Positive Solutions for a Higher-Order Semipositone...
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Research Article Positive Solutions for a Higher-Order Semipositone Nonlocal Fractional Differential Equation with Singularities on Both Time and Space Variable Kemei Zhang School of Mathematics Sciences, Qufu Normal University, Qufu , Shandong, China Correspondence should be addressed to Kemei Zhang; [email protected] Received 6 August 2018; Accepted 15 January 2019; Published 3 February 2019 Academic Editor: Shanhe Wu Copyright © 2019 Kemei Zhang. is is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited. In this paper, we consider the following higher-order semipositone nonlocal Riemann-Liouville fractional differential equation 0+ () + (, (), 0+ ()) + () = 0, 0 < < 1, 0+ (0) = +1 0+ (0) = ⋅ ⋅ ⋅ = +−2 0+ (0) = 0, and 0+ (1) = ∑ −2 =1 0+ ( ), where 0 + and 0 + are the standard Riemann-Liouville fractional derivatives. e existence results of positive solution are given by Guo-krasnosel’skii fixed point theorem and Schauder’s fixed point theorem. 1. Introduction In this paper, we devote to the investigation of the following nonlinear fractional differential equation 0+ () + (, () , 0+ ()) + () = 0, 0 < < 1, 0+ (0) = +1 0+ (0) = ⋅ ⋅ ⋅ = +−2 0+ (0) = 0, 0+ (1) = −2 ∑ =1 0+ ( ), (1) where 0+ and 0+ are the standard Riemann-Liouville derivatives, ≥ 2,1 ≤ − ≤ − 1, − 1 < ≤ ,0 < 1 < 2 < ⋅⋅⋅ < −2 <1, 0<∑ =1 −−1 <1. e nonlinear term (, , V) is continuous and may be singular on both = 0, 1 and V =0; () ∈ 1 ([0, 1], R) permits sign- changing. Differential equation models can describe many nonlin- ear phenomena in applied mathematics, economics, finance, engineering, and physical and biological processes [1, 2]. In recent years, there has been a great deal of research on the existence and/or uniqueness of solution in studying FDEs nonlocal problems for their wide applications in modeling some important physical laws (see [3–16], for instance). In [3], the authors were concerned with the existence of monotone positive solutions to the following fractional-order multipoint boundary value problems 0+ () + () (, () , ()) = 0, ∈ (0, 1) , (0) = (0) = 0, (1) = ∑ =0 ( ), (2) where 0< 1 < 2 < ⋅⋅⋅ < <1, 2<<3, ≥0, and ∑ =1 −2 <1. e authors obtained the existence of monotone positive solutions and establish iterative schemes for approximating the solutions. In [4], the authors investigated the existence of positive solutions of the following fractional differential equation multipoint boundary value problems with changing sign nonlinearity Hindawi Journal of Function Spaces Volume 2019, Article ID 7161894, 10 pages https://doi.org/10.1155/2019/7161894
Transcript of Positive Solutions for a Higher-Order Semipositone...
Research ArticlePositive Solutions for a Higher-Order Semipositone NonlocalFractional Differential Equation with Singularities on Both Timeand Space Variable
Kemei Zhang
School of Mathematics Sciences Qufu Normal University Qufu 273165 Shandong China
Correspondence should be addressed to Kemei Zhang zhkm90126com
Received 6 August 2018 Accepted 15 January 2019 Published 3 February 2019
Academic Editor Shanhe Wu
Copyright copy 2019 Kemei ZhangThis is an open access article distributed under the Creative Commons Attribution License whichpermits unrestricted use distribution and reproduction in any medium provided the original work is properly cited
In this paper we consider the following higher-order semipositone nonlocal Riemann-Liouville fractional differential equation1198631205720+119909(119905) + 119891(119905 119909(119905) 1198631205730+119909(119905)) + 119890(119905) = 0 0 lt 119905 lt 11198631205730+119909(0) = 119863120573+10+ 119909(0) = sdot sdot sdot = 119863119899+120573minus20+ 119909(0) = 0 and 1198631205730+119909(1) = sum119898minus2119894=1 1205781198941198631205730+119909(120585119894)where 1198631205720+ and 1198631205730+ are the standard Riemann-Liouville fractional derivatives The existence results of positive solution are givenby Guo-krasnoselrsquoskii fixed point theorem and Schauderrsquos fixed point theorem
1 Introduction
In this paper we devote to the investigation of the followingnonlinear fractional differential equation
1198631205720+119909 (119905) + 119891 (119905 119909 (119905) 1198631205730+119909 (119905)) + 119890 (119905) = 00 lt 119905 lt 1
1198631205730+119909 (0) = 119863120573+10+ 119909 (0) = sdot sdot sdot = 119863119899+120573minus20+ 119909 (0) = 01198631205730+119909 (1) =
119898minus2sum119894=1
1205781198941198631205730+119909 (120585119894)
(1)
where 1198631205720+ and 1198631205730+ are the standard Riemann-Liouvillederivatives 120572 ge 2 1 le 120572 minus 120573 le 119899 minus 1 119899 minus 1 lt 120572 le 119899 0 lt1205851 lt 1205852 lt sdot sdot sdot lt 120585119898minus2 lt 1 0 lt sum119898119894=1 120578119894120585119894120572minus120573minus1 lt 1 Thenonlinear term 119891(119905 119906 V) is continuous and may be singularon both 119905 = 0 1 and V = 0 119890(119905) isin 1198711([0 1]R) permits sign-changing
Differential equation models can describe many nonlin-ear phenomena in applied mathematics economics financeengineering and physical and biological processes [1 2] In
recent years there has been a great deal of research on theexistence andor uniqueness of solution in studying FDEsnonlocal problems for their wide applications in modelingsome important physical laws (see [3ndash16] for instance)
In [3] the authors were concerned with the existence ofmonotone positive solutions to the following fractional-ordermultipoint boundary value problems
1198631205720+119906 (119905) + 119902 (119905) 119891 (119905 119906 (119905) 1199061015840 (119905)) = 0 119905 isin (0 1) 119906 (0) = 1199061015840 (0) = 01199061015840 (1) = 119898sum
119894=0
1205781198941199061015840 (120585119894) (2)
where 0 lt 1205851 lt 1205852 lt sdot sdot sdot lt 120585119898 lt 1 2 lt 120572 lt 3 120578119894 ge 0and sum119898119894=1 120578119894120585119894120572minus2 lt 1 The authors obtained the existence ofmonotone positive solutions and establish iterative schemesfor approximating the solutions
In [4] the authors investigated the existence of positivesolutions of the following fractional differential equationmultipoint boundary value problems with changing signnonlinearity
HindawiJournal of Function SpacesVolume 2019 Article ID 7161894 10 pageshttpsdoiorg10115520197161894
2 Journal of Function Spaces
1198631205720+119906 (119905) + 120582119891 (119905 119906 (119905)) = 0 119905 isin (0 1) 119906 (0) = 1199061015840 (0) = sdot sdot sdot = 119906(119899minus2) = 0
119906(119894) (1) = 119898minus2sum119895=0
1205781198951199061015840 (120585119895) (3)
where 120582 is a positive parameter 120572 ge 2 119899 minus 1 lt 120572 le 119899 119894 isinN 1 le 119894 le 119899 minus 2 120578119895 ge 0 (119895 = 1 2 sdot sdot sdot 119898 minus 2) 0 lt 1205851 lt 1205852 ltsdot sdot sdot lt 120585119898minus2 lt 1 and 119891may change sign and may be singular at119905 = 0 1 By employing the cone expansion and compressionfixed point theorem the existence of positive solutions wasobtained
In [5] the authors established the uniqueness of a positivesolution to the following higher-order fractional differentialequation
1198631205720+119906 (119905) + 119902 (119905) 119891 (119905 119906 (119905) 11986312058310+119906 (119905) sdot sdot sdot 119863120583119899minus20+ 119906 (119905))= 0 119905 isin (0 1)
119906 (0) = 1199061205831 (0) = sdot sdot sdot = 119906120583119899minus2 (0) = 01198631205830+119906 (1) =
119898minus2sum119894=1
1205781198941198631205730+119909 (120585119894) (4)
where 119891 [0 1] times (0 +infin)119899minus1 997888rarr [0 +infin) is continuous and119891(119905 1199061 119906119899minus1)may be singular at 1199061 = 0 119906119899minus1 = 0 and119902(119905) (0 1) 997888rarr [0 +infin) is continuous and may be singularat 119905 = 0 andor 119905 = 1 By using the fixed point theoremfor the mixed monotone operator the existence of uniquepositive solutions for above singular nonlocal boundary valueproblems of fractional differential equations is establishedThe nonlinear term 119891 in [11] is nonnegative
In [11] the authors studied the existence of positive solu-tions for the following nonlocal fractional-order differentialequations with sign-changing singular perturbation
minus 119863120572+20+ 119910 (119905) + 1198631205720+119910 (119905) = 119891 (119905 119910 (119905) 1198631205720+119910 (119905)) + 119890 (119905) 119905 isin (0 1)
1198861198631205720+119910 (0) minus 119887119863120572+10+ 119910 (0) =119898minus2sum119895=1
1198861198951198631205720+119910 (120585119895)
1198881198631205720+119910 (1) minus 119889119863120572+10+ 119910 (1) =119898minus2sum119895=1
1198871198951198631205720+119910 (120585119895)
(5)
where 0 lt 120572 le 1 119886 119888 ge 0 119887 119889 gt 0 0 lt 120585119895 lt 1 119886119895 119887119895 isin[0 +infin) 119891 (0 1) times [0infin) times (0infin) 997888rarr (0infin) is continuousand may be singular near the zero for the third argumentand 119890(119905) isin 1198711([0 1]R) may be sign-changing By meansof Schauderrsquos fixed point theorem the conditions for theexistence of positive solutions are established respectivelyfor the cases where the nonlinearity is positive negative andsemipositone
Motivated by the work mentioned above we consider thefractional-order singular nonlocal BVP (1) and establish the
existence results of positive solutions for (1) The main toolsused in this paper are Guo-krasnoselrsquoskii fixed point theoremand Schauderrsquos fixed point theorem For the concepts andproperties about the cone theory and the fixed point theoremone can refer to [17ndash21]
The rest of this paper is organized as follows in Section 2we present some useful preliminaries and lemmas The mainresults are given in Section 3 and Section 4 in which thesingular cases with respect to the time variables and spacevariables are discussed respectively
2 Preliminaries and Some Lemmas
Definition 1 (see [1 2]) The Riemann-Liouville fractionalintegral of order 120572 gt 0 of a function 119909(119905) (0infin) 997888rarr R
is given by
1198681205720+119909 (119905) = 1Γ (120572) int119905
0(119905 minus 119904)120572minus1 119909 (119904) 119889119904 (6)
provided the right-hand side is pointwisely defined on (0 infin)Definition 2 (see [1 2]) The Riemann-Liouville fractionalderivative of order 120572 gt 0 of a function 119909(119905) (0infin) 997888rarr R isgiven by
1198631205720+119909 (119905) = 1Γ (119899 minus 120572) ( 119889119889119905)119899 int1199050
119909 (119904)(119905 minus 119904)120572minus119899+1119889119904 (7)
provided the right-hand side is pointwisely defined on (0 infin)where 119899 = [120572] + 1 [120572] denotes the integer part of the number120572Lemma 3 (see [1 2]) e unique solution of the followinglinear Riemann-Liouville fractional differential equation oforder 120572 gt 0
1198631205720+119909 (119905) = 0 (8)
is
119909 (119905) = 1198881119905120572minus1 + 1198882119905120572minus2 + sdot sdot sdot + 119888119899119905120572minus119899 (9)
where 119888119894 = 1 2 119899 119899 = [120572] + 1 [120572] denotes the integer partof the number 120572Lemma 4 (see [1 2]) If 119909 isin 1198711(0 1) and 1198631205720+119909 isin 1198711(0 1)then
1198681205720+1198631205720+119909 (119905) = 119909 (119905) + 1198881119905120572minus1 + 1198882119905120572minus2 + sdot sdot sdot + 119888119899119905120572minus119899 (10)
where 119888119894 isin 119877 119894 = 1 2 119899 119899 = [120572] + 1The similar proof of the following three lemmas can be
traced to [5 12] in order to be convenient for readers to readwe now give the detailed process of proof for Lemma 5 theproofs for other two lemmas are omitted here
Journal of Function Spaces 3
Lemma 5 Let ℎ isin 1198711(0 1) and 0 lt sum119898119894=1 120578119894120585119894120572minus120573minus1 lt 1 thenthe unique solution of the problem
119863120572minus1205730+ 119909 (119905) + ℎ (119905) = 0 0 lt 119905 lt 1119909 (0) = 1199091015840 (0) = sdot sdot sdot = 119909(119899minus2) (0) = 0119909 (1) = 119898minus2sum
119894=1
120578119894119909 (120585119894)(11)
can be expressed uniquely by
119909 (119905) = int10119866 (119905 119904) ℎ (119904) 119889119904 (12)
where
119866 (119905 119904) = 1Γ (120572 minus 120573) 119892 (0)sdot
119892 (119904) [119905 (1 minus 119904)]120572minus120573minus1 0 le 119905 le 119904 le 1119892 (119904) [119905 (1 minus 119904)]120572minus120573minus1 minus (119905 minus 119904)120572minus120573minus1 119892 (0) 0 le 119904 le 119905 le 1
(13)
119892 (119904) = 1 minus sum119904le120585119894
120578119894 (120585119894 minus 1199041 minus 119904 )120572minus120573minus1 (14)
Proof By Lemma 4 the solution of (13) can be written as
119909 (119905) = minus 1Γ (120572 minus 120573) int119905
0(119905 minus 119904)120572minus120573minus1 ℎ (119904) 119889119904 + 1198621119905120572minus120573minus1
+ 1198622119905120572minus120573minus2 + sdot sdot sdot + 119862119899119905120572minus120573minus119899(15)
It follows from 119909(0) = 1199091015840(0) = sdot sdot sdot = 119909(119899minus2)(0) = 0 that 1198622 =sdot sdot sdot = 119862119899 = 0 ie119909 (119905) = minus 1Γ (120572 minus 120573) int
119905
0(119905 minus 119904)120572minus120573minus1 ℎ (119904) 119889119904 + 1198621119905120572minus120573minus1 (16)
thus
119909 (1) = minus 1Γ (120572 minus 120573) int1
0(1 minus 119904)120572minus120573minus1 ℎ (119904) 119889119904 + 1198621
119909 (120585119894) = minus 1Γ (120572 minus 120573) int120585119894
0(120585119894 minus 119904)120572minus120573minus1 ℎ (119904) 119889119904
+ 1198621120585119894120572minus120573minus1(17)
which together with the boundary value condition 119909(1) =sum119898minus2119894=1 120578119894119909(120585119894) implies that
minus 1Γ (120572 minus 120573) int1
0(1 minus 119904)120572minus120573minus1 ℎ (119904) 119889119904 + 1198621
= 119898minus2sum119894=1
120578119894 (minus 1Γ (120572 minus 120573) int120585119894
0(120585119894 minus 119904)120572minus120573minus1 ℎ (119904) 119889119904
+ 1198621120585119894120572minus120573minus1)
1198621(1 minus 119898minus2sum119894=1
120578119894120585120572minus120573minus1119894 ) = 1Γ (120572 minus 120573) int1
0(1 minus 119904)120572minus120573minus1
sdot ℎ (119904) 119889119904 minus 119898minus2sum119894=1
120578119894 int1205851198940 (120585119894 minus 119904)120572minus120573minus1 ℎ (119904) 119889119904Γ (120572 minus 120573)
1198621 = int10
(1 minus 119904)120572minus120573minus1(1 minus sum119898minus2119894=1 120578119894120585120572minus120573minus1119894 ) Γ (120572 minus 120573)ℎ (119904) 119889119904
minus 119898minus2sum119894=1
120578119894 int1205851198940(120585119894 minus 119904)120572minus120573minus1 ℎ (119904) 119889119904
Γ (120572 minus 120573) (1 minus sum119898minus2119894=1 120578119894120585120572minus120573minus1119894 ) (18)
ie
1198621 = 1119892 (0) Γ (120572 minus 120573) [int1
0(1 minus 119904)120572minus120573minus1 ℎ (119904) 119889119904
minus 119898minus2sum119894=1
120578119894 int1205851198940(120585119894 minus 119904)120572minus120573minus1 ℎ (119904) 119889119904]
= 1119892 (0) Γ (120572 minus 120573) [int1
0(1 minus 119904)120572minus120573minus1 119892 (119904) ℎ (119904) 119889119904
(19)
thus
119909 (119905) = minus 1Γ (120572 minus 120573) int119905
0(119905 minus 119904)120572minus120573minus1 ℎ (119904) 119889119904 + 1198621119905120572minus120573minus1
= 1119892 (0) Γ (120572 minus 120573)sdot int1199050[119905120572minus120573minus1 (1 minus 119904)120572minus120573minus1 119892 (119904) minus (119905 minus 119904)120572minus120573minus1 119892 (0)]
sdot ℎ (119904) 119889119904 minus 1119892 (0) Γ (120572 minus 120573)sdot int1119905119905120572minus120573minus1 (1 minus 119904)120572minus120573minus1 ℎ (119904) 119889119904 = int1
0119866 (119905 119904)
sdot ℎ (119904) 119889119904
(20)
Lemma 6 If 0 lt sum119898119894=1 120578119894120585119894120572minus120573minus1 lt 1 then the function 119892satisfies the following conditions
(1) 119892 is a nondecreasing function on [0 1](2) there exist1198721 ge 1198981 ge 0 such that1198981119905+119892(0) le 119892(119905) le1198721119905 + 119892(0) for any 119905 isin [0 1] where 1198721 = sup0lt119905le1((119892(119905) minus119892(0))119905)1198981 = inf0lt119905le1((119892(119905) minus 119892(0))119905)
Remark 7 It is easy to prove that 1198981 gt 0Lemma 8 e function 119866(119905 119904) defined by (13) has the follow-ing properties
(1) 119866(119905 119904) gt 0 for any (119905 119904) isin (0 1)
4 Journal of Function Spaces
(2) 119866(119905 119904) ge (1198981119892(0)Γ(120572minus120573))119904(1 minus 119904)120572minus120573minus1119905120572minus120573minus1 for any(119905 119904) isin [0 1](3) 119866(119905 119904) le ((1198721 + 119892(0)[120572 minus 120573])119892(0)Γ(120572 minus120573))119904(1 minus 119904)120572minus120573minus1119905120572minus120573minus1 for any (119905 119904) isin [0 1](4) 119866(119905 119904) le 119872119904(1 minus 119904)120572minus120573minus1
where [120572 minus 120573] denotes the integer part of the number 120572 minus 120573119872 = 1198721 + 119892 (0) (120572 minus 120573 minus 1)119892 (0) Γ (120572 minus 120573) (21)
Set V(119905) = 1198631205730+119909(119905) then (1) can be transformed into the
following form
119863120572minus1205730+ V (119905) + 119891 (119905 1198681205730+V (119905) V (119905)) + 119890 (119905) = 00 lt 119905 lt 1
V (0) = V1015840 (0) = sdot sdot sdot = V(119899minus2) (0) = 0V (1) = 119898minus2sum
119894=1
120578119894V (120585119894) (22)
FromLemma 5we know that the solution V(119905) of (22) satisfiesV (119905) = int1
0119866 (119905 119904) [119891 (119904 1198681205730+V (119904) V (119904)) + 119890 (119904)] 119889119904 (23)
Lemma 9 (see [17 18]) Suppose that 119864 is a Banach space and119863 sub 119864 is a bounded convex closed set the operator 119860 119863 997888rarr119863 is completely continuous then 119860 has one fixed point on119863
Lemma 10 (see [20] (Guo-krasnoselrsquoskii fixed point theo-rem)) Let Ω1 and Ω1 be two bounded open sets in Banachspace119864 such that 120579 isin Ω1 andΩ1 sub Ω2119860 119875⋂(Ω2Ω1) 997888rarr 119875a completely continuous operator where 120579 denotes the zeroelement of119864 and119875 a cone of119864 Suppose that one of the followingconditions
(i) 119860119909 ge 119909 forall119909 isin 119875 cap 120597Ω1 and 119860119909 le 119909 forall119909 isin119875 cap 120597Ω2(ii) 119860119909 le 119909 forall119909 isin 119875 cap 120597Ω1 and 119860119909 ge 119909 forall119909 isin119875 cap 120597Ω2
holds en 119860 has at least one fixed point in 119875 cap (Ω2 Ω1)3 Main Result I 119891 Is Singular with Respect tothe Time Variables
Let119864 = 119862[0 1] 119906 = max0le119905le1|119906(119905)| then (119864 sdot) is a Banachspace Set
119875 = V isin 119864 V (119905) ge 11989811198721 + 119892 (0) [120572 minus 120573] 119905120572minus120573minus1 V 119905
isin [0 1] (24)
where [120572 minus 120573] denotes the integer part of the number 120572 minus 120573Then 119875 sub 119864 is a positive cone of 119864 For convenience we listsome conditions which will be used in this section
(1198671) For any (119905 119909 119910) isin (0 1) times [0 +infin) times [0 +infin)0 le 119891 (119905 119909 119910) le 120601 (119905) (120588 (119909) + ℎ (119910)) (25)
where 120601 isin 119862(0 1) 120601(119905) gt 0 on (0 1) 120588(119909) gt 0 is continuousand increasing on [0 +infin) ℎ(119909) gt 0 is continuous anddecreasing on [0 +infin)
(1198672)0 lt int10119904 (1 minus 119904)120572minus120573minus1 (120601 (119904) + 119890+ (119904) + 119890minus (119904)) 119889119904
lt +infin(26)
where 119890+ = max119890(119905) 0 119890minus = minusmin119890(119905) 0 are the positivepart and negative part of 119890(119905) respectively
(1198673) There exists 1199031 gt 0 such that
120588(1199031 + 119908Γ (120573) )int10119904 (1 minus 119904)120572minus120573minus1 120601 (119904) 119889119904
+ int10119904 (1 minus 119904)120572minus120573minus1 [ℎ (0) 120601 (119904) + 119890+ (119904)] 119889119904
lt 1199031119872(27)
where119908(119905) is the solution of the following linear equation
119863120572minus1205730+ V (119905) + 119890minus (119905) = 0 0 lt 119905 lt 1V (0) = V1015840 (0) = sdot sdot sdot = V(119899minus2) (0) = 0V (1) = 119898minus2sum
119894=1
120578119894V (120585119894) (28)
ie 119908(119905) = int10119866(119905 119904)119890minus(119904)119889119904
(1198674) There exists [119886 119887] sub (0 1) such that
lim119909+119910997888rarr+infin
119891 (119905 119909 119910)119909 + 119910 gt 119871 (29)
uniformly holds for 119905 isin [119886 119887] where119871 = 2Γ (120572 minus 120573) 119892 (0) (1198721 + 119892 (0) [120572 minus 120573])
11989812119886120572minus120573minus1 int119887119886 119904120572minus120573 (1 minus 119904)120572minus120573minus1 119889119904 (30)
For any V isin 119875 let[V (119905) minus 119908 (119905)]lowast = max V (119905) minus 119908 (119905) 0 (31)
and define operator
119865V (119905) = int10119866 (119905 119904)
sdot [119891 (119904 1198681205730+ [V (119904) minus 119908 (119904)]lowast [V (119904) minus 119908 (119904)]lowast)+ 119890+ (119904)] 119889119904
(32)
From condition (1198671) and (1198672) it is easy to know that 119865 is welldefined
Journal of Function Spaces 5
Lemma 11 119865 119875 997888rarr 119875 is a completely continuous operator
Proof For any V isin 119875 it follows from Lemma 8 that
119865V (119905) = int10119866 (119905 119904)
sdot [119891 (119904 1198681205730+ [V (119904) minus 119908 (119904)]lowast [V (119904) minus 119908 (119904)]lowast)+ 119890+ (119904)] 119889119904 le 1198721 + 119892 (0) [120572 minus 120573]119892 (0) Γ (120572 minus 120573)sdot 119905120572minus120573minus1 int1
0119904 (1 minus 119904)120572minus120573minus1
sdot [119891 (119904 1198681205730+ [V (119904) minus 119908 (119904)]lowast [V (119904) minus 119908 (119904)]lowast)+ 119890+ (119904)] 119889119904 le 1198721 + 119892 (0) [120572 minus 120573]119892 (0) Γ (120572 minus 120573) int1
0119904 (1
minus 119904)120572minus120573minus1sdot [119891 (119904 1198681205730+ [V (119904) minus 119908 (119904)]lowast [V (119904) minus 119908 (119904)]lowast)+ 119890+ (119904)] 119889119904
119865V (119905) = int10119866 (119905 119904)
sdot [119891 (119904 1198681205730+ [V (119904) minus 119908 (119904)]lowast [V (119904) minus 119908 (119904)]lowast)+ 119890+ (119904)] 119889119904 ge 1198981119892 (0) Γ (120572 minus 120573)119905120572minus120573minus1 int
1
0119904 (1
minus 119904)120572minus120573minus1sdot [119891 (119904 1198681205730+ [V (119904) minus 119908 (119904)]lowast [V (119904) minus 119908 (119904)]lowast)+ 119890+ (119904)] 119889119904
(33)
which deduce that 119865V(119905) ge (1198981(1198721 + 119892(0)[120572 minus120573]))119905120572minus120573minus1119865V ie 119865 119875 997888rarr 119875Let 119861 sub 119875 be a bounded set ie there exists 1198711 gt 0 such
that V le 1198711 for any V isin 119861 then0 le [V (119905) minus 119908 (119905)]lowast le 1198711 + int1
0119866 (119905 119904) 119890minus (119904) 119889119904 le 1198711
+ int10119872119904 (1 minus 119904)120572minus120573minus1 119890minus (119904) 119889119904 ≐
(34)
119865V (119905) = int10119866 (119905 119904)
sdot [119891 (119904 1198681205730+ [V (119904) minus 119908 (119904)]lowast [V (119904) minus 119908 (119904)]lowast)+ 119890+ (119904)] 119889119904 le int1
0119866 (119905 119904)
sdot [120601 (119904) (120588 (1198681205730+) + ℎ (0)) + 119890+ (119904)] 119889119904
le 119872(120588( 120573Γ (120573)) + ℎ (0))int10119904 (1 minus 119904)120572minus120573minus1
sdot 120601 (119904) 119889119904 +119872int10119904 (1 minus 119904)120572minus120573minus1 119890+ (119904) 119889119904
lt +infin(35)
therefore
119865V le 119872(120588( 120573Γ (120573)) + ℎ (0))sdot int10119904 (1 minus 119904)120572minus120573minus1 120601 (119904) 119889119904
+119872int10119904 (1 minus 119904)120572minus120573minus1 119890+ (119904) 119889119904
(36)
for any V isin 119861 which implies that 119865 is uniformly boundedFrom (1198672) the absolutely continuity of integral and the
uniformly continuity of119866(119905 119904) on [0 1] we know that for any120576 gt 0 exist120575 gt 0 such that
int1205750119904 (1 minus 119904)120572minus120573minus1 120601 (119904) 119889119904lt 1205766119872(120588 (120573Γ (120573)) + ℎ (0))
(37)
int1205751minus120575
119904 (1 minus 119904)120572minus120573minus1 120601 (119904) 119889119904lt 1205766119872(120588 (120573Γ (120573)) + ℎ (0))
(38)
and 1003816100381610038161003816119866 (1199051 119904) minus 119866 (1199052 119904)1003816100381610038161003816 lt 1205763119897 (120588 (120573Γ (120573)) + ℎ (0)) (39)
for any 1199051 1199052 119904 isin [0 1] with |1199051 minus 1199052| lt 120575(37)-(39) together with Lemma 8 imply that
1003816100381610038161003816119865V (1199051) minus 119865V (1199052)1003816100381610038161003816 le int10
1003816100381610038161003816119866 (1199051 119904) minus 119866 (1199052 119904)1003816100381610038161003816sdot 119891 (119904 1198681205730+ [V (119904) minus 119908 (119904)]lowast [V (119904) minus 119908 (119904)]lowast) 119889119904= int1205750
1003816100381610038161003816119866 (1199051 119904) minus 119866 (1199052 119904)1003816100381610038161003816sdot 119891 (119904 1198681205730+ [V (119904) minus 119908 (119904)]lowast [V (119904) minus 119908 (119904)]lowast) 119889119904+ int11minus120575
1003816100381610038161003816119866 (1199051 119904) minus 119866 (1199052 119904)1003816100381610038161003816sdot 119891 (119904 1198681205730+ [V (119904) minus 119908 (119904)]lowast [V (119904) minus 119908 (119904)]lowast) 119889119904+ int1minus120575120575
1003816100381610038161003816119866 (1199051 119904) minus 119866 (1199052 119904)1003816100381610038161003816sdot 119891 (119904 1198681205730+ [V (119904) minus 119908 (119904)]lowast [V (119904) minus 119908 (119904)]lowast) 119889119904
6 Journal of Function Spaces
le 2119872int1205750119904 (1 minus 119904)120572minus120573minus1 120601 (119904)
sdot (120588 (1198681205730+) + ℎ (0)) 119889119904 + 2119872int11minus120575
119904 (1 minus 119904)120572minus120573minus1sdot 120601 (119904) (120588 (1198681205730+) + ℎ (0)) 119889119904+ int1minus120575120575
1003816100381610038161003816119866 (1199051 119904) minus 119866 (1199052 119904)1003816100381610038161003816sdot 119897 (120588 (1198681205730+) + ℎ (0)) 119889119904 lt 1205763 + 1205763 + 1205763 = 120576
(40)
for any 1199051 1199052 isin [0 1] with |1199051 minus 1199052| lt 120575 and any V isin 119861 where119897 = max120575le119905le1minus120575120601(119905) which deduces that 119865 is equicontinuouson [0 1]Thus according to Ascoli-Arzela theorem we knowthat 119865119861 is a relatively compact set and that 119865 is a completelycontinuous operator
Theorem 12 Suppose that (1198671) minus (1198674) hold then the FVP (1)has at least one positive solution
Proof For any V isin 1205971198751199031 where 1198751199031 = V isin 119875 | V lt 1199031 by(32) Lemma 8 and condition (1198673) one can get that
119865V (119905) = int10119866 (119905 119904)
sdot [119891 (119904 1198681205730+ [V (119904) minus 119908 (119904)]lowast [V (119904) minus 119908 (119904)]lowast)+ 119890+ (119904)] 119889119904 le int1
0119866 (119905 119904)
sdot [120601 (119904) (120588 (1198681205730+ (V (119904) minus 119908 (119904)) + ℎ (0))+ 119890+ (119904)] 119889119904 le int1
0119866 (119905 119904)
sdot [120601 (119904) (120588 (1198681205730+ (V + 119908)) + ℎ (0)) + 119890+ (119904)] 119889119904= 119872int1
0119904 (1 minus 119904)120572minus120573minus1
sdot [120601 (119904) (120588 (1198681205730+ (1199031 + 119908)) + ℎ (0)) + 119890+ (119904)] 119889119904= 119872int1
0119904 (1 minus 119904)120572minus120573minus1
sdot [120601 (119904) (120588(1199031 + 119908Γ (120573) ) + ℎ (0)) + 119890+ (119904)] 119889119904lt 1199031 = V
(41)
ie 119865V le V V isin 1205971198751199031 By condition (1198674) exist119883 gt 0 such that
119891 (119905 119909 119910) gt 119871 (119909 + 119910) (42)
for any 119909 gt 119883 and any 119905 isin [119886 119887] Choose 1199032 such that
1199032 gt max1199031 21198881 2 (1198721 + 119892 (0) [120572 minus 120573])1198831198981119886120572minus120573minus1 (43)
where
1198881 = (1198721 + 119892 (0) [120572 minus 120573])2 sdot int10 (1 minus 119904)120572minus120573minus1 119890minus (119904) 1198891199041198981Γ (120572 minus 120573) 119892 (0) (44)
For any V isin 1205971198751199032 where 1198751199032 = V isin 119875 | V lt 1199032 Because119908 (119905) = int1
0119866 (119905 119904) 119890minus (119904) 119889119904
le 1198721 + 119892 (0) [120572 minus 120573]Γ (120572 minus 120573) 119892 (0)sdot 119905120572minus120573minus1 int1
0(1 minus 119904)120572minus120573minus1 119890minus (119904) 119889119904
le (1198721 + 119892 (0) [120572 minus 120573])2 sdot int10 (1 minus 119904)120572minus120573minus1 119890minus (119904) 1198891199041198981Γ (120572 minus 120573) 119892 (0) 1199032sdot V (119905) = 11988811199032 V (119905)
(45)
so we have
