Research Article Existence and Multiple Positive Solutions ...value problems for the fractional di...
Transcript of Research Article Existence and Multiple Positive Solutions ...value problems for the fractional di...
Research ArticleExistence and Multiple Positive Solutions forBoundary Value Problem of Fractional Differential Equationwith 119901-Laplacian Operator
Min Jiang and Shouming Zhong
School of Mathematics Science University of Electronic Science and Technology of China Chengdu 611731 China
Correspondence should be addressed to Min Jiang jiangmin10701126com
Received 28 February 2014 Accepted 26 April 2014 Published 15 May 2014
Academic Editor Bashir Ahmad
Copyright copy 2014 M Jiang and S Zhong This is an open access article distributed under the Creative Commons AttributionLicense which permits unrestricted use distribution and reproduction in any medium provided the original work is properlycited
This paper investigates the existence multiplicity nonexistence and uniqueness of positive solutions to a kind of two-pointboundary value problem for nonlinear fractional differential equations with 119901-Laplacian operator By using fixed point techniquescombining with partially ordered structure of Banach space we establish some criteria for existence and uniqueness of positivesolution of fractional differential equations with 119901-Laplacian operator in terms of different value of parameter In particular thedependence of positive solution on the parameter was obtained Finally several illustrative examples are given to support theobtained new results The study of illustrative examples shows that the obtained results are applicable
1 Introduction
In this paper we consider boundary value problem for afractional differential equation with 119901-Laplacian operator
(120601119901 (119863120572
0+119909 (119905)))
1015840
+ 120582120596 (119905) 119891 (119905 119909 (119905)) = 0 0 lt 119905 lt 1
1199091015840(0) = 119909
10158401015840(0) = 0
119909 (1) = 120578int
1
0
119892 (119904) 119909 (119904) 119889119904
(1)
where 2 lt 120572 le 3 0 lt 120578 lt 1 is a constant and 120582 gt 0 is aparameter 119901 gt 1 120601119901(119904) is the p-Laplacian operator that is120601119901(119904) = |119904|
119901minus2119904 120601minus1
119901(119904) = 120601119902(119904) 1119901 + 1119902 = 1 119863 denotes
the Caputo fractional derivativeFractional differential equations have been of great inter-
est recently It is caused both by the intensive developmentof the theory of fractional calculus itself and by the appli-cations of such constructions in various fields of sciencesand engineering such as physics chemistry aerodynamicselectrodynamics of complex medium electrical circuits andbiology (see [1ndash7] and their references) Differential equa-tions with p-Laplacian arise naturally in non-Newtonian
mechanics nonlinear elasticity glaciology population biol-ogy combustion theory and nonlinear flow laws Since thep-Laplacian operator and fractional calculus arise from somany applied fields the fractional p-Laplacian differentialequations are worth studying Recently there have appeareda very large number of papers which are devoted to theexistence of solutions of boundary value problems and initialvalue problems for the fractional differential equations (see[8ndash12]) and the existence of solutions of boundary valueproblems for the fractional p-Laplacian differential equationshas just begun in recent years (see [13ndash22]) On the otherhand there are few papers that consider the eigenvalueintervals of fractional boundary value problems (see [23 24])In [23] the author discussed the following system
119863120572
0+119909 (119905) + 120582ℎ (119905) 119891 (119909 (119905)) = 0 0 lt 119905 lt 1
2 lt 120572 le 3
119909 (0) = 1199091015840(0) = 119909
10158401015840(0) = 119909
10158401015840(1) = 0
(2)
By the use of appropriate conditions with respect tolim119905rarr0(119891(119905)119905) = 1198910 and lim119905rarrinfin(119891(119905)119905) = 119891infin the authorproved that the above problem has at least one or two positive
Hindawi Publishing CorporationAbstract and Applied AnalysisVolume 2014 Article ID 512426 18 pageshttpdxdoiorg1011552014512426
2 Abstract and Applied Analysis
solutions for some 120582 where 1198910 = 0 119897infin 119891infin = 0 119897infin 0 lt
119897 lt infin But almost all the results which the author obtaineddepend on both 1198910 and 119891infin the case depends on one of 1198910and 119891infin the author only discussed 1198910 = 0infin and 119891infin = 0infinthe case 1198910 = 119897 or 119891infin = 119897 has not been discussed On theother hand there exist several results on the existence of onesolution to fractional p-Laplacian boundary value problems(BVPs) there are to the best of our knowledge relatively fewresults on the nonexistence and the uniqueness of positivesolutions to fractional p-Laplacian differential equation withparameter
Motivated by the above questions in this paper we willestablish several sufficient conditions for the existence ofpositive solutions of (1) by using fixed point theorem andfixed point index theory
The work is organized in the following fashion In Sec-tion 2 we provide some necessary background In particularwe will introduce some lemmas and definitions associatedwith fixed point index theory The main results will be statedand proved in Section 3 Two examples are given in Section 4
2 Preliminaries
In this section we introduce definitions and preliminary factswhich are used throughout this paper
Definition 1 (see [1]) The Riemann-Liouville fractional inte-gral of order 120572 gt 0 of a function 119909 (0 +infin) rarr R is givenby
119868120572
0+119909 (119905) =
1
Γ (120572)int
119905
0
(119905 minus 119904)120572minus1
119909 (119904) 119889119904 (3)
provided that the right side is point-wise defined on (0 +infin)
and Γ is the Gamma function
Definition 2 (see [1]) The Caputo fractional derivative oforder 120572 gt 0 of a continuous function 119909 (0 +infin) rarr R isgiven by
119863120572
0+119909 (119905) =
1
Γ (119899 minus 120572)int
119905
0
(119905 minus 119904)119899minus120572minus1
119909(119899)
(119904) 119889119904 (4)
where 119899 = [120572] + 1 provided that the right side is point-wisedefined on (0 +infin)
Lemma 3 (see [1]) Let 120572 gt 0 and assume that 119909 isin 119862[0 1]and then the fractional differential equation 119863
120572
0119909(119905) = 0 has
unique solutions
119906 (119905) = 1198880 + 1198881119905 + 11988821199052+ sdot sdot sdot + 119888119899minus1119905
119899minus1
119888119894 isin R 119894 = 0 1 2 119899 minus 1 119899 minus 1 lt 120572 le 119899
(5)
Lemma 4 (see [1]) Let 120572 gt 0 Then
119868120572
0+119863120572
0+119909 (119905)
= 119909 (119905) + 1198880 + 1198881119905 + 11988821199052+ sdot sdot sdot + 119888119899minus1119905
119899minus1
(6)
where 119888119894 isin R 119894 = 0 1 2 119899 minus 1 119899 minus 1 lt 120572 le 119899
Lemma 5 (see [25]) Let 119875 be a cone in a real Banach space119864 and letΩ be a bounded open set of 119864 Assume that operator119860 119875 cap Ω rarr 119875 is completely continuous If there exists a1199090 gt 0 such that
119909 minus 119860119909 = 1199051199090 forall119909 isin 119875 cap 120597Ω 119905 ge 0 (7)
then 119894(119860 119875 cap Ω 119875) = 0
Lemma 6 (see [26]) Let 119864 be a Banach space and 119875 sube 119864 acone and Ω1 and Ω2 are open set with 0 isin Ω1 Ω1 sube Ω2 andlet 119879 119875 cap (Ω2 Ω1) rarr 119875 be completely continuous operatorsuch that either
(i) 119879119906 le 119906 119906 isin 119875cap120597Ω1 and 119879119906 ge 119906 119906 isin 119875cap120597Ω2
or
(ii) 119879119906 ge 119906 119906 isin 119875cap120597Ω1 and 119879119906 le 119906 119906 isin 119875cap120597Ω2
holds Then 119879 has a fixed point in 119875 cap (Ω2 Ω1)
3 Main Results
In this section we present some new results on the existencemultiplicity nonexistence and the uniqueness of positivesolution of problem (1) and dependence of the positivesolution 119909120582(119905) on the parameter 120582
Let 119864 = 119862[0 1] then 119864 is a real Banach space with thenorm sdot defined by 119909 = max0le119905le1|119909(119905)|
Lemma 7 Let 119910(119905) isin 119862(0 1) cap 1198711(0 1) and 119910(119905) ge 0 then the
solution of the problem
(120601119901 (119863120572
0+119909 (119905)))
1015840
+ 120582119910 (119905) = 0 0 lt 119905 lt 1
1199091015840(0) = 119909
10158401015840(0) = 0
119909 (1) = 120578int
1
0
119892 (119904) 119909 (119904) 119889119904
(8)
is given by
119909 (119905) =1
Γ (120572)int
1
0
(1 minus 119904)120572minus1
120601119902 (120582int
119904
0
119910 (120591) 119889120591) 119889119904
minusint
119905
0
(119905 minus 119904)120572minus1
120601119902 (120582int
119904
0
119910 (120591) 119889120591) 119889119904
+120578
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
119892 (119904)
times (int
1
0
(1 minus 120591)120572minus1
120601119902 (120582int
120591
0
119910 (120583) 119889120583)119889120591
minus int
119904
0
(119904 minus 120591)120572minus1
times 120601119902 (120582int
120591
0
119910 (120583) 119889120583)119889120591)119889119904
(9)
Abstract and Applied Analysis 3
119909 (119905) ge 0 min0le119905le1
119909 (119905) ge 120590 119909 (10)
where 2 lt 120572 le 3 0 lt 120578 lt 1 1 minus 120578 int1
0119892(119904)119889119904 gt 0 and 120590 =
120578 int1
0(1 minus 119904)119892(119904)119889119904(1 minus 120578 int
1
0119904119892(119904)119889119904)
Proof It is easy to see by integration of (8) that
120601119901 (119863120572
0+119909 (119905)) minus 120601119901 (119863
120572
0+119909 (0)) = minus120582int
119905
0
119910 (119904) 119889119904 (11)
By the boundary condition 1199091015840(0) = 119909
10158401015840(0) = 0 we can easily
get that1198631205720+
119909(0) = 0 and we obtain that
120601119901 (119863120572
0+119909 (119905)) = minus120582int
119905
0
119910 (119904) 119889119904 (12)
that is
119863120572
0+119909 (119905) = minus120601119902 (120582int
119905
0
119910 (119904) 119889119904) (13)
By Lemmas 3 and 4 we get that
119909 (119905) = minus119868120572120601119902 (120582int
119905
0
119910 (119904) 119889119904) + 1198880 + 1198881119905 + 11988821199052 (14)
Using the boundary condition 1199091015840(0) = 119909
10158401015840(0) = 0 119909(1) =
120578 int1
0119892(119904)119909(119904)119889119904 we get 1198881 = 1198882 = 0 and
1198880 =1
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
(1 minus 119904)120572minus1
120601119902 (120582int
119904
0
119910 (120591) 119889120591) 119889119904
minus120578
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
119892 (119905) (int
119905
0
(119905 minus 119904)120572minus1
120601119902 (120582int
119904
0
119910 (120591) 119889120591) 119889119904) 119889119905
(15)
Thus
119909 (119905) = minus1
Γ (120572)int
119905
0
(119905 minus 119904)120572minus1
120601119902 (120582int
119904
0
119910 (120591) 119889120591) 119889119904
+1
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
(1 minus 119904)120572minus1
120601119902 (120582int
119904
0
119910 (120591) 119889120591) 119889119904
minus120578
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
119892 (119905) (int
119905
0
(119905 minus 119904)120572minus1
120601119902 (120582int
119904
0
119910 (120591) 119889120591) 119889119904) 119889119905
=1
Γ (120572)int
1
0
(1 minus 119904)120572minus1
120601119902 (120582int
119904
0
119910 (120591) 119889120591) 119889119904
minusint
119905
0
(119905 minus 119904)120572minus1
120601119902 (120582int
119904
0
119910 (120591) 119889120591) 119889119904
+120578
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
119892 (119904) (int
1
0
(1 minus 120591)120572minus1
120601119902 (120582int
120591
0
119910 (120583) 119889120583)119889120591
minus int
119904
0
(119904 minus 120591)120572minus1
times 120601119902 (120582int
120591
0
119910 (120583) 119889120583)119889120591)119889119904
(16)
Direct differentiation of (9) implies
1199091015840(119905)
= minus1
Γ (120572 minus 1)int
119905
0
(119905 minus 119904)120572minus2
120601119902 (120582int
119904
0
119910 (120591) 119889120591) 119889119904 le 0
(17)
By differentiation of (17) we get
11990910158401015840(119905)
= minus1
Γ (120572 minus 2)int
119905
0
(119905 minus 119904)120572minus3
120601119902 (120582int
119904
0
119910 (120591) 119889120591) 119889119904 le 0
(18)
Thus the solution of problem (8) is nonincreasing andconcave on [0 1] and
119909 = 119909 (0) min0le119905le1
119909 (119905) = 119909 (1) (19)
On the other hand as we assume that 119910(119905) ge 0 we see that
119909 (1) =120578
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
119892 (119904) (int
1
0
(1 minus 120591)120572minus1
120601119902 (120582int
120591
0
119910 (120583) 119889120583)119889120591
minus int
119904
0
(119904 minus 120591)120572minus1
times 120601119902 (120582int
120591
0
119910 (120583) 119889120583)119889120591)119889119904
=120578
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
119892 (119904) (int
119904
0
[(1 minus 120591)120572minus1
minus (119904 minus 120591)120572minus1
]
4 Abstract and Applied Analysis
times 120601119902 (120582int
120591
0
119910 (120583) 119889120583)119889120591
+int
1
119904
(1 minus 120591)120572minus1
120601119902 (120582int
120591
0
119910 (120583) 119889120583)119889120591)119889119904
ge 0
(20)
Therefore 119909(119905) ge 0 and is concave for 119905 isin [0 1] So for every119905 isin [0 1]
119909 (119905) ge 119905119909 (1) + (1 minus 119905) 119909 (0) (21)
Therefore
120578int
1
0
119909 (119905) 119892 (119905) 119889119905 ge 120578119909 (1) int
1
0
119905119892 (119905) 119889119905
+ 120578119909 (0) int
1
0
(1 minus 119905) 119892 (119905) 119889119905
(22)
Since 119909(1) = 120578 int1
0119909(119905)119892(119905)119889119905 1 minus 120578 int
1
0119892(119905)119889119905 gt 0 we have
119909 (1) ge120578 int1
0(1 minus 119905) 119892 (119905) 119889119905
1 minus 120578 int1
0119905119892 (119905) 119889119905
119909 (0) (23)
that is
min0le119905le1
119909 (119905) = 119909 (1) ge120578 int1
0(1 minus 119905) 119892 (119905) 119889119905
1 minus 120578 int1
0119905119892 (119905) 119889119905
119909 (0) = 120590 119909 (24)
The Lemma is proved
We construct a cone 119875 in 119864 by
119875 = 119909 isin 119864 119909 ge 0 min0le119905le1
119909 (119905) ge 120590 119909 (25)
where 120590 is defined in Lemma 7 It is easy to see 119875 is a closedconvex cone of 119864
Define 119879 119875 rarr 119864 by
(119879119909) (119905)
=1
Γ (120572)int
1
0
(1 minus 119904)120572minus1
120601119902 (int
119904
0
120596 (120591) 119891 (120591 119909 (120591)) 119889120591) 119889119904
minusint
119905
0
(119905 minus 119904)120572minus1
120601119902 (int
119904
0
120596 (120591) 119891 (120591 119909 (120591)) 119889120591) 119889119904
+120578
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
119892 (119904)
times (int
1
0
(1 minus 120591)120572minus1
120601119902 (int
120591
0
120596 (120583) 119891 (120583 119909 (120583)) 119889120583)119889120591
minus int
119904
0
(119904 minus 120591)120572minus1
times 120601119902 (int
120591
0
120596 (120583) 119891 (120583 119909 (120583)) 119889120583)119889120591)119889119904
(26)
It is clear that the fixed points of the operator 120601119902(120582)119879 arethe solutions of the boundary value problems (1)
We make the following hypotheses
(H1) 119891 [0 1] times [0infin) rarr [0infin) is continuous(H2) 120596 (0 1) rarr [0infin) is continuous and not identical
zero on any closed subinterval of (0 1) with 0 lt
int1
0120596(119905)119889119905 lt +infin
(H3) 1 minus 120578 int1
0119892(119904)119889119904 gt 0
(H4) 1198910
120601isin [0infin] where 119891
0
120601= lim119909rarr0(119891(119905 119909)120601119901(119909))
uniformly for 119905 isin [0 1](H5) 119891
infin
120601isin [0infin] where 119891
infin
120601= lim119909rarrinfin(119891(119905 119909)120601119901(119909))
uniformly for 119905 isin [0 1](H6) 119891(119905 119909) gt 0 for any 119905 isin [0 1] and 119909 gt 0
Lemma 8 Assume that (H1)ndash(H3) hold Then 119879 119875 rarr 119864 iscompletely continuous
Proof First we show that 119879 is continuous It is easy to see119879(119875) sub 119875 For 119909119899(119905) isin 119864 and 119909119899(119905) rarr 119909(119905) as 119899 rarr infin bythe continuous of 119891(119905 119909(119905)) we get 119891(119905 119909119899(119905)) rarr 119891(119905 119909(119905))as 119899 rarr infin This implies that
120601119902 (int
119905
0
119908 (119904) 119891 (119904 119909119899 (119904)) 119889119904)
997888rarr 120601119902 (int
119905
0
119908 (119904) 119891 (119904 119909 (119904)) 119889119904)
(27)
So we have
sup119905isin[01]
10038161003816100381610038161003816100381610038161003816120601119902 (int
119905
0
119908 (119904) 119891 (119904 119909119899 (119904)) 119889119904)
minus 120601119902 (int
119905
0
119908 (119904) 119891 (119904 119909 (119904)) 119889119904)
10038161003816100381610038161003816100381610038161003816997888rarr 0 119899 997888rarr infin
(28)
Denote Π = |(119879119909119899)(119905) minus (119879119909)(119905)| then
Π le1
Γ (120572)int
1
0
(1 minus 119904)120572minus1
10038161003816100381610038161003816100381610038161003816120601119902 (int
119904
0
120596 (120591) 119891 (120591 119909119899 (120591)) 119889120591)
minus 120601119902 (int
119904
0
120596 (120591) 119891 (120591 119909 (120591)) 119889120591)
10038161003816100381610038161003816100381610038161003816119889119904
Abstract and Applied Analysis 5
+int
119905
0
(119905 minus 119904)120572minus1
10038161003816100381610038161003816100381610038161003816120601119902 (int
119904
0
120596 (120591) 119891 (120591 119909119899 (120591)) 119889120591)minus120601119902
times (int
119904
0
120596 (120591) 119891 (120591 119909 (120591)) 119889120591)
10038161003816100381610038161003816100381610038161003816119889119904
+120578
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
119892 (119904)
times (int
1
0
(1 minus 120591)120572minus1
times
10038161003816100381610038161003816100381610038161003816120601119902 (int
120591
0
120596 (120583) 119891 (120583 119909119899 (120583)) 119889120583)
minus 120601119902 (int
120591
0
120596 (120583) 119891 (120583 119909 (120583)) 119889120583)
10038161003816100381610038161003816100381610038161003816119889120591
+ int
119904
0
(119904 minus 120591)120572minus1
times
10038161003816100381610038161003816100381610038161003816120601119902 (int
120591
0
120596 (120583) 119891 (120583 119909119899 (120583)) 119889120583) minus 120601119902
times (int
120591
0
120596 (120583) 119891 (120583 119909 (120583)) 119889120583)
10038161003816100381610038161003816100381610038161003816119889120591) 119889119904
le sup119904isin[01]
10038161003816100381610038161003816100381610038161003816120601119902 (int
119904
0
119908 (120591) 119891 (120591 119909119899 (120591)) 119889120591)
minus 120601119902 (int
119904
0
119908 (120591) 119891 (120591 119909 (120591)) 119889120591)
10038161003816100381610038161003816100381610038161003816
times2
Γ (120572 + 1)(1 +
120578 int1
0119892 (119904) 119889119904
1 minus 120578 int1
0119892 (119904) 119889119904
) 997888rarr 0
119899 997888rarr infin
(29)
Hence 119879 is continuousSecond we show that 119879 is compactFor 119909(119905) isin 119861119903 = 119909 isin 119875 119909 le 119903 by condition (H1)
119891(119905 119909(119905)) lt infin Denote119872 = max0le119905le1119909le119903119891(119905 119909(119905))Then
|119879119909 (119905)|
le1
Γ (120572)int
1
0
(1 minus 119904)120572minus1
10038161003816100381610038161003816100381610038161003816120601119902 (int
119904
0
120596 (120591) 119891 (120591 119909 (120591)) 119889120591)
10038161003816100381610038161003816100381610038161003816119889119904
+int
119905
0
(119905 minus 119904)120572minus1
10038161003816100381610038161003816100381610038161003816120601119902 (int
119904
0
120596 (120591) 119891 (120591 119909 (120591)) 119889120591)
10038161003816100381610038161003816100381610038161003816119889119904
+120578
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
119892 (119904)
times (int
1
0
(1 minus 120591)120572minus1
times
10038161003816100381610038161003816100381610038161003816120601119902 (int
120591
0
120596 (120583) 119891 (120583 119909 (120583)) 119889120583)
10038161003816100381610038161003816100381610038161003816119889120591
+ int
119904
0
(119904 minus 120591)120572minus1
times
10038161003816100381610038161003816100381610038161003816120601119902 (int
120591
0
120596 (120583) 119891 (120583 119909 (120583)) 119889120583)
10038161003816100381610038161003816100381610038161003816119889120591) 119889119904
le1
Γ (120572)
times int
1
0
(1 minus 119904)120572minus1
10038161003816100381610038161003816100381610038161003816120601119902 (119872int
119904
0
120596 (120591) 119889120591)
10038161003816100381610038161003816100381610038161003816119889119904
+int
119905
0
(119905 minus 119904)120572minus1
10038161003816100381610038161003816100381610038161003816120601119902 (119872int
119904
0
120596 (120591) 119889120591)
10038161003816100381610038161003816100381610038161003816119889119904
+120578
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
119892 (119904) (int
1
0
(1 minus 120591)120572minus1
times
10038161003816100381610038161003816100381610038161003816120601119902 (119872int
120591
0
120596 (120583) 119889120583)119889120591
10038161003816100381610038161003816100381610038161003816
+ int
119904
0
(119904 minus 120591)120572minus1
times
10038161003816100381610038161003816100381610038161003816120601119902 (119872int
120591
0
120596 (120583) 119889120583)
10038161003816100381610038161003816100381610038161003816119889120591) 119889119904
le120578
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
120601119902 (119872int
1
0
120596 (120591) 119889120591)
times int
1
0
119892 (119904) (int
1
0
(1 minus 120591)120572minus1
119889120591 + int
119904
0
(119904 minus 120591)120572minus1
119889120591)119889119904
+1
Γ (120572)120601119902 (119872int
1
0
120596 (120591) 119889120591)
times int
1
0
(1 minus 119904)120572minus1
119889119904 + int
119905
0
(119905 minus 119904)120572minus1
119889119904
le
2120601119902 (119872int1
0120596 (120591) 119889120591)
Γ (120572 + 1) (1 minus 120578 int1
0119892 (120591) 119889120591)
lt infin
(30)
that is 119879 maps bounded sets into bounded sets in 119875For 119909(119905) isin 119861119903 0 lt 1199051 lt 1199052 lt 1
1003816100381610038161003816119879119909 (1199052) minus 119879119909 (1199051)1003816100381610038161003816
=1
Γ (120572)
10038161003816100381610038161003816100381610038161003816int
1199052
0
(1199052 minus 119904)120572minus1
120601119902 (int
119904
0
120596 (120591) 119891 (120591 119909 (120591)) 119889120591) 119889119904
6 Abstract and Applied Analysis
minus int
1199051
0
(1199051 minus 119904)120572minus1
times 120601119902 (int
119904
0
120596 (120591) 119891 (120591 119909 (120591)) 119889120591) 119889119904
10038161003816100381610038161003816100381610038161003816
le1
Γ (120572)120601119902 (119872int
1
0
120596 (120591) 119889120591)
times
100381610038161003816100381610038161003816100381610038161003816
int
1199051
0
((1199052 minus 119904)120572minus1
minus (1199051 minus 119904)120572minus1
) 119889119904
+int
1199052
1199051
(1199052 minus 119904)120572minus1
119889119904
100381610038161003816100381610038161003816100381610038161003816
(31)
By mean value theorem we obtain that
int
1199051
0
((1199052 minus 119904)120572minus1
minus (1199051 minus 119904)120572minus1
) 119889119904
le int
1199051
0
(120572 minus 1) (1 minus 119904)120572minus2
(1199052 minus 1199051) 119889119904
le int
1
0
(120572 minus 1) (1 minus 119904)120572minus2
(1199052 minus 1199051) 119889119904
= 1199052 minus 1199051
(32)
Thus1003816100381610038161003816119879119909 (1199052) minus 119879119909 (1199051)
1003816100381610038161003816
le1
Γ (120572)120601119902 (119872int
1
0
120596 (120591) 119889120591) [(1199052 minus 1199051) +1
120572(1199052 minus 1199051)
120572]
(33)
This shows that |119879119909(1199052) minus 119879119909(1199051)| rarr 0 as 1199052 minus 1199051 rarr 0so 119879119909 119909 isin 119861119903 is equicontinuous Therefore the operator119879 119875 rarr 119875 is completely continuous by the Arzela-Ascolitheorem
Now for convenience we introduce the following nota-tions Let
int
1
0
(1 minus 119904)120572minus1
120601119902 (int
119904
0
120596 (120591) 119889120591) 119889119904 = 1205741
120601119902 (int
1
0
120596 (120591) 119889120591) = 1205742
119898 (119903) = minmin119905
119891 (119905 119909)
120601119901 (119903) 119909 isin [120590119903 119903]
(34)
Theorem 9 Assume that (H1)ndash(H5) hold
(i) If 0 lt 1198910
120601lt +infin then there exists 1205850 gt 0 such that for
every 0 lt 119903 lt 1205850 problem (1) has a positive solution119909119903(119905) satisfying 119909119903 = 119903 associated with
120582 = 120582119903 isin [1205820 1205820] (35)
where 1205820 and 1205820 are two positive finite numbers
(ii) If 0 lt 119891infin
120601lt +infin then there exists 120585lowast
0gt 0 such that for
every 119877 gt 120585lowast
0 problem (1) has a positive solution 119909119877(119905)
satisfying 119909119877 = 119877 for any
120582 = 120582119877 isin [120582lowast
0 120582lowast
0] (36)
where 120582lowast
0and 120582
lowast
0are two positive finite numbers
(iii) If 1198910120601
= +infin then there exists 1205851 gt 0 such that for any0 lt 119903lowast
lt 1205851 problem (1) has a positive solution 119909119903lowast(119905)
satisfying 119909119903lowast = 119903lowast for any
120582 = 120582119903lowast isin (0 120582lowast] (37)
where 120582lowast is a positive finite number
(iv) If 119891infin
120601= +infin then there exists 1205851 gt 0 such that
for every 119877lowast gt 1205851 problem (1) has a positive solution119909119877lowast
(119905) satisfying 119909119877lowast
= 119877lowast for any
120582 = 120582119877lowast
isin (0 120582lowast] (38)
where 120582lowast is a positive finite number(v) If there exists 120585 gt 0 such that 119898(119903) gt 120585 then problem
(1) has positive solution 119909119903(119905) satisfying 119909119903 = 119903 forany
120582 = 120582119903 isin (0 ] (39)
where is a positive finite number
Proof (i) It follows from 0 lt 1198910
120601lt +infin that there exists
1198971 1198972 120588 gt 0 such that
1198971120601119901 (119909) le 119891 (119905 119909) le 1198972120601119901 (119909)
(forall119905 isin [0 1] 0 le 119909 le 120588)
(40)
Let 1205850 = 120588120590 we show that 1205850 is required When 119909 isin 119875 cap
120597Ω119903 we have
119909 (119905) ge 120590 119909 = 120590119903 (41)
we need 119909(119905) lt 120588 this implies that 120590119903 lt 120588 as 119903 lt 1205850 so120590119903 lt 1205901205850 = 120588
Let 1205820 = (11198971)(Γ(120572)1205901205741)119901minus1 Then we may assume that
119909 minus 120601119902 (1205820) 119879119909 = 0 (forall119909 isin 119875 cap 120597Ω119903) (42)
if not then there exists119909119903 isin 119875cap120597Ω119903 such that120601119902(1205820)119879119909119903 = 119909119903and then (35) already holds for 120582119903 = 1205820
Define 120595(119905) equiv 1 forall119905 isin [0 1] then 120595(119905) isin 119875 and 120595 equiv 1We now show that
119909 minus 120601119902 (1205820) 119879119909 = 120581120595 (forall119909 isin 119875 cap 120597Ω119903 120581 ge 0) (43)
In fact if there exists 1199091 isin 119875 cap 120597Ω119903 1205811 ge 0 such that 1199091 minus120601119902(1205820)1198791199091 = 1205811120595 then (42) implies that 1205811 gt 0 On the otherhand 1199091 = 120601119902(1205820)1198791199091 + 1205811120595 ge 1205811120595 we may choose 120581
lowast=
max120581 | 1199091 ge 120581120595 then 1205811 le 120581lowastlt +infin 1199091 ge 120581
lowast120595 Therefore
120581lowast= 120581lowast 1003817100381710038171003817120595
1003817100381710038171003817 le10038171003817100381710038171199091
1003817100381710038171003817 = 119903 (44)
Abstract and Applied Analysis 7
Consequently for any 119905 isin [0 1] (26) and (44) imply that
1199091 (119905)
= 120601119902 (1205820)1
Γ (120572)
times int
1
0
(1 minus 119904)120572minus1
times 120601119902 (int
119904
0
120596 (120591) 119891 (120591 1199091 (120591)) 119889120591) 119889119904
minus int
119905
0
(119905 minus 119904)120572minus1
times 120601119902 (int
119904
0
120596 (120591) 119891 (120591 1199091 (120591)) 119889120591) 119889119904
+120578120601119902 (1205820)
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
119892 (119904)
times (int
1
0
(1 minus 120591)120572minus1
times 120601119902 (int
120591
0
120596 (120583) 119891 (120583 1199091 (120583)) 119889120583)119889120591
minus int
119904
0
(119904 minus 120591)120572minus1
times 120601119902 (int
120591
0
120596 (120583) 119891 (120583 1199091 (120583)) 119889120583)119889120591)119889119904
+ 1205811120595
ge120578120590120601119902 (1205820)
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
119892 (119904)
times (int
1
0
(1 minus 120591)120572minus1
times 120601119902 (int
120591
0
120596 (120583) 119891 (120583 1199091 (120583)) 119889120583)119889120591
minus int
119904
0
(119904 minus 120591)120572minus1
120601119902
times (int
120591
0
120596 (120583) 119891 (120583 1199091 (120583)) 119889120583)119889120591)119889119904
+120590120601119902 (1205820)
Γ (120572)
times int
1
0
(1 minus 119904)120572minus1
120601119902 (int
119904
0
120596 (120591) 119891 (120591 1199091 (120591)) 119889120591) 119889119904
+ 1205811120595
ge120590120601119902 (1205820)
Γ (120572)
times int
1
0
(1 minus 119904)120572minus1
120601119902 (int
119904
0
120596 (120591) 119891 (120591 1199091 (120591)) 119889120591) 119889119904
+ 1205811120595
ge120590120601119902 (1205820)
Γ (120572)
times int
1
0
(1 minus 119904)120572minus1
120601119902 (1198971 int
119904
0
120596 (120591) 120601119901 (1199091 (120591)) 119889120591) 119889119904
+ 1205811120595
ge120590120601119902 (1205820)
Γ (120572)
times int
1
0
(1 minus 119904)120572minus1
120601119902 (1198971 int
119904
0
120596 (120591) 120601119901 (120581lowast120595 (120591)) 119889120591) 119889119904
+ 1205811120595
=120590120581lowast119897119902minus1
1120582119902minus1
0
Γ (120572)int
1
0
(1 minus 119904)120572minus1
120601119902 (int
119904
0
120596 (120591) 119889120591) 119889119904 + 1205811120595
=120590120581lowast119897119902minus1
1120582119902minus1
01205741
Γ (120572)+ 1205811120595
= 120581lowast+ 120581120595
(45)
which implies that 1199091(119905) ge (120581lowast+ 120581)120595(119905) 119905 isin [0 1] which is a
contradiction to the definition of 120581lowast Thus (42) holds and byLemma 5 the fixed point index
119894 (120601119902 (1205820) 119879 119875 cap 120597Ω119903 119875) = 0 (46)
On the other hand by the fact that the fixed point index ofconstants operator is 1 so
119894 (120599 119875 cap 120597Ω119903 119875) = 1 (47)
where 120599 is the zero operator It follows therefore from (46)and (47) and the homotopy invariance property that thereexist 119909119903 isin 119875 cap 120597Ω119903 and 0 lt ]0 lt 1 such that ]0120601119902(1205820)119879119909119903 =119909119903 but 119909119903 = 120601119902(120582119903)119879119909119903 which implies that ]0120601119902(1205820)119879119909119903 =
120601119902(120582119903)119879119909119903 it follows that we get ]0120601119902(1205820) = 120601119902(120582119903)) that is
]0120582119902minus1
0= 120582119902minus1
119903 (48)
and then
120582119903 = ]1(119902minus1)0
1205820 = ]119901minus10
1205820 lt 1205820 (49)
From the proof above for any 119903 lt 1205850 there exists a positivesolution 119909119903 isin 119875 cap 120597Ω119903 associated with 120582 = 120582119903 gt 0
8 Abstract and Applied Analysis
Thus
119909119903 (119905)
=120601119902 (120582119903)
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902 (int
119904
0
120596 (120591) 119891 (120591 119909119903 (120591)) 119889120591) 119889119904
minus int
119905
0
(119905 minus 119904)120572minus1
times 120601119902 (int
119904
0
120596 (120591) 119891 (120591 119909119903 (120591)) 119889120591) 119889119904
+120578120601119902 (120582119903)
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
119892 (119904)
times (int
1
0
(1 minus 120591)120572minus1
times 120601119902 (int
120591
0
120596 (120583) 119891 (120583 119909119903 (120583)) 119889120583)119889120591
minus int
119904
0
(119904 minus 120591)120572minus1
times 120601119902 (int
120591
0
120596 (120583) 119891 (120583 119909119903 (120583)) 119889120583)119889120591)119889119904
(50)
with 119909119903 = 119903Next we show that 120582119903 ge 1205820 In fact
119909119903 (119905)
=120601119902 (120582119903)
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902 (int
119904
0
120596 (120591) 119891 (120591 119909119903 (120591)) 119889120591) 119889119904
minus int
119905
0
(119905 minus 119904)120572minus1
times 120601119902 (int
119904
0
120596 (120591) 119891 (120591 119909119903 (120591)) 119889120591) 119889119904
+120578120601119902 (120582119903)
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
119892 (119904)
times (int
1
0
(1 minus 120591)120572minus1
times 120601119902 (int
120591
0
120596 (120583) 119891 (120583 119909119903 (120583)) 119889120583)119889120591
minus int
119904
0
(119904 minus 120591)120572minus1
times 120601119902 (int
120591
0
120596 (120583) 119891 (120583 119909119903 (120583)) 119889120583)119889120591)119889119904
le120578120601119902 (120582119903)
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
119892 (119904) int
1
0
(1 minus 120591)120572minus1
times 120601119902 (int
120591
0
120596 (120583) 119891 (120583 119909119903 (120583)) 119889120583)119889120591 119889119904
+120601119902 (120582119903)
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902 (int
119904
0
120596 (120591) 119891 (120591 119909119903 (120591)) 119889120591) 119889119904
=120601119902 (120582119903) 120578 int
1
0119892 (119904) 119889119904
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
(1 minus 119904)120572minus1
times 120601119902 (int
119904
0
120596 (120591) 119891 (120591 119909119903 (120591)) 119889120591) 119889119904
+120601119902 (120582119903)
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902 (int
119904
0
120596 (120591) 119891 (120591 119909119903 (120591)) 119889120591) 119889119904
le120601119902 (120582119903) 120578 int
1
0119892 (119904) 119889119904
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
(1 minus 119904)120572minus1
times 120601119902 (1198972 int
119904
0
120596 (120591) 119909119903 (120591) 119889120591) 119889119904
+120601119902 (120582119903)
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902 (1198972120582119903 int
119904
0
120596 (120591) 119909119903 (120591) 119889120591) 119889119904
=120601119902 (120582119903)
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
(1 minus 119904)120572minus1
times 120601119902 (1198972 int
119904
0
120596 (120591) 119909119903 (120591) 119889120591) 119889119904
Abstract and Applied Analysis 9
le
10038171003817100381710038171199091199031003817100381710038171003817 120601119902 (1205821199031198972)
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
(1 minus 119904)120572minus1
times 120601119902 (int
1
0
120596 (120591) 119889120591) 119889119904
=
10038171003817100381710038171199091199031003817100381710038171003817 120601119902 (1205821199031198972) 1205742
Γ (120572 + 1) (1 minus 120578 int1
0119892 (119904) 119889119904)
(51)
which implies that
10038171003817100381710038171199091199031003817100381710038171003817 le
10038171003817100381710038171199091199031003817100381710038171003817 120601119902 (1205821199031198972) 1205742
Γ (120572 + 1) (1 minus 120578 int1
0119892 (119904) 119889119904)
(52)
that is
120601119902 (1205821199031198972) 1205742
Γ (120572 + 1) (1 minus 120578 int1
0119892 (119904) 119889119904)
ge 1 (53)
it follows that we get
120582119903 ge1
1198972(
Γ (120572 + 1) (1 minus 120578 int1
0119892 (119904) 119889119904)
1205742)
119901minus1
= 1205820(54)
In conclusion 120582119903 isin [1205820 1205820](ii) It follows from 0 lt 119891
infin
120601lt infin that there exist
119897lowast
1 119897lowast
2 120588lowastgt 0 such that
119897lowast
1120601119901 (119909) le 119891 (119905 119909) le 119897
lowast
2120601119901 (119909)
(forall119905 isin [0 1] 119909 ge 120588lowast)
(55)
Let 120585lowast0= 120588lowast120590 the following proof are similar to (i)
(iii) It follows from 1198910
120601= +infin there exists 119897lowast gt 0 120588
lowastgt 0
such that
1198910
120601ge 119897lowast120601119901 (119909) (forall119905 isin [0 1] 0 le 119909 le 120588
lowast) (56)
Let 1205851 = 120588lowast120590 we show that 1205851 is required When 119909 isin 119875 cap
120597Ω119903lowast we have 119909(119905) ge 120590119909 = 120590119903lowast since 119909 le 120588
lowast then we need120590119903lowastlt 120588lowast as 119903lowast lt 1205851 so 120590119903
lowastlt 120588lowast holds
Let 120582lowast = (1119897lowast)(Γ(120572)1205901205741)
1(119902minus1) we proceed in the sameway as in the proof of (i) replacing (42) we may assume that119909 minus 120601119902(120582
lowast)119879119909 = 0 (forall119909 isin 119875 cap 120597Ω119903lowast) and replacing (43) we
can prove 119909 minus 120601119902(120582lowast)119879119909 = 120581120595 (forall119909 isin 119875 cap 120597Ω119903lowast 120581 ge 0) It
follows from Lemma 5 that 119894(120601119902(120582lowast)119879 119875 cap Ω119903lowast 119875) = 0 Note
that 119894(120579 119875 cap Ω119903lowast 119875) = 1 we can easily show that there exists119909119903lowast isin 119875cap120597Ω119903lowast and 0 lt ]119903lowast lt 1 such that ]119903lowast120601119902(120582
lowast)119879119909119903lowast = 119909119903lowast
Hence (37) holds for 120582119903lowast = 120582lowast]1minus119901119903lowast
lt 120582lowast
The proof ofTheorem 9(iv) follows by themethod similarto Theorem 9(iii) we omit it here
(v) It follows that 119898(119903) gt 120585 for any 119909 isin 119875 cap 120597Ω119903 we have120590119903 le 119909 le 119903 and 119891(119905 119909)120601119901(119909) ge 119891(119905 119909)120601119901(119903) ge 119898(119903) ge 120585then
119891 (119905 119909) ge 120585120601119901 (119909) forall119909 isin [120590119903 119903] 119905 isin [0 1] (57)
Let = (1120585)(Γ(120572)1205901205741)119901minus1 The following proof is similar to
that of (iv) This finished the proof of (v)
Remark 10 InTheorem 9 all the criteria obtained depend onone of 1198910
120601and 119891
infin
120601
Let Ψ(119909) = 120601119902(120582)119879119909 the following theorems give out themultiply nonexistence and the dependence of parameter
Theorem 11 Assume that (H1)ndash(H5) hold
(i) If 1198910
120601= 0 and 119891
infin
120601= 0 lim
119909rarrinfin119891(119905 119909) = 119891infin isin
(0 +infin] uniformly for 119905 isin [0 1] then there exists120582lowast
gt 0 such that the problem (1) has two solutions forany 120582 gt 120582
lowast
(ii) If 1198910120601
= 0 and 119891infin
120601= 0 then there exists 120582 gt 0 such
that for any 120582 lt 120582 problem (1) has no solution
Proof (i) It follows from 119891infin isin (0 +infin] that for any 120577 gt 0there exists 120588 gt 0 such that
119891 (119905 119909) gt 120577 (forall119905 isin [0 1] 119909 ge 120588) (58)
Since 1198910
120601= 0 there exists 1205761 gt 0 120588
1gt 0 such that
119891 (119905 119909) le 1205761120601119901 (119909) (forall119905 isin [0 1] 0 le 119909 le 1205881) (59)
where 1205761 satisfied 120601119902(1205821205761)1205742Γ(120572 + 1)(1 minus 120578 int1
0119892(119904)119889119904) le 1
For forall119909 isin 119875 cap 120597Ω1205881
we have
Ψ (119909) = (120601119902 (120582) 119879119909) (119905)
=1
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902 (120582int
119904
0
120596 (120591) 119891 (120591 119909 (120591)) 119889120591) 119889119904
minus int
119905
0
(119905 minus 119904)120572minus1
times 120601119902 (120582int
119904
0
120596 (120591) 119891 (120591 119909 (120591)) 119889120591) 119889119904
+120578
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
119892 (119904)
10 Abstract and Applied Analysis
times (int
1
0
(1 minus 120591)120572minus1
times 120601119902 (120582int
120591
0
120596 (120583) 119891 (120583 119909 (120583)) 119889120583)119889120591
minus int
119904
0
(119904 minus 120591)120572minus1
times 120601119902 (120582int
120591
0
120596 (120583) 119891 (120583 119909 (120583)) 119889120583)119889120591)119889119904
le120578
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
119892 (119904)
times (int
1
0
(1 minus 120591)120572minus1
times 120601119902(120582int
120591
0
120596 (120583) 119891 (120583 119909 (120583)) 119889120583)119889120591)119889119904
+1
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902 (120582int
119904
0
120596 (120591) 1205761119891 (120591 119909 (120591)) 119889120591) 119889119904
le120578 int1
0119892 (119904) 119889119904
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
(1 minus 120591)120572minus1
times 120601119902 (120582int
1
0
120596 (120583) 1205761120601119901 (119909 (120583)) 119889120583)119889120591
+1
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902(120582int
1
0
120596 (120591) 1205761120601119901 (119909 (120591)) 119889120591) 119889119904
le1
Γ (120572 + 1) (1 minus 120578 int1
0119892 (119904) 119889119904)
times 120601119902 (120582int
1
0
120596 (120591) 1205761120601119901 (119909) 119889120591)
=120601119902 (1205821205761) 1205742
Γ (120572 + 1) (1 minus 120578 int1
0119892 (119904) 119889119904)
119909 le 119909
(60)
This implied
Ψ119909 le 119909 forall119909 isin 119875 cap 120597Ω1205881
(61)
Let 1205883gt max120588120590 120588
1 and 120582
lowast= (1120577)(120588
3Γ(120572)1205901205741)
119901minus1 thenforall119909 isin 119875 cap 120597Ω120588
3
we have 120590119909 le 119909 le 1205883and
(Ψ119909) (119905)
=1
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902 (120582int
119904
0
120596 (120591) 119891 (120591 119909 (120591)) 119889120591) 119889119904
minus int
119905
0
(119905 minus 119904)120572minus1
times 120601119902 (120582int
119904
0
120596 (120591) 119891 (120591 119909 (120591)) 119889120591) 119889119904
+120578
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
119892 (119904)
times (int
1
0
(1 minus 120591)120572minus1
times 120601119902 (120582int
120591
0
120596 (120583) 119891 (120583 119909 (120583)) 119889120583)119889120591
minus int
119904
0
(119904 minus 120591)120572minus1
times 120601119902(120582int
120591
0
120596 (120583) 119891 (120583 119909 (120583)) 119889120583)119889120591)119889119904
ge120590120578
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
119892 (119904)
times (int
1
0
(1 minus 120591)120572minus1
times 120601119902 (120582int
120591
0
120596 (120583) 119891 (120583 119909 (120583)) 119889120583)119889120591
minus int
119904
0
(119904 minus 120591)120572minus1
times 120601119902(120582int
120591
0
120596 (120583) 119891 (120583 119909 (120583)) 119889120583)119889120591)119889119904
+120590
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902 (120582int
119904
0
120596 (120591) 119891 (120591 119909 (120591)) 119889120591) 119889119904
ge120590
Γ (120572)int
1
0
(1 minus 119904)120572minus1
Abstract and Applied Analysis 11
times 120601119902 (120582int
119904
0
120596 (120591) 119891 (120591 119909 (120591)) 119889120591) 119889119904
gt120590
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902 (120582int
119904
0
120596 (120591) 120577 119889120591) 119889119904
gt120590120601119902 (120582120577)
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902 (int
119904
0
120596 (120591) 119889120591) 119889119904
=120590120601119902 (120582120577) 1205741
Γ (120572)= 1205883
(62)
which implies
Ψ119909 gt 119909 forall119909 isin 119875 cap 120597Ω1205883
(63)
Next for 119891infin
120601= 0 there exist 1205762 gt 0 120588
2gt 1205883gt 0 such
that
119891 (119905 119909) le 1205762120601119901 (119909) (forall119905 isin [0 1] 119909 ge 1205882) (64)
where 1205762 satisfied 120601119902(1205821205762)1205742Γ(120572 + 1)(1 minus 120578 int1
0119892(119904)119889119904) le 1
Similar to the above proof we get
Ψ119909 le 119909 forall119909 isin 119875 cap 120597Ω1205882
(65)
Applying Lemma 7 to (61) (65) and (63) yields that Ψ
has two fixed points 1199091 1199092 such that 1199091 isin 119875cap (Ω1205883
Ω1205881
) and1199092 isin 119875 cap (Ω120588
2
Ω1205883
)(ii) It follows from 119891
0
120601= 0 and 119891
infin
120601= 0 that there exists
1205761 1205762 1205881 1205882 gt 0 such that
119891 (119905 119909) le 1205761120601119901 (119909) (forall119905 isin [0 1] 0 le 119909 le 1205881)
119891 (119905 119909) le 1205762120601119901 (119909) (forall119905 isin [0 1] 119909 ge 1205882)
(66)
then for 119909 isin [1205881 1205882] by the continuous of 119891(119905 119909(119905)) there
exists 119909 isin [1205881 1205882] such that
max1205881le119909le120588
2
119891 (119905 119909)
120601119901 (119909)=
119891 (119905 119909)
120601119901 (119909) (67)
Let Υ = max1205761 1205762 119891(119905 119909)120601119901(119909) gt 0 then
119891 (119905 119909) le Υ120601119901 (119909) (forall0 le 119905 le 1 119909 ge 0) (68)
Assuming 119909(119905) is a positive solution of problem (1) we willshow that this leads to a contradiction for 120582 lt 120582 where
120582 = (1Υ)(Γ(120572 + 1)(1 minus 120578 int1
0119892(119904)119889119904)1205742)
119901minus1 It follows from(26) that
119909 = 119909 (0)
=120578
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
119892 (119904)
times (int
1
0
(1 minus 120591)120572minus1
times 120601119902 (120582int
120591
0
120596 (120583) 119891 (120583 119909 (120583)) 119889120583)119889120591
minus int
119904
0
(119904 minus 120591)120572minus1
times 120601119902(120582int
120591
0
120596 (120583) 119891 (120583 119909 (120583)) 119889120583)119889120591)119889119904
+1
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902 (120582int
119904
0
120596 (120591) 119891 (120591 119909 (120591)) 119889120591) 119889119904
le120578
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
119892 (119904)
times int
1
0
(1 minus 120591)120572minus1
times 120601119902(120582int
120591
0
120596 (120583) 119891 (120583 119909 (120583)) 119889120583)119889120591 119889119904
+1
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902 (120582int
119904
0
120596 (120591) 119891 (120591 119909 (120591)) 119889120591) 119889119904
=1
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
(1 minus 119904)120572minus1
times 120601119902 (120582int
119904
0
120596 (120591) 119891 (120591 119909 (120591)) 119889120591) 119889119904
le1
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
(1 minus 119904)120572minus1
12 Abstract and Applied Analysis
times 120601119902 (120582int
1
0
120596 (120591) Υ120601119901 (119909 (120591)) 119889120591) 119889119904
=120601119902 (120582Υ) 1205742
Γ (120572 + 1) (1 minus 120578 int1
0119892 (119904) 119889119904)
119909
(69)
This implies that 1 le 120601119902(120582Υ)1205742Γ(120572 + 1)(1 minus 120578 int1
0119892(119904)119889119904) =
(120582Υ)119902minus1
1205742Γ(120572+1)(1minus120578 int1
0119892(119904)119889119904) lt (120582 Υ)
119902minus11205742Γ(120572+1)(1minus
120578 int1
0119892(119904)119889119904) = 1 which is a contradiction This finishes the
proof of (ii)
Remark 12 In (i) the condition lim119909rarrinfin
119891(119905 119909) = 119891infin isin
(0 +infin] uniformly for 119905 isin [0 1] can be replaced withcondition (H6) The same methods can be used to prove theresults
Remark 13 From the above proof the condition 119891infin isin
(0 +infin] is important to keep the problem (1) having at leasttwo positive solutions if this condition is not satisfied thenthere exists 120582 gt 0 small enough such that the problem (1) hasno positive solutions for all 120582 lt 120582
Theorem 14 Assume that (H1)ndash(H6) hold(i) If 1198910
120601= +infin and 119891
infin
120601= +infin then there exists 120582lowast0 gt 0
such that problem (1) has at least two solutions for 120582 isin
(0 120582lowast0)(ii) If 1198910
120601= +infin and 119891
infin
120601= +infin then there exists 120582 gt 0
such that problem (1) has no solution for all 120582 gt 120582
Proof (i) It follows from 1198910
120601= +infin that there exists 1199031 gt 0
such that
119891 (119905 119909) ge 119871lowast1120601119901 (119909) (forall119905 isin [0 1] 0 le 119909 le 1199031) (70)
where 119871lowast1 satisfies 1205902120601119902(120582119871lowast1)1205741Γ(120572) ge 1
Then for 119909 isin 119875cap120597Ω1199031
we have 120590119909 le 119909 le 1199031 for 119905 isin [0 1]
and
(Ψ119909) (119905)
=1
Γ (120572)
times int
1
0
(1 minus 119904 )120572minus1
times 120601119902 (120582int
119904
0
120596 (120591) 119891 (120591 119909 (120591)) 119889120591) 119889119904
minus int
119905
0
(119905 minus 119904)120572minus1
times 120601119902 (120582int
119904
0
120596 (120591) 119891 (120591 119909 (120591)) 119889120591) 119889119904
+120578
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
119892 (119904)
times (int
1
0
(1 minus 120591)120572minus1
times 120601119902 (120582int
120591
0
120596 (120583) 119891 (120583 119909 (120583)) 119889120583)119889120591
minus int
119904
0
(119904 minus 120591)120572minus1
times 120601119902 (120582int
120591
0
120596 (120583) 119891 (120583 119909 (120583)) 119889120583)119889120591)119889119904
ge120590120578
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
119892 (119904)
times (int
1
0
(1 minus 120591)120572minus1
times 120601119902 (120582int
120591
0
120596 (120583) 119891 (120583 119909 (120583)) 119889120583)119889120591
minus int
119904
0
(119904 minus 120591)120572minus1
times120601119902 (120582int
120591
0
120596 (120583) 119891 (120583 119909 (120583)) 119889120583)119889120591)119889119904
+120590
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902 (120582int
119904
0
120596 (120591) 119891 (120591 119909 (120591)) 119889120591) 119889119904
ge120590
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902 (120582int
119904
0
120596 (120591) 119891 (120591 119909 (120591)) 119889120591) 119889119904
ge120590
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902 (120582int
119904
0
120596 (120591) 119871lowast1120601119901 (119909 (120591)) 119889120591) 119889119904
ge1205902120601119902 (120582119871lowast1) 119909
Γ (120572)
times int
1
0
(1 minus 119904)120572minus1
times 120601119902 (int
119904
0
120596 (120591) 119889120591) 119889119904
=1205902120601119902 (120582119871lowast1) 1205742
Γ (120572)119909
(71)
Abstract and Applied Analysis 13
which implies that
Ψ119909 ge 119909 forall119909 isin 119875 cap 120597Ω1199031
(72)
Next from 119891infin
120601= +infin there exists 1199032 gt 0 such that
119891 (119905 119909) ge 119871lowast2120601119901 (119909) forall119909 ge 1199032 (73)
where 119871lowast2 satisfies
1205902120601119902 (120582119871lowast2) 1205742
Γ (120572)ge 1 (74)
Let 1199032 gt max1199031 1199032120590 then for 119909 isin 119875 cap 120597Ω1199032
we have
119909 (119905) ge 120590 119909 = 1205901199032 gt 1199032 (75)
Similar to the above proof we can get
Ψ119909 ge 119909 forall119909 isin 119875 cap 120597Ω1199032
(76)
Let 1199031 lt 1199033 lt 11990321198721199033
= max119891(119905 119909) | 119905 isin [0 1] 119909 le 1199033+1 gt
0 and 120582lowast0 = (11198721199033
)(Γ(120572+1)1199033(1minus120578 int1
0119892(119904)119889119904)1205742)
119901minus1Thenfor 119909 isin 119875 cap 120597Ω119903
3
we have
119891 (119905 119909) le 1198721199033
forall119905 isin [0 1] (77)
Hence
(Ψ119909) (119905)
=1
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902 (120582int
119904
0
120596 (120591) 119891 (120591 119909 (120591)) 119889120591) 119889119904
minus int
119905
0
(119905 minus 119904)120572minus1
times 120601119902 (120582int
119904
0
120596 (120591) 119891 (120591 119909 (120591)) 119889120591) 119889119904
+120578
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
119892 (119904)
times (int
1
0
(1 minus 120591)120572minus1
times 120601119902 (120582int
120591
0
120596 (120583) 119891 (120583 119909 (120583)) 119889120583)119889120591
minus int
119904
0
(119904 minus 120591)120572minus1
times120601119902 (120582int
120591
0
120596 (120583) 119891 (120583 119909 (120583)) 119889120583)119889120591)119889119904
le120578
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
119892 (119904)
times int
1
0
(1 minus 120591)120572minus1
120601119902
times (120582int
120591
0
120596 (120583) 119891 (120583 119909 (120583)) 119889120583)119889120591 119889119904
+1
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902 (120582int
119904
0
120596 (120591) 119891 (120591 119909 (120591)) 119889120591) 119889119904
le120578
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
119892 (119904)
times int
1
0
(1 minus 120591)120572minus1
120601119902
times (120582int
120591
0
120596 (120583)1198721199033
119889120583)119889120591119889119904
+1
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902 (120582int
119904
0
120596 (120591)1198721199033
119889120591)119889119904
le120578120601119902 (120582119872119903
3
) int1
0119892 (119904) 119889119904
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
(1 minus 120591)120572minus1
times 120601119902 (int
1
0
120596 (120583) 119889120583)119889120591
+120601119902 (120582119872119903
3
)
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902 (int
1
0
120596 (120591) 119889120591) 119889119904
=1205742120601119902 (120582119872119903
3
)
Γ (120572 + 1) (1 minus 120578 int1
0119892 (119904) 119889119904)
lt1205742120601119902 (120582lowast0119872119903
3
)
Γ (120572 + 1) (1 minus 120578 int1
0119892 (119904) 119889119904)
= 1199033
(78)
14 Abstract and Applied Analysis
This implies that
Ψ119909 lt 119909 forall119909 isin 119875 cap 120597Ω1199033
(79)
Applying Lemma 7 to (72) (76) and (79) yields that Ψ hastwo fixed points 1199091 1199092 such that 1199091 isin 119875 cap (Ω119903
3
Ω1199031
) and1199092 isin 119875 cap (Ω119903
2
Ω1199033
)(ii) By the proof of (i) and the continuity of 119891(119905 119909) there
exists 119909 isin [1199031 1199032] such that min119909isin[11990311199032](119891(119905 119909)120601119901(119909)) =
119891(119905 119909)120601119901(119909)Let Υ = min119871lowast1 119871lowast2 119891(119905 119909)120601119901(119909) since (H6) holds
then Υ gt 0 and
119891 (119905 119909) ge Υ120601119901 (119909) forall119909 ge 0 119905 isin [0 1] (80)
Assuming 119909(119905) is a positive solution of problem (1) we willshow that this leads to contradiction for all 120582 gt 120582 where 120582 =
(1Υ)(Γ(120572)1205741120590)119901minus1 It follows form (26) that
119909 = 119909 (0)
=120578
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
119892 (119904)
times (int
1
0
(1 minus 120591)120572minus1
times 120601119902 (120582int
120591
0
120596 (120583) 119891 (120583 119909 (120583)) 119889120583)119889120591
minus int
119904
0
(119904 minus 120591)120572minus1
times120601119902 (120582int
120591
0
120596 (120583) 119891 (120583 119909 (120583)) 119889120583)119889120591)119889119904
+1
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902 (120582int
119904
0
120596 (120591) 119891 (120591 119909 (120591)) 119889120591) 119889119904
ge1
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902 (120582int
119904
0
120596 (120591) 119891 (120591 119909 (120591)) 119889120591) 119889119904
ge1
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902 (120582int
119904
0
120596 (120591) Υ120601119901 (119909 (120591)) 119889120591) 119889119904
ge1
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902 (120582int
119904
0
120596 (120591) Υ120601119901 (120590 119909 119889120591)) 119889119904
ge120601119902 (120582Υ) 120590 119909
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902 (int
119904
0
120596 (120591) 119889120591) 119889119904
=120601119902 (120582Υ) 1205741120590
Γ (120572)119909
(81)
This implies that 1 ge 120601119902(120582Υ)1205741120590Γ(120572) = (120582Υ)119902minus1
1205741120590Γ(120572) gt
(120582Υ)119902minus1
1205741120590Γ(120572) = 1 which is a contradiction This com-pletes the proof of (ii)
Corollary 15 Assume that (H1)ndash(H5) holds
(i) If lim119909rarrinfin
119891(119905 119909) = 119891infin isin (0 +infin] uniformly for 119905 isin
[0 1] and one of 1198910120601
= 0 and 119891infin
120601= 0 is satisfied then
there exists 120582lowast gt 0 such that the problem (1) has at leastone positive solution for any 120582 gt 120582
lowast(ii) If one of 1198910
120601= +infin and 119891
infin
120601= +infin holds then there
exists 120582lowast0 gt 0 such that problem (1) has at least onepositive solutions for any 120582 isin (0 120582lowast0)
Proof (i) The conclusion is a direct consequence of Theo-rem 11(i)
(ii) The conclusion is a direct consequence of Theo-rem 14(i)
Theorem 16 Assume that (H1)ndash(H5) holdThen the followingconclusions hold
(i) If 1198910120601
= 0 and 119891infin
120601= infin then the problem (1) has a
positive solution 119909120582(119905) for all 120582 gt 0(ii) If 1198910
120601= infin and 119891
infin
120601= 0 then the problem (1) has a
positive solution 119909120582(119905) for all 120582 gt 0
Proof Weonly prove (i) the proof of (ii) is similar sowe omitit here
Let 120582 gt 0 since 1198910
120601= 0 there exists 119903 gt 0 119897 gt 0 such that
119891 (119905 119909) le 119897120601119901 (119909) (forall119905 isin [0 1] 119909 isin [0 119903]) (82)
where 119897 satisfied
1205742120582119902minus1
119897119902minus1
Γ (120572 + 1) (1 minus 120578 int1
0119892 (119904) 119889119904)
le 1 (83)
Similar to the proof of Theorem 14 we obtain that
Ψ119909 le 119909 forall119909 isin 119875 cap 120597Ω119903 (84)
Abstract and Applied Analysis 15
From 119891infin
120601= infin there exists 119877 gt 119903 gt 0 119871 gt 0 such that
119891 (119905 119909) ge 119871120601119901 (119909) (forall119905 isin [0 1] 119909 ge 119877) (85)where 119871 satisfied
1205902120582119902minus1
119871119902minus1
1205741
Γ (120572)ge 1 (86)
We get thatΨ119909 ge 119909 forall119909 isin 119875 cap 120597Ω119877 (87)
Applying Lemma 7 to (84) (87) yields thatΨ has a fixed point119909 such that 119909 isin 119875 cap (Ω119877 Ω119903)
Theorem 17 Assume that (H1)ndash(H6) holdThen the followingconclusions hold
If 0 lt 1198910
120601lt infin and 0 lt 119891
infin
120601lt infin then there exists 1 2 gt
0 such that problem (1) has no positive solution for any 120582 gt 1
and 0 lt 120582 lt 2
Proof It follows from 0 lt 1198910
120601lt infin and 0 lt 119891
infin
120601lt infin that
there exist 1 gt 0 2 gt 0 1 gt 0 2 gt 0 and 1199032 gt 1199031 gt 0
such that1120601119901 (119909) le 119891 (119905 119909) le 1120601119901 (119909)
(forall119905 isin [0 1] 0 le 119909 le 1199031)
2120601119901 (119909) le 119891 (119905 119909) le 2120601119901 (119909)
(forall119905 isin [0 1] 119909 ge 1199032)
(88)
Let = max119891(119905 119909)120601119901(119909) 1199031 le 119909 le 1199032 0 le 119905 le 1 =
min119891(119905 119909)120601119901(119909) 1199031 le 119909 le 1199032 0 le 119905 le 1 By the condition(H6) gt 0 Then forall119909 ge 0 we have
120601119901 (119909) le 119891 (119905 119909) le 120601119901 (119909) (89)
where = max1 2 gt 0 and = min1198971 2 gt 0Let 1 = (1)(Γ(120572)1205741120590)
119901minus1 and 2 =
(1)(Γ(120572 + 1)(1 minus 120578 int1
0119892(119904)119889119904)1205742)
119901minus1
If problem (1)has positive solution 119909(119905) in 119875 then
119909 = 119909 (0)
=120578
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
119892 (119904)
times (int
1
0
(1 minus 120591)120572minus1
times 120601119902 (120582int
120591
0
120596 (120583) 119891 (120583 119909 (120583)) 119889120583)119889120591
minus int
119904
0
(119904 minus 120591)120572minus1
times120601119902 (120582int
120591
0
120596 (120583) 119891 (120583 119909 (120583)) 119889120583)119889120591)119889119904
+1
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902 (120582int
119904
0
120596 (120591) 119891 (120591 119909 (120591)) 119889120591) 119889119904
ge1
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902 (120582int
119904
0
120596 (120591) 119891 (120591 119909 (120591)) 119889120591) 119889119904
ge1
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902 (120582int
119904
0
120596 (120591) 120601119901 (119909 (120591)) 119889120591) 119889119904
ge1
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902 (120582int
119904
0
120596 (120591) 120601119901 (120590 119909 119889120591)) 119889119904
ge120601119902 (120582) 120590 119909
Γ (120572)
times int
1
0
(1 minus 119904)120572minus1
120601119902 (int
119904
0
120596 (120591) 119889120591) 119889119904
=120601119902 (120582) 1205741120590
Γ (120572)119909
gt120601119902 (1 ) 1205741120590
Γ (120572)119909 = 119909
119909 = 119909 (0)
=120578
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
119892 (119904)
times (int
1
0
(1 minus 120591)120572minus1
120601119902
times (120582int
120591
0
120596 (120583) 119891 (120583 119909 (120583)) 119889120583)119889120591
minus int
119904
0
(119904 minus 120591)120572minus1
120601119902
times(120582int
120591
0
120596 (120583) 119891 (120583 119909 (120583)) 119889120583)119889120591)119889119904
+1
Γ (120572)int
1
0
(1 minus 119904)120572minus1
120601119902
times (120582int
119904
0
120596 (120591) 119891 (120591 119909 (120591)) 119889120591) 119889119904
16 Abstract and Applied Analysis
le120578
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
119892 (119904)
times int
1
0
(1 minus 120591)120572minus1
times 120601119902 (120582int
120591
0
120596 (120583) 119891 (120583 119909 (120583)) 119889120583)119889120591 119889119904
+1
Γ (120572)int
1
0
(1 minus 119904)120572minus1
120601119902
times (120582int
119904
0
120596 (120591) 119891 (120591 119909 (120591)) 119889120591) 119889119904
=1
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
(1 minus 119904)120572minus1
120601119902
times (120582int
119904
0
120596 (120591) 119891 (120591 119909 (120591)) 119889120591) 119889119904
le1
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
(1 minus 119904)120572minus1
120601119902
