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Research Article Positive Solutions for Nonlinear -Fractional Difference Eigenvalue Problem with Nonlocal Conditions Wafa Shammakh and Maryam Al-Yami Sciences Faculty for Girls, King Abdulaziz University, Jeddah, Saudi Arabia Correspondence should be addressed to Maryam Al-Yami; [email protected] Received 16 August 2015; Revised 5 November 2015; Accepted 26 November 2015 Academic Editor: Svatoslav Stanˇ ek Copyright © 2015 W. Shammakh and M. Al-Yami. is is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited. e problem of positive solutions for nonlinear -fractional difference eigenvalue problem with nonlocal boundary conditions is investigated. Based on the fixed point index theory in cones, sufficient existence of positive solutions conditions is derived for the problem. 1. Introduction e fractional -calculus is the -extension of ordinary fractional calculus. It has been used by many researchers to adequately describe the evolution of a variety of engineering, economical, physical, and biological processes. We consider a nonlinear -fractional difference eigen- value problem with nonlocal boundary conditions given by () + () (, ()) = 0, 0 ≤ ≤ 1, 0 < < 1, (1) (0) = 0, (0) = 0, 2 ≤ ≤ − 1, (1) = [] , (2) where denote the fractional -derivative of the Caputo type, − 1 < ≤ , > 2, > 0 is a parameter, and [] is given by a Riemann-Stieltjes integral [] = ∫ 1 0 () (). is type of BC includes, as particular cases, multipoint problems when [] = ∑ −2 =1 ( ), (see [1]) and a contin- uously distributed case when [] = ∫ 1 0 ()() (see [2– 4]). More recently, many people pay attention to BVPs involv- ing nonlinear -difference equations [5–12]. In [13], Yuan and Yang dealt with some existence and uniqueness results for nonlinear boundary value problems for delayed -fractional difference systems based on a con- traction mapping principle and Krasnoselskii’s fixed-point theorem. In [14], Yang investigated the sufficient conditions for the existence and nonexistence positive solutions for BVP involving nonlinear -fractional difference equations. Ferreira [4] studied the existence of positive solutions to the nonlinear -fractional BVPs by means of Krasnoselskii’s fixed point theorem in cones. In this paper, we obtain the results on the existence of one and two positive solutions by utilizing the results of Webb and Lan [15] involving comparison with the principle characteristic value of a related linear problem to the - fractional case. We then use the theory worked out by Webb and Infante in [16–19] to study the general nonlocal BCs. 2. Preliminaries In this section, we will present some definitions and lemmas that will be used in the proof of our main results. Let ∈ (0, 1) defined by [20] [] = −1 −1 = −1 + ⋅ ⋅ ⋅ + 1, R. (3) Hindawi Publishing Corporation Abstract and Applied Analysis Volume 2015, Article ID 759378, 10 pages http://dx.doi.org/10.1155/2015/759378
Transcript
Page 1: Research Article Positive Solutions for Nonlinear -Fractional …downloads.hindawi.com/journals/aaa/2015/759378.pdf · 2019. 7. 31. · Research Article Positive Solutions for Nonlinear

Research ArticlePositive Solutions for Nonlinear 119902-Fractional DifferenceEigenvalue Problem with Nonlocal Conditions

Wafa Shammakh and Maryam Al-Yami

Sciences Faculty for Girls King Abdulaziz University Jeddah Saudi Arabia

Correspondence should be addressed to Maryam Al-Yami malyamikauedusa

Received 16 August 2015 Revised 5 November 2015 Accepted 26 November 2015

Academic Editor Svatoslav Stanek

Copyright copy 2015 W Shammakh and M Al-Yami This is an open access article distributed under the Creative CommonsAttribution License which permits unrestricted use distribution and reproduction in any medium provided the original work isproperly cited

The problem of positive solutions for nonlinear 119902-fractional difference eigenvalue problem with nonlocal boundary conditions isinvestigated Based on the fixed point index theory in cones sufficient existence of positive solutions conditions is derived for theproblem

1 Introduction

The fractional 119902-calculus is the 119902-extension of ordinaryfractional calculus It has been used by many researchers toadequately describe the evolution of a variety of engineeringeconomical physical and biological processes

We consider a nonlinear 119902-fractional difference eigen-value problem with nonlocal boundary conditions given by119862119863120572

119902119906 (119905) + 120582119892 (119905) 119891 (119905 119906 (119905)) = 0

0 le 119905 le 1 0 lt 119902 lt 1

(1)

119863119896

119902119906 (0) = 0

119906 (0) = 0 2 le 119896 le 119899 minus 1

119863119902119906 (1) = 120579 [119906]

(2)

where 119862119863120572119902denote the fractional 119902-derivative of the Caputo

type 119899 minus 1 lt 120572 le 119899 119899 gt 2 120582 gt 0 is a parameter and 120579[119906] isgiven by a Riemann-Stieltjes integral 120579[119906] = int1

0119906(119905)119889119902119860(119905)

This type of BC includes as particular cases multipointproblems when 120579[119906] = sum119898minus2

119894=1120572119894119906(120577119894) (see [1]) and a contin-

uously distributed case when 120579[119906] = int10120572(119904)119906(119904)119889

119902119904 (see [2ndash

4])More recently many people pay attention to BVPs involv-

ing nonlinear 119902-difference equations [5ndash12]

In [13] Yuan and Yang dealt with some existence anduniqueness results for nonlinear boundary value problemsfor delayed 119902-fractional difference systems based on a con-traction mapping principle and Krasnoselskiirsquos fixed-pointtheorem

In [14] Yang investigated the sufficient conditions forthe existence and nonexistence positive solutions for BVPinvolving nonlinear 119902-fractional difference equations

Ferreira [4] studied the existence of positive solutions tothe nonlinear 119902-fractional BVPs by means of Krasnoselskiirsquosfixed point theorem in cones

In this paper we obtain the results on the existence ofone and two positive solutions by utilizing the results ofWebb and Lan [15] involving comparison with the principlecharacteristic value of a related linear problem to the 119902-fractional case We then use the theory worked out by Webband Infante in [16ndash19] to study the general nonlocal BCs

2 Preliminaries

In this section we will present some definitions and lemmasthat will be used in the proof of our main results

Let 119902 isin (0 1) defined by [20]

[119886]119902 =119902119886minus 1

119902 minus 1= 119902119886minus1+ sdot sdot sdot + 1 119886 isin R (3)

Hindawi Publishing CorporationAbstract and Applied AnalysisVolume 2015 Article ID 759378 10 pageshttpdxdoiorg1011552015759378

2 Abstract and Applied Analysis

The 119902-analogue of the power function (119886 minus 119887)119899 with 119899 isin N is

(119886 minus 119887)0= 1

(119886 minus 119887)(119899)=

119899minus1

prod

119896=0

(119886 minus 119887119902119896) 119886 119887 isin R 119899 isin N

(4)

More generally if 120572 isin R then

(119886 minus 119887)(120572)= 119886120572

infin

prod

119894=0

(119886 minus 119887119902119894)

(119886 minus 119887119902120572+119894) (5)

Note that if 119887 = 0 then 119886(120572) = 119886120572The 119902-gamma function isdefined by

Γ119902 (119909) =

(1 minus 119902)(119909minus1)

(1 minus 119902)119909minus1

119909 isin R 0 minus1 minus2 0 lt 119902 lt 1

(6)

and satisfies Γ119902(119909 + 1) = [119909]

119902Γ119902(119909)

The 119902-derivative of a function 119891(119909) is here defined by

119863119902119891 (119909) =

119889119902119891 (119909)

119889119902119909=119891 (119902119909) minus 119891 (119909)

(119902 minus 1) 119909 (7)

and 119902-derivatives of higher order are defined by

119863119899

119902119891 (119909) =

119891 (119909) if 119899 = 0

119863119902119863119899minus1

119902119891 (119909) if 119899 isin N

(8)

The 119902-integral of a function 119891 defined in the interval [0 119887] isgiven by

int

119909

0

119891 (119905) 119889119902119905 = 119909 (1 minus 119902)

infin

sum

119899=0

119891 (119909119902119899) 119902119899

0 le10038161003816100381610038161199021003816100381610038161003816 lt 1 119909 isin [0 119887]

(9)

If 119886 isin [0 119887] and 119891 is defined in the interval [0 119887] its integralfrom 119886 to 119887 is defined by

int

119887

119886

119891 (119905) 119889119902119905 = int

119887

0

119891 (119905) 119889119902119905 minus int

119886

0

119891 (119905) 119889119902119905 (10)

Similarly as done for derivatives it can be defined an operator119868119899

119902 namely

(1198680

119902119891) (119909) = 119891 (119909)

(119868119899

119902119891) (119909) = 119868119902 (119868

119899minus1

119902119891) (119909) 119899 isin N

(11)

The fundamental theorem of calculus applies to these opera-tors 119868119902and119863

119902 that is

(119863119902119868119902119891) (119909) = 119891 (119909) (12)

and if 119891 is continuous at 119909 = 0 then

(119868119902119863119902119891) (119909) = 119891 (119909) minus 119891 (0) (13)

Basic properties of the two operators can be found in the book[20] We now point out four formulas that will be used later

[119886 (119905 minus 119904)](120572)= 119886120572(119905 minus 119904)

(120572)

119905119863119902 (119905 minus 119904)

(120572)= [120572]119902 (119905 minus 119904)

(120572minus1)

119904119863119902 (119905 minus 119904)

(120572)= minus [120572]119902 (119905 minus 119902119904)

(120572minus1)

(119909119863119902 int

119909

0

119891 (119909 119905) 119889119902119905) (119909)

= int

119909

0119909119863119902 119891 (119909 119905) 119889119902119905 + 119891 (119902119909 119909)

(14)

where119894119863119902denotes the 119902-derivative with respect to variable 119894

[21]

Remark 1 (see [21]) We note that if 120572 gt 0 and 119886 le 119887 le 119905 then(119905 minus 119886)

(120572)ge (119905 minus 119887)

(120572)

Definition 2 (see [22]) Let 120572 ge 0 and let 119891 be a functiondefined on [0 1]The fractional 119902-integral of the Riemann-Liouville type is (

RL1198680

119902119891)(119909) = 119891(119909) and

(RL119868120572

119902119891) (119909) =

1

Γ119902 (120572)

int

119909

0

(119909 minus 119902119905)(120572minus1)

119891 (119905) 119889119902119905

120572 isin R+ 119909 isin [0 1]

(15)

Definition 3 (see [22]) The fractional 119902-derivative of theRiemann-Liouville type of order 120572 ge 0 is defined by(RL1198630

119902119891)(119909) = 119891(119909) and

(RL119863120572

119902119891) (119909) = (119863

[120572]

119902119868[120572]minus120572

119902119891) (119909) 120572 gt 0 (16)

where [120572] is the smallest integer greater than or equal to 120572

Definition 4 (see [22]) The fractional 119902-derivative of theCaputo type of order 120572 ge 0 is defined by

(119862119863120572

119902119891) (119909) = (119868

[120572]minus120572

119902119863[120572]

119902119891) (119909) 120572 gt 0 (17)

Lemma 5 (see [22]) Let 120572 120573 ge 0 and let 119891 be a functiondefined on [0 1]Then the next formulas hold

(1) (119868120573119902119868120572

119902119891)(119909) = (119868

120572+120573

119902119891)(119909)

(2) (119862119863120572

119902119868120572

119902119891)(119909) = 119891(119909)

Abstract and Applied Analysis 3

Lemma 6 (see [22]) Let 120572 isin R+ N 120582 isin (minus1infin) Then thenext formulas hold

(1) 119868120572

119902119909120582=

Γ119902 (120582 + 1)

Γ119902 (120582 + 1 + 120572)

119909120582+120572

(2)119877119871119863120572

119902119909120582=

Γ119902 (120582 + 1)

Γ119902 (120582 + 120572 + 1)119909120582minus120572

(3)119862119863120572

119902119909120582=

0 119894119891 120582 isin N0 120572 gt 120582

119877119871119863120572

119902119909120582119894119891 119900119905ℎ119890119903119908119894119904119890

(18)

Theorem 7 (see [23]) Let 119909 gt 0 and 120572 isin R+ NThen thefollowing equality holds

(119868120572

119902 119862119863120572

119902119891) (119909) = 119891 (119909) minus

[120572]minus1

sum

119896=0

119909119896

Γ119902 (119896 + 1)

(119863119896

119902119891) (0) (19)

Lemma 8 (see [24]) Suppose 119879 119870 rarr 119870 is a completelycontinuous operator and has no fixed points on 120597119870

120588cap119870 Then

the following are true(i) If 119879119906 le 119906 for all 119906 isin 120597119870

120588cap 119870 then 119894(119879 119870

120588cap

119870119870) = 1 where 119894 is the fixed point index on 119870(ii) If 119879119906 ge 119906 for all 119906 isin 120597119870

120588cap 119870 then 119894(119879 119870

120588cap

119870119870) = 0

Lemma 9 (see [24]) Let 119870 be a cone in Banach space 119864Suppose that119879 119870

120588rarr 119870 is a completely continuous operator

There exists 1199060isin 1198700 such that 119906minus119879119906 = 120583119906

0for any 119906 isin 120597119870

119903

and 120583 ge 0 119894(119879 119870120588 119870) = 0

Lemma 10 (see [24]) Let 119870 be a cone in Banach space 119864Suppose that119879 119870

120588rarr 119870 is a completely continuous operator

If 119879119906 = 120583119906 for any 119906 isin 120597119870119903and 120583 ge 1 then 119894(119879 119870

120588 119870) = 1

Lemma 11 Let 119910 isin 119862[0 1] be a given function and 119899 minus 1 lt120572 le 119899 then 119906 is a solution of BVP (1)-(2) if and only if 119906 is asolution of the integral equation

119906 (119905) = 120574 (119905) 120579 [119906] + int

1

0

1198660 (119905 119902119904) 119910 (119904) 119889119902119904 (20)

where120574 (119905) = 119905

1198660(119905 119902119904)

=

[120572 minus 1]119902 119905 (1 minus 119902119904)(120572minus2)

minus (119905 minus 119902119904)(120572minus1)

Γ119902 (120572)

0 le 119902119904 le 119905 le 1

[120572 minus 1]119902 119905 (1 minus 119902119904)(120572minus2)

Γ119902 (120572) 0 le 119905 le 119902119904 le 1

(21)

Proof Assume that 119906 is a solution of BVP (1)-(2)Applying Theorem 7 (1) can be reduced to an equivalent

integral equation

119906 (119905) = minus1

Γ119902 (120572)

int

119905

0

(119905 minus 119902119904)(120572minus1)

119910 (119904) 119889119902119904 + 1198880 + 1198881119905

+ 11988821199052+ sdot sdot sdot + 119888

119899minus1119905119899minus1

(22)

By (2) we obtain

1198880= 0

1198882 = sdot sdot sdot = 119888119899minus1 = 0

1198881= 120579 [119906] +

[120572 minus 1]119902

Γ119902 (120572)

int

1

0

(1 minus 119902119904)(120572minus2)

119910 (119904) 119889119902119904

(23)

Therefore we obtain

119906 (119905) = 120574 (119905) 120579 [119906] +119905 [120572 minus 1]119902

Γ119902 (120572)

int

1

0

(1 minus 119902119904)(120572minus2)

sdot 119910 (119904) 119889119902119904 minus1

Γ119902 (120572)

int

119905

0

(119905 minus 119902119904)(120572minus1)

119910 (119904) 119889119902119904

= 120574 (119905) 120579 [119906]

+ int

119905

0

[

[

[120572 minus 1]119902 119905 (1 minus 119902119904)(120572minus2)

minus (119905 minus 119902119904)(120572minus1)

Γ119902 (120572)

]

]

sdot 119910 (119904) 119889119902119904 + int

1

119905

[

[

[120572 minus 1]119902 119905 (1 minus 119902119904)(120572minus2)

Γ119902 (120572)

]

]

sdot 119910 (119904) 119889119902119904 = 120574 (119905) 120579 [119906] + int

1

0

1198660 (119905 119902119904) 119910 (119904) 119889119902119904

(24)

Conversely if 119906 is a solution of the integral equation (20)using Lemmas 5 and 6 we have

119862119863120572

119902119906 (119905)

=119862119863120572

119902119905120579 [119906]

+119862119863120572

119902119905(int

1

0

[120572 minus 1]119902 (1 minus 119902119904)(120572minus2)

Γ119902 (120572)

119910 (119904) 119889119902119904)

minus119862119863120572

119902(int

119905

0

(119905 minus 119902119904)(120572minus1)

Γ119902 (120572)

119910 (119904) 119889119902119904)

= minus119862119863120572

119902119868120572

119902119910 (119905) = minus119910 (119905)

(25)

A simple computation shows 119906(0) = 0119863119896119902119906(0) = 0 2 le 119896 le

119899 minus 1119863119902119906(1) = 120579[119906]

Remark 12 1198660(119905 119902119904) is Greenrsquos function for the local BVP

119862119863120572

119902119906 (119905) + 120582119892 (119905) 119891 (119905 119906 (119905)) = 0

119905 isin [0 1] 119899 minus 1 lt 120572 le 119899 0 lt 119902 lt 1

119863119896

119902119906 (0) = 0

119906 (0) = 0 2 le 119896 le 119899 minus 1

119863119902119906 (1) = 0

(26)

4 Abstract and Applied Analysis

Lemma 13 Function 1198660(119905 119902119904) defined in (20) satisfies the

following conditions

(H1) 1198660(119905 119902119904) ge 0 is continuous and 119866

0(119905 119904) le Φ

0(119902119904) for

all 0 le 119905 119904 le 1(H2) 119866

0(119905 119902119904) ge 119888

0(119905)Φ0(119902119904) for all 0 le 119905 119904 le 1 where

Φ0(119902119904) = 119866

0(1 119902119904)

=[120572 minus 1]119902 (1 minus 119902119904)

(120572minus2)minus (1 minus 119902119904)

(120572minus1)

Γ119902 (120572)

1198880 (119905) = 119905

120572minus1

(27)

Proof It is obvious that1198660(119905 119902119904) is nonnegative and continu-

ous(H1) For 0 le 119902119904 le 119905 le 1

1198660(119905 119902119904) =

1

Γ119902 (120572)[[120572 minus 1]119902 119905 (1 minus 119902119904)

(120572minus2)

minus (119905 minus 119902119904)(120572minus1)

] =1

Γ119902 (120572)

[[120572 minus 1]119902 119905 (1 minus 119902119904)(120572minus2)

minus 119905120572minus1(1 minus 119902

119904

119905)

(120572minus1)

]

ge1

Γ119902 (120572)[[120572 minus 1]119902 119905

120572minus1(1 minus 119902119904)

(120572minus2)

minus 119905120572minus1(1 minus 119902119904)

(120572minus1)]

=119905120572minus1

Γ119902 (120572)

[[120572 minus 1]119902 (1 minus 119902119904)(120572minus2)

minus (1 minus 119902119904)(120572minus1)

]

ge 0

(28)

and for 0 le 119905 le 119902119904 le 1

1198660 (119905 119902119904) =

[120572 minus 1]119902 119905 (1 minus 119902119904)(120572minus2)

Γ119902 (120572)

(29)

and it is clear that 1198660(119905 119902119904) ge 0 and 119866

0(0 119902119904) = 0Therefore

1198660(119905 119902119904) ge 0

For fixed 119904 isin [0 1] and 119905 ge 119902119904 we have

119905119863119902 1198660 (119905 119902119904)

=[120572 minus 1]119902 (1 minus 119902119904)

(120572minus2)minus [120572 minus 1]119902 (119905 minus 119902119904)

(120572minus2)

Γ119902 (120572)

=(1 minus 119902119904)

(120572minus2)minus (119905 minus 119902119904)

(120572minus2)

Γ119902 (120572 minus 1)

ge 0

(30)

that is 1198660(119905 119902119904) is an increasing function of 119905 Obviously

1198660(119905 119902119904) 119905 le 119902119904 is increasing in 119905 therefore 119866

0(119905 119902119904) is an

increasing function of 119905 for fixed 119904 isin [0 1]Thus (H1) holds

(H2) Suppose now that 119905 ge 119902119904

1198660(119905 119902119904)

Φ0 (119902119904)=[120572 minus 1]119902 119905 (1 minus 119902119904)

(120572minus2)minus (119905 minus 119902119904)

(120572minus1)

[120572 minus 1]119902 (1 minus 119902119904)(120572minus2)

minus (1 minus 119902119904)(120572minus1)

ge

119905120572minus1[[120572 minus 1]119902 (1 minus 119902119904)

(120572minus2)minus (1 minus 119902119904)

(120572minus1)]

[120572 minus 1]119902 (1 minus 119902119904)(120572minus2)

minus (1 minus 119902119904)(120572minus1)

= 119905120572minus1

(31)

On the other hand if 119905 le 119902119904 then we have

1198660(119905 119902119904)

Φ0(119902119904)

=[120572 minus 1]119902 119905 (1 minus 119902119904)

(120572minus2)

[120572 minus 1]119902 (1 minus 119902119904)(120572minus2)

minus (1 minus 119902119904)(120572minus1)

ge[120572 minus 1]119902 119905

120572minus1(1 minus 119902119904)

(120572minus2)minus 119905120572minus1(1 minus 119902119904)

(120572minus1)

[120572 minus 1]119902 (1 minus 119902119904)(120572minus2)

minus (1 minus 119902119904)(120572minus1)

= 119905120572minus1

(32)

and this finished the proof of (H2)

Defining G119860(119902119904) = int

1

01198660(119905 119902119904)119889

119902119860(119905) Greenrsquos function

for nonlocal BVP (1)-(2) is given by

119866 (119905 119902119904) =120574 (119905)

[1 minus 120579 [120574]]G119860(119902119904) + 119866

0(119905 119902119904) (33)

Throughout the paper we assume the following(H3) A is a function of bounded variation andG119860(119902119904) =

int1

01198660(119905 119902119904)119889

119902119860(119905) satisfies G

119860(119902119904) ge 0 for almost every 119904 isin

[0 1] Note thatG119860(119902119904) exists for almost every 119904 by (H1)

(H4)The functions 119892Φ satisfy 119892 ge 0 almost everywhere119892Φ isin 119871

1[0 1] and

int

119887

119886

Φ(119902119904) 119892 (119904) 119889119902119904 gt 0 (34)

(H5) 119891 [0 1] times [0infin) rarr [0infin) satisfies Caratheodoryconditions that is 119891(sdot 119906) is measurable for each fixed 119906 isin[0infin) and 119891(119905 sdot) is continuous for almost every 119905 isin [0 1]and for each 119903 gt 0 there exists 120601

119903isin 119871infin[0 1] such that 0 le

119891(119905 119906) le 120601119903for all 119906 isin [0 119903] and almost all 119905 isin [0 1]

(H6) One has the following 120574 isin 119862[0 1] 120574(119905) ge 0 0 le120579[120574] lt 1

Lemma 14 If 1198660satisfies (H1) (H2) then 119866 satisfies (H1)

(H2) for a function Φ the same interval [119886 119887] and the sameconstant 119888 where Φ satisfies (H4) and 119888 = min1198880(119905) 119905 isin[119886 119887]

Proof We have

119866 (119905 119902119904) =120574 (119905)

[1 minus 120579 [120574]]G119860(119902119904) + 119866

0(119905 119902119904)

le

10038171003817100381710038171205741003817100381710038171003817

[1 minus 120579 [120574]]G119860(119902119904) + Φ

0(119902119904) = Φ (119902119904)

(35)

Abstract and Applied Analysis 5

and for 119905 isin [119886 119887]

119866 (119905 119902119904) ge11988810038171003817100381710038171205741003817100381710038171003817

[1 minus 120579 [120574]]G119860 (119902119904) + 119888Φ0 (119902119904) = 119888Φ (119902119904) (36)

Note that 119892Φ isin 119871infin because 119860 has finite variation and

G119860(119902119904) le Φ(119902119904) var(119860)Thus Greenrsquos function 119866(119905 119902119904) satisfies (H1) (H2) for a

functionΦ and the constant 119888

3 Main Result

Set 119864 = 119862[0 1] as a Banach space with the norm 119906 =

sup119905isin[01]

|119906(119905)| Let 119875 = 119906 isin 119864 119906 ge 0 denote the standardcone of nonnegative functions Define

119870 = 119906 isin 119875 min119886le119905le119887

119906 (119905) ge 119888 119906 (37)

where [119886 119887] is some subset of [0 1]Note that 120574 isin 119870 so 119870 = 0 For any 0 lt 119903 lt 119877 lt +infin

let 119870119903= 119906 isin 119870 119906 lt 119903 120597119870

119903= 119906 isin 119870 119906 = 119903

119870119903 = 119906 isin 119870 119906 le 119903 119870119877 119870119903 = 119906 isin 119870 119903 le 119906 le 119877and 119881119903 = 119906 isin 119870 min119905isin[119886119887]119906(119905) lt 119903 and 119881119903 is bounded

Define a nonlinear operator 119879 119875 rarr 119870 and a linearoperator 119871 119875 rarr 119870 by

119879119906 (119905) = 120582int

1

0

119866 (119905 119902119904) 119892 (119904) 119891 (119904 119906 (119904)) 119889119902119904 (38)

119871119906 (119905) fl int1

0

119866 (119905 119902119904) 119892 (119904) 119906 (119904) 119889119902119904 (39)

Lemma 15 (see [18]) Under hypotheses (H1)ndash(H6) the maps119879 119875 rarr 119864 defined in (38) are compact

Theorem 16 Under hypotheses (H1)ndash(H6) the maps are 119879 119875 rarr 119870

Proof For 119906 isin 119875 and 119905 isin [0 1] we have

119879119906 (119905) le 120582int

1

0

Φ(119902119904) 119892 (119904) 119891 (119904 119906 (119904)) 119889119902119904 (40)

Hence

119879119906 le 120582int

1

0

Φ(119902119904) 119892 (119904) 119891 (119904 119906 (119904)) 119889119902119904 (41)

Also for 119905 isin [119886 119887] we have

119879119906 (119905) ge 119888120582int

1

0

Φ(119902119904) 119892 (119904) 119891 (119904 119906 (119904)) 119889119902119904 ge 119888 119879119906 (42)

Similar to the proofs of Lemma 15 and Theorem 16 119871119906(119905) iscompact and maps 119875 into119870

We will use the Krein-Rutman theorem We recall that 120582is an eigenvalue of 119871 with corresponding eigenfunction 120601 if120601 = 0 and 120582120601 = 119871120601 The reciprocals of eigenvalues are calledcharacteristic values of 119871 The radius of the spectrum of 119871denoted by 119903(119871) is given by the well-known spectral radiusformula 119903(119871) = lim

119899rarrinfin1198711198991119899

Theorem 17 (see [15]) Let 119870 be a total cone in a real Banachspace 119864 and let 119864 rarr 119864 be a compact linear operator with(119870) sube 119870 If 119903() gt 0 then there is 120601

1isin 119870 0 such that

1206011= 119903()120601

1

Thus 1205821 fl 119903() is an eigenvalue of the largest possiblereal eigenvalue and 1205831 = 1120582

1is the smallest positive

characteristic value

Lemma 18 (see [15]) Assume that (H1)ndash(H3) hold and let 119871be as defined in (39) Then 119903(119871) gt 0

Theorem 19 (see [15]) When (H1)ndash(H3) hold 119903(119871) is aneigenvalue of 119871 with eigenfunction 120601

1in 119870

Theorem 20 (see [15]) Let 1205831= 1119903(119871) and 120601

1(119905) be a

corresponding eigenfunction in119875 of norm 1Then119898 le 1205831le 119872

where

119898 = ( sup119905isin[01]

int

1

0

119866 (119905 119902119904) 119892 (119904) 119889119902119904)

minus1

119872 = ( inf119905isin[119886119887]

int

119887

119886

119866 (119905 119902119904) 119892 (119904) 119889119902119904)

minus1

(43)

If 119892(119905) gt 0 for 119905 isin [0 1] and 119866(119905 119902119904) gt 0 for 119905 119904 isin [0 1] thefirst inequality is strict unless 1206011(119905) is constant for 119905 isin [0 1] If119892(119905)120601(119905) gt 0 for 119905 isin [119886 119887] the second inequality is strict unless1206011(119905) is constant for 119905 isin [119886 119887]

Proof (for the local BVP (1)-(2) if 119892(119905) equiv 1) We now computethe constant 119898 and the optimal value of119872(119886 119887) that is wedetermine 119886 119887 so that119872(119886 119887) is minimal

For 119902119904 le 119905 we have by direct integration

int

119905

0

1198660 (119905 119902119904) 119889119902119904

= int

119905

0

[

[

[120572 minus 1]119902 119905 (1 minus 119902119904)(120572minus2)

minus (119905 minus 119902119904)(120572minus1)

Γ119902 (120572)

]

]

119889119902119904

=119905 minus 119905 (1 minus 119905)

(120572minus1)

Γ119902 (120572)

minus119905120572

[120572]119902 Γ119902 (120572)

(44)

For 119902119904 ge 119905

int

1

119905

1198660(119905 119902119904) 119889

119902119904 = int

1

119905

[120572 minus 1]119902 119905 (1 minus 119902119904)(120572minus2)

Γ119902 (120572)119889119902119904

=119905 (1 minus 119905)

(120572minus1)

Γ119902 (120572)

(45)

Then we have

int

1

0

1198660 (119905 119902119904) 119889119902119904 =119905

Γ119902 (120572)

minus119905120572

[120572]119902 Γ119902 (120572) (46)

6 Abstract and Applied Analysis

And the maximum of this expression occurs when 119905 = 1hence

sup119905isin[01]

int

1

0

1198660 (119905 119902119904) 119889119902119904 =1

Γ119902 (120572)

minus1

[120572]119902 Γ119902 (120572)

=[120572]119902 minus 1

[120572]119902 Γ119902 (120572)

(47)

Then119898 = [120572]119902Γ119902(120572)([120572]

119902minus 1)

For 119886 lt 119887 we have by direct integration

int

119905

119886

1198660(119905 119902119904) 119889

119902119904 = minus

119905 (1 minus 119905)(120572minus1)

Γ119902 (120572)

+119905 (1 minus 119886)

(120572minus1)

Γ119902 (120572)

minus(119905 minus 119886)

(120572)

[120572]119902 Γ119902 (120572)

int

119887

119905

1198660 (119905 119904) 119889119904 = minus

119905 (1 minus 119887)(120572minus1)

Γ119902 (120572)+119905 (1 minus 119905)

(120572minus1)

Γ119902 (120572)

(48)

Then

int

119887

119886

1198660(119905 119902119904) 119889

119902119904 =119905 (1 minus 119886)

(120572minus1)

Γ119902 (120572)minus(119905 minus 119886)

(120572)

[120572]119902 Γ119902 (120572)

minus119905 (1 minus 119887)

(120572minus1)

Γ119902 (120572)

=119905 [(1 minus 119886)

(120572minus1)minus (1 minus 119887)

(120572minus1)]

Γ119902 (120572)

minus(119905 minus 119886)

(120572)

[120572]119902 Γ119902 (120572)= 119877 (119905 119886 119887)

119905119863119902119877 (119905 119886 119887) =

[(1 minus 119886)(120572minus1)

minus (1 minus 119887)(120572minus1)

]

Γ119902 (120572)

minus(119905 minus 119886)

(120572minus1)

Γ119902 (120572)

(49)

The sign of derivative119905119863119902119877 shows that this is an increasing

function of 119905 so the minimum occurs at 119905 = 119886 Let

119877 (119886 119887) =119886

Γ119902 (120572)

[(1 minus 119886)(120572minus1)

minus (1 minus 119887)(120572minus1)

] (50)

The minimal value of 119872(119886 119887) corresponds to the maximalvalue of 119877(119886 119887) Consider

119887119863119902119877 (119886 119887) =

119886 [120572 minus 1]119902 (1 minus 119902119887)(120572minus2)

Γ119902 (120572)gt 0 (51)

The quantity 119877(119886 119887) is an increasing function of 119887 so itsmaximum occurs when 119887 = 1 Let

119877 (119886) =119886 (1 minus 119886)

(120572minus1)

Γ119902 (120572)

(52)

Then the maximum of 119877(119886) occurs when 119886 = 1(1+ [120572minus1]119902)

Consider

min119905isin[119886119887]

int

119887

119886

1198660(119905 119902119904) 119889

119902119904 = 119877(

1

1 + [120572 minus 1]119902

1) (53)

Hence the minimal value of119872(119886 119887) is

119872(1

1 + [120572 minus 1]119902

1) = (119877(1

1 + [120572 minus 1]119902

1))

minus1

(54)

4 The Existence of at Least OnePositive Solution

For convenience we introduce the following notations

119891 (119906) fl sup119905isin[01]

119891 (119905 119906)

119891 (119906) fl inf119905isin[01]