V (119905) minus 119908 (119905) ge (1 minus 11988811199032) V (119905) ge 12V (119905) (46)
and then for 119905 isin [119886 119887]V (119905) minus 119908 (119905) ge 12V (119905) ge 12 sdot 1198981119886120572minus120573minus11198721 + 119892 (0) [120572 minus 120573] sdot V
= 12 sdot 1198981119886120572minus120573minus11198721 + 119892 (0) [120572 minus 120573] sdot 1199032 gt 119883(47)
follows from (43) and the definition of cone 119875From (42) (43) and (47) one can obtain that
119865V (119905) = int10119866 (119905 119904)
sdot [119891 (119904 1198681205730+ [V (119904) minus 119908 (119904)]lowast [V (119904) minus 119908 (119904)]lowast)+ 119890+ (119904)] 119889119904 ge int119887
119886119866 (119905 119904) 119871 (1198681205730+ (V (119904) minus 119908 (119904))
+ (V (119904) minus 119908 (119904))) + 119890+ (119904) ]119889119904 ge 119871int119887119886119866 (119905 119904)
sdot 12V (119904) 119889119904 ge 1198712 sdot 1198981Γ (120572 minus 120573) 119892 (0) sdot int119887
119886119904 (1
minus 119904)120572minus120573minus1 119905120572minus120573minus1V (119904) 119889119904 ge 1198712 sdot 1198981119886120572minus120573minus1Γ (120572 minus 120573) 119892 (0)
Journal of Function Spaces 7
sdot 11989811198721 + 119892 (0) [120572 minus 120573] sdot int119887
119886119904 (1 minus 119904)120572minus120573minus1 119904120572minus120573minus1119889119904
sdot V = 1198712 sdot 1198981119886120572minus120573minus1Γ (120572 minus 120573) 119892 (0) sdot11989811198721 + 119892 (0) [120572 minus 120573]
sdot int119887119886(1 minus 119904)120572minus120573minus1 119904120572minus120573119889119904 sdot V ge 1198712 sdot 1198981119886120572minus120573minus1Γ (120572 minus 120573) 119892 (0)
sdot 11989811198721 + 119892 (0) [120572 minus 120573] sdot int119887
119886(1 minus 119904)120572minus120573minus1 119904120572minus120573119889119904 sdot 1199032
= 1199032 = V (48)
ie 119865V ge V V isin 1205971198751199032 It follows from Lemma 10 that 119865 has at least fixed point
V1 isin 1198751199032 1198751199031 ie V1 satisfies119863120572minus1205730+ V1 (119905) + 119891 (119905 1198681205730+ (V1 (119905) minus 119908 (119905)) V1 (119905) minus 119908 (119905))
+ 119890 (119905) = 0 0 lt 119905 lt 1V1 (0) = V10158401 (0) = sdot sdot sdot = V(119899minus2)1 (0) = 0V1 (1) = 119898minus2sum
119894=1
120578119894V1 (120585119894) (49)
Set V1(119905) = V1(119905) minus 119908(119905) noticing that V1(119905) 119908(119905) are thesolutions of BVP (32) and (49) respectively therefore we canconclude that V1(119905) is a positive solution of (22) Let 1199091(119905) =1198681205730+V1(119905) then 1199091(119905) is a positive solution of the nonlinearfractional differential equations (1)
4 Main Result II 119891 Is Singular withRespect to Both the Time Variables andthe Space Variable
In this section we always suppose that the following condi-tion holds
(1198675) 119891(119905 119906 V) (0 1) times [0infin) times (0infin) 997888rarr [0infin)is continuous there exist 120576 isin (0 1) and 1205831 1205832 isin 119862+[0 1]1205831(119905) equiv 0 for 119905 isin [0 1] such that
1205831 (119905)(119909 + 119910)120576 le 119891 (119905 119909 119910) le1205832 (119905)(119909 + 119910)120576 (50)
for any (119909 119910) isin [0infin) times (0infin) 119905 isin (0 1) where119862+ [0 1] = 119909 (119905) isin 119862 [0 1] | 119909 (119905) ge 0 119905 isin [0 1] (51)
Set
120593 (119905) = int10119866 (119905 119904) 119890 (119904) 119889119904 119905 isin [0 1] (52)
it follows from Lemma 5 that 120593(119905) is the solution of thefollowing linear equation
119863120572minus1205730+ V (119905) + 119890 (119905) = 0 0 lt 119905 lt 1V (0) = V1015840 (0) = sdot sdot sdot = V(119899minus2) (0) = 0V (1) = 119898minus2sum
119894=1
120578119894V (120585119894) (53)
Denote
1198861 (119905) = int10119866 (119905 119904) 1205831 (119904) 119889119904
1198862 (119905) = int10119866 (119905 119904) 1205832 (119904) 119889119904
120593lowast = inf0le119905le1
120593 (119905) 120593lowast = sup0le119905le1
120593 (119905) 1198861lowast = min
0le119905le11198861 (119905)
119886lowast1 = max0le119905le1
1198861 (119905) 1198862lowast = min
0le119905le11198862 (119905)
119886lowast2 = max0le119905le1
1198862 (119905)
(54)
Clearly 119886lowast119895 ge 119886119895lowast gt 0 119895 = 1 2Theorem 13 Suppose that the condition (H5) holds and 120593lowast ge0 en the FVP (1) has at least one positive solution
Proof Because 120593lowast ge 0 so we can choose 119877 gt 0 large enoughsuch that
119903119877120576 + 120593lowast ge 1119877119877120576 (119886lowast2 + 120593lowast) le 119877
(55)
where 119903 = 1198861lowast[1 + 1(1 + 120573Γ(120573))]120576 In fact since
lim119909997888rarr+infin
119909120576119909 = 0 lt min119903 1119886lowast2 + 120593lowast (56)
there exists 1198831 gt 0 such that 119909120576119909 lt min119903 1(119886lowast2 + 120593lowast) forany 119909 gt 1198831 ie
119903119909120576 ge 1119909 119909120576 (119886lowast2 + 120593lowast) le 119909 (57)
for any 119909 gt 1198831 If 120593lowast gt 0 then from lim119909997888rarr+infin[1119909 minus 119903119909120576] =0 lt 120593lowast one can get that there exists 1198832 gt 0 such that 1119909 minus119903119909120576 lt 120593lowast for any 119909 gt 1198832 ie119903119909120576 + 120593lowast ge 1119909 (58)
8 Journal of Function Spaces
for any 119909 gt 1198832 By (57) (58) we can choose 119877 gt max1198831 1198832such that 119877 satisfies (55) Set
119863 = V isin 119862+ [0 1] 1119877 le V (119905) le 119877 119905 isin [0 1] (59)
For any V isin 1198631 from (23) we have
1198681205730+V (119905) = 1Γ (120573) int119905
0(119905 minus 119904)120573minus1 V (119904) 119889119904
le 119877Γ (120573) int119905
0(119905 minus 119904)120573minus1 119889119904 = 119877119905120573120573Γ (120573) le 119877120573Γ (120573)
(60)
1198681205730+V (119905) = 1Γ (120573) int119905
0(119905 minus 119904)120573minus1 V (119904) 119889119904
ge 1119877Γ (120573) int119905
0(119905 minus 119904)120573minus1 119889119904 = 119905120573119877120573Γ (120573)
(61)
It follows from (60) (61) and (H5) that
1205831 (119905)119877120576 (1 + 1120573Γ (120573))120576 le1205831 (119905)
(V (119905) + 1198681205730+V (119905))120576le 119891 (119905 1198681205730+V (119905) V (119905))
119891 (119905 1198681205730+V (119905) V (119905)) le 1205832 (119905)(V (119905) + 1198681205730+V (119905))120576
le 1205832 (119905)(1119877 + 119905120573119877120573Γ (120573))120576le 1198771205761205832 (119905)
(62)
ie
1205831 (119905)119877120576 (1 + 1120573Γ (120573))120576 le 119891 (119905 1198681205730+V (119905) V (119905)) le 1198771205761205832 (119905) (63)
And then
119879V (119905) = int10119866 (119905 119904) [119891 (119904 1198681205730+V (119904) V (119904)) + 119890 (119904)] 119889119904
= int10119866 (119905 119904) 119891 (119904 1198681205730+V (119904) V (119904)) 119889119904 + 120593 (119905)
le int10119866 (119905 119904) 119891 (119904 1198681205730+V (119904) V (119904)) 119889119904 + 120593lowast
le 119877120576 int10119866 (119905 119904) 1205832 (119904) 119889119904 + 120593lowast lt +infin
(64)
which deduces that the operator 119879 is well definedNow we shall prove that 119879 119863 997888rarr 119863 For V isin 119863 it
is easy to see that 119879V(119905) isin 119862+[0 1] and by (55) (63) we canobtain that
119879V (119905) = int10119866 (119905 119904) [119891 (119904 1198681205730+V (119904) V (119904)) + 119890 (119904)] 119889119904
= int10119866 (119905 119904) 119891 (119904 1198681205730+V (119904) V (119904)) 119889119904 + 120593 (119905)
ge 1119877120576 (1 + 1120573Γ (120573))120576 int
1
0119866 (119905 119904) 1205831 (119904) 119889119904
+ 120593 (119905) ge 1119877120576 (1 + 1120573Γ (120573))120576 1198861lowast + 120593lowast ge
1119877
(65)
119879V (119905) = int10119866 (119905 119904) [119891 (119904 1198681205730+V (119904) V (119904)) + 119890 (119904)] 119889119904
= int10119866 (119905 119904) 119891 (119904 1198681205730+V (119904) V (119904)) 119889119904 + 120593 (119905)
le 119877120576 int10119866 (119905 119904) 1205832 (119904) 119889119904 + 120593lowast le 119877120576119886lowast2 + 120593lowast
le 119877120576 (119886lowast2 + 120593lowast) le 119877
(66)
ie 119879 119863 997888rarr 119863Next let us prove that 119879 119863 997888rarr 119863 is completely
continuousFor any V119899 sub 119863 V0 isin 119863 and V119899 997888rarr V0The continuity
of 119891 deduces that
119891 (119905 1198681205730+V119899 (119905) V119899 (119905)) 997888rarr 119891(119905 1198681205730+V0 (119905) V0 (119905)) 119899 997888rarr infin (67)
and it follows from (63) that100381610038161003816100381610038161003816119891 (119905 1198681205730+V119899 (119905) V119899 (119905)) minus 119891 (119905 1198681205730+V0 (119905) V0 (119905))100381610038161003816100381610038161003816le 2119877120598max max
0le119905le11205831 (119905) max
0le119905le11205832 (119905) (68)
By using the Lebesgue dominated convergence theorem weobtain that
lim119899997888rarrinfin
1003817100381710038171003817119879V119899 minus 119879V01003817100381710038171003817 = lim119899997888rarrinfin
max0le119905le1
100381610038161003816100381610038161003816100381610038161003816int1
0119866 (119905 119904) [119891 (119904 1198681205730+V119899 (119904) V119899 (119904)) minus 119891 (119904 1198681205730+V0 (119904) V0 (119904))] 119889119904
100381610038161003816100381610038161003816100381610038161003816le lim119899997888rarrinfin
max0le119905le1
int10119866 (119905 119904) 100381610038161003816100381610038161003816119891 (119904 1198681205730+V119899 (119904) V119899 (119904)) minus 119891 (119904 1198681205730+V0 (119904) V0 (119904))100381610038161003816100381610038161003816 119889119904
Journal of Function Spaces 9
= max0le119905119904le1
119866 (119905 119904) sdot int10
lim119899997888rarrinfin
100381610038161003816100381610038161003816119891 (119904 1198681205730+V119899 (119904) V119899 (119904)) minus 119891 (119904 1198681205730+V0 (119904) V0 (119904))100381610038161003816100381610038161003816 119889119904 = 0(69)
and this implies that 119879 is a continuous operatorNow we shall prove that 119879 119863 997888rarr 119863 is compact For any
V isin 119863 119879V isin 119863 which deduces that 1119877 le 119879V(119905) le 119877 for119905 isin [0 1] ie 119879 is uniformly boundedSince 119866(119905 119904) isin 119862([0 1] times [0 1]) it is also uniformly
continuous on [0 1] times [0 1] and then for any 120576 gt 0 exist120575 gt 0st for any (1199051 119904) (1199052 119904) isin [0 1] times [0 1] with |1199051 minus 1199052| lt 120575 wealways have
1003816100381610038161003816119866 (1199051 119904) minus 119866 (1199052 119904)1003816100381610038161003816 lt 120576119877120576 int101205832 (119904) 119889119904 + int10 |119890 (119904)| 119889119904
(70)
Thus one can obtain by virtue of (63) (70) that
1003816100381610038161003816119879V (1199051) minus 119879V (1199052)1003816100381610038161003816 =100381610038161003816100381610038161003816100381610038161003816int1
0[119866 (1199051 119904) minus 119866 (1199052 119904)]
sdot [119891 (119904 119868120573V (119904) V (119904)) + 119890 (119904)] 119889119904100381610038161003816100381610038161003816100381610038161003816 le int1
0
1003816100381610038161003816119866 (1199051 119904)minus 119866 (1199052 119904)1003816100381610038161003816 [100381610038161003816100381610038161003816119891 (119904 1198681205730+V (119904) V (119904))100381610038161003816100381610038161003816 + |119890 (119904)|] 119889119904lt [119877120576 int1
01205832 (119904) 119889119904 + int1
0|119890 (119904)| 119889119904]
sdot 120576119877120576 int101205832 (119904) 119889119904 + int10 |119890 (119904)| 119889119904
= 120576
(71)
for any 1199051 1199052 isin [0 1] with |1199051 minus 1199052| lt 120575 and any V isin 1198631 whichshows that 119879 119863 997888rarr 119863 is equicontinuous Thus Arzela-Ascoli theorem guarantees that 119879 119863 997888rarr 119863 is completelycontinuous Existence of at least one fixed point V1 isin 119863follows from Lemma 6 ie 1199061(119905) = 1198681205730+V1(119905) is a positivesolution of differential equation (1) which satisfies
1119877Γ (120572)119905120572 le 1199061 (119905) = 1Γ (120572) int119905
0(119905 minus 119904)120572minus1 V1 (119904) 119889119904
le 119877Γ (120572) 119905120572(72)
Data Availability
The data used to support the findings of this study areavailable from the corresponding author upon request
Conflicts of Interest
The author declares that there are no conflicts of interestregarding the publication of this paper
Acknowledgments
This research was supported by Shandong Provincial Nat-ural Science Foundation of China (no ZR2016FM10) andNational Natural Science Foundation of China (no 11571197)
References
[1] I Podlubny Fractional Differential Equations Mathematics inScience and Engineering Academic Press NewYork NY USA1999
[2] A A Kilbas H M Srivastava and J J Nietoeory and Appli-cational Differential Equations Elsevier Amsterdam Nether-lands 2006
[3] T Hu Y Sun and W Sun ldquoExistence of positive solutions for athird-order multipoint boundary value problem and extensionto fractional caserdquo Boundary Value Problems vol 2016 article197 2016
[4] Y Jia and X Zhang ldquoPositive solutions for a class of fractionaldifferential equation multi-point boundary value problemswith changing sign nonlinearityrdquo Applied Mathematics andComputation vol 47 no 1-2 pp 15ndash31 2015
[5] X Hao ldquoPositive solution for singular fractional differentialequations involving derivativesrdquo Advances in Difference Equa-tions vol 2016 article 139 2016
[6] Y Zou and G He ldquoOn the uniqueness of solutions for a class offractional differential equationsrdquo Applied Mathematics Lettersvol 74 pp 68ndash73 2017
[7] X Hao H Wang L Liu and Y Cui ldquoPositive solutionsfor a system of nonlinear fractional nonlocal boundary valueproblems with parameters and p-Laplacian operatorrdquoBoundaryValue Problems vol 2017 article 182 2017
[8] X Zhang ldquoPositive solutions for a class of singular fractionaldifferential equation with infinite-point boundary value condi-tionsrdquo Applied Mathematics Letters vol 39 pp 22ndash27 2015
[9] K M Zhang ldquoOn a sign-changing solution for some fractionaldifferential equationsrdquo Boundary Value Problems vol 2017article 59 2017
[10] Y L Guan Z Q Zhao and X L Lin ldquoOn the existence ofpositive solutions and negative solutions of singular fractionaldifferential equations via global bifurcation techniquesrdquoBound-ary Value Problems vol 2016 article 141 2016
[11] X G Zhang Y H Wu and L Caccetta ldquoNonlocal fractionalorder differential equations with changing-sign singular pertur-bationrdquo Applied Mathematical Modelling vol 56 pp 116ndash1262015
[12] YWang and L S Liu ldquoPositive solutions for fractionalm-pointboundary value problems in Banach spacesrdquo ActaMathematicaScientia vol 32 no 1 pp 246ndash256 2012
[13] Y Cui and Y Zou ldquoAn existence and uniqueness theorem for asecond order nonlinear system with coupled integral boundaryvalue conditionsrdquo Applied Mathematics and Computation vol256 pp 438ndash444 2015
10 Journal of Function Spaces
[14] X Q Zhang L Wang and Q S Wang ldquoExistence of positivesolutions for a class of nonlinear fractional differential equa-tions with integral boundary conditionsrdquo Applied Mathematicsand Computation vol 226 pp 708ndash718 2014
[15] XG Zhang L S Liu YHWu andBWiwatanapataphee ldquoThespectral analysis for a singular fractional differential equationwith a signedmeasurerdquoApplied Mathematics and Computationvol 257 pp 252ndash263 2015
[16] Y Wang and L Liu ldquoPositive solutions for a class of fractional3-point boundary value problems at resonancerdquo Advances inDifference Equations vol 13 article 13 2017
[17] M A Krasnoselskii and P P Zabreiko Geometrical Methods ofNonlinear Analysis Springer New York NY USA 1984
[18] K Deimling Nonlinear Functional Analysis Springer BerlinGermany 1985
[19] K Zhang ldquoNontrivial solutions of fourth-order singularboundary value problems with sign-changing nonlinear termsrdquoTopological Methods in Nonlinear Analysis vol 40 no 1 pp 53ndash70 2012
[20] D J Guo and V Lakshmikantham Nonlinear Problems inAbstract Cones Academic Press Massachusetts Mass USA1988
[21] J Liu and Z Zhao ldquoMultiple solutions for impulsive problemswith non-autonomous perturbationsrdquo Applied MathematicsLetters vol 64 pp 143ndash149 2017
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2 Journal of Function Spaces
1198631205720+119906 (119905) + 120582119891 (119905 119906 (119905)) = 0 119905 isin (0 1) 119906 (0) = 1199061015840 (0) = sdot sdot sdot = 119906(119899minus2) = 0
119906(119894) (1) = 119898minus2sum119895=0
1205781198951199061015840 (120585119895) (3)
where 120582 is a positive parameter 120572 ge 2 119899 minus 1 lt 120572 le 119899 119894 isinN 1 le 119894 le 119899 minus 2 120578119895 ge 0 (119895 = 1 2 sdot sdot sdot 119898 minus 2) 0 lt 1205851 lt 1205852 ltsdot sdot sdot lt 120585119898minus2 lt 1 and 119891may change sign and may be singular at119905 = 0 1 By employing the cone expansion and compressionfixed point theorem the existence of positive solutions wasobtained
In [5] the authors established the uniqueness of a positivesolution to the following higher-order fractional differentialequation
1198631205720+119906 (119905) + 119902 (119905) 119891 (119905 119906 (119905) 11986312058310+119906 (119905) sdot sdot sdot 119863120583119899minus20+ 119906 (119905))= 0 119905 isin (0 1)
119906 (0) = 1199061205831 (0) = sdot sdot sdot = 119906120583119899minus2 (0) = 01198631205830+119906 (1) =
119898minus2sum119894=1
1205781198941198631205730+119909 (120585119894) (4)
where 119891 [0 1] times (0 +infin)119899minus1 997888rarr [0 +infin) is continuous and119891(119905 1199061 119906119899minus1)may be singular at 1199061 = 0 119906119899minus1 = 0 and119902(119905) (0 1) 997888rarr [0 +infin) is continuous and may be singularat 119905 = 0 andor 119905 = 1 By using the fixed point theoremfor the mixed monotone operator the existence of uniquepositive solutions for above singular nonlocal boundary valueproblems of fractional differential equations is establishedThe nonlinear term 119891 in [11] is nonnegative
In [11] the authors studied the existence of positive solu-tions for the following nonlocal fractional-order differentialequations with sign-changing singular perturbation
minus 119863120572+20+ 119910 (119905) + 1198631205720+119910 (119905) = 119891 (119905 119910 (119905) 1198631205720+119910 (119905)) + 119890 (119905) 119905 isin (0 1)
1198861198631205720+119910 (0) minus 119887119863120572+10+ 119910 (0) =119898minus2sum119895=1
1198861198951198631205720+119910 (120585119895)
1198881198631205720+119910 (1) minus 119889119863120572+10+ 119910 (1) =119898minus2sum119895=1
1198871198951198631205720+119910 (120585119895)
(5)
where 0 lt 120572 le 1 119886 119888 ge 0 119887 119889 gt 0 0 lt 120585119895 lt 1 119886119895 119887119895 isin[0 +infin) 119891 (0 1) times [0infin) times (0infin) 997888rarr (0infin) is continuousand may be singular near the zero for the third argumentand 119890(119905) isin 1198711([0 1]R) may be sign-changing By meansof Schauderrsquos fixed point theorem the conditions for theexistence of positive solutions are established respectivelyfor the cases where the nonlinearity is positive negative andsemipositone
Motivated by the work mentioned above we consider thefractional-order singular nonlocal BVP (1) and establish the
existence results of positive solutions for (1) The main toolsused in this paper are Guo-krasnoselrsquoskii fixed point theoremand Schauderrsquos fixed point theorem For the concepts andproperties about the cone theory and the fixed point theoremone can refer to [17ndash21]
The rest of this paper is organized as follows in Section 2we present some useful preliminaries and lemmas The mainresults are given in Section 3 and Section 4 in which thesingular cases with respect to the time variables and spacevariables are discussed respectively
2 Preliminaries and Some Lemmas
Definition 1 (see [1 2]) The Riemann-Liouville fractionalintegral of order 120572 gt 0 of a function 119909(119905) (0infin) 997888rarr R
is given by
1198681205720+119909 (119905) = 1Γ (120572) int119905
0(119905 minus 119904)120572minus1 119909 (119904) 119889119904 (6)
provided the right-hand side is pointwisely defined on (0 infin)Definition 2 (see [1 2]) The Riemann-Liouville fractionalderivative of order 120572 gt 0 of a function 119909(119905) (0infin) 997888rarr R isgiven by
1198631205720+119909 (119905) = 1Γ (119899 minus 120572) ( 119889119889119905)119899 int1199050
119909 (119904)(119905 minus 119904)120572minus119899+1119889119904 (7)
provided the right-hand side is pointwisely defined on (0 infin)where 119899 = [120572] + 1 [120572] denotes the integer part of the number120572Lemma 3 (see [1 2]) e unique solution of the followinglinear Riemann-Liouville fractional differential equation oforder 120572 gt 0
1198631205720+119909 (119905) = 0 (8)
is
119909 (119905) = 1198881119905120572minus1 + 1198882119905120572minus2 + sdot sdot sdot + 119888119899119905120572minus119899 (9)
where 119888119894 = 1 2 119899 119899 = [120572] + 1 [120572] denotes the integer partof the number 120572Lemma 4 (see [1 2]) If 119909 isin 1198711(0 1) and 1198631205720+119909 isin 1198711(0 1)then
1198681205720+1198631205720+119909 (119905) = 119909 (119905) + 1198881119905120572minus1 + 1198882119905120572minus2 + sdot sdot sdot + 119888119899119905120572minus119899 (10)
where 119888119894 isin 119877 119894 = 1 2 119899 119899 = [120572] + 1The similar proof of the following three lemmas can be
traced to [5 12] in order to be convenient for readers to readwe now give the detailed process of proof for Lemma 5 theproofs for other two lemmas are omitted here
Journal of Function Spaces 3
Lemma 5 Let ℎ isin 1198711(0 1) and 0 lt sum119898119894=1 120578119894120585119894120572minus120573minus1 lt 1 thenthe unique solution of the problem
119863120572minus1205730+ 119909 (119905) + ℎ (119905) = 0 0 lt 119905 lt 1119909 (0) = 1199091015840 (0) = sdot sdot sdot = 119909(119899minus2) (0) = 0119909 (1) = 119898minus2sum
119894=1
120578119894119909 (120585119894)(11)
can be expressed uniquely by
119909 (119905) = int10119866 (119905 119904) ℎ (119904) 119889119904 (12)
where
119866 (119905 119904) = 1Γ (120572 minus 120573) 119892 (0)sdot
119892 (119904) [119905 (1 minus 119904)]120572minus120573minus1 0 le 119905 le 119904 le 1119892 (119904) [119905 (1 minus 119904)]120572minus120573minus1 minus (119905 minus 119904)120572minus120573minus1 119892 (0) 0 le 119904 le 119905 le 1
(13)
119892 (119904) = 1 minus sum119904le120585119894
120578119894 (120585119894 minus 1199041 minus 119904 )120572minus120573minus1 (14)
Proof By Lemma 4 the solution of (13) can be written as
119909 (119905) = minus 1Γ (120572 minus 120573) int119905
0(119905 minus 119904)120572minus120573minus1 ℎ (119904) 119889119904 + 1198621119905120572minus120573minus1
+ 1198622119905120572minus120573minus2 + sdot sdot sdot + 119862119899119905120572minus120573minus119899(15)
It follows from 119909(0) = 1199091015840(0) = sdot sdot sdot = 119909(119899minus2)(0) = 0 that 1198622 =sdot sdot sdot = 119862119899 = 0 ie119909 (119905) = minus 1Γ (120572 minus 120573) int
119905
0(119905 minus 119904)120572minus120573minus1 ℎ (119904) 119889119904 + 1198621119905120572minus120573minus1 (16)
thus
119909 (1) = minus 1Γ (120572 minus 120573) int1
0(1 minus 119904)120572minus120573minus1 ℎ (119904) 119889119904 + 1198621
119909 (120585119894) = minus 1Γ (120572 minus 120573) int120585119894
0(120585119894 minus 119904)120572minus120573minus1 ℎ (119904) 119889119904
+ 1198621120585119894120572minus120573minus1(17)
which together with the boundary value condition 119909(1) =sum119898minus2119894=1 120578119894119909(120585119894) implies that
minus 1Γ (120572 minus 120573) int1
0(1 minus 119904)120572minus120573minus1 ℎ (119904) 119889119904 + 1198621
= 119898minus2sum119894=1
120578119894 (minus 1Γ (120572 minus 120573) int120585119894
0(120585119894 minus 119904)120572minus120573minus1 ℎ (119904) 119889119904
+ 1198621120585119894120572minus120573minus1)
1198621(1 minus 119898minus2sum119894=1
120578119894120585120572minus120573minus1119894 ) = 1Γ (120572 minus 120573) int1
0(1 minus 119904)120572minus120573minus1
sdot ℎ (119904) 119889119904 minus 119898minus2sum119894=1
120578119894 int1205851198940 (120585119894 minus 119904)120572minus120573minus1 ℎ (119904) 119889119904Γ (120572 minus 120573)
1198621 = int10
(1 minus 119904)120572minus120573minus1(1 minus sum119898minus2119894=1 120578119894120585120572minus120573minus1119894 ) Γ (120572 minus 120573)ℎ (119904) 119889119904
minus 119898minus2sum119894=1
120578119894 int1205851198940(120585119894 minus 119904)120572minus120573minus1 ℎ (119904) 119889119904
Γ (120572 minus 120573) (1 minus sum119898minus2119894=1 120578119894120585120572minus120573minus1119894 ) (18)
ie
1198621 = 1119892 (0) Γ (120572 minus 120573) [int1
0(1 minus 119904)120572minus120573minus1 ℎ (119904) 119889119904
minus 119898minus2sum119894=1
120578119894 int1205851198940(120585119894 minus 119904)120572minus120573minus1 ℎ (119904) 119889119904]
= 1119892 (0) Γ (120572 minus 120573) [int1
0(1 minus 119904)120572minus120573minus1 119892 (119904) ℎ (119904) 119889119904
(19)
thus
119909 (119905) = minus 1Γ (120572 minus 120573) int119905
0(119905 minus 119904)120572minus120573minus1 ℎ (119904) 119889119904 + 1198621119905120572minus120573minus1
= 1119892 (0) Γ (120572 minus 120573)sdot int1199050[119905120572minus120573minus1 (1 minus 119904)120572minus120573minus1 119892 (119904) minus (119905 minus 119904)120572minus120573minus1 119892 (0)]
sdot ℎ (119904) 119889119904 minus 1119892 (0) Γ (120572 minus 120573)sdot int1119905119905120572minus120573minus1 (1 minus 119904)120572minus120573minus1 ℎ (119904) 119889119904 = int1
0119866 (119905 119904)
sdot ℎ (119904) 119889119904
(20)
Lemma 6 If 0 lt sum119898119894=1 120578119894120585119894120572minus120573minus1 lt 1 then the function 119892satisfies the following conditions
(1) 119892 is a nondecreasing function on [0 1](2) there exist1198721 ge 1198981 ge 0 such that1198981119905+119892(0) le 119892(119905) le1198721119905 + 119892(0) for any 119905 isin [0 1] where 1198721 = sup0lt119905le1((119892(119905) minus119892(0))119905)1198981 = inf0lt119905le1((119892(119905) minus 119892(0))119905)
Remark 7 It is easy to prove that 1198981 gt 0Lemma 8 e function 119866(119905 119904) defined by (13) has the follow-ing properties
(1) 119866(119905 119904) gt 0 for any (119905 119904) isin (0 1)
4 Journal of Function Spaces
(2) 119866(119905 119904) ge (1198981119892(0)Γ(120572minus120573))119904(1 minus 119904)120572minus120573minus1119905120572minus120573minus1 for any(119905 119904) isin [0 1](3) 119866(119905 119904) le ((1198721 + 119892(0)[120572 minus 120573])119892(0)Γ(120572 minus120573))119904(1 minus 119904)120572minus120573minus1119905120572minus120573minus1 for any (119905 119904) isin [0 1](4) 119866(119905 119904) le 119872119904(1 minus 119904)120572minus120573minus1
where [120572 minus 120573] denotes the integer part of the number 120572 minus 120573119872 = 1198721 + 119892 (0) (120572 minus 120573 minus 1)119892 (0) Γ (120572 minus 120573) (21)
Set V(119905) = 1198631205730+119909(119905) then (1) can be transformed into the
following form
119863120572minus1205730+ V (119905) + 119891 (119905 1198681205730+V (119905) V (119905)) + 119890 (119905) = 00 lt 119905 lt 1
V (0) = V1015840 (0) = sdot sdot sdot = V(119899minus2) (0) = 0V (1) = 119898minus2sum
119894=1
120578119894V (120585119894) (22)
FromLemma 5we know that the solution V(119905) of (22) satisfiesV (119905) = int1
0119866 (119905 119904) [119891 (119904 1198681205730+V (119904) V (119904)) + 119890 (119904)] 119889119904 (23)
Lemma 9 (see [17 18]) Suppose that 119864 is a Banach space and119863 sub 119864 is a bounded convex closed set the operator 119860 119863 997888rarr119863 is completely continuous then 119860 has one fixed point on119863
Lemma 10 (see [20] (Guo-krasnoselrsquoskii fixed point theo-rem)) Let Ω1 and Ω1 be two bounded open sets in Banachspace119864 such that 120579 isin Ω1 andΩ1 sub Ω2119860 119875⋂(Ω2Ω1) 997888rarr 119875a completely continuous operator where 120579 denotes the zeroelement of119864 and119875 a cone of119864 Suppose that one of the followingconditions
(i) 119860119909 ge 119909 forall119909 isin 119875 cap 120597Ω1 and 119860119909 le 119909 forall119909 isin119875 cap 120597Ω2(ii) 119860119909 le 119909 forall119909 isin 119875 cap 120597Ω1 and 119860119909 ge 119909 forall119909 isin119875 cap 120597Ω2
holds en 119860 has at least one fixed point in 119875 cap (Ω2 Ω1)3 Main Result I 119891 Is Singular with Respect tothe Time Variables
Let119864 = 119862[0 1] 119906 = max0le119905le1|119906(119905)| then (119864 sdot) is a Banachspace Set
119875 = V isin 119864 V (119905) ge 11989811198721 + 119892 (0) [120572 minus 120573] 119905120572minus120573minus1 V 119905
isin [0 1] (24)