times (120582int
1
0
120596 (120591) 120601119901 (119909 (120591)) 119889120591) 119889119904
=120601119902 (120582) 1205742
Γ (120572 + 1) (1 minus 120578 int1
0119892 (119904) 119889119904)
119909
lt120601119902 (2) 1205742
Γ (120572 + 1) (1 minus 120578 int1
0119892 (119904) 119889119904)
119909 = 119909
(90)
which is a contradiction This finishes the proof of Theo-rem 17
4 Examples
In this section we give out same examples to illustrate ourmain results can be used in practice
Example 1 Consider the following fractional differentialequation
(120601119901 (11986352
0+119909 (119905)))
1015840
+ 120582119905(1 + 1199052)12
119909119901minus1 arctan119909 = 0
0 lt 119905 lt 1
1199091015840(0) = 119909
10158401015840(0) = 0
119909 (1) =1
2int
1
0
1
2119909 (119904) 119889119904
(91)
where 120572 = 52 119901 = 32 119902 = 3 120596(119905) = 119905 119891(119905 119909) = (1 +
1199052)12
119909119901minus1 arctan119909 120578 = 12 and 119892(119905) = 12
By simple computation 119897lowast1= 1205872 119897
lowast
2= radic21205872 1205741 = int
1
0(1minus
119904)32
1206013(int119904
0120591 119889120591)119889119904 asymp 00043 1205742 = 1206013(int
1
0120591 119889120591) = 025 120590 asymp
01429 120582lowast
0= (1119897
lowast
1)(Γ(120572)1205901205741)
119901minus1asymp 25149 120582
lowast
0= (1119897
lowast
2)(Γ(120572+
1)(1 minus 120578 int1
0119892(119904)119889119904)1205742)
119901minus1asymp 14222 Then it is easy to see
all the conditions of Theorem 9(ii) are satisfied thus forall120582 isin
[14222 25149]] the problem (91) has at least one positivesolution
Example 2 Consider the boundary value problem
(120601119901 (11986352
0+119909 (119905)))
1015840
+ 120582119905(1 + 1199052)12
119909119901119890minus119909
= 0
0 lt 119905 lt 1
1199091015840(0) = 119909
10158401015840(0) = 0
119909 (1) =1
2int
1
0
1
2119909 (119904) 119889119904
(92)
where 120572 = 52 119901 = 32 119902 = 3 120596(119905) = 119905 119891(119905 119909) = (1 +
1199052)12
119909119901119890minus119909
120578 = 12 and 119892(119905) = 12
1198910
120601= lim119909rarr0
119891 (119905 119909)
120601119901 (119909)= lim119909rarr0
(1 + 1199052)12
119909119901119890minus119909
119909119901minus1= 0
forall119905 isin [0 1]
1198910
120601= lim119909rarrinfin
119891 (119905 119909)
120601119901 (119909)= lim119909rarrinfin
(1 + 1199052)12
119909119901119890minus119909
119909119901minus1= 0
forall119905 isin [0 1]
(93)
Let 1205881= 12 120588
2= 4 then
119891 (119905 119909)
120601119901 (119909)le (1 + 119905
2)12 1
2radic119890le
1
radic2119890
forall119905 isin [0 1] 119909 isin [01
2]
119891 (119905 119909)
120601119901 (119909)le (1 + 119905
2)12 4
1198904le
4radic2
1198904
forall119905 isin [0 1] 119909 isin [4infin)
(94)
so 1205761 = 1radic2119890 1205762 = 4radic21198904 and forall119909 isin
[12 4] max12le119909le4(119891(119905 119909)120601119901(119909)) = radic2119890minus1 We
have Υ = max1radic2119890 4radic21198904 radic2119890minus1
= radic2119890minus1 1205742 = 025
Abstract and Applied Analysis 17
and 120582 = (1Υ)(Γ(120572+1)(1minus120578 int1
0119892(119904)119889119904)1205742)
119901minus1= 3radic5120587
141198904
By Theorem 11(ii) for any 120582 lt 120582 = 3radic512058714
1198904 the problemhas no positive solution
5 Conclusions
Recently differential equations with p-Laplacian operatorwere widely discussed by several authors In this paperby using fixed point techniques combining with partiallyordered structure of Banach space we obtained the existencemultiply and the dependence of parameter These newresults we presented can be used in numerical computationand analyze mathematical models of physical phenomenamechanics nonlinear dynamics and many other relatedfields
Conflict of Interests
The authors declare that there is no conflict of interestsregarding the publication of this paper
Acknowledgments
The authors are very grateful to anonymous referees for theirvaluable and detailed suggestions and comments to improvethe original paper This work is supported by the Programfor New Century Excellent Talents in University (NCET-10-0097)
References
[1] I Podlubny Fractional Differential Equations vol 198 Aca-demic Press San Diego Calif USA 1999
[2] A A Kilbas H M Srivastava and J J Trujillo Theoryand Applications of Fractional Differential Equations vol 204of North-Holland Mathematics Studies Elsevier Science BVAmsterdam The Netherlands 2006
[3] Y A Rossikhin and M V Shitikova ldquoApplication of fractionalderivatives to the analysis of damped vibrations of viscoelasticsingle mass systemsrdquoActa Mechanica vol 120 no 1ndash4 pp 109ndash125 1997
[4] J Sabatier O Agrawal and J T Machado Advances in Frac-tional Calculus Theoretical developments and applications inphysics and engineering Springer Berlin Germany 2007
[5] N Engheta ldquoOn fractional calculus and fractional multipolesin electromagnetismrdquo IEEE Transactions on Antennas andPropagation vol 44 no 4 pp 554ndash566 1996
[6] R Hilfer Applications of Fractional Calculus in Physics WorldScientific Singapore 2000
[7] R L Bagley and R A Calico ldquoFractional order state equationsfor the control of viscoelastically damped structuresrdquo Journalof Guidance Control and Dynamics vol 14 no 2 pp 304ndash3111991
[8] W-X Zhou andY-D Chu ldquoExistence of solutions for fractionaldifferential equations with multi-point boundary conditionsrdquoCommunications in Nonlinear Science and Numerical Simula-tion vol 17 no 3 pp 1142ndash1148 2012
[9] B Ahmad and G Wang ldquoA study of an impulsive four-pointnonlocal boundary value problemof nonlinear fractional differ-ential equationsrdquo Computers amp Mathematics with Applicationsvol 62 no 3 pp 1341ndash1349 2011
[10] B Ahmad and S Sivasundaram ldquoOn four-point nonlocalboundary value problems of nonlinear integro-differentialequations of fractional orderrdquo Applied Mathematics and Com-putation vol 217 no 2 pp 480ndash487 2010
[11] N Nyamoradi ldquoPositive solutions for multi-point boundaryvalue problems for nonlinear fractional differential equationsrdquoJournal of Contemporary Mathematical Analysis vol 48 no 4pp 145ndash157 2013
[12] B Ahmad and J J Nieto ldquoExistence of solutions for nonlocalboundary value problems of higher-order nonlinear fractionaldifferential equationsrdquo Abstract and Applied Analysis vol 2009Article ID 494720 9 pages 2009
[13] X Liu M Jia and X Xiang ldquoOn the solvability of a fractionaldifferential equationmodel involving the119901-Laplacian operatorrdquoComputers ampMathematics with Applications vol 64 no 10 pp3267ndash3275 2012
[14] J Wang H Xiang and Z Liu ldquoExistence of concave positivesolutions for boundary value problem of nonlinear fractionaldifferential equation with 119901-Laplacian operatorrdquo InternationalJournal of Mathematics and Mathematical Sciences vol 2010Article ID 495138 17 pages 2010
[15] J Wang and H Xiang ldquoUpper and lower solutions method fora class of singular fractional boundary value problems with 119901-Laplacian operatorrdquo Abstract and Applied Analysis vol 2010Article ID 971824 12 pages 2010
[16] H Lu andZHan ldquoExistence of positive solutions for boundary-value problem of fractional differential equation with 119901-laplacian operatorrdquo The American Jouranl of Engineering andTechnology Research vol 11 pp 3757ndash3764 2011
[17] G Chai ldquoPositive solutions for boundary value problem offractional differential equation with 119901-Laplacian operatorrdquoBoundary Value Problems vol 2012 article 18 2012
[18] T Chen W Liu and Z Hu ldquoA boundary value problem forfractional differential equation with 119901-Laplacian operator atresonancerdquoNonlinear AnalysisTheoryMethodsampApplicationsvol 75 no 6 pp 3210ndash3217 2012
[19] T Chen andW Liu ldquoAn anti-periodic boundary value problemfor the fractional differential equation with a 119901-LaplacianoperatorrdquoAppliedMathematics Letters of Rapid Publication vol25 no 11 pp 1671ndash1675 2012
[20] S J Li X G Zhang Y H Wu and L Caccetta ldquoExtremalsolutions for 119901-Laplacian differential systems via iterative com-putationrdquo Applied Mathematics and Computation vol 26 pp1151ndash1158 2013
[21] Z Liu and L Lu ldquoA class of BVPs for nonlinear fractionaldifferential equations with 119901-Laplacian operatorrdquo ElectronicJournal of Qualitative Theory of Differential Equations vol 70pp 1ndash16 2012
[22] Z Liu L Lu and I Szanto ldquoExistence of solutions for fractionalimpulsive differential equations with 119901-Laplacian operatorrdquoActa Mathematica Hungarica vol 141 no 3 pp 203ndash219 2013
[23] Z Bai ldquoEigenvalue intervals for a class of fractional boundaryvalue problemrdquo Computers amp Mathematics with Applicationsvol 64 no 10 pp 3253ndash3257 2012
[24] X Zhang L Wang and Q Sun ldquoExistence of positive solutionsfor a class of nonlinear fractional differential equations withintegral boundary conditions and a parameterrdquo Applied Mathe-matics and Computation vol 226 pp 708ndash718 2014
18 Abstract and Applied Analysis
[25] D J Guo and V Lakshmikantham Nonlinear Problems inAbstract Cones vol 5 Academic Press New York NY USA1988
[26] M A Krasnoselrsquoskii Topological Methods in the Theory ofNonlinear Integral Equations Translated by A H ArmstrongElmsford Pergamon Turkey 1964
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Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Stochastic AnalysisInternational Journal of
2 Abstract and Applied Analysis
solutions for some 120582 where 1198910 = 0 119897infin 119891infin = 0 119897infin 0 lt
119897 lt infin But almost all the results which the author obtaineddepend on both 1198910 and 119891infin the case depends on one of 1198910and 119891infin the author only discussed 1198910 = 0infin and 119891infin = 0infinthe case 1198910 = 119897 or 119891infin = 119897 has not been discussed On theother hand there exist several results on the existence of onesolution to fractional p-Laplacian boundary value problems(BVPs) there are to the best of our knowledge relatively fewresults on the nonexistence and the uniqueness of positivesolutions to fractional p-Laplacian differential equation withparameter
Motivated by the above questions in this paper we willestablish several sufficient conditions for the existence ofpositive solutions of (1) by using fixed point theorem andfixed point index theory
The work is organized in the following fashion In Sec-tion 2 we provide some necessary background In particularwe will introduce some lemmas and definitions associatedwith fixed point index theory The main results will be statedand proved in Section 3 Two examples are given in Section 4
2 Preliminaries
In this section we introduce definitions and preliminary factswhich are used throughout this paper
Definition 1 (see [1]) The Riemann-Liouville fractional inte-gral of order 120572 gt 0 of a function 119909 (0 +infin) rarr R is givenby
119868120572
0+119909 (119905) =
1
Γ (120572)int
119905
0
(119905 minus 119904)120572minus1
119909 (119904) 119889119904 (3)
provided that the right side is point-wise defined on (0 +infin)
and Γ is the Gamma function
Definition 2 (see [1]) The Caputo fractional derivative oforder 120572 gt 0 of a continuous function 119909 (0 +infin) rarr R isgiven by
119863120572
0+119909 (119905) =
1
Γ (119899 minus 120572)int
119905
0
(119905 minus 119904)119899minus120572minus1
119909(119899)
(119904) 119889119904 (4)
where 119899 = [120572] + 1 provided that the right side is point-wisedefined on (0 +infin)
Lemma 3 (see [1]) Let 120572 gt 0 and assume that 119909 isin 119862[0 1]and then the fractional differential equation 119863
120572
0119909(119905) = 0 has
unique solutions
119906 (119905) = 1198880 + 1198881119905 + 11988821199052+ sdot sdot sdot + 119888119899minus1119905
119899minus1
119888119894 isin R 119894 = 0 1 2 119899 minus 1 119899 minus 1 lt 120572 le 119899
(5)
Lemma 4 (see [1]) Let 120572 gt 0 Then
119868120572
0+119863120572
0+119909 (119905)
= 119909 (119905) + 1198880 + 1198881119905 + 11988821199052+ sdot sdot sdot + 119888119899minus1119905
119899minus1
(6)
where 119888119894 isin R 119894 = 0 1 2 119899 minus 1 119899 minus 1 lt 120572 le 119899
Lemma 5 (see [25]) Let 119875 be a cone in a real Banach space119864 and letΩ be a bounded open set of 119864 Assume that operator119860 119875 cap Ω rarr 119875 is completely continuous If there exists a1199090 gt 0 such that
119909 minus 119860119909 = 1199051199090 forall119909 isin 119875 cap 120597Ω 119905 ge 0 (7)
then 119894(119860 119875 cap Ω 119875) = 0
Lemma 6 (see [26]) Let 119864 be a Banach space and 119875 sube 119864 acone and Ω1 and Ω2 are open set with 0 isin Ω1 Ω1 sube Ω2 andlet 119879 119875 cap (Ω2 Ω1) rarr 119875 be completely continuous operatorsuch that either
(i) 119879119906 le 119906 119906 isin 119875cap120597Ω1 and 119879119906 ge 119906 119906 isin 119875cap120597Ω2
or
(ii) 119879119906 ge 119906 119906 isin 119875cap120597Ω1 and 119879119906 le 119906 119906 isin 119875cap120597Ω2
holds Then 119879 has a fixed point in 119875 cap (Ω2 Ω1)
3 Main Results
In this section we present some new results on the existencemultiplicity nonexistence and the uniqueness of positivesolution of problem (1) and dependence of the positivesolution 119909120582(119905) on the parameter 120582
Let 119864 = 119862[0 1] then 119864 is a real Banach space with thenorm sdot defined by 119909 = max0le119905le1|119909(119905)|
Lemma 7 Let 119910(119905) isin 119862(0 1) cap 1198711(0 1) and 119910(119905) ge 0 then the
solution of the problem
(120601119901 (119863120572
0+119909 (119905)))
1015840
+ 120582119910 (119905) = 0 0 lt 119905 lt 1
1199091015840(0) = 119909
10158401015840(0) = 0
119909 (1) = 120578int
1
0
119892 (119904) 119909 (119904) 119889119904
(8)
is given by
119909 (119905) =1
Γ (120572)int
1
0
(1 minus 119904)120572minus1
120601119902 (120582int
119904
0
119910 (120591) 119889120591) 119889119904
minusint
119905
0
(119905 minus 119904)120572minus1
120601119902 (120582int
119904
0
119910 (120591) 119889120591) 119889119904
+120578
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
119892 (119904)
times (int
1
0
(1 minus 120591)120572minus1
120601119902 (120582int
120591
0
119910 (120583) 119889120583)119889120591
minus int
119904
0
(119904 minus 120591)120572minus1
times 120601119902 (120582int
120591
0
119910 (120583) 119889120583)119889120591)119889119904
(9)
Abstract and Applied Analysis 3
119909 (119905) ge 0 min0le119905le1
119909 (119905) ge 120590 119909 (10)
where 2 lt 120572 le 3 0 lt 120578 lt 1 1 minus 120578 int1
0119892(119904)119889119904 gt 0 and 120590 =
120578 int1
0(1 minus 119904)119892(119904)119889119904(1 minus 120578 int
1
0119904119892(119904)119889119904)
Proof It is easy to see by integration of (8) that
120601119901 (119863120572
0+119909 (119905)) minus 120601119901 (119863
120572
0+119909 (0)) = minus120582int
119905
0
119910 (119904) 119889119904 (11)
By the boundary condition 1199091015840(0) = 119909
10158401015840(0) = 0 we can easily
get that1198631205720+
119909(0) = 0 and we obtain that
120601119901 (119863120572
0+119909 (119905)) = minus120582int
119905
0
119910 (119904) 119889119904 (12)
that is
119863120572
0+119909 (119905) = minus120601119902 (120582int
119905
0
119910 (119904) 119889119904) (13)
By Lemmas 3 and 4 we get that
119909 (119905) = minus119868120572120601119902 (120582int
119905
0
119910 (119904) 119889119904) + 1198880 + 1198881119905 + 11988821199052 (14)
Using the boundary condition 1199091015840(0) = 119909
10158401015840(0) = 0 119909(1) =
120578 int1
0119892(119904)119909(119904)119889119904 we get 1198881 = 1198882 = 0 and
1198880 =1
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
(1 minus 119904)120572minus1
120601119902 (120582int
119904
0
119910 (120591) 119889120591) 119889119904
minus120578
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
119892 (119905) (int
119905
0
(119905 minus 119904)120572minus1
120601119902 (120582int
119904
0
119910 (120591) 119889120591) 119889119904) 119889119905
(15)
Thus
119909 (119905) = minus1
Γ (120572)int
119905
0
(119905 minus 119904)120572minus1
120601119902 (120582int
119904
0
119910 (120591) 119889120591) 119889119904
+1
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
(1 minus 119904)120572minus1
120601119902 (120582int
119904
0
119910 (120591) 119889120591) 119889119904
minus120578
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
119892 (119905) (int
119905
0
(119905 minus 119904)120572minus1
120601119902 (120582int
119904
0
119910 (120591) 119889120591) 119889119904) 119889119905
=1
Γ (120572)int
1
0
(1 minus 119904)120572minus1
120601119902 (120582int
119904
0
119910 (120591) 119889120591) 119889119904
minusint
119905
0
(119905 minus 119904)120572minus1
120601119902 (120582int
119904
0
119910 (120591) 119889120591) 119889119904
+120578
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
119892 (119904) (int
1
0
(1 minus 120591)120572minus1
120601119902 (120582int
120591
0
119910 (120583) 119889120583)119889120591
minus int
119904
0
(119904 minus 120591)120572minus1
times 120601119902 (120582int
120591
0
119910 (120583) 119889120583)119889120591)119889119904
(16)
Direct differentiation of (9) implies
1199091015840(119905)
= minus1
Γ (120572 minus 1)int
119905
0
(119905 minus 119904)120572minus2
120601119902 (120582int
119904
0
119910 (120591) 119889120591) 119889119904 le 0
(17)
By differentiation of (17) we get
11990910158401015840(119905)
= minus1
Γ (120572 minus 2)int
119905
0
(119905 minus 119904)120572minus3
120601119902 (120582int
119904
0
119910 (120591) 119889120591) 119889119904 le 0
(18)
Thus the solution of problem (8) is nonincreasing andconcave on [0 1] and
119909 = 119909 (0) min0le119905le1
119909 (119905) = 119909 (1) (19)
On the other hand as we assume that 119910(119905) ge 0 we see that
119909 (1) =120578
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
119892 (119904) (int
1
0
(1 minus 120591)120572minus1
120601119902 (120582int
120591
0
119910 (120583) 119889120583)119889120591
minus int
119904
0
(119904 minus 120591)120572minus1
times 120601119902 (120582int
120591
0
119910 (120583) 119889120583)119889120591)119889119904
=120578
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
119892 (119904) (int
119904
0
[(1 minus 120591)120572minus1
minus (119904 minus 120591)120572minus1
]
4 Abstract and Applied Analysis
times 120601119902 (120582int
120591
0
119910 (120583) 119889120583)119889120591
+int
1
119904
(1 minus 120591)120572minus1
120601119902 (120582int
120591
0
119910 (120583) 119889120583)119889120591)119889119904
ge 0
(20)
Therefore 119909(119905) ge 0 and is concave for 119905 isin [0 1] So for every119905 isin [0 1]
119909 (119905) ge 119905119909 (1) + (1 minus 119905) 119909 (0) (21)
Therefore
120578int
1
0
119909 (119905) 119892 (119905) 119889119905 ge 120578119909 (1) int
1
0
119905119892 (119905) 119889119905
+ 120578119909 (0) int
1
0
(1 minus 119905) 119892 (119905) 119889119905
(22)
Since 119909(1) = 120578 int1
0119909(119905)119892(119905)119889119905 1 minus 120578 int
1
0119892(119905)119889119905 gt 0 we have
119909 (1) ge120578 int1
0(1 minus 119905) 119892 (119905) 119889119905
1 minus 120578 int1
0119905119892 (119905) 119889119905
119909 (0) (23)
that is
min0le119905le1
119909 (119905) = 119909 (1) ge120578 int1
0(1 minus 119905) 119892 (119905) 119889119905
1 minus 120578 int1
0119905119892 (119905) 119889119905
119909 (0) = 120590 119909 (24)
The Lemma is proved
We construct a cone 119875 in 119864 by
119875 = 119909 isin 119864 119909 ge 0 min0le119905le1
119909 (119905) ge 120590 119909 (25)
where 120590 is defined in Lemma 7 It is easy to see 119875 is a closedconvex cone of 119864
Define 119879 119875 rarr 119864 by
(119879119909) (119905)
=1
Γ (120572)int
1
0
(1 minus 119904)120572minus1
120601119902 (int
119904
0
120596 (120591) 119891 (120591 119909 (120591)) 119889120591) 119889119904
minusint
119905
0
(119905 minus 119904)120572minus1
120601119902 (int
119904
0
120596 (120591) 119891 (120591 119909 (120591)) 119889120591) 119889119904
+120578
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
119892 (119904)
times (int
1
0
(1 minus 120591)120572minus1
120601119902 (int
120591
0
120596 (120583) 119891 (120583 119909 (120583)) 119889120583)119889120591
minus int
119904
0
(119904 minus 120591)120572minus1
times 120601119902 (int
120591
0
120596 (120583) 119891 (120583 119909 (120583)) 119889120583)119889120591)119889119904
(26)
It is clear that the fixed points of the operator 120601119902(120582)119879 arethe solutions of the boundary value problems (1)
We make the following hypotheses
(H1) 119891 [0 1] times [0infin) rarr [0infin) is continuous(H2) 120596 (0 1) rarr [0infin) is continuous and not identical
zero on any closed subinterval of (0 1) with 0 lt
int1
0120596(119905)119889119905 lt +infin
(H3) 1 minus 120578 int1
0119892(119904)119889119904 gt 0
(H4) 1198910
120601isin [0infin] where 119891
0
120601= lim119909rarr0(119891(119905 119909)120601119901(119909))
uniformly for 119905 isin [0 1](H5) 119891
infin
120601isin [0infin] where 119891
infin
120601= lim119909rarrinfin(119891(119905 119909)120601119901(119909))
uniformly for 119905 isin [0 1](H6) 119891(119905 119909) gt 0 for any 119905 isin [0 1] and 119909 gt 0
Lemma 8 Assume that (H1)ndash(H3) hold Then 119879 119875 rarr 119864 iscompletely continuous
Proof First we show that 119879 is continuous It is easy to see119879(119875) sub 119875 For 119909119899(119905) isin 119864 and 119909119899(119905) rarr 119909(119905) as 119899 rarr infin bythe continuous of 119891(119905 119909(119905)) we get 119891(119905 119909119899(119905)) rarr 119891(119905 119909(119905))as 119899 rarr infin This implies that
120601119902 (int
119905
0
119908 (119904) 119891 (119904 119909119899 (119904)) 119889119904)
997888rarr 120601119902 (int
119905
0
119908 (119904) 119891 (119904 119909 (119904)) 119889119904)
(27)
So we have
sup119905isin[01]
10038161003816100381610038161003816100381610038161003816120601119902 (int
119905
0
119908 (119904) 119891 (119904 119909119899 (119904)) 119889119904)
minus 120601119902 (int
119905
0
119908 (119904) 119891 (119904 119909 (119904)) 119889119904)
10038161003816100381610038161003816100381610038161003816997888rarr 0 119899 997888rarr infin
(28)
Denote Π = |(119879119909119899)(119905) minus (119879119909)(119905)| then
Π le1
Γ (120572)int
1
0
(1 minus 119904)120572minus1
10038161003816100381610038161003816100381610038161003816120601119902 (int
119904
0
120596 (120591) 119891 (120591 119909119899 (120591)) 119889120591)
minus 120601119902 (int
119904
0
120596 (120591) 119891 (120591 119909 (120591)) 119889120591)
10038161003816100381610038161003816100381610038161003816119889119904
Abstract and Applied Analysis 5
+int
119905
0
(119905 minus 119904)120572minus1
10038161003816100381610038161003816100381610038161003816120601119902 (int
119904
0
120596 (120591) 119891 (120591 119909119899 (120591)) 119889120591)minus120601119902
times (int
119904
0
120596 (120591) 119891 (120591 119909 (120591)) 119889120591)
10038161003816100381610038161003816100381610038161003816119889119904
+120578
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
119892 (119904)
times (int
1
0
(1 minus 120591)120572minus1
times
10038161003816100381610038161003816100381610038161003816120601119902 (int
120591
0
120596 (120583) 119891 (120583 119909119899 (120583)) 119889120583)
minus 120601119902 (int
120591
0
120596 (120583) 119891 (120583 119909 (120583)) 119889120583)
10038161003816100381610038161003816100381610038161003816119889120591
+ int
119904
0
(119904 minus 120591)120572minus1
times
10038161003816100381610038161003816100381610038161003816120601119902 (int
120591
0
120596 (120583) 119891 (120583 119909119899 (120583)) 119889120583) minus 120601119902
times (int
120591
0
120596 (120583) 119891 (120583 119909 (120583)) 119889120583)
10038161003816100381610038161003816100381610038161003816119889120591) 119889119904
le sup119904isin[01]
10038161003816100381610038161003816100381610038161003816120601119902 (int
119904
0
119908 (120591) 119891 (120591 119909119899 (120591)) 119889120591)
minus 120601119902 (int
119904
0
119908 (120591) 119891 (120591 119909 (120591)) 119889120591)
10038161003816100381610038161003816100381610038161003816
times2
Γ (120572 + 1)(1 +
120578 int1
0119892 (119904) 119889119904
1 minus 120578 int1
0119892 (119904) 119889119904
) 997888rarr 0
119899 997888rarr infin
(29)
Hence 119879 is continuousSecond we show that 119879 is compactFor 119909(119905) isin 119861119903 = 119909 isin 119875 119909 le 119903 by condition (H1)
119891(119905 119909(119905)) lt infin Denote119872 = max0le119905le1119909le119903119891(119905 119909(119905))Then
|119879119909 (119905)|
le1
Γ (120572)int
1
0
(1 minus 119904)120572minus1
10038161003816100381610038161003816100381610038161003816120601119902 (int
119904
0
120596 (120591) 119891 (120591 119909 (120591)) 119889120591)
10038161003816100381610038161003816100381610038161003816119889119904
+int
119905
0
(119905 minus 119904)120572minus1
10038161003816100381610038161003816100381610038161003816120601119902 (int
119904
0
120596 (120591) 119891 (120591 119909 (120591)) 119889120591)
10038161003816100381610038161003816100381610038161003816119889119904
+120578
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
119892 (119904)
times (int
1
0
(1 minus 120591)120572minus1
times
10038161003816100381610038161003816100381610038161003816120601119902 (int
120591
0
120596 (120583) 119891 (120583 119909 (120583)) 119889120583)
10038161003816100381610038161003816100381610038161003816119889120591
+ int
119904
0
(119904 minus 120591)120572minus1
times
10038161003816100381610038161003816100381610038161003816120601119902 (int
120591
0
120596 (120583) 119891 (120583 119909 (120583)) 119889120583)
10038161003816100381610038161003816100381610038161003816119889120591) 119889119904
le1
Γ (120572)
times int
1
0
(1 minus 119904)120572minus1
10038161003816100381610038161003816100381610038161003816120601119902 (119872int
119904
0
120596 (120591) 119889120591)
10038161003816100381610038161003816100381610038161003816119889119904
+int
119905
0
(119905 minus 119904)120572minus1
10038161003816100381610038161003816100381610038161003816120601119902 (119872int
119904
0
120596 (120591) 119889120591)
10038161003816100381610038161003816100381610038161003816119889119904
+120578
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
119892 (119904) (int
1
0
(1 minus 120591)120572minus1
times
10038161003816100381610038161003816100381610038161003816120601119902 (119872int
120591
0
120596 (120583) 119889120583)119889120591
10038161003816100381610038161003816100381610038161003816
+ int
119904
0
(119904 minus 120591)120572minus1
times
10038161003816100381610038161003816100381610038161003816120601119902 (119872int
120591
0
120596 (120583) 119889120583)
10038161003816100381610038161003816100381610038161003816119889120591) 119889119904
le120578
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
120601119902 (119872int
1
0
120596 (120591) 119889120591)
times int
1
0
119892 (119904) (int
1
0
(1 minus 120591)120572minus1
119889120591 + int
119904
0
(119904 minus 120591)120572minus1
119889120591)119889119904
+1
Γ (120572)120601119902 (119872int
1
0
120596 (120591) 119889120591)
times int
1
0
(1 minus 119904)120572minus1
119889119904 + int
119905
0
(119905 minus 119904)120572minus1
119889119904
le
2120601119902 (119872int1
0120596 (120591) 119889120591)
Γ (120572 + 1) (1 minus 120578 int1
0119892 (120591) 119889120591)
lt infin
(30)
that is 119879 maps bounded sets into bounded sets in 119875For 119909(119905) isin 119861119903 0 lt 1199051 lt 1199052 lt 1
1003816100381610038161003816119879119909 (1199052) minus 119879119909 (1199051)1003816100381610038161003816
=1
Γ (120572)
10038161003816100381610038161003816100381610038161003816int
1199052
0
(1199052 minus 119904)120572minus1
120601119902 (int
119904
0
120596 (120591) 119891 (120591 119909 (120591)) 119889120591) 119889119904
6 Abstract and Applied Analysis
minus int
1199051
0
(1199051 minus 119904)120572minus1
times 120601119902 (int
119904
0
120596 (120591) 119891 (120591 119909 (120591)) 119889120591) 119889119904
10038161003816100381610038161003816100381610038161003816
le1
Γ (120572)120601119902 (119872int
1
0
120596 (120591) 119889120591)
times
100381610038161003816100381610038161003816100381610038161003816
int
1199051
0
((1199052 minus 119904)120572minus1
minus (1199051 minus 119904)120572minus1
) 119889119904
+int
1199052
1199051
(1199052 minus 119904)120572minus1
119889119904
100381610038161003816100381610038161003816100381610038161003816
(31)
By mean value theorem we obtain that
int
1199051
0
((1199052 minus 119904)120572minus1
minus (1199051 minus 119904)120572minus1
) 119889119904
le int
1199051
0
(120572 minus 1) (1 minus 119904)120572minus2
(1199052 minus 1199051) 119889119904
le int
1
0
(120572 minus 1) (1 minus 119904)120572minus2
(1199052 minus 1199051) 119889119904
= 1199052 minus 1199051
(32)
Thus1003816100381610038161003816119879119909 (1199052) minus 119879119909 (1199051)
1003816100381610038161003816
le1
Γ (120572)120601119902 (119872int
1
0
120596 (120591) 119889120591) [(1199052 minus 1199051) +1
120572(1199052 minus 1199051)
120572]
(33)
This shows that |119879119909(1199052) minus 119879119909(1199051)| rarr 0 as 1199052 minus 1199051 rarr 0so 119879119909 119909 isin 119861119903 is equicontinuous Therefore the operator119879 119875 rarr 119875 is completely continuous by the Arzela-Ascolitheorem
Now for convenience we introduce the following nota-tions Let
int
1
0
(1 minus 119904)120572minus1
120601119902 (int
119904
0
120596 (120591) 119889120591) 119889119904 = 1205741
120601119902 (int
1
0
120596 (120591) 119889120591) = 1205742
119898 (119903) = minmin119905
119891 (119905 119909)
120601119901 (119903) 119909 isin [120590119903 119903]
(34)
Theorem 9 Assume that (H1)ndash(H5) hold
(i) If 0 lt 1198910
120601lt +infin then there exists 1205850 gt 0 such that for
every 0 lt 119903 lt 1205850 problem (1) has a positive solution119909119903(119905) satisfying 119909119903 = 119903 associated with
120582 = 120582119903 isin [1205820 1205820] (35)
where 1205820 and 1205820 are two positive finite numbers
(ii) If 0 lt 119891infin
120601lt +infin then there exists 120585lowast
0gt 0 such that for
every 119877 gt 120585lowast
0 problem (1) has a positive solution 119909119877(119905)
satisfying 119909119877 = 119877 for any
120582 = 120582119877 isin [120582lowast
0 120582lowast
0] (36)
where 120582lowast
0and 120582
lowast
0are two positive finite numbers
(iii) If 1198910120601
= +infin then there exists 1205851 gt 0 such that for any0 lt 119903lowast
lt 1205851 problem (1) has a positive solution 119909119903lowast(119905)
satisfying 119909119903lowast = 119903lowast for any
120582 = 120582119903lowast isin (0 120582lowast] (37)
where 120582lowast is a positive finite number
(iv) If 119891infin
120601= +infin then there exists 1205851 gt 0 such that
for every 119877lowast gt 1205851 problem (1) has a positive solution119909119877lowast
(119905) satisfying 119909119877lowast
= 119877lowast for any
120582 = 120582119877lowast
isin (0 120582lowast] (38)
where 120582lowast is a positive finite number(v) If there exists 120585 gt 0 such that 119898(119903) gt 120585 then problem
(1) has positive solution 119909119903(119905) satisfying 119909119903 = 119903 forany
120582 = 120582119903 isin (0 ] (39)
where is a positive finite number
Proof (i) It follows from 0 lt 1198910
120601lt +infin that there exists
1198971 1198972 120588 gt 0 such that
1198971120601119901 (119909) le 119891 (119905 119909) le 1198972120601119901 (119909)
(forall119905 isin [0 1] 0 le 119909 le 120588)
(40)
Let 1205850 = 120588120590 we show that 1205850 is required When 119909 isin 119875 cap
120597Ω119903 we have
119909 (119905) ge 120590 119909 = 120590119903 (41)
we need 119909(119905) lt 120588 this implies that 120590119903 lt 120588 as 119903 lt 1205850 so120590119903 lt 1205901205850 = 120588
Let 1205820 = (11198971)(Γ(120572)1205901205741)119901minus1 Then we may assume that
119909 minus 120601119902 (1205820) 119879119909 = 0 (forall119909 isin 119875 cap 120597Ω119903) (42)
if not then there exists119909119903 isin 119875cap120597Ω119903 such that120601119902(1205820)119879119909119903 = 119909119903and then (35) already holds for 120582119903 = 1205820
Define 120595(119905) equiv 1 forall119905 isin [0 1] then 120595(119905) isin 119875 and 120595 equiv 1We now show that
119909 minus 120601119902 (1205820) 119879119909 = 120581120595 (forall119909 isin 119875 cap 120597Ω119903 120581 ge 0) (43)
In fact if there exists 1199091 isin 119875 cap 120597Ω119903 1205811 ge 0 such that 1199091 minus120601119902(1205820)1198791199091 = 1205811120595 then (42) implies that 1205811 gt 0 On the otherhand 1199091 = 120601119902(1205820)1198791199091 + 1205811120595 ge 1205811120595 we may choose 120581
lowast=
max120581 | 1199091 ge 120581120595 then 1205811 le 120581lowastlt +infin 1199091 ge 120581
lowast120595 Therefore
120581lowast= 120581lowast 1003817100381710038171003817120595
1003817100381710038171003817 le10038171003817100381710038171199091
1003817100381710038171003817 = 119903 (44)
Abstract and Applied Analysis 7
Consequently for any 119905 isin [0 1] (26) and (44) imply that
1199091 (119905)
= 120601119902 (1205820)1
Γ (120572)
times int
1
0
(1 minus 119904)120572minus1
times 120601119902 (int
119904
0
120596 (120591) 119891 (120591 1199091 (120591)) 119889120591) 119889119904
minus int
119905
0
(119905 minus 119904)120572minus1
times 120601119902 (int
119904
0
120596 (120591) 119891 (120591 1199091 (120591)) 119889120591) 119889119904
+120578120601119902 (1205820)
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
119892 (119904)
times (int
1
0
(1 minus 120591)120572minus1
times 120601119902 (int
120591
0
120596 (120583) 119891 (120583 1199091 (120583)) 119889120583)119889120591
minus int
119904
0
(119904 minus 120591)120572minus1
times 120601119902 (int
120591
0
120596 (120583) 119891 (120583 1199091 (120583)) 119889120583)119889120591)119889119904
+ 1205811120595
ge120578120590120601119902 (1205820)
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
119892 (119904)
times (int
1
0
(1 minus 120591)120572minus1
times 120601119902 (int
120591
0
120596 (120583) 119891 (120583 1199091 (120583)) 119889120583)119889120591
minus int
119904
0
(119904 minus 120591)120572minus1
120601119902
times (int
120591
0
120596 (120583) 119891 (120583 1199091 (120583)) 119889120583)119889120591)119889119904
+120590120601119902 (1205820)
Γ (120572)
times int
1
0
(1 minus 119904)120572minus1
120601119902 (int
119904
0
120596 (120591) 119891 (120591 1199091 (120591)) 119889120591) 119889119904
+ 1205811120595
ge120590120601119902 (1205820)
Γ (120572)
times int
1
0
(1 minus 119904)120572minus1
120601119902 (int
119904
0
120596 (120591) 119891 (120591 1199091 (120591)) 119889120591) 119889119904
+ 1205811120595
ge120590120601119902 (1205820)
Γ (120572)
times int
1
0
(1 minus 119904)120572minus1
120601119902 (1198971 int
119904
0
120596 (120591) 120601119901 (1199091 (120591)) 119889120591) 119889119904
+ 1205811120595
ge120590120601119902 (1205820)
Γ (120572)
times int
1
0
(1 minus 119904)120572minus1
120601119902 (1198971 int
119904
0
120596 (120591) 120601119901 (120581lowast120595 (120591)) 119889120591) 119889119904
+ 1205811120595
=120590120581lowast119897119902minus1
1120582119902minus1
0
Γ (120572)int
1
0
(1 minus 119904)120572minus1
120601119902 (int
119904
0
120596 (120591) 119889120591) 119889119904 + 1205811120595
=120590120581lowast119897119902minus1
1120582119902minus1
01205741
Γ (120572)+ 1205811120595
= 120581lowast+ 120581120595
(45)
which implies that 1199091(119905) ge (120581lowast+ 120581)120595(119905) 119905 isin [0 1] which is a
contradiction to the definition of 120581lowast Thus (42) holds and byLemma 5 the fixed point index
119894 (120601119902 (1205820) 119879 119875 cap 120597Ω119903 119875) = 0 (46)
On the other hand by the fact that the fixed point index ofconstants operator is 1 so
119894 (120599 119875 cap 120597Ω119903 119875) = 1 (47)
where 120599 is the zero operator It follows therefore from (46)and (47) and the homotopy invariance property that thereexist 119909119903 isin 119875 cap 120597Ω119903 and 0 lt ]0 lt 1 such that ]0120601119902(1205820)119879119909119903 =119909119903 but 119909119903 = 120601119902(120582119903)119879119909119903 which implies that ]0120601119902(1205820)119879119909119903 =
120601119902(120582119903)119879119909119903 it follows that we get ]0120601119902(1205820) = 120601119902(120582119903)) that is
]0120582119902minus1
0= 120582119902minus1
119903 (48)
and then
120582119903 = ]1(119902minus1)0
1205820 = ]119901minus10
1205820 lt 1205820 (49)
From the proof above for any 119903 lt 1205850 there exists a positivesolution 119909119903 isin 119875 cap 120597Ω119903 associated with 120582 = 120582119903 gt 0
8 Abstract and Applied Analysis
Thus
119909119903 (119905)
=120601119902 (120582119903)
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902 (int
119904
0
120596 (120591) 119891 (120591 119909119903 (120591)) 119889120591) 119889119904
minus int
119905
0
(119905 minus 119904)120572minus1
times 120601119902 (int
119904
0
120596 (120591) 119891 (120591 119909119903 (120591)) 119889120591) 119889119904
+120578120601119902 (120582119903)
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
119892 (119904)
times (int
1
0
(1 minus 120591)120572minus1
times 120601119902 (int
120591
0
120596 (120583) 119891 (120583 119909119903 (120583)) 119889120583)119889120591
minus int
119904
0
(119904 minus 120591)120572minus1
times 120601119902 (int
120591
0
120596 (120583) 119891 (120583 119909119903 (120583)) 119889120583)119889120591)119889119904
(50)
with 119909119903 = 119903Next we show that 120582119903 ge 1205820 In fact
119909119903 (119905)
=120601119902 (120582119903)
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902 (int
119904
0
120596 (120591) 119891 (120591 119909119903 (120591)) 119889120591) 119889119904
minus int
119905
0
(119905 minus 119904)120572minus1
times 120601119902 (int
119904
0
120596 (120591) 119891 (120591 119909119903 (120591)) 119889120591) 119889119904
+120578120601119902 (120582119903)
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
119892 (119904)
times (int
1
0
(1 minus 120591)120572minus1
times 120601119902 (int
120591
0
120596 (120583) 119891 (120583 119909119903 (120583)) 119889120583)119889120591
minus int
119904
0
(119904 minus 120591)120572minus1
times 120601119902 (int
120591
0
120596 (120583) 119891 (120583 119909119903 (120583)) 119889120583)119889120591)119889119904
le120578120601119902 (120582119903)
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
119892 (119904) int
1
0
(1 minus 120591)120572minus1
times 120601119902 (int
120591
0
120596 (120583) 119891 (120583 119909119903 (120583)) 119889120583)119889120591 119889119904
+120601119902 (120582119903)
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902 (int
119904
0
120596 (120591) 119891 (120591 119909119903 (120591)) 119889120591) 119889119904
=120601119902 (120582119903) 120578 int
1
0119892 (119904) 119889119904
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
(1 minus 119904)120572minus1
times 120601119902 (int
119904
0
120596 (120591) 119891 (120591 119909119903 (120591)) 119889120591) 119889119904
+120601119902 (120582119903)
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902 (int
119904
0
120596 (120591) 119891 (120591 119909119903 (120591)) 119889120591) 119889119904
le120601119902 (120582119903) 120578 int
1
0119892 (119904) 119889119904
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
(1 minus 119904)120572minus1
times 120601119902 (1198972 int
119904
0
120596 (120591) 119909119903 (120591) 119889120591) 119889119904
+120601119902 (120582119903)
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902 (1198972120582119903 int
119904
0
120596 (120591) 119909119903 (120591) 119889120591) 119889119904
=120601119902 (120582119903)
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
(1 minus 119904)120572minus1
times 120601119902 (1198972 int
119904
0
120596 (120591) 119909119903 (120591) 119889120591) 119889119904
Abstract and Applied Analysis 9
le
10038171003817100381710038171199091199031003817100381710038171003817 120601119902 (1205821199031198972)
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
(1 minus 119904)120572minus1
times 120601119902 (int
1
0
120596 (120591) 119889120591) 119889119904
=
10038171003817100381710038171199091199031003817100381710038171003817 120601119902 (1205821199031198972) 1205742
Γ (120572 + 1) (1 minus 120578 int1
0119892 (119904) 119889119904)
(51)
which implies that
10038171003817100381710038171199091199031003817100381710038171003817 le
10038171003817100381710038171199091199031003817100381710038171003817 120601119902 (1205821199031198972) 1205742
Γ (120572 + 1) (1 minus 120578 int1
0119892 (119904) 119889119904)
(52)
that is
120601119902 (1205821199031198972) 1205742
Γ (120572 + 1) (1 minus 120578 int1
0119892 (119904) 119889119904)
ge 1 (53)
it follows that we get
120582119903 ge1
1198972(
Γ (120572 + 1) (1 minus 120578 int1
0119892 (119904) 119889119904)
1205742)
119901minus1
= 1205820(54)
In conclusion 120582119903 isin [1205820 1205820](ii) It follows from 0 lt 119891
infin
120601lt infin that there exist
119897lowast
1 119897lowast
2 120588lowastgt 0 such that
119897lowast
1120601119901 (119909) le 119891 (119905 119909) le 119897
lowast
2120601119901 (119909)
(forall119905 isin [0 1] 119909 ge 120588lowast)
(55)
Let 120585lowast0= 120588lowast120590 the following proof are similar to (i)
(iii) It follows from 1198910
120601= +infin there exists 119897lowast gt 0 120588
lowastgt 0
such that
1198910
120601ge 119897lowast120601119901 (119909) (forall119905 isin [0 1] 0 le 119909 le 120588
lowast) (56)
Let 1205851 = 120588lowast120590 we show that 1205851 is required When 119909 isin 119875 cap
120597Ω119903lowast we have 119909(119905) ge 120590119909 = 120590119903lowast since 119909 le 120588
lowast then we need120590119903lowastlt 120588lowast as 119903lowast lt 1205851 so 120590119903
lowastlt 120588lowast holds
Let 120582lowast = (1119897lowast)(Γ(120572)1205901205741)
1(119902minus1) we proceed in the sameway as in the proof of (i) replacing (42) we may assume that119909 minus 120601119902(120582
lowast)119879119909 = 0 (forall119909 isin 119875 cap 120597Ω119903lowast) and replacing (43) we
can prove 119909 minus 120601119902(120582lowast)119879119909 = 120581120595 (forall119909 isin 119875 cap 120597Ω119903lowast 120581 ge 0) It
follows from Lemma 5 that 119894(120601119902(120582lowast)119879 119875 cap Ω119903lowast 119875) = 0 Note
that 119894(120579 119875 cap Ω119903lowast 119875) = 1 we can easily show that there exists119909119903lowast isin 119875cap120597Ω119903lowast and 0 lt ]119903lowast lt 1 such that ]119903lowast120601119902(120582
lowast)119879119909119903lowast = 119909119903lowast
Hence (37) holds for 120582119903lowast = 120582lowast]1minus119901119903lowast
lt 120582lowast
The proof ofTheorem 9(iv) follows by themethod similarto Theorem 9(iii) we omit it here
(v) It follows that 119898(119903) gt 120585 for any 119909 isin 119875 cap 120597Ω119903 we have120590119903 le 119909 le 119903 and 119891(119905 119909)120601119901(119909) ge 119891(119905 119909)120601119901(119903) ge 119898(119903) ge 120585then
119891 (119905 119909) ge 120585120601119901 (119909) forall119909 isin [120590119903 119903] 119905 isin [0 1] (57)
Let = (1120585)(Γ(120572)1205901205741)119901minus1 The following proof is similar to
that of (iv) This finished the proof of (v)
Remark 10 InTheorem 9 all the criteria obtained depend onone of 1198910
120601and 119891
infin
120601
Let Ψ(119909) = 120601119902(120582)119879119909 the following theorems give out themultiply nonexistence and the dependence of parameter
Theorem 11 Assume that (H1)ndash(H5) hold
(i) If 1198910
120601= 0 and 119891
infin
120601= 0 lim
119909rarrinfin119891(119905 119909) = 119891infin isin
(0 +infin] uniformly for 119905 isin [0 1] then there exists120582lowast
gt 0 such that the problem (1) has two solutions forany 120582 gt 120582
lowast
(ii) If 1198910120601
= 0 and 119891infin
120601= 0 then there exists 120582 gt 0 such
that for any 120582 lt 120582 problem (1) has no solution
Proof (i) It follows from 119891infin isin (0 +infin] that for any 120577 gt 0there exists 120588 gt 0 such that
119891 (119905 119909) gt 120577 (forall119905 isin [0 1] 119909 ge 120588) (58)
Since 1198910
120601= 0 there exists 1205761 gt 0 120588
1gt 0 such that
119891 (119905 119909) le 1205761120601119901 (119909) (forall119905 isin [0 1] 0 le 119909 le 1205881) (59)
where 1205761 satisfied 120601119902(1205821205761)1205742Γ(120572 + 1)(1 minus 120578 int1
0119892(119904)119889119904) le 1
For forall119909 isin 119875 cap 120597Ω1205881
we have
Ψ (119909) = (120601119902 (120582) 119879119909) (119905)
=1
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902 (120582int
119904
0
120596 (120591) 119891 (120591 119909 (120591)) 119889120591) 119889119904
minus int
119905
0
(119905 minus 119904)120572minus1
times 120601119902 (120582int
119904
0
120596 (120591) 119891 (120591 119909 (120591)) 119889120591) 119889119904
+120578
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
119892 (119904)
10 Abstract and Applied Analysis
times (int
1
0
(1 minus 120591)120572minus1
times 120601119902 (120582int
120591
0
120596 (120583) 119891 (120583 119909 (120583)) 119889120583)119889120591
minus int
119904
0
(119904 minus 120591)120572minus1
times 120601119902 (120582int
120591
0
120596 (120583) 119891 (120583 119909 (120583)) 119889120583)119889120591)119889119904
le120578
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
119892 (119904)
times (int
1
0
(1 minus 120591)120572minus1
times 120601119902(120582int
120591
0
120596 (120583) 119891 (120583 119909 (120583)) 119889120583)119889120591)119889119904
+1
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902 (120582int
119904
0
120596 (120591) 1205761119891 (120591 119909 (120591)) 119889120591) 119889119904
le120578 int1
0119892 (119904) 119889119904
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
(1 minus 120591)120572minus1
times 120601119902 (120582int
1
0
120596 (120583) 1205761120601119901 (119909 (120583)) 119889120583)119889120591
+1
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902(120582int
1
0
120596 (120591) 1205761120601119901 (119909 (120591)) 119889120591) 119889119904
le1
Γ (120572 + 1) (1 minus 120578 int1
0119892 (119904) 119889119904)
times 120601119902 (120582int
1
0
120596 (120591) 1205761120601119901 (119909) 119889120591)
=120601119902 (1205821205761) 1205742
Γ (120572 + 1) (1 minus 120578 int1
0119892 (119904) 119889119904)
119909 le 119909
(60)
This implied
Ψ119909 le 119909 forall119909 isin 119875 cap 120597Ω1205881
(61)
Let 1205883gt max120588120590 120588
1 and 120582
lowast= (1120577)(120588
3Γ(120572)1205901205741)
119901minus1 thenforall119909 isin 119875 cap 120597Ω120588
3
we have 120590119909 le 119909 le 1205883and
(Ψ119909) (119905)
=1
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902 (120582int
119904
0
120596 (120591) 119891 (120591 119909 (120591)) 119889120591) 119889119904
minus int
119905
0
(119905 minus 119904)120572minus1
times 120601119902 (120582int
119904
0
120596 (120591) 119891 (120591 119909 (120591)) 119889120591) 119889119904
+120578
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
119892 (119904)
times (int
1
0
(1 minus 120591)120572minus1
times 120601119902 (120582int
120591
0
120596 (120583) 119891 (120583 119909 (120583)) 119889120583)119889120591
minus int
119904
0
(119904 minus 120591)120572minus1
times 120601119902(120582int
120591
0
120596 (120583) 119891 (120583 119909 (120583)) 119889120583)119889120591)119889119904
ge120590120578
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
119892 (119904)
times (int
1
0
(1 minus 120591)120572minus1
times 120601119902 (120582int
120591
0
120596 (120583) 119891 (120583 119909 (120583)) 119889120583)119889120591
minus int
119904
0
(119904 minus 120591)120572minus1
times 120601119902(120582int
120591
0
120596 (120583) 119891 (120583 119909 (120583)) 119889120583)119889120591)119889119904
+120590
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902 (120582int
119904
0
120596 (120591) 119891 (120591 119909 (120591)) 119889120591) 119889119904
ge120590
Γ (120572)int
1
0
(1 minus 119904)120572minus1
Abstract and Applied Analysis 11
times 120601119902 (120582int
119904
0
120596 (120591) 119891 (120591 119909 (120591)) 119889120591) 119889119904
gt120590
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902 (120582int
119904
0
120596 (120591) 120577 119889120591) 119889119904
gt120590120601119902 (120582120577)
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902 (int
119904
0
120596 (120591) 119889120591) 119889119904
=120590120601119902 (120582120577) 1205741
Γ (120572)= 1205883
(62)
which implies
Ψ119909 gt 119909 forall119909 isin 119875 cap 120597Ω1205883
(63)
Next for 119891infin
120601= 0 there exist 1205762 gt 0 120588
2gt 1205883gt 0 such
that
119891 (119905 119909) le 1205762120601119901 (119909) (forall119905 isin [0 1] 119909 ge 1205882) (64)
where 1205762 satisfied 120601119902(1205821205762)1205742Γ(120572 + 1)(1 minus 120578 int1
0119892(119904)119889119904) le 1
Similar to the above proof we get
Ψ119909 le 119909 forall119909 isin 119875 cap 120597Ω1205882
(65)
Applying Lemma 7 to (61) (65) and (63) yields that Ψ
has two fixed points 1199091 1199092 such that 1199091 isin 119875cap (Ω1205883
Ω1205881
) and1199092 isin 119875 cap (Ω120588
2
Ω1205883
)(ii) It follows from 119891
0
120601= 0 and 119891
infin
120601= 0 that there exists
1205761 1205762 1205881 1205882 gt 0 such that
119891 (119905 119909) le 1205761120601119901 (119909) (forall119905 isin [0 1] 0 le 119909 le 1205881)
119891 (119905 119909) le 1205762120601119901 (119909) (forall119905 isin [0 1] 119909 ge 1205882)
(66)
then for 119909 isin [1205881 1205882] by the continuous of 119891(119905 119909(119905)) there
exists 119909 isin [1205881 1205882] such that
max1205881le119909le120588
2
119891 (119905 119909)
120601119901 (119909)=
119891 (119905 119909)
120601119901 (119909) (67)
Let Υ = max1205761 1205762 119891(119905 119909)120601119901(119909) gt 0 then
119891 (119905 119909) le Υ120601119901 (119909) (forall0 le 119905 le 1 119909 ge 0) (68)
Assuming 119909(119905) is a positive solution of problem (1) we willshow that this leads to a contradiction for 120582 lt 120582 where
120582 = (1Υ)(Γ(120572 + 1)(1 minus 120578 int1
0119892(119904)119889119904)1205742)
119901minus1 It follows from(26) that
119909 = 119909 (0)
=120578
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
119892 (119904)
times (int
1
0
(1 minus 120591)120572minus1
times 120601119902 (120582int
120591
0
120596 (120583) 119891 (120583 119909 (120583)) 119889120583)119889120591
minus int
119904
0
(119904 minus 120591)120572minus1
times 120601119902(120582int
120591
0
120596 (120583) 119891 (120583 119909 (120583)) 119889120583)119889120591)119889119904
+1
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902 (120582int
119904
0
120596 (120591) 119891 (120591 119909 (120591)) 119889120591) 119889119904
le120578
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
119892 (119904)
times int
1
0
(1 minus 120591)120572minus1
times 120601119902(120582int
120591
0
120596 (120583) 119891 (120583 119909 (120583)) 119889120583)119889120591 119889119904
+1
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902 (120582int
119904
0
120596 (120591) 119891 (120591 119909 (120591)) 119889120591) 119889119904
=1
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
(1 minus 119904)120572minus1
times 120601119902 (120582int
119904
0
120596 (120591) 119891 (120591 119909 (120591)) 119889120591) 119889119904
le1
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
(1 minus 119904)120572minus1
12 Abstract and Applied Analysis
times 120601119902 (120582int
1
0
120596 (120591) Υ120601119901 (119909 (120591)) 119889120591) 119889119904
=120601119902 (120582Υ) 1205742
Γ (120572 + 1) (1 minus 120578 int1
0119892 (119904) 119889119904)
119909
(69)
This implies that 1 le 120601119902(120582Υ)1205742Γ(120572 + 1)(1 minus 120578 int1
0119892(119904)119889119904) =
(120582Υ)119902minus1
1205742Γ(120572+1)(1minus120578 int1
0119892(119904)119889119904) lt (120582 Υ)
119902minus11205742Γ(120572+1)(1minus
120578 int1
0119892(119904)119889119904) = 1 which is a contradiction This finishes the
proof of (ii)
Remark 12 In (i) the condition lim119909rarrinfin
119891(119905 119909) = 119891infin isin
(0 +infin] uniformly for 119905 isin [0 1] can be replaced withcondition (H6) The same methods can be used to prove theresults
Remark 13 From the above proof the condition 119891infin isin
(0 +infin] is important to keep the problem (1) having at leasttwo positive solutions if this condition is not satisfied thenthere exists 120582 gt 0 small enough such that the problem (1) hasno positive solutions for all 120582 lt 120582
Theorem 14 Assume that (H1)ndash(H6) hold(i) If 1198910
120601= +infin and 119891
infin
120601= +infin then there exists 120582lowast0 gt 0
such that problem (1) has at least two solutions for 120582 isin
(0 120582lowast0)(ii) If 1198910
120601= +infin and 119891
infin
120601= +infin then there exists 120582 gt 0
such that problem (1) has no solution for all 120582 gt 120582
Proof (i) It follows from 1198910
120601= +infin that there exists 1199031 gt 0
such that
119891 (119905 119909) ge 119871lowast1120601119901 (119909) (forall119905 isin [0 1] 0 le 119909 le 1199031) (70)
where 119871lowast1 satisfies 1205902120601119902(120582119871lowast1)1205741Γ(120572) ge 1
Then for 119909 isin 119875cap120597Ω1199031
we have 120590119909 le 119909 le 1199031 for 119905 isin [0 1]
and
(Ψ119909) (119905)
=1
Γ (120572)
times int
1
0
(1 minus 119904 )120572minus1
times 120601119902 (120582int
119904
0
120596 (120591) 119891 (120591 119909 (120591)) 119889120591) 119889119904
minus int
119905
0
(119905 minus 119904)120572minus1
times 120601119902 (120582int
119904
0
120596 (120591) 119891 (120591 119909 (120591)) 119889120591) 119889119904
+120578
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
119892 (119904)
times (int
1
0
(1 minus 120591)120572minus1
times 120601119902 (120582int
120591
0
120596 (120583) 119891 (120583 119909 (120583)) 119889120583)119889120591
minus int
119904
0
(119904 minus 120591)120572minus1
times 120601119902 (120582int
120591
0
120596 (120583) 119891 (120583 119909 (120583)) 119889120583)119889120591)119889119904
ge120590120578
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
119892 (119904)
times (int
1
0
(1 minus 120591)120572minus1
times 120601119902 (120582int
120591
0
120596 (120583) 119891 (120583 119909 (120583)) 119889120583)119889120591
minus int
119904
0
(119904 minus 120591)120572minus1
times120601119902 (120582int
120591
0
120596 (120583) 119891 (120583 119909 (120583)) 119889120583)119889120591)119889119904
+120590
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902 (120582int
119904
0
120596 (120591) 119891 (120591 119909 (120591)) 119889120591) 119889119904
ge120590
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902 (120582int
119904
0
120596 (120591) 119891 (120591 119909 (120591)) 119889120591) 119889119904
ge120590
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902 (120582int
119904
0
120596 (120591) 119871lowast1120601119901 (119909 (120591)) 119889120591) 119889119904
ge1205902120601119902 (120582119871lowast1) 119909
Γ (120572)
times int
1
0
(1 minus 119904)120572minus1
times 120601119902 (int
119904
0
120596 (120591) 119889120591) 119889119904
=1205902120601119902 (120582119871lowast1) 1205742
Γ (120572)119909
(71)
Abstract and Applied Analysis 13
which implies that
Ψ119909 ge 119909 forall119909 isin 119875 cap 120597Ω1199031
(72)
Next from 119891infin
120601= +infin there exists 1199032 gt 0 such that
119891 (119905 119909) ge 119871lowast2120601119901 (119909) forall119909 ge 1199032 (73)
where 119871lowast2 satisfies
1205902120601119902 (120582119871lowast2) 1205742
Γ (120572)ge 1 (74)
Let 1199032 gt max1199031 1199032120590 then for 119909 isin 119875 cap 120597Ω1199032
we have
119909 (119905) ge 120590 119909 = 1205901199032 gt 1199032 (75)
Similar to the above proof we can get
Ψ119909 ge 119909 forall119909 isin 119875 cap 120597Ω1199032
(76)
Let 1199031 lt 1199033 lt 11990321198721199033
= max119891(119905 119909) | 119905 isin [0 1] 119909 le 1199033+1 gt
0 and 120582lowast0 = (11198721199033
)(Γ(120572+1)1199033(1minus120578 int1
0119892(119904)119889119904)1205742)
119901minus1Thenfor 119909 isin 119875 cap 120597Ω119903
3
we have
119891 (119905 119909) le 1198721199033
forall119905 isin [0 1] (77)
Hence
(Ψ119909) (119905)
=1
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902 (120582int
119904
0
120596 (120591) 119891 (120591 119909 (120591)) 119889120591) 119889119904
minus int
119905
0
(119905 minus 119904)120572minus1
times 120601119902 (120582int
119904
0
120596 (120591) 119891 (120591 119909 (120591)) 119889120591) 119889119904
+120578
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
119892 (119904)
times (int
1
0
(1 minus 120591)120572minus1
times 120601119902 (120582int
120591
0
120596 (120583) 119891 (120583 119909 (120583)) 119889120583)119889120591
minus int
119904
0
(119904 minus 120591)120572minus1
times120601119902 (120582int
120591
0
120596 (120583) 119891 (120583 119909 (120583)) 119889120583)119889120591)119889119904
le120578
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
119892 (119904)
times int
1
0
(1 minus 120591)120572minus1
120601119902
times (120582int
120591
0
120596 (120583) 119891 (120583 119909 (120583)) 119889120583)119889120591 119889119904
+1
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902 (120582int
119904
0
120596 (120591) 119891 (120591 119909 (120591)) 119889120591) 119889119904
le120578
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
119892 (119904)
times int
1
0
(1 minus 120591)120572minus1
120601119902
times (120582int
120591
0
120596 (120583)1198721199033
119889120583)119889120591119889119904
+1
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902 (120582int
119904
0
120596 (120591)1198721199033
119889120591)119889119904
le120578120601119902 (120582119872119903
3
) int1
0119892 (119904) 119889119904
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
(1 minus 120591)120572minus1
times 120601119902 (int
1
0
120596 (120583) 119889120583)119889120591
+120601119902 (120582119872119903
3
)
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902 (int
1
0
120596 (120591) 119889120591) 119889119904
=1205742120601119902 (120582119872119903
3
)
Γ (120572 + 1) (1 minus 120578 int1
0119892 (119904) 119889119904)
lt1205742120601119902 (120582lowast0119872119903
3
)
Γ (120572 + 1) (1 minus 120578 int1
0119892 (119904) 119889119904)
= 1199033
(78)
14 Abstract and Applied Analysis
This implies that
Ψ119909 lt 119909 forall119909 isin 119875 cap 120597Ω1199033
(79)
Applying Lemma 7 to (72) (76) and (79) yields that Ψ hastwo fixed points 1199091 1199092 such that 1199091 isin 119875 cap (Ω119903
3
Ω1199031
) and1199092 isin 119875 cap (Ω119903
2
Ω1199033
)(ii) By the proof of (i) and the continuity of 119891(119905 119909) there
exists 119909 isin [1199031 1199032] such that min119909isin[11990311199032](119891(119905 119909)120601119901(119909)) =
119891(119905 119909)120601119901(119909)Let Υ = min119871lowast1 119871lowast2 119891(119905 119909)120601119901(119909) since (H6) holds
then Υ gt 0 and
119891 (119905 119909) ge Υ120601119901 (119909) forall119909 ge 0 119905 isin [0 1] (80)
Assuming 119909(119905) is a positive solution of problem (1) we willshow that this leads to contradiction for all 120582 gt 120582 where 120582 =
(1Υ)(Γ(120572)1205741120590)119901minus1 It follows form (26) that
119909 = 119909 (0)
=120578
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
119892 (119904)
times (int
1
0
(1 minus 120591)120572minus1
times 120601119902 (120582int
120591
0
120596 (120583) 119891 (120583 119909 (120583)) 119889120583)119889120591
minus int
119904
0
(119904 minus 120591)120572minus1
times120601119902 (120582int
120591
0
120596 (120583) 119891 (120583 119909 (120583)) 119889120583)119889120591)119889119904
+1
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902 (120582int
119904
0
120596 (120591) 119891 (120591 119909 (120591)) 119889120591) 119889119904
ge1
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902 (120582int
119904
0
120596 (120591) 119891 (120591 119909 (120591)) 119889120591) 119889119904
ge1
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902 (120582int
119904
0
120596 (120591) Υ120601119901 (119909 (120591)) 119889120591) 119889119904
ge1
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902 (120582int
119904
0
120596 (120591) Υ120601119901 (120590 119909 119889120591)) 119889119904
ge120601119902 (120582Υ) 120590 119909
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902 (int
119904
0
120596 (120591) 119889120591) 119889119904
=120601119902 (120582Υ) 1205741120590
Γ (120572)119909
(81)
This implies that 1 ge 120601119902(120582Υ)1205741120590Γ(120572) = (120582Υ)119902minus1
1205741120590Γ(120572) gt
(120582Υ)119902minus1
1205741120590Γ(120572) = 1 which is a contradiction This com-pletes the proof of (ii)
Corollary 15 Assume that (H1)ndash(H5) holds
(i) If lim119909rarrinfin
119891(119905 119909) = 119891infin isin (0 +infin] uniformly for 119905 isin
[0 1] and one of 1198910120601
= 0 and 119891infin
120601= 0 is satisfied then
there exists 120582lowast gt 0 such that the problem (1) has at leastone positive solution for any 120582 gt 120582
lowast(ii) If one of 1198910
120601= +infin and 119891
infin
120601= +infin holds then there
exists 120582lowast0 gt 0 such that problem (1) has at least onepositive solutions for any 120582 isin (0 120582lowast0)
Proof (i) The conclusion is a direct consequence of Theo-rem 11(i)
(ii) The conclusion is a direct consequence of Theo-rem 14(i)
Theorem 16 Assume that (H1)ndash(H5) holdThen the followingconclusions hold
(i) If 1198910120601
= 0 and 119891infin
120601= infin then the problem (1) has a