119891 (119905 119906)

1198910 fl lim sup119906rarr0

+

119891 (119906)

119906

1198910fl lim inf119906rarr0

+

119891 (119906)

119906

119891infin fl lim sup

119906rarrinfin

119891 (119906)

119906

119891infin

fl lim inf119906rarrinfin

119891 (119906)

119906

1198910119903 fl sup0le119905le1 0le119906le119903

119891 (119905 119906)

119903

119891119903119903119888

fl inf119886le119905le119887 119903le119906le119903119888

119891 (119905 119906)

119903

(55)

Under hypotheses (H1)ndash(H4) let be defined by

119906 (119905) = int

119887

119886

119866 (119905 119902119904) 119892 (119904) 119906 (119904) 119889119902119904 (56)

Then is a compact linear operator and (119875) sube 119870Hence 119903() is an eigenvalue of with an eigenfunction 120601

1

in119870 Let 1205831fl 1119903()Note that 120583

1ge 1205831 hence the condition

in the following theorem ismore stringent comparedwith thecase if 119903(119871) could be used

Theorem 21 Assume that

(A1) 0 le 1205821198910 lt 1205831(A2) 120583

1lt 120582119891infinle infin

Then (1)-(2) had at least one positive solution

Abstract and Applied Analysis 7

Proof Let 120576 gt 0 be such that 1198910 le (1120582)(1205831minus 120576)Then there

exists 1205880gt 0 such that

119891 (119905 119906) le1

120582(1205831 minus 120576) 119906

forall119906 isin [0 1205880] and almost all 119905 isin [0 1]

(57)

Let 120588 isin (0 1205880]We prove that

119879119906 = 120573119906 for 119906 isin 120597119870120588 120573 ge 1 (58)

which implies the result In fact if (58) does not hold thenthere exist 119906 isin 120597119870

120588and 120573 ge 1 such that 119879119906 = 120573119906

This implies

120573119906 (119905) = 120582int

1

0

119866 (119905 119902119904) 119892 (119904) 119891 (119904 119906 (119904)) 119889119902119904

le (1205831minus 120576) int

1

0

119866 (119905 119902119904) 119892 (119904) 119906 (119904) 119889119902119904

= (1205831minus 120576) 119871119906 (119905)

(59)

Thus we have shown 119906(119905) le (1205831 minus 120576)119871119906(119905)This gives

119906 (119905) le (1205831 minus 120576) 119871 [(1205831 minus 120576) 119871119906 (119905)]

= (1205831minus 120576)21198712119906 (119905)

(60)

And by iterating

119906 (119905) le (1205831 minus 120576)119899119871119899119906 (119905) for 119899 isin 119873 (61)

Therefore

119906 le (1205831 minus 120576)119899 10038171003817100381710038171198711198991003817100381710038171003817 119906

1 le (1205831 minus 120576)119899 10038171003817100381710038171198711198991003817100381710038171003817

(62)

and we have

1 le (1205831 minus 120576) lim

119899rarr+infin

100381710038171003817100381711987111989910038171003817100381710038171119899= (1205831 minus 120576)

1

1205831

lt 1 (63)

a contradiction It follows that

119894119896 (119879119870120588) = 1 for each 120588 isin (0 1205880] (64)

Let 1205881gt 0 120588

1gt 120588 be chosen so that 119891(119905 119906) gt (120583

1120582)119906 for all

119906 ge 1198881205881 119888 as in (H2) and almost all 119905 isin [0 1]

We claim that 119906 = 119879119906 + 1205731206011 for all 120573 gt 0 and 119906 isin 120597119870120588lowast

when 120588lowast gt 1205881 Note that 119906 isin 119870 with 119906 = 120588lowast ge 120588

1

We have 119906(119905) ge 1198881205881for all 119905 isin [119886 119887]

Now if our claim is false then we have

119906 (119905) = 120582int

1

0

119866 (119905 119902119904) 119892 (119904) 119891 (119904 119906 (119904)) 119889119902119904 + 1205731206011 (119905) (65)

Therefore

119906 (119905) ge 1205831 int

119887

119886

119866 (119905 119902119904) 119892 (119904) 119906 (119904) 119889119902119904 + 1205731206011 (119905)

= 1205831119906 (119905) + 1205731206011 (119905)

(66)

From (66) we firstly deduce that 119906(119905) ge 1205731206011(119905) on [0 1]

Then we have

1205831119906 (119905) ge 1205831 (1205731206011 (119905)) = 1205731206011 (119905) (67)

Inserting this into (66) we obtain 119906(119905) ge 21205731206011(119905) for 119905 isin [0 1]

Repeating this process gives

119906 (119905) ge 1198991205731206011 (119905) for 119905 isin [0 1] 119899 isin 119873 (68)

Since 1206011(119905) is strictly positive on [0 1] this is a contradiction

then

119894119870(119879119870120588lowast) = 0 for 119906 isin 120597119870

120588lowast (69)

By (64) and (69) one has

119894119870(119879119870120588lowast 119870120588) = 119894119870(119879119870120588lowast) minus 119894119870(119879119870120588) = minus1 (70)

Therefore 119879 has at least one fixed point 1199060isin 119870120588lowast 119870120588 and

1199060is a positive solution of BVP (1)-(2)

Theorem 22 Assume that

(A3) 1205831 lt 1205821198910 le infin(A4) 0 le 120582119891infin lt 1205831

Then (1)-(2) had at least one positive solution

Proof Let 120576 gt 0 satisfy 1198910gt (1120582)(120583

1+ 120576)Then there exists

1198771 gt 0 such that

119891 (119905 119906) ge1

120582(1205831+ 120576) 119906 forall119905 isin [0 1] 119906 isin [0 1198771] (71)

For any 119906 isin 1205971198701198771

we have by (71) that

119879119906 (119905) = 120582int

1

0

119866 (119905 119902119904) 119892 (119904) 119891 (119904 119906 (119904)) 119889119902119904

ge (1205831 + 120576) int

1

0

119866 (119905 119902119904) 119892 (119904) 119906 (119904) 119889119902119904

ge 1205831119871119906 (119905) forall119905 isin [0 1]

(72)

Let 1 be the positive eigenfunction of 119871 corresponding to 1205831

that is 1 = 12058311198711 Wemay suppose that119879 has no fixed pointon 120597119870119877

1

otherwise the proof is finished In the following wewill show that

119906 minus 119879119906 = 1205731 forall119906 isin 1205971198701198771

120573 ge 0 (73)

If (73) is not true then there is 0isin 1205971198701198771

and 1205730ge 0 such that

0minus1198790= 12057301 It is clear that 120573

0gt 0 and

0= 1198790+12057301ge

12057301Set

120573lowast= sup 120573 0 ge 1205731 (74)

Obviously 120573lowast ge 1205730gt 0 It follows from 119871(119875) sub 119875 that

12058311198710ge 1205831119871120573lowast1= 120573lowast12058311198711= 120573lowast1 (75)

8 Abstract and Applied Analysis

and using this and (72) we have

0 = 1198790 + 12057301 ge 12058311198710 + 12057301 ge 120573lowast1 + 12057301 (76)

which contradicts (74) Thus (73) holdsBy Lemma 9 we have

119894119870(1198791198701198771

) = 0 (77)

On the other hand let 120576 gt 0 satisfy 119891infin lt (1120582)(1205831minus 120576) Then

there exists 1198772gt 1198771such that

119891 (119905 119906) le1

120582(1205831 minus 120576) 119906 forall119905 isin [0 1] 119906 ge 1198772 (78)

By (H5) there exists an 119871infin function 1205931such that

119891 (119905 119906) le1

1205821205931 (119905) forall119906 isin [0 1198772] 119905 isin [0 1] (79)

Hence we have

119891 (119905 119906) le1

120582[(1205831 minus 120576) 119906 + 1205931 (119905)]

forall119906 isin 119877+ 119905 isin [0 1]

(80)

Since 11205831is the radius of the spectrum of 119871 (119868(120583

1minus120576)minus119871)

minus1

existsLet

119862 = int

1

0

1205931 (119904) Φ (119904) 119892 (119904) 119889119902119904

1198770= (

119868

(1205831minus 120576)

minus 119871)

minus1

(119888

(1205831minus 120576)

)

(81)

We prove that for each 119877 gt 1198770

119879119906 = 120573119906 forall119906 isin 120597119870119877 120573 ge 1 (82)

In fact if not there exist 119906 isin 120597119870119877and 120573 ge 1 such that 119879119906 =

120573119906This together with (80) implies

119906 (119905) le int

1

0

119866 (119905 119902119904) 119892 (119904) ((1205831 minus 120576) 119906 (119904) + 1205931 (119904)) 119889119902119904

= (1205831minus 120576) int

1

0

119866 (119905 119902119904) 119892 (119904) 119906 (119904) 119889119902119904

+ int

1

0

119866 (119905 119902119904) 119892 (119904) 1205931 (119904) 119889119902119904

le (1205831 minus 120576) 119871119906 (119905) + 119862

(83)

This implies

(119868

1205831minus 120576minus 119871)119906 (119905) le

119862

1205831minus 120576

119906 (119905) le (119868

1205831minus 120576minus 119871)

minus1

(119862

1205831minus 120576)

= 1198770

(84)

Therefore we have 119906 le 1198770lt 119877 a contradiction Taking

119877 gt 1198772 it follows from (74) and properties of index that

119894119870(119879119870119877) = 1 forall119877 gt 119877

0 (85)

Now (77) and (85) combined imply

119894119870(119879119870119877 1198701198771

) = 119894119870(119879119870119877) minus 119894119870(1198791198701198771

) = 1 (86)

Therefore 119879 has at least one fixed point 1199060isin 1198701198771198701198771

and 1199060

is a positive solution of BVP (1)-(2)

5 The Existence of Two Positive Solutions

Theorem 23 Suppose (A2) (A3) and

(A5) 12058211989101205881015840

le 119898 for some 1205881015840 gt 0

Then (1)-(2) had at least two positive solutions

Proof By (A5) we have

119879119906 (119905) = 120582int

1

0

119866 (119905 119902119904) 119892 (119904) 119891 (119904 119906 (119904)) 119889119902119904

le int

1

0

119866 (119905 119902119904) 119892 (119904) 1205881015840119898119889119902119904

(87)

so that 119879119906 le 1205881015840 = 119906 for all 119906 isin 1205971198811205881015840 Now Lemma 8 yields

119894119896(119879 1198811205881015840) = 1 (88)

On the other hand in view of (A2) we may take 120588lowast gt 1205881015840so that (69) holds (see the proof of Theorem 21) From (A3)we may take 119877

1isin (0 120588

1015840) so that (77) holds (see the proof of

Theorem 22)Combining (88) (69) and (77) we arrive at

119894119896 (119879119870120588lowast 1198811205881015840) = 0 minus 1 = minus1

119894119896(119879 1198811205881015840 1198701198771

) = 1 minus 0 = 1

(89)

Consequently 119879 has at least two fixed points with one on119870120588lowast 1198811205881015840 and the other on119881

1205881015840 1198701198771

Therefore (1)-(2) had atleast two positive solutions

Theorem 24 Suppose (A1) (A4) and

(A6) 12058211989112058810158401205881015840119888 ge 119872 for some 1205881015840 gt 0

Then (1)-(2) had at least two positive solutions

Proof By (A6) we have

119879119906 (119905) = 120582int

1

0

119866 (119905 119902119904) 119892 (119904) 119891 (119904 119906 (119904)) 119889119902119904

ge 120582int

119887

119886

119866 (119905 119902119904) 119892 (119904) 119891 (119904 119906 (119904)) 119889119902119904

ge int

119887

119886

119866 (119905 119902119904) 119892 (119904)1198721205881015840119889119902119904

(90)

Abstract and Applied Analysis 9

so that 119879119906 ge 1205881015840 = 119906 for all 119906 isin 1205971198811205881015840 and by Lemma 8

this yields

119894119896 (119879 1198811205881015840) = 0 (91)

On the other hand in view of (A1) we may take 120588 isin (0 1205881015840)so that (64) holds (see the proof of Theorem 21) In additionfrom (A4) we may take 119877 gt 1205881015840 so that (85) holds (see theproof of Theorem 22)

Combining (91) (64) and (85) we arrive at

119894119896(119879119870119877 1198811205881015840) = 1 minus 0 = 1

119894119896(119879 1198811205881015840 119870120588) = 0 minus 1 = minus1

(92)

Hence 119879 has at least two fixed points with one on 1198811205881015840 119870120588

and the other on 119870119877 1198811205881015840 Therefore (1)-(2) had at least two

positive solutions

We illustrate the applicability of these results with someexamples

Example 25 Consider the problem

11986305(25)119906 (119905) + 120582 (5119905 + 3) (

71199062+ 119906

119906 + 1) (2 + cos 119906) = 0

119905 isin (0 1)

1198632

05119906 (0) = 0

119906 (0) = 0

11986305119906 (1) = 0

(93)

Herewe have119892(119905) = 5119905+3119891(119906) = (2+cos 119906)((71199062+119906)(119906+1))and 2 lt 120572 le 3

It is readily shown that 1198910 = 1198910= 3 119891infin = 21 119891

infin= 7

Also 3119906 le 119891(119906) le 21119906 for 119906 ge 0 By calculation wefind119898 = 019722 and the smallest119872 calculated is119872(119886 119887) asymp119872(0484405 1) asymp 074665 We find 120583

1asymp 030366 Hence

by Theorem 21 there is at least one positive solution if 3120582 lt1205831and 7120582 gt 120583

1 that is there is a positive solution if 120582 isin

(047047 109773)

ByTheorem 22 there does not exist a positive solution ifeither 3120582 gt 120583

1or 21120582 lt 120583

1 that is if 120582 lt 109773 or 120582 gt

015682 no positive solution exists

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

References

[1] N Pongarm S Asawasamrit J Tariboon and S K NtouyasldquoMulti-strip fractional q-integral boundary value problemsfor nonlinear fractional q-difference equationsrdquo Advances inDifference Equations vol 2014 no 1 article 193 2014

[2] R Almeida and N Martins ldquoExistence results for fractional q-difference equations of order120572 isin ]2 3[with three-point bound-ary conditionsrdquo Communications in Nonlinear Science andNumerical Simulation vol 19 no 6 pp 1675ndash1685 2014

[3] Y ZhaoHChen andQZhang ldquoExistence results for fractionalq-difference equations with nonlocal q-integral boundary con-ditionsrdquo Advances in Difference Equations vol 2013 article 482013

[4] R A Ferreira ldquoPositive solutions of a nonlinear q-fractionaldifference equation with integral boundary conditionsrdquo Inter-national Journal of Difference Equations vol 9 no 2 pp 135ndash145 2014

[5] B Ahmad and S K Ntouyas ldquoFractional 119902-difference hybridequations and inclusions with Dirichlet boundary conditionsrdquoAdvances in Difference Equations vol 2014 article 199 2014

[6] B Ahmad S KNtouyas andAAlsaedi ldquoExistence of solutionsfor fractional q-integro-difference inclusions with fractional q-integral boundary conditionsrdquo Advances in Difference Equa-tions vol 2014 article 257 2014

[7] R P Agarwal B Ahmad A Alsaedi and H Al-Hutami ldquoOnnonlinear fractional q-difference equations involving two frac-tional orders with three-point nonlocal boundary conditionsrdquoDynamics of Continuous Discrete amp Impulsive Systems Series AMathematical Analysis vol 21 no 1 pp 135ndash151 2014

[8] M El-Shahed andM A Al-Yami ldquoPositive solutions of bound-ary value problems for nth order q-differential equationsrdquoInternational Journal of Mathematical Archive vol 2 pp 521ndash532 2011

[9] J Ma and J Yang ldquoExistence of solutions for multi-pointboundary value problem of fractional q-difference equationrdquoElectronic Journal ofQualitativeTheory ofDifferential Equationsvol 92 pp 1ndash10 2011

[10] P M Rajkovic S D Marinkovic and M S Stankovic ldquoFrac-tional integrals and derivatives in q-calculusrdquo Applicable Anal-ysis and Discrete Mathematics vol 1 no 1 pp 311ndash323 2007

[11] Y Zhao H Chen and B Qin ldquoMultiple solutions for acoupled systemof nonlinear fractional differential equations viavariational methodsrdquo Applied Mathematics and Computationvol 257 pp 417ndash427 2015

[12] W-X Zhou X Liu and J-G Zhang ldquoSome new existenceand uniqueness results of solutions to semilinear impulsivefractional integro-differential equationsrdquoAdvances inDifferenceEquations vol 2015 article 38 2015

[13] Q Yuan andW Yang ldquoPositive solutions of nonlinear boundaryvalue problems for delayed fractional q-difference systemsrdquoAdvances in Difference Equations vol 2014 no 1 article 51 16pages 2014

[14] W Yang ldquoPositive solutions for nonlinear semipositone frac-tional q-difference system with coupled integral boundaryconditionsrdquo Applied Mathematics and Computation vol 244pp 702ndash725 2014

[15] J R L Webb and K Q Lan ldquoEigenvalue criteria for existenceof multiple positive solutions of nonlinear boundary valueproblems of local and nonlocal typerdquo Topological Methods inNonlinear Analysis vol 27 no 1 pp 91ndash116 2006

[16] J R L Webb ldquoNonlocal conjugate type boundary value prob-lems of higher orderrdquo Nonlinear Analysis Theory Methods ampApplications vol 71 no 5-6 pp 1933ndash1940 2009

[17] J R L Webb and G Infante ldquoNon-local boundary value prob-lems of arbitrary orderrdquo Journal of the London MathematicalSociety vol 79 no 1 pp 238ndash258 2009

10 Abstract and Applied Analysis

[18] J R Webb and G Infante ldquoPositive solutions of nonlocalboundary value problems involving integral conditionsrdquo Non-linear Differential Equations and Applications NoDEA vol 15no 1-2 pp 45ndash67 2008

[19] J R L Webb and G Infante ldquoPositive solutions of nonlocalboundary value problems a unified approachrdquo Journal of theLondon Mathematical Society vol 74 no 3 pp 673ndash693 2006

[20] V Kac and P Cheung Quantum Calculus Springer Science ampBusiness Media 2002

[21] R A C Ferreira ldquoNontrivial solutions for fractional q-difference boundary value problemsrdquo Electronic Journal ofQualitative Theory of Differential Equations no 70 pp 1ndash102010

[22] M S Stankovic P M Rajkovic and S D Marinkovic ldquoOn q-fractional derivatives of Riemann-Liouville and Caputo typerdquohttparxivorgabs09090387

[23] M H Annaby and Z S Mansour q-Fractional Calculus andEquations vol 2056 of Lecture Notes in Mathematics SpringerBerlin Germany 2012

[24] M El-Shahed and W M Shammakh ldquoMultiple positive solu-tions for nonlinear fractional eigenvalue problemwith non localconditionsrdquo Fractional Calculus and Applied Analysis vol 3 pp1ndash13 2012

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

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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

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Mathematical PhysicsAdvances in

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Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

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CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

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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

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Page 2: Research Article Positive Solutions for Nonlinear -Fractional …downloads.hindawi.com/journals/aaa/2015/759378.pdf · 2019. 7. 31. · Research Article Positive Solutions for Nonlinear

2 Abstract and Applied Analysis

The 119902-analogue of the power function (119886 minus 119887)119899 with 119899 isin N is

(119886 minus 119887)0= 1

(119886 minus 119887)(119899)=

119899minus1

prod

119896=0

(119886 minus 119887119902119896) 119886 119887 isin R 119899 isin N

(4)

More generally if 120572 isin R then

(119886 minus 119887)(120572)= 119886120572

infin

prod

119894=0

(119886 minus 119887119902119894)

(119886 minus 119887119902120572+119894) (5)

Note that if 119887 = 0 then 119886(120572) = 119886120572The 119902-gamma function isdefined by

Γ119902 (119909) =

(1 minus 119902)(119909minus1)

(1 minus 119902)119909minus1

119909 isin R 0 minus1 minus2 0 lt 119902 lt 1

(6)

and satisfies Γ119902(119909 + 1) = [119909]

119902Γ119902(119909)

The 119902-derivative of a function 119891(119909) is here defined by

119863119902119891 (119909) =

119889119902119891 (119909)

119889119902119909=119891 (119902119909) minus 119891 (119909)

(119902 minus 1) 119909 (7)

and 119902-derivatives of higher order are defined by

119863119899

119902119891 (119909) =

119891 (119909) if 119899 = 0

119863119902119863119899minus1

119902119891 (119909) if 119899 isin N

(8)

The 119902-integral of a function 119891 defined in the interval [0 119887] isgiven by

int

119909

0

119891 (119905) 119889119902119905 = 119909 (1 minus 119902)

infin

sum

119899=0

119891 (119909119902119899) 119902119899

0 le10038161003816100381610038161199021003816100381610038161003816 lt 1 119909 isin [0 119887]

(9)

If 119886 isin [0 119887] and 119891 is defined in the interval [0 119887] its integralfrom 119886 to 119887 is defined by

int

119887

119886

119891 (119905) 119889119902119905 = int

119887

0

119891 (119905) 119889119902119905 minus int

119886

0

119891 (119905) 119889119902119905 (10)

Similarly as done for derivatives it can be defined an operator119868119899

119902 namely

(1198680

119902119891) (119909) = 119891 (119909)

(119868119899

119902119891) (119909) = 119868119902 (119868

119899minus1

119902119891) (119909) 119899 isin N

(11)

The fundamental theorem of calculus applies to these opera-tors 119868119902and119863

119902 that is

(119863119902119868119902119891) (119909) = 119891 (119909) (12)

and if 119891 is continuous at 119909 = 0 then

(119868119902119863119902119891) (119909) = 119891 (119909) minus 119891 (0) (13)

Basic properties of the two operators can be found in the book[20] We now point out four formulas that will be used later

[119886 (119905 minus 119904)](120572)= 119886120572(119905 minus 119904)

(120572)

119905119863119902 (119905 minus 119904)

(120572)= [120572]119902 (119905 minus 119904)

(120572minus1)

119904119863119902 (119905 minus 119904)

(120572)= minus [120572]119902 (119905 minus 119902119904)

(120572minus1)

(119909119863119902 int

119909

0

119891 (119909 119905) 119889119902119905) (119909)

= int

119909

0119909119863119902 119891 (119909 119905) 119889119902119905 + 119891 (119902119909 119909)

(14)

where119894119863119902denotes the 119902-derivative with respect to variable 119894

[21]

Remark 1 (see [21]) We note that if 120572 gt 0 and 119886 le 119887 le 119905 then(119905 minus 119886)

(120572)ge (119905 minus 119887)

(120572)

Definition 2 (see [22]) Let 120572 ge 0 and let 119891 be a functiondefined on [0 1]The fractional 119902-integral of the Riemann-Liouville type is (

RL1198680

119902119891)(119909) = 119891(119909) and

(RL119868120572

119902119891) (119909) =

1

Γ119902 (120572)

int

119909

0

(119909 minus 119902119905)(120572minus1)

119891 (119905) 119889119902119905

120572 isin R+ 119909 isin [0 1]

(15)

Definition 3 (see [22]) The fractional 119902-derivative of theRiemann-Liouville type of order 120572 ge 0 is defined by(RL1198630

119902119891)(119909) = 119891(119909) and

(RL119863120572

119902119891) (119909) = (119863

[120572]

119902119868[120572]minus120572

119902119891) (119909) 120572 gt 0 (16)

where [120572] is the smallest integer greater than or equal to 120572

Definition 4 (see [22]) The fractional 119902-derivative of theCaputo type of order 120572 ge 0 is defined by

(119862119863120572

119902119891) (119909) = (119868

[120572]minus120572

119902119863[120572]

119902119891) (119909) 120572 gt 0 (17)

Lemma 5 (see [22]) Let 120572 120573 ge 0 and let 119891 be a functiondefined on [0 1]Then the next formulas hold

(1) (119868120573119902119868120572

119902119891)(119909) = (119868

120572+120573

119902119891)(119909)

(2) (119862119863120572

119902119868120572

119902119891)(119909) = 119891(119909)

Abstract and Applied Analysis 3

Lemma 6 (see [22]) Let 120572 isin R+ N 120582 isin (minus1infin) Then thenext formulas hold

(1) 119868120572

119902119909120582=

Γ119902 (120582 + 1)

Γ119902 (120582 + 1 + 120572)

119909120582+120572

(2)119877119871119863120572

119902119909120582=

Γ119902 (120582 + 1)

Γ119902 (120582 + 120572 + 1)119909120582minus120572

(3)119862119863120572

119902119909120582=

0 119894119891 120582 isin N0 120572 gt 120582

119877119871119863120572

119902119909120582119894119891 119900119905ℎ119890119903119908119894119904119890

(18)

Theorem 7 (see [23]) Let 119909 gt 0 and 120572 isin R+ NThen thefollowing equality holds

(119868120572

119902 119862119863120572

119902119891) (119909) = 119891 (119909) minus

[120572]minus1

sum

119896=0

119909119896

Γ119902 (119896 + 1)

(119863119896

119902119891) (0) (19)

Lemma 8 (see [24]) Suppose 119879 119870 rarr 119870 is a completelycontinuous operator and has no fixed points on 120597119870

120588cap119870 Then

the following are true(i) If 119879119906 le 119906 for all 119906 isin 120597119870

120588cap 119870 then 119894(119879 119870

120588cap

119870119870) = 1 where 119894 is the fixed point index on 119870(ii) If 119879119906 ge 119906 for all 119906 isin 120597119870

120588cap 119870 then 119894(119879 119870

120588cap

119870119870) = 0

Lemma 9 (see [24]) Let 119870 be a cone in Banach space 119864Suppose that119879 119870

120588rarr 119870 is a completely continuous operator

There exists 1199060isin 1198700 such that 119906minus119879119906 = 120583119906

0for any 119906 isin 120597119870

119903

and 120583 ge 0 119894(119879 119870120588 119870) = 0

Lemma 10 (see [24]) Let 119870 be a cone in Banach space 119864Suppose that119879 119870

120588rarr 119870 is a completely continuous operator

If 119879119906 = 120583119906 for any 119906 isin 120597119870119903and 120583 ge 1 then 119894(119879 119870

120588 119870) = 1

Lemma 11 Let 119910 isin 119862[0 1] be a given function and 119899 minus 1 lt120572 le 119899 then 119906 is a solution of BVP (1)-(2) if and only if 119906 is asolution of the integral equation

119906 (119905) = 120574 (119905) 120579 [119906] + int

1

0

1198660 (119905 119902119904) 119910 (119904) 119889119902119904 (20)

where120574 (119905) = 119905

1198660(119905 119902119904)

=

[120572 minus 1]119902 119905 (1 minus 119902119904)(120572minus2)

minus (119905 minus 119902119904)(120572minus1)

Γ119902 (120572)

0 le 119902119904 le 119905 le 1

[120572 minus 1]119902 119905 (1 minus 119902119904)(120572minus2)

Γ119902 (120572) 0 le 119905 le 119902119904 le 1

(21)

Proof Assume that 119906 is a solution of BVP (1)-(2)Applying Theorem 7 (1) can be reduced to an equivalent

integral equation

119906 (119905) = minus1

Γ119902 (120572)

int

119905

0

(119905 minus 119902119904)(120572minus1)

119910 (119904) 119889119902119904 + 1198880 + 1198881119905

+ 11988821199052+ sdot sdot sdot + 119888

119899minus1119905119899minus1

(22)

By (2) we obtain

1198880= 0

1198882 = sdot sdot sdot = 119888119899minus1 = 0

1198881= 120579 [119906] +

[120572 minus 1]119902

Γ119902 (120572)

int

1

0

(1 minus 119902119904)(120572minus2)

119910 (119904) 119889119902119904

(23)

Therefore we obtain

119906 (119905) = 120574 (119905) 120579 [119906] +119905 [120572 minus 1]119902

Γ119902 (120572)

int

1

0

(1 minus 119902119904)(120572minus2)

sdot 119910 (119904) 119889119902119904 minus1

Γ119902 (120572)

int

119905

0

(119905 minus 119902119904)(120572minus1)

119910 (119904) 119889119902119904

= 120574 (119905) 120579 [119906]

+ int

119905

0

[

[

[120572 minus 1]119902 119905 (1 minus 119902119904)(120572minus2)

minus (119905 minus 119902119904)(120572minus1)

Γ119902 (120572)

]

]

sdot 119910 (119904) 119889119902119904 + int

1

119905

[

[

[120572 minus 1]119902 119905 (1 minus 119902119904)(120572minus2)

Γ119902 (120572)

]

]

sdot 119910 (119904) 119889119902119904 = 120574 (119905) 120579 [119906] + int

1

0

1198660 (119905 119902119904) 119910 (119904) 119889119902119904

(24)

Conversely if 119906 is a solution of the integral equation (20)using Lemmas 5 and 6 we have

119862119863120572

119902119906 (119905)

=119862119863120572

119902119905120579 [119906]

+119862119863120572

119902119905(int

1

0

[120572 minus 1]119902 (1 minus 119902119904)(120572minus2)

Γ119902 (120572)

119910 (119904) 119889119902119904)

minus119862119863120572

119902(int

119905

0

(119905 minus 119902119904)(120572minus1)

Γ119902 (120572)

119910 (119904) 119889119902119904)

= minus119862119863120572

119902119868120572

119902119910 (119905) = minus119910 (119905)

(25)

A simple computation shows 119906(0) = 0119863119896119902119906(0) = 0 2 le 119896 le

119899 minus 1119863119902119906(1) = 120579[119906]

Remark 12 1198660(119905 119902119904) is Greenrsquos function for the local BVP

119862119863120572

119902119906 (119905) + 120582119892 (119905) 119891 (119905 119906 (119905)) = 0

119905 isin [0 1] 119899 minus 1 lt 120572 le 119899 0 lt 119902 lt 1

119863119896

119902119906 (0) = 0

119906 (0) = 0 2 le 119896 le 119899 minus 1

119863119902119906 (1) = 0

(26)

4 Abstract and Applied Analysis

Lemma 13 Function 1198660(119905 119902119904) defined in (20) satisfies the

following conditions

(H1) 1198660(119905 119902119904) ge 0 is continuous and 119866

0(119905 119904) le Φ

0(119902119904) for

all 0 le 119905 119904 le 1(H2) 119866

0(119905 119902119904) ge 119888

0(119905)Φ0(119902119904) for all 0 le 119905 119904 le 1 where

Φ0(119902119904) = 119866

0(1 119902119904)

=[120572 minus 1]119902 (1 minus 119902119904)

(120572minus2)minus (1 minus 119902119904)

(120572minus1)

Γ119902 (120572)

1198880 (119905) = 119905

120572minus1

(27)

Proof It is obvious that1198660(119905 119902119904) is nonnegative and continu-

ous(H1) For 0 le 119902119904 le 119905 le 1

1198660(119905 119902119904) =

1

Γ119902 (120572)[[120572 minus 1]119902 119905 (1 minus 119902119904)

(120572minus2)

minus (119905 minus 119902119904)(120572minus1)

] =1

Γ119902 (120572)

[[120572 minus 1]119902 119905 (1 minus 119902119904)(120572minus2)

minus 119905120572minus1(1 minus 119902

119904

119905)

(120572minus1)

]

ge1

Γ119902 (120572)[[120572 minus 1]119902 119905

120572minus1(1 minus 119902119904)

(120572minus2)

minus 119905120572minus1(1 minus 119902119904)

(120572minus1)]

=119905120572minus1

Γ119902 (120572)

[[120572 minus 1]119902 (1 minus 119902119904)(120572minus2)

minus (1 minus 119902119904)(120572minus1)

]

ge 0

(28)

and for 0 le 119905 le 119902119904 le 1

1198660 (119905 119902119904) =

[120572 minus 1]119902 119905 (1 minus 119902119904)(120572minus2)

Γ119902 (120572)

(29)

and it is clear that 1198660(119905 119902119904) ge 0 and 119866

0(0 119902119904) = 0Therefore

1198660(119905 119902119904) ge 0

For fixed 119904 isin [0 1] and 119905 ge 119902119904 we have

119905119863119902 1198660 (119905 119902119904)

=[120572 minus 1]119902 (1 minus 119902119904)

(120572minus2)minus [120572 minus 1]119902 (119905 minus 119902119904)

(120572minus2)

Γ119902 (120572)

=(1 minus 119902119904)

(120572minus2)minus (119905 minus 119902119904)

(120572minus2)

Γ119902 (120572 minus 1)

ge 0

(30)

that is 1198660(119905 119902119904) is an increasing function of 119905 Obviously

1198660(119905 119902119904) 119905 le 119902119904 is increasing in 119905 therefore 119866

0(119905 119902119904) is an

increasing function of 119905 for fixed 119904 isin [0 1]Thus (H1) holds

(H2) Suppose now that 119905 ge 119902119904

1198660(119905 119902119904)