where [120572 minus 120573] denotes the integer part of the number 120572 minus 120573Then 119875 sub 119864 is a positive cone of 119864 For convenience we listsome conditions which will be used in this section
(1198671) For any (119905 119909 119910) isin (0 1) times [0 +infin) times [0 +infin)0 le 119891 (119905 119909 119910) le 120601 (119905) (120588 (119909) + ℎ (119910)) (25)
where 120601 isin 119862(0 1) 120601(119905) gt 0 on (0 1) 120588(119909) gt 0 is continuousand increasing on [0 +infin) ℎ(119909) gt 0 is continuous anddecreasing on [0 +infin)
(1198672)0 lt int10119904 (1 minus 119904)120572minus120573minus1 (120601 (119904) + 119890+ (119904) + 119890minus (119904)) 119889119904
lt +infin(26)
where 119890+ = max119890(119905) 0 119890minus = minusmin119890(119905) 0 are the positivepart and negative part of 119890(119905) respectively
(1198673) There exists 1199031 gt 0 such that
120588(1199031 + 119908Γ (120573) )int10119904 (1 minus 119904)120572minus120573minus1 120601 (119904) 119889119904
+ int10119904 (1 minus 119904)120572minus120573minus1 [ℎ (0) 120601 (119904) + 119890+ (119904)] 119889119904
lt 1199031119872(27)
where119908(119905) is the solution of the following linear equation
119863120572minus1205730+ V (119905) + 119890minus (119905) = 0 0 lt 119905 lt 1V (0) = V1015840 (0) = sdot sdot sdot = V(119899minus2) (0) = 0V (1) = 119898minus2sum
119894=1
120578119894V (120585119894) (28)
ie 119908(119905) = int10119866(119905 119904)119890minus(119904)119889119904
(1198674) There exists [119886 119887] sub (0 1) such that
lim119909+119910997888rarr+infin
119891 (119905 119909 119910)119909 + 119910 gt 119871 (29)
uniformly holds for 119905 isin [119886 119887] where119871 = 2Γ (120572 minus 120573) 119892 (0) (1198721 + 119892 (0) [120572 minus 120573])
11989812119886120572minus120573minus1 int119887119886 119904120572minus120573 (1 minus 119904)120572minus120573minus1 119889119904 (30)
For any V isin 119875 let[V (119905) minus 119908 (119905)]lowast = max V (119905) minus 119908 (119905) 0 (31)
and define operator
119865V (119905) = int10119866 (119905 119904)
sdot [119891 (119904 1198681205730+ [V (119904) minus 119908 (119904)]lowast [V (119904) minus 119908 (119904)]lowast)+ 119890+ (119904)] 119889119904
(32)
From condition (1198671) and (1198672) it is easy to know that 119865 is welldefined
Journal of Function Spaces 5
Lemma 11 119865 119875 997888rarr 119875 is a completely continuous operator
Proof For any V isin 119875 it follows from Lemma 8 that
119865V (119905) = int10119866 (119905 119904)
sdot [119891 (119904 1198681205730+ [V (119904) minus 119908 (119904)]lowast [V (119904) minus 119908 (119904)]lowast)+ 119890+ (119904)] 119889119904 le 1198721 + 119892 (0) [120572 minus 120573]119892 (0) Γ (120572 minus 120573)sdot 119905120572minus120573minus1 int1
0119904 (1 minus 119904)120572minus120573minus1
sdot [119891 (119904 1198681205730+ [V (119904) minus 119908 (119904)]lowast [V (119904) minus 119908 (119904)]lowast)+ 119890+ (119904)] 119889119904 le 1198721 + 119892 (0) [120572 minus 120573]119892 (0) Γ (120572 minus 120573) int1
0119904 (1
minus 119904)120572minus120573minus1sdot [119891 (119904 1198681205730+ [V (119904) minus 119908 (119904)]lowast [V (119904) minus 119908 (119904)]lowast)+ 119890+ (119904)] 119889119904
119865V (119905) = int10119866 (119905 119904)
sdot [119891 (119904 1198681205730+ [V (119904) minus 119908 (119904)]lowast [V (119904) minus 119908 (119904)]lowast)+ 119890+ (119904)] 119889119904 ge 1198981119892 (0) Γ (120572 minus 120573)119905120572minus120573minus1 int
1
0119904 (1
minus 119904)120572minus120573minus1sdot [119891 (119904 1198681205730+ [V (119904) minus 119908 (119904)]lowast [V (119904) minus 119908 (119904)]lowast)+ 119890+ (119904)] 119889119904
(33)
which deduce that 119865V(119905) ge (1198981(1198721 + 119892(0)[120572 minus120573]))119905120572minus120573minus1119865V ie 119865 119875 997888rarr 119875Let 119861 sub 119875 be a bounded set ie there exists 1198711 gt 0 such
that V le 1198711 for any V isin 119861 then0 le [V (119905) minus 119908 (119905)]lowast le 1198711 + int1
0119866 (119905 119904) 119890minus (119904) 119889119904 le 1198711
+ int10119872119904 (1 minus 119904)120572minus120573minus1 119890minus (119904) 119889119904 ≐
(34)
119865V (119905) = int10119866 (119905 119904)
sdot [119891 (119904 1198681205730+ [V (119904) minus 119908 (119904)]lowast [V (119904) minus 119908 (119904)]lowast)+ 119890+ (119904)] 119889119904 le int1
0119866 (119905 119904)
sdot [120601 (119904) (120588 (1198681205730+) + ℎ (0)) + 119890+ (119904)] 119889119904
le 119872(120588( 120573Γ (120573)) + ℎ (0))int10119904 (1 minus 119904)120572minus120573minus1
sdot 120601 (119904) 119889119904 +119872int10119904 (1 minus 119904)120572minus120573minus1 119890+ (119904) 119889119904
lt +infin(35)
therefore
119865V le 119872(120588( 120573Γ (120573)) + ℎ (0))sdot int10119904 (1 minus 119904)120572minus120573minus1 120601 (119904) 119889119904
+119872int10119904 (1 minus 119904)120572minus120573minus1 119890+ (119904) 119889119904
(36)
for any V isin 119861 which implies that 119865 is uniformly boundedFrom (1198672) the absolutely continuity of integral and the
uniformly continuity of119866(119905 119904) on [0 1] we know that for any120576 gt 0 exist120575 gt 0 such that
int1205750119904 (1 minus 119904)120572minus120573minus1 120601 (119904) 119889119904lt 1205766119872(120588 (120573Γ (120573)) + ℎ (0))
(37)
int1205751minus120575
119904 (1 minus 119904)120572minus120573minus1 120601 (119904) 119889119904lt 1205766119872(120588 (120573Γ (120573)) + ℎ (0))
(38)
and 1003816100381610038161003816119866 (1199051 119904) minus 119866 (1199052 119904)1003816100381610038161003816 lt 1205763119897 (120588 (120573Γ (120573)) + ℎ (0)) (39)
for any 1199051 1199052 119904 isin [0 1] with |1199051 minus 1199052| lt 120575(37)-(39) together with Lemma 8 imply that
1003816100381610038161003816119865V (1199051) minus 119865V (1199052)1003816100381610038161003816 le int10
1003816100381610038161003816119866 (1199051 119904) minus 119866 (1199052 119904)1003816100381610038161003816sdot 119891 (119904 1198681205730+ [V (119904) minus 119908 (119904)]lowast [V (119904) minus 119908 (119904)]lowast) 119889119904= int1205750
1003816100381610038161003816119866 (1199051 119904) minus 119866 (1199052 119904)1003816100381610038161003816sdot 119891 (119904 1198681205730+ [V (119904) minus 119908 (119904)]lowast [V (119904) minus 119908 (119904)]lowast) 119889119904+ int11minus120575
1003816100381610038161003816119866 (1199051 119904) minus 119866 (1199052 119904)1003816100381610038161003816sdot 119891 (119904 1198681205730+ [V (119904) minus 119908 (119904)]lowast [V (119904) minus 119908 (119904)]lowast) 119889119904+ int1minus120575120575
1003816100381610038161003816119866 (1199051 119904) minus 119866 (1199052 119904)1003816100381610038161003816sdot 119891 (119904 1198681205730+ [V (119904) minus 119908 (119904)]lowast [V (119904) minus 119908 (119904)]lowast) 119889119904
6 Journal of Function Spaces
le 2119872int1205750119904 (1 minus 119904)120572minus120573minus1 120601 (119904)
sdot (120588 (1198681205730+) + ℎ (0)) 119889119904 + 2119872int11minus120575
119904 (1 minus 119904)120572minus120573minus1sdot 120601 (119904) (120588 (1198681205730+) + ℎ (0)) 119889119904+ int1minus120575120575
1003816100381610038161003816119866 (1199051 119904) minus 119866 (1199052 119904)1003816100381610038161003816sdot 119897 (120588 (1198681205730+) + ℎ (0)) 119889119904 lt 1205763 + 1205763 + 1205763 = 120576
(40)
for any 1199051 1199052 isin [0 1] with |1199051 minus 1199052| lt 120575 and any V isin 119861 where119897 = max120575le119905le1minus120575120601(119905) which deduces that 119865 is equicontinuouson [0 1]Thus according to Ascoli-Arzela theorem we knowthat 119865119861 is a relatively compact set and that 119865 is a completelycontinuous operator
Theorem 12 Suppose that (1198671) minus (1198674) hold then the FVP (1)has at least one positive solution
Proof For any V isin 1205971198751199031 where 1198751199031 = V isin 119875 | V lt 1199031 by(32) Lemma 8 and condition (1198673) one can get that
119865V (119905) = int10119866 (119905 119904)
sdot [119891 (119904 1198681205730+ [V (119904) minus 119908 (119904)]lowast [V (119904) minus 119908 (119904)]lowast)+ 119890+ (119904)] 119889119904 le int1
0119866 (119905 119904)
sdot [120601 (119904) (120588 (1198681205730+ (V (119904) minus 119908 (119904)) + ℎ (0))+ 119890+ (119904)] 119889119904 le int1
0119866 (119905 119904)
sdot [120601 (119904) (120588 (1198681205730+ (V + 119908)) + ℎ (0)) + 119890+ (119904)] 119889119904= 119872int1
0119904 (1 minus 119904)120572minus120573minus1
sdot [120601 (119904) (120588 (1198681205730+ (1199031 + 119908)) + ℎ (0)) + 119890+ (119904)] 119889119904= 119872int1
0119904 (1 minus 119904)120572minus120573minus1
sdot [120601 (119904) (120588(1199031 + 119908Γ (120573) ) + ℎ (0)) + 119890+ (119904)] 119889119904lt 1199031 = V
(41)
ie 119865V le V V isin 1205971198751199031 By condition (1198674) exist119883 gt 0 such that
119891 (119905 119909 119910) gt 119871 (119909 + 119910) (42)
for any 119909 gt 119883 and any 119905 isin [119886 119887] Choose 1199032 such that
1199032 gt max1199031 21198881 2 (1198721 + 119892 (0) [120572 minus 120573])1198831198981119886120572minus120573minus1 (43)
where
1198881 = (1198721 + 119892 (0) [120572 minus 120573])2 sdot int10 (1 minus 119904)120572minus120573minus1 119890minus (119904) 1198891199041198981Γ (120572 minus 120573) 119892 (0) (44)
For any V isin 1205971198751199032 where 1198751199032 = V isin 119875 | V lt 1199032 Because119908 (119905) = int1
0119866 (119905 119904) 119890minus (119904) 119889119904
le 1198721 + 119892 (0) [120572 minus 120573]Γ (120572 minus 120573) 119892 (0)sdot 119905120572minus120573minus1 int1
0(1 minus 119904)120572minus120573minus1 119890minus (119904) 119889119904
le (1198721 + 119892 (0) [120572 minus 120573])2 sdot int10 (1 minus 119904)120572minus120573minus1 119890minus (119904) 1198891199041198981Γ (120572 minus 120573) 119892 (0) 1199032sdot V (119905) = 11988811199032 V (119905)
(45)
so we have
V (119905) minus 119908 (119905) ge (1 minus 11988811199032) V (119905) ge 12V (119905) (46)
and then for 119905 isin [119886 119887]V (119905) minus 119908 (119905) ge 12V (119905) ge 12 sdot 1198981119886120572minus120573minus11198721 + 119892 (0) [120572 minus 120573] sdot V
= 12 sdot 1198981119886120572minus120573minus11198721 + 119892 (0) [120572 minus 120573] sdot 1199032 gt 119883(47)
follows from (43) and the definition of cone 119875From (42) (43) and (47) one can obtain that
119865V (119905) = int10119866 (119905 119904)
sdot [119891 (119904 1198681205730+ [V (119904) minus 119908 (119904)]lowast [V (119904) minus 119908 (119904)]lowast)+ 119890+ (119904)] 119889119904 ge int119887
119886119866 (119905 119904) 119871 (1198681205730+ (V (119904) minus 119908 (119904))
+ (V (119904) minus 119908 (119904))) + 119890+ (119904) ]119889119904 ge 119871int119887119886119866 (119905 119904)
sdot 12V (119904) 119889119904 ge 1198712 sdot 1198981Γ (120572 minus 120573) 119892 (0) sdot int119887
119886119904 (1
minus 119904)120572minus120573minus1 119905120572minus120573minus1V (119904) 119889119904 ge 1198712 sdot 1198981119886120572minus120573minus1Γ (120572 minus 120573) 119892 (0)
Journal of Function Spaces 7
sdot 11989811198721 + 119892 (0) [120572 minus 120573] sdot int119887
119886119904 (1 minus 119904)120572minus120573minus1 119904120572minus120573minus1119889119904
sdot V = 1198712 sdot 1198981119886120572minus120573minus1Γ (120572 minus 120573) 119892 (0) sdot11989811198721 + 119892 (0) [120572 minus 120573]
sdot int119887119886(1 minus 119904)120572minus120573minus1 119904120572minus120573119889119904 sdot V ge 1198712 sdot 1198981119886120572minus120573minus1Γ (120572 minus 120573) 119892 (0)
sdot 11989811198721 + 119892 (0) [120572 minus 120573] sdot int119887
119886(1 minus 119904)120572minus120573minus1 119904120572minus120573119889119904 sdot 1199032
= 1199032 = V (48)
ie 119865V ge V V isin 1205971198751199032 It follows from Lemma 10 that 119865 has at least fixed point
V1 isin 1198751199032 1198751199031 ie V1 satisfies119863120572minus1205730+ V1 (119905) + 119891 (119905 1198681205730+ (V1 (119905) minus 119908 (119905)) V1 (119905) minus 119908 (119905))
+ 119890 (119905) = 0 0 lt 119905 lt 1V1 (0) = V10158401 (0) = sdot sdot sdot = V(119899minus2)1 (0) = 0V1 (1) = 119898minus2sum
119894=1
120578119894V1 (120585119894) (49)
Set V1(119905) = V1(119905) minus 119908(119905) noticing that V1(119905) 119908(119905) are thesolutions of BVP (32) and (49) respectively therefore we canconclude that V1(119905) is a positive solution of (22) Let 1199091(119905) =1198681205730+V1(119905) then 1199091(119905) is a positive solution of the nonlinearfractional differential equations (1)
4 Main Result II 119891 Is Singular withRespect to Both the Time Variables andthe Space Variable
In this section we always suppose that the following condi-tion holds
(1198675) 119891(119905 119906 V) (0 1) times [0infin) times (0infin) 997888rarr [0infin)is continuous there exist 120576 isin (0 1) and 1205831 1205832 isin 119862+[0 1]1205831(119905) equiv 0 for 119905 isin [0 1] such that
1205831 (119905)(119909 + 119910)120576 le 119891 (119905 119909 119910) le1205832 (119905)(119909 + 119910)120576 (50)
for any (119909 119910) isin [0infin) times (0infin) 119905 isin (0 1) where119862+ [0 1] = 119909 (119905) isin 119862 [0 1] | 119909 (119905) ge 0 119905 isin [0 1] (51)
Set
120593 (119905) = int10119866 (119905 119904) 119890 (119904) 119889119904 119905 isin [0 1] (52)
it follows from Lemma 5 that 120593(119905) is the solution of thefollowing linear equation
119863120572minus1205730+ V (119905) + 119890 (119905) = 0 0 lt 119905 lt 1V (0) = V1015840 (0) = sdot sdot sdot = V(119899minus2) (0) = 0V (1) = 119898minus2sum
119894=1
120578119894V (120585119894) (53)
Denote
1198861 (119905) = int10119866 (119905 119904) 1205831 (119904) 119889119904
1198862 (119905) = int10119866 (119905 119904) 1205832 (119904) 119889119904
120593lowast = inf0le119905le1
120593 (119905) 120593lowast = sup0le119905le1
120593 (119905) 1198861lowast = min
0le119905le11198861 (119905)
119886lowast1 = max0le119905le1
1198861 (119905) 1198862lowast = min
0le119905le11198862 (119905)
119886lowast2 = max0le119905le1
1198862 (119905)
(54)
Clearly 119886lowast119895 ge 119886119895lowast gt 0 119895 = 1 2Theorem 13 Suppose that the condition (H5) holds and 120593lowast ge0 en the FVP (1) has at least one positive solution
Proof Because 120593lowast ge 0 so we can choose 119877 gt 0 large enoughsuch that
119903119877120576 + 120593lowast ge 1119877119877120576 (119886lowast2 + 120593lowast) le 119877
(55)
where 119903 = 1198861lowast[1 + 1(1 + 120573Γ(120573))]120576 In fact since
lim119909997888rarr+infin
119909120576119909 = 0 lt min119903 1119886lowast2 + 120593lowast (56)
there exists 1198831 gt 0 such that 119909120576119909 lt min119903 1(119886lowast2 + 120593lowast) forany 119909 gt 1198831 ie
119903119909120576 ge 1119909 119909120576 (119886lowast2 + 120593lowast) le 119909 (57)
for any 119909 gt 1198831 If 120593lowast gt 0 then from lim119909997888rarr+infin[1119909 minus 119903119909120576] =0 lt 120593lowast one can get that there exists 1198832 gt 0 such that 1119909 minus119903119909120576 lt 120593lowast for any 119909 gt 1198832 ie119903119909120576 + 120593lowast ge 1119909 (58)
8 Journal of Function Spaces
for any 119909 gt 1198832 By (57) (58) we can choose 119877 gt max1198831 1198832such that 119877 satisfies (55) Set
119863 = V isin 119862+ [0 1] 1119877 le V (119905) le 119877 119905 isin [0 1] (59)
For any V isin 1198631 from (23) we have
1198681205730+V (119905) = 1Γ (120573) int119905
0(119905 minus 119904)120573minus1 V (119904) 119889119904
le 119877Γ (120573) int119905
0(119905 minus 119904)120573minus1 119889119904 = 119877119905120573120573Γ (120573) le 119877120573Γ (120573)
(60)
1198681205730+V (119905) = 1Γ (120573) int119905
0(119905 minus 119904)120573minus1 V (119904) 119889119904
ge 1119877Γ (120573) int119905
0(119905 minus 119904)120573minus1 119889119904 = 119905120573119877120573Γ (120573)
(61)
It follows from (60) (61) and (H5) that
1205831 (119905)119877120576 (1 + 1120573Γ (120573))120576 le1205831 (119905)
(V (119905) + 1198681205730+V (119905))120576le 119891 (119905 1198681205730+V (119905) V (119905))
119891 (119905 1198681205730+V (119905) V (119905)) le 1205832 (119905)(V (119905) + 1198681205730+V (119905))120576
le 1205832 (119905)(1119877 + 119905120573119877120573Γ (120573))120576le 1198771205761205832 (119905)
(62)
ie
1205831 (119905)119877120576 (1 + 1120573Γ (120573))120576 le 119891 (119905 1198681205730+V (119905) V (119905)) le 1198771205761205832 (119905) (63)
And then
119879V (119905) = int10119866 (119905 119904) [119891 (119904 1198681205730+V (119904) V (119904)) + 119890 (119904)] 119889119904
= int10119866 (119905 119904) 119891 (119904 1198681205730+V (119904) V (119904)) 119889119904 + 120593 (119905)
le int10119866 (119905 119904) 119891 (119904 1198681205730+V (119904) V (119904)) 119889119904 + 120593lowast
le 119877120576 int10119866 (119905 119904) 1205832 (119904) 119889119904 + 120593lowast lt +infin
(64)
which deduces that the operator 119879 is well definedNow we shall prove that 119879 119863 997888rarr 119863 For V isin 119863 it
is easy to see that 119879V(119905) isin 119862+[0 1] and by (55) (63) we canobtain that
119879V (119905) = int10119866 (119905 119904) [119891 (119904 1198681205730+V (119904) V (119904)) + 119890 (119904)] 119889119904
= int10119866 (119905 119904) 119891 (119904 1198681205730+V (119904) V (119904)) 119889119904 + 120593 (119905)
ge 1119877120576 (1 + 1120573Γ (120573))120576 int
1
0119866 (119905 119904) 1205831 (119904) 119889119904
+ 120593 (119905) ge 1119877120576 (1 + 1120573Γ (120573))120576 1198861lowast + 120593lowast ge
1119877
(65)
119879V (119905) = int10119866 (119905 119904) [119891 (119904 1198681205730+V (119904) V (119904)) + 119890 (119904)] 119889119904
= int10119866 (119905 119904) 119891 (119904 1198681205730+V (119904) V (119904)) 119889119904 + 120593 (119905)
le 119877120576 int10119866 (119905 119904) 1205832 (119904) 119889119904 + 120593lowast le 119877120576119886lowast2 + 120593lowast
le 119877120576 (119886lowast2 + 120593lowast) le 119877
(66)
ie 119879 119863 997888rarr 119863Next let us prove that 119879 119863 997888rarr 119863 is completely
continuousFor any V119899 sub 119863 V0 isin 119863 and V119899 997888rarr V0The continuity
of 119891 deduces that
119891 (119905 1198681205730+V119899 (119905) V119899 (119905)) 997888rarr 119891(119905 1198681205730+V0 (119905) V0 (119905)) 119899 997888rarr infin (67)
and it follows from (63) that100381610038161003816100381610038161003816119891 (119905 1198681205730+V119899 (119905) V119899 (119905)) minus 119891 (119905 1198681205730+V0 (119905) V0 (119905))100381610038161003816100381610038161003816le 2119877120598max max
0le119905le11205831 (119905) max
0le119905le11205832 (119905) (68)
By using the Lebesgue dominated convergence theorem weobtain that
lim119899997888rarrinfin
1003817100381710038171003817119879V119899 minus 119879V01003817100381710038171003817 = lim119899997888rarrinfin
max0le119905le1
100381610038161003816100381610038161003816100381610038161003816int1
0119866 (119905 119904) [119891 (119904 1198681205730+V119899 (119904) V119899 (119904)) minus 119891 (119904 1198681205730+V0 (119904) V0 (119904))] 119889119904
100381610038161003816100381610038161003816100381610038161003816le lim119899997888rarrinfin
max0le119905le1
int10119866 (119905 119904) 100381610038161003816100381610038161003816119891 (119904 1198681205730+V119899 (119904) V119899 (119904)) minus 119891 (119904 1198681205730+V0 (119904) V0 (119904))100381610038161003816100381610038161003816 119889119904
Journal of Function Spaces 9
= max0le119905119904le1
119866 (119905 119904) sdot int10
lim119899997888rarrinfin
100381610038161003816100381610038161003816119891 (119904 1198681205730+V119899 (119904) V119899 (119904)) minus 119891 (119904 1198681205730+V0 (119904) V0 (119904))100381610038161003816100381610038161003816 119889119904 = 0(69)
and this implies that 119879 is a continuous operatorNow we shall prove that 119879 119863 997888rarr 119863 is compact For any
V isin 119863 119879V isin 119863 which deduces that 1119877 le 119879V(119905) le 119877 for119905 isin [0 1] ie 119879 is uniformly boundedSince 119866(119905 119904) isin 119862([0 1] times [0 1]) it is also uniformly
continuous on [0 1] times [0 1] and then for any 120576 gt 0 exist120575 gt 0st for any (1199051 119904) (1199052 119904) isin [0 1] times [0 1] with |1199051 minus 1199052| lt 120575 wealways have
1003816100381610038161003816119866 (1199051 119904) minus 119866 (1199052 119904)1003816100381610038161003816 lt 120576119877120576 int101205832 (119904) 119889119904 + int10 |119890 (119904)| 119889119904
(70)
Thus one can obtain by virtue of (63) (70) that
1003816100381610038161003816119879V (1199051) minus 119879V (1199052)1003816100381610038161003816 =100381610038161003816100381610038161003816100381610038161003816int1
0[119866 (1199051 119904) minus 119866 (1199052 119904)]
sdot [119891 (119904 119868120573V (119904) V (119904)) + 119890 (119904)] 119889119904100381610038161003816100381610038161003816100381610038161003816 le int1
0
1003816100381610038161003816119866 (1199051 119904)minus 119866 (1199052 119904)1003816100381610038161003816 [100381610038161003816100381610038161003816119891 (119904 1198681205730+V (119904) V (119904))100381610038161003816100381610038161003816 + |119890 (119904)|] 119889119904lt [119877120576 int1
01205832 (119904) 119889119904 + int1
0|119890 (119904)| 119889119904]
sdot 120576119877120576 int101205832 (119904) 119889119904 + int10 |119890 (119904)| 119889119904
= 120576
(71)
for any 1199051 1199052 isin [0 1] with |1199051 minus 1199052| lt 120575 and any V isin 1198631 whichshows that 119879 119863 997888rarr 119863 is equicontinuous Thus Arzela-Ascoli theorem guarantees that 119879 119863 997888rarr 119863 is completelycontinuous Existence of at least one fixed point V1 isin 119863follows from Lemma 6 ie 1199061(119905) = 1198681205730+V1(119905) is a positivesolution of differential equation (1) which satisfies
1119877Γ (120572)119905120572 le 1199061 (119905) = 1Γ (120572) int119905
0(119905 minus 119904)120572minus1 V1 (119904) 119889119904
le 119877Γ (120572) 119905120572(72)
Data Availability
The data used to support the findings of this study areavailable from the corresponding author upon request
Conflicts of Interest
The author declares that there are no conflicts of interestregarding the publication of this paper
Acknowledgments
This research was supported by Shandong Provincial Nat-ural Science Foundation of China (no ZR2016FM10) andNational Natural Science Foundation of China (no 11571197)
References
[1] I Podlubny Fractional Differential Equations Mathematics inScience and Engineering Academic Press NewYork NY USA1999
[2] A A Kilbas H M Srivastava and J J Nietoeory and Appli-cational Differential Equations Elsevier Amsterdam Nether-lands 2006
[3] T Hu Y Sun and W Sun ldquoExistence of positive solutions for athird-order multipoint boundary value problem and extensionto fractional caserdquo Boundary Value Problems vol 2016 article197 2016
[4] Y Jia and X Zhang ldquoPositive solutions for a class of fractionaldifferential equation multi-point boundary value problemswith changing sign nonlinearityrdquo Applied Mathematics andComputation vol 47 no 1-2 pp 15ndash31 2015
[5] X Hao ldquoPositive solution for singular fractional differentialequations involving derivativesrdquo Advances in Difference Equa-tions vol 2016 article 139 2016
[6] Y Zou and G He ldquoOn the uniqueness of solutions for a class offractional differential equationsrdquo Applied Mathematics Lettersvol 74 pp 68ndash73 2017
[7] X Hao H Wang L Liu and Y Cui ldquoPositive solutionsfor a system of nonlinear fractional nonlocal boundary valueproblems with parameters and p-Laplacian operatorrdquoBoundaryValue Problems vol 2017 article 182 2017
[8] X Zhang ldquoPositive solutions for a class of singular fractionaldifferential equation with infinite-point boundary value condi-tionsrdquo Applied Mathematics Letters vol 39 pp 22ndash27 2015
[9] K M Zhang ldquoOn a sign-changing solution for some fractionaldifferential equationsrdquo Boundary Value Problems vol 2017article 59 2017
[10] Y L Guan Z Q Zhao and X L Lin ldquoOn the existence ofpositive solutions and negative solutions of singular fractionaldifferential equations via global bifurcation techniquesrdquoBound-ary Value Problems vol 2016 article 141 2016
[11] X G Zhang Y H Wu and L Caccetta ldquoNonlocal fractionalorder differential equations with changing-sign singular pertur-bationrdquo Applied Mathematical Modelling vol 56 pp 116ndash1262015
[12] YWang and L S Liu ldquoPositive solutions for fractionalm-pointboundary value problems in Banach spacesrdquo ActaMathematicaScientia vol 32 no 1 pp 246ndash256 2012
[13] Y Cui and Y Zou ldquoAn existence and uniqueness theorem for asecond order nonlinear system with coupled integral boundaryvalue conditionsrdquo Applied Mathematics and Computation vol256 pp 438ndash444 2015
10 Journal of Function Spaces
[14] X Q Zhang L Wang and Q S Wang ldquoExistence of positivesolutions for a class of nonlinear fractional differential equa-tions with integral boundary conditionsrdquo Applied Mathematicsand Computation vol 226 pp 708ndash718 2014
[15] XG Zhang L S Liu YHWu andBWiwatanapataphee ldquoThespectral analysis for a singular fractional differential equationwith a signedmeasurerdquoApplied Mathematics and Computationvol 257 pp 252ndash263 2015
[16] Y Wang and L Liu ldquoPositive solutions for a class of fractional3-point boundary value problems at resonancerdquo Advances inDifference Equations vol 13 article 13 2017
[17] M A Krasnoselskii and P P Zabreiko Geometrical Methods ofNonlinear Analysis Springer New York NY USA 1984
[18] K Deimling Nonlinear Functional Analysis Springer BerlinGermany 1985
[19] K Zhang ldquoNontrivial solutions of fourth-order singularboundary value problems with sign-changing nonlinear termsrdquoTopological Methods in Nonlinear Analysis vol 40 no 1 pp 53ndash70 2012
[20] D J Guo and V Lakshmikantham Nonlinear Problems inAbstract Cones Academic Press Massachusetts Mass USA1988
[21] J Liu and Z Zhao ldquoMultiple solutions for impulsive problemswith non-autonomous perturbationsrdquo Applied MathematicsLetters vol 64 pp 143ndash149 2017
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Journal of Function Spaces 3
Lemma 5 Let ℎ isin 1198711(0 1) and 0 lt sum119898119894=1 120578119894120585119894120572minus120573minus1 lt 1 thenthe unique solution of the problem
119863120572minus1205730+ 119909 (119905) + ℎ (119905) = 0 0 lt 119905 lt 1119909 (0) = 1199091015840 (0) = sdot sdot sdot = 119909(119899minus2) (0) = 0119909 (1) = 119898minus2sum
119894=1
120578119894119909 (120585119894)(11)
can be expressed uniquely by
119909 (119905) = int10119866 (119905 119904) ℎ (119904) 119889119904 (12)
where
119866 (119905 119904) = 1Γ (120572 minus 120573) 119892 (0)sdot
119892 (119904) [119905 (1 minus 119904)]120572minus120573minus1 0 le 119905 le 119904 le 1119892 (119904) [119905 (1 minus 119904)]120572minus120573minus1 minus (119905 minus 119904)120572minus120573minus1 119892 (0) 0 le 119904 le 119905 le 1
(13)
119892 (119904) = 1 minus sum119904le120585119894
120578119894 (120585119894 minus 1199041 minus 119904 )120572minus120573minus1 (14)
Proof By Lemma 4 the solution of (13) can be written as
119909 (119905) = minus 1Γ (120572 minus 120573) int119905
0(119905 minus 119904)120572minus120573minus1 ℎ (119904) 119889119904 + 1198621119905120572minus120573minus1
+ 1198622119905120572minus120573minus2 + sdot sdot sdot + 119862119899119905120572minus120573minus119899(15)
It follows from 119909(0) = 1199091015840(0) = sdot sdot sdot = 119909(119899minus2)(0) = 0 that 1198622 =sdot sdot sdot = 119862119899 = 0 ie119909 (119905) = minus 1Γ (120572 minus 120573) int
119905
0(119905 minus 119904)120572minus120573minus1 ℎ (119904) 119889119904 + 1198621119905120572minus120573minus1 (16)
thus
119909 (1) = minus 1Γ (120572 minus 120573) int1
0(1 minus 119904)120572minus120573minus1 ℎ (119904) 119889119904 + 1198621
119909 (120585119894) = minus 1Γ (120572 minus 120573) int120585119894
0(120585119894 minus 119904)120572minus120573minus1 ℎ (119904) 119889119904
+ 1198621120585119894120572minus120573minus1(17)
which together with the boundary value condition 119909(1) =sum119898minus2119894=1 120578119894119909(120585119894) implies that
minus 1Γ (120572 minus 120573) int1
0(1 minus 119904)120572minus120573minus1 ℎ (119904) 119889119904 + 1198621
= 119898minus2sum119894=1
120578119894 (minus 1Γ (120572 minus 120573) int120585119894
0(120585119894 minus 119904)120572minus120573minus1 ℎ (119904) 119889119904
+ 1198621120585119894120572minus120573minus1)
1198621(1 minus 119898minus2sum119894=1
120578119894120585120572minus120573minus1119894 ) = 1Γ (120572 minus 120573) int1
0(1 minus 119904)120572minus120573minus1
sdot ℎ (119904) 119889119904 minus 119898minus2sum119894=1
