Research ArticleThe Existence of Solutions for Four-PointCoupled Boundary Value Problems of FractionalDifferential Equations at Resonance

Yumei Zou1 Lishan Liu2 and Yujun Cui3

1 Department of Statistics and Finance Shandong University of Science and Technology Qingdao 266590 China2 School of Mathematical Sciences Qufu Normal University Qufu Shandong 273165 China3Department of Mathematics Shandong University of Science and Technology Qingdao Shandong 266590 China

Correspondence should be addressed to Yujun Cui cyj720201163com

Received 18 December 2013 Accepted 14 February 2014 Published 23 March 2014

Academic Editor Xinguang Zhang

Copyright copy 2014 Yumei Zou et al This is an open access article distributed under the Creative Commons Attribution Licensewhich permits unrestricted use distribution and reproduction in any medium provided the original work is properly cited

A four-point coupled boundary value problem of fractional differential equations is studied Based onMawhinrsquos coincidence degreetheory some existence theorems are obtained in the case of resonance

1 Introduction

In this paper we are concerned with the following four-pointcoupled boundary value problem for nonlinear fractionaldifferential equation Consider

119863120572

0+119906 (119905) = 119891 (119905 119906 (119905) 119863

120572minus1

0+119906 (119905) V (119905) 119863120573minus1

0+V (119905))

119863120573

0+V (119905) = 119892 (119905 119906 (119905) 119863

120572minus1

0+119906 (119905) V (119905) 119863120573minus1

0+V (119905))

1198682minus120572

0+119906 (119905) |119905=0

= 0 119906 (1) = 119886V (120585)

1198682minus120573

0+V (119905) |119905=0

= 0 V (1) = 119887119906 (120578)

(1)

where 1 lt 120572 120573 lt 2 1198631205720+

and 1198681205720+

are the standard Riemann-Liouville differentiation and integration 119891 119892 isin 119862([0 1] times

R4R) 119886 119887 isin R 120585 120578 isin (0 1) and

119886119887120585120573minus1

120578120572minus1

= 1 (2)

The subject of fractional calculus has gained considerablepopularity and importance because of its intensive develop-ment of the theory of fractional calculus itself and its variedapplications in many fields of science and engineering As

a result the subject of fractional differential equations hasattracted much attention see [1ndash11] for a good overview

At the same time we notice that coupled boundary valueproblems which arise in the study of reaction-diffusion equa-tions and Sturm-Liouville problems have wide applicationsin various fields of sciences and engineering for examplethe heat equation [12ndash14] and mathematical biology [15 16]In [17] Asif and Khan used the Guo-Krasnoselrsquoskii fixed-point theorem to show the existence of positive solutionsto the nonlinear differential system with coupled four-pointboundary value conditions

minus11990910158401015840

(119905) = 119891 (119905 119909 (119905) 119910 (119905)) 119905 isin (0 1)

minus11991010158401015840

(119905) = 119892 (119905 119909 (119905) 119910 (119905)) 119905 isin (0 1)

119909 (0) = 119910 (0) = 0

119909 (1) = 120572119910 (120585) 119910 (1) = 120573119909 (120578)

(3)

where 120585 120578 isin (0 1) 0 lt 120572120573120585120578 lt 1 and 119891 119892 (0 1) times [0 +infin)times

[0 +infin) rarr [0 +infin) are two given continuous functionsIn [18] the authors considered the existence of positive

solutions of four-point coupled boundary value problem for

Hindawi Publishing CorporationAbstract and Applied AnalysisVolume 2014 Article ID 314083 8 pageshttpdxdoiorg1011552014314083

2 Abstract and Applied Analysis

systems of the nonlinear semipositone fractional differentialequation

119863120572

0+119906 + 120582119891 (119905 119906 V) = 0 119905 isin (0 1) 120582 gt 0

119863120572

0+V + 120582119892 (119905 119906 V) = 0

119906(119894)

(0) = V(119894) (0) = 0 0 le 119894 le 119899 minus 2

119906 (1) = 119886V (120585) V (1) = 119887119906 (120578)

(4)

where 120582 is a parameter 119886 119887 120585 120578 satisfy 120585 120578 isin (0 1) 0 lt

119886119887120585120578 lt 1 120572 isin (119899 minus 1 119899] is a real number and 119899 ge 3 and119863120572

0+119906 is Riemann-Liouvillersquos fractional derivativeRecently Cui and Sun [19] showed the existence of

positive solutions of singular superlinear coupled integralboundary value problems for differential systems

minus11990910158401015840

(119905) = 1198911(119905 119909 (119905) 119910 (119905)) 119905 isin (0 1)

minus11991010158401015840

(119905) = 1198912(119905 119909 (119905) 119910 (119905)) 119905 isin (0 1)

119909 (0) = 119910 (0) = 0 119909 (1) = 120572 [119910] 119910 (1) = 120573 [119909]

(5)

where 120572[119909] 120573[119909] are bounded linear functionals on 119862[0 1]

given by

120572 [119909] = int

1

0

119909 (119905) 119889119860 (119905) 120573 [119909] = int

1

0

119909 (119905) 119889119861 (119905) (6)

with 119860 119861 being functions of bounded variation with positivemeasures

A key assumption in the above papers is that the casestudied is not at resonance that is the associated fractional(or ordinary) linear differential operators are invertible Inthis paper instead we are interested in the resonance casedue to the critical condition (2) Boundary value problemsat resonance have been studied by several authors includingthemost recent works [20ndash31] In this paper we establish newresults on the existence of a solution for the couple boundaryvalue problems at resonance Our method is based on thecoincidence degree theorem of Mawhin [32 33]

Now we briefly recall some notations and an abstractexistence result

Let 119884119885 be real Banach spaces and let 119871 dom 119871 sub 119884 rarr

119885 be a Fredholm operator of index zero 119875 119884 rarr 119884 and119876 119885 rarr 119885 are continuous projectors such that

Im119875 = Ker 119871 Ker119876 = Im 119871

119884 = Ker 119871 oplus Ker119875 119885 = Im 119871 oplus Im119876

(7)

It follows that 119871|dom119871capKer119875 dom 119871 cap Ker119875 rarr Im 119871

is invertible We denote the inverse of the mapping by 119870119875

(generalized inverse operator of 119871) If Ω is an open boundedsubset of119884 such that dom 119871capΩ = 0 themapping119873 119884 rarr 119885

will be called 119871-compact on Ω if 119876119873(Ω) is bounded and119870119875(119868 minus 119876)119873 Ω rarr 119884 is compact

Theorem 1 (see [32 33]) Let 119871 be a Fredholm operator ofindex zero and let 119873 be 119871-compact on Ω Assume that thefollowing conditions are satisfied

(i) 119871119909 = 120582119873119909 for every (119909 120582) isin [(dom 119871 Ker 119871) cap 120597Ω] times(0 1)

(ii) 119873119909 notin Im 119871 for every 119909 isin Ker 119871 cap 120597Ω(iii) deg(119876119873|Ker119871Ker 119871 cap Ω 0) = 0 where 119876 119885 rarr 119885 is

a projector as above with Im 119871 = Ker119876

Then the equation119871119909 = 119873119909 has at least one solution indom119871cap

Ω

For convenience let us set the following notations

Δ1= max1 + 1

Γ (120573) (1 +

1

Γ (120572))

times (1+119886120585120573minus1

Γ (120572) [119887

Γ (120572)+

1

Γ (120573)])

Δ2=

Γ (120573 + 1) Γ (120572 + 1)

120572120585120573minus1Γ (120572 + 1) (120585 minus 1) + Γ (120573 + 1) (120578 minus 1)

Δ3=

1

Γ (120573)max 119886120585120573minus1 (Γ (120572) + 1) Γ (120573) + 1

Δ4= max Δ

1

1003817100381710038171003817119886110038171003817100381710038171+ Δ3

1003817100381710038171003817119886210038171003817100381710038171 Δ1

1003817100381710038171003817119887110038171003817100381710038171+ Δ3

1003817100381710038171003817119886210038171003817100381710038171

Δ5= max Δ

1

1003817100381710038171003817119888110038171003817100381710038171+ Δ3

1003817100381710038171003817119888210038171003817100381710038171 Δ1

1003817100381710038171003817119889110038171003817100381710038171+ Δ3

1003817100381710038171003817119889210038171003817100381710038171

(8)

2 Preliminaries and Lemmas

In this section first we provide recall of some basic definitionsand lemmas of the fractional calculus which will be used inthis paper For more details we refer to books [1 2 4]

Definition 2 (see [1 4]) The Riemann-Liouville fractionalintegral of order 120572 gt 0 of a function 119906 (0infin) rarr R isgiven by

119868120572

0+119906 (119905) =

1

Γ (120572)int

119905

0

(119905 minus 119904)120572minus1

119906 (119904) 119889119904 (9)

provided that the right-hand side is pointwise defined on(0infin)

Definition 3 (see [1 4]) The Riemann-Liouville fractionalderivative of order 120572 gt 0 of a continuous function 119906

(0infin) rarr R is given by

119863120572

0+119906 (119905) =

1

Γ (119899 minus 120572)(119889

119889119905)

119899

int

119905

0

119906 (119904)

(119905 minus 119904)120572minus119899+1

119889119904 (10)

where 119899 minus 1 le 120572 lt 119899 provided that the right-hand side ispointwise defined on (0infin)

We use the classical Banach space 119862[0 1] with the norm119906infin

= max119905isin[01]

|119906(119905)| and 1198711

[0 1] with the norm 1199061=

int1

0

|119906(119905)|119889119905 We also use the space 119860119862119899[0 1] defined by

119860119862119899

[0 1] = 119906 [0 1] 997888rarr R | 119906(119899minus1)

are absolutely continuous on [0 1]

(11)

Abstract and Applied Analysis 3

and the Banach space 119862120583[0 1] (120583 gt 0)

119862120583

[0 1]

= 119906 (119905) | 119906 (119905) = 119868120583

0+119909 (119905) + 119888

1119905120583minus1

+ 1198882119905120583minus2

+ sdot sdot sdot

+ 119888119873minus1

119905120583minus(119873minus1)

119909 isin 119862 [0 1] 119905 isin [0 1]

119888119894isin R 119894 = 1 2 119873 = [120583] + 1

(12)

with the norm 119906119862120583 = 119863

120583

0+119906infin+sdot sdot sdot+119863

120583minus(119873minus1)

0+119906infin+119906infin

Lemma 4 (see [1]) Let 120572 gt 0 119899 = [120572] + 1 Assume that 119906 isin

1198711

(0 1)with a fractional integration of order 119899minus120572 that belongsto 119860119862119899[0 1] Then the equality

(119868120572

0+119863120572

0+119906) (119905) = 119906 (119905) minus

119899

sum

119894=1

((119868119899minus120572

0+119906) (119905))

(119899minus119894)

|119905=0

Γ (120572 minus 119894 + 1)119905120572minus119894 (13)

holds almost everywhere on [0 1]

In the following lemma we use the unified notationof both for fractional integrals and fractional derivativesassuming that 119868120572

0+= 119863minus120572

0+for 120572 lt 0

Lemma 5 (see [1]) Assume that 120572 gt 0 then

(i) let 119896 isin N If119863120572119886+119906(119905) and (119863120572+119896

119886+119906)(119905) exist then

(119863119896

119863120572

119886+) 119906 (119905) = (119863

120572+119896

119886+119906) (119905) (14)

(ii) if 120573 gt 0 120572 + 120573 gt 1 then

(119868120572

119886+119868120573

119886+) 119906 (119905) = (119868

120572+120573

119886+119906) (119905) (15)

satisfies at any point on [119886 119887] for 119906 isin 119871119901(119886 119887) and 1 le

119901 le +infin(iii) let 119906 isin 119862[119886 119887] Then (119863

120572

119886+119868120572

119886+)119906(119905) = 119906(119905) holds on

[119886 119887](iv) note that for 120582 gt minus1 120582 = 120572 minus 1 120572 minus 2 120572 minus 119899 one

has

119863120572

119905120582

=Γ (120582 + 1)

Γ (120582 minus 120572 + 1)119905120582minus120572

119863120572

119905120572minus119894

= 0 119894 = 1 2 119899

(16)

Remark 6 If 1 lt 120572 lt 2 and 119906 satisfies1198631205720+119906 = 119891(119905) isin 119871

1

(0 1)

and 1198682minus1205720+

119906|119905=0

= 0 then 119906 isin 119862120572minus1[0 1] In fact with Lemma 4one has

119906 (119905) = 119868120572

0+119891 (119905) + 119888

1119905120572minus1

+ 1198882119905120572minus2

(17)

Combined with 1198682minus1205720+

119906|119905=0

= 0 there is 1198882= 0 So

119906 (119905) = 119868120572

0+119891 (119905) + 119888

1119905120572minus1

= 119868120572minus1

0+[1198681

0+119891 (119905) + 119888

1Γ (120572)] (18)

Lemma 7 (see [34]) 119865 sub 119862120583

[0 1] is a sequentially compactset if and only if 119865 is uniformly bounded and equicontinuousHere to be uniformly bounded means that there exists119872 gt 0

such that for every 119906 isin 119865

119906119862120583 =

1003817100381710038171003817119863120583

0+1199061003817100381710038171003817infin

+ sdot sdot sdot +100381710038171003817100381710038171003817119863120583minus(119873minus1)

0+119906100381710038171003817100381710038171003817infin

+ 119906infinlt 119872 (19)

and to be equicontinuous means that for all 120598 gt 0 exist120575 gt 0 andfor all 119905

1 1199052isin [0 1] |119905

1minus 1199052| lt 120575 119906 isin 119865 and 119894 isin 1 [120583]

the following holds1003816100381610038161003816119906 (1199051) minus 119906 (1199052)

1003816100381610038161003816 lt 12057610038161003816100381610038161003816119863120583minus119894

0+119906 (1199051) minus 119863120583minus119894

0+119906 (1199052)10038161003816100381610038161003816lt 120576

(20)

We also use the following two Banach spaces 119884 =

119862120572minus1

[0 1] times 119862120573minus1

[0 1] with the norm1003817100381710038171003817(119909 119910)

1003817100381710038171003817119884= max 119909

119862120572minus1

10038171003817100381710038171199101003817100381710038171003817119862120573minus1

(21)

and 119885 = 1198711

[0 1] times 1198711

[0 1] with the norm1003817100381710038171003817(119909 119910)

1003817100381710038171003817119885= max 119909

1100381710038171003817100381711991010038171003817100381710038171 (22)

Let the linear operator 119871 dom 119871 sub 119884 rarr 119885 with

dom 119871 = (119906 V) isin 119884 1198682minus120572

0+119906 (119905) |119905=0

= 0 119906 (1) = 119886V (120585)

1198682minus120573

0+V (119905) |119905=0

= 0 V (1) = 119887119906 (120578)

(23)

be defined by

119871 (119906 V) = (1198711119906 1198712V) (24)

where 1198711 119862120572minus1

[0 1] rarr 1198711

[0 1] and 1198712 119862120573minus1

[0 1] rarr

1198711

[0 1] are defined by

1198711119906 = 119863

120572

0+119906 (119905) 119871

2V = 119863

120573

0+V (119905) (25)

Let the nonlinear operator119873 119884 rarr 119885 be defined by

(119873 (119906 V)) (119905) = (1198731(119906 V) (119905) 119873

2(119906 V) (119905)) (26)

where1198731 1198732 119884 rarr 119871

1

[0 1] are defined by

1198731(119906 V) (119905) = 119891 (119905 119906 (119905) 119863

120572minus1

0+119906 (119905) V (119905) 119863120573minus1

0+V (119905))

1198732(119909 119910) (119905) = 119892 (119905 119906 (119905) 119863

120572minus1

0+119906 (119905) V (119905) 119863120573minus1

0+V (119905))

(27)

Then four-point coupled boundary value problems (1) can bewritten as

119871 (119906 V) = 119873 (119906 V) (28)

Lemma 8 Let 119871 be the linear operator defined as above If (2)holds then

Ker 119871 = (119906 V) isin dom 119871 (119906 V) = 119888 (119886120585120573minus1

119905120572minus1

119905120573minus1

)

119888 isin R 119905 isin [0 1]

(29)

Im 119871 = (119909 119910) isin 119885 119886119868120573

0+119910 (120585) minus 119868

120572

0+119909 (1)

+119886120585120573minus1

[119887119868120572

0+119909 (120578) minus 119868

120573

0+119910 (1)] = 0

(30)

4 Abstract and Applied Analysis

Proof Let 119906(119905) = 119886120585120573minus1

119905120572minus1and let V(119905) = 119905

120573minus1 Then byLemma 5 we have 1198682minus120572

0+119906(119905)|119905=0

= 1198682minus120573

0+V(119905)|119905=0

= 0 119906(1) =

119886120585120573minus1

= 119886V(120585) V(1) = 1 = 119887119906(120578) and 1198631205720+119906(119905) = 119863

120573

0+V(119905) = 0

So

(119906 V) isin dom 119871 (119906 V) = 119888 (119886120585120573minus1

119905120572minus1

119905120573minus1

) 119888 isin R

sub Ker 119871(31)

For every (119906 V) isin Ker 119871 if1198631205720+119906(119905) = 119863

120573

0+V(119905) = 0 then

119906 (119905) = 1198881119905120572minus1

+ 1198882119905120572minus2

V (119905) = 1198883119905120573minus1

+ 1198884119905120573minus2

(32)

Considering that 1198682minus1205720+

119906(119905)|119905=0

= 1198682minus120573

0+V(119905)|119905=0

= 0 119906(1) = 119886V(120585)and V(1) = 119887119906(120578) we can obtain that 119888

2= 1198884= 0 and 119888

1 1198883=

119886120585120573minus1 It yields the following

Ker 119871 sub (119906 V) isin dom 119871 (119906 V)

= 119888 (119886120585120573minus1

119905120572minus1

119905120573minus1

) 119888 isin R

(33)

Let (119909 119910) isin Im 119871 then there is (119906 V) isin dom 119871 such that(119909 119910) = 119871(119906 V) that is 119906 isin 119862

120572minus1

[0 1] 1198631205720+119906(119905) = 119909(119905) and

V isin 119862120573minus1[0 1]1198631205730+V(119905) = 119910(119905) By Lemma 4

119868120572

0+119909 (119905) = 119906 (119905) minus

((119863120572minus1

0+119906) (119905))

10038161003816100381610038161003816119905=0

Γ (120572)119905120572minus1

minus

((1198682minus120572

0+119906) (119905))

10038161003816100381610038161003816119905=0

Γ (120572 minus 1)119905120572minus2

119868120573

0+119910 (119905) = V (119905) minus

((119863120573minus1

0+V) (119905))

100381610038161003816100381610038161003816119905=0

Γ (120573)119905120573minus1

minus

((1198682minus120573

0+V) (119905))

100381610038161003816100381610038161003816119905=0

Γ (120573 minus 1)119905120573minus2

(34)

and by the couple boundary conditions we have

119906 (119905) = 119868120572

0+119909 (119905) + 119888

1119905120572minus1

V (119905) = 119868120573

0+119910 (119905) + 119888

2119905120573minus1

119906 (1) = 119868120572

0+119909 (1) + 119888

1= 119886V (120585) = 119886 [119868

120573

0+119910 (120585) + 119888

2120585120573minus1

]

V (1) = 119868120573

0+119910 (1) + 119888

2= 119887119906 (120578) = 119887 [119868

120572

0+119910 (120578) + 119888

1120578120572minus1

]

(35)

It yields the following

119886119868120573

0+119910 (120585) minus 119868

120572

0+119909 (1) + 119886120585

120573minus1

[119887119868120572

0+119909 (120578) minus 119868

120573

0+119910 (1)] = 0

(36)

On the other hand suppose that (119909 119910) isin 119885 satisfy (36) Let119906(119905) = 119868

120572

0+119909(119905) + 119886120585

120573minus1

119905120572minus1 and V(119905) = 119868

120573

0+119910(119905) + [119887119868

120572

0+119909(120578) minus

119868120573

0+119910(1) + 1]119905

120573minus1 and then 119863120572

0+119906(119905) = 119909(119905) 119863120573

0+V(119905) = 119910(119905)

and

1198682minus120572

0+119906 (119905) |119905=0

= 1198682minus120573

0+V (119905) |119905=0

= 0

119906 (1) = 119868120572

0+119909 (1) + 119886120585

120573minus1

= 119886119868120573

0+119910 (120585) + 119886 [119887119868

120572

0+119909 (120578) minus 119868

120573

0+119910 (1) + 1] 120585

120573minus1

= 119886V (120585)

V (1) = 119868120573

0+119910 (1) + [119887119868

120572

0+119909 (120578) minus 119868

120573

0+119910 (1) + 1]

= 119887119868120572

0+119909 (120578) + 119886119887120585

120573minus1

120578120572minus1

= 119887119906 (120578)

(37)

Therefore (30) holds

Lemma 9 If (2) holds then 119871 is a Fredholm operator of indexzero and dim Ker 119871 = codim Im 119871 = 1 Furthermore thelinear operator119870

119901 Im 119871 rarr dom 119871capKer119875 can be defined by

119870119875(119906 V) (119905)

= 119868120572

0+119906 (119905) minus 119886120585

120573minus1

119905120572minus1

[119887119868120572

0+119906 (120578) minus 119868

120573

0+V (1)] 119868120573

0+V (119905)

(38)

Also10038171003817100381710038171003817119870119901(119906 V)

10038171003817100381710038171003817119884le Δ1(119906 V)

119885 (39)

Proof Define operator 119876 119885 rarr 119885 as follows

119876 (119906 V) = Δ21198761(119906 V) (1 1) (40)

where 1198761 119885 rarr R is defined by

1198761(119906 V) = 119886119868

120573

0+V (120585) minus 119868120572

0+119906 (1)

+ 119886120585120573minus1

[119887119868120572

0+119906 (120578) minus 119868

120573

0+V (1)]

Δ2=

Γ (120573 + 1) Γ (120572 + 1)

120572120585120573minus1Γ (120572 + 1) (120585 minus 1) + Γ (120573 + 1) (120578 minus 1)= 0

(41)

It is easy to see that 1198762(119906 V) = 119876(119906 V) that is 119876 119885 rarr 119885

is a continuous linear projector Furthermore Ker119876 = Im 119871For (119906 V) isin 119885 set (119906 V) = [(119906 V) minus 119876(119906 V)] + 119876(119906 V) Then(119906 V) minus 119876(119906 V) isin Ker119876 and 119876(119906 V) isin Im119876 It follows fromKer119876 = Im 119871 and1198762(119906 V) = 119876(119906 V) that Im 119871capIm119876 = (0 0)So we have

119885 = Im 119871 oplus Im119876 (42)

Now Ind119871 = dim Ker 119871 minus codim Im 119871 = dim Ker 119871 minus

dim Im119876 = 0 and so 119871 is a Fredholm operator of index 0

Abstract and Applied Analysis 5

Let 119875 119884 rarr 119884 be continuous linear operator defined by

119875 (119906 V) =119863120573minus1

0+V (0)

Γ (120573)(119886120585120573minus1

119905120572minus1

119905120573minus1

) (43)

Obviously 119875 is a linear projector and

Ker119875 = (119906 V) isin 119884 119863120573minus1

0+V (0) = 0 (44)

It is easy to know that 119884 = Ker119875 oplus Ker 119871Define119870

119875 Im 119871 rarr dom 119871 cap Ker119875 by

119870119875(119906 V) (119905)

= (119868120572

0+119906 (119905) minus 119886120585

120573minus1

119905120572minus1

[119887119868120572

0+119906 (120578) minus 119868

120573

0+V (1)] 119868120573

0+V (119905))

(45)

Since

100381610038161003816100381610038161003816119887119868120572

0+119906 (120578) minus 119868

120573

0+V (1)

100381610038161003816100381610038161003816le

10038161003816100381610038161003816100381610038161003816

119887

Γ (120572)int

120578

0

(120578 minus 119904)120572minus1

119906 (119904) 119889119904

10038161003816100381610038161003816100381610038161003816

+

100381610038161003816100381610038161003816100381610038161003816

1

Γ (120573)int

1

0

(1 minus 119904)120572minus1V (119904) 119889119904

100381610038161003816100381610038161003816100381610038161003816

le119887

Γ (120572)1199061+

1

Γ (120573)V1

119863120572minus1

0+119868120572

0+119906 (119905) = int

119905

0

119906 (119904) 119889119904

119868120572

0+119906 (119905) =

1

Γ (120572)int

119905

0

(119905 minus 119904)120572minus1

119906 (119904) 119889119904

119863120573minus1

0+119868120573

0+V (119905) = int

119905

0

V (119904) 119889119904

119868120573

0+V (119905) =

1

Γ (120573)int

119905

0

(119905 minus 119904)120573minus1V (119904) 119889119904

(46)

then1003817100381710038171003817119870119875 (119906 V)

1003817100381710038171003817119884le Δ1(119906 V)

119885 (47)

In fact if (119906 V) isin Im 119871 then

119871119870119875(119906 V) (119905)

= 119863120572

0+[119868120572

0+119906 (119905) minus 119886120585

120573minus1

119905120572minus1

[119887119868120572

0+119906 (120578) minus 119868

120573

0+V (1)]]

119863120573

0+119868120573

0+V (119905) = (119906 V)

(48)

By Lemma 4 for (119906 V) isin dom 119871 cap Ker119875

119870119875119871 (119906 V) (119905)

= (119868120572

0+119863120572

0+119906 (119905)minus119886120585

120572minus1

119905120572minus1

[119887119868120572

0+119863120572

0+119906 (120578)minus119868

120573

0+119863120573

0+V (1)]

119868120573

0+119863120573

0+V (119905))

= (119906 (119905) minus119863120572minus1

0+119906 (0)

Γ (120572)119905120572minus1

minus1198682minus120572

0+119906 (0)

Γ (120572 minus 1)119905120572minus2

minus 119886120585120573minus1

119905120572minus1

times [119887119906 (120578) minus 119887119863120572minus1

0+119906 (0)

Γ (120572)120578120572minus1

minus1198682minus120572

0+119906 (0)

Γ (120572 minus 1)120578120572minus2

minusV (1)+119863120573minus1

0+V (0)

Γ (120573)] V (119905)minus

119863120573minus1

0+V (0)

Γ (120573)119905120573minus1

)

= (119906 (119905) minus119863120572minus1

0+119906 (0)

Γ (120572)119905120572minus1

+119886120585120573minus1

119905120572minus1

119887119863120572minus1

0+119906 (0)

Γ (120572)120578120572minus1

V (119905))

= (119906 (119905) V (119905))(49)

(119863120573minus10+

V(0) = 0 since (119906 V) isin Ker119875 and 1198682minus120572

0+119906(0) = 0 since

(119906 V) isin dom 119871) Hence

119870119901= (119871|dom119871capKer119875)

minus1

(50)

The proof is complete

3 Main Results

In this section we will use Theorem 1 to prove the existenceof solutions to BVP (1) To obtain our main theorem we usethe following assumptions

(H1) There exist functions 119886119894 119887119894 119888119894 119889119894 119890119894isin 1198711

[0 1] (119894 =

1 2) such that for all (119909 119910 119911 119908) isin R4 119905 isin [0 1]

1003816100381610038161003816119891 (119905 119909 119910 119911 119908)1003816100381610038161003816 le 1198901(119905) + 119886

1(119905) |119909| + 119887

1(119905)

10038161003816100381610038161199101003816100381610038161003816

+ 1198881(119905) |119911| + 119889

1(119905) |119908|

1003816100381610038161003816119892 (119905 119909 119910 119911 119908)1003816100381610038161003816 le 1198902(119905) + 119886