positive solution 119909120582(119905) for all 120582 gt 0(ii) If 1198910
120601= infin and 119891
infin
120601= 0 then the problem (1) has a
positive solution 119909120582(119905) for all 120582 gt 0
Proof Weonly prove (i) the proof of (ii) is similar sowe omitit here
Let 120582 gt 0 since 1198910
120601= 0 there exists 119903 gt 0 119897 gt 0 such that
119891 (119905 119909) le 119897120601119901 (119909) (forall119905 isin [0 1] 119909 isin [0 119903]) (82)
where 119897 satisfied
1205742120582119902minus1
119897119902minus1
Γ (120572 + 1) (1 minus 120578 int1
0119892 (119904) 119889119904)
le 1 (83)
Similar to the proof of Theorem 14 we obtain that
Ψ119909 le 119909 forall119909 isin 119875 cap 120597Ω119903 (84)
Abstract and Applied Analysis 15
From 119891infin
120601= infin there exists 119877 gt 119903 gt 0 119871 gt 0 such that
119891 (119905 119909) ge 119871120601119901 (119909) (forall119905 isin [0 1] 119909 ge 119877) (85)where 119871 satisfied
1205902120582119902minus1
119871119902minus1
1205741
Γ (120572)ge 1 (86)
We get thatΨ119909 ge 119909 forall119909 isin 119875 cap 120597Ω119877 (87)
Applying Lemma 7 to (84) (87) yields thatΨ has a fixed point119909 such that 119909 isin 119875 cap (Ω119877 Ω119903)
Theorem 17 Assume that (H1)ndash(H6) holdThen the followingconclusions hold
If 0 lt 1198910
120601lt infin and 0 lt 119891
infin
120601lt infin then there exists 1 2 gt
0 such that problem (1) has no positive solution for any 120582 gt 1
and 0 lt 120582 lt 2
Proof It follows from 0 lt 1198910
120601lt infin and 0 lt 119891
infin
120601lt infin that
there exist 1 gt 0 2 gt 0 1 gt 0 2 gt 0 and 1199032 gt 1199031 gt 0
such that1120601119901 (119909) le 119891 (119905 119909) le 1120601119901 (119909)
(forall119905 isin [0 1] 0 le 119909 le 1199031)
2120601119901 (119909) le 119891 (119905 119909) le 2120601119901 (119909)
(forall119905 isin [0 1] 119909 ge 1199032)
(88)
Let = max119891(119905 119909)120601119901(119909) 1199031 le 119909 le 1199032 0 le 119905 le 1 =
min119891(119905 119909)120601119901(119909) 1199031 le 119909 le 1199032 0 le 119905 le 1 By the condition(H6) gt 0 Then forall119909 ge 0 we have
120601119901 (119909) le 119891 (119905 119909) le 120601119901 (119909) (89)
where = max1 2 gt 0 and = min1198971 2 gt 0Let 1 = (1)(Γ(120572)1205741120590)
119901minus1 and 2 =
(1)(Γ(120572 + 1)(1 minus 120578 int1
0119892(119904)119889119904)1205742)
119901minus1
If problem (1)has positive solution 119909(119905) in 119875 then
119909 = 119909 (0)
=120578
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
119892 (119904)
times (int
1
0
(1 minus 120591)120572minus1
times 120601119902 (120582int
120591
0
120596 (120583) 119891 (120583 119909 (120583)) 119889120583)119889120591
minus int
119904
0
(119904 minus 120591)120572minus1
times120601119902 (120582int
120591
0
120596 (120583) 119891 (120583 119909 (120583)) 119889120583)119889120591)119889119904
+1
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902 (120582int
119904
0
120596 (120591) 119891 (120591 119909 (120591)) 119889120591) 119889119904
ge1
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902 (120582int
119904
0
120596 (120591) 119891 (120591 119909 (120591)) 119889120591) 119889119904
ge1
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902 (120582int
119904
0
120596 (120591) 120601119901 (119909 (120591)) 119889120591) 119889119904
ge1
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902 (120582int
119904
0
120596 (120591) 120601119901 (120590 119909 119889120591)) 119889119904
ge120601119902 (120582) 120590 119909
Γ (120572)
times int
1
0
(1 minus 119904)120572minus1
120601119902 (int
119904
0
120596 (120591) 119889120591) 119889119904
=120601119902 (120582) 1205741120590
Γ (120572)119909
gt120601119902 (1 ) 1205741120590
Γ (120572)119909 = 119909
119909 = 119909 (0)
=120578
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
119892 (119904)
times (int
1
0
(1 minus 120591)120572minus1
120601119902
times (120582int
120591
0
120596 (120583) 119891 (120583 119909 (120583)) 119889120583)119889120591
minus int
119904
0
(119904 minus 120591)120572minus1
120601119902
times(120582int
120591
0
120596 (120583) 119891 (120583 119909 (120583)) 119889120583)119889120591)119889119904
+1
Γ (120572)int
1
0
(1 minus 119904)120572minus1
120601119902
times (120582int
119904
0
120596 (120591) 119891 (120591 119909 (120591)) 119889120591) 119889119904
16 Abstract and Applied Analysis
le120578
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
119892 (119904)
times int
1
0
(1 minus 120591)120572minus1
times 120601119902 (120582int
120591
0
120596 (120583) 119891 (120583 119909 (120583)) 119889120583)119889120591 119889119904
+1
Γ (120572)int
1
0
(1 minus 119904)120572minus1
120601119902
times (120582int
119904
0
120596 (120591) 119891 (120591 119909 (120591)) 119889120591) 119889119904
=1
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
(1 minus 119904)120572minus1
120601119902
times (120582int
119904
0
120596 (120591) 119891 (120591 119909 (120591)) 119889120591) 119889119904
le1
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
(1 minus 119904)120572minus1
120601119902
times (120582int
1
0
120596 (120591) 120601119901 (119909 (120591)) 119889120591) 119889119904
=120601119902 (120582) 1205742
Γ (120572 + 1) (1 minus 120578 int1
0119892 (119904) 119889119904)
119909
lt120601119902 (2) 1205742
Γ (120572 + 1) (1 minus 120578 int1
0119892 (119904) 119889119904)
119909 = 119909
(90)
which is a contradiction This finishes the proof of Theo-rem 17
4 Examples
In this section we give out same examples to illustrate ourmain results can be used in practice
Example 1 Consider the following fractional differentialequation
(120601119901 (11986352
0+119909 (119905)))
1015840
+ 120582119905(1 + 1199052)12
119909119901minus1 arctan119909 = 0
0 lt 119905 lt 1
1199091015840(0) = 119909
10158401015840(0) = 0
119909 (1) =1
2int
1
0
1
2119909 (119904) 119889119904
(91)
where 120572 = 52 119901 = 32 119902 = 3 120596(119905) = 119905 119891(119905 119909) = (1 +
1199052)12
119909119901minus1 arctan119909 120578 = 12 and 119892(119905) = 12
By simple computation 119897lowast1= 1205872 119897
lowast
2= radic21205872 1205741 = int
1
0(1minus
119904)32
1206013(int119904
0120591 119889120591)119889119904 asymp 00043 1205742 = 1206013(int
1
0120591 119889120591) = 025 120590 asymp
01429 120582lowast
0= (1119897
lowast
1)(Γ(120572)1205901205741)
119901minus1asymp 25149 120582
lowast
0= (1119897
lowast
2)(Γ(120572+
1)(1 minus 120578 int1
0119892(119904)119889119904)1205742)
119901minus1asymp 14222 Then it is easy to see
all the conditions of Theorem 9(ii) are satisfied thus forall120582 isin
[14222 25149]] the problem (91) has at least one positivesolution
Example 2 Consider the boundary value problem
(120601119901 (11986352
0+119909 (119905)))
1015840
+ 120582119905(1 + 1199052)12
119909119901119890minus119909
= 0
0 lt 119905 lt 1
1199091015840(0) = 119909
10158401015840(0) = 0
119909 (1) =1
2int
1
0
1
2119909 (119904) 119889119904
(92)
where 120572 = 52 119901 = 32 119902 = 3 120596(119905) = 119905 119891(119905 119909) = (1 +
1199052)12
119909119901119890minus119909
120578 = 12 and 119892(119905) = 12
1198910
120601= lim119909rarr0
119891 (119905 119909)
120601119901 (119909)= lim119909rarr0
(1 + 1199052)12
119909119901119890minus119909
119909119901minus1= 0
forall119905 isin [0 1]
1198910
120601= lim119909rarrinfin
119891 (119905 119909)
120601119901 (119909)= lim119909rarrinfin
(1 + 1199052)12
119909119901119890minus119909
119909119901minus1= 0
forall119905 isin [0 1]
(93)
Let 1205881= 12 120588
2= 4 then
119891 (119905 119909)
120601119901 (119909)le (1 + 119905
2)12 1
2radic119890le
1
radic2119890
forall119905 isin [0 1] 119909 isin [01
2]
119891 (119905 119909)
120601119901 (119909)le (1 + 119905
2)12 4
1198904le
4radic2
1198904
forall119905 isin [0 1] 119909 isin [4infin)
(94)
so 1205761 = 1radic2119890 1205762 = 4radic21198904 and forall119909 isin
[12 4] max12le119909le4(119891(119905 119909)120601119901(119909)) = radic2119890minus1 We
have Υ = max1radic2119890 4radic21198904 radic2119890minus1
= radic2119890minus1 1205742 = 025
Abstract and Applied Analysis 17
and 120582 = (1Υ)(Γ(120572+1)(1minus120578 int1
0119892(119904)119889119904)1205742)
119901minus1= 3radic5120587
141198904
By Theorem 11(ii) for any 120582 lt 120582 = 3radic512058714
1198904 the problemhas no positive solution
5 Conclusions
Recently differential equations with p-Laplacian operatorwere widely discussed by several authors In this paperby using fixed point techniques combining with partiallyordered structure of Banach space we obtained the existencemultiply and the dependence of parameter These newresults we presented can be used in numerical computationand analyze mathematical models of physical phenomenamechanics nonlinear dynamics and many other relatedfields
Conflict of Interests
The authors declare that there is no conflict of interestsregarding the publication of this paper
Acknowledgments
The authors are very grateful to anonymous referees for theirvaluable and detailed suggestions and comments to improvethe original paper This work is supported by the Programfor New Century Excellent Talents in University (NCET-10-0097)
References
[1] I Podlubny Fractional Differential Equations vol 198 Aca-demic Press San Diego Calif USA 1999
[2] A A Kilbas H M Srivastava and J J Trujillo Theoryand Applications of Fractional Differential Equations vol 204of North-Holland Mathematics Studies Elsevier Science BVAmsterdam The Netherlands 2006
[3] Y A Rossikhin and M V Shitikova ldquoApplication of fractionalderivatives to the analysis of damped vibrations of viscoelasticsingle mass systemsrdquoActa Mechanica vol 120 no 1ndash4 pp 109ndash125 1997
[4] J Sabatier O Agrawal and J T Machado Advances in Frac-tional Calculus Theoretical developments and applications inphysics and engineering Springer Berlin Germany 2007
[5] N Engheta ldquoOn fractional calculus and fractional multipolesin electromagnetismrdquo IEEE Transactions on Antennas andPropagation vol 44 no 4 pp 554ndash566 1996
[6] R Hilfer Applications of Fractional Calculus in Physics WorldScientific Singapore 2000
[7] R L Bagley and R A Calico ldquoFractional order state equationsfor the control of viscoelastically damped structuresrdquo Journalof Guidance Control and Dynamics vol 14 no 2 pp 304ndash3111991
[8] W-X Zhou andY-D Chu ldquoExistence of solutions for fractionaldifferential equations with multi-point boundary conditionsrdquoCommunications in Nonlinear Science and Numerical Simula-tion vol 17 no 3 pp 1142ndash1148 2012
[9] B Ahmad and G Wang ldquoA study of an impulsive four-pointnonlocal boundary value problemof nonlinear fractional differ-ential equationsrdquo Computers amp Mathematics with Applicationsvol 62 no 3 pp 1341ndash1349 2011
[10] B Ahmad and S Sivasundaram ldquoOn four-point nonlocalboundary value problems of nonlinear integro-differentialequations of fractional orderrdquo Applied Mathematics and Com-putation vol 217 no 2 pp 480ndash487 2010
[11] N Nyamoradi ldquoPositive solutions for multi-point boundaryvalue problems for nonlinear fractional differential equationsrdquoJournal of Contemporary Mathematical Analysis vol 48 no 4pp 145ndash157 2013
[12] B Ahmad and J J Nieto ldquoExistence of solutions for nonlocalboundary value problems of higher-order nonlinear fractionaldifferential equationsrdquo Abstract and Applied Analysis vol 2009Article ID 494720 9 pages 2009
[13] X Liu M Jia and X Xiang ldquoOn the solvability of a fractionaldifferential equationmodel involving the119901-Laplacian operatorrdquoComputers ampMathematics with Applications vol 64 no 10 pp3267ndash3275 2012
[14] J Wang H Xiang and Z Liu ldquoExistence of concave positivesolutions for boundary value problem of nonlinear fractionaldifferential equation with 119901-Laplacian operatorrdquo InternationalJournal of Mathematics and Mathematical Sciences vol 2010Article ID 495138 17 pages 2010
[15] J Wang and H Xiang ldquoUpper and lower solutions method fora class of singular fractional boundary value problems with 119901-Laplacian operatorrdquo Abstract and Applied Analysis vol 2010Article ID 971824 12 pages 2010
[16] H Lu andZHan ldquoExistence of positive solutions for boundary-value problem of fractional differential equation with 119901-laplacian operatorrdquo The American Jouranl of Engineering andTechnology Research vol 11 pp 3757ndash3764 2011
[17] G Chai ldquoPositive solutions for boundary value problem offractional differential equation with 119901-Laplacian operatorrdquoBoundary Value Problems vol 2012 article 18 2012
[18] T Chen W Liu and Z Hu ldquoA boundary value problem forfractional differential equation with 119901-Laplacian operator atresonancerdquoNonlinear AnalysisTheoryMethodsampApplicationsvol 75 no 6 pp 3210ndash3217 2012
[19] T Chen andW Liu ldquoAn anti-periodic boundary value problemfor the fractional differential equation with a 119901-LaplacianoperatorrdquoAppliedMathematics Letters of Rapid Publication vol25 no 11 pp 1671ndash1675 2012
[20] S J Li X G Zhang Y H Wu and L Caccetta ldquoExtremalsolutions for 119901-Laplacian differential systems via iterative com-putationrdquo Applied Mathematics and Computation vol 26 pp1151ndash1158 2013
[21] Z Liu and L Lu ldquoA class of BVPs for nonlinear fractionaldifferential equations with 119901-Laplacian operatorrdquo ElectronicJournal of Qualitative Theory of Differential Equations vol 70pp 1ndash16 2012
[22] Z Liu L Lu and I Szanto ldquoExistence of solutions for fractionalimpulsive differential equations with 119901-Laplacian operatorrdquoActa Mathematica Hungarica vol 141 no 3 pp 203ndash219 2013
[23] Z Bai ldquoEigenvalue intervals for a class of fractional boundaryvalue problemrdquo Computers amp Mathematics with Applicationsvol 64 no 10 pp 3253ndash3257 2012
[24] X Zhang L Wang and Q Sun ldquoExistence of positive solutionsfor a class of nonlinear fractional differential equations withintegral boundary conditions and a parameterrdquo Applied Mathe-matics and Computation vol 226 pp 708ndash718 2014
18 Abstract and Applied Analysis
[25] D J Guo and V Lakshmikantham Nonlinear Problems inAbstract Cones vol 5 Academic Press New York NY USA1988
[26] M A Krasnoselrsquoskii Topological Methods in the Theory ofNonlinear Integral Equations Translated by A H ArmstrongElmsford Pergamon Turkey 1964
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Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
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Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
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Algebra
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Decision SciencesAdvances in
Discrete MathematicsJournal of
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Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Stochastic AnalysisInternational Journal of
Abstract and Applied Analysis 3
119909 (119905) ge 0 min0le119905le1
119909 (119905) ge 120590 119909 (10)
where 2 lt 120572 le 3 0 lt 120578 lt 1 1 minus 120578 int1
0119892(119904)119889119904 gt 0 and 120590 =
120578 int1
0(1 minus 119904)119892(119904)119889119904(1 minus 120578 int
1
0119904119892(119904)119889119904)
Proof It is easy to see by integration of (8) that
120601119901 (119863120572
0+119909 (119905)) minus 120601119901 (119863
120572
0+119909 (0)) = minus120582int
119905
0
119910 (119904) 119889119904 (11)
By the boundary condition 1199091015840(0) = 119909
10158401015840(0) = 0 we can easily
get that1198631205720+
119909(0) = 0 and we obtain that
120601119901 (119863120572
0+119909 (119905)) = minus120582int
119905
0
119910 (119904) 119889119904 (12)
that is
119863120572
0+119909 (119905) = minus120601119902 (120582int
119905
0
119910 (119904) 119889119904) (13)
By Lemmas 3 and 4 we get that
119909 (119905) = minus119868120572120601119902 (120582int
119905
0
119910 (119904) 119889119904) + 1198880 + 1198881119905 + 11988821199052 (14)
Using the boundary condition 1199091015840(0) = 119909
10158401015840(0) = 0 119909(1) =
120578 int1
0119892(119904)119909(119904)119889119904 we get 1198881 = 1198882 = 0 and
1198880 =1
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
(1 minus 119904)120572minus1
120601119902 (120582int
119904
0
119910 (120591) 119889120591) 119889119904
minus120578
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
119892 (119905) (int
119905
0
(119905 minus 119904)120572minus1
120601119902 (120582int
119904
0
119910 (120591) 119889120591) 119889119904) 119889119905
(15)
Thus
119909 (119905) = minus1
Γ (120572)int
119905
0
(119905 minus 119904)120572minus1
120601119902 (120582int
119904
0
119910 (120591) 119889120591) 119889119904
+1
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
(1 minus 119904)120572minus1
120601119902 (120582int
119904
0
119910 (120591) 119889120591) 119889119904
minus120578
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
119892 (119905) (int
119905
0
(119905 minus 119904)120572minus1
120601119902 (120582int
119904
0
119910 (120591) 119889120591) 119889119904) 119889119905
=1
Γ (120572)int
1
0
(1 minus 119904)120572minus1
120601119902 (120582int
119904
0
119910 (120591) 119889120591) 119889119904
minusint
119905
0
(119905 minus 119904)120572minus1
120601119902 (120582int
119904
0
119910 (120591) 119889120591) 119889119904
+120578
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
119892 (119904) (int
1
0
(1 minus 120591)120572minus1
120601119902 (120582int
120591
0
119910 (120583) 119889120583)119889120591
minus int
119904
0
(119904 minus 120591)120572minus1
times 120601119902 (120582int
120591
0
119910 (120583) 119889120583)119889120591)119889119904
(16)
Direct differentiation of (9) implies
1199091015840(119905)
= minus1
Γ (120572 minus 1)int
119905
0
(119905 minus 119904)120572minus2
120601119902 (120582int
119904
0
119910 (120591) 119889120591) 119889119904 le 0
(17)
By differentiation of (17) we get
11990910158401015840(119905)
= minus1
Γ (120572 minus 2)int
119905
0
(119905 minus 119904)120572minus3
120601119902 (120582int
119904
0
119910 (120591) 119889120591) 119889119904 le 0
(18)
Thus the solution of problem (8) is nonincreasing andconcave on [0 1] and
119909 = 119909 (0) min0le119905le1
119909 (119905) = 119909 (1) (19)
On the other hand as we assume that 119910(119905) ge 0 we see that
119909 (1) =120578
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
119892 (119904) (int
1
0
(1 minus 120591)120572minus1
120601119902 (120582int
120591
0
119910 (120583) 119889120583)119889120591
minus int
119904
0
(119904 minus 120591)120572minus1
times 120601119902 (120582int
120591
0
119910 (120583) 119889120583)119889120591)119889119904
=120578
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
119892 (119904) (int
119904
0
[(1 minus 120591)120572minus1
minus (119904 minus 120591)120572minus1
]
4 Abstract and Applied Analysis
times 120601119902 (120582int
120591
0
119910 (120583) 119889120583)119889120591
+int
1
119904
(1 minus 120591)120572minus1
120601119902 (120582int
120591
0
119910 (120583) 119889120583)119889120591)119889119904
ge 0
(20)
Therefore 119909(119905) ge 0 and is concave for 119905 isin [0 1] So for every119905 isin [0 1]
119909 (119905) ge 119905119909 (1) + (1 minus 119905) 119909 (0) (21)
Therefore
120578int
1
0
119909 (119905) 119892 (119905) 119889119905 ge 120578119909 (1) int
1
0
119905119892 (119905) 119889119905
+ 120578119909 (0) int
1
0
(1 minus 119905) 119892 (119905) 119889119905
(22)
Since 119909(1) = 120578 int1
0119909(119905)119892(119905)119889119905 1 minus 120578 int
1
0119892(119905)119889119905 gt 0 we have
119909 (1) ge120578 int1
0(1 minus 119905) 119892 (119905) 119889119905
1 minus 120578 int1
0119905119892 (119905) 119889119905
119909 (0) (23)
that is
min0le119905le1
119909 (119905) = 119909 (1) ge120578 int1
0(1 minus 119905) 119892 (119905) 119889119905
1 minus 120578 int1
0119905119892 (119905) 119889119905
119909 (0) = 120590 119909 (24)
The Lemma is proved
We construct a cone 119875 in 119864 by
119875 = 119909 isin 119864 119909 ge 0 min0le119905le1
119909 (119905) ge 120590 119909 (25)
where 120590 is defined in Lemma 7 It is easy to see 119875 is a closedconvex cone of 119864
Define 119879 119875 rarr 119864 by
(119879119909) (119905)
=1
Γ (120572)int
1
0
(1 minus 119904)120572minus1
120601119902 (int
119904
0
120596 (120591) 119891 (120591 119909 (120591)) 119889120591) 119889119904
minusint
119905
0
(119905 minus 119904)120572minus1
120601119902 (int
119904
0
120596 (120591) 119891 (120591 119909 (120591)) 119889120591) 119889119904
+120578
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
119892 (119904)
times (int
1
0
(1 minus 120591)120572minus1
120601119902 (int
120591
0
120596 (120583) 119891 (120583 119909 (120583)) 119889120583)119889120591
minus int
119904
0
(119904 minus 120591)120572minus1
times 120601119902 (int
120591
0
120596 (120583) 119891 (120583 119909 (120583)) 119889120583)119889120591)119889119904
(26)
It is clear that the fixed points of the operator 120601119902(120582)119879 arethe solutions of the boundary value problems (1)
We make the following hypotheses
(H1) 119891 [0 1] times [0infin) rarr [0infin) is continuous(H2) 120596 (0 1) rarr [0infin) is continuous and not identical
zero on any closed subinterval of (0 1) with 0 lt
int1
0120596(119905)119889119905 lt +infin
(H3) 1 minus 120578 int1
0119892(119904)119889119904 gt 0
(H4) 1198910
120601isin [0infin] where 119891
0
120601= lim119909rarr0(119891(119905 119909)120601119901(119909))
uniformly for 119905 isin [0 1](H5) 119891
infin
120601isin [0infin] where 119891
infin
120601= lim119909rarrinfin(119891(119905 119909)120601119901(119909))
uniformly for 119905 isin [0 1](H6) 119891(119905 119909) gt 0 for any 119905 isin [0 1] and 119909 gt 0
Lemma 8 Assume that (H1)ndash(H3) hold Then 119879 119875 rarr 119864 iscompletely continuous
Proof First we show that 119879 is continuous It is easy to see119879(119875) sub 119875 For 119909119899(119905) isin 119864 and 119909119899(119905) rarr 119909(119905) as 119899 rarr infin bythe continuous of 119891(119905 119909(119905)) we get 119891(119905 119909119899(119905)) rarr 119891(119905 119909(119905))as 119899 rarr infin This implies that
120601119902 (int
119905
0
119908 (119904) 119891 (119904 119909119899 (119904)) 119889119904)
997888rarr 120601119902 (int
119905
0
119908 (119904) 119891 (119904 119909 (119904)) 119889119904)
(27)
So we have
sup119905isin[01]
10038161003816100381610038161003816100381610038161003816120601119902 (int
119905
0
119908 (119904) 119891 (119904 119909119899 (119904)) 119889119904)
minus 120601119902 (int
119905
0
119908 (119904) 119891 (119904 119909 (119904)) 119889119904)
10038161003816100381610038161003816100381610038161003816997888rarr 0 119899 997888rarr infin
(28)
Denote Π = |(119879119909119899)(119905) minus (119879119909)(119905)| then
Π le1
Γ (120572)int
1
0
(1 minus 119904)120572minus1
10038161003816100381610038161003816100381610038161003816120601119902 (int
119904
0
120596 (120591) 119891 (120591 119909119899 (120591)) 119889120591)
minus 120601119902 (int
119904
0
120596 (120591) 119891 (120591 119909 (120591)) 119889120591)
10038161003816100381610038161003816100381610038161003816119889119904
Abstract and Applied Analysis 5
+int
119905
0
(119905 minus 119904)120572minus1
10038161003816100381610038161003816100381610038161003816120601119902 (int
119904
0
120596 (120591) 119891 (120591 119909119899 (120591)) 119889120591)minus120601119902
times (int
119904
0
120596 (120591) 119891 (120591 119909 (120591)) 119889120591)
10038161003816100381610038161003816100381610038161003816119889119904
+120578
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
119892 (119904)
times (int
1
0
(1 minus 120591)120572minus1
times
10038161003816100381610038161003816100381610038161003816120601119902 (int
120591
0
120596 (120583) 119891 (120583 119909119899 (120583)) 119889120583)
minus 120601119902 (int
120591
0
120596 (120583) 119891 (120583 119909 (120583)) 119889120583)
10038161003816100381610038161003816100381610038161003816119889120591
+ int
119904
0
(119904 minus 120591)120572minus1
times
10038161003816100381610038161003816100381610038161003816120601119902 (int
120591
0
120596 (120583) 119891 (120583 119909119899 (120583)) 119889120583) minus 120601119902
times (int
120591
0
120596 (120583) 119891 (120583 119909 (120583)) 119889120583)
10038161003816100381610038161003816100381610038161003816119889120591) 119889119904
le sup119904isin[01]
10038161003816100381610038161003816100381610038161003816120601119902 (int
119904
0
119908 (120591) 119891 (120591 119909119899 (120591)) 119889120591)
minus 120601119902 (int
119904
0
119908 (120591) 119891 (120591 119909 (120591)) 119889120591)
10038161003816100381610038161003816100381610038161003816
times2
Γ (120572 + 1)(1 +
120578 int1
0119892 (119904) 119889119904
1 minus 120578 int1
0119892 (119904) 119889119904
) 997888rarr 0
119899 997888rarr infin
(29)
Hence 119879 is continuousSecond we show that 119879 is compactFor 119909(119905) isin 119861119903 = 119909 isin 119875 119909 le 119903 by condition (H1)
119891(119905 119909(119905)) lt infin Denote119872 = max0le119905le1119909le119903119891(119905 119909(119905))Then
|119879119909 (119905)|
le1
Γ (120572)int
1
0
(1 minus 119904)120572minus1
10038161003816100381610038161003816100381610038161003816120601119902 (int
119904
0
120596 (120591) 119891 (120591 119909 (120591)) 119889120591)
10038161003816100381610038161003816100381610038161003816119889119904
+int
119905
0
(119905 minus 119904)120572minus1
10038161003816100381610038161003816100381610038161003816120601119902 (int
119904
0
120596 (120591) 119891 (120591 119909 (120591)) 119889120591)
10038161003816100381610038161003816100381610038161003816119889119904
+120578
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
119892 (119904)
times (int
1
0
(1 minus 120591)120572minus1
times
10038161003816100381610038161003816100381610038161003816120601119902 (int
120591
0
120596 (120583) 119891 (120583 119909 (120583)) 119889120583)
10038161003816100381610038161003816100381610038161003816119889120591
+ int
119904
0
(119904 minus 120591)120572minus1
times
10038161003816100381610038161003816100381610038161003816120601119902 (int
120591
0
120596 (120583) 119891 (120583 119909 (120583)) 119889120583)
10038161003816100381610038161003816100381610038161003816119889120591) 119889119904
le1
Γ (120572)
times int
1
0
(1 minus 119904)120572minus1
10038161003816100381610038161003816100381610038161003816120601119902 (119872int
119904
0
120596 (120591) 119889120591)
10038161003816100381610038161003816100381610038161003816119889119904
+int
119905
0
(119905 minus 119904)120572minus1
10038161003816100381610038161003816100381610038161003816120601119902 (119872int
119904
0
120596 (120591) 119889120591)
10038161003816100381610038161003816100381610038161003816119889119904
+120578
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
119892 (119904) (int
1
0
(1 minus 120591)120572minus1
times
10038161003816100381610038161003816100381610038161003816120601119902 (119872int
120591
0
120596 (120583) 119889120583)119889120591
10038161003816100381610038161003816100381610038161003816
+ int
119904
0
(119904 minus 120591)120572minus1
times
10038161003816100381610038161003816100381610038161003816120601119902 (119872int
120591
0
120596 (120583) 119889120583)
10038161003816100381610038161003816100381610038161003816119889120591) 119889119904
le120578
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
120601119902 (119872int
1
0
120596 (120591) 119889120591)
times int
1
0
119892 (119904) (int
1
0
(1 minus 120591)120572minus1
119889120591 + int
119904
0
(119904 minus 120591)120572minus1
119889120591)119889119904
+1
Γ (120572)120601119902 (119872int
1
0
120596 (120591) 119889120591)
times int
1
0
(1 minus 119904)120572minus1
119889119904 + int
119905
0
(119905 minus 119904)120572minus1
119889119904
le
2120601119902 (119872int1
0120596 (120591) 119889120591)
Γ (120572 + 1) (1 minus 120578 int1
0119892 (120591) 119889120591)
lt infin
(30)
that is 119879 maps bounded sets into bounded sets in 119875For 119909(119905) isin 119861119903 0 lt 1199051 lt 1199052 lt 1
1003816100381610038161003816119879119909 (1199052) minus 119879119909 (1199051)1003816100381610038161003816
=1
Γ (120572)
10038161003816100381610038161003816100381610038161003816int
1199052
0
(1199052 minus 119904)120572minus1
120601119902 (int
119904
0
120596 (120591) 119891 (120591 119909 (120591)) 119889120591) 119889119904
6 Abstract and Applied Analysis
minus int
1199051
0
(1199051 minus 119904)120572minus1
times 120601119902 (int
119904
0
120596 (120591) 119891 (120591 119909 (120591)) 119889120591) 119889119904
10038161003816100381610038161003816100381610038161003816
le1
Γ (120572)120601119902 (119872int
1
0
120596 (120591) 119889120591)
times
100381610038161003816100381610038161003816100381610038161003816
int
1199051
0
((1199052 minus 119904)120572minus1
minus (1199051 minus 119904)120572minus1
) 119889119904
+int
1199052
1199051
(1199052 minus 119904)120572minus1
119889119904
100381610038161003816100381610038161003816100381610038161003816
(31)
By mean value theorem we obtain that
int
1199051
0
((1199052 minus 119904)120572minus1
minus (1199051 minus 119904)120572minus1
) 119889119904
le int
1199051
0
(120572 minus 1) (1 minus 119904)120572minus2
(1199052 minus 1199051) 119889119904
le int
1
0
(120572 minus 1) (1 minus 119904)120572minus2
(1199052 minus 1199051) 119889119904
= 1199052 minus 1199051
(32)
Thus1003816100381610038161003816119879119909 (1199052) minus 119879119909 (1199051)
1003816100381610038161003816
le1
Γ (120572)120601119902 (119872int
1
0
120596 (120591) 119889120591) [(1199052 minus 1199051) +1
120572(1199052 minus 1199051)
120572]
(33)
This shows that |119879119909(1199052) minus 119879119909(1199051)| rarr 0 as 1199052 minus 1199051 rarr 0so 119879119909 119909 isin 119861119903 is equicontinuous Therefore the operator119879 119875 rarr 119875 is completely continuous by the Arzela-Ascolitheorem
Now for convenience we introduce the following nota-tions Let
int
1
0
(1 minus 119904)120572minus1
120601119902 (int
119904
0
120596 (120591) 119889120591) 119889119904 = 1205741
120601119902 (int
1
0
120596 (120591) 119889120591) = 1205742
119898 (119903) = minmin119905
119891 (119905 119909)
120601119901 (119903) 119909 isin [120590119903 119903]
(34)
Theorem 9 Assume that (H1)ndash(H5) hold
(i) If 0 lt 1198910
120601lt +infin then there exists 1205850 gt 0 such that for
every 0 lt 119903 lt 1205850 problem (1) has a positive solution119909119903(119905) satisfying 119909119903 = 119903 associated with
120582 = 120582119903 isin [1205820 1205820] (35)
where 1205820 and 1205820 are two positive finite numbers
(ii) If 0 lt 119891infin
120601lt +infin then there exists 120585lowast
0gt 0 such that for
every 119877 gt 120585lowast
0 problem (1) has a positive solution 119909119877(119905)
satisfying 119909119877 = 119877 for any
120582 = 120582119877 isin [120582lowast
0 120582lowast
0] (36)
where 120582lowast
0and 120582
lowast
0are two positive finite numbers
(iii) If 1198910120601
= +infin then there exists 1205851 gt 0 such that for any0 lt 119903lowast
lt 1205851 problem (1) has a positive solution 119909119903lowast(119905)
satisfying 119909119903lowast = 119903lowast for any
120582 = 120582119903lowast isin (0 120582lowast] (37)
where 120582lowast is a positive finite number
(iv) If 119891infin
120601= +infin then there exists 1205851 gt 0 such that
for every 119877lowast gt 1205851 problem (1) has a positive solution119909119877lowast
(119905) satisfying 119909119877lowast
= 119877lowast for any
120582 = 120582119877lowast
isin (0 120582lowast] (38)
where 120582lowast is a positive finite number(v) If there exists 120585 gt 0 such that 119898(119903) gt 120585 then problem
(1) has positive solution 119909119903(119905) satisfying 119909119903 = 119903 forany
120582 = 120582119903 isin (0 ] (39)
where is a positive finite number
Proof (i) It follows from 0 lt 1198910
120601lt +infin that there exists
1198971 1198972 120588 gt 0 such that
1198971120601119901 (119909) le 119891 (119905 119909) le 1198972120601119901 (119909)
(forall119905 isin [0 1] 0 le 119909 le 120588)
(40)
Let 1205850 = 120588120590 we show that 1205850 is required When 119909 isin 119875 cap
120597Ω119903 we have
119909 (119905) ge 120590 119909 = 120590119903 (41)
we need 119909(119905) lt 120588 this implies that 120590119903 lt 120588 as 119903 lt 1205850 so120590119903 lt 1205901205850 = 120588
Let 1205820 = (11198971)(Γ(120572)1205901205741)119901minus1 Then we may assume that
119909 minus 120601119902 (1205820) 119879119909 = 0 (forall119909 isin 119875 cap 120597Ω119903) (42)
if not then there exists119909119903 isin 119875cap120597Ω119903 such that120601119902(1205820)119879119909119903 = 119909119903and then (35) already holds for 120582119903 = 1205820
Define 120595(119905) equiv 1 forall119905 isin [0 1] then 120595(119905) isin 119875 and 120595 equiv 1We now show that
119909 minus 120601119902 (1205820) 119879119909 = 120581120595 (forall119909 isin 119875 cap 120597Ω119903 120581 ge 0) (43)
In fact if there exists 1199091 isin 119875 cap 120597Ω119903 1205811 ge 0 such that 1199091 minus120601119902(1205820)1198791199091 = 1205811120595 then (42) implies that 1205811 gt 0 On the otherhand 1199091 = 120601119902(1205820)1198791199091 + 1205811120595 ge 1205811120595 we may choose 120581
lowast=
max120581 | 1199091 ge 120581120595 then 1205811 le 120581lowastlt +infin 1199091 ge 120581
lowast120595 Therefore
120581lowast= 120581lowast 1003817100381710038171003817120595
1003817100381710038171003817 le10038171003817100381710038171199091
1003817100381710038171003817 = 119903 (44)
Abstract and Applied Analysis 7
Consequently for any 119905 isin [0 1] (26) and (44) imply that
1199091 (119905)
= 120601119902 (1205820)1
Γ (120572)
times int
1
0
(1 minus 119904)120572minus1
times 120601119902 (int
119904
0
120596 (120591) 119891 (120591 1199091 (120591)) 119889120591) 119889119904
minus int
119905
0
(119905 minus 119904)120572minus1
times 120601119902 (int
119904
0
120596 (120591) 119891 (120591 1199091 (120591)) 119889120591) 119889119904
+120578120601119902 (1205820)
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
119892 (119904)
times (int
1
0
(1 minus 120591)120572minus1
times 120601119902 (int
120591
0
120596 (120583) 119891 (120583 1199091 (120583)) 119889120583)119889120591
minus int
119904
0
(119904 minus 120591)120572minus1
times 120601119902 (int
120591
0
120596 (120583) 119891 (120583 1199091 (120583)) 119889120583)119889120591)119889119904
+ 1205811120595
ge120578120590120601119902 (1205820)
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
119892 (119904)
times (int
1
0
(1 minus 120591)120572minus1
times 120601119902 (int
120591
0
120596 (120583) 119891 (120583 1199091 (120583)) 119889120583)119889120591
minus int
119904
0
(119904 minus 120591)120572minus1
120601119902
times (int
120591
0
120596 (120583) 119891 (120583 1199091 (120583)) 119889120583)119889120591)119889119904
+120590120601119902 (1205820)
Γ (120572)
times int
1
0
(1 minus 119904)120572minus1
120601119902 (int
119904
0
120596 (120591) 119891 (120591 1199091 (120591)) 119889120591) 119889119904
+ 1205811120595
ge120590120601119902 (1205820)
Γ (120572)
times int
1
0
(1 minus 119904)120572minus1
120601119902 (int
119904
0
120596 (120591) 119891 (120591 1199091 (120591)) 119889120591) 119889119904
+ 1205811120595
ge120590120601119902 (1205820)
Γ (120572)
times int
1
0
(1 minus 119904)120572minus1
120601119902 (1198971 int
119904
0
120596 (120591) 120601119901 (1199091 (120591)) 119889120591) 119889119904
+ 1205811120595
ge120590120601119902 (1205820)
Γ (120572)
times int
1
0
(1 minus 119904)120572minus1
120601119902 (1198971 int
119904
0
120596 (120591) 120601119901 (120581lowast120595 (120591)) 119889120591) 119889119904
+ 1205811120595
=120590120581lowast119897119902minus1
1120582119902minus1
0
Γ (120572)int
1
0
(1 minus 119904)120572minus1
120601119902 (int
119904
0
120596 (120591) 119889120591) 119889119904 + 1205811120595
=120590120581lowast119897119902minus1
1120582119902minus1
01205741
Γ (120572)+ 1205811120595
= 120581lowast+ 120581120595
(45)
which implies that 1199091(119905) ge (120581lowast+ 120581)120595(119905) 119905 isin [0 1] which is a
contradiction to the definition of 120581lowast Thus (42) holds and byLemma 5 the fixed point index
119894 (120601119902 (1205820) 119879 119875 cap 120597Ω119903 119875) = 0 (46)
On the other hand by the fact that the fixed point index ofconstants operator is 1 so
119894 (120599 119875 cap 120597Ω119903 119875) = 1 (47)
where 120599 is the zero operator It follows therefore from (46)and (47) and the homotopy invariance property that thereexist 119909119903 isin 119875 cap 120597Ω119903 and 0 lt ]0 lt 1 such that ]0120601119902(1205820)119879119909119903 =119909119903 but 119909119903 = 120601119902(120582119903)119879119909119903 which implies that ]0120601119902(1205820)119879119909119903 =
120601119902(120582119903)119879119909119903 it follows that we get ]0120601119902(1205820) = 120601119902(120582119903)) that is
]0120582119902minus1
0= 120582119902minus1
119903 (48)
and then
120582119903 = ]1(119902minus1)0
1205820 = ]119901minus10
1205820 lt 1205820 (49)
From the proof above for any 119903 lt 1205850 there exists a positivesolution 119909119903 isin 119875 cap 120597Ω119903 associated with 120582 = 120582119903 gt 0
8 Abstract and Applied Analysis
Thus
119909119903 (119905)
=120601119902 (120582119903)
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902 (int
119904
0
120596 (120591) 119891 (120591 119909119903 (120591)) 119889120591) 119889119904
minus int
119905
0
(119905 minus 119904)120572minus1
times 120601119902 (int
119904
0
120596 (120591) 119891 (120591 119909119903 (120591)) 119889120591) 119889119904
+120578120601119902 (120582119903)
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
119892 (119904)
times (int
1
0
(1 minus 120591)120572minus1
times 120601119902 (int
120591
0
120596 (120583) 119891 (120583 119909119903 (120583)) 119889120583)119889120591
minus int
119904
0
(119904 minus 120591)120572minus1
times 120601119902 (int
120591
0
120596 (120583) 119891 (120583 119909119903 (120583)) 119889120583)119889120591)119889119904
(50)
with 119909119903 = 119903Next we show that 120582119903 ge 1205820 In fact
119909119903 (119905)
=120601119902 (120582119903)
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902 (int
119904
0
120596 (120591) 119891 (120591 119909119903 (120591)) 119889120591) 119889119904
minus int
119905
0
(119905 minus 119904)120572minus1
times 120601119902 (int
119904
0
120596 (120591) 119891 (120591 119909119903 (120591)) 119889120591) 119889119904
+120578120601119902 (120582119903)
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
119892 (119904)
times (int
1
0
(1 minus 120591)120572minus1
times 120601119902 (int
120591
0
120596 (120583) 119891 (120583 119909119903 (120583)) 119889120583)119889120591
minus int
119904
0
(119904 minus 120591)120572minus1
times 120601119902 (int
120591
0
120596 (120583) 119891 (120583 119909119903 (120583)) 119889120583)119889120591)119889119904
le120578120601119902 (120582119903)
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
119892 (119904) int
1
0
(1 minus 120591)120572minus1
times 120601119902 (int
120591
0
120596 (120583) 119891 (120583 119909119903 (120583)) 119889120583)119889120591 119889119904
+120601119902 (120582119903)
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902 (int
119904
0
120596 (120591) 119891 (120591 119909119903 (120591)) 119889120591) 119889119904
=120601119902 (120582119903) 120578 int
1
0119892 (119904) 119889119904
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
(1 minus 119904)120572minus1
times 120601119902 (int
119904
0
120596 (120591) 119891 (120591 119909119903 (120591)) 119889120591) 119889119904
+120601119902 (120582119903)
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902 (int
119904
0
120596 (120591) 119891 (120591 119909119903 (120591)) 119889120591) 119889119904
le120601119902 (120582119903) 120578 int
1
0119892 (119904) 119889119904
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
(1 minus 119904)120572minus1
times 120601119902 (1198972 int
119904
0
120596 (120591) 119909119903 (120591) 119889120591) 119889119904
+120601119902 (120582119903)
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902 (1198972120582119903 int
119904
0
120596 (120591) 119909119903 (120591) 119889120591) 119889119904
=120601119902 (120582119903)
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
(1 minus 119904)120572minus1
times 120601119902 (1198972 int
119904
0
120596 (120591) 119909119903 (120591) 119889120591) 119889119904
Abstract and Applied Analysis 9
le
10038171003817100381710038171199091199031003817100381710038171003817 120601119902 (1205821199031198972)
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
(1 minus 119904)120572minus1
times 120601119902 (int
1
0
120596 (120591) 119889120591) 119889119904
=
10038171003817100381710038171199091199031003817100381710038171003817 120601119902 (1205821199031198972) 1205742
Γ (120572 + 1) (1 minus 120578 int1
0119892 (119904) 119889119904)
(51)
which implies that
10038171003817100381710038171199091199031003817100381710038171003817 le
10038171003817100381710038171199091199031003817100381710038171003817 120601119902 (1205821199031198972) 1205742
Γ (120572 + 1) (1 minus 120578 int1
0119892 (119904) 119889119904)
(52)
that is
120601119902 (1205821199031198972) 1205742
Γ (120572 + 1) (1 minus 120578 int1
0119892 (119904) 119889119904)
ge 1 (53)
it follows that we get
120582119903 ge1
1198972(
Γ (120572 + 1) (1 minus 120578 int1
0119892 (119904) 119889119904)
1205742)
119901minus1
= 1205820(54)
In conclusion 120582119903 isin [1205820 1205820](ii) It follows from 0 lt 119891
infin
120601lt infin that there exist
119897lowast
1 119897lowast
2 120588lowastgt 0 such that
119897lowast
1120601119901 (119909) le 119891 (119905 119909) le 119897
lowast
2120601119901 (119909)
(forall119905 isin [0 1] 119909 ge 120588lowast)
(55)
Let 120585lowast0= 120588lowast120590 the following proof are similar to (i)
(iii) It follows from 1198910
120601= +infin there exists 119897lowast gt 0 120588
lowastgt 0
such that
1198910
120601ge 119897lowast120601119901 (119909) (forall119905 isin [0 1] 0 le 119909 le 120588
lowast) (56)
Let 1205851 = 120588lowast120590 we show that 1205851 is required When 119909 isin 119875 cap
120597Ω119903lowast we have 119909(119905) ge 120590119909 = 120590119903lowast since 119909 le 120588
lowast then we need120590119903lowastlt 120588lowast as 119903lowast lt 1205851 so 120590119903
lowastlt 120588lowast holds
Let 120582lowast = (1119897lowast)(Γ(120572)1205901205741)
1(119902minus1) we proceed in the sameway as in the proof of (i) replacing (42) we may assume that119909 minus 120601119902(120582
lowast)119879119909 = 0 (forall119909 isin 119875 cap 120597Ω119903lowast) and replacing (43) we
can prove 119909 minus 120601119902(120582lowast)119879119909 = 120581120595 (forall119909 isin 119875 cap 120597Ω119903lowast 120581 ge 0) It
follows from Lemma 5 that 119894(120601119902(120582lowast)119879 119875 cap Ω119903lowast 119875) = 0 Note
that 119894(120579 119875 cap Ω119903lowast 119875) = 1 we can easily show that there exists119909119903lowast isin 119875cap120597Ω119903lowast and 0 lt ]119903lowast lt 1 such that ]119903lowast120601119902(120582
lowast)119879119909119903lowast = 119909119903lowast
Hence (37) holds for 120582119903lowast = 120582lowast]1minus119901119903lowast
lt 120582lowast
The proof ofTheorem 9(iv) follows by themethod similarto Theorem 9(iii) we omit it here
(v) It follows that 119898(119903) gt 120585 for any 119909 isin 119875 cap 120597Ω119903 we have120590119903 le 119909 le 119903 and 119891(119905 119909)120601119901(119909) ge 119891(119905 119909)120601119901(119903) ge 119898(119903) ge 120585then
119891 (119905 119909) ge 120585120601119901 (119909) forall119909 isin [120590119903 119903] 119905 isin [0 1] (57)
Let = (1120585)(Γ(120572)1205901205741)119901minus1 The following proof is similar to
that of (iv) This finished the proof of (v)
Remark 10 InTheorem 9 all the criteria obtained depend onone of 1198910
120601and 119891
infin
120601
Let Ψ(119909) = 120601119902(120582)119879119909 the following theorems give out themultiply nonexistence and the dependence of parameter
Theorem 11 Assume that (H1)ndash(H5) hold
(i) If 1198910
120601= 0 and 119891
infin
120601= 0 lim
119909rarrinfin119891(119905 119909) = 119891infin isin
(0 +infin] uniformly for 119905 isin [0 1] then there exists120582lowast
gt 0 such that the problem (1) has two solutions forany 120582 gt 120582
lowast
(ii) If 1198910120601
= 0 and 119891infin
120601= 0 then there exists 120582 gt 0 such
that for any 120582 lt 120582 problem (1) has no solution
Proof (i) It follows from 119891infin isin (0 +infin] that for any 120577 gt 0there exists 120588 gt 0 such that
119891 (119905 119909) gt 120577 (forall119905 isin [0 1] 119909 ge 120588) (58)
Since 1198910
120601= 0 there exists 1205761 gt 0 120588
1gt 0 such that
119891 (119905 119909) le 1205761120601119901 (119909) (forall119905 isin [0 1] 0 le 119909 le 1205881) (59)
where 1205761 satisfied 120601119902(1205821205761)1205742Γ(120572 + 1)(1 minus 120578 int1
0119892(119904)119889119904) le 1
For forall119909 isin 119875 cap 120597Ω1205881
we have
Ψ (119909) = (120601119902 (120582) 119879119909) (119905)
=1
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902 (120582int
119904
0
120596 (120591) 119891 (120591 119909 (120591)) 119889120591) 119889119904
minus int
119905
0
(119905 minus 119904)120572minus1
times 120601119902 (120582int
119904
0
120596 (120591) 119891 (120591 119909 (120591)) 119889120591) 119889119904
+120578
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
119892 (119904)
10 Abstract and Applied Analysis
times (int
1
0
(1 minus 120591)120572minus1
times 120601119902 (120582int
120591
0
120596 (120583) 119891 (120583 119909 (120583)) 119889120583)119889120591
minus int
119904
0
(119904 minus 120591)120572minus1
times 120601119902 (120582int
120591
0
120596 (120583) 119891 (120583 119909 (120583)) 119889120583)119889120591)119889119904
le120578
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
119892 (119904)
times (int
1
0
(1 minus 120591)120572minus1
times 120601119902(120582int
120591
0
120596 (120583) 119891 (120583 119909 (120583)) 119889120583)119889120591)119889119904
+1
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902 (120582int
119904
0
120596 (120591) 1205761119891 (120591 119909 (120591)) 119889120591) 119889119904
le120578 int1
0119892 (119904) 119889119904
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
(1 minus 120591)120572minus1
times 120601119902 (120582int
1
0
120596 (120583) 1205761120601119901 (119909 (120583)) 119889120583)119889120591
+1
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902(120582int
1
0
120596 (120591) 1205761120601119901 (119909 (120591)) 119889120591) 119889119904
le1
Γ (120572 + 1) (1 minus 120578 int1
0119892 (119904) 119889119904)
times 120601119902 (120582int
1
0
120596 (120591) 1205761120601119901 (119909) 119889120591)
=120601119902 (1205821205761) 1205742
Γ (120572 + 1) (1 minus 120578 int1
0119892 (119904) 119889119904)
119909 le 119909
(60)
This implied
Ψ119909 le 119909 forall119909 isin 119875 cap 120597Ω1205881
(61)
Let 1205883gt max120588120590 120588
1 and 120582
lowast= (1120577)(120588
3Γ(120572)1205901205741)
119901minus1 thenforall119909 isin 119875 cap 120597Ω120588
3
we have 120590119909 le 119909 le 1205883and
(Ψ119909) (119905)
=1
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902 (120582int
119904
0
120596 (120591) 119891 (120591 119909 (120591)) 119889120591) 119889119904
minus int
119905
0
(119905 minus 119904)120572minus1
times 120601119902 (120582int
119904
0
120596 (120591) 119891 (120591 119909 (120591)) 119889120591) 119889119904
+120578
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
119892 (119904)
times (int
1
0
(1 minus 120591)120572minus1
times 120601119902 (120582int
120591
0
120596 (120583) 119891 (120583 119909 (120583)) 119889120583)119889120591
minus int
119904
0
(119904 minus 120591)120572minus1
times 120601119902(120582int
120591
0
120596 (120583) 119891 (120583 119909 (120583)) 119889120583)119889120591)119889119904
ge120590120578
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
119892 (119904)
times (int
1
0
(1 minus 120591)120572minus1
times 120601119902 (120582int
120591
0
120596 (120583) 119891 (120583 119909 (120583)) 119889120583)119889120591
minus int
119904
0
(119904 minus 120591)120572minus1
times 120601119902(120582int
120591
0
120596 (120583) 119891 (120583 119909 (120583)) 119889120583)119889120591)119889119904
+120590
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902 (120582int
119904
0
120596 (120591) 119891 (120591 119909 (120591)) 119889120591) 119889119904
ge120590
Γ (120572)int
1
0
(1 minus 119904)120572minus1
Abstract and Applied Analysis 11
times 120601119902 (120582int
119904
0
120596 (120591) 119891 (120591 119909 (120591)) 119889120591) 119889119904
gt120590
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902 (120582int
119904
0
120596 (120591) 120577 119889120591) 119889119904
gt120590120601119902 (120582120577)
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902 (int
119904
0
120596 (120591) 119889120591) 119889119904
=120590120601119902 (120582120577) 1205741
Γ (120572)= 1205883
(62)
which implies
Ψ119909 gt 119909 forall119909 isin 119875 cap 120597Ω1205883
(63)
Next for 119891infin
120601= 0 there exist 1205762 gt 0 120588
2gt 1205883gt 0 such
that
119891 (119905 119909) le 1205762120601119901 (119909) (forall119905 isin [0 1] 119909 ge 1205882) (64)
where 1205762 satisfied 120601119902(1205821205762)1205742Γ(120572 + 1)(1 minus 120578 int1
0119892(119904)119889119904) le 1
Similar to the above proof we get
Ψ119909 le 119909 forall119909 isin 119875 cap 120597Ω1205882
(65)
Applying Lemma 7 to (61) (65) and (63) yields that Ψ
has two fixed points 1199091 1199092 such that 1199091 isin 119875cap (Ω1205883
Ω1205881
) and1199092 isin 119875 cap (Ω120588
2
Ω1205883
)(ii) It follows from 119891
0
120601= 0 and 119891
infin
120601= 0 that there exists
1205761 1205762 1205881 1205882 gt 0 such that
119891 (119905 119909) le 1205761120601119901 (119909) (forall119905 isin [0 1] 0 le 119909 le 1205881)
119891 (119905 119909) le 1205762120601119901 (119909) (forall119905 isin [0 1] 119909 ge 1205882)
(66)
then for 119909 isin [1205881 1205882] by the continuous of 119891(119905 119909(119905)) there
exists 119909 isin [1205881 1205882] such that
max1205881le119909le120588
2
119891 (119905 119909)
120601119901 (119909)=
119891 (119905 119909)
120601119901 (119909) (67)
Let Υ = max1205761 1205762 119891(119905 119909)120601119901(119909) gt 0 then
119891 (119905 119909) le Υ120601119901 (119909) (forall0 le 119905 le 1 119909 ge 0) (68)
Assuming 119909(119905) is a positive solution of problem (1) we willshow that this leads to a contradiction for 120582 lt 120582 where
120582 = (1Υ)(Γ(120572 + 1)(1 minus 120578 int1
0119892(119904)119889119904)1205742)
119901minus1 It follows from(26) that
119909 = 119909 (0)
=120578
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
119892 (119904)
times (int
1
0
(1 minus 120591)120572minus1
times 120601119902 (120582int
120591
0
120596 (120583) 119891 (120583 119909 (120583)) 119889120583)119889120591
minus int
119904
0
(119904 minus 120591)120572minus1
times 120601119902(120582int
120591
0
120596 (120583) 119891 (120583 119909 (120583)) 119889120583)119889120591)119889119904
+1
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902 (120582int
119904
0
120596 (120591) 119891 (120591 119909 (120591)) 119889120591) 119889119904
le120578
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
119892 (119904)
times int
1
0
(1 minus 120591)120572minus1
times 120601119902(120582int
120591
0
120596 (120583) 119891 (120583 119909 (120583)) 119889120583)119889120591 119889119904
+1
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902 (120582int
119904
0
120596 (120591) 119891 (120591 119909 (120591)) 119889120591) 119889119904
=1
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
(1 minus 119904)120572minus1
times 120601119902 (120582int
119904
0
120596 (120591) 119891 (120591 119909 (120591)) 119889120591) 119889119904
le1
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
(1 minus 119904)120572minus1
12 Abstract and Applied Analysis
times 120601119902 (120582int
1
0
120596 (120591) Υ120601119901 (119909 (120591)) 119889120591) 119889119904
=120601119902 (120582Υ) 1205742
Γ (120572 + 1) (1 minus 120578 int1
0119892 (119904) 119889119904)
119909
(69)
This implies that 1 le 120601119902(120582Υ)1205742Γ(120572 + 1)(1 minus 120578 int1
0119892(119904)119889119904) =
(120582Υ)119902minus1
1205742Γ(120572+1)(1minus120578 int1
0119892(119904)119889119904) lt (120582 Υ)
119902minus11205742Γ(120572+1)(1minus
120578 int1
0119892(119904)119889119904) = 1 which is a contradiction This finishes the
proof of (ii)
Remark 12 In (i) the condition lim119909rarrinfin
119891(119905 119909) = 119891infin isin
(0 +infin] uniformly for 119905 isin [0 1] can be replaced withcondition (H6) The same methods can be used to prove theresults
Remark 13 From the above proof the condition 119891infin isin
(0 +infin] is important to keep the problem (1) having at leasttwo positive solutions if this condition is not satisfied thenthere exists 120582 gt 0 small enough such that the problem (1) hasno positive solutions for all 120582 lt 120582
Theorem 14 Assume that (H1)ndash(H6) hold(i) If 1198910
120601= +infin and 119891
infin
120601= +infin then there exists 120582lowast0 gt 0
such that problem (1) has at least two solutions for 120582 isin
(0 120582lowast0)(ii) If 1198910
120601= +infin and 119891
infin
120601= +infin then there exists 120582 gt 0
such that problem (1) has no solution for all 120582 gt 120582
Proof (i) It follows from 1198910
120601= +infin that there exists 1199031 gt 0
such that
119891 (119905 119909) ge 119871lowast1120601119901 (119909) (forall119905 isin [0 1] 0 le 119909 le 1199031) (70)
where 119871lowast1 satisfies 1205902120601119902(120582119871lowast1)1205741Γ(120572) ge 1
Then for 119909 isin 119875cap120597Ω1199031
we have 120590119909 le 119909 le 1199031 for 119905 isin [0 1]
and
(Ψ119909) (119905)
=1
Γ (120572)
times int
1
0
(1 minus 119904 )120572minus1
times 120601119902 (120582int
119904
0
120596 (120591) 119891 (120591 119909 (120591)) 119889120591) 119889119904
minus int
119905
0
(119905 minus 119904)120572minus1
times 120601119902 (120582int
119904
0
120596 (120591) 119891 (120591 119909 (120591)) 119889120591) 119889119904
+120578
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
119892 (119904)
times (int
1
0
(1 minus 120591)120572minus1
times 120601119902 (120582int
120591
0
120596 (120583) 119891 (120583 119909 (120583)) 119889120583)119889120591
minus int
119904
0
(119904 minus 120591)120572minus1
times 120601119902 (120582int
120591
0
120596 (120583) 119891 (120583 119909 (120583)) 119889120583)119889120591)119889119904
ge120590120578
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
119892 (119904)
times (int
1
0
(1 minus 120591)120572minus1
times 120601119902 (120582int
120591
0
120596 (120583) 119891 (120583 119909 (120583)) 119889120583)119889120591
minus int
119904
0
(119904 minus 120591)120572minus1
times120601119902 (120582int
120591
0
120596 (120583) 119891 (120583 119909 (120583)) 119889120583)119889120591)119889119904
+120590
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902 (120582int
119904
0
120596 (120591) 119891 (120591 119909 (120591)) 119889120591) 119889119904
ge120590
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902 (120582int
119904
0
120596 (120591) 119891 (120591 119909 (120591)) 119889120591) 119889119904
ge120590
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902 (120582int
119904
0
120596 (120591) 119871lowast1120601119901 (119909 (120591)) 119889120591) 119889119904
ge1205902120601119902 (120582119871lowast1) 119909
Γ (120572)
times int
1
0
(1 minus 119904)120572minus1
times 120601119902 (int
119904
0
120596 (120591) 119889120591) 119889119904
=1205902120601119902 (120582119871lowast1) 1205742
Γ (120572)119909
(71)
Abstract and Applied Analysis 13
which implies that
Ψ119909 ge 119909 forall119909 isin 119875 cap 120597Ω1199031
(72)
Next from 119891infin
120601= +infin there exists 1199032 gt 0 such that
119891 (119905 119909) ge 119871lowast2120601119901 (119909) forall119909 ge 1199032 (73)
where 119871lowast2 satisfies
1205902120601119902 (120582119871lowast2) 1205742
Γ (120572)ge 1 (74)
Let 1199032 gt max1199031 1199032120590 then for 119909 isin 119875 cap 120597Ω1199032
we have
119909 (119905) ge 120590 119909 = 1205901199032 gt 1199032 (75)
Similar to the above proof we can get
Ψ119909 ge 119909 forall119909 isin 119875 cap 120597Ω1199032
(76)
Let 1199031 lt 1199033 lt 11990321198721199033
= max119891(119905 119909) | 119905 isin [0 1] 119909 le 1199033+1 gt
0 and 120582lowast0 = (11198721199033
)(Γ(120572+1)1199033(1minus120578 int1
0119892(119904)119889119904)1205742)
119901minus1Thenfor 119909 isin 119875 cap 120597Ω119903
3
we have
119891 (119905 119909) le 1198721199033
forall119905 isin [0 1] (77)
Hence
(Ψ119909) (119905)
=1
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902 (120582int
119904
0
120596 (120591) 119891 (120591 119909 (120591)) 119889120591) 119889119904
minus int
119905
0
(119905 minus 119904)120572minus1
times 120601119902 (120582int
119904
0
120596 (120591) 119891 (120591 119909 (120591)) 119889120591) 119889119904
+120578
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
119892 (119904)
times (int
1
0
(1 minus 120591)120572minus1
times 120601119902 (120582int
120591
0
120596 (120583) 119891 (120583 119909 (120583)) 119889120583)119889120591
minus int
119904
0
(119904 minus 120591)120572minus1
times120601119902 (120582int
120591
0
120596 (120583) 119891 (120583 119909 (120583)) 119889120583)119889120591)119889119904
le120578
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
119892 (119904)
times int
1
0
(1 minus 120591)120572minus1
120601119902
times (120582int
120591
0
120596 (120583) 119891 (120583 119909 (120583)) 119889120583)119889120591 119889119904
+1
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902 (120582int
119904
0
120596 (120591) 119891 (120591 119909 (120591)) 119889120591) 119889119904
le120578
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
119892 (119904)
times int
1
0
(1 minus 120591)120572minus1
120601119902
times (120582int
120591
0
120596 (120583)1198721199033
119889120583)119889120591119889119904
+1
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902 (120582int
119904
0
120596 (120591)1198721199033
119889120591)119889119904
le120578120601119902 (120582119872119903
3
) int1
0119892 (119904) 119889119904
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
(1 minus 120591)120572minus1
times 120601119902 (int
1
0
120596 (120583) 119889120583)119889120591
+120601119902 (120582119872119903
3
)
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902 (int
1
0
120596 (120591) 119889120591) 119889119904
=1205742120601119902 (120582119872119903
3
)
Γ (120572 + 1) (1 minus 120578 int1
0119892 (119904) 119889119904)
lt1205742120601119902 (120582lowast0119872119903
3
)
Γ (120572 + 1) (1 minus 120578 int1
0119892 (119904) 119889119904)
= 1199033
(78)
14 Abstract and Applied Analysis
This implies that
Ψ119909 lt 119909 forall119909 isin 119875 cap 120597Ω1199033
(79)
Applying Lemma 7 to (72) (76) and (79) yields that Ψ hastwo fixed points 1199091 1199092 such that 1199091 isin 119875 cap (Ω119903
3
Ω1199031
) and1199092 isin 119875 cap (Ω119903
2
Ω1199033
)(ii) By the proof of (i) and the continuity of 119891(119905 119909) there
exists 119909 isin [1199031 1199032] such that min119909isin[11990311199032](119891(119905 119909)120601119901(119909)) =
119891(119905 119909)120601119901(119909)Let Υ = min119871lowast1 119871lowast2 119891(119905 119909)120601119901(119909) since (H6) holds
then Υ gt 0 and
119891 (119905 119909) ge Υ120601119901 (119909) forall119909 ge 0 119905 isin [0 1] (80)
Assuming 119909(119905) is a positive solution of problem (1) we willshow that this leads to contradiction for all 120582 gt 120582 where 120582 =
(1Υ)(Γ(120572)1205741120590)119901minus1 It follows form (26) that
119909 = 119909 (0)
=120578
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
119892 (119904)
times (int
1
0
(1 minus 120591)120572minus1
times 120601119902 (120582int
120591
0
120596 (120583) 119891 (120583 119909 (120583)) 119889120583)119889120591
minus int
119904
0
(119904 minus 120591)120572minus1
times120601119902 (120582int
120591
0
120596 (120583) 119891 (120583 119909 (120583)) 119889120583)119889120591)119889119904
+1
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902 (120582int
119904
0
120596 (120591) 119891 (120591 119909 (120591)) 119889120591) 119889119904
ge1
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902 (120582int
119904
0
120596 (120591) 119891 (120591 119909 (120591)) 119889120591) 119889119904
ge1
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902 (120582int
119904
0
120596 (120591) Υ120601119901 (119909 (120591)) 119889120591) 119889119904
ge1
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902 (120582int
119904
0
120596 (120591) Υ120601119901 (120590 119909 119889120591)) 119889119904
ge120601119902 (120582Υ) 120590 119909
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902 (int
119904
0
120596 (120591) 119889120591) 119889119904
=120601119902 (120582Υ) 1205741120590
Γ (120572)119909
(81)
This implies that 1 ge 120601119902(120582Υ)1205741120590Γ(120572) = (120582Υ)119902minus1
1205741120590Γ(120572) gt
(120582Υ)119902minus1
1205741120590Γ(120572) = 1 which is a contradiction This com-pletes the proof of (ii)
Corollary 15 Assume that (H1)ndash(H5) holds
(i) If lim119909rarrinfin
119891(119905 119909) = 119891infin isin (0 +infin] uniformly for 119905 isin
[0 1] and one of 1198910120601
= 0 and 119891infin
120601= 0 is satisfied then