120578119894 int1205851198940 (120585119894 minus 119904)120572minus120573minus1 ℎ (119904) 119889119904Γ (120572 minus 120573)
1198621 = int10
(1 minus 119904)120572minus120573minus1(1 minus sum119898minus2119894=1 120578119894120585120572minus120573minus1119894 ) Γ (120572 minus 120573)ℎ (119904) 119889119904
minus 119898minus2sum119894=1
120578119894 int1205851198940(120585119894 minus 119904)120572minus120573minus1 ℎ (119904) 119889119904
Γ (120572 minus 120573) (1 minus sum119898minus2119894=1 120578119894120585120572minus120573minus1119894 ) (18)
ie
1198621 = 1119892 (0) Γ (120572 minus 120573) [int1
0(1 minus 119904)120572minus120573minus1 ℎ (119904) 119889119904
minus 119898minus2sum119894=1
120578119894 int1205851198940(120585119894 minus 119904)120572minus120573minus1 ℎ (119904) 119889119904]
= 1119892 (0) Γ (120572 minus 120573) [int1
0(1 minus 119904)120572minus120573minus1 119892 (119904) ℎ (119904) 119889119904
(19)
thus
119909 (119905) = minus 1Γ (120572 minus 120573) int119905
0(119905 minus 119904)120572minus120573minus1 ℎ (119904) 119889119904 + 1198621119905120572minus120573minus1
= 1119892 (0) Γ (120572 minus 120573)sdot int1199050[119905120572minus120573minus1 (1 minus 119904)120572minus120573minus1 119892 (119904) minus (119905 minus 119904)120572minus120573minus1 119892 (0)]
sdot ℎ (119904) 119889119904 minus 1119892 (0) Γ (120572 minus 120573)sdot int1119905119905120572minus120573minus1 (1 minus 119904)120572minus120573minus1 ℎ (119904) 119889119904 = int1
0119866 (119905 119904)
sdot ℎ (119904) 119889119904
(20)
Lemma 6 If 0 lt sum119898119894=1 120578119894120585119894120572minus120573minus1 lt 1 then the function 119892satisfies the following conditions
(1) 119892 is a nondecreasing function on [0 1](2) there exist1198721 ge 1198981 ge 0 such that1198981119905+119892(0) le 119892(119905) le1198721119905 + 119892(0) for any 119905 isin [0 1] where 1198721 = sup0lt119905le1((119892(119905) minus119892(0))119905)1198981 = inf0lt119905le1((119892(119905) minus 119892(0))119905)
Remark 7 It is easy to prove that 1198981 gt 0Lemma 8 e function 119866(119905 119904) defined by (13) has the follow-ing properties
(1) 119866(119905 119904) gt 0 for any (119905 119904) isin (0 1)
4 Journal of Function Spaces
(2) 119866(119905 119904) ge (1198981119892(0)Γ(120572minus120573))119904(1 minus 119904)120572minus120573minus1119905120572minus120573minus1 for any(119905 119904) isin [0 1](3) 119866(119905 119904) le ((1198721 + 119892(0)[120572 minus 120573])119892(0)Γ(120572 minus120573))119904(1 minus 119904)120572minus120573minus1119905120572minus120573minus1 for any (119905 119904) isin [0 1](4) 119866(119905 119904) le 119872119904(1 minus 119904)120572minus120573minus1
where [120572 minus 120573] denotes the integer part of the number 120572 minus 120573119872 = 1198721 + 119892 (0) (120572 minus 120573 minus 1)119892 (0) Γ (120572 minus 120573) (21)
Set V(119905) = 1198631205730+119909(119905) then (1) can be transformed into the
following form
119863120572minus1205730+ V (119905) + 119891 (119905 1198681205730+V (119905) V (119905)) + 119890 (119905) = 00 lt 119905 lt 1
V (0) = V1015840 (0) = sdot sdot sdot = V(119899minus2) (0) = 0V (1) = 119898minus2sum
119894=1
120578119894V (120585119894) (22)
FromLemma 5we know that the solution V(119905) of (22) satisfiesV (119905) = int1
0119866 (119905 119904) [119891 (119904 1198681205730+V (119904) V (119904)) + 119890 (119904)] 119889119904 (23)
Lemma 9 (see [17 18]) Suppose that 119864 is a Banach space and119863 sub 119864 is a bounded convex closed set the operator 119860 119863 997888rarr119863 is completely continuous then 119860 has one fixed point on119863
Lemma 10 (see [20] (Guo-krasnoselrsquoskii fixed point theo-rem)) Let Ω1 and Ω1 be two bounded open sets in Banachspace119864 such that 120579 isin Ω1 andΩ1 sub Ω2119860 119875⋂(Ω2Ω1) 997888rarr 119875a completely continuous operator where 120579 denotes the zeroelement of119864 and119875 a cone of119864 Suppose that one of the followingconditions
(i) 119860119909 ge 119909 forall119909 isin 119875 cap 120597Ω1 and 119860119909 le 119909 forall119909 isin119875 cap 120597Ω2(ii) 119860119909 le 119909 forall119909 isin 119875 cap 120597Ω1 and 119860119909 ge 119909 forall119909 isin119875 cap 120597Ω2
holds en 119860 has at least one fixed point in 119875 cap (Ω2 Ω1)3 Main Result I 119891 Is Singular with Respect tothe Time Variables
Let119864 = 119862[0 1] 119906 = max0le119905le1|119906(119905)| then (119864 sdot) is a Banachspace Set
119875 = V isin 119864 V (119905) ge 11989811198721 + 119892 (0) [120572 minus 120573] 119905120572minus120573minus1 V 119905
isin [0 1] (24)
where [120572 minus 120573] denotes the integer part of the number 120572 minus 120573Then 119875 sub 119864 is a positive cone of 119864 For convenience we listsome conditions which will be used in this section
(1198671) For any (119905 119909 119910) isin (0 1) times [0 +infin) times [0 +infin)0 le 119891 (119905 119909 119910) le 120601 (119905) (120588 (119909) + ℎ (119910)) (25)
where 120601 isin 119862(0 1) 120601(119905) gt 0 on (0 1) 120588(119909) gt 0 is continuousand increasing on [0 +infin) ℎ(119909) gt 0 is continuous anddecreasing on [0 +infin)
(1198672)0 lt int10119904 (1 minus 119904)120572minus120573minus1 (120601 (119904) + 119890+ (119904) + 119890minus (119904)) 119889119904
lt +infin(26)
where 119890+ = max119890(119905) 0 119890minus = minusmin119890(119905) 0 are the positivepart and negative part of 119890(119905) respectively
(1198673) There exists 1199031 gt 0 such that
120588(1199031 + 119908Γ (120573) )int10119904 (1 minus 119904)120572minus120573minus1 120601 (119904) 119889119904
+ int10119904 (1 minus 119904)120572minus120573minus1 [ℎ (0) 120601 (119904) + 119890+ (119904)] 119889119904
lt 1199031119872(27)
where119908(119905) is the solution of the following linear equation
119863120572minus1205730+ V (119905) + 119890minus (119905) = 0 0 lt 119905 lt 1V (0) = V1015840 (0) = sdot sdot sdot = V(119899minus2) (0) = 0V (1) = 119898minus2sum
119894=1
120578119894V (120585119894) (28)
ie 119908(119905) = int10119866(119905 119904)119890minus(119904)119889119904
(1198674) There exists [119886 119887] sub (0 1) such that
lim119909+119910997888rarr+infin
119891 (119905 119909 119910)119909 + 119910 gt 119871 (29)
uniformly holds for 119905 isin [119886 119887] where119871 = 2Γ (120572 minus 120573) 119892 (0) (1198721 + 119892 (0) [120572 minus 120573])
11989812119886120572minus120573minus1 int119887119886 119904120572minus120573 (1 minus 119904)120572minus120573minus1 119889119904 (30)
For any V isin 119875 let[V (119905) minus 119908 (119905)]lowast = max V (119905) minus 119908 (119905) 0 (31)
and define operator
119865V (119905) = int10119866 (119905 119904)
sdot [119891 (119904 1198681205730+ [V (119904) minus 119908 (119904)]lowast [V (119904) minus 119908 (119904)]lowast)+ 119890+ (119904)] 119889119904
(32)
From condition (1198671) and (1198672) it is easy to know that 119865 is welldefined
Journal of Function Spaces 5
Lemma 11 119865 119875 997888rarr 119875 is a completely continuous operator
Proof For any V isin 119875 it follows from Lemma 8 that
119865V (119905) = int10119866 (119905 119904)
sdot [119891 (119904 1198681205730+ [V (119904) minus 119908 (119904)]lowast [V (119904) minus 119908 (119904)]lowast)+ 119890+ (119904)] 119889119904 le 1198721 + 119892 (0) [120572 minus 120573]119892 (0) Γ (120572 minus 120573)sdot 119905120572minus120573minus1 int1
0119904 (1 minus 119904)120572minus120573minus1
sdot [119891 (119904 1198681205730+ [V (119904) minus 119908 (119904)]lowast [V (119904) minus 119908 (119904)]lowast)+ 119890+ (119904)] 119889119904 le 1198721 + 119892 (0) [120572 minus 120573]119892 (0) Γ (120572 minus 120573) int1
0119904 (1
minus 119904)120572minus120573minus1sdot [119891 (119904 1198681205730+ [V (119904) minus 119908 (119904)]lowast [V (119904) minus 119908 (119904)]lowast)+ 119890+ (119904)] 119889119904
119865V (119905) = int10119866 (119905 119904)
sdot [119891 (119904 1198681205730+ [V (119904) minus 119908 (119904)]lowast [V (119904) minus 119908 (119904)]lowast)+ 119890+ (119904)] 119889119904 ge 1198981119892 (0) Γ (120572 minus 120573)119905120572minus120573minus1 int
1
0119904 (1
minus 119904)120572minus120573minus1sdot [119891 (119904 1198681205730+ [V (119904) minus 119908 (119904)]lowast [V (119904) minus 119908 (119904)]lowast)+ 119890+ (119904)] 119889119904
(33)
which deduce that 119865V(119905) ge (1198981(1198721 + 119892(0)[120572 minus120573]))119905120572minus120573minus1119865V ie 119865 119875 997888rarr 119875Let 119861 sub 119875 be a bounded set ie there exists 1198711 gt 0 such
that V le 1198711 for any V isin 119861 then0 le [V (119905) minus 119908 (119905)]lowast le 1198711 + int1
0119866 (119905 119904) 119890minus (119904) 119889119904 le 1198711
+ int10119872119904 (1 minus 119904)120572minus120573minus1 119890minus (119904) 119889119904 ≐
(34)
119865V (119905) = int10119866 (119905 119904)
sdot [119891 (119904 1198681205730+ [V (119904) minus 119908 (119904)]lowast [V (119904) minus 119908 (119904)]lowast)+ 119890+ (119904)] 119889119904 le int1
0119866 (119905 119904)
sdot [120601 (119904) (120588 (1198681205730+) + ℎ (0)) + 119890+ (119904)] 119889119904
le 119872(120588( 120573Γ (120573)) + ℎ (0))int10119904 (1 minus 119904)120572minus120573minus1
sdot 120601 (119904) 119889119904 +119872int10119904 (1 minus 119904)120572minus120573minus1 119890+ (119904) 119889119904
lt +infin(35)
therefore
119865V le 119872(120588( 120573Γ (120573)) + ℎ (0))sdot int10119904 (1 minus 119904)120572minus120573minus1 120601 (119904) 119889119904
+119872int10119904 (1 minus 119904)120572minus120573minus1 119890+ (119904) 119889119904
(36)
for any V isin 119861 which implies that 119865 is uniformly boundedFrom (1198672) the absolutely continuity of integral and the
uniformly continuity of119866(119905 119904) on [0 1] we know that for any120576 gt 0 exist120575 gt 0 such that
int1205750119904 (1 minus 119904)120572minus120573minus1 120601 (119904) 119889119904lt 1205766119872(120588 (120573Γ (120573)) + ℎ (0))
(37)
int1205751minus120575
119904 (1 minus 119904)120572minus120573minus1 120601 (119904) 119889119904lt 1205766119872(120588 (120573Γ (120573)) + ℎ (0))
(38)
and 1003816100381610038161003816119866 (1199051 119904) minus 119866 (1199052 119904)1003816100381610038161003816 lt 1205763119897 (120588 (120573Γ (120573)) + ℎ (0)) (39)
for any 1199051 1199052 119904 isin [0 1] with |1199051 minus 1199052| lt 120575(37)-(39) together with Lemma 8 imply that
1003816100381610038161003816119865V (1199051) minus 119865V (1199052)1003816100381610038161003816 le int10
1003816100381610038161003816119866 (1199051 119904) minus 119866 (1199052 119904)1003816100381610038161003816sdot 119891 (119904 1198681205730+ [V (119904) minus 119908 (119904)]lowast [V (119904) minus 119908 (119904)]lowast) 119889119904= int1205750
1003816100381610038161003816119866 (1199051 119904) minus 119866 (1199052 119904)1003816100381610038161003816sdot 119891 (119904 1198681205730+ [V (119904) minus 119908 (119904)]lowast [V (119904) minus 119908 (119904)]lowast) 119889119904+ int11minus120575
1003816100381610038161003816119866 (1199051 119904) minus 119866 (1199052 119904)1003816100381610038161003816sdot 119891 (119904 1198681205730+ [V (119904) minus 119908 (119904)]lowast [V (119904) minus 119908 (119904)]lowast) 119889119904+ int1minus120575120575
1003816100381610038161003816119866 (1199051 119904) minus 119866 (1199052 119904)1003816100381610038161003816sdot 119891 (119904 1198681205730+ [V (119904) minus 119908 (119904)]lowast [V (119904) minus 119908 (119904)]lowast) 119889119904
6 Journal of Function Spaces
le 2119872int1205750119904 (1 minus 119904)120572minus120573minus1 120601 (119904)
sdot (120588 (1198681205730+) + ℎ (0)) 119889119904 + 2119872int11minus120575
119904 (1 minus 119904)120572minus120573minus1sdot 120601 (119904) (120588 (1198681205730+) + ℎ (0)) 119889119904+ int1minus120575120575
1003816100381610038161003816119866 (1199051 119904) minus 119866 (1199052 119904)1003816100381610038161003816sdot 119897 (120588 (1198681205730+) + ℎ (0)) 119889119904 lt 1205763 + 1205763 + 1205763 = 120576
(40)
for any 1199051 1199052 isin [0 1] with |1199051 minus 1199052| lt 120575 and any V isin 119861 where119897 = max120575le119905le1minus120575120601(119905) which deduces that 119865 is equicontinuouson [0 1]Thus according to Ascoli-Arzela theorem we knowthat 119865119861 is a relatively compact set and that 119865 is a completelycontinuous operator
Theorem 12 Suppose that (1198671) minus (1198674) hold then the FVP (1)has at least one positive solution
Proof For any V isin 1205971198751199031 where 1198751199031 = V isin 119875 | V lt 1199031 by(32) Lemma 8 and condition (1198673) one can get that
119865V (119905) = int10119866 (119905 119904)
sdot [119891 (119904 1198681205730+ [V (119904) minus 119908 (119904)]lowast [V (119904) minus 119908 (119904)]lowast)+ 119890+ (119904)] 119889119904 le int1
0119866 (119905 119904)
sdot [120601 (119904) (120588 (1198681205730+ (V (119904) minus 119908 (119904)) + ℎ (0))+ 119890+ (119904)] 119889119904 le int1
0119866 (119905 119904)
sdot [120601 (119904) (120588 (1198681205730+ (V + 119908)) + ℎ (0)) + 119890+ (119904)] 119889119904= 119872int1
0119904 (1 minus 119904)120572minus120573minus1
sdot [120601 (119904) (120588 (1198681205730+ (1199031 + 119908)) + ℎ (0)) + 119890+ (119904)] 119889119904= 119872int1
0119904 (1 minus 119904)120572minus120573minus1
sdot [120601 (119904) (120588(1199031 + 119908Γ (120573) ) + ℎ (0)) + 119890+ (119904)] 119889119904lt 1199031 = V
(41)
ie 119865V le V V isin 1205971198751199031 By condition (1198674) exist119883 gt 0 such that
119891 (119905 119909 119910) gt 119871 (119909 + 119910) (42)
for any 119909 gt 119883 and any 119905 isin [119886 119887] Choose 1199032 such that
1199032 gt max1199031 21198881 2 (1198721 + 119892 (0) [120572 minus 120573])1198831198981119886120572minus120573minus1 (43)
where
1198881 = (1198721 + 119892 (0) [120572 minus 120573])2 sdot int10 (1 minus 119904)120572minus120573minus1 119890minus (119904) 1198891199041198981Γ (120572 minus 120573) 119892 (0) (44)
For any V isin 1205971198751199032 where 1198751199032 = V isin 119875 | V lt 1199032 Because119908 (119905) = int1
0119866 (119905 119904) 119890minus (119904) 119889119904
le 1198721 + 119892 (0) [120572 minus 120573]Γ (120572 minus 120573) 119892 (0)sdot 119905120572minus120573minus1 int1
0(1 minus 119904)120572minus120573minus1 119890minus (119904) 119889119904
le (1198721 + 119892 (0) [120572 minus 120573])2 sdot int10 (1 minus 119904)120572minus120573minus1 119890minus (119904) 1198891199041198981Γ (120572 minus 120573) 119892 (0) 1199032sdot V (119905) = 11988811199032 V (119905)
(45)
so we have
V (119905) minus 119908 (119905) ge (1 minus 11988811199032) V (119905) ge 12V (119905) (46)
and then for 119905 isin [119886 119887]V (119905) minus 119908 (119905) ge 12V (119905) ge 12 sdot 1198981119886120572minus120573minus11198721 + 119892 (0) [120572 minus 120573] sdot V
= 12 sdot 1198981119886120572minus120573minus11198721 + 119892 (0) [120572 minus 120573] sdot 1199032 gt 119883(47)
follows from (43) and the definition of cone 119875From (42) (43) and (47) one can obtain that
119865V (119905) = int10119866 (119905 119904)
sdot [119891 (119904 1198681205730+ [V (119904) minus 119908 (119904)]lowast [V (119904) minus 119908 (119904)]lowast)+ 119890+ (119904)] 119889119904 ge int119887
119886119866 (119905 119904) 119871 (1198681205730+ (V (119904) minus 119908 (119904))
+ (V (119904) minus 119908 (119904))) + 119890+ (119904) ]119889119904 ge 119871int119887119886119866 (119905 119904)
sdot 12V (119904) 119889119904 ge 1198712 sdot 1198981Γ (120572 minus 120573) 119892 (0) sdot int119887
119886119904 (1
minus 119904)120572minus120573minus1 119905120572minus120573minus1V (119904) 119889119904 ge 1198712 sdot 1198981119886120572minus120573minus1Γ (120572 minus 120573) 119892 (0)
Journal of Function Spaces 7
sdot 11989811198721 + 119892 (0) [120572 minus 120573] sdot int119887
119886119904 (1 minus 119904)120572minus120573minus1 119904120572minus120573minus1119889119904
sdot V = 1198712 sdot 1198981119886120572minus120573minus1Γ (120572 minus 120573) 119892 (0) sdot11989811198721 + 119892 (0) [120572 minus 120573]
sdot int119887119886(1 minus 119904)120572minus120573minus1 119904120572minus120573119889119904 sdot V ge 1198712 sdot 1198981119886120572minus120573minus1Γ (120572 minus 120573) 119892 (0)
sdot 11989811198721 + 119892 (0) [120572 minus 120573] sdot int119887
119886(1 minus 119904)120572minus120573minus1 119904120572minus120573119889119904 sdot 1199032
= 1199032 = V (48)
ie 119865V ge V V isin 1205971198751199032 It follows from Lemma 10 that 119865 has at least fixed point
V1 isin 1198751199032 1198751199031 ie V1 satisfies119863120572minus1205730+ V1 (119905) + 119891 (119905 1198681205730+ (V1 (119905) minus 119908 (119905)) V1 (119905) minus 119908 (119905))
+ 119890 (119905) = 0 0 lt 119905 lt 1V1 (0) = V10158401 (0) = sdot sdot sdot = V(119899minus2)1 (0) = 0V1 (1) = 119898minus2sum
119894=1
120578119894V1 (120585119894) (49)
Set V1(119905) = V1(119905) minus 119908(119905) noticing that V1(119905) 119908(119905) are thesolutions of BVP (32) and (49) respectively therefore we canconclude that V1(119905) is a positive solution of (22) Let 1199091(119905) =1198681205730+V1(119905) then 1199091(119905) is a positive solution of the nonlinearfractional differential equations (1)
4 Main Result II 119891 Is Singular withRespect to Both the Time Variables andthe Space Variable
In this section we always suppose that the following condi-tion holds
(1198675) 119891(119905 119906 V) (0 1) times [0infin) times (0infin) 997888rarr [0infin)is continuous there exist 120576 isin (0 1) and 1205831 1205832 isin 119862+[0 1]1205831(119905) equiv 0 for 119905 isin [0 1] such that
1205831 (119905)(119909 + 119910)120576 le 119891 (119905 119909 119910) le1205832 (119905)(119909 + 119910)120576 (50)
for any (119909 119910) isin [0infin) times (0infin) 119905 isin (0 1) where119862+ [0 1] = 119909 (119905) isin 119862 [0 1] | 119909 (119905) ge 0 119905 isin [0 1] (51)
Set
120593 (119905) = int10119866 (119905 119904) 119890 (119904) 119889119904 119905 isin [0 1] (52)
it follows from Lemma 5 that 120593(119905) is the solution of thefollowing linear equation
119863120572minus1205730+ V (119905) + 119890 (119905) = 0 0 lt 119905 lt 1V (0) = V1015840 (0) = sdot sdot sdot = V(119899minus2) (0) = 0V (1) = 119898minus2sum
119894=1
120578119894V (120585119894) (53)
Denote
1198861 (119905) = int10119866 (119905 119904) 1205831 (119904) 119889119904
1198862 (119905) = int10119866 (119905 119904) 1205832 (119904) 119889119904
120593lowast = inf0le119905le1
120593 (119905) 120593lowast = sup0le119905le1
120593 (119905) 1198861lowast = min
0le119905le11198861 (119905)
119886lowast1 = max0le119905le1
1198861 (119905) 1198862lowast = min
0le119905le11198862 (119905)
119886lowast2 = max0le119905le1
1198862 (119905)
(54)
Clearly 119886lowast119895 ge 119886119895lowast gt 0 119895 = 1 2Theorem 13 Suppose that the condition (H5) holds and 120593lowast ge0 en the FVP (1) has at least one positive solution
Proof Because 120593lowast ge 0 so we can choose 119877 gt 0 large enoughsuch that
119903119877120576 + 120593lowast ge 1119877119877120576 (119886lowast2 + 120593lowast) le 119877
(55)
where 119903 = 1198861lowast[1 + 1(1 + 120573Γ(120573))]120576 In fact since
lim119909997888rarr+infin
119909120576119909 = 0 lt min119903 1119886lowast2 + 120593lowast (56)
there exists 1198831 gt 0 such that 119909120576119909 lt min119903 1(119886lowast2 + 120593lowast) forany 119909 gt 1198831 ie
119903119909120576 ge 1119909 119909120576 (119886lowast2 + 120593lowast) le 119909 (57)
for any 119909 gt 1198831 If 120593lowast gt 0 then from lim119909997888rarr+infin[1119909 minus 119903119909120576] =0 lt 120593lowast one can get that there exists 1198832 gt 0 such that 1119909 minus119903119909120576 lt 120593lowast for any 119909 gt 1198832 ie119903119909120576 + 120593lowast ge 1119909 (58)
8 Journal of Function Spaces
for any 119909 gt 1198832 By (57) (58) we can choose 119877 gt max1198831 1198832such that 119877 satisfies (55) Set
119863 = V isin 119862+ [0 1] 1119877 le V (119905) le 119877 119905 isin [0 1] (59)
For any V isin 1198631 from (23) we have
1198681205730+V (119905) = 1Γ (120573) int119905
0(119905 minus 119904)120573minus1 V (119904) 119889119904
le 119877Γ (120573) int119905
0(119905 minus 119904)120573minus1 119889119904 = 119877119905120573120573Γ (120573) le 119877120573Γ (120573)
(60)
1198681205730+V (119905) = 1Γ (120573) int119905
0(119905 minus 119904)120573minus1 V (119904) 119889119904
ge 1119877Γ (120573) int119905
0(119905 minus 119904)120573minus1 119889119904 = 119905120573119877120573Γ (120573)
(61)
It follows from (60) (61) and (H5) that
1205831 (119905)119877120576 (1 + 1120573Γ (120573))120576 le1205831 (119905)
(V (119905) + 1198681205730+V (119905))120576le 119891 (119905 1198681205730+V (119905) V (119905))
119891 (119905 1198681205730+V (119905) V (119905)) le 1205832 (119905)(V (119905) + 1198681205730+V (119905))120576
le 1205832 (119905)(1119877 + 119905120573119877120573Γ (120573))120576le 1198771205761205832 (119905)
(62)
ie
1205831 (119905)119877120576 (1 + 1120573Γ (120573))120576 le 119891 (119905 1198681205730+V (119905) V (119905)) le 1198771205761205832 (119905) (63)
And then
119879V (119905) = int10119866 (119905 119904) [119891 (119904 1198681205730+V (119904) V (119904)) + 119890 (119904)] 119889119904
= int10119866 (119905 119904) 119891 (119904 1198681205730+V (119904) V (119904)) 119889119904 + 120593 (119905)
le int10119866 (119905 119904) 119891 (119904 1198681205730+V (119904) V (119904)) 119889119904 + 120593lowast
le 119877120576 int10119866 (119905 119904) 1205832 (119904) 119889119904 + 120593lowast lt +infin
(64)
which deduces that the operator 119879 is well definedNow we shall prove that 119879 119863 997888rarr 119863 For V isin 119863 it
is easy to see that 119879V(119905) isin 119862+[0 1] and by (55) (63) we canobtain that
119879V (119905) = int10119866 (119905 119904) [119891 (119904 1198681205730+V (119904) V (119904)) + 119890 (119904)] 119889119904
= int10119866 (119905 119904) 119891 (119904 1198681205730+V (119904) V (119904)) 119889119904 + 120593 (119905)
ge 1119877120576 (1 + 1120573Γ (120573))120576 int
1
0119866 (119905 119904) 1205831 (119904) 119889119904
+ 120593 (119905) ge 1119877120576 (1 + 1120573Γ (120573))120576 1198861lowast + 120593lowast ge
1119877
(65)
119879V (119905) = int10119866 (119905 119904) [119891 (119904 1198681205730+V (119904) V (119904)) + 119890 (119904)] 119889119904
= int10119866 (119905 119904) 119891 (119904 1198681205730+V (119904) V (119904)) 119889119904 + 120593 (119905)
le 119877120576 int10119866 (119905 119904) 1205832 (119904) 119889119904 + 120593lowast le 119877120576119886lowast2 + 120593lowast
le 119877120576 (119886lowast2 + 120593lowast) le 119877
(66)
ie 119879 119863 997888rarr 119863Next let us prove that 119879 119863 997888rarr 119863 is completely
continuousFor any V119899 sub 119863 V0 isin 119863 and V119899 997888rarr V0The continuity
of 119891 deduces that
119891 (119905 1198681205730+V119899 (119905) V119899 (119905)) 997888rarr 119891(119905 1198681205730+V0 (119905) V0 (119905)) 119899 997888rarr infin (67)
and it follows from (63) that100381610038161003816100381610038161003816119891 (119905 1198681205730+V119899 (119905) V119899 (119905)) minus 119891 (119905 1198681205730+V0 (119905) V0 (119905))100381610038161003816100381610038161003816le 2119877120598max max
0le119905le11205831 (119905) max
0le119905le11205832 (119905) (68)
By using the Lebesgue dominated convergence theorem weobtain that
lim119899997888rarrinfin
1003817100381710038171003817119879V119899 minus 119879V01003817100381710038171003817 = lim119899997888rarrinfin
max0le119905le1
100381610038161003816100381610038161003816100381610038161003816int1
0119866 (119905 119904) [119891 (119904 1198681205730+V119899 (119904) V119899 (119904)) minus 119891 (119904 1198681205730+V0 (119904) V0 (119904))] 119889119904
100381610038161003816100381610038161003816100381610038161003816le lim119899997888rarrinfin
max0le119905le1
int10119866 (119905 119904) 100381610038161003816100381610038161003816119891 (119904 1198681205730+V119899 (119904) V119899 (119904)) minus 119891 (119904 1198681205730+V0 (119904) V0 (119904))100381610038161003816100381610038161003816 119889119904
Journal of Function Spaces 9
= max0le119905119904le1
119866 (119905 119904) sdot int10
lim119899997888rarrinfin
100381610038161003816100381610038161003816119891 (119904 1198681205730+V119899 (119904) V119899 (119904)) minus 119891 (119904 1198681205730+V0 (119904) V0 (119904))100381610038161003816100381610038161003816 119889119904 = 0(69)
and this implies that 119879 is a continuous operatorNow we shall prove that 119879 119863 997888rarr 119863 is compact For any
V isin 119863 119879V isin 119863 which deduces that 1119877 le 119879V(119905) le 119877 for119905 isin [0 1] ie 119879 is uniformly boundedSince 119866(119905 119904) isin 119862([0 1] times [0 1]) it is also uniformly
continuous on [0 1] times [0 1] and then for any 120576 gt 0 exist120575 gt 0st for any (1199051 119904) (1199052 119904) isin [0 1] times [0 1] with |1199051 minus 1199052| lt 120575 wealways have
1003816100381610038161003816119866 (1199051 119904) minus 119866 (1199052 119904)1003816100381610038161003816 lt 120576119877120576 int101205832 (119904) 119889119904 + int10 |119890 (119904)| 119889119904
(70)
Thus one can obtain by virtue of (63) (70) that
1003816100381610038161003816119879V (1199051) minus 119879V (1199052)1003816100381610038161003816 =100381610038161003816100381610038161003816100381610038161003816int1
0[119866 (1199051 119904) minus 119866 (1199052 119904)]
sdot [119891 (119904 119868120573V (119904) V (119904)) + 119890 (119904)] 119889119904100381610038161003816100381610038161003816100381610038161003816 le int1
0
1003816100381610038161003816119866 (1199051 119904)minus 119866 (1199052 119904)1003816100381610038161003816 [100381610038161003816100381610038161003816119891 (119904 1198681205730+V (119904) V (119904))100381610038161003816100381610038161003816 + |119890 (119904)|] 119889119904lt [119877120576 int1
01205832 (119904) 119889119904 + int1
0|119890 (119904)| 119889119904]
sdot 120576119877120576 int101205832 (119904) 119889119904 + int10 |119890 (119904)| 119889119904
= 120576
(71)
for any 1199051 1199052 isin [0 1] with |1199051 minus 1199052| lt 120575 and any V isin 1198631 whichshows that 119879 119863 997888rarr 119863 is equicontinuous Thus Arzela-Ascoli theorem guarantees that 119879 119863 997888rarr 119863 is completelycontinuous Existence of at least one fixed point V1 isin 119863follows from Lemma 6 ie 1199061(119905) = 1198681205730+V1(119905) is a positivesolution of differential equation (1) which satisfies
1119877Γ (120572)119905120572 le 1199061 (119905) = 1Γ (120572) int119905
0(119905 minus 119904)120572minus1 V1 (119904) 119889119904
le 119877Γ (120572) 119905120572(72)
Data Availability
The data used to support the findings of this study areavailable from the corresponding author upon request
Conflicts of Interest
The author declares that there are no conflicts of interestregarding the publication of this paper
Acknowledgments
This research was supported by Shandong Provincial Nat-ural Science Foundation of China (no ZR2016FM10) andNational Natural Science Foundation of China (no 11571197)
References
[1] I Podlubny Fractional Differential Equations Mathematics inScience and Engineering Academic Press NewYork NY USA1999
[2] A A Kilbas H M Srivastava and J J Nietoeory and Appli-cational Differential Equations Elsevier Amsterdam Nether-lands 2006
[3] T Hu Y Sun and W Sun ldquoExistence of positive solutions for athird-order multipoint boundary value problem and extensionto fractional caserdquo Boundary Value Problems vol 2016 article197 2016
[4] Y Jia and X Zhang ldquoPositive solutions for a class of fractionaldifferential equation multi-point boundary value problemswith changing sign nonlinearityrdquo Applied Mathematics andComputation vol 47 no 1-2 pp 15ndash31 2015
[5] X Hao ldquoPositive solution for singular fractional differentialequations involving derivativesrdquo Advances in Difference Equa-tions vol 2016 article 139 2016
[6] Y Zou and G He ldquoOn the uniqueness of solutions for a class offractional differential equationsrdquo Applied Mathematics Lettersvol 74 pp 68ndash73 2017
[7] X Hao H Wang L Liu and Y Cui ldquoPositive solutionsfor a system of nonlinear fractional nonlocal boundary valueproblems with parameters and p-Laplacian operatorrdquoBoundaryValue Problems vol 2017 article 182 2017
[8] X Zhang ldquoPositive solutions for a class of singular fractionaldifferential equation with infinite-point boundary value condi-tionsrdquo Applied Mathematics Letters vol 39 pp 22ndash27 2015
[9] K M Zhang ldquoOn a sign-changing solution for some fractionaldifferential equationsrdquo Boundary Value Problems vol 2017article 59 2017