Φ0 (119902119904)=[120572 minus 1]119902 119905 (1 minus 119902119904)

(120572minus2)minus (119905 minus 119902119904)

(120572minus1)

[120572 minus 1]119902 (1 minus 119902119904)(120572minus2)

minus (1 minus 119902119904)(120572minus1)

ge

119905120572minus1[[120572 minus 1]119902 (1 minus 119902119904)

(120572minus2)minus (1 minus 119902119904)

(120572minus1)]

[120572 minus 1]119902 (1 minus 119902119904)(120572minus2)

minus (1 minus 119902119904)(120572minus1)

= 119905120572minus1

(31)

On the other hand if 119905 le 119902119904 then we have

1198660(119905 119902119904)

Φ0(119902119904)

=[120572 minus 1]119902 119905 (1 minus 119902119904)

(120572minus2)

[120572 minus 1]119902 (1 minus 119902119904)(120572minus2)

minus (1 minus 119902119904)(120572minus1)

ge[120572 minus 1]119902 119905

120572minus1(1 minus 119902119904)

(120572minus2)minus 119905120572minus1(1 minus 119902119904)

(120572minus1)

[120572 minus 1]119902 (1 minus 119902119904)(120572minus2)

minus (1 minus 119902119904)(120572minus1)

= 119905120572minus1

(32)

and this finished the proof of (H2)

Defining G119860(119902119904) = int

1

01198660(119905 119902119904)119889

119902119860(119905) Greenrsquos function

for nonlocal BVP (1)-(2) is given by

119866 (119905 119902119904) =120574 (119905)

[1 minus 120579 [120574]]G119860(119902119904) + 119866

0(119905 119902119904) (33)

Throughout the paper we assume the following(H3) A is a function of bounded variation andG119860(119902119904) =

int1

01198660(119905 119902119904)119889

119902119860(119905) satisfies G

119860(119902119904) ge 0 for almost every 119904 isin

[0 1] Note thatG119860(119902119904) exists for almost every 119904 by (H1)

(H4)The functions 119892Φ satisfy 119892 ge 0 almost everywhere119892Φ isin 119871

1[0 1] and

int

119887

119886

Φ(119902119904) 119892 (119904) 119889119902119904 gt 0 (34)

(H5) 119891 [0 1] times [0infin) rarr [0infin) satisfies Caratheodoryconditions that is 119891(sdot 119906) is measurable for each fixed 119906 isin[0infin) and 119891(119905 sdot) is continuous for almost every 119905 isin [0 1]and for each 119903 gt 0 there exists 120601

119903isin 119871infin[0 1] such that 0 le

119891(119905 119906) le 120601119903for all 119906 isin [0 119903] and almost all 119905 isin [0 1]

(H6) One has the following 120574 isin 119862[0 1] 120574(119905) ge 0 0 le120579[120574] lt 1

Lemma 14 If 1198660satisfies (H1) (H2) then 119866 satisfies (H1)

(H2) for a function Φ the same interval [119886 119887] and the sameconstant 119888 where Φ satisfies (H4) and 119888 = min1198880(119905) 119905 isin[119886 119887]

Proof We have

119866 (119905 119902119904) =120574 (119905)

[1 minus 120579 [120574]]G119860(119902119904) + 119866

0(119905 119902119904)

le

10038171003817100381710038171205741003817100381710038171003817

[1 minus 120579 [120574]]G119860(119902119904) + Φ

0(119902119904) = Φ (119902119904)

(35)

Abstract and Applied Analysis 5

and for 119905 isin [119886 119887]

119866 (119905 119902119904) ge11988810038171003817100381710038171205741003817100381710038171003817

[1 minus 120579 [120574]]G119860 (119902119904) + 119888Φ0 (119902119904) = 119888Φ (119902119904) (36)

Note that 119892Φ isin 119871infin because 119860 has finite variation and

G119860(119902119904) le Φ(119902119904) var(119860)Thus Greenrsquos function 119866(119905 119902119904) satisfies (H1) (H2) for a

functionΦ and the constant 119888

3 Main Result

Set 119864 = 119862[0 1] as a Banach space with the norm 119906 =

sup119905isin[01]

|119906(119905)| Let 119875 = 119906 isin 119864 119906 ge 0 denote the standardcone of nonnegative functions Define

119870 = 119906 isin 119875 min119886le119905le119887

119906 (119905) ge 119888 119906 (37)

where [119886 119887] is some subset of [0 1]Note that 120574 isin 119870 so 119870 = 0 For any 0 lt 119903 lt 119877 lt +infin

let 119870119903= 119906 isin 119870 119906 lt 119903 120597119870

119903= 119906 isin 119870 119906 = 119903

119870119903 = 119906 isin 119870 119906 le 119903 119870119877 119870119903 = 119906 isin 119870 119903 le 119906 le 119877and 119881119903 = 119906 isin 119870 min119905isin[119886119887]119906(119905) lt 119903 and 119881119903 is bounded

Define a nonlinear operator 119879 119875 rarr 119870 and a linearoperator 119871 119875 rarr 119870 by

119879119906 (119905) = 120582int

1

0

119866 (119905 119902119904) 119892 (119904) 119891 (119904 119906 (119904)) 119889119902119904 (38)

119871119906 (119905) fl int1

0

119866 (119905 119902119904) 119892 (119904) 119906 (119904) 119889119902119904 (39)

Lemma 15 (see [18]) Under hypotheses (H1)ndash(H6) the maps119879 119875 rarr 119864 defined in (38) are compact

Theorem 16 Under hypotheses (H1)ndash(H6) the maps are 119879 119875 rarr 119870

Proof For 119906 isin 119875 and 119905 isin [0 1] we have

119879119906 (119905) le 120582int

1

0

Φ(119902119904) 119892 (119904) 119891 (119904 119906 (119904)) 119889119902119904 (40)

Hence

119879119906 le 120582int

1

0

Φ(119902119904) 119892 (119904) 119891 (119904 119906 (119904)) 119889119902119904 (41)

Also for 119905 isin [119886 119887] we have

119879119906 (119905) ge 119888120582int

1

0

Φ(119902119904) 119892 (119904) 119891 (119904 119906 (119904)) 119889119902119904 ge 119888 119879119906 (42)

Similar to the proofs of Lemma 15 and Theorem 16 119871119906(119905) iscompact and maps 119875 into119870

We will use the Krein-Rutman theorem We recall that 120582is an eigenvalue of 119871 with corresponding eigenfunction 120601 if120601 = 0 and 120582120601 = 119871120601 The reciprocals of eigenvalues are calledcharacteristic values of 119871 The radius of the spectrum of 119871denoted by 119903(119871) is given by the well-known spectral radiusformula 119903(119871) = lim

119899rarrinfin1198711198991119899

Theorem 17 (see [15]) Let 119870 be a total cone in a real Banachspace 119864 and let 119864 rarr 119864 be a compact linear operator with(119870) sube 119870 If 119903() gt 0 then there is 120601

1isin 119870 0 such that

1206011= 119903()120601

1

Thus 1205821 fl 119903() is an eigenvalue of the largest possiblereal eigenvalue and 1205831 = 1120582

1is the smallest positive

characteristic value

Lemma 18 (see [15]) Assume that (H1)ndash(H3) hold and let 119871be as defined in (39) Then 119903(119871) gt 0

Theorem 19 (see [15]) When (H1)ndash(H3) hold 119903(119871) is aneigenvalue of 119871 with eigenfunction 120601

1in 119870

Theorem 20 (see [15]) Let 1205831= 1119903(119871) and 120601

1(119905) be a

corresponding eigenfunction in119875 of norm 1Then119898 le 1205831le 119872

where

119898 = ( sup119905isin[01]

int

1

0

119866 (119905 119902119904) 119892 (119904) 119889119902119904)

minus1

119872 = ( inf119905isin[119886119887]

int

119887

119886

119866 (119905 119902119904) 119892 (119904) 119889119902119904)

minus1

(43)

If 119892(119905) gt 0 for 119905 isin [0 1] and 119866(119905 119902119904) gt 0 for 119905 119904 isin [0 1] thefirst inequality is strict unless 1206011(119905) is constant for 119905 isin [0 1] If119892(119905)120601(119905) gt 0 for 119905 isin [119886 119887] the second inequality is strict unless1206011(119905) is constant for 119905 isin [119886 119887]

Proof (for the local BVP (1)-(2) if 119892(119905) equiv 1) We now computethe constant 119898 and the optimal value of119872(119886 119887) that is wedetermine 119886 119887 so that119872(119886 119887) is minimal

For 119902119904 le 119905 we have by direct integration

int

119905

0

1198660 (119905 119902119904) 119889119902119904

= int

119905

0

[

[

[120572 minus 1]119902 119905 (1 minus 119902119904)(120572minus2)

minus (119905 minus 119902119904)(120572minus1)

Γ119902 (120572)

]

]

119889119902119904

=119905 minus 119905 (1 minus 119905)

(120572minus1)

Γ119902 (120572)

minus119905120572

[120572]119902 Γ119902 (120572)

(44)

For 119902119904 ge 119905

int

1

119905

1198660(119905 119902119904) 119889

119902119904 = int

1

119905

[120572 minus 1]119902 119905 (1 minus 119902119904)(120572minus2)

Γ119902 (120572)119889119902119904

=119905 (1 minus 119905)

(120572minus1)

Γ119902 (120572)

(45)

Then we have

int

1

0

1198660 (119905 119902119904) 119889119902119904 =119905

Γ119902 (120572)

minus119905120572

[120572]119902 Γ119902 (120572) (46)

6 Abstract and Applied Analysis

And the maximum of this expression occurs when 119905 = 1hence

sup119905isin[01]

int

1

0

1198660 (119905 119902119904) 119889119902119904 =1

Γ119902 (120572)

minus1

[120572]119902 Γ119902 (120572)

=[120572]119902 minus 1

[120572]119902 Γ119902 (120572)

(47)

Then119898 = [120572]119902Γ119902(120572)([120572]

119902minus 1)

For 119886 lt 119887 we have by direct integration

int

119905

119886

1198660(119905 119902119904) 119889

119902119904 = minus

119905 (1 minus 119905)(120572minus1)

Γ119902 (120572)

+119905 (1 minus 119886)

(120572minus1)

Γ119902 (120572)

minus(119905 minus 119886)

(120572)

[120572]119902 Γ119902 (120572)

int

119887

119905

1198660 (119905 119904) 119889119904 = minus

119905 (1 minus 119887)(120572minus1)

Γ119902 (120572)+119905 (1 minus 119905)

(120572minus1)

Γ119902 (120572)

(48)

Then

int

119887

119886

1198660(119905 119902119904) 119889

119902119904 =119905 (1 minus 119886)

(120572minus1)

Γ119902 (120572)minus(119905 minus 119886)

(120572)

[120572]119902 Γ119902 (120572)

minus119905 (1 minus 119887)

(120572minus1)

Γ119902 (120572)

=119905 [(1 minus 119886)

(120572minus1)minus (1 minus 119887)

(120572minus1)]

Γ119902 (120572)

minus(119905 minus 119886)

(120572)

[120572]119902 Γ119902 (120572)= 119877 (119905 119886 119887)

119905119863119902119877 (119905 119886 119887) =

[(1 minus 119886)(120572minus1)

minus (1 minus 119887)(120572minus1)

]

Γ119902 (120572)

minus(119905 minus 119886)

(120572minus1)

Γ119902 (120572)

(49)

The sign of derivative119905119863119902119877 shows that this is an increasing

function of 119905 so the minimum occurs at 119905 = 119886 Let

119877 (119886 119887) =119886

Γ119902 (120572)

[(1 minus 119886)(120572minus1)

minus (1 minus 119887)(120572minus1)

] (50)

The minimal value of 119872(119886 119887) corresponds to the maximalvalue of 119877(119886 119887) Consider

119887119863119902119877 (119886 119887) =

119886 [120572 minus 1]119902 (1 minus 119902119887)(120572minus2)

Γ119902 (120572)gt 0 (51)

The quantity 119877(119886 119887) is an increasing function of 119887 so itsmaximum occurs when 119887 = 1 Let

119877 (119886) =119886 (1 minus 119886)

(120572minus1)

Γ119902 (120572)

(52)

Then the maximum of 119877(119886) occurs when 119886 = 1(1+ [120572minus1]119902)

Consider

min119905isin[119886119887]

int

119887

119886

1198660(119905 119902119904) 119889

119902119904 = 119877(

1

1 + [120572 minus 1]119902

1) (53)

Hence the minimal value of119872(119886 119887) is

119872(1

1 + [120572 minus 1]119902

1) = (119877(1

1 + [120572 minus 1]119902

1))

minus1

(54)

4 The Existence of at Least OnePositive Solution

For convenience we introduce the following notations

119891 (119906) fl sup119905isin[01]

119891 (119905 119906)

119891 (119906) fl inf119905isin[01]

119891 (119905 119906)

1198910 fl lim sup119906rarr0

+

119891 (119906)

119906

1198910fl lim inf119906rarr0

+

119891 (119906)

119906

119891infin fl lim sup

119906rarrinfin

119891 (119906)

119906

119891infin

fl lim inf119906rarrinfin

119891 (119906)

119906

1198910119903 fl sup0le119905le1 0le119906le119903

119891 (119905 119906)

119903

119891119903119903119888

fl inf119886le119905le119887 119903le119906le119903119888

119891 (119905 119906)

119903

(55)

Under hypotheses (H1)ndash(H4) let be defined by

119906 (119905) = int

119887

119886

119866 (119905 119902119904) 119892 (119904) 119906 (119904) 119889119902119904 (56)

Then is a compact linear operator and (119875) sube 119870Hence 119903() is an eigenvalue of with an eigenfunction 120601

1

in119870 Let 1205831fl 1119903()Note that 120583

1ge 1205831 hence the condition

in the following theorem ismore stringent comparedwith thecase if 119903(119871) could be used

Theorem 21 Assume that

(A1) 0 le 1205821198910 lt 1205831(A2) 120583

1lt 120582119891infinle infin

Then (1)-(2) had at least one positive solution

Abstract and Applied Analysis 7

Proof Let 120576 gt 0 be such that 1198910 le (1120582)(1205831minus 120576)Then there

exists 1205880gt 0 such that

119891 (119905 119906) le1

120582(1205831 minus 120576) 119906

forall119906 isin [0 1205880] and almost all 119905 isin [0 1]

(57)

Let 120588 isin (0 1205880]We prove that

119879119906 = 120573119906 for 119906 isin 120597119870120588 120573 ge 1 (58)

which implies the result In fact if (58) does not hold thenthere exist 119906 isin 120597119870

120588and 120573 ge 1 such that 119879119906 = 120573119906

This implies

120573119906 (119905) = 120582int

1

0

119866 (119905 119902119904) 119892 (119904) 119891 (119904 119906 (119904)) 119889119902119904

le (1205831minus 120576) int

1

0

119866 (119905 119902119904) 119892 (119904) 119906 (119904) 119889119902119904

= (1205831minus 120576) 119871119906 (119905)

(59)

Thus we have shown 119906(119905) le (1205831 minus 120576)119871119906(119905)This gives

119906 (119905) le (1205831 minus 120576) 119871 [(1205831 minus 120576) 119871119906 (119905)]

= (1205831minus 120576)21198712119906 (119905)

(60)

And by iterating

119906 (119905) le (1205831 minus 120576)119899119871119899119906 (119905) for 119899 isin 119873 (61)

Therefore

119906 le (1205831 minus 120576)119899 10038171003817100381710038171198711198991003817100381710038171003817 119906

1 le (1205831 minus 120576)119899 10038171003817100381710038171198711198991003817100381710038171003817

(62)

and we have

1 le (1205831 minus 120576) lim

119899rarr+infin

100381710038171003817100381711987111989910038171003817100381710038171119899= (1205831 minus 120576)

1

1205831

lt 1 (63)

a contradiction It follows that

119894119896 (119879119870120588) = 1 for each 120588 isin (0 1205880] (64)

Let 1205881gt 0 120588

1gt 120588 be chosen so that 119891(119905 119906) gt (120583

1120582)119906 for all

119906 ge 1198881205881 119888 as in (H2) and almost all 119905 isin [0 1]

We claim that 119906 = 119879119906 + 1205731206011 for all 120573 gt 0 and 119906 isin 120597119870120588lowast

when 120588lowast gt 1205881 Note that 119906 isin 119870 with 119906 = 120588lowast ge 120588

1

We have 119906(119905) ge 1198881205881for all 119905 isin [119886 119887]

Now if our claim is false then we have

119906 (119905) = 120582int

1

0

119866 (119905 119902119904) 119892 (119904) 119891 (119904 119906 (119904)) 119889119902119904 + 1205731206011 (119905) (65)

Therefore

119906 (119905) ge 1205831 int

119887

119886

119866 (119905 119902119904) 119892 (119904) 119906 (119904) 119889119902119904 + 1205731206011 (119905)

= 1205831119906 (119905) + 1205731206011 (119905)

(66)

From (66) we firstly deduce that 119906(119905) ge 1205731206011(119905) on [0 1]

Then we have

1205831119906 (119905) ge 1205831 (1205731206011 (119905)) = 1205731206011 (119905) (67)

Inserting this into (66) we obtain 119906(119905) ge 21205731206011(119905) for 119905 isin [0 1]

Repeating this process gives

119906 (119905) ge 1198991205731206011 (119905) for 119905 isin [0 1] 119899 isin 119873 (68)

Since 1206011(119905) is strictly positive on [0 1] this is a contradiction

then

119894119870(119879119870120588lowast) = 0 for 119906 isin 120597119870

120588lowast (69)

By (64) and (69) one has

119894119870(119879119870120588lowast 119870120588) = 119894119870(119879119870120588lowast) minus 119894119870(119879119870120588) = minus1 (70)

Therefore 119879 has at least one fixed point 1199060isin 119870120588lowast 119870120588 and

1199060is a positive solution of BVP (1)-(2)

Theorem 22 Assume that

(A3) 1205831 lt 1205821198910 le infin(A4) 0 le 120582119891infin lt 1205831

Then (1)-(2) had at least one positive solution

Proof Let 120576 gt 0 satisfy 1198910gt (1120582)(120583

1+ 120576)Then there exists

1198771 gt 0 such that

119891 (119905 119906) ge1

120582(1205831+ 120576) 119906 forall119905 isin [0 1] 119906 isin [0 1198771] (71)

For any 119906 isin 1205971198701198771

we have by (71) that

119879119906 (119905) = 120582int

1

0

119866 (119905 119902119904) 119892 (119904) 119891 (119904 119906 (119904)) 119889119902119904

ge (1205831 + 120576) int

1

0

119866 (119905 119902119904) 119892 (119904) 119906 (119904) 119889119902119904

ge 1205831119871119906 (119905) forall119905 isin [0 1]

(72)

Let 1 be the positive eigenfunction of 119871 corresponding to 1205831

that is 1 = 12058311198711 Wemay suppose that119879 has no fixed pointon 120597119870119877

1

otherwise the proof is finished In the following wewill show that

119906 minus 119879119906 = 1205731 forall119906 isin 1205971198701198771

120573 ge 0 (73)

If (73) is not true then there is 0isin 1205971198701198771

and 1205730ge 0 such that

0minus1198790= 12057301 It is clear that 120573

0gt 0 and

0= 1198790+12057301ge

12057301Set

120573lowast= sup 120573 0 ge 1205731 (74)

Obviously 120573lowast ge 1205730gt 0 It follows from 119871(119875) sub 119875 that

12058311198710ge 1205831119871120573lowast1= 120573lowast12058311198711= 120573lowast1 (75)

8 Abstract and Applied Analysis

and using this and (72) we have

0 = 1198790 + 12057301 ge 12058311198710 + 12057301 ge 120573lowast1 + 12057301 (76)

which contradicts (74) Thus (73) holdsBy Lemma 9 we have

119894119870(1198791198701198771

) = 0 (77)

On the other hand let 120576 gt 0 satisfy 119891infin lt (1120582)(1205831minus 120576) Then

there exists 1198772gt 1198771such that

119891 (119905 119906) le1

120582(1205831 minus 120576) 119906 forall119905 isin [0 1] 119906 ge 1198772 (78)

By (H5) there exists an 119871infin function 1205931such that

119891 (119905 119906) le1

1205821205931 (119905) forall119906 isin [0 1198772] 119905 isin [0 1] (79)

Hence we have

119891 (119905 119906) le1

120582[(1205831 minus 120576) 119906 + 1205931 (119905)]

forall119906 isin 119877+ 119905 isin [0 1]

(80)

Since 11205831is the radius of the spectrum of 119871 (119868(120583

1minus120576)minus119871)

minus1

existsLet

119862 = int

1

0

1205931 (119904) Φ (119904) 119892 (119904) 119889119902119904

1198770= (

119868

(1205831minus 120576)

minus 119871)

minus1

(119888

(1205831minus 120576)

)

(81)

We prove that for each 119877 gt 1198770

119879119906 = 120573119906 forall119906 isin 120597119870119877 120573 ge 1 (82)

In fact if not there exist 119906 isin 120597119870119877and 120573 ge 1 such that 119879119906 =

120573119906This together with (80) implies

119906 (119905) le int

1

0

119866 (119905 119902119904) 119892 (119904) ((1205831 minus 120576) 119906 (119904) + 1205931 (119904)) 119889119902119904

= (1205831minus 120576) int

1

0

119866 (119905 119902119904) 119892 (119904) 119906 (119904) 119889119902119904

+ int

1

0

119866 (119905 119902119904) 119892 (119904) 1205931 (119904) 119889119902119904

le (1205831 minus 120576) 119871119906 (119905) + 119862

(83)

This implies

(119868

1205831minus 120576minus 119871)119906 (119905) le

119862

1205831minus 120576

119906 (119905) le (119868

1205831minus 120576minus 119871)

minus1

(119862

1205831minus 120576)

= 1198770

(84)

Therefore we have 119906 le 1198770lt 119877 a contradiction Taking

119877 gt 1198772 it follows from (74) and properties of index that

119894119870(119879119870119877) = 1 forall119877 gt 119877

0 (85)

Now (77) and (85) combined imply

119894119870(119879119870119877 1198701198771

) = 119894119870(119879119870119877) minus 119894119870(1198791198701198771

) = 1 (86)

Therefore 119879 has at least one fixed point 1199060isin 1198701198771198701198771

and 1199060

is a positive solution of BVP (1)-(2)

5 The Existence of Two Positive Solutions

Theorem 23 Suppose (A2) (A3) and

(A5) 12058211989101205881015840

le 119898 for some 1205881015840 gt 0

Then (1)-(2) had at least two positive solutions

Proof By (A5) we have

119879119906 (119905) = 120582int

1

0

119866 (119905 119902119904) 119892 (119904) 119891 (119904 119906 (119904)) 119889119902119904

le int

1

0

119866 (119905 119902119904) 119892 (119904) 1205881015840119898119889119902119904

(87)

so that 119879119906 le 1205881015840 = 119906 for all 119906 isin 1205971198811205881015840 Now Lemma 8 yields

119894119896(119879 1198811205881015840) = 1 (88)

On the other hand in view of (A2) we may take 120588lowast gt 1205881015840so that (69) holds (see the proof of Theorem 21) From (A3)we may take 119877

1isin (0 120588

1015840) so that (77) holds (see the proof of

Theorem 22)Combining (88) (69) and (77) we arrive at

119894119896 (119879119870120588lowast 1198811205881015840) = 0 minus 1 = minus1

119894119896(119879 1198811205881015840 1198701198771

) = 1 minus 0 = 1

(89)

Consequently 119879 has at least two fixed points with one on119870120588lowast 1198811205881015840 and the other on119881

1205881015840 1198701198771

Therefore (1)-(2) had atleast two positive solutions

Theorem 24 Suppose (A1) (A4) and

(A6) 12058211989112058810158401205881015840119888 ge 119872 for some 1205881015840 gt 0

Then (1)-(2) had at least two positive solutions

Proof By (A6) we have

119879119906 (119905) = 120582int

1

0

119866 (119905 119902119904) 119892 (119904) 119891 (119904 119906 (119904)) 119889119902119904

ge 120582int

119887

119886

119866 (119905 119902119904) 119892 (119904) 119891 (119904 119906 (119904)) 119889119902119904

ge int

119887

119886

119866 (119905 119902119904) 119892 (119904)1198721205881015840119889119902119904

(90)

Abstract and Applied Analysis 9

so that 119879119906 ge 1205881015840 = 119906 for all 119906 isin 1205971198811205881015840 and by Lemma 8

this yields

119894119896 (119879 1198811205881015840) = 0 (91)

On the other hand in view of (A1) we may take 120588 isin (0 1205881015840)so that (64) holds (see the proof of Theorem 21) In additionfrom (A4) we may take 119877 gt 1205881015840 so that (85) holds (see theproof of Theorem 22)

Combining (91) (64) and (85) we arrive at

119894119896(119879119870119877 1198811205881015840) = 1 minus 0 = 1

119894119896(119879 1198811205881015840 119870120588) = 0 minus 1 = minus1

(92)

Hence 119879 has at least two fixed points with one on 1198811205881015840 119870120588

and the other on 119870119877 1198811205881015840 Therefore (1)-(2) had at least two

positive solutions

We illustrate the applicability of these results with someexamples

Example 25 Consider the problem

11986305(25)119906 (119905) + 120582 (5119905 + 3) (

71199062+ 119906

119906 + 1) (2 + cos 119906) = 0

119905 isin (0 1)

1198632

05119906 (0) = 0

119906 (0) = 0

11986305119906 (1) = 0

(93)

Herewe have119892(119905) = 5119905+3119891(119906) = (2+cos 119906)((71199062+119906)(119906+1))and 2 lt 120572 le 3

It is readily shown that 1198910 = 1198910= 3 119891infin = 21 119891

infin= 7

Also 3119906 le 119891(119906) le 21119906 for 119906 ge 0 By calculation wefind119898 = 019722 and the smallest119872 calculated is119872(119886 119887) asymp119872(0484405 1) asymp 074665 We find 120583

1asymp 030366 Hence

by Theorem 21 there is at least one positive solution if 3120582 lt1205831and 7120582 gt 120583

1 that is there is a positive solution if 120582 isin

(047047 109773)

ByTheorem 22 there does not exist a positive solution ifeither 3120582 gt 120583

1or 21120582 lt 120583

1 that is if 120582 lt 109773 or 120582 gt

015682 no positive solution exists

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

References

[1] N Pongarm S Asawasamrit J Tariboon and S K NtouyasldquoMulti-strip fractional q-integral boundary value problemsfor nonlinear fractional q-difference equationsrdquo Advances inDifference Equations vol 2014 no 1 article 193 2014

[2] R Almeida and N Martins ldquoExistence results for fractional q-difference equations of order120572 isin ]2 3[with three-point bound-ary conditionsrdquo Communications in Nonlinear Science andNumerical Simulation vol 19 no 6 pp 1675ndash1685 2014

[3] Y ZhaoHChen andQZhang ldquoExistence results for fractionalq-difference equations with nonlocal q-integral boundary con-ditionsrdquo Advances in Difference Equations vol 2013 article 482013

[4] R A Ferreira ldquoPositive solutions of a nonlinear q-fractionaldifference equation with integral boundary conditionsrdquo Inter-national Journal of Difference Equations vol 9 no 2 pp 135ndash145 2014

[5] B Ahmad and S K Ntouyas ldquoFractional 119902-difference hybridequations and inclusions with Dirichlet boundary conditionsrdquoAdvances in Difference Equations vol 2014 article 199 2014

[6] B Ahmad S KNtouyas andAAlsaedi ldquoExistence of solutionsfor fractional q-integro-difference inclusions with fractional q-integral boundary conditionsrdquo Advances in Difference Equa-tions vol 2014 article 257 2014

[7] R P Agarwal B Ahmad A Alsaedi and H Al-Hutami ldquoOnnonlinear fractional q-difference equations involving two frac-tional orders with three-point nonlocal boundary conditionsrdquoDynamics of Continuous Discrete amp Impulsive Systems Series AMathematical Analysis vol 21 no 1 pp 135ndash151 2014

[8] M El-Shahed andM A Al-Yami ldquoPositive solutions of bound-ary value problems for nth order q-differential equationsrdquoInternational Journal of Mathematical Archive vol 2 pp 521ndash532 2011

[9] J Ma and J Yang ldquoExistence of solutions for multi-pointboundary value problem of fractional q-difference equationrdquoElectronic Journal ofQualitativeTheory ofDifferential Equationsvol 92 pp 1ndash10 2011

[10] P M Rajkovic S D Marinkovic and M S Stankovic ldquoFrac-tional integrals and derivatives in q-calculusrdquo Applicable Anal-ysis and Discrete Mathematics vol 1 no 1 pp 311ndash323 2007

[11] Y Zhao H Chen and B Qin ldquoMultiple solutions for acoupled systemof nonlinear fractional differential equations viavariational methodsrdquo Applied Mathematics and Computationvol 257 pp 417ndash427 2015

[12] W-X Zhou X Liu and J-G Zhang ldquoSome new existenceand uniqueness results of solutions to semilinear impulsivefractional integro-differential equationsrdquoAdvances inDifferenceEquations vol 2015 article 38 2015

[13] Q Yuan andW Yang ldquoPositive solutions of nonlinear boundaryvalue problems for delayed fractional q-difference systemsrdquoAdvances in Difference Equations vol 2014 no 1 article 51 16pages 2014

[14] W Yang ldquoPositive solutions for nonlinear semipositone frac-tional q-difference system with coupled integral boundaryconditionsrdquo Applied Mathematics and Computation vol 244pp 702ndash725 2014

[15] J R L Webb and K Q Lan ldquoEigenvalue criteria for existenceof multiple positive solutions of nonlinear boundary valueproblems of local and nonlocal typerdquo Topological Methods inNonlinear Analysis vol 27 no 1 pp 91ndash116 2006

[16] J R L Webb ldquoNonlocal conjugate type boundary value prob-lems of higher orderrdquo Nonlinear Analysis Theory Methods ampApplications vol 71 no 5-6 pp 1933ndash1940 2009

[17] J R L Webb and G Infante ldquoNon-local boundary value prob-lems of arbitrary orderrdquo Journal of the London MathematicalSociety vol 79 no 1 pp 238ndash258 2009

10 Abstract and Applied Analysis

[18] J R Webb and G Infante ldquoPositive solutions of nonlocalboundary value problems involving integral conditionsrdquo Non-linear Differential Equations and Applications NoDEA vol 15no 1-2 pp 45ndash67 2008

[19] J R L Webb and G Infante ldquoPositive solutions of nonlocalboundary value problems a unified approachrdquo Journal of theLondon Mathematical Society vol 74 no 3 pp 673ndash693 2006

[20] V Kac and P Cheung Quantum Calculus Springer Science ampBusiness Media 2002

[21] R A C Ferreira ldquoNontrivial solutions for fractional q-difference boundary value problemsrdquo Electronic Journal ofQualitative Theory of Differential Equations no 70 pp 1ndash102010

[22] M S Stankovic P M Rajkovic and S D Marinkovic ldquoOn q-fractional derivatives of Riemann-Liouville and Caputo typerdquohttparxivorgabs09090387

[23] M H Annaby and Z S Mansour q-Fractional Calculus andEquations vol 2056 of Lecture Notes in Mathematics SpringerBerlin Germany 2012

[24] M El-Shahed and W M Shammakh ldquoMultiple positive solu-tions for nonlinear fractional eigenvalue problemwith non localconditionsrdquo Fractional Calculus and Applied Analysis vol 3 pp1ndash13 2012

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

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Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

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Hindawi 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

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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

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

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Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Page 3: Research Article Positive Solutions for Nonlinear -Fractional …downloads.hindawi.com/journals/aaa/2015/759378.pdf · 2019. 7. 31. · Research Article Positive Solutions for Nonlinear

Abstract and Applied Analysis 3

Lemma 6 (see [22]) Let 120572 isin R+ N 120582 isin (minus1infin) Then thenext formulas hold

(1) 119868120572

119902119909120582=

Γ119902 (120582 + 1)

Γ119902 (120582 + 1 + 120572)

119909120582+120572

(2)119877119871119863120572

119902119909120582=

Γ119902 (120582 + 1)

Γ119902 (120582 + 120572 + 1)119909120582minus120572

(3)119862119863120572

119902119909120582=

0 119894119891 120582 isin N0 120572 gt 120582

119877119871119863120572

119902119909120582119894119891 119900119905ℎ119890119903119908119894119904119890

(18)

Theorem 7 (see [23]) Let 119909 gt 0 and 120572 isin R+ NThen thefollowing equality holds

(119868120572

119902 119862119863120572

119902119891) (119909) = 119891 (119909) minus

[120572]minus1

sum

119896=0

119909119896

Γ119902 (119896 + 1)

(119863119896

119902119891) (0) (19)