there exists 120582lowast gt 0 such that the problem (1) has at leastone positive solution for any 120582 gt 120582
lowast(ii) If one of 1198910
120601= +infin and 119891
infin
120601= +infin holds then there
exists 120582lowast0 gt 0 such that problem (1) has at least onepositive solutions for any 120582 isin (0 120582lowast0)
Proof (i) The conclusion is a direct consequence of Theo-rem 11(i)
(ii) The conclusion is a direct consequence of Theo-rem 14(i)
Theorem 16 Assume that (H1)ndash(H5) holdThen the followingconclusions hold
(i) If 1198910120601
= 0 and 119891infin
120601= infin then the problem (1) has a
positive solution 119909120582(119905) for all 120582 gt 0(ii) If 1198910
120601= infin and 119891
infin
120601= 0 then the problem (1) has a
positive solution 119909120582(119905) for all 120582 gt 0
Proof Weonly prove (i) the proof of (ii) is similar sowe omitit here
Let 120582 gt 0 since 1198910
120601= 0 there exists 119903 gt 0 119897 gt 0 such that
119891 (119905 119909) le 119897120601119901 (119909) (forall119905 isin [0 1] 119909 isin [0 119903]) (82)
where 119897 satisfied
1205742120582119902minus1
119897119902minus1
Γ (120572 + 1) (1 minus 120578 int1
0119892 (119904) 119889119904)
le 1 (83)
Similar to the proof of Theorem 14 we obtain that
Ψ119909 le 119909 forall119909 isin 119875 cap 120597Ω119903 (84)
Abstract and Applied Analysis 15
From 119891infin
120601= infin there exists 119877 gt 119903 gt 0 119871 gt 0 such that
119891 (119905 119909) ge 119871120601119901 (119909) (forall119905 isin [0 1] 119909 ge 119877) (85)where 119871 satisfied
1205902120582119902minus1
119871119902minus1
1205741
Γ (120572)ge 1 (86)
We get thatΨ119909 ge 119909 forall119909 isin 119875 cap 120597Ω119877 (87)
Applying Lemma 7 to (84) (87) yields thatΨ has a fixed point119909 such that 119909 isin 119875 cap (Ω119877 Ω119903)
Theorem 17 Assume that (H1)ndash(H6) holdThen the followingconclusions hold
If 0 lt 1198910
120601lt infin and 0 lt 119891
infin
120601lt infin then there exists 1 2 gt
0 such that problem (1) has no positive solution for any 120582 gt 1
and 0 lt 120582 lt 2
Proof It follows from 0 lt 1198910
120601lt infin and 0 lt 119891
infin
120601lt infin that
there exist 1 gt 0 2 gt 0 1 gt 0 2 gt 0 and 1199032 gt 1199031 gt 0
such that1120601119901 (119909) le 119891 (119905 119909) le 1120601119901 (119909)
(forall119905 isin [0 1] 0 le 119909 le 1199031)
2120601119901 (119909) le 119891 (119905 119909) le 2120601119901 (119909)
(forall119905 isin [0 1] 119909 ge 1199032)
(88)
Let = max119891(119905 119909)120601119901(119909) 1199031 le 119909 le 1199032 0 le 119905 le 1 =
min119891(119905 119909)120601119901(119909) 1199031 le 119909 le 1199032 0 le 119905 le 1 By the condition(H6) gt 0 Then forall119909 ge 0 we have
120601119901 (119909) le 119891 (119905 119909) le 120601119901 (119909) (89)
where = max1 2 gt 0 and = min1198971 2 gt 0Let 1 = (1)(Γ(120572)1205741120590)
119901minus1 and 2 =
(1)(Γ(120572 + 1)(1 minus 120578 int1
0119892(119904)119889119904)1205742)
119901minus1
If problem (1)has positive solution 119909(119905) in 119875 then
119909 = 119909 (0)
=120578
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
119892 (119904)
times (int
1
0
(1 minus 120591)120572minus1
times 120601119902 (120582int
120591
0
120596 (120583) 119891 (120583 119909 (120583)) 119889120583)119889120591
minus int
119904
0
(119904 minus 120591)120572minus1
times120601119902 (120582int
120591
0
120596 (120583) 119891 (120583 119909 (120583)) 119889120583)119889120591)119889119904
+1
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902 (120582int
119904
0
120596 (120591) 119891 (120591 119909 (120591)) 119889120591) 119889119904
ge1
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902 (120582int
119904
0
120596 (120591) 119891 (120591 119909 (120591)) 119889120591) 119889119904
ge1
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902 (120582int
119904
0
120596 (120591) 120601119901 (119909 (120591)) 119889120591) 119889119904
ge1
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902 (120582int
119904
0
120596 (120591) 120601119901 (120590 119909 119889120591)) 119889119904
ge120601119902 (120582) 120590 119909
Γ (120572)
times int
1
0
(1 minus 119904)120572minus1
120601119902 (int
119904
0
120596 (120591) 119889120591) 119889119904
=120601119902 (120582) 1205741120590
Γ (120572)119909
gt120601119902 (1 ) 1205741120590
Γ (120572)119909 = 119909
119909 = 119909 (0)
=120578
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
119892 (119904)
times (int
1
0
(1 minus 120591)120572minus1
120601119902
times (120582int
120591
0
120596 (120583) 119891 (120583 119909 (120583)) 119889120583)119889120591
minus int
119904
0
(119904 minus 120591)120572minus1
120601119902
times(120582int
120591
0
120596 (120583) 119891 (120583 119909 (120583)) 119889120583)119889120591)119889119904
+1
Γ (120572)int
1
0
(1 minus 119904)120572minus1
120601119902
times (120582int
119904
0
120596 (120591) 119891 (120591 119909 (120591)) 119889120591) 119889119904
16 Abstract and Applied Analysis
le120578
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
119892 (119904)
times int
1
0
(1 minus 120591)120572minus1
times 120601119902 (120582int
120591
0
120596 (120583) 119891 (120583 119909 (120583)) 119889120583)119889120591 119889119904
+1
Γ (120572)int
1
0
(1 minus 119904)120572minus1
120601119902
times (120582int
119904
0
120596 (120591) 119891 (120591 119909 (120591)) 119889120591) 119889119904
=1
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
(1 minus 119904)120572minus1
120601119902
times (120582int
119904
0
120596 (120591) 119891 (120591 119909 (120591)) 119889120591) 119889119904
le1
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
(1 minus 119904)120572minus1
120601119902
times (120582int
1
0
120596 (120591) 120601119901 (119909 (120591)) 119889120591) 119889119904
=120601119902 (120582) 1205742
Γ (120572 + 1) (1 minus 120578 int1
0119892 (119904) 119889119904)
119909
lt120601119902 (2) 1205742
Γ (120572 + 1) (1 minus 120578 int1
0119892 (119904) 119889119904)
119909 = 119909
(90)
which is a contradiction This finishes the proof of Theo-rem 17
4 Examples
In this section we give out same examples to illustrate ourmain results can be used in practice
Example 1 Consider the following fractional differentialequation
(120601119901 (11986352
0+119909 (119905)))
1015840
+ 120582119905(1 + 1199052)12
119909119901minus1 arctan119909 = 0
0 lt 119905 lt 1
1199091015840(0) = 119909
10158401015840(0) = 0
119909 (1) =1
2int
1
0
1
2119909 (119904) 119889119904
(91)
where 120572 = 52 119901 = 32 119902 = 3 120596(119905) = 119905 119891(119905 119909) = (1 +
1199052)12
119909119901minus1 arctan119909 120578 = 12 and 119892(119905) = 12
By simple computation 119897lowast1= 1205872 119897
lowast
2= radic21205872 1205741 = int
1
0(1minus
119904)32
1206013(int119904
0120591 119889120591)119889119904 asymp 00043 1205742 = 1206013(int
1
0120591 119889120591) = 025 120590 asymp
01429 120582lowast
0= (1119897
lowast
1)(Γ(120572)1205901205741)
119901minus1asymp 25149 120582
lowast
0= (1119897
lowast
2)(Γ(120572+
1)(1 minus 120578 int1
0119892(119904)119889119904)1205742)
119901minus1asymp 14222 Then it is easy to see
all the conditions of Theorem 9(ii) are satisfied thus forall120582 isin
[14222 25149]] the problem (91) has at least one positivesolution
Example 2 Consider the boundary value problem
(120601119901 (11986352
0+119909 (119905)))
1015840
+ 120582119905(1 + 1199052)12
119909119901119890minus119909
= 0
0 lt 119905 lt 1
1199091015840(0) = 119909
10158401015840(0) = 0
119909 (1) =1
2int
1
0
1
2119909 (119904) 119889119904
(92)
where 120572 = 52 119901 = 32 119902 = 3 120596(119905) = 119905 119891(119905 119909) = (1 +
1199052)12
119909119901119890minus119909
120578 = 12 and 119892(119905) = 12
1198910
120601= lim119909rarr0
119891 (119905 119909)
120601119901 (119909)= lim119909rarr0
(1 + 1199052)12
119909119901119890minus119909
119909119901minus1= 0
forall119905 isin [0 1]
1198910
120601= lim119909rarrinfin
119891 (119905 119909)
120601119901 (119909)= lim119909rarrinfin
(1 + 1199052)12
119909119901119890minus119909
119909119901minus1= 0
forall119905 isin [0 1]
(93)
Let 1205881= 12 120588
2= 4 then
119891 (119905 119909)
120601119901 (119909)le (1 + 119905
2)12 1
2radic119890le
1
radic2119890
forall119905 isin [0 1] 119909 isin [01
2]
119891 (119905 119909)
120601119901 (119909)le (1 + 119905
2)12 4
1198904le
4radic2
1198904
forall119905 isin [0 1] 119909 isin [4infin)
(94)
so 1205761 = 1radic2119890 1205762 = 4radic21198904 and forall119909 isin
[12 4] max12le119909le4(119891(119905 119909)120601119901(119909)) = radic2119890minus1 We
have Υ = max1radic2119890 4radic21198904 radic2119890minus1
= radic2119890minus1 1205742 = 025
Abstract and Applied Analysis 17
and 120582 = (1Υ)(Γ(120572+1)(1minus120578 int1
0119892(119904)119889119904)1205742)
119901minus1= 3radic5120587
141198904
By Theorem 11(ii) for any 120582 lt 120582 = 3radic512058714
1198904 the problemhas no positive solution
5 Conclusions
Recently differential equations with p-Laplacian operatorwere widely discussed by several authors In this paperby using fixed point techniques combining with partiallyordered structure of Banach space we obtained the existencemultiply and the dependence of parameter These newresults we presented can be used in numerical computationand analyze mathematical models of physical phenomenamechanics nonlinear dynamics and many other relatedfields
Conflict of Interests
The authors declare that there is no conflict of interestsregarding the publication of this paper
Acknowledgments
The authors are very grateful to anonymous referees for theirvaluable and detailed suggestions and comments to improvethe original paper This work is supported by the Programfor New Century Excellent Talents in University (NCET-10-0097)
References
[1] I Podlubny Fractional Differential Equations vol 198 Aca-demic Press San Diego Calif USA 1999
[2] A A Kilbas H M Srivastava and J J Trujillo Theoryand Applications of Fractional Differential Equations vol 204of North-Holland Mathematics Studies Elsevier Science BVAmsterdam The Netherlands 2006
[3] Y A Rossikhin and M V Shitikova ldquoApplication of fractionalderivatives to the analysis of damped vibrations of viscoelasticsingle mass systemsrdquoActa Mechanica vol 120 no 1ndash4 pp 109ndash125 1997
[4] J Sabatier O Agrawal and J T Machado Advances in Frac-tional Calculus Theoretical developments and applications inphysics and engineering Springer Berlin Germany 2007
[5] N Engheta ldquoOn fractional calculus and fractional multipolesin electromagnetismrdquo IEEE Transactions on Antennas andPropagation vol 44 no 4 pp 554ndash566 1996
[6] R Hilfer Applications of Fractional Calculus in Physics WorldScientific Singapore 2000
[7] R L Bagley and R A Calico ldquoFractional order state equationsfor the control of viscoelastically damped structuresrdquo Journalof Guidance Control and Dynamics vol 14 no 2 pp 304ndash3111991
[8] W-X Zhou andY-D Chu ldquoExistence of solutions for fractionaldifferential equations with multi-point boundary conditionsrdquoCommunications in Nonlinear Science and Numerical Simula-tion vol 17 no 3 pp 1142ndash1148 2012
[9] B Ahmad and G Wang ldquoA study of an impulsive four-pointnonlocal boundary value problemof nonlinear fractional differ-ential equationsrdquo Computers amp Mathematics with Applicationsvol 62 no 3 pp 1341ndash1349 2011
[10] B Ahmad and S Sivasundaram ldquoOn four-point nonlocalboundary value problems of nonlinear integro-differentialequations of fractional orderrdquo Applied Mathematics and Com-putation vol 217 no 2 pp 480ndash487 2010
[11] N Nyamoradi ldquoPositive solutions for multi-point boundaryvalue problems for nonlinear fractional differential equationsrdquoJournal of Contemporary Mathematical Analysis vol 48 no 4pp 145ndash157 2013
[12] B Ahmad and J J Nieto ldquoExistence of solutions for nonlocalboundary value problems of higher-order nonlinear fractionaldifferential equationsrdquo Abstract and Applied Analysis vol 2009Article ID 494720 9 pages 2009
[13] X Liu M Jia and X Xiang ldquoOn the solvability of a fractionaldifferential equationmodel involving the119901-Laplacian operatorrdquoComputers ampMathematics with Applications vol 64 no 10 pp3267ndash3275 2012
[14] J Wang H Xiang and Z Liu ldquoExistence of concave positivesolutions for boundary value problem of nonlinear fractionaldifferential equation with 119901-Laplacian operatorrdquo InternationalJournal of Mathematics and Mathematical Sciences vol 2010Article ID 495138 17 pages 2010
[15] J Wang and H Xiang ldquoUpper and lower solutions method fora class of singular fractional boundary value problems with 119901-Laplacian operatorrdquo Abstract and Applied Analysis vol 2010Article ID 971824 12 pages 2010
[16] H Lu andZHan ldquoExistence of positive solutions for boundary-value problem of fractional differential equation with 119901-laplacian operatorrdquo The American Jouranl of Engineering andTechnology Research vol 11 pp 3757ndash3764 2011
[17] G Chai ldquoPositive solutions for boundary value problem offractional differential equation with 119901-Laplacian operatorrdquoBoundary Value Problems vol 2012 article 18 2012
[18] T Chen W Liu and Z Hu ldquoA boundary value problem forfractional differential equation with 119901-Laplacian operator atresonancerdquoNonlinear AnalysisTheoryMethodsampApplicationsvol 75 no 6 pp 3210ndash3217 2012
[19] T Chen andW Liu ldquoAn anti-periodic boundary value problemfor the fractional differential equation with a 119901-LaplacianoperatorrdquoAppliedMathematics Letters of Rapid Publication vol25 no 11 pp 1671ndash1675 2012
[20] S J Li X G Zhang Y H Wu and L Caccetta ldquoExtremalsolutions for 119901-Laplacian differential systems via iterative com-putationrdquo Applied Mathematics and Computation vol 26 pp1151ndash1158 2013
[21] Z Liu and L Lu ldquoA class of BVPs for nonlinear fractionaldifferential equations with 119901-Laplacian operatorrdquo ElectronicJournal of Qualitative Theory of Differential Equations vol 70pp 1ndash16 2012
[22] Z Liu L Lu and I Szanto ldquoExistence of solutions for fractionalimpulsive differential equations with 119901-Laplacian operatorrdquoActa Mathematica Hungarica vol 141 no 3 pp 203ndash219 2013
[23] Z Bai ldquoEigenvalue intervals for a class of fractional boundaryvalue problemrdquo Computers amp Mathematics with Applicationsvol 64 no 10 pp 3253ndash3257 2012
[24] X Zhang L Wang and Q Sun ldquoExistence of positive solutionsfor a class of nonlinear fractional differential equations withintegral boundary conditions and a parameterrdquo Applied Mathe-matics and Computation vol 226 pp 708ndash718 2014
18 Abstract and Applied Analysis
[25] D J Guo and V Lakshmikantham Nonlinear Problems inAbstract Cones vol 5 Academic Press New York NY USA1988
[26] M A Krasnoselrsquoskii Topological Methods in the Theory ofNonlinear Integral Equations Translated by A H ArmstrongElmsford Pergamon Turkey 1964
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Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
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Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Stochastic AnalysisInternational Journal of
4 Abstract and Applied Analysis
times 120601119902 (120582int
120591
0
119910 (120583) 119889120583)119889120591
+int
1
119904
(1 minus 120591)120572minus1
120601119902 (120582int
120591
0
119910 (120583) 119889120583)119889120591)119889119904
ge 0
(20)
Therefore 119909(119905) ge 0 and is concave for 119905 isin [0 1] So for every119905 isin [0 1]
119909 (119905) ge 119905119909 (1) + (1 minus 119905) 119909 (0) (21)
Therefore
120578int
1
0
119909 (119905) 119892 (119905) 119889119905 ge 120578119909 (1) int
1
0
119905119892 (119905) 119889119905
+ 120578119909 (0) int
1
0
(1 minus 119905) 119892 (119905) 119889119905
(22)
Since 119909(1) = 120578 int1
0119909(119905)119892(119905)119889119905 1 minus 120578 int
1
0119892(119905)119889119905 gt 0 we have
119909 (1) ge120578 int1
0(1 minus 119905) 119892 (119905) 119889119905
1 minus 120578 int1
0119905119892 (119905) 119889119905
119909 (0) (23)
that is
min0le119905le1
119909 (119905) = 119909 (1) ge120578 int1
0(1 minus 119905) 119892 (119905) 119889119905
1 minus 120578 int1
0119905119892 (119905) 119889119905
119909 (0) = 120590 119909 (24)
The Lemma is proved
We construct a cone 119875 in 119864 by
119875 = 119909 isin 119864 119909 ge 0 min0le119905le1
119909 (119905) ge 120590 119909 (25)
where 120590 is defined in Lemma 7 It is easy to see 119875 is a closedconvex cone of 119864
Define 119879 119875 rarr 119864 by
(119879119909) (119905)
=1
Γ (120572)int
1
0
(1 minus 119904)120572minus1
120601119902 (int
119904
0
120596 (120591) 119891 (120591 119909 (120591)) 119889120591) 119889119904
minusint
119905
0
(119905 minus 119904)120572minus1
120601119902 (int
119904
0
120596 (120591) 119891 (120591 119909 (120591)) 119889120591) 119889119904
+120578
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
119892 (119904)
times (int
1
0
(1 minus 120591)120572minus1
120601119902 (int
120591
0
120596 (120583) 119891 (120583 119909 (120583)) 119889120583)119889120591
minus int
119904
0
(119904 minus 120591)120572minus1
times 120601119902 (int
120591
0
120596 (120583) 119891 (120583 119909 (120583)) 119889120583)119889120591)119889119904
(26)
It is clear that the fixed points of the operator 120601119902(120582)119879 arethe solutions of the boundary value problems (1)
We make the following hypotheses
(H1) 119891 [0 1] times [0infin) rarr [0infin) is continuous(H2) 120596 (0 1) rarr [0infin) is continuous and not identical
zero on any closed subinterval of (0 1) with 0 lt
int1
0120596(119905)119889119905 lt +infin
(H3) 1 minus 120578 int1
0119892(119904)119889119904 gt 0
(H4) 1198910
120601isin [0infin] where 119891
0
120601= lim119909rarr0(119891(119905 119909)120601119901(119909))
uniformly for 119905 isin [0 1](H5) 119891
infin
120601isin [0infin] where 119891
infin
120601= lim119909rarrinfin(119891(119905 119909)120601119901(119909))
uniformly for 119905 isin [0 1](H6) 119891(119905 119909) gt 0 for any 119905 isin [0 1] and 119909 gt 0
Lemma 8 Assume that (H1)ndash(H3) hold Then 119879 119875 rarr 119864 iscompletely continuous
Proof First we show that 119879 is continuous It is easy to see119879(119875) sub 119875 For 119909119899(119905) isin 119864 and 119909119899(119905) rarr 119909(119905) as 119899 rarr infin bythe continuous of 119891(119905 119909(119905)) we get 119891(119905 119909119899(119905)) rarr 119891(119905 119909(119905))as 119899 rarr infin This implies that
120601119902 (int
119905
0
119908 (119904) 119891 (119904 119909119899 (119904)) 119889119904)
997888rarr 120601119902 (int
119905
0
119908 (119904) 119891 (119904 119909 (119904)) 119889119904)
(27)
So we have
sup119905isin[01]
10038161003816100381610038161003816100381610038161003816120601119902 (int
119905
0
119908 (119904) 119891 (119904 119909119899 (119904)) 119889119904)
minus 120601119902 (int
119905
0
119908 (119904) 119891 (119904 119909 (119904)) 119889119904)
10038161003816100381610038161003816100381610038161003816997888rarr 0 119899 997888rarr infin
(28)
Denote Π = |(119879119909119899)(119905) minus (119879119909)(119905)| then
Π le1
Γ (120572)int
1
0
(1 minus 119904)120572minus1
10038161003816100381610038161003816100381610038161003816120601119902 (int
119904
0
120596 (120591) 119891 (120591 119909119899 (120591)) 119889120591)
minus 120601119902 (int
119904
0
120596 (120591) 119891 (120591 119909 (120591)) 119889120591)
10038161003816100381610038161003816100381610038161003816119889119904
Abstract and Applied Analysis 5
+int
119905
0
(119905 minus 119904)120572minus1
10038161003816100381610038161003816100381610038161003816120601119902 (int
119904
0
120596 (120591) 119891 (120591 119909119899 (120591)) 119889120591)minus120601119902
times (int
119904
0
120596 (120591) 119891 (120591 119909 (120591)) 119889120591)
10038161003816100381610038161003816100381610038161003816119889119904
+120578
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
119892 (119904)
times (int
1
0
(1 minus 120591)120572minus1
times
10038161003816100381610038161003816100381610038161003816120601119902 (int
120591
0
120596 (120583) 119891 (120583 119909119899 (120583)) 119889120583)
minus 120601119902 (int
120591
0
120596 (120583) 119891 (120583 119909 (120583)) 119889120583)
10038161003816100381610038161003816100381610038161003816119889120591
+ int
119904
0
(119904 minus 120591)120572minus1
times
10038161003816100381610038161003816100381610038161003816120601119902 (int
120591
0
120596 (120583) 119891 (120583 119909119899 (120583)) 119889120583) minus 120601119902
times (int
120591
0
120596 (120583) 119891 (120583 119909 (120583)) 119889120583)
10038161003816100381610038161003816100381610038161003816119889120591) 119889119904
le sup119904isin[01]
10038161003816100381610038161003816100381610038161003816120601119902 (int
119904
0
119908 (120591) 119891 (120591 119909119899 (120591)) 119889120591)
minus 120601119902 (int
119904
0
119908 (120591) 119891 (120591 119909 (120591)) 119889120591)
10038161003816100381610038161003816100381610038161003816
times2
Γ (120572 + 1)(1 +
120578 int1
0119892 (119904) 119889119904
1 minus 120578 int1
0119892 (119904) 119889119904
) 997888rarr 0
119899 997888rarr infin
(29)
Hence 119879 is continuousSecond we show that 119879 is compactFor 119909(119905) isin 119861119903 = 119909 isin 119875 119909 le 119903 by condition (H1)
119891(119905 119909(119905)) lt infin Denote119872 = max0le119905le1119909le119903119891(119905 119909(119905))Then
|119879119909 (119905)|
le1
Γ (120572)int
1
0
(1 minus 119904)120572minus1
10038161003816100381610038161003816100381610038161003816120601119902 (int
119904
0
120596 (120591) 119891 (120591 119909 (120591)) 119889120591)
10038161003816100381610038161003816100381610038161003816119889119904
+int
119905
0
(119905 minus 119904)120572minus1
10038161003816100381610038161003816100381610038161003816120601119902 (int
119904
0
120596 (120591) 119891 (120591 119909 (120591)) 119889120591)
10038161003816100381610038161003816100381610038161003816119889119904
+120578
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
119892 (119904)
times (int
1
0
(1 minus 120591)120572minus1
times
10038161003816100381610038161003816100381610038161003816120601119902 (int
120591
0
120596 (120583) 119891 (120583 119909 (120583)) 119889120583)
10038161003816100381610038161003816100381610038161003816119889120591
+ int
119904
0
(119904 minus 120591)120572minus1
times
10038161003816100381610038161003816100381610038161003816120601119902 (int
120591
0
120596 (120583) 119891 (120583 119909 (120583)) 119889120583)
10038161003816100381610038161003816100381610038161003816119889120591) 119889119904
le1
Γ (120572)
times int
1
0
(1 minus 119904)120572minus1
10038161003816100381610038161003816100381610038161003816120601119902 (119872int
119904
0
120596 (120591) 119889120591)
10038161003816100381610038161003816100381610038161003816119889119904
+int
119905
0
(119905 minus 119904)120572minus1
10038161003816100381610038161003816100381610038161003816120601119902 (119872int
119904
0
120596 (120591) 119889120591)
10038161003816100381610038161003816100381610038161003816119889119904
+120578
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
119892 (119904) (int
1
0
(1 minus 120591)120572minus1
times
10038161003816100381610038161003816100381610038161003816120601119902 (119872int
120591
0
120596 (120583) 119889120583)119889120591
10038161003816100381610038161003816100381610038161003816
+ int
119904
0
(119904 minus 120591)120572minus1
times
10038161003816100381610038161003816100381610038161003816120601119902 (119872int
120591
0
120596 (120583) 119889120583)
10038161003816100381610038161003816100381610038161003816119889120591) 119889119904
le120578
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
120601119902 (119872int
1
0
120596 (120591) 119889120591)
times int
1
0
119892 (119904) (int
1
0
(1 minus 120591)120572minus1
119889120591 + int
119904
0
(119904 minus 120591)120572minus1
119889120591)119889119904
+1
Γ (120572)120601119902 (119872int
1
0
120596 (120591) 119889120591)
times int
1
0
(1 minus 119904)120572minus1
119889119904 + int
119905
0
(119905 minus 119904)120572minus1
119889119904
le
2120601119902 (119872int1
0120596 (120591) 119889120591)
Γ (120572 + 1) (1 minus 120578 int1
0119892 (120591) 119889120591)
lt infin
(30)
that is 119879 maps bounded sets into bounded sets in 119875For 119909(119905) isin 119861119903 0 lt 1199051 lt 1199052 lt 1
1003816100381610038161003816119879119909 (1199052) minus 119879119909 (1199051)1003816100381610038161003816
=1
Γ (120572)
10038161003816100381610038161003816100381610038161003816int
1199052
0
(1199052 minus 119904)120572minus1
120601119902 (int
119904
0
120596 (120591) 119891 (120591 119909 (120591)) 119889120591) 119889119904
6 Abstract and Applied Analysis
minus int
1199051
0
(1199051 minus 119904)120572minus1
times 120601119902 (int
119904
0
120596 (120591) 119891 (120591 119909 (120591)) 119889120591) 119889119904
10038161003816100381610038161003816100381610038161003816
le1
Γ (120572)120601119902 (119872int
1
0
120596 (120591) 119889120591)
times
100381610038161003816100381610038161003816100381610038161003816
int
1199051
0
((1199052 minus 119904)120572minus1
minus (1199051 minus 119904)120572minus1
) 119889119904
+int
1199052
1199051
(1199052 minus 119904)120572minus1
119889119904
100381610038161003816100381610038161003816100381610038161003816
(31)
By mean value theorem we obtain that
int
1199051
0
((1199052 minus 119904)120572minus1
minus (1199051 minus 119904)120572minus1
) 119889119904
le int
1199051
0
(120572 minus 1) (1 minus 119904)120572minus2
(1199052 minus 1199051) 119889119904
le int
1
0
(120572 minus 1) (1 minus 119904)120572minus2
(1199052 minus 1199051) 119889119904
= 1199052 minus 1199051
(32)
Thus1003816100381610038161003816119879119909 (1199052) minus 119879119909 (1199051)
1003816100381610038161003816
le1
Γ (120572)120601119902 (119872int
1
0
120596 (120591) 119889120591) [(1199052 minus 1199051) +1
120572(1199052 minus 1199051)
120572]
(33)
This shows that |119879119909(1199052) minus 119879119909(1199051)| rarr 0 as 1199052 minus 1199051 rarr 0so 119879119909 119909 isin 119861119903 is equicontinuous Therefore the operator119879 119875 rarr 119875 is completely continuous by the Arzela-Ascolitheorem
Now for convenience we introduce the following nota-tions Let
int
1
0
(1 minus 119904)120572minus1
120601119902 (int
119904
0
120596 (120591) 119889120591) 119889119904 = 1205741
120601119902 (int
1
0
120596 (120591) 119889120591) = 1205742
119898 (119903) = minmin119905
119891 (119905 119909)
120601119901 (119903) 119909 isin [120590119903 119903]
(34)
Theorem 9 Assume that (H1)ndash(H5) hold
(i) If 0 lt 1198910
120601lt +infin then there exists 1205850 gt 0 such that for
every 0 lt 119903 lt 1205850 problem (1) has a positive solution119909119903(119905) satisfying 119909119903 = 119903 associated with
120582 = 120582119903 isin [1205820 1205820] (35)
where 1205820 and 1205820 are two positive finite numbers
(ii) If 0 lt 119891infin
120601lt +infin then there exists 120585lowast
0gt 0 such that for
every 119877 gt 120585lowast
0 problem (1) has a positive solution 119909119877(119905)
satisfying 119909119877 = 119877 for any
120582 = 120582119877 isin [120582lowast
0 120582lowast
0] (36)
where 120582lowast
0and 120582
lowast
0are two positive finite numbers
(iii) If 1198910120601
= +infin then there exists 1205851 gt 0 such that for any0 lt 119903lowast
lt 1205851 problem (1) has a positive solution 119909119903lowast(119905)
satisfying 119909119903lowast = 119903lowast for any
120582 = 120582119903lowast isin (0 120582lowast] (37)
where 120582lowast is a positive finite number
(iv) If 119891infin
120601= +infin then there exists 1205851 gt 0 such that
for every 119877lowast gt 1205851 problem (1) has a positive solution119909119877lowast
(119905) satisfying 119909119877lowast
= 119877lowast for any
120582 = 120582119877lowast
isin (0 120582lowast] (38)
where 120582lowast is a positive finite number(v) If there exists 120585 gt 0 such that 119898(119903) gt 120585 then problem
(1) has positive solution 119909119903(119905) satisfying 119909119903 = 119903 forany
120582 = 120582119903 isin (0 ] (39)
where is a positive finite number
Proof (i) It follows from 0 lt 1198910
120601lt +infin that there exists
1198971 1198972 120588 gt 0 such that
1198971120601119901 (119909) le 119891 (119905 119909) le 1198972120601119901 (119909)
(forall119905 isin [0 1] 0 le 119909 le 120588)
(40)
Let 1205850 = 120588120590 we show that 1205850 is required When 119909 isin 119875 cap
120597Ω119903 we have
119909 (119905) ge 120590 119909 = 120590119903 (41)
we need 119909(119905) lt 120588 this implies that 120590119903 lt 120588 as 119903 lt 1205850 so120590119903 lt 1205901205850 = 120588
Let 1205820 = (11198971)(Γ(120572)1205901205741)119901minus1 Then we may assume that
119909 minus 120601119902 (1205820) 119879119909 = 0 (forall119909 isin 119875 cap 120597Ω119903) (42)
if not then there exists119909119903 isin 119875cap120597Ω119903 such that120601119902(1205820)119879119909119903 = 119909119903and then (35) already holds for 120582119903 = 1205820
Define 120595(119905) equiv 1 forall119905 isin [0 1] then 120595(119905) isin 119875 and 120595 equiv 1We now show that
119909 minus 120601119902 (1205820) 119879119909 = 120581120595 (forall119909 isin 119875 cap 120597Ω119903 120581 ge 0) (43)
In fact if there exists 1199091 isin 119875 cap 120597Ω119903 1205811 ge 0 such that 1199091 minus120601119902(1205820)1198791199091 = 1205811120595 then (42) implies that 1205811 gt 0 On the otherhand 1199091 = 120601119902(1205820)1198791199091 + 1205811120595 ge 1205811120595 we may choose 120581
lowast=
max120581 | 1199091 ge 120581120595 then 1205811 le 120581lowastlt +infin 1199091 ge 120581
lowast120595 Therefore
120581lowast= 120581lowast 1003817100381710038171003817120595
1003817100381710038171003817 le10038171003817100381710038171199091
1003817100381710038171003817 = 119903 (44)
Abstract and Applied Analysis 7
Consequently for any 119905 isin [0 1] (26) and (44) imply that
1199091 (119905)
= 120601119902 (1205820)1
Γ (120572)
times int
1
0
(1 minus 119904)120572minus1
times 120601119902 (int
119904
0
120596 (120591) 119891 (120591 1199091 (120591)) 119889120591) 119889119904
minus int
119905
0
(119905 minus 119904)120572minus1
times 120601119902 (int
119904
0
120596 (120591) 119891 (120591 1199091 (120591)) 119889120591) 119889119904
+120578120601119902 (1205820)
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
119892 (119904)
times (int
1
0
(1 minus 120591)120572minus1
times 120601119902 (int
120591
0
120596 (120583) 119891 (120583 1199091 (120583)) 119889120583)119889120591
minus int
119904
0
(119904 minus 120591)120572minus1
times 120601119902 (int
120591
0
120596 (120583) 119891 (120583 1199091 (120583)) 119889120583)119889120591)119889119904
+ 1205811120595
ge120578120590120601119902 (1205820)
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
119892 (119904)
times (int
1
0
(1 minus 120591)120572minus1
times 120601119902 (int
120591
0
120596 (120583) 119891 (120583 1199091 (120583)) 119889120583)119889120591
minus int
119904
0
(119904 minus 120591)120572minus1
120601119902
times (int
120591
0
120596 (120583) 119891 (120583 1199091 (120583)) 119889120583)119889120591)119889119904
+120590120601119902 (1205820)
Γ (120572)
times int
1
0
(1 minus 119904)120572minus1
120601119902 (int
119904
0
120596 (120591) 119891 (120591 1199091 (120591)) 119889120591) 119889119904
+ 1205811120595
ge120590120601119902 (1205820)
Γ (120572)
times int
1
0
(1 minus 119904)120572minus1
120601119902 (int
119904
0
120596 (120591) 119891 (120591 1199091 (120591)) 119889120591) 119889119904
+ 1205811120595
ge120590120601119902 (1205820)
Γ (120572)
times int
1
0
(1 minus 119904)120572minus1
120601119902 (1198971 int
119904
0
120596 (120591) 120601119901 (1199091 (120591)) 119889120591) 119889119904
+ 1205811120595
ge120590120601119902 (1205820)
Γ (120572)
times int
1
0
(1 minus 119904)120572minus1
120601119902 (1198971 int
119904
0
120596 (120591) 120601119901 (120581lowast120595 (120591)) 119889120591) 119889119904
+ 1205811120595
=120590120581lowast119897119902minus1
1120582119902minus1
0
Γ (120572)int
1
0
(1 minus 119904)120572minus1
120601119902 (int
119904
0
120596 (120591) 119889120591) 119889119904 + 1205811120595
=120590120581lowast119897119902minus1
1120582119902minus1
01205741
Γ (120572)+ 1205811120595
= 120581lowast+ 120581120595
(45)
which implies that 1199091(119905) ge (120581lowast+ 120581)120595(119905) 119905 isin [0 1] which is a
contradiction to the definition of 120581lowast Thus (42) holds and byLemma 5 the fixed point index
119894 (120601119902 (1205820) 119879 119875 cap 120597Ω119903 119875) = 0 (46)
On the other hand by the fact that the fixed point index ofconstants operator is 1 so
119894 (120599 119875 cap 120597Ω119903 119875) = 1 (47)
where 120599 is the zero operator It follows therefore from (46)and (47) and the homotopy invariance property that thereexist 119909119903 isin 119875 cap 120597Ω119903 and 0 lt ]0 lt 1 such that ]0120601119902(1205820)119879119909119903 =119909119903 but 119909119903 = 120601119902(120582119903)119879119909119903 which implies that ]0120601119902(1205820)119879119909119903 =
120601119902(120582119903)119879119909119903 it follows that we get ]0120601119902(1205820) = 120601119902(120582119903)) that is
]0120582119902minus1
0= 120582119902minus1
119903 (48)
and then
120582119903 = ]1(119902minus1)0
1205820 = ]119901minus10
1205820 lt 1205820 (49)
From the proof above for any 119903 lt 1205850 there exists a positivesolution 119909119903 isin 119875 cap 120597Ω119903 associated with 120582 = 120582119903 gt 0
8 Abstract and Applied Analysis
Thus
119909119903 (119905)
=120601119902 (120582119903)
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902 (int
119904
0
120596 (120591) 119891 (120591 119909119903 (120591)) 119889120591) 119889119904
minus int
119905
0
(119905 minus 119904)120572minus1
times 120601119902 (int
119904
0
120596 (120591) 119891 (120591 119909119903 (120591)) 119889120591) 119889119904
+120578120601119902 (120582119903)
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
119892 (119904)
times (int
1
0
(1 minus 120591)120572minus1
times 120601119902 (int
120591
0
120596 (120583) 119891 (120583 119909119903 (120583)) 119889120583)119889120591
minus int
119904
0
(119904 minus 120591)120572minus1
times 120601119902 (int
120591
0
120596 (120583) 119891 (120583 119909119903 (120583)) 119889120583)119889120591)119889119904
(50)
with 119909119903 = 119903Next we show that 120582119903 ge 1205820 In fact
119909119903 (119905)
=120601119902 (120582119903)
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902 (int
119904
0
120596 (120591) 119891 (120591 119909119903 (120591)) 119889120591) 119889119904
minus int
119905
0
(119905 minus 119904)120572minus1
times 120601119902 (int
119904
0
120596 (120591) 119891 (120591 119909119903 (120591)) 119889120591) 119889119904
+120578120601119902 (120582119903)
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
119892 (119904)
times (int
1
0
(1 minus 120591)120572minus1
times 120601119902 (int
120591
0
120596 (120583) 119891 (120583 119909119903 (120583)) 119889120583)119889120591
minus int
119904
0
(119904 minus 120591)120572minus1
times 120601119902 (int
120591
0
120596 (120583) 119891 (120583 119909119903 (120583)) 119889120583)119889120591)119889119904
le120578120601119902 (120582119903)
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
119892 (119904) int
1
0
(1 minus 120591)120572minus1
times 120601119902 (int
120591
0
120596 (120583) 119891 (120583 119909119903 (120583)) 119889120583)119889120591 119889119904
+120601119902 (120582119903)
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902 (int
119904
0
120596 (120591) 119891 (120591 119909119903 (120591)) 119889120591) 119889119904
=120601119902 (120582119903) 120578 int
1
0119892 (119904) 119889119904
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
(1 minus 119904)120572minus1
times 120601119902 (int
119904
0
120596 (120591) 119891 (120591 119909119903 (120591)) 119889120591) 119889119904
+120601119902 (120582119903)
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902 (int
119904
0
120596 (120591) 119891 (120591 119909119903 (120591)) 119889120591) 119889119904
le120601119902 (120582119903) 120578 int
1
0119892 (119904) 119889119904
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
(1 minus 119904)120572minus1
times 120601119902 (1198972 int
119904
0
120596 (120591) 119909119903 (120591) 119889120591) 119889119904
+120601119902 (120582119903)
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902 (1198972120582119903 int
119904
0
120596 (120591) 119909119903 (120591) 119889120591) 119889119904
=120601119902 (120582119903)
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
(1 minus 119904)120572minus1
times 120601119902 (1198972 int
119904
0
120596 (120591) 119909119903 (120591) 119889120591) 119889119904
Abstract and Applied Analysis 9
le
10038171003817100381710038171199091199031003817100381710038171003817 120601119902 (1205821199031198972)
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
(1 minus 119904)120572minus1
times 120601119902 (int
1
0
120596 (120591) 119889120591) 119889119904
=
10038171003817100381710038171199091199031003817100381710038171003817 120601119902 (1205821199031198972) 1205742
Γ (120572 + 1) (1 minus 120578 int1
0119892 (119904) 119889119904)
(51)
which implies that
10038171003817100381710038171199091199031003817100381710038171003817 le
10038171003817100381710038171199091199031003817100381710038171003817 120601119902 (1205821199031198972) 1205742
Γ (120572 + 1) (1 minus 120578 int1
0119892 (119904) 119889119904)
(52)
that is
120601119902 (1205821199031198972) 1205742
Γ (120572 + 1) (1 minus 120578 int1
0119892 (119904) 119889119904)
ge 1 (53)
it follows that we get
120582119903 ge1
1198972(
Γ (120572 + 1) (1 minus 120578 int1
0119892 (119904) 119889119904)
1205742)
119901minus1
= 1205820(54)
In conclusion 120582119903 isin [1205820 1205820](ii) It follows from 0 lt 119891
infin
120601lt infin that there exist
119897lowast
1 119897lowast
2 120588lowastgt 0 such that
119897lowast
1120601119901 (119909) le 119891 (119905 119909) le 119897
lowast
2120601119901 (119909)
(forall119905 isin [0 1] 119909 ge 120588lowast)
(55)
Let 120585lowast0= 120588lowast120590 the following proof are similar to (i)
(iii) It follows from 1198910
120601= +infin there exists 119897lowast gt 0 120588
lowastgt 0
such that
1198910
120601ge 119897lowast120601119901 (119909) (forall119905 isin [0 1] 0 le 119909 le 120588
lowast) (56)
Let 1205851 = 120588lowast120590 we show that 1205851 is required When 119909 isin 119875 cap
120597Ω119903lowast we have 119909(119905) ge 120590119909 = 120590119903lowast since 119909 le 120588
lowast then we need120590119903lowastlt 120588lowast as 119903lowast lt 1205851 so 120590119903
lowastlt 120588lowast holds
Let 120582lowast = (1119897lowast)(Γ(120572)1205901205741)
1(119902minus1) we proceed in the sameway as in the proof of (i) replacing (42) we may assume that119909 minus 120601119902(120582
lowast)119879119909 = 0 (forall119909 isin 119875 cap 120597Ω119903lowast) and replacing (43) we
can prove 119909 minus 120601119902(120582lowast)119879119909 = 120581120595 (forall119909 isin 119875 cap 120597Ω119903lowast 120581 ge 0) It
follows from Lemma 5 that 119894(120601119902(120582lowast)119879 119875 cap Ω119903lowast 119875) = 0 Note
that 119894(120579 119875 cap Ω119903lowast 119875) = 1 we can easily show that there exists119909119903lowast isin 119875cap120597Ω119903lowast and 0 lt ]119903lowast lt 1 such that ]119903lowast120601119902(120582
lowast)119879119909119903lowast = 119909119903lowast
Hence (37) holds for 120582119903lowast = 120582lowast]1minus119901119903lowast
lt 120582lowast
The proof ofTheorem 9(iv) follows by themethod similarto Theorem 9(iii) we omit it here
(v) It follows that 119898(119903) gt 120585 for any 119909 isin 119875 cap 120597Ω119903 we have120590119903 le 119909 le 119903 and 119891(119905 119909)120601119901(119909) ge 119891(119905 119909)120601119901(119903) ge 119898(119903) ge 120585then
119891 (119905 119909) ge 120585120601119901 (119909) forall119909 isin [120590119903 119903] 119905 isin [0 1] (57)
Let = (1120585)(Γ(120572)1205901205741)119901minus1 The following proof is similar to
that of (iv) This finished the proof of (v)
Remark 10 InTheorem 9 all the criteria obtained depend onone of 1198910
120601and 119891
infin
120601
Let Ψ(119909) = 120601119902(120582)119879119909 the following theorems give out themultiply nonexistence and the dependence of parameter
Theorem 11 Assume that (H1)ndash(H5) hold
(i) If 1198910
120601= 0 and 119891
infin
120601= 0 lim
119909rarrinfin119891(119905 119909) = 119891infin isin
(0 +infin] uniformly for 119905 isin [0 1] then there exists120582lowast
gt 0 such that the problem (1) has two solutions forany 120582 gt 120582
lowast
(ii) If 1198910120601
= 0 and 119891infin
120601= 0 then there exists 120582 gt 0 such
that for any 120582 lt 120582 problem (1) has no solution
Proof (i) It follows from 119891infin isin (0 +infin] that for any 120577 gt 0there exists 120588 gt 0 such that
119891 (119905 119909) gt 120577 (forall119905 isin [0 1] 119909 ge 120588) (58)
Since 1198910
120601= 0 there exists 1205761 gt 0 120588
1gt 0 such that
119891 (119905 119909) le 1205761120601119901 (119909) (forall119905 isin [0 1] 0 le 119909 le 1205881) (59)
where 1205761 satisfied 120601119902(1205821205761)1205742Γ(120572 + 1)(1 minus 120578 int1
0119892(119904)119889119904) le 1
For forall119909 isin 119875 cap 120597Ω1205881
we have
Ψ (119909) = (120601119902 (120582) 119879119909) (119905)
=1
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902 (120582int
119904
0
120596 (120591) 119891 (120591 119909 (120591)) 119889120591) 119889119904
minus int
119905
0
(119905 minus 119904)120572minus1
times 120601119902 (120582int
119904
0
120596 (120591) 119891 (120591 119909 (120591)) 119889120591) 119889119904
+120578
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
119892 (119904)
10 Abstract and Applied Analysis
times (int
1
0
(1 minus 120591)120572minus1
times 120601119902 (120582int
120591
0
120596 (120583) 119891 (120583 119909 (120583)) 119889120583)119889120591
minus int
119904
0
(119904 minus 120591)120572minus1
times 120601119902 (120582int
120591
0
120596 (120583) 119891 (120583 119909 (120583)) 119889120583)119889120591)119889119904
le120578
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
119892 (119904)
times (int
1
0
(1 minus 120591)120572minus1
times 120601119902(120582int
120591
0
120596 (120583) 119891 (120583 119909 (120583)) 119889120583)119889120591)119889119904
+1
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902 (120582int
119904
0
120596 (120591) 1205761119891 (120591 119909 (120591)) 119889120591) 119889119904
le120578 int1
0119892 (119904) 119889119904
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
(1 minus 120591)120572minus1
times 120601119902 (120582int
1
0
120596 (120583) 1205761120601119901 (119909 (120583)) 119889120583)119889120591
+1
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902(120582int
1
0
120596 (120591) 1205761120601119901 (119909 (120591)) 119889120591) 119889119904
le1
Γ (120572 + 1) (1 minus 120578 int1
0119892 (119904) 119889119904)
times 120601119902 (120582int
1
0
120596 (120591) 1205761120601119901 (119909) 119889120591)
=120601119902 (1205821205761) 1205742
Γ (120572 + 1) (1 minus 120578 int1
0119892 (119904) 119889119904)
119909 le 119909
(60)
This implied
Ψ119909 le 119909 forall119909 isin 119875 cap 120597Ω1205881
(61)
Let 1205883gt max120588120590 120588
1 and 120582
lowast= (1120577)(120588
3Γ(120572)1205901205741)
119901minus1 thenforall119909 isin 119875 cap 120597Ω120588
3
we have 120590119909 le 119909 le 1205883and
(Ψ119909) (119905)
=1
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902 (120582int
119904
0
120596 (120591) 119891 (120591 119909 (120591)) 119889120591) 119889119904
minus int
119905
0
(119905 minus 119904)120572minus1
times 120601119902 (120582int
119904
0
120596 (120591) 119891 (120591 119909 (120591)) 119889120591) 119889119904
+120578
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
119892 (119904)
times (int
1
0
(1 minus 120591)120572minus1
times 120601119902 (120582int
120591
0
120596 (120583) 119891 (120583 119909 (120583)) 119889120583)119889120591
minus int
119904
0
(119904 minus 120591)120572minus1
times 120601119902(120582int
120591
0
120596 (120583) 119891 (120583 119909 (120583)) 119889120583)119889120591)119889119904
ge120590120578
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
119892 (119904)
times (int
1
0
(1 minus 120591)120572minus1
times 120601119902 (120582int
120591
0
120596 (120583) 119891 (120583 119909 (120583)) 119889120583)119889120591
minus int
119904
0
(119904 minus 120591)120572minus1
times 120601119902(120582int
120591
0
120596 (120583) 119891 (120583 119909 (120583)) 119889120583)119889120591)119889119904
+120590
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902 (120582int
119904
0
120596 (120591) 119891 (120591 119909 (120591)) 119889120591) 119889119904
ge120590
Γ (120572)int
1
0
(1 minus 119904)120572minus1
Abstract and Applied Analysis 11
times 120601119902 (120582int
119904
0
120596 (120591) 119891 (120591 119909 (120591)) 119889120591) 119889119904
gt120590
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902 (120582int
119904
0
120596 (120591) 120577 119889120591) 119889119904
gt120590120601119902 (120582120577)
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902 (int
119904
0
120596 (120591) 119889120591) 119889119904
=120590120601119902 (120582120577) 1205741
Γ (120572)= 1205883
(62)
which implies
Ψ119909 gt 119909 forall119909 isin 119875 cap 120597Ω1205883
(63)
Next for 119891infin
120601= 0 there exist 1205762 gt 0 120588
2gt 1205883gt 0 such
that
119891 (119905 119909) le 1205762120601119901 (119909) (forall119905 isin [0 1] 119909 ge 1205882) (64)
where 1205762 satisfied 120601119902(1205821205762)1205742Γ(120572 + 1)(1 minus 120578 int1
0119892(119904)119889119904) le 1
Similar to the above proof we get
Ψ119909 le 119909 forall119909 isin 119875 cap 120597Ω1205882
(65)
Applying Lemma 7 to (61) (65) and (63) yields that Ψ
has two fixed points 1199091 1199092 such that 1199091 isin 119875cap (Ω1205883
Ω1205881
) and1199092 isin 119875 cap (Ω120588
2
Ω1205883
)(ii) It follows from 119891
0
120601= 0 and 119891
infin
120601= 0 that there exists
1205761 1205762 1205881 1205882 gt 0 such that
119891 (119905 119909) le 1205761120601119901 (119909) (forall119905 isin [0 1] 0 le 119909 le 1205881)
119891 (119905 119909) le 1205762120601119901 (119909) (forall119905 isin [0 1] 119909 ge 1205882)
(66)
then for 119909 isin [1205881 1205882] by the continuous of 119891(119905 119909(119905)) there
exists 119909 isin [1205881 1205882] such that
max1205881le119909le120588
2
119891 (119905 119909)
120601119901 (119909)=
119891 (119905 119909)
120601119901 (119909) (67)
Let Υ = max1205761 1205762 119891(119905 119909)120601119901(119909) gt 0 then
119891 (119905 119909) le Υ120601119901 (119909) (forall0 le 119905 le 1 119909 ge 0) (68)
Assuming 119909(119905) is a positive solution of problem (1) we willshow that this leads to a contradiction for 120582 lt 120582 where
120582 = (1Υ)(Γ(120572 + 1)(1 minus 120578 int1
0119892(119904)119889119904)1205742)
119901minus1 It follows from(26) that
119909 = 119909 (0)
=120578
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
119892 (119904)
times (int
1
0
(1 minus 120591)120572minus1
times 120601119902 (120582int
120591
0
120596 (120583) 119891 (120583 119909 (120583)) 119889120583)119889120591
minus int
119904
0
(119904 minus 120591)120572minus1
times 120601119902(120582int
120591
0
120596 (120583) 119891 (120583 119909 (120583)) 119889120583)119889120591)119889119904
+1
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902 (120582int
119904
0
120596 (120591) 119891 (120591 119909 (120591)) 119889120591) 119889119904
le120578
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
119892 (119904)
times int
1
0
(1 minus 120591)120572minus1
times 120601119902(120582int
120591
0
120596 (120583) 119891 (120583 119909 (120583)) 119889120583)119889120591 119889119904
+1
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902 (120582int
119904
0
120596 (120591) 119891 (120591 119909 (120591)) 119889120591) 119889119904
=1
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
(1 minus 119904)120572minus1
times 120601119902 (120582int
119904
0
120596 (120591) 119891 (120591 119909 (120591)) 119889120591) 119889119904
le1
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
(1 minus 119904)120572minus1
12 Abstract and Applied Analysis
times 120601119902 (120582int
1
0
120596 (120591) Υ120601119901 (119909 (120591)) 119889120591) 119889119904
=120601119902 (120582Υ) 1205742
Γ (120572 + 1) (1 minus 120578 int1
0119892 (119904) 119889119904)
119909
(69)
This implies that 1 le 120601119902(120582Υ)1205742Γ(120572 + 1)(1 minus 120578 int1
0119892(119904)119889119904) =
(120582Υ)119902minus1
1205742Γ(120572+1)(1minus120578 int1
0119892(119904)119889119904) lt (120582 Υ)
119902minus11205742Γ(120572+1)(1minus
120578 int1
0119892(119904)119889119904) = 1 which is a contradiction This finishes the
proof of (ii)
Remark 12 In (i) the condition lim119909rarrinfin
119891(119905 119909) = 119891infin isin
(0 +infin] uniformly for 119905 isin [0 1] can be replaced withcondition (H6) The same methods can be used to prove theresults
Remark 13 From the above proof the condition 119891infin isin
(0 +infin] is important to keep the problem (1) having at leasttwo positive solutions if this condition is not satisfied thenthere exists 120582 gt 0 small enough such that the problem (1) hasno positive solutions for all 120582 lt 120582
Theorem 14 Assume that (H1)ndash(H6) hold(i) If 1198910
120601= +infin and 119891
infin
120601= +infin then there exists 120582lowast0 gt 0
such that problem (1) has at least two solutions for 120582 isin
(0 120582lowast0)(ii) If 1198910
120601= +infin and 119891
infin
120601= +infin then there exists 120582 gt 0
such that problem (1) has no solution for all 120582 gt 120582
Proof (i) It follows from 1198910
120601= +infin that there exists 1199031 gt 0
such that
119891 (119905 119909) ge 119871lowast1120601119901 (119909) (forall119905 isin [0 1] 0 le 119909 le 1199031) (70)
where 119871lowast1 satisfies 1205902120601119902(120582119871lowast1)1205741Γ(120572) ge 1
Then for 119909 isin 119875cap120597Ω1199031
we have 120590119909 le 119909 le 1199031 for 119905 isin [0 1]
and
(Ψ119909) (119905)
=1
Γ (120572)
times int
1
0
(1 minus 119904 )120572minus1
times 120601119902 (120582int
119904
0
120596 (120591) 119891 (120591 119909 (120591)) 119889120591) 119889119904
minus int
119905
0
(119905 minus 119904)120572minus1
times 120601119902 (120582int
119904
0
120596 (120591) 119891 (120591 119909 (120591)) 119889120591) 119889119904
+120578
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
119892 (119904)
times (int
1
0
(1 minus 120591)120572minus1
times 120601119902 (120582int
120591
0
120596 (120583) 119891 (120583 119909 (120583)) 119889120583)119889120591
minus int
119904
0
(119904 minus 120591)120572minus1
times 120601119902 (120582int
120591
0
120596 (120583) 119891 (120583 119909 (120583)) 119889120583)119889120591)119889119904
ge120590120578
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
119892 (119904)
times (int
1
0
(1 minus 120591)120572minus1
times 120601119902 (120582int
120591
0
120596 (120583) 119891 (120583 119909 (120583)) 119889120583)119889120591
minus int
119904
0
(119904 minus 120591)120572minus1
times120601119902 (120582int
120591
0
120596 (120583) 119891 (120583 119909 (120583)) 119889120583)119889120591)119889119904
+120590
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902 (120582int
119904
0
120596 (120591) 119891 (120591 119909 (120591)) 119889120591) 119889119904
ge120590
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902 (120582int
119904
0
120596 (120591) 119891 (120591 119909 (120591)) 119889120591) 119889119904
ge120590
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902 (120582int
119904
0
120596 (120591) 119871lowast1120601119901 (119909 (120591)) 119889120591) 119889119904
ge1205902120601119902 (120582119871lowast1) 119909
Γ (120572)
times int
1
0
(1 minus 119904)120572minus1
times 120601119902 (int
119904
0
120596 (120591) 119889120591) 119889119904
=1205902120601119902 (120582119871lowast1) 1205742
Γ (120572)119909
(71)
Abstract and Applied Analysis 13
which implies that
Ψ119909 ge 119909 forall119909 isin 119875 cap 120597Ω1199031
(72)
Next from 119891infin
120601= +infin there exists 1199032 gt 0 such that
119891 (119905 119909) ge 119871lowast2120601119901 (119909) forall119909 ge 1199032 (73)
where 119871lowast2 satisfies
1205902120601119902 (120582119871lowast2) 1205742
Γ (120572)ge 1 (74)
Let 1199032 gt max1199031 1199032120590 then for 119909 isin 119875 cap 120597Ω1199032
we have
119909 (119905) ge 120590 119909 = 1205901199032 gt 1199032 (75)
Similar to the above proof we can get
Ψ119909 ge 119909 forall119909 isin 119875 cap 120597Ω1199032
(76)
Let 1199031 lt 1199033 lt 11990321198721199033
= max119891(119905 119909) | 119905 isin [0 1] 119909 le 1199033+1 gt
0 and 120582lowast0 = (11198721199033
)(Γ(120572+1)1199033(1minus120578 int1
0119892(119904)119889119904)1205742)
119901minus1Thenfor 119909 isin 119875 cap 120597Ω119903
3
we have
119891 (119905 119909) le 1198721199033
forall119905 isin [0 1] (77)
Hence
(Ψ119909) (119905)
=1
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902 (120582int
119904
0
120596 (120591) 119891 (120591 119909 (120591)) 119889120591) 119889119904
minus int
119905
0
(119905 minus 119904)120572minus1
times 120601119902 (120582int
119904
0
120596 (120591) 119891 (120591 119909 (120591)) 119889120591) 119889119904
+120578
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
119892 (119904)
times (int
1
0
(1 minus 120591)120572minus1
times 120601119902 (120582int
120591
0
120596 (120583) 119891 (120583 119909 (120583)) 119889120583)119889120591
minus int
119904
0
(119904 minus 120591)120572minus1
times120601119902 (120582int
120591
0
120596 (120583) 119891 (120583 119909 (120583)) 119889120583)119889120591)119889119904
le120578
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
119892 (119904)
times int
1
0
(1 minus 120591)120572minus1
120601119902
times (120582int
120591
0
120596 (120583) 119891 (120583 119909 (120583)) 119889120583)119889120591 119889119904
+1
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902 (120582int
119904
0
120596 (120591) 119891 (120591 119909 (120591)) 119889120591) 119889119904
le120578
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
119892 (119904)
times int
1
0
(1 minus 120591)120572minus1
120601119902
times (120582int
120591
0
120596 (120583)1198721199033
119889120583)119889120591119889119904
+1
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902 (120582int
119904
0
120596 (120591)1198721199033
119889120591)119889119904
le120578120601119902 (120582119872119903
3
) int1
0119892 (119904) 119889119904
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
(1 minus 120591)120572minus1
times 120601119902 (int
1
0
120596 (120583) 119889120583)119889120591
+120601119902 (120582119872119903
3
)
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902 (int
1
0
120596 (120591) 119889120591) 119889119904
=1205742120601119902 (120582119872119903
3
)
Γ (120572 + 1) (1 minus 120578 int1
0119892 (119904) 119889119904)
lt1205742120601119902 (120582lowast0119872119903
3
)
Γ (120572 + 1) (1 minus 120578 int1
0119892 (119904) 119889119904)
= 1199033
(78)
14 Abstract and Applied Analysis
This implies that
Ψ119909 lt 119909 forall119909 isin 119875 cap 120597Ω1199033
(79)
Applying Lemma 7 to (72) (76) and (79) yields that Ψ hastwo fixed points 1199091 1199092 such that 1199091 isin 119875 cap (Ω119903
3
Ω1199031
) and1199092 isin 119875 cap (Ω119903
2
Ω1199033
)(ii) By the proof of (i) and the continuity of 119891(119905 119909) there
exists 119909 isin [1199031 1199032] such that min119909isin[11990311199032](119891(119905 119909)120601119901(119909)) =
119891(119905 119909)120601119901(119909)Let Υ = min119871lowast1 119871lowast2 119891(119905 119909)120601119901(119909) since (H6) holds
then Υ gt 0 and
119891 (119905 119909) ge Υ120601119901 (119909) forall119909 ge 0 119905 isin [0 1] (80)
Assuming 119909(119905) is a positive solution of problem (1) we willshow that this leads to contradiction for all 120582 gt 120582 where 120582 =
(1Υ)(Γ(120572)1205741120590)119901minus1 It follows form (26) that
119909 = 119909 (0)
=120578
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
119892 (119904)
times (int
1
0
(1 minus 120591)120572minus1
times 120601119902 (120582int
120591
0
120596 (120583) 119891 (120583 119909 (120583)) 119889120583)119889120591
minus int
119904
0
(119904 minus 120591)120572minus1
times120601119902 (120582int
120591
0
120596 (120583) 119891 (120583 119909 (120583)) 119889120583)119889120591)119889119904
+1
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902 (120582int
119904
0
120596 (120591) 119891 (120591 119909 (120591)) 119889120591) 119889119904
ge1
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902 (120582int
119904
0
120596 (120591) 119891 (120591 119909 (120591)) 119889120591) 119889119904
ge1
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902 (120582int
119904
0
120596 (120591) Υ120601119901 (119909 (120591)) 119889120591) 119889119904
ge1
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902 (120582int
119904
0
120596 (120591) Υ120601119901 (120590 119909 119889120591)) 119889119904
ge120601119902 (120582Υ) 120590 119909
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902 (int
119904
0
120596 (120591) 119889120591) 119889119904
=120601119902 (120582Υ) 1205741120590
Γ (120572)119909
(81)
This implies that 1 ge 120601119902(120582Υ)1205741120590Γ(120572) = (120582Υ)119902minus1
1205741120590Γ(120572) gt
(120582Υ)119902minus1
1205741120590Γ(120572) = 1 which is a contradiction This com-pletes the proof of (ii)
Corollary 15 Assume that (H1)ndash(H5) holds
(i) If lim119909rarrinfin
119891(119905 119909) = 119891infin isin (0 +infin] uniformly for 119905 isin
[0 1] and one of 1198910120601
= 0 and 119891infin
120601= 0 is satisfied then
there exists 120582lowast gt 0 such that the problem (1) has at leastone positive solution for any 120582 gt 120582
lowast(ii) If one of 1198910
120601= +infin and 119891
infin
120601= +infin holds then there
exists 120582lowast0 gt 0 such that problem (1) has at least onepositive solutions for any 120582 isin (0 120582lowast0)
Proof (i) The conclusion is a direct consequence of Theo-rem 11(i)
(ii) The conclusion is a direct consequence of Theo-rem 14(i)
Theorem 16 Assume that (H1)ndash(H5) holdThen the followingconclusions hold
(i) If 1198910120601
= 0 and 119891infin
120601= infin then the problem (1) has a
positive solution 119909120582(119905) for all 120582 gt 0(ii) If 1198910
120601= infin and 119891
infin
120601= 0 then the problem (1) has a
positive solution 119909120582(119905) for all 120582 gt 0
Proof Weonly prove (i) the proof of (ii) is similar sowe omitit here
Let 120582 gt 0 since 1198910
120601= 0 there exists 119903 gt 0 119897 gt 0 such that
119891 (119905 119909) le 119897120601119901 (119909) (forall119905 isin [0 1] 119909 isin [0 119903]) (82)
where 119897 satisfied
1205742120582119902minus1
119897119902minus1
Γ (120572 + 1) (1 minus 120578 int1
0119892 (119904) 119889119904)
le 1 (83)
Similar to the proof of Theorem 14 we obtain that
Ψ119909 le 119909 forall119909 isin 119875 cap 120597Ω119903 (84)
Abstract and Applied Analysis 15
From 119891infin
120601= infin there exists 119877 gt 119903 gt 0 119871 gt 0 such that
119891 (119905 119909) ge 119871120601119901 (119909) (forall119905 isin [0 1] 119909 ge 119877) (85)where 119871 satisfied
1205902120582119902minus1
119871119902minus1
1205741
Γ (120572)ge 1 (86)
We get thatΨ119909 ge 119909 forall119909 isin 119875 cap 120597Ω119877 (87)
Applying Lemma 7 to (84) (87) yields thatΨ has a fixed point119909 such that 119909 isin 119875 cap (Ω119877 Ω119903)
Theorem 17 Assume that (H1)ndash(H6) holdThen the followingconclusions hold
If 0 lt 1198910
120601lt infin and 0 lt 119891
infin
120601lt infin then there exists 1 2 gt
0 such that problem (1) has no positive solution for any 120582 gt 1
and 0 lt 120582 lt 2
Proof It follows from 0 lt 1198910
120601lt infin and 0 lt 119891
infin
120601lt infin that
there exist 1 gt 0 2 gt 0 1 gt 0 2 gt 0 and 1199032 gt 1199031 gt 0
such that1120601119901 (119909) le 119891 (119905 119909) le 1120601119901 (119909)
(forall119905 isin [0 1] 0 le 119909 le 1199031)
2120601119901 (119909) le 119891 (119905 119909) le 2120601119901 (119909)
(forall119905 isin [0 1] 119909 ge 1199032)
(88)
Let = max119891(119905 119909)120601119901(119909) 1199031 le 119909 le 1199032 0 le 119905 le 1 =
min119891(119905 119909)120601119901(119909) 1199031 le 119909 le 1199032 0 le 119905 le 1 By the condition(H6) gt 0 Then forall119909 ge 0 we have
120601119901 (119909) le 119891 (119905 119909) le 120601119901 (119909) (89)
where = max1 2 gt 0 and = min1198971 2 gt 0Let 1 = (1)(Γ(120572)1205741120590)
119901minus1 and 2 =
(1)(Γ(120572 + 1)(1 minus 120578 int1
0119892(119904)119889119904)1205742)
119901minus1
If problem (1)has positive solution 119909(119905) in 119875 then
119909 = 119909 (0)
=120578
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
119892 (119904)
times (int
1
0
(1 minus 120591)120572minus1
times 120601119902 (120582int
120591
0
120596 (120583) 119891 (120583 119909 (120583)) 119889120583)119889120591
minus int
119904
0
(119904 minus 120591)120572minus1
times120601119902 (120582int
120591
0
120596 (120583) 119891 (120583 119909 (120583)) 119889120583)119889120591)119889119904
+1
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902 (120582int
119904
0