[10] Y L Guan Z Q Zhao and X L Lin ldquoOn the existence ofpositive solutions and negative solutions of singular fractionaldifferential equations via global bifurcation techniquesrdquoBound-ary Value Problems vol 2016 article 141 2016
[11] X G Zhang Y H Wu and L Caccetta ldquoNonlocal fractionalorder differential equations with changing-sign singular pertur-bationrdquo Applied Mathematical Modelling vol 56 pp 116ndash1262015
[12] YWang and L S Liu ldquoPositive solutions for fractionalm-pointboundary value problems in Banach spacesrdquo ActaMathematicaScientia vol 32 no 1 pp 246ndash256 2012
[13] Y Cui and Y Zou ldquoAn existence and uniqueness theorem for asecond order nonlinear system with coupled integral boundaryvalue conditionsrdquo Applied Mathematics and Computation vol256 pp 438ndash444 2015
10 Journal of Function Spaces
[14] X Q Zhang L Wang and Q S Wang ldquoExistence of positivesolutions for a class of nonlinear fractional differential equa-tions with integral boundary conditionsrdquo Applied Mathematicsand Computation vol 226 pp 708ndash718 2014
[15] XG Zhang L S Liu YHWu andBWiwatanapataphee ldquoThespectral analysis for a singular fractional differential equationwith a signedmeasurerdquoApplied Mathematics and Computationvol 257 pp 252ndash263 2015
[16] Y Wang and L Liu ldquoPositive solutions for a class of fractional3-point boundary value problems at resonancerdquo Advances inDifference Equations vol 13 article 13 2017
[17] M A Krasnoselskii and P P Zabreiko Geometrical Methods ofNonlinear Analysis Springer New York NY USA 1984
[18] K Deimling Nonlinear Functional Analysis Springer BerlinGermany 1985
[19] K Zhang ldquoNontrivial solutions of fourth-order singularboundary value problems with sign-changing nonlinear termsrdquoTopological Methods in Nonlinear Analysis vol 40 no 1 pp 53ndash70 2012
[20] D J Guo and V Lakshmikantham Nonlinear Problems inAbstract Cones Academic Press Massachusetts Mass USA1988
[21] J Liu and Z Zhao ldquoMultiple solutions for impulsive problemswith non-autonomous perturbationsrdquo Applied MathematicsLetters vol 64 pp 143ndash149 2017
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4 Journal of Function Spaces
(2) 119866(119905 119904) ge (1198981119892(0)Γ(120572minus120573))119904(1 minus 119904)120572minus120573minus1119905120572minus120573minus1 for any(119905 119904) isin [0 1](3) 119866(119905 119904) le ((1198721 + 119892(0)[120572 minus 120573])119892(0)Γ(120572 minus120573))119904(1 minus 119904)120572minus120573minus1119905120572minus120573minus1 for any (119905 119904) isin [0 1](4) 119866(119905 119904) le 119872119904(1 minus 119904)120572minus120573minus1
where [120572 minus 120573] denotes the integer part of the number 120572 minus 120573119872 = 1198721 + 119892 (0) (120572 minus 120573 minus 1)119892 (0) Γ (120572 minus 120573) (21)
Set V(119905) = 1198631205730+119909(119905) then (1) can be transformed into the
following form
119863120572minus1205730+ V (119905) + 119891 (119905 1198681205730+V (119905) V (119905)) + 119890 (119905) = 00 lt 119905 lt 1
V (0) = V1015840 (0) = sdot sdot sdot = V(119899minus2) (0) = 0V (1) = 119898minus2sum
119894=1
120578119894V (120585119894) (22)
FromLemma 5we know that the solution V(119905) of (22) satisfiesV (119905) = int1
0119866 (119905 119904) [119891 (119904 1198681205730+V (119904) V (119904)) + 119890 (119904)] 119889119904 (23)
Lemma 9 (see [17 18]) Suppose that 119864 is a Banach space and119863 sub 119864 is a bounded convex closed set the operator 119860 119863 997888rarr119863 is completely continuous then 119860 has one fixed point on119863
Lemma 10 (see [20] (Guo-krasnoselrsquoskii fixed point theo-rem)) Let Ω1 and Ω1 be two bounded open sets in Banachspace119864 such that 120579 isin Ω1 andΩ1 sub Ω2119860 119875⋂(Ω2Ω1) 997888rarr 119875a completely continuous operator where 120579 denotes the zeroelement of119864 and119875 a cone of119864 Suppose that one of the followingconditions
(i) 119860119909 ge 119909 forall119909 isin 119875 cap 120597Ω1 and 119860119909 le 119909 forall119909 isin119875 cap 120597Ω2(ii) 119860119909 le 119909 forall119909 isin 119875 cap 120597Ω1 and 119860119909 ge 119909 forall119909 isin119875 cap 120597Ω2
holds en 119860 has at least one fixed point in 119875 cap (Ω2 Ω1)3 Main Result I 119891 Is Singular with Respect tothe Time Variables
Let119864 = 119862[0 1] 119906 = max0le119905le1|119906(119905)| then (119864 sdot) is a Banachspace Set
119875 = V isin 119864 V (119905) ge 11989811198721 + 119892 (0) [120572 minus 120573] 119905120572minus120573minus1 V 119905
isin [0 1] (24)
where [120572 minus 120573] denotes the integer part of the number 120572 minus 120573Then 119875 sub 119864 is a positive cone of 119864 For convenience we listsome conditions which will be used in this section
(1198671) For any (119905 119909 119910) isin (0 1) times [0 +infin) times [0 +infin)0 le 119891 (119905 119909 119910) le 120601 (119905) (120588 (119909) + ℎ (119910)) (25)
where 120601 isin 119862(0 1) 120601(119905) gt 0 on (0 1) 120588(119909) gt 0 is continuousand increasing on [0 +infin) ℎ(119909) gt 0 is continuous anddecreasing on [0 +infin)
(1198672)0 lt int10119904 (1 minus 119904)120572minus120573minus1 (120601 (119904) + 119890+ (119904) + 119890minus (119904)) 119889119904
lt +infin(26)
where 119890+ = max119890(119905) 0 119890minus = minusmin119890(119905) 0 are the positivepart and negative part of 119890(119905) respectively
(1198673) There exists 1199031 gt 0 such that
120588(1199031 + 119908Γ (120573) )int10119904 (1 minus 119904)120572minus120573minus1 120601 (119904) 119889119904
+ int10119904 (1 minus 119904)120572minus120573minus1 [ℎ (0) 120601 (119904) + 119890+ (119904)] 119889119904
lt 1199031119872(27)
where119908(119905) is the solution of the following linear equation
119863120572minus1205730+ V (119905) + 119890minus (119905) = 0 0 lt 119905 lt 1V (0) = V1015840 (0) = sdot sdot sdot = V(119899minus2) (0) = 0V (1) = 119898minus2sum
119894=1
120578119894V (120585119894) (28)
ie 119908(119905) = int10119866(119905 119904)119890minus(119904)119889119904
(1198674) There exists [119886 119887] sub (0 1) such that
lim119909+119910997888rarr+infin
119891 (119905 119909 119910)119909 + 119910 gt 119871 (29)
uniformly holds for 119905 isin [119886 119887] where119871 = 2Γ (120572 minus 120573) 119892 (0) (1198721 + 119892 (0) [120572 minus 120573])
11989812119886120572minus120573minus1 int119887119886 119904120572minus120573 (1 minus 119904)120572minus120573minus1 119889119904 (30)
For any V isin 119875 let[V (119905) minus 119908 (119905)]lowast = max V (119905) minus 119908 (119905) 0 (31)
and define operator
119865V (119905) = int10119866 (119905 119904)
sdot [119891 (119904 1198681205730+ [V (119904) minus 119908 (119904)]lowast [V (119904) minus 119908 (119904)]lowast)+ 119890+ (119904)] 119889119904
(32)
From condition (1198671) and (1198672) it is easy to know that 119865 is welldefined
Journal of Function Spaces 5
Lemma 11 119865 119875 997888rarr 119875 is a completely continuous operator
Proof For any V isin 119875 it follows from Lemma 8 that
119865V (119905) = int10119866 (119905 119904)
sdot [119891 (119904 1198681205730+ [V (119904) minus 119908 (119904)]lowast [V (119904) minus 119908 (119904)]lowast)+ 119890+ (119904)] 119889119904 le 1198721 + 119892 (0) [120572 minus 120573]119892 (0) Γ (120572 minus 120573)sdot 119905120572minus120573minus1 int1
0119904 (1 minus 119904)120572minus120573minus1
sdot [119891 (119904 1198681205730+ [V (119904) minus 119908 (119904)]lowast [V (119904) minus 119908 (119904)]lowast)+ 119890+ (119904)] 119889119904 le 1198721 + 119892 (0) [120572 minus 120573]119892 (0) Γ (120572 minus 120573) int1
0119904 (1
minus 119904)120572minus120573minus1sdot [119891 (119904 1198681205730+ [V (119904) minus 119908 (119904)]lowast [V (119904) minus 119908 (119904)]lowast)+ 119890+ (119904)] 119889119904
119865V (119905) = int10119866 (119905 119904)
sdot [119891 (119904 1198681205730+ [V (119904) minus 119908 (119904)]lowast [V (119904) minus 119908 (119904)]lowast)+ 119890+ (119904)] 119889119904 ge 1198981119892 (0) Γ (120572 minus 120573)119905120572minus120573minus1 int
1
0119904 (1
minus 119904)120572minus120573minus1sdot [119891 (119904 1198681205730+ [V (119904) minus 119908 (119904)]lowast [V (119904) minus 119908 (119904)]lowast)+ 119890+ (119904)] 119889119904
(33)
which deduce that 119865V(119905) ge (1198981(1198721 + 119892(0)[120572 minus120573]))119905120572minus120573minus1119865V ie 119865 119875 997888rarr 119875Let 119861 sub 119875 be a bounded set ie there exists 1198711 gt 0 such
that V le 1198711 for any V isin 119861 then0 le [V (119905) minus 119908 (119905)]lowast le 1198711 + int1
0119866 (119905 119904) 119890minus (119904) 119889119904 le 1198711
+ int10119872119904 (1 minus 119904)120572minus120573minus1 119890minus (119904) 119889119904 ≐
(34)
119865V (119905) = int10119866 (119905 119904)
sdot [119891 (119904 1198681205730+ [V (119904) minus 119908 (119904)]lowast [V (119904) minus 119908 (119904)]lowast)+ 119890+ (119904)] 119889119904 le int1
0119866 (119905 119904)
sdot [120601 (119904) (120588 (1198681205730+) + ℎ (0)) + 119890+ (119904)] 119889119904
le 119872(120588( 120573Γ (120573)) + ℎ (0))int10119904 (1 minus 119904)120572minus120573minus1
sdot 120601 (119904) 119889119904 +119872int10119904 (1 minus 119904)120572minus120573minus1 119890+ (119904) 119889119904
lt +infin(35)
therefore
119865V le 119872(120588( 120573Γ (120573)) + ℎ (0))sdot int10119904 (1 minus 119904)120572minus120573minus1 120601 (119904) 119889119904
+119872int10119904 (1 minus 119904)120572minus120573minus1 119890+ (119904) 119889119904
(36)
for any V isin 119861 which implies that 119865 is uniformly boundedFrom (1198672) the absolutely continuity of integral and the
uniformly continuity of119866(119905 119904) on [0 1] we know that for any120576 gt 0 exist120575 gt 0 such that
int1205750119904 (1 minus 119904)120572minus120573minus1 120601 (119904) 119889119904lt 1205766119872(120588 (120573Γ (120573)) + ℎ (0))
(37)
int1205751minus120575
119904 (1 minus 119904)120572minus120573minus1 120601 (119904) 119889119904lt 1205766119872(120588 (120573Γ (120573)) + ℎ (0))
(38)
and 1003816100381610038161003816119866 (1199051 119904) minus 119866 (1199052 119904)1003816100381610038161003816 lt 1205763119897 (120588 (120573Γ (120573)) + ℎ (0)) (39)
for any 1199051 1199052 119904 isin [0 1] with |1199051 minus 1199052| lt 120575(37)-(39) together with Lemma 8 imply that
1003816100381610038161003816119865V (1199051) minus 119865V (1199052)1003816100381610038161003816 le int10
1003816100381610038161003816119866 (1199051 119904) minus 119866 (1199052 119904)1003816100381610038161003816sdot 119891 (119904 1198681205730+ [V (119904) minus 119908 (119904)]lowast [V (119904) minus 119908 (119904)]lowast) 119889119904= int1205750
1003816100381610038161003816119866 (1199051 119904) minus 119866 (1199052 119904)1003816100381610038161003816sdot 119891 (119904 1198681205730+ [V (119904) minus 119908 (119904)]lowast [V (119904) minus 119908 (119904)]lowast) 119889119904+ int11minus120575
1003816100381610038161003816119866 (1199051 119904) minus 119866 (1199052 119904)1003816100381610038161003816sdot 119891 (119904 1198681205730+ [V (119904) minus 119908 (119904)]lowast [V (119904) minus 119908 (119904)]lowast) 119889119904+ int1minus120575120575
1003816100381610038161003816119866 (1199051 119904) minus 119866 (1199052 119904)1003816100381610038161003816sdot 119891 (119904 1198681205730+ [V (119904) minus 119908 (119904)]lowast [V (119904) minus 119908 (119904)]lowast) 119889119904
6 Journal of Function Spaces
le 2119872int1205750119904 (1 minus 119904)120572minus120573minus1 120601 (119904)
sdot (120588 (1198681205730+) + ℎ (0)) 119889119904 + 2119872int11minus120575
119904 (1 minus 119904)120572minus120573minus1sdot 120601 (119904) (120588 (1198681205730+) + ℎ (0)) 119889119904+ int1minus120575120575
1003816100381610038161003816119866 (1199051 119904) minus 119866 (1199052 119904)1003816100381610038161003816sdot 119897 (120588 (1198681205730+) + ℎ (0)) 119889119904 lt 1205763 + 1205763 + 1205763 = 120576
(40)
for any 1199051 1199052 isin [0 1] with |1199051 minus 1199052| lt 120575 and any V isin 119861 where119897 = max120575le119905le1minus120575120601(119905) which deduces that 119865 is equicontinuouson [0 1]Thus according to Ascoli-Arzela theorem we knowthat 119865119861 is a relatively compact set and that 119865 is a completelycontinuous operator
Theorem 12 Suppose that (1198671) minus (1198674) hold then the FVP (1)has at least one positive solution
Proof For any V isin 1205971198751199031 where 1198751199031 = V isin 119875 | V lt 1199031 by(32) Lemma 8 and condition (1198673) one can get that
119865V (119905) = int10119866 (119905 119904)
sdot [119891 (119904 1198681205730+ [V (119904) minus 119908 (119904)]lowast [V (119904) minus 119908 (119904)]lowast)+ 119890+ (119904)] 119889119904 le int1
0119866 (119905 119904)
sdot [120601 (119904) (120588 (1198681205730+ (V (119904) minus 119908 (119904)) + ℎ (0))+ 119890+ (119904)] 119889119904 le int1
0119866 (119905 119904)
sdot [120601 (119904) (120588 (1198681205730+ (V + 119908)) + ℎ (0)) + 119890+ (119904)] 119889119904= 119872int1
0119904 (1 minus 119904)120572minus120573minus1
sdot [120601 (119904) (120588 (1198681205730+ (1199031 + 119908)) + ℎ (0)) + 119890+ (119904)] 119889119904= 119872int1
0119904 (1 minus 119904)120572minus120573minus1
sdot [120601 (119904) (120588(1199031 + 119908Γ (120573) ) + ℎ (0)) + 119890+ (119904)] 119889119904lt 1199031 = V
(41)
ie 119865V le V V isin 1205971198751199031 By condition (1198674) exist119883 gt 0 such that
119891 (119905 119909 119910) gt 119871 (119909 + 119910) (42)
for any 119909 gt 119883 and any 119905 isin [119886 119887] Choose 1199032 such that
1199032 gt max1199031 21198881 2 (1198721 + 119892 (0) [120572 minus 120573])1198831198981119886120572minus120573minus1 (43)
where
1198881 = (1198721 + 119892 (0) [120572 minus 120573])2 sdot int10 (1 minus 119904)120572minus120573minus1 119890minus (119904) 1198891199041198981Γ (120572 minus 120573) 119892 (0) (44)
For any V isin 1205971198751199032 where 1198751199032 = V isin 119875 | V lt 1199032 Because119908 (119905) = int1
0119866 (119905 119904) 119890minus (119904) 119889119904
le 1198721 + 119892 (0) [120572 minus 120573]Γ (120572 minus 120573) 119892 (0)sdot 119905120572minus120573minus1 int1
0(1 minus 119904)120572minus120573minus1 119890minus (119904) 119889119904
le (1198721 + 119892 (0) [120572 minus 120573])2 sdot int10 (1 minus 119904)120572minus120573minus1 119890minus (119904) 1198891199041198981Γ (120572 minus 120573) 119892 (0) 1199032sdot V (119905) = 11988811199032 V (119905)
(45)
so we have
V (119905) minus 119908 (119905) ge (1 minus 11988811199032) V (119905) ge 12V (119905) (46)
and then for 119905 isin [119886 119887]V (119905) minus 119908 (119905) ge 12V (119905) ge 12 sdot 1198981119886120572minus120573minus11198721 + 119892 (0) [120572 minus 120573] sdot V
= 12 sdot 1198981119886120572minus120573minus11198721 + 119892 (0) [120572 minus 120573] sdot 1199032 gt 119883(47)
follows from (43) and the definition of cone 119875From (42) (43) and (47) one can obtain that
119865V (119905) = int10119866 (119905 119904)
sdot [119891 (119904 1198681205730+ [V (119904) minus 119908 (119904)]lowast [V (119904) minus 119908 (119904)]lowast)+ 119890+ (119904)] 119889119904 ge int119887
119886119866 (119905 119904) 119871 (1198681205730+ (V (119904) minus 119908 (119904))
+ (V (119904) minus 119908 (119904))) + 119890+ (119904) ]119889119904 ge 119871int119887119886119866 (119905 119904)
sdot 12V (119904) 119889119904 ge 1198712 sdot 1198981Γ (120572 minus 120573) 119892 (0) sdot int119887
119886119904 (1
minus 119904)120572minus120573minus1 119905120572minus120573minus1V (119904) 119889119904 ge 1198712 sdot 1198981119886120572minus120573minus1Γ (120572 minus 120573) 119892 (0)
Journal of Function Spaces 7
sdot 11989811198721 + 119892 (0) [120572 minus 120573] sdot int119887
119886119904 (1 minus 119904)120572minus120573minus1 119904120572minus120573minus1119889119904
sdot V = 1198712 sdot 1198981119886120572minus120573minus1Γ (120572 minus 120573) 119892 (0) sdot11989811198721 + 119892 (0) [120572 minus 120573]
sdot int119887119886(1 minus 119904)120572minus120573minus1 119904120572minus120573119889119904 sdot V ge 1198712 sdot 1198981119886120572minus120573minus1Γ (120572 minus 120573) 119892 (0)
sdot 11989811198721 + 119892 (0) [120572 minus 120573] sdot int119887
119886(1 minus 119904)120572minus120573minus1 119904120572minus120573119889119904 sdot 1199032
= 1199032 = V (48)
ie 119865V ge V V isin 1205971198751199032 It follows from Lemma 10 that 119865 has at least fixed point
V1 isin 1198751199032 1198751199031 ie V1 satisfies119863120572minus1205730+ V1 (119905) + 119891 (119905 1198681205730+ (V1 (119905) minus 119908 (119905)) V1 (119905) minus 119908 (119905))
+ 119890 (119905) = 0 0 lt 119905 lt 1V1 (0) = V10158401 (0) = sdot sdot sdot = V(119899minus2)1 (0) = 0V1 (1) = 119898minus2sum
119894=1
120578119894V1 (120585119894) (49)
Set V1(119905) = V1(119905) minus 119908(119905) noticing that V1(119905) 119908(119905) are thesolutions of BVP (32) and (49) respectively therefore we canconclude that V1(119905) is a positive solution of (22) Let 1199091(119905) =1198681205730+V1(119905) then 1199091(119905) is a positive solution of the nonlinearfractional differential equations (1)
4 Main Result II 119891 Is Singular withRespect to Both the Time Variables andthe Space Variable
In this section we always suppose that the following condi-tion holds
(1198675) 119891(119905 119906 V) (0 1) times [0infin) times (0infin) 997888rarr [0infin)is continuous there exist 120576 isin (0 1) and 1205831 1205832 isin 119862+[0 1]1205831(119905) equiv 0 for 119905 isin [0 1] such that
1205831 (119905)(119909 + 119910)120576 le 119891 (119905 119909 119910) le1205832 (119905)(119909 + 119910)120576 (50)
for any (119909 119910) isin [0infin) times (0infin) 119905 isin (0 1) where119862+ [0 1] = 119909 (119905) isin 119862 [0 1] | 119909 (119905) ge 0 119905 isin [0 1] (51)
Set
120593 (119905) = int10119866 (119905 119904) 119890 (119904) 119889119904 119905 isin [0 1] (52)
it follows from Lemma 5 that 120593(119905) is the solution of thefollowing linear equation
119863120572minus1205730+ V (119905) + 119890 (119905) = 0 0 lt 119905 lt 1V (0) = V1015840 (0) = sdot sdot sdot = V(119899minus2) (0) = 0V (1) = 119898minus2sum
119894=1
120578119894V (120585119894) (53)
Denote
1198861 (119905) = int10119866 (119905 119904) 1205831 (119904) 119889119904
1198862 (119905) = int10119866 (119905 119904) 1205832 (119904) 119889119904
120593lowast = inf0le119905le1
120593 (119905) 120593lowast = sup0le119905le1
120593 (119905) 1198861lowast = min
0le119905le11198861 (119905)
119886lowast1 = max0le119905le1
1198861 (119905) 1198862lowast = min
0le119905le11198862 (119905)
119886lowast2 = max0le119905le1
1198862 (119905)
(54)
Clearly 119886lowast119895 ge 119886119895lowast gt 0 119895 = 1 2Theorem 13 Suppose that the condition (H5) holds and 120593lowast ge0 en the FVP (1) has at least one positive solution
Proof Because 120593lowast ge 0 so we can choose 119877 gt 0 large enoughsuch that
119903119877120576 + 120593lowast ge 1119877119877120576 (119886lowast2 + 120593lowast) le 119877
(55)
where 119903 = 1198861lowast[1 + 1(1 + 120573Γ(120573))]120576 In fact since
lim119909997888rarr+infin
119909120576119909 = 0 lt min119903 1119886lowast2 + 120593lowast (56)
there exists 1198831 gt 0 such that 119909120576119909 lt min119903 1(119886lowast2 + 120593lowast) forany 119909 gt 1198831 ie
119903119909120576 ge 1119909 119909120576 (119886lowast2 + 120593lowast) le 119909 (57)
for any 119909 gt 1198831 If 120593lowast gt 0 then from lim119909997888rarr+infin[1119909 minus 119903119909120576] =0 lt 120593lowast one can get that there exists 1198832 gt 0 such that 1119909 minus119903119909120576 lt 120593lowast for any 119909 gt 1198832 ie119903119909120576 + 120593lowast ge 1119909 (58)
8 Journal of Function Spaces
for any 119909 gt 1198832 By (57) (58) we can choose 119877 gt max1198831 1198832such that 119877 satisfies (55) Set
119863 = V isin 119862+ [0 1] 1119877 le V (119905) le 119877 119905 isin [0 1] (59)
For any V isin 1198631 from (23) we have
1198681205730+V (119905) = 1Γ (120573) int119905
0(119905 minus 119904)120573minus1 V (119904) 119889119904
le 119877Γ (120573) int119905
0(119905 minus 119904)120573minus1 119889119904 = 119877119905120573120573Γ (120573) le 119877120573Γ (120573)
(60)
1198681205730+V (119905) = 1Γ (120573) int119905
0(119905 minus 119904)120573minus1 V (119904) 119889119904
ge 1119877Γ (120573) int119905
0(119905 minus 119904)120573minus1 119889119904 = 119905120573119877120573Γ (120573)
(61)
It follows from (60) (61) and (H5) that
1205831 (119905)119877120576 (1 + 1120573Γ (120573))120576 le1205831 (119905)
(V (119905) + 1198681205730+V (119905))120576le 119891 (119905 1198681205730+V (119905) V (119905))
119891 (119905 1198681205730+V (119905) V (119905)) le 1205832 (119905)(V (119905) + 1198681205730+V (119905))120576
le 1205832 (119905)(1119877 + 119905120573119877120573Γ (120573))120576le 1198771205761205832 (119905)
(62)
ie
1205831 (119905)119877120576 (1 + 1120573Γ (120573))120576 le 119891 (119905 1198681205730+V (119905) V (119905)) le 1198771205761205832 (119905) (63)
And then
119879V (119905) = int10119866 (119905 119904) [119891 (119904 1198681205730+V (119904) V (119904)) + 119890 (119904)] 119889119904
= int10119866 (119905 119904) 119891 (119904 1198681205730+V (119904) V (119904)) 119889119904 + 120593 (119905)
le int10119866 (119905 119904) 119891 (119904 1198681205730+V (119904) V (119904)) 119889119904 + 120593lowast
le 119877120576 int10119866 (119905 119904) 1205832 (119904) 119889119904 + 120593lowast lt +infin
(64)
which deduces that the operator 119879 is well definedNow we shall prove that 119879 119863 997888rarr 119863 For V isin 119863 it
is easy to see that 119879V(119905) isin 119862+[0 1] and by (55) (63) we canobtain that
119879V (119905) = int10119866 (119905 119904) [119891 (119904 1198681205730+V (119904) V (119904)) + 119890 (119904)] 119889119904
= int10119866 (119905 119904) 119891 (119904 1198681205730+V (119904) V (119904)) 119889119904 + 120593 (119905)
ge 1119877120576 (1 + 1120573Γ (120573))120576 int
1
0119866 (119905 119904) 1205831 (119904) 119889119904
+ 120593 (119905) ge 1119877120576 (1 + 1120573Γ (120573))120576 1198861lowast + 120593lowast ge
1119877
(65)
119879V (119905) = int10119866 (119905 119904) [119891 (119904 1198681205730+V (119904) V (119904)) + 119890 (119904)] 119889119904
= int10119866 (119905 119904) 119891 (119904 1198681205730+V (119904) V (119904)) 119889119904 + 120593 (119905)
le 119877120576 int10119866 (119905 119904) 1205832 (119904) 119889119904 + 120593lowast le 119877120576119886lowast2 + 120593lowast
le 119877120576 (119886lowast2 + 120593lowast) le 119877
(66)
ie 119879 119863 997888rarr 119863Next let us prove that 119879 119863 997888rarr 119863 is completely
continuousFor any V119899 sub 119863 V0 isin 119863 and V119899 997888rarr V0The continuity
of 119891 deduces that
119891 (119905 1198681205730+V119899 (119905) V119899 (119905)) 997888rarr 119891(119905 1198681205730+V0 (119905) V0 (119905)) 119899 997888rarr infin (67)
and it follows from (63) that100381610038161003816100381610038161003816119891 (119905 1198681205730+V119899 (119905) V119899 (119905)) minus 119891 (119905 1198681205730+V0 (119905) V0 (119905))100381610038161003816100381610038161003816le 2119877120598max max
0le119905le11205831 (119905) max
0le119905le11205832 (119905) (68)
By using the Lebesgue dominated convergence theorem weobtain that
lim119899997888rarrinfin
1003817100381710038171003817119879V119899 minus 119879V01003817100381710038171003817 = lim119899997888rarrinfin
max0le119905le1
100381610038161003816100381610038161003816100381610038161003816int1
0119866 (119905 119904) [119891 (119904 1198681205730+V119899 (119904) V119899 (119904)) minus 119891 (119904 1198681205730+V0 (119904) V0 (119904))] 119889119904
100381610038161003816100381610038161003816100381610038161003816le lim119899997888rarrinfin
max0le119905le1
int10119866 (119905 119904) 100381610038161003816100381610038161003816119891 (119904 1198681205730+V119899 (119904) V119899 (119904)) minus 119891 (119904 1198681205730+V0 (119904) V0 (119904))100381610038161003816100381610038161003816 119889119904
Journal of Function Spaces 9
= max0le119905119904le1
119866 (119905 119904) sdot int10
lim119899997888rarrinfin
100381610038161003816100381610038161003816119891 (119904 1198681205730+V119899 (119904) V119899 (119904)) minus 119891 (119904 1198681205730+V0 (119904) V0 (119904))100381610038161003816100381610038161003816 119889119904 = 0(69)
and this implies that 119879 is a continuous operatorNow we shall prove that 119879 119863 997888rarr 119863 is compact For any
V isin 119863 119879V isin 119863 which deduces that 1119877 le 119879V(119905) le 119877 for119905 isin [0 1] ie 119879 is uniformly boundedSince 119866(119905 119904) isin 119862([0 1] times [0 1]) it is also uniformly
continuous on [0 1] times [0 1] and then for any 120576 gt 0 exist120575 gt 0st for any (1199051 119904) (1199052 119904) isin [0 1] times [0 1] with |1199051 minus 1199052| lt 120575 wealways have
1003816100381610038161003816119866 (1199051 119904) minus 119866 (1199052 119904)1003816100381610038161003816 lt 120576119877120576 int101205832 (119904) 119889119904 + int10 |119890 (119904)| 119889119904
(70)
Thus one can obtain by virtue of (63) (70) that
1003816100381610038161003816119879V (1199051) minus 119879V (1199052)1003816100381610038161003816 =100381610038161003816100381610038161003816100381610038161003816int1
0[119866 (1199051 119904) minus 119866 (1199052 119904)]
sdot [119891 (119904 119868120573V (119904) V (119904)) + 119890 (119904)] 119889119904100381610038161003816100381610038161003816100381610038161003816 le int1
0
1003816100381610038161003816119866 (1199051 119904)minus 119866 (1199052 119904)1003816100381610038161003816 [100381610038161003816100381610038161003816119891 (119904 1198681205730+V (119904) V (119904))100381610038161003816100381610038161003816 + |119890 (119904)|] 119889119904lt [119877120576 int1
01205832 (119904) 119889119904 + int1
0|119890 (119904)| 119889119904]
sdot 120576119877120576 int101205832 (119904) 119889119904 + int10 |119890 (119904)| 119889119904
= 120576
(71)
for any 1199051 1199052 isin [0 1] with |1199051 minus 1199052| lt 120575 and any V isin 1198631 whichshows that 119879 119863 997888rarr 119863 is equicontinuous Thus Arzela-Ascoli theorem guarantees that 119879 119863 997888rarr 119863 is completelycontinuous Existence of at least one fixed point V1 isin 119863follows from Lemma 6 ie 1199061(119905) = 1198681205730+V1(119905) is a positivesolution of differential equation (1) which satisfies
1119877Γ (120572)119905120572 le 1199061 (119905) = 1Γ (120572) int119905
0(119905 minus 119904)120572minus1 V1 (119904) 119889119904
le 119877Γ (120572) 119905120572(72)
Data Availability
The data used to support the findings of this study areavailable from the corresponding author upon request
Conflicts of Interest
The author declares that there are no conflicts of interestregarding the publication of this paper
Acknowledgments
This research was supported by Shandong Provincial Nat-ural Science Foundation of China (no ZR2016FM10) andNational Natural Science Foundation of China (no 11571197)
References
[1] I Podlubny Fractional Differential Equations Mathematics inScience and Engineering Academic Press NewYork NY USA1999
[2] A A Kilbas H M Srivastava and J J Nietoeory and Appli-cational Differential Equations Elsevier Amsterdam Nether-lands 2006
[3] T Hu Y Sun and W Sun ldquoExistence of positive solutions for athird-order multipoint boundary value problem and extensionto fractional caserdquo Boundary Value Problems vol 2016 article197 2016
[4] Y Jia and X Zhang ldquoPositive solutions for a class of fractionaldifferential equation multi-point boundary value problemswith changing sign nonlinearityrdquo Applied Mathematics andComputation vol 47 no 1-2 pp 15ndash31 2015
[5] X Hao ldquoPositive solution for singular fractional differentialequations involving derivativesrdquo Advances in Difference Equa-tions vol 2016 article 139 2016
[6] Y Zou and G He ldquoOn the uniqueness of solutions for a class offractional differential equationsrdquo Applied Mathematics Lettersvol 74 pp 68ndash73 2017
[7] X Hao H Wang L Liu and Y Cui ldquoPositive solutionsfor a system of nonlinear fractional nonlocal boundary valueproblems with parameters and p-Laplacian operatorrdquoBoundaryValue Problems vol 2017 article 182 2017
[8] X Zhang ldquoPositive solutions for a class of singular fractionaldifferential equation with infinite-point boundary value condi-tionsrdquo Applied Mathematics Letters vol 39 pp 22ndash27 2015
[9] K M Zhang ldquoOn a sign-changing solution for some fractionaldifferential equationsrdquo Boundary Value Problems vol 2017article 59 2017
[10] Y L Guan Z Q Zhao and X L Lin ldquoOn the existence ofpositive solutions and negative solutions of singular fractionaldifferential equations via global bifurcation techniquesrdquoBound-ary Value Problems vol 2016 article 141 2016