2(119905) |119909| + 119887

2(119905)

10038161003816100381610038161199101003816100381610038161003816

+ 1198882(119905) |119911| + 119889

2(119905) |119908|

(51)

(H2) There exists a constant 119860 gt 0 such that for (119906 V) isindom 119871 if |119863120573minus1

0+V(119905)| gt 119860 for all 119905 isin [0 1] then

1198761(1198731(119906 V)) = 0 or 119876

1(1198732(119906 V)) = 0

(H3) There exists a constant 119861 gt 0 such that either for each119888 isin R |119888| gt 119861

1198881198731(119888119886120585120573minus1

119905120572minus1

119888119905120573minus1

) gt 0 1198881198732(119888119886120585120573minus1

119905120572minus1

119888119905120573minus1

) gt 0

(52)

or for each 119886 isin R |119886| gt 119861

1198881198731(119888119886120585120573minus1

119905120572minus1

119888119905120573minus1

) lt 0

1198881198732(119888119886120585120573minus1

119905120572minus1

119888119905120573minus1

) lt 0

(53)

6 Abstract and Applied Analysis

Theorem 10 Suppose (2) and (1198671)ndash(1198673) hold Then (1) hasat least one solution in Y provided that

max (Δ1+ Δ3)max 10038171003817100381710038171198862

1003817100381710038171003817110038171003817100381710038171198872

1003817100381710038171003817110038171003817100381710038171198882

1003817100381710038171003817110038171003817100381710038171198892

10038171003817100381710038171 Δ4 Δ5

lt 1

(54)

Proof Set

Ω1= (119906 V) isin dom 119871 Ker 119871 119871 (119906 V) = 120582119873 (119906 V)

for some 120582 isin [0 1] (55)

Take (119906 V) isin Ω1 Since 119871119906 = 120582119873119906 so 120582 = 0 and 119873(119906 V) isin

Im 119871 = Ker119876 hence

119876119873(119906 V) = 1198761(1198731(119906 V) 119873

2(119906 V)) (1 1) = 0 (56)

Thus from (H2) there exist 1199050isin [0 1] such that

100381610038161003816100381610038161003816119863120573minus1

0+V (1199050)100381610038161003816100381610038161003816le 119860 (57)

Noticing that

119863120573minus1

0+V (119905) = 119863

120573minus1

0+V (1199050) + int

119905

1199050

119863120573

0+V (119904) 119889119904 (58)

so100381610038161003816100381610038161003816119863120573minus1

0+V (0)

100381610038161003816100381610038161003816le100381710038171003817100381710038171003817119863120573minus1

0+V (119905)

100381710038171003817100381710038171003817infinle100381610038161003816100381610038161003816119863120573minus2

0+V (1199050)100381610038161003816100381610038161003816+100381710038171003817100381710038171003817119863120573

0+V (119905)

1003817100381710038171003817100381710038171

le 119860 +10038171003817100381710038171198712V

10038171003817100381710038171le 119860 +

10038171003817100381710038171198732 (119906 V)10038171003817100381710038171

(59)

Thus

119875 (119906 V)119884=

1003817100381710038171003817100381710038171003817100381710038171003817

119863120573minus1

0+V (0)

Γ (120573)(119886120585120573minus1

119905120572minus1

119905120573minus1

)

1003817100381710038171003817100381710038171003817100381710038171003817119884

= Δ3

100381610038161003816100381610038161003816119863120573minus1

0+V (0)

100381610038161003816100381610038161003816le Δ3(119860 +

10038171003817100381710038171198732 (119906 V)10038171003817100381710038171)

(60)

For all (119906 V) isin Ω1 (119868minus119875)(119906 V) isin dom 119871capKer119875 Considering

Lemma 9 we get 119871119875(119906 V) = (0 0) Together with (39) wehave

(119868 minus 119875) (119906 V)119884=1003817100381710038171003817119870119875119871 (119868 minus 119875) (119906 V)

1003817100381710038171003817119884

le Δ1119871 (119868 minus 119875) (119906 V)

119885

le Δ1119873 (119906 V)

119885

(61)

From (60) and (61) we have

(119906 V)119884

le 119875 (119906 V)119884+ (119868 minus 119875) (119906 V)

119884

le Δ3(119860 +

10038171003817100381710038171198732 (119906 V)10038171003817100381710038171) + Δ1119873 (119906 V)

119885

= 119860Δ3+max (Δ

1+ Δ3)10038171003817100381710038171198732 (119906 V)

10038171003817100381710038171 Δ1

10038171003817100381710038171198731 (119906 V)10038171003817100381710038171

+Δ3

10038171003817100381710038171198732 (119906 V)10038171003817100381710038171

(62)

From (62) we discuss various cases

Case 1 ((119906 V)119884le 119860Δ

3+ (Δ1+ Δ3)1198732(119906 V)

1) From (H1)

we have

(119906 V)119884

le 119860Δ3+ (Δ1+ Δ3)

times (10038171003817100381710038171198862

10038171003817100381710038171119906infin+10038171003817100381710038171198872

10038171003817100381710038171

10038171003817100381710038171003817119863120572minus1

0+11990610038171003817100381710038171003817infin

+10038171003817100381710038171198882

10038171003817100381710038171Vinfin

+10038171003817100381710038171198892

10038171003817100381710038171

100381710038171003817100381710038171003817119863120573minus1

0+V100381710038171003817100381710038171003817infin

+10038171003817100381710038171198902

10038171003817100381710038171)

le 119860Δ3+ (Δ1+ Δ3)10038171003817100381710038171198902

10038171003817100381710038171

+ (Δ1+ Δ3)max 10038171003817100381710038171198862

1003817100381710038171003817110038171003817100381710038171198872

10038171003817100381710038171 119906119862120572minus1

+ (Δ1+ Δ3)max 10038171003817100381710038171198882

1003817100381710038171003817110038171003817100381710038171198892

10038171003817100381710038171 V119862120573minus1

le 119860Δ3+ (Δ1+ Δ3)10038171003817100381710038171198902

10038171003817100381710038171

+ (Δ1+ Δ3)max 10038171003817100381710038171198862

1003817100381710038171003817110038171003817100381710038171198872

1003817100381710038171003817110038171003817100381710038171198882

1003817100381710038171003817110038171003817100381710038171198892

10038171003817100381710038171 (119906 V)

119884

(63)

which yield

(119906 V)119884

le119860Δ2+ (Δ1+ Δ3)10038171003817100381710038171198902

10038171003817100381710038171

1 minus (Δ1+ Δ3)max 10038171003817100381710038171198862

1003817100381710038171003817110038171003817100381710038171198872

1003817100381710038171003817110038171003817100381710038171198882

1003817100381710038171003817110038171003817100381710038171198892

10038171003817100381710038171

(64)

ThusΩ1is bounded

Case 2 ((119906 V)119884le 119860Δ

3+ Δ11198731(119906 V)

1+ Δ31198732(119906 V)

1)

From (H1) we have

(119906 V)119884

le 119860Δ3+ Δ1(10038171003817100381710038171198861

10038171003817100381710038171119906infin+10038171003817100381710038171198871

10038171003817100381710038171

10038171003817100381710038171003817119863120572minus1

0+11990610038171003817100381710038171003817infin

+10038171003817100381710038171198881

10038171003817100381710038171Vinfin

+10038171003817100381710038171198891

10038171003817100381710038171

100381710038171003817100381710038171003817119863120573minus1

0+V100381710038171003817100381710038171003817infin

+10038171003817100381710038171198901

10038171003817100381710038171)

+ Δ3(10038171003817100381710038171198862

10038171003817100381710038171119906infin+10038171003817100381710038171198872

10038171003817100381710038171

10038171003817100381710038171003817119863120572minus1

0+11990610038171003817100381710038171003817infin

+10038171003817100381710038171198882

10038171003817100381710038171Vinfin

+10038171003817100381710038171198892

10038171003817100381710038171

100381710038171003817100381710038171003817119863120573minus1

0+V100381710038171003817100381710038171003817infin

+10038171003817100381710038171198902

10038171003817100381710038171)

le 119860Δ3+ Δ1

1003817100381710038171003817119890110038171003817100381710038171+ Δ3

1003817100381710038171003817119890210038171003817100381710038171+ Δ4119906119862120572minus1 + Δ

5V119862120573minus1

le 119860Δ3+ Δ1

1003817100381710038171003817119890110038171003817100381710038171+ Δ3

1003817100381710038171003817119890210038171003817100381710038171+max Δ

4 Δ5 (119906 V)

119884

(65)

which yield

(119906 V)119884le119860Δ3+ Δ1

1003817100381710038171003817119890110038171003817100381710038171+ Δ3

1003817100381710038171003817119890210038171003817100381710038171

1 minusmax Δ4 Δ5

(66)

ThusΩ1is bounded Let

Ω2= (119906 V) isin Ker 119871 119873 (119906 V) isin Im 119871 (67)

For (119906 V) isin Ω2and (119906 V) isin Ker 119871 so (119906 V) = 119888(119886120585

120573minus1

119905120572minus1

119905120573minus1

) 119905 isin [0 1] 119888 isin R Noticing that Im 119871 = Ker119876 then we

Abstract and Applied Analysis 7

get 119876119873(119906 V) = 0 and thus 11987611198731(119888119886120585120573minus1

119905120572minus1

119888119905120573minus1

) = 0 and11987611198732(119888119886120585120573minus1

119905120572minus1

119888119905120573minus1

) = 0 From (H2) we get |119888| le 119860Γ(120573)and thusΩ

2is bounded

We define the isomorphism 119869 Ker 119871 rarr Im119876 by

119869 (119888119886120585120573minus1

119905120572minus1

119888119905120573minus1

) = (119888 119888) (68)

If the first part of (H3) is satisfied then let

Ω3= (119906 V) isin ker 119871 120582119869 (119906 V)

+ (1 minus 120582)119876119873 (119906 V) = 0 120582 isin [0 1]

(69)

For (119906 V) = (119888119886120585120573minus1

119905120572minus1

119888119905120573minus1

) isin Ω3

120582 (119888 119888) = minus (1 minus 120582) Δ21198761119873(119888119886120585

120573minus1

119905120572minus1

119888119905120573minus1

) (1 1) (70)

If 120582 = 1 then 119888 = 0 Otherwise if |119888| gt 119861 in view of (H3) andΔ2lt 0 one has

(1 minus 120582) Δ31198761119873(119888119886120585

120573minus1

119905120572minus1

119888119905120573minus1

) lt 0 (71)

which contradict 1205821198882 ge 0ThusΩ3sub (119906 V) isin Ker 119871 | (119906 V) =

(119888119886120585120573minus1

119905120572minus1

119888119905120573minus1

) |119888| le 119861 is boundedIf the second part of (H3) holds then define the set

Ω3= (119906 V) isin Ker 119871 minus120582119869 (119906 V)

+ (1 minus 120582)119876119873 (119906 V) = 0 120582 isin [0 1]

(72)

and here 119869 is as above Similar to the above argument we canshow thatΩ

3is bounded too

In the following we will prove that all conditions ofTheorem 1 are satisfied LetΩ be a bounded open subset of 119884such that cup3

119894=1Ω119894sub Ω By standard arguments we can prove

that 119870119875(119868 minus 119876)119873 Ω rarr 119884 is compact and thus 119873 is 119871-

compact onΩ Then by the above argument we have

(i) 119871119906 = 120582119873119906 for every (119906 120582) isin [(dom 119871 Ker 119871) cap 120597Ω]times(0 1)

(ii) 119873119906 notin Im 119871 for 119906 isin Ker 119871 cap 120597Ω

Finally we will prove that (iii) of Theorem 1 is satisfied Let119867(119906 120582) = plusmn120582119869119906 + (1 minus 120582)119876119873119906 According to the aboveargument we know that

119867(119906 120582) = 0 for 119906 isin Ker 119871 cap 120597Ω (73)

Thus by the homotopy property of degree

deg (119876119873|Ker119871Ker 119871 cap Ω 0)

= deg (119867 (sdot 0) Ker 119871 cap Ω 0)

= deg (119867 (sdot 1) Ker 119871 cap Ω 0)

= deg (plusmn119869Ker 119871 cap Ω 0) = 0

(74)

Then byTheorem 1 119871(119906 V) = 119873(119906 V) has at least one solutionin dom 119871 cap Ω so that BVP (1) has a solution in 119862

120572minus1

[0 1] times

119862120573minus1

[0 1] The proof is complete

Conflict of Interests

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

Acknowledgments

The authors are very grateful to the anonymous refereesfor many valuable comments and suggestions which helpedto improve the presentation of the paper The project wassupported by the National Natural Science Foundation ofChina (11371221 61304074) the Specialized Research Foun-dation for the Doctoral Program of Higher Education ofChina (20123705110001) the Program for Scientific ResearchInnovation Team in Colleges and Universities of ShandongProvince the Postdoctoral Science Foundation of ShandongProvince (201303074) and Foundation of SDUST

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[1] A A Kilbas H M Srivastava and J J Trujillo Theoryand Applications of Fractional Differential Equations vol 204of North-Holland Mathematics Studies Elsevier Science BVAmsterdam The Netherlands 2006

[2] V Lakshmikantham S Leela and J Vasundhara DeviTheory ofFractional Dynamic Systems Cambridge Academic PublishersCambridge UK 2009

[3] J J Nieto and J Pimentel ldquoPositive solutions of a fractionalthermostat modelrdquo Boundary Value Problems vol 2013 article5 2013

[4] I Podlubny Fractional Differential Equations vol 198 Aca-demic Press San Diego Calif USA 1999

[5] W Wu and X Zhou ldquoEigenvalue of fractional differentialequations with P-Laplacian operatorrdquo Discrete Dynamics inNature and Society vol 2013 Article ID 137890 8 pages 2013

[6] X Zhang L Liu Y Wu and Y Lu ldquoThe iterative solutions ofnonlinear fractional differential equationsrdquo Applied Mathemat-ics and Computation vol 219 no 9 pp 4680ndash4691 2013

[7] X Zhang L Liu and Y Wu ldquoThe uniqueness of positivesolution for a singular fractional differential system involvingderivativesrdquo Communications in Nonlinear Science and Numer-ical Simulation vol 18 no 6 pp 1400ndash1409 2013

[8] X Zhang L Liu and Y Wu ldquoThe eigenvalue problem for asingular higher order fractional differential equation involvingfractional derivativesrdquo Applied Mathematics and Computationvol 218 no 17 pp 8526ndash8536 2012

[9] X Zhang L Liu and YWu ldquoExistence results for multiple pos-itive solutions of nonlinear higher order perturbed fractionaldifferential equations with derivativesrdquo Applied Mathematicsand Computation vol 219 no 4 pp 1420ndash1433 2012

[10] X Zhang L Liu B Wiwatanapataphee and Y Wu ldquoPositivesolutions of eigenvalue problems for a class of fractionaldifferential equations with derivativesrdquo Abstract and AppliedAnalysis vol 2012 Article ID 512127 16 pages 2012

[11] X Zhang L Liu and Y Wu ldquoMultiple positive solutionsof a singular fractional differential equation with negativelyperturbed termrdquo Mathematical and Computer Modelling vol55 no 3-4 pp 1263ndash1274 2012

[12] K Deng ldquoBlow-up rates for parabolic systemsrdquo Zeitschrift furAngewandte Mathematik und Physik vol 47 no 1 pp 132ndash1431996

8 Abstract and Applied Analysis

[13] K Deng ldquoGlobal existence and blow-up for a system of heatequations with non-linear boundary conditionsrdquoMathematicalMethods in the Applied Sciences vol 18 no 4 pp 307ndash315 1995

[14] L Zhigui and X Chunhong ldquoThe blow-up rate for a system ofheat equations with nonlinear boundary conditionsrdquoNonlinearAnalysisTheory Methods ampApplications vol 34 no 5 pp 767ndash778 1998

[15] D G Aronson ldquoA comparison method for stability analysis ofnonlinear parabolic problemsrdquo SIAM Review vol 20 no 2 pp245ndash264 1978

[16] M Pedersen and Z Lin ldquoBlow-up analysis for a system of heatequations coupled through a nonlinear boundary conditionrdquoApplied Mathematics Letters vol 14 no 2 pp 171ndash176 2001

[17] N A Asif and R A Khan ldquoPositive solutions to singularsystem with four-point coupled boundary conditionsrdquo Journalof Mathematical Analysis and Applications vol 386 no 2 pp848ndash861 2012

[18] C Yuan D Jiang D OrsquoRegan and R P Agarwal ldquoMultiple pos-itive solutions to systems of nonlinear semipositone fractionaldifferential equations with coupled boundary conditionsrdquo Elec-tronic Journal of QualitativeTheory of Differential Equations no13 p 17 2012

[19] Y Cui and J Sun ldquoOn existence of positive solutions of coupledintegral boundary value problems for a nonlinear singularsuperlinear differential systemrdquo Electronic Journal of QualitativeTheory of Differential Equations no 41 pp 1ndash13 2012

[20] Z Bai and Y Zhang ldquoSolvability of fractional three-pointboundary value problems with nonlinear growthrdquo AppliedMathematics and Computation vol 218 no 5 pp 1719ndash17252011

[21] Y Cui ldquoSolvability of second-order boundary-value problemsat resonance involving integral conditionsrdquo Electronic Journalof Differential Equations no 45 pp 1ndash9 2012

[22] Z Hu and W Liu ldquoSolvability for fractional order boundaryvalue problems at resonancerdquo Boundary Value Problems vol2011 article 20 2011

[23] W Jiang ldquoThe existence of solutions to boundary value prob-lems of fractional differential equations at resonancerdquoNonlinearAnalysis Theory Methods amp Applications vol 74 no 5 pp1987ndash1994 2011

[24] N Kosmatov ldquoMulti-point boundary value problems on timescales at resonancerdquo Journal of Mathematical Analysis andApplications vol 323 no 1 pp 253ndash266 2006

[25] G Wang W Liu S Zhu and T Zheng ldquoExistence results fora coupled system of nonlinear fractional 2m-point boundaryvalue problems at resonancerdquo Advances in Difference Equationsvol 2011 article 80 2011

[26] J R L Webb ldquoRemarks on nonlocal boundary value problemsat resonancerdquo Applied Mathematics and Computation vol 216no 2 pp 497ndash500 2010

[27] J R L Webb and M Zima ldquoMultiple positive solutions ofresonant and non-resonant non-local fourth-order boundaryvalue problemsrdquo Glasgow Mathematical Journal vol 54 no 1pp 225ndash240 2012

[28] X ZhangM Feng andWGe ldquoExistence result of second-orderdifferential equations with integral boundary conditions atresonancerdquo Journal of Mathematical Analysis and Applicationsvol 353 no 1 pp 311ndash319 2009

[29] X Zhang C Zhu and Z Wu ldquoSolvability for a coupled systemof fractional differential equations with impulses at resonancerdquoBoundary Value Problems vol 2013 artice 80 2013

[30] Z Zhao and J Liang ldquoExistence of solutions to functionalboundary value problem of second-order nonlinear differentialequationrdquo Journal of Mathematical Analysis and Applicationsvol 373 no 2 pp 614ndash634 2011

[31] Y Zou and Y Cui ldquoExistence results for a functional boundaryvalue problem of fractional differential equationsrdquo Advances inDifference Equations vol 2013 article 25 2013

[32] J Mawhin ldquoTopological degree and boundary value problemsfor nonlinear differential equationsrdquo in Topological Methods forOrdinary Differential Equations P M Fitzpertrick M MartelliJ Mawhin and R Nussbaum Eds vol 1537 of Lecture Notes inMathematics pp 74ndash142 Springer Berlin Germany 1993

[33] J Mawhin Topological Degree Methods in Nonlinear BoundaryValue Problems vol 40 of CBMS Regional Conference Series inMathematics American Mathematical Society Providence RIUSA 1979

[34] Y Zhang and Z Bai ldquoExistence of solutions for nonlinearfractional three-point boundary value problems at resonancerdquoJournal of Applied Mathematics and Computing vol 36 no 1-2pp 417ndash440 2011

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Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

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Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

2 Abstract and Applied Analysis

systems of the nonlinear semipositone fractional differentialequation

119863120572

0+119906 + 120582119891 (119905 119906 V) = 0 119905 isin (0 1) 120582 gt 0

119863120572

0+V + 120582119892 (119905 119906 V) = 0

119906(119894)

(0) = V(119894) (0) = 0 0 le 119894 le 119899 minus 2

119906 (1) = 119886V (120585) V (1) = 119887119906 (120578)

(4)

where 120582 is a parameter 119886 119887 120585 120578 satisfy 120585 120578 isin (0 1) 0 lt

119886119887120585120578 lt 1 120572 isin (119899 minus 1 119899] is a real number and 119899 ge 3 and119863120572

0+119906 is Riemann-Liouvillersquos fractional derivativeRecently Cui and Sun [19] showed the existence of

positive solutions of singular superlinear coupled integralboundary value problems for differential systems

minus11990910158401015840

(119905) = 1198911(119905 119909 (119905) 119910 (119905)) 119905 isin (0 1)

minus11991010158401015840

(119905) = 1198912(119905 119909 (119905) 119910 (119905)) 119905 isin (0 1)

119909 (0) = 119910 (0) = 0 119909 (1) = 120572 [119910] 119910 (1) = 120573 [119909]

(5)

where 120572[119909] 120573[119909] are bounded linear functionals on 119862[0 1]

given by

120572 [119909] = int

1

0

119909 (119905) 119889119860 (119905) 120573 [119909] = int

1

0

119909 (119905) 119889119861 (119905) (6)

with 119860 119861 being functions of bounded variation with positivemeasures

A key assumption in the above papers is that the casestudied is not at resonance that is the associated fractional(or ordinary) linear differential operators are invertible Inthis paper instead we are interested in the resonance casedue to the critical condition (2) Boundary value problemsat resonance have been studied by several authors includingthemost recent works [20ndash31] In this paper we establish newresults on the existence of a solution for the couple boundaryvalue problems at resonance Our method is based on thecoincidence degree theorem of Mawhin [32 33]

Now we briefly recall some notations and an abstractexistence result

Let 119884119885 be real Banach spaces and let 119871 dom 119871 sub 119884 rarr

119885 be a Fredholm operator of index zero 119875 119884 rarr 119884 and119876 119885 rarr 119885 are continuous projectors such that

Im119875 = Ker 119871 Ker119876 = Im 119871

119884 = Ker 119871 oplus Ker119875 119885 = Im 119871 oplus Im119876

(7)

It follows that 119871|dom119871capKer119875 dom 119871 cap Ker119875 rarr Im 119871

is invertible We denote the inverse of the mapping by 119870119875

(generalized inverse operator of 119871) If Ω is an open boundedsubset of119884 such that dom 119871capΩ = 0 themapping119873 119884 rarr 119885

will be called 119871-compact on Ω if 119876119873(Ω) is bounded and119870119875(119868 minus 119876)119873 Ω rarr 119884 is compact

Theorem 1 (see [32 33]) Let 119871 be a Fredholm operator ofindex zero and let 119873 be 119871-compact on Ω Assume that thefollowing conditions are satisfied

(i) 119871119909 = 120582119873119909 for every (119909 120582) isin [(dom 119871 Ker 119871) cap 120597Ω] times(0 1)

(ii) 119873119909 notin Im 119871 for every 119909 isin Ker 119871 cap 120597Ω(iii) deg(119876119873|Ker119871Ker 119871 cap Ω 0) = 0 where 119876 119885 rarr 119885 is

a projector as above with Im 119871 = Ker119876

Then the equation119871119909 = 119873119909 has at least one solution indom119871cap

Ω

For convenience let us set the following notations

Δ1= max1 + 1

Γ (120573) (1 +

1

Γ (120572))

times (1+119886120585120573minus1

Γ (120572) [119887

Γ (120572)+

1

Γ (120573)])

Δ2=

Γ (120573 + 1) Γ (120572 + 1)

120572120585120573minus1Γ (120572 + 1) (120585 minus 1) + Γ (120573 + 1) (120578 minus 1)

Δ3=

1

Γ (120573)max 119886120585120573minus1 (Γ (120572) + 1) Γ (120573) + 1

Δ4= max Δ

1

1003817100381710038171003817119886110038171003817100381710038171+ Δ3

1003817100381710038171003817119886210038171003817100381710038171 Δ1

1003817100381710038171003817119887110038171003817100381710038171+ Δ3

1003817100381710038171003817119886210038171003817100381710038171

Δ5= max Δ

1

1003817100381710038171003817119888110038171003817100381710038171+ Δ3

1003817100381710038171003817119888210038171003817100381710038171 Δ1

1003817100381710038171003817119889110038171003817100381710038171+ Δ3

1003817100381710038171003817119889210038171003817100381710038171

(8)

2 Preliminaries and Lemmas

In this section first we provide recall of some basic definitionsand lemmas of the fractional calculus which will be used inthis paper For more details we refer to books [1 2 4]

Definition 2 (see [1 4]) The Riemann-Liouville fractionalintegral of order 120572 gt 0 of a function 119906 (0infin) rarr R isgiven by

119868120572

0+119906 (119905) =

1

Γ (120572)int

119905

0

(119905 minus 119904)120572minus1

119906 (119904) 119889119904 (9)

provided that the right-hand side is pointwise defined on(0infin)

Definition 3 (see [1 4]) The Riemann-Liouville fractionalderivative of order 120572 gt 0 of a continuous function 119906

(0infin) rarr R is given by

119863120572

0+119906 (119905) =

1

Γ (119899 minus 120572)(119889

119889119905)

119899

int

119905

0

119906 (119904)

(119905 minus 119904)120572minus119899+1

119889119904 (10)

where 119899 minus 1 le 120572 lt 119899 provided that the right-hand side ispointwise defined on (0infin)

We use the classical Banach space 119862[0 1] with the norm119906infin

= max119905isin[01]

|119906(119905)| and 1198711

[0 1] with the norm 1199061=

int1

0

|119906(119905)|119889119905 We also use the space 119860119862119899[0 1] defined by

119860119862119899

[0 1] = 119906 [0 1] 997888rarr R | 119906(119899minus1)

are absolutely continuous on [0 1]

(11)

Abstract and Applied Analysis 3

and the Banach space 119862120583[0 1] (120583 gt 0)

119862120583

[0 1]

= 119906 (119905) | 119906 (119905) = 119868120583

0+119909 (119905) + 119888

1119905120583minus1

+ 1198882119905120583minus2

+ sdot sdot sdot

+ 119888119873minus1

119905120583minus(119873minus1)

119909 isin 119862 [0 1] 119905 isin [0 1]

119888119894isin R 119894 = 1 2 119873 = [120583] + 1

(12)

with the norm 119906119862120583 = 119863

120583

0+119906infin+sdot sdot sdot+119863

120583minus(119873minus1)

0+119906infin+119906infin

Lemma 4 (see [1]) Let 120572 gt 0 119899 = [120572] + 1 Assume that 119906 isin

1198711

(0 1)with a fractional integration of order 119899minus120572 that belongsto 119860119862119899[0 1] Then the equality

(119868120572

0+119863120572

0+119906) (119905) = 119906 (119905) minus

119899

sum

119894=1

((119868119899minus120572

0+119906) (119905))

(119899minus119894)

|119905=0

Γ (120572 minus 119894 + 1)119905120572minus119894 (13)

holds almost everywhere on [0 1]

In the following lemma we use the unified notationof both for fractional integrals and fractional derivativesassuming that 119868120572

0+= 119863minus120572

0+for 120572 lt 0

Lemma 5 (see [1]) Assume that 120572 gt 0 then

(i) let 119896 isin N If119863120572119886+119906(119905) and (119863120572+119896