120596 (120591) 119891 (120591 119909 (120591)) 119889120591) 119889119904
ge1
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902 (120582int
119904
0
120596 (120591) 119891 (120591 119909 (120591)) 119889120591) 119889119904
ge1
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902 (120582int
119904
0
120596 (120591) 120601119901 (119909 (120591)) 119889120591) 119889119904
ge1
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902 (120582int
119904
0
120596 (120591) 120601119901 (120590 119909 119889120591)) 119889119904
ge120601119902 (120582) 120590 119909
Γ (120572)
times int
1
0
(1 minus 119904)120572minus1
120601119902 (int
119904
0
120596 (120591) 119889120591) 119889119904
=120601119902 (120582) 1205741120590
Γ (120572)119909
gt120601119902 (1 ) 1205741120590
Γ (120572)119909 = 119909
119909 = 119909 (0)
=120578
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
119892 (119904)
times (int
1
0
(1 minus 120591)120572minus1
120601119902
times (120582int
120591
0
120596 (120583) 119891 (120583 119909 (120583)) 119889120583)119889120591
minus int
119904
0
(119904 minus 120591)120572minus1
120601119902
times(120582int
120591
0
120596 (120583) 119891 (120583 119909 (120583)) 119889120583)119889120591)119889119904
+1
Γ (120572)int
1
0
(1 minus 119904)120572minus1
120601119902
times (120582int
119904
0
120596 (120591) 119891 (120591 119909 (120591)) 119889120591) 119889119904
16 Abstract and Applied Analysis
le120578
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
119892 (119904)
times int
1
0
(1 minus 120591)120572minus1
times 120601119902 (120582int
120591
0
120596 (120583) 119891 (120583 119909 (120583)) 119889120583)119889120591 119889119904
+1
Γ (120572)int
1
0
(1 minus 119904)120572minus1
120601119902
times (120582int
119904
0
120596 (120591) 119891 (120591 119909 (120591)) 119889120591) 119889119904
=1
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
(1 minus 119904)120572minus1
120601119902
times (120582int
119904
0
120596 (120591) 119891 (120591 119909 (120591)) 119889120591) 119889119904
le1
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
(1 minus 119904)120572minus1
120601119902
times (120582int
1
0
120596 (120591) 120601119901 (119909 (120591)) 119889120591) 119889119904
=120601119902 (120582) 1205742
Γ (120572 + 1) (1 minus 120578 int1
0119892 (119904) 119889119904)
119909
lt120601119902 (2) 1205742
Γ (120572 + 1) (1 minus 120578 int1
0119892 (119904) 119889119904)
119909 = 119909
(90)
which is a contradiction This finishes the proof of Theo-rem 17
4 Examples
In this section we give out same examples to illustrate ourmain results can be used in practice
Example 1 Consider the following fractional differentialequation
(120601119901 (11986352
0+119909 (119905)))
1015840
+ 120582119905(1 + 1199052)12
119909119901minus1 arctan119909 = 0
0 lt 119905 lt 1
1199091015840(0) = 119909
10158401015840(0) = 0
119909 (1) =1
2int
1
0
1
2119909 (119904) 119889119904
(91)
where 120572 = 52 119901 = 32 119902 = 3 120596(119905) = 119905 119891(119905 119909) = (1 +
1199052)12
119909119901minus1 arctan119909 120578 = 12 and 119892(119905) = 12
By simple computation 119897lowast1= 1205872 119897
lowast
2= radic21205872 1205741 = int
1
0(1minus
119904)32
1206013(int119904
0120591 119889120591)119889119904 asymp 00043 1205742 = 1206013(int
1
0120591 119889120591) = 025 120590 asymp
01429 120582lowast
0= (1119897
lowast
1)(Γ(120572)1205901205741)
119901minus1asymp 25149 120582
lowast
0= (1119897
lowast
2)(Γ(120572+
1)(1 minus 120578 int1
0119892(119904)119889119904)1205742)
119901minus1asymp 14222 Then it is easy to see
all the conditions of Theorem 9(ii) are satisfied thus forall120582 isin
[14222 25149]] the problem (91) has at least one positivesolution
Example 2 Consider the boundary value problem
(120601119901 (11986352
0+119909 (119905)))
1015840
+ 120582119905(1 + 1199052)12
119909119901119890minus119909
= 0
0 lt 119905 lt 1
1199091015840(0) = 119909
10158401015840(0) = 0
119909 (1) =1
2int
1
0
1
2119909 (119904) 119889119904
(92)
where 120572 = 52 119901 = 32 119902 = 3 120596(119905) = 119905 119891(119905 119909) = (1 +
1199052)12
119909119901119890minus119909
120578 = 12 and 119892(119905) = 12
1198910
120601= lim119909rarr0
119891 (119905 119909)
120601119901 (119909)= lim119909rarr0
(1 + 1199052)12
119909119901119890minus119909
119909119901minus1= 0
forall119905 isin [0 1]
1198910
120601= lim119909rarrinfin
119891 (119905 119909)
120601119901 (119909)= lim119909rarrinfin
(1 + 1199052)12
119909119901119890minus119909
119909119901minus1= 0
forall119905 isin [0 1]
(93)
Let 1205881= 12 120588
2= 4 then
119891 (119905 119909)
120601119901 (119909)le (1 + 119905
2)12 1
2radic119890le
1
radic2119890
forall119905 isin [0 1] 119909 isin [01
2]
119891 (119905 119909)
120601119901 (119909)le (1 + 119905
2)12 4
1198904le
4radic2
1198904
forall119905 isin [0 1] 119909 isin [4infin)
(94)
so 1205761 = 1radic2119890 1205762 = 4radic21198904 and forall119909 isin
[12 4] max12le119909le4(119891(119905 119909)120601119901(119909)) = radic2119890minus1 We
have Υ = max1radic2119890 4radic21198904 radic2119890minus1
= radic2119890minus1 1205742 = 025
Abstract and Applied Analysis 17
and 120582 = (1Υ)(Γ(120572+1)(1minus120578 int1
0119892(119904)119889119904)1205742)
119901minus1= 3radic5120587
141198904
By Theorem 11(ii) for any 120582 lt 120582 = 3radic512058714
1198904 the problemhas no positive solution
5 Conclusions
Recently differential equations with p-Laplacian operatorwere widely discussed by several authors In this paperby using fixed point techniques combining with partiallyordered structure of Banach space we obtained the existencemultiply and the dependence of parameter These newresults we presented can be used in numerical computationand analyze mathematical models of physical phenomenamechanics nonlinear dynamics and many other relatedfields
Conflict of Interests
The authors declare that there is no conflict of interestsregarding the publication of this paper
Acknowledgments
The authors are very grateful to anonymous referees for theirvaluable and detailed suggestions and comments to improvethe original paper This work is supported by the Programfor New Century Excellent Talents in University (NCET-10-0097)
References
[1] I Podlubny Fractional Differential Equations vol 198 Aca-demic Press San Diego Calif USA 1999
[2] A A Kilbas H M Srivastava and J J Trujillo Theoryand Applications of Fractional Differential Equations vol 204of North-Holland Mathematics Studies Elsevier Science BVAmsterdam The Netherlands 2006
[3] Y A Rossikhin and M V Shitikova ldquoApplication of fractionalderivatives to the analysis of damped vibrations of viscoelasticsingle mass systemsrdquoActa Mechanica vol 120 no 1ndash4 pp 109ndash125 1997
[4] J Sabatier O Agrawal and J T Machado Advances in Frac-tional Calculus Theoretical developments and applications inphysics and engineering Springer Berlin Germany 2007
[5] N Engheta ldquoOn fractional calculus and fractional multipolesin electromagnetismrdquo IEEE Transactions on Antennas andPropagation vol 44 no 4 pp 554ndash566 1996
[6] R Hilfer Applications of Fractional Calculus in Physics WorldScientific Singapore 2000
[7] R L Bagley and R A Calico ldquoFractional order state equationsfor the control of viscoelastically damped structuresrdquo Journalof Guidance Control and Dynamics vol 14 no 2 pp 304ndash3111991
[8] W-X Zhou andY-D Chu ldquoExistence of solutions for fractionaldifferential equations with multi-point boundary conditionsrdquoCommunications in Nonlinear Science and Numerical Simula-tion vol 17 no 3 pp 1142ndash1148 2012
[9] B Ahmad and G Wang ldquoA study of an impulsive four-pointnonlocal boundary value problemof nonlinear fractional differ-ential equationsrdquo Computers amp Mathematics with Applicationsvol 62 no 3 pp 1341ndash1349 2011
[10] B Ahmad and S Sivasundaram ldquoOn four-point nonlocalboundary value problems of nonlinear integro-differentialequations of fractional orderrdquo Applied Mathematics and Com-putation vol 217 no 2 pp 480ndash487 2010
[11] N Nyamoradi ldquoPositive solutions for multi-point boundaryvalue problems for nonlinear fractional differential equationsrdquoJournal of Contemporary Mathematical Analysis vol 48 no 4pp 145ndash157 2013
[12] B Ahmad and J J Nieto ldquoExistence of solutions for nonlocalboundary value problems of higher-order nonlinear fractionaldifferential equationsrdquo Abstract and Applied Analysis vol 2009Article ID 494720 9 pages 2009
[13] X Liu M Jia and X Xiang ldquoOn the solvability of a fractionaldifferential equationmodel involving the119901-Laplacian operatorrdquoComputers ampMathematics with Applications vol 64 no 10 pp3267ndash3275 2012
[14] J Wang H Xiang and Z Liu ldquoExistence of concave positivesolutions for boundary value problem of nonlinear fractionaldifferential equation with 119901-Laplacian operatorrdquo InternationalJournal of Mathematics and Mathematical Sciences vol 2010Article ID 495138 17 pages 2010
[15] J Wang and H Xiang ldquoUpper and lower solutions method fora class of singular fractional boundary value problems with 119901-Laplacian operatorrdquo Abstract and Applied Analysis vol 2010Article ID 971824 12 pages 2010
[16] H Lu andZHan ldquoExistence of positive solutions for boundary-value problem of fractional differential equation with 119901-laplacian operatorrdquo The American Jouranl of Engineering andTechnology Research vol 11 pp 3757ndash3764 2011
[17] G Chai ldquoPositive solutions for boundary value problem offractional differential equation with 119901-Laplacian operatorrdquoBoundary Value Problems vol 2012 article 18 2012
[18] T Chen W Liu and Z Hu ldquoA boundary value problem forfractional differential equation with 119901-Laplacian operator atresonancerdquoNonlinear AnalysisTheoryMethodsampApplicationsvol 75 no 6 pp 3210ndash3217 2012
[19] T Chen andW Liu ldquoAn anti-periodic boundary value problemfor the fractional differential equation with a 119901-LaplacianoperatorrdquoAppliedMathematics Letters of Rapid Publication vol25 no 11 pp 1671ndash1675 2012
[20] S J Li X G Zhang Y H Wu and L Caccetta ldquoExtremalsolutions for 119901-Laplacian differential systems via iterative com-putationrdquo Applied Mathematics and Computation vol 26 pp1151ndash1158 2013
[21] Z Liu and L Lu ldquoA class of BVPs for nonlinear fractionaldifferential equations with 119901-Laplacian operatorrdquo ElectronicJournal of Qualitative Theory of Differential Equations vol 70pp 1ndash16 2012
[22] Z Liu L Lu and I Szanto ldquoExistence of solutions for fractionalimpulsive differential equations with 119901-Laplacian operatorrdquoActa Mathematica Hungarica vol 141 no 3 pp 203ndash219 2013
[23] Z Bai ldquoEigenvalue intervals for a class of fractional boundaryvalue problemrdquo Computers amp Mathematics with Applicationsvol 64 no 10 pp 3253ndash3257 2012
[24] X Zhang L Wang and Q Sun ldquoExistence of positive solutionsfor a class of nonlinear fractional differential equations withintegral boundary conditions and a parameterrdquo Applied Mathe-matics and Computation vol 226 pp 708ndash718 2014
18 Abstract and Applied Analysis
[25] D J Guo and V Lakshmikantham Nonlinear Problems inAbstract Cones vol 5 Academic Press New York NY USA1988
[26] M A Krasnoselrsquoskii Topological Methods in the Theory ofNonlinear Integral Equations Translated by A H ArmstrongElmsford Pergamon Turkey 1964
Submit your manuscripts athttpwwwhindawicom
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
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Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
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Differential EquationsInternational Journal of
Volume 2014
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OptimizationJournal of
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International Journal of
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Function Spaces
Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of Mathematics and Mathematical Sciences
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Algebra
Discrete Dynamics in Nature and Society
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Decision SciencesAdvances in
Discrete MathematicsJournal of
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Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Stochastic AnalysisInternational Journal of
Abstract and Applied Analysis 5
+int
119905
0
(119905 minus 119904)120572minus1
10038161003816100381610038161003816100381610038161003816120601119902 (int
119904
0
120596 (120591) 119891 (120591 119909119899 (120591)) 119889120591)minus120601119902
times (int
119904
0
120596 (120591) 119891 (120591 119909 (120591)) 119889120591)
10038161003816100381610038161003816100381610038161003816119889119904
+120578
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
119892 (119904)
times (int
1
0
(1 minus 120591)120572minus1
times
10038161003816100381610038161003816100381610038161003816120601119902 (int
120591
0
120596 (120583) 119891 (120583 119909119899 (120583)) 119889120583)
minus 120601119902 (int
120591
0
120596 (120583) 119891 (120583 119909 (120583)) 119889120583)
10038161003816100381610038161003816100381610038161003816119889120591
+ int
119904
0
(119904 minus 120591)120572minus1
times
10038161003816100381610038161003816100381610038161003816120601119902 (int
120591
0
120596 (120583) 119891 (120583 119909119899 (120583)) 119889120583) minus 120601119902
times (int
120591
0
120596 (120583) 119891 (120583 119909 (120583)) 119889120583)
10038161003816100381610038161003816100381610038161003816119889120591) 119889119904
le sup119904isin[01]
10038161003816100381610038161003816100381610038161003816120601119902 (int
119904
0
119908 (120591) 119891 (120591 119909119899 (120591)) 119889120591)
minus 120601119902 (int
119904
0
119908 (120591) 119891 (120591 119909 (120591)) 119889120591)
10038161003816100381610038161003816100381610038161003816
times2
Γ (120572 + 1)(1 +
120578 int1
0119892 (119904) 119889119904
1 minus 120578 int1
0119892 (119904) 119889119904
) 997888rarr 0
119899 997888rarr infin
(29)
Hence 119879 is continuousSecond we show that 119879 is compactFor 119909(119905) isin 119861119903 = 119909 isin 119875 119909 le 119903 by condition (H1)
119891(119905 119909(119905)) lt infin Denote119872 = max0le119905le1119909le119903119891(119905 119909(119905))Then
|119879119909 (119905)|
le1
Γ (120572)int
1
0
(1 minus 119904)120572minus1
10038161003816100381610038161003816100381610038161003816120601119902 (int
119904
0
120596 (120591) 119891 (120591 119909 (120591)) 119889120591)
10038161003816100381610038161003816100381610038161003816119889119904
+int
119905
0
(119905 minus 119904)120572minus1
10038161003816100381610038161003816100381610038161003816120601119902 (int
119904
0
120596 (120591) 119891 (120591 119909 (120591)) 119889120591)
10038161003816100381610038161003816100381610038161003816119889119904
+120578
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
119892 (119904)
times (int
1
0
(1 minus 120591)120572minus1
times
10038161003816100381610038161003816100381610038161003816120601119902 (int
120591
0
120596 (120583) 119891 (120583 119909 (120583)) 119889120583)
10038161003816100381610038161003816100381610038161003816119889120591
+ int
119904
0
(119904 minus 120591)120572minus1
times
10038161003816100381610038161003816100381610038161003816120601119902 (int
120591
0
120596 (120583) 119891 (120583 119909 (120583)) 119889120583)
10038161003816100381610038161003816100381610038161003816119889120591) 119889119904
le1
Γ (120572)
times int
1
0
(1 minus 119904)120572minus1
10038161003816100381610038161003816100381610038161003816120601119902 (119872int
119904
0
120596 (120591) 119889120591)
10038161003816100381610038161003816100381610038161003816119889119904
+int
119905
0
(119905 minus 119904)120572minus1
10038161003816100381610038161003816100381610038161003816120601119902 (119872int
119904
0
120596 (120591) 119889120591)
10038161003816100381610038161003816100381610038161003816119889119904
+120578
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
119892 (119904) (int
1
0
(1 minus 120591)120572minus1
times
10038161003816100381610038161003816100381610038161003816120601119902 (119872int
120591
0
120596 (120583) 119889120583)119889120591
10038161003816100381610038161003816100381610038161003816
+ int
119904
0
(119904 minus 120591)120572minus1
times
10038161003816100381610038161003816100381610038161003816120601119902 (119872int
120591
0
120596 (120583) 119889120583)
10038161003816100381610038161003816100381610038161003816119889120591) 119889119904
le120578
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
120601119902 (119872int
1
0
120596 (120591) 119889120591)
times int
1
0
119892 (119904) (int
1
0
(1 minus 120591)120572minus1
119889120591 + int
119904
0
(119904 minus 120591)120572minus1
119889120591)119889119904
+1
Γ (120572)120601119902 (119872int
1
0
120596 (120591) 119889120591)
times int
1
0
(1 minus 119904)120572minus1
119889119904 + int
119905
0
(119905 minus 119904)120572minus1
119889119904
le
2120601119902 (119872int1
0120596 (120591) 119889120591)
Γ (120572 + 1) (1 minus 120578 int1
0119892 (120591) 119889120591)
lt infin
(30)
that is 119879 maps bounded sets into bounded sets in 119875For 119909(119905) isin 119861119903 0 lt 1199051 lt 1199052 lt 1
1003816100381610038161003816119879119909 (1199052) minus 119879119909 (1199051)1003816100381610038161003816
=1
Γ (120572)
10038161003816100381610038161003816100381610038161003816int
1199052
0
(1199052 minus 119904)120572minus1
120601119902 (int
119904
0
120596 (120591) 119891 (120591 119909 (120591)) 119889120591) 119889119904
6 Abstract and Applied Analysis
minus int
1199051
0
(1199051 minus 119904)120572minus1
times 120601119902 (int
119904
0
120596 (120591) 119891 (120591 119909 (120591)) 119889120591) 119889119904
10038161003816100381610038161003816100381610038161003816
le1
Γ (120572)120601119902 (119872int
1
0
120596 (120591) 119889120591)
times
100381610038161003816100381610038161003816100381610038161003816
int
1199051
0
((1199052 minus 119904)120572minus1
minus (1199051 minus 119904)120572minus1
) 119889119904
+int
1199052
1199051
(1199052 minus 119904)120572minus1
119889119904
100381610038161003816100381610038161003816100381610038161003816
(31)
By mean value theorem we obtain that
int
1199051
0
((1199052 minus 119904)120572minus1
minus (1199051 minus 119904)120572minus1
) 119889119904
le int
1199051
0
(120572 minus 1) (1 minus 119904)120572minus2
(1199052 minus 1199051) 119889119904
le int
1
0
(120572 minus 1) (1 minus 119904)120572minus2
(1199052 minus 1199051) 119889119904
= 1199052 minus 1199051
(32)
Thus1003816100381610038161003816119879119909 (1199052) minus 119879119909 (1199051)
1003816100381610038161003816
le1
Γ (120572)120601119902 (119872int
1
0
120596 (120591) 119889120591) [(1199052 minus 1199051) +1
120572(1199052 minus 1199051)
120572]
(33)
This shows that |119879119909(1199052) minus 119879119909(1199051)| rarr 0 as 1199052 minus 1199051 rarr 0so 119879119909 119909 isin 119861119903 is equicontinuous Therefore the operator119879 119875 rarr 119875 is completely continuous by the Arzela-Ascolitheorem
Now for convenience we introduce the following nota-tions Let
int
1
0
(1 minus 119904)120572minus1
120601119902 (int
119904
0
120596 (120591) 119889120591) 119889119904 = 1205741
120601119902 (int
1
0
120596 (120591) 119889120591) = 1205742
119898 (119903) = minmin119905
119891 (119905 119909)
120601119901 (119903) 119909 isin [120590119903 119903]
(34)
Theorem 9 Assume that (H1)ndash(H5) hold
(i) If 0 lt 1198910
120601lt +infin then there exists 1205850 gt 0 such that for
every 0 lt 119903 lt 1205850 problem (1) has a positive solution119909119903(119905) satisfying 119909119903 = 119903 associated with
120582 = 120582119903 isin [1205820 1205820] (35)
where 1205820 and 1205820 are two positive finite numbers
(ii) If 0 lt 119891infin
120601lt +infin then there exists 120585lowast
0gt 0 such that for
every 119877 gt 120585lowast
0 problem (1) has a positive solution 119909119877(119905)
satisfying 119909119877 = 119877 for any
120582 = 120582119877 isin [120582lowast
0 120582lowast
0] (36)
where 120582lowast
0and 120582
lowast
0are two positive finite numbers
(iii) If 1198910120601
= +infin then there exists 1205851 gt 0 such that for any0 lt 119903lowast
lt 1205851 problem (1) has a positive solution 119909119903lowast(119905)
satisfying 119909119903lowast = 119903lowast for any
120582 = 120582119903lowast isin (0 120582lowast] (37)
where 120582lowast is a positive finite number
(iv) If 119891infin
120601= +infin then there exists 1205851 gt 0 such that
for every 119877lowast gt 1205851 problem (1) has a positive solution119909119877lowast
(119905) satisfying 119909119877lowast
= 119877lowast for any
120582 = 120582119877lowast
isin (0 120582lowast] (38)
where 120582lowast is a positive finite number(v) If there exists 120585 gt 0 such that 119898(119903) gt 120585 then problem
(1) has positive solution 119909119903(119905) satisfying 119909119903 = 119903 forany
120582 = 120582119903 isin (0 ] (39)
where is a positive finite number
Proof (i) It follows from 0 lt 1198910
120601lt +infin that there exists
1198971 1198972 120588 gt 0 such that
1198971120601119901 (119909) le 119891 (119905 119909) le 1198972120601119901 (119909)
(forall119905 isin [0 1] 0 le 119909 le 120588)
(40)
Let 1205850 = 120588120590 we show that 1205850 is required When 119909 isin 119875 cap
120597Ω119903 we have
119909 (119905) ge 120590 119909 = 120590119903 (41)
we need 119909(119905) lt 120588 this implies that 120590119903 lt 120588 as 119903 lt 1205850 so120590119903 lt 1205901205850 = 120588
Let 1205820 = (11198971)(Γ(120572)1205901205741)119901minus1 Then we may assume that
119909 minus 120601119902 (1205820) 119879119909 = 0 (forall119909 isin 119875 cap 120597Ω119903) (42)
if not then there exists119909119903 isin 119875cap120597Ω119903 such that120601119902(1205820)119879119909119903 = 119909119903and then (35) already holds for 120582119903 = 1205820
Define 120595(119905) equiv 1 forall119905 isin [0 1] then 120595(119905) isin 119875 and 120595 equiv 1We now show that
119909 minus 120601119902 (1205820) 119879119909 = 120581120595 (forall119909 isin 119875 cap 120597Ω119903 120581 ge 0) (43)
In fact if there exists 1199091 isin 119875 cap 120597Ω119903 1205811 ge 0 such that 1199091 minus120601119902(1205820)1198791199091 = 1205811120595 then (42) implies that 1205811 gt 0 On the otherhand 1199091 = 120601119902(1205820)1198791199091 + 1205811120595 ge 1205811120595 we may choose 120581
lowast=
max120581 | 1199091 ge 120581120595 then 1205811 le 120581lowastlt +infin 1199091 ge 120581
lowast120595 Therefore
120581lowast= 120581lowast 1003817100381710038171003817120595
1003817100381710038171003817 le10038171003817100381710038171199091
1003817100381710038171003817 = 119903 (44)
Abstract and Applied Analysis 7
Consequently for any 119905 isin [0 1] (26) and (44) imply that
1199091 (119905)
= 120601119902 (1205820)1
Γ (120572)
times int
1
0
(1 minus 119904)120572minus1
times 120601119902 (int
119904
0
120596 (120591) 119891 (120591 1199091 (120591)) 119889120591) 119889119904
minus int
119905
0
(119905 minus 119904)120572minus1
times 120601119902 (int
119904
0
120596 (120591) 119891 (120591 1199091 (120591)) 119889120591) 119889119904
+120578120601119902 (1205820)
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
119892 (119904)
times (int
1
0
(1 minus 120591)120572minus1
times 120601119902 (int
120591
0
120596 (120583) 119891 (120583 1199091 (120583)) 119889120583)119889120591
minus int
119904
0
(119904 minus 120591)120572minus1
times 120601119902 (int
120591
0
120596 (120583) 119891 (120583 1199091 (120583)) 119889120583)119889120591)119889119904
+ 1205811120595
ge120578120590120601119902 (1205820)
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
119892 (119904)
times (int
1
0
(1 minus 120591)120572minus1
times 120601119902 (int
120591
0
120596 (120583) 119891 (120583 1199091 (120583)) 119889120583)119889120591
minus int
119904
0
(119904 minus 120591)120572minus1
120601119902
times (int
120591
0
120596 (120583) 119891 (120583 1199091 (120583)) 119889120583)119889120591)119889119904
+120590120601119902 (1205820)
Γ (120572)
times int
1
0
(1 minus 119904)120572minus1
120601119902 (int
119904
0
120596 (120591) 119891 (120591 1199091 (120591)) 119889120591) 119889119904
+ 1205811120595
ge120590120601119902 (1205820)
Γ (120572)
times int
1
0
(1 minus 119904)120572minus1
120601119902 (int
119904
0
120596 (120591) 119891 (120591 1199091 (120591)) 119889120591) 119889119904
+ 1205811120595
ge120590120601119902 (1205820)
Γ (120572)
times int
1
0
(1 minus 119904)120572minus1
120601119902 (1198971 int
119904
0
120596 (120591) 120601119901 (1199091 (120591)) 119889120591) 119889119904
+ 1205811120595
ge120590120601119902 (1205820)
Γ (120572)
times int
1
0
(1 minus 119904)120572minus1
120601119902 (1198971 int
119904
0
120596 (120591) 120601119901 (120581lowast120595 (120591)) 119889120591) 119889119904
+ 1205811120595
=120590120581lowast119897119902minus1
1120582119902minus1
0
Γ (120572)int
1
0
(1 minus 119904)120572minus1
120601119902 (int
119904
0
120596 (120591) 119889120591) 119889119904 + 1205811120595
=120590120581lowast119897119902minus1
1120582119902minus1
01205741
Γ (120572)+ 1205811120595
= 120581lowast+ 120581120595
(45)
which implies that 1199091(119905) ge (120581lowast+ 120581)120595(119905) 119905 isin [0 1] which is a
contradiction to the definition of 120581lowast Thus (42) holds and byLemma 5 the fixed point index
119894 (120601119902 (1205820) 119879 119875 cap 120597Ω119903 119875) = 0 (46)
On the other hand by the fact that the fixed point index ofconstants operator is 1 so
119894 (120599 119875 cap 120597Ω119903 119875) = 1 (47)
where 120599 is the zero operator It follows therefore from (46)and (47) and the homotopy invariance property that thereexist 119909119903 isin 119875 cap 120597Ω119903 and 0 lt ]0 lt 1 such that ]0120601119902(1205820)119879119909119903 =119909119903 but 119909119903 = 120601119902(120582119903)119879119909119903 which implies that ]0120601119902(1205820)119879119909119903 =
120601119902(120582119903)119879119909119903 it follows that we get ]0120601119902(1205820) = 120601119902(120582119903)) that is
]0120582119902minus1
0= 120582119902minus1
119903 (48)
and then
120582119903 = ]1(119902minus1)0
1205820 = ]119901minus10
1205820 lt 1205820 (49)
From the proof above for any 119903 lt 1205850 there exists a positivesolution 119909119903 isin 119875 cap 120597Ω119903 associated with 120582 = 120582119903 gt 0
8 Abstract and Applied Analysis
Thus
119909119903 (119905)
=120601119902 (120582119903)
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902 (int
119904
0
120596 (120591) 119891 (120591 119909119903 (120591)) 119889120591) 119889119904
minus int
119905
0
(119905 minus 119904)120572minus1
times 120601119902 (int
119904
0
120596 (120591) 119891 (120591 119909119903 (120591)) 119889120591) 119889119904
+120578120601119902 (120582119903)
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
119892 (119904)
times (int
1
0
(1 minus 120591)120572minus1
times 120601119902 (int
120591
0
120596 (120583) 119891 (120583 119909119903 (120583)) 119889120583)119889120591
minus int
119904
0
(119904 minus 120591)120572minus1
times 120601119902 (int
120591
0
120596 (120583) 119891 (120583 119909119903 (120583)) 119889120583)119889120591)119889119904
(50)
with 119909119903 = 119903Next we show that 120582119903 ge 1205820 In fact
119909119903 (119905)
=120601119902 (120582119903)
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902 (int
119904
0
120596 (120591) 119891 (120591 119909119903 (120591)) 119889120591) 119889119904
minus int
119905
0
(119905 minus 119904)120572minus1
times 120601119902 (int
119904
0
120596 (120591) 119891 (120591 119909119903 (120591)) 119889120591) 119889119904
+120578120601119902 (120582119903)
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
119892 (119904)
times (int
1
0
(1 minus 120591)120572minus1
times 120601119902 (int
120591
0
120596 (120583) 119891 (120583 119909119903 (120583)) 119889120583)119889120591
minus int
119904
0
(119904 minus 120591)120572minus1
times 120601119902 (int
120591
0
120596 (120583) 119891 (120583 119909119903 (120583)) 119889120583)119889120591)119889119904
le120578120601119902 (120582119903)
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
119892 (119904) int
1
0
(1 minus 120591)120572minus1
times 120601119902 (int
120591
0
120596 (120583) 119891 (120583 119909119903 (120583)) 119889120583)119889120591 119889119904
+120601119902 (120582119903)
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902 (int
119904
0
120596 (120591) 119891 (120591 119909119903 (120591)) 119889120591) 119889119904
=120601119902 (120582119903) 120578 int
1
0119892 (119904) 119889119904
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
(1 minus 119904)120572minus1
times 120601119902 (int
119904
0
120596 (120591) 119891 (120591 119909119903 (120591)) 119889120591) 119889119904
+120601119902 (120582119903)
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902 (int
119904
0
120596 (120591) 119891 (120591 119909119903 (120591)) 119889120591) 119889119904
le120601119902 (120582119903) 120578 int
1
0119892 (119904) 119889119904
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
(1 minus 119904)120572minus1
times 120601119902 (1198972 int
119904
0
120596 (120591) 119909119903 (120591) 119889120591) 119889119904
+120601119902 (120582119903)
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902 (1198972120582119903 int
119904
0
120596 (120591) 119909119903 (120591) 119889120591) 119889119904
=120601119902 (120582119903)
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
(1 minus 119904)120572minus1
times 120601119902 (1198972 int
119904
0
120596 (120591) 119909119903 (120591) 119889120591) 119889119904
Abstract and Applied Analysis 9
le
10038171003817100381710038171199091199031003817100381710038171003817 120601119902 (1205821199031198972)
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
(1 minus 119904)120572minus1
times 120601119902 (int
1
0
120596 (120591) 119889120591) 119889119904
=
10038171003817100381710038171199091199031003817100381710038171003817 120601119902 (1205821199031198972) 1205742
Γ (120572 + 1) (1 minus 120578 int1
0119892 (119904) 119889119904)
(51)
which implies that
10038171003817100381710038171199091199031003817100381710038171003817 le
10038171003817100381710038171199091199031003817100381710038171003817 120601119902 (1205821199031198972) 1205742
Γ (120572 + 1) (1 minus 120578 int1
0119892 (119904) 119889119904)
(52)
that is
120601119902 (1205821199031198972) 1205742
Γ (120572 + 1) (1 minus 120578 int1
0119892 (119904) 119889119904)
ge 1 (53)
it follows that we get
120582119903 ge1
1198972(
Γ (120572 + 1) (1 minus 120578 int1
0119892 (119904) 119889119904)
1205742)
119901minus1
= 1205820(54)
In conclusion 120582119903 isin [1205820 1205820](ii) It follows from 0 lt 119891
infin
120601lt infin that there exist
119897lowast
1 119897lowast
2 120588lowastgt 0 such that
119897lowast
1120601119901 (119909) le 119891 (119905 119909) le 119897
lowast
2120601119901 (119909)
(forall119905 isin [0 1] 119909 ge 120588lowast)
(55)
Let 120585lowast0= 120588lowast120590 the following proof are similar to (i)
(iii) It follows from 1198910
120601= +infin there exists 119897lowast gt 0 120588
lowastgt 0
such that
1198910
120601ge 119897lowast120601119901 (119909) (forall119905 isin [0 1] 0 le 119909 le 120588
lowast) (56)
Let 1205851 = 120588lowast120590 we show that 1205851 is required When 119909 isin 119875 cap
120597Ω119903lowast we have 119909(119905) ge 120590119909 = 120590119903lowast since 119909 le 120588
lowast then we need120590119903lowastlt 120588lowast as 119903lowast lt 1205851 so 120590119903
lowastlt 120588lowast holds
Let 120582lowast = (1119897lowast)(Γ(120572)1205901205741)
1(119902minus1) we proceed in the sameway as in the proof of (i) replacing (42) we may assume that119909 minus 120601119902(120582
lowast)119879119909 = 0 (forall119909 isin 119875 cap 120597Ω119903lowast) and replacing (43) we
can prove 119909 minus 120601119902(120582lowast)119879119909 = 120581120595 (forall119909 isin 119875 cap 120597Ω119903lowast 120581 ge 0) It
follows from Lemma 5 that 119894(120601119902(120582lowast)119879 119875 cap Ω119903lowast 119875) = 0 Note
that 119894(120579 119875 cap Ω119903lowast 119875) = 1 we can easily show that there exists119909119903lowast isin 119875cap120597Ω119903lowast and 0 lt ]119903lowast lt 1 such that ]119903lowast120601119902(120582
lowast)119879119909119903lowast = 119909119903lowast
Hence (37) holds for 120582119903lowast = 120582lowast]1minus119901119903lowast
lt 120582lowast
The proof ofTheorem 9(iv) follows by themethod similarto Theorem 9(iii) we omit it here
(v) It follows that 119898(119903) gt 120585 for any 119909 isin 119875 cap 120597Ω119903 we have120590119903 le 119909 le 119903 and 119891(119905 119909)120601119901(119909) ge 119891(119905 119909)120601119901(119903) ge 119898(119903) ge 120585then
119891 (119905 119909) ge 120585120601119901 (119909) forall119909 isin [120590119903 119903] 119905 isin [0 1] (57)
Let = (1120585)(Γ(120572)1205901205741)119901minus1 The following proof is similar to
that of (iv) This finished the proof of (v)
Remark 10 InTheorem 9 all the criteria obtained depend onone of 1198910
120601and 119891
infin
120601
Let Ψ(119909) = 120601119902(120582)119879119909 the following theorems give out themultiply nonexistence and the dependence of parameter
Theorem 11 Assume that (H1)ndash(H5) hold
(i) If 1198910
120601= 0 and 119891
infin
120601= 0 lim
119909rarrinfin119891(119905 119909) = 119891infin isin
(0 +infin] uniformly for 119905 isin [0 1] then there exists120582lowast
gt 0 such that the problem (1) has two solutions forany 120582 gt 120582
lowast
(ii) If 1198910120601
= 0 and 119891infin
120601= 0 then there exists 120582 gt 0 such
that for any 120582 lt 120582 problem (1) has no solution
Proof (i) It follows from 119891infin isin (0 +infin] that for any 120577 gt 0there exists 120588 gt 0 such that
119891 (119905 119909) gt 120577 (forall119905 isin [0 1] 119909 ge 120588) (58)
Since 1198910
120601= 0 there exists 1205761 gt 0 120588
1gt 0 such that
119891 (119905 119909) le 1205761120601119901 (119909) (forall119905 isin [0 1] 0 le 119909 le 1205881) (59)
where 1205761 satisfied 120601119902(1205821205761)1205742Γ(120572 + 1)(1 minus 120578 int1
0119892(119904)119889119904) le 1
For forall119909 isin 119875 cap 120597Ω1205881
we have
Ψ (119909) = (120601119902 (120582) 119879119909) (119905)
=1
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902 (120582int
119904
0
120596 (120591) 119891 (120591 119909 (120591)) 119889120591) 119889119904
minus int
119905
0
(119905 minus 119904)120572minus1
times 120601119902 (120582int
119904
0
120596 (120591) 119891 (120591 119909 (120591)) 119889120591) 119889119904
+120578
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
119892 (119904)
10 Abstract and Applied Analysis
times (int
1
0
(1 minus 120591)120572minus1
times 120601119902 (120582int
120591
0
120596 (120583) 119891 (120583 119909 (120583)) 119889120583)119889120591
minus int
119904
0
(119904 minus 120591)120572minus1
times 120601119902 (120582int
120591
0
120596 (120583) 119891 (120583 119909 (120583)) 119889120583)119889120591)119889119904
le120578
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
119892 (119904)
times (int
1
0
(1 minus 120591)120572minus1
times 120601119902(120582int
120591
0
120596 (120583) 119891 (120583 119909 (120583)) 119889120583)119889120591)119889119904
+1
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902 (120582int
119904
0
120596 (120591) 1205761119891 (120591 119909 (120591)) 119889120591) 119889119904
le120578 int1
0119892 (119904) 119889119904
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
(1 minus 120591)120572minus1
times 120601119902 (120582int
1
0
120596 (120583) 1205761120601119901 (119909 (120583)) 119889120583)119889120591
+1
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902(120582int
1
0
120596 (120591) 1205761120601119901 (119909 (120591)) 119889120591) 119889119904
le1
Γ (120572 + 1) (1 minus 120578 int1
0119892 (119904) 119889119904)
times 120601119902 (120582int
1
0
120596 (120591) 1205761120601119901 (119909) 119889120591)
=120601119902 (1205821205761) 1205742
Γ (120572 + 1) (1 minus 120578 int1
0119892 (119904) 119889119904)
119909 le 119909
(60)
This implied
Ψ119909 le 119909 forall119909 isin 119875 cap 120597Ω1205881
(61)
Let 1205883gt max120588120590 120588
1 and 120582
lowast= (1120577)(120588
3Γ(120572)1205901205741)
119901minus1 thenforall119909 isin 119875 cap 120597Ω120588
3
we have 120590119909 le 119909 le 1205883and
(Ψ119909) (119905)
=1
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902 (120582int
119904
0
120596 (120591) 119891 (120591 119909 (120591)) 119889120591) 119889119904
minus int
119905
0
(119905 minus 119904)120572minus1
times 120601119902 (120582int
119904
0
120596 (120591) 119891 (120591 119909 (120591)) 119889120591) 119889119904
+120578
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
119892 (119904)
times (int
1
0
(1 minus 120591)120572minus1
times 120601119902 (120582int
120591
0
120596 (120583) 119891 (120583 119909 (120583)) 119889120583)119889120591
minus int
119904
0
(119904 minus 120591)120572minus1
times 120601119902(120582int
120591
0
120596 (120583) 119891 (120583 119909 (120583)) 119889120583)119889120591)119889119904
ge120590120578
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
119892 (119904)
times (int
1
0
(1 minus 120591)120572minus1
times 120601119902 (120582int
120591
0
120596 (120583) 119891 (120583 119909 (120583)) 119889120583)119889120591
minus int
119904
0
(119904 minus 120591)120572minus1
times 120601119902(120582int
120591
0
120596 (120583) 119891 (120583 119909 (120583)) 119889120583)119889120591)119889119904
+120590
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902 (120582int
119904
0
120596 (120591) 119891 (120591 119909 (120591)) 119889120591) 119889119904
ge120590
Γ (120572)int
1
0
(1 minus 119904)120572minus1
Abstract and Applied Analysis 11
times 120601119902 (120582int
119904
0
120596 (120591) 119891 (120591 119909 (120591)) 119889120591) 119889119904
gt120590
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902 (120582int
119904
0
120596 (120591) 120577 119889120591) 119889119904
gt120590120601119902 (120582120577)
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902 (int
119904
0
120596 (120591) 119889120591) 119889119904
=120590120601119902 (120582120577) 1205741
Γ (120572)= 1205883
(62)
which implies
Ψ119909 gt 119909 forall119909 isin 119875 cap 120597Ω1205883
(63)
Next for 119891infin
120601= 0 there exist 1205762 gt 0 120588
2gt 1205883gt 0 such
that
119891 (119905 119909) le 1205762120601119901 (119909) (forall119905 isin [0 1] 119909 ge 1205882) (64)
where 1205762 satisfied 120601119902(1205821205762)1205742Γ(120572 + 1)(1 minus 120578 int1
0119892(119904)119889119904) le 1
Similar to the above proof we get
Ψ119909 le 119909 forall119909 isin 119875 cap 120597Ω1205882
(65)
Applying Lemma 7 to (61) (65) and (63) yields that Ψ
has two fixed points 1199091 1199092 such that 1199091 isin 119875cap (Ω1205883
Ω1205881
) and1199092 isin 119875 cap (Ω120588
2
Ω1205883
)(ii) It follows from 119891
0
120601= 0 and 119891
infin
120601= 0 that there exists
1205761 1205762 1205881 1205882 gt 0 such that
119891 (119905 119909) le 1205761120601119901 (119909) (forall119905 isin [0 1] 0 le 119909 le 1205881)
119891 (119905 119909) le 1205762120601119901 (119909) (forall119905 isin [0 1] 119909 ge 1205882)
(66)
then for 119909 isin [1205881 1205882] by the continuous of 119891(119905 119909(119905)) there
exists 119909 isin [1205881 1205882] such that
max1205881le119909le120588
2
119891 (119905 119909)
120601119901 (119909)=
119891 (119905 119909)
120601119901 (119909) (67)
Let Υ = max1205761 1205762 119891(119905 119909)120601119901(119909) gt 0 then
119891 (119905 119909) le Υ120601119901 (119909) (forall0 le 119905 le 1 119909 ge 0) (68)
Assuming 119909(119905) is a positive solution of problem (1) we willshow that this leads to a contradiction for 120582 lt 120582 where
120582 = (1Υ)(Γ(120572 + 1)(1 minus 120578 int1
0119892(119904)119889119904)1205742)
119901minus1 It follows from(26) that
119909 = 119909 (0)
=120578
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
119892 (119904)
times (int
1
0
(1 minus 120591)120572minus1
times 120601119902 (120582int
120591
0
120596 (120583) 119891 (120583 119909 (120583)) 119889120583)119889120591
minus int
119904
0
(119904 minus 120591)120572minus1
times 120601119902(120582int
120591
0
120596 (120583) 119891 (120583 119909 (120583)) 119889120583)119889120591)119889119904
+1
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902 (120582int
119904
0
120596 (120591) 119891 (120591 119909 (120591)) 119889120591) 119889119904
le120578
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
119892 (119904)
times int
1
0
(1 minus 120591)120572minus1
times 120601119902(120582int
120591
0
120596 (120583) 119891 (120583 119909 (120583)) 119889120583)119889120591 119889119904
+1
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902 (120582int
119904
0
120596 (120591) 119891 (120591 119909 (120591)) 119889120591) 119889119904
=1
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
(1 minus 119904)120572minus1
times 120601119902 (120582int
119904
0
120596 (120591) 119891 (120591 119909 (120591)) 119889120591) 119889119904
le1
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
(1 minus 119904)120572minus1
12 Abstract and Applied Analysis
times 120601119902 (120582int
1
0
120596 (120591) Υ120601119901 (119909 (120591)) 119889120591) 119889119904
=120601119902 (120582Υ) 1205742
Γ (120572 + 1) (1 minus 120578 int1
0119892 (119904) 119889119904)
119909
(69)
This implies that 1 le 120601119902(120582Υ)1205742Γ(120572 + 1)(1 minus 120578 int1
0119892(119904)119889119904) =
(120582Υ)119902minus1
1205742Γ(120572+1)(1minus120578 int1
0119892(119904)119889119904) lt (120582 Υ)
119902minus11205742Γ(120572+1)(1minus
120578 int1
0119892(119904)119889119904) = 1 which is a contradiction This finishes the
proof of (ii)
Remark 12 In (i) the condition lim119909rarrinfin
119891(119905 119909) = 119891infin isin
(0 +infin] uniformly for 119905 isin [0 1] can be replaced withcondition (H6) The same methods can be used to prove theresults
Remark 13 From the above proof the condition 119891infin isin
(0 +infin] is important to keep the problem (1) having at leasttwo positive solutions if this condition is not satisfied thenthere exists 120582 gt 0 small enough such that the problem (1) hasno positive solutions for all 120582 lt 120582
Theorem 14 Assume that (H1)ndash(H6) hold(i) If 1198910
120601= +infin and 119891
infin
120601= +infin then there exists 120582lowast0 gt 0
such that problem (1) has at least two solutions for 120582 isin
(0 120582lowast0)(ii) If 1198910
120601= +infin and 119891
infin
120601= +infin then there exists 120582 gt 0
such that problem (1) has no solution for all 120582 gt 120582
Proof (i) It follows from 1198910
120601= +infin that there exists 1199031 gt 0
such that
119891 (119905 119909) ge 119871lowast1120601119901 (119909) (forall119905 isin [0 1] 0 le 119909 le 1199031) (70)
where 119871lowast1 satisfies 1205902120601119902(120582119871lowast1)1205741Γ(120572) ge 1
Then for 119909 isin 119875cap120597Ω1199031
we have 120590119909 le 119909 le 1199031 for 119905 isin [0 1]
and
(Ψ119909) (119905)
=1
Γ (120572)
times int
1
0
(1 minus 119904 )120572minus1
times 120601119902 (120582int
119904
0
120596 (120591) 119891 (120591 119909 (120591)) 119889120591) 119889119904
minus int
119905
0
(119905 minus 119904)120572minus1
times 120601119902 (120582int
119904
0
120596 (120591) 119891 (120591 119909 (120591)) 119889120591) 119889119904
+120578
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
119892 (119904)
times (int
1
0
(1 minus 120591)120572minus1
times 120601119902 (120582int
120591
0
120596 (120583) 119891 (120583 119909 (120583)) 119889120583)119889120591
minus int
119904
0
(119904 minus 120591)120572minus1
times 120601119902 (120582int
120591
0
120596 (120583) 119891 (120583 119909 (120583)) 119889120583)119889120591)119889119904
ge120590120578
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
119892 (119904)
times (int
1
0
(1 minus 120591)120572minus1
times 120601119902 (120582int
120591
0
120596 (120583) 119891 (120583 119909 (120583)) 119889120583)119889120591
minus int
119904
0
(119904 minus 120591)120572minus1
times120601119902 (120582int
120591
0
120596 (120583) 119891 (120583 119909 (120583)) 119889120583)119889120591)119889119904
+120590
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902 (120582int
119904
0
120596 (120591) 119891 (120591 119909 (120591)) 119889120591) 119889119904
ge120590
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902 (120582int
119904
0
120596 (120591) 119891 (120591 119909 (120591)) 119889120591) 119889119904
ge120590
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902 (120582int
119904
0
120596 (120591) 119871lowast1120601119901 (119909 (120591)) 119889120591) 119889119904
ge1205902120601119902 (120582119871lowast1) 119909
Γ (120572)
times int
1
0
(1 minus 119904)120572minus1
times 120601119902 (int
119904
0
120596 (120591) 119889120591) 119889119904
=1205902120601119902 (120582119871lowast1) 1205742
Γ (120572)119909
(71)
Abstract and Applied Analysis 13
which implies that
Ψ119909 ge 119909 forall119909 isin 119875 cap 120597Ω1199031
(72)
Next from 119891infin
120601= +infin there exists 1199032 gt 0 such that
119891 (119905 119909) ge 119871lowast2120601119901 (119909) forall119909 ge 1199032 (73)
where 119871lowast2 satisfies
1205902120601119902 (120582119871lowast2) 1205742
Γ (120572)ge 1 (74)
Let 1199032 gt max1199031 1199032120590 then for 119909 isin 119875 cap 120597Ω1199032
we have
119909 (119905) ge 120590 119909 = 1205901199032 gt 1199032 (75)
Similar to the above proof we can get
Ψ119909 ge 119909 forall119909 isin 119875 cap 120597Ω1199032
(76)
Let 1199031 lt 1199033 lt 11990321198721199033
= max119891(119905 119909) | 119905 isin [0 1] 119909 le 1199033+1 gt
0 and 120582lowast0 = (11198721199033
)(Γ(120572+1)1199033(1minus120578 int1
0119892(119904)119889119904)1205742)
119901minus1Thenfor 119909 isin 119875 cap 120597Ω119903
3
we have
119891 (119905 119909) le 1198721199033
forall119905 isin [0 1] (77)
Hence
(Ψ119909) (119905)
=1
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902 (120582int
119904
0
120596 (120591) 119891 (120591 119909 (120591)) 119889120591) 119889119904
minus int
119905
0
(119905 minus 119904)120572minus1
times 120601119902 (120582int
119904
0
120596 (120591) 119891 (120591 119909 (120591)) 119889120591) 119889119904
+120578
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
119892 (119904)
times (int
1
0
(1 minus 120591)120572minus1
times 120601119902 (120582int
120591
0
120596 (120583) 119891 (120583 119909 (120583)) 119889120583)119889120591
minus int
119904
0
(119904 minus 120591)120572minus1
times120601119902 (120582int
120591
0
120596 (120583) 119891 (120583 119909 (120583)) 119889120583)119889120591)119889119904
le120578
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
119892 (119904)
times int
1
0
(1 minus 120591)120572minus1
120601119902
times (120582int
120591
0
120596 (120583) 119891 (120583 119909 (120583)) 119889120583)119889120591 119889119904
+1
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902 (120582int
119904
0
120596 (120591) 119891 (120591 119909 (120591)) 119889120591) 119889119904
le120578
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
119892 (119904)
times int
1
0
(1 minus 120591)120572minus1
120601119902
times (120582int
120591
0
120596 (120583)1198721199033
119889120583)119889120591119889119904
+1
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902 (120582int
119904
0
120596 (120591)1198721199033
119889120591)119889119904
le120578120601119902 (120582119872119903
3
) int1
0119892 (119904) 119889119904
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
(1 minus 120591)120572minus1
times 120601119902 (int
1
0
120596 (120583) 119889120583)119889120591
+120601119902 (120582119872119903
3
)
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902 (int
1
0
120596 (120591) 119889120591) 119889119904
=1205742120601119902 (120582119872119903
3
)
Γ (120572 + 1) (1 minus 120578 int1
0119892 (119904) 119889119904)
lt1205742120601119902 (120582lowast0119872119903
3
)
Γ (120572 + 1) (1 minus 120578 int1
0119892 (119904) 119889119904)
= 1199033
(78)
14 Abstract and Applied Analysis
This implies that
Ψ119909 lt 119909 forall119909 isin 119875 cap 120597Ω1199033
(79)
Applying Lemma 7 to (72) (76) and (79) yields that Ψ hastwo fixed points 1199091 1199092 such that 1199091 isin 119875 cap (Ω119903
3
Ω1199031
) and1199092 isin 119875 cap (Ω119903
2
Ω1199033
)(ii) By the proof of (i) and the continuity of 119891(119905 119909) there
exists 119909 isin [1199031 1199032] such that min119909isin[11990311199032](119891(119905 119909)120601119901(119909)) =
119891(119905 119909)120601119901(119909)Let Υ = min119871lowast1 119871lowast2 119891(119905 119909)120601119901(119909) since (H6) holds
then Υ gt 0 and
119891 (119905 119909) ge Υ120601119901 (119909) forall119909 ge 0 119905 isin [0 1] (80)
Assuming 119909(119905) is a positive solution of problem (1) we willshow that this leads to contradiction for all 120582 gt 120582 where 120582 =
(1Υ)(Γ(120572)1205741120590)119901minus1 It follows form (26) that
119909 = 119909 (0)
=120578
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
119892 (119904)
times (int
1
0
(1 minus 120591)120572minus1
times 120601119902 (120582int
120591
0
120596 (120583) 119891 (120583 119909 (120583)) 119889120583)119889120591
minus int
119904
0
(119904 minus 120591)120572minus1
times120601119902 (120582int
120591
0
120596 (120583) 119891 (120583 119909 (120583)) 119889120583)119889120591)119889119904
+1
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902 (120582int
119904
0
120596 (120591) 119891 (120591 119909 (120591)) 119889120591) 119889119904
ge1
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902 (120582int
119904
0
120596 (120591) 119891 (120591 119909 (120591)) 119889120591) 119889119904
ge1
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902 (120582int
119904
0
120596 (120591) Υ120601119901 (119909 (120591)) 119889120591) 119889119904
ge1
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902 (120582int
119904
0
120596 (120591) Υ120601119901 (120590 119909 119889120591)) 119889119904
ge120601119902 (120582Υ) 120590 119909
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902 (int
119904
0
120596 (120591) 119889120591) 119889119904
=120601119902 (120582Υ) 1205741120590
Γ (120572)119909
(81)
This implies that 1 ge 120601119902(120582Υ)1205741120590Γ(120572) = (120582Υ)119902minus1
1205741120590Γ(120572) gt
(120582Υ)119902minus1
1205741120590Γ(120572) = 1 which is a contradiction This com-pletes the proof of (ii)
Corollary 15 Assume that (H1)ndash(H5) holds
(i) If lim119909rarrinfin
119891(119905 119909) = 119891infin isin (0 +infin] uniformly for 119905 isin
[0 1] and one of 1198910120601
= 0 and 119891infin
120601= 0 is satisfied then
there exists 120582lowast gt 0 such that the problem (1) has at leastone positive solution for any 120582 gt 120582
lowast(ii) If one of 1198910
120601= +infin and 119891
infin
120601= +infin holds then there
exists 120582lowast0 gt 0 such that problem (1) has at least onepositive solutions for any 120582 isin (0 120582lowast0)
Proof (i) The conclusion is a direct consequence of Theo-rem 11(i)
(ii) The conclusion is a direct consequence of Theo-rem 14(i)
Theorem 16 Assume that (H1)ndash(H5) holdThen the followingconclusions hold
(i) If 1198910120601
= 0 and 119891infin
120601= infin then the problem (1) has a
positive solution 119909120582(119905) for all 120582 gt 0(ii) If 1198910
120601= infin and 119891
infin
120601= 0 then the problem (1) has a
positive solution 119909120582(119905) for all 120582 gt 0
Proof Weonly prove (i) the proof of (ii) is similar sowe omitit here
Let 120582 gt 0 since 1198910
120601= 0 there exists 119903 gt 0 119897 gt 0 such that
119891 (119905 119909) le 119897120601119901 (119909) (forall119905 isin [0 1] 119909 isin [0 119903]) (82)
where 119897 satisfied
1205742120582119902minus1
119897119902minus1
Γ (120572 + 1) (1 minus 120578 int1
0119892 (119904) 119889119904)
le 1 (83)
Similar to the proof of Theorem 14 we obtain that
Ψ119909 le 119909 forall119909 isin 119875 cap 120597Ω119903 (84)
Abstract and Applied Analysis 15
From 119891infin
120601= infin there exists 119877 gt 119903 gt 0 119871 gt 0 such that
119891 (119905 119909) ge 119871120601119901 (119909) (forall119905 isin [0 1] 119909 ge 119877) (85)where 119871 satisfied
1205902120582119902minus1
119871119902minus1
1205741
Γ (120572)ge 1 (86)
We get thatΨ119909 ge 119909 forall119909 isin 119875 cap 120597Ω119877 (87)
Applying Lemma 7 to (84) (87) yields thatΨ has a fixed point119909 such that 119909 isin 119875 cap (Ω119877 Ω119903)
Theorem 17 Assume that (H1)ndash(H6) holdThen the followingconclusions hold
If 0 lt 1198910
120601lt infin and 0 lt 119891
infin
120601lt infin then there exists 1 2 gt
0 such that problem (1) has no positive solution for any 120582 gt 1
and 0 lt 120582 lt 2
Proof It follows from 0 lt 1198910
120601lt infin and 0 lt 119891
infin
120601lt infin that
there exist 1 gt 0 2 gt 0 1 gt 0 2 gt 0 and 1199032 gt 1199031 gt 0
such that1120601119901 (119909) le 119891 (119905 119909) le 1120601119901 (119909)
(forall119905 isin [0 1] 0 le 119909 le 1199031)
2120601119901 (119909) le 119891 (119905 119909) le 2120601119901 (119909)
(forall119905 isin [0 1] 119909 ge 1199032)
(88)
Let = max119891(119905 119909)120601119901(119909) 1199031 le 119909 le 1199032 0 le 119905 le 1 =
min119891(119905 119909)120601119901(119909) 1199031 le 119909 le 1199032 0 le 119905 le 1 By the condition(H6) gt 0 Then forall119909 ge 0 we have
120601119901 (119909) le 119891 (119905 119909) le 120601119901 (119909) (89)
where = max1 2 gt 0 and = min1198971 2 gt 0Let 1 = (1)(Γ(120572)1205741120590)
119901minus1 and 2 =
(1)(Γ(120572 + 1)(1 minus 120578 int1
0119892(119904)119889119904)1205742)
119901minus1
If problem (1)has positive solution 119909(119905) in 119875 then
119909 = 119909 (0)
=120578
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
119892 (119904)
times (int
1
0
(1 minus 120591)120572minus1
times 120601119902 (120582int
120591
0
120596 (120583) 119891 (120583 119909 (120583)) 119889120583)119889120591
minus int
119904
0
(119904 minus 120591)120572minus1
times120601119902 (120582int
120591
0
120596 (120583) 119891 (120583 119909 (120583)) 119889120583)119889120591)119889119904
+1
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902 (120582int
119904
0
120596 (120591) 119891 (120591 119909 (120591)) 119889120591) 119889119904
ge1
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902 (120582int
119904
0
120596 (120591) 119891 (120591 119909 (120591)) 119889120591) 119889119904
ge1
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902 (120582int
119904
0
120596 (120591) 120601119901 (119909 (120591)) 119889120591) 119889119904
ge1
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902 (120582int
119904
0
120596 (120591) 120601119901 (120590 119909 119889120591)) 119889119904
ge120601119902 (120582) 120590 119909
Γ (120572)
times int
1
0
(1 minus 119904)120572minus1
120601119902 (int
119904
0
120596 (120591) 119889120591) 119889119904
=120601119902 (120582) 1205741120590
Γ (120572)119909
gt120601119902 (1 ) 1205741120590
Γ (120572)119909 = 119909
119909 = 119909 (0)
=120578
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
119892 (119904)
times (int
1
0
(1 minus 120591)120572minus1
120601119902
times (120582int
120591
0
120596 (120583) 119891 (120583 119909 (120583)) 119889120583)119889120591
minus int
119904
0
(119904 minus 120591)120572minus1
120601119902
times(120582int
120591
0
120596 (120583) 119891 (120583 119909 (120583)) 119889120583)119889120591)119889119904
+1
Γ (120572)int
1
0
(1 minus 119904)120572minus1
120601119902
times (120582int
119904
0
120596 (120591) 119891 (120591 119909 (120591)) 119889120591) 119889119904
16 Abstract and Applied Analysis
le120578
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
119892 (119904)
times int
1
0
(1 minus 120591)120572minus1
times 120601119902 (120582int
120591
0
120596 (120583) 119891 (120583 119909 (120583)) 119889120583)119889120591 119889119904
+1
Γ (120572)int
1
0
(1 minus 119904)120572minus1
120601119902
times (120582int
119904
0
120596 (120591) 119891 (120591 119909 (120591)) 119889120591) 119889119904
=1
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
(1 minus 119904)120572minus1
120601119902
times (120582int
119904
0
120596 (120591) 119891 (120591 119909 (120591)) 119889120591) 119889119904
le1
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
(1 minus 119904)120572minus1
120601119902
times (120582int
1
0
120596 (120591) 120601119901 (119909 (120591)) 119889120591) 119889119904
=120601119902 (120582) 1205742
Γ (120572 + 1) (1 minus 120578 int1
0119892 (119904) 119889119904)
119909
lt120601119902 (2) 1205742
Γ (120572 + 1) (1 minus 120578 int1
0119892 (119904) 119889119904)
119909 = 119909
(90)
which is a contradiction This finishes the proof of Theo-rem 17
4 Examples
In this section we give out same examples to illustrate ourmain results can be used in practice
Example 1 Consider the following fractional differentialequation
(120601119901 (11986352
0+119909 (119905)))
1015840
+ 120582119905(1 + 1199052)12
119909119901minus1 arctan119909 = 0
0 lt 119905 lt 1
1199091015840(0) = 119909
10158401015840(0) = 0
119909 (1) =1
2int
1
0
1
2119909 (119904) 119889119904
(91)
where 120572 = 52 119901 = 32 119902 = 3 120596(119905) = 119905 119891(119905 119909) = (1 +
1199052)12
119909119901minus1 arctan119909 120578 = 12 and 119892(119905) = 12
By simple computation 119897lowast1= 1205872 119897
lowast
2= radic21205872 1205741 = int
1
0(1minus
119904)32
1206013(int119904
0120591 119889120591)119889119904 asymp 00043 1205742 = 1206013(int
1
0120591 119889120591) = 025 120590 asymp
01429 120582lowast
0= (1119897
lowast
1)(Γ(120572)1205901205741)
119901minus1asymp 25149 120582
lowast
0= (1119897
lowast
2)(Γ(120572+
1)(1 minus 120578 int1
0119892(119904)119889119904)1205742)
119901minus1asymp 14222 Then it is easy to see
all the conditions of Theorem 9(ii) are satisfied thus forall120582 isin
[14222 25149]] the problem (91) has at least one positivesolution
Example 2 Consider the boundary value problem
(120601119901 (11986352
0+119909 (119905)))
1015840
+ 120582119905(1 + 1199052)12
119909119901119890minus119909
= 0
0 lt 119905 lt 1
1199091015840(0) = 119909
10158401015840(0) = 0
119909 (1) =1
2int
1
0
1
2119909 (119904) 119889119904
(92)
where 120572 = 52 119901 = 32 119902 = 3 120596(119905) = 119905 119891(119905 119909) = (1 +
1199052)12
119909119901119890minus119909
120578 = 12 and 119892(119905) = 12
1198910
120601= lim119909rarr0
119891 (119905 119909)
120601119901 (119909)= lim119909rarr0
(1 + 1199052)12
119909119901119890minus119909
119909119901minus1= 0
forall119905 isin [0 1]
1198910
120601= lim119909rarrinfin
119891 (119905 119909)
120601119901 (119909)= lim119909rarrinfin
(1 + 1199052)12
119909119901119890minus119909
119909119901minus1= 0
forall119905 isin [0 1]
(93)
Let 1205881= 12 120588
2= 4 then
119891 (119905 119909)
120601119901 (119909)le (1 + 119905
2)12 1
2radic119890le
1
radic2119890
forall119905 isin [0 1] 119909 isin [01
2]
119891 (119905 119909)
120601119901 (119909)le (1 + 119905
2)12 4
1198904le
4radic2
1198904
forall119905 isin [0 1] 119909 isin [4infin)
(94)
so 1205761 = 1radic2119890 1205762 = 4radic21198904 and forall119909 isin
[12 4] max12le119909le4(119891(119905 119909)120601119901(119909)) = radic2119890minus1 We
have Υ = max1radic2119890 4radic21198904 radic2119890minus1
= radic2119890minus1 1205742 = 025
Abstract and Applied Analysis 17
and 120582 = (1Υ)(Γ(120572+1)(1minus120578 int1
0119892(119904)119889119904)1205742)
119901minus1= 3radic5120587
141198904
By Theorem 11(ii) for any 120582 lt 120582 = 3radic512058714
1198904 the problemhas no positive solution
5 Conclusions
Recently differential equations with p-Laplacian operatorwere widely discussed by several authors In this paperby using fixed point techniques combining with partiallyordered structure of Banach space we obtained the existencemultiply and the dependence of parameter These newresults we presented can be used in numerical computationand analyze mathematical models of physical phenomenamechanics nonlinear dynamics and many other relatedfields
Conflict of Interests
The authors declare that there is no conflict of interestsregarding the publication of this paper
Acknowledgments
The authors are very grateful to anonymous referees for theirvaluable and detailed suggestions and comments to improvethe original paper This work is supported by the Programfor New Century Excellent Talents in University (NCET-10-0097)
References
[1] I Podlubny Fractional Differential Equations vol 198 Aca-demic Press San Diego Calif USA 1999
[2] A A Kilbas H M Srivastava and J J Trujillo Theoryand Applications of Fractional Differential Equations vol 204of North-Holland Mathematics Studies Elsevier Science BVAmsterdam The Netherlands 2006
[3] Y A Rossikhin and M V Shitikova ldquoApplication of fractionalderivatives to the analysis of damped vibrations of viscoelasticsingle mass systemsrdquoActa Mechanica vol 120 no 1ndash4 pp 109ndash125 1997
[4] J Sabatier O Agrawal and J T Machado Advances in Frac-tional Calculus Theoretical developments and applications inphysics and engineering Springer Berlin Germany 2007
[5] N Engheta ldquoOn fractional calculus and fractional multipolesin electromagnetismrdquo IEEE Transactions on Antennas andPropagation vol 44 no 4 pp 554ndash566 1996
[6] R Hilfer Applications of Fractional Calculus in Physics WorldScientific Singapore 2000
[7] R L Bagley and R A Calico ldquoFractional order state equationsfor the control of viscoelastically damped structuresrdquo Journalof Guidance Control and Dynamics vol 14 no 2 pp 304ndash3111991