[11] X G Zhang Y H Wu and L Caccetta ldquoNonlocal fractionalorder differential equations with changing-sign singular pertur-bationrdquo Applied Mathematical Modelling vol 56 pp 116ndash1262015
[12] YWang and L S Liu ldquoPositive solutions for fractionalm-pointboundary value problems in Banach spacesrdquo ActaMathematicaScientia vol 32 no 1 pp 246ndash256 2012
[13] Y Cui and Y Zou ldquoAn existence and uniqueness theorem for asecond order nonlinear system with coupled integral boundaryvalue conditionsrdquo Applied Mathematics and Computation vol256 pp 438ndash444 2015
10 Journal of Function Spaces
[14] X Q Zhang L Wang and Q S Wang ldquoExistence of positivesolutions for a class of nonlinear fractional differential equa-tions with integral boundary conditionsrdquo Applied Mathematicsand Computation vol 226 pp 708ndash718 2014
[15] XG Zhang L S Liu YHWu andBWiwatanapataphee ldquoThespectral analysis for a singular fractional differential equationwith a signedmeasurerdquoApplied Mathematics and Computationvol 257 pp 252ndash263 2015
[16] Y Wang and L Liu ldquoPositive solutions for a class of fractional3-point boundary value problems at resonancerdquo Advances inDifference Equations vol 13 article 13 2017
[17] M A Krasnoselskii and P P Zabreiko Geometrical Methods ofNonlinear Analysis Springer New York NY USA 1984
[18] K Deimling Nonlinear Functional Analysis Springer BerlinGermany 1985
[19] K Zhang ldquoNontrivial solutions of fourth-order singularboundary value problems with sign-changing nonlinear termsrdquoTopological Methods in Nonlinear Analysis vol 40 no 1 pp 53ndash70 2012
[20] D J Guo and V Lakshmikantham Nonlinear Problems inAbstract Cones Academic Press Massachusetts Mass USA1988
[21] J Liu and Z Zhao ldquoMultiple solutions for impulsive problemswith non-autonomous perturbationsrdquo Applied MathematicsLetters vol 64 pp 143ndash149 2017
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Journal of Function Spaces 5
Lemma 11 119865 119875 997888rarr 119875 is a completely continuous operator
Proof For any V isin 119875 it follows from Lemma 8 that
119865V (119905) = int10119866 (119905 119904)
sdot [119891 (119904 1198681205730+ [V (119904) minus 119908 (119904)]lowast [V (119904) minus 119908 (119904)]lowast)+ 119890+ (119904)] 119889119904 le 1198721 + 119892 (0) [120572 minus 120573]119892 (0) Γ (120572 minus 120573)sdot 119905120572minus120573minus1 int1
0119904 (1 minus 119904)120572minus120573minus1
sdot [119891 (119904 1198681205730+ [V (119904) minus 119908 (119904)]lowast [V (119904) minus 119908 (119904)]lowast)+ 119890+ (119904)] 119889119904 le 1198721 + 119892 (0) [120572 minus 120573]119892 (0) Γ (120572 minus 120573) int1
0119904 (1
minus 119904)120572minus120573minus1sdot [119891 (119904 1198681205730+ [V (119904) minus 119908 (119904)]lowast [V (119904) minus 119908 (119904)]lowast)+ 119890+ (119904)] 119889119904
119865V (119905) = int10119866 (119905 119904)
sdot [119891 (119904 1198681205730+ [V (119904) minus 119908 (119904)]lowast [V (119904) minus 119908 (119904)]lowast)+ 119890+ (119904)] 119889119904 ge 1198981119892 (0) Γ (120572 minus 120573)119905120572minus120573minus1 int
1
0119904 (1
minus 119904)120572minus120573minus1sdot [119891 (119904 1198681205730+ [V (119904) minus 119908 (119904)]lowast [V (119904) minus 119908 (119904)]lowast)+ 119890+ (119904)] 119889119904
(33)
which deduce that 119865V(119905) ge (1198981(1198721 + 119892(0)[120572 minus120573]))119905120572minus120573minus1119865V ie 119865 119875 997888rarr 119875Let 119861 sub 119875 be a bounded set ie there exists 1198711 gt 0 such
that V le 1198711 for any V isin 119861 then0 le [V (119905) minus 119908 (119905)]lowast le 1198711 + int1
0119866 (119905 119904) 119890minus (119904) 119889119904 le 1198711
+ int10119872119904 (1 minus 119904)120572minus120573minus1 119890minus (119904) 119889119904 ≐
(34)
119865V (119905) = int10119866 (119905 119904)
sdot [119891 (119904 1198681205730+ [V (119904) minus 119908 (119904)]lowast [V (119904) minus 119908 (119904)]lowast)+ 119890+ (119904)] 119889119904 le int1
0119866 (119905 119904)
sdot [120601 (119904) (120588 (1198681205730+) + ℎ (0)) + 119890+ (119904)] 119889119904
le 119872(120588( 120573Γ (120573)) + ℎ (0))int10119904 (1 minus 119904)120572minus120573minus1
sdot 120601 (119904) 119889119904 +119872int10119904 (1 minus 119904)120572minus120573minus1 119890+ (119904) 119889119904
lt +infin(35)
therefore
119865V le 119872(120588( 120573Γ (120573)) + ℎ (0))sdot int10119904 (1 minus 119904)120572minus120573minus1 120601 (119904) 119889119904
+119872int10119904 (1 minus 119904)120572minus120573minus1 119890+ (119904) 119889119904
(36)
for any V isin 119861 which implies that 119865 is uniformly boundedFrom (1198672) the absolutely continuity of integral and the
uniformly continuity of119866(119905 119904) on [0 1] we know that for any120576 gt 0 exist120575 gt 0 such that
int1205750119904 (1 minus 119904)120572minus120573minus1 120601 (119904) 119889119904lt 1205766119872(120588 (120573Γ (120573)) + ℎ (0))
(37)
int1205751minus120575
119904 (1 minus 119904)120572minus120573minus1 120601 (119904) 119889119904lt 1205766119872(120588 (120573Γ (120573)) + ℎ (0))
(38)
and 1003816100381610038161003816119866 (1199051 119904) minus 119866 (1199052 119904)1003816100381610038161003816 lt 1205763119897 (120588 (120573Γ (120573)) + ℎ (0)) (39)
for any 1199051 1199052 119904 isin [0 1] with |1199051 minus 1199052| lt 120575(37)-(39) together with Lemma 8 imply that
1003816100381610038161003816119865V (1199051) minus 119865V (1199052)1003816100381610038161003816 le int10
1003816100381610038161003816119866 (1199051 119904) minus 119866 (1199052 119904)1003816100381610038161003816sdot 119891 (119904 1198681205730+ [V (119904) minus 119908 (119904)]lowast [V (119904) minus 119908 (119904)]lowast) 119889119904= int1205750
1003816100381610038161003816119866 (1199051 119904) minus 119866 (1199052 119904)1003816100381610038161003816sdot 119891 (119904 1198681205730+ [V (119904) minus 119908 (119904)]lowast [V (119904) minus 119908 (119904)]lowast) 119889119904+ int11minus120575
1003816100381610038161003816119866 (1199051 119904) minus 119866 (1199052 119904)1003816100381610038161003816sdot 119891 (119904 1198681205730+ [V (119904) minus 119908 (119904)]lowast [V (119904) minus 119908 (119904)]lowast) 119889119904+ int1minus120575120575
1003816100381610038161003816119866 (1199051 119904) minus 119866 (1199052 119904)1003816100381610038161003816sdot 119891 (119904 1198681205730+ [V (119904) minus 119908 (119904)]lowast [V (119904) minus 119908 (119904)]lowast) 119889119904
6 Journal of Function Spaces
le 2119872int1205750119904 (1 minus 119904)120572minus120573minus1 120601 (119904)
sdot (120588 (1198681205730+) + ℎ (0)) 119889119904 + 2119872int11minus120575
119904 (1 minus 119904)120572minus120573minus1sdot 120601 (119904) (120588 (1198681205730+) + ℎ (0)) 119889119904+ int1minus120575120575
1003816100381610038161003816119866 (1199051 119904) minus 119866 (1199052 119904)1003816100381610038161003816sdot 119897 (120588 (1198681205730+) + ℎ (0)) 119889119904 lt 1205763 + 1205763 + 1205763 = 120576
(40)
for any 1199051 1199052 isin [0 1] with |1199051 minus 1199052| lt 120575 and any V isin 119861 where119897 = max120575le119905le1minus120575120601(119905) which deduces that 119865 is equicontinuouson [0 1]Thus according to Ascoli-Arzela theorem we knowthat 119865119861 is a relatively compact set and that 119865 is a completelycontinuous operator
Theorem 12 Suppose that (1198671) minus (1198674) hold then the FVP (1)has at least one positive solution
Proof For any V isin 1205971198751199031 where 1198751199031 = V isin 119875 | V lt 1199031 by(32) Lemma 8 and condition (1198673) one can get that
119865V (119905) = int10119866 (119905 119904)
sdot [119891 (119904 1198681205730+ [V (119904) minus 119908 (119904)]lowast [V (119904) minus 119908 (119904)]lowast)+ 119890+ (119904)] 119889119904 le int1
0119866 (119905 119904)
sdot [120601 (119904) (120588 (1198681205730+ (V (119904) minus 119908 (119904)) + ℎ (0))+ 119890+ (119904)] 119889119904 le int1
0119866 (119905 119904)
sdot [120601 (119904) (120588 (1198681205730+ (V + 119908)) + ℎ (0)) + 119890+ (119904)] 119889119904= 119872int1
0119904 (1 minus 119904)120572minus120573minus1
sdot [120601 (119904) (120588 (1198681205730+ (1199031 + 119908)) + ℎ (0)) + 119890+ (119904)] 119889119904= 119872int1
0119904 (1 minus 119904)120572minus120573minus1
sdot [120601 (119904) (120588(1199031 + 119908Γ (120573) ) + ℎ (0)) + 119890+ (119904)] 119889119904lt 1199031 = V
(41)
ie 119865V le V V isin 1205971198751199031 By condition (1198674) exist119883 gt 0 such that
119891 (119905 119909 119910) gt 119871 (119909 + 119910) (42)
for any 119909 gt 119883 and any 119905 isin [119886 119887] Choose 1199032 such that
1199032 gt max1199031 21198881 2 (1198721 + 119892 (0) [120572 minus 120573])1198831198981119886120572minus120573minus1 (43)
where
1198881 = (1198721 + 119892 (0) [120572 minus 120573])2 sdot int10 (1 minus 119904)120572minus120573minus1 119890minus (119904) 1198891199041198981Γ (120572 minus 120573) 119892 (0) (44)
For any V isin 1205971198751199032 where 1198751199032 = V isin 119875 | V lt 1199032 Because119908 (119905) = int1
0119866 (119905 119904) 119890minus (119904) 119889119904
le 1198721 + 119892 (0) [120572 minus 120573]Γ (120572 minus 120573) 119892 (0)sdot 119905120572minus120573minus1 int1
0(1 minus 119904)120572minus120573minus1 119890minus (119904) 119889119904
le (1198721 + 119892 (0) [120572 minus 120573])2 sdot int10 (1 minus 119904)120572minus120573minus1 119890minus (119904) 1198891199041198981Γ (120572 minus 120573) 119892 (0) 1199032sdot V (119905) = 11988811199032 V (119905)
(45)
so we have
V (119905) minus 119908 (119905) ge (1 minus 11988811199032) V (119905) ge 12V (119905) (46)
and then for 119905 isin [119886 119887]V (119905) minus 119908 (119905) ge 12V (119905) ge 12 sdot 1198981119886120572minus120573minus11198721 + 119892 (0) [120572 minus 120573] sdot V
= 12 sdot 1198981119886120572minus120573minus11198721 + 119892 (0) [120572 minus 120573] sdot 1199032 gt 119883(47)
follows from (43) and the definition of cone 119875From (42) (43) and (47) one can obtain that
119865V (119905) = int10119866 (119905 119904)
sdot [119891 (119904 1198681205730+ [V (119904) minus 119908 (119904)]lowast [V (119904) minus 119908 (119904)]lowast)+ 119890+ (119904)] 119889119904 ge int119887
119886119866 (119905 119904) 119871 (1198681205730+ (V (119904) minus 119908 (119904))
+ (V (119904) minus 119908 (119904))) + 119890+ (119904) ]119889119904 ge 119871int119887119886119866 (119905 119904)
sdot 12V (119904) 119889119904 ge 1198712 sdot 1198981Γ (120572 minus 120573) 119892 (0) sdot int119887
119886119904 (1
minus 119904)120572minus120573minus1 119905120572minus120573minus1V (119904) 119889119904 ge 1198712 sdot 1198981119886120572minus120573minus1Γ (120572 minus 120573) 119892 (0)
Journal of Function Spaces 7
sdot 11989811198721 + 119892 (0) [120572 minus 120573] sdot int119887
119886119904 (1 minus 119904)120572minus120573minus1 119904120572minus120573minus1119889119904
sdot V = 1198712 sdot 1198981119886120572minus120573minus1Γ (120572 minus 120573) 119892 (0) sdot11989811198721 + 119892 (0) [120572 minus 120573]
sdot int119887119886(1 minus 119904)120572minus120573minus1 119904120572minus120573119889119904 sdot V ge 1198712 sdot 1198981119886120572minus120573minus1Γ (120572 minus 120573) 119892 (0)
sdot 11989811198721 + 119892 (0) [120572 minus 120573] sdot int119887
119886(1 minus 119904)120572minus120573minus1 119904120572minus120573119889119904 sdot 1199032
= 1199032 = V (48)
ie 119865V ge V V isin 1205971198751199032 It follows from Lemma 10 that 119865 has at least fixed point
V1 isin 1198751199032 1198751199031 ie V1 satisfies119863120572minus1205730+ V1 (119905) + 119891 (119905 1198681205730+ (V1 (119905) minus 119908 (119905)) V1 (119905) minus 119908 (119905))
+ 119890 (119905) = 0 0 lt 119905 lt 1V1 (0) = V10158401 (0) = sdot sdot sdot = V(119899minus2)1 (0) = 0V1 (1) = 119898minus2sum
119894=1
120578119894V1 (120585119894) (49)
Set V1(119905) = V1(119905) minus 119908(119905) noticing that V1(119905) 119908(119905) are thesolutions of BVP (32) and (49) respectively therefore we canconclude that V1(119905) is a positive solution of (22) Let 1199091(119905) =1198681205730+V1(119905) then 1199091(119905) is a positive solution of the nonlinearfractional differential equations (1)
4 Main Result II 119891 Is Singular withRespect to Both the Time Variables andthe Space Variable
In this section we always suppose that the following condi-tion holds
(1198675) 119891(119905 119906 V) (0 1) times [0infin) times (0infin) 997888rarr [0infin)is continuous there exist 120576 isin (0 1) and 1205831 1205832 isin 119862+[0 1]1205831(119905) equiv 0 for 119905 isin [0 1] such that
1205831 (119905)(119909 + 119910)120576 le 119891 (119905 119909 119910) le1205832 (119905)(119909 + 119910)120576 (50)
for any (119909 119910) isin [0infin) times (0infin) 119905 isin (0 1) where119862+ [0 1] = 119909 (119905) isin 119862 [0 1] | 119909 (119905) ge 0 119905 isin [0 1] (51)
Set
120593 (119905) = int10119866 (119905 119904) 119890 (119904) 119889119904 119905 isin [0 1] (52)
it follows from Lemma 5 that 120593(119905) is the solution of thefollowing linear equation
119863120572minus1205730+ V (119905) + 119890 (119905) = 0 0 lt 119905 lt 1V (0) = V1015840 (0) = sdot sdot sdot = V(119899minus2) (0) = 0V (1) = 119898minus2sum
119894=1
120578119894V (120585119894) (53)
Denote
1198861 (119905) = int10119866 (119905 119904) 1205831 (119904) 119889119904
1198862 (119905) = int10119866 (119905 119904) 1205832 (119904) 119889119904
120593lowast = inf0le119905le1
120593 (119905) 120593lowast = sup0le119905le1
120593 (119905) 1198861lowast = min
0le119905le11198861 (119905)
119886lowast1 = max0le119905le1
1198861 (119905) 1198862lowast = min
0le119905le11198862 (119905)
119886lowast2 = max0le119905le1
1198862 (119905)
(54)
Clearly 119886lowast119895 ge 119886119895lowast gt 0 119895 = 1 2Theorem 13 Suppose that the condition (H5) holds and 120593lowast ge0 en the FVP (1) has at least one positive solution
Proof Because 120593lowast ge 0 so we can choose 119877 gt 0 large enoughsuch that
119903119877120576 + 120593lowast ge 1119877119877120576 (119886lowast2 + 120593lowast) le 119877
(55)
where 119903 = 1198861lowast[1 + 1(1 + 120573Γ(120573))]120576 In fact since
lim119909997888rarr+infin
119909120576119909 = 0 lt min119903 1119886lowast2 + 120593lowast (56)
there exists 1198831 gt 0 such that 119909120576119909 lt min119903 1(119886lowast2 + 120593lowast) forany 119909 gt 1198831 ie
119903119909120576 ge 1119909 119909120576 (119886lowast2 + 120593lowast) le 119909 (57)
for any 119909 gt 1198831 If 120593lowast gt 0 then from lim119909997888rarr+infin[1119909 minus 119903119909120576] =0 lt 120593lowast one can get that there exists 1198832 gt 0 such that 1119909 minus119903119909120576 lt 120593lowast for any 119909 gt 1198832 ie119903119909120576 + 120593lowast ge 1119909 (58)
8 Journal of Function Spaces
for any 119909 gt 1198832 By (57) (58) we can choose 119877 gt max1198831 1198832such that 119877 satisfies (55) Set
119863 = V isin 119862+ [0 1] 1119877 le V (119905) le 119877 119905 isin [0 1] (59)
For any V isin 1198631 from (23) we have
1198681205730+V (119905) = 1Γ (120573) int119905
0(119905 minus 119904)120573minus1 V (119904) 119889119904
le 119877Γ (120573) int119905
0(119905 minus 119904)120573minus1 119889119904 = 119877119905120573120573Γ (120573) le 119877120573Γ (120573)
(60)
1198681205730+V (119905) = 1Γ (120573) int119905
0(119905 minus 119904)120573minus1 V (119904) 119889119904
ge 1119877Γ (120573) int119905
0(119905 minus 119904)120573minus1 119889119904 = 119905120573119877120573Γ (120573)
(61)
It follows from (60) (61) and (H5) that
1205831 (119905)119877120576 (1 + 1120573Γ (120573))120576 le1205831 (119905)
(V (119905) + 1198681205730+V (119905))120576le 119891 (119905 1198681205730+V (119905) V (119905))
119891 (119905 1198681205730+V (119905) V (119905)) le 1205832 (119905)(V (119905) + 1198681205730+V (119905))120576
le 1205832 (119905)(1119877 + 119905120573119877120573Γ (120573))120576le 1198771205761205832 (119905)
(62)
ie
1205831 (119905)119877120576 (1 + 1120573Γ (120573))120576 le 119891 (119905 1198681205730+V (119905) V (119905)) le 1198771205761205832 (119905) (63)
And then
119879V (119905) = int10119866 (119905 119904) [119891 (119904 1198681205730+V (119904) V (119904)) + 119890 (119904)] 119889119904
= int10119866 (119905 119904) 119891 (119904 1198681205730+V (119904) V (119904)) 119889119904 + 120593 (119905)
le int10119866 (119905 119904) 119891 (119904 1198681205730+V (119904) V (119904)) 119889119904 + 120593lowast
le 119877120576 int10119866 (119905 119904) 1205832 (119904) 119889119904 + 120593lowast lt +infin
(64)
which deduces that the operator 119879 is well definedNow we shall prove that 119879 119863 997888rarr 119863 For V isin 119863 it
is easy to see that 119879V(119905) isin 119862+[0 1] and by (55) (63) we canobtain that
119879V (119905) = int10119866 (119905 119904) [119891 (119904 1198681205730+V (119904) V (119904)) + 119890 (119904)] 119889119904
= int10119866 (119905 119904) 119891 (119904 1198681205730+V (119904) V (119904)) 119889119904 + 120593 (119905)
ge 1119877120576 (1 + 1120573Γ (120573))120576 int
1
0119866 (119905 119904) 1205831 (119904) 119889119904
+ 120593 (119905) ge 1119877120576 (1 + 1120573Γ (120573))120576 1198861lowast + 120593lowast ge
1119877
(65)
119879V (119905) = int10119866 (119905 119904) [119891 (119904 1198681205730+V (119904) V (119904)) + 119890 (119904)] 119889119904
= int10119866 (119905 119904) 119891 (119904 1198681205730+V (119904) V (119904)) 119889119904 + 120593 (119905)
le 119877120576 int10119866 (119905 119904) 1205832 (119904) 119889119904 + 120593lowast le 119877120576119886lowast2 + 120593lowast
le 119877120576 (119886lowast2 + 120593lowast) le 119877
(66)
ie 119879 119863 997888rarr 119863Next let us prove that 119879 119863 997888rarr 119863 is completely
continuousFor any V119899 sub 119863 V0 isin 119863 and V119899 997888rarr V0The continuity
of 119891 deduces that
119891 (119905 1198681205730+V119899 (119905) V119899 (119905)) 997888rarr 119891(119905 1198681205730+V0 (119905) V0 (119905)) 119899 997888rarr infin (67)
and it follows from (63) that100381610038161003816100381610038161003816119891 (119905 1198681205730+V119899 (119905) V119899 (119905)) minus 119891 (119905 1198681205730+V0 (119905) V0 (119905))100381610038161003816100381610038161003816le 2119877120598max max
0le119905le11205831 (119905) max
0le119905le11205832 (119905) (68)
By using the Lebesgue dominated convergence theorem weobtain that
lim119899997888rarrinfin
1003817100381710038171003817119879V119899 minus 119879V01003817100381710038171003817 = lim119899997888rarrinfin
max0le119905le1
100381610038161003816100381610038161003816100381610038161003816int1
0119866 (119905 119904) [119891 (119904 1198681205730+V119899 (119904) V119899 (119904)) minus 119891 (119904 1198681205730+V0 (119904) V0 (119904))] 119889119904
100381610038161003816100381610038161003816100381610038161003816le lim119899997888rarrinfin
max0le119905le1
int10119866 (119905 119904) 100381610038161003816100381610038161003816119891 (119904 1198681205730+V119899 (119904) V119899 (119904)) minus 119891 (119904 1198681205730+V0 (119904) V0 (119904))100381610038161003816100381610038161003816 119889119904
Journal of Function Spaces 9
= max0le119905119904le1
119866 (119905 119904) sdot int10
lim119899997888rarrinfin
100381610038161003816100381610038161003816119891 (119904 1198681205730+V119899 (119904) V119899 (119904)) minus 119891 (119904 1198681205730+V0 (119904) V0 (119904))100381610038161003816100381610038161003816 119889119904 = 0(69)
and this implies that 119879 is a continuous operatorNow we shall prove that 119879 119863 997888rarr 119863 is compact For any
V isin 119863 119879V isin 119863 which deduces that 1119877 le 119879V(119905) le 119877 for119905 isin [0 1] ie 119879 is uniformly boundedSince 119866(119905 119904) isin 119862([0 1] times [0 1]) it is also uniformly
continuous on [0 1] times [0 1] and then for any 120576 gt 0 exist120575 gt 0st for any (1199051 119904) (1199052 119904) isin [0 1] times [0 1] with |1199051 minus 1199052| lt 120575 wealways have
1003816100381610038161003816119866 (1199051 119904) minus 119866 (1199052 119904)1003816100381610038161003816 lt 120576119877120576 int101205832 (119904) 119889119904 + int10 |119890 (119904)| 119889119904
(70)
Thus one can obtain by virtue of (63) (70) that
1003816100381610038161003816119879V (1199051) minus 119879V (1199052)1003816100381610038161003816 =100381610038161003816100381610038161003816100381610038161003816int1
0[119866 (1199051 119904) minus 119866 (1199052 119904)]
sdot [119891 (119904 119868120573V (119904) V (119904)) + 119890 (119904)] 119889119904100381610038161003816100381610038161003816100381610038161003816 le int1
0
1003816100381610038161003816119866 (1199051 119904)minus 119866 (1199052 119904)1003816100381610038161003816 [100381610038161003816100381610038161003816119891 (119904 1198681205730+V (119904) V (119904))100381610038161003816100381610038161003816 + |119890 (119904)|] 119889119904lt [119877120576 int1
01205832 (119904) 119889119904 + int1
0|119890 (119904)| 119889119904]
sdot 120576119877120576 int101205832 (119904) 119889119904 + int10 |119890 (119904)| 119889119904
= 120576
(71)
for any 1199051 1199052 isin [0 1] with |1199051 minus 1199052| lt 120575 and any V isin 1198631 whichshows that 119879 119863 997888rarr 119863 is equicontinuous Thus Arzela-Ascoli theorem guarantees that 119879 119863 997888rarr 119863 is completelycontinuous Existence of at least one fixed point V1 isin 119863follows from Lemma 6 ie 1199061(119905) = 1198681205730+V1(119905) is a positivesolution of differential equation (1) which satisfies
1119877Γ (120572)119905120572 le 1199061 (119905) = 1Γ (120572) int119905
0(119905 minus 119904)120572minus1 V1 (119904) 119889119904
le 119877Γ (120572) 119905120572(72)
Data Availability
The data used to support the findings of this study areavailable from the corresponding author upon request
Conflicts of Interest
The author declares that there are no conflicts of interestregarding the publication of this paper
Acknowledgments
This research was supported by Shandong Provincial Nat-ural Science Foundation of China (no ZR2016FM10) andNational Natural Science Foundation of China (no 11571197)
References
[1] I Podlubny Fractional Differential Equations Mathematics inScience and Engineering Academic Press NewYork NY USA1999
[2] A A Kilbas H M Srivastava and J J Nietoeory and Appli-cational Differential Equations Elsevier Amsterdam Nether-lands 2006
[3] T Hu Y Sun and W Sun ldquoExistence of positive solutions for athird-order multipoint boundary value problem and extensionto fractional caserdquo Boundary Value Problems vol 2016 article197 2016
[4] Y Jia and X Zhang ldquoPositive solutions for a class of fractionaldifferential equation multi-point boundary value problemswith changing sign nonlinearityrdquo Applied Mathematics andComputation vol 47 no 1-2 pp 15ndash31 2015
[5] X Hao ldquoPositive solution for singular fractional differentialequations involving derivativesrdquo Advances in Difference Equa-tions vol 2016 article 139 2016
[6] Y Zou and G He ldquoOn the uniqueness of solutions for a class offractional differential equationsrdquo Applied Mathematics Lettersvol 74 pp 68ndash73 2017
[7] X Hao H Wang L Liu and Y Cui ldquoPositive solutionsfor a system of nonlinear fractional nonlocal boundary valueproblems with parameters and p-Laplacian operatorrdquoBoundaryValue Problems vol 2017 article 182 2017
[8] X Zhang ldquoPositive solutions for a class of singular fractionaldifferential equation with infinite-point boundary value condi-tionsrdquo Applied Mathematics Letters vol 39 pp 22ndash27 2015
[9] K M Zhang ldquoOn a sign-changing solution for some fractionaldifferential equationsrdquo Boundary Value Problems vol 2017article 59 2017
[10] Y L Guan Z Q Zhao and X L Lin ldquoOn the existence ofpositive solutions and negative solutions of singular fractionaldifferential equations via global bifurcation techniquesrdquoBound-ary Value Problems vol 2016 article 141 2016
[11] X G Zhang Y H Wu and L Caccetta ldquoNonlocal fractionalorder differential equations with changing-sign singular pertur-bationrdquo Applied Mathematical Modelling vol 56 pp 116ndash1262015
[12] YWang and L S Liu ldquoPositive solutions for fractionalm-pointboundary value problems in Banach spacesrdquo ActaMathematicaScientia vol 32 no 1 pp 246ndash256 2012
[13] Y Cui and Y Zou ldquoAn existence and uniqueness theorem for asecond order nonlinear system with coupled integral boundaryvalue conditionsrdquo Applied Mathematics and Computation vol256 pp 438ndash444 2015
10 Journal of Function Spaces
[14] X Q Zhang L Wang and Q S Wang ldquoExistence of positivesolutions for a class of nonlinear fractional differential equa-tions with integral boundary conditionsrdquo Applied Mathematicsand Computation vol 226 pp 708ndash718 2014
[15] XG Zhang L S Liu YHWu andBWiwatanapataphee ldquoThespectral analysis for a singular fractional differential equationwith a signedmeasurerdquoApplied Mathematics and Computationvol 257 pp 252ndash263 2015
[16] Y Wang and L Liu ldquoPositive solutions for a class of fractional3-point boundary value problems at resonancerdquo Advances inDifference Equations vol 13 article 13 2017
[17] M A Krasnoselskii and P P Zabreiko Geometrical Methods ofNonlinear Analysis Springer New York NY USA 1984
[18] K Deimling Nonlinear Functional Analysis Springer BerlinGermany 1985
[19] K Zhang ldquoNontrivial solutions of fourth-order singularboundary value problems with sign-changing nonlinear termsrdquoTopological Methods in Nonlinear Analysis vol 40 no 1 pp 53ndash70 2012
[20] D J Guo and V Lakshmikantham Nonlinear Problems inAbstract Cones Academic Press Massachusetts Mass USA1988
[21] J Liu and Z Zhao ldquoMultiple solutions for impulsive problemswith non-autonomous perturbationsrdquo Applied MathematicsLetters vol 64 pp 143ndash149 2017
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6 Journal of Function Spaces
le 2119872int1205750119904 (1 minus 119904)120572minus120573minus1 120601 (119904)
sdot (120588 (1198681205730+) + ℎ (0)) 119889119904 + 2119872int11minus120575
119904 (1 minus 119904)120572minus120573minus1sdot 120601 (119904) (120588 (1198681205730+) + ℎ (0)) 119889119904+ int1minus120575120575
1003816100381610038161003816119866 (1199051 119904) minus 119866 (1199052 119904)1003816100381610038161003816sdot 119897 (120588 (1198681205730+) + ℎ (0)) 119889119904 lt 1205763 + 1205763 + 1205763 = 120576
(40)
for any 1199051 1199052 isin [0 1] with |1199051 minus 1199052| lt 120575 and any V isin 119861 where119897 = max120575le119905le1minus120575120601(119905) which deduces that 119865 is equicontinuouson [0 1]Thus according to Ascoli-Arzela theorem we knowthat 119865119861 is a relatively compact set and that 119865 is a completelycontinuous operator
Theorem 12 Suppose that (1198671) minus (1198674) hold then the FVP (1)has at least one positive solution
Proof For any V isin 1205971198751199031 where 1198751199031 = V isin 119875 | V lt 1199031 by(32) Lemma 8 and condition (1198673) one can get that
119865V (119905) = int10119866 (119905 119904)
sdot [119891 (119904 1198681205730+ [V (119904) minus 119908 (119904)]lowast [V (119904) minus 119908 (119904)]lowast)+ 119890+ (119904)] 119889119904 le int1
0119866 (119905 119904)
sdot [120601 (119904) (120588 (1198681205730+ (V (119904) minus 119908 (119904)) + ℎ (0))+ 119890+ (119904)] 119889119904 le int1
0119866 (119905 119904)
sdot [120601 (119904) (120588 (1198681205730+ (V + 119908)) + ℎ (0)) + 119890+ (119904)] 119889119904= 119872int1
0119904 (1 minus 119904)120572minus120573minus1
sdot [120601 (119904) (120588 (1198681205730+ (1199031 + 119908)) + ℎ (0)) + 119890+ (119904)] 119889119904= 119872int1
0119904 (1 minus 119904)120572minus120573minus1
sdot [120601 (119904) (120588(1199031 + 119908Γ (120573) ) + ℎ (0)) + 119890+ (119904)] 119889119904lt 1199031 = V
(41)
ie 119865V le V V isin 1205971198751199031 By condition (1198674) exist119883 gt 0 such that
119891 (119905 119909 119910) gt 119871 (119909 + 119910) (42)
for any 119909 gt 119883 and any 119905 isin [119886 119887] Choose 1199032 such that
1199032 gt max1199031 21198881 2 (1198721 + 119892 (0) [120572 minus 120573])1198831198981119886120572minus120573minus1 (43)
where
1198881 = (1198721 + 119892 (0) [120572 minus 120573])2 sdot int10 (1 minus 119904)120572minus120573minus1 119890minus (119904) 1198891199041198981Γ (120572 minus 120573) 119892 (0) (44)
For any V isin 1205971198751199032 where 1198751199032 = V isin 119875 | V lt 1199032 Because119908 (119905) = int1
0119866 (119905 119904) 119890minus (119904) 119889119904
le 1198721 + 119892 (0) [120572 minus 120573]Γ (120572 minus 120573) 119892 (0)sdot 119905120572minus120573minus1 int1
0(1 minus 119904)120572minus120573minus1 119890minus (119904) 119889119904