119886+119906)(119905) exist then

(119863119896

119863120572

119886+) 119906 (119905) = (119863

120572+119896

119886+119906) (119905) (14)

(ii) if 120573 gt 0 120572 + 120573 gt 1 then

(119868120572

119886+119868120573

119886+) 119906 (119905) = (119868

120572+120573

119886+119906) (119905) (15)

satisfies at any point on [119886 119887] for 119906 isin 119871119901(119886 119887) and 1 le

119901 le +infin(iii) let 119906 isin 119862[119886 119887] Then (119863

120572

119886+119868120572

119886+)119906(119905) = 119906(119905) holds on

[119886 119887](iv) note that for 120582 gt minus1 120582 = 120572 minus 1 120572 minus 2 120572 minus 119899 one

has

119863120572

119905120582

=Γ (120582 + 1)

Γ (120582 minus 120572 + 1)119905120582minus120572

119863120572

119905120572minus119894

= 0 119894 = 1 2 119899

(16)

Remark 6 If 1 lt 120572 lt 2 and 119906 satisfies1198631205720+119906 = 119891(119905) isin 119871

1

(0 1)

and 1198682minus1205720+

119906|119905=0

= 0 then 119906 isin 119862120572minus1[0 1] In fact with Lemma 4one has

119906 (119905) = 119868120572

0+119891 (119905) + 119888

1119905120572minus1

+ 1198882119905120572minus2

(17)

Combined with 1198682minus1205720+

119906|119905=0

= 0 there is 1198882= 0 So

119906 (119905) = 119868120572

0+119891 (119905) + 119888

1119905120572minus1

= 119868120572minus1

0+[1198681

0+119891 (119905) + 119888

1Γ (120572)] (18)

Lemma 7 (see [34]) 119865 sub 119862120583

[0 1] is a sequentially compactset if and only if 119865 is uniformly bounded and equicontinuousHere to be uniformly bounded means that there exists119872 gt 0

such that for every 119906 isin 119865

119906119862120583 =

1003817100381710038171003817119863120583

0+1199061003817100381710038171003817infin

+ sdot sdot sdot +100381710038171003817100381710038171003817119863120583minus(119873minus1)

0+119906100381710038171003817100381710038171003817infin

+ 119906infinlt 119872 (19)

and to be equicontinuous means that for all 120598 gt 0 exist120575 gt 0 andfor all 119905

1 1199052isin [0 1] |119905

1minus 1199052| lt 120575 119906 isin 119865 and 119894 isin 1 [120583]

the following holds1003816100381610038161003816119906 (1199051) minus 119906 (1199052)

1003816100381610038161003816 lt 12057610038161003816100381610038161003816119863120583minus119894

0+119906 (1199051) minus 119863120583minus119894

0+119906 (1199052)10038161003816100381610038161003816lt 120576

(20)

We also use the following two Banach spaces 119884 =

119862120572minus1

[0 1] times 119862120573minus1

[0 1] with the norm1003817100381710038171003817(119909 119910)

1003817100381710038171003817119884= max 119909

119862120572minus1

10038171003817100381710038171199101003817100381710038171003817119862120573minus1

(21)

and 119885 = 1198711

[0 1] times 1198711

[0 1] with the norm1003817100381710038171003817(119909 119910)

1003817100381710038171003817119885= max 119909

1100381710038171003817100381711991010038171003817100381710038171 (22)

Let the linear operator 119871 dom 119871 sub 119884 rarr 119885 with

dom 119871 = (119906 V) isin 119884 1198682minus120572

0+119906 (119905) |119905=0

= 0 119906 (1) = 119886V (120585)

1198682minus120573

0+V (119905) |119905=0

= 0 V (1) = 119887119906 (120578)

(23)

be defined by

119871 (119906 V) = (1198711119906 1198712V) (24)

where 1198711 119862120572minus1

[0 1] rarr 1198711

[0 1] and 1198712 119862120573minus1

[0 1] rarr

1198711

[0 1] are defined by

1198711119906 = 119863

120572

0+119906 (119905) 119871

2V = 119863

120573

0+V (119905) (25)

Let the nonlinear operator119873 119884 rarr 119885 be defined by

(119873 (119906 V)) (119905) = (1198731(119906 V) (119905) 119873

2(119906 V) (119905)) (26)

where1198731 1198732 119884 rarr 119871

1

[0 1] are defined by

1198731(119906 V) (119905) = 119891 (119905 119906 (119905) 119863

120572minus1

0+119906 (119905) V (119905) 119863120573minus1

0+V (119905))

1198732(119909 119910) (119905) = 119892 (119905 119906 (119905) 119863

120572minus1

0+119906 (119905) V (119905) 119863120573minus1

0+V (119905))

(27)

Then four-point coupled boundary value problems (1) can bewritten as

119871 (119906 V) = 119873 (119906 V) (28)

Lemma 8 Let 119871 be the linear operator defined as above If (2)holds then

Ker 119871 = (119906 V) isin dom 119871 (119906 V) = 119888 (119886120585120573minus1

119905120572minus1

119905120573minus1

)

119888 isin R 119905 isin [0 1]

(29)

Im 119871 = (119909 119910) isin 119885 119886119868120573

0+119910 (120585) minus 119868

120572

0+119909 (1)

+119886120585120573minus1

[119887119868120572

0+119909 (120578) minus 119868

120573

0+119910 (1)] = 0

(30)

4 Abstract and Applied Analysis

Proof Let 119906(119905) = 119886120585120573minus1

119905120572minus1and let V(119905) = 119905

120573minus1 Then byLemma 5 we have 1198682minus120572

0+119906(119905)|119905=0

= 1198682minus120573

0+V(119905)|119905=0

= 0 119906(1) =

119886120585120573minus1

= 119886V(120585) V(1) = 1 = 119887119906(120578) and 1198631205720+119906(119905) = 119863

120573

0+V(119905) = 0

So

(119906 V) isin dom 119871 (119906 V) = 119888 (119886120585120573minus1

119905120572minus1

119905120573minus1

) 119888 isin R

sub Ker 119871(31)

For every (119906 V) isin Ker 119871 if1198631205720+119906(119905) = 119863

120573

0+V(119905) = 0 then

119906 (119905) = 1198881119905120572minus1

+ 1198882119905120572minus2

V (119905) = 1198883119905120573minus1

+ 1198884119905120573minus2

(32)

Considering that 1198682minus1205720+

119906(119905)|119905=0

= 1198682minus120573

0+V(119905)|119905=0

= 0 119906(1) = 119886V(120585)and V(1) = 119887119906(120578) we can obtain that 119888

2= 1198884= 0 and 119888

1 1198883=

119886120585120573minus1 It yields the following

Ker 119871 sub (119906 V) isin dom 119871 (119906 V)

= 119888 (119886120585120573minus1

119905120572minus1

119905120573minus1

) 119888 isin R

(33)

Let (119909 119910) isin Im 119871 then there is (119906 V) isin dom 119871 such that(119909 119910) = 119871(119906 V) that is 119906 isin 119862

120572minus1

[0 1] 1198631205720+119906(119905) = 119909(119905) and

V isin 119862120573minus1[0 1]1198631205730+V(119905) = 119910(119905) By Lemma 4

119868120572

0+119909 (119905) = 119906 (119905) minus

((119863120572minus1

0+119906) (119905))

10038161003816100381610038161003816119905=0

Γ (120572)119905120572minus1

minus

((1198682minus120572

0+119906) (119905))

10038161003816100381610038161003816119905=0

Γ (120572 minus 1)119905120572minus2

119868120573

0+119910 (119905) = V (119905) minus

((119863120573minus1

0+V) (119905))

100381610038161003816100381610038161003816119905=0

Γ (120573)119905120573minus1

minus

((1198682minus120573

0+V) (119905))

100381610038161003816100381610038161003816119905=0

Γ (120573 minus 1)119905120573minus2

(34)

and by the couple boundary conditions we have

119906 (119905) = 119868120572

0+119909 (119905) + 119888

1119905120572minus1

V (119905) = 119868120573

0+119910 (119905) + 119888

2119905120573minus1

119906 (1) = 119868120572

0+119909 (1) + 119888

1= 119886V (120585) = 119886 [119868

120573

0+119910 (120585) + 119888

2120585120573minus1

]

V (1) = 119868120573

0+119910 (1) + 119888

2= 119887119906 (120578) = 119887 [119868

120572

0+119910 (120578) + 119888

1120578120572minus1

]

(35)

It yields the following

119886119868120573

0+119910 (120585) minus 119868

120572

0+119909 (1) + 119886120585

120573minus1

[119887119868120572

0+119909 (120578) minus 119868

120573

0+119910 (1)] = 0

(36)

On the other hand suppose that (119909 119910) isin 119885 satisfy (36) Let119906(119905) = 119868

120572

0+119909(119905) + 119886120585

120573minus1

119905120572minus1 and V(119905) = 119868

120573

0+119910(119905) + [119887119868

120572

0+119909(120578) minus

119868120573

0+119910(1) + 1]119905

120573minus1 and then 119863120572

0+119906(119905) = 119909(119905) 119863120573

0+V(119905) = 119910(119905)

and

1198682minus120572

0+119906 (119905) |119905=0

= 1198682minus120573

0+V (119905) |119905=0

= 0

119906 (1) = 119868120572

0+119909 (1) + 119886120585

120573minus1

= 119886119868120573

0+119910 (120585) + 119886 [119887119868

120572

0+119909 (120578) minus 119868

120573

0+119910 (1) + 1] 120585

120573minus1

= 119886V (120585)

V (1) = 119868120573

0+119910 (1) + [119887119868

120572

0+119909 (120578) minus 119868

120573

0+119910 (1) + 1]

= 119887119868120572

0+119909 (120578) + 119886119887120585

120573minus1

120578120572minus1

= 119887119906 (120578)

(37)

Therefore (30) holds

Lemma 9 If (2) holds then 119871 is a Fredholm operator of indexzero and dim Ker 119871 = codim Im 119871 = 1 Furthermore thelinear operator119870

119901 Im 119871 rarr dom 119871capKer119875 can be defined by

119870119875(119906 V) (119905)

= 119868120572

0+119906 (119905) minus 119886120585

120573minus1

119905120572minus1

[119887119868120572

0+119906 (120578) minus 119868

120573

0+V (1)] 119868120573

0+V (119905)

(38)

Also10038171003817100381710038171003817119870119901(119906 V)

10038171003817100381710038171003817119884le Δ1(119906 V)

119885 (39)

Proof Define operator 119876 119885 rarr 119885 as follows

119876 (119906 V) = Δ21198761(119906 V) (1 1) (40)

where 1198761 119885 rarr R is defined by

1198761(119906 V) = 119886119868

120573

0+V (120585) minus 119868120572

0+119906 (1)

+ 119886120585120573minus1

[119887119868120572

0+119906 (120578) minus 119868

120573

0+V (1)]

Δ2=

Γ (120573 + 1) Γ (120572 + 1)

120572120585120573minus1Γ (120572 + 1) (120585 minus 1) + Γ (120573 + 1) (120578 minus 1)= 0

(41)

It is easy to see that 1198762(119906 V) = 119876(119906 V) that is 119876 119885 rarr 119885

is a continuous linear projector Furthermore Ker119876 = Im 119871For (119906 V) isin 119885 set (119906 V) = [(119906 V) minus 119876(119906 V)] + 119876(119906 V) Then(119906 V) minus 119876(119906 V) isin Ker119876 and 119876(119906 V) isin Im119876 It follows fromKer119876 = Im 119871 and1198762(119906 V) = 119876(119906 V) that Im 119871capIm119876 = (0 0)So we have

119885 = Im 119871 oplus Im119876 (42)

Now Ind119871 = dim Ker 119871 minus codim Im 119871 = dim Ker 119871 minus

dim Im119876 = 0 and so 119871 is a Fredholm operator of index 0

Abstract and Applied Analysis 5

Let 119875 119884 rarr 119884 be continuous linear operator defined by

119875 (119906 V) =119863120573minus1

0+V (0)

Γ (120573)(119886120585120573minus1

119905120572minus1

119905120573minus1

) (43)

Obviously 119875 is a linear projector and

Ker119875 = (119906 V) isin 119884 119863120573minus1

0+V (0) = 0 (44)

It is easy to know that 119884 = Ker119875 oplus Ker 119871Define119870

119875 Im 119871 rarr dom 119871 cap Ker119875 by

119870119875(119906 V) (119905)

= (119868120572

0+119906 (119905) minus 119886120585

120573minus1

119905120572minus1

[119887119868120572

0+119906 (120578) minus 119868

120573

0+V (1)] 119868120573

0+V (119905))

(45)

Since

100381610038161003816100381610038161003816119887119868120572

0+119906 (120578) minus 119868

120573

0+V (1)

100381610038161003816100381610038161003816le

10038161003816100381610038161003816100381610038161003816

119887

Γ (120572)int

120578

0

(120578 minus 119904)120572minus1

119906 (119904) 119889119904

10038161003816100381610038161003816100381610038161003816

+

100381610038161003816100381610038161003816100381610038161003816

1

Γ (120573)int

1

0

(1 minus 119904)120572minus1V (119904) 119889119904

100381610038161003816100381610038161003816100381610038161003816

le119887

Γ (120572)1199061+

1

Γ (120573)V1

119863120572minus1

0+119868120572

0+119906 (119905) = int

119905

0

119906 (119904) 119889119904

119868120572

0+119906 (119905) =

1

Γ (120572)int

119905

0

(119905 minus 119904)120572minus1

119906 (119904) 119889119904

119863120573minus1

0+119868120573

0+V (119905) = int

119905

0

V (119904) 119889119904

119868120573

0+V (119905) =

1

Γ (120573)int

119905

0

(119905 minus 119904)120573minus1V (119904) 119889119904

(46)

then1003817100381710038171003817119870119875 (119906 V)

1003817100381710038171003817119884le Δ1(119906 V)

119885 (47)

In fact if (119906 V) isin Im 119871 then

119871119870119875(119906 V) (119905)

= 119863120572

0+[119868120572

0+119906 (119905) minus 119886120585

120573minus1

119905120572minus1

[119887119868120572

0+119906 (120578) minus 119868

120573

0+V (1)]]

119863120573

0+119868120573

0+V (119905) = (119906 V)

(48)

By Lemma 4 for (119906 V) isin dom 119871 cap Ker119875

119870119875119871 (119906 V) (119905)

= (119868120572

0+119863120572

0+119906 (119905)minus119886120585

120572minus1

119905120572minus1

[119887119868120572

0+119863120572

0+119906 (120578)minus119868

120573

0+119863120573

0+V (1)]

119868120573

0+119863120573

0+V (119905))

= (119906 (119905) minus119863120572minus1

0+119906 (0)

Γ (120572)119905120572minus1

minus1198682minus120572

0+119906 (0)

Γ (120572 minus 1)119905120572minus2

minus 119886120585120573minus1

119905120572minus1

times [119887119906 (120578) minus 119887119863120572minus1

0+119906 (0)

Γ (120572)120578120572minus1

minus1198682minus120572

0+119906 (0)

Γ (120572 minus 1)120578120572minus2

minusV (1)+119863120573minus1

0+V (0)

Γ (120573)] V (119905)minus

119863120573minus1

0+V (0)

Γ (120573)119905120573minus1

)

= (119906 (119905) minus119863120572minus1

0+119906 (0)

Γ (120572)119905120572minus1

+119886120585120573minus1

119905120572minus1

119887119863120572minus1

0+119906 (0)

Γ (120572)120578120572minus1

V (119905))

= (119906 (119905) V (119905))(49)

(119863120573minus10+

V(0) = 0 since (119906 V) isin Ker119875 and 1198682minus120572

0+119906(0) = 0 since

(119906 V) isin dom 119871) Hence

119870119901= (119871|dom119871capKer119875)

minus1

(50)

The proof is complete

3 Main Results

In this section we will use Theorem 1 to prove the existenceof solutions to BVP (1) To obtain our main theorem we usethe following assumptions

(H1) There exist functions 119886119894 119887119894 119888119894 119889119894 119890119894isin 1198711

[0 1] (119894 =

1 2) such that for all (119909 119910 119911 119908) isin R4 119905 isin [0 1]

1003816100381610038161003816119891 (119905 119909 119910 119911 119908)1003816100381610038161003816 le 1198901(119905) + 119886

1(119905) |119909| + 119887

1(119905)

10038161003816100381610038161199101003816100381610038161003816

+ 1198881(119905) |119911| + 119889

1(119905) |119908|

1003816100381610038161003816119892 (119905 119909 119910 119911 119908)1003816100381610038161003816 le 1198902(119905) + 119886

2(119905) |119909| + 119887

2(119905)

10038161003816100381610038161199101003816100381610038161003816

+ 1198882(119905) |119911| + 119889

2(119905) |119908|

(51)

(H2) There exists a constant 119860 gt 0 such that for (119906 V) isindom 119871 if |119863120573minus1

0+V(119905)| gt 119860 for all 119905 isin [0 1] then

1198761(1198731(119906 V)) = 0 or 119876

1(1198732(119906 V)) = 0

(H3) There exists a constant 119861 gt 0 such that either for each119888 isin R |119888| gt 119861

1198881198731(119888119886120585120573minus1

119905120572minus1

119888119905120573minus1

) gt 0 1198881198732(119888119886120585120573minus1

119905120572minus1

119888119905120573minus1

) gt 0

(52)

or for each 119886 isin R |119886| gt 119861

1198881198731(119888119886120585120573minus1

119905120572minus1

119888119905120573minus1

) lt 0

1198881198732(119888119886120585120573minus1

119905120572minus1

119888119905120573minus1

) lt 0

(53)

6 Abstract and Applied Analysis

Theorem 10 Suppose (2) and (1198671)ndash(1198673) hold Then (1) hasat least one solution in Y provided that

max (Δ1+ Δ3)max 10038171003817100381710038171198862

1003817100381710038171003817110038171003817100381710038171198872

1003817100381710038171003817110038171003817100381710038171198882

1003817100381710038171003817110038171003817100381710038171198892

10038171003817100381710038171 Δ4 Δ5

lt 1

(54)

Proof Set

Ω1= (119906 V) isin dom 119871 Ker 119871 119871 (119906 V) = 120582119873 (119906 V)

for some 120582 isin [0 1] (55)

Take (119906 V) isin Ω1 Since 119871119906 = 120582119873119906 so 120582 = 0 and 119873(119906 V) isin

Im 119871 = Ker119876 hence

119876119873(119906 V) = 1198761(1198731(119906 V) 119873

2(119906 V)) (1 1) = 0 (56)

Thus from (H2) there exist 1199050isin [0 1] such that

100381610038161003816100381610038161003816119863120573minus1

0+V (1199050)100381610038161003816100381610038161003816le 119860 (57)

Noticing that

119863120573minus1

0+V (119905) = 119863

120573minus1

0+V (1199050) + int

119905

1199050

119863120573

0+V (119904) 119889119904 (58)

so100381610038161003816100381610038161003816119863120573minus1

0+V (0)

100381610038161003816100381610038161003816le100381710038171003817100381710038171003817119863120573minus1

0+V (119905)

100381710038171003817100381710038171003817infinle100381610038161003816100381610038161003816119863120573minus2

0+V (1199050)100381610038161003816100381610038161003816+100381710038171003817100381710038171003817119863120573

0+V (119905)

1003817100381710038171003817100381710038171

le 119860 +10038171003817100381710038171198712V

10038171003817100381710038171le 119860 +

10038171003817100381710038171198732 (119906 V)10038171003817100381710038171

(59)

Thus

119875 (119906 V)119884=

1003817100381710038171003817100381710038171003817100381710038171003817

119863120573minus1

0+V (0)

Γ (120573)(119886120585120573minus1

119905120572minus1

119905120573minus1

)

1003817100381710038171003817100381710038171003817100381710038171003817119884

= Δ3

100381610038161003816100381610038161003816119863120573minus1

0+V (0)

100381610038161003816100381610038161003816le Δ3(119860 +

10038171003817100381710038171198732 (119906 V)10038171003817100381710038171)

(60)

For all (119906 V) isin Ω1 (119868minus119875)(119906 V) isin dom 119871capKer119875 Considering

Lemma 9 we get 119871119875(119906 V) = (0 0) Together with (39) wehave

(119868 minus 119875) (119906 V)119884=1003817100381710038171003817119870119875119871 (119868 minus 119875) (119906 V)

1003817100381710038171003817119884

le Δ1119871 (119868 minus 119875) (119906 V)

119885

le Δ1119873 (119906 V)

119885

(61)

From (60) and (61) we have

(119906 V)119884

le 119875 (119906 V)119884+ (119868 minus 119875) (119906 V)

119884

le Δ3(119860 +

10038171003817100381710038171198732 (119906 V)10038171003817100381710038171) + Δ1119873 (119906 V)

119885

= 119860Δ3+max (Δ

1+ Δ3)10038171003817100381710038171198732 (119906 V)

10038171003817100381710038171 Δ1

10038171003817100381710038171198731 (119906 V)10038171003817100381710038171

+Δ3

10038171003817100381710038171198732 (119906 V)10038171003817100381710038171

(62)

From (62) we discuss various cases

Case 1 ((119906 V)119884le 119860Δ

3+ (Δ1+ Δ3)1198732(119906 V)

1) From (H1)

we have

(119906 V)119884

le 119860Δ3+ (Δ1+ Δ3)

times (10038171003817100381710038171198862

10038171003817100381710038171119906infin+10038171003817100381710038171198872

10038171003817100381710038171

10038171003817100381710038171003817119863120572minus1

0+11990610038171003817100381710038171003817infin

+10038171003817100381710038171198882

10038171003817100381710038171Vinfin

+10038171003817100381710038171198892

10038171003817100381710038171

100381710038171003817100381710038171003817119863120573minus1

0+V100381710038171003817100381710038171003817infin

+10038171003817100381710038171198902

10038171003817100381710038171)

le 119860Δ3+ (Δ1+ Δ3)10038171003817100381710038171198902

10038171003817100381710038171

+ (Δ1+ Δ3)max 10038171003817100381710038171198862

1003817100381710038171003817110038171003817100381710038171198872

10038171003817100381710038171 119906119862120572minus1

+ (Δ1+ Δ3)max 10038171003817100381710038171198882

1003817100381710038171003817110038171003817100381710038171198892

10038171003817100381710038171 V119862120573minus1

le 119860Δ3+ (Δ1+ Δ3)10038171003817100381710038171198902

10038171003817100381710038171

+ (Δ1+ Δ3)max 10038171003817100381710038171198862

1003817100381710038171003817110038171003817100381710038171198872

1003817100381710038171003817110038171003817100381710038171198882

1003817100381710038171003817110038171003817100381710038171198892

10038171003817100381710038171 (119906 V)

119884

(63)

which yield

(119906 V)119884

le119860Δ2+ (Δ1+ Δ3)10038171003817100381710038171198902

10038171003817100381710038171

1 minus (Δ1+ Δ3)max 10038171003817100381710038171198862

1003817100381710038171003817110038171003817100381710038171198872

1003817100381710038171003817110038171003817100381710038171198882

1003817100381710038171003817110038171003817100381710038171198892

10038171003817100381710038171

(64)

ThusΩ1is bounded

Case 2 ((119906 V)119884le 119860Δ

3+ Δ11198731(119906 V)

1+ Δ31198732(119906 V)

1)

From (H1) we have

(119906 V)119884

le 119860Δ3+ Δ1(10038171003817100381710038171198861

10038171003817100381710038171119906infin+10038171003817100381710038171198871

10038171003817100381710038171

10038171003817100381710038171003817119863120572minus1

0+11990610038171003817100381710038171003817infin

+10038171003817100381710038171198881

10038171003817100381710038171Vinfin

+10038171003817100381710038171198891

10038171003817100381710038171

100381710038171003817100381710038171003817119863120573minus1

0+V100381710038171003817100381710038171003817infin

+10038171003817100381710038171198901

10038171003817100381710038171)

+ Δ3(10038171003817100381710038171198862

10038171003817100381710038171119906infin+10038171003817100381710038171198872

10038171003817100381710038171

10038171003817100381710038171003817119863120572minus1

0+11990610038171003817100381710038171003817infin

+10038171003817100381710038171198882

10038171003817100381710038171Vinfin

+10038171003817100381710038171198892

10038171003817100381710038171

100381710038171003817100381710038171003817119863120573minus1

0+V100381710038171003817100381710038171003817infin

+10038171003817100381710038171198902

10038171003817100381710038171)

le 119860Δ3+ Δ1

1003817100381710038171003817119890110038171003817100381710038171+ Δ3

1003817100381710038171003817119890210038171003817100381710038171+ Δ4119906119862120572minus1 + Δ

5V119862120573minus1

le 119860Δ3+ Δ1

1003817100381710038171003817119890110038171003817100381710038171+ Δ3

1003817100381710038171003817119890210038171003817100381710038171+max Δ

4 Δ5 (119906 V)

119884

(65)

which yield

(119906 V)119884le119860Δ3+ Δ1

1003817100381710038171003817119890110038171003817100381710038171+ Δ3

1003817100381710038171003817119890210038171003817100381710038171

1 minusmax Δ4 Δ5

(66)

ThusΩ1is bounded Let

Ω2= (119906 V) isin Ker 119871 119873 (119906 V) isin Im 119871 (67)

For (119906 V) isin Ω2and (119906 V) isin Ker 119871 so (119906 V) = 119888(119886120585

120573minus1

119905120572minus1

119905120573minus1

) 119905 isin [0 1] 119888 isin R Noticing that Im 119871 = Ker119876 then we

Abstract and Applied Analysis 7

get 119876119873(119906 V) = 0 and thus 11987611198731(119888119886120585120573minus1

119905120572minus1

119888119905120573minus1

) = 0 and11987611198732(119888119886120585120573minus1

119905120572minus1

119888119905120573minus1

) = 0 From (H2) we get |119888| le 119860Γ(120573)and thusΩ

2is bounded

We define the isomorphism 119869 Ker 119871 rarr Im119876 by

119869 (119888119886120585120573minus1

119905120572minus1

119888119905120573minus1

) = (119888 119888) (68)

If the first part of (H3) is satisfied then let

Ω3= (119906 V) isin ker 119871 120582119869 (119906 V)

+ (1 minus 120582)119876119873 (119906 V) = 0 120582 isin [0 1]

(69)

For (119906 V) = (119888119886120585120573minus1

119905120572minus1

119888119905120573minus1

) isin Ω3

120582 (119888 119888) = minus (1 minus 120582) Δ21198761119873(119888119886120585

120573minus1

119905120572minus1

119888119905120573minus1

) (1 1) (70)

If 120582 = 1 then 119888 = 0 Otherwise if |119888| gt 119861 in view of (H3) andΔ2lt 0 one has

(1 minus 120582) Δ31198761119873(119888119886120585

120573minus1

119905120572minus1

119888119905120573minus1

) lt 0 (71)

which contradict 1205821198882 ge 0ThusΩ3sub (119906 V) isin Ker 119871 | (119906 V) =

(119888119886120585120573minus1

119905120572minus1

119888119905120573minus1

) |119888| le 119861 is boundedIf the second part of (H3) holds then define the set

Ω3= (119906 V) isin Ker 119871 minus120582119869 (119906 V)

+ (1 minus 120582)119876119873 (119906 V) = 0 120582 isin [0 1]

(72)

and here 119869 is as above Similar to the above argument we canshow thatΩ

3is bounded too

In the following we will prove that all conditions ofTheorem 1 are satisfied LetΩ be a bounded open subset of 119884such that cup3