[8] W-X Zhou andY-D Chu ldquoExistence of solutions for fractionaldifferential equations with multi-point boundary conditionsrdquoCommunications in Nonlinear Science and Numerical Simula-tion vol 17 no 3 pp 1142ndash1148 2012
[9] B Ahmad and G Wang ldquoA study of an impulsive four-pointnonlocal boundary value problemof nonlinear fractional differ-ential equationsrdquo Computers amp Mathematics with Applicationsvol 62 no 3 pp 1341ndash1349 2011
[10] B Ahmad and S Sivasundaram ldquoOn four-point nonlocalboundary value problems of nonlinear integro-differentialequations of fractional orderrdquo Applied Mathematics and Com-putation vol 217 no 2 pp 480ndash487 2010
[11] N Nyamoradi ldquoPositive solutions for multi-point boundaryvalue problems for nonlinear fractional differential equationsrdquoJournal of Contemporary Mathematical Analysis vol 48 no 4pp 145ndash157 2013
[12] B Ahmad and J J Nieto ldquoExistence of solutions for nonlocalboundary value problems of higher-order nonlinear fractionaldifferential equationsrdquo Abstract and Applied Analysis vol 2009Article ID 494720 9 pages 2009
[13] X Liu M Jia and X Xiang ldquoOn the solvability of a fractionaldifferential equationmodel involving the119901-Laplacian operatorrdquoComputers ampMathematics with Applications vol 64 no 10 pp3267ndash3275 2012
[14] J Wang H Xiang and Z Liu ldquoExistence of concave positivesolutions for boundary value problem of nonlinear fractionaldifferential equation with 119901-Laplacian operatorrdquo InternationalJournal of Mathematics and Mathematical Sciences vol 2010Article ID 495138 17 pages 2010
[15] J Wang and H Xiang ldquoUpper and lower solutions method fora class of singular fractional boundary value problems with 119901-Laplacian operatorrdquo Abstract and Applied Analysis vol 2010Article ID 971824 12 pages 2010
[16] H Lu andZHan ldquoExistence of positive solutions for boundary-value problem of fractional differential equation with 119901-laplacian operatorrdquo The American Jouranl of Engineering andTechnology Research vol 11 pp 3757ndash3764 2011
[17] G Chai ldquoPositive solutions for boundary value problem offractional differential equation with 119901-Laplacian operatorrdquoBoundary Value Problems vol 2012 article 18 2012
[18] T Chen W Liu and Z Hu ldquoA boundary value problem forfractional differential equation with 119901-Laplacian operator atresonancerdquoNonlinear AnalysisTheoryMethodsampApplicationsvol 75 no 6 pp 3210ndash3217 2012
[19] T Chen andW Liu ldquoAn anti-periodic boundary value problemfor the fractional differential equation with a 119901-LaplacianoperatorrdquoAppliedMathematics Letters of Rapid Publication vol25 no 11 pp 1671ndash1675 2012
[20] S J Li X G Zhang Y H Wu and L Caccetta ldquoExtremalsolutions for 119901-Laplacian differential systems via iterative com-putationrdquo Applied Mathematics and Computation vol 26 pp1151ndash1158 2013
[21] Z Liu and L Lu ldquoA class of BVPs for nonlinear fractionaldifferential equations with 119901-Laplacian operatorrdquo ElectronicJournal of Qualitative Theory of Differential Equations vol 70pp 1ndash16 2012
[22] Z Liu L Lu and I Szanto ldquoExistence of solutions for fractionalimpulsive differential equations with 119901-Laplacian operatorrdquoActa Mathematica Hungarica vol 141 no 3 pp 203ndash219 2013
[23] Z Bai ldquoEigenvalue intervals for a class of fractional boundaryvalue problemrdquo Computers amp Mathematics with Applicationsvol 64 no 10 pp 3253ndash3257 2012
[24] X Zhang L Wang and Q Sun ldquoExistence of positive solutionsfor a class of nonlinear fractional differential equations withintegral boundary conditions and a parameterrdquo Applied Mathe-matics and Computation vol 226 pp 708ndash718 2014
18 Abstract and Applied Analysis
[25] D J Guo and V Lakshmikantham Nonlinear Problems inAbstract Cones vol 5 Academic Press New York NY USA1988
[26] M A Krasnoselrsquoskii Topological Methods in the Theory ofNonlinear Integral Equations Translated by A H ArmstrongElmsford Pergamon Turkey 1964
Submit your manuscripts athttpwwwhindawicom
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical Problems in Engineering
Hindawi Publishing Corporationhttpwwwhindawicom
Differential EquationsInternational Journal of
Volume 2014
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Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical PhysicsAdvances in
Complex AnalysisJournal of
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OptimizationJournal of
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CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of
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Operations ResearchAdvances in
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Function Spaces
Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of Mathematics and Mathematical Sciences
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014
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Algebra
Discrete Dynamics in Nature and Society
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
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Decision SciencesAdvances in
Discrete MathematicsJournal of
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Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Stochastic AnalysisInternational Journal of
6 Abstract and Applied Analysis
minus int
1199051
0
(1199051 minus 119904)120572minus1
times 120601119902 (int
119904
0
120596 (120591) 119891 (120591 119909 (120591)) 119889120591) 119889119904
10038161003816100381610038161003816100381610038161003816
le1
Γ (120572)120601119902 (119872int
1
0
120596 (120591) 119889120591)
times
100381610038161003816100381610038161003816100381610038161003816
int
1199051
0
((1199052 minus 119904)120572minus1
minus (1199051 minus 119904)120572minus1
) 119889119904
+int
1199052
1199051
(1199052 minus 119904)120572minus1
119889119904
100381610038161003816100381610038161003816100381610038161003816
(31)
By mean value theorem we obtain that
int
1199051
0
((1199052 minus 119904)120572minus1
minus (1199051 minus 119904)120572minus1
) 119889119904
le int
1199051
0
(120572 minus 1) (1 minus 119904)120572minus2
(1199052 minus 1199051) 119889119904
le int
1
0
(120572 minus 1) (1 minus 119904)120572minus2
(1199052 minus 1199051) 119889119904
= 1199052 minus 1199051
(32)
Thus1003816100381610038161003816119879119909 (1199052) minus 119879119909 (1199051)
1003816100381610038161003816
le1
Γ (120572)120601119902 (119872int
1
0
120596 (120591) 119889120591) [(1199052 minus 1199051) +1
120572(1199052 minus 1199051)
120572]
(33)
This shows that |119879119909(1199052) minus 119879119909(1199051)| rarr 0 as 1199052 minus 1199051 rarr 0so 119879119909 119909 isin 119861119903 is equicontinuous Therefore the operator119879 119875 rarr 119875 is completely continuous by the Arzela-Ascolitheorem
Now for convenience we introduce the following nota-tions Let
int
1
0
(1 minus 119904)120572minus1
120601119902 (int
119904
0
120596 (120591) 119889120591) 119889119904 = 1205741
120601119902 (int
1
0
120596 (120591) 119889120591) = 1205742
119898 (119903) = minmin119905
119891 (119905 119909)
120601119901 (119903) 119909 isin [120590119903 119903]
(34)
Theorem 9 Assume that (H1)ndash(H5) hold
(i) If 0 lt 1198910
120601lt +infin then there exists 1205850 gt 0 such that for
every 0 lt 119903 lt 1205850 problem (1) has a positive solution119909119903(119905) satisfying 119909119903 = 119903 associated with
120582 = 120582119903 isin [1205820 1205820] (35)
where 1205820 and 1205820 are two positive finite numbers
(ii) If 0 lt 119891infin
120601lt +infin then there exists 120585lowast
0gt 0 such that for
every 119877 gt 120585lowast
0 problem (1) has a positive solution 119909119877(119905)
satisfying 119909119877 = 119877 for any
120582 = 120582119877 isin [120582lowast
0 120582lowast
0] (36)
where 120582lowast
0and 120582
lowast
0are two positive finite numbers
(iii) If 1198910120601
= +infin then there exists 1205851 gt 0 such that for any0 lt 119903lowast
lt 1205851 problem (1) has a positive solution 119909119903lowast(119905)
satisfying 119909119903lowast = 119903lowast for any
120582 = 120582119903lowast isin (0 120582lowast] (37)
where 120582lowast is a positive finite number
(iv) If 119891infin
120601= +infin then there exists 1205851 gt 0 such that
for every 119877lowast gt 1205851 problem (1) has a positive solution119909119877lowast
(119905) satisfying 119909119877lowast
= 119877lowast for any
120582 = 120582119877lowast
isin (0 120582lowast] (38)
where 120582lowast is a positive finite number(v) If there exists 120585 gt 0 such that 119898(119903) gt 120585 then problem
(1) has positive solution 119909119903(119905) satisfying 119909119903 = 119903 forany
120582 = 120582119903 isin (0 ] (39)
where is a positive finite number
Proof (i) It follows from 0 lt 1198910
120601lt +infin that there exists
1198971 1198972 120588 gt 0 such that
1198971120601119901 (119909) le 119891 (119905 119909) le 1198972120601119901 (119909)
(forall119905 isin [0 1] 0 le 119909 le 120588)
(40)
Let 1205850 = 120588120590 we show that 1205850 is required When 119909 isin 119875 cap
120597Ω119903 we have
119909 (119905) ge 120590 119909 = 120590119903 (41)
we need 119909(119905) lt 120588 this implies that 120590119903 lt 120588 as 119903 lt 1205850 so120590119903 lt 1205901205850 = 120588
Let 1205820 = (11198971)(Γ(120572)1205901205741)119901minus1 Then we may assume that
119909 minus 120601119902 (1205820) 119879119909 = 0 (forall119909 isin 119875 cap 120597Ω119903) (42)
if not then there exists119909119903 isin 119875cap120597Ω119903 such that120601119902(1205820)119879119909119903 = 119909119903and then (35) already holds for 120582119903 = 1205820
Define 120595(119905) equiv 1 forall119905 isin [0 1] then 120595(119905) isin 119875 and 120595 equiv 1We now show that
119909 minus 120601119902 (1205820) 119879119909 = 120581120595 (forall119909 isin 119875 cap 120597Ω119903 120581 ge 0) (43)
In fact if there exists 1199091 isin 119875 cap 120597Ω119903 1205811 ge 0 such that 1199091 minus120601119902(1205820)1198791199091 = 1205811120595 then (42) implies that 1205811 gt 0 On the otherhand 1199091 = 120601119902(1205820)1198791199091 + 1205811120595 ge 1205811120595 we may choose 120581
lowast=
max120581 | 1199091 ge 120581120595 then 1205811 le 120581lowastlt +infin 1199091 ge 120581
lowast120595 Therefore
120581lowast= 120581lowast 1003817100381710038171003817120595
1003817100381710038171003817 le10038171003817100381710038171199091
1003817100381710038171003817 = 119903 (44)
Abstract and Applied Analysis 7
Consequently for any 119905 isin [0 1] (26) and (44) imply that
1199091 (119905)
= 120601119902 (1205820)1
Γ (120572)
times int
1
0
(1 minus 119904)120572minus1
times 120601119902 (int
119904
0
120596 (120591) 119891 (120591 1199091 (120591)) 119889120591) 119889119904
minus int
119905
0
(119905 minus 119904)120572minus1
times 120601119902 (int
119904
0
120596 (120591) 119891 (120591 1199091 (120591)) 119889120591) 119889119904
+120578120601119902 (1205820)
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
119892 (119904)
times (int
1
0
(1 minus 120591)120572minus1
times 120601119902 (int
120591
0
120596 (120583) 119891 (120583 1199091 (120583)) 119889120583)119889120591
minus int
119904
0
(119904 minus 120591)120572minus1
times 120601119902 (int
120591
0
120596 (120583) 119891 (120583 1199091 (120583)) 119889120583)119889120591)119889119904
+ 1205811120595
ge120578120590120601119902 (1205820)
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
119892 (119904)
times (int
1
0
(1 minus 120591)120572minus1
times 120601119902 (int
120591
0
120596 (120583) 119891 (120583 1199091 (120583)) 119889120583)119889120591
minus int
119904
0
(119904 minus 120591)120572minus1
120601119902
times (int
120591
0
120596 (120583) 119891 (120583 1199091 (120583)) 119889120583)119889120591)119889119904
+120590120601119902 (1205820)
Γ (120572)
times int
1
0
(1 minus 119904)120572minus1
120601119902 (int
119904
0
120596 (120591) 119891 (120591 1199091 (120591)) 119889120591) 119889119904
+ 1205811120595
ge120590120601119902 (1205820)
Γ (120572)
times int
1
0
(1 minus 119904)120572minus1
120601119902 (int
119904
0
120596 (120591) 119891 (120591 1199091 (120591)) 119889120591) 119889119904
+ 1205811120595
ge120590120601119902 (1205820)
Γ (120572)
times int
1
0
(1 minus 119904)120572minus1
120601119902 (1198971 int
119904
0
120596 (120591) 120601119901 (1199091 (120591)) 119889120591) 119889119904
+ 1205811120595
ge120590120601119902 (1205820)
Γ (120572)
times int
1
0
(1 minus 119904)120572minus1
120601119902 (1198971 int
119904
0
120596 (120591) 120601119901 (120581lowast120595 (120591)) 119889120591) 119889119904
+ 1205811120595
=120590120581lowast119897119902minus1
1120582119902minus1
0
Γ (120572)int
1
0
(1 minus 119904)120572minus1
120601119902 (int
119904
0
120596 (120591) 119889120591) 119889119904 + 1205811120595
=120590120581lowast119897119902minus1
1120582119902minus1
01205741
Γ (120572)+ 1205811120595
= 120581lowast+ 120581120595
(45)
which implies that 1199091(119905) ge (120581lowast+ 120581)120595(119905) 119905 isin [0 1] which is a
contradiction to the definition of 120581lowast Thus (42) holds and byLemma 5 the fixed point index
119894 (120601119902 (1205820) 119879 119875 cap 120597Ω119903 119875) = 0 (46)
On the other hand by the fact that the fixed point index ofconstants operator is 1 so
119894 (120599 119875 cap 120597Ω119903 119875) = 1 (47)
where 120599 is the zero operator It follows therefore from (46)and (47) and the homotopy invariance property that thereexist 119909119903 isin 119875 cap 120597Ω119903 and 0 lt ]0 lt 1 such that ]0120601119902(1205820)119879119909119903 =119909119903 but 119909119903 = 120601119902(120582119903)119879119909119903 which implies that ]0120601119902(1205820)119879119909119903 =
120601119902(120582119903)119879119909119903 it follows that we get ]0120601119902(1205820) = 120601119902(120582119903)) that is
]0120582119902minus1
0= 120582119902minus1
119903 (48)
and then
120582119903 = ]1(119902minus1)0
1205820 = ]119901minus10
1205820 lt 1205820 (49)
From the proof above for any 119903 lt 1205850 there exists a positivesolution 119909119903 isin 119875 cap 120597Ω119903 associated with 120582 = 120582119903 gt 0
8 Abstract and Applied Analysis
Thus
119909119903 (119905)
=120601119902 (120582119903)
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902 (int
119904
0
120596 (120591) 119891 (120591 119909119903 (120591)) 119889120591) 119889119904
minus int
119905
0
(119905 minus 119904)120572minus1
times 120601119902 (int
119904
0
120596 (120591) 119891 (120591 119909119903 (120591)) 119889120591) 119889119904
+120578120601119902 (120582119903)
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
119892 (119904)
times (int
1
0
(1 minus 120591)120572minus1
times 120601119902 (int
120591
0
120596 (120583) 119891 (120583 119909119903 (120583)) 119889120583)119889120591
minus int
119904
0
(119904 minus 120591)120572minus1
times 120601119902 (int
120591
0
120596 (120583) 119891 (120583 119909119903 (120583)) 119889120583)119889120591)119889119904
(50)
with 119909119903 = 119903Next we show that 120582119903 ge 1205820 In fact
119909119903 (119905)
=120601119902 (120582119903)
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902 (int
119904
0
120596 (120591) 119891 (120591 119909119903 (120591)) 119889120591) 119889119904
minus int
119905
0
(119905 minus 119904)120572minus1
times 120601119902 (int
119904
0
120596 (120591) 119891 (120591 119909119903 (120591)) 119889120591) 119889119904
+120578120601119902 (120582119903)
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
119892 (119904)
times (int
1
0
(1 minus 120591)120572minus1
times 120601119902 (int
120591
0
120596 (120583) 119891 (120583 119909119903 (120583)) 119889120583)119889120591
minus int
119904
0
(119904 minus 120591)120572minus1
times 120601119902 (int
120591
0
120596 (120583) 119891 (120583 119909119903 (120583)) 119889120583)119889120591)119889119904
le120578120601119902 (120582119903)
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
119892 (119904) int
1
0
(1 minus 120591)120572minus1
times 120601119902 (int
120591
0
120596 (120583) 119891 (120583 119909119903 (120583)) 119889120583)119889120591 119889119904
+120601119902 (120582119903)
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902 (int
119904
0
120596 (120591) 119891 (120591 119909119903 (120591)) 119889120591) 119889119904
=120601119902 (120582119903) 120578 int
1
0119892 (119904) 119889119904
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
(1 minus 119904)120572minus1
times 120601119902 (int
119904
0
120596 (120591) 119891 (120591 119909119903 (120591)) 119889120591) 119889119904
+120601119902 (120582119903)
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902 (int
119904
0
120596 (120591) 119891 (120591 119909119903 (120591)) 119889120591) 119889119904
le120601119902 (120582119903) 120578 int
1
0119892 (119904) 119889119904
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
(1 minus 119904)120572minus1
times 120601119902 (1198972 int
119904
0
120596 (120591) 119909119903 (120591) 119889120591) 119889119904
+120601119902 (120582119903)
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902 (1198972120582119903 int
119904
0
120596 (120591) 119909119903 (120591) 119889120591) 119889119904
=120601119902 (120582119903)
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
(1 minus 119904)120572minus1
times 120601119902 (1198972 int
119904
0
120596 (120591) 119909119903 (120591) 119889120591) 119889119904
Abstract and Applied Analysis 9
le
10038171003817100381710038171199091199031003817100381710038171003817 120601119902 (1205821199031198972)
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
(1 minus 119904)120572minus1
times 120601119902 (int
1
0
120596 (120591) 119889120591) 119889119904
=
10038171003817100381710038171199091199031003817100381710038171003817 120601119902 (1205821199031198972) 1205742
Γ (120572 + 1) (1 minus 120578 int1
0119892 (119904) 119889119904)
(51)
which implies that
10038171003817100381710038171199091199031003817100381710038171003817 le
10038171003817100381710038171199091199031003817100381710038171003817 120601119902 (1205821199031198972) 1205742
Γ (120572 + 1) (1 minus 120578 int1
0119892 (119904) 119889119904)
(52)
that is
120601119902 (1205821199031198972) 1205742
Γ (120572 + 1) (1 minus 120578 int1
0119892 (119904) 119889119904)
ge 1 (53)
it follows that we get
120582119903 ge1
1198972(
Γ (120572 + 1) (1 minus 120578 int1
0119892 (119904) 119889119904)
1205742)
119901minus1
= 1205820(54)
In conclusion 120582119903 isin [1205820 1205820](ii) It follows from 0 lt 119891
infin
120601lt infin that there exist
119897lowast
1 119897lowast
2 120588lowastgt 0 such that
119897lowast
1120601119901 (119909) le 119891 (119905 119909) le 119897
lowast
2120601119901 (119909)
(forall119905 isin [0 1] 119909 ge 120588lowast)
(55)
Let 120585lowast0= 120588lowast120590 the following proof are similar to (i)
(iii) It follows from 1198910
120601= +infin there exists 119897lowast gt 0 120588
lowastgt 0
such that
1198910
120601ge 119897lowast120601119901 (119909) (forall119905 isin [0 1] 0 le 119909 le 120588
lowast) (56)
Let 1205851 = 120588lowast120590 we show that 1205851 is required When 119909 isin 119875 cap
120597Ω119903lowast we have 119909(119905) ge 120590119909 = 120590119903lowast since 119909 le 120588
lowast then we need120590119903lowastlt 120588lowast as 119903lowast lt 1205851 so 120590119903
lowastlt 120588lowast holds
Let 120582lowast = (1119897lowast)(Γ(120572)1205901205741)
1(119902minus1) we proceed in the sameway as in the proof of (i) replacing (42) we may assume that119909 minus 120601119902(120582
lowast)119879119909 = 0 (forall119909 isin 119875 cap 120597Ω119903lowast) and replacing (43) we
can prove 119909 minus 120601119902(120582lowast)119879119909 = 120581120595 (forall119909 isin 119875 cap 120597Ω119903lowast 120581 ge 0) It
follows from Lemma 5 that 119894(120601119902(120582lowast)119879 119875 cap Ω119903lowast 119875) = 0 Note
that 119894(120579 119875 cap Ω119903lowast 119875) = 1 we can easily show that there exists119909119903lowast isin 119875cap120597Ω119903lowast and 0 lt ]119903lowast lt 1 such that ]119903lowast120601119902(120582
lowast)119879119909119903lowast = 119909119903lowast
Hence (37) holds for 120582119903lowast = 120582lowast]1minus119901119903lowast
lt 120582lowast
The proof ofTheorem 9(iv) follows by themethod similarto Theorem 9(iii) we omit it here
(v) It follows that 119898(119903) gt 120585 for any 119909 isin 119875 cap 120597Ω119903 we have120590119903 le 119909 le 119903 and 119891(119905 119909)120601119901(119909) ge 119891(119905 119909)120601119901(119903) ge 119898(119903) ge 120585then
119891 (119905 119909) ge 120585120601119901 (119909) forall119909 isin [120590119903 119903] 119905 isin [0 1] (57)
Let = (1120585)(Γ(120572)1205901205741)119901minus1 The following proof is similar to
that of (iv) This finished the proof of (v)
Remark 10 InTheorem 9 all the criteria obtained depend onone of 1198910
120601and 119891
infin
120601
Let Ψ(119909) = 120601119902(120582)119879119909 the following theorems give out themultiply nonexistence and the dependence of parameter
Theorem 11 Assume that (H1)ndash(H5) hold
(i) If 1198910
120601= 0 and 119891
infin
120601= 0 lim
119909rarrinfin119891(119905 119909) = 119891infin isin
(0 +infin] uniformly for 119905 isin [0 1] then there exists120582lowast
gt 0 such that the problem (1) has two solutions forany 120582 gt 120582
lowast
(ii) If 1198910120601
= 0 and 119891infin
120601= 0 then there exists 120582 gt 0 such
that for any 120582 lt 120582 problem (1) has no solution
Proof (i) It follows from 119891infin isin (0 +infin] that for any 120577 gt 0there exists 120588 gt 0 such that
119891 (119905 119909) gt 120577 (forall119905 isin [0 1] 119909 ge 120588) (58)
Since 1198910
120601= 0 there exists 1205761 gt 0 120588
1gt 0 such that
119891 (119905 119909) le 1205761120601119901 (119909) (forall119905 isin [0 1] 0 le 119909 le 1205881) (59)
where 1205761 satisfied 120601119902(1205821205761)1205742Γ(120572 + 1)(1 minus 120578 int1
0119892(119904)119889119904) le 1
For forall119909 isin 119875 cap 120597Ω1205881
we have
Ψ (119909) = (120601119902 (120582) 119879119909) (119905)
=1
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902 (120582int
119904
0
120596 (120591) 119891 (120591 119909 (120591)) 119889120591) 119889119904
minus int
119905
0
(119905 minus 119904)120572minus1
times 120601119902 (120582int
119904
0
120596 (120591) 119891 (120591 119909 (120591)) 119889120591) 119889119904
+120578
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
119892 (119904)
10 Abstract and Applied Analysis
times (int
1
0
(1 minus 120591)120572minus1
times 120601119902 (120582int
120591
0
120596 (120583) 119891 (120583 119909 (120583)) 119889120583)119889120591
minus int
119904
0
(119904 minus 120591)120572minus1
times 120601119902 (120582int
120591
0
120596 (120583) 119891 (120583 119909 (120583)) 119889120583)119889120591)119889119904
le120578
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
119892 (119904)
times (int
1
0
(1 minus 120591)120572minus1
times 120601119902(120582int
120591
0
120596 (120583) 119891 (120583 119909 (120583)) 119889120583)119889120591)119889119904
+1
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902 (120582int
119904
0
120596 (120591) 1205761119891 (120591 119909 (120591)) 119889120591) 119889119904
le120578 int1
0119892 (119904) 119889119904
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
(1 minus 120591)120572minus1
times 120601119902 (120582int
1
0
120596 (120583) 1205761120601119901 (119909 (120583)) 119889120583)119889120591
+1
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902(120582int
1
0
120596 (120591) 1205761120601119901 (119909 (120591)) 119889120591) 119889119904
le1
Γ (120572 + 1) (1 minus 120578 int1
0119892 (119904) 119889119904)
times 120601119902 (120582int
1
0
120596 (120591) 1205761120601119901 (119909) 119889120591)
=120601119902 (1205821205761) 1205742
Γ (120572 + 1) (1 minus 120578 int1
0119892 (119904) 119889119904)
119909 le 119909
(60)
This implied
Ψ119909 le 119909 forall119909 isin 119875 cap 120597Ω1205881
(61)
Let 1205883gt max120588120590 120588
1 and 120582
lowast= (1120577)(120588
3Γ(120572)1205901205741)
119901minus1 thenforall119909 isin 119875 cap 120597Ω120588
3
we have 120590119909 le 119909 le 1205883and
(Ψ119909) (119905)
=1
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902 (120582int
119904
0
120596 (120591) 119891 (120591 119909 (120591)) 119889120591) 119889119904
minus int
119905
0
(119905 minus 119904)120572minus1
times 120601119902 (120582int
119904
0
120596 (120591) 119891 (120591 119909 (120591)) 119889120591) 119889119904
+120578
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
119892 (119904)
times (int
1
0
(1 minus 120591)120572minus1
times 120601119902 (120582int
120591
0
120596 (120583) 119891 (120583 119909 (120583)) 119889120583)119889120591
minus int
119904
0
(119904 minus 120591)120572minus1
times 120601119902(120582int
120591
0
120596 (120583) 119891 (120583 119909 (120583)) 119889120583)119889120591)119889119904
ge120590120578
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
119892 (119904)
times (int
1
0
(1 minus 120591)120572minus1
times 120601119902 (120582int
120591
0
120596 (120583) 119891 (120583 119909 (120583)) 119889120583)119889120591
minus int
119904
0
(119904 minus 120591)120572minus1
times 120601119902(120582int
120591
0
120596 (120583) 119891 (120583 119909 (120583)) 119889120583)119889120591)119889119904
+120590
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902 (120582int
119904
0
120596 (120591) 119891 (120591 119909 (120591)) 119889120591) 119889119904
ge120590
Γ (120572)int
1
0
(1 minus 119904)120572minus1
Abstract and Applied Analysis 11
times 120601119902 (120582int
119904
0
120596 (120591) 119891 (120591 119909 (120591)) 119889120591) 119889119904
gt120590
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902 (120582int
119904
0
120596 (120591) 120577 119889120591) 119889119904
gt120590120601119902 (120582120577)
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902 (int
119904
0
120596 (120591) 119889120591) 119889119904
=120590120601119902 (120582120577) 1205741
Γ (120572)= 1205883
(62)
which implies
Ψ119909 gt 119909 forall119909 isin 119875 cap 120597Ω1205883
(63)
Next for 119891infin
120601= 0 there exist 1205762 gt 0 120588
2gt 1205883gt 0 such
that
119891 (119905 119909) le 1205762120601119901 (119909) (forall119905 isin [0 1] 119909 ge 1205882) (64)
where 1205762 satisfied 120601119902(1205821205762)1205742Γ(120572 + 1)(1 minus 120578 int1
0119892(119904)119889119904) le 1
Similar to the above proof we get
Ψ119909 le 119909 forall119909 isin 119875 cap 120597Ω1205882
(65)
Applying Lemma 7 to (61) (65) and (63) yields that Ψ
has two fixed points 1199091 1199092 such that 1199091 isin 119875cap (Ω1205883
Ω1205881
) and1199092 isin 119875 cap (Ω120588
2
Ω1205883
)(ii) It follows from 119891
0
120601= 0 and 119891
infin
120601= 0 that there exists
1205761 1205762 1205881 1205882 gt 0 such that
119891 (119905 119909) le 1205761120601119901 (119909) (forall119905 isin [0 1] 0 le 119909 le 1205881)
119891 (119905 119909) le 1205762120601119901 (119909) (forall119905 isin [0 1] 119909 ge 1205882)
(66)
then for 119909 isin [1205881 1205882] by the continuous of 119891(119905 119909(119905)) there
exists 119909 isin [1205881 1205882] such that
max1205881le119909le120588
2
119891 (119905 119909)
120601119901 (119909)=
119891 (119905 119909)
120601119901 (119909) (67)
Let Υ = max1205761 1205762 119891(119905 119909)120601119901(119909) gt 0 then
119891 (119905 119909) le Υ120601119901 (119909) (forall0 le 119905 le 1 119909 ge 0) (68)
Assuming 119909(119905) is a positive solution of problem (1) we willshow that this leads to a contradiction for 120582 lt 120582 where
120582 = (1Υ)(Γ(120572 + 1)(1 minus 120578 int1
0119892(119904)119889119904)1205742)
119901minus1 It follows from(26) that
119909 = 119909 (0)
=120578
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
119892 (119904)
times (int
1
0
(1 minus 120591)120572minus1
times 120601119902 (120582int
120591
0
120596 (120583) 119891 (120583 119909 (120583)) 119889120583)119889120591
minus int
119904
0
(119904 minus 120591)120572minus1
times 120601119902(120582int
120591
0
120596 (120583) 119891 (120583 119909 (120583)) 119889120583)119889120591)119889119904
+1
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902 (120582int
119904
0
120596 (120591) 119891 (120591 119909 (120591)) 119889120591) 119889119904
le120578
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
119892 (119904)
times int
1
0
(1 minus 120591)120572minus1
times 120601119902(120582int
120591
0
120596 (120583) 119891 (120583 119909 (120583)) 119889120583)119889120591 119889119904
+1
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902 (120582int
119904
0
120596 (120591) 119891 (120591 119909 (120591)) 119889120591) 119889119904
=1
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
(1 minus 119904)120572minus1
times 120601119902 (120582int
119904
0
120596 (120591) 119891 (120591 119909 (120591)) 119889120591) 119889119904
le1
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
(1 minus 119904)120572minus1
12 Abstract and Applied Analysis
times 120601119902 (120582int
1
0
120596 (120591) Υ120601119901 (119909 (120591)) 119889120591) 119889119904
=120601119902 (120582Υ) 1205742
Γ (120572 + 1) (1 minus 120578 int1
0119892 (119904) 119889119904)
119909
(69)
This implies that 1 le 120601119902(120582Υ)1205742Γ(120572 + 1)(1 minus 120578 int1
0119892(119904)119889119904) =
(120582Υ)119902minus1
1205742Γ(120572+1)(1minus120578 int1
0119892(119904)119889119904) lt (120582 Υ)
119902minus11205742Γ(120572+1)(1minus
120578 int1
0119892(119904)119889119904) = 1 which is a contradiction This finishes the
proof of (ii)
Remark 12 In (i) the condition lim119909rarrinfin
119891(119905 119909) = 119891infin isin
(0 +infin] uniformly for 119905 isin [0 1] can be replaced withcondition (H6) The same methods can be used to prove theresults
Remark 13 From the above proof the condition 119891infin isin
(0 +infin] is important to keep the problem (1) having at leasttwo positive solutions if this condition is not satisfied thenthere exists 120582 gt 0 small enough such that the problem (1) hasno positive solutions for all 120582 lt 120582
Theorem 14 Assume that (H1)ndash(H6) hold(i) If 1198910
120601= +infin and 119891
infin
120601= +infin then there exists 120582lowast0 gt 0
such that problem (1) has at least two solutions for 120582 isin
(0 120582lowast0)(ii) If 1198910
120601= +infin and 119891
infin
120601= +infin then there exists 120582 gt 0
such that problem (1) has no solution for all 120582 gt 120582
Proof (i) It follows from 1198910
120601= +infin that there exists 1199031 gt 0
such that
119891 (119905 119909) ge 119871lowast1120601119901 (119909) (forall119905 isin [0 1] 0 le 119909 le 1199031) (70)
where 119871lowast1 satisfies 1205902120601119902(120582119871lowast1)1205741Γ(120572) ge 1
Then for 119909 isin 119875cap120597Ω1199031
we have 120590119909 le 119909 le 1199031 for 119905 isin [0 1]
and
(Ψ119909) (119905)
=1
Γ (120572)
times int
1
0
(1 minus 119904 )120572minus1
times 120601119902 (120582int
119904
0
120596 (120591) 119891 (120591 119909 (120591)) 119889120591) 119889119904
minus int
119905
0
(119905 minus 119904)120572minus1
times 120601119902 (120582int
119904
0
120596 (120591) 119891 (120591 119909 (120591)) 119889120591) 119889119904
+120578
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
119892 (119904)
times (int
1
0
(1 minus 120591)120572minus1
times 120601119902 (120582int
120591
0
120596 (120583) 119891 (120583 119909 (120583)) 119889120583)119889120591
minus int
119904
0
(119904 minus 120591)120572minus1
times 120601119902 (120582int
120591
0
120596 (120583) 119891 (120583 119909 (120583)) 119889120583)119889120591)119889119904
ge120590120578
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
119892 (119904)
times (int
1
0
(1 minus 120591)120572minus1
times 120601119902 (120582int
120591
0
120596 (120583) 119891 (120583 119909 (120583)) 119889120583)119889120591
minus int
119904
0
(119904 minus 120591)120572minus1
times120601119902 (120582int
120591
0
120596 (120583) 119891 (120583 119909 (120583)) 119889120583)119889120591)119889119904
+120590
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902 (120582int
119904
0
120596 (120591) 119891 (120591 119909 (120591)) 119889120591) 119889119904
ge120590
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902 (120582int
119904
0
120596 (120591) 119891 (120591 119909 (120591)) 119889120591) 119889119904
ge120590
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902 (120582int
119904
0
120596 (120591) 119871lowast1120601119901 (119909 (120591)) 119889120591) 119889119904
ge1205902120601119902 (120582119871lowast1) 119909
Γ (120572)
times int
1
0
(1 minus 119904)120572minus1
times 120601119902 (int
119904
0
120596 (120591) 119889120591) 119889119904
=1205902120601119902 (120582119871lowast1) 1205742
Γ (120572)119909
(71)
Abstract and Applied Analysis 13
which implies that
Ψ119909 ge 119909 forall119909 isin 119875 cap 120597Ω1199031
(72)
Next from 119891infin
120601= +infin there exists 1199032 gt 0 such that
119891 (119905 119909) ge 119871lowast2120601119901 (119909) forall119909 ge 1199032 (73)
where 119871lowast2 satisfies
1205902120601119902 (120582119871lowast2) 1205742
Γ (120572)ge 1 (74)
Let 1199032 gt max1199031 1199032120590 then for 119909 isin 119875 cap 120597Ω1199032
we have
119909 (119905) ge 120590 119909 = 1205901199032 gt 1199032 (75)
Similar to the above proof we can get
Ψ119909 ge 119909 forall119909 isin 119875 cap 120597Ω1199032
(76)
Let 1199031 lt 1199033 lt 11990321198721199033
= max119891(119905 119909) | 119905 isin [0 1] 119909 le 1199033+1 gt
0 and 120582lowast0 = (11198721199033
)(Γ(120572+1)1199033(1minus120578 int1
0119892(119904)119889119904)1205742)
119901minus1Thenfor 119909 isin 119875 cap 120597Ω119903
3
we have
119891 (119905 119909) le 1198721199033
forall119905 isin [0 1] (77)
Hence
(Ψ119909) (119905)
=1
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902 (120582int
119904
0
120596 (120591) 119891 (120591 119909 (120591)) 119889120591) 119889119904
minus int
119905
0
(119905 minus 119904)120572minus1
times 120601119902 (120582int
119904
0
120596 (120591) 119891 (120591 119909 (120591)) 119889120591) 119889119904
+120578
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
119892 (119904)
times (int
1
0
(1 minus 120591)120572minus1
times 120601119902 (120582int
120591
0
120596 (120583) 119891 (120583 119909 (120583)) 119889120583)119889120591
minus int
119904
0
(119904 minus 120591)120572minus1
times120601119902 (120582int
120591
0
120596 (120583) 119891 (120583 119909 (120583)) 119889120583)119889120591)119889119904
le120578
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
119892 (119904)
times int
1
0
(1 minus 120591)120572minus1
120601119902
times (120582int
120591
0
120596 (120583) 119891 (120583 119909 (120583)) 119889120583)119889120591 119889119904
+1
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902 (120582int
119904
0
120596 (120591) 119891 (120591 119909 (120591)) 119889120591) 119889119904
le120578
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
119892 (119904)
times int
1
0
(1 minus 120591)120572minus1
120601119902
times (120582int
120591
0
120596 (120583)1198721199033
119889120583)119889120591119889119904
+1
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902 (120582int
119904
0
120596 (120591)1198721199033
119889120591)119889119904
le120578120601119902 (120582119872119903
3
) int1
0119892 (119904) 119889119904
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
(1 minus 120591)120572minus1
times 120601119902 (int
1
0
120596 (120583) 119889120583)119889120591
+120601119902 (120582119872119903
3
)
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902 (int
1
0
120596 (120591) 119889120591) 119889119904
=1205742120601119902 (120582119872119903
3
)
Γ (120572 + 1) (1 minus 120578 int1
0119892 (119904) 119889119904)
lt1205742120601119902 (120582lowast0119872119903
3
)
Γ (120572 + 1) (1 minus 120578 int1
0119892 (119904) 119889119904)
= 1199033
(78)
14 Abstract and Applied Analysis
This implies that
Ψ119909 lt 119909 forall119909 isin 119875 cap 120597Ω1199033
(79)
Applying Lemma 7 to (72) (76) and (79) yields that Ψ hastwo fixed points 1199091 1199092 such that 1199091 isin 119875 cap (Ω119903
3
Ω1199031
) and1199092 isin 119875 cap (Ω119903
2
Ω1199033
)(ii) By the proof of (i) and the continuity of 119891(119905 119909) there
exists 119909 isin [1199031 1199032] such that min119909isin[11990311199032](119891(119905 119909)120601119901(119909)) =
119891(119905 119909)120601119901(119909)Let Υ = min119871lowast1 119871lowast2 119891(119905 119909)120601119901(119909) since (H6) holds
then Υ gt 0 and
119891 (119905 119909) ge Υ120601119901 (119909) forall119909 ge 0 119905 isin [0 1] (80)
Assuming 119909(119905) is a positive solution of problem (1) we willshow that this leads to contradiction for all 120582 gt 120582 where 120582 =
(1Υ)(Γ(120572)1205741120590)119901minus1 It follows form (26) that
119909 = 119909 (0)
=120578
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
119892 (119904)
times (int
1
0
(1 minus 120591)120572minus1
times 120601119902 (120582int
120591
0
120596 (120583) 119891 (120583 119909 (120583)) 119889120583)119889120591
minus int
119904
0
(119904 minus 120591)120572minus1
times120601119902 (120582int
120591
0
120596 (120583) 119891 (120583 119909 (120583)) 119889120583)119889120591)119889119904
+1
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902 (120582int
119904
0
120596 (120591) 119891 (120591 119909 (120591)) 119889120591) 119889119904
ge1
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902 (120582int
119904
0
120596 (120591) 119891 (120591 119909 (120591)) 119889120591) 119889119904
ge1
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902 (120582int
119904
0
120596 (120591) Υ120601119901 (119909 (120591)) 119889120591) 119889119904
ge1
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902 (120582int
119904
0
120596 (120591) Υ120601119901 (120590 119909 119889120591)) 119889119904
ge120601119902 (120582Υ) 120590 119909
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902 (int
119904
0
120596 (120591) 119889120591) 119889119904
=120601119902 (120582Υ) 1205741120590
Γ (120572)119909
(81)
This implies that 1 ge 120601119902(120582Υ)1205741120590Γ(120572) = (120582Υ)119902minus1
1205741120590Γ(120572) gt
(120582Υ)119902minus1
1205741120590Γ(120572) = 1 which is a contradiction This com-pletes the proof of (ii)
Corollary 15 Assume that (H1)ndash(H5) holds
(i) If lim119909rarrinfin
119891(119905 119909) = 119891infin isin (0 +infin] uniformly for 119905 isin
[0 1] and one of 1198910120601
= 0 and 119891infin
120601= 0 is satisfied then
there exists 120582lowast gt 0 such that the problem (1) has at leastone positive solution for any 120582 gt 120582
lowast(ii) If one of 1198910
120601= +infin and 119891
infin
120601= +infin holds then there
exists 120582lowast0 gt 0 such that problem (1) has at least onepositive solutions for any 120582 isin (0 120582lowast0)
Proof (i) The conclusion is a direct consequence of Theo-rem 11(i)
(ii) The conclusion is a direct consequence of Theo-rem 14(i)
Theorem 16 Assume that (H1)ndash(H5) holdThen the followingconclusions hold
(i) If 1198910120601
= 0 and 119891infin
120601= infin then the problem (1) has a
positive solution 119909120582(119905) for all 120582 gt 0(ii) If 1198910
120601= infin and 119891
infin
120601= 0 then the problem (1) has a
positive solution 119909120582(119905) for all 120582 gt 0
Proof Weonly prove (i) the proof of (ii) is similar sowe omitit here
Let 120582 gt 0 since 1198910
120601= 0 there exists 119903 gt 0 119897 gt 0 such that
119891 (119905 119909) le 119897120601119901 (119909) (forall119905 isin [0 1] 119909 isin [0 119903]) (82)
where 119897 satisfied
1205742120582119902minus1
119897119902minus1
Γ (120572 + 1) (1 minus 120578 int1
0119892 (119904) 119889119904)
le 1 (83)
Similar to the proof of Theorem 14 we obtain that
Ψ119909 le 119909 forall119909 isin 119875 cap 120597Ω119903 (84)
Abstract and Applied Analysis 15
From 119891infin
120601= infin there exists 119877 gt 119903 gt 0 119871 gt 0 such that
119891 (119905 119909) ge 119871120601119901 (119909) (forall119905 isin [0 1] 119909 ge 119877) (85)where 119871 satisfied
1205902120582119902minus1
119871119902minus1
1205741
Γ (120572)ge 1 (86)
We get thatΨ119909 ge 119909 forall119909 isin 119875 cap 120597Ω119877 (87)
Applying Lemma 7 to (84) (87) yields thatΨ has a fixed point119909 such that 119909 isin 119875 cap (Ω119877 Ω119903)
Theorem 17 Assume that (H1)ndash(H6) holdThen the followingconclusions hold
If 0 lt 1198910
120601lt infin and 0 lt 119891
infin
120601lt infin then there exists 1 2 gt
0 such that problem (1) has no positive solution for any 120582 gt 1
and 0 lt 120582 lt 2
Proof It follows from 0 lt 1198910
120601lt infin and 0 lt 119891
infin
120601lt infin that
there exist 1 gt 0 2 gt 0 1 gt 0 2 gt 0 and 1199032 gt 1199031 gt 0
such that1120601119901 (119909) le 119891 (119905 119909) le 1120601119901 (119909)
(forall119905 isin [0 1] 0 le 119909 le 1199031)
2120601119901 (119909) le 119891 (119905 119909) le 2120601119901 (119909)
(forall119905 isin [0 1] 119909 ge 1199032)
(88)
Let = max119891(119905 119909)120601119901(119909) 1199031 le 119909 le 1199032 0 le 119905 le 1 =
min119891(119905 119909)120601119901(119909) 1199031 le 119909 le 1199032 0 le 119905 le 1 By the condition(H6) gt 0 Then forall119909 ge 0 we have
120601119901 (119909) le 119891 (119905 119909) le 120601119901 (119909) (89)
where = max1 2 gt 0 and = min1198971 2 gt 0Let 1 = (1)(Γ(120572)1205741120590)
119901minus1 and 2 =
(1)(Γ(120572 + 1)(1 minus 120578 int1
0119892(119904)119889119904)1205742)
119901minus1
If problem (1)has positive solution 119909(119905) in 119875 then
119909 = 119909 (0)
=120578
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
119892 (119904)
times (int
1
0
(1 minus 120591)120572minus1
times 120601119902 (120582int
120591
0
120596 (120583) 119891 (120583 119909 (120583)) 119889120583)119889120591
minus int
119904
0
(119904 minus 120591)120572minus1
times120601119902 (120582int
120591
0
120596 (120583) 119891 (120583 119909 (120583)) 119889120583)119889120591)119889119904
+1
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902 (120582int
119904
0
120596 (120591) 119891 (120591 119909 (120591)) 119889120591) 119889119904
ge1
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902 (120582int
119904
0
120596 (120591) 119891 (120591 119909 (120591)) 119889120591) 119889119904
ge1
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902 (120582int
119904
0
120596 (120591) 120601119901 (119909 (120591)) 119889120591) 119889119904
ge1
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902 (120582int
119904
0
120596 (120591) 120601119901 (120590 119909 119889120591)) 119889119904
ge120601119902 (120582) 120590 119909
Γ (120572)
times int
1
0
(1 minus 119904)120572minus1
120601119902 (int
119904
0
120596 (120591) 119889120591) 119889119904
=120601119902 (120582) 1205741120590
Γ (120572)119909
gt120601119902 (1 ) 1205741120590
Γ (120572)119909 = 119909
119909 = 119909 (0)
=120578
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
119892 (119904)
times (int
1
0
(1 minus 120591)120572minus1
120601119902
times (120582int
120591
0
120596 (120583) 119891 (120583 119909 (120583)) 119889120583)119889120591
minus int
119904
0
(119904 minus 120591)120572minus1
120601119902
times(120582int
120591
0
120596 (120583) 119891 (120583 119909 (120583)) 119889120583)119889120591)119889119904
+1
Γ (120572)int
1
0
(1 minus 119904)120572minus1
120601119902
times (120582int
119904
0
120596 (120591) 119891 (120591 119909 (120591)) 119889120591) 119889119904
16 Abstract and Applied Analysis
le120578
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
119892 (119904)
times int
1
0
(1 minus 120591)120572minus1
times 120601119902 (120582int
120591
0
120596 (120583) 119891 (120583 119909 (120583)) 119889120583)119889120591 119889119904
+1
Γ (120572)int
1
0
(1 minus 119904)120572minus1
120601119902
times (120582int
119904
0
120596 (120591) 119891 (120591 119909 (120591)) 119889120591) 119889119904
=1
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
(1 minus 119904)120572minus1
120601119902
times (120582int
119904
0
120596 (120591) 119891 (120591 119909 (120591)) 119889120591) 119889119904
le1
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
(1 minus 119904)120572minus1
120601119902
times (120582int
1
0
120596 (120591) 120601119901 (119909 (120591)) 119889120591) 119889119904
=120601119902 (120582) 1205742
Γ (120572 + 1) (1 minus 120578 int1
0119892 (119904) 119889119904)
119909
lt120601119902 (2) 1205742
Γ (120572 + 1) (1 minus 120578 int1
0119892 (119904) 119889119904)
119909 = 119909
(90)
which is a contradiction This finishes the proof of Theo-rem 17
4 Examples
In this section we give out same examples to illustrate ourmain results can be used in practice
Example 1 Consider the following fractional differentialequation
(120601119901 (11986352
0+119909 (119905)))
1015840
+ 120582119905(1 + 1199052)12
119909119901minus1 arctan119909 = 0
0 lt 119905 lt 1
1199091015840(0) = 119909
10158401015840(0) = 0
119909 (1) =1
2int
1
0
1
2119909 (119904) 119889119904
(91)
where 120572 = 52 119901 = 32 119902 = 3 120596(119905) = 119905 119891(119905 119909) = (1 +
1199052)12
119909119901minus1 arctan119909 120578 = 12 and 119892(119905) = 12
By simple computation 119897lowast1= 1205872 119897
lowast
2= radic21205872 1205741 = int
1
0(1minus
119904)32
1206013(int119904
0120591 119889120591)119889119904 asymp 00043 1205742 = 1206013(int
1
0120591 119889120591) = 025 120590 asymp
01429 120582lowast
0= (1119897
lowast
1)(Γ(120572)1205901205741)
119901minus1asymp 25149 120582
lowast
0= (1119897
lowast
2)(Γ(120572+
1)(1 minus 120578 int1
0119892(119904)119889119904)1205742)
119901minus1asymp 14222 Then it is easy to see
all the conditions of Theorem 9(ii) are satisfied thus forall120582 isin
[14222 25149]] the problem (91) has at least one positivesolution
Example 2 Consider the boundary value problem
(120601119901 (11986352
0+119909 (119905)))
1015840
+ 120582119905(1 + 1199052)12
119909119901119890minus119909
= 0
0 lt 119905 lt 1
1199091015840(0) = 119909
10158401015840(0) = 0
119909 (1) =1
2int
1
0
1
2119909 (119904) 119889119904
(92)
where 120572 = 52 119901 = 32 119902 = 3 120596(119905) = 119905 119891(119905 119909) = (1 +
1199052)12
119909119901119890minus119909
120578 = 12 and 119892(119905) = 12
1198910
120601= lim119909rarr0
119891 (119905 119909)
120601119901 (119909)= lim119909rarr0
(1 + 1199052)12
119909119901119890minus119909
119909119901minus1= 0
forall119905 isin [0 1]
1198910
120601= lim119909rarrinfin
119891 (119905 119909)
120601119901 (119909)= lim119909rarrinfin
(1 + 1199052)12
119909119901119890minus119909
119909119901minus1= 0
forall119905 isin [0 1]
(93)
Let 1205881= 12 120588
2= 4 then
119891 (119905 119909)
120601119901 (119909)le (1 + 119905
2)12 1
2radic119890le
1
radic2119890
forall119905 isin [0 1] 119909 isin [01
2]
119891 (119905 119909)
120601119901 (119909)le (1 + 119905
2)12 4
1198904le
4radic2
1198904
forall119905 isin [0 1] 119909 isin [4infin)
(94)
so 1205761 = 1radic2119890 1205762 = 4radic21198904 and forall119909 isin
[12 4] max12le119909le4(119891(119905 119909)120601119901(119909)) = radic2119890minus1 We
have Υ = max1radic2119890 4radic21198904 radic2119890minus1
= radic2119890minus1 1205742 = 025
Abstract and Applied Analysis 17
and 120582 = (1Υ)(Γ(120572+1)(1minus120578 int1
0119892(119904)119889119904)1205742)
119901minus1= 3radic5120587
141198904
By Theorem 11(ii) for any 120582 lt 120582 = 3radic512058714
1198904 the problemhas no positive solution
5 Conclusions
Recently differential equations with p-Laplacian operatorwere widely discussed by several authors In this paperby using fixed point techniques combining with partiallyordered structure of Banach space we obtained the existencemultiply and the dependence of parameter These newresults we presented can be used in numerical computationand analyze mathematical models of physical phenomenamechanics nonlinear dynamics and many other relatedfields
Conflict of Interests
The authors declare that there is no conflict of interestsregarding the publication of this paper
Acknowledgments
The authors are very grateful to anonymous referees for theirvaluable and detailed suggestions and comments to improvethe original paper This work is supported by the Programfor New Century Excellent Talents in University (NCET-10-0097)
References
[1] I Podlubny Fractional Differential Equations vol 198 Aca-demic Press San Diego Calif USA 1999
[2] A A Kilbas H M Srivastava and J J Trujillo Theoryand Applications of Fractional Differential Equations vol 204of North-Holland Mathematics Studies Elsevier Science BVAmsterdam The Netherlands 2006
[3] Y A Rossikhin and M V Shitikova ldquoApplication of fractionalderivatives to the analysis of damped vibrations of viscoelasticsingle mass systemsrdquoActa Mechanica vol 120 no 1ndash4 pp 109ndash125 1997
[4] J Sabatier O Agrawal and J T Machado Advances in Frac-tional Calculus Theoretical developments and applications inphysics and engineering Springer Berlin Germany 2007
[5] N Engheta ldquoOn fractional calculus and fractional multipolesin electromagnetismrdquo IEEE Transactions on Antennas andPropagation vol 44 no 4 pp 554ndash566 1996
[6] R Hilfer Applications of Fractional Calculus in Physics WorldScientific Singapore 2000
[7] R L Bagley and R A Calico ldquoFractional order state equationsfor the control of viscoelastically damped structuresrdquo Journalof Guidance Control and Dynamics vol 14 no 2 pp 304ndash3111991
[8] W-X Zhou andY-D Chu ldquoExistence of solutions for fractionaldifferential equations with multi-point boundary conditionsrdquoCommunications in Nonlinear Science and Numerical Simula-tion vol 17 no 3 pp 1142ndash1148 2012
[9] B Ahmad and G Wang ldquoA study of an impulsive four-pointnonlocal boundary value problemof nonlinear fractional differ-ential equationsrdquo Computers amp Mathematics with Applicationsvol 62 no 3 pp 1341ndash1349 2011
[10] B Ahmad and S Sivasundaram ldquoOn four-point nonlocalboundary value problems of nonlinear integro-differentialequations of fractional orderrdquo Applied Mathematics and Com-putation vol 217 no 2 pp 480ndash487 2010
[11] N Nyamoradi ldquoPositive solutions for multi-point boundaryvalue problems for nonlinear fractional differential equationsrdquoJournal of Contemporary Mathematical Analysis vol 48 no 4pp 145ndash157 2013
[12] B Ahmad and J J Nieto ldquoExistence of solutions for nonlocalboundary value problems of higher-order nonlinear fractionaldifferential equationsrdquo Abstract and Applied Analysis vol 2009Article ID 494720 9 pages 2009
[13] X Liu M Jia and X Xiang ldquoOn the solvability of a fractionaldifferential equationmodel involving the119901-Laplacian operatorrdquoComputers ampMathematics with Applications vol 64 no 10 pp3267ndash3275 2012
[14] J Wang H Xiang and Z Liu ldquoExistence of concave positivesolutions for boundary value problem of nonlinear fractionaldifferential equation with 119901-Laplacian operatorrdquo InternationalJournal of Mathematics and Mathematical Sciences vol 2010Article ID 495138 17 pages 2010
[15] J Wang and H Xiang ldquoUpper and lower solutions method fora class of singular fractional boundary value problems with 119901-Laplacian operatorrdquo Abstract and Applied Analysis vol 2010Article ID 971824 12 pages 2010
[16] H Lu andZHan ldquoExistence of positive solutions for boundary-value problem of fractional differential equation with 119901-laplacian operatorrdquo The American Jouranl of Engineering andTechnology Research vol 11 pp 3757ndash3764 2011
[17] G Chai ldquoPositive solutions for boundary value problem offractional differential equation with 119901-Laplacian operatorrdquoBoundary Value Problems vol 2012 article 18 2012
[18] T Chen W Liu and Z Hu ldquoA boundary value problem forfractional differential equation with 119901-Laplacian operator atresonancerdquoNonlinear AnalysisTheoryMethodsampApplicationsvol 75 no 6 pp 3210ndash3217 2012
[19] T Chen andW Liu ldquoAn anti-periodic boundary value problemfor the fractional differential equation with a 119901-LaplacianoperatorrdquoAppliedMathematics Letters of Rapid Publication vol25 no 11 pp 1671ndash1675 2012
[20] S J Li X G Zhang Y H Wu and L Caccetta ldquoExtremalsolutions for 119901-Laplacian differential systems via iterative com-putationrdquo Applied Mathematics and Computation vol 26 pp1151ndash1158 2013
[21] Z Liu and L Lu ldquoA class of BVPs for nonlinear fractionaldifferential equations with 119901-Laplacian operatorrdquo ElectronicJournal of Qualitative Theory of Differential Equations vol 70pp 1ndash16 2012
[22] Z Liu L Lu and I Szanto ldquoExistence of solutions for fractionalimpulsive differential equations with 119901-Laplacian operatorrdquoActa Mathematica Hungarica vol 141 no 3 pp 203ndash219 2013
[23] Z Bai ldquoEigenvalue intervals for a class of fractional boundaryvalue problemrdquo Computers amp Mathematics with Applicationsvol 64 no 10 pp 3253ndash3257 2012
[24] X Zhang L Wang and Q Sun ldquoExistence of positive solutionsfor a class of nonlinear fractional differential equations withintegral boundary conditions and a parameterrdquo Applied Mathe-matics and Computation vol 226 pp 708ndash718 2014
18 Abstract and Applied Analysis
[25] D J Guo and V Lakshmikantham Nonlinear Problems inAbstract Cones vol 5 Academic Press New York NY USA1988
[26] M A Krasnoselrsquoskii Topological Methods in the Theory ofNonlinear Integral Equations Translated by A H ArmstrongElmsford Pergamon Turkey 1964
Submit your manuscripts athttpwwwhindawicom
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
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Differential EquationsInternational Journal of
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Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
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Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Stochastic AnalysisInternational Journal of
Abstract and Applied Analysis 7
Consequently for any 119905 isin [0 1] (26) and (44) imply that
1199091 (119905)
= 120601119902 (1205820)1
Γ (120572)
times int
1
0
(1 minus 119904)120572minus1
times 120601119902 (int
119904
0
120596 (120591) 119891 (120591 1199091 (120591)) 119889120591) 119889119904
minus int
119905
0
(119905 minus 119904)120572minus1
times 120601119902 (int
119904
0
120596 (120591) 119891 (120591 1199091 (120591)) 119889120591) 119889119904
+120578120601119902 (1205820)
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
119892 (119904)
times (int
1
0
(1 minus 120591)120572minus1
times 120601119902 (int
120591
0
120596 (120583) 119891 (120583 1199091 (120583)) 119889120583)119889120591
minus int
119904
0
(119904 minus 120591)120572minus1
times 120601119902 (int
120591
0
120596 (120583) 119891 (120583 1199091 (120583)) 119889120583)119889120591)119889119904
+ 1205811120595
ge120578120590120601119902 (1205820)
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
119892 (119904)
times (int
1
0
(1 minus 120591)120572minus1
times 120601119902 (int
120591
0
120596 (120583) 119891 (120583 1199091 (120583)) 119889120583)119889120591
minus int
119904
0
(119904 minus 120591)120572minus1
120601119902
times (int
120591
0
120596 (120583) 119891 (120583 1199091 (120583)) 119889120583)119889120591)119889119904
+120590120601119902 (1205820)
Γ (120572)
times int
1
0
(1 minus 119904)120572minus1
120601119902 (int
119904
0
120596 (120591) 119891 (120591 1199091 (120591)) 119889120591) 119889119904
+ 1205811120595
ge120590120601119902 (1205820)
Γ (120572)
times int
1
0
(1 minus 119904)120572minus1
120601119902 (int
119904
0
120596 (120591) 119891 (120591 1199091 (120591)) 119889120591) 119889119904
+ 1205811120595
ge120590120601119902 (1205820)
Γ (120572)
times int
1
0
(1 minus 119904)120572minus1
120601119902 (1198971 int
119904
0
120596 (120591) 120601119901 (1199091 (120591)) 119889120591) 119889119904
+ 1205811120595
ge120590120601119902 (1205820)
Γ (120572)
times int
1
0
(1 minus 119904)120572minus1
120601119902 (1198971 int
119904
0
120596 (120591) 120601119901 (120581lowast120595 (120591)) 119889120591) 119889119904
+ 1205811120595
=120590120581lowast119897119902minus1
1120582119902minus1
0
Γ (120572)int
1
0
(1 minus 119904)120572minus1
120601119902 (int
119904
0
120596 (120591) 119889120591) 119889119904 + 1205811120595
=120590120581lowast119897119902minus1
1120582119902minus1
01205741
Γ (120572)+ 1205811120595
= 120581lowast+ 120581120595
(45)
which implies that 1199091(119905) ge (120581lowast+ 120581)120595(119905) 119905 isin [0 1] which is a
contradiction to the definition of 120581lowast Thus (42) holds and byLemma 5 the fixed point index
119894 (120601119902 (1205820) 119879 119875 cap 120597Ω119903 119875) = 0 (46)
On the other hand by the fact that the fixed point index ofconstants operator is 1 so
119894 (120599 119875 cap 120597Ω119903 119875) = 1 (47)
where 120599 is the zero operator It follows therefore from (46)and (47) and the homotopy invariance property that thereexist 119909119903 isin 119875 cap 120597Ω119903 and 0 lt ]0 lt 1 such that ]0120601119902(1205820)119879119909119903 =119909119903 but 119909119903 = 120601119902(120582119903)119879119909119903 which implies that ]0120601119902(1205820)119879119909119903 =
120601119902(120582119903)119879119909119903 it follows that we get ]0120601119902(1205820) = 120601119902(120582119903)) that is
]0120582119902minus1
0= 120582119902minus1
119903 (48)
and then
120582119903 = ]1(119902minus1)0
1205820 = ]119901minus10
1205820 lt 1205820 (49)
From the proof above for any 119903 lt 1205850 there exists a positivesolution 119909119903 isin 119875 cap 120597Ω119903 associated with 120582 = 120582119903 gt 0
8 Abstract and Applied Analysis
Thus
119909119903 (119905)
=120601119902 (120582119903)
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902 (int
119904
0
120596 (120591) 119891 (120591 119909119903 (120591)) 119889120591) 119889119904
minus int
119905
0
(119905 minus 119904)120572minus1
times 120601119902 (int
119904
0
120596 (120591) 119891 (120591 119909119903 (120591)) 119889120591) 119889119904
+120578120601119902 (120582119903)
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
119892 (119904)
times (int
1
0
(1 minus 120591)120572minus1
times 120601119902 (int
120591
0
120596 (120583) 119891 (120583 119909119903 (120583)) 119889120583)119889120591
minus int
119904
0
(119904 minus 120591)120572minus1
times 120601119902 (int
120591
0
120596 (120583) 119891 (120583 119909119903 (120583)) 119889120583)119889120591)119889119904
(50)
with 119909119903 = 119903Next we show that 120582119903 ge 1205820 In fact
119909119903 (119905)
=120601119902 (120582119903)
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902 (int
119904
0
120596 (120591) 119891 (120591 119909119903 (120591)) 119889120591) 119889119904
minus int
119905
0
(119905 minus 119904)120572minus1
times 120601119902 (int
119904
0
120596 (120591) 119891 (120591 119909119903 (120591)) 119889120591) 119889119904
+120578120601119902 (120582119903)
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
119892 (119904)
times (int
1
0
(1 minus 120591)120572minus1
times 120601119902 (int
120591
0
120596 (120583) 119891 (120583 119909119903 (120583)) 119889120583)119889120591
minus int
119904
0
(119904 minus 120591)120572minus1
times 120601119902 (int
120591
0
120596 (120583) 119891 (120583 119909119903 (120583)) 119889120583)119889120591)119889119904
le120578120601119902 (120582119903)
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
119892 (119904) int
1
0
(1 minus 120591)120572minus1
times 120601119902 (int
120591
0
120596 (120583) 119891 (120583 119909119903 (120583)) 119889120583)119889120591 119889119904
+120601119902 (120582119903)
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902 (int
119904
0
120596 (120591) 119891 (120591 119909119903 (120591)) 119889120591) 119889119904
=120601119902 (120582119903) 120578 int
1
0119892 (119904) 119889119904
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
(1 minus 119904)120572minus1
times 120601119902 (int
119904
0
120596 (120591) 119891 (120591 119909119903 (120591)) 119889120591) 119889119904
+120601119902 (120582119903)
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902 (int
119904
0
120596 (120591) 119891 (120591 119909119903 (120591)) 119889120591) 119889119904
le120601119902 (120582119903) 120578 int
1
0119892 (119904) 119889119904
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
(1 minus 119904)120572minus1
times 120601119902 (1198972 int
119904
0
120596 (120591) 119909119903 (120591) 119889120591) 119889119904
+120601119902 (120582119903)
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902 (1198972120582119903 int
119904
0
120596 (120591) 119909119903 (120591) 119889120591) 119889119904
=120601119902 (120582119903)
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
(1 minus 119904)120572minus1
times 120601119902 (1198972 int
119904
0
120596 (120591) 119909119903 (120591) 119889120591) 119889119904
Abstract and Applied Analysis 9
le
10038171003817100381710038171199091199031003817100381710038171003817 120601119902 (1205821199031198972)
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
(1 minus 119904)120572minus1
times 120601119902 (int
1
0
120596 (120591) 119889120591) 119889119904
=
10038171003817100381710038171199091199031003817100381710038171003817 120601119902 (1205821199031198972) 1205742
Γ (120572 + 1) (1 minus 120578 int1
0119892 (119904) 119889119904)
(51)
which implies that
10038171003817100381710038171199091199031003817100381710038171003817 le
10038171003817100381710038171199091199031003817100381710038171003817 120601119902 (1205821199031198972) 1205742