le (1198721 + 119892 (0) [120572 minus 120573])2 sdot int10 (1 minus 119904)120572minus120573minus1 119890minus (119904) 1198891199041198981Γ (120572 minus 120573) 119892 (0) 1199032sdot V (119905) = 11988811199032 V (119905)
(45)
so we have
V (119905) minus 119908 (119905) ge (1 minus 11988811199032) V (119905) ge 12V (119905) (46)
and then for 119905 isin [119886 119887]V (119905) minus 119908 (119905) ge 12V (119905) ge 12 sdot 1198981119886120572minus120573minus11198721 + 119892 (0) [120572 minus 120573] sdot V
= 12 sdot 1198981119886120572minus120573minus11198721 + 119892 (0) [120572 minus 120573] sdot 1199032 gt 119883(47)
follows from (43) and the definition of cone 119875From (42) (43) and (47) one can obtain that
119865V (119905) = int10119866 (119905 119904)
sdot [119891 (119904 1198681205730+ [V (119904) minus 119908 (119904)]lowast [V (119904) minus 119908 (119904)]lowast)+ 119890+ (119904)] 119889119904 ge int119887
119886119866 (119905 119904) 119871 (1198681205730+ (V (119904) minus 119908 (119904))
+ (V (119904) minus 119908 (119904))) + 119890+ (119904) ]119889119904 ge 119871int119887119886119866 (119905 119904)
sdot 12V (119904) 119889119904 ge 1198712 sdot 1198981Γ (120572 minus 120573) 119892 (0) sdot int119887
119886119904 (1
minus 119904)120572minus120573minus1 119905120572minus120573minus1V (119904) 119889119904 ge 1198712 sdot 1198981119886120572minus120573minus1Γ (120572 minus 120573) 119892 (0)
Journal of Function Spaces 7
sdot 11989811198721 + 119892 (0) [120572 minus 120573] sdot int119887
119886119904 (1 minus 119904)120572minus120573minus1 119904120572minus120573minus1119889119904
sdot V = 1198712 sdot 1198981119886120572minus120573minus1Γ (120572 minus 120573) 119892 (0) sdot11989811198721 + 119892 (0) [120572 minus 120573]
sdot int119887119886(1 minus 119904)120572minus120573minus1 119904120572minus120573119889119904 sdot V ge 1198712 sdot 1198981119886120572minus120573minus1Γ (120572 minus 120573) 119892 (0)
sdot 11989811198721 + 119892 (0) [120572 minus 120573] sdot int119887
119886(1 minus 119904)120572minus120573minus1 119904120572minus120573119889119904 sdot 1199032
= 1199032 = V (48)
ie 119865V ge V V isin 1205971198751199032 It follows from Lemma 10 that 119865 has at least fixed point
V1 isin 1198751199032 1198751199031 ie V1 satisfies119863120572minus1205730+ V1 (119905) + 119891 (119905 1198681205730+ (V1 (119905) minus 119908 (119905)) V1 (119905) minus 119908 (119905))
+ 119890 (119905) = 0 0 lt 119905 lt 1V1 (0) = V10158401 (0) = sdot sdot sdot = V(119899minus2)1 (0) = 0V1 (1) = 119898minus2sum
119894=1
120578119894V1 (120585119894) (49)
Set V1(119905) = V1(119905) minus 119908(119905) noticing that V1(119905) 119908(119905) are thesolutions of BVP (32) and (49) respectively therefore we canconclude that V1(119905) is a positive solution of (22) Let 1199091(119905) =1198681205730+V1(119905) then 1199091(119905) is a positive solution of the nonlinearfractional differential equations (1)
4 Main Result II 119891 Is Singular withRespect to Both the Time Variables andthe Space Variable
In this section we always suppose that the following condi-tion holds
(1198675) 119891(119905 119906 V) (0 1) times [0infin) times (0infin) 997888rarr [0infin)is continuous there exist 120576 isin (0 1) and 1205831 1205832 isin 119862+[0 1]1205831(119905) equiv 0 for 119905 isin [0 1] such that
1205831 (119905)(119909 + 119910)120576 le 119891 (119905 119909 119910) le1205832 (119905)(119909 + 119910)120576 (50)
for any (119909 119910) isin [0infin) times (0infin) 119905 isin (0 1) where119862+ [0 1] = 119909 (119905) isin 119862 [0 1] | 119909 (119905) ge 0 119905 isin [0 1] (51)
Set
120593 (119905) = int10119866 (119905 119904) 119890 (119904) 119889119904 119905 isin [0 1] (52)
it follows from Lemma 5 that 120593(119905) is the solution of thefollowing linear equation
119863120572minus1205730+ V (119905) + 119890 (119905) = 0 0 lt 119905 lt 1V (0) = V1015840 (0) = sdot sdot sdot = V(119899minus2) (0) = 0V (1) = 119898minus2sum
119894=1
120578119894V (120585119894) (53)
Denote
1198861 (119905) = int10119866 (119905 119904) 1205831 (119904) 119889119904
1198862 (119905) = int10119866 (119905 119904) 1205832 (119904) 119889119904
120593lowast = inf0le119905le1
120593 (119905) 120593lowast = sup0le119905le1
120593 (119905) 1198861lowast = min
0le119905le11198861 (119905)
119886lowast1 = max0le119905le1
1198861 (119905) 1198862lowast = min
0le119905le11198862 (119905)
119886lowast2 = max0le119905le1
1198862 (119905)
(54)
Clearly 119886lowast119895 ge 119886119895lowast gt 0 119895 = 1 2Theorem 13 Suppose that the condition (H5) holds and 120593lowast ge0 en the FVP (1) has at least one positive solution
Proof Because 120593lowast ge 0 so we can choose 119877 gt 0 large enoughsuch that
119903119877120576 + 120593lowast ge 1119877119877120576 (119886lowast2 + 120593lowast) le 119877
(55)
where 119903 = 1198861lowast[1 + 1(1 + 120573Γ(120573))]120576 In fact since
lim119909997888rarr+infin
119909120576119909 = 0 lt min119903 1119886lowast2 + 120593lowast (56)
there exists 1198831 gt 0 such that 119909120576119909 lt min119903 1(119886lowast2 + 120593lowast) forany 119909 gt 1198831 ie
119903119909120576 ge 1119909 119909120576 (119886lowast2 + 120593lowast) le 119909 (57)
for any 119909 gt 1198831 If 120593lowast gt 0 then from lim119909997888rarr+infin[1119909 minus 119903119909120576] =0 lt 120593lowast one can get that there exists 1198832 gt 0 such that 1119909 minus119903119909120576 lt 120593lowast for any 119909 gt 1198832 ie119903119909120576 + 120593lowast ge 1119909 (58)
8 Journal of Function Spaces
for any 119909 gt 1198832 By (57) (58) we can choose 119877 gt max1198831 1198832such that 119877 satisfies (55) Set
119863 = V isin 119862+ [0 1] 1119877 le V (119905) le 119877 119905 isin [0 1] (59)
For any V isin 1198631 from (23) we have
1198681205730+V (119905) = 1Γ (120573) int119905
0(119905 minus 119904)120573minus1 V (119904) 119889119904
le 119877Γ (120573) int119905
0(119905 minus 119904)120573minus1 119889119904 = 119877119905120573120573Γ (120573) le 119877120573Γ (120573)
(60)
1198681205730+V (119905) = 1Γ (120573) int119905
0(119905 minus 119904)120573minus1 V (119904) 119889119904
ge 1119877Γ (120573) int119905
0(119905 minus 119904)120573minus1 119889119904 = 119905120573119877120573Γ (120573)
(61)
It follows from (60) (61) and (H5) that
1205831 (119905)119877120576 (1 + 1120573Γ (120573))120576 le1205831 (119905)
(V (119905) + 1198681205730+V (119905))120576le 119891 (119905 1198681205730+V (119905) V (119905))
119891 (119905 1198681205730+V (119905) V (119905)) le 1205832 (119905)(V (119905) + 1198681205730+V (119905))120576
le 1205832 (119905)(1119877 + 119905120573119877120573Γ (120573))120576le 1198771205761205832 (119905)
(62)
ie
1205831 (119905)119877120576 (1 + 1120573Γ (120573))120576 le 119891 (119905 1198681205730+V (119905) V (119905)) le 1198771205761205832 (119905) (63)
And then
119879V (119905) = int10119866 (119905 119904) [119891 (119904 1198681205730+V (119904) V (119904)) + 119890 (119904)] 119889119904
= int10119866 (119905 119904) 119891 (119904 1198681205730+V (119904) V (119904)) 119889119904 + 120593 (119905)
le int10119866 (119905 119904) 119891 (119904 1198681205730+V (119904) V (119904)) 119889119904 + 120593lowast
le 119877120576 int10119866 (119905 119904) 1205832 (119904) 119889119904 + 120593lowast lt +infin
(64)
which deduces that the operator 119879 is well definedNow we shall prove that 119879 119863 997888rarr 119863 For V isin 119863 it
is easy to see that 119879V(119905) isin 119862+[0 1] and by (55) (63) we canobtain that
119879V (119905) = int10119866 (119905 119904) [119891 (119904 1198681205730+V (119904) V (119904)) + 119890 (119904)] 119889119904
= int10119866 (119905 119904) 119891 (119904 1198681205730+V (119904) V (119904)) 119889119904 + 120593 (119905)
ge 1119877120576 (1 + 1120573Γ (120573))120576 int
1
0119866 (119905 119904) 1205831 (119904) 119889119904
+ 120593 (119905) ge 1119877120576 (1 + 1120573Γ (120573))120576 1198861lowast + 120593lowast ge
1119877
(65)
119879V (119905) = int10119866 (119905 119904) [119891 (119904 1198681205730+V (119904) V (119904)) + 119890 (119904)] 119889119904
= int10119866 (119905 119904) 119891 (119904 1198681205730+V (119904) V (119904)) 119889119904 + 120593 (119905)
le 119877120576 int10119866 (119905 119904) 1205832 (119904) 119889119904 + 120593lowast le 119877120576119886lowast2 + 120593lowast
le 119877120576 (119886lowast2 + 120593lowast) le 119877
(66)
ie 119879 119863 997888rarr 119863Next let us prove that 119879 119863 997888rarr 119863 is completely
continuousFor any V119899 sub 119863 V0 isin 119863 and V119899 997888rarr V0The continuity
of 119891 deduces that
119891 (119905 1198681205730+V119899 (119905) V119899 (119905)) 997888rarr 119891(119905 1198681205730+V0 (119905) V0 (119905)) 119899 997888rarr infin (67)
and it follows from (63) that100381610038161003816100381610038161003816119891 (119905 1198681205730+V119899 (119905) V119899 (119905)) minus 119891 (119905 1198681205730+V0 (119905) V0 (119905))100381610038161003816100381610038161003816le 2119877120598max max
0le119905le11205831 (119905) max
0le119905le11205832 (119905) (68)
By using the Lebesgue dominated convergence theorem weobtain that
lim119899997888rarrinfin
1003817100381710038171003817119879V119899 minus 119879V01003817100381710038171003817 = lim119899997888rarrinfin
max0le119905le1
100381610038161003816100381610038161003816100381610038161003816int1
0119866 (119905 119904) [119891 (119904 1198681205730+V119899 (119904) V119899 (119904)) minus 119891 (119904 1198681205730+V0 (119904) V0 (119904))] 119889119904
100381610038161003816100381610038161003816100381610038161003816le lim119899997888rarrinfin
max0le119905le1
int10119866 (119905 119904) 100381610038161003816100381610038161003816119891 (119904 1198681205730+V119899 (119904) V119899 (119904)) minus 119891 (119904 1198681205730+V0 (119904) V0 (119904))100381610038161003816100381610038161003816 119889119904
Journal of Function Spaces 9
= max0le119905119904le1
119866 (119905 119904) sdot int10
lim119899997888rarrinfin
100381610038161003816100381610038161003816119891 (119904 1198681205730+V119899 (119904) V119899 (119904)) minus 119891 (119904 1198681205730+V0 (119904) V0 (119904))100381610038161003816100381610038161003816 119889119904 = 0(69)
and this implies that 119879 is a continuous operatorNow we shall prove that 119879 119863 997888rarr 119863 is compact For any
V isin 119863 119879V isin 119863 which deduces that 1119877 le 119879V(119905) le 119877 for119905 isin [0 1] ie 119879 is uniformly boundedSince 119866(119905 119904) isin 119862([0 1] times [0 1]) it is also uniformly
continuous on [0 1] times [0 1] and then for any 120576 gt 0 exist120575 gt 0st for any (1199051 119904) (1199052 119904) isin [0 1] times [0 1] with |1199051 minus 1199052| lt 120575 wealways have
1003816100381610038161003816119866 (1199051 119904) minus 119866 (1199052 119904)1003816100381610038161003816 lt 120576119877120576 int101205832 (119904) 119889119904 + int10 |119890 (119904)| 119889119904
(70)
Thus one can obtain by virtue of (63) (70) that
1003816100381610038161003816119879V (1199051) minus 119879V (1199052)1003816100381610038161003816 =100381610038161003816100381610038161003816100381610038161003816int1
0[119866 (1199051 119904) minus 119866 (1199052 119904)]
sdot [119891 (119904 119868120573V (119904) V (119904)) + 119890 (119904)] 119889119904100381610038161003816100381610038161003816100381610038161003816 le int1
0
1003816100381610038161003816119866 (1199051 119904)minus 119866 (1199052 119904)1003816100381610038161003816 [100381610038161003816100381610038161003816119891 (119904 1198681205730+V (119904) V (119904))100381610038161003816100381610038161003816 + |119890 (119904)|] 119889119904lt [119877120576 int1
01205832 (119904) 119889119904 + int1
0|119890 (119904)| 119889119904]
sdot 120576119877120576 int101205832 (119904) 119889119904 + int10 |119890 (119904)| 119889119904
= 120576
(71)
for any 1199051 1199052 isin [0 1] with |1199051 minus 1199052| lt 120575 and any V isin 1198631 whichshows that 119879 119863 997888rarr 119863 is equicontinuous Thus Arzela-Ascoli theorem guarantees that 119879 119863 997888rarr 119863 is completelycontinuous Existence of at least one fixed point V1 isin 119863follows from Lemma 6 ie 1199061(119905) = 1198681205730+V1(119905) is a positivesolution of differential equation (1) which satisfies
1119877Γ (120572)119905120572 le 1199061 (119905) = 1Γ (120572) int119905
0(119905 minus 119904)120572minus1 V1 (119904) 119889119904
le 119877Γ (120572) 119905120572(72)
Data Availability
The data used to support the findings of this study areavailable from the corresponding author upon request
Conflicts of Interest
The author declares that there are no conflicts of interestregarding the publication of this paper
Acknowledgments
This research was supported by Shandong Provincial Nat-ural Science Foundation of China (no ZR2016FM10) andNational Natural Science Foundation of China (no 11571197)
References
[1] I Podlubny Fractional Differential Equations Mathematics inScience and Engineering Academic Press NewYork NY USA1999
[2] A A Kilbas H M Srivastava and J J Nietoeory and Appli-cational Differential Equations Elsevier Amsterdam Nether-lands 2006
[3] T Hu Y Sun and W Sun ldquoExistence of positive solutions for athird-order multipoint boundary value problem and extensionto fractional caserdquo Boundary Value Problems vol 2016 article197 2016
[4] Y Jia and X Zhang ldquoPositive solutions for a class of fractionaldifferential equation multi-point boundary value problemswith changing sign nonlinearityrdquo Applied Mathematics andComputation vol 47 no 1-2 pp 15ndash31 2015
[5] X Hao ldquoPositive solution for singular fractional differentialequations involving derivativesrdquo Advances in Difference Equa-tions vol 2016 article 139 2016
[6] Y Zou and G He ldquoOn the uniqueness of solutions for a class offractional differential equationsrdquo Applied Mathematics Lettersvol 74 pp 68ndash73 2017
[7] X Hao H Wang L Liu and Y Cui ldquoPositive solutionsfor a system of nonlinear fractional nonlocal boundary valueproblems with parameters and p-Laplacian operatorrdquoBoundaryValue Problems vol 2017 article 182 2017
[8] X Zhang ldquoPositive solutions for a class of singular fractionaldifferential equation with infinite-point boundary value condi-tionsrdquo Applied Mathematics Letters vol 39 pp 22ndash27 2015
[9] K M Zhang ldquoOn a sign-changing solution for some fractionaldifferential equationsrdquo Boundary Value Problems vol 2017article 59 2017
[10] Y L Guan Z Q Zhao and X L Lin ldquoOn the existence ofpositive solutions and negative solutions of singular fractionaldifferential equations via global bifurcation techniquesrdquoBound-ary Value Problems vol 2016 article 141 2016
[11] X G Zhang Y H Wu and L Caccetta ldquoNonlocal fractionalorder differential equations with changing-sign singular pertur-bationrdquo Applied Mathematical Modelling vol 56 pp 116ndash1262015
[12] YWang and L S Liu ldquoPositive solutions for fractionalm-pointboundary value problems in Banach spacesrdquo ActaMathematicaScientia vol 32 no 1 pp 246ndash256 2012
[13] Y Cui and Y Zou ldquoAn existence and uniqueness theorem for asecond order nonlinear system with coupled integral boundaryvalue conditionsrdquo Applied Mathematics and Computation vol256 pp 438ndash444 2015
10 Journal of Function Spaces
[14] X Q Zhang L Wang and Q S Wang ldquoExistence of positivesolutions for a class of nonlinear fractional differential equa-tions with integral boundary conditionsrdquo Applied Mathematicsand Computation vol 226 pp 708ndash718 2014
[15] XG Zhang L S Liu YHWu andBWiwatanapataphee ldquoThespectral analysis for a singular fractional differential equationwith a signedmeasurerdquoApplied Mathematics and Computationvol 257 pp 252ndash263 2015
[16] Y Wang and L Liu ldquoPositive solutions for a class of fractional3-point boundary value problems at resonancerdquo Advances inDifference Equations vol 13 article 13 2017
[17] M A Krasnoselskii and P P Zabreiko Geometrical Methods ofNonlinear Analysis Springer New York NY USA 1984
[18] K Deimling Nonlinear Functional Analysis Springer BerlinGermany 1985
[19] K Zhang ldquoNontrivial solutions of fourth-order singularboundary value problems with sign-changing nonlinear termsrdquoTopological Methods in Nonlinear Analysis vol 40 no 1 pp 53ndash70 2012
[20] D J Guo and V Lakshmikantham Nonlinear Problems inAbstract Cones Academic Press Massachusetts Mass USA1988
[21] J Liu and Z Zhao ldquoMultiple solutions for impulsive problemswith non-autonomous perturbationsrdquo Applied MathematicsLetters vol 64 pp 143ndash149 2017
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Journal of Function Spaces 7
sdot 11989811198721 + 119892 (0) [120572 minus 120573] sdot int119887
119886119904 (1 minus 119904)120572minus120573minus1 119904120572minus120573minus1119889119904
sdot V = 1198712 sdot 1198981119886120572minus120573minus1Γ (120572 minus 120573) 119892 (0) sdot11989811198721 + 119892 (0) [120572 minus 120573]
sdot int119887119886(1 minus 119904)120572minus120573minus1 119904120572minus120573119889119904 sdot V ge 1198712 sdot 1198981119886120572minus120573minus1Γ (120572 minus 120573) 119892 (0)
sdot 11989811198721 + 119892 (0) [120572 minus 120573] sdot int119887
119886(1 minus 119904)120572minus120573minus1 119904120572minus120573119889119904 sdot 1199032
= 1199032 = V (48)
ie 119865V ge V V isin 1205971198751199032 It follows from Lemma 10 that 119865 has at least fixed point
V1 isin 1198751199032 1198751199031 ie V1 satisfies119863120572minus1205730+ V1 (119905) + 119891 (119905 1198681205730+ (V1 (119905) minus 119908 (119905)) V1 (119905) minus 119908 (119905))
+ 119890 (119905) = 0 0 lt 119905 lt 1V1 (0) = V10158401 (0) = sdot sdot sdot = V(119899minus2)1 (0) = 0V1 (1) = 119898minus2sum
119894=1
120578119894V1 (120585119894) (49)
Set V1(119905) = V1(119905) minus 119908(119905) noticing that V1(119905) 119908(119905) are thesolutions of BVP (32) and (49) respectively therefore we canconclude that V1(119905) is a positive solution of (22) Let 1199091(119905) =1198681205730+V1(119905) then 1199091(119905) is a positive solution of the nonlinearfractional differential equations (1)
4 Main Result II 119891 Is Singular withRespect to Both the Time Variables andthe Space Variable
In this section we always suppose that the following condi-tion holds
(1198675) 119891(119905 119906 V) (0 1) times [0infin) times (0infin) 997888rarr [0infin)is continuous there exist 120576 isin (0 1) and 1205831 1205832 isin 119862+[0 1]1205831(119905) equiv 0 for 119905 isin [0 1] such that
1205831 (119905)(119909 + 119910)120576 le 119891 (119905 119909 119910) le1205832 (119905)(119909 + 119910)120576 (50)
for any (119909 119910) isin [0infin) times (0infin) 119905 isin (0 1) where119862+ [0 1] = 119909 (119905) isin 119862 [0 1] | 119909 (119905) ge 0 119905 isin [0 1] (51)
Set
120593 (119905) = int10119866 (119905 119904) 119890 (119904) 119889119904 119905 isin [0 1] (52)
it follows from Lemma 5 that 120593(119905) is the solution of thefollowing linear equation
119863120572minus1205730+ V (119905) + 119890 (119905) = 0 0 lt 119905 lt 1V (0) = V1015840 (0) = sdot sdot sdot = V(119899minus2) (0) = 0V (1) = 119898minus2sum
119894=1
120578119894V (120585119894) (53)
Denote
1198861 (119905) = int10119866 (119905 119904) 1205831 (119904) 119889119904
1198862 (119905) = int10119866 (119905 119904) 1205832 (119904) 119889119904
120593lowast = inf0le119905le1
120593 (119905) 120593lowast = sup0le119905le1
120593 (119905) 1198861lowast = min
0le119905le11198861 (119905)
119886lowast1 = max0le119905le1
1198861 (119905) 1198862lowast = min
0le119905le11198862 (119905)
119886lowast2 = max0le119905le1
1198862 (119905)
(54)
Clearly 119886lowast119895 ge 119886119895lowast gt 0 119895 = 1 2Theorem 13 Suppose that the condition (H5) holds and 120593lowast ge0 en the FVP (1) has at least one positive solution
Proof Because 120593lowast ge 0 so we can choose 119877 gt 0 large enoughsuch that
119903119877120576 + 120593lowast ge 1119877119877120576 (119886lowast2 + 120593lowast) le 119877
(55)
where 119903 = 1198861lowast[1 + 1(1 + 120573Γ(120573))]120576 In fact since
lim119909997888rarr+infin
119909120576119909 = 0 lt min119903 1119886lowast2 + 120593lowast (56)
there exists 1198831 gt 0 such that 119909120576119909 lt min119903 1(119886lowast2 + 120593lowast) forany 119909 gt 1198831 ie
119903119909120576 ge 1119909 119909120576 (119886lowast2 + 120593lowast) le 119909 (57)
for any 119909 gt 1198831 If 120593lowast gt 0 then from lim119909997888rarr+infin[1119909 minus 119903119909120576] =0 lt 120593lowast one can get that there exists 1198832 gt 0 such that 1119909 minus119903119909120576 lt 120593lowast for any 119909 gt 1198832 ie119903119909120576 + 120593lowast ge 1119909 (58)
8 Journal of Function Spaces
for any 119909 gt 1198832 By (57) (58) we can choose 119877 gt max1198831 1198832such that 119877 satisfies (55) Set
119863 = V isin 119862+ [0 1] 1119877 le V (119905) le 119877 119905 isin [0 1] (59)
For any V isin 1198631 from (23) we have
1198681205730+V (119905) = 1Γ (120573) int119905
0(119905 minus 119904)120573minus1 V (119904) 119889119904
le 119877Γ (120573) int119905
0(119905 minus 119904)120573minus1 119889119904 = 119877119905120573120573Γ (120573) le 119877120573Γ (120573)
(60)
1198681205730+V (119905) = 1Γ (120573) int119905
0(119905 minus 119904)120573minus1 V (119904) 119889119904
ge 1119877Γ (120573) int119905
0(119905 minus 119904)120573minus1 119889119904 = 119905120573119877120573Γ (120573)
(61)
It follows from (60) (61) and (H5) that
1205831 (119905)119877120576 (1 + 1120573Γ (120573))120576 le1205831 (119905)
(V (119905) + 1198681205730+V (119905))120576le 119891 (119905 1198681205730+V (119905) V (119905))
119891 (119905 1198681205730+V (119905) V (119905)) le 1205832 (119905)(V (119905) + 1198681205730+V (119905))120576
le 1205832 (119905)(1119877 + 119905120573119877120573Γ (120573))120576le 1198771205761205832 (119905)
(62)
ie
1205831 (119905)119877120576 (1 + 1120573Γ (120573))120576 le 119891 (119905 1198681205730+V (119905) V (119905)) le 1198771205761205832 (119905) (63)
And then
119879V (119905) = int10119866 (119905 119904) [119891 (119904 1198681205730+V (119904) V (119904)) + 119890 (119904)] 119889119904
= int10119866 (119905 119904) 119891 (119904 1198681205730+V (119904) V (119904)) 119889119904 + 120593 (119905)
le int10119866 (119905 119904) 119891 (119904 1198681205730+V (119904) V (119904)) 119889119904 + 120593lowast
le 119877120576 int10119866 (119905 119904) 1205832 (119904) 119889119904 + 120593lowast lt +infin
(64)
which deduces that the operator 119879 is well definedNow we shall prove that 119879 119863 997888rarr 119863 For V isin 119863 it
is easy to see that 119879V(119905) isin 119862+[0 1] and by (55) (63) we canobtain that
119879V (119905) = int10119866 (119905 119904) [119891 (119904 1198681205730+V (119904) V (119904)) + 119890 (119904)] 119889119904
= int10119866 (119905 119904) 119891 (119904 1198681205730+V (119904) V (119904)) 119889119904 + 120593 (119905)
ge 1119877120576 (1 + 1120573Γ (120573))120576 int
1
0119866 (119905 119904) 1205831 (119904) 119889119904
+ 120593 (119905) ge 1119877120576 (1 + 1120573Γ (120573))120576 1198861lowast + 120593lowast ge
1119877
(65)
119879V (119905) = int10119866 (119905 119904) [119891 (119904 1198681205730+V (119904) V (119904)) + 119890 (119904)] 119889119904
= int10119866 (119905 119904) 119891 (119904 1198681205730+V (119904) V (119904)) 119889119904 + 120593 (119905)
le 119877120576 int10119866 (119905 119904) 1205832 (119904) 119889119904 + 120593lowast le 119877120576119886lowast2 + 120593lowast
le 119877120576 (119886lowast2 + 120593lowast) le 119877
(66)
ie 119879 119863 997888rarr 119863Next let us prove that 119879 119863 997888rarr 119863 is completely
continuousFor any V119899 sub 119863 V0 isin 119863 and V119899 997888rarr V0The continuity
of 119891 deduces that
119891 (119905 1198681205730+V119899 (119905) V119899 (119905)) 997888rarr 119891(119905 1198681205730+V0 (119905) V0 (119905)) 119899 997888rarr infin (67)
and it follows from (63) that100381610038161003816100381610038161003816119891 (119905 1198681205730+V119899 (119905) V119899 (119905)) minus 119891 (119905 1198681205730+V0 (119905) V0 (119905))100381610038161003816100381610038161003816le 2119877120598max max
0le119905le11205831 (119905) max
0le119905le11205832 (119905) (68)
By using the Lebesgue dominated convergence theorem weobtain that
lim119899997888rarrinfin
1003817100381710038171003817119879V119899 minus 119879V01003817100381710038171003817 = lim119899997888rarrinfin
max0le119905le1
100381610038161003816100381610038161003816100381610038161003816int1
0119866 (119905 119904) [119891 (119904 1198681205730+V119899 (119904) V119899 (119904)) minus 119891 (119904 1198681205730+V0 (119904) V0 (119904))] 119889119904
100381610038161003816100381610038161003816100381610038161003816le lim119899997888rarrinfin
max0le119905le1
int10119866 (119905 119904) 100381610038161003816100381610038161003816119891 (119904 1198681205730+V119899 (119904) V119899 (119904)) minus 119891 (119904 1198681205730+V0 (119904) V0 (119904))100381610038161003816100381610038161003816 119889119904
Journal of Function Spaces 9
= max0le119905119904le1
119866 (119905 119904) sdot int10
lim119899997888rarrinfin
100381610038161003816100381610038161003816119891 (119904 1198681205730+V119899 (119904) V119899 (119904)) minus 119891 (119904 1198681205730+V0 (119904) V0 (119904))100381610038161003816100381610038161003816 119889119904 = 0(69)
and this implies that 119879 is a continuous operatorNow we shall prove that 119879 119863 997888rarr 119863 is compact For any
V isin 119863 119879V isin 119863 which deduces that 1119877 le 119879V(119905) le 119877 for119905 isin [0 1] ie 119879 is uniformly boundedSince 119866(119905 119904) isin 119862([0 1] times [0 1]) it is also uniformly
continuous on [0 1] times [0 1] and then for any 120576 gt 0 exist120575 gt 0st for any (1199051 119904) (1199052 119904) isin [0 1] times [0 1] with |1199051 minus 1199052| lt 120575 wealways have
1003816100381610038161003816119866 (1199051 119904) minus 119866 (1199052 119904)1003816100381610038161003816 lt 120576119877120576 int101205832 (119904) 119889119904 + int10 |119890 (119904)| 119889119904
(70)
Thus one can obtain by virtue of (63) (70) that
1003816100381610038161003816119879V (1199051) minus 119879V (1199052)1003816100381610038161003816 =100381610038161003816100381610038161003816100381610038161003816int1
0[119866 (1199051 119904) minus 119866 (1199052 119904)]
sdot [119891 (119904 119868120573V (119904) V (119904)) + 119890 (119904)] 119889119904100381610038161003816100381610038161003816100381610038161003816 le int1
0
1003816100381610038161003816119866 (1199051 119904)minus 119866 (1199052 119904)1003816100381610038161003816 [100381610038161003816100381610038161003816119891 (119904 1198681205730+V (119904) V (119904))100381610038161003816100381610038161003816 + |119890 (119904)|] 119889119904lt [119877120576 int1
01205832 (119904) 119889119904 + int1
0|119890 (119904)| 119889119904]
sdot 120576119877120576 int101205832 (119904) 119889119904 + int10 |119890 (119904)| 119889119904
= 120576
(71)
for any 1199051 1199052 isin [0 1] with |1199051 minus 1199052| lt 120575 and any V isin 1198631 whichshows that 119879 119863 997888rarr 119863 is equicontinuous Thus Arzela-Ascoli theorem guarantees that 119879 119863 997888rarr 119863 is completelycontinuous Existence of at least one fixed point V1 isin 119863follows from Lemma 6 ie 1199061(119905) = 1198681205730+V1(119905) is a positivesolution of differential equation (1) which satisfies
1119877Γ (120572)119905120572 le 1199061 (119905) = 1Γ (120572) int119905
0(119905 minus 119904)120572minus1 V1 (119904) 119889119904
le 119877Γ (120572) 119905120572(72)
Data Availability
The data used to support the findings of this study areavailable from the corresponding author upon request
Conflicts of Interest
The author declares that there are no conflicts of interestregarding the publication of this paper
Acknowledgments
This research was supported by Shandong Provincial Nat-ural Science Foundation of China (no ZR2016FM10) andNational Natural Science Foundation of China (no 11571197)
References
[1] I Podlubny Fractional Differential Equations Mathematics inScience and Engineering Academic Press NewYork NY USA1999
[2] A A Kilbas H M Srivastava and J J Nietoeory and Appli-cational Differential Equations Elsevier Amsterdam Nether-lands 2006
[3] T Hu Y Sun and W Sun ldquoExistence of positive solutions for athird-order multipoint boundary value problem and extensionto fractional caserdquo Boundary Value Problems vol 2016 article197 2016
[4] Y Jia and X Zhang ldquoPositive solutions for a class of fractionaldifferential equation multi-point boundary value problemswith changing sign nonlinearityrdquo Applied Mathematics andComputation vol 47 no 1-2 pp 15ndash31 2015
[5] X Hao ldquoPositive solution for singular fractional differentialequations involving derivativesrdquo Advances in Difference Equa-tions vol 2016 article 139 2016