119894=1Ω119894sub Ω By standard arguments we can prove

that 119870119875(119868 minus 119876)119873 Ω rarr 119884 is compact and thus 119873 is 119871-

compact onΩ Then by the above argument we have

(i) 119871119906 = 120582119873119906 for every (119906 120582) isin [(dom 119871 Ker 119871) cap 120597Ω]times(0 1)

(ii) 119873119906 notin Im 119871 for 119906 isin Ker 119871 cap 120597Ω

Finally we will prove that (iii) of Theorem 1 is satisfied Let119867(119906 120582) = plusmn120582119869119906 + (1 minus 120582)119876119873119906 According to the aboveargument we know that

119867(119906 120582) = 0 for 119906 isin Ker 119871 cap 120597Ω (73)

Thus by the homotopy property of degree

deg (119876119873|Ker119871Ker 119871 cap Ω 0)

= deg (119867 (sdot 0) Ker 119871 cap Ω 0)

= deg (119867 (sdot 1) Ker 119871 cap Ω 0)

= deg (plusmn119869Ker 119871 cap Ω 0) = 0

(74)

Then byTheorem 1 119871(119906 V) = 119873(119906 V) has at least one solutionin dom 119871 cap Ω so that BVP (1) has a solution in 119862

120572minus1

[0 1] times

119862120573minus1

[0 1] The proof is complete

Conflict of Interests

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

Acknowledgments

The authors are very grateful to the anonymous refereesfor many valuable comments and suggestions which helpedto improve the presentation of the paper The project wassupported by the National Natural Science Foundation ofChina (11371221 61304074) the Specialized Research Foun-dation for the Doctoral Program of Higher Education ofChina (20123705110001) the Program for Scientific ResearchInnovation Team in Colleges and Universities of ShandongProvince the Postdoctoral Science Foundation of ShandongProvince (201303074) and Foundation of SDUST

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[1] A A Kilbas H M Srivastava and J J Trujillo Theoryand Applications of Fractional Differential Equations vol 204of North-Holland Mathematics Studies Elsevier Science BVAmsterdam The Netherlands 2006

[2] V Lakshmikantham S Leela and J Vasundhara DeviTheory ofFractional Dynamic Systems Cambridge Academic PublishersCambridge UK 2009

[3] J J Nieto and J Pimentel ldquoPositive solutions of a fractionalthermostat modelrdquo Boundary Value Problems vol 2013 article5 2013

[4] I Podlubny Fractional Differential Equations vol 198 Aca-demic Press San Diego Calif USA 1999

[5] W Wu and X Zhou ldquoEigenvalue of fractional differentialequations with P-Laplacian operatorrdquo Discrete Dynamics inNature and Society vol 2013 Article ID 137890 8 pages 2013

[6] X Zhang L Liu Y Wu and Y Lu ldquoThe iterative solutions ofnonlinear fractional differential equationsrdquo Applied Mathemat-ics and Computation vol 219 no 9 pp 4680ndash4691 2013

[7] X Zhang L Liu and Y Wu ldquoThe uniqueness of positivesolution for a singular fractional differential system involvingderivativesrdquo Communications in Nonlinear Science and Numer-ical Simulation vol 18 no 6 pp 1400ndash1409 2013

[8] X Zhang L Liu and Y Wu ldquoThe eigenvalue problem for asingular higher order fractional differential equation involvingfractional derivativesrdquo Applied Mathematics and Computationvol 218 no 17 pp 8526ndash8536 2012

[9] X Zhang L Liu and YWu ldquoExistence results for multiple pos-itive solutions of nonlinear higher order perturbed fractionaldifferential equations with derivativesrdquo Applied Mathematicsand Computation vol 219 no 4 pp 1420ndash1433 2012

[10] X Zhang L Liu B Wiwatanapataphee and Y Wu ldquoPositivesolutions of eigenvalue problems for a class of fractionaldifferential equations with derivativesrdquo Abstract and AppliedAnalysis vol 2012 Article ID 512127 16 pages 2012

[11] X Zhang L Liu and Y Wu ldquoMultiple positive solutionsof a singular fractional differential equation with negativelyperturbed termrdquo Mathematical and Computer Modelling vol55 no 3-4 pp 1263ndash1274 2012

[12] K Deng ldquoBlow-up rates for parabolic systemsrdquo Zeitschrift furAngewandte Mathematik und Physik vol 47 no 1 pp 132ndash1431996

8 Abstract and Applied Analysis

[13] K Deng ldquoGlobal existence and blow-up for a system of heatequations with non-linear boundary conditionsrdquoMathematicalMethods in the Applied Sciences vol 18 no 4 pp 307ndash315 1995

[14] L Zhigui and X Chunhong ldquoThe blow-up rate for a system ofheat equations with nonlinear boundary conditionsrdquoNonlinearAnalysisTheory Methods ampApplications vol 34 no 5 pp 767ndash778 1998

[15] D G Aronson ldquoA comparison method for stability analysis ofnonlinear parabolic problemsrdquo SIAM Review vol 20 no 2 pp245ndash264 1978

[16] M Pedersen and Z Lin ldquoBlow-up analysis for a system of heatequations coupled through a nonlinear boundary conditionrdquoApplied Mathematics Letters vol 14 no 2 pp 171ndash176 2001

[17] N A Asif and R A Khan ldquoPositive solutions to singularsystem with four-point coupled boundary conditionsrdquo Journalof Mathematical Analysis and Applications vol 386 no 2 pp848ndash861 2012

[18] C Yuan D Jiang D OrsquoRegan and R P Agarwal ldquoMultiple pos-itive solutions to systems of nonlinear semipositone fractionaldifferential equations with coupled boundary conditionsrdquo Elec-tronic Journal of QualitativeTheory of Differential Equations no13 p 17 2012

[19] Y Cui and J Sun ldquoOn existence of positive solutions of coupledintegral boundary value problems for a nonlinear singularsuperlinear differential systemrdquo Electronic Journal of QualitativeTheory of Differential Equations no 41 pp 1ndash13 2012

[20] Z Bai and Y Zhang ldquoSolvability of fractional three-pointboundary value problems with nonlinear growthrdquo AppliedMathematics and Computation vol 218 no 5 pp 1719ndash17252011

[21] Y Cui ldquoSolvability of second-order boundary-value problemsat resonance involving integral conditionsrdquo Electronic Journalof Differential Equations no 45 pp 1ndash9 2012

[22] Z Hu and W Liu ldquoSolvability for fractional order boundaryvalue problems at resonancerdquo Boundary Value Problems vol2011 article 20 2011

[23] W Jiang ldquoThe existence of solutions to boundary value prob-lems of fractional differential equations at resonancerdquoNonlinearAnalysis Theory Methods amp Applications vol 74 no 5 pp1987ndash1994 2011

[24] N Kosmatov ldquoMulti-point boundary value problems on timescales at resonancerdquo Journal of Mathematical Analysis andApplications vol 323 no 1 pp 253ndash266 2006

[25] G Wang W Liu S Zhu and T Zheng ldquoExistence results fora coupled system of nonlinear fractional 2m-point boundaryvalue problems at resonancerdquo Advances in Difference Equationsvol 2011 article 80 2011

[26] J R L Webb ldquoRemarks on nonlocal boundary value problemsat resonancerdquo Applied Mathematics and Computation vol 216no 2 pp 497ndash500 2010

[27] J R L Webb and M Zima ldquoMultiple positive solutions ofresonant and non-resonant non-local fourth-order boundaryvalue problemsrdquo Glasgow Mathematical Journal vol 54 no 1pp 225ndash240 2012

[28] X ZhangM Feng andWGe ldquoExistence result of second-orderdifferential equations with integral boundary conditions atresonancerdquo Journal of Mathematical Analysis and Applicationsvol 353 no 1 pp 311ndash319 2009

[29] X Zhang C Zhu and Z Wu ldquoSolvability for a coupled systemof fractional differential equations with impulses at resonancerdquoBoundary Value Problems vol 2013 artice 80 2013

[30] Z Zhao and J Liang ldquoExistence of solutions to functionalboundary value problem of second-order nonlinear differentialequationrdquo Journal of Mathematical Analysis and Applicationsvol 373 no 2 pp 614ndash634 2011

[31] Y Zou and Y Cui ldquoExistence results for a functional boundaryvalue problem of fractional differential equationsrdquo Advances inDifference Equations vol 2013 article 25 2013

[32] J Mawhin ldquoTopological degree and boundary value problemsfor nonlinear differential equationsrdquo in Topological Methods forOrdinary Differential Equations P M Fitzpertrick M MartelliJ Mawhin and R Nussbaum Eds vol 1537 of Lecture Notes inMathematics pp 74ndash142 Springer Berlin Germany 1993

[33] J Mawhin Topological Degree Methods in Nonlinear BoundaryValue Problems vol 40 of CBMS Regional Conference Series inMathematics American Mathematical Society Providence RIUSA 1979

[34] Y Zhang and Z Bai ldquoExistence of solutions for nonlinearfractional three-point boundary value problems at resonancerdquoJournal of Applied Mathematics and Computing vol 36 no 1-2pp 417ndash440 2011

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

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Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

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Decision SciencesAdvances in

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Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Abstract and Applied Analysis 3

and the Banach space 119862120583[0 1] (120583 gt 0)

119862120583

[0 1]

= 119906 (119905) | 119906 (119905) = 119868120583

0+119909 (119905) + 119888

1119905120583minus1

+ 1198882119905120583minus2

+ sdot sdot sdot

+ 119888119873minus1

119905120583minus(119873minus1)

119909 isin 119862 [0 1] 119905 isin [0 1]

119888119894isin R 119894 = 1 2 119873 = [120583] + 1

(12)

with the norm 119906119862120583 = 119863

120583

0+119906infin+sdot sdot sdot+119863

120583minus(119873minus1)

0+119906infin+119906infin

Lemma 4 (see [1]) Let 120572 gt 0 119899 = [120572] + 1 Assume that 119906 isin

1198711

(0 1)with a fractional integration of order 119899minus120572 that belongsto 119860119862119899[0 1] Then the equality

(119868120572

0+119863120572

0+119906) (119905) = 119906 (119905) minus

119899

sum

119894=1

((119868119899minus120572

0+119906) (119905))

(119899minus119894)

|119905=0

Γ (120572 minus 119894 + 1)119905120572minus119894 (13)

holds almost everywhere on [0 1]

In the following lemma we use the unified notationof both for fractional integrals and fractional derivativesassuming that 119868120572

0+= 119863minus120572

0+for 120572 lt 0

Lemma 5 (see [1]) Assume that 120572 gt 0 then

(i) let 119896 isin N If119863120572119886+119906(119905) and (119863120572+119896

119886+119906)(119905) exist then

(119863119896

119863120572

119886+) 119906 (119905) = (119863

120572+119896

119886+119906) (119905) (14)

(ii) if 120573 gt 0 120572 + 120573 gt 1 then

(119868120572

119886+119868120573

119886+) 119906 (119905) = (119868

120572+120573

119886+119906) (119905) (15)

satisfies at any point on [119886 119887] for 119906 isin 119871119901(119886 119887) and 1 le

119901 le +infin(iii) let 119906 isin 119862[119886 119887] Then (119863

120572

119886+119868120572

119886+)119906(119905) = 119906(119905) holds on

[119886 119887](iv) note that for 120582 gt minus1 120582 = 120572 minus 1 120572 minus 2 120572 minus 119899 one

has

119863120572

119905120582

=Γ (120582 + 1)

Γ (120582 minus 120572 + 1)119905120582minus120572

119863120572

119905120572minus119894

= 0 119894 = 1 2 119899

(16)

Remark 6 If 1 lt 120572 lt 2 and 119906 satisfies1198631205720+119906 = 119891(119905) isin 119871

1

(0 1)

and 1198682minus1205720+

119906|119905=0

= 0 then 119906 isin 119862120572minus1[0 1] In fact with Lemma 4one has

119906 (119905) = 119868120572

0+119891 (119905) + 119888

1119905120572minus1

+ 1198882119905120572minus2

(17)

Combined with 1198682minus1205720+

119906|119905=0

= 0 there is 1198882= 0 So

119906 (119905) = 119868120572

0+119891 (119905) + 119888

1119905120572minus1

= 119868120572minus1

0+[1198681

0+119891 (119905) + 119888

1Γ (120572)] (18)

Lemma 7 (see [34]) 119865 sub 119862120583

[0 1] is a sequentially compactset if and only if 119865 is uniformly bounded and equicontinuousHere to be uniformly bounded means that there exists119872 gt 0

such that for every 119906 isin 119865

119906119862120583 =

1003817100381710038171003817119863120583

0+1199061003817100381710038171003817infin

+ sdot sdot sdot +100381710038171003817100381710038171003817119863120583minus(119873minus1)

0+119906100381710038171003817100381710038171003817infin

+ 119906infinlt 119872 (19)

and to be equicontinuous means that for all 120598 gt 0 exist120575 gt 0 andfor all 119905

1 1199052isin [0 1] |119905

1minus 1199052| lt 120575 119906 isin 119865 and 119894 isin 1 [120583]

the following holds1003816100381610038161003816119906 (1199051) minus 119906 (1199052)

1003816100381610038161003816 lt 12057610038161003816100381610038161003816119863120583minus119894

0+119906 (1199051) minus 119863120583minus119894

0+119906 (1199052)10038161003816100381610038161003816lt 120576

(20)

We also use the following two Banach spaces 119884 =

119862120572minus1

[0 1] times 119862120573minus1

[0 1] with the norm1003817100381710038171003817(119909 119910)

1003817100381710038171003817119884= max 119909

119862120572minus1

10038171003817100381710038171199101003817100381710038171003817119862120573minus1

(21)

and 119885 = 1198711

[0 1] times 1198711

[0 1] with the norm1003817100381710038171003817(119909 119910)

1003817100381710038171003817119885= max 119909

1100381710038171003817100381711991010038171003817100381710038171 (22)

Let the linear operator 119871 dom 119871 sub 119884 rarr 119885 with

dom 119871 = (119906 V) isin 119884 1198682minus120572

0+119906 (119905) |119905=0

= 0 119906 (1) = 119886V (120585)

1198682minus120573

0+V (119905) |119905=0

= 0 V (1) = 119887119906 (120578)

(23)

be defined by

119871 (119906 V) = (1198711119906 1198712V) (24)

where 1198711 119862120572minus1

[0 1] rarr 1198711

[0 1] and 1198712 119862120573minus1

[0 1] rarr

1198711

[0 1] are defined by

1198711119906 = 119863

120572

0+119906 (119905) 119871

2V = 119863

120573

0+V (119905) (25)

Let the nonlinear operator119873 119884 rarr 119885 be defined by

(119873 (119906 V)) (119905) = (1198731(119906 V) (119905) 119873

2(119906 V) (119905)) (26)

where1198731 1198732 119884 rarr 119871

1

[0 1] are defined by

1198731(119906 V) (119905) = 119891 (119905 119906 (119905) 119863

120572minus1

0+119906 (119905) V (119905) 119863120573minus1

0+V (119905))

1198732(119909 119910) (119905) = 119892 (119905 119906 (119905) 119863

120572minus1

0+119906 (119905) V (119905) 119863120573minus1

0+V (119905))

(27)

Then four-point coupled boundary value problems (1) can bewritten as

119871 (119906 V) = 119873 (119906 V) (28)

Lemma 8 Let 119871 be the linear operator defined as above If (2)holds then

Ker 119871 = (119906 V) isin dom 119871 (119906 V) = 119888 (119886120585120573minus1

119905120572minus1

119905120573minus1

)

119888 isin R 119905 isin [0 1]

(29)

Im 119871 = (119909 119910) isin 119885 119886119868120573

0+119910 (120585) minus 119868

120572

0+119909 (1)

+119886120585120573minus1

[119887119868120572

0+119909 (120578) minus 119868

120573

0+119910 (1)] = 0

(30)

4 Abstract and Applied Analysis

Proof Let 119906(119905) = 119886120585120573minus1

119905120572minus1and let V(119905) = 119905

120573minus1 Then byLemma 5 we have 1198682minus120572

0+119906(119905)|119905=0

= 1198682minus120573

0+V(119905)|119905=0

= 0 119906(1) =

119886120585120573minus1

= 119886V(120585) V(1) = 1 = 119887119906(120578) and 1198631205720+119906(119905) = 119863

120573

0+V(119905) = 0

So

(119906 V) isin dom 119871 (119906 V) = 119888 (119886120585120573minus1

119905120572minus1

119905120573minus1

) 119888 isin R

sub Ker 119871(31)

For every (119906 V) isin Ker 119871 if1198631205720+119906(119905) = 119863

120573

0+V(119905) = 0 then

119906 (119905) = 1198881119905120572minus1

+ 1198882119905120572minus2

V (119905) = 1198883119905120573minus1

+ 1198884119905120573minus2

(32)

Considering that 1198682minus1205720+

119906(119905)|119905=0

= 1198682minus120573

0+V(119905)|119905=0

= 0 119906(1) = 119886V(120585)and V(1) = 119887119906(120578) we can obtain that 119888

2= 1198884= 0 and 119888

1 1198883=

119886120585120573minus1 It yields the following

Ker 119871 sub (119906 V) isin dom 119871 (119906 V)

= 119888 (119886120585120573minus1

119905120572minus1

119905120573minus1

) 119888 isin R

(33)

Let (119909 119910) isin Im 119871 then there is (119906 V) isin dom 119871 such that(119909 119910) = 119871(119906 V) that is 119906 isin 119862

120572minus1

[0 1] 1198631205720+119906(119905) = 119909(119905) and

V isin 119862120573minus1[0 1]1198631205730+V(119905) = 119910(119905) By Lemma 4

119868120572

0+119909 (119905) = 119906 (119905) minus

((119863120572minus1

0+119906) (119905))

10038161003816100381610038161003816119905=0

Γ (120572)119905120572minus1

minus

((1198682minus120572

0+119906) (119905))

10038161003816100381610038161003816119905=0

Γ (120572 minus 1)119905120572minus2

119868120573

0+119910 (119905) = V (119905) minus

((119863120573minus1

0+V) (119905))

100381610038161003816100381610038161003816119905=0

Γ (120573)119905120573minus1

minus

((1198682minus120573

0+V) (119905))

100381610038161003816100381610038161003816119905=0

Γ (120573 minus 1)119905120573minus2

(34)

and by the couple boundary conditions we have

119906 (119905) = 119868120572

0+119909 (119905) + 119888

1119905120572minus1

V (119905) = 119868120573

0+119910 (119905) + 119888

2119905120573minus1

119906 (1) = 119868120572

0+119909 (1) + 119888

1= 119886V (120585) = 119886 [119868

120573

0+119910 (120585) + 119888

2120585120573minus1

]

V (1) = 119868120573

0+119910 (1) + 119888

2= 119887119906 (120578) = 119887 [119868

120572

0+119910 (120578) + 119888

1120578120572minus1

]

(35)

It yields the following

119886119868120573

0+119910 (120585) minus 119868

120572

0+119909 (1) + 119886120585

120573minus1

[119887119868120572

0+119909 (120578) minus 119868

120573

0+119910 (1)] = 0

(36)

On the other hand suppose that (119909 119910) isin 119885 satisfy (36) Let119906(119905) = 119868

120572

0+119909(119905) + 119886120585

120573minus1

119905120572minus1 and V(119905) = 119868

120573

0+119910(119905) + [119887119868

120572

0+119909(120578) minus

119868120573

0+119910(1) + 1]119905

120573minus1 and then 119863120572

0+119906(119905) = 119909(119905) 119863120573

0+V(119905) = 119910(119905)

and

1198682minus120572

0+119906 (119905) |119905=0

= 1198682minus120573

0+V (119905) |119905=0

= 0

119906 (1) = 119868120572

0+119909 (1) + 119886120585

120573minus1

= 119886119868120573

0+119910 (120585) + 119886 [119887119868

120572

0+119909 (120578) minus 119868

120573

0+119910 (1) + 1] 120585

120573minus1

= 119886V (120585)

V (1) = 119868120573

0+119910 (1) + [119887119868

120572

0+119909 (120578) minus 119868

120573

0+119910 (1) + 1]

= 119887119868120572

0+119909 (120578) + 119886119887120585

120573minus1

120578120572minus1

= 119887119906 (120578)

(37)

Therefore (30) holds

Lemma 9 If (2) holds then 119871 is a Fredholm operator of indexzero and dim Ker 119871 = codim Im 119871 = 1 Furthermore thelinear operator119870

119901 Im 119871 rarr dom 119871capKer119875 can be defined by

119870119875(119906 V) (119905)

= 119868120572

0+119906 (119905) minus 119886120585

120573minus1

119905120572minus1

[119887119868120572

0+119906 (120578) minus 119868

120573

0+V (1)] 119868120573

0+V (119905)

(38)

Also10038171003817100381710038171003817119870119901(119906 V)

10038171003817100381710038171003817119884le Δ1(119906 V)

119885 (39)

Proof Define operator 119876 119885 rarr 119885 as follows

119876 (119906 V) = Δ21198761(119906 V) (1 1) (40)

where 1198761 119885 rarr R is defined by

1198761(119906 V) = 119886119868

120573

0+V (120585) minus 119868120572

0+119906 (1)

+ 119886120585120573minus1

[119887119868120572

0+119906 (120578) minus 119868

120573

0+V (1)]

Δ2=

Γ (120573 + 1) Γ (120572 + 1)

120572120585120573minus1Γ (120572 + 1) (120585 minus 1) + Γ (120573 + 1) (120578 minus 1)= 0

(41)

It is easy to see that 1198762(119906 V) = 119876(119906 V) that is 119876 119885 rarr 119885

is a continuous linear projector Furthermore Ker119876 = Im 119871For (119906 V) isin 119885 set (119906 V) = [(119906 V) minus 119876(119906 V)] + 119876(119906 V) Then(119906 V) minus 119876(119906 V) isin Ker119876 and 119876(119906 V) isin Im119876 It follows fromKer119876 = Im 119871 and1198762(119906 V) = 119876(119906 V) that Im 119871capIm119876 = (0 0)So we have

119885 = Im 119871 oplus Im119876 (42)

Now Ind119871 = dim Ker 119871 minus codim Im 119871 = dim Ker 119871 minus

dim Im119876 = 0 and so 119871 is a Fredholm operator of index 0

Abstract and Applied Analysis 5

Let 119875 119884 rarr 119884 be continuous linear operator defined by

119875 (119906 V) =119863120573minus1

0+V (0)

Γ (120573)(119886120585120573minus1

119905120572minus1

119905120573minus1

) (43)

Obviously 119875 is a linear projector and

Ker119875 = (119906 V) isin 119884 119863120573minus1

0+V (0) = 0 (44)

It is easy to know that 119884 = Ker119875 oplus Ker 119871Define119870

119875 Im 119871 rarr dom 119871 cap Ker119875 by

119870119875(119906 V) (119905)

= (119868120572

0+119906 (119905) minus 119886120585

120573minus1

119905120572minus1

[119887119868120572

0+119906 (120578) minus 119868

120573

0+V (1)] 119868120573

0+V (119905))

(45)

Since

100381610038161003816100381610038161003816119887119868120572

0+119906 (120578) minus 119868

120573

0+V (1)

100381610038161003816100381610038161003816le

10038161003816100381610038161003816100381610038161003816

119887

Γ (120572)int

120578

0

(120578 minus 119904)120572minus1

119906 (119904) 119889119904

10038161003816100381610038161003816100381610038161003816

+

100381610038161003816100381610038161003816100381610038161003816

1

Γ (120573)int

1

0

(1 minus 119904)120572minus1V (119904) 119889119904

100381610038161003816100381610038161003816100381610038161003816

le119887

Γ (120572)1199061+

1

Γ (120573)V1

119863120572minus1

0+119868120572

0+119906 (119905) = int

119905

0

119906 (119904) 119889119904

119868120572

0+119906 (119905) =

1

Γ (120572)int

119905

0

(119905 minus 119904)120572minus1

119906 (119904) 119889119904

119863120573minus1

0+119868120573

0+V (119905) = int

119905

0

V (119904) 119889119904

119868120573

0+V (119905) =

1

Γ (120573)int

119905

0

(119905 minus 119904)120573minus1V (119904) 119889119904

(46)

then1003817100381710038171003817119870119875 (119906 V)

1003817100381710038171003817119884le Δ1(119906 V)

119885 (47)

In fact if (119906 V) isin Im 119871 then

119871119870119875(119906 V) (119905)

= 119863120572

0+[119868120572

0+119906 (119905) minus 119886120585

120573minus1

119905120572minus1

[119887119868120572

0+119906 (120578) minus 119868

120573

0+V (1)]]

119863120573

0+119868120573

0+V (119905) = (119906 V)

(48)

By Lemma 4 for (119906 V) isin dom 119871 cap Ker119875

119870119875119871 (119906 V) (119905)

= (119868120572

0+119863120572

0+119906 (119905)minus119886120585

120572minus1

119905120572minus1

[119887119868120572

0+119863120572

0+119906 (120578)minus119868

120573

0+119863120573

0+V (1)]

119868120573

0+119863120573

0+V (119905))

= (119906 (119905) minus119863120572minus1

0+119906 (0)

Γ (120572)119905120572minus1

minus1198682minus120572

0+119906 (0)

Γ (120572 minus 1)119905120572minus2

minus 119886120585120573minus1

119905120572minus1

times [119887119906 (120578) minus 119887119863120572minus1

0+119906 (0)

Γ (120572)120578120572minus1

minus1198682minus120572

0+119906 (0)

Γ (120572 minus 1)120578120572minus2

minusV (1)+119863120573minus1

0+V (0)

Γ (120573)] V (119905)minus

119863120573minus1

0+V (0)

Γ (120573)119905120573minus1

)

= (119906 (119905) minus119863120572minus1

0+119906 (0)

Γ (120572)119905120572minus1

+119886120585120573minus1

119905120572minus1

119887119863120572minus1

0+119906 (0)

Γ (120572)120578120572minus1

V (119905))

= (119906 (119905) V (119905))(49)

(119863120573minus10+

V(0) = 0 since (119906 V) isin Ker119875 and 1198682minus120572

0+119906(0) = 0 since

(119906 V) isin dom 119871) Hence

119870119901= (119871|dom119871capKer119875)

minus1

(50)

The proof is complete

3 Main Results

In this section we will use Theorem 1 to prove the existenceof solutions to BVP (1) To obtain our main theorem we usethe following assumptions

(H1) There exist functions 119886119894 119887119894 119888119894 119889119894 119890119894isin 1198711

[0 1] (119894 =

1 2) such that for all (119909 119910 119911 119908) isin R4 119905 isin [0 1]

1003816100381610038161003816119891 (119905 119909 119910 119911 119908)1003816100381610038161003816 le 1198901(119905) + 119886

1(119905) |119909| + 119887

1(119905)

10038161003816100381610038161199101003816100381610038161003816

+ 1198881(119905) |119911| + 119889

1(119905) |119908|

1003816100381610038161003816119892 (119905 119909 119910 119911 119908)1003816100381610038161003816 le 1198902(119905) + 119886

2(119905) |119909| + 119887

2(119905)

10038161003816100381610038161199101003816100381610038161003816

+ 1198882(119905) |119911| + 119889

2(119905) |119908|

(51)

(H2) There exists a constant 119860 gt 0 such that for (119906 V) isindom 119871 if |119863120573minus1

0+V(119905)| gt 119860 for all 119905 isin [0 1] then

1198761(1198731(119906 V)) = 0 or 119876

1(1198732(119906 V)) = 0

(H3) There exists a constant 119861 gt 0 such that either for each119888 isin R |119888| gt 119861