Lemma 8 (see [24]) Suppose 119879 119870 rarr 119870 is a completelycontinuous operator and has no fixed points on 120597119870

120588cap119870 Then

the following are true(i) If 119879119906 le 119906 for all 119906 isin 120597119870

120588cap 119870 then 119894(119879 119870

120588cap

119870119870) = 1 where 119894 is the fixed point index on 119870(ii) If 119879119906 ge 119906 for all 119906 isin 120597119870

120588cap 119870 then 119894(119879 119870

120588cap

119870119870) = 0

Lemma 9 (see [24]) Let 119870 be a cone in Banach space 119864Suppose that119879 119870

120588rarr 119870 is a completely continuous operator

There exists 1199060isin 1198700 such that 119906minus119879119906 = 120583119906

0for any 119906 isin 120597119870

119903

and 120583 ge 0 119894(119879 119870120588 119870) = 0

Lemma 10 (see [24]) Let 119870 be a cone in Banach space 119864Suppose that119879 119870

120588rarr 119870 is a completely continuous operator

If 119879119906 = 120583119906 for any 119906 isin 120597119870119903and 120583 ge 1 then 119894(119879 119870

120588 119870) = 1

Lemma 11 Let 119910 isin 119862[0 1] be a given function and 119899 minus 1 lt120572 le 119899 then 119906 is a solution of BVP (1)-(2) if and only if 119906 is asolution of the integral equation

119906 (119905) = 120574 (119905) 120579 [119906] + int

1

0

1198660 (119905 119902119904) 119910 (119904) 119889119902119904 (20)

where120574 (119905) = 119905

1198660(119905 119902119904)

=

[120572 minus 1]119902 119905 (1 minus 119902119904)(120572minus2)

minus (119905 minus 119902119904)(120572minus1)

Γ119902 (120572)

0 le 119902119904 le 119905 le 1

[120572 minus 1]119902 119905 (1 minus 119902119904)(120572minus2)

Γ119902 (120572) 0 le 119905 le 119902119904 le 1

(21)

Proof Assume that 119906 is a solution of BVP (1)-(2)Applying Theorem 7 (1) can be reduced to an equivalent

integral equation

119906 (119905) = minus1

Γ119902 (120572)

int

119905

0

(119905 minus 119902119904)(120572minus1)

119910 (119904) 119889119902119904 + 1198880 + 1198881119905

+ 11988821199052+ sdot sdot sdot + 119888

119899minus1119905119899minus1

(22)

By (2) we obtain

1198880= 0

1198882 = sdot sdot sdot = 119888119899minus1 = 0

1198881= 120579 [119906] +

[120572 minus 1]119902

Γ119902 (120572)

int

1

0

(1 minus 119902119904)(120572minus2)

119910 (119904) 119889119902119904

(23)

Therefore we obtain

119906 (119905) = 120574 (119905) 120579 [119906] +119905 [120572 minus 1]119902

Γ119902 (120572)

int

1

0

(1 minus 119902119904)(120572minus2)

sdot 119910 (119904) 119889119902119904 minus1

Γ119902 (120572)

int

119905

0

(119905 minus 119902119904)(120572minus1)

119910 (119904) 119889119902119904

= 120574 (119905) 120579 [119906]

+ int

119905

0

[

[

[120572 minus 1]119902 119905 (1 minus 119902119904)(120572minus2)

minus (119905 minus 119902119904)(120572minus1)

Γ119902 (120572)

]

]

sdot 119910 (119904) 119889119902119904 + int

1

119905

[

[

[120572 minus 1]119902 119905 (1 minus 119902119904)(120572minus2)

Γ119902 (120572)

]

]

sdot 119910 (119904) 119889119902119904 = 120574 (119905) 120579 [119906] + int

1

0

1198660 (119905 119902119904) 119910 (119904) 119889119902119904

(24)

Conversely if 119906 is a solution of the integral equation (20)using Lemmas 5 and 6 we have

119862119863120572

119902119906 (119905)

=119862119863120572

119902119905120579 [119906]

+119862119863120572

119902119905(int

1

0

[120572 minus 1]119902 (1 minus 119902119904)(120572minus2)

Γ119902 (120572)

119910 (119904) 119889119902119904)

minus119862119863120572

119902(int

119905

0

(119905 minus 119902119904)(120572minus1)

Γ119902 (120572)

119910 (119904) 119889119902119904)

= minus119862119863120572

119902119868120572

119902119910 (119905) = minus119910 (119905)

(25)

A simple computation shows 119906(0) = 0119863119896119902119906(0) = 0 2 le 119896 le

119899 minus 1119863119902119906(1) = 120579[119906]

Remark 12 1198660(119905 119902119904) is Greenrsquos function for the local BVP

119862119863120572

119902119906 (119905) + 120582119892 (119905) 119891 (119905 119906 (119905)) = 0

119905 isin [0 1] 119899 minus 1 lt 120572 le 119899 0 lt 119902 lt 1

119863119896

119902119906 (0) = 0

119906 (0) = 0 2 le 119896 le 119899 minus 1

119863119902119906 (1) = 0

(26)

4 Abstract and Applied Analysis

Lemma 13 Function 1198660(119905 119902119904) defined in (20) satisfies the

following conditions

(H1) 1198660(119905 119902119904) ge 0 is continuous and 119866

0(119905 119904) le Φ

0(119902119904) for

all 0 le 119905 119904 le 1(H2) 119866

0(119905 119902119904) ge 119888

0(119905)Φ0(119902119904) for all 0 le 119905 119904 le 1 where

Φ0(119902119904) = 119866

0(1 119902119904)

=[120572 minus 1]119902 (1 minus 119902119904)

(120572minus2)minus (1 minus 119902119904)

(120572minus1)

Γ119902 (120572)

1198880 (119905) = 119905

120572minus1

(27)

Proof It is obvious that1198660(119905 119902119904) is nonnegative and continu-

ous(H1) For 0 le 119902119904 le 119905 le 1

1198660(119905 119902119904) =

1

Γ119902 (120572)[[120572 minus 1]119902 119905 (1 minus 119902119904)

(120572minus2)

minus (119905 minus 119902119904)(120572minus1)

] =1

Γ119902 (120572)

[[120572 minus 1]119902 119905 (1 minus 119902119904)(120572minus2)

minus 119905120572minus1(1 minus 119902

119904

119905)

(120572minus1)

]

ge1

Γ119902 (120572)[[120572 minus 1]119902 119905

120572minus1(1 minus 119902119904)

(120572minus2)

minus 119905120572minus1(1 minus 119902119904)

(120572minus1)]

=119905120572minus1

Γ119902 (120572)

[[120572 minus 1]119902 (1 minus 119902119904)(120572minus2)

minus (1 minus 119902119904)(120572minus1)

]

ge 0

(28)

and for 0 le 119905 le 119902119904 le 1

1198660 (119905 119902119904) =

[120572 minus 1]119902 119905 (1 minus 119902119904)(120572minus2)

Γ119902 (120572)

(29)

and it is clear that 1198660(119905 119902119904) ge 0 and 119866

0(0 119902119904) = 0Therefore

1198660(119905 119902119904) ge 0

For fixed 119904 isin [0 1] and 119905 ge 119902119904 we have

119905119863119902 1198660 (119905 119902119904)

=[120572 minus 1]119902 (1 minus 119902119904)

(120572minus2)minus [120572 minus 1]119902 (119905 minus 119902119904)

(120572minus2)

Γ119902 (120572)

=(1 minus 119902119904)

(120572minus2)minus (119905 minus 119902119904)

(120572minus2)

Γ119902 (120572 minus 1)

ge 0

(30)

that is 1198660(119905 119902119904) is an increasing function of 119905 Obviously

1198660(119905 119902119904) 119905 le 119902119904 is increasing in 119905 therefore 119866

0(119905 119902119904) is an

increasing function of 119905 for fixed 119904 isin [0 1]Thus (H1) holds

(H2) Suppose now that 119905 ge 119902119904

1198660(119905 119902119904)

Φ0 (119902119904)=[120572 minus 1]119902 119905 (1 minus 119902119904)

(120572minus2)minus (119905 minus 119902119904)

(120572minus1)

[120572 minus 1]119902 (1 minus 119902119904)(120572minus2)

minus (1 minus 119902119904)(120572minus1)

ge

119905120572minus1[[120572 minus 1]119902 (1 minus 119902119904)

(120572minus2)minus (1 minus 119902119904)

(120572minus1)]

[120572 minus 1]119902 (1 minus 119902119904)(120572minus2)

minus (1 minus 119902119904)(120572minus1)

= 119905120572minus1

(31)

On the other hand if 119905 le 119902119904 then we have

1198660(119905 119902119904)

Φ0(119902119904)

=[120572 minus 1]119902 119905 (1 minus 119902119904)

(120572minus2)

[120572 minus 1]119902 (1 minus 119902119904)(120572minus2)

minus (1 minus 119902119904)(120572minus1)

ge[120572 minus 1]119902 119905

120572minus1(1 minus 119902119904)

(120572minus2)minus 119905120572minus1(1 minus 119902119904)

(120572minus1)

[120572 minus 1]119902 (1 minus 119902119904)(120572minus2)

minus (1 minus 119902119904)(120572minus1)

= 119905120572minus1

(32)

and this finished the proof of (H2)

Defining G119860(119902119904) = int

1

01198660(119905 119902119904)119889

119902119860(119905) Greenrsquos function

for nonlocal BVP (1)-(2) is given by

119866 (119905 119902119904) =120574 (119905)

[1 minus 120579 [120574]]G119860(119902119904) + 119866

0(119905 119902119904) (33)

Throughout the paper we assume the following(H3) A is a function of bounded variation andG119860(119902119904) =

int1

01198660(119905 119902119904)119889

119902119860(119905) satisfies G

119860(119902119904) ge 0 for almost every 119904 isin

[0 1] Note thatG119860(119902119904) exists for almost every 119904 by (H1)

(H4)The functions 119892Φ satisfy 119892 ge 0 almost everywhere119892Φ isin 119871

1[0 1] and

int

119887

119886

Φ(119902119904) 119892 (119904) 119889119902119904 gt 0 (34)

(H5) 119891 [0 1] times [0infin) rarr [0infin) satisfies Caratheodoryconditions that is 119891(sdot 119906) is measurable for each fixed 119906 isin[0infin) and 119891(119905 sdot) is continuous for almost every 119905 isin [0 1]and for each 119903 gt 0 there exists 120601

119903isin 119871infin[0 1] such that 0 le

119891(119905 119906) le 120601119903for all 119906 isin [0 119903] and almost all 119905 isin [0 1]

(H6) One has the following 120574 isin 119862[0 1] 120574(119905) ge 0 0 le120579[120574] lt 1

Lemma 14 If 1198660satisfies (H1) (H2) then 119866 satisfies (H1)

(H2) for a function Φ the same interval [119886 119887] and the sameconstant 119888 where Φ satisfies (H4) and 119888 = min1198880(119905) 119905 isin[119886 119887]

Proof We have

119866 (119905 119902119904) =120574 (119905)

[1 minus 120579 [120574]]G119860(119902119904) + 119866

0(119905 119902119904)

le

10038171003817100381710038171205741003817100381710038171003817

[1 minus 120579 [120574]]G119860(119902119904) + Φ

0(119902119904) = Φ (119902119904)

(35)

Abstract and Applied Analysis 5

and for 119905 isin [119886 119887]

119866 (119905 119902119904) ge11988810038171003817100381710038171205741003817100381710038171003817

[1 minus 120579 [120574]]G119860 (119902119904) + 119888Φ0 (119902119904) = 119888Φ (119902119904) (36)

Note that 119892Φ isin 119871infin because 119860 has finite variation and

G119860(119902119904) le Φ(119902119904) var(119860)Thus Greenrsquos function 119866(119905 119902119904) satisfies (H1) (H2) for a

functionΦ and the constant 119888

3 Main Result

Set 119864 = 119862[0 1] as a Banach space with the norm 119906 =

sup119905isin[01]

|119906(119905)| Let 119875 = 119906 isin 119864 119906 ge 0 denote the standardcone of nonnegative functions Define

119870 = 119906 isin 119875 min119886le119905le119887

119906 (119905) ge 119888 119906 (37)

where [119886 119887] is some subset of [0 1]Note that 120574 isin 119870 so 119870 = 0 For any 0 lt 119903 lt 119877 lt +infin

let 119870119903= 119906 isin 119870 119906 lt 119903 120597119870

119903= 119906 isin 119870 119906 = 119903

119870119903 = 119906 isin 119870 119906 le 119903 119870119877 119870119903 = 119906 isin 119870 119903 le 119906 le 119877and 119881119903 = 119906 isin 119870 min119905isin[119886119887]119906(119905) lt 119903 and 119881119903 is bounded

Define a nonlinear operator 119879 119875 rarr 119870 and a linearoperator 119871 119875 rarr 119870 by

119879119906 (119905) = 120582int

1

0

119866 (119905 119902119904) 119892 (119904) 119891 (119904 119906 (119904)) 119889119902119904 (38)

119871119906 (119905) fl int1

0

119866 (119905 119902119904) 119892 (119904) 119906 (119904) 119889119902119904 (39)

Lemma 15 (see [18]) Under hypotheses (H1)ndash(H6) the maps119879 119875 rarr 119864 defined in (38) are compact

Theorem 16 Under hypotheses (H1)ndash(H6) the maps are 119879 119875 rarr 119870

Proof For 119906 isin 119875 and 119905 isin [0 1] we have

119879119906 (119905) le 120582int

1

0

Φ(119902119904) 119892 (119904) 119891 (119904 119906 (119904)) 119889119902119904 (40)

Hence

119879119906 le 120582int

1

0

Φ(119902119904) 119892 (119904) 119891 (119904 119906 (119904)) 119889119902119904 (41)

Also for 119905 isin [119886 119887] we have

119879119906 (119905) ge 119888120582int

1

0

Φ(119902119904) 119892 (119904) 119891 (119904 119906 (119904)) 119889119902119904 ge 119888 119879119906 (42)

Similar to the proofs of Lemma 15 and Theorem 16 119871119906(119905) iscompact and maps 119875 into119870

We will use the Krein-Rutman theorem We recall that 120582is an eigenvalue of 119871 with corresponding eigenfunction 120601 if120601 = 0 and 120582120601 = 119871120601 The reciprocals of eigenvalues are calledcharacteristic values of 119871 The radius of the spectrum of 119871denoted by 119903(119871) is given by the well-known spectral radiusformula 119903(119871) = lim

119899rarrinfin1198711198991119899

Theorem 17 (see [15]) Let 119870 be a total cone in a real Banachspace 119864 and let 119864 rarr 119864 be a compact linear operator with(119870) sube 119870 If 119903() gt 0 then there is 120601

1isin 119870 0 such that

1206011= 119903()120601

1

Thus 1205821 fl 119903() is an eigenvalue of the largest possiblereal eigenvalue and 1205831 = 1120582

1is the smallest positive

characteristic value

Lemma 18 (see [15]) Assume that (H1)ndash(H3) hold and let 119871be as defined in (39) Then 119903(119871) gt 0

Theorem 19 (see [15]) When (H1)ndash(H3) hold 119903(119871) is aneigenvalue of 119871 with eigenfunction 120601

1in 119870

Theorem 20 (see [15]) Let 1205831= 1119903(119871) and 120601

1(119905) be a

corresponding eigenfunction in119875 of norm 1Then119898 le 1205831le 119872

where

119898 = ( sup119905isin[01]

int

1

0

119866 (119905 119902119904) 119892 (119904) 119889119902119904)

minus1

119872 = ( inf119905isin[119886119887]

int

119887

119886

119866 (119905 119902119904) 119892 (119904) 119889119902119904)

minus1

(43)

If 119892(119905) gt 0 for 119905 isin [0 1] and 119866(119905 119902119904) gt 0 for 119905 119904 isin [0 1] thefirst inequality is strict unless 1206011(119905) is constant for 119905 isin [0 1] If119892(119905)120601(119905) gt 0 for 119905 isin [119886 119887] the second inequality is strict unless1206011(119905) is constant for 119905 isin [119886 119887]

Proof (for the local BVP (1)-(2) if 119892(119905) equiv 1) We now computethe constant 119898 and the optimal value of119872(119886 119887) that is wedetermine 119886 119887 so that119872(119886 119887) is minimal

For 119902119904 le 119905 we have by direct integration

int

119905

0

1198660 (119905 119902119904) 119889119902119904

= int

119905

0

[

[

[120572 minus 1]119902 119905 (1 minus 119902119904)(120572minus2)

minus (119905 minus 119902119904)(120572minus1)

Γ119902 (120572)

]

]

119889119902119904

=119905 minus 119905 (1 minus 119905)

(120572minus1)

Γ119902 (120572)

minus119905120572

[120572]119902 Γ119902 (120572)

(44)

For 119902119904 ge 119905

int

1

119905

1198660(119905 119902119904) 119889

119902119904 = int

1

119905

[120572 minus 1]119902 119905 (1 minus 119902119904)(120572minus2)

Γ119902 (120572)119889119902119904

=119905 (1 minus 119905)

(120572minus1)

Γ119902 (120572)

(45)

Then we have

int

1

0

1198660 (119905 119902119904) 119889119902119904 =119905

Γ119902 (120572)

minus119905120572

[120572]119902 Γ119902 (120572) (46)

6 Abstract and Applied Analysis

And the maximum of this expression occurs when 119905 = 1hence

sup119905isin[01]

int

1

0

1198660 (119905 119902119904) 119889119902119904 =1

Γ119902 (120572)

minus1

[120572]119902 Γ119902 (120572)

=[120572]119902 minus 1

[120572]119902 Γ119902 (120572)

(47)

Then119898 = [120572]119902Γ119902(120572)([120572]

119902minus 1)

For 119886 lt 119887 we have by direct integration

int

119905

119886

1198660(119905 119902119904) 119889

119902119904 = minus

119905 (1 minus 119905)(120572minus1)

Γ119902 (120572)

+119905 (1 minus 119886)

(120572minus1)

Γ119902 (120572)

minus(119905 minus 119886)

(120572)

[120572]119902 Γ119902 (120572)

int

119887

119905

1198660 (119905 119904) 119889119904 = minus

119905 (1 minus 119887)(120572minus1)

Γ119902 (120572)+119905 (1 minus 119905)

(120572minus1)

Γ119902 (120572)

(48)

Then

int

119887

119886

1198660(119905 119902119904) 119889

119902119904 =119905 (1 minus 119886)

(120572minus1)

Γ119902 (120572)minus(119905 minus 119886)

(120572)

[120572]119902 Γ119902 (120572)

minus119905 (1 minus 119887)

(120572minus1)

Γ119902 (120572)

=119905 [(1 minus 119886)

(120572minus1)minus (1 minus 119887)

(120572minus1)]

Γ119902 (120572)

minus(119905 minus 119886)

(120572)

[120572]119902 Γ119902 (120572)= 119877 (119905 119886 119887)

119905119863119902119877 (119905 119886 119887) =

[(1 minus 119886)(120572minus1)

minus (1 minus 119887)(120572minus1)

]

Γ119902 (120572)

minus(119905 minus 119886)

(120572minus1)

Γ119902 (120572)

(49)

The sign of derivative119905119863119902119877 shows that this is an increasing

function of 119905 so the minimum occurs at 119905 = 119886 Let

119877 (119886 119887) =119886

Γ119902 (120572)

[(1 minus 119886)(120572minus1)

minus (1 minus 119887)(120572minus1)

] (50)

The minimal value of 119872(119886 119887) corresponds to the maximalvalue of 119877(119886 119887) Consider

119887119863119902119877 (119886 119887) =

119886 [120572 minus 1]119902 (1 minus 119902119887)(120572minus2)

Γ119902 (120572)gt 0 (51)

The quantity 119877(119886 119887) is an increasing function of 119887 so itsmaximum occurs when 119887 = 1 Let

119877 (119886) =119886 (1 minus 119886)

(120572minus1)

Γ119902 (120572)

(52)

Then the maximum of 119877(119886) occurs when 119886 = 1(1+ [120572minus1]119902)

Consider

min119905isin[119886119887]

int

119887

119886

1198660(119905 119902119904) 119889

119902119904 = 119877(

1

1 + [120572 minus 1]119902

1) (53)

Hence the minimal value of119872(119886 119887) is

119872(1

1 + [120572 minus 1]119902

1) = (119877(1

1 + [120572 minus 1]119902

1))

minus1

(54)

4 The Existence of at Least OnePositive Solution

For convenience we introduce the following notations

119891 (119906) fl sup119905isin[01]

119891 (119905 119906)

119891 (119906) fl inf119905isin[01]

119891 (119905 119906)

1198910 fl lim sup119906rarr0

+

119891 (119906)

119906

1198910fl lim inf119906rarr0

+

119891 (119906)

119906

119891infin fl lim sup

119906rarrinfin

119891 (119906)

119906

119891infin

fl lim inf119906rarrinfin

119891 (119906)

119906

1198910119903 fl sup0le119905le1 0le119906le119903

119891 (119905 119906)

119903

119891119903119903119888

fl inf119886le119905le119887 119903le119906le119903119888

119891 (119905 119906)

119903

(55)

Under hypotheses (H1)ndash(H4) let be defined by

119906 (119905) = int

119887

119886

119866 (119905 119902119904) 119892 (119904) 119906 (119904) 119889119902119904 (56)

Then is a compact linear operator and (119875) sube 119870Hence 119903() is an eigenvalue of with an eigenfunction 120601

1

in119870 Let 1205831fl 1119903()Note that 120583

1ge 1205831 hence the condition

in the following theorem ismore stringent comparedwith thecase if 119903(119871) could be used

Theorem 21 Assume that

(A1) 0 le 1205821198910 lt 1205831(A2) 120583

1lt 120582119891infinle infin

Then (1)-(2) had at least one positive solution

Abstract and Applied Analysis 7

Proof Let 120576 gt 0 be such that 1198910 le (1120582)(1205831minus 120576)Then there

exists 1205880gt 0 such that

119891 (119905 119906) le1

120582(1205831 minus 120576) 119906

forall119906 isin [0 1205880] and almost all 119905 isin [0 1]

(57)

Let 120588 isin (0 1205880]We prove that

119879119906 = 120573119906 for 119906 isin 120597119870120588 120573 ge 1 (58)

which implies the result In fact if (58) does not hold thenthere exist 119906 isin 120597119870

120588and 120573 ge 1 such that 119879119906 = 120573119906

This implies

120573119906 (119905) = 120582int

1

0

119866 (119905 119902119904) 119892 (119904) 119891 (119904 119906 (119904)) 119889119902119904

le (1205831minus 120576) int

1

0

119866 (119905 119902119904) 119892 (119904) 119906 (119904) 119889119902119904

= (1205831minus 120576) 119871119906 (119905)

(59)

Thus we have shown 119906(119905) le (1205831 minus 120576)119871119906(119905)This gives

119906 (119905) le (1205831 minus 120576) 119871 [(1205831 minus 120576) 119871119906 (119905)]

= (1205831minus 120576)21198712119906 (119905)

(60)

And by iterating

119906 (119905) le (1205831 minus 120576)119899119871119899119906 (119905) for 119899 isin 119873 (61)

Therefore

119906 le (1205831 minus 120576)119899 10038171003817100381710038171198711198991003817100381710038171003817 119906

1 le (1205831 minus 120576)119899 10038171003817100381710038171198711198991003817100381710038171003817

(62)

and we have

1 le (1205831 minus 120576) lim

119899rarr+infin

100381710038171003817100381711987111989910038171003817100381710038171119899= (1205831 minus 120576)

1

1205831

lt 1 (63)

a contradiction It follows that

119894119896 (119879119870120588) = 1 for each 120588 isin (0 1205880] (64)

Let 1205881gt 0 120588

1gt 120588 be chosen so that 119891(119905 119906) gt (120583

1120582)119906 for all

119906 ge 1198881205881 119888 as in (H2) and almost all 119905 isin [0 1]

We claim that 119906 = 119879119906 + 1205731206011 for all 120573 gt 0 and 119906 isin 120597119870120588lowast

when 120588lowast gt 1205881 Note that 119906 isin 119870 with 119906 = 120588lowast ge 120588

1

We have 119906(119905) ge 1198881205881for all 119905 isin [119886 119887]

Now if our claim is false then we have

119906 (119905) = 120582int

1

0

119866 (119905 119902119904) 119892 (119904) 119891 (119904 119906 (119904)) 119889119902119904 + 1205731206011 (119905) (65)

Therefore

119906 (119905) ge 1205831 int

119887

119886

119866 (119905 119902119904) 119892 (119904) 119906 (119904) 119889119902119904 + 1205731206011 (119905)

= 1205831119906 (119905) + 1205731206011 (119905)

(66)

From (66) we firstly deduce that 119906(119905) ge 1205731206011(119905) on [0 1]

Then we have

1205831119906 (119905) ge 1205831 (1205731206011 (119905)) = 1205731206011 (119905) (67)

Inserting this into (66) we obtain 119906(119905) ge 21205731206011(119905) for 119905 isin [0 1]

Repeating this process gives

119906 (119905) ge 1198991205731206011 (119905) for 119905 isin [0 1] 119899 isin 119873 (68)

Since 1206011(119905) is strictly positive on [0 1] this is a contradiction

then

119894119870(119879119870120588lowast) = 0 for 119906 isin 120597119870

120588lowast (69)

By (64) and (69) one has

119894119870(119879119870120588lowast 119870120588) = 119894119870(119879119870120588lowast) minus 119894119870(119879119870120588) = minus1 (70)

Therefore 119879 has at least one fixed point 1199060isin 119870120588lowast 119870120588 and

1199060is a positive solution of BVP (1)-(2)

Theorem 22 Assume that

(A3) 1205831 lt 1205821198910 le infin(A4) 0 le 120582119891infin lt 1205831

Then (1)-(2) had at least one positive solution

Proof Let 120576 gt 0 satisfy 1198910gt (1120582)(120583

1+ 120576)Then there exists

1198771 gt 0 such that

119891 (119905 119906) ge1

120582(1205831+ 120576) 119906 forall119905 isin [0 1] 119906 isin [0 1198771] (71)

For any 119906 isin 1205971198701198771

we have by (71) that

119879119906 (119905) = 120582int

1

0

119866 (119905 119902119904) 119892 (119904) 119891 (119904 119906 (119904)) 119889119902119904

ge (1205831 + 120576) int

1

0

119866 (119905 119902119904) 119892 (119904) 119906 (119904) 119889119902119904

ge 1205831119871119906 (119905) forall119905 isin [0 1]

(72)

Let 1 be the positive eigenfunction of 119871 corresponding to 1205831

that is 1 = 12058311198711 Wemay suppose that119879 has no fixed pointon 120597119870119877

1

otherwise the proof is finished In the following wewill show that

119906 minus 119879119906 = 1205731 forall119906 isin 1205971198701198771

120573 ge 0 (73)

If (73) is not true then there is 0isin 1205971198701198771

and 1205730ge 0 such that

0minus1198790= 12057301 It is clear that 120573

0gt 0 and

0= 1198790+12057301ge

12057301Set

120573lowast= sup 120573 0 ge 1205731 (74)

Obviously 120573lowast ge 1205730gt 0 It follows from 119871(119875) sub 119875 that

12058311198710ge 1205831119871120573lowast1= 120573lowast12058311198711= 120573lowast1 (75)

8 Abstract and Applied Analysis

and using this and (72) we have

0 = 1198790 + 12057301 ge 12058311198710 + 12057301 ge 120573lowast1 + 12057301 (76)

which contradicts (74) Thus (73) holdsBy Lemma 9 we have

119894119870(1198791198701198771

) = 0 (77)

On the other hand let 120576 gt 0 satisfy 119891infin lt (1120582)(1205831minus 120576) Then

there exists 1198772gt 1198771such that

119891 (119905 119906) le1

120582(1205831 minus 120576) 119906 forall119905 isin [0 1] 119906 ge 1198772 (78)

By (H5) there exists an 119871infin function 1205931such that

119891 (119905 119906) le1

1205821205931 (119905) forall119906 isin [0 1198772] 119905 isin [0 1] (79)

Hence we have

119891 (119905 119906) le1

120582[(1205831 minus 120576) 119906 + 1205931 (119905)]

forall119906 isin 119877+ 119905 isin [0 1]

(80)

Since 11205831is the radius of the spectrum of 119871 (119868(120583

1minus120576)minus119871)

minus1

existsLet

119862 = int

1

0

1205931 (119904) Φ (119904) 119892 (119904) 119889119902119904

1198770= (

119868

(1205831minus 120576)

minus 119871)

minus1

(119888

(1205831minus 120576)

)

(81)

We prove that for each 119877 gt 1198770

119879119906 = 120573119906 forall119906 isin 120597119870119877 120573 ge 1 (82)

In fact if not there exist 119906 isin 120597119870119877and 120573 ge 1 such that 119879119906 =

120573119906This together with (80) implies

119906 (119905) le int

1

0

119866 (119905 119902119904) 119892 (119904) ((1205831 minus 120576) 119906 (119904) + 1205931 (119904)) 119889119902119904

= (1205831minus 120576) int

1

0

119866 (119905 119902119904) 119892 (119904) 119906 (119904) 119889119902119904

+ int

1

0

119866 (119905 119902119904) 119892 (119904) 1205931 (119904) 119889119902119904

le (1205831 minus 120576) 119871119906 (119905) + 119862

(83)

This implies

(119868

1205831minus 120576minus 119871)119906 (119905) le

119862

1205831minus 120576

119906 (119905) le (119868

1205831minus 120576minus 119871)

minus1

(119862

1205831minus 120576)

= 1198770

(84)

Therefore we have 119906 le 1198770lt 119877 a contradiction Taking

119877 gt 1198772 it follows from (74) and properties of index that

119894119870(119879119870119877) = 1 forall119877 gt 119877

0 (85)

Now (77) and (85) combined imply

119894119870(119879119870119877 1198701198771

) = 119894119870(119879119870119877) minus 119894119870(1198791198701198771

) = 1 (86)

Therefore 119879 has at least one fixed point 1199060isin 1198701198771198701198771

and 1199060

is a positive solution of BVP (1)-(2)

5 The Existence of Two Positive Solutions

Theorem 23 Suppose (A2) (A3) and

(A5) 12058211989101205881015840

le 119898 for some 1205881015840 gt 0

Then (1)-(2) had at least two positive solutions

Proof By (A5) we have

119879119906 (119905) = 120582int

1

0

119866 (119905 119902119904) 119892 (119904) 119891 (119904 119906 (119904)) 119889119902119904

le int

1

0

119866 (119905 119902119904) 119892 (119904) 1205881015840119898119889119902119904

(87)

so that 119879119906 le 1205881015840 = 119906 for all 119906 isin 1205971198811205881015840 Now Lemma 8 yields

119894119896(119879 1198811205881015840) = 1 (88)

On the other hand in view of (A2) we may take 120588lowast gt 1205881015840so that (69) holds (see the proof of Theorem 21) From (A3)we may take 119877

1isin (0 120588

1015840) so that (77) holds (see the proof of

Theorem 22)Combining (88) (69) and (77) we arrive at

119894119896 (119879119870120588lowast 1198811205881015840) = 0 minus 1 = minus1

119894119896(119879 1198811205881015840 1198701198771

) = 1 minus 0 = 1

(89)

Consequently 119879 has at least two fixed points with one on119870120588lowast 1198811205881015840 and the other on119881

1205881015840 1198701198771

Therefore (1)-(2) had atleast two positive solutions

Theorem 24 Suppose (A1) (A4) and

(A6) 12058211989112058810158401205881015840119888 ge 119872 for some 1205881015840 gt 0

Then (1)-(2) had at least two positive solutions

Proof By (A6) we have

119879119906 (119905) = 120582int

1

0

119866 (119905 119902119904) 119892 (119904) 119891 (119904 119906 (119904)) 119889119902119904

ge 120582int

119887

119886

119866 (119905 119902119904) 119892 (119904) 119891 (119904 119906 (119904)) 119889119902119904

ge int

119887

119886

119866 (119905 119902119904) 119892 (119904)1198721205881015840119889119902119904

(90)

Abstract and Applied Analysis 9

so that 119879119906 ge 1205881015840 = 119906 for all 119906 isin 1205971198811205881015840 and by Lemma 8

this yields

119894119896 (119879 1198811205881015840) = 0 (91)

On the other hand in view of (A1) we may take 120588 isin (0 1205881015840)so that (64) holds (see the proof of Theorem 21) In additionfrom (A4) we may take 119877 gt 1205881015840 so that (85) holds (see theproof of Theorem 22)

Combining (91) (64) and (85) we arrive at

119894119896(119879119870119877 1198811205881015840) = 1 minus 0 = 1

119894119896(119879 1198811205881015840 119870120588) = 0 minus 1 = minus1

(92)