[6] Y Zou and G He ldquoOn the uniqueness of solutions for a class offractional differential equationsrdquo Applied Mathematics Lettersvol 74 pp 68ndash73 2017
[7] X Hao H Wang L Liu and Y Cui ldquoPositive solutionsfor a system of nonlinear fractional nonlocal boundary valueproblems with parameters and p-Laplacian operatorrdquoBoundaryValue Problems vol 2017 article 182 2017
[8] X Zhang ldquoPositive solutions for a class of singular fractionaldifferential equation with infinite-point boundary value condi-tionsrdquo Applied Mathematics Letters vol 39 pp 22ndash27 2015
[9] K M Zhang ldquoOn a sign-changing solution for some fractionaldifferential equationsrdquo Boundary Value Problems vol 2017article 59 2017
[10] Y L Guan Z Q Zhao and X L Lin ldquoOn the existence ofpositive solutions and negative solutions of singular fractionaldifferential equations via global bifurcation techniquesrdquoBound-ary Value Problems vol 2016 article 141 2016
[11] X G Zhang Y H Wu and L Caccetta ldquoNonlocal fractionalorder differential equations with changing-sign singular pertur-bationrdquo Applied Mathematical Modelling vol 56 pp 116ndash1262015
[12] YWang and L S Liu ldquoPositive solutions for fractionalm-pointboundary value problems in Banach spacesrdquo ActaMathematicaScientia vol 32 no 1 pp 246ndash256 2012
[13] Y Cui and Y Zou ldquoAn existence and uniqueness theorem for asecond order nonlinear system with coupled integral boundaryvalue conditionsrdquo Applied Mathematics and Computation vol256 pp 438ndash444 2015
10 Journal of Function Spaces
[14] X Q Zhang L Wang and Q S Wang ldquoExistence of positivesolutions for a class of nonlinear fractional differential equa-tions with integral boundary conditionsrdquo Applied Mathematicsand Computation vol 226 pp 708ndash718 2014
[15] XG Zhang L S Liu YHWu andBWiwatanapataphee ldquoThespectral analysis for a singular fractional differential equationwith a signedmeasurerdquoApplied Mathematics and Computationvol 257 pp 252ndash263 2015
[16] Y Wang and L Liu ldquoPositive solutions for a class of fractional3-point boundary value problems at resonancerdquo Advances inDifference Equations vol 13 article 13 2017
[17] M A Krasnoselskii and P P Zabreiko Geometrical Methods ofNonlinear Analysis Springer New York NY USA 1984
[18] K Deimling Nonlinear Functional Analysis Springer BerlinGermany 1985
[19] K Zhang ldquoNontrivial solutions of fourth-order singularboundary value problems with sign-changing nonlinear termsrdquoTopological Methods in Nonlinear Analysis vol 40 no 1 pp 53ndash70 2012
[20] D J Guo and V Lakshmikantham Nonlinear Problems inAbstract Cones Academic Press Massachusetts Mass USA1988
[21] J Liu and Z Zhao ldquoMultiple solutions for impulsive problemswith non-autonomous perturbationsrdquo Applied MathematicsLetters vol 64 pp 143ndash149 2017
Hindawiwwwhindawicom Volume 2018
MathematicsJournal of
Hindawiwwwhindawicom Volume 2018
Mathematical Problems in Engineering
Applied MathematicsJournal of
Hindawiwwwhindawicom Volume 2018
Probability and StatisticsHindawiwwwhindawicom Volume 2018
Journal of
Hindawiwwwhindawicom Volume 2018
Mathematical PhysicsAdvances in
Complex AnalysisJournal of
Hindawiwwwhindawicom Volume 2018
OptimizationJournal of
Hindawiwwwhindawicom Volume 2018
Hindawiwwwhindawicom Volume 2018
Engineering Mathematics
International Journal of
Hindawiwwwhindawicom Volume 2018
Operations ResearchAdvances in
Journal of
Hindawiwwwhindawicom Volume 2018
Function SpacesAbstract and Applied AnalysisHindawiwwwhindawicom Volume 2018
International Journal of Mathematics and Mathematical Sciences
Hindawiwwwhindawicom Volume 2018
Hindawi Publishing Corporation httpwwwhindawicom Volume 2013Hindawiwwwhindawicom
The Scientific World Journal
Volume 2018
Hindawiwwwhindawicom Volume 2018Volume 2018
Numerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisAdvances inAdvances in Discrete Dynamics in
Nature and SocietyHindawiwwwhindawicom Volume 2018
Hindawiwwwhindawicom
Dierential EquationsInternational Journal of
Volume 2018
Hindawiwwwhindawicom Volume 2018
Decision SciencesAdvances in
Hindawiwwwhindawicom Volume 2018
AnalysisInternational Journal of
Hindawiwwwhindawicom Volume 2018
Stochastic AnalysisInternational Journal of
Submit your manuscripts atwwwhindawicom
8 Journal of Function Spaces
for any 119909 gt 1198832 By (57) (58) we can choose 119877 gt max1198831 1198832such that 119877 satisfies (55) Set
119863 = V isin 119862+ [0 1] 1119877 le V (119905) le 119877 119905 isin [0 1] (59)
For any V isin 1198631 from (23) we have
1198681205730+V (119905) = 1Γ (120573) int119905
0(119905 minus 119904)120573minus1 V (119904) 119889119904
le 119877Γ (120573) int119905
0(119905 minus 119904)120573minus1 119889119904 = 119877119905120573120573Γ (120573) le 119877120573Γ (120573)
(60)
1198681205730+V (119905) = 1Γ (120573) int119905
0(119905 minus 119904)120573minus1 V (119904) 119889119904
ge 1119877Γ (120573) int119905
0(119905 minus 119904)120573minus1 119889119904 = 119905120573119877120573Γ (120573)
(61)
It follows from (60) (61) and (H5) that
1205831 (119905)119877120576 (1 + 1120573Γ (120573))120576 le1205831 (119905)
(V (119905) + 1198681205730+V (119905))120576le 119891 (119905 1198681205730+V (119905) V (119905))
119891 (119905 1198681205730+V (119905) V (119905)) le 1205832 (119905)(V (119905) + 1198681205730+V (119905))120576
le 1205832 (119905)(1119877 + 119905120573119877120573Γ (120573))120576le 1198771205761205832 (119905)
(62)
ie
1205831 (119905)119877120576 (1 + 1120573Γ (120573))120576 le 119891 (119905 1198681205730+V (119905) V (119905)) le 1198771205761205832 (119905) (63)
And then
119879V (119905) = int10119866 (119905 119904) [119891 (119904 1198681205730+V (119904) V (119904)) + 119890 (119904)] 119889119904
= int10119866 (119905 119904) 119891 (119904 1198681205730+V (119904) V (119904)) 119889119904 + 120593 (119905)
le int10119866 (119905 119904) 119891 (119904 1198681205730+V (119904) V (119904)) 119889119904 + 120593lowast
le 119877120576 int10119866 (119905 119904) 1205832 (119904) 119889119904 + 120593lowast lt +infin
(64)
which deduces that the operator 119879 is well definedNow we shall prove that 119879 119863 997888rarr 119863 For V isin 119863 it
is easy to see that 119879V(119905) isin 119862+[0 1] and by (55) (63) we canobtain that
119879V (119905) = int10119866 (119905 119904) [119891 (119904 1198681205730+V (119904) V (119904)) + 119890 (119904)] 119889119904
= int10119866 (119905 119904) 119891 (119904 1198681205730+V (119904) V (119904)) 119889119904 + 120593 (119905)
ge 1119877120576 (1 + 1120573Γ (120573))120576 int
1
0119866 (119905 119904) 1205831 (119904) 119889119904
+ 120593 (119905) ge 1119877120576 (1 + 1120573Γ (120573))120576 1198861lowast + 120593lowast ge
1119877
(65)
119879V (119905) = int10119866 (119905 119904) [119891 (119904 1198681205730+V (119904) V (119904)) + 119890 (119904)] 119889119904
= int10119866 (119905 119904) 119891 (119904 1198681205730+V (119904) V (119904)) 119889119904 + 120593 (119905)
le 119877120576 int10119866 (119905 119904) 1205832 (119904) 119889119904 + 120593lowast le 119877120576119886lowast2 + 120593lowast
le 119877120576 (119886lowast2 + 120593lowast) le 119877
(66)
ie 119879 119863 997888rarr 119863Next let us prove that 119879 119863 997888rarr 119863 is completely
continuousFor any V119899 sub 119863 V0 isin 119863 and V119899 997888rarr V0The continuity
of 119891 deduces that
119891 (119905 1198681205730+V119899 (119905) V119899 (119905)) 997888rarr 119891(119905 1198681205730+V0 (119905) V0 (119905)) 119899 997888rarr infin (67)
and it follows from (63) that100381610038161003816100381610038161003816119891 (119905 1198681205730+V119899 (119905) V119899 (119905)) minus 119891 (119905 1198681205730+V0 (119905) V0 (119905))100381610038161003816100381610038161003816le 2119877120598max max
0le119905le11205831 (119905) max
0le119905le11205832 (119905) (68)
By using the Lebesgue dominated convergence theorem weobtain that
lim119899997888rarrinfin
1003817100381710038171003817119879V119899 minus 119879V01003817100381710038171003817 = lim119899997888rarrinfin
max0le119905le1
100381610038161003816100381610038161003816100381610038161003816int1
0119866 (119905 119904) [119891 (119904 1198681205730+V119899 (119904) V119899 (119904)) minus 119891 (119904 1198681205730+V0 (119904) V0 (119904))] 119889119904
100381610038161003816100381610038161003816100381610038161003816le lim119899997888rarrinfin
max0le119905le1
int10119866 (119905 119904) 100381610038161003816100381610038161003816119891 (119904 1198681205730+V119899 (119904) V119899 (119904)) minus 119891 (119904 1198681205730+V0 (119904) V0 (119904))100381610038161003816100381610038161003816 119889119904
Journal of Function Spaces 9
= max0le119905119904le1
119866 (119905 119904) sdot int10
lim119899997888rarrinfin
100381610038161003816100381610038161003816119891 (119904 1198681205730+V119899 (119904) V119899 (119904)) minus 119891 (119904 1198681205730+V0 (119904) V0 (119904))100381610038161003816100381610038161003816 119889119904 = 0(69)
and this implies that 119879 is a continuous operatorNow we shall prove that 119879 119863 997888rarr 119863 is compact For any
V isin 119863 119879V isin 119863 which deduces that 1119877 le 119879V(119905) le 119877 for119905 isin [0 1] ie 119879 is uniformly boundedSince 119866(119905 119904) isin 119862([0 1] times [0 1]) it is also uniformly
continuous on [0 1] times [0 1] and then for any 120576 gt 0 exist120575 gt 0st for any (1199051 119904) (1199052 119904) isin [0 1] times [0 1] with |1199051 minus 1199052| lt 120575 wealways have
1003816100381610038161003816119866 (1199051 119904) minus 119866 (1199052 119904)1003816100381610038161003816 lt 120576119877120576 int101205832 (119904) 119889119904 + int10 |119890 (119904)| 119889119904
(70)
Thus one can obtain by virtue of (63) (70) that
1003816100381610038161003816119879V (1199051) minus 119879V (1199052)1003816100381610038161003816 =100381610038161003816100381610038161003816100381610038161003816int1
0[119866 (1199051 119904) minus 119866 (1199052 119904)]
sdot [119891 (119904 119868120573V (119904) V (119904)) + 119890 (119904)] 119889119904100381610038161003816100381610038161003816100381610038161003816 le int1
0
1003816100381610038161003816119866 (1199051 119904)minus 119866 (1199052 119904)1003816100381610038161003816 [100381610038161003816100381610038161003816119891 (119904 1198681205730+V (119904) V (119904))100381610038161003816100381610038161003816 + |119890 (119904)|] 119889119904lt [119877120576 int1
01205832 (119904) 119889119904 + int1
0|119890 (119904)| 119889119904]
sdot 120576119877120576 int101205832 (119904) 119889119904 + int10 |119890 (119904)| 119889119904
= 120576
(71)
for any 1199051 1199052 isin [0 1] with |1199051 minus 1199052| lt 120575 and any V isin 1198631 whichshows that 119879 119863 997888rarr 119863 is equicontinuous Thus Arzela-Ascoli theorem guarantees that 119879 119863 997888rarr 119863 is completelycontinuous Existence of at least one fixed point V1 isin 119863follows from Lemma 6 ie 1199061(119905) = 1198681205730+V1(119905) is a positivesolution of differential equation (1) which satisfies
1119877Γ (120572)119905120572 le 1199061 (119905) = 1Γ (120572) int119905
0(119905 minus 119904)120572minus1 V1 (119904) 119889119904
le 119877Γ (120572) 119905120572(72)
Data Availability
The data used to support the findings of this study areavailable from the corresponding author upon request
Conflicts of Interest
The author declares that there are no conflicts of interestregarding the publication of this paper
Acknowledgments
This research was supported by Shandong Provincial Nat-ural Science Foundation of China (no ZR2016FM10) andNational Natural Science Foundation of China (no 11571197)
References
[1] I Podlubny Fractional Differential Equations Mathematics inScience and Engineering Academic Press NewYork NY USA1999
[2] A A Kilbas H M Srivastava and J J Nietoeory and Appli-cational Differential Equations Elsevier Amsterdam Nether-lands 2006
[3] T Hu Y Sun and W Sun ldquoExistence of positive solutions for athird-order multipoint boundary value problem and extensionto fractional caserdquo Boundary Value Problems vol 2016 article197 2016
[4] Y Jia and X Zhang ldquoPositive solutions for a class of fractionaldifferential equation multi-point boundary value problemswith changing sign nonlinearityrdquo Applied Mathematics andComputation vol 47 no 1-2 pp 15ndash31 2015
[5] X Hao ldquoPositive solution for singular fractional differentialequations involving derivativesrdquo Advances in Difference Equa-tions vol 2016 article 139 2016
[6] Y Zou and G He ldquoOn the uniqueness of solutions for a class offractional differential equationsrdquo Applied Mathematics Lettersvol 74 pp 68ndash73 2017
[7] X Hao H Wang L Liu and Y Cui ldquoPositive solutionsfor a system of nonlinear fractional nonlocal boundary valueproblems with parameters and p-Laplacian operatorrdquoBoundaryValue Problems vol 2017 article 182 2017
[8] X Zhang ldquoPositive solutions for a class of singular fractionaldifferential equation with infinite-point boundary value condi-tionsrdquo Applied Mathematics Letters vol 39 pp 22ndash27 2015
[9] K M Zhang ldquoOn a sign-changing solution for some fractionaldifferential equationsrdquo Boundary Value Problems vol 2017article 59 2017
[10] Y L Guan Z Q Zhao and X L Lin ldquoOn the existence ofpositive solutions and negative solutions of singular fractionaldifferential equations via global bifurcation techniquesrdquoBound-ary Value Problems vol 2016 article 141 2016
[11] X G Zhang Y H Wu and L Caccetta ldquoNonlocal fractionalorder differential equations with changing-sign singular pertur-bationrdquo Applied Mathematical Modelling vol 56 pp 116ndash1262015
[12] YWang and L S Liu ldquoPositive solutions for fractionalm-pointboundary value problems in Banach spacesrdquo ActaMathematicaScientia vol 32 no 1 pp 246ndash256 2012
[13] Y Cui and Y Zou ldquoAn existence and uniqueness theorem for asecond order nonlinear system with coupled integral boundaryvalue conditionsrdquo Applied Mathematics and Computation vol256 pp 438ndash444 2015
10 Journal of Function Spaces
[14] X Q Zhang L Wang and Q S Wang ldquoExistence of positivesolutions for a class of nonlinear fractional differential equa-tions with integral boundary conditionsrdquo Applied Mathematicsand Computation vol 226 pp 708ndash718 2014
[15] XG Zhang L S Liu YHWu andBWiwatanapataphee ldquoThespectral analysis for a singular fractional differential equationwith a signedmeasurerdquoApplied Mathematics and Computationvol 257 pp 252ndash263 2015
[16] Y Wang and L Liu ldquoPositive solutions for a class of fractional3-point boundary value problems at resonancerdquo Advances inDifference Equations vol 13 article 13 2017
[17] M A Krasnoselskii and P P Zabreiko Geometrical Methods ofNonlinear Analysis Springer New York NY USA 1984
[18] K Deimling Nonlinear Functional Analysis Springer BerlinGermany 1985
[19] K Zhang ldquoNontrivial solutions of fourth-order singularboundary value problems with sign-changing nonlinear termsrdquoTopological Methods in Nonlinear Analysis vol 40 no 1 pp 53ndash70 2012
[20] D J Guo and V Lakshmikantham Nonlinear Problems inAbstract Cones Academic Press Massachusetts Mass USA1988
[21] J Liu and Z Zhao ldquoMultiple solutions for impulsive problemswith non-autonomous perturbationsrdquo Applied MathematicsLetters vol 64 pp 143ndash149 2017
Hindawiwwwhindawicom Volume 2018
MathematicsJournal of
Hindawiwwwhindawicom Volume 2018
Mathematical Problems in Engineering
Applied MathematicsJournal of
Hindawiwwwhindawicom Volume 2018
Probability and StatisticsHindawiwwwhindawicom Volume 2018
Journal of
Hindawiwwwhindawicom Volume 2018
Mathematical PhysicsAdvances in
Complex AnalysisJournal of
Hindawiwwwhindawicom Volume 2018
OptimizationJournal of
Hindawiwwwhindawicom Volume 2018
Hindawiwwwhindawicom Volume 2018
Engineering Mathematics
International Journal of
Hindawiwwwhindawicom Volume 2018
Operations ResearchAdvances in
Journal of
Hindawiwwwhindawicom Volume 2018
Function SpacesAbstract and Applied AnalysisHindawiwwwhindawicom Volume 2018
International Journal of Mathematics and Mathematical Sciences
Hindawiwwwhindawicom Volume 2018
Hindawi Publishing Corporation httpwwwhindawicom Volume 2013Hindawiwwwhindawicom
The Scientific World Journal
Volume 2018
Hindawiwwwhindawicom Volume 2018Volume 2018
Numerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisAdvances inAdvances in Discrete Dynamics in
Nature and SocietyHindawiwwwhindawicom Volume 2018
Hindawiwwwhindawicom
Dierential EquationsInternational Journal of
Volume 2018
Hindawiwwwhindawicom Volume 2018
Decision SciencesAdvances in
Hindawiwwwhindawicom Volume 2018
AnalysisInternational Journal of
Hindawiwwwhindawicom Volume 2018
Stochastic AnalysisInternational Journal of
Submit your manuscripts atwwwhindawicom
Journal of Function Spaces 9
= max0le119905119904le1
119866 (119905 119904) sdot int10
lim119899997888rarrinfin
100381610038161003816100381610038161003816119891 (119904 1198681205730+V119899 (119904) V119899 (119904)) minus 119891 (119904 1198681205730+V0 (119904) V0 (119904))100381610038161003816100381610038161003816 119889119904 = 0(69)
and this implies that 119879 is a continuous operatorNow we shall prove that 119879 119863 997888rarr 119863 is compact For any
V isin 119863 119879V isin 119863 which deduces that 1119877 le 119879V(119905) le 119877 for119905 isin [0 1] ie 119879 is uniformly boundedSince 119866(119905 119904) isin 119862([0 1] times [0 1]) it is also uniformly
continuous on [0 1] times [0 1] and then for any 120576 gt 0 exist120575 gt 0st for any (1199051 119904) (1199052 119904) isin [0 1] times [0 1] with |1199051 minus 1199052| lt 120575 wealways have
1003816100381610038161003816119866 (1199051 119904) minus 119866 (1199052 119904)1003816100381610038161003816 lt 120576119877120576 int101205832 (119904) 119889119904 + int10 |119890 (119904)| 119889119904
(70)
Thus one can obtain by virtue of (63) (70) that
1003816100381610038161003816119879V (1199051) minus 119879V (1199052)1003816100381610038161003816 =100381610038161003816100381610038161003816100381610038161003816int1
0[119866 (1199051 119904) minus 119866 (1199052 119904)]
sdot [119891 (119904 119868120573V (119904) V (119904)) + 119890 (119904)] 119889119904100381610038161003816100381610038161003816100381610038161003816 le int1
0
1003816100381610038161003816119866 (1199051 119904)minus 119866 (1199052 119904)1003816100381610038161003816 [100381610038161003816100381610038161003816119891 (119904 1198681205730+V (119904) V (119904))100381610038161003816100381610038161003816 + |119890 (119904)|] 119889119904lt [119877120576 int1
01205832 (119904) 119889119904 + int1
0|119890 (119904)| 119889119904]
sdot 120576119877120576 int101205832 (119904) 119889119904 + int10 |119890 (119904)| 119889119904
= 120576
(71)
for any 1199051 1199052 isin [0 1] with |1199051 minus 1199052| lt 120575 and any V isin 1198631 whichshows that 119879 119863 997888rarr 119863 is equicontinuous Thus Arzela-Ascoli theorem guarantees that 119879 119863 997888rarr 119863 is completelycontinuous Existence of at least one fixed point V1 isin 119863follows from Lemma 6 ie 1199061(119905) = 1198681205730+V1(119905) is a positivesolution of differential equation (1) which satisfies
1119877Γ (120572)119905120572 le 1199061 (119905) = 1Γ (120572) int119905
0(119905 minus 119904)120572minus1 V1 (119904) 119889119904
le 119877Γ (120572) 119905120572(72)
Data Availability
The data used to support the findings of this study areavailable from the corresponding author upon request
Conflicts of Interest
The author declares that there are no conflicts of interestregarding the publication of this paper
Acknowledgments
This research was supported by Shandong Provincial Nat-ural Science Foundation of China (no ZR2016FM10) andNational Natural Science Foundation of China (no 11571197)
References
[1] I Podlubny Fractional Differential Equations Mathematics inScience and Engineering Academic Press NewYork NY USA1999
[2] A A Kilbas H M Srivastava and J J Nietoeory and Appli-cational Differential Equations Elsevier Amsterdam Nether-lands 2006
[3] T Hu Y Sun and W Sun ldquoExistence of positive solutions for athird-order multipoint boundary value problem and extensionto fractional caserdquo Boundary Value Problems vol 2016 article197 2016
[4] Y Jia and X Zhang ldquoPositive solutions for a class of fractionaldifferential equation multi-point boundary value problemswith changing sign nonlinearityrdquo Applied Mathematics andComputation vol 47 no 1-2 pp 15ndash31 2015
[5] X Hao ldquoPositive solution for singular fractional differentialequations involving derivativesrdquo Advances in Difference Equa-tions vol 2016 article 139 2016
[6] Y Zou and G He ldquoOn the uniqueness of solutions for a class offractional differential equationsrdquo Applied Mathematics Lettersvol 74 pp 68ndash73 2017
[7] X Hao H Wang L Liu and Y Cui ldquoPositive solutionsfor a system of nonlinear fractional nonlocal boundary valueproblems with parameters and p-Laplacian operatorrdquoBoundaryValue Problems vol 2017 article 182 2017
[8] X Zhang ldquoPositive solutions for a class of singular fractionaldifferential equation with infinite-point boundary value condi-tionsrdquo Applied Mathematics Letters vol 39 pp 22ndash27 2015
[9] K M Zhang ldquoOn a sign-changing solution for some fractionaldifferential equationsrdquo Boundary Value Problems vol 2017article 59 2017
[10] Y L Guan Z Q Zhao and X L Lin ldquoOn the existence ofpositive solutions and negative solutions of singular fractionaldifferential equations via global bifurcation techniquesrdquoBound-ary Value Problems vol 2016 article 141 2016
[11] X G Zhang Y H Wu and L Caccetta ldquoNonlocal fractionalorder differential equations with changing-sign singular pertur-bationrdquo Applied Mathematical Modelling vol 56 pp 116ndash1262015
[12] YWang and L S Liu ldquoPositive solutions for fractionalm-pointboundary value problems in Banach spacesrdquo ActaMathematicaScientia vol 32 no 1 pp 246ndash256 2012
[13] Y Cui and Y Zou ldquoAn existence and uniqueness theorem for asecond order nonlinear system with coupled integral boundaryvalue conditionsrdquo Applied Mathematics and Computation vol256 pp 438ndash444 2015
10 Journal of Function Spaces
[14] X Q Zhang L Wang and Q S Wang ldquoExistence of positivesolutions for a class of nonlinear fractional differential equa-tions with integral boundary conditionsrdquo Applied Mathematicsand Computation vol 226 pp 708ndash718 2014
[15] XG Zhang L S Liu YHWu andBWiwatanapataphee ldquoThespectral analysis for a singular fractional differential equationwith a signedmeasurerdquoApplied Mathematics and Computationvol 257 pp 252ndash263 2015
[16] Y Wang and L Liu ldquoPositive solutions for a class of fractional3-point boundary value problems at resonancerdquo Advances inDifference Equations vol 13 article 13 2017
[17] M A Krasnoselskii and P P Zabreiko Geometrical Methods ofNonlinear Analysis Springer New York NY USA 1984
[18] K Deimling Nonlinear Functional Analysis Springer BerlinGermany 1985
[19] K Zhang ldquoNontrivial solutions of fourth-order singularboundary value problems with sign-changing nonlinear termsrdquoTopological Methods in Nonlinear Analysis vol 40 no 1 pp 53ndash70 2012
[20] D J Guo and V Lakshmikantham Nonlinear Problems inAbstract Cones Academic Press Massachusetts Mass USA1988
[21] J Liu and Z Zhao ldquoMultiple solutions for impulsive problemswith non-autonomous perturbationsrdquo Applied MathematicsLetters vol 64 pp 143ndash149 2017
Hindawiwwwhindawicom Volume 2018
MathematicsJournal of
Hindawiwwwhindawicom Volume 2018
Mathematical Problems in Engineering
Applied MathematicsJournal of
Hindawiwwwhindawicom Volume 2018
Probability and StatisticsHindawiwwwhindawicom Volume 2018
Journal of
Hindawiwwwhindawicom Volume 2018
Mathematical PhysicsAdvances in
Complex AnalysisJournal of
Hindawiwwwhindawicom Volume 2018
OptimizationJournal of
Hindawiwwwhindawicom Volume 2018
Hindawiwwwhindawicom Volume 2018
Engineering Mathematics
International Journal of
Hindawiwwwhindawicom Volume 2018
Operations ResearchAdvances in
Journal of
Hindawiwwwhindawicom Volume 2018
Function SpacesAbstract and Applied AnalysisHindawiwwwhindawicom Volume 2018
International Journal of Mathematics and Mathematical Sciences
Hindawiwwwhindawicom Volume 2018
Hindawi Publishing Corporation httpwwwhindawicom Volume 2013Hindawiwwwhindawicom
The Scientific World Journal
Volume 2018
Hindawiwwwhindawicom Volume 2018Volume 2018
Numerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisAdvances inAdvances in Discrete Dynamics in
Nature and SocietyHindawiwwwhindawicom Volume 2018
Hindawiwwwhindawicom
Dierential EquationsInternational Journal of
Volume 2018
Hindawiwwwhindawicom Volume 2018
Decision SciencesAdvances in
Hindawiwwwhindawicom Volume 2018
AnalysisInternational Journal of
Hindawiwwwhindawicom Volume 2018
Stochastic AnalysisInternational Journal of
Submit your manuscripts atwwwhindawicom
10 Journal of Function Spaces
[14] X Q Zhang L Wang and Q S Wang ldquoExistence of positivesolutions for a class of nonlinear fractional differential equa-tions with integral boundary conditionsrdquo Applied Mathematicsand Computation vol 226 pp 708ndash718 2014
[15] XG Zhang L S Liu YHWu andBWiwatanapataphee ldquoThespectral analysis for a singular fractional differential equationwith a signedmeasurerdquoApplied Mathematics and Computationvol 257 pp 252ndash263 2015
[16] Y Wang and L Liu ldquoPositive solutions for a class of fractional3-point boundary value problems at resonancerdquo Advances inDifference Equations vol 13 article 13 2017
[17] M A Krasnoselskii and P P Zabreiko Geometrical Methods ofNonlinear Analysis Springer New York NY USA 1984
[18] K Deimling Nonlinear Functional Analysis Springer BerlinGermany 1985
[19] K Zhang ldquoNontrivial solutions of fourth-order singularboundary value problems with sign-changing nonlinear termsrdquoTopological Methods in Nonlinear Analysis vol 40 no 1 pp 53ndash70 2012
[20] D J Guo and V Lakshmikantham Nonlinear Problems inAbstract Cones Academic Press Massachusetts Mass USA1988
[21] J Liu and Z Zhao ldquoMultiple solutions for impulsive problemswith non-autonomous perturbationsrdquo Applied MathematicsLetters vol 64 pp 143ndash149 2017
Hindawiwwwhindawicom Volume 2018
MathematicsJournal of
Hindawiwwwhindawicom Volume 2018
Mathematical Problems in Engineering
Applied MathematicsJournal of
Hindawiwwwhindawicom Volume 2018
Probability and StatisticsHindawiwwwhindawicom Volume 2018
Journal of
Hindawiwwwhindawicom Volume 2018
Mathematical PhysicsAdvances in
Complex AnalysisJournal of
Hindawiwwwhindawicom Volume 2018
OptimizationJournal of
Hindawiwwwhindawicom Volume 2018
Hindawiwwwhindawicom Volume 2018
Engineering Mathematics
International Journal of
Hindawiwwwhindawicom Volume 2018
Operations ResearchAdvances in
Journal of
Hindawiwwwhindawicom Volume 2018
Function SpacesAbstract and Applied AnalysisHindawiwwwhindawicom Volume 2018
International Journal of Mathematics and Mathematical Sciences
Hindawiwwwhindawicom Volume 2018
Hindawi Publishing Corporation httpwwwhindawicom Volume 2013Hindawiwwwhindawicom
The Scientific World Journal
Volume 2018
Hindawiwwwhindawicom Volume 2018Volume 2018
Numerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisAdvances inAdvances in Discrete Dynamics in
Nature and SocietyHindawiwwwhindawicom Volume 2018
Hindawiwwwhindawicom
Dierential EquationsInternational Journal of
Volume 2018
Hindawiwwwhindawicom Volume 2018
Decision SciencesAdvances in
Hindawiwwwhindawicom Volume 2018
AnalysisInternational Journal of
Hindawiwwwhindawicom Volume 2018
Stochastic AnalysisInternational Journal of
Submit your manuscripts atwwwhindawicom
Hindawiwwwhindawicom Volume 2018
MathematicsJournal of
Hindawiwwwhindawicom Volume 2018
Mathematical Problems in Engineering
Applied MathematicsJournal of
Hindawiwwwhindawicom Volume 2018
Probability and StatisticsHindawiwwwhindawicom Volume 2018
Journal of
Hindawiwwwhindawicom Volume 2018
Mathematical PhysicsAdvances in
Complex AnalysisJournal of
Hindawiwwwhindawicom Volume 2018
OptimizationJournal of
Hindawiwwwhindawicom Volume 2018
Hindawiwwwhindawicom Volume 2018
Engineering Mathematics
International Journal of
Hindawiwwwhindawicom Volume 2018
Operations ResearchAdvances in
Journal of
Hindawiwwwhindawicom Volume 2018
Function SpacesAbstract and Applied AnalysisHindawiwwwhindawicom Volume 2018
International Journal of Mathematics and Mathematical Sciences
Hindawiwwwhindawicom Volume 2018
Hindawi Publishing Corporation httpwwwhindawicom Volume 2013Hindawiwwwhindawicom
The Scientific World Journal
Volume 2018
Hindawiwwwhindawicom Volume 2018Volume 2018
Numerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisNumerical AnalysisAdvances inAdvances in Discrete Dynamics in
Nature and SocietyHindawiwwwhindawicom Volume 2018
Hindawiwwwhindawicom
Dierential EquationsInternational Journal of
Volume 2018
Hindawiwwwhindawicom Volume 2018
Decision SciencesAdvances in
Hindawiwwwhindawicom Volume 2018
AnalysisInternational Journal of
Hindawiwwwhindawicom Volume 2018
Stochastic AnalysisInternational Journal of
Submit your manuscripts atwwwhindawicom