1198881198731(119888119886120585120573minus1

119905120572minus1

119888119905120573minus1

) gt 0 1198881198732(119888119886120585120573minus1

119905120572minus1

119888119905120573minus1

) gt 0

(52)

or for each 119886 isin R |119886| gt 119861

1198881198731(119888119886120585120573minus1

119905120572minus1

119888119905120573minus1

) lt 0

1198881198732(119888119886120585120573minus1

119905120572minus1

119888119905120573minus1

) lt 0

(53)

6 Abstract and Applied Analysis

Theorem 10 Suppose (2) and (1198671)ndash(1198673) hold Then (1) hasat least one solution in Y provided that

max (Δ1+ Δ3)max 10038171003817100381710038171198862

1003817100381710038171003817110038171003817100381710038171198872

1003817100381710038171003817110038171003817100381710038171198882

1003817100381710038171003817110038171003817100381710038171198892

10038171003817100381710038171 Δ4 Δ5

lt 1

(54)

Proof Set

Ω1= (119906 V) isin dom 119871 Ker 119871 119871 (119906 V) = 120582119873 (119906 V)

for some 120582 isin [0 1] (55)

Take (119906 V) isin Ω1 Since 119871119906 = 120582119873119906 so 120582 = 0 and 119873(119906 V) isin

Im 119871 = Ker119876 hence

119876119873(119906 V) = 1198761(1198731(119906 V) 119873

2(119906 V)) (1 1) = 0 (56)

Thus from (H2) there exist 1199050isin [0 1] such that

100381610038161003816100381610038161003816119863120573minus1

0+V (1199050)100381610038161003816100381610038161003816le 119860 (57)

Noticing that

119863120573minus1

0+V (119905) = 119863

120573minus1

0+V (1199050) + int

119905

1199050

119863120573

0+V (119904) 119889119904 (58)

so100381610038161003816100381610038161003816119863120573minus1

0+V (0)

100381610038161003816100381610038161003816le100381710038171003817100381710038171003817119863120573minus1

0+V (119905)

100381710038171003817100381710038171003817infinle100381610038161003816100381610038161003816119863120573minus2

0+V (1199050)100381610038161003816100381610038161003816+100381710038171003817100381710038171003817119863120573

0+V (119905)

1003817100381710038171003817100381710038171

le 119860 +10038171003817100381710038171198712V

10038171003817100381710038171le 119860 +

10038171003817100381710038171198732 (119906 V)10038171003817100381710038171

(59)

Thus

119875 (119906 V)119884=

1003817100381710038171003817100381710038171003817100381710038171003817

119863120573minus1

0+V (0)

Γ (120573)(119886120585120573minus1

119905120572minus1

119905120573minus1

)

1003817100381710038171003817100381710038171003817100381710038171003817119884

= Δ3

100381610038161003816100381610038161003816119863120573minus1

0+V (0)

100381610038161003816100381610038161003816le Δ3(119860 +

10038171003817100381710038171198732 (119906 V)10038171003817100381710038171)

(60)

For all (119906 V) isin Ω1 (119868minus119875)(119906 V) isin dom 119871capKer119875 Considering

Lemma 9 we get 119871119875(119906 V) = (0 0) Together with (39) wehave

(119868 minus 119875) (119906 V)119884=1003817100381710038171003817119870119875119871 (119868 minus 119875) (119906 V)

1003817100381710038171003817119884

le Δ1119871 (119868 minus 119875) (119906 V)

119885

le Δ1119873 (119906 V)

119885

(61)

From (60) and (61) we have

(119906 V)119884

le 119875 (119906 V)119884+ (119868 minus 119875) (119906 V)

119884

le Δ3(119860 +

10038171003817100381710038171198732 (119906 V)10038171003817100381710038171) + Δ1119873 (119906 V)

119885

= 119860Δ3+max (Δ

1+ Δ3)10038171003817100381710038171198732 (119906 V)

10038171003817100381710038171 Δ1

10038171003817100381710038171198731 (119906 V)10038171003817100381710038171

+Δ3

10038171003817100381710038171198732 (119906 V)10038171003817100381710038171

(62)

From (62) we discuss various cases

Case 1 ((119906 V)119884le 119860Δ

3+ (Δ1+ Δ3)1198732(119906 V)

1) From (H1)

we have

(119906 V)119884

le 119860Δ3+ (Δ1+ Δ3)

times (10038171003817100381710038171198862

10038171003817100381710038171119906infin+10038171003817100381710038171198872

10038171003817100381710038171

10038171003817100381710038171003817119863120572minus1

0+11990610038171003817100381710038171003817infin

+10038171003817100381710038171198882

10038171003817100381710038171Vinfin

+10038171003817100381710038171198892

10038171003817100381710038171

100381710038171003817100381710038171003817119863120573minus1

0+V100381710038171003817100381710038171003817infin

+10038171003817100381710038171198902

10038171003817100381710038171)

le 119860Δ3+ (Δ1+ Δ3)10038171003817100381710038171198902

10038171003817100381710038171

+ (Δ1+ Δ3)max 10038171003817100381710038171198862

1003817100381710038171003817110038171003817100381710038171198872

10038171003817100381710038171 119906119862120572minus1

+ (Δ1+ Δ3)max 10038171003817100381710038171198882

1003817100381710038171003817110038171003817100381710038171198892

10038171003817100381710038171 V119862120573minus1

le 119860Δ3+ (Δ1+ Δ3)10038171003817100381710038171198902

10038171003817100381710038171

+ (Δ1+ Δ3)max 10038171003817100381710038171198862

1003817100381710038171003817110038171003817100381710038171198872

1003817100381710038171003817110038171003817100381710038171198882

1003817100381710038171003817110038171003817100381710038171198892

10038171003817100381710038171 (119906 V)

119884

(63)

which yield

(119906 V)119884

le119860Δ2+ (Δ1+ Δ3)10038171003817100381710038171198902

10038171003817100381710038171

1 minus (Δ1+ Δ3)max 10038171003817100381710038171198862

1003817100381710038171003817110038171003817100381710038171198872

1003817100381710038171003817110038171003817100381710038171198882

1003817100381710038171003817110038171003817100381710038171198892

10038171003817100381710038171

(64)

ThusΩ1is bounded

Case 2 ((119906 V)119884le 119860Δ

3+ Δ11198731(119906 V)

1+ Δ31198732(119906 V)

1)

From (H1) we have

(119906 V)119884

le 119860Δ3+ Δ1(10038171003817100381710038171198861

10038171003817100381710038171119906infin+10038171003817100381710038171198871

10038171003817100381710038171

10038171003817100381710038171003817119863120572minus1

0+11990610038171003817100381710038171003817infin

+10038171003817100381710038171198881

10038171003817100381710038171Vinfin

+10038171003817100381710038171198891

10038171003817100381710038171

100381710038171003817100381710038171003817119863120573minus1

0+V100381710038171003817100381710038171003817infin

+10038171003817100381710038171198901

10038171003817100381710038171)

+ Δ3(10038171003817100381710038171198862

10038171003817100381710038171119906infin+10038171003817100381710038171198872

10038171003817100381710038171

10038171003817100381710038171003817119863120572minus1

0+11990610038171003817100381710038171003817infin

+10038171003817100381710038171198882

10038171003817100381710038171Vinfin

+10038171003817100381710038171198892

10038171003817100381710038171

100381710038171003817100381710038171003817119863120573minus1

0+V100381710038171003817100381710038171003817infin

+10038171003817100381710038171198902

10038171003817100381710038171)

le 119860Δ3+ Δ1

1003817100381710038171003817119890110038171003817100381710038171+ Δ3

1003817100381710038171003817119890210038171003817100381710038171+ Δ4119906119862120572minus1 + Δ

5V119862120573minus1

le 119860Δ3+ Δ1

1003817100381710038171003817119890110038171003817100381710038171+ Δ3

1003817100381710038171003817119890210038171003817100381710038171+max Δ

4 Δ5 (119906 V)

119884

(65)

which yield

(119906 V)119884le119860Δ3+ Δ1

1003817100381710038171003817119890110038171003817100381710038171+ Δ3

1003817100381710038171003817119890210038171003817100381710038171

1 minusmax Δ4 Δ5

(66)

ThusΩ1is bounded Let

Ω2= (119906 V) isin Ker 119871 119873 (119906 V) isin Im 119871 (67)

For (119906 V) isin Ω2and (119906 V) isin Ker 119871 so (119906 V) = 119888(119886120585

120573minus1

119905120572minus1

119905120573minus1

) 119905 isin [0 1] 119888 isin R Noticing that Im 119871 = Ker119876 then we

Abstract and Applied Analysis 7

get 119876119873(119906 V) = 0 and thus 11987611198731(119888119886120585120573minus1

119905120572minus1

119888119905120573minus1

) = 0 and11987611198732(119888119886120585120573minus1

119905120572minus1

119888119905120573minus1

) = 0 From (H2) we get |119888| le 119860Γ(120573)and thusΩ

2is bounded

We define the isomorphism 119869 Ker 119871 rarr Im119876 by

119869 (119888119886120585120573minus1

119905120572minus1

119888119905120573minus1

) = (119888 119888) (68)

If the first part of (H3) is satisfied then let

Ω3= (119906 V) isin ker 119871 120582119869 (119906 V)

+ (1 minus 120582)119876119873 (119906 V) = 0 120582 isin [0 1]

(69)

For (119906 V) = (119888119886120585120573minus1

119905120572minus1

119888119905120573minus1

) isin Ω3

120582 (119888 119888) = minus (1 minus 120582) Δ21198761119873(119888119886120585

120573minus1

119905120572minus1

119888119905120573minus1

) (1 1) (70)

If 120582 = 1 then 119888 = 0 Otherwise if |119888| gt 119861 in view of (H3) andΔ2lt 0 one has

(1 minus 120582) Δ31198761119873(119888119886120585

120573minus1

119905120572minus1

119888119905120573minus1

) lt 0 (71)

which contradict 1205821198882 ge 0ThusΩ3sub (119906 V) isin Ker 119871 | (119906 V) =

(119888119886120585120573minus1

119905120572minus1

119888119905120573minus1

) |119888| le 119861 is boundedIf the second part of (H3) holds then define the set

Ω3= (119906 V) isin Ker 119871 minus120582119869 (119906 V)

+ (1 minus 120582)119876119873 (119906 V) = 0 120582 isin [0 1]

(72)

and here 119869 is as above Similar to the above argument we canshow thatΩ

3is bounded too

In the following we will prove that all conditions ofTheorem 1 are satisfied LetΩ be a bounded open subset of 119884such that cup3

119894=1Ω119894sub Ω By standard arguments we can prove

that 119870119875(119868 minus 119876)119873 Ω rarr 119884 is compact and thus 119873 is 119871-

compact onΩ Then by the above argument we have

(i) 119871119906 = 120582119873119906 for every (119906 120582) isin [(dom 119871 Ker 119871) cap 120597Ω]times(0 1)

(ii) 119873119906 notin Im 119871 for 119906 isin Ker 119871 cap 120597Ω

Finally we will prove that (iii) of Theorem 1 is satisfied Let119867(119906 120582) = plusmn120582119869119906 + (1 minus 120582)119876119873119906 According to the aboveargument we know that

119867(119906 120582) = 0 for 119906 isin Ker 119871 cap 120597Ω (73)

Thus by the homotopy property of degree

deg (119876119873|Ker119871Ker 119871 cap Ω 0)

= deg (119867 (sdot 0) Ker 119871 cap Ω 0)

= deg (119867 (sdot 1) Ker 119871 cap Ω 0)

= deg (plusmn119869Ker 119871 cap Ω 0) = 0

(74)

Then byTheorem 1 119871(119906 V) = 119873(119906 V) has at least one solutionin dom 119871 cap Ω so that BVP (1) has a solution in 119862

120572minus1

[0 1] times

119862120573minus1

[0 1] The proof is complete

Conflict of Interests

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

Acknowledgments

The authors are very grateful to the anonymous refereesfor many valuable comments and suggestions which helpedto improve the presentation of the paper The project wassupported by the National Natural Science Foundation ofChina (11371221 61304074) the Specialized Research Foun-dation for the Doctoral Program of Higher Education ofChina (20123705110001) the Program for Scientific ResearchInnovation Team in Colleges and Universities of ShandongProvince the Postdoctoral Science Foundation of ShandongProvince (201303074) and Foundation of SDUST

References

[1] A A Kilbas H M Srivastava and J J Trujillo Theoryand Applications of Fractional Differential Equations vol 204of North-Holland Mathematics Studies Elsevier Science BVAmsterdam The Netherlands 2006

[2] V Lakshmikantham S Leela and J Vasundhara DeviTheory ofFractional Dynamic Systems Cambridge Academic PublishersCambridge UK 2009

[3] J J Nieto and J Pimentel ldquoPositive solutions of a fractionalthermostat modelrdquo Boundary Value Problems vol 2013 article5 2013

[4] I Podlubny Fractional Differential Equations vol 198 Aca-demic Press San Diego Calif USA 1999

[5] W Wu and X Zhou ldquoEigenvalue of fractional differentialequations with P-Laplacian operatorrdquo Discrete Dynamics inNature and Society vol 2013 Article ID 137890 8 pages 2013

[6] X Zhang L Liu Y Wu and Y Lu ldquoThe iterative solutions ofnonlinear fractional differential equationsrdquo Applied Mathemat-ics and Computation vol 219 no 9 pp 4680ndash4691 2013

[7] X Zhang L Liu and Y Wu ldquoThe uniqueness of positivesolution for a singular fractional differential system involvingderivativesrdquo Communications in Nonlinear Science and Numer-ical Simulation vol 18 no 6 pp 1400ndash1409 2013

[8] X Zhang L Liu and Y Wu ldquoThe eigenvalue problem for asingular higher order fractional differential equation involvingfractional derivativesrdquo Applied Mathematics and Computationvol 218 no 17 pp 8526ndash8536 2012

[9] X Zhang L Liu and YWu ldquoExistence results for multiple pos-itive solutions of nonlinear higher order perturbed fractionaldifferential equations with derivativesrdquo Applied Mathematicsand Computation vol 219 no 4 pp 1420ndash1433 2012

[10] X Zhang L Liu B Wiwatanapataphee and Y Wu ldquoPositivesolutions of eigenvalue problems for a class of fractionaldifferential equations with derivativesrdquo Abstract and AppliedAnalysis vol 2012 Article ID 512127 16 pages 2012

[11] X Zhang L Liu and Y Wu ldquoMultiple positive solutionsof a singular fractional differential equation with negativelyperturbed termrdquo Mathematical and Computer Modelling vol55 no 3-4 pp 1263ndash1274 2012

[12] K Deng ldquoBlow-up rates for parabolic systemsrdquo Zeitschrift furAngewandte Mathematik und Physik vol 47 no 1 pp 132ndash1431996

8 Abstract and Applied Analysis

[13] K Deng ldquoGlobal existence and blow-up for a system of heatequations with non-linear boundary conditionsrdquoMathematicalMethods in the Applied Sciences vol 18 no 4 pp 307ndash315 1995

[14] L Zhigui and X Chunhong ldquoThe blow-up rate for a system ofheat equations with nonlinear boundary conditionsrdquoNonlinearAnalysisTheory Methods ampApplications vol 34 no 5 pp 767ndash778 1998

[15] D G Aronson ldquoA comparison method for stability analysis ofnonlinear parabolic problemsrdquo SIAM Review vol 20 no 2 pp245ndash264 1978

[16] M Pedersen and Z Lin ldquoBlow-up analysis for a system of heatequations coupled through a nonlinear boundary conditionrdquoApplied Mathematics Letters vol 14 no 2 pp 171ndash176 2001

[17] N A Asif and R A Khan ldquoPositive solutions to singularsystem with four-point coupled boundary conditionsrdquo Journalof Mathematical Analysis and Applications vol 386 no 2 pp848ndash861 2012

[18] C Yuan D Jiang D OrsquoRegan and R P Agarwal ldquoMultiple pos-itive solutions to systems of nonlinear semipositone fractionaldifferential equations with coupled boundary conditionsrdquo Elec-tronic Journal of QualitativeTheory of Differential Equations no13 p 17 2012

[19] Y Cui and J Sun ldquoOn existence of positive solutions of coupledintegral boundary value problems for a nonlinear singularsuperlinear differential systemrdquo Electronic Journal of QualitativeTheory of Differential Equations no 41 pp 1ndash13 2012

[20] Z Bai and Y Zhang ldquoSolvability of fractional three-pointboundary value problems with nonlinear growthrdquo AppliedMathematics and Computation vol 218 no 5 pp 1719ndash17252011

[21] Y Cui ldquoSolvability of second-order boundary-value problemsat resonance involving integral conditionsrdquo Electronic Journalof Differential Equations no 45 pp 1ndash9 2012

[22] Z Hu and W Liu ldquoSolvability for fractional order boundaryvalue problems at resonancerdquo Boundary Value Problems vol2011 article 20 2011

[23] W Jiang ldquoThe existence of solutions to boundary value prob-lems of fractional differential equations at resonancerdquoNonlinearAnalysis Theory Methods amp Applications vol 74 no 5 pp1987ndash1994 2011

[24] N Kosmatov ldquoMulti-point boundary value problems on timescales at resonancerdquo Journal of Mathematical Analysis andApplications vol 323 no 1 pp 253ndash266 2006

[25] G Wang W Liu S Zhu and T Zheng ldquoExistence results fora coupled system of nonlinear fractional 2m-point boundaryvalue problems at resonancerdquo Advances in Difference Equationsvol 2011 article 80 2011

[26] J R L Webb ldquoRemarks on nonlocal boundary value problemsat resonancerdquo Applied Mathematics and Computation vol 216no 2 pp 497ndash500 2010

[27] J R L Webb and M Zima ldquoMultiple positive solutions ofresonant and non-resonant non-local fourth-order boundaryvalue problemsrdquo Glasgow Mathematical Journal vol 54 no 1pp 225ndash240 2012

[28] X ZhangM Feng andWGe ldquoExistence result of second-orderdifferential equations with integral boundary conditions atresonancerdquo Journal of Mathematical Analysis and Applicationsvol 353 no 1 pp 311ndash319 2009

[29] X Zhang C Zhu and Z Wu ldquoSolvability for a coupled systemof fractional differential equations with impulses at resonancerdquoBoundary Value Problems vol 2013 artice 80 2013

[30] Z Zhao and J Liang ldquoExistence of solutions to functionalboundary value problem of second-order nonlinear differentialequationrdquo Journal of Mathematical Analysis and Applicationsvol 373 no 2 pp 614ndash634 2011

[31] Y Zou and Y Cui ldquoExistence results for a functional boundaryvalue problem of fractional differential equationsrdquo Advances inDifference Equations vol 2013 article 25 2013

[32] J Mawhin ldquoTopological degree and boundary value problemsfor nonlinear differential equationsrdquo in Topological Methods forOrdinary Differential Equations P M Fitzpertrick M MartelliJ Mawhin and R Nussbaum Eds vol 1537 of Lecture Notes inMathematics pp 74ndash142 Springer Berlin Germany 1993

[33] J Mawhin Topological Degree Methods in Nonlinear BoundaryValue Problems vol 40 of CBMS Regional Conference Series inMathematics American Mathematical Society Providence RIUSA 1979

[34] Y Zhang and Z Bai ldquoExistence of solutions for nonlinearfractional three-point boundary value problems at resonancerdquoJournal of Applied Mathematics and Computing vol 36 no 1-2pp 417ndash440 2011

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

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Mathematical Problems in Engineering

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Differential EquationsInternational Journal of

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Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom 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

4 Abstract and Applied Analysis

Proof Let 119906(119905) = 119886120585120573minus1

119905120572minus1and let V(119905) = 119905

120573minus1 Then byLemma 5 we have 1198682minus120572

0+119906(119905)|119905=0

= 1198682minus120573

0+V(119905)|119905=0

= 0 119906(1) =

119886120585120573minus1

= 119886V(120585) V(1) = 1 = 119887119906(120578) and 1198631205720+119906(119905) = 119863

120573

0+V(119905) = 0

So

(119906 V) isin dom 119871 (119906 V) = 119888 (119886120585120573minus1

119905120572minus1

119905120573minus1

) 119888 isin R

sub Ker 119871(31)

For every (119906 V) isin Ker 119871 if1198631205720+119906(119905) = 119863

120573

0+V(119905) = 0 then

119906 (119905) = 1198881119905120572minus1

+ 1198882119905120572minus2

V (119905) = 1198883119905120573minus1

+ 1198884119905120573minus2

(32)

Considering that 1198682minus1205720+

119906(119905)|119905=0

= 1198682minus120573

0+V(119905)|119905=0

= 0 119906(1) = 119886V(120585)and V(1) = 119887119906(120578) we can obtain that 119888

2= 1198884= 0 and 119888

1 1198883=

119886120585120573minus1 It yields the following

Ker 119871 sub (119906 V) isin dom 119871 (119906 V)

= 119888 (119886120585120573minus1

119905120572minus1

119905120573minus1

) 119888 isin R

(33)

Let (119909 119910) isin Im 119871 then there is (119906 V) isin dom 119871 such that(119909 119910) = 119871(119906 V) that is 119906 isin 119862

120572minus1

[0 1] 1198631205720+119906(119905) = 119909(119905) and

V isin 119862120573minus1[0 1]1198631205730+V(119905) = 119910(119905) By Lemma 4

119868120572

0+119909 (119905) = 119906 (119905) minus

((119863120572minus1

0+119906) (119905))

10038161003816100381610038161003816119905=0

Γ (120572)119905120572minus1

minus

((1198682minus120572

0+119906) (119905))

10038161003816100381610038161003816119905=0

Γ (120572 minus 1)119905120572minus2

119868120573

0+119910 (119905) = V (119905) minus

((119863120573minus1

0+V) (119905))

100381610038161003816100381610038161003816119905=0

Γ (120573)119905120573minus1

minus

((1198682minus120573

0+V) (119905))

100381610038161003816100381610038161003816119905=0

Γ (120573 minus 1)119905120573minus2

(34)

and by the couple boundary conditions we have

119906 (119905) = 119868120572

0+119909 (119905) + 119888

1119905120572minus1

V (119905) = 119868120573

0+119910 (119905) + 119888

2119905120573minus1

119906 (1) = 119868120572

0+119909 (1) + 119888

1= 119886V (120585) = 119886 [119868

120573

0+119910 (120585) + 119888

2120585120573minus1

]

V (1) = 119868120573

0+119910 (1) + 119888

2= 119887119906 (120578) = 119887 [119868

120572

0+119910 (120578) + 119888

1120578120572minus1

]

(35)

It yields the following

119886119868120573

0+119910 (120585) minus 119868

120572

0+119909 (1) + 119886120585

120573minus1

[119887119868120572

0+119909 (120578) minus 119868

120573

0+119910 (1)] = 0

(36)

On the other hand suppose that (119909 119910) isin 119885 satisfy (36) Let119906(119905) = 119868

120572

0+119909(119905) + 119886120585

120573minus1

119905120572minus1 and V(119905) = 119868

120573

0+119910(119905) + [119887119868

120572

0+119909(120578) minus

119868120573

0+119910(1) + 1]119905

120573minus1 and then 119863120572

0+119906(119905) = 119909(119905) 119863120573

0+V(119905) = 119910(119905)

and

1198682minus120572

0+119906 (119905) |119905=0

= 1198682minus120573

0+V (119905) |119905=0

= 0

119906 (1) = 119868120572

0+119909 (1) + 119886120585

120573minus1

= 119886119868120573

0+119910 (120585) + 119886 [119887119868

120572

0+119909 (120578) minus 119868

120573

0+119910 (1) + 1] 120585

120573minus1

= 119886V (120585)

V (1) = 119868120573

0+119910 (1) + [119887119868

120572

0+119909 (120578) minus 119868

120573

0+119910 (1) + 1]

= 119887119868120572

0+119909 (120578) + 119886119887120585

120573minus1

120578120572minus1

= 119887119906 (120578)

(37)

Therefore (30) holds

Lemma 9 If (2) holds then 119871 is a Fredholm operator of indexzero and dim Ker 119871 = codim Im 119871 = 1 Furthermore thelinear operator119870

119901 Im 119871 rarr dom 119871capKer119875 can be defined by

119870119875(119906 V) (119905)

= 119868120572

0+119906 (119905) minus 119886120585

120573minus1

119905120572minus1

[119887119868120572

0+119906 (120578) minus 119868

120573

0+V (1)] 119868120573

0+V (119905)

(38)

Also10038171003817100381710038171003817119870119901(119906 V)

10038171003817100381710038171003817119884le Δ1(119906 V)

119885 (39)

Proof Define operator 119876 119885 rarr 119885 as follows

119876 (119906 V) = Δ21198761(119906 V) (1 1) (40)

where 1198761 119885 rarr R is defined by

1198761(119906 V) = 119886119868

120573

0+V (120585) minus 119868120572

0+119906 (1)

+ 119886120585120573minus1

[119887119868120572

0+119906 (120578) minus 119868

120573

0+V (1)]

Δ2=

Γ (120573 + 1) Γ (120572 + 1)

120572120585120573minus1Γ (120572 + 1) (120585 minus 1) + Γ (120573 + 1) (120578 minus 1)= 0

(41)

It is easy to see that 1198762(119906 V) = 119876(119906 V) that is 119876 119885 rarr 119885

is a continuous linear projector Furthermore Ker119876 = Im 119871For (119906 V) isin 119885 set (119906 V) = [(119906 V) minus 119876(119906 V)] + 119876(119906 V) Then(119906 V) minus 119876(119906 V) isin Ker119876 and 119876(119906 V) isin Im119876 It follows fromKer119876 = Im 119871 and1198762(119906 V) = 119876(119906 V) that Im 119871capIm119876 = (0 0)So we have

119885 = Im 119871 oplus Im119876 (42)

Now Ind119871 = dim Ker 119871 minus codim Im 119871 = dim Ker 119871 minus

dim Im119876 = 0 and so 119871 is a Fredholm operator of index 0

Abstract and Applied Analysis 5

Let 119875 119884 rarr 119884 be continuous linear operator defined by

119875 (119906 V) =119863120573minus1

0+V (0)

Γ (120573)(119886120585120573minus1

119905120572minus1

119905120573minus1

) (43)

Obviously 119875 is a linear projector and

Ker119875 = (119906 V) isin 119884 119863120573minus1

0+V (0) = 0 (44)

It is easy to know that 119884 = Ker119875 oplus Ker 119871Define119870

119875 Im 119871 rarr dom 119871 cap Ker119875 by

119870119875(119906 V) (119905)

= (119868120572

0+119906 (119905) minus 119886120585

120573minus1

119905120572minus1

[119887119868120572

0+119906 (120578) minus 119868

120573

0+V (1)] 119868120573

0+V (119905))

(45)

Since

100381610038161003816100381610038161003816119887119868120572

0+119906 (120578) minus 119868

120573

0+V (1)

100381610038161003816100381610038161003816le

10038161003816100381610038161003816100381610038161003816

119887

Γ (120572)int

120578

0

(120578 minus 119904)120572minus1

119906 (119904) 119889119904

10038161003816100381610038161003816100381610038161003816

+

100381610038161003816100381610038161003816100381610038161003816

1

Γ (120573)int

1

0

(1 minus 119904)120572minus1V (119904) 119889119904

100381610038161003816100381610038161003816100381610038161003816

le119887

Γ (120572)1199061+

1

Γ (120573)V1

119863120572minus1

0+119868120572

0+119906 (119905) = int

119905

0

119906 (119904) 119889119904

119868120572

0+119906 (119905) =

1

Γ (120572)int

119905

0

(119905 minus 119904)120572minus1

119906 (119904) 119889119904

119863120573minus1

0+119868120573

0+V (119905) = int

119905

0

V (119904) 119889119904

119868120573

0+V (119905) =

1

Γ (120573)int

119905

0

(119905 minus 119904)120573minus1V (119904) 119889119904

(46)

then1003817100381710038171003817119870119875 (119906 V)