Hence 119879 has at least two fixed points with one on 1198811205881015840 119870120588

and the other on 119870119877 1198811205881015840 Therefore (1)-(2) had at least two

positive solutions

We illustrate the applicability of these results with someexamples

Example 25 Consider the problem

11986305(25)119906 (119905) + 120582 (5119905 + 3) (

71199062+ 119906

119906 + 1) (2 + cos 119906) = 0

119905 isin (0 1)

1198632

05119906 (0) = 0

119906 (0) = 0

11986305119906 (1) = 0

(93)

Herewe have119892(119905) = 5119905+3119891(119906) = (2+cos 119906)((71199062+119906)(119906+1))and 2 lt 120572 le 3

It is readily shown that 1198910 = 1198910= 3 119891infin = 21 119891

infin= 7

Also 3119906 le 119891(119906) le 21119906 for 119906 ge 0 By calculation wefind119898 = 019722 and the smallest119872 calculated is119872(119886 119887) asymp119872(0484405 1) asymp 074665 We find 120583

1asymp 030366 Hence

by Theorem 21 there is at least one positive solution if 3120582 lt1205831and 7120582 gt 120583

1 that is there is a positive solution if 120582 isin

(047047 109773)

ByTheorem 22 there does not exist a positive solution ifeither 3120582 gt 120583

1or 21120582 lt 120583

1 that is if 120582 lt 109773 or 120582 gt

015682 no positive solution exists

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

References

[1] N Pongarm S Asawasamrit J Tariboon and S K NtouyasldquoMulti-strip fractional q-integral boundary value problemsfor nonlinear fractional q-difference equationsrdquo Advances inDifference Equations vol 2014 no 1 article 193 2014

[2] R Almeida and N Martins ldquoExistence results for fractional q-difference equations of order120572 isin ]2 3[with three-point bound-ary conditionsrdquo Communications in Nonlinear Science andNumerical Simulation vol 19 no 6 pp 1675ndash1685 2014

[3] Y ZhaoHChen andQZhang ldquoExistence results for fractionalq-difference equations with nonlocal q-integral boundary con-ditionsrdquo Advances in Difference Equations vol 2013 article 482013

[4] R A Ferreira ldquoPositive solutions of a nonlinear q-fractionaldifference equation with integral boundary conditionsrdquo Inter-national Journal of Difference Equations vol 9 no 2 pp 135ndash145 2014

[5] B Ahmad and S K Ntouyas ldquoFractional 119902-difference hybridequations and inclusions with Dirichlet boundary conditionsrdquoAdvances in Difference Equations vol 2014 article 199 2014

[6] B Ahmad S KNtouyas andAAlsaedi ldquoExistence of solutionsfor fractional q-integro-difference inclusions with fractional q-integral boundary conditionsrdquo Advances in Difference Equa-tions vol 2014 article 257 2014

[7] R P Agarwal B Ahmad A Alsaedi and H Al-Hutami ldquoOnnonlinear fractional q-difference equations involving two frac-tional orders with three-point nonlocal boundary conditionsrdquoDynamics of Continuous Discrete amp Impulsive Systems Series AMathematical Analysis vol 21 no 1 pp 135ndash151 2014

[8] M El-Shahed andM A Al-Yami ldquoPositive solutions of bound-ary value problems for nth order q-differential equationsrdquoInternational Journal of Mathematical Archive vol 2 pp 521ndash532 2011

[9] J Ma and J Yang ldquoExistence of solutions for multi-pointboundary value problem of fractional q-difference equationrdquoElectronic Journal ofQualitativeTheory ofDifferential Equationsvol 92 pp 1ndash10 2011

[10] P M Rajkovic S D Marinkovic and M S Stankovic ldquoFrac-tional integrals and derivatives in q-calculusrdquo Applicable Anal-ysis and Discrete Mathematics vol 1 no 1 pp 311ndash323 2007

[11] Y Zhao H Chen and B Qin ldquoMultiple solutions for acoupled systemof nonlinear fractional differential equations viavariational methodsrdquo Applied Mathematics and Computationvol 257 pp 417ndash427 2015

[12] W-X Zhou X Liu and J-G Zhang ldquoSome new existenceand uniqueness results of solutions to semilinear impulsivefractional integro-differential equationsrdquoAdvances inDifferenceEquations vol 2015 article 38 2015

[13] Q Yuan andW Yang ldquoPositive solutions of nonlinear boundaryvalue problems for delayed fractional q-difference systemsrdquoAdvances in Difference Equations vol 2014 no 1 article 51 16pages 2014

[14] W Yang ldquoPositive solutions for nonlinear semipositone frac-tional q-difference system with coupled integral boundaryconditionsrdquo Applied Mathematics and Computation vol 244pp 702ndash725 2014

[15] J R L Webb and K Q Lan ldquoEigenvalue criteria for existenceof multiple positive solutions of nonlinear boundary valueproblems of local and nonlocal typerdquo Topological Methods inNonlinear Analysis vol 27 no 1 pp 91ndash116 2006

[16] J R L Webb ldquoNonlocal conjugate type boundary value prob-lems of higher orderrdquo Nonlinear Analysis Theory Methods ampApplications vol 71 no 5-6 pp 1933ndash1940 2009

[17] J R L Webb and G Infante ldquoNon-local boundary value prob-lems of arbitrary orderrdquo Journal of the London MathematicalSociety vol 79 no 1 pp 238ndash258 2009

10 Abstract and Applied Analysis

[18] J R Webb and G Infante ldquoPositive solutions of nonlocalboundary value problems involving integral conditionsrdquo Non-linear Differential Equations and Applications NoDEA vol 15no 1-2 pp 45ndash67 2008

[19] J R L Webb and G Infante ldquoPositive solutions of nonlocalboundary value problems a unified approachrdquo Journal of theLondon Mathematical Society vol 74 no 3 pp 673ndash693 2006

[20] V Kac and P Cheung Quantum Calculus Springer Science ampBusiness Media 2002

[21] R A C Ferreira ldquoNontrivial solutions for fractional q-difference boundary value problemsrdquo Electronic Journal ofQualitative Theory of Differential Equations no 70 pp 1ndash102010

[22] M S Stankovic P M Rajkovic and S D Marinkovic ldquoOn q-fractional derivatives of Riemann-Liouville and Caputo typerdquohttparxivorgabs09090387

[23] M H Annaby and Z S Mansour q-Fractional Calculus andEquations vol 2056 of Lecture Notes in Mathematics SpringerBerlin Germany 2012

[24] M El-Shahed and W M Shammakh ldquoMultiple positive solu-tions for nonlinear fractional eigenvalue problemwith non localconditionsrdquo Fractional Calculus and Applied Analysis vol 3 pp1ndash13 2012

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

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Page 4: Research Article Positive Solutions for Nonlinear -Fractional …downloads.hindawi.com/journals/aaa/2015/759378.pdf · 2019. 7. 31. · Research Article Positive Solutions for Nonlinear

4 Abstract and Applied Analysis

Lemma 13 Function 1198660(119905 119902119904) defined in (20) satisfies the

following conditions

(H1) 1198660(119905 119902119904) ge 0 is continuous and 119866

0(119905 119904) le Φ

0(119902119904) for

all 0 le 119905 119904 le 1(H2) 119866

0(119905 119902119904) ge 119888

0(119905)Φ0(119902119904) for all 0 le 119905 119904 le 1 where

Φ0(119902119904) = 119866

0(1 119902119904)

=[120572 minus 1]119902 (1 minus 119902119904)

(120572minus2)minus (1 minus 119902119904)

(120572minus1)

Γ119902 (120572)

1198880 (119905) = 119905

120572minus1

(27)

Proof It is obvious that1198660(119905 119902119904) is nonnegative and continu-

ous(H1) For 0 le 119902119904 le 119905 le 1

1198660(119905 119902119904) =

1

Γ119902 (120572)[[120572 minus 1]119902 119905 (1 minus 119902119904)

(120572minus2)

minus (119905 minus 119902119904)(120572minus1)

] =1

Γ119902 (120572)

[[120572 minus 1]119902 119905 (1 minus 119902119904)(120572minus2)

minus 119905120572minus1(1 minus 119902

119904

119905)

(120572minus1)

]

ge1

Γ119902 (120572)[[120572 minus 1]119902 119905

120572minus1(1 minus 119902119904)

(120572minus2)

minus 119905120572minus1(1 minus 119902119904)

(120572minus1)]

=119905120572minus1

Γ119902 (120572)

[[120572 minus 1]119902 (1 minus 119902119904)(120572minus2)

minus (1 minus 119902119904)(120572minus1)

]

ge 0

(28)

and for 0 le 119905 le 119902119904 le 1

1198660 (119905 119902119904) =

[120572 minus 1]119902 119905 (1 minus 119902119904)(120572minus2)

Γ119902 (120572)

(29)

and it is clear that 1198660(119905 119902119904) ge 0 and 119866

0(0 119902119904) = 0Therefore

1198660(119905 119902119904) ge 0

For fixed 119904 isin [0 1] and 119905 ge 119902119904 we have

119905119863119902 1198660 (119905 119902119904)

=[120572 minus 1]119902 (1 minus 119902119904)

(120572minus2)minus [120572 minus 1]119902 (119905 minus 119902119904)

(120572minus2)

Γ119902 (120572)

=(1 minus 119902119904)

(120572minus2)minus (119905 minus 119902119904)

(120572minus2)

Γ119902 (120572 minus 1)

ge 0

(30)

that is 1198660(119905 119902119904) is an increasing function of 119905 Obviously

1198660(119905 119902119904) 119905 le 119902119904 is increasing in 119905 therefore 119866

0(119905 119902119904) is an

increasing function of 119905 for fixed 119904 isin [0 1]Thus (H1) holds

(H2) Suppose now that 119905 ge 119902119904

1198660(119905 119902119904)

Φ0 (119902119904)=[120572 minus 1]119902 119905 (1 minus 119902119904)

(120572minus2)minus (119905 minus 119902119904)

(120572minus1)

[120572 minus 1]119902 (1 minus 119902119904)(120572minus2)

minus (1 minus 119902119904)(120572minus1)

ge

119905120572minus1[[120572 minus 1]119902 (1 minus 119902119904)

(120572minus2)minus (1 minus 119902119904)

(120572minus1)]

[120572 minus 1]119902 (1 minus 119902119904)(120572minus2)

minus (1 minus 119902119904)(120572minus1)

= 119905120572minus1

(31)

On the other hand if 119905 le 119902119904 then we have

1198660(119905 119902119904)

Φ0(119902119904)

=[120572 minus 1]119902 119905 (1 minus 119902119904)

(120572minus2)

[120572 minus 1]119902 (1 minus 119902119904)(120572minus2)

minus (1 minus 119902119904)(120572minus1)

ge[120572 minus 1]119902 119905

120572minus1(1 minus 119902119904)

(120572minus2)minus 119905120572minus1(1 minus 119902119904)

(120572minus1)

[120572 minus 1]119902 (1 minus 119902119904)(120572minus2)

minus (1 minus 119902119904)(120572minus1)

= 119905120572minus1

(32)

and this finished the proof of (H2)

Defining G119860(119902119904) = int

1

01198660(119905 119902119904)119889

119902119860(119905) Greenrsquos function

for nonlocal BVP (1)-(2) is given by

119866 (119905 119902119904) =120574 (119905)

[1 minus 120579 [120574]]G119860(119902119904) + 119866

0(119905 119902119904) (33)

Throughout the paper we assume the following(H3) A is a function of bounded variation andG119860(119902119904) =

int1

01198660(119905 119902119904)119889

119902119860(119905) satisfies G

119860(119902119904) ge 0 for almost every 119904 isin

[0 1] Note thatG119860(119902119904) exists for almost every 119904 by (H1)

(H4)The functions 119892Φ satisfy 119892 ge 0 almost everywhere119892Φ isin 119871

1[0 1] and

int

119887

119886

Φ(119902119904) 119892 (119904) 119889119902119904 gt 0 (34)

(H5) 119891 [0 1] times [0infin) rarr [0infin) satisfies Caratheodoryconditions that is 119891(sdot 119906) is measurable for each fixed 119906 isin[0infin) and 119891(119905 sdot) is continuous for almost every 119905 isin [0 1]and for each 119903 gt 0 there exists 120601

119903isin 119871infin[0 1] such that 0 le

119891(119905 119906) le 120601119903for all 119906 isin [0 119903] and almost all 119905 isin [0 1]

(H6) One has the following 120574 isin 119862[0 1] 120574(119905) ge 0 0 le120579[120574] lt 1

Lemma 14 If 1198660satisfies (H1) (H2) then 119866 satisfies (H1)

(H2) for a function Φ the same interval [119886 119887] and the sameconstant 119888 where Φ satisfies (H4) and 119888 = min1198880(119905) 119905 isin[119886 119887]

Proof We have

119866 (119905 119902119904) =120574 (119905)

[1 minus 120579 [120574]]G119860(119902119904) + 119866

0(119905 119902119904)

le

10038171003817100381710038171205741003817100381710038171003817

[1 minus 120579 [120574]]G119860(119902119904) + Φ

0(119902119904) = Φ (119902119904)

(35)

Abstract and Applied Analysis 5

and for 119905 isin [119886 119887]

119866 (119905 119902119904) ge11988810038171003817100381710038171205741003817100381710038171003817

[1 minus 120579 [120574]]G119860 (119902119904) + 119888Φ0 (119902119904) = 119888Φ (119902119904) (36)

Note that 119892Φ isin 119871infin because 119860 has finite variation and

G119860(119902119904) le Φ(119902119904) var(119860)Thus Greenrsquos function 119866(119905 119902119904) satisfies (H1) (H2) for a

functionΦ and the constant 119888

3 Main Result

Set 119864 = 119862[0 1] as a Banach space with the norm 119906 =

sup119905isin[01]

|119906(119905)| Let 119875 = 119906 isin 119864 119906 ge 0 denote the standardcone of nonnegative functions Define

119870 = 119906 isin 119875 min119886le119905le119887

119906 (119905) ge 119888 119906 (37)

where [119886 119887] is some subset of [0 1]Note that 120574 isin 119870 so 119870 = 0 For any 0 lt 119903 lt 119877 lt +infin

let 119870119903= 119906 isin 119870 119906 lt 119903 120597119870

119903= 119906 isin 119870 119906 = 119903

119870119903 = 119906 isin 119870 119906 le 119903 119870119877 119870119903 = 119906 isin 119870 119903 le 119906 le 119877and 119881119903 = 119906 isin 119870 min119905isin[119886119887]119906(119905) lt 119903 and 119881119903 is bounded

Define a nonlinear operator 119879 119875 rarr 119870 and a linearoperator 119871 119875 rarr 119870 by

119879119906 (119905) = 120582int

1

0

119866 (119905 119902119904) 119892 (119904) 119891 (119904 119906 (119904)) 119889119902119904 (38)

119871119906 (119905) fl int1

0

119866 (119905 119902119904) 119892 (119904) 119906 (119904) 119889119902119904 (39)

Lemma 15 (see [18]) Under hypotheses (H1)ndash(H6) the maps119879 119875 rarr 119864 defined in (38) are compact

Theorem 16 Under hypotheses (H1)ndash(H6) the maps are 119879 119875 rarr 119870

Proof For 119906 isin 119875 and 119905 isin [0 1] we have

119879119906 (119905) le 120582int

1

0

Φ(119902119904) 119892 (119904) 119891 (119904 119906 (119904)) 119889119902119904 (40)

Hence

119879119906 le 120582int

1

0

Φ(119902119904) 119892 (119904) 119891 (119904 119906 (119904)) 119889119902119904 (41)

Also for 119905 isin [119886 119887] we have

119879119906 (119905) ge 119888120582int

1

0

Φ(119902119904) 119892 (119904) 119891 (119904 119906 (119904)) 119889119902119904 ge 119888 119879119906 (42)

Similar to the proofs of Lemma 15 and Theorem 16 119871119906(119905) iscompact and maps 119875 into119870

We will use the Krein-Rutman theorem We recall that 120582is an eigenvalue of 119871 with corresponding eigenfunction 120601 if120601 = 0 and 120582120601 = 119871120601 The reciprocals of eigenvalues are calledcharacteristic values of 119871 The radius of the spectrum of 119871denoted by 119903(119871) is given by the well-known spectral radiusformula 119903(119871) = lim

119899rarrinfin1198711198991119899

Theorem 17 (see [15]) Let 119870 be a total cone in a real Banachspace 119864 and let 119864 rarr 119864 be a compact linear operator with(119870) sube 119870 If 119903() gt 0 then there is 120601

1isin 119870 0 such that

1206011= 119903()120601

1

Thus 1205821 fl 119903() is an eigenvalue of the largest possiblereal eigenvalue and 1205831 = 1120582

1is the smallest positive

characteristic value

Lemma 18 (see [15]) Assume that (H1)ndash(H3) hold and let 119871be as defined in (39) Then 119903(119871) gt 0

Theorem 19 (see [15]) When (H1)ndash(H3) hold 119903(119871) is aneigenvalue of 119871 with eigenfunction 120601

1in 119870

Theorem 20 (see [15]) Let 1205831= 1119903(119871) and 120601

1(119905) be a

corresponding eigenfunction in119875 of norm 1Then119898 le 1205831le 119872

where

119898 = ( sup119905isin[01]

int

1

0

119866 (119905 119902119904) 119892 (119904) 119889119902119904)

minus1

119872 = ( inf119905isin[119886119887]

int

119887

119886

119866 (119905 119902119904) 119892 (119904) 119889119902119904)

minus1

(43)

If 119892(119905) gt 0 for 119905 isin [0 1] and 119866(119905 119902119904) gt 0 for 119905 119904 isin [0 1] thefirst inequality is strict unless 1206011(119905) is constant for 119905 isin [0 1] If119892(119905)120601(119905) gt 0 for 119905 isin [119886 119887] the second inequality is strict unless1206011(119905) is constant for 119905 isin [119886 119887]

Proof (for the local BVP (1)-(2) if 119892(119905) equiv 1) We now computethe constant 119898 and the optimal value of119872(119886 119887) that is wedetermine 119886 119887 so that119872(119886 119887) is minimal

For 119902119904 le 119905 we have by direct integration

int

119905

0

1198660 (119905 119902119904) 119889119902119904

= int

119905

0

[

[

[120572 minus 1]119902 119905 (1 minus 119902119904)(120572minus2)

minus (119905 minus 119902119904)(120572minus1)

Γ119902 (120572)

]

]

119889119902119904

=119905 minus 119905 (1 minus 119905)

(120572minus1)

Γ119902 (120572)

minus119905120572

[120572]119902 Γ119902 (120572)

(44)

For 119902119904 ge 119905

int

1

119905

1198660(119905 119902119904) 119889

119902119904 = int

1

119905

[120572 minus 1]119902 119905 (1 minus 119902119904)(120572minus2)

Γ119902 (120572)119889119902119904

=119905 (1 minus 119905)

(120572minus1)

Γ119902 (120572)

(45)

Then we have

int

1

0

1198660 (119905 119902119904) 119889119902119904 =119905

Γ119902 (120572)

minus119905120572

[120572]119902 Γ119902 (120572) (46)

6 Abstract and Applied Analysis

And the maximum of this expression occurs when 119905 = 1hence

sup119905isin[01]

int

1

0

1198660 (119905 119902119904) 119889119902119904 =1

Γ119902 (120572)

minus1

[120572]119902 Γ119902 (120572)

=[120572]119902 minus 1

[120572]119902 Γ119902 (120572)

(47)

Then119898 = [120572]119902Γ119902(120572)([120572]

119902minus 1)

For 119886 lt 119887 we have by direct integration

int

119905

119886

1198660(119905 119902119904) 119889

119902119904 = minus

119905 (1 minus 119905)(120572minus1)

Γ119902 (120572)

+119905 (1 minus 119886)

(120572minus1)

Γ119902 (120572)

minus(119905 minus 119886)

(120572)

[120572]119902 Γ119902 (120572)

int

119887

119905

1198660 (119905 119904) 119889119904 = minus

119905 (1 minus 119887)(120572minus1)

Γ119902 (120572)+119905 (1 minus 119905)

(120572minus1)

Γ119902 (120572)

(48)

Then

int

119887

119886

1198660(119905 119902119904) 119889

119902119904 =119905 (1 minus 119886)

(120572minus1)

Γ119902 (120572)minus(119905 minus 119886)

(120572)

[120572]119902 Γ119902 (120572)

minus119905 (1 minus 119887)

(120572minus1)

Γ119902 (120572)

=119905 [(1 minus 119886)

(120572minus1)minus (1 minus 119887)

(120572minus1)]

Γ119902 (120572)

minus(119905 minus 119886)

(120572)

[120572]119902 Γ119902 (120572)= 119877 (119905 119886 119887)

119905119863119902119877 (119905 119886 119887) =

[(1 minus 119886)(120572minus1)

minus (1 minus 119887)(120572minus1)

]

Γ119902 (120572)

minus(119905 minus 119886)

(120572minus1)

Γ119902 (120572)

(49)

The sign of derivative119905119863119902119877 shows that this is an increasing

function of 119905 so the minimum occurs at 119905 = 119886 Let

119877 (119886 119887) =119886

Γ119902 (120572)

[(1 minus 119886)(120572minus1)

minus (1 minus 119887)(120572minus1)

] (50)

The minimal value of 119872(119886 119887) corresponds to the maximalvalue of 119877(119886 119887) Consider

119887119863119902119877 (119886 119887) =

119886 [120572 minus 1]119902 (1 minus 119902119887)(120572minus2)

Γ119902 (120572)gt 0 (51)

The quantity 119877(119886 119887) is an increasing function of 119887 so itsmaximum occurs when 119887 = 1 Let

119877 (119886) =119886 (1 minus 119886)

(120572minus1)

Γ119902 (120572)

(52)

Then the maximum of 119877(119886) occurs when 119886 = 1(1+ [120572minus1]119902)

Consider

min119905isin[119886119887]

int

119887

119886

1198660(119905 119902119904) 119889

119902119904 = 119877(

1

1 + [120572 minus 1]119902

1) (53)

Hence the minimal value of119872(119886 119887) is

119872(1

1 + [120572 minus 1]119902

1) = (119877(1

1 + [120572 minus 1]119902

1))

minus1

(54)

4 The Existence of at Least OnePositive Solution

For convenience we introduce the following notations

119891 (119906) fl sup119905isin[01]

119891 (119905 119906)

119891 (119906) fl inf119905isin[01]

119891 (119905 119906)

1198910 fl lim sup119906rarr0

+

119891 (119906)

119906

1198910fl lim inf119906rarr0

+

119891 (119906)

119906

119891infin fl lim sup

119906rarrinfin

119891 (119906)

119906

119891infin

fl lim inf119906rarrinfin

119891 (119906)

119906

1198910119903 fl sup0le119905le1 0le119906le119903

119891 (119905 119906)

119903

119891119903119903119888

fl inf119886le119905le119887 119903le119906le119903119888

119891 (119905 119906)

119903

(55)

Under hypotheses (H1)ndash(H4) let be defined by

119906 (119905) = int

119887

119886

119866 (119905 119902119904) 119892 (119904) 119906 (119904) 119889119902119904 (56)

Then is a compact linear operator and (119875) sube 119870Hence 119903() is an eigenvalue of with an eigenfunction 120601

1

in119870 Let 1205831fl 1119903()Note that 120583

1ge 1205831 hence the condition

in the following theorem ismore stringent comparedwith thecase if 119903(119871) could be used

Theorem 21 Assume that

(A1) 0 le 1205821198910 lt 1205831(A2) 120583

1lt 120582119891infinle infin

Then (1)-(2) had at least one positive solution

Abstract and Applied Analysis 7

Proof Let 120576 gt 0 be such that 1198910 le (1120582)(1205831minus 120576)Then there

exists 1205880gt 0 such that

119891 (119905 119906) le1

120582(1205831 minus 120576) 119906

forall119906 isin [0 1205880] and almost all 119905 isin [0 1]

(57)

Let 120588 isin (0 1205880]We prove that

119879119906 = 120573119906 for 119906 isin 120597119870120588 120573 ge 1 (58)

which implies the result In fact if (58) does not hold thenthere exist 119906 isin 120597119870

120588and 120573 ge 1 such that 119879119906 = 120573119906

This implies

120573119906 (119905) = 120582int

1

0

119866 (119905 119902119904) 119892 (119904) 119891 (119904 119906 (119904)) 119889119902119904

le (1205831minus 120576) int

1

0

119866 (119905 119902119904) 119892 (119904) 119906 (119904) 119889119902119904

= (1205831minus 120576) 119871119906 (119905)

(59)

Thus we have shown 119906(119905) le (1205831 minus 120576)119871119906(119905)This gives

119906 (119905) le (1205831 minus 120576) 119871 [(1205831 minus 120576) 119871119906 (119905)]

= (1205831minus 120576)21198712119906 (119905)

(60)

And by iterating

119906 (119905) le (1205831 minus 120576)119899119871119899119906 (119905) for 119899 isin 119873 (61)

Therefore

119906 le (1205831 minus 120576)119899 10038171003817100381710038171198711198991003817100381710038171003817 119906

1 le (1205831 minus 120576)119899 10038171003817100381710038171198711198991003817100381710038171003817

(62)

and we have

1 le (1205831 minus 120576) lim

119899rarr+infin

100381710038171003817100381711987111989910038171003817100381710038171119899= (1205831 minus 120576)

1

1205831

lt 1 (63)

a contradiction It follows that

119894119896 (119879119870120588) = 1 for each 120588 isin (0 1205880] (64)

Let 1205881gt 0 120588

1gt 120588 be chosen so that 119891(119905 119906) gt (120583

1120582)119906 for all

119906 ge 1198881205881 119888 as in (H2) and almost all 119905 isin [0 1]

We claim that 119906 = 119879119906 + 1205731206011 for all 120573 gt 0 and 119906 isin 120597119870120588lowast

when 120588lowast gt 1205881 Note that 119906 isin 119870 with 119906 = 120588lowast ge 120588

1

We have 119906(119905) ge 1198881205881for all 119905 isin [119886 119887]

Now if our claim is false then we have

119906 (119905) = 120582int

1

0

119866 (119905 119902119904) 119892 (119904) 119891 (119904 119906 (119904)) 119889119902119904 + 1205731206011 (119905) (65)

Therefore

119906 (119905) ge 1205831 int

119887

119886

119866 (119905 119902119904) 119892 (119904) 119906 (119904) 119889119902119904 + 1205731206011 (119905)

= 1205831119906 (119905) + 1205731206011 (119905)

(66)

From (66) we firstly deduce that 119906(119905) ge 1205731206011(119905) on [0 1]

Then we have

1205831119906 (119905) ge 1205831 (1205731206011 (119905)) = 1205731206011 (119905) (67)

Inserting this into (66) we obtain 119906(119905) ge 21205731206011(119905) for 119905 isin [0 1]

Repeating this process gives

119906 (119905) ge 1198991205731206011 (119905) for 119905 isin [0 1] 119899 isin 119873 (68)

Since 1206011(119905) is strictly positive on [0 1] this is a contradiction

then

119894119870(119879119870120588lowast) = 0 for 119906 isin 120597119870

120588lowast (69)

By (64) and (69) one has

119894119870(119879119870120588lowast 119870120588) = 119894119870(119879119870120588lowast) minus 119894119870(119879119870120588) = minus1 (70)

Therefore 119879 has at least one fixed point 1199060isin 119870120588lowast 119870120588 and

1199060is a positive solution of BVP (1)-(2)

Theorem 22 Assume that

(A3) 1205831 lt 1205821198910 le infin(A4) 0 le 120582119891infin lt 1205831

Then (1)-(2) had at least one positive solution

Proof Let 120576 gt 0 satisfy 1198910gt (1120582)(120583

1+ 120576)Then there exists

1198771 gt 0 such that

119891 (119905 119906) ge1

120582(1205831+ 120576) 119906 forall119905 isin [0 1] 119906 isin [0 1198771] (71)

For any 119906 isin 1205971198701198771

we have by (71) that

119879119906 (119905) = 120582int

1

0

119866 (119905 119902119904) 119892 (119904) 119891 (119904 119906 (119904)) 119889119902119904

ge (1205831 + 120576) int

1

0

119866 (119905 119902119904) 119892 (119904) 119906 (119904) 119889119902119904

ge 1205831119871119906 (119905) forall119905 isin [0 1]

(72)

Let 1 be the positive eigenfunction of 119871 corresponding to 1205831

that is 1 = 12058311198711 Wemay suppose that119879 has no fixed pointon 120597119870119877

1

otherwise the proof is finished In the following wewill show that

119906 minus 119879119906 = 1205731 forall119906 isin 1205971198701198771

120573 ge 0 (73)

If (73) is not true then there is 0isin 1205971198701198771

and 1205730ge 0 such that

0minus1198790= 12057301 It is clear that 120573

0gt 0 and

0= 1198790+12057301ge

12057301Set

120573lowast= sup 120573 0 ge 1205731 (74)

Obviously 120573lowast ge 1205730gt 0 It follows from 119871(119875) sub 119875 that

12058311198710ge 1205831119871120573lowast1= 120573lowast12058311198711= 120573lowast1 (75)

8 Abstract and Applied Analysis

and using this and (72) we have

0 = 1198790 + 12057301 ge 12058311198710 + 12057301 ge 120573lowast1 + 12057301 (76)

which contradicts (74) Thus (73) holdsBy Lemma 9 we have

119894119870(1198791198701198771

) = 0 (77)

On the other hand let 120576 gt 0 satisfy 119891infin lt (1120582)(1205831minus 120576) Then

there exists 1198772gt 1198771such that

119891 (119905 119906) le1

120582(1205831 minus 120576) 119906 forall119905 isin [0 1] 119906 ge 1198772 (78)

By (H5) there exists an 119871infin function 1205931such that

119891 (119905 119906) le1

1205821205931 (119905) forall119906 isin [0 1198772] 119905 isin [0 1] (79)

Hence we have

119891 (119905 119906) le1

120582[(1205831 minus 120576) 119906 + 1205931 (119905)]

forall119906 isin 119877+ 119905 isin [0 1]

(80)

Since 11205831is the radius of the spectrum of 119871 (119868(120583

1minus120576)minus119871)

minus1

existsLet

119862 = int

1

0

1205931 (119904) Φ (119904) 119892 (119904) 119889119902119904

1198770= (

119868

(1205831minus 120576)

minus 119871)

minus1

(119888

(1205831minus 120576)

)

(81)

We prove that for each 119877 gt 1198770

119879119906 = 120573119906 forall119906 isin 120597119870119877 120573 ge 1 (82)

In fact if not there exist 119906 isin 120597119870119877and 120573 ge 1 such that 119879119906 =

120573119906This together with (80) implies

119906 (119905) le int

1

0

119866 (119905 119902119904) 119892 (119904) ((1205831 minus 120576) 119906 (119904) + 1205931 (119904)) 119889119902119904

= (1205831minus 120576) int

1

0

119866 (119905 119902119904) 119892 (119904) 119906 (119904) 119889119902119904

+ int

1

0

119866 (119905 119902119904) 119892 (119904) 1205931 (119904) 119889119902119904

le (1205831 minus 120576) 119871119906 (119905) + 119862

(83)

This implies

(119868

1205831minus 120576minus 119871)119906 (119905) le

119862

1205831minus 120576

119906 (119905) le (119868

1205831minus 120576minus 119871)

minus1

(119862

1205831minus 120576)

= 1198770

(84)

Therefore we have 119906 le 1198770lt 119877 a contradiction Taking

119877 gt 1198772 it follows from (74) and properties of index that

119894119870(119879119870119877) = 1 forall119877 gt 119877

0 (85)

Now (77) and (85) combined imply

119894119870(119879119870119877 1198701198771

) = 119894119870(119879119870119877) minus 119894119870(1198791198701198771

) = 1 (86)

Therefore 119879 has at least one fixed point 1199060isin 1198701198771198701198771

and 1199060

is a positive solution of BVP (1)-(2)

5 The Existence of Two Positive Solutions

Theorem 23 Suppose (A2) (A3) and

(A5) 12058211989101205881015840

le 119898 for some 1205881015840 gt 0

Then (1)-(2) had at least two positive solutions

Proof By (A5) we have

119879119906 (119905) = 120582int

1

0

119866 (119905 119902119904) 119892 (119904) 119891 (119904 119906 (119904)) 119889119902119904

le int

1

0

119866 (119905 119902119904) 119892 (119904) 1205881015840119898119889119902119904

(87)

so that 119879119906 le 1205881015840 = 119906 for all 119906 isin 1205971198811205881015840 Now Lemma 8 yields

119894119896(119879 1198811205881015840) = 1 (88)

On the other hand in view of (A2) we may take 120588lowast gt 1205881015840so that (69) holds (see the proof of Theorem 21) From (A3)we may take 119877