Γ (120572 + 1) (1 minus 120578 int1
0119892 (119904) 119889119904)
(52)
that is
120601119902 (1205821199031198972) 1205742
Γ (120572 + 1) (1 minus 120578 int1
0119892 (119904) 119889119904)
ge 1 (53)
it follows that we get
120582119903 ge1
1198972(
Γ (120572 + 1) (1 minus 120578 int1
0119892 (119904) 119889119904)
1205742)
119901minus1
= 1205820(54)
In conclusion 120582119903 isin [1205820 1205820](ii) It follows from 0 lt 119891
infin
120601lt infin that there exist
119897lowast
1 119897lowast
2 120588lowastgt 0 such that
119897lowast
1120601119901 (119909) le 119891 (119905 119909) le 119897
lowast
2120601119901 (119909)
(forall119905 isin [0 1] 119909 ge 120588lowast)
(55)
Let 120585lowast0= 120588lowast120590 the following proof are similar to (i)
(iii) It follows from 1198910
120601= +infin there exists 119897lowast gt 0 120588
lowastgt 0
such that
1198910
120601ge 119897lowast120601119901 (119909) (forall119905 isin [0 1] 0 le 119909 le 120588
lowast) (56)
Let 1205851 = 120588lowast120590 we show that 1205851 is required When 119909 isin 119875 cap
120597Ω119903lowast we have 119909(119905) ge 120590119909 = 120590119903lowast since 119909 le 120588
lowast then we need120590119903lowastlt 120588lowast as 119903lowast lt 1205851 so 120590119903
lowastlt 120588lowast holds
Let 120582lowast = (1119897lowast)(Γ(120572)1205901205741)
1(119902minus1) we proceed in the sameway as in the proof of (i) replacing (42) we may assume that119909 minus 120601119902(120582
lowast)119879119909 = 0 (forall119909 isin 119875 cap 120597Ω119903lowast) and replacing (43) we
can prove 119909 minus 120601119902(120582lowast)119879119909 = 120581120595 (forall119909 isin 119875 cap 120597Ω119903lowast 120581 ge 0) It
follows from Lemma 5 that 119894(120601119902(120582lowast)119879 119875 cap Ω119903lowast 119875) = 0 Note
that 119894(120579 119875 cap Ω119903lowast 119875) = 1 we can easily show that there exists119909119903lowast isin 119875cap120597Ω119903lowast and 0 lt ]119903lowast lt 1 such that ]119903lowast120601119902(120582
lowast)119879119909119903lowast = 119909119903lowast
Hence (37) holds for 120582119903lowast = 120582lowast]1minus119901119903lowast
lt 120582lowast
The proof ofTheorem 9(iv) follows by themethod similarto Theorem 9(iii) we omit it here
(v) It follows that 119898(119903) gt 120585 for any 119909 isin 119875 cap 120597Ω119903 we have120590119903 le 119909 le 119903 and 119891(119905 119909)120601119901(119909) ge 119891(119905 119909)120601119901(119903) ge 119898(119903) ge 120585then
119891 (119905 119909) ge 120585120601119901 (119909) forall119909 isin [120590119903 119903] 119905 isin [0 1] (57)
Let = (1120585)(Γ(120572)1205901205741)119901minus1 The following proof is similar to
that of (iv) This finished the proof of (v)
Remark 10 InTheorem 9 all the criteria obtained depend onone of 1198910
120601and 119891
infin
120601
Let Ψ(119909) = 120601119902(120582)119879119909 the following theorems give out themultiply nonexistence and the dependence of parameter
Theorem 11 Assume that (H1)ndash(H5) hold
(i) If 1198910
120601= 0 and 119891
infin
120601= 0 lim
119909rarrinfin119891(119905 119909) = 119891infin isin
(0 +infin] uniformly for 119905 isin [0 1] then there exists120582lowast
gt 0 such that the problem (1) has two solutions forany 120582 gt 120582
lowast
(ii) If 1198910120601
= 0 and 119891infin
120601= 0 then there exists 120582 gt 0 such
that for any 120582 lt 120582 problem (1) has no solution
Proof (i) It follows from 119891infin isin (0 +infin] that for any 120577 gt 0there exists 120588 gt 0 such that
119891 (119905 119909) gt 120577 (forall119905 isin [0 1] 119909 ge 120588) (58)
Since 1198910
120601= 0 there exists 1205761 gt 0 120588
1gt 0 such that
119891 (119905 119909) le 1205761120601119901 (119909) (forall119905 isin [0 1] 0 le 119909 le 1205881) (59)
where 1205761 satisfied 120601119902(1205821205761)1205742Γ(120572 + 1)(1 minus 120578 int1
0119892(119904)119889119904) le 1
For forall119909 isin 119875 cap 120597Ω1205881
we have
Ψ (119909) = (120601119902 (120582) 119879119909) (119905)
=1
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902 (120582int
119904
0
120596 (120591) 119891 (120591 119909 (120591)) 119889120591) 119889119904
minus int
119905
0
(119905 minus 119904)120572minus1
times 120601119902 (120582int
119904
0
120596 (120591) 119891 (120591 119909 (120591)) 119889120591) 119889119904
+120578
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
119892 (119904)
10 Abstract and Applied Analysis
times (int
1
0
(1 minus 120591)120572minus1
times 120601119902 (120582int
120591
0
120596 (120583) 119891 (120583 119909 (120583)) 119889120583)119889120591
minus int
119904
0
(119904 minus 120591)120572minus1
times 120601119902 (120582int
120591
0
120596 (120583) 119891 (120583 119909 (120583)) 119889120583)119889120591)119889119904
le120578
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
119892 (119904)
times (int
1
0
(1 minus 120591)120572minus1
times 120601119902(120582int
120591
0
120596 (120583) 119891 (120583 119909 (120583)) 119889120583)119889120591)119889119904
+1
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902 (120582int
119904
0
120596 (120591) 1205761119891 (120591 119909 (120591)) 119889120591) 119889119904
le120578 int1
0119892 (119904) 119889119904
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
(1 minus 120591)120572minus1
times 120601119902 (120582int
1
0
120596 (120583) 1205761120601119901 (119909 (120583)) 119889120583)119889120591
+1
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902(120582int
1
0
120596 (120591) 1205761120601119901 (119909 (120591)) 119889120591) 119889119904
le1
Γ (120572 + 1) (1 minus 120578 int1
0119892 (119904) 119889119904)
times 120601119902 (120582int
1
0
120596 (120591) 1205761120601119901 (119909) 119889120591)
=120601119902 (1205821205761) 1205742
Γ (120572 + 1) (1 minus 120578 int1
0119892 (119904) 119889119904)
119909 le 119909
(60)
This implied
Ψ119909 le 119909 forall119909 isin 119875 cap 120597Ω1205881
(61)
Let 1205883gt max120588120590 120588
1 and 120582
lowast= (1120577)(120588
3Γ(120572)1205901205741)
119901minus1 thenforall119909 isin 119875 cap 120597Ω120588
3
we have 120590119909 le 119909 le 1205883and
(Ψ119909) (119905)
=1
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902 (120582int
119904
0
120596 (120591) 119891 (120591 119909 (120591)) 119889120591) 119889119904
minus int
119905
0
(119905 minus 119904)120572minus1
times 120601119902 (120582int
119904
0
120596 (120591) 119891 (120591 119909 (120591)) 119889120591) 119889119904
+120578
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
119892 (119904)
times (int
1
0
(1 minus 120591)120572minus1
times 120601119902 (120582int
120591
0
120596 (120583) 119891 (120583 119909 (120583)) 119889120583)119889120591
minus int
119904
0
(119904 minus 120591)120572minus1
times 120601119902(120582int
120591
0
120596 (120583) 119891 (120583 119909 (120583)) 119889120583)119889120591)119889119904
ge120590120578
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
119892 (119904)
times (int
1
0
(1 minus 120591)120572minus1
times 120601119902 (120582int
120591
0
120596 (120583) 119891 (120583 119909 (120583)) 119889120583)119889120591
minus int
119904
0
(119904 minus 120591)120572minus1
times 120601119902(120582int
120591
0
120596 (120583) 119891 (120583 119909 (120583)) 119889120583)119889120591)119889119904
+120590
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902 (120582int
119904
0
120596 (120591) 119891 (120591 119909 (120591)) 119889120591) 119889119904
ge120590
Γ (120572)int
1
0
(1 minus 119904)120572minus1
Abstract and Applied Analysis 11
times 120601119902 (120582int
119904
0
120596 (120591) 119891 (120591 119909 (120591)) 119889120591) 119889119904
gt120590
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902 (120582int
119904
0
120596 (120591) 120577 119889120591) 119889119904
gt120590120601119902 (120582120577)
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902 (int
119904
0
120596 (120591) 119889120591) 119889119904
=120590120601119902 (120582120577) 1205741
Γ (120572)= 1205883
(62)
which implies
Ψ119909 gt 119909 forall119909 isin 119875 cap 120597Ω1205883
(63)
Next for 119891infin
120601= 0 there exist 1205762 gt 0 120588
2gt 1205883gt 0 such
that
119891 (119905 119909) le 1205762120601119901 (119909) (forall119905 isin [0 1] 119909 ge 1205882) (64)
where 1205762 satisfied 120601119902(1205821205762)1205742Γ(120572 + 1)(1 minus 120578 int1
0119892(119904)119889119904) le 1
Similar to the above proof we get
Ψ119909 le 119909 forall119909 isin 119875 cap 120597Ω1205882
(65)
Applying Lemma 7 to (61) (65) and (63) yields that Ψ
has two fixed points 1199091 1199092 such that 1199091 isin 119875cap (Ω1205883
Ω1205881
) and1199092 isin 119875 cap (Ω120588
2
Ω1205883
)(ii) It follows from 119891
0
120601= 0 and 119891
infin
120601= 0 that there exists
1205761 1205762 1205881 1205882 gt 0 such that
119891 (119905 119909) le 1205761120601119901 (119909) (forall119905 isin [0 1] 0 le 119909 le 1205881)
119891 (119905 119909) le 1205762120601119901 (119909) (forall119905 isin [0 1] 119909 ge 1205882)
(66)
then for 119909 isin [1205881 1205882] by the continuous of 119891(119905 119909(119905)) there
exists 119909 isin [1205881 1205882] such that
max1205881le119909le120588
2
119891 (119905 119909)
120601119901 (119909)=
119891 (119905 119909)
120601119901 (119909) (67)
Let Υ = max1205761 1205762 119891(119905 119909)120601119901(119909) gt 0 then
119891 (119905 119909) le Υ120601119901 (119909) (forall0 le 119905 le 1 119909 ge 0) (68)
Assuming 119909(119905) is a positive solution of problem (1) we willshow that this leads to a contradiction for 120582 lt 120582 where
120582 = (1Υ)(Γ(120572 + 1)(1 minus 120578 int1
0119892(119904)119889119904)1205742)
119901minus1 It follows from(26) that
119909 = 119909 (0)
=120578
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
119892 (119904)
times (int
1
0
(1 minus 120591)120572minus1
times 120601119902 (120582int
120591
0
120596 (120583) 119891 (120583 119909 (120583)) 119889120583)119889120591
minus int
119904
0
(119904 minus 120591)120572minus1
times 120601119902(120582int
120591
0
120596 (120583) 119891 (120583 119909 (120583)) 119889120583)119889120591)119889119904
+1
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902 (120582int
119904
0
120596 (120591) 119891 (120591 119909 (120591)) 119889120591) 119889119904
le120578
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
119892 (119904)
times int
1
0
(1 minus 120591)120572minus1
times 120601119902(120582int
120591
0
120596 (120583) 119891 (120583 119909 (120583)) 119889120583)119889120591 119889119904
+1
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902 (120582int
119904
0
120596 (120591) 119891 (120591 119909 (120591)) 119889120591) 119889119904
=1
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
(1 minus 119904)120572minus1
times 120601119902 (120582int
119904
0
120596 (120591) 119891 (120591 119909 (120591)) 119889120591) 119889119904
le1
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
(1 minus 119904)120572minus1
12 Abstract and Applied Analysis
times 120601119902 (120582int
1
0
120596 (120591) Υ120601119901 (119909 (120591)) 119889120591) 119889119904
=120601119902 (120582Υ) 1205742
Γ (120572 + 1) (1 minus 120578 int1
0119892 (119904) 119889119904)
119909
(69)
This implies that 1 le 120601119902(120582Υ)1205742Γ(120572 + 1)(1 minus 120578 int1
0119892(119904)119889119904) =
(120582Υ)119902minus1
1205742Γ(120572+1)(1minus120578 int1
0119892(119904)119889119904) lt (120582 Υ)
119902minus11205742Γ(120572+1)(1minus
120578 int1
0119892(119904)119889119904) = 1 which is a contradiction This finishes the
proof of (ii)
Remark 12 In (i) the condition lim119909rarrinfin
119891(119905 119909) = 119891infin isin
(0 +infin] uniformly for 119905 isin [0 1] can be replaced withcondition (H6) The same methods can be used to prove theresults
Remark 13 From the above proof the condition 119891infin isin
(0 +infin] is important to keep the problem (1) having at leasttwo positive solutions if this condition is not satisfied thenthere exists 120582 gt 0 small enough such that the problem (1) hasno positive solutions for all 120582 lt 120582
Theorem 14 Assume that (H1)ndash(H6) hold(i) If 1198910
120601= +infin and 119891
infin
120601= +infin then there exists 120582lowast0 gt 0
such that problem (1) has at least two solutions for 120582 isin
(0 120582lowast0)(ii) If 1198910
120601= +infin and 119891
infin
120601= +infin then there exists 120582 gt 0
such that problem (1) has no solution for all 120582 gt 120582
Proof (i) It follows from 1198910
120601= +infin that there exists 1199031 gt 0
such that
119891 (119905 119909) ge 119871lowast1120601119901 (119909) (forall119905 isin [0 1] 0 le 119909 le 1199031) (70)
where 119871lowast1 satisfies 1205902120601119902(120582119871lowast1)1205741Γ(120572) ge 1
Then for 119909 isin 119875cap120597Ω1199031
we have 120590119909 le 119909 le 1199031 for 119905 isin [0 1]
and
(Ψ119909) (119905)
=1
Γ (120572)
times int
1
0
(1 minus 119904 )120572minus1
times 120601119902 (120582int
119904
0
120596 (120591) 119891 (120591 119909 (120591)) 119889120591) 119889119904
minus int
119905
0
(119905 minus 119904)120572minus1
times 120601119902 (120582int
119904
0
120596 (120591) 119891 (120591 119909 (120591)) 119889120591) 119889119904
+120578
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
119892 (119904)
times (int
1
0
(1 minus 120591)120572minus1
times 120601119902 (120582int
120591
0
120596 (120583) 119891 (120583 119909 (120583)) 119889120583)119889120591
minus int
119904
0
(119904 minus 120591)120572minus1
times 120601119902 (120582int
120591
0
120596 (120583) 119891 (120583 119909 (120583)) 119889120583)119889120591)119889119904
ge120590120578
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
119892 (119904)
times (int
1
0
(1 minus 120591)120572minus1
times 120601119902 (120582int
120591
0
120596 (120583) 119891 (120583 119909 (120583)) 119889120583)119889120591
minus int
119904
0
(119904 minus 120591)120572minus1
times120601119902 (120582int
120591
0
120596 (120583) 119891 (120583 119909 (120583)) 119889120583)119889120591)119889119904
+120590
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902 (120582int
119904
0
120596 (120591) 119891 (120591 119909 (120591)) 119889120591) 119889119904
ge120590
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902 (120582int
119904
0
120596 (120591) 119891 (120591 119909 (120591)) 119889120591) 119889119904
ge120590
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902 (120582int
119904
0
120596 (120591) 119871lowast1120601119901 (119909 (120591)) 119889120591) 119889119904
ge1205902120601119902 (120582119871lowast1) 119909
Γ (120572)
times int
1
0
(1 minus 119904)120572minus1
times 120601119902 (int
119904
0
120596 (120591) 119889120591) 119889119904
=1205902120601119902 (120582119871lowast1) 1205742
Γ (120572)119909
(71)
Abstract and Applied Analysis 13
which implies that
Ψ119909 ge 119909 forall119909 isin 119875 cap 120597Ω1199031
(72)
Next from 119891infin
120601= +infin there exists 1199032 gt 0 such that
119891 (119905 119909) ge 119871lowast2120601119901 (119909) forall119909 ge 1199032 (73)
where 119871lowast2 satisfies
1205902120601119902 (120582119871lowast2) 1205742
Γ (120572)ge 1 (74)
Let 1199032 gt max1199031 1199032120590 then for 119909 isin 119875 cap 120597Ω1199032
we have
119909 (119905) ge 120590 119909 = 1205901199032 gt 1199032 (75)
Similar to the above proof we can get
Ψ119909 ge 119909 forall119909 isin 119875 cap 120597Ω1199032
(76)
Let 1199031 lt 1199033 lt 11990321198721199033
= max119891(119905 119909) | 119905 isin [0 1] 119909 le 1199033+1 gt
0 and 120582lowast0 = (11198721199033
)(Γ(120572+1)1199033(1minus120578 int1
0119892(119904)119889119904)1205742)
119901minus1Thenfor 119909 isin 119875 cap 120597Ω119903
3
we have
119891 (119905 119909) le 1198721199033
forall119905 isin [0 1] (77)
Hence
(Ψ119909) (119905)
=1
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902 (120582int
119904
0
120596 (120591) 119891 (120591 119909 (120591)) 119889120591) 119889119904
minus int
119905
0
(119905 minus 119904)120572minus1
times 120601119902 (120582int
119904
0
120596 (120591) 119891 (120591 119909 (120591)) 119889120591) 119889119904
+120578
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
119892 (119904)
times (int
1
0
(1 minus 120591)120572minus1
times 120601119902 (120582int
120591
0
120596 (120583) 119891 (120583 119909 (120583)) 119889120583)119889120591
minus int
119904
0
(119904 minus 120591)120572minus1
times120601119902 (120582int
120591
0
120596 (120583) 119891 (120583 119909 (120583)) 119889120583)119889120591)119889119904
le120578
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
119892 (119904)
times int
1
0
(1 minus 120591)120572minus1
120601119902
times (120582int
120591
0
120596 (120583) 119891 (120583 119909 (120583)) 119889120583)119889120591 119889119904
+1
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902 (120582int
119904
0
120596 (120591) 119891 (120591 119909 (120591)) 119889120591) 119889119904
le120578
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
119892 (119904)
times int
1
0
(1 minus 120591)120572minus1
120601119902
times (120582int
120591
0
120596 (120583)1198721199033
119889120583)119889120591119889119904
+1
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902 (120582int
119904
0
120596 (120591)1198721199033
119889120591)119889119904
le120578120601119902 (120582119872119903
3
) int1
0119892 (119904) 119889119904
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
(1 minus 120591)120572minus1
times 120601119902 (int
1
0
120596 (120583) 119889120583)119889120591
+120601119902 (120582119872119903
3
)
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902 (int
1
0
120596 (120591) 119889120591) 119889119904
=1205742120601119902 (120582119872119903
3
)
Γ (120572 + 1) (1 minus 120578 int1
0119892 (119904) 119889119904)
lt1205742120601119902 (120582lowast0119872119903
3
)
Γ (120572 + 1) (1 minus 120578 int1
0119892 (119904) 119889119904)
= 1199033
(78)
14 Abstract and Applied Analysis
This implies that
Ψ119909 lt 119909 forall119909 isin 119875 cap 120597Ω1199033
(79)
Applying Lemma 7 to (72) (76) and (79) yields that Ψ hastwo fixed points 1199091 1199092 such that 1199091 isin 119875 cap (Ω119903
3
Ω1199031
) and1199092 isin 119875 cap (Ω119903
2
Ω1199033
)(ii) By the proof of (i) and the continuity of 119891(119905 119909) there
exists 119909 isin [1199031 1199032] such that min119909isin[11990311199032](119891(119905 119909)120601119901(119909)) =
119891(119905 119909)120601119901(119909)Let Υ = min119871lowast1 119871lowast2 119891(119905 119909)120601119901(119909) since (H6) holds
then Υ gt 0 and
119891 (119905 119909) ge Υ120601119901 (119909) forall119909 ge 0 119905 isin [0 1] (80)
Assuming 119909(119905) is a positive solution of problem (1) we willshow that this leads to contradiction for all 120582 gt 120582 where 120582 =
(1Υ)(Γ(120572)1205741120590)119901minus1 It follows form (26) that
119909 = 119909 (0)
=120578
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
119892 (119904)
times (int
1
0
(1 minus 120591)120572minus1
times 120601119902 (120582int
120591
0
120596 (120583) 119891 (120583 119909 (120583)) 119889120583)119889120591
minus int
119904
0
(119904 minus 120591)120572minus1
times120601119902 (120582int
120591
0
120596 (120583) 119891 (120583 119909 (120583)) 119889120583)119889120591)119889119904
+1
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902 (120582int
119904
0
120596 (120591) 119891 (120591 119909 (120591)) 119889120591) 119889119904
ge1
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902 (120582int
119904
0
120596 (120591) 119891 (120591 119909 (120591)) 119889120591) 119889119904
ge1
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902 (120582int
119904
0
120596 (120591) Υ120601119901 (119909 (120591)) 119889120591) 119889119904
ge1
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902 (120582int
119904
0
120596 (120591) Υ120601119901 (120590 119909 119889120591)) 119889119904
ge120601119902 (120582Υ) 120590 119909
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902 (int
119904
0
120596 (120591) 119889120591) 119889119904
=120601119902 (120582Υ) 1205741120590
Γ (120572)119909
(81)
This implies that 1 ge 120601119902(120582Υ)1205741120590Γ(120572) = (120582Υ)119902minus1
1205741120590Γ(120572) gt
(120582Υ)119902minus1
1205741120590Γ(120572) = 1 which is a contradiction This com-pletes the proof of (ii)
Corollary 15 Assume that (H1)ndash(H5) holds
(i) If lim119909rarrinfin
119891(119905 119909) = 119891infin isin (0 +infin] uniformly for 119905 isin
[0 1] and one of 1198910120601
= 0 and 119891infin
120601= 0 is satisfied then
there exists 120582lowast gt 0 such that the problem (1) has at leastone positive solution for any 120582 gt 120582
lowast(ii) If one of 1198910
120601= +infin and 119891
infin
120601= +infin holds then there
exists 120582lowast0 gt 0 such that problem (1) has at least onepositive solutions for any 120582 isin (0 120582lowast0)
Proof (i) The conclusion is a direct consequence of Theo-rem 11(i)
(ii) The conclusion is a direct consequence of Theo-rem 14(i)
Theorem 16 Assume that (H1)ndash(H5) holdThen the followingconclusions hold
(i) If 1198910120601
= 0 and 119891infin
120601= infin then the problem (1) has a
positive solution 119909120582(119905) for all 120582 gt 0(ii) If 1198910
120601= infin and 119891
infin
120601= 0 then the problem (1) has a
positive solution 119909120582(119905) for all 120582 gt 0
Proof Weonly prove (i) the proof of (ii) is similar sowe omitit here
Let 120582 gt 0 since 1198910
120601= 0 there exists 119903 gt 0 119897 gt 0 such that
119891 (119905 119909) le 119897120601119901 (119909) (forall119905 isin [0 1] 119909 isin [0 119903]) (82)
where 119897 satisfied
1205742120582119902minus1
119897119902minus1
Γ (120572 + 1) (1 minus 120578 int1
0119892 (119904) 119889119904)
le 1 (83)
Similar to the proof of Theorem 14 we obtain that
Ψ119909 le 119909 forall119909 isin 119875 cap 120597Ω119903 (84)
Abstract and Applied Analysis 15
From 119891infin
120601= infin there exists 119877 gt 119903 gt 0 119871 gt 0 such that
119891 (119905 119909) ge 119871120601119901 (119909) (forall119905 isin [0 1] 119909 ge 119877) (85)where 119871 satisfied
1205902120582119902minus1
119871119902minus1
1205741
Γ (120572)ge 1 (86)
We get thatΨ119909 ge 119909 forall119909 isin 119875 cap 120597Ω119877 (87)
Applying Lemma 7 to (84) (87) yields thatΨ has a fixed point119909 such that 119909 isin 119875 cap (Ω119877 Ω119903)
Theorem 17 Assume that (H1)ndash(H6) holdThen the followingconclusions hold
If 0 lt 1198910
120601lt infin and 0 lt 119891
infin
120601lt infin then there exists 1 2 gt
0 such that problem (1) has no positive solution for any 120582 gt 1
and 0 lt 120582 lt 2
Proof It follows from 0 lt 1198910
120601lt infin and 0 lt 119891
infin
120601lt infin that
there exist 1 gt 0 2 gt 0 1 gt 0 2 gt 0 and 1199032 gt 1199031 gt 0
such that1120601119901 (119909) le 119891 (119905 119909) le 1120601119901 (119909)
(forall119905 isin [0 1] 0 le 119909 le 1199031)
2120601119901 (119909) le 119891 (119905 119909) le 2120601119901 (119909)
(forall119905 isin [0 1] 119909 ge 1199032)
(88)
Let = max119891(119905 119909)120601119901(119909) 1199031 le 119909 le 1199032 0 le 119905 le 1 =
min119891(119905 119909)120601119901(119909) 1199031 le 119909 le 1199032 0 le 119905 le 1 By the condition(H6) gt 0 Then forall119909 ge 0 we have
120601119901 (119909) le 119891 (119905 119909) le 120601119901 (119909) (89)
where = max1 2 gt 0 and = min1198971 2 gt 0Let 1 = (1)(Γ(120572)1205741120590)
119901minus1 and 2 =
(1)(Γ(120572 + 1)(1 minus 120578 int1
0119892(119904)119889119904)1205742)
119901minus1
If problem (1)has positive solution 119909(119905) in 119875 then
119909 = 119909 (0)
=120578
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
119892 (119904)
times (int
1
0
(1 minus 120591)120572minus1
times 120601119902 (120582int
120591
0
120596 (120583) 119891 (120583 119909 (120583)) 119889120583)119889120591
minus int
119904
0
(119904 minus 120591)120572minus1
times120601119902 (120582int
120591
0
120596 (120583) 119891 (120583 119909 (120583)) 119889120583)119889120591)119889119904
+1
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902 (120582int
119904
0
120596 (120591) 119891 (120591 119909 (120591)) 119889120591) 119889119904
ge1
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902 (120582int
119904
0
120596 (120591) 119891 (120591 119909 (120591)) 119889120591) 119889119904
ge1
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902 (120582int
119904
0
120596 (120591) 120601119901 (119909 (120591)) 119889120591) 119889119904
ge1
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902 (120582int
119904
0
120596 (120591) 120601119901 (120590 119909 119889120591)) 119889119904
ge120601119902 (120582) 120590 119909
Γ (120572)
times int
1
0
(1 minus 119904)120572minus1
120601119902 (int
119904
0
120596 (120591) 119889120591) 119889119904
=120601119902 (120582) 1205741120590
Γ (120572)119909
gt120601119902 (1 ) 1205741120590
Γ (120572)119909 = 119909
119909 = 119909 (0)
=120578
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
119892 (119904)
times (int
1
0
(1 minus 120591)120572minus1
120601119902
times (120582int
120591
0
120596 (120583) 119891 (120583 119909 (120583)) 119889120583)119889120591
minus int
119904
0
(119904 minus 120591)120572minus1
120601119902
times(120582int
120591
0
120596 (120583) 119891 (120583 119909 (120583)) 119889120583)119889120591)119889119904
+1
Γ (120572)int
1
0
(1 minus 119904)120572minus1
120601119902
times (120582int
119904
0
120596 (120591) 119891 (120591 119909 (120591)) 119889120591) 119889119904
16 Abstract and Applied Analysis
le120578
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
119892 (119904)
times int
1
0
(1 minus 120591)120572minus1
times 120601119902 (120582int
120591
0
120596 (120583) 119891 (120583 119909 (120583)) 119889120583)119889120591 119889119904
+1
Γ (120572)int
1
0
(1 minus 119904)120572minus1
120601119902
times (120582int
119904
0
120596 (120591) 119891 (120591 119909 (120591)) 119889120591) 119889119904
=1
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
(1 minus 119904)120572minus1
120601119902
times (120582int
119904
0
120596 (120591) 119891 (120591 119909 (120591)) 119889120591) 119889119904
le1
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
(1 minus 119904)120572minus1
120601119902
times (120582int
1
0
120596 (120591) 120601119901 (119909 (120591)) 119889120591) 119889119904
=120601119902 (120582) 1205742
Γ (120572 + 1) (1 minus 120578 int1
0119892 (119904) 119889119904)
119909
lt120601119902 (2) 1205742
Γ (120572 + 1) (1 minus 120578 int1
0119892 (119904) 119889119904)
119909 = 119909
(90)
which is a contradiction This finishes the proof of Theo-rem 17
4 Examples
In this section we give out same examples to illustrate ourmain results can be used in practice
Example 1 Consider the following fractional differentialequation
(120601119901 (11986352
0+119909 (119905)))
1015840
+ 120582119905(1 + 1199052)12
119909119901minus1 arctan119909 = 0
0 lt 119905 lt 1
1199091015840(0) = 119909
10158401015840(0) = 0
119909 (1) =1
2int
1
0
1
2119909 (119904) 119889119904
(91)
where 120572 = 52 119901 = 32 119902 = 3 120596(119905) = 119905 119891(119905 119909) = (1 +
1199052)12
119909119901minus1 arctan119909 120578 = 12 and 119892(119905) = 12
By simple computation 119897lowast1= 1205872 119897
lowast
2= radic21205872 1205741 = int
1
0(1minus
119904)32
1206013(int119904
0120591 119889120591)119889119904 asymp 00043 1205742 = 1206013(int
1
0120591 119889120591) = 025 120590 asymp
01429 120582lowast
0= (1119897
lowast
1)(Γ(120572)1205901205741)
119901minus1asymp 25149 120582
lowast
0= (1119897
lowast
2)(Γ(120572+
1)(1 minus 120578 int1
0119892(119904)119889119904)1205742)
119901minus1asymp 14222 Then it is easy to see
all the conditions of Theorem 9(ii) are satisfied thus forall120582 isin
[14222 25149]] the problem (91) has at least one positivesolution
Example 2 Consider the boundary value problem
(120601119901 (11986352
0+119909 (119905)))
1015840
+ 120582119905(1 + 1199052)12
119909119901119890minus119909
= 0
0 lt 119905 lt 1
1199091015840(0) = 119909
10158401015840(0) = 0
119909 (1) =1
2int
1
0
1
2119909 (119904) 119889119904
(92)
where 120572 = 52 119901 = 32 119902 = 3 120596(119905) = 119905 119891(119905 119909) = (1 +
1199052)12
119909119901119890minus119909
120578 = 12 and 119892(119905) = 12
1198910
120601= lim119909rarr0
119891 (119905 119909)
120601119901 (119909)= lim119909rarr0
(1 + 1199052)12
119909119901119890minus119909
119909119901minus1= 0
forall119905 isin [0 1]
1198910
120601= lim119909rarrinfin
119891 (119905 119909)
120601119901 (119909)= lim119909rarrinfin
(1 + 1199052)12
119909119901119890minus119909
119909119901minus1= 0
forall119905 isin [0 1]
(93)
Let 1205881= 12 120588
2= 4 then
119891 (119905 119909)
120601119901 (119909)le (1 + 119905
2)12 1
2radic119890le
1
radic2119890
forall119905 isin [0 1] 119909 isin [01
2]
119891 (119905 119909)
120601119901 (119909)le (1 + 119905
2)12 4
1198904le
4radic2
1198904
forall119905 isin [0 1] 119909 isin [4infin)
(94)
so 1205761 = 1radic2119890 1205762 = 4radic21198904 and forall119909 isin
[12 4] max12le119909le4(119891(119905 119909)120601119901(119909)) = radic2119890minus1 We
have Υ = max1radic2119890 4radic21198904 radic2119890minus1
= radic2119890minus1 1205742 = 025
Abstract and Applied Analysis 17
and 120582 = (1Υ)(Γ(120572+1)(1minus120578 int1
0119892(119904)119889119904)1205742)
119901minus1= 3radic5120587
141198904
By Theorem 11(ii) for any 120582 lt 120582 = 3radic512058714
1198904 the problemhas no positive solution
5 Conclusions
Recently differential equations with p-Laplacian operatorwere widely discussed by several authors In this paperby using fixed point techniques combining with partiallyordered structure of Banach space we obtained the existencemultiply and the dependence of parameter These newresults we presented can be used in numerical computationand analyze mathematical models of physical phenomenamechanics nonlinear dynamics and many other relatedfields
Conflict of Interests
The authors declare that there is no conflict of interestsregarding the publication of this paper
Acknowledgments
The authors are very grateful to anonymous referees for theirvaluable and detailed suggestions and comments to improvethe original paper This work is supported by the Programfor New Century Excellent Talents in University (NCET-10-0097)
References
[1] I Podlubny Fractional Differential Equations vol 198 Aca-demic Press San Diego Calif USA 1999
[2] A A Kilbas H M Srivastava and J J Trujillo Theoryand Applications of Fractional Differential Equations vol 204of North-Holland Mathematics Studies Elsevier Science BVAmsterdam The Netherlands 2006
[3] Y A Rossikhin and M V Shitikova ldquoApplication of fractionalderivatives to the analysis of damped vibrations of viscoelasticsingle mass systemsrdquoActa Mechanica vol 120 no 1ndash4 pp 109ndash125 1997
[4] J Sabatier O Agrawal and J T Machado Advances in Frac-tional Calculus Theoretical developments and applications inphysics and engineering Springer Berlin Germany 2007
[5] N Engheta ldquoOn fractional calculus and fractional multipolesin electromagnetismrdquo IEEE Transactions on Antennas andPropagation vol 44 no 4 pp 554ndash566 1996
[6] R Hilfer Applications of Fractional Calculus in Physics WorldScientific Singapore 2000
[7] R L Bagley and R A Calico ldquoFractional order state equationsfor the control of viscoelastically damped structuresrdquo Journalof Guidance Control and Dynamics vol 14 no 2 pp 304ndash3111991
[8] W-X Zhou andY-D Chu ldquoExistence of solutions for fractionaldifferential equations with multi-point boundary conditionsrdquoCommunications in Nonlinear Science and Numerical Simula-tion vol 17 no 3 pp 1142ndash1148 2012
[9] B Ahmad and G Wang ldquoA study of an impulsive four-pointnonlocal boundary value problemof nonlinear fractional differ-ential equationsrdquo Computers amp Mathematics with Applicationsvol 62 no 3 pp 1341ndash1349 2011
[10] B Ahmad and S Sivasundaram ldquoOn four-point nonlocalboundary value problems of nonlinear integro-differentialequations of fractional orderrdquo Applied Mathematics and Com-putation vol 217 no 2 pp 480ndash487 2010
[11] N Nyamoradi ldquoPositive solutions for multi-point boundaryvalue problems for nonlinear fractional differential equationsrdquoJournal of Contemporary Mathematical Analysis vol 48 no 4pp 145ndash157 2013
[12] B Ahmad and J J Nieto ldquoExistence of solutions for nonlocalboundary value problems of higher-order nonlinear fractionaldifferential equationsrdquo Abstract and Applied Analysis vol 2009Article ID 494720 9 pages 2009
[13] X Liu M Jia and X Xiang ldquoOn the solvability of a fractionaldifferential equationmodel involving the119901-Laplacian operatorrdquoComputers ampMathematics with Applications vol 64 no 10 pp3267ndash3275 2012
[14] J Wang H Xiang and Z Liu ldquoExistence of concave positivesolutions for boundary value problem of nonlinear fractionaldifferential equation with 119901-Laplacian operatorrdquo InternationalJournal of Mathematics and Mathematical Sciences vol 2010Article ID 495138 17 pages 2010
[15] J Wang and H Xiang ldquoUpper and lower solutions method fora class of singular fractional boundary value problems with 119901-Laplacian operatorrdquo Abstract and Applied Analysis vol 2010Article ID 971824 12 pages 2010
[16] H Lu andZHan ldquoExistence of positive solutions for boundary-value problem of fractional differential equation with 119901-laplacian operatorrdquo The American Jouranl of Engineering andTechnology Research vol 11 pp 3757ndash3764 2011
[17] G Chai ldquoPositive solutions for boundary value problem offractional differential equation with 119901-Laplacian operatorrdquoBoundary Value Problems vol 2012 article 18 2012
[18] T Chen W Liu and Z Hu ldquoA boundary value problem forfractional differential equation with 119901-Laplacian operator atresonancerdquoNonlinear AnalysisTheoryMethodsampApplicationsvol 75 no 6 pp 3210ndash3217 2012
[19] T Chen andW Liu ldquoAn anti-periodic boundary value problemfor the fractional differential equation with a 119901-LaplacianoperatorrdquoAppliedMathematics Letters of Rapid Publication vol25 no 11 pp 1671ndash1675 2012
[20] S J Li X G Zhang Y H Wu and L Caccetta ldquoExtremalsolutions for 119901-Laplacian differential systems via iterative com-putationrdquo Applied Mathematics and Computation vol 26 pp1151ndash1158 2013
[21] Z Liu and L Lu ldquoA class of BVPs for nonlinear fractionaldifferential equations with 119901-Laplacian operatorrdquo ElectronicJournal of Qualitative Theory of Differential Equations vol 70pp 1ndash16 2012
[22] Z Liu L Lu and I Szanto ldquoExistence of solutions for fractionalimpulsive differential equations with 119901-Laplacian operatorrdquoActa Mathematica Hungarica vol 141 no 3 pp 203ndash219 2013
[23] Z Bai ldquoEigenvalue intervals for a class of fractional boundaryvalue problemrdquo Computers amp Mathematics with Applicationsvol 64 no 10 pp 3253ndash3257 2012
[24] X Zhang L Wang and Q Sun ldquoExistence of positive solutionsfor a class of nonlinear fractional differential equations withintegral boundary conditions and a parameterrdquo Applied Mathe-matics and Computation vol 226 pp 708ndash718 2014
18 Abstract and Applied Analysis
[25] D J Guo and V Lakshmikantham Nonlinear Problems inAbstract Cones vol 5 Academic Press New York NY USA1988
[26] M A Krasnoselrsquoskii Topological Methods in the Theory ofNonlinear Integral Equations Translated by A H ArmstrongElmsford Pergamon Turkey 1964
Submit your manuscripts athttpwwwhindawicom
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
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Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical Problems in Engineering
Hindawi Publishing Corporationhttpwwwhindawicom
Differential EquationsInternational Journal of
Volume 2014
Applied MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical PhysicsAdvances in
Complex AnalysisJournal of
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OptimizationJournal of
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CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of
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Operations ResearchAdvances in
Journal of
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Function Spaces
Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of Mathematics and Mathematical Sciences
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Algebra
Discrete Dynamics in Nature and Society
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Decision SciencesAdvances in
Discrete MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom
Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Stochastic AnalysisInternational Journal of
8 Abstract and Applied Analysis
Thus
119909119903 (119905)
=120601119902 (120582119903)
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902 (int
119904
0
120596 (120591) 119891 (120591 119909119903 (120591)) 119889120591) 119889119904
minus int
119905
0
(119905 minus 119904)120572minus1
times 120601119902 (int
119904
0
120596 (120591) 119891 (120591 119909119903 (120591)) 119889120591) 119889119904
+120578120601119902 (120582119903)
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
119892 (119904)
times (int
1
0
(1 minus 120591)120572minus1
times 120601119902 (int
120591
0
120596 (120583) 119891 (120583 119909119903 (120583)) 119889120583)119889120591
minus int
119904
0
(119904 minus 120591)120572minus1
times 120601119902 (int
120591
0
120596 (120583) 119891 (120583 119909119903 (120583)) 119889120583)119889120591)119889119904
(50)
with 119909119903 = 119903Next we show that 120582119903 ge 1205820 In fact
119909119903 (119905)
=120601119902 (120582119903)
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902 (int
119904
0
120596 (120591) 119891 (120591 119909119903 (120591)) 119889120591) 119889119904
minus int
119905
0
(119905 minus 119904)120572minus1
times 120601119902 (int
119904
0
120596 (120591) 119891 (120591 119909119903 (120591)) 119889120591) 119889119904
+120578120601119902 (120582119903)
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
119892 (119904)
times (int
1
0
(1 minus 120591)120572minus1
times 120601119902 (int
120591
0
120596 (120583) 119891 (120583 119909119903 (120583)) 119889120583)119889120591
minus int
119904
0
(119904 minus 120591)120572minus1
times 120601119902 (int
120591
0
120596 (120583) 119891 (120583 119909119903 (120583)) 119889120583)119889120591)119889119904
le120578120601119902 (120582119903)
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
119892 (119904) int
1
0
(1 minus 120591)120572minus1
times 120601119902 (int
120591
0
120596 (120583) 119891 (120583 119909119903 (120583)) 119889120583)119889120591 119889119904
+120601119902 (120582119903)
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902 (int
119904
0
120596 (120591) 119891 (120591 119909119903 (120591)) 119889120591) 119889119904
=120601119902 (120582119903) 120578 int
1
0119892 (119904) 119889119904
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
(1 minus 119904)120572minus1
times 120601119902 (int
119904
0
120596 (120591) 119891 (120591 119909119903 (120591)) 119889120591) 119889119904
+120601119902 (120582119903)
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902 (int
119904
0
120596 (120591) 119891 (120591 119909119903 (120591)) 119889120591) 119889119904
le120601119902 (120582119903) 120578 int
1
0119892 (119904) 119889119904
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
(1 minus 119904)120572minus1
times 120601119902 (1198972 int
119904
0
120596 (120591) 119909119903 (120591) 119889120591) 119889119904
+120601119902 (120582119903)
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902 (1198972120582119903 int
119904
0
120596 (120591) 119909119903 (120591) 119889120591) 119889119904
=120601119902 (120582119903)
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
(1 minus 119904)120572minus1
times 120601119902 (1198972 int
119904
0
120596 (120591) 119909119903 (120591) 119889120591) 119889119904
Abstract and Applied Analysis 9
le
10038171003817100381710038171199091199031003817100381710038171003817 120601119902 (1205821199031198972)
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
(1 minus 119904)120572minus1
times 120601119902 (int
1
0
120596 (120591) 119889120591) 119889119904
=
10038171003817100381710038171199091199031003817100381710038171003817 120601119902 (1205821199031198972) 1205742
Γ (120572 + 1) (1 minus 120578 int1
0119892 (119904) 119889119904)
(51)
which implies that
10038171003817100381710038171199091199031003817100381710038171003817 le
10038171003817100381710038171199091199031003817100381710038171003817 120601119902 (1205821199031198972) 1205742
Γ (120572 + 1) (1 minus 120578 int1
0119892 (119904) 119889119904)
(52)
that is
120601119902 (1205821199031198972) 1205742
Γ (120572 + 1) (1 minus 120578 int1
0119892 (119904) 119889119904)
ge 1 (53)
it follows that we get
120582119903 ge1
1198972(
Γ (120572 + 1) (1 minus 120578 int1
0119892 (119904) 119889119904)
1205742)
119901minus1
= 1205820(54)
In conclusion 120582119903 isin [1205820 1205820](ii) It follows from 0 lt 119891
infin
120601lt infin that there exist
119897lowast
1 119897lowast
2 120588lowastgt 0 such that
119897lowast
1120601119901 (119909) le 119891 (119905 119909) le 119897
lowast
2120601119901 (119909)
(forall119905 isin [0 1] 119909 ge 120588lowast)
(55)
Let 120585lowast0= 120588lowast120590 the following proof are similar to (i)
(iii) It follows from 1198910
120601= +infin there exists 119897lowast gt 0 120588
lowastgt 0
such that
1198910
120601ge 119897lowast120601119901 (119909) (forall119905 isin [0 1] 0 le 119909 le 120588
lowast) (56)
Let 1205851 = 120588lowast120590 we show that 1205851 is required When 119909 isin 119875 cap
120597Ω119903lowast we have 119909(119905) ge 120590119909 = 120590119903lowast since 119909 le 120588
lowast then we need120590119903lowastlt 120588lowast as 119903lowast lt 1205851 so 120590119903
lowastlt 120588lowast holds
Let 120582lowast = (1119897lowast)(Γ(120572)1205901205741)
1(119902minus1) we proceed in the sameway as in the proof of (i) replacing (42) we may assume that119909 minus 120601119902(120582
lowast)119879119909 = 0 (forall119909 isin 119875 cap 120597Ω119903lowast) and replacing (43) we
can prove 119909 minus 120601119902(120582lowast)119879119909 = 120581120595 (forall119909 isin 119875 cap 120597Ω119903lowast 120581 ge 0) It
follows from Lemma 5 that 119894(120601119902(120582lowast)119879 119875 cap Ω119903lowast 119875) = 0 Note
that 119894(120579 119875 cap Ω119903lowast 119875) = 1 we can easily show that there exists119909119903lowast isin 119875cap120597Ω119903lowast and 0 lt ]119903lowast lt 1 such that ]119903lowast120601119902(120582
lowast)119879119909119903lowast = 119909119903lowast
Hence (37) holds for 120582119903lowast = 120582lowast]1minus119901119903lowast
lt 120582lowast
The proof ofTheorem 9(iv) follows by themethod similarto Theorem 9(iii) we omit it here
(v) It follows that 119898(119903) gt 120585 for any 119909 isin 119875 cap 120597Ω119903 we have120590119903 le 119909 le 119903 and 119891(119905 119909)120601119901(119909) ge 119891(119905 119909)120601119901(119903) ge 119898(119903) ge 120585then
119891 (119905 119909) ge 120585120601119901 (119909) forall119909 isin [120590119903 119903] 119905 isin [0 1] (57)
Let = (1120585)(Γ(120572)1205901205741)119901minus1 The following proof is similar to
that of (iv) This finished the proof of (v)
Remark 10 InTheorem 9 all the criteria obtained depend onone of 1198910
120601and 119891
infin
120601
Let Ψ(119909) = 120601119902(120582)119879119909 the following theorems give out themultiply nonexistence and the dependence of parameter
Theorem 11 Assume that (H1)ndash(H5) hold
(i) If 1198910
120601= 0 and 119891
infin
120601= 0 lim
119909rarrinfin119891(119905 119909) = 119891infin isin
(0 +infin] uniformly for 119905 isin [0 1] then there exists120582lowast
gt 0 such that the problem (1) has two solutions forany 120582 gt 120582
lowast
(ii) If 1198910120601
= 0 and 119891infin
120601= 0 then there exists 120582 gt 0 such
that for any 120582 lt 120582 problem (1) has no solution
Proof (i) It follows from 119891infin isin (0 +infin] that for any 120577 gt 0there exists 120588 gt 0 such that
119891 (119905 119909) gt 120577 (forall119905 isin [0 1] 119909 ge 120588) (58)
Since 1198910
120601= 0 there exists 1205761 gt 0 120588
1gt 0 such that
119891 (119905 119909) le 1205761120601119901 (119909) (forall119905 isin [0 1] 0 le 119909 le 1205881) (59)
where 1205761 satisfied 120601119902(1205821205761)1205742Γ(120572 + 1)(1 minus 120578 int1
0119892(119904)119889119904) le 1
For forall119909 isin 119875 cap 120597Ω1205881
we have
Ψ (119909) = (120601119902 (120582) 119879119909) (119905)
=1
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902 (120582int
119904
0
120596 (120591) 119891 (120591 119909 (120591)) 119889120591) 119889119904
minus int
119905
0
(119905 minus 119904)120572minus1
times 120601119902 (120582int
119904
0
120596 (120591) 119891 (120591 119909 (120591)) 119889120591) 119889119904
+120578
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
119892 (119904)
10 Abstract and Applied Analysis
times (int
1
0
(1 minus 120591)120572minus1
times 120601119902 (120582int
120591
0
120596 (120583) 119891 (120583 119909 (120583)) 119889120583)119889120591
minus int
119904
0
(119904 minus 120591)120572minus1
times 120601119902 (120582int
120591
0
120596 (120583) 119891 (120583 119909 (120583)) 119889120583)119889120591)119889119904
le120578
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
119892 (119904)
times (int
1
0
(1 minus 120591)120572minus1
times 120601119902(120582int
120591
0
120596 (120583) 119891 (120583 119909 (120583)) 119889120583)119889120591)119889119904
+1
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902 (120582int
119904
0
120596 (120591) 1205761119891 (120591 119909 (120591)) 119889120591) 119889119904
le120578 int1
0119892 (119904) 119889119904
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
(1 minus 120591)120572minus1
times 120601119902 (120582int
1
0
120596 (120583) 1205761120601119901 (119909 (120583)) 119889120583)119889120591
+1
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902(120582int
1
0
120596 (120591) 1205761120601119901 (119909 (120591)) 119889120591) 119889119904
le1
Γ (120572 + 1) (1 minus 120578 int1
0119892 (119904) 119889119904)
times 120601119902 (120582int
1
0
120596 (120591) 1205761120601119901 (119909) 119889120591)
=120601119902 (1205821205761) 1205742
Γ (120572 + 1) (1 minus 120578 int1
0119892 (119904) 119889119904)
119909 le 119909
(60)
This implied
Ψ119909 le 119909 forall119909 isin 119875 cap 120597Ω1205881
(61)
Let 1205883gt max120588120590 120588
1 and 120582
lowast= (1120577)(120588
3Γ(120572)1205901205741)
119901minus1 thenforall119909 isin 119875 cap 120597Ω120588
3
we have 120590119909 le 119909 le 1205883and
(Ψ119909) (119905)
=1
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902 (120582int
119904
0
120596 (120591) 119891 (120591 119909 (120591)) 119889120591) 119889119904
minus int
119905
0
(119905 minus 119904)120572minus1
times 120601119902 (120582int
119904
0
120596 (120591) 119891 (120591 119909 (120591)) 119889120591) 119889119904
+120578
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
119892 (119904)
times (int
1
0
(1 minus 120591)120572minus1
times 120601119902 (120582int
120591
0
120596 (120583) 119891 (120583 119909 (120583)) 119889120583)119889120591
minus int
119904
0
(119904 minus 120591)120572minus1
times 120601119902(120582int
120591
0
120596 (120583) 119891 (120583 119909 (120583)) 119889120583)119889120591)119889119904
ge120590120578
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
119892 (119904)
times (int
1
0
(1 minus 120591)120572minus1
times 120601119902 (120582int
120591
0
120596 (120583) 119891 (120583 119909 (120583)) 119889120583)119889120591
minus int
119904
0
(119904 minus 120591)120572minus1
times 120601119902(120582int
120591
0
120596 (120583) 119891 (120583 119909 (120583)) 119889120583)119889120591)119889119904
+120590
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902 (120582int
119904
0
120596 (120591) 119891 (120591 119909 (120591)) 119889120591) 119889119904
ge120590
Γ (120572)int
1
0
(1 minus 119904)120572minus1
Abstract and Applied Analysis 11
times 120601119902 (120582int
119904
0
120596 (120591) 119891 (120591 119909 (120591)) 119889120591) 119889119904
gt120590
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902 (120582int
119904
0
120596 (120591) 120577 119889120591) 119889119904
gt120590120601119902 (120582120577)
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902 (int
119904
0
120596 (120591) 119889120591) 119889119904
=120590120601119902 (120582120577) 1205741
Γ (120572)= 1205883
(62)
which implies
Ψ119909 gt 119909 forall119909 isin 119875 cap 120597Ω1205883
(63)
Next for 119891infin
120601= 0 there exist 1205762 gt 0 120588
2gt 1205883gt 0 such
that
119891 (119905 119909) le 1205762120601119901 (119909) (forall119905 isin [0 1] 119909 ge 1205882) (64)
where 1205762 satisfied 120601119902(1205821205762)1205742Γ(120572 + 1)(1 minus 120578 int1
0119892(119904)119889119904) le 1
Similar to the above proof we get
Ψ119909 le 119909 forall119909 isin 119875 cap 120597Ω1205882
(65)
Applying Lemma 7 to (61) (65) and (63) yields that Ψ
has two fixed points 1199091 1199092 such that 1199091 isin 119875cap (Ω1205883
Ω1205881
) and1199092 isin 119875 cap (Ω120588
2
Ω1205883
)(ii) It follows from 119891
0
120601= 0 and 119891
infin
120601= 0 that there exists
1205761 1205762 1205881 1205882 gt 0 such that
119891 (119905 119909) le 1205761120601119901 (119909) (forall119905 isin [0 1] 0 le 119909 le 1205881)
119891 (119905 119909) le 1205762120601119901 (119909) (forall119905 isin [0 1] 119909 ge 1205882)
(66)
then for 119909 isin [1205881 1205882] by the continuous of 119891(119905 119909(119905)) there
exists 119909 isin [1205881 1205882] such that
max1205881le119909le120588
2
119891 (119905 119909)
120601119901 (119909)=
119891 (119905 119909)
120601119901 (119909) (67)
Let Υ = max1205761 1205762 119891(119905 119909)120601119901(119909) gt 0 then
119891 (119905 119909) le Υ120601119901 (119909) (forall0 le 119905 le 1 119909 ge 0) (68)
Assuming 119909(119905) is a positive solution of problem (1) we willshow that this leads to a contradiction for 120582 lt 120582 where
120582 = (1Υ)(Γ(120572 + 1)(1 minus 120578 int1
0119892(119904)119889119904)1205742)
119901minus1 It follows from(26) that
119909 = 119909 (0)
=120578
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
119892 (119904)
times (int
1
0
(1 minus 120591)120572minus1
times 120601119902 (120582int
120591
0
120596 (120583) 119891 (120583 119909 (120583)) 119889120583)119889120591
minus int
119904
0
(119904 minus 120591)120572minus1
times 120601119902(120582int
120591
0
120596 (120583) 119891 (120583 119909 (120583)) 119889120583)119889120591)119889119904
+1
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902 (120582int
119904
0
120596 (120591) 119891 (120591 119909 (120591)) 119889120591) 119889119904
le120578
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
119892 (119904)
times int
1
0
(1 minus 120591)120572minus1
times 120601119902(120582int
120591
0
120596 (120583) 119891 (120583 119909 (120583)) 119889120583)119889120591 119889119904
+1
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902 (120582int
119904
0
120596 (120591) 119891 (120591 119909 (120591)) 119889120591) 119889119904
=1
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
(1 minus 119904)120572minus1
times 120601119902 (120582int
119904
0
120596 (120591) 119891 (120591 119909 (120591)) 119889120591) 119889119904
le1
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
(1 minus 119904)120572minus1
12 Abstract and Applied Analysis
times 120601119902 (120582int
1
0
120596 (120591) Υ120601119901 (119909 (120591)) 119889120591) 119889119904
=120601119902 (120582Υ) 1205742
Γ (120572 + 1) (1 minus 120578 int1
0119892 (119904) 119889119904)
119909
(69)
This implies that 1 le 120601119902(120582Υ)1205742Γ(120572 + 1)(1 minus 120578 int1
0119892(119904)119889119904) =
(120582Υ)119902minus1
1205742Γ(120572+1)(1minus120578 int1
0119892(119904)119889119904) lt (120582 Υ)
119902minus11205742Γ(120572+1)(1minus
120578 int1
0119892(119904)119889119904) = 1 which is a contradiction This finishes the
proof of (ii)
Remark 12 In (i) the condition lim119909rarrinfin
119891(119905 119909) = 119891infin isin
(0 +infin] uniformly for 119905 isin [0 1] can be replaced withcondition (H6) The same methods can be used to prove theresults
Remark 13 From the above proof the condition 119891infin isin
(0 +infin] is important to keep the problem (1) having at leasttwo positive solutions if this condition is not satisfied thenthere exists 120582 gt 0 small enough such that the problem (1) hasno positive solutions for all 120582 lt 120582
Theorem 14 Assume that (H1)ndash(H6) hold(i) If 1198910
120601= +infin and 119891
infin
120601= +infin then there exists 120582lowast0 gt 0
such that problem (1) has at least two solutions for 120582 isin
(0 120582lowast0)(ii) If 1198910
120601= +infin and 119891
infin
120601= +infin then there exists 120582 gt 0
such that problem (1) has no solution for all 120582 gt 120582
Proof (i) It follows from 1198910
120601= +infin that there exists 1199031 gt 0
such that
119891 (119905 119909) ge 119871lowast1120601119901 (119909) (forall119905 isin [0 1] 0 le 119909 le 1199031) (70)
where 119871lowast1 satisfies 1205902120601119902(120582119871lowast1)1205741Γ(120572) ge 1
Then for 119909 isin 119875cap120597Ω1199031
we have 120590119909 le 119909 le 1199031 for 119905 isin [0 1]
and
(Ψ119909) (119905)
=1
Γ (120572)
times int
1
0
(1 minus 119904 )120572minus1
times 120601119902 (120582int
119904
0
120596 (120591) 119891 (120591 119909 (120591)) 119889120591) 119889119904
minus int
119905
0
(119905 minus 119904)120572minus1
times 120601119902 (120582int
119904
0
120596 (120591) 119891 (120591 119909 (120591)) 119889120591) 119889119904
+120578
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
119892 (119904)
times (int
1
0
(1 minus 120591)120572minus1
times 120601119902 (120582int
120591
0
120596 (120583) 119891 (120583 119909 (120583)) 119889120583)119889120591
minus int
119904
0
(119904 minus 120591)120572minus1
times 120601119902 (120582int
120591
0
120596 (120583) 119891 (120583 119909 (120583)) 119889120583)119889120591)119889119904
ge120590120578
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
119892 (119904)
times (int
1
0
(1 minus 120591)120572minus1
times 120601119902 (120582int
120591
0
120596 (120583) 119891 (120583 119909 (120583)) 119889120583)119889120591
minus int
119904
0
(119904 minus 120591)120572minus1
times120601119902 (120582int
120591
0
120596 (120583) 119891 (120583 119909 (120583)) 119889120583)119889120591)119889119904
+120590
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902 (120582int
119904
0
120596 (120591) 119891 (120591 119909 (120591)) 119889120591) 119889119904
ge120590
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902 (120582int
119904
0
120596 (120591) 119891 (120591 119909 (120591)) 119889120591) 119889119904
ge120590
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902 (120582int
119904
0
120596 (120591) 119871lowast1120601119901 (119909 (120591)) 119889120591) 119889119904
ge1205902120601119902 (120582119871lowast1) 119909
Γ (120572)
times int
1
0
(1 minus 119904)120572minus1
times 120601119902 (int
119904
0
120596 (120591) 119889120591) 119889119904
=1205902120601119902 (120582119871lowast1) 1205742
Γ (120572)119909
(71)
Abstract and Applied Analysis 13
which implies that
Ψ119909 ge 119909 forall119909 isin 119875 cap 120597Ω1199031
(72)
Next from 119891infin
120601= +infin there exists 1199032 gt 0 such that
119891 (119905 119909) ge 119871lowast2120601119901 (119909) forall119909 ge 1199032 (73)
where 119871lowast2 satisfies
1205902120601119902 (120582119871lowast2) 1205742
Γ (120572)ge 1 (74)
Let 1199032 gt max1199031 1199032120590 then for 119909 isin 119875 cap 120597Ω1199032
we have
119909 (119905) ge 120590 119909 = 1205901199032 gt 1199032 (75)
Similar to the above proof we can get
Ψ119909 ge 119909 forall119909 isin 119875 cap 120597Ω1199032
(76)
Let 1199031 lt 1199033 lt 11990321198721199033
= max119891(119905 119909) | 119905 isin [0 1] 119909 le 1199033+1 gt
0 and 120582lowast0 = (11198721199033
)(Γ(120572+1)1199033(1minus120578 int1
0119892(119904)119889119904)1205742)
119901minus1Thenfor 119909 isin 119875 cap 120597Ω119903
3
we have
119891 (119905 119909) le 1198721199033
forall119905 isin [0 1] (77)
Hence
(Ψ119909) (119905)
=1
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902 (120582int
119904
0
120596 (120591) 119891 (120591 119909 (120591)) 119889120591) 119889119904
minus int
119905
0
(119905 minus 119904)120572minus1
times 120601119902 (120582int
119904
0
120596 (120591) 119891 (120591 119909 (120591)) 119889120591) 119889119904
+120578
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
119892 (119904)
times (int
1
0
(1 minus 120591)120572minus1
times 120601119902 (120582int
120591
0
120596 (120583) 119891 (120583 119909 (120583)) 119889120583)119889120591
minus int
119904
0
(119904 minus 120591)120572minus1
times120601119902 (120582int
120591
0
120596 (120583) 119891 (120583 119909 (120583)) 119889120583)119889120591)119889119904
le120578
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
119892 (119904)
times int
1
0
(1 minus 120591)120572minus1
120601119902
times (120582int
120591
0
120596 (120583) 119891 (120583 119909 (120583)) 119889120583)119889120591 119889119904
+1
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902 (120582int
119904
0
120596 (120591) 119891 (120591 119909 (120591)) 119889120591) 119889119904
le120578
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
119892 (119904)
times int
1
0
(1 minus 120591)120572minus1
120601119902
times (120582int
120591
0
120596 (120583)1198721199033
119889120583)119889120591119889119904
+1
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902 (120582int
119904
0
120596 (120591)1198721199033
119889120591)119889119904
le120578120601119902 (120582119872119903
3
) int1
0119892 (119904) 119889119904
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
(1 minus 120591)120572minus1
times 120601119902 (int
1
0
120596 (120583) 119889120583)119889120591
+120601119902 (120582119872119903
3
)
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902 (int
1
0
120596 (120591) 119889120591) 119889119904
=1205742120601119902 (120582119872119903
3
)
Γ (120572 + 1) (1 minus 120578 int1
0119892 (119904) 119889119904)
lt1205742120601119902 (120582lowast0119872119903
3
)
Γ (120572 + 1) (1 minus 120578 int1
0119892 (119904) 119889119904)
= 1199033
(78)
14 Abstract and Applied Analysis
This implies that
Ψ119909 lt 119909 forall119909 isin 119875 cap 120597Ω1199033
(79)
Applying Lemma 7 to (72) (76) and (79) yields that Ψ hastwo fixed points 1199091 1199092 such that 1199091 isin 119875 cap (Ω119903
3
Ω1199031
) and1199092 isin 119875 cap (Ω119903
2
Ω1199033
)(ii) By the proof of (i) and the continuity of 119891(119905 119909) there
exists 119909 isin [1199031 1199032] such that min119909isin[11990311199032](119891(119905 119909)120601119901(119909)) =
119891(119905 119909)120601119901(119909)Let Υ = min119871lowast1 119871lowast2 119891(119905 119909)120601119901(119909) since (H6) holds
then Υ gt 0 and
119891 (119905 119909) ge Υ120601119901 (119909) forall119909 ge 0 119905 isin [0 1] (80)
Assuming 119909(119905) is a positive solution of problem (1) we willshow that this leads to contradiction for all 120582 gt 120582 where 120582 =
(1Υ)(Γ(120572)1205741120590)119901minus1 It follows form (26) that
119909 = 119909 (0)
=120578
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
119892 (119904)
times (int
1
0
(1 minus 120591)120572minus1
times 120601119902 (120582int
120591
0
120596 (120583) 119891 (120583 119909 (120583)) 119889120583)119889120591
minus int
119904
0
(119904 minus 120591)120572minus1
times120601119902 (120582int
120591
0
120596 (120583) 119891 (120583 119909 (120583)) 119889120583)119889120591)119889119904
+1
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902 (120582int
119904
0
120596 (120591) 119891 (120591 119909 (120591)) 119889120591) 119889119904
ge1
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902 (120582int
119904
0
120596 (120591) 119891 (120591 119909 (120591)) 119889120591) 119889119904
ge1
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902 (120582int
119904
0
120596 (120591) Υ120601119901 (119909 (120591)) 119889120591) 119889119904
ge1
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902 (120582int
119904
0
120596 (120591) Υ120601119901 (120590 119909 119889120591)) 119889119904
ge120601119902 (120582Υ) 120590 119909
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902 (int
119904
0
120596 (120591) 119889120591) 119889119904
=120601119902 (120582Υ) 1205741120590
Γ (120572)119909
(81)
This implies that 1 ge 120601119902(120582Υ)1205741120590Γ(120572) = (120582Υ)119902minus1
1205741120590Γ(120572) gt
(120582Υ)119902minus1
1205741120590Γ(120572) = 1 which is a contradiction This com-pletes the proof of (ii)
Corollary 15 Assume that (H1)ndash(H5) holds
(i) If lim119909rarrinfin
119891(119905 119909) = 119891infin isin (0 +infin] uniformly for 119905 isin
[0 1] and one of 1198910120601
= 0 and 119891infin
120601= 0 is satisfied then
there exists 120582lowast gt 0 such that the problem (1) has at leastone positive solution for any 120582 gt 120582
lowast(ii) If one of 1198910
120601= +infin and 119891
infin
120601= +infin holds then there
exists 120582lowast0 gt 0 such that problem (1) has at least onepositive solutions for any 120582 isin (0 120582lowast0)
Proof (i) The conclusion is a direct consequence of Theo-rem 11(i)
(ii) The conclusion is a direct consequence of Theo-rem 14(i)