1003817100381710038171003817119884le Δ1(119906 V)

119885 (47)

In fact if (119906 V) isin Im 119871 then

119871119870119875(119906 V) (119905)

= 119863120572

0+[119868120572

0+119906 (119905) minus 119886120585

120573minus1

119905120572minus1

[119887119868120572

0+119906 (120578) minus 119868

120573

0+V (1)]]

119863120573

0+119868120573

0+V (119905) = (119906 V)

(48)

By Lemma 4 for (119906 V) isin dom 119871 cap Ker119875

119870119875119871 (119906 V) (119905)

= (119868120572

0+119863120572

0+119906 (119905)minus119886120585

120572minus1

119905120572minus1

[119887119868120572

0+119863120572

0+119906 (120578)minus119868

120573

0+119863120573

0+V (1)]

119868120573

0+119863120573

0+V (119905))

= (119906 (119905) minus119863120572minus1

0+119906 (0)

Γ (120572)119905120572minus1

minus1198682minus120572

0+119906 (0)

Γ (120572 minus 1)119905120572minus2

minus 119886120585120573minus1

119905120572minus1

times [119887119906 (120578) minus 119887119863120572minus1

0+119906 (0)

Γ (120572)120578120572minus1

minus1198682minus120572

0+119906 (0)

Γ (120572 minus 1)120578120572minus2

minusV (1)+119863120573minus1

0+V (0)

Γ (120573)] V (119905)minus

119863120573minus1

0+V (0)

Γ (120573)119905120573minus1

)

= (119906 (119905) minus119863120572minus1

0+119906 (0)

Γ (120572)119905120572minus1

+119886120585120573minus1

119905120572minus1

119887119863120572minus1

0+119906 (0)

Γ (120572)120578120572minus1

V (119905))

= (119906 (119905) V (119905))(49)

(119863120573minus10+

V(0) = 0 since (119906 V) isin Ker119875 and 1198682minus120572

0+119906(0) = 0 since

(119906 V) isin dom 119871) Hence

119870119901= (119871|dom119871capKer119875)

minus1

(50)

The proof is complete

3 Main Results

In this section we will use Theorem 1 to prove the existenceof solutions to BVP (1) To obtain our main theorem we usethe following assumptions

(H1) There exist functions 119886119894 119887119894 119888119894 119889119894 119890119894isin 1198711

[0 1] (119894 =

1 2) such that for all (119909 119910 119911 119908) isin R4 119905 isin [0 1]

1003816100381610038161003816119891 (119905 119909 119910 119911 119908)1003816100381610038161003816 le 1198901(119905) + 119886

1(119905) |119909| + 119887

1(119905)

10038161003816100381610038161199101003816100381610038161003816

+ 1198881(119905) |119911| + 119889

1(119905) |119908|

1003816100381610038161003816119892 (119905 119909 119910 119911 119908)1003816100381610038161003816 le 1198902(119905) + 119886

2(119905) |119909| + 119887

2(119905)

10038161003816100381610038161199101003816100381610038161003816

+ 1198882(119905) |119911| + 119889

2(119905) |119908|

(51)

(H2) There exists a constant 119860 gt 0 such that for (119906 V) isindom 119871 if |119863120573minus1

0+V(119905)| gt 119860 for all 119905 isin [0 1] then

1198761(1198731(119906 V)) = 0 or 119876

1(1198732(119906 V)) = 0

(H3) There exists a constant 119861 gt 0 such that either for each119888 isin R |119888| gt 119861

1198881198731(119888119886120585120573minus1

119905120572minus1

119888119905120573minus1

) gt 0 1198881198732(119888119886120585120573minus1

119905120572minus1

119888119905120573minus1

) gt 0

(52)

or for each 119886 isin R |119886| gt 119861

1198881198731(119888119886120585120573minus1

119905120572minus1

119888119905120573minus1

) lt 0

1198881198732(119888119886120585120573minus1

119905120572minus1

119888119905120573minus1

) lt 0

(53)

6 Abstract and Applied Analysis

Theorem 10 Suppose (2) and (1198671)ndash(1198673) hold Then (1) hasat least one solution in Y provided that

max (Δ1+ Δ3)max 10038171003817100381710038171198862

1003817100381710038171003817110038171003817100381710038171198872

1003817100381710038171003817110038171003817100381710038171198882

1003817100381710038171003817110038171003817100381710038171198892

10038171003817100381710038171 Δ4 Δ5

lt 1

(54)

Proof Set

Ω1= (119906 V) isin dom 119871 Ker 119871 119871 (119906 V) = 120582119873 (119906 V)

for some 120582 isin [0 1] (55)

Take (119906 V) isin Ω1 Since 119871119906 = 120582119873119906 so 120582 = 0 and 119873(119906 V) isin

Im 119871 = Ker119876 hence

119876119873(119906 V) = 1198761(1198731(119906 V) 119873

2(119906 V)) (1 1) = 0 (56)

Thus from (H2) there exist 1199050isin [0 1] such that

100381610038161003816100381610038161003816119863120573minus1

0+V (1199050)100381610038161003816100381610038161003816le 119860 (57)

Noticing that

119863120573minus1

0+V (119905) = 119863

120573minus1

0+V (1199050) + int

119905

1199050

119863120573

0+V (119904) 119889119904 (58)

so100381610038161003816100381610038161003816119863120573minus1

0+V (0)

100381610038161003816100381610038161003816le100381710038171003817100381710038171003817119863120573minus1

0+V (119905)

100381710038171003817100381710038171003817infinle100381610038161003816100381610038161003816119863120573minus2

0+V (1199050)100381610038161003816100381610038161003816+100381710038171003817100381710038171003817119863120573

0+V (119905)

1003817100381710038171003817100381710038171

le 119860 +10038171003817100381710038171198712V

10038171003817100381710038171le 119860 +

10038171003817100381710038171198732 (119906 V)10038171003817100381710038171

(59)

Thus

119875 (119906 V)119884=

1003817100381710038171003817100381710038171003817100381710038171003817

119863120573minus1

0+V (0)

Γ (120573)(119886120585120573minus1

119905120572minus1

119905120573minus1

)

1003817100381710038171003817100381710038171003817100381710038171003817119884

= Δ3

100381610038161003816100381610038161003816119863120573minus1

0+V (0)

100381610038161003816100381610038161003816le Δ3(119860 +

10038171003817100381710038171198732 (119906 V)10038171003817100381710038171)

(60)

For all (119906 V) isin Ω1 (119868minus119875)(119906 V) isin dom 119871capKer119875 Considering

Lemma 9 we get 119871119875(119906 V) = (0 0) Together with (39) wehave

(119868 minus 119875) (119906 V)119884=1003817100381710038171003817119870119875119871 (119868 minus 119875) (119906 V)

1003817100381710038171003817119884

le Δ1119871 (119868 minus 119875) (119906 V)

119885

le Δ1119873 (119906 V)

119885

(61)

From (60) and (61) we have

(119906 V)119884

le 119875 (119906 V)119884+ (119868 minus 119875) (119906 V)

119884

le Δ3(119860 +

10038171003817100381710038171198732 (119906 V)10038171003817100381710038171) + Δ1119873 (119906 V)

119885

= 119860Δ3+max (Δ

1+ Δ3)10038171003817100381710038171198732 (119906 V)

10038171003817100381710038171 Δ1

10038171003817100381710038171198731 (119906 V)10038171003817100381710038171

+Δ3

10038171003817100381710038171198732 (119906 V)10038171003817100381710038171

(62)

From (62) we discuss various cases

Case 1 ((119906 V)119884le 119860Δ

3+ (Δ1+ Δ3)1198732(119906 V)

1) From (H1)

we have

(119906 V)119884

le 119860Δ3+ (Δ1+ Δ3)

times (10038171003817100381710038171198862

10038171003817100381710038171119906infin+10038171003817100381710038171198872

10038171003817100381710038171

10038171003817100381710038171003817119863120572minus1

0+11990610038171003817100381710038171003817infin

+10038171003817100381710038171198882

10038171003817100381710038171Vinfin

+10038171003817100381710038171198892

10038171003817100381710038171

100381710038171003817100381710038171003817119863120573minus1

0+V100381710038171003817100381710038171003817infin

+10038171003817100381710038171198902

10038171003817100381710038171)

le 119860Δ3+ (Δ1+ Δ3)10038171003817100381710038171198902

10038171003817100381710038171

+ (Δ1+ Δ3)max 10038171003817100381710038171198862

1003817100381710038171003817110038171003817100381710038171198872

10038171003817100381710038171 119906119862120572minus1

+ (Δ1+ Δ3)max 10038171003817100381710038171198882

1003817100381710038171003817110038171003817100381710038171198892

10038171003817100381710038171 V119862120573minus1

le 119860Δ3+ (Δ1+ Δ3)10038171003817100381710038171198902

10038171003817100381710038171

+ (Δ1+ Δ3)max 10038171003817100381710038171198862

1003817100381710038171003817110038171003817100381710038171198872

1003817100381710038171003817110038171003817100381710038171198882

1003817100381710038171003817110038171003817100381710038171198892

10038171003817100381710038171 (119906 V)

119884

(63)

which yield

(119906 V)119884

le119860Δ2+ (Δ1+ Δ3)10038171003817100381710038171198902

10038171003817100381710038171

1 minus (Δ1+ Δ3)max 10038171003817100381710038171198862

1003817100381710038171003817110038171003817100381710038171198872

1003817100381710038171003817110038171003817100381710038171198882

1003817100381710038171003817110038171003817100381710038171198892

10038171003817100381710038171

(64)

ThusΩ1is bounded

Case 2 ((119906 V)119884le 119860Δ

3+ Δ11198731(119906 V)

1+ Δ31198732(119906 V)

1)

From (H1) we have

(119906 V)119884

le 119860Δ3+ Δ1(10038171003817100381710038171198861

10038171003817100381710038171119906infin+10038171003817100381710038171198871

10038171003817100381710038171

10038171003817100381710038171003817119863120572minus1

0+11990610038171003817100381710038171003817infin

+10038171003817100381710038171198881

10038171003817100381710038171Vinfin

+10038171003817100381710038171198891

10038171003817100381710038171

100381710038171003817100381710038171003817119863120573minus1

0+V100381710038171003817100381710038171003817infin

+10038171003817100381710038171198901

10038171003817100381710038171)

+ Δ3(10038171003817100381710038171198862

10038171003817100381710038171119906infin+10038171003817100381710038171198872

10038171003817100381710038171

10038171003817100381710038171003817119863120572minus1

0+11990610038171003817100381710038171003817infin

+10038171003817100381710038171198882

10038171003817100381710038171Vinfin

+10038171003817100381710038171198892

10038171003817100381710038171

100381710038171003817100381710038171003817119863120573minus1

0+V100381710038171003817100381710038171003817infin

+10038171003817100381710038171198902

10038171003817100381710038171)

le 119860Δ3+ Δ1

1003817100381710038171003817119890110038171003817100381710038171+ Δ3

1003817100381710038171003817119890210038171003817100381710038171+ Δ4119906119862120572minus1 + Δ

5V119862120573minus1

le 119860Δ3+ Δ1

1003817100381710038171003817119890110038171003817100381710038171+ Δ3

1003817100381710038171003817119890210038171003817100381710038171+max Δ

4 Δ5 (119906 V)

119884

(65)

which yield

(119906 V)119884le119860Δ3+ Δ1

1003817100381710038171003817119890110038171003817100381710038171+ Δ3

1003817100381710038171003817119890210038171003817100381710038171

1 minusmax Δ4 Δ5

(66)

ThusΩ1is bounded Let

Ω2= (119906 V) isin Ker 119871 119873 (119906 V) isin Im 119871 (67)

For (119906 V) isin Ω2and (119906 V) isin Ker 119871 so (119906 V) = 119888(119886120585

120573minus1

119905120572minus1

119905120573minus1

) 119905 isin [0 1] 119888 isin R Noticing that Im 119871 = Ker119876 then we

Abstract and Applied Analysis 7

get 119876119873(119906 V) = 0 and thus 11987611198731(119888119886120585120573minus1

119905120572minus1

119888119905120573minus1

) = 0 and11987611198732(119888119886120585120573minus1

119905120572minus1

119888119905120573minus1

) = 0 From (H2) we get |119888| le 119860Γ(120573)and thusΩ

2is bounded

We define the isomorphism 119869 Ker 119871 rarr Im119876 by

119869 (119888119886120585120573minus1

119905120572minus1

119888119905120573minus1

) = (119888 119888) (68)

If the first part of (H3) is satisfied then let

Ω3= (119906 V) isin ker 119871 120582119869 (119906 V)

+ (1 minus 120582)119876119873 (119906 V) = 0 120582 isin [0 1]

(69)

For (119906 V) = (119888119886120585120573minus1

119905120572minus1

119888119905120573minus1

) isin Ω3

120582 (119888 119888) = minus (1 minus 120582) Δ21198761119873(119888119886120585

120573minus1

119905120572minus1

119888119905120573minus1

) (1 1) (70)

If 120582 = 1 then 119888 = 0 Otherwise if |119888| gt 119861 in view of (H3) andΔ2lt 0 one has

(1 minus 120582) Δ31198761119873(119888119886120585

120573minus1

119905120572minus1

119888119905120573minus1

) lt 0 (71)

which contradict 1205821198882 ge 0ThusΩ3sub (119906 V) isin Ker 119871 | (119906 V) =

(119888119886120585120573minus1

119905120572minus1

119888119905120573minus1

) |119888| le 119861 is boundedIf the second part of (H3) holds then define the set

Ω3= (119906 V) isin Ker 119871 minus120582119869 (119906 V)

+ (1 minus 120582)119876119873 (119906 V) = 0 120582 isin [0 1]

(72)

and here 119869 is as above Similar to the above argument we canshow thatΩ

3is bounded too

In the following we will prove that all conditions ofTheorem 1 are satisfied LetΩ be a bounded open subset of 119884such that cup3

119894=1Ω119894sub Ω By standard arguments we can prove

that 119870119875(119868 minus 119876)119873 Ω rarr 119884 is compact and thus 119873 is 119871-

compact onΩ Then by the above argument we have

(i) 119871119906 = 120582119873119906 for every (119906 120582) isin [(dom 119871 Ker 119871) cap 120597Ω]times(0 1)

(ii) 119873119906 notin Im 119871 for 119906 isin Ker 119871 cap 120597Ω

Finally we will prove that (iii) of Theorem 1 is satisfied Let119867(119906 120582) = plusmn120582119869119906 + (1 minus 120582)119876119873119906 According to the aboveargument we know that

119867(119906 120582) = 0 for 119906 isin Ker 119871 cap 120597Ω (73)

Thus by the homotopy property of degree

deg (119876119873|Ker119871Ker 119871 cap Ω 0)

= deg (119867 (sdot 0) Ker 119871 cap Ω 0)

= deg (119867 (sdot 1) Ker 119871 cap Ω 0)

= deg (plusmn119869Ker 119871 cap Ω 0) = 0

(74)

Then byTheorem 1 119871(119906 V) = 119873(119906 V) has at least one solutionin dom 119871 cap Ω so that BVP (1) has a solution in 119862

120572minus1

[0 1] times

119862120573minus1

[0 1] The proof is complete

Conflict of Interests

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

Acknowledgments

The authors are very grateful to the anonymous refereesfor many valuable comments and suggestions which helpedto improve the presentation of the paper The project wassupported by the National Natural Science Foundation ofChina (11371221 61304074) the Specialized Research Foun-dation for the Doctoral Program of Higher Education ofChina (20123705110001) the Program for Scientific ResearchInnovation Team in Colleges and Universities of ShandongProvince the Postdoctoral Science Foundation of ShandongProvince (201303074) and Foundation of SDUST

References

[1] A A Kilbas H M Srivastava and J J Trujillo Theoryand Applications of Fractional Differential Equations vol 204of North-Holland Mathematics Studies Elsevier Science BVAmsterdam The Netherlands 2006

[2] V Lakshmikantham S Leela and J Vasundhara DeviTheory ofFractional Dynamic Systems Cambridge Academic PublishersCambridge UK 2009

[3] J J Nieto and J Pimentel ldquoPositive solutions of a fractionalthermostat modelrdquo Boundary Value Problems vol 2013 article5 2013

[4] I Podlubny Fractional Differential Equations vol 198 Aca-demic Press San Diego Calif USA 1999

[5] W Wu and X Zhou ldquoEigenvalue of fractional differentialequations with P-Laplacian operatorrdquo Discrete Dynamics inNature and Society vol 2013 Article ID 137890 8 pages 2013

[6] X Zhang L Liu Y Wu and Y Lu ldquoThe iterative solutions ofnonlinear fractional differential equationsrdquo Applied Mathemat-ics and Computation vol 219 no 9 pp 4680ndash4691 2013

[7] X Zhang L Liu and Y Wu ldquoThe uniqueness of positivesolution for a singular fractional differential system involvingderivativesrdquo Communications in Nonlinear Science and Numer-ical Simulation vol 18 no 6 pp 1400ndash1409 2013

[8] X Zhang L Liu and Y Wu ldquoThe eigenvalue problem for asingular higher order fractional differential equation involvingfractional derivativesrdquo Applied Mathematics and Computationvol 218 no 17 pp 8526ndash8536 2012

[9] X Zhang L Liu and YWu ldquoExistence results for multiple pos-itive solutions of nonlinear higher order perturbed fractionaldifferential equations with derivativesrdquo Applied Mathematicsand Computation vol 219 no 4 pp 1420ndash1433 2012

[10] X Zhang L Liu B Wiwatanapataphee and Y Wu ldquoPositivesolutions of eigenvalue problems for a class of fractionaldifferential equations with derivativesrdquo Abstract and AppliedAnalysis vol 2012 Article ID 512127 16 pages 2012

[11] X Zhang L Liu and Y Wu ldquoMultiple positive solutionsof a singular fractional differential equation with negativelyperturbed termrdquo Mathematical and Computer Modelling vol55 no 3-4 pp 1263ndash1274 2012

[12] K Deng ldquoBlow-up rates for parabolic systemsrdquo Zeitschrift furAngewandte Mathematik und Physik vol 47 no 1 pp 132ndash1431996

8 Abstract and Applied Analysis

[13] K Deng ldquoGlobal existence and blow-up for a system of heatequations with non-linear boundary conditionsrdquoMathematicalMethods in the Applied Sciences vol 18 no 4 pp 307ndash315 1995

[14] L Zhigui and X Chunhong ldquoThe blow-up rate for a system ofheat equations with nonlinear boundary conditionsrdquoNonlinearAnalysisTheory Methods ampApplications vol 34 no 5 pp 767ndash778 1998

[15] D G Aronson ldquoA comparison method for stability analysis ofnonlinear parabolic problemsrdquo SIAM Review vol 20 no 2 pp245ndash264 1978

[16] M Pedersen and Z Lin ldquoBlow-up analysis for a system of heatequations coupled through a nonlinear boundary conditionrdquoApplied Mathematics Letters vol 14 no 2 pp 171ndash176 2001

[17] N A Asif and R A Khan ldquoPositive solutions to singularsystem with four-point coupled boundary conditionsrdquo Journalof Mathematical Analysis and Applications vol 386 no 2 pp848ndash861 2012

[18] C Yuan D Jiang D OrsquoRegan and R P Agarwal ldquoMultiple pos-itive solutions to systems of nonlinear semipositone fractionaldifferential equations with coupled boundary conditionsrdquo Elec-tronic Journal of QualitativeTheory of Differential Equations no13 p 17 2012

[19] Y Cui and J Sun ldquoOn existence of positive solutions of coupledintegral boundary value problems for a nonlinear singularsuperlinear differential systemrdquo Electronic Journal of QualitativeTheory of Differential Equations no 41 pp 1ndash13 2012

[20] Z Bai and Y Zhang ldquoSolvability of fractional three-pointboundary value problems with nonlinear growthrdquo AppliedMathematics and Computation vol 218 no 5 pp 1719ndash17252011

[21] Y Cui ldquoSolvability of second-order boundary-value problemsat resonance involving integral conditionsrdquo Electronic Journalof Differential Equations no 45 pp 1ndash9 2012

[22] Z Hu and W Liu ldquoSolvability for fractional order boundaryvalue problems at resonancerdquo Boundary Value Problems vol2011 article 20 2011

[23] W Jiang ldquoThe existence of solutions to boundary value prob-lems of fractional differential equations at resonancerdquoNonlinearAnalysis Theory Methods amp Applications vol 74 no 5 pp1987ndash1994 2011

[24] N Kosmatov ldquoMulti-point boundary value problems on timescales at resonancerdquo Journal of Mathematical Analysis andApplications vol 323 no 1 pp 253ndash266 2006

[25] G Wang W Liu S Zhu and T Zheng ldquoExistence results fora coupled system of nonlinear fractional 2m-point boundaryvalue problems at resonancerdquo Advances in Difference Equationsvol 2011 article 80 2011

[26] J R L Webb ldquoRemarks on nonlocal boundary value problemsat resonancerdquo Applied Mathematics and Computation vol 216no 2 pp 497ndash500 2010

[27] J R L Webb and M Zima ldquoMultiple positive solutions ofresonant and non-resonant non-local fourth-order boundaryvalue problemsrdquo Glasgow Mathematical Journal vol 54 no 1pp 225ndash240 2012

[28] X ZhangM Feng andWGe ldquoExistence result of second-orderdifferential equations with integral boundary conditions atresonancerdquo Journal of Mathematical Analysis and Applicationsvol 353 no 1 pp 311ndash319 2009

[29] X Zhang C Zhu and Z Wu ldquoSolvability for a coupled systemof fractional differential equations with impulses at resonancerdquoBoundary Value Problems vol 2013 artice 80 2013

[30] Z Zhao and J Liang ldquoExistence of solutions to functionalboundary value problem of second-order nonlinear differentialequationrdquo Journal of Mathematical Analysis and Applicationsvol 373 no 2 pp 614ndash634 2011

[31] Y Zou and Y Cui ldquoExistence results for a functional boundaryvalue problem of fractional differential equationsrdquo Advances inDifference Equations vol 2013 article 25 2013

[32] J Mawhin ldquoTopological degree and boundary value problemsfor nonlinear differential equationsrdquo in Topological Methods forOrdinary Differential Equations P M Fitzpertrick M MartelliJ Mawhin and R Nussbaum Eds vol 1537 of Lecture Notes inMathematics pp 74ndash142 Springer Berlin Germany 1993

[33] J Mawhin Topological Degree Methods in Nonlinear BoundaryValue Problems vol 40 of CBMS Regional Conference Series inMathematics American Mathematical Society Providence RIUSA 1979

[34] Y Zhang and Z Bai ldquoExistence of solutions for nonlinearfractional three-point boundary value problems at resonancerdquoJournal of Applied Mathematics and Computing vol 36 no 1-2pp 417ndash440 2011

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

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OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

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

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

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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

Abstract and Applied Analysis 5

Let 119875 119884 rarr 119884 be continuous linear operator defined by

119875 (119906 V) =119863120573minus1

0+V (0)

Γ (120573)(119886120585120573minus1

119905120572minus1

119905120573minus1

) (43)

Obviously 119875 is a linear projector and

Ker119875 = (119906 V) isin 119884 119863120573minus1

0+V (0) = 0 (44)

It is easy to know that 119884 = Ker119875 oplus Ker 119871Define119870

119875 Im 119871 rarr dom 119871 cap Ker119875 by

119870119875(119906 V) (119905)

= (119868120572

0+119906 (119905) minus 119886120585

120573minus1

119905120572minus1

[119887119868120572

0+119906 (120578) minus 119868

120573

0+V (1)] 119868120573

0+V (119905))

(45)

Since

100381610038161003816100381610038161003816119887119868120572

0+119906 (120578) minus 119868

120573

0+V (1)

100381610038161003816100381610038161003816le

10038161003816100381610038161003816100381610038161003816

119887

Γ (120572)int

120578

0

(120578 minus 119904)120572minus1

119906 (119904) 119889119904

10038161003816100381610038161003816100381610038161003816

+

100381610038161003816100381610038161003816100381610038161003816

1

Γ (120573)int

1

0

(1 minus 119904)120572minus1V (119904) 119889119904

100381610038161003816100381610038161003816100381610038161003816

le119887

Γ (120572)1199061+

1

Γ (120573)V1

119863120572minus1

0+119868120572

0+119906 (119905) = int

119905

0

119906 (119904) 119889119904

119868120572

0+119906 (119905) =

1

Γ (120572)int

119905

0

(119905 minus 119904)120572minus1

119906 (119904) 119889119904

119863120573minus1

0+119868120573

0+V (119905) = int

119905

0

V (119904) 119889119904

119868120573

0+V (119905) =

1

Γ (120573)int

119905

0

(119905 minus 119904)120573minus1V (119904) 119889119904

(46)

then1003817100381710038171003817119870119875 (119906 V)

1003817100381710038171003817119884le Δ1(119906 V)

119885 (47)

In fact if (119906 V) isin Im 119871 then

119871119870119875(119906 V) (119905)

= 119863120572

0+[119868120572

0+119906 (119905) minus 119886120585

120573minus1

119905120572minus1

[119887119868120572

0+119906 (120578) minus 119868

120573

0+V (1)]]

119863120573

0+119868120573

0+V (119905) = (119906 V)

(48)

By Lemma 4 for (119906 V) isin dom 119871 cap Ker119875

119870119875119871 (119906 V) (119905)

= (119868120572

0+119863120572

0+119906 (119905)minus119886120585

120572minus1

119905120572minus1

[119887119868120572

0+119863120572

0+119906 (120578)minus119868

120573

0+119863120573

0+V (1)]

119868120573

0+119863120573

0+V (119905))

= (119906 (119905) minus119863120572minus1

0+119906 (0)

Γ (120572)119905120572minus1

minus1198682minus120572

0+119906 (0)

Γ (120572 minus 1)119905120572minus2

minus 119886120585120573minus1

119905120572minus1

times [119887119906 (120578) minus 119887119863120572minus1

0+119906 (0)

Γ (120572)120578120572minus1

minus1198682minus120572

0+119906 (0)

Γ (120572 minus 1)120578120572minus2

minusV (1)+119863120573minus1

0+V (0)

Γ (120573)] V (119905)minus

119863120573minus1

0+V (0)

Γ (120573)119905120573minus1

)

= (119906 (119905) minus119863120572minus1

0+119906 (0)

Γ (120572)119905120572minus1

+119886120585120573minus1

119905120572minus1

119887119863120572minus1

0+119906 (0)

Γ (120572)120578120572minus1

V (119905))

= (119906 (119905) V (119905))(49)

(119863120573minus10+

V(0) = 0 since (119906 V) isin Ker119875 and 1198682minus120572

0+119906(0) = 0 since

(119906 V) isin dom 119871) Hence

119870119901= (119871|dom119871capKer119875)

minus1

(50)

The proof is complete

3 Main Results

In this section we will use Theorem 1 to prove the existenceof solutions to BVP (1) To obtain our main theorem we usethe following assumptions

(H1) There exist functions 119886119894 119887119894 119888119894 119889119894 119890119894isin 1198711