1isin (0 120588

1015840) so that (77) holds (see the proof of

Theorem 22)Combining (88) (69) and (77) we arrive at

119894119896 (119879119870120588lowast 1198811205881015840) = 0 minus 1 = minus1

119894119896(119879 1198811205881015840 1198701198771

) = 1 minus 0 = 1

(89)

Consequently 119879 has at least two fixed points with one on119870120588lowast 1198811205881015840 and the other on119881

1205881015840 1198701198771

Therefore (1)-(2) had atleast two positive solutions

Theorem 24 Suppose (A1) (A4) and

(A6) 12058211989112058810158401205881015840119888 ge 119872 for some 1205881015840 gt 0

Then (1)-(2) had at least two positive solutions

Proof By (A6) we have

119879119906 (119905) = 120582int

1

0

119866 (119905 119902119904) 119892 (119904) 119891 (119904 119906 (119904)) 119889119902119904

ge 120582int

119887

119886

119866 (119905 119902119904) 119892 (119904) 119891 (119904 119906 (119904)) 119889119902119904

ge int

119887

119886

119866 (119905 119902119904) 119892 (119904)1198721205881015840119889119902119904

(90)

Abstract and Applied Analysis 9

so that 119879119906 ge 1205881015840 = 119906 for all 119906 isin 1205971198811205881015840 and by Lemma 8

this yields

119894119896 (119879 1198811205881015840) = 0 (91)

On the other hand in view of (A1) we may take 120588 isin (0 1205881015840)so that (64) holds (see the proof of Theorem 21) In additionfrom (A4) we may take 119877 gt 1205881015840 so that (85) holds (see theproof of Theorem 22)

Combining (91) (64) and (85) we arrive at

119894119896(119879119870119877 1198811205881015840) = 1 minus 0 = 1

119894119896(119879 1198811205881015840 119870120588) = 0 minus 1 = minus1

(92)

Hence 119879 has at least two fixed points with one on 1198811205881015840 119870120588

and the other on 119870119877 1198811205881015840 Therefore (1)-(2) had at least two

positive solutions

We illustrate the applicability of these results with someexamples

Example 25 Consider the problem

11986305(25)119906 (119905) + 120582 (5119905 + 3) (

71199062+ 119906

119906 + 1) (2 + cos 119906) = 0

119905 isin (0 1)

1198632

05119906 (0) = 0

119906 (0) = 0

11986305119906 (1) = 0

(93)

Herewe have119892(119905) = 5119905+3119891(119906) = (2+cos 119906)((71199062+119906)(119906+1))and 2 lt 120572 le 3

It is readily shown that 1198910 = 1198910= 3 119891infin = 21 119891

infin= 7

Also 3119906 le 119891(119906) le 21119906 for 119906 ge 0 By calculation wefind119898 = 019722 and the smallest119872 calculated is119872(119886 119887) asymp119872(0484405 1) asymp 074665 We find 120583

1asymp 030366 Hence

by Theorem 21 there is at least one positive solution if 3120582 lt1205831and 7120582 gt 120583

1 that is there is a positive solution if 120582 isin

(047047 109773)

ByTheorem 22 there does not exist a positive solution ifeither 3120582 gt 120583

1or 21120582 lt 120583

1 that is if 120582 lt 109773 or 120582 gt

015682 no positive solution exists

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

References

[1] N Pongarm S Asawasamrit J Tariboon and S K NtouyasldquoMulti-strip fractional q-integral boundary value problemsfor nonlinear fractional q-difference equationsrdquo Advances inDifference Equations vol 2014 no 1 article 193 2014

[2] R Almeida and N Martins ldquoExistence results for fractional q-difference equations of order120572 isin ]2 3[with three-point bound-ary conditionsrdquo Communications in Nonlinear Science andNumerical Simulation vol 19 no 6 pp 1675ndash1685 2014

[3] Y ZhaoHChen andQZhang ldquoExistence results for fractionalq-difference equations with nonlocal q-integral boundary con-ditionsrdquo Advances in Difference Equations vol 2013 article 482013

[4] R A Ferreira ldquoPositive solutions of a nonlinear q-fractionaldifference equation with integral boundary conditionsrdquo Inter-national Journal of Difference Equations vol 9 no 2 pp 135ndash145 2014

[5] B Ahmad and S K Ntouyas ldquoFractional 119902-difference hybridequations and inclusions with Dirichlet boundary conditionsrdquoAdvances in Difference Equations vol 2014 article 199 2014

[6] B Ahmad S KNtouyas andAAlsaedi ldquoExistence of solutionsfor fractional q-integro-difference inclusions with fractional q-integral boundary conditionsrdquo Advances in Difference Equa-tions vol 2014 article 257 2014

[7] R P Agarwal B Ahmad A Alsaedi and H Al-Hutami ldquoOnnonlinear fractional q-difference equations involving two frac-tional orders with three-point nonlocal boundary conditionsrdquoDynamics of Continuous Discrete amp Impulsive Systems Series AMathematical Analysis vol 21 no 1 pp 135ndash151 2014

[8] M El-Shahed andM A Al-Yami ldquoPositive solutions of bound-ary value problems for nth order q-differential equationsrdquoInternational Journal of Mathematical Archive vol 2 pp 521ndash532 2011

[9] J Ma and J Yang ldquoExistence of solutions for multi-pointboundary value problem of fractional q-difference equationrdquoElectronic Journal ofQualitativeTheory ofDifferential Equationsvol 92 pp 1ndash10 2011

[10] P M Rajkovic S D Marinkovic and M S Stankovic ldquoFrac-tional integrals and derivatives in q-calculusrdquo Applicable Anal-ysis and Discrete Mathematics vol 1 no 1 pp 311ndash323 2007

[11] Y Zhao H Chen and B Qin ldquoMultiple solutions for acoupled systemof nonlinear fractional differential equations viavariational methodsrdquo Applied Mathematics and Computationvol 257 pp 417ndash427 2015

[12] W-X Zhou X Liu and J-G Zhang ldquoSome new existenceand uniqueness results of solutions to semilinear impulsivefractional integro-differential equationsrdquoAdvances inDifferenceEquations vol 2015 article 38 2015

[13] Q Yuan andW Yang ldquoPositive solutions of nonlinear boundaryvalue problems for delayed fractional q-difference systemsrdquoAdvances in Difference Equations vol 2014 no 1 article 51 16pages 2014

[14] W Yang ldquoPositive solutions for nonlinear semipositone frac-tional q-difference system with coupled integral boundaryconditionsrdquo Applied Mathematics and Computation vol 244pp 702ndash725 2014

[15] J R L Webb and K Q Lan ldquoEigenvalue criteria for existenceof multiple positive solutions of nonlinear boundary valueproblems of local and nonlocal typerdquo Topological Methods inNonlinear Analysis vol 27 no 1 pp 91ndash116 2006

[16] J R L Webb ldquoNonlocal conjugate type boundary value prob-lems of higher orderrdquo Nonlinear Analysis Theory Methods ampApplications vol 71 no 5-6 pp 1933ndash1940 2009

[17] J R L Webb and G Infante ldquoNon-local boundary value prob-lems of arbitrary orderrdquo Journal of the London MathematicalSociety vol 79 no 1 pp 238ndash258 2009

10 Abstract and Applied Analysis

[18] J R Webb and G Infante ldquoPositive solutions of nonlocalboundary value problems involving integral conditionsrdquo Non-linear Differential Equations and Applications NoDEA vol 15no 1-2 pp 45ndash67 2008

[19] J R L Webb and G Infante ldquoPositive solutions of nonlocalboundary value problems a unified approachrdquo Journal of theLondon Mathematical Society vol 74 no 3 pp 673ndash693 2006

[20] V Kac and P Cheung Quantum Calculus Springer Science ampBusiness Media 2002

[21] R A C Ferreira ldquoNontrivial solutions for fractional q-difference boundary value problemsrdquo Electronic Journal ofQualitative Theory of Differential Equations no 70 pp 1ndash102010

[22] M S Stankovic P M Rajkovic and S D Marinkovic ldquoOn q-fractional derivatives of Riemann-Liouville and Caputo typerdquohttparxivorgabs09090387

[23] M H Annaby and Z S Mansour q-Fractional Calculus andEquations vol 2056 of Lecture Notes in Mathematics SpringerBerlin Germany 2012

[24] M El-Shahed and W M Shammakh ldquoMultiple positive solu-tions for nonlinear fractional eigenvalue problemwith non localconditionsrdquo Fractional Calculus and Applied Analysis vol 3 pp1ndash13 2012

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

Page 5: Research Article Positive Solutions for Nonlinear -Fractional …downloads.hindawi.com/journals/aaa/2015/759378.pdf · 2019. 7. 31. · Research Article Positive Solutions for Nonlinear

Abstract and Applied Analysis 5

and for 119905 isin [119886 119887]

119866 (119905 119902119904) ge11988810038171003817100381710038171205741003817100381710038171003817

[1 minus 120579 [120574]]G119860 (119902119904) + 119888Φ0 (119902119904) = 119888Φ (119902119904) (36)

Note that 119892Φ isin 119871infin because 119860 has finite variation and

G119860(119902119904) le Φ(119902119904) var(119860)Thus Greenrsquos function 119866(119905 119902119904) satisfies (H1) (H2) for a

functionΦ and the constant 119888

3 Main Result

Set 119864 = 119862[0 1] as a Banach space with the norm 119906 =

sup119905isin[01]

|119906(119905)| Let 119875 = 119906 isin 119864 119906 ge 0 denote the standardcone of nonnegative functions Define

119870 = 119906 isin 119875 min119886le119905le119887

119906 (119905) ge 119888 119906 (37)

where [119886 119887] is some subset of [0 1]Note that 120574 isin 119870 so 119870 = 0 For any 0 lt 119903 lt 119877 lt +infin

let 119870119903= 119906 isin 119870 119906 lt 119903 120597119870

119903= 119906 isin 119870 119906 = 119903

119870119903 = 119906 isin 119870 119906 le 119903 119870119877 119870119903 = 119906 isin 119870 119903 le 119906 le 119877and 119881119903 = 119906 isin 119870 min119905isin[119886119887]119906(119905) lt 119903 and 119881119903 is bounded

Define a nonlinear operator 119879 119875 rarr 119870 and a linearoperator 119871 119875 rarr 119870 by

119879119906 (119905) = 120582int

1

0

119866 (119905 119902119904) 119892 (119904) 119891 (119904 119906 (119904)) 119889119902119904 (38)

119871119906 (119905) fl int1

0

119866 (119905 119902119904) 119892 (119904) 119906 (119904) 119889119902119904 (39)

Lemma 15 (see [18]) Under hypotheses (H1)ndash(H6) the maps119879 119875 rarr 119864 defined in (38) are compact

Theorem 16 Under hypotheses (H1)ndash(H6) the maps are 119879 119875 rarr 119870

Proof For 119906 isin 119875 and 119905 isin [0 1] we have

119879119906 (119905) le 120582int

1

0

Φ(119902119904) 119892 (119904) 119891 (119904 119906 (119904)) 119889119902119904 (40)

Hence

119879119906 le 120582int

1

0

Φ(119902119904) 119892 (119904) 119891 (119904 119906 (119904)) 119889119902119904 (41)

Also for 119905 isin [119886 119887] we have

119879119906 (119905) ge 119888120582int

1

0

Φ(119902119904) 119892 (119904) 119891 (119904 119906 (119904)) 119889119902119904 ge 119888 119879119906 (42)

Similar to the proofs of Lemma 15 and Theorem 16 119871119906(119905) iscompact and maps 119875 into119870

We will use the Krein-Rutman theorem We recall that 120582is an eigenvalue of 119871 with corresponding eigenfunction 120601 if120601 = 0 and 120582120601 = 119871120601 The reciprocals of eigenvalues are calledcharacteristic values of 119871 The radius of the spectrum of 119871denoted by 119903(119871) is given by the well-known spectral radiusformula 119903(119871) = lim

119899rarrinfin1198711198991119899

Theorem 17 (see [15]) Let 119870 be a total cone in a real Banachspace 119864 and let 119864 rarr 119864 be a compact linear operator with(119870) sube 119870 If 119903() gt 0 then there is 120601

1isin 119870 0 such that

1206011= 119903()120601

1

Thus 1205821 fl 119903() is an eigenvalue of the largest possiblereal eigenvalue and 1205831 = 1120582

1is the smallest positive

characteristic value

Lemma 18 (see [15]) Assume that (H1)ndash(H3) hold and let 119871be as defined in (39) Then 119903(119871) gt 0

Theorem 19 (see [15]) When (H1)ndash(H3) hold 119903(119871) is aneigenvalue of 119871 with eigenfunction 120601

1in 119870

Theorem 20 (see [15]) Let 1205831= 1119903(119871) and 120601

1(119905) be a

corresponding eigenfunction in119875 of norm 1Then119898 le 1205831le 119872

where

119898 = ( sup119905isin[01]

int

1

0

119866 (119905 119902119904) 119892 (119904) 119889119902119904)

minus1

119872 = ( inf119905isin[119886119887]

int

119887

119886

119866 (119905 119902119904) 119892 (119904) 119889119902119904)

minus1

(43)

If 119892(119905) gt 0 for 119905 isin [0 1] and 119866(119905 119902119904) gt 0 for 119905 119904 isin [0 1] thefirst inequality is strict unless 1206011(119905) is constant for 119905 isin [0 1] If119892(119905)120601(119905) gt 0 for 119905 isin [119886 119887] the second inequality is strict unless1206011(119905) is constant for 119905 isin [119886 119887]

Proof (for the local BVP (1)-(2) if 119892(119905) equiv 1) We now computethe constant 119898 and the optimal value of119872(119886 119887) that is wedetermine 119886 119887 so that119872(119886 119887) is minimal

For 119902119904 le 119905 we have by direct integration

int

119905

0

1198660 (119905 119902119904) 119889119902119904

= int

119905

0

[

[

[120572 minus 1]119902 119905 (1 minus 119902119904)(120572minus2)

minus (119905 minus 119902119904)(120572minus1)

Γ119902 (120572)

]

]

119889119902119904

=119905 minus 119905 (1 minus 119905)

(120572minus1)

Γ119902 (120572)

minus119905120572

[120572]119902 Γ119902 (120572)

(44)

For 119902119904 ge 119905

int

1

119905

1198660(119905 119902119904) 119889

119902119904 = int

1

119905

[120572 minus 1]119902 119905 (1 minus 119902119904)(120572minus2)

Γ119902 (120572)119889119902119904

=119905 (1 minus 119905)

(120572minus1)

Γ119902 (120572)

(45)

Then we have

int

1

0

1198660 (119905 119902119904) 119889119902119904 =119905

Γ119902 (120572)

minus119905120572

[120572]119902 Γ119902 (120572) (46)

6 Abstract and Applied Analysis

And the maximum of this expression occurs when 119905 = 1hence

sup119905isin[01]

int

1

0

1198660 (119905 119902119904) 119889119902119904 =1

Γ119902 (120572)

minus1

[120572]119902 Γ119902 (120572)

=[120572]119902 minus 1

[120572]119902 Γ119902 (120572)

(47)

Then119898 = [120572]119902Γ119902(120572)([120572]

119902minus 1)

For 119886 lt 119887 we have by direct integration

int

119905

119886

1198660(119905 119902119904) 119889

119902119904 = minus

119905 (1 minus 119905)(120572minus1)

Γ119902 (120572)

+119905 (1 minus 119886)

(120572minus1)

Γ119902 (120572)

minus(119905 minus 119886)

(120572)

[120572]119902 Γ119902 (120572)

int

119887

119905

1198660 (119905 119904) 119889119904 = minus

119905 (1 minus 119887)(120572minus1)

Γ119902 (120572)+119905 (1 minus 119905)

(120572minus1)

Γ119902 (120572)

(48)

Then

int

119887

119886

1198660(119905 119902119904) 119889

119902119904 =119905 (1 minus 119886)

(120572minus1)

Γ119902 (120572)minus(119905 minus 119886)

(120572)

[120572]119902 Γ119902 (120572)

minus119905 (1 minus 119887)

(120572minus1)

Γ119902 (120572)

=119905 [(1 minus 119886)

(120572minus1)minus (1 minus 119887)

(120572minus1)]

Γ119902 (120572)

minus(119905 minus 119886)

(120572)

[120572]119902 Γ119902 (120572)= 119877 (119905 119886 119887)

119905119863119902119877 (119905 119886 119887) =

[(1 minus 119886)(120572minus1)

minus (1 minus 119887)(120572minus1)

]

Γ119902 (120572)

minus(119905 minus 119886)

(120572minus1)

Γ119902 (120572)

(49)

The sign of derivative119905119863119902119877 shows that this is an increasing

function of 119905 so the minimum occurs at 119905 = 119886 Let

119877 (119886 119887) =119886

Γ119902 (120572)

[(1 minus 119886)(120572minus1)

minus (1 minus 119887)(120572minus1)

] (50)

The minimal value of 119872(119886 119887) corresponds to the maximalvalue of 119877(119886 119887) Consider

119887119863119902119877 (119886 119887) =

119886 [120572 minus 1]119902 (1 minus 119902119887)(120572minus2)

Γ119902 (120572)gt 0 (51)

The quantity 119877(119886 119887) is an increasing function of 119887 so itsmaximum occurs when 119887 = 1 Let

119877 (119886) =119886 (1 minus 119886)

(120572minus1)

Γ119902 (120572)

(52)

Then the maximum of 119877(119886) occurs when 119886 = 1(1+ [120572minus1]119902)

Consider

min119905isin[119886119887]

int

119887

119886

1198660(119905 119902119904) 119889

119902119904 = 119877(

1

1 + [120572 minus 1]119902

1) (53)

Hence the minimal value of119872(119886 119887) is

119872(1

1 + [120572 minus 1]119902

1) = (119877(1

1 + [120572 minus 1]119902

1))

minus1

(54)

4 The Existence of at Least OnePositive Solution

For convenience we introduce the following notations

119891 (119906) fl sup119905isin[01]

119891 (119905 119906)

119891 (119906) fl inf119905isin[01]

119891 (119905 119906)

1198910 fl lim sup119906rarr0

+

119891 (119906)

119906

1198910fl lim inf119906rarr0

+

119891 (119906)

119906

119891infin fl lim sup

119906rarrinfin

119891 (119906)

119906

119891infin

fl lim inf119906rarrinfin

119891 (119906)

119906

1198910119903 fl sup0le119905le1 0le119906le119903

119891 (119905 119906)

119903

119891119903119903119888

fl inf119886le119905le119887 119903le119906le119903119888

119891 (119905 119906)

119903

(55)

Under hypotheses (H1)ndash(H4) let be defined by

119906 (119905) = int

119887

119886

119866 (119905 119902119904) 119892 (119904) 119906 (119904) 119889119902119904 (56)

Then is a compact linear operator and (119875) sube 119870Hence 119903() is an eigenvalue of with an eigenfunction 120601

1

in119870 Let 1205831fl 1119903()Note that 120583

1ge 1205831 hence the condition

in the following theorem ismore stringent comparedwith thecase if 119903(119871) could be used

Theorem 21 Assume that

(A1) 0 le 1205821198910 lt 1205831(A2) 120583

1lt 120582119891infinle infin

Then (1)-(2) had at least one positive solution

Abstract and Applied Analysis 7

Proof Let 120576 gt 0 be such that 1198910 le (1120582)(1205831minus 120576)Then there

exists 1205880gt 0 such that

119891 (119905 119906) le1

120582(1205831 minus 120576) 119906

forall119906 isin [0 1205880] and almost all 119905 isin [0 1]

(57)

Let 120588 isin (0 1205880]We prove that

119879119906 = 120573119906 for 119906 isin 120597119870120588 120573 ge 1 (58)

which implies the result In fact if (58) does not hold thenthere exist 119906 isin 120597119870

120588and 120573 ge 1 such that 119879119906 = 120573119906

This implies

120573119906 (119905) = 120582int

1

0

119866 (119905 119902119904) 119892 (119904) 119891 (119904 119906 (119904)) 119889119902119904

le (1205831minus 120576) int

1

0

119866 (119905 119902119904) 119892 (119904) 119906 (119904) 119889119902119904

= (1205831minus 120576) 119871119906 (119905)

(59)

Thus we have shown 119906(119905) le (1205831 minus 120576)119871119906(119905)This gives

119906 (119905) le (1205831 minus 120576) 119871 [(1205831 minus 120576) 119871119906 (119905)]

= (1205831minus 120576)21198712119906 (119905)

(60)

And by iterating

119906 (119905) le (1205831 minus 120576)119899119871119899119906 (119905) for 119899 isin 119873 (61)

Therefore

119906 le (1205831 minus 120576)119899 10038171003817100381710038171198711198991003817100381710038171003817 119906

1 le (1205831 minus 120576)119899 10038171003817100381710038171198711198991003817100381710038171003817

(62)

and we have

1 le (1205831 minus 120576) lim

119899rarr+infin

100381710038171003817100381711987111989910038171003817100381710038171119899= (1205831 minus 120576)

1

1205831

lt 1 (63)

a contradiction It follows that

119894119896 (119879119870120588) = 1 for each 120588 isin (0 1205880] (64)

Let 1205881gt 0 120588

1gt 120588 be chosen so that 119891(119905 119906) gt (120583

1120582)119906 for all

119906 ge 1198881205881 119888 as in (H2) and almost all 119905 isin [0 1]

We claim that 119906 = 119879119906 + 1205731206011 for all 120573 gt 0 and 119906 isin 120597119870120588lowast

when 120588lowast gt 1205881 Note that 119906 isin 119870 with 119906 = 120588lowast ge 120588

1

We have 119906(119905) ge 1198881205881for all 119905 isin [119886 119887]

Now if our claim is false then we have

119906 (119905) = 120582int

1

0

119866 (119905 119902119904) 119892 (119904) 119891 (119904 119906 (119904)) 119889119902119904 + 1205731206011 (119905) (65)

Therefore

119906 (119905) ge 1205831 int

119887

119886

119866 (119905 119902119904) 119892 (119904) 119906 (119904) 119889119902119904 + 1205731206011 (119905)

= 1205831119906 (119905) + 1205731206011 (119905)

(66)

From (66) we firstly deduce that 119906(119905) ge 1205731206011(119905) on [0 1]

Then we have

1205831119906 (119905) ge 1205831 (1205731206011 (119905)) = 1205731206011 (119905) (67)

Inserting this into (66) we obtain 119906(119905) ge 21205731206011(119905) for 119905 isin [0 1]

Repeating this process gives

119906 (119905) ge 1198991205731206011 (119905) for 119905 isin [0 1] 119899 isin 119873 (68)

Since 1206011(119905) is strictly positive on [0 1] this is a contradiction

then

119894119870(119879119870120588lowast) = 0 for 119906 isin 120597119870

120588lowast (69)

By (64) and (69) one has

119894119870(119879119870120588lowast 119870120588) = 119894119870(119879119870120588lowast) minus 119894119870(119879119870120588) = minus1 (70)

Therefore 119879 has at least one fixed point 1199060isin 119870120588lowast 119870120588 and

1199060is a positive solution of BVP (1)-(2)

Theorem 22 Assume that

(A3) 1205831 lt 1205821198910 le infin(A4) 0 le 120582119891infin lt 1205831

Then (1)-(2) had at least one positive solution

Proof Let 120576 gt 0 satisfy 1198910gt (1120582)(120583

1+ 120576)Then there exists

1198771 gt 0 such that

119891 (119905 119906) ge1

120582(1205831+ 120576) 119906 forall119905 isin [0 1] 119906 isin [0 1198771] (71)

For any 119906 isin 1205971198701198771

we have by (71) that

119879119906 (119905) = 120582int

1

0

119866 (119905 119902119904) 119892 (119904) 119891 (119904 119906 (119904)) 119889119902119904

ge (1205831 + 120576) int

1

0

119866 (119905 119902119904) 119892 (119904) 119906 (119904) 119889119902119904

ge 1205831119871119906 (119905) forall119905 isin [0 1]

(72)

Let 1 be the positive eigenfunction of 119871 corresponding to 1205831

that is 1 = 12058311198711 Wemay suppose that119879 has no fixed pointon 120597119870119877

1

otherwise the proof is finished In the following wewill show that

119906 minus 119879119906 = 1205731 forall119906 isin 1205971198701198771

120573 ge 0 (73)

If (73) is not true then there is 0isin 1205971198701198771

and 1205730ge 0 such that

0minus1198790= 12057301 It is clear that 120573

0gt 0 and

0= 1198790+12057301ge

12057301Set

120573lowast= sup 120573 0 ge 1205731 (74)

Obviously 120573lowast ge 1205730gt 0 It follows from 119871(119875) sub 119875 that

12058311198710ge 1205831119871120573lowast1= 120573lowast12058311198711= 120573lowast1 (75)

8 Abstract and Applied Analysis

and using this and (72) we have

0 = 1198790 + 12057301 ge 12058311198710 + 12057301 ge 120573lowast1 + 12057301 (76)

which contradicts (74) Thus (73) holdsBy Lemma 9 we have

119894119870(1198791198701198771

) = 0 (77)

On the other hand let 120576 gt 0 satisfy 119891infin lt (1120582)(1205831minus 120576) Then

there exists 1198772gt 1198771such that

119891 (119905 119906) le1

120582(1205831 minus 120576) 119906 forall119905 isin [0 1] 119906 ge 1198772 (78)

By (H5) there exists an 119871infin function 1205931such that

119891 (119905 119906) le1

1205821205931 (119905) forall119906 isin [0 1198772] 119905 isin [0 1] (79)

Hence we have

119891 (119905 119906) le1

120582[(1205831 minus 120576) 119906 + 1205931 (119905)]

forall119906 isin 119877+ 119905 isin [0 1]

(80)

Since 11205831is the radius of the spectrum of 119871 (119868(120583

1minus120576)minus119871)

minus1

existsLet

119862 = int

1

0

1205931 (119904) Φ (119904) 119892 (119904) 119889119902119904

1198770= (

119868

(1205831minus 120576)

minus 119871)

minus1

(119888

(1205831minus 120576)

)

(81)

We prove that for each 119877 gt 1198770

119879119906 = 120573119906 forall119906 isin 120597119870119877 120573 ge 1 (82)

In fact if not there exist 119906 isin 120597119870119877and 120573 ge 1 such that 119879119906 =

120573119906This together with (80) implies

119906 (119905) le int

1

0

119866 (119905 119902119904) 119892 (119904) ((1205831 minus 120576) 119906 (119904) + 1205931 (119904)) 119889119902119904

= (1205831minus 120576) int

1

0

119866 (119905 119902119904) 119892 (119904) 119906 (119904) 119889119902119904

+ int

1

0

119866 (119905 119902119904) 119892 (119904) 1205931 (119904) 119889119902119904

le (1205831 minus 120576) 119871119906 (119905) + 119862

(83)

This implies

(119868

1205831minus 120576minus 119871)119906 (119905) le

119862

1205831minus 120576

119906 (119905) le (119868

1205831minus 120576minus 119871)

minus1

(119862

1205831minus 120576)

= 1198770

(84)

Therefore we have 119906 le 1198770lt 119877 a contradiction Taking

119877 gt 1198772 it follows from (74) and properties of index that

119894119870(119879119870119877) = 1 forall119877 gt 119877

0 (85)

Now (77) and (85) combined imply

119894119870(119879119870119877 1198701198771

) = 119894119870(119879119870119877) minus 119894119870(1198791198701198771

) = 1 (86)

Therefore 119879 has at least one fixed point 1199060isin 1198701198771198701198771

and 1199060

is a positive solution of BVP (1)-(2)

5 The Existence of Two Positive Solutions

Theorem 23 Suppose (A2) (A3) and

(A5) 12058211989101205881015840

le 119898 for some 1205881015840 gt 0

Then (1)-(2) had at least two positive solutions

Proof By (A5) we have

119879119906 (119905) = 120582int

1

0

119866 (119905 119902119904) 119892 (119904) 119891 (119904 119906 (119904)) 119889119902119904

le int

1

0

119866 (119905 119902119904) 119892 (119904) 1205881015840119898119889119902119904

(87)

so that 119879119906 le 1205881015840 = 119906 for all 119906 isin 1205971198811205881015840 Now Lemma 8 yields

119894119896(119879 1198811205881015840) = 1 (88)

On the other hand in view of (A2) we may take 120588lowast gt 1205881015840so that (69) holds (see the proof of Theorem 21) From (A3)we may take 119877

1isin (0 120588

1015840) so that (77) holds (see the proof of

Theorem 22)Combining (88) (69) and (77) we arrive at

119894119896 (119879119870120588lowast 1198811205881015840) = 0 minus 1 = minus1

119894119896(119879 1198811205881015840 1198701198771

) = 1 minus 0 = 1

(89)

Consequently 119879 has at least two fixed points with one on119870120588lowast 1198811205881015840 and the other on119881

1205881015840 1198701198771

Therefore (1)-(2) had atleast two positive solutions

Theorem 24 Suppose (A1) (A4) and

(A6) 12058211989112058810158401205881015840119888 ge 119872 for some 1205881015840 gt 0

Then (1)-(2) had at least two positive solutions

Proof By (A6) we have

119879119906 (119905) = 120582int

1

0

119866 (119905 119902119904) 119892 (119904) 119891 (119904 119906 (119904)) 119889119902119904

ge 120582int

119887

119886

119866 (119905 119902119904) 119892 (119904) 119891 (119904 119906 (119904)) 119889119902119904

ge int

119887

119886

119866 (119905 119902119904) 119892 (119904)1198721205881015840119889119902119904

(90)

Abstract and Applied Analysis 9

so that 119879119906 ge 1205881015840 = 119906 for all 119906 isin 1205971198811205881015840 and by Lemma 8

this yields

119894119896 (119879 1198811205881015840) = 0 (91)

On the other hand in view of (A1) we may take 120588 isin (0 1205881015840)so that (64) holds (see the proof of Theorem 21) In additionfrom (A4) we may take 119877 gt 1205881015840 so that (85) holds (see theproof of Theorem 22)

Combining (91) (64) and (85) we arrive at

119894119896(119879119870119877 1198811205881015840) = 1 minus 0 = 1

119894119896(119879 1198811205881015840 119870120588) = 0 minus 1 = minus1

(92)

Hence 119879 has at least two fixed points with one on 1198811205881015840 119870120588

and the other on 119870119877 1198811205881015840 Therefore (1)-(2) had at least two

positive solutions

We illustrate the applicability of these results with someexamples

Example 25 Consider the problem

11986305(25)119906 (119905) + 120582 (5119905 + 3) (

71199062+ 119906

119906 + 1) (2 + cos 119906) = 0

119905 isin (0 1)

1198632

05119906 (0) = 0

119906 (0) = 0

11986305119906 (1) = 0

(93)

Herewe have119892(119905) = 5119905+3119891(119906) = (2+cos 119906)((71199062+119906)(119906+1))and 2 lt 120572 le 3

It is readily shown that 1198910 = 1198910= 3 119891infin = 21 119891

infin= 7

Also 3119906 le 119891(119906) le 21119906 for 119906 ge 0 By calculation wefind119898 = 019722 and the smallest119872 calculated is119872(119886 119887) asymp119872(0484405 1) asymp 074665 We find 120583

1asymp 030366 Hence

by Theorem 21 there is at least one positive solution if 3120582 lt1205831and 7120582 gt 120583

1 that is there is a positive solution if 120582 isin

(047047 109773)

ByTheorem 22 there does not exist a positive solution ifeither 3120582 gt 120583

1or 21120582 lt 120583

1 that is if 120582 lt 109773 or 120582 gt

015682 no positive solution exists

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

References

[1] N Pongarm S Asawasamrit J Tariboon and S K NtouyasldquoMulti-strip fractional q-integral boundary value problemsfor nonlinear fractional q-difference equationsrdquo Advances inDifference Equations vol 2014 no 1 article 193 2014

[2] R Almeida and N Martins ldquoExistence results for fractional q-difference equations of order120572 isin ]2 3[with three-point bound-ary conditionsrdquo Communications in Nonlinear Science andNumerical Simulation vol 19 no 6 pp 1675ndash1685 2014

[3] Y ZhaoHChen andQZhang ldquoExistence results for fractionalq-difference equations with nonlocal q-integral boundary con-ditionsrdquo Advances in Difference Equations vol 2013 article 482013

[4] R A Ferreira ldquoPositive solutions of a nonlinear q-fractionaldifference equation with integral boundary conditionsrdquo Inter-national Journal of Difference Equations vol 9 no 2 pp 135ndash145 2014

[5] B Ahmad and S K Ntouyas ldquoFractional 119902-difference hybridequations and inclusions with Dirichlet boundary conditionsrdquoAdvances in Difference Equations vol 2014 article 199 2014