[0 1] (119894 =

1 2) such that for all (119909 119910 119911 119908) isin R4 119905 isin [0 1]

1003816100381610038161003816119891 (119905 119909 119910 119911 119908)1003816100381610038161003816 le 1198901(119905) + 119886

1(119905) |119909| + 119887

1(119905)

10038161003816100381610038161199101003816100381610038161003816

+ 1198881(119905) |119911| + 119889

1(119905) |119908|

1003816100381610038161003816119892 (119905 119909 119910 119911 119908)1003816100381610038161003816 le 1198902(119905) + 119886

2(119905) |119909| + 119887

2(119905)

10038161003816100381610038161199101003816100381610038161003816

+ 1198882(119905) |119911| + 119889

2(119905) |119908|

(51)

(H2) There exists a constant 119860 gt 0 such that for (119906 V) isindom 119871 if |119863120573minus1

0+V(119905)| gt 119860 for all 119905 isin [0 1] then

1198761(1198731(119906 V)) = 0 or 119876

1(1198732(119906 V)) = 0

(H3) There exists a constant 119861 gt 0 such that either for each119888 isin R |119888| gt 119861

1198881198731(119888119886120585120573minus1

119905120572minus1

119888119905120573minus1

) gt 0 1198881198732(119888119886120585120573minus1

119905120572minus1

119888119905120573minus1

) gt 0

(52)

or for each 119886 isin R |119886| gt 119861

1198881198731(119888119886120585120573minus1

119905120572minus1

119888119905120573minus1

) lt 0

1198881198732(119888119886120585120573minus1

119905120572minus1

119888119905120573minus1

) lt 0

(53)

6 Abstract and Applied Analysis

Theorem 10 Suppose (2) and (1198671)ndash(1198673) hold Then (1) hasat least one solution in Y provided that

max (Δ1+ Δ3)max 10038171003817100381710038171198862

1003817100381710038171003817110038171003817100381710038171198872

1003817100381710038171003817110038171003817100381710038171198882

1003817100381710038171003817110038171003817100381710038171198892

10038171003817100381710038171 Δ4 Δ5

lt 1

(54)

Proof Set

Ω1= (119906 V) isin dom 119871 Ker 119871 119871 (119906 V) = 120582119873 (119906 V)

for some 120582 isin [0 1] (55)

Take (119906 V) isin Ω1 Since 119871119906 = 120582119873119906 so 120582 = 0 and 119873(119906 V) isin

Im 119871 = Ker119876 hence

119876119873(119906 V) = 1198761(1198731(119906 V) 119873

2(119906 V)) (1 1) = 0 (56)

Thus from (H2) there exist 1199050isin [0 1] such that

100381610038161003816100381610038161003816119863120573minus1

0+V (1199050)100381610038161003816100381610038161003816le 119860 (57)

Noticing that

119863120573minus1

0+V (119905) = 119863

120573minus1

0+V (1199050) + int

119905

1199050

119863120573

0+V (119904) 119889119904 (58)

so100381610038161003816100381610038161003816119863120573minus1

0+V (0)

100381610038161003816100381610038161003816le100381710038171003817100381710038171003817119863120573minus1

0+V (119905)

100381710038171003817100381710038171003817infinle100381610038161003816100381610038161003816119863120573minus2

0+V (1199050)100381610038161003816100381610038161003816+100381710038171003817100381710038171003817119863120573

0+V (119905)

1003817100381710038171003817100381710038171

le 119860 +10038171003817100381710038171198712V

10038171003817100381710038171le 119860 +

10038171003817100381710038171198732 (119906 V)10038171003817100381710038171

(59)

Thus

119875 (119906 V)119884=

1003817100381710038171003817100381710038171003817100381710038171003817

119863120573minus1

0+V (0)

Γ (120573)(119886120585120573minus1

119905120572minus1

119905120573minus1

)

1003817100381710038171003817100381710038171003817100381710038171003817119884

= Δ3

100381610038161003816100381610038161003816119863120573minus1

0+V (0)

100381610038161003816100381610038161003816le Δ3(119860 +

10038171003817100381710038171198732 (119906 V)10038171003817100381710038171)

(60)

For all (119906 V) isin Ω1 (119868minus119875)(119906 V) isin dom 119871capKer119875 Considering

Lemma 9 we get 119871119875(119906 V) = (0 0) Together with (39) wehave

(119868 minus 119875) (119906 V)119884=1003817100381710038171003817119870119875119871 (119868 minus 119875) (119906 V)

1003817100381710038171003817119884

le Δ1119871 (119868 minus 119875) (119906 V)

119885

le Δ1119873 (119906 V)

119885

(61)

From (60) and (61) we have

(119906 V)119884

le 119875 (119906 V)119884+ (119868 minus 119875) (119906 V)

119884

le Δ3(119860 +

10038171003817100381710038171198732 (119906 V)10038171003817100381710038171) + Δ1119873 (119906 V)

119885

= 119860Δ3+max (Δ

1+ Δ3)10038171003817100381710038171198732 (119906 V)

10038171003817100381710038171 Δ1

10038171003817100381710038171198731 (119906 V)10038171003817100381710038171

+Δ3

10038171003817100381710038171198732 (119906 V)10038171003817100381710038171

(62)

From (62) we discuss various cases

Case 1 ((119906 V)119884le 119860Δ

3+ (Δ1+ Δ3)1198732(119906 V)

1) From (H1)

we have

(119906 V)119884

le 119860Δ3+ (Δ1+ Δ3)

times (10038171003817100381710038171198862

10038171003817100381710038171119906infin+10038171003817100381710038171198872

10038171003817100381710038171

10038171003817100381710038171003817119863120572minus1

0+11990610038171003817100381710038171003817infin

+10038171003817100381710038171198882

10038171003817100381710038171Vinfin

+10038171003817100381710038171198892

10038171003817100381710038171

100381710038171003817100381710038171003817119863120573minus1

0+V100381710038171003817100381710038171003817infin

+10038171003817100381710038171198902

10038171003817100381710038171)

le 119860Δ3+ (Δ1+ Δ3)10038171003817100381710038171198902

10038171003817100381710038171

+ (Δ1+ Δ3)max 10038171003817100381710038171198862

1003817100381710038171003817110038171003817100381710038171198872

10038171003817100381710038171 119906119862120572minus1

+ (Δ1+ Δ3)max 10038171003817100381710038171198882

1003817100381710038171003817110038171003817100381710038171198892

10038171003817100381710038171 V119862120573minus1

le 119860Δ3+ (Δ1+ Δ3)10038171003817100381710038171198902

10038171003817100381710038171

+ (Δ1+ Δ3)max 10038171003817100381710038171198862

1003817100381710038171003817110038171003817100381710038171198872

1003817100381710038171003817110038171003817100381710038171198882

1003817100381710038171003817110038171003817100381710038171198892

10038171003817100381710038171 (119906 V)

119884

(63)

which yield

(119906 V)119884

le119860Δ2+ (Δ1+ Δ3)10038171003817100381710038171198902

10038171003817100381710038171

1 minus (Δ1+ Δ3)max 10038171003817100381710038171198862

1003817100381710038171003817110038171003817100381710038171198872

1003817100381710038171003817110038171003817100381710038171198882

1003817100381710038171003817110038171003817100381710038171198892

10038171003817100381710038171

(64)

ThusΩ1is bounded

Case 2 ((119906 V)119884le 119860Δ

3+ Δ11198731(119906 V)

1+ Δ31198732(119906 V)

1)

From (H1) we have

(119906 V)119884

le 119860Δ3+ Δ1(10038171003817100381710038171198861

10038171003817100381710038171119906infin+10038171003817100381710038171198871

10038171003817100381710038171

10038171003817100381710038171003817119863120572minus1

0+11990610038171003817100381710038171003817infin

+10038171003817100381710038171198881

10038171003817100381710038171Vinfin

+10038171003817100381710038171198891

10038171003817100381710038171

100381710038171003817100381710038171003817119863120573minus1

0+V100381710038171003817100381710038171003817infin

+10038171003817100381710038171198901

10038171003817100381710038171)

+ Δ3(10038171003817100381710038171198862

10038171003817100381710038171119906infin+10038171003817100381710038171198872

10038171003817100381710038171

10038171003817100381710038171003817119863120572minus1

0+11990610038171003817100381710038171003817infin

+10038171003817100381710038171198882

10038171003817100381710038171Vinfin

+10038171003817100381710038171198892

10038171003817100381710038171

100381710038171003817100381710038171003817119863120573minus1

0+V100381710038171003817100381710038171003817infin

+10038171003817100381710038171198902

10038171003817100381710038171)

le 119860Δ3+ Δ1

1003817100381710038171003817119890110038171003817100381710038171+ Δ3

1003817100381710038171003817119890210038171003817100381710038171+ Δ4119906119862120572minus1 + Δ

5V119862120573minus1

le 119860Δ3+ Δ1

1003817100381710038171003817119890110038171003817100381710038171+ Δ3

1003817100381710038171003817119890210038171003817100381710038171+max Δ

4 Δ5 (119906 V)

119884

(65)

which yield

(119906 V)119884le119860Δ3+ Δ1

1003817100381710038171003817119890110038171003817100381710038171+ Δ3

1003817100381710038171003817119890210038171003817100381710038171

1 minusmax Δ4 Δ5

(66)

ThusΩ1is bounded Let

Ω2= (119906 V) isin Ker 119871 119873 (119906 V) isin Im 119871 (67)

For (119906 V) isin Ω2and (119906 V) isin Ker 119871 so (119906 V) = 119888(119886120585

120573minus1

119905120572minus1

119905120573minus1

) 119905 isin [0 1] 119888 isin R Noticing that Im 119871 = Ker119876 then we

Abstract and Applied Analysis 7

get 119876119873(119906 V) = 0 and thus 11987611198731(119888119886120585120573minus1

119905120572minus1

119888119905120573minus1

) = 0 and11987611198732(119888119886120585120573minus1

119905120572minus1

119888119905120573minus1

) = 0 From (H2) we get |119888| le 119860Γ(120573)and thusΩ

2is bounded

We define the isomorphism 119869 Ker 119871 rarr Im119876 by

119869 (119888119886120585120573minus1

119905120572minus1

119888119905120573minus1

) = (119888 119888) (68)

If the first part of (H3) is satisfied then let

Ω3= (119906 V) isin ker 119871 120582119869 (119906 V)

+ (1 minus 120582)119876119873 (119906 V) = 0 120582 isin [0 1]

(69)

For (119906 V) = (119888119886120585120573minus1

119905120572minus1

119888119905120573minus1

) isin Ω3

120582 (119888 119888) = minus (1 minus 120582) Δ21198761119873(119888119886120585

120573minus1

119905120572minus1

119888119905120573minus1

) (1 1) (70)

If 120582 = 1 then 119888 = 0 Otherwise if |119888| gt 119861 in view of (H3) andΔ2lt 0 one has

(1 minus 120582) Δ31198761119873(119888119886120585

120573minus1

119905120572minus1

119888119905120573minus1

) lt 0 (71)

which contradict 1205821198882 ge 0ThusΩ3sub (119906 V) isin Ker 119871 | (119906 V) =

(119888119886120585120573minus1

119905120572minus1

119888119905120573minus1

) |119888| le 119861 is boundedIf the second part of (H3) holds then define the set

Ω3= (119906 V) isin Ker 119871 minus120582119869 (119906 V)

+ (1 minus 120582)119876119873 (119906 V) = 0 120582 isin [0 1]

(72)

and here 119869 is as above Similar to the above argument we canshow thatΩ

3is bounded too

In the following we will prove that all conditions ofTheorem 1 are satisfied LetΩ be a bounded open subset of 119884such that cup3

119894=1Ω119894sub Ω By standard arguments we can prove

that 119870119875(119868 minus 119876)119873 Ω rarr 119884 is compact and thus 119873 is 119871-

compact onΩ Then by the above argument we have

(i) 119871119906 = 120582119873119906 for every (119906 120582) isin [(dom 119871 Ker 119871) cap 120597Ω]times(0 1)

(ii) 119873119906 notin Im 119871 for 119906 isin Ker 119871 cap 120597Ω

Finally we will prove that (iii) of Theorem 1 is satisfied Let119867(119906 120582) = plusmn120582119869119906 + (1 minus 120582)119876119873119906 According to the aboveargument we know that

119867(119906 120582) = 0 for 119906 isin Ker 119871 cap 120597Ω (73)

Thus by the homotopy property of degree

deg (119876119873|Ker119871Ker 119871 cap Ω 0)

= deg (119867 (sdot 0) Ker 119871 cap Ω 0)

= deg (119867 (sdot 1) Ker 119871 cap Ω 0)

= deg (plusmn119869Ker 119871 cap Ω 0) = 0

(74)

Then byTheorem 1 119871(119906 V) = 119873(119906 V) has at least one solutionin dom 119871 cap Ω so that BVP (1) has a solution in 119862

120572minus1

[0 1] times

119862120573minus1

[0 1] The proof is complete

Conflict of Interests

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

Acknowledgments

The authors are very grateful to the anonymous refereesfor many valuable comments and suggestions which helpedto improve the presentation of the paper The project wassupported by the National Natural Science Foundation ofChina (11371221 61304074) the Specialized Research Foun-dation for the Doctoral Program of Higher Education ofChina (20123705110001) the Program for Scientific ResearchInnovation Team in Colleges and Universities of ShandongProvince the Postdoctoral Science Foundation of ShandongProvince (201303074) and Foundation of SDUST

References

[1] A A Kilbas H M Srivastava and J J Trujillo Theoryand Applications of Fractional Differential Equations vol 204of North-Holland Mathematics Studies Elsevier Science BVAmsterdam The Netherlands 2006

[2] V Lakshmikantham S Leela and J Vasundhara DeviTheory ofFractional Dynamic Systems Cambridge Academic PublishersCambridge UK 2009

[3] J J Nieto and J Pimentel ldquoPositive solutions of a fractionalthermostat modelrdquo Boundary Value Problems vol 2013 article5 2013

[4] I Podlubny Fractional Differential Equations vol 198 Aca-demic Press San Diego Calif USA 1999

[5] W Wu and X Zhou ldquoEigenvalue of fractional differentialequations with P-Laplacian operatorrdquo Discrete Dynamics inNature and Society vol 2013 Article ID 137890 8 pages 2013

[6] X Zhang L Liu Y Wu and Y Lu ldquoThe iterative solutions ofnonlinear fractional differential equationsrdquo Applied Mathemat-ics and Computation vol 219 no 9 pp 4680ndash4691 2013

[7] X Zhang L Liu and Y Wu ldquoThe uniqueness of positivesolution for a singular fractional differential system involvingderivativesrdquo Communications in Nonlinear Science and Numer-ical Simulation vol 18 no 6 pp 1400ndash1409 2013

[8] X Zhang L Liu and Y Wu ldquoThe eigenvalue problem for asingular higher order fractional differential equation involvingfractional derivativesrdquo Applied Mathematics and Computationvol 218 no 17 pp 8526ndash8536 2012

[9] X Zhang L Liu and YWu ldquoExistence results for multiple pos-itive solutions of nonlinear higher order perturbed fractionaldifferential equations with derivativesrdquo Applied Mathematicsand Computation vol 219 no 4 pp 1420ndash1433 2012

[10] X Zhang L Liu B Wiwatanapataphee and Y Wu ldquoPositivesolutions of eigenvalue problems for a class of fractionaldifferential equations with derivativesrdquo Abstract and AppliedAnalysis vol 2012 Article ID 512127 16 pages 2012

[11] X Zhang L Liu and Y Wu ldquoMultiple positive solutionsof a singular fractional differential equation with negativelyperturbed termrdquo Mathematical and Computer Modelling vol55 no 3-4 pp 1263ndash1274 2012

[12] K Deng ldquoBlow-up rates for parabolic systemsrdquo Zeitschrift furAngewandte Mathematik und Physik vol 47 no 1 pp 132ndash1431996

8 Abstract and Applied Analysis

[13] K Deng ldquoGlobal existence and blow-up for a system of heatequations with non-linear boundary conditionsrdquoMathematicalMethods in the Applied Sciences vol 18 no 4 pp 307ndash315 1995

[14] L Zhigui and X Chunhong ldquoThe blow-up rate for a system ofheat equations with nonlinear boundary conditionsrdquoNonlinearAnalysisTheory Methods ampApplications vol 34 no 5 pp 767ndash778 1998

[15] D G Aronson ldquoA comparison method for stability analysis ofnonlinear parabolic problemsrdquo SIAM Review vol 20 no 2 pp245ndash264 1978

[16] M Pedersen and Z Lin ldquoBlow-up analysis for a system of heatequations coupled through a nonlinear boundary conditionrdquoApplied Mathematics Letters vol 14 no 2 pp 171ndash176 2001

[17] N A Asif and R A Khan ldquoPositive solutions to singularsystem with four-point coupled boundary conditionsrdquo Journalof Mathematical Analysis and Applications vol 386 no 2 pp848ndash861 2012

[18] C Yuan D Jiang D OrsquoRegan and R P Agarwal ldquoMultiple pos-itive solutions to systems of nonlinear semipositone fractionaldifferential equations with coupled boundary conditionsrdquo Elec-tronic Journal of QualitativeTheory of Differential Equations no13 p 17 2012

[19] Y Cui and J Sun ldquoOn existence of positive solutions of coupledintegral boundary value problems for a nonlinear singularsuperlinear differential systemrdquo Electronic Journal of QualitativeTheory of Differential Equations no 41 pp 1ndash13 2012

[20] Z Bai and Y Zhang ldquoSolvability of fractional three-pointboundary value problems with nonlinear growthrdquo AppliedMathematics and Computation vol 218 no 5 pp 1719ndash17252011

[21] Y Cui ldquoSolvability of second-order boundary-value problemsat resonance involving integral conditionsrdquo Electronic Journalof Differential Equations no 45 pp 1ndash9 2012

[22] Z Hu and W Liu ldquoSolvability for fractional order boundaryvalue problems at resonancerdquo Boundary Value Problems vol2011 article 20 2011

[23] W Jiang ldquoThe existence of solutions to boundary value prob-lems of fractional differential equations at resonancerdquoNonlinearAnalysis Theory Methods amp Applications vol 74 no 5 pp1987ndash1994 2011

[24] N Kosmatov ldquoMulti-point boundary value problems on timescales at resonancerdquo Journal of Mathematical Analysis andApplications vol 323 no 1 pp 253ndash266 2006

[25] G Wang W Liu S Zhu and T Zheng ldquoExistence results fora coupled system of nonlinear fractional 2m-point boundaryvalue problems at resonancerdquo Advances in Difference Equationsvol 2011 article 80 2011

[26] J R L Webb ldquoRemarks on nonlocal boundary value problemsat resonancerdquo Applied Mathematics and Computation vol 216no 2 pp 497ndash500 2010

[27] J R L Webb and M Zima ldquoMultiple positive solutions ofresonant and non-resonant non-local fourth-order boundaryvalue problemsrdquo Glasgow Mathematical Journal vol 54 no 1pp 225ndash240 2012

[28] X ZhangM Feng andWGe ldquoExistence result of second-orderdifferential equations with integral boundary conditions atresonancerdquo Journal of Mathematical Analysis and Applicationsvol 353 no 1 pp 311ndash319 2009

[29] X Zhang C Zhu and Z Wu ldquoSolvability for a coupled systemof fractional differential equations with impulses at resonancerdquoBoundary Value Problems vol 2013 artice 80 2013

[30] Z Zhao and J Liang ldquoExistence of solutions to functionalboundary value problem of second-order nonlinear differentialequationrdquo Journal of Mathematical Analysis and Applicationsvol 373 no 2 pp 614ndash634 2011

[31] Y Zou and Y Cui ldquoExistence results for a functional boundaryvalue problem of fractional differential equationsrdquo Advances inDifference Equations vol 2013 article 25 2013

[32] J Mawhin ldquoTopological degree and boundary value problemsfor nonlinear differential equationsrdquo in Topological Methods forOrdinary Differential Equations P M Fitzpertrick M MartelliJ Mawhin and R Nussbaum Eds vol 1537 of Lecture Notes inMathematics pp 74ndash142 Springer Berlin Germany 1993

[33] J Mawhin Topological Degree Methods in Nonlinear BoundaryValue Problems vol 40 of CBMS Regional Conference Series inMathematics American Mathematical Society Providence RIUSA 1979

[34] Y Zhang and Z Bai ldquoExistence of solutions for nonlinearfractional three-point boundary value problems at resonancerdquoJournal of Applied Mathematics and Computing vol 36 no 1-2pp 417ndash440 2011

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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OptimizationJournal of

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CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

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Journal of

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Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

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The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

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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

6 Abstract and Applied Analysis

Theorem 10 Suppose (2) and (1198671)ndash(1198673) hold Then (1) hasat least one solution in Y provided that

max (Δ1+ Δ3)max 10038171003817100381710038171198862

1003817100381710038171003817110038171003817100381710038171198872

1003817100381710038171003817110038171003817100381710038171198882

1003817100381710038171003817110038171003817100381710038171198892

10038171003817100381710038171 Δ4 Δ5

lt 1

(54)

Proof Set

Ω1= (119906 V) isin dom 119871 Ker 119871 119871 (119906 V) = 120582119873 (119906 V)

for some 120582 isin [0 1] (55)

Take (119906 V) isin Ω1 Since 119871119906 = 120582119873119906 so 120582 = 0 and 119873(119906 V) isin

Im 119871 = Ker119876 hence

119876119873(119906 V) = 1198761(1198731(119906 V) 119873

2(119906 V)) (1 1) = 0 (56)

Thus from (H2) there exist 1199050isin [0 1] such that

100381610038161003816100381610038161003816119863120573minus1

0+V (1199050)100381610038161003816100381610038161003816le 119860 (57)

Noticing that

119863120573minus1

0+V (119905) = 119863

120573minus1

0+V (1199050) + int

119905

1199050

119863120573

0+V (119904) 119889119904 (58)

so100381610038161003816100381610038161003816119863120573minus1

0+V (0)

100381610038161003816100381610038161003816le100381710038171003817100381710038171003817119863120573minus1

0+V (119905)

100381710038171003817100381710038171003817infinle100381610038161003816100381610038161003816119863120573minus2

0+V (1199050)100381610038161003816100381610038161003816+100381710038171003817100381710038171003817119863120573

0+V (119905)

1003817100381710038171003817100381710038171

le 119860 +10038171003817100381710038171198712V

10038171003817100381710038171le 119860 +

10038171003817100381710038171198732 (119906 V)10038171003817100381710038171

(59)

Thus

119875 (119906 V)119884=

1003817100381710038171003817100381710038171003817100381710038171003817

119863120573minus1

0+V (0)

Γ (120573)(119886120585120573minus1

119905120572minus1

119905120573minus1

)

1003817100381710038171003817100381710038171003817100381710038171003817119884

= Δ3

100381610038161003816100381610038161003816119863120573minus1

0+V (0)

100381610038161003816100381610038161003816le Δ3(119860 +

10038171003817100381710038171198732 (119906 V)10038171003817100381710038171)

(60)

For all (119906 V) isin Ω1 (119868minus119875)(119906 V) isin dom 119871capKer119875 Considering

Lemma 9 we get 119871119875(119906 V) = (0 0) Together with (39) wehave

(119868 minus 119875) (119906 V)119884=1003817100381710038171003817119870119875119871 (119868 minus 119875) (119906 V)

1003817100381710038171003817119884

le Δ1119871 (119868 minus 119875) (119906 V)

119885

le Δ1119873 (119906 V)

119885

(61)

From (60) and (61) we have

(119906 V)119884

le 119875 (119906 V)119884+ (119868 minus 119875) (119906 V)

119884

le Δ3(119860 +

10038171003817100381710038171198732 (119906 V)10038171003817100381710038171) + Δ1119873 (119906 V)

119885

= 119860Δ3+max (Δ

1+ Δ3)10038171003817100381710038171198732 (119906 V)

10038171003817100381710038171 Δ1

10038171003817100381710038171198731 (119906 V)10038171003817100381710038171

+Δ3

10038171003817100381710038171198732 (119906 V)10038171003817100381710038171

(62)

From (62) we discuss various cases

Case 1 ((119906 V)119884le 119860Δ

3+ (Δ1+ Δ3)1198732(119906 V)

1) From (H1)

we have

(119906 V)119884

le 119860Δ3+ (Δ1+ Δ3)

times (10038171003817100381710038171198862

10038171003817100381710038171119906infin+10038171003817100381710038171198872

10038171003817100381710038171

10038171003817100381710038171003817119863120572minus1

0+11990610038171003817100381710038171003817infin

+10038171003817100381710038171198882

10038171003817100381710038171Vinfin

+10038171003817100381710038171198892

10038171003817100381710038171

100381710038171003817100381710038171003817119863120573minus1

0+V100381710038171003817100381710038171003817infin

+10038171003817100381710038171198902

10038171003817100381710038171)

le 119860Δ3+ (Δ1+ Δ3)10038171003817100381710038171198902

10038171003817100381710038171

+ (Δ1+ Δ3)max 10038171003817100381710038171198862

1003817100381710038171003817110038171003817100381710038171198872

10038171003817100381710038171 119906119862120572minus1

+ (Δ1+ Δ3)max 10038171003817100381710038171198882

1003817100381710038171003817110038171003817100381710038171198892

10038171003817100381710038171 V119862120573minus1

le 119860Δ3+ (Δ1+ Δ3)10038171003817100381710038171198902

10038171003817100381710038171

+ (Δ1+ Δ3)max 10038171003817100381710038171198862

1003817100381710038171003817110038171003817100381710038171198872

1003817100381710038171003817110038171003817100381710038171198882

1003817100381710038171003817110038171003817100381710038171198892

10038171003817100381710038171 (119906 V)

119884

(63)

which yield

(119906 V)119884

le119860Δ2+ (Δ1+ Δ3)10038171003817100381710038171198902

10038171003817100381710038171

1 minus (Δ1+ Δ3)max 10038171003817100381710038171198862

1003817100381710038171003817110038171003817100381710038171198872

1003817100381710038171003817110038171003817100381710038171198882

1003817100381710038171003817110038171003817100381710038171198892

10038171003817100381710038171

(64)

ThusΩ1is bounded

Case 2 ((119906 V)119884le 119860Δ

3+ Δ11198731(119906 V)

1+ Δ31198732(119906 V)

1)

From (H1) we have

(119906 V)119884

le 119860Δ3+ Δ1(10038171003817100381710038171198861

10038171003817100381710038171119906infin+10038171003817100381710038171198871

10038171003817100381710038171

10038171003817100381710038171003817119863120572minus1

0+11990610038171003817100381710038171003817infin

+10038171003817100381710038171198881

10038171003817100381710038171Vinfin

+10038171003817100381710038171198891

10038171003817100381710038171

100381710038171003817100381710038171003817119863120573minus1

0+V100381710038171003817100381710038171003817infin

+10038171003817100381710038171198901

10038171003817100381710038171)

+ Δ3(10038171003817100381710038171198862

10038171003817100381710038171119906infin+10038171003817100381710038171198872

10038171003817100381710038171

10038171003817100381710038171003817119863120572minus1

0+11990610038171003817100381710038171003817infin

+10038171003817100381710038171198882

10038171003817100381710038171Vinfin

+10038171003817100381710038171198892

10038171003817100381710038171

100381710038171003817100381710038171003817119863120573minus1

0+V100381710038171003817100381710038171003817infin

+10038171003817100381710038171198902

10038171003817100381710038171)

le 119860Δ3+ Δ1

1003817100381710038171003817119890110038171003817100381710038171+ Δ3

1003817100381710038171003817119890210038171003817100381710038171+ Δ4119906119862120572minus1 + Δ

5V119862120573minus1

le 119860Δ3+ Δ1

1003817100381710038171003817119890110038171003817100381710038171+ Δ3

1003817100381710038171003817119890210038171003817100381710038171+max Δ

4 Δ5 (119906 V)

119884

(65)

which yield

(119906 V)119884le119860Δ3+ Δ1

1003817100381710038171003817119890110038171003817100381710038171+ Δ3

1003817100381710038171003817119890210038171003817100381710038171

1 minusmax Δ4 Δ5

(66)

ThusΩ1is bounded Let

Ω2= (119906 V) isin Ker 119871 119873 (119906 V) isin Im 119871 (67)

For (119906 V) isin Ω2and (119906 V) isin Ker 119871 so (119906 V) = 119888(119886120585

120573minus1

119905120572minus1

119905120573minus1

) 119905 isin [0 1] 119888 isin R Noticing that Im 119871 = Ker119876 then we

Abstract and Applied Analysis 7

get 119876119873(119906 V) = 0 and thus 11987611198731(119888119886120585120573minus1

119905120572minus1

119888119905120573minus1

) = 0 and11987611198732(119888119886120585120573minus1

119905120572minus1

119888119905120573minus1

) = 0 From (H2) we get |119888| le 119860Γ(120573)and thusΩ

2is bounded

We define the isomorphism 119869 Ker 119871 rarr Im119876 by

119869 (119888119886120585120573minus1

119905120572minus1

119888119905120573minus1

) = (119888 119888) (68)

If the first part of (H3) is satisfied then let

Ω3= (119906 V) isin ker 119871 120582119869 (119906 V)

+ (1 minus 120582)119876119873 (119906 V) = 0 120582 isin [0 1]

(69)

For (119906 V) = (119888119886120585120573minus1

119905120572minus1

119888119905120573minus1

) isin Ω3

120582 (119888 119888) = minus (1 minus 120582) Δ21198761119873(119888119886120585

120573minus1

119905120572minus1

119888119905120573minus1

) (1 1) (70)

If 120582 = 1 then 119888 = 0 Otherwise if |119888| gt 119861 in view of (H3) andΔ2lt 0 one has

(1 minus 120582) Δ31198761119873(119888119886120585

120573minus1

119905120572minus1

119888119905120573minus1

) lt 0 (71)

which contradict 1205821198882 ge 0ThusΩ3sub (119906 V) isin Ker 119871 | (119906 V) =

(119888119886120585120573minus1

119905120572minus1

119888119905120573minus1

) |119888| le 119861 is boundedIf the second part of (H3) holds then define the set

Ω3= (119906 V) isin Ker 119871 minus120582119869 (119906 V)

+ (1 minus 120582)119876119873 (119906 V) = 0 120582 isin [0 1]

(72)

and here 119869 is as above Similar to the above argument we canshow thatΩ

3is bounded too

In the following we will prove that all conditions ofTheorem 1 are satisfied LetΩ be a bounded open subset of 119884such that cup3

119894=1Ω119894sub Ω By standard arguments we can prove

that 119870119875(119868 minus 119876)119873 Ω rarr 119884 is compact and thus 119873 is 119871-

compact onΩ Then by the above argument we have

(i) 119871119906 = 120582119873119906 for every (119906 120582) isin [(dom 119871 Ker 119871) cap 120597Ω]times(0 1)

(ii) 119873119906 notin Im 119871 for 119906 isin Ker 119871 cap 120597Ω

Finally we will prove that (iii) of Theorem 1 is satisfied Let119867(119906 120582) = plusmn120582119869119906 + (1 minus 120582)119876119873119906 According to the aboveargument we know that

119867(119906 120582) = 0 for 119906 isin Ker 119871 cap 120597Ω (73)

Thus by the homotopy property of degree

deg (119876119873|Ker119871Ker 119871 cap Ω 0)

= deg (119867 (sdot 0) Ker 119871 cap Ω 0)

= deg (119867 (sdot 1) Ker 119871 cap Ω 0)

= deg (plusmn119869Ker 119871 cap Ω 0) = 0

(74)

Then byTheorem 1 119871(119906 V) = 119873(119906 V) has at least one solutionin dom 119871 cap Ω so that BVP (1) has a solution in 119862

120572minus1

[0 1] times

119862120573minus1

[0 1] The proof is complete

Conflict of Interests

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

Acknowledgments

The authors are very grateful to the anonymous refereesfor many valuable comments and suggestions which helpedto improve the presentation of the paper The project wassupported by the National Natural Science Foundation ofChina (11371221 61304074) the Specialized Research Foun-dation for the Doctoral Program of Higher Education ofChina (20123705110001) the Program for Scientific ResearchInnovation Team in Colleges and Universities of ShandongProvince the Postdoctoral Science Foundation of ShandongProvince (201303074) and Foundation of SDUST

References

[1] A A Kilbas H M Srivastava and J J Trujillo Theoryand Applications of Fractional Differential Equations vol 204of North-Holland Mathematics Studies Elsevier Science BVAmsterdam The Netherlands 2006

[2] V Lakshmikantham S Leela and J Vasundhara DeviTheory ofFractional Dynamic Systems Cambridge Academic PublishersCambridge UK 2009

[3] J J Nieto and J Pimentel ldquoPositive solutions of a fractionalthermostat modelrdquo Boundary Value Problems vol 2013 article5 2013

[4] I Podlubny Fractional Differential Equations vol 198 Aca-demic Press San Diego Calif USA 1999

[5] W Wu and X Zhou ldquoEigenvalue of fractional differentialequations with P-Laplacian operatorrdquo Discrete Dynamics inNature and Society vol 2013 Article ID 137890 8 pages 2013

[6] X Zhang L Liu Y Wu and Y Lu ldquoThe iterative solutions ofnonlinear fractional differential equationsrdquo Applied Mathemat-ics and Computation vol 219 no 9 pp 4680ndash4691 2013

[7] X Zhang L Liu and Y Wu ldquoThe uniqueness of positivesolution for a singular fractional differential system involvingderivativesrdquo Communications in Nonlinear Science and Numer-ical Simulation vol 18 no 6 pp 1400ndash1409 2013

[8] X Zhang L Liu and Y Wu ldquoThe eigenvalue problem for asingular higher order fractional differential equation involvingfractional derivativesrdquo Applied Mathematics and Computationvol 218 no 17 pp 8526ndash8536 2012

[9] X Zhang L Liu and YWu ldquoExistence results for multiple pos-itive solutions of nonlinear higher order perturbed fractionaldifferential equations with derivativesrdquo Applied Mathematicsand Computation vol 219 no 4 pp 1420ndash1433 2012

[10] X Zhang L Liu B Wiwatanapataphee and Y Wu ldquoPositivesolutions of eigenvalue problems for a class of fractionaldifferential equations with derivativesrdquo Abstract and AppliedAnalysis vol 2012 Article ID 512127 16 pages 2012

[11] X Zhang L Liu and Y Wu ldquoMultiple positive solutionsof a singular fractional differential equation with negativelyperturbed termrdquo Mathematical and Computer Modelling vol55 no 3-4 pp 1263ndash1274 2012

[12] K Deng ldquoBlow-up rates for parabolic systemsrdquo Zeitschrift furAngewandte Mathematik und Physik vol 47 no 1 pp 132ndash1431996

8 Abstract and Applied Analysis

[13] K Deng ldquoGlobal existence and blow-up for a system of heatequations with non-linear boundary conditionsrdquoMathematicalMethods in the Applied Sciences vol 18 no 4 pp 307ndash315 1995

[14] L Zhigui and X Chunhong ldquoThe blow-up rate for a system ofheat equations with nonlinear boundary conditionsrdquoNonlinearAnalysisTheory Methods ampApplications vol 34 no 5 pp 767ndash778 1998

[15] D G Aronson ldquoA comparison method for stability analysis ofnonlinear parabolic problemsrdquo SIAM Review vol 20 no 2 pp245ndash264 1978

[16] M Pedersen and Z Lin ldquoBlow-up analysis for a system of heatequations coupled through a nonlinear boundary conditionrdquoApplied Mathematics Letters vol 14 no 2 pp 171ndash176 2001

[17] N A Asif and R A Khan ldquoPositive solutions to singularsystem with four-point coupled boundary conditionsrdquo Journalof Mathematical Analysis and Applications vol 386 no 2 pp848ndash861 2012

[18] C Yuan D Jiang D OrsquoRegan and R P Agarwal ldquoMultiple pos-itive solutions to systems of nonlinear semipositone fractionaldifferential equations with coupled boundary conditionsrdquo Elec-tronic Journal of QualitativeTheory of Differential Equations no13 p 17 2012

[19] Y Cui and J Sun ldquoOn existence of positive solutions of coupledintegral boundary value problems for a nonlinear singularsuperlinear differential systemrdquo Electronic Journal of QualitativeTheory of Differential Equations no 41 pp 1ndash13 2012

[20] Z Bai and Y Zhang ldquoSolvability of fractional three-pointboundary value problems with nonlinear growthrdquo AppliedMathematics and Computation vol 218 no 5 pp 1719ndash17252011

[21] Y Cui ldquoSolvability of second-order boundary-value problemsat resonance involving integral conditionsrdquo Electronic Journalof Differential Equations no 45 pp 1ndash9 2012

[22] Z Hu and W Liu ldquoSolvability for fractional order boundaryvalue problems at resonancerdquo Boundary Value Problems vol2011 article 20 2011

[23] W Jiang ldquoThe existence of solutions to boundary value prob-lems of fractional differential equations at resonancerdquoNonlinearAnalysis Theory Methods amp Applications vol 74 no 5 pp1987ndash1994 2011

[24] N Kosmatov ldquoMulti-point boundary value problems on timescales at resonancerdquo Journal of Mathematical Analysis andApplications vol 323 no 1 pp 253ndash266 2006

[25] G Wang W Liu S Zhu and T Zheng ldquoExistence results fora coupled system of nonlinear fractional 2m-point boundaryvalue problems at resonancerdquo Advances in Difference Equationsvol 2011 article 80 2011

[26] J R L Webb ldquoRemarks on nonlocal boundary value problemsat resonancerdquo Applied Mathematics and Computation vol 216no 2 pp 497ndash500 2010

[27] J R L Webb and M Zima ldquoMultiple positive solutions ofresonant and non-resonant non-local fourth-order boundaryvalue problemsrdquo Glasgow Mathematical Journal vol 54 no 1pp 225ndash240 2012

[28] X ZhangM Feng andWGe ldquoExistence result of second-orderdifferential equations with integral boundary conditions atresonancerdquo Journal of Mathematical Analysis and Applicationsvol 353 no 1 pp 311ndash319 2009

[29] X Zhang C Zhu and Z Wu ldquoSolvability for a coupled systemof fractional differential equations with impulses at resonancerdquoBoundary Value Problems vol 2013 artice 80 2013

[30] Z Zhao and J Liang ldquoExistence of solutions to functionalboundary value problem of second-order nonlinear differentialequationrdquo Journal of Mathematical Analysis and Applicationsvol 373 no 2 pp 614ndash634 2011

[31] Y Zou and Y Cui ldquoExistence results for a functional boundaryvalue problem of fractional differential equationsrdquo Advances inDifference Equations vol 2013 article 25 2013

[32] J Mawhin ldquoTopological degree and boundary value problemsfor nonlinear differential equationsrdquo in Topological Methods forOrdinary Differential Equations P M Fitzpertrick M MartelliJ Mawhin and R Nussbaum Eds vol 1537 of Lecture Notes inMathematics pp 74ndash142 Springer Berlin Germany 1993

[33] J Mawhin Topological Degree Methods in Nonlinear BoundaryValue Problems vol 40 of CBMS Regional Conference Series inMathematics American Mathematical Society Providence RIUSA 1979

[34] Y Zhang and Z Bai ldquoExistence of solutions for nonlinearfractional three-point boundary value problems at resonancerdquoJournal of Applied Mathematics and Computing vol 36 no 1-2pp 417ndash440 2011

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

Abstract and Applied Analysis 7

get 119876119873(119906 V) = 0 and thus 11987611198731(119888119886120585120573minus1

119905120572minus1

119888119905120573minus1

) = 0 and11987611198732(119888119886120585120573minus1

119905120572minus1

119888119905120573minus1

) = 0 From (H2) we get |119888| le 119860Γ(120573)and thusΩ

2is bounded

We define the isomorphism 119869 Ker 119871 rarr Im119876 by

119869 (119888119886120585120573minus1

119905120572minus1

119888119905120573minus1

) = (119888 119888) (68)

If the first part of (H3) is satisfied then let

Ω3= (119906 V) isin ker 119871 120582119869 (119906 V)

+ (1 minus 120582)119876119873 (119906 V) = 0 120582 isin [0 1]

(69)

For (119906 V) = (119888119886120585120573minus1

119905120572minus1

119888119905120573minus1

) isin Ω3

120582 (119888 119888) = minus (1 minus 120582) Δ21198761119873(119888119886120585

120573minus1

119905120572minus1

119888119905120573minus1

) (1 1) (70)

If 120582 = 1 then 119888 = 0 Otherwise if |119888| gt 119861 in view of (H3) andΔ2lt 0 one has

(1 minus 120582) Δ31198761119873(119888119886120585

120573minus1

119905120572minus1

119888119905120573minus1

) lt 0 (71)

which contradict 1205821198882 ge 0ThusΩ3sub (119906 V) isin Ker 119871 | (119906 V) =

(119888119886120585120573minus1

119905120572minus1

119888119905120573minus1

) |119888| le 119861 is boundedIf the second part of (H3) holds then define the set

Ω3= (119906 V) isin Ker 119871 minus120582119869 (119906 V)

+ (1 minus 120582)119876119873 (119906 V) = 0 120582 isin [0 1]

(72)

and here 119869 is as above Similar to the above argument we canshow thatΩ

3is bounded too

In the following we will prove that all conditions ofTheorem 1 are satisfied LetΩ be a bounded open subset of 119884such that cup3

119894=1Ω119894sub Ω By standard arguments we can prove

that 119870119875(119868 minus 119876)119873 Ω rarr 119884 is compact and thus 119873 is 119871-

compact onΩ Then by the above argument we have

(i) 119871119906 = 120582119873119906 for every (119906 120582) isin [(dom 119871 Ker 119871) cap 120597Ω]times(0 1)

(ii) 119873119906 notin Im 119871 for 119906 isin Ker 119871 cap 120597Ω

Finally we will prove that (iii) of Theorem 1 is satisfied Let119867(119906 120582) = plusmn120582119869119906 + (1 minus 120582)119876119873119906 According to the aboveargument we know that

119867(119906 120582) = 0 for 119906 isin Ker 119871 cap 120597Ω (73)

Thus by the homotopy property of degree

deg (119876119873|Ker119871Ker 119871 cap Ω 0)

= deg (119867 (sdot 0) Ker 119871 cap Ω 0)

= deg (119867 (sdot 1) Ker 119871 cap Ω 0)

= deg (plusmn119869Ker 119871 cap Ω 0) = 0

(74)

Then byTheorem 1 119871(119906 V) = 119873(119906 V) has at least one solutionin dom 119871 cap Ω so that BVP (1) has a solution in 119862

120572minus1

[0 1] times

119862120573minus1

[0 1] The proof is complete

Conflict of Interests

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

Acknowledgments

The authors are very grateful to the anonymous refereesfor many valuable comments and suggestions which helpedto improve the presentation of the paper The project wassupported by the National Natural Science Foundation ofChina (11371221 61304074) the Specialized Research Foun-dation for the Doctoral Program of Higher Education ofChina (20123705110001) the Program for Scientific ResearchInnovation Team in Colleges and Universities of ShandongProvince the Postdoctoral Science Foundation of ShandongProvince (201303074) and Foundation of SDUST

References

[1] A A Kilbas H M Srivastava and J J Trujillo Theoryand Applications of Fractional Differential Equations vol 204of North-Holland Mathematics Studies Elsevier Science BVAmsterdam The Netherlands 2006

[2] V Lakshmikantham S Leela and J Vasundhara DeviTheory ofFractional Dynamic Systems Cambridge Academic PublishersCambridge UK 2009

[3] J J Nieto and J Pimentel ldquoPositive solutions of a fractionalthermostat modelrdquo Boundary Value Problems vol 2013 article5 2013

[4] I Podlubny Fractional Differential Equations vol 198 Aca-demic Press San Diego Calif USA 1999

[5] W Wu and X Zhou ldquoEigenvalue of fractional differentialequations with P-Laplacian operatorrdquo Discrete Dynamics inNature and Society vol 2013 Article ID 137890 8 pages 2013

[6] X Zhang L Liu Y Wu and Y Lu ldquoThe iterative solutions ofnonlinear fractional differential equationsrdquo Applied Mathemat-ics and Computation vol 219 no 9 pp 4680ndash4691 2013

[7] X Zhang L Liu and Y Wu ldquoThe uniqueness of positivesolution for a singular fractional differential system involvingderivativesrdquo Communications in Nonlinear Science and Numer-ical Simulation vol 18 no 6 pp 1400ndash1409 2013

[8] X Zhang L Liu and Y Wu ldquoThe eigenvalue problem for asingular higher order fractional differential equation involvingfractional derivativesrdquo Applied Mathematics and Computationvol 218 no 17 pp 8526ndash8536 2012

[9] X Zhang L Liu and YWu ldquoExistence results for multiple pos-itive solutions of nonlinear higher order perturbed fractionaldifferential equations with derivativesrdquo Applied Mathematicsand Computation vol 219 no 4 pp 1420ndash1433 2012

[10] X Zhang L Liu B Wiwatanapataphee and Y Wu ldquoPositivesolutions of eigenvalue problems for a class of fractionaldifferential equations with derivativesrdquo Abstract and AppliedAnalysis vol 2012 Article ID 512127 16 pages 2012

[11] X Zhang L Liu and Y Wu ldquoMultiple positive solutionsof a singular fractional differential equation with negativelyperturbed termrdquo Mathematical and Computer Modelling vol55 no 3-4 pp 1263ndash1274 2012

[12] K Deng ldquoBlow-up rates for parabolic systemsrdquo Zeitschrift furAngewandte Mathematik und Physik vol 47 no 1 pp 132ndash1431996

8 Abstract and Applied Analysis

[13] K Deng ldquoGlobal existence and blow-up for a system of heatequations with non-linear boundary conditionsrdquoMathematicalMethods in the Applied Sciences vol 18 no 4 pp 307ndash315 1995

[14] L Zhigui and X Chunhong ldquoThe blow-up rate for a system ofheat equations with nonlinear boundary conditionsrdquoNonlinearAnalysisTheory Methods ampApplications vol 34 no 5 pp 767ndash778 1998

[15] D G Aronson ldquoA comparison method for stability analysis ofnonlinear parabolic problemsrdquo SIAM Review vol 20 no 2 pp245ndash264 1978

[16] M Pedersen and Z Lin ldquoBlow-up analysis for a system of heatequations coupled through a nonlinear boundary conditionrdquoApplied Mathematics Letters vol 14 no 2 pp 171ndash176 2001

[17] N A Asif and R A Khan ldquoPositive solutions to singularsystem with four-point coupled boundary conditionsrdquo Journalof Mathematical Analysis and Applications vol 386 no 2 pp848ndash861 2012

[18] C Yuan D Jiang D OrsquoRegan and R P Agarwal ldquoMultiple pos-itive solutions to systems of nonlinear semipositone fractionaldifferential equations with coupled boundary conditionsrdquo Elec-tronic Journal of QualitativeTheory of Differential Equations no13 p 17 2012

[19] Y Cui and J Sun ldquoOn existence of positive solutions of coupledintegral boundary value problems for a nonlinear singularsuperlinear differential systemrdquo Electronic Journal of QualitativeTheory of Differential Equations no 41 pp 1ndash13 2012

[20] Z Bai and Y Zhang ldquoSolvability of fractional three-pointboundary value problems with nonlinear growthrdquo AppliedMathematics and Computation vol 218 no 5 pp 1719ndash17252011

[21] Y Cui ldquoSolvability of second-order boundary-value problemsat resonance involving integral conditionsrdquo Electronic Journalof Differential Equations no 45 pp 1ndash9 2012

[22] Z Hu and W Liu ldquoSolvability for fractional order boundaryvalue problems at resonancerdquo Boundary Value Problems vol2011 article 20 2011

[23] W Jiang ldquoThe existence of solutions to boundary value prob-lems of fractional differential equations at resonancerdquoNonlinearAnalysis Theory Methods amp Applications vol 74 no 5 pp1987ndash1994 2011

[24] N Kosmatov ldquoMulti-point boundary value problems on timescales at resonancerdquo Journal of Mathematical Analysis andApplications vol 323 no 1 pp 253ndash266 2006

[25] G Wang W Liu S Zhu and T Zheng ldquoExistence results fora coupled system of nonlinear fractional 2m-point boundaryvalue problems at resonancerdquo Advances in Difference Equationsvol 2011 article 80 2011

[26] J R L Webb ldquoRemarks on nonlocal boundary value problemsat resonancerdquo Applied Mathematics and Computation vol 216no 2 pp 497ndash500 2010

[27] J R L Webb and M Zima ldquoMultiple positive solutions ofresonant and non-resonant non-local fourth-order boundaryvalue problemsrdquo Glasgow Mathematical Journal vol 54 no 1pp 225ndash240 2012

[28] X ZhangM Feng andWGe ldquoExistence result of second-orderdifferential equations with integral boundary conditions atresonancerdquo Journal of Mathematical Analysis and Applicationsvol 353 no 1 pp 311ndash319 2009

[29] X Zhang C Zhu and Z Wu ldquoSolvability for a coupled systemof fractional differential equations with impulses at resonancerdquoBoundary Value Problems vol 2013 artice 80 2013

[30] Z Zhao and J Liang ldquoExistence of solutions to functionalboundary value problem of second-order nonlinear differentialequationrdquo Journal of Mathematical Analysis and Applicationsvol 373 no 2 pp 614ndash634 2011

[31] Y Zou and Y Cui ldquoExistence results for a functional boundaryvalue problem of fractional differential equationsrdquo Advances inDifference Equations vol 2013 article 25 2013

[32] J Mawhin ldquoTopological degree and boundary value problemsfor nonlinear differential equationsrdquo in Topological Methods forOrdinary Differential Equations P M Fitzpertrick M MartelliJ Mawhin and R Nussbaum Eds vol 1537 of Lecture Notes inMathematics pp 74ndash142 Springer Berlin Germany 1993

[33] J Mawhin Topological Degree Methods in Nonlinear BoundaryValue Problems vol 40 of CBMS Regional Conference Series inMathematics American Mathematical Society Providence RIUSA 1979

[34] Y Zhang and Z Bai ldquoExistence of solutions for nonlinearfractional three-point boundary value problems at resonancerdquoJournal of Applied Mathematics and Computing vol 36 no 1-2pp 417ndash440 2011

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

8 Abstract and Applied Analysis

[13] K Deng ldquoGlobal existence and blow-up for a system of heatequations with non-linear boundary conditionsrdquoMathematicalMethods in the Applied Sciences vol 18 no 4 pp 307ndash315 1995

[14] L Zhigui and X Chunhong ldquoThe blow-up rate for a system ofheat equations with nonlinear boundary conditionsrdquoNonlinearAnalysisTheory Methods ampApplications vol 34 no 5 pp 767ndash778 1998

[15] D G Aronson ldquoA comparison method for stability analysis ofnonlinear parabolic problemsrdquo SIAM Review vol 20 no 2 pp245ndash264 1978

[16] M Pedersen and Z Lin ldquoBlow-up analysis for a system of heatequations coupled through a nonlinear boundary conditionrdquoApplied Mathematics Letters vol 14 no 2 pp 171ndash176 2001

[17] N A Asif and R A Khan ldquoPositive solutions to singularsystem with four-point coupled boundary conditionsrdquo Journalof Mathematical Analysis and Applications vol 386 no 2 pp848ndash861 2012

[18] C Yuan D Jiang D OrsquoRegan and R P Agarwal ldquoMultiple pos-itive solutions to systems of nonlinear semipositone fractionaldifferential equations with coupled boundary conditionsrdquo Elec-tronic Journal of QualitativeTheory of Differential Equations no13 p 17 2012

[19] Y Cui and J Sun ldquoOn existence of positive solutions of coupledintegral boundary value problems for a nonlinear singularsuperlinear differential systemrdquo Electronic Journal of QualitativeTheory of Differential Equations no 41 pp 1ndash13 2012

[20] Z Bai and Y Zhang ldquoSolvability of fractional three-pointboundary value problems with nonlinear growthrdquo AppliedMathematics and Computation vol 218 no 5 pp 1719ndash17252011

[21] Y Cui ldquoSolvability of second-order boundary-value problemsat resonance involving integral conditionsrdquo Electronic Journalof Differential Equations no 45 pp 1ndash9 2012

[22] Z Hu and W Liu ldquoSolvability for fractional order boundaryvalue problems at resonancerdquo Boundary Value Problems vol2011 article 20 2011

[23] W Jiang ldquoThe existence of solutions to boundary value prob-lems of fractional differential equations at resonancerdquoNonlinearAnalysis Theory Methods amp Applications vol 74 no 5 pp1987ndash1994 2011

[24] N Kosmatov ldquoMulti-point boundary value problems on timescales at resonancerdquo Journal of Mathematical Analysis andApplications vol 323 no 1 pp 253ndash266 2006

[25] G Wang W Liu S Zhu and T Zheng ldquoExistence results fora coupled system of nonlinear fractional 2m-point boundaryvalue problems at resonancerdquo Advances in Difference Equationsvol 2011 article 80 2011

[26] J R L Webb ldquoRemarks on nonlocal boundary value problemsat resonancerdquo Applied Mathematics and Computation vol 216no 2 pp 497ndash500 2010

[27] J R L Webb and M Zima ldquoMultiple positive solutions ofresonant and non-resonant non-local fourth-order boundaryvalue problemsrdquo Glasgow Mathematical Journal vol 54 no 1pp 225ndash240 2012

[28] X ZhangM Feng andWGe ldquoExistence result of second-orderdifferential equations with integral boundary conditions atresonancerdquo Journal of Mathematical Analysis and Applicationsvol 353 no 1 pp 311ndash319 2009

[29] X Zhang C Zhu and Z Wu ldquoSolvability for a coupled systemof fractional differential equations with impulses at resonancerdquoBoundary Value Problems vol 2013 artice 80 2013

[30] Z Zhao and J Liang ldquoExistence of solutions to functionalboundary value problem of second-order nonlinear differentialequationrdquo Journal of Mathematical Analysis and Applicationsvol 373 no 2 pp 614ndash634 2011

[31] Y Zou and Y Cui ldquoExistence results for a functional boundaryvalue problem of fractional differential equationsrdquo Advances inDifference Equations vol 2013 article 25 2013

[32] J Mawhin ldquoTopological degree and boundary value problemsfor nonlinear differential equationsrdquo in Topological Methods forOrdinary Differential Equations P M Fitzpertrick M MartelliJ Mawhin and R Nussbaum Eds vol 1537 of Lecture Notes inMathematics pp 74ndash142 Springer Berlin Germany 1993

[33] J Mawhin Topological Degree Methods in Nonlinear BoundaryValue Problems vol 40 of CBMS Regional Conference Series inMathematics American Mathematical Society Providence RIUSA 1979

[34] Y Zhang and Z Bai ldquoExistence of solutions for nonlinearfractional three-point boundary value problems at resonancerdquoJournal of Applied Mathematics and Computing vol 36 no 1-2pp 417ndash440 2011

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

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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