[6] B Ahmad S KNtouyas andAAlsaedi ldquoExistence of solutionsfor fractional q-integro-difference inclusions with fractional q-integral boundary conditionsrdquo Advances in Difference Equa-tions vol 2014 article 257 2014

[7] R P Agarwal B Ahmad A Alsaedi and H Al-Hutami ldquoOnnonlinear fractional q-difference equations involving two frac-tional orders with three-point nonlocal boundary conditionsrdquoDynamics of Continuous Discrete amp Impulsive Systems Series AMathematical Analysis vol 21 no 1 pp 135ndash151 2014

[8] M El-Shahed andM A Al-Yami ldquoPositive solutions of bound-ary value problems for nth order q-differential equationsrdquoInternational Journal of Mathematical Archive vol 2 pp 521ndash532 2011

[9] J Ma and J Yang ldquoExistence of solutions for multi-pointboundary value problem of fractional q-difference equationrdquoElectronic Journal ofQualitativeTheory ofDifferential Equationsvol 92 pp 1ndash10 2011

[10] P M Rajkovic S D Marinkovic and M S Stankovic ldquoFrac-tional integrals and derivatives in q-calculusrdquo Applicable Anal-ysis and Discrete Mathematics vol 1 no 1 pp 311ndash323 2007

[11] Y Zhao H Chen and B Qin ldquoMultiple solutions for acoupled systemof nonlinear fractional differential equations viavariational methodsrdquo Applied Mathematics and Computationvol 257 pp 417ndash427 2015

[12] W-X Zhou X Liu and J-G Zhang ldquoSome new existenceand uniqueness results of solutions to semilinear impulsivefractional integro-differential equationsrdquoAdvances inDifferenceEquations vol 2015 article 38 2015

[13] Q Yuan andW Yang ldquoPositive solutions of nonlinear boundaryvalue problems for delayed fractional q-difference systemsrdquoAdvances in Difference Equations vol 2014 no 1 article 51 16pages 2014

[14] W Yang ldquoPositive solutions for nonlinear semipositone frac-tional q-difference system with coupled integral boundaryconditionsrdquo Applied Mathematics and Computation vol 244pp 702ndash725 2014

[15] J R L Webb and K Q Lan ldquoEigenvalue criteria for existenceof multiple positive solutions of nonlinear boundary valueproblems of local and nonlocal typerdquo Topological Methods inNonlinear Analysis vol 27 no 1 pp 91ndash116 2006

[16] J R L Webb ldquoNonlocal conjugate type boundary value prob-lems of higher orderrdquo Nonlinear Analysis Theory Methods ampApplications vol 71 no 5-6 pp 1933ndash1940 2009

[17] J R L Webb and G Infante ldquoNon-local boundary value prob-lems of arbitrary orderrdquo Journal of the London MathematicalSociety vol 79 no 1 pp 238ndash258 2009

10 Abstract and Applied Analysis

[18] J R Webb and G Infante ldquoPositive solutions of nonlocalboundary value problems involving integral conditionsrdquo Non-linear Differential Equations and Applications NoDEA vol 15no 1-2 pp 45ndash67 2008

[19] J R L Webb and G Infante ldquoPositive solutions of nonlocalboundary value problems a unified approachrdquo Journal of theLondon Mathematical Society vol 74 no 3 pp 673ndash693 2006

[20] V Kac and P Cheung Quantum Calculus Springer Science ampBusiness Media 2002

[21] R A C Ferreira ldquoNontrivial solutions for fractional q-difference boundary value problemsrdquo Electronic Journal ofQualitative Theory of Differential Equations no 70 pp 1ndash102010

[22] M S Stankovic P M Rajkovic and S D Marinkovic ldquoOn q-fractional derivatives of Riemann-Liouville and Caputo typerdquohttparxivorgabs09090387

[23] M H Annaby and Z S Mansour q-Fractional Calculus andEquations vol 2056 of Lecture Notes in Mathematics SpringerBerlin Germany 2012

[24] M El-Shahed and W M Shammakh ldquoMultiple positive solu-tions for nonlinear fractional eigenvalue problemwith non localconditionsrdquo Fractional Calculus and Applied Analysis vol 3 pp1ndash13 2012

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

Page 6: Research Article Positive Solutions for Nonlinear -Fractional …downloads.hindawi.com/journals/aaa/2015/759378.pdf · 2019. 7. 31. · Research Article Positive Solutions for Nonlinear

6 Abstract and Applied Analysis

And the maximum of this expression occurs when 119905 = 1hence

sup119905isin[01]

int

1

0

1198660 (119905 119902119904) 119889119902119904 =1

Γ119902 (120572)

minus1

[120572]119902 Γ119902 (120572)

=[120572]119902 minus 1

[120572]119902 Γ119902 (120572)

(47)

Then119898 = [120572]119902Γ119902(120572)([120572]

119902minus 1)

For 119886 lt 119887 we have by direct integration

int

119905

119886

1198660(119905 119902119904) 119889

119902119904 = minus

119905 (1 minus 119905)(120572minus1)

Γ119902 (120572)

+119905 (1 minus 119886)

(120572minus1)

Γ119902 (120572)

minus(119905 minus 119886)

(120572)

[120572]119902 Γ119902 (120572)

int

119887

119905

1198660 (119905 119904) 119889119904 = minus

119905 (1 minus 119887)(120572minus1)

Γ119902 (120572)+119905 (1 minus 119905)

(120572minus1)

Γ119902 (120572)

(48)

Then

int

119887

119886

1198660(119905 119902119904) 119889

119902119904 =119905 (1 minus 119886)

(120572minus1)

Γ119902 (120572)minus(119905 minus 119886)

(120572)

[120572]119902 Γ119902 (120572)

minus119905 (1 minus 119887)

(120572minus1)

Γ119902 (120572)

=119905 [(1 minus 119886)

(120572minus1)minus (1 minus 119887)

(120572minus1)]

Γ119902 (120572)

minus(119905 minus 119886)

(120572)

[120572]119902 Γ119902 (120572)= 119877 (119905 119886 119887)

119905119863119902119877 (119905 119886 119887) =

[(1 minus 119886)(120572minus1)

minus (1 minus 119887)(120572minus1)

]

Γ119902 (120572)

minus(119905 minus 119886)

(120572minus1)

Γ119902 (120572)

(49)

The sign of derivative119905119863119902119877 shows that this is an increasing

function of 119905 so the minimum occurs at 119905 = 119886 Let

119877 (119886 119887) =119886

Γ119902 (120572)

[(1 minus 119886)(120572minus1)

minus (1 minus 119887)(120572minus1)

] (50)

The minimal value of 119872(119886 119887) corresponds to the maximalvalue of 119877(119886 119887) Consider

119887119863119902119877 (119886 119887) =

119886 [120572 minus 1]119902 (1 minus 119902119887)(120572minus2)

Γ119902 (120572)gt 0 (51)

The quantity 119877(119886 119887) is an increasing function of 119887 so itsmaximum occurs when 119887 = 1 Let

119877 (119886) =119886 (1 minus 119886)

(120572minus1)

Γ119902 (120572)

(52)

Then the maximum of 119877(119886) occurs when 119886 = 1(1+ [120572minus1]119902)

Consider

min119905isin[119886119887]

int

119887

119886

1198660(119905 119902119904) 119889

119902119904 = 119877(

1

1 + [120572 minus 1]119902

1) (53)

Hence the minimal value of119872(119886 119887) is

119872(1

1 + [120572 minus 1]119902

1) = (119877(1

1 + [120572 minus 1]119902

1))

minus1

(54)

4 The Existence of at Least OnePositive Solution

For convenience we introduce the following notations

119891 (119906) fl sup119905isin[01]

119891 (119905 119906)

119891 (119906) fl inf119905isin[01]

119891 (119905 119906)

1198910 fl lim sup119906rarr0

+

119891 (119906)

119906

1198910fl lim inf119906rarr0

+

119891 (119906)

119906

119891infin fl lim sup

119906rarrinfin

119891 (119906)

119906

119891infin

fl lim inf119906rarrinfin

119891 (119906)

119906

1198910119903 fl sup0le119905le1 0le119906le119903

119891 (119905 119906)

119903

119891119903119903119888

fl inf119886le119905le119887 119903le119906le119903119888

119891 (119905 119906)

119903

(55)

Under hypotheses (H1)ndash(H4) let be defined by

119906 (119905) = int

119887

119886

119866 (119905 119902119904) 119892 (119904) 119906 (119904) 119889119902119904 (56)

Then is a compact linear operator and (119875) sube 119870Hence 119903() is an eigenvalue of with an eigenfunction 120601

1

in119870 Let 1205831fl 1119903()Note that 120583

1ge 1205831 hence the condition

in the following theorem ismore stringent comparedwith thecase if 119903(119871) could be used

Theorem 21 Assume that

(A1) 0 le 1205821198910 lt 1205831(A2) 120583

1lt 120582119891infinle infin

Then (1)-(2) had at least one positive solution

Abstract and Applied Analysis 7

Proof Let 120576 gt 0 be such that 1198910 le (1120582)(1205831minus 120576)Then there

exists 1205880gt 0 such that

119891 (119905 119906) le1

120582(1205831 minus 120576) 119906

forall119906 isin [0 1205880] and almost all 119905 isin [0 1]

(57)

Let 120588 isin (0 1205880]We prove that

119879119906 = 120573119906 for 119906 isin 120597119870120588 120573 ge 1 (58)

which implies the result In fact if (58) does not hold thenthere exist 119906 isin 120597119870

120588and 120573 ge 1 such that 119879119906 = 120573119906

This implies

120573119906 (119905) = 120582int

1

0

119866 (119905 119902119904) 119892 (119904) 119891 (119904 119906 (119904)) 119889119902119904

le (1205831minus 120576) int

1

0

119866 (119905 119902119904) 119892 (119904) 119906 (119904) 119889119902119904

= (1205831minus 120576) 119871119906 (119905)

(59)

Thus we have shown 119906(119905) le (1205831 minus 120576)119871119906(119905)This gives

119906 (119905) le (1205831 minus 120576) 119871 [(1205831 minus 120576) 119871119906 (119905)]

= (1205831minus 120576)21198712119906 (119905)

(60)

And by iterating

119906 (119905) le (1205831 minus 120576)119899119871119899119906 (119905) for 119899 isin 119873 (61)

Therefore

119906 le (1205831 minus 120576)119899 10038171003817100381710038171198711198991003817100381710038171003817 119906

1 le (1205831 minus 120576)119899 10038171003817100381710038171198711198991003817100381710038171003817

(62)

and we have

1 le (1205831 minus 120576) lim

119899rarr+infin

100381710038171003817100381711987111989910038171003817100381710038171119899= (1205831 minus 120576)

1

1205831

lt 1 (63)

a contradiction It follows that

119894119896 (119879119870120588) = 1 for each 120588 isin (0 1205880] (64)

Let 1205881gt 0 120588

1gt 120588 be chosen so that 119891(119905 119906) gt (120583

1120582)119906 for all

119906 ge 1198881205881 119888 as in (H2) and almost all 119905 isin [0 1]

We claim that 119906 = 119879119906 + 1205731206011 for all 120573 gt 0 and 119906 isin 120597119870120588lowast

when 120588lowast gt 1205881 Note that 119906 isin 119870 with 119906 = 120588lowast ge 120588

1

We have 119906(119905) ge 1198881205881for all 119905 isin [119886 119887]

Now if our claim is false then we have

119906 (119905) = 120582int

1

0

119866 (119905 119902119904) 119892 (119904) 119891 (119904 119906 (119904)) 119889119902119904 + 1205731206011 (119905) (65)

Therefore

119906 (119905) ge 1205831 int

119887

119886

119866 (119905 119902119904) 119892 (119904) 119906 (119904) 119889119902119904 + 1205731206011 (119905)

= 1205831119906 (119905) + 1205731206011 (119905)

(66)

From (66) we firstly deduce that 119906(119905) ge 1205731206011(119905) on [0 1]

Then we have

1205831119906 (119905) ge 1205831 (1205731206011 (119905)) = 1205731206011 (119905) (67)

Inserting this into (66) we obtain 119906(119905) ge 21205731206011(119905) for 119905 isin [0 1]

Repeating this process gives

119906 (119905) ge 1198991205731206011 (119905) for 119905 isin [0 1] 119899 isin 119873 (68)

Since 1206011(119905) is strictly positive on [0 1] this is a contradiction

then

119894119870(119879119870120588lowast) = 0 for 119906 isin 120597119870

120588lowast (69)

By (64) and (69) one has

119894119870(119879119870120588lowast 119870120588) = 119894119870(119879119870120588lowast) minus 119894119870(119879119870120588) = minus1 (70)

Therefore 119879 has at least one fixed point 1199060isin 119870120588lowast 119870120588 and

1199060is a positive solution of BVP (1)-(2)

Theorem 22 Assume that

(A3) 1205831 lt 1205821198910 le infin(A4) 0 le 120582119891infin lt 1205831

Then (1)-(2) had at least one positive solution

Proof Let 120576 gt 0 satisfy 1198910gt (1120582)(120583

1+ 120576)Then there exists

1198771 gt 0 such that

119891 (119905 119906) ge1

120582(1205831+ 120576) 119906 forall119905 isin [0 1] 119906 isin [0 1198771] (71)

For any 119906 isin 1205971198701198771

we have by (71) that

119879119906 (119905) = 120582int

1

0

119866 (119905 119902119904) 119892 (119904) 119891 (119904 119906 (119904)) 119889119902119904

ge (1205831 + 120576) int

1

0

119866 (119905 119902119904) 119892 (119904) 119906 (119904) 119889119902119904

ge 1205831119871119906 (119905) forall119905 isin [0 1]

(72)

Let 1 be the positive eigenfunction of 119871 corresponding to 1205831

that is 1 = 12058311198711 Wemay suppose that119879 has no fixed pointon 120597119870119877

1

otherwise the proof is finished In the following wewill show that

119906 minus 119879119906 = 1205731 forall119906 isin 1205971198701198771

120573 ge 0 (73)

If (73) is not true then there is 0isin 1205971198701198771

and 1205730ge 0 such that

0minus1198790= 12057301 It is clear that 120573

0gt 0 and

0= 1198790+12057301ge

12057301Set

120573lowast= sup 120573 0 ge 1205731 (74)

Obviously 120573lowast ge 1205730gt 0 It follows from 119871(119875) sub 119875 that

12058311198710ge 1205831119871120573lowast1= 120573lowast12058311198711= 120573lowast1 (75)

8 Abstract and Applied Analysis

and using this and (72) we have

0 = 1198790 + 12057301 ge 12058311198710 + 12057301 ge 120573lowast1 + 12057301 (76)

which contradicts (74) Thus (73) holdsBy Lemma 9 we have

119894119870(1198791198701198771

) = 0 (77)

On the other hand let 120576 gt 0 satisfy 119891infin lt (1120582)(1205831minus 120576) Then

there exists 1198772gt 1198771such that

119891 (119905 119906) le1

120582(1205831 minus 120576) 119906 forall119905 isin [0 1] 119906 ge 1198772 (78)

By (H5) there exists an 119871infin function 1205931such that

119891 (119905 119906) le1

1205821205931 (119905) forall119906 isin [0 1198772] 119905 isin [0 1] (79)

Hence we have

119891 (119905 119906) le1

120582[(1205831 minus 120576) 119906 + 1205931 (119905)]

forall119906 isin 119877+ 119905 isin [0 1]

(80)

Since 11205831is the radius of the spectrum of 119871 (119868(120583

1minus120576)minus119871)

minus1

existsLet

119862 = int

1

0

1205931 (119904) Φ (119904) 119892 (119904) 119889119902119904

1198770= (

119868

(1205831minus 120576)

minus 119871)

minus1

(119888

(1205831minus 120576)

)

(81)

We prove that for each 119877 gt 1198770

119879119906 = 120573119906 forall119906 isin 120597119870119877 120573 ge 1 (82)

In fact if not there exist 119906 isin 120597119870119877and 120573 ge 1 such that 119879119906 =

120573119906This together with (80) implies

119906 (119905) le int

1

0

119866 (119905 119902119904) 119892 (119904) ((1205831 minus 120576) 119906 (119904) + 1205931 (119904)) 119889119902119904

= (1205831minus 120576) int

1

0

119866 (119905 119902119904) 119892 (119904) 119906 (119904) 119889119902119904

+ int

1

0

119866 (119905 119902119904) 119892 (119904) 1205931 (119904) 119889119902119904

le (1205831 minus 120576) 119871119906 (119905) + 119862

(83)

This implies

(119868

1205831minus 120576minus 119871)119906 (119905) le

119862

1205831minus 120576

119906 (119905) le (119868

1205831minus 120576minus 119871)

minus1

(119862

1205831minus 120576)

= 1198770

(84)

Therefore we have 119906 le 1198770lt 119877 a contradiction Taking

119877 gt 1198772 it follows from (74) and properties of index that

119894119870(119879119870119877) = 1 forall119877 gt 119877

0 (85)

Now (77) and (85) combined imply

119894119870(119879119870119877 1198701198771

) = 119894119870(119879119870119877) minus 119894119870(1198791198701198771

) = 1 (86)

Therefore 119879 has at least one fixed point 1199060isin 1198701198771198701198771

and 1199060

is a positive solution of BVP (1)-(2)

5 The Existence of Two Positive Solutions

Theorem 23 Suppose (A2) (A3) and

(A5) 12058211989101205881015840

le 119898 for some 1205881015840 gt 0

Then (1)-(2) had at least two positive solutions

Proof By (A5) we have

119879119906 (119905) = 120582int

1

0

119866 (119905 119902119904) 119892 (119904) 119891 (119904 119906 (119904)) 119889119902119904

le int

1

0

119866 (119905 119902119904) 119892 (119904) 1205881015840119898119889119902119904

(87)

so that 119879119906 le 1205881015840 = 119906 for all 119906 isin 1205971198811205881015840 Now Lemma 8 yields

119894119896(119879 1198811205881015840) = 1 (88)

On the other hand in view of (A2) we may take 120588lowast gt 1205881015840so that (69) holds (see the proof of Theorem 21) From (A3)we may take 119877

1isin (0 120588

1015840) so that (77) holds (see the proof of

Theorem 22)Combining (88) (69) and (77) we arrive at

119894119896 (119879119870120588lowast 1198811205881015840) = 0 minus 1 = minus1

119894119896(119879 1198811205881015840 1198701198771

) = 1 minus 0 = 1

(89)

Consequently 119879 has at least two fixed points with one on119870120588lowast 1198811205881015840 and the other on119881

1205881015840 1198701198771

Therefore (1)-(2) had atleast two positive solutions

Theorem 24 Suppose (A1) (A4) and

(A6) 12058211989112058810158401205881015840119888 ge 119872 for some 1205881015840 gt 0

Then (1)-(2) had at least two positive solutions

Proof By (A6) we have

119879119906 (119905) = 120582int

1

0

119866 (119905 119902119904) 119892 (119904) 119891 (119904 119906 (119904)) 119889119902119904

ge 120582int

119887

119886

119866 (119905 119902119904) 119892 (119904) 119891 (119904 119906 (119904)) 119889119902119904

ge int

119887

119886

119866 (119905 119902119904) 119892 (119904)1198721205881015840119889119902119904

(90)

Abstract and Applied Analysis 9

so that 119879119906 ge 1205881015840 = 119906 for all 119906 isin 1205971198811205881015840 and by Lemma 8

this yields

119894119896 (119879 1198811205881015840) = 0 (91)

On the other hand in view of (A1) we may take 120588 isin (0 1205881015840)so that (64) holds (see the proof of Theorem 21) In additionfrom (A4) we may take 119877 gt 1205881015840 so that (85) holds (see theproof of Theorem 22)

Combining (91) (64) and (85) we arrive at

119894119896(119879119870119877 1198811205881015840) = 1 minus 0 = 1

119894119896(119879 1198811205881015840 119870120588) = 0 minus 1 = minus1

(92)

Hence 119879 has at least two fixed points with one on 1198811205881015840 119870120588

and the other on 119870119877 1198811205881015840 Therefore (1)-(2) had at least two

positive solutions

We illustrate the applicability of these results with someexamples

Example 25 Consider the problem

11986305(25)119906 (119905) + 120582 (5119905 + 3) (

71199062+ 119906

119906 + 1) (2 + cos 119906) = 0

119905 isin (0 1)

1198632

05119906 (0) = 0

119906 (0) = 0

11986305119906 (1) = 0

(93)

Herewe have119892(119905) = 5119905+3119891(119906) = (2+cos 119906)((71199062+119906)(119906+1))and 2 lt 120572 le 3

It is readily shown that 1198910 = 1198910= 3 119891infin = 21 119891

infin= 7

Also 3119906 le 119891(119906) le 21119906 for 119906 ge 0 By calculation wefind119898 = 019722 and the smallest119872 calculated is119872(119886 119887) asymp119872(0484405 1) asymp 074665 We find 120583

1asymp 030366 Hence

by Theorem 21 there is at least one positive solution if 3120582 lt1205831and 7120582 gt 120583

1 that is there is a positive solution if 120582 isin

(047047 109773)

ByTheorem 22 there does not exist a positive solution ifeither 3120582 gt 120583

1or 21120582 lt 120583

1 that is if 120582 lt 109773 or 120582 gt

015682 no positive solution exists

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

References

[1] N Pongarm S Asawasamrit J Tariboon and S K NtouyasldquoMulti-strip fractional q-integral boundary value problemsfor nonlinear fractional q-difference equationsrdquo Advances inDifference Equations vol 2014 no 1 article 193 2014

[2] R Almeida and N Martins ldquoExistence results for fractional q-difference equations of order120572 isin ]2 3[with three-point bound-ary conditionsrdquo Communications in Nonlinear Science andNumerical Simulation vol 19 no 6 pp 1675ndash1685 2014

[3] Y ZhaoHChen andQZhang ldquoExistence results for fractionalq-difference equations with nonlocal q-integral boundary con-ditionsrdquo Advances in Difference Equations vol 2013 article 482013

[4] R A Ferreira ldquoPositive solutions of a nonlinear q-fractionaldifference equation with integral boundary conditionsrdquo Inter-national Journal of Difference Equations vol 9 no 2 pp 135ndash145 2014

[5] B Ahmad and S K Ntouyas ldquoFractional 119902-difference hybridequations and inclusions with Dirichlet boundary conditionsrdquoAdvances in Difference Equations vol 2014 article 199 2014

[6] B Ahmad S KNtouyas andAAlsaedi ldquoExistence of solutionsfor fractional q-integro-difference inclusions with fractional q-integral boundary conditionsrdquo Advances in Difference Equa-tions vol 2014 article 257 2014

[7] R P Agarwal B Ahmad A Alsaedi and H Al-Hutami ldquoOnnonlinear fractional q-difference equations involving two frac-tional orders with three-point nonlocal boundary conditionsrdquoDynamics of Continuous Discrete amp Impulsive Systems Series AMathematical Analysis vol 21 no 1 pp 135ndash151 2014

[8] M El-Shahed andM A Al-Yami ldquoPositive solutions of bound-ary value problems for nth order q-differential equationsrdquoInternational Journal of Mathematical Archive vol 2 pp 521ndash532 2011

[9] J Ma and J Yang ldquoExistence of solutions for multi-pointboundary value problem of fractional q-difference equationrdquoElectronic Journal ofQualitativeTheory ofDifferential Equationsvol 92 pp 1ndash10 2011

[10] P M Rajkovic S D Marinkovic and M S Stankovic ldquoFrac-tional integrals and derivatives in q-calculusrdquo Applicable Anal-ysis and Discrete Mathematics vol 1 no 1 pp 311ndash323 2007

[11] Y Zhao H Chen and B Qin ldquoMultiple solutions for acoupled systemof nonlinear fractional differential equations viavariational methodsrdquo Applied Mathematics and Computationvol 257 pp 417ndash427 2015

[12] W-X Zhou X Liu and J-G Zhang ldquoSome new existenceand uniqueness results of solutions to semilinear impulsivefractional integro-differential equationsrdquoAdvances inDifferenceEquations vol 2015 article 38 2015

[13] Q Yuan andW Yang ldquoPositive solutions of nonlinear boundaryvalue problems for delayed fractional q-difference systemsrdquoAdvances in Difference Equations vol 2014 no 1 article 51 16pages 2014

[14] W Yang ldquoPositive solutions for nonlinear semipositone frac-tional q-difference system with coupled integral boundaryconditionsrdquo Applied Mathematics and Computation vol 244pp 702ndash725 2014

[15] J R L Webb and K Q Lan ldquoEigenvalue criteria for existenceof multiple positive solutions of nonlinear boundary valueproblems of local and nonlocal typerdquo Topological Methods inNonlinear Analysis vol 27 no 1 pp 91ndash116 2006

[16] J R L Webb ldquoNonlocal conjugate type boundary value prob-lems of higher orderrdquo Nonlinear Analysis Theory Methods ampApplications vol 71 no 5-6 pp 1933ndash1940 2009

[17] J R L Webb and G Infante ldquoNon-local boundary value prob-lems of arbitrary orderrdquo Journal of the London MathematicalSociety vol 79 no 1 pp 238ndash258 2009

10 Abstract and Applied Analysis

[18] J R Webb and G Infante ldquoPositive solutions of nonlocalboundary value problems involving integral conditionsrdquo Non-linear Differential Equations and Applications NoDEA vol 15no 1-2 pp 45ndash67 2008

[19] J R L Webb and G Infante ldquoPositive solutions of nonlocalboundary value problems a unified approachrdquo Journal of theLondon Mathematical Society vol 74 no 3 pp 673ndash693 2006

[20] V Kac and P Cheung Quantum Calculus Springer Science ampBusiness Media 2002

[21] R A C Ferreira ldquoNontrivial solutions for fractional q-difference boundary value problemsrdquo Electronic Journal ofQualitative Theory of Differential Equations no 70 pp 1ndash102010

[22] M S Stankovic P M Rajkovic and S D Marinkovic ldquoOn q-fractional derivatives of Riemann-Liouville and Caputo typerdquohttparxivorgabs09090387

[23] M H Annaby and Z S Mansour q-Fractional Calculus andEquations vol 2056 of Lecture Notes in Mathematics SpringerBerlin Germany 2012

[24] M El-Shahed and W M Shammakh ldquoMultiple positive solu-tions for nonlinear fractional eigenvalue problemwith non localconditionsrdquo Fractional Calculus and Applied Analysis vol 3 pp1ndash13 2012

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

Page 7: Research Article Positive Solutions for Nonlinear -Fractional …downloads.hindawi.com/journals/aaa/2015/759378.pdf · 2019. 7. 31. · Research Article Positive Solutions for Nonlinear

Abstract and Applied Analysis 7

Proof Let 120576 gt 0 be such that 1198910 le (1120582)(1205831minus 120576)Then there

exists 1205880gt 0 such that

119891 (119905 119906) le1

120582(1205831 minus 120576) 119906

forall119906 isin [0 1205880] and almost all 119905 isin [0 1]

(57)

Let 120588 isin (0 1205880]We prove that

119879119906 = 120573119906 for 119906 isin 120597119870120588 120573 ge 1 (58)

which implies the result In fact if (58) does not hold thenthere exist 119906 isin 120597119870

120588and 120573 ge 1 such that 119879119906 = 120573119906

This implies

120573119906 (119905) = 120582int

1

0

119866 (119905 119902119904) 119892 (119904) 119891 (119904 119906 (119904)) 119889119902119904

le (1205831minus 120576) int

1

0

119866 (119905 119902119904) 119892 (119904) 119906 (119904) 119889119902119904

= (1205831minus 120576) 119871119906 (119905)

(59)

Thus we have shown 119906(119905) le (1205831 minus 120576)119871119906(119905)This gives

119906 (119905) le (1205831 minus 120576) 119871 [(1205831 minus 120576) 119871119906 (119905)]

= (1205831minus 120576)21198712119906 (119905)

(60)

And by iterating

119906 (119905) le (1205831 minus 120576)119899119871119899119906 (119905) for 119899 isin 119873 (61)

Therefore

119906 le (1205831 minus 120576)119899 10038171003817100381710038171198711198991003817100381710038171003817 119906

1 le (1205831 minus 120576)119899 10038171003817100381710038171198711198991003817100381710038171003817

(62)

and we have

1 le (1205831 minus 120576) lim

119899rarr+infin

100381710038171003817100381711987111989910038171003817100381710038171119899= (1205831 minus 120576)

1

1205831

lt 1 (63)

a contradiction It follows that

119894119896 (119879119870120588) = 1 for each 120588 isin (0 1205880] (64)

Let 1205881gt 0 120588

1gt 120588 be chosen so that 119891(119905 119906) gt (120583

1120582)119906 for all

119906 ge 1198881205881 119888 as in (H2) and almost all 119905 isin [0 1]

We claim that 119906 = 119879119906 + 1205731206011 for all 120573 gt 0 and 119906 isin 120597119870120588lowast

when 120588lowast gt 1205881 Note that 119906 isin 119870 with 119906 = 120588lowast ge 120588

1

We have 119906(119905) ge 1198881205881for all 119905 isin [119886 119887]

Now if our claim is false then we have

119906 (119905) = 120582int

1

0

119866 (119905 119902119904) 119892 (119904) 119891 (119904 119906 (119904)) 119889119902119904 + 1205731206011 (119905) (65)

Therefore

119906 (119905) ge 1205831 int

119887

119886

119866 (119905 119902119904) 119892 (119904) 119906 (119904) 119889119902119904 + 1205731206011 (119905)

= 1205831119906 (119905) + 1205731206011 (119905)

(66)

From (66) we firstly deduce that 119906(119905) ge 1205731206011(119905) on [0 1]

Then we have

1205831119906 (119905) ge 1205831 (1205731206011 (119905)) = 1205731206011 (119905) (67)

Inserting this into (66) we obtain 119906(119905) ge 21205731206011(119905) for 119905 isin [0 1]

Repeating this process gives

119906 (119905) ge 1198991205731206011 (119905) for 119905 isin [0 1] 119899 isin 119873 (68)

Since 1206011(119905) is strictly positive on [0 1] this is a contradiction

then

119894119870(119879119870120588lowast) = 0 for 119906 isin 120597119870

120588lowast (69)

By (64) and (69) one has

119894119870(119879119870120588lowast 119870120588) = 119894119870(119879119870120588lowast) minus 119894119870(119879119870120588) = minus1 (70)

Therefore 119879 has at least one fixed point 1199060isin 119870120588lowast 119870120588 and

1199060is a positive solution of BVP (1)-(2)

Theorem 22 Assume that

(A3) 1205831 lt 1205821198910 le infin(A4) 0 le 120582119891infin lt 1205831

Then (1)-(2) had at least one positive solution

Proof Let 120576 gt 0 satisfy 1198910gt (1120582)(120583

1+ 120576)Then there exists

1198771 gt 0 such that

119891 (119905 119906) ge1

120582(1205831+ 120576) 119906 forall119905 isin [0 1] 119906 isin [0 1198771] (71)

For any 119906 isin 1205971198701198771

we have by (71) that

119879119906 (119905) = 120582int

1

0

119866 (119905 119902119904) 119892 (119904) 119891 (119904 119906 (119904)) 119889119902119904

ge (1205831 + 120576) int

1

0

119866 (119905 119902119904) 119892 (119904) 119906 (119904) 119889119902119904

ge 1205831119871119906 (119905) forall119905 isin [0 1]

(72)

Let 1 be the positive eigenfunction of 119871 corresponding to 1205831

that is 1 = 12058311198711 Wemay suppose that119879 has no fixed pointon 120597119870119877

1

otherwise the proof is finished In the following wewill show that

119906 minus 119879119906 = 1205731 forall119906 isin 1205971198701198771

120573 ge 0 (73)

If (73) is not true then there is 0isin 1205971198701198771

and 1205730ge 0 such that

0minus1198790= 12057301 It is clear that 120573

0gt 0 and

0= 1198790+12057301ge

12057301Set

120573lowast= sup 120573 0 ge 1205731 (74)

Obviously 120573lowast ge 1205730gt 0 It follows from 119871(119875) sub 119875 that

12058311198710ge 1205831119871120573lowast1= 120573lowast12058311198711= 120573lowast1 (75)

8 Abstract and Applied Analysis

and using this and (72) we have

0 = 1198790 + 12057301 ge 12058311198710 + 12057301 ge 120573lowast1 + 12057301 (76)

which contradicts (74) Thus (73) holdsBy Lemma 9 we have

119894119870(1198791198701198771

) = 0 (77)

On the other hand let 120576 gt 0 satisfy 119891infin lt (1120582)(1205831minus 120576) Then

there exists 1198772gt 1198771such that

119891 (119905 119906) le1

120582(1205831 minus 120576) 119906 forall119905 isin [0 1] 119906 ge 1198772 (78)

By (H5) there exists an 119871infin function 1205931such that

119891 (119905 119906) le1

1205821205931 (119905) forall119906 isin [0 1198772] 119905 isin [0 1] (79)

Hence we have

119891 (119905 119906) le1

120582[(1205831 minus 120576) 119906 + 1205931 (119905)]

forall119906 isin 119877+ 119905 isin [0 1]

(80)

Since 11205831is the radius of the spectrum of 119871 (119868(120583

1minus120576)minus119871)

minus1

existsLet

119862 = int

1

0

1205931 (119904) Φ (119904) 119892 (119904) 119889119902119904

1198770= (

119868

(1205831minus 120576)

minus 119871)

minus1

(119888

(1205831minus 120576)

)

(81)

We prove that for each 119877 gt 1198770

119879119906 = 120573119906 forall119906 isin 120597119870119877 120573 ge 1 (82)

In fact if not there exist 119906 isin 120597119870119877and 120573 ge 1 such that 119879119906 =

120573119906This together with (80) implies

119906 (119905) le int

1

0

119866 (119905 119902119904) 119892 (119904) ((1205831 minus 120576) 119906 (119904) + 1205931 (119904)) 119889119902119904

= (1205831minus 120576) int

1

0

119866 (119905 119902119904) 119892 (119904) 119906 (119904) 119889119902119904

+ int

1

0

119866 (119905 119902119904) 119892 (119904) 1205931 (119904) 119889119902119904

le (1205831 minus 120576) 119871119906 (119905) + 119862

(83)

This implies

(119868

1205831minus 120576minus 119871)119906 (119905) le

119862

1205831minus 120576

119906 (119905) le (119868

1205831minus 120576minus 119871)

minus1

(119862

1205831minus 120576)

= 1198770

(84)

Therefore we have 119906 le 1198770lt 119877 a contradiction Taking

119877 gt 1198772 it follows from (74) and properties of index that

119894119870(119879119870119877) = 1 forall119877 gt 119877

0 (85)

Now (77) and (85) combined imply

119894119870(119879119870119877 1198701198771

) = 119894119870(119879119870119877) minus 119894119870(1198791198701198771

) = 1 (86)

Therefore 119879 has at least one fixed point 1199060isin 1198701198771198701198771

and 1199060

is a positive solution of BVP (1)-(2)

5 The Existence of Two Positive Solutions

Theorem 23 Suppose (A2) (A3) and

(A5) 12058211989101205881015840

le 119898 for some 1205881015840 gt 0

Then (1)-(2) had at least two positive solutions

Proof By (A5) we have

119879119906 (119905) = 120582int

1

0

119866 (119905 119902119904) 119892 (119904) 119891 (119904 119906 (119904)) 119889119902119904

le int

1

0

119866 (119905 119902119904) 119892 (119904) 1205881015840119898119889119902119904

(87)

so that 119879119906 le 1205881015840 = 119906 for all 119906 isin 1205971198811205881015840 Now Lemma 8 yields

119894119896(119879 1198811205881015840) = 1 (88)

On the other hand in view of (A2) we may take 120588lowast gt 1205881015840so that (69) holds (see the proof of Theorem 21) From (A3)we may take 119877

1isin (0 120588

1015840) so that (77) holds (see the proof of

Theorem 22)Combining (88) (69) and (77) we arrive at

119894119896 (119879119870120588lowast 1198811205881015840) = 0 minus 1 = minus1

119894119896(119879 1198811205881015840 1198701198771

) = 1 minus 0 = 1

(89)

Consequently 119879 has at least two fixed points with one on119870120588lowast 1198811205881015840 and the other on119881

1205881015840 1198701198771

Therefore (1)-(2) had atleast two positive solutions

Theorem 24 Suppose (A1) (A4) and

(A6) 12058211989112058810158401205881015840119888 ge 119872 for some 1205881015840 gt 0

Then (1)-(2) had at least two positive solutions

Proof By (A6) we have

119879119906 (119905) = 120582int

1

0

119866 (119905 119902119904) 119892 (119904) 119891 (119904 119906 (119904)) 119889119902119904

ge 120582int

119887

119886

119866 (119905 119902119904) 119892 (119904) 119891 (119904 119906 (119904)) 119889119902119904

ge int

119887

119886

119866 (119905 119902119904) 119892 (119904)1198721205881015840119889119902119904

(90)

Abstract and Applied Analysis 9

so that 119879119906 ge 1205881015840 = 119906 for all 119906 isin 1205971198811205881015840 and by Lemma 8

this yields

119894119896 (119879 1198811205881015840) = 0 (91)

On the other hand in view of (A1) we may take 120588 isin (0 1205881015840)so that (64) holds (see the proof of Theorem 21) In additionfrom (A4) we may take 119877 gt 1205881015840 so that (85) holds (see theproof of Theorem 22)

Combining (91) (64) and (85) we arrive at

119894119896(119879119870119877 1198811205881015840) = 1 minus 0 = 1

119894119896(119879 1198811205881015840 119870120588) = 0 minus 1 = minus1

(92)

Hence 119879 has at least two fixed points with one on 1198811205881015840 119870120588

and the other on 119870119877 1198811205881015840 Therefore (1)-(2) had at least two

positive solutions

We illustrate the applicability of these results with someexamples

Example 25 Consider the problem

11986305(25)119906 (119905) + 120582 (5119905 + 3) (

71199062+ 119906

119906 + 1) (2 + cos 119906) = 0

119905 isin (0 1)

1198632

05119906 (0) = 0

119906 (0) = 0

11986305119906 (1) = 0

(93)

Herewe have119892(119905) = 5119905+3119891(119906) = (2+cos 119906)((71199062+119906)(119906+1))and 2 lt 120572 le 3

It is readily shown that 1198910 = 1198910= 3 119891infin = 21 119891

infin= 7

Also 3119906 le 119891(119906) le 21119906 for 119906 ge 0 By calculation wefind119898 = 019722 and the smallest119872 calculated is119872(119886 119887) asymp119872(0484405 1) asymp 074665 We find 120583

1asymp 030366 Hence

by Theorem 21 there is at least one positive solution if 3120582 lt1205831and 7120582 gt 120583

1 that is there is a positive solution if 120582 isin

(047047 109773)

ByTheorem 22 there does not exist a positive solution ifeither 3120582 gt 120583

1or 21120582 lt 120583

1 that is if 120582 lt 109773 or 120582 gt

015682 no positive solution exists

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

References

[1] N Pongarm S Asawasamrit J Tariboon and S K NtouyasldquoMulti-strip fractional q-integral boundary value problemsfor nonlinear fractional q-difference equationsrdquo Advances inDifference Equations vol 2014 no 1 article 193 2014

[2] R Almeida and N Martins ldquoExistence results for fractional q-difference equations of order120572 isin ]2 3[with three-point bound-ary conditionsrdquo Communications in Nonlinear Science andNumerical Simulation vol 19 no 6 pp 1675ndash1685 2014

[3] Y ZhaoHChen andQZhang ldquoExistence results for fractionalq-difference equations with nonlocal q-integral boundary con-ditionsrdquo Advances in Difference Equations vol 2013 article 482013

[4] R A Ferreira ldquoPositive solutions of a nonlinear q-fractionaldifference equation with integral boundary conditionsrdquo Inter-national Journal of Difference Equations vol 9 no 2 pp 135ndash145 2014

[5] B Ahmad and S K Ntouyas ldquoFractional 119902-difference hybridequations and inclusions with Dirichlet boundary conditionsrdquoAdvances in Difference Equations vol 2014 article 199 2014

[6] B Ahmad S KNtouyas andAAlsaedi ldquoExistence of solutionsfor fractional q-integro-difference inclusions with fractional q-integral boundary conditionsrdquo Advances in Difference Equa-tions vol 2014 article 257 2014

[7] R P Agarwal B Ahmad A Alsaedi and H Al-Hutami ldquoOnnonlinear fractional q-difference equations involving two frac-tional orders with three-point nonlocal boundary conditionsrdquoDynamics of Continuous Discrete amp Impulsive Systems Series AMathematical Analysis vol 21 no 1 pp 135ndash151 2014

[8] M El-Shahed andM A Al-Yami ldquoPositive solutions of bound-ary value problems for nth order q-differential equationsrdquoInternational Journal of Mathematical Archive vol 2 pp 521ndash532 2011

[9] J Ma and J Yang ldquoExistence of solutions for multi-pointboundary value problem of fractional q-difference equationrdquoElectronic Journal ofQualitativeTheory ofDifferential Equationsvol 92 pp 1ndash10 2011

[10] P M Rajkovic S D Marinkovic and M S Stankovic ldquoFrac-tional integrals and derivatives in q-calculusrdquo Applicable Anal-ysis and Discrete Mathematics vol 1 no 1 pp 311ndash323 2007

[11] Y Zhao H Chen and B Qin ldquoMultiple solutions for acoupled systemof nonlinear fractional differential equations viavariational methodsrdquo Applied Mathematics and Computationvol 257 pp 417ndash427 2015

[12] W-X Zhou X Liu and J-G Zhang ldquoSome new existenceand uniqueness results of solutions to semilinear impulsivefractional integro-differential equationsrdquoAdvances inDifferenceEquations vol 2015 article 38 2015

[13] Q Yuan andW Yang ldquoPositive solutions of nonlinear boundaryvalue problems for delayed fractional q-difference systemsrdquoAdvances in Difference Equations vol 2014 no 1 article 51 16pages 2014

[14] W Yang ldquoPositive solutions for nonlinear semipositone frac-tional q-difference system with coupled integral boundaryconditionsrdquo Applied Mathematics and Computation vol 244pp 702ndash725 2014

[15] J R L Webb and K Q Lan ldquoEigenvalue criteria for existenceof multiple positive solutions of nonlinear boundary valueproblems of local and nonlocal typerdquo Topological Methods inNonlinear Analysis vol 27 no 1 pp 91ndash116 2006

[16] J R L Webb ldquoNonlocal conjugate type boundary value prob-lems of higher orderrdquo Nonlinear Analysis Theory Methods ampApplications vol 71 no 5-6 pp 1933ndash1940 2009

[17] J R L Webb and G Infante ldquoNon-local boundary value prob-lems of arbitrary orderrdquo Journal of the London MathematicalSociety vol 79 no 1 pp 238ndash258 2009

10 Abstract and Applied Analysis

[18] J R Webb and G Infante ldquoPositive solutions of nonlocalboundary value problems involving integral conditionsrdquo Non-linear Differential Equations and Applications NoDEA vol 15no 1-2 pp 45ndash67 2008

[19] J R L Webb and G Infante ldquoPositive solutions of nonlocalboundary value problems a unified approachrdquo Journal of theLondon Mathematical Society vol 74 no 3 pp 673ndash693 2006

[20] V Kac and P Cheung Quantum Calculus Springer Science ampBusiness Media 2002

[21] R A C Ferreira ldquoNontrivial solutions for fractional q-difference boundary value problemsrdquo Electronic Journal ofQualitative Theory of Differential Equations no 70 pp 1ndash102010

[22] M S Stankovic P M Rajkovic and S D Marinkovic ldquoOn q-fractional derivatives of Riemann-Liouville and Caputo typerdquohttparxivorgabs09090387

[23] M H Annaby and Z S Mansour q-Fractional Calculus andEquations vol 2056 of Lecture Notes in Mathematics SpringerBerlin Germany 2012

[24] M El-Shahed and W M Shammakh ldquoMultiple positive solu-tions for nonlinear fractional eigenvalue problemwith non localconditionsrdquo Fractional Calculus and Applied Analysis vol 3 pp1ndash13 2012

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

Page 8: Research Article Positive Solutions for Nonlinear -Fractional …downloads.hindawi.com/journals/aaa/2015/759378.pdf · 2019. 7. 31. · Research Article Positive Solutions for Nonlinear

8 Abstract and Applied Analysis

and using this and (72) we have

0 = 1198790 + 12057301 ge 12058311198710 + 12057301 ge 120573lowast1 + 12057301 (76)

which contradicts (74) Thus (73) holdsBy Lemma 9 we have

119894119870(1198791198701198771

) = 0 (77)

On the other hand let 120576 gt 0 satisfy 119891infin lt (1120582)(1205831minus 120576) Then

there exists 1198772gt 1198771such that

119891 (119905 119906) le1

120582(1205831 minus 120576) 119906 forall119905 isin [0 1] 119906 ge 1198772 (78)

By (H5) there exists an 119871infin function 1205931such that

119891 (119905 119906) le1

1205821205931 (119905) forall119906 isin [0 1198772] 119905 isin [0 1] (79)

Hence we have

119891 (119905 119906) le1

120582[(1205831 minus 120576) 119906 + 1205931 (119905)]

forall119906 isin 119877+ 119905 isin [0 1]

(80)

Since 11205831is the radius of the spectrum of 119871 (119868(120583

1minus120576)minus119871)

minus1

existsLet

119862 = int

1

0

1205931 (119904) Φ (119904) 119892 (119904) 119889119902119904

1198770= (

119868

(1205831minus 120576)

minus 119871)

minus1

(119888

(1205831minus 120576)

)

(81)

We prove that for each 119877 gt 1198770

119879119906 = 120573119906 forall119906 isin 120597119870119877 120573 ge 1 (82)

In fact if not there exist 119906 isin 120597119870119877and 120573 ge 1 such that 119879119906 =

120573119906This together with (80) implies

119906 (119905) le int

1

0

119866 (119905 119902119904) 119892 (119904) ((1205831 minus 120576) 119906 (119904) + 1205931 (119904)) 119889119902119904

= (1205831minus 120576) int

1

0

119866 (119905 119902119904) 119892 (119904) 119906 (119904) 119889119902119904

+ int

1

0

119866 (119905 119902119904) 119892 (119904) 1205931 (119904) 119889119902119904

le (1205831 minus 120576) 119871119906 (119905) + 119862

(83)

This implies

(119868

1205831minus 120576minus 119871)119906 (119905) le

119862

1205831minus 120576

119906 (119905) le (119868

1205831minus 120576minus 119871)

minus1

(119862

1205831minus 120576)

= 1198770

(84)

Therefore we have 119906 le 1198770lt 119877 a contradiction Taking

119877 gt 1198772 it follows from (74) and properties of index that

119894119870(119879119870119877) = 1 forall119877 gt 119877

0 (85)

Now (77) and (85) combined imply

119894119870(119879119870119877 1198701198771

) = 119894119870(119879119870119877) minus 119894119870(1198791198701198771

) = 1 (86)

Therefore 119879 has at least one fixed point 1199060isin 1198701198771198701198771

and 1199060

is a positive solution of BVP (1)-(2)

5 The Existence of Two Positive Solutions

Theorem 23 Suppose (A2) (A3) and

(A5) 12058211989101205881015840

le 119898 for some 1205881015840 gt 0

Then (1)-(2) had at least two positive solutions

Proof By (A5) we have

119879119906 (119905) = 120582int

1

0

119866 (119905 119902119904) 119892 (119904) 119891 (119904 119906 (119904)) 119889119902119904

le int

1

0

119866 (119905 119902119904) 119892 (119904) 1205881015840119898119889119902119904

(87)

so that 119879119906 le 1205881015840 = 119906 for all 119906 isin 1205971198811205881015840 Now Lemma 8 yields

119894119896(119879 1198811205881015840) = 1 (88)

On the other hand in view of (A2) we may take 120588lowast gt 1205881015840so that (69) holds (see the proof of Theorem 21) From (A3)we may take 119877

1isin (0 120588

1015840) so that (77) holds (see the proof of

Theorem 22)Combining (88) (69) and (77) we arrive at

119894119896 (119879119870120588lowast 1198811205881015840) = 0 minus 1 = minus1

119894119896(119879 1198811205881015840 1198701198771

) = 1 minus 0 = 1

(89)

Consequently 119879 has at least two fixed points with one on119870120588lowast 1198811205881015840 and the other on119881

1205881015840 1198701198771

Therefore (1)-(2) had atleast two positive solutions

Theorem 24 Suppose (A1) (A4) and

(A6) 12058211989112058810158401205881015840119888 ge 119872 for some 1205881015840 gt 0

Then (1)-(2) had at least two positive solutions

Proof By (A6) we have

119879119906 (119905) = 120582int

1

0

119866 (119905 119902119904) 119892 (119904) 119891 (119904 119906 (119904)) 119889119902119904

ge 120582int

119887

119886

119866 (119905 119902119904) 119892 (119904) 119891 (119904 119906 (119904)) 119889119902119904

ge int

119887

119886

119866 (119905 119902119904) 119892 (119904)1198721205881015840119889119902119904

(90)

Abstract and Applied Analysis 9

so that 119879119906 ge 1205881015840 = 119906 for all 119906 isin 1205971198811205881015840 and by Lemma 8

this yields

119894119896 (119879 1198811205881015840) = 0 (91)

On the other hand in view of (A1) we may take 120588 isin (0 1205881015840)so that (64) holds (see the proof of Theorem 21) In additionfrom (A4) we may take 119877 gt 1205881015840 so that (85) holds (see theproof of Theorem 22)

Combining (91) (64) and (85) we arrive at

119894119896(119879119870119877 1198811205881015840) = 1 minus 0 = 1

119894119896(119879 1198811205881015840 119870120588) = 0 minus 1 = minus1

(92)

Hence 119879 has at least two fixed points with one on 1198811205881015840 119870120588

and the other on 119870119877 1198811205881015840 Therefore (1)-(2) had at least two

positive solutions

We illustrate the applicability of these results with someexamples

Example 25 Consider the problem

11986305(25)119906 (119905) + 120582 (5119905 + 3) (

71199062+ 119906

119906 + 1) (2 + cos 119906) = 0

119905 isin (0 1)

1198632

05119906 (0) = 0

119906 (0) = 0

11986305119906 (1) = 0

(93)

Herewe have119892(119905) = 5119905+3119891(119906) = (2+cos 119906)((71199062+119906)(119906+1))and 2 lt 120572 le 3

It is readily shown that 1198910 = 1198910= 3 119891infin = 21 119891

infin= 7

Also 3119906 le 119891(119906) le 21119906 for 119906 ge 0 By calculation wefind119898 = 019722 and the smallest119872 calculated is119872(119886 119887) asymp119872(0484405 1) asymp 074665 We find 120583

1asymp 030366 Hence

by Theorem 21 there is at least one positive solution if 3120582 lt1205831and 7120582 gt 120583

1 that is there is a positive solution if 120582 isin

(047047 109773)

ByTheorem 22 there does not exist a positive solution ifeither 3120582 gt 120583

1or 21120582 lt 120583

1 that is if 120582 lt 109773 or 120582 gt

015682 no positive solution exists

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

References

[1] N Pongarm S Asawasamrit J Tariboon and S K NtouyasldquoMulti-strip fractional q-integral boundary value problemsfor nonlinear fractional q-difference equationsrdquo Advances inDifference Equations vol 2014 no 1 article 193 2014

[2] R Almeida and N Martins ldquoExistence results for fractional q-difference equations of order120572 isin ]2 3[with three-point bound-ary conditionsrdquo Communications in Nonlinear Science andNumerical Simulation vol 19 no 6 pp 1675ndash1685 2014

[3] Y ZhaoHChen andQZhang ldquoExistence results for fractionalq-difference equations with nonlocal q-integral boundary con-ditionsrdquo Advances in Difference Equations vol 2013 article 482013

[4] R A Ferreira ldquoPositive solutions of a nonlinear q-fractionaldifference equation with integral boundary conditionsrdquo Inter-national Journal of Difference Equations vol 9 no 2 pp 135ndash145 2014

[5] B Ahmad and S K Ntouyas ldquoFractional 119902-difference hybridequations and inclusions with Dirichlet boundary conditionsrdquoAdvances in Difference Equations vol 2014 article 199 2014

[6] B Ahmad S KNtouyas andAAlsaedi ldquoExistence of solutionsfor fractional q-integro-difference inclusions with fractional q-integral boundary conditionsrdquo Advances in Difference Equa-tions vol 2014 article 257 2014

[7] R P Agarwal B Ahmad A Alsaedi and H Al-Hutami ldquoOnnonlinear fractional q-difference equations involving two frac-tional orders with three-point nonlocal boundary conditionsrdquoDynamics of Continuous Discrete amp Impulsive Systems Series AMathematical Analysis vol 21 no 1 pp 135ndash151 2014

[8] M El-Shahed andM A Al-Yami ldquoPositive solutions of bound-ary value problems for nth order q-differential equationsrdquoInternational Journal of Mathematical Archive vol 2 pp 521ndash532 2011

[9] J Ma and J Yang ldquoExistence of solutions for multi-pointboundary value problem of fractional q-difference equationrdquoElectronic Journal ofQualitativeTheory ofDifferential Equationsvol 92 pp 1ndash10 2011

[10] P M Rajkovic S D Marinkovic and M S Stankovic ldquoFrac-tional integrals and derivatives in q-calculusrdquo Applicable Anal-ysis and Discrete Mathematics vol 1 no 1 pp 311ndash323 2007

[11] Y Zhao H Chen and B Qin ldquoMultiple solutions for acoupled systemof nonlinear fractional differential equations viavariational methodsrdquo Applied Mathematics and Computationvol 257 pp 417ndash427 2015

[12] W-X Zhou X Liu and J-G Zhang ldquoSome new existenceand uniqueness results of solutions to semilinear impulsivefractional integro-differential equationsrdquoAdvances inDifferenceEquations vol 2015 article 38 2015

[13] Q Yuan andW Yang ldquoPositive solutions of nonlinear boundaryvalue problems for delayed fractional q-difference systemsrdquoAdvances in Difference Equations vol 2014 no 1 article 51 16pages 2014

[14] W Yang ldquoPositive solutions for nonlinear semipositone frac-tional q-difference system with coupled integral boundaryconditionsrdquo Applied Mathematics and Computation vol 244pp 702ndash725 2014

[15] J R L Webb and K Q Lan ldquoEigenvalue criteria for existenceof multiple positive solutions of nonlinear boundary valueproblems of local and nonlocal typerdquo Topological Methods inNonlinear Analysis vol 27 no 1 pp 91ndash116 2006

[16] J R L Webb ldquoNonlocal conjugate type boundary value prob-lems of higher orderrdquo Nonlinear Analysis Theory Methods ampApplications vol 71 no 5-6 pp 1933ndash1940 2009

[17] J R L Webb and G Infante ldquoNon-local boundary value prob-lems of arbitrary orderrdquo Journal of the London MathematicalSociety vol 79 no 1 pp 238ndash258 2009

10 Abstract and Applied Analysis

[18] J R Webb and G Infante ldquoPositive solutions of nonlocalboundary value problems involving integral conditionsrdquo Non-linear Differential Equations and Applications NoDEA vol 15no 1-2 pp 45ndash67 2008

[19] J R L Webb and G Infante ldquoPositive solutions of nonlocalboundary value problems a unified approachrdquo Journal of theLondon Mathematical Society vol 74 no 3 pp 673ndash693 2006

[20] V Kac and P Cheung Quantum Calculus Springer Science ampBusiness Media 2002

[21] R A C Ferreira ldquoNontrivial solutions for fractional q-difference boundary value problemsrdquo Electronic Journal ofQualitative Theory of Differential Equations no 70 pp 1ndash102010

[22] M S Stankovic P M Rajkovic and S D Marinkovic ldquoOn q-fractional derivatives of Riemann-Liouville and Caputo typerdquohttparxivorgabs09090387

[23] M H Annaby and Z S Mansour q-Fractional Calculus andEquations vol 2056 of Lecture Notes in Mathematics SpringerBerlin Germany 2012

[24] M El-Shahed and W M Shammakh ldquoMultiple positive solu-tions for nonlinear fractional eigenvalue problemwith non localconditionsrdquo Fractional Calculus and Applied Analysis vol 3 pp1ndash13 2012

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

Page 9: Research Article Positive Solutions for Nonlinear -Fractional …downloads.hindawi.com/journals/aaa/2015/759378.pdf · 2019. 7. 31. · Research Article Positive Solutions for Nonlinear

Abstract and Applied Analysis 9

so that 119879119906 ge 1205881015840 = 119906 for all 119906 isin 1205971198811205881015840 and by Lemma 8

this yields

119894119896 (119879 1198811205881015840) = 0 (91)

On the other hand in view of (A1) we may take 120588 isin (0 1205881015840)so that (64) holds (see the proof of Theorem 21) In additionfrom (A4) we may take 119877 gt 1205881015840 so that (85) holds (see theproof of Theorem 22)

Combining (91) (64) and (85) we arrive at

119894119896(119879119870119877 1198811205881015840) = 1 minus 0 = 1

119894119896(119879 1198811205881015840 119870120588) = 0 minus 1 = minus1

(92)

Hence 119879 has at least two fixed points with one on 1198811205881015840 119870120588

and the other on 119870119877 1198811205881015840 Therefore (1)-(2) had at least two

positive solutions

We illustrate the applicability of these results with someexamples

Example 25 Consider the problem

11986305(25)119906 (119905) + 120582 (5119905 + 3) (

71199062+ 119906

119906 + 1) (2 + cos 119906) = 0

119905 isin (0 1)

1198632

05119906 (0) = 0

119906 (0) = 0

11986305119906 (1) = 0

(93)

Herewe have119892(119905) = 5119905+3119891(119906) = (2+cos 119906)((71199062+119906)(119906+1))and 2 lt 120572 le 3

It is readily shown that 1198910 = 1198910= 3 119891infin = 21 119891

infin= 7

Also 3119906 le 119891(119906) le 21119906 for 119906 ge 0 By calculation wefind119898 = 019722 and the smallest119872 calculated is119872(119886 119887) asymp119872(0484405 1) asymp 074665 We find 120583

1asymp 030366 Hence

by Theorem 21 there is at least one positive solution if 3120582 lt1205831and 7120582 gt 120583

1 that is there is a positive solution if 120582 isin

(047047 109773)

ByTheorem 22 there does not exist a positive solution ifeither 3120582 gt 120583

1or 21120582 lt 120583

1 that is if 120582 lt 109773 or 120582 gt

015682 no positive solution exists

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

References

[1] N Pongarm S Asawasamrit J Tariboon and S K NtouyasldquoMulti-strip fractional q-integral boundary value problemsfor nonlinear fractional q-difference equationsrdquo Advances inDifference Equations vol 2014 no 1 article 193 2014

[2] R Almeida and N Martins ldquoExistence results for fractional q-difference equations of order120572 isin ]2 3[with three-point bound-ary conditionsrdquo Communications in Nonlinear Science andNumerical Simulation vol 19 no 6 pp 1675ndash1685 2014

[3] Y ZhaoHChen andQZhang ldquoExistence results for fractionalq-difference equations with nonlocal q-integral boundary con-ditionsrdquo Advances in Difference Equations vol 2013 article 482013

[4] R A Ferreira ldquoPositive solutions of a nonlinear q-fractionaldifference equation with integral boundary conditionsrdquo Inter-national Journal of Difference Equations vol 9 no 2 pp 135ndash145 2014

[5] B Ahmad and S K Ntouyas ldquoFractional 119902-difference hybridequations and inclusions with Dirichlet boundary conditionsrdquoAdvances in Difference Equations vol 2014 article 199 2014

[6] B Ahmad S KNtouyas andAAlsaedi ldquoExistence of solutionsfor fractional q-integro-difference inclusions with fractional q-integral boundary conditionsrdquo Advances in Difference Equa-tions vol 2014 article 257 2014

[7] R P Agarwal B Ahmad A Alsaedi and H Al-Hutami ldquoOnnonlinear fractional q-difference equations involving two frac-tional orders with three-point nonlocal boundary conditionsrdquoDynamics of Continuous Discrete amp Impulsive Systems Series AMathematical Analysis vol 21 no 1 pp 135ndash151 2014

[8] M El-Shahed andM A Al-Yami ldquoPositive solutions of bound-ary value problems for nth order q-differential equationsrdquoInternational Journal of Mathematical Archive vol 2 pp 521ndash532 2011

[9] J Ma and J Yang ldquoExistence of solutions for multi-pointboundary value problem of fractional q-difference equationrdquoElectronic Journal ofQualitativeTheory ofDifferential Equationsvol 92 pp 1ndash10 2011

[10] P M Rajkovic S D Marinkovic and M S Stankovic ldquoFrac-tional integrals and derivatives in q-calculusrdquo Applicable Anal-ysis and Discrete Mathematics vol 1 no 1 pp 311ndash323 2007

[11] Y Zhao H Chen and B Qin ldquoMultiple solutions for acoupled systemof nonlinear fractional differential equations viavariational methodsrdquo Applied Mathematics and Computationvol 257 pp 417ndash427 2015

[12] W-X Zhou X Liu and J-G Zhang ldquoSome new existenceand uniqueness results of solutions to semilinear impulsivefractional integro-differential equationsrdquoAdvances inDifferenceEquations vol 2015 article 38 2015

[13] Q Yuan andW Yang ldquoPositive solutions of nonlinear boundaryvalue problems for delayed fractional q-difference systemsrdquoAdvances in Difference Equations vol 2014 no 1 article 51 16pages 2014

[14] W Yang ldquoPositive solutions for nonlinear semipositone frac-tional q-difference system with coupled integral boundaryconditionsrdquo Applied Mathematics and Computation vol 244pp 702ndash725 2014

[15] J R L Webb and K Q Lan ldquoEigenvalue criteria for existenceof multiple positive solutions of nonlinear boundary valueproblems of local and nonlocal typerdquo Topological Methods inNonlinear Analysis vol 27 no 1 pp 91ndash116 2006

[16] J R L Webb ldquoNonlocal conjugate type boundary value prob-lems of higher orderrdquo Nonlinear Analysis Theory Methods ampApplications vol 71 no 5-6 pp 1933ndash1940 2009

[17] J R L Webb and G Infante ldquoNon-local boundary value prob-lems of arbitrary orderrdquo Journal of the London MathematicalSociety vol 79 no 1 pp 238ndash258 2009

10 Abstract and Applied Analysis

[18] J R Webb and G Infante ldquoPositive solutions of nonlocalboundary value problems involving integral conditionsrdquo Non-linear Differential Equations and Applications NoDEA vol 15no 1-2 pp 45ndash67 2008

[19] J R L Webb and G Infante ldquoPositive solutions of nonlocalboundary value problems a unified approachrdquo Journal of theLondon Mathematical Society vol 74 no 3 pp 673ndash693 2006

[20] V Kac and P Cheung Quantum Calculus Springer Science ampBusiness Media 2002

[21] R A C Ferreira ldquoNontrivial solutions for fractional q-difference boundary value problemsrdquo Electronic Journal ofQualitative Theory of Differential Equations no 70 pp 1ndash102010

[22] M S Stankovic P M Rajkovic and S D Marinkovic ldquoOn q-fractional derivatives of Riemann-Liouville and Caputo typerdquohttparxivorgabs09090387

[23] M H Annaby and Z S Mansour q-Fractional Calculus andEquations vol 2056 of Lecture Notes in Mathematics SpringerBerlin Germany 2012

[24] M El-Shahed and W M Shammakh ldquoMultiple positive solu-tions for nonlinear fractional eigenvalue problemwith non localconditionsrdquo Fractional Calculus and Applied Analysis vol 3 pp1ndash13 2012

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

Page 10: Research Article Positive Solutions for Nonlinear -Fractional …downloads.hindawi.com/journals/aaa/2015/759378.pdf · 2019. 7. 31. · Research Article Positive Solutions for Nonlinear

10 Abstract and Applied Analysis

[18] J R Webb and G Infante ldquoPositive solutions of nonlocalboundary value problems involving integral conditionsrdquo Non-linear Differential Equations and Applications NoDEA vol 15no 1-2 pp 45ndash67 2008

[19] J R L Webb and G Infante ldquoPositive solutions of nonlocalboundary value problems a unified approachrdquo Journal of theLondon Mathematical Society vol 74 no 3 pp 673ndash693 2006

[20] V Kac and P Cheung Quantum Calculus Springer Science ampBusiness Media 2002

[21] R A C Ferreira ldquoNontrivial solutions for fractional q-difference boundary value problemsrdquo Electronic Journal ofQualitative Theory of Differential Equations no 70 pp 1ndash102010

[22] M S Stankovic P M Rajkovic and S D Marinkovic ldquoOn q-fractional derivatives of Riemann-Liouville and Caputo typerdquohttparxivorgabs09090387

[23] M H Annaby and Z S Mansour q-Fractional Calculus andEquations vol 2056 of Lecture Notes in Mathematics SpringerBerlin Germany 2012

[24] M El-Shahed and W M Shammakh ldquoMultiple positive solu-tions for nonlinear fractional eigenvalue problemwith non localconditionsrdquo Fractional Calculus and Applied Analysis vol 3 pp1ndash13 2012

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

Page 11: Research Article Positive Solutions for Nonlinear -Fractional …downloads.hindawi.com/journals/aaa/2015/759378.pdf · 2019. 7. 31. · Research Article Positive Solutions for Nonlinear

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


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