Theorem 16 Assume that (H1)ndash(H5) holdThen the followingconclusions hold
(i) If 1198910120601
= 0 and 119891infin
120601= infin then the problem (1) has a
positive solution 119909120582(119905) for all 120582 gt 0(ii) If 1198910
120601= infin and 119891
infin
120601= 0 then the problem (1) has a
positive solution 119909120582(119905) for all 120582 gt 0
Proof Weonly prove (i) the proof of (ii) is similar sowe omitit here
Let 120582 gt 0 since 1198910
120601= 0 there exists 119903 gt 0 119897 gt 0 such that
119891 (119905 119909) le 119897120601119901 (119909) (forall119905 isin [0 1] 119909 isin [0 119903]) (82)
where 119897 satisfied
1205742120582119902minus1
119897119902minus1
Γ (120572 + 1) (1 minus 120578 int1
0119892 (119904) 119889119904)
le 1 (83)
Similar to the proof of Theorem 14 we obtain that
Ψ119909 le 119909 forall119909 isin 119875 cap 120597Ω119903 (84)
Abstract and Applied Analysis 15
From 119891infin
120601= infin there exists 119877 gt 119903 gt 0 119871 gt 0 such that
119891 (119905 119909) ge 119871120601119901 (119909) (forall119905 isin [0 1] 119909 ge 119877) (85)where 119871 satisfied
1205902120582119902minus1
119871119902minus1
1205741
Γ (120572)ge 1 (86)
We get thatΨ119909 ge 119909 forall119909 isin 119875 cap 120597Ω119877 (87)
Applying Lemma 7 to (84) (87) yields thatΨ has a fixed point119909 such that 119909 isin 119875 cap (Ω119877 Ω119903)
Theorem 17 Assume that (H1)ndash(H6) holdThen the followingconclusions hold
If 0 lt 1198910
120601lt infin and 0 lt 119891
infin
120601lt infin then there exists 1 2 gt
0 such that problem (1) has no positive solution for any 120582 gt 1
and 0 lt 120582 lt 2
Proof It follows from 0 lt 1198910
120601lt infin and 0 lt 119891
infin
120601lt infin that
there exist 1 gt 0 2 gt 0 1 gt 0 2 gt 0 and 1199032 gt 1199031 gt 0
such that1120601119901 (119909) le 119891 (119905 119909) le 1120601119901 (119909)
(forall119905 isin [0 1] 0 le 119909 le 1199031)
2120601119901 (119909) le 119891 (119905 119909) le 2120601119901 (119909)
(forall119905 isin [0 1] 119909 ge 1199032)
(88)
Let = max119891(119905 119909)120601119901(119909) 1199031 le 119909 le 1199032 0 le 119905 le 1 =
min119891(119905 119909)120601119901(119909) 1199031 le 119909 le 1199032 0 le 119905 le 1 By the condition(H6) gt 0 Then forall119909 ge 0 we have
120601119901 (119909) le 119891 (119905 119909) le 120601119901 (119909) (89)
where = max1 2 gt 0 and = min1198971 2 gt 0Let 1 = (1)(Γ(120572)1205741120590)
119901minus1 and 2 =
(1)(Γ(120572 + 1)(1 minus 120578 int1
0119892(119904)119889119904)1205742)
119901minus1
If problem (1)has positive solution 119909(119905) in 119875 then
119909 = 119909 (0)
=120578
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
119892 (119904)
times (int
1
0
(1 minus 120591)120572minus1
times 120601119902 (120582int
120591
0
120596 (120583) 119891 (120583 119909 (120583)) 119889120583)119889120591
minus int
119904
0
(119904 minus 120591)120572minus1
times120601119902 (120582int
120591
0
120596 (120583) 119891 (120583 119909 (120583)) 119889120583)119889120591)119889119904
+1
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902 (120582int
119904
0
120596 (120591) 119891 (120591 119909 (120591)) 119889120591) 119889119904
ge1
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902 (120582int
119904
0
120596 (120591) 119891 (120591 119909 (120591)) 119889120591) 119889119904
ge1
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902 (120582int
119904
0
120596 (120591) 120601119901 (119909 (120591)) 119889120591) 119889119904
ge1
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902 (120582int
119904
0
120596 (120591) 120601119901 (120590 119909 119889120591)) 119889119904
ge120601119902 (120582) 120590 119909
Γ (120572)
times int
1
0
(1 minus 119904)120572minus1
120601119902 (int
119904
0
120596 (120591) 119889120591) 119889119904
=120601119902 (120582) 1205741120590
Γ (120572)119909
gt120601119902 (1 ) 1205741120590
Γ (120572)119909 = 119909
119909 = 119909 (0)
=120578
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
119892 (119904)
times (int
1
0
(1 minus 120591)120572minus1
120601119902
times (120582int
120591
0
120596 (120583) 119891 (120583 119909 (120583)) 119889120583)119889120591
minus int
119904
0
(119904 minus 120591)120572minus1
120601119902
times(120582int
120591
0
120596 (120583) 119891 (120583 119909 (120583)) 119889120583)119889120591)119889119904
+1
Γ (120572)int
1
0
(1 minus 119904)120572minus1
120601119902
times (120582int
119904
0
120596 (120591) 119891 (120591 119909 (120591)) 119889120591) 119889119904
16 Abstract and Applied Analysis
le120578
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
119892 (119904)
times int
1
0
(1 minus 120591)120572minus1
times 120601119902 (120582int
120591
0
120596 (120583) 119891 (120583 119909 (120583)) 119889120583)119889120591 119889119904
+1
Γ (120572)int
1
0
(1 minus 119904)120572minus1
120601119902
times (120582int
119904
0
120596 (120591) 119891 (120591 119909 (120591)) 119889120591) 119889119904
=1
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
(1 minus 119904)120572minus1
120601119902
times (120582int
119904
0
120596 (120591) 119891 (120591 119909 (120591)) 119889120591) 119889119904
le1
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
(1 minus 119904)120572minus1
120601119902
times (120582int
1
0
120596 (120591) 120601119901 (119909 (120591)) 119889120591) 119889119904
=120601119902 (120582) 1205742
Γ (120572 + 1) (1 minus 120578 int1
0119892 (119904) 119889119904)
119909
lt120601119902 (2) 1205742
Γ (120572 + 1) (1 minus 120578 int1
0119892 (119904) 119889119904)
119909 = 119909
(90)
which is a contradiction This finishes the proof of Theo-rem 17
4 Examples
In this section we give out same examples to illustrate ourmain results can be used in practice
Example 1 Consider the following fractional differentialequation
(120601119901 (11986352
0+119909 (119905)))
1015840
+ 120582119905(1 + 1199052)12
119909119901minus1 arctan119909 = 0
0 lt 119905 lt 1
1199091015840(0) = 119909
10158401015840(0) = 0
119909 (1) =1
2int
1
0
1
2119909 (119904) 119889119904
(91)
where 120572 = 52 119901 = 32 119902 = 3 120596(119905) = 119905 119891(119905 119909) = (1 +
1199052)12
119909119901minus1 arctan119909 120578 = 12 and 119892(119905) = 12
By simple computation 119897lowast1= 1205872 119897
lowast
2= radic21205872 1205741 = int
1
0(1minus
119904)32
1206013(int119904
0120591 119889120591)119889119904 asymp 00043 1205742 = 1206013(int
1
0120591 119889120591) = 025 120590 asymp
01429 120582lowast
0= (1119897
lowast
1)(Γ(120572)1205901205741)
119901minus1asymp 25149 120582
lowast
0= (1119897
lowast
2)(Γ(120572+
1)(1 minus 120578 int1
0119892(119904)119889119904)1205742)
119901minus1asymp 14222 Then it is easy to see
all the conditions of Theorem 9(ii) are satisfied thus forall120582 isin
[14222 25149]] the problem (91) has at least one positivesolution
Example 2 Consider the boundary value problem
(120601119901 (11986352
0+119909 (119905)))
1015840
+ 120582119905(1 + 1199052)12
119909119901119890minus119909
= 0
0 lt 119905 lt 1
1199091015840(0) = 119909
10158401015840(0) = 0
119909 (1) =1
2int
1
0
1
2119909 (119904) 119889119904
(92)
where 120572 = 52 119901 = 32 119902 = 3 120596(119905) = 119905 119891(119905 119909) = (1 +
1199052)12
119909119901119890minus119909
120578 = 12 and 119892(119905) = 12
1198910
120601= lim119909rarr0
119891 (119905 119909)
120601119901 (119909)= lim119909rarr0
(1 + 1199052)12
119909119901119890minus119909
119909119901minus1= 0
forall119905 isin [0 1]
1198910
120601= lim119909rarrinfin
119891 (119905 119909)
120601119901 (119909)= lim119909rarrinfin
(1 + 1199052)12
119909119901119890minus119909
119909119901minus1= 0
forall119905 isin [0 1]
(93)
Let 1205881= 12 120588
2= 4 then
119891 (119905 119909)
120601119901 (119909)le (1 + 119905
2)12 1
2radic119890le
1
radic2119890
forall119905 isin [0 1] 119909 isin [01
2]
119891 (119905 119909)
120601119901 (119909)le (1 + 119905
2)12 4
1198904le
4radic2
1198904
forall119905 isin [0 1] 119909 isin [4infin)
(94)
so 1205761 = 1radic2119890 1205762 = 4radic21198904 and forall119909 isin
[12 4] max12le119909le4(119891(119905 119909)120601119901(119909)) = radic2119890minus1 We
have Υ = max1radic2119890 4radic21198904 radic2119890minus1
= radic2119890minus1 1205742 = 025
Abstract and Applied Analysis 17
and 120582 = (1Υ)(Γ(120572+1)(1minus120578 int1
0119892(119904)119889119904)1205742)
119901minus1= 3radic5120587
141198904
By Theorem 11(ii) for any 120582 lt 120582 = 3radic512058714
1198904 the problemhas no positive solution
5 Conclusions
Recently differential equations with p-Laplacian operatorwere widely discussed by several authors In this paperby using fixed point techniques combining with partiallyordered structure of Banach space we obtained the existencemultiply and the dependence of parameter These newresults we presented can be used in numerical computationand analyze mathematical models of physical phenomenamechanics nonlinear dynamics and many other relatedfields
Conflict of Interests
The authors declare that there is no conflict of interestsregarding the publication of this paper
Acknowledgments
The authors are very grateful to anonymous referees for theirvaluable and detailed suggestions and comments to improvethe original paper This work is supported by the Programfor New Century Excellent Talents in University (NCET-10-0097)
References
[1] I Podlubny Fractional Differential Equations vol 198 Aca-demic Press San Diego Calif USA 1999
[2] A A Kilbas H M Srivastava and J J Trujillo Theoryand Applications of Fractional Differential Equations vol 204of North-Holland Mathematics Studies Elsevier Science BVAmsterdam The Netherlands 2006
[3] Y A Rossikhin and M V Shitikova ldquoApplication of fractionalderivatives to the analysis of damped vibrations of viscoelasticsingle mass systemsrdquoActa Mechanica vol 120 no 1ndash4 pp 109ndash125 1997
[4] J Sabatier O Agrawal and J T Machado Advances in Frac-tional Calculus Theoretical developments and applications inphysics and engineering Springer Berlin Germany 2007
[5] N Engheta ldquoOn fractional calculus and fractional multipolesin electromagnetismrdquo IEEE Transactions on Antennas andPropagation vol 44 no 4 pp 554ndash566 1996
[6] R Hilfer Applications of Fractional Calculus in Physics WorldScientific Singapore 2000
[7] R L Bagley and R A Calico ldquoFractional order state equationsfor the control of viscoelastically damped structuresrdquo Journalof Guidance Control and Dynamics vol 14 no 2 pp 304ndash3111991
[8] W-X Zhou andY-D Chu ldquoExistence of solutions for fractionaldifferential equations with multi-point boundary conditionsrdquoCommunications in Nonlinear Science and Numerical Simula-tion vol 17 no 3 pp 1142ndash1148 2012
[9] B Ahmad and G Wang ldquoA study of an impulsive four-pointnonlocal boundary value problemof nonlinear fractional differ-ential equationsrdquo Computers amp Mathematics with Applicationsvol 62 no 3 pp 1341ndash1349 2011
[10] B Ahmad and S Sivasundaram ldquoOn four-point nonlocalboundary value problems of nonlinear integro-differentialequations of fractional orderrdquo Applied Mathematics and Com-putation vol 217 no 2 pp 480ndash487 2010
[11] N Nyamoradi ldquoPositive solutions for multi-point boundaryvalue problems for nonlinear fractional differential equationsrdquoJournal of Contemporary Mathematical Analysis vol 48 no 4pp 145ndash157 2013
[12] B Ahmad and J J Nieto ldquoExistence of solutions for nonlocalboundary value problems of higher-order nonlinear fractionaldifferential equationsrdquo Abstract and Applied Analysis vol 2009Article ID 494720 9 pages 2009
[13] X Liu M Jia and X Xiang ldquoOn the solvability of a fractionaldifferential equationmodel involving the119901-Laplacian operatorrdquoComputers ampMathematics with Applications vol 64 no 10 pp3267ndash3275 2012
[14] J Wang H Xiang and Z Liu ldquoExistence of concave positivesolutions for boundary value problem of nonlinear fractionaldifferential equation with 119901-Laplacian operatorrdquo InternationalJournal of Mathematics and Mathematical Sciences vol 2010Article ID 495138 17 pages 2010
[15] J Wang and H Xiang ldquoUpper and lower solutions method fora class of singular fractional boundary value problems with 119901-Laplacian operatorrdquo Abstract and Applied Analysis vol 2010Article ID 971824 12 pages 2010
[16] H Lu andZHan ldquoExistence of positive solutions for boundary-value problem of fractional differential equation with 119901-laplacian operatorrdquo The American Jouranl of Engineering andTechnology Research vol 11 pp 3757ndash3764 2011
[17] G Chai ldquoPositive solutions for boundary value problem offractional differential equation with 119901-Laplacian operatorrdquoBoundary Value Problems vol 2012 article 18 2012
[18] T Chen W Liu and Z Hu ldquoA boundary value problem forfractional differential equation with 119901-Laplacian operator atresonancerdquoNonlinear AnalysisTheoryMethodsampApplicationsvol 75 no 6 pp 3210ndash3217 2012
[19] T Chen andW Liu ldquoAn anti-periodic boundary value problemfor the fractional differential equation with a 119901-LaplacianoperatorrdquoAppliedMathematics Letters of Rapid Publication vol25 no 11 pp 1671ndash1675 2012
[20] S J Li X G Zhang Y H Wu and L Caccetta ldquoExtremalsolutions for 119901-Laplacian differential systems via iterative com-putationrdquo Applied Mathematics and Computation vol 26 pp1151ndash1158 2013
[21] Z Liu and L Lu ldquoA class of BVPs for nonlinear fractionaldifferential equations with 119901-Laplacian operatorrdquo ElectronicJournal of Qualitative Theory of Differential Equations vol 70pp 1ndash16 2012
[22] Z Liu L Lu and I Szanto ldquoExistence of solutions for fractionalimpulsive differential equations with 119901-Laplacian operatorrdquoActa Mathematica Hungarica vol 141 no 3 pp 203ndash219 2013
[23] Z Bai ldquoEigenvalue intervals for a class of fractional boundaryvalue problemrdquo Computers amp Mathematics with Applicationsvol 64 no 10 pp 3253ndash3257 2012
[24] X Zhang L Wang and Q Sun ldquoExistence of positive solutionsfor a class of nonlinear fractional differential equations withintegral boundary conditions and a parameterrdquo Applied Mathe-matics and Computation vol 226 pp 708ndash718 2014
18 Abstract and Applied Analysis
[25] D J Guo and V Lakshmikantham Nonlinear Problems inAbstract Cones vol 5 Academic Press New York NY USA1988
[26] M A Krasnoselrsquoskii Topological Methods in the Theory ofNonlinear Integral Equations Translated by A H ArmstrongElmsford Pergamon Turkey 1964
Submit your manuscripts athttpwwwhindawicom
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Differential EquationsInternational Journal of
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International Journal of
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Function Spaces
Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
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Algebra
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Decision SciencesAdvances in
Discrete MathematicsJournal of
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Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Stochastic AnalysisInternational Journal of
Abstract and Applied Analysis 9
le
10038171003817100381710038171199091199031003817100381710038171003817 120601119902 (1205821199031198972)
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
(1 minus 119904)120572minus1
times 120601119902 (int
1
0
120596 (120591) 119889120591) 119889119904
=
10038171003817100381710038171199091199031003817100381710038171003817 120601119902 (1205821199031198972) 1205742
Γ (120572 + 1) (1 minus 120578 int1
0119892 (119904) 119889119904)
(51)
which implies that
10038171003817100381710038171199091199031003817100381710038171003817 le
10038171003817100381710038171199091199031003817100381710038171003817 120601119902 (1205821199031198972) 1205742
Γ (120572 + 1) (1 minus 120578 int1
0119892 (119904) 119889119904)
(52)
that is
120601119902 (1205821199031198972) 1205742
Γ (120572 + 1) (1 minus 120578 int1
0119892 (119904) 119889119904)
ge 1 (53)
it follows that we get
120582119903 ge1
1198972(
Γ (120572 + 1) (1 minus 120578 int1
0119892 (119904) 119889119904)
1205742)
119901minus1
= 1205820(54)
In conclusion 120582119903 isin [1205820 1205820](ii) It follows from 0 lt 119891
infin
120601lt infin that there exist
119897lowast
1 119897lowast
2 120588lowastgt 0 such that
119897lowast
1120601119901 (119909) le 119891 (119905 119909) le 119897
lowast
2120601119901 (119909)
(forall119905 isin [0 1] 119909 ge 120588lowast)
(55)
Let 120585lowast0= 120588lowast120590 the following proof are similar to (i)
(iii) It follows from 1198910
120601= +infin there exists 119897lowast gt 0 120588
lowastgt 0
such that
1198910
120601ge 119897lowast120601119901 (119909) (forall119905 isin [0 1] 0 le 119909 le 120588
lowast) (56)
Let 1205851 = 120588lowast120590 we show that 1205851 is required When 119909 isin 119875 cap
120597Ω119903lowast we have 119909(119905) ge 120590119909 = 120590119903lowast since 119909 le 120588
lowast then we need120590119903lowastlt 120588lowast as 119903lowast lt 1205851 so 120590119903
lowastlt 120588lowast holds
Let 120582lowast = (1119897lowast)(Γ(120572)1205901205741)
1(119902minus1) we proceed in the sameway as in the proof of (i) replacing (42) we may assume that119909 minus 120601119902(120582
lowast)119879119909 = 0 (forall119909 isin 119875 cap 120597Ω119903lowast) and replacing (43) we
can prove 119909 minus 120601119902(120582lowast)119879119909 = 120581120595 (forall119909 isin 119875 cap 120597Ω119903lowast 120581 ge 0) It
follows from Lemma 5 that 119894(120601119902(120582lowast)119879 119875 cap Ω119903lowast 119875) = 0 Note
that 119894(120579 119875 cap Ω119903lowast 119875) = 1 we can easily show that there exists119909119903lowast isin 119875cap120597Ω119903lowast and 0 lt ]119903lowast lt 1 such that ]119903lowast120601119902(120582
lowast)119879119909119903lowast = 119909119903lowast
Hence (37) holds for 120582119903lowast = 120582lowast]1minus119901119903lowast
lt 120582lowast
The proof ofTheorem 9(iv) follows by themethod similarto Theorem 9(iii) we omit it here
(v) It follows that 119898(119903) gt 120585 for any 119909 isin 119875 cap 120597Ω119903 we have120590119903 le 119909 le 119903 and 119891(119905 119909)120601119901(119909) ge 119891(119905 119909)120601119901(119903) ge 119898(119903) ge 120585then
119891 (119905 119909) ge 120585120601119901 (119909) forall119909 isin [120590119903 119903] 119905 isin [0 1] (57)
Let = (1120585)(Γ(120572)1205901205741)119901minus1 The following proof is similar to
that of (iv) This finished the proof of (v)
Remark 10 InTheorem 9 all the criteria obtained depend onone of 1198910
120601and 119891
infin
120601
Let Ψ(119909) = 120601119902(120582)119879119909 the following theorems give out themultiply nonexistence and the dependence of parameter
Theorem 11 Assume that (H1)ndash(H5) hold
(i) If 1198910
120601= 0 and 119891
infin
120601= 0 lim
119909rarrinfin119891(119905 119909) = 119891infin isin
(0 +infin] uniformly for 119905 isin [0 1] then there exists120582lowast
gt 0 such that the problem (1) has two solutions forany 120582 gt 120582
lowast
(ii) If 1198910120601
= 0 and 119891infin
120601= 0 then there exists 120582 gt 0 such
that for any 120582 lt 120582 problem (1) has no solution
Proof (i) It follows from 119891infin isin (0 +infin] that for any 120577 gt 0there exists 120588 gt 0 such that
119891 (119905 119909) gt 120577 (forall119905 isin [0 1] 119909 ge 120588) (58)
Since 1198910
120601= 0 there exists 1205761 gt 0 120588
1gt 0 such that
119891 (119905 119909) le 1205761120601119901 (119909) (forall119905 isin [0 1] 0 le 119909 le 1205881) (59)
where 1205761 satisfied 120601119902(1205821205761)1205742Γ(120572 + 1)(1 minus 120578 int1
0119892(119904)119889119904) le 1
For forall119909 isin 119875 cap 120597Ω1205881
we have
Ψ (119909) = (120601119902 (120582) 119879119909) (119905)
=1
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902 (120582int
119904
0
120596 (120591) 119891 (120591 119909 (120591)) 119889120591) 119889119904
minus int
119905
0
(119905 minus 119904)120572minus1
times 120601119902 (120582int
119904
0
120596 (120591) 119891 (120591 119909 (120591)) 119889120591) 119889119904
+120578
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
119892 (119904)
10 Abstract and Applied Analysis
times (int
1
0
(1 minus 120591)120572minus1
times 120601119902 (120582int
120591
0
120596 (120583) 119891 (120583 119909 (120583)) 119889120583)119889120591
minus int
119904
0
(119904 minus 120591)120572minus1
times 120601119902 (120582int
120591
0
120596 (120583) 119891 (120583 119909 (120583)) 119889120583)119889120591)119889119904
le120578
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
119892 (119904)
times (int
1
0
(1 minus 120591)120572minus1
times 120601119902(120582int
120591
0
120596 (120583) 119891 (120583 119909 (120583)) 119889120583)119889120591)119889119904
+1
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902 (120582int
119904
0
120596 (120591) 1205761119891 (120591 119909 (120591)) 119889120591) 119889119904
le120578 int1
0119892 (119904) 119889119904
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
(1 minus 120591)120572minus1
times 120601119902 (120582int
1
0
120596 (120583) 1205761120601119901 (119909 (120583)) 119889120583)119889120591
+1
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902(120582int
1
0
120596 (120591) 1205761120601119901 (119909 (120591)) 119889120591) 119889119904
le1
Γ (120572 + 1) (1 minus 120578 int1
0119892 (119904) 119889119904)
times 120601119902 (120582int
1
0
120596 (120591) 1205761120601119901 (119909) 119889120591)
=120601119902 (1205821205761) 1205742
Γ (120572 + 1) (1 minus 120578 int1
0119892 (119904) 119889119904)
119909 le 119909
(60)
This implied
Ψ119909 le 119909 forall119909 isin 119875 cap 120597Ω1205881
(61)
Let 1205883gt max120588120590 120588
1 and 120582
lowast= (1120577)(120588
3Γ(120572)1205901205741)
119901minus1 thenforall119909 isin 119875 cap 120597Ω120588
3
we have 120590119909 le 119909 le 1205883and
(Ψ119909) (119905)
=1
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902 (120582int
119904
0
120596 (120591) 119891 (120591 119909 (120591)) 119889120591) 119889119904
minus int
119905
0
(119905 minus 119904)120572minus1
times 120601119902 (120582int
119904
0
120596 (120591) 119891 (120591 119909 (120591)) 119889120591) 119889119904
+120578
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
119892 (119904)
times (int
1
0
(1 minus 120591)120572minus1
times 120601119902 (120582int
120591
0
120596 (120583) 119891 (120583 119909 (120583)) 119889120583)119889120591
minus int
119904
0
(119904 minus 120591)120572minus1
times 120601119902(120582int
120591
0
120596 (120583) 119891 (120583 119909 (120583)) 119889120583)119889120591)119889119904
ge120590120578
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
119892 (119904)
times (int
1
0
(1 minus 120591)120572minus1
times 120601119902 (120582int
120591
0
120596 (120583) 119891 (120583 119909 (120583)) 119889120583)119889120591
minus int
119904
0
(119904 minus 120591)120572minus1
times 120601119902(120582int
120591
0
120596 (120583) 119891 (120583 119909 (120583)) 119889120583)119889120591)119889119904
+120590
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902 (120582int
119904
0
120596 (120591) 119891 (120591 119909 (120591)) 119889120591) 119889119904
ge120590
Γ (120572)int
1
0
(1 minus 119904)120572minus1
Abstract and Applied Analysis 11
times 120601119902 (120582int
119904
0
120596 (120591) 119891 (120591 119909 (120591)) 119889120591) 119889119904
gt120590
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902 (120582int
119904
0
120596 (120591) 120577 119889120591) 119889119904
gt120590120601119902 (120582120577)
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902 (int
119904
0
120596 (120591) 119889120591) 119889119904
=120590120601119902 (120582120577) 1205741
Γ (120572)= 1205883
(62)
which implies
Ψ119909 gt 119909 forall119909 isin 119875 cap 120597Ω1205883
(63)
Next for 119891infin
120601= 0 there exist 1205762 gt 0 120588
2gt 1205883gt 0 such
that
119891 (119905 119909) le 1205762120601119901 (119909) (forall119905 isin [0 1] 119909 ge 1205882) (64)
where 1205762 satisfied 120601119902(1205821205762)1205742Γ(120572 + 1)(1 minus 120578 int1
0119892(119904)119889119904) le 1
Similar to the above proof we get
Ψ119909 le 119909 forall119909 isin 119875 cap 120597Ω1205882
(65)
Applying Lemma 7 to (61) (65) and (63) yields that Ψ
has two fixed points 1199091 1199092 such that 1199091 isin 119875cap (Ω1205883
Ω1205881
) and1199092 isin 119875 cap (Ω120588
2
Ω1205883
)(ii) It follows from 119891
0
120601= 0 and 119891
infin
120601= 0 that there exists
1205761 1205762 1205881 1205882 gt 0 such that
119891 (119905 119909) le 1205761120601119901 (119909) (forall119905 isin [0 1] 0 le 119909 le 1205881)
119891 (119905 119909) le 1205762120601119901 (119909) (forall119905 isin [0 1] 119909 ge 1205882)
(66)
then for 119909 isin [1205881 1205882] by the continuous of 119891(119905 119909(119905)) there
exists 119909 isin [1205881 1205882] such that
max1205881le119909le120588
2
119891 (119905 119909)
120601119901 (119909)=
119891 (119905 119909)
120601119901 (119909) (67)
Let Υ = max1205761 1205762 119891(119905 119909)120601119901(119909) gt 0 then
119891 (119905 119909) le Υ120601119901 (119909) (forall0 le 119905 le 1 119909 ge 0) (68)
Assuming 119909(119905) is a positive solution of problem (1) we willshow that this leads to a contradiction for 120582 lt 120582 where
120582 = (1Υ)(Γ(120572 + 1)(1 minus 120578 int1
0119892(119904)119889119904)1205742)
119901minus1 It follows from(26) that
119909 = 119909 (0)
=120578
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
119892 (119904)
times (int
1
0
(1 minus 120591)120572minus1
times 120601119902 (120582int
120591
0
120596 (120583) 119891 (120583 119909 (120583)) 119889120583)119889120591
minus int
119904
0
(119904 minus 120591)120572minus1
times 120601119902(120582int
120591
0
120596 (120583) 119891 (120583 119909 (120583)) 119889120583)119889120591)119889119904
+1
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902 (120582int
119904
0
120596 (120591) 119891 (120591 119909 (120591)) 119889120591) 119889119904
le120578
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
119892 (119904)
times int
1
0
(1 minus 120591)120572minus1
times 120601119902(120582int
120591
0
120596 (120583) 119891 (120583 119909 (120583)) 119889120583)119889120591 119889119904
+1
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902 (120582int
119904
0
120596 (120591) 119891 (120591 119909 (120591)) 119889120591) 119889119904
=1
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
(1 minus 119904)120572minus1
times 120601119902 (120582int
119904
0
120596 (120591) 119891 (120591 119909 (120591)) 119889120591) 119889119904
le1
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
(1 minus 119904)120572minus1
12 Abstract and Applied Analysis
times 120601119902 (120582int
1
0
120596 (120591) Υ120601119901 (119909 (120591)) 119889120591) 119889119904
=120601119902 (120582Υ) 1205742
Γ (120572 + 1) (1 minus 120578 int1
0119892 (119904) 119889119904)
119909
(69)
This implies that 1 le 120601119902(120582Υ)1205742Γ(120572 + 1)(1 minus 120578 int1
0119892(119904)119889119904) =
(120582Υ)119902minus1
1205742Γ(120572+1)(1minus120578 int1
0119892(119904)119889119904) lt (120582 Υ)
119902minus11205742Γ(120572+1)(1minus
120578 int1
0119892(119904)119889119904) = 1 which is a contradiction This finishes the
proof of (ii)
Remark 12 In (i) the condition lim119909rarrinfin
119891(119905 119909) = 119891infin isin
(0 +infin] uniformly for 119905 isin [0 1] can be replaced withcondition (H6) The same methods can be used to prove theresults
Remark 13 From the above proof the condition 119891infin isin
(0 +infin] is important to keep the problem (1) having at leasttwo positive solutions if this condition is not satisfied thenthere exists 120582 gt 0 small enough such that the problem (1) hasno positive solutions for all 120582 lt 120582
Theorem 14 Assume that (H1)ndash(H6) hold(i) If 1198910
120601= +infin and 119891
infin
120601= +infin then there exists 120582lowast0 gt 0
such that problem (1) has at least two solutions for 120582 isin
(0 120582lowast0)(ii) If 1198910
120601= +infin and 119891
infin
120601= +infin then there exists 120582 gt 0
such that problem (1) has no solution for all 120582 gt 120582
Proof (i) It follows from 1198910
120601= +infin that there exists 1199031 gt 0
such that
119891 (119905 119909) ge 119871lowast1120601119901 (119909) (forall119905 isin [0 1] 0 le 119909 le 1199031) (70)
where 119871lowast1 satisfies 1205902120601119902(120582119871lowast1)1205741Γ(120572) ge 1
Then for 119909 isin 119875cap120597Ω1199031
we have 120590119909 le 119909 le 1199031 for 119905 isin [0 1]
and
(Ψ119909) (119905)
=1
Γ (120572)
times int
1
0
(1 minus 119904 )120572minus1
times 120601119902 (120582int
119904
0
120596 (120591) 119891 (120591 119909 (120591)) 119889120591) 119889119904
minus int
119905
0
(119905 minus 119904)120572minus1
times 120601119902 (120582int
119904
0
120596 (120591) 119891 (120591 119909 (120591)) 119889120591) 119889119904
+120578
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
119892 (119904)
times (int
1
0
(1 minus 120591)120572minus1
times 120601119902 (120582int
120591
0
120596 (120583) 119891 (120583 119909 (120583)) 119889120583)119889120591
minus int
119904
0
(119904 minus 120591)120572minus1
times 120601119902 (120582int
120591
0
120596 (120583) 119891 (120583 119909 (120583)) 119889120583)119889120591)119889119904
ge120590120578
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
119892 (119904)
times (int
1
0
(1 minus 120591)120572minus1
times 120601119902 (120582int
120591
0
120596 (120583) 119891 (120583 119909 (120583)) 119889120583)119889120591
minus int
119904
0
(119904 minus 120591)120572minus1
times120601119902 (120582int
120591
0
120596 (120583) 119891 (120583 119909 (120583)) 119889120583)119889120591)119889119904
+120590
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902 (120582int
119904
0
120596 (120591) 119891 (120591 119909 (120591)) 119889120591) 119889119904
ge120590
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902 (120582int
119904
0
120596 (120591) 119891 (120591 119909 (120591)) 119889120591) 119889119904
ge120590
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902 (120582int
119904
0
120596 (120591) 119871lowast1120601119901 (119909 (120591)) 119889120591) 119889119904
ge1205902120601119902 (120582119871lowast1) 119909
Γ (120572)
times int
1
0
(1 minus 119904)120572minus1
times 120601119902 (int
119904
0
120596 (120591) 119889120591) 119889119904
=1205902120601119902 (120582119871lowast1) 1205742
Γ (120572)119909
(71)
Abstract and Applied Analysis 13
which implies that
Ψ119909 ge 119909 forall119909 isin 119875 cap 120597Ω1199031
(72)
Next from 119891infin
120601= +infin there exists 1199032 gt 0 such that
119891 (119905 119909) ge 119871lowast2120601119901 (119909) forall119909 ge 1199032 (73)
where 119871lowast2 satisfies
1205902120601119902 (120582119871lowast2) 1205742
Γ (120572)ge 1 (74)
Let 1199032 gt max1199031 1199032120590 then for 119909 isin 119875 cap 120597Ω1199032
we have
119909 (119905) ge 120590 119909 = 1205901199032 gt 1199032 (75)
Similar to the above proof we can get
Ψ119909 ge 119909 forall119909 isin 119875 cap 120597Ω1199032
(76)
Let 1199031 lt 1199033 lt 11990321198721199033
= max119891(119905 119909) | 119905 isin [0 1] 119909 le 1199033+1 gt
0 and 120582lowast0 = (11198721199033
)(Γ(120572+1)1199033(1minus120578 int1
0119892(119904)119889119904)1205742)
119901minus1Thenfor 119909 isin 119875 cap 120597Ω119903
3
we have
119891 (119905 119909) le 1198721199033
forall119905 isin [0 1] (77)
Hence
(Ψ119909) (119905)
=1
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902 (120582int
119904
0
120596 (120591) 119891 (120591 119909 (120591)) 119889120591) 119889119904
minus int
119905
0
(119905 minus 119904)120572minus1
times 120601119902 (120582int
119904
0
120596 (120591) 119891 (120591 119909 (120591)) 119889120591) 119889119904
+120578
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
119892 (119904)
times (int
1
0
(1 minus 120591)120572minus1
times 120601119902 (120582int
120591
0
120596 (120583) 119891 (120583 119909 (120583)) 119889120583)119889120591
minus int
119904
0
(119904 minus 120591)120572minus1
times120601119902 (120582int
120591
0
120596 (120583) 119891 (120583 119909 (120583)) 119889120583)119889120591)119889119904
le120578
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
119892 (119904)
times int
1
0
(1 minus 120591)120572minus1
120601119902
times (120582int
120591
0
120596 (120583) 119891 (120583 119909 (120583)) 119889120583)119889120591 119889119904
+1
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902 (120582int
119904
0
120596 (120591) 119891 (120591 119909 (120591)) 119889120591) 119889119904
le120578
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
119892 (119904)
times int
1
0
(1 minus 120591)120572minus1
120601119902
times (120582int
120591
0
120596 (120583)1198721199033
119889120583)119889120591119889119904
+1
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902 (120582int
119904
0
120596 (120591)1198721199033
119889120591)119889119904
le120578120601119902 (120582119872119903
3
) int1
0119892 (119904) 119889119904
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
(1 minus 120591)120572minus1
times 120601119902 (int
1
0
120596 (120583) 119889120583)119889120591
+120601119902 (120582119872119903
3
)
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902 (int
1
0
120596 (120591) 119889120591) 119889119904
=1205742120601119902 (120582119872119903
3
)
Γ (120572 + 1) (1 minus 120578 int1
0119892 (119904) 119889119904)
lt1205742120601119902 (120582lowast0119872119903
3
)
Γ (120572 + 1) (1 minus 120578 int1
0119892 (119904) 119889119904)
= 1199033
(78)
14 Abstract and Applied Analysis
This implies that
Ψ119909 lt 119909 forall119909 isin 119875 cap 120597Ω1199033
(79)
Applying Lemma 7 to (72) (76) and (79) yields that Ψ hastwo fixed points 1199091 1199092 such that 1199091 isin 119875 cap (Ω119903
3
Ω1199031
) and1199092 isin 119875 cap (Ω119903
2
Ω1199033
)(ii) By the proof of (i) and the continuity of 119891(119905 119909) there
exists 119909 isin [1199031 1199032] such that min119909isin[11990311199032](119891(119905 119909)120601119901(119909)) =
119891(119905 119909)120601119901(119909)Let Υ = min119871lowast1 119871lowast2 119891(119905 119909)120601119901(119909) since (H6) holds
then Υ gt 0 and
119891 (119905 119909) ge Υ120601119901 (119909) forall119909 ge 0 119905 isin [0 1] (80)
Assuming 119909(119905) is a positive solution of problem (1) we willshow that this leads to contradiction for all 120582 gt 120582 where 120582 =
(1Υ)(Γ(120572)1205741120590)119901minus1 It follows form (26) that
119909 = 119909 (0)
=120578
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
119892 (119904)
times (int
1
0
(1 minus 120591)120572minus1
times 120601119902 (120582int
120591
0
120596 (120583) 119891 (120583 119909 (120583)) 119889120583)119889120591
minus int
119904
0
(119904 minus 120591)120572minus1
times120601119902 (120582int
120591
0
120596 (120583) 119891 (120583 119909 (120583)) 119889120583)119889120591)119889119904
+1
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902 (120582int
119904
0
120596 (120591) 119891 (120591 119909 (120591)) 119889120591) 119889119904
ge1
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902 (120582int
119904
0
120596 (120591) 119891 (120591 119909 (120591)) 119889120591) 119889119904
ge1
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902 (120582int
119904
0
120596 (120591) Υ120601119901 (119909 (120591)) 119889120591) 119889119904
ge1
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902 (120582int
119904
0
120596 (120591) Υ120601119901 (120590 119909 119889120591)) 119889119904
ge120601119902 (120582Υ) 120590 119909
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902 (int
119904
0
120596 (120591) 119889120591) 119889119904
=120601119902 (120582Υ) 1205741120590
Γ (120572)119909
(81)
This implies that 1 ge 120601119902(120582Υ)1205741120590Γ(120572) = (120582Υ)119902minus1
1205741120590Γ(120572) gt
(120582Υ)119902minus1
1205741120590Γ(120572) = 1 which is a contradiction This com-pletes the proof of (ii)
Corollary 15 Assume that (H1)ndash(H5) holds
(i) If lim119909rarrinfin
119891(119905 119909) = 119891infin isin (0 +infin] uniformly for 119905 isin
[0 1] and one of 1198910120601
= 0 and 119891infin
120601= 0 is satisfied then
there exists 120582lowast gt 0 such that the problem (1) has at leastone positive solution for any 120582 gt 120582
lowast(ii) If one of 1198910
120601= +infin and 119891
infin
120601= +infin holds then there
exists 120582lowast0 gt 0 such that problem (1) has at least onepositive solutions for any 120582 isin (0 120582lowast0)
Proof (i) The conclusion is a direct consequence of Theo-rem 11(i)
(ii) The conclusion is a direct consequence of Theo-rem 14(i)
Theorem 16 Assume that (H1)ndash(H5) holdThen the followingconclusions hold
(i) If 1198910120601
= 0 and 119891infin
120601= infin then the problem (1) has a
positive solution 119909120582(119905) for all 120582 gt 0(ii) If 1198910
120601= infin and 119891
infin
120601= 0 then the problem (1) has a
positive solution 119909120582(119905) for all 120582 gt 0
Proof Weonly prove (i) the proof of (ii) is similar sowe omitit here
Let 120582 gt 0 since 1198910
120601= 0 there exists 119903 gt 0 119897 gt 0 such that
119891 (119905 119909) le 119897120601119901 (119909) (forall119905 isin [0 1] 119909 isin [0 119903]) (82)
where 119897 satisfied
1205742120582119902minus1
119897119902minus1
Γ (120572 + 1) (1 minus 120578 int1
0119892 (119904) 119889119904)
le 1 (83)
Similar to the proof of Theorem 14 we obtain that
Ψ119909 le 119909 forall119909 isin 119875 cap 120597Ω119903 (84)
Abstract and Applied Analysis 15
From 119891infin
120601= infin there exists 119877 gt 119903 gt 0 119871 gt 0 such that
119891 (119905 119909) ge 119871120601119901 (119909) (forall119905 isin [0 1] 119909 ge 119877) (85)where 119871 satisfied
1205902120582119902minus1
119871119902minus1
1205741
Γ (120572)ge 1 (86)
We get thatΨ119909 ge 119909 forall119909 isin 119875 cap 120597Ω119877 (87)
Applying Lemma 7 to (84) (87) yields thatΨ has a fixed point119909 such that 119909 isin 119875 cap (Ω119877 Ω119903)
Theorem 17 Assume that (H1)ndash(H6) holdThen the followingconclusions hold
If 0 lt 1198910
120601lt infin and 0 lt 119891
infin
120601lt infin then there exists 1 2 gt
0 such that problem (1) has no positive solution for any 120582 gt 1
and 0 lt 120582 lt 2
Proof It follows from 0 lt 1198910
120601lt infin and 0 lt 119891
infin
120601lt infin that
there exist 1 gt 0 2 gt 0 1 gt 0 2 gt 0 and 1199032 gt 1199031 gt 0
such that1120601119901 (119909) le 119891 (119905 119909) le 1120601119901 (119909)
(forall119905 isin [0 1] 0 le 119909 le 1199031)
2120601119901 (119909) le 119891 (119905 119909) le 2120601119901 (119909)
(forall119905 isin [0 1] 119909 ge 1199032)
(88)
Let = max119891(119905 119909)120601119901(119909) 1199031 le 119909 le 1199032 0 le 119905 le 1 =
min119891(119905 119909)120601119901(119909) 1199031 le 119909 le 1199032 0 le 119905 le 1 By the condition(H6) gt 0 Then forall119909 ge 0 we have
120601119901 (119909) le 119891 (119905 119909) le 120601119901 (119909) (89)
where = max1 2 gt 0 and = min1198971 2 gt 0Let 1 = (1)(Γ(120572)1205741120590)
119901minus1 and 2 =
(1)(Γ(120572 + 1)(1 minus 120578 int1
0119892(119904)119889119904)1205742)
119901minus1
If problem (1)has positive solution 119909(119905) in 119875 then
119909 = 119909 (0)
=120578
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
119892 (119904)
times (int
1
0
(1 minus 120591)120572minus1
times 120601119902 (120582int
120591
0
120596 (120583) 119891 (120583 119909 (120583)) 119889120583)119889120591
minus int
119904
0
(119904 minus 120591)120572minus1
times120601119902 (120582int
120591
0
120596 (120583) 119891 (120583 119909 (120583)) 119889120583)119889120591)119889119904
+1
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902 (120582int
119904
0
120596 (120591) 119891 (120591 119909 (120591)) 119889120591) 119889119904
ge1
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902 (120582int
119904
0
120596 (120591) 119891 (120591 119909 (120591)) 119889120591) 119889119904
ge1
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902 (120582int
119904
0
120596 (120591) 120601119901 (119909 (120591)) 119889120591) 119889119904
ge1
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902 (120582int
119904
0
120596 (120591) 120601119901 (120590 119909 119889120591)) 119889119904
ge120601119902 (120582) 120590 119909
Γ (120572)
times int
1
0
(1 minus 119904)120572minus1
120601119902 (int
119904
0
120596 (120591) 119889120591) 119889119904
=120601119902 (120582) 1205741120590
Γ (120572)119909
gt120601119902 (1 ) 1205741120590
Γ (120572)119909 = 119909
119909 = 119909 (0)
=120578
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
119892 (119904)
times (int
1
0
(1 minus 120591)120572minus1
120601119902
times (120582int
120591
0
120596 (120583) 119891 (120583 119909 (120583)) 119889120583)119889120591
minus int
119904
0
(119904 minus 120591)120572minus1
120601119902
times(120582int
120591
0
120596 (120583) 119891 (120583 119909 (120583)) 119889120583)119889120591)119889119904
+1
Γ (120572)int
1
0
(1 minus 119904)120572minus1
120601119902
times (120582int
119904
0
120596 (120591) 119891 (120591 119909 (120591)) 119889120591) 119889119904
16 Abstract and Applied Analysis
le120578
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
119892 (119904)
times int
1
0
(1 minus 120591)120572minus1
times 120601119902 (120582int
120591
0
120596 (120583) 119891 (120583 119909 (120583)) 119889120583)119889120591 119889119904
+1
Γ (120572)int
1
0
(1 minus 119904)120572minus1
120601119902
times (120582int
119904
0
120596 (120591) 119891 (120591 119909 (120591)) 119889120591) 119889119904
=1
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
(1 minus 119904)120572minus1
120601119902
times (120582int
119904
0
120596 (120591) 119891 (120591 119909 (120591)) 119889120591) 119889119904
le1
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
(1 minus 119904)120572minus1
120601119902
times (120582int
1
0
120596 (120591) 120601119901 (119909 (120591)) 119889120591) 119889119904
=120601119902 (120582) 1205742
Γ (120572 + 1) (1 minus 120578 int1
0119892 (119904) 119889119904)
119909
lt120601119902 (2) 1205742
Γ (120572 + 1) (1 minus 120578 int1
0119892 (119904) 119889119904)
119909 = 119909
(90)
which is a contradiction This finishes the proof of Theo-rem 17
4 Examples
In this section we give out same examples to illustrate ourmain results can be used in practice
Example 1 Consider the following fractional differentialequation
(120601119901 (11986352
0+119909 (119905)))
1015840
+ 120582119905(1 + 1199052)12
119909119901minus1 arctan119909 = 0
0 lt 119905 lt 1
1199091015840(0) = 119909
10158401015840(0) = 0
119909 (1) =1
2int
1
0
1
2119909 (119904) 119889119904
(91)
where 120572 = 52 119901 = 32 119902 = 3 120596(119905) = 119905 119891(119905 119909) = (1 +
1199052)12
119909119901minus1 arctan119909 120578 = 12 and 119892(119905) = 12
By simple computation 119897lowast1= 1205872 119897
lowast
2= radic21205872 1205741 = int
1
0(1minus
119904)32
1206013(int119904
0120591 119889120591)119889119904 asymp 00043 1205742 = 1206013(int
1
0120591 119889120591) = 025 120590 asymp
01429 120582lowast
0= (1119897
lowast
1)(Γ(120572)1205901205741)
119901minus1asymp 25149 120582
lowast
0= (1119897
lowast
2)(Γ(120572+
1)(1 minus 120578 int1
0119892(119904)119889119904)1205742)
119901minus1asymp 14222 Then it is easy to see
all the conditions of Theorem 9(ii) are satisfied thus forall120582 isin
[14222 25149]] the problem (91) has at least one positivesolution
Example 2 Consider the boundary value problem
(120601119901 (11986352
0+119909 (119905)))
1015840
+ 120582119905(1 + 1199052)12
119909119901119890minus119909
= 0
0 lt 119905 lt 1
1199091015840(0) = 119909
10158401015840(0) = 0
119909 (1) =1
2int
1
0
1
2119909 (119904) 119889119904
(92)
where 120572 = 52 119901 = 32 119902 = 3 120596(119905) = 119905 119891(119905 119909) = (1 +
1199052)12
119909119901119890minus119909
120578 = 12 and 119892(119905) = 12
1198910
120601= lim119909rarr0
119891 (119905 119909)
120601119901 (119909)= lim119909rarr0
(1 + 1199052)12
119909119901119890minus119909
119909119901minus1= 0
forall119905 isin [0 1]
1198910
120601= lim119909rarrinfin
119891 (119905 119909)
120601119901 (119909)= lim119909rarrinfin
(1 + 1199052)12
119909119901119890minus119909
119909119901minus1= 0
forall119905 isin [0 1]
(93)
Let 1205881= 12 120588
2= 4 then
119891 (119905 119909)
120601119901 (119909)le (1 + 119905
2)12 1
2radic119890le
1
radic2119890
forall119905 isin [0 1] 119909 isin [01
2]
119891 (119905 119909)
120601119901 (119909)le (1 + 119905
2)12 4
1198904le
4radic2
1198904
forall119905 isin [0 1] 119909 isin [4infin)
(94)
so 1205761 = 1radic2119890 1205762 = 4radic21198904 and forall119909 isin
[12 4] max12le119909le4(119891(119905 119909)120601119901(119909)) = radic2119890minus1 We
have Υ = max1radic2119890 4radic21198904 radic2119890minus1
= radic2119890minus1 1205742 = 025
Abstract and Applied Analysis 17
and 120582 = (1Υ)(Γ(120572+1)(1minus120578 int1
0119892(119904)119889119904)1205742)
119901minus1= 3radic5120587
141198904
By Theorem 11(ii) for any 120582 lt 120582 = 3radic512058714
1198904 the problemhas no positive solution
5 Conclusions
Recently differential equations with p-Laplacian operatorwere widely discussed by several authors In this paperby using fixed point techniques combining with partiallyordered structure of Banach space we obtained the existencemultiply and the dependence of parameter These newresults we presented can be used in numerical computationand analyze mathematical models of physical phenomenamechanics nonlinear dynamics and many other relatedfields
Conflict of Interests
The authors declare that there is no conflict of interestsregarding the publication of this paper
Acknowledgments
The authors are very grateful to anonymous referees for theirvaluable and detailed suggestions and comments to improvethe original paper This work is supported by the Programfor New Century Excellent Talents in University (NCET-10-0097)
References
[1] I Podlubny Fractional Differential Equations vol 198 Aca-demic Press San Diego Calif USA 1999
[2] A A Kilbas H M Srivastava and J J Trujillo Theoryand Applications of Fractional Differential Equations vol 204of North-Holland Mathematics Studies Elsevier Science BVAmsterdam The Netherlands 2006
[3] Y A Rossikhin and M V Shitikova ldquoApplication of fractionalderivatives to the analysis of damped vibrations of viscoelasticsingle mass systemsrdquoActa Mechanica vol 120 no 1ndash4 pp 109ndash125 1997
[4] J Sabatier O Agrawal and J T Machado Advances in Frac-tional Calculus Theoretical developments and applications inphysics and engineering Springer Berlin Germany 2007
[5] N Engheta ldquoOn fractional calculus and fractional multipolesin electromagnetismrdquo IEEE Transactions on Antennas andPropagation vol 44 no 4 pp 554ndash566 1996
[6] R Hilfer Applications of Fractional Calculus in Physics WorldScientific Singapore 2000
[7] R L Bagley and R A Calico ldquoFractional order state equationsfor the control of viscoelastically damped structuresrdquo Journalof Guidance Control and Dynamics vol 14 no 2 pp 304ndash3111991
[8] W-X Zhou andY-D Chu ldquoExistence of solutions for fractionaldifferential equations with multi-point boundary conditionsrdquoCommunications in Nonlinear Science and Numerical Simula-tion vol 17 no 3 pp 1142ndash1148 2012
[9] B Ahmad and G Wang ldquoA study of an impulsive four-pointnonlocal boundary value problemof nonlinear fractional differ-ential equationsrdquo Computers amp Mathematics with Applicationsvol 62 no 3 pp 1341ndash1349 2011
[10] B Ahmad and S Sivasundaram ldquoOn four-point nonlocalboundary value problems of nonlinear integro-differentialequations of fractional orderrdquo Applied Mathematics and Com-putation vol 217 no 2 pp 480ndash487 2010
[11] N Nyamoradi ldquoPositive solutions for multi-point boundaryvalue problems for nonlinear fractional differential equationsrdquoJournal of Contemporary Mathematical Analysis vol 48 no 4pp 145ndash157 2013
[12] B Ahmad and J J Nieto ldquoExistence of solutions for nonlocalboundary value problems of higher-order nonlinear fractionaldifferential equationsrdquo Abstract and Applied Analysis vol 2009Article ID 494720 9 pages 2009
[13] X Liu M Jia and X Xiang ldquoOn the solvability of a fractionaldifferential equationmodel involving the119901-Laplacian operatorrdquoComputers ampMathematics with Applications vol 64 no 10 pp3267ndash3275 2012
[14] J Wang H Xiang and Z Liu ldquoExistence of concave positivesolutions for boundary value problem of nonlinear fractionaldifferential equation with 119901-Laplacian operatorrdquo InternationalJournal of Mathematics and Mathematical Sciences vol 2010Article ID 495138 17 pages 2010
[15] J Wang and H Xiang ldquoUpper and lower solutions method fora class of singular fractional boundary value problems with 119901-Laplacian operatorrdquo Abstract and Applied Analysis vol 2010Article ID 971824 12 pages 2010
[16] H Lu andZHan ldquoExistence of positive solutions for boundary-value problem of fractional differential equation with 119901-laplacian operatorrdquo The American Jouranl of Engineering andTechnology Research vol 11 pp 3757ndash3764 2011
[17] G Chai ldquoPositive solutions for boundary value problem offractional differential equation with 119901-Laplacian operatorrdquoBoundary Value Problems vol 2012 article 18 2012
[18] T Chen W Liu and Z Hu ldquoA boundary value problem forfractional differential equation with 119901-Laplacian operator atresonancerdquoNonlinear AnalysisTheoryMethodsampApplicationsvol 75 no 6 pp 3210ndash3217 2012
[19] T Chen andW Liu ldquoAn anti-periodic boundary value problemfor the fractional differential equation with a 119901-LaplacianoperatorrdquoAppliedMathematics Letters of Rapid Publication vol25 no 11 pp 1671ndash1675 2012
[20] S J Li X G Zhang Y H Wu and L Caccetta ldquoExtremalsolutions for 119901-Laplacian differential systems via iterative com-putationrdquo Applied Mathematics and Computation vol 26 pp1151ndash1158 2013
[21] Z Liu and L Lu ldquoA class of BVPs for nonlinear fractionaldifferential equations with 119901-Laplacian operatorrdquo ElectronicJournal of Qualitative Theory of Differential Equations vol 70pp 1ndash16 2012
[22] Z Liu L Lu and I Szanto ldquoExistence of solutions for fractionalimpulsive differential equations with 119901-Laplacian operatorrdquoActa Mathematica Hungarica vol 141 no 3 pp 203ndash219 2013
[23] Z Bai ldquoEigenvalue intervals for a class of fractional boundaryvalue problemrdquo Computers amp Mathematics with Applicationsvol 64 no 10 pp 3253ndash3257 2012
[24] X Zhang L Wang and Q Sun ldquoExistence of positive solutionsfor a class of nonlinear fractional differential equations withintegral boundary conditions and a parameterrdquo Applied Mathe-matics and Computation vol 226 pp 708ndash718 2014
18 Abstract and Applied Analysis
[25] D J Guo and V Lakshmikantham Nonlinear Problems inAbstract Cones vol 5 Academic Press New York NY USA1988
[26] M A Krasnoselrsquoskii Topological Methods in the Theory ofNonlinear Integral Equations Translated by A H ArmstrongElmsford Pergamon Turkey 1964
Submit your manuscripts athttpwwwhindawicom
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Differential EquationsInternational Journal of
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International Journal of
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Function Spaces
Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
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Algebra
Discrete Dynamics in Nature and Society
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Decision SciencesAdvances in
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Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Stochastic AnalysisInternational Journal of
10 Abstract and Applied Analysis
times (int
1
0
(1 minus 120591)120572minus1
times 120601119902 (120582int
120591
0
120596 (120583) 119891 (120583 119909 (120583)) 119889120583)119889120591
minus int
119904
0
(119904 minus 120591)120572minus1
times 120601119902 (120582int
120591
0
120596 (120583) 119891 (120583 119909 (120583)) 119889120583)119889120591)119889119904
le120578
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
119892 (119904)
times (int
1
0
(1 minus 120591)120572minus1
times 120601119902(120582int
120591
0
120596 (120583) 119891 (120583 119909 (120583)) 119889120583)119889120591)119889119904
+1
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902 (120582int
119904
0
120596 (120591) 1205761119891 (120591 119909 (120591)) 119889120591) 119889119904
le120578 int1
0119892 (119904) 119889119904
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
(1 minus 120591)120572minus1
times 120601119902 (120582int
1
0
120596 (120583) 1205761120601119901 (119909 (120583)) 119889120583)119889120591
+1
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902(120582int
1
0
120596 (120591) 1205761120601119901 (119909 (120591)) 119889120591) 119889119904
le1
Γ (120572 + 1) (1 minus 120578 int1
0119892 (119904) 119889119904)
times 120601119902 (120582int
1
0
120596 (120591) 1205761120601119901 (119909) 119889120591)
=120601119902 (1205821205761) 1205742
Γ (120572 + 1) (1 minus 120578 int1
0119892 (119904) 119889119904)
119909 le 119909
(60)
This implied
Ψ119909 le 119909 forall119909 isin 119875 cap 120597Ω1205881
(61)
Let 1205883gt max120588120590 120588
1 and 120582
lowast= (1120577)(120588
3Γ(120572)1205901205741)
119901minus1 thenforall119909 isin 119875 cap 120597Ω120588
3
we have 120590119909 le 119909 le 1205883and
(Ψ119909) (119905)
=1
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902 (120582int
119904
0
120596 (120591) 119891 (120591 119909 (120591)) 119889120591) 119889119904
minus int
119905
0
(119905 minus 119904)120572minus1
times 120601119902 (120582int
119904
0
120596 (120591) 119891 (120591 119909 (120591)) 119889120591) 119889119904
+120578
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
119892 (119904)
times (int
1
0
(1 minus 120591)120572minus1
times 120601119902 (120582int
120591
0
120596 (120583) 119891 (120583 119909 (120583)) 119889120583)119889120591
minus int
119904
0
(119904 minus 120591)120572minus1
times 120601119902(120582int
120591
0
120596 (120583) 119891 (120583 119909 (120583)) 119889120583)119889120591)119889119904
ge120590120578
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
119892 (119904)
times (int
1
0
(1 minus 120591)120572minus1
times 120601119902 (120582int
120591
0
120596 (120583) 119891 (120583 119909 (120583)) 119889120583)119889120591
minus int
119904
0
(119904 minus 120591)120572minus1
times 120601119902(120582int
120591
0
120596 (120583) 119891 (120583 119909 (120583)) 119889120583)119889120591)119889119904
+120590
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902 (120582int
119904
0
120596 (120591) 119891 (120591 119909 (120591)) 119889120591) 119889119904
ge120590
Γ (120572)int
1
0
(1 minus 119904)120572minus1
Abstract and Applied Analysis 11
times 120601119902 (120582int
119904
0
120596 (120591) 119891 (120591 119909 (120591)) 119889120591) 119889119904
gt120590
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902 (120582int
119904
0
120596 (120591) 120577 119889120591) 119889119904
gt120590120601119902 (120582120577)
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902 (int
119904
0
120596 (120591) 119889120591) 119889119904
=120590120601119902 (120582120577) 1205741
Γ (120572)= 1205883
(62)
which implies
Ψ119909 gt 119909 forall119909 isin 119875 cap 120597Ω1205883
(63)
Next for 119891infin
120601= 0 there exist 1205762 gt 0 120588
2gt 1205883gt 0 such
that
119891 (119905 119909) le 1205762120601119901 (119909) (forall119905 isin [0 1] 119909 ge 1205882) (64)
where 1205762 satisfied 120601119902(1205821205762)1205742Γ(120572 + 1)(1 minus 120578 int1
0119892(119904)119889119904) le 1
Similar to the above proof we get
Ψ119909 le 119909 forall119909 isin 119875 cap 120597Ω1205882
(65)
Applying Lemma 7 to (61) (65) and (63) yields that Ψ
has two fixed points 1199091 1199092 such that 1199091 isin 119875cap (Ω1205883
Ω1205881
) and1199092 isin 119875 cap (Ω120588
2
Ω1205883
)(ii) It follows from 119891
0
120601= 0 and 119891
infin
120601= 0 that there exists
1205761 1205762 1205881 1205882 gt 0 such that
119891 (119905 119909) le 1205761120601119901 (119909) (forall119905 isin [0 1] 0 le 119909 le 1205881)
119891 (119905 119909) le 1205762120601119901 (119909) (forall119905 isin [0 1] 119909 ge 1205882)
(66)
then for 119909 isin [1205881 1205882] by the continuous of 119891(119905 119909(119905)) there
exists 119909 isin [1205881 1205882] such that
max1205881le119909le120588
2
119891 (119905 119909)
120601119901 (119909)=
119891 (119905 119909)
120601119901 (119909) (67)
Let Υ = max1205761 1205762 119891(119905 119909)120601119901(119909) gt 0 then
119891 (119905 119909) le Υ120601119901 (119909) (forall0 le 119905 le 1 119909 ge 0) (68)
Assuming 119909(119905) is a positive solution of problem (1) we willshow that this leads to a contradiction for 120582 lt 120582 where
120582 = (1Υ)(Γ(120572 + 1)(1 minus 120578 int1
0119892(119904)119889119904)1205742)
119901minus1 It follows from(26) that
119909 = 119909 (0)
=120578
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
119892 (119904)
times (int
1
0
(1 minus 120591)120572minus1
times 120601119902 (120582int
120591
0
120596 (120583) 119891 (120583 119909 (120583)) 119889120583)119889120591
minus int
119904
0
(119904 minus 120591)120572minus1
times 120601119902(120582int
120591
0
120596 (120583) 119891 (120583 119909 (120583)) 119889120583)119889120591)119889119904
+1
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902 (120582int
119904
0
120596 (120591) 119891 (120591 119909 (120591)) 119889120591) 119889119904
le120578
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
119892 (119904)
times int
1
0
(1 minus 120591)120572minus1
times 120601119902(120582int
120591
0
120596 (120583) 119891 (120583 119909 (120583)) 119889120583)119889120591 119889119904
+1
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902 (120582int
119904
0
120596 (120591) 119891 (120591 119909 (120591)) 119889120591) 119889119904
=1
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
(1 minus 119904)120572minus1
times 120601119902 (120582int
119904
0
120596 (120591) 119891 (120591 119909 (120591)) 119889120591) 119889119904
le1
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
(1 minus 119904)120572minus1
12 Abstract and Applied Analysis
times 120601119902 (120582int
1
0
120596 (120591) Υ120601119901 (119909 (120591)) 119889120591) 119889119904
=120601119902 (120582Υ) 1205742
Γ (120572 + 1) (1 minus 120578 int1
0119892 (119904) 119889119904)
119909
(69)
This implies that 1 le 120601119902(120582Υ)1205742Γ(120572 + 1)(1 minus 120578 int1
0119892(119904)119889119904) =
(120582Υ)119902minus1
1205742Γ(120572+1)(1minus120578 int1
0119892(119904)119889119904) lt (120582 Υ)
119902minus11205742Γ(120572+1)(1minus
120578 int1
0119892(119904)119889119904) = 1 which is a contradiction This finishes the
proof of (ii)
Remark 12 In (i) the condition lim119909rarrinfin
119891(119905 119909) = 119891infin isin
(0 +infin] uniformly for 119905 isin [0 1] can be replaced withcondition (H6) The same methods can be used to prove theresults
Remark 13 From the above proof the condition 119891infin isin
(0 +infin] is important to keep the problem (1) having at leasttwo positive solutions if this condition is not satisfied thenthere exists 120582 gt 0 small enough such that the problem (1) hasno positive solutions for all 120582 lt 120582
Theorem 14 Assume that (H1)ndash(H6) hold(i) If 1198910
120601= +infin and 119891
infin
120601= +infin then there exists 120582lowast0 gt 0
such that problem (1) has at least two solutions for 120582 isin
(0 120582lowast0)(ii) If 1198910
120601= +infin and 119891
infin
120601= +infin then there exists 120582 gt 0
such that problem (1) has no solution for all 120582 gt 120582
Proof (i) It follows from 1198910
120601= +infin that there exists 1199031 gt 0
such that
119891 (119905 119909) ge 119871lowast1120601119901 (119909) (forall119905 isin [0 1] 0 le 119909 le 1199031) (70)
where 119871lowast1 satisfies 1205902120601119902(120582119871lowast1)1205741Γ(120572) ge 1
Then for 119909 isin 119875cap120597Ω1199031
we have 120590119909 le 119909 le 1199031 for 119905 isin [0 1]
and
(Ψ119909) (119905)
=1
Γ (120572)
times int
1
0
(1 minus 119904 )120572minus1
times 120601119902 (120582int
119904
0
120596 (120591) 119891 (120591 119909 (120591)) 119889120591) 119889119904
minus int
119905
0
(119905 minus 119904)120572minus1
times 120601119902 (120582int
119904
0
120596 (120591) 119891 (120591 119909 (120591)) 119889120591) 119889119904
+120578
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
119892 (119904)
times (int
1
0
(1 minus 120591)120572minus1
times 120601119902 (120582int
120591
0
120596 (120583) 119891 (120583 119909 (120583)) 119889120583)119889120591
minus int
119904
0
(119904 minus 120591)120572minus1
times 120601119902 (120582int
120591
0
120596 (120583) 119891 (120583 119909 (120583)) 119889120583)119889120591)119889119904
ge120590120578
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
119892 (119904)
times (int
1
0
(1 minus 120591)120572minus1
times 120601119902 (120582int
120591
0
120596 (120583) 119891 (120583 119909 (120583)) 119889120583)119889120591
minus int
119904
0
(119904 minus 120591)120572minus1
times120601119902 (120582int
120591
0
120596 (120583) 119891 (120583 119909 (120583)) 119889120583)119889120591)119889119904
+120590
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902 (120582int
119904
0
120596 (120591) 119891 (120591 119909 (120591)) 119889120591) 119889119904
ge120590
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902 (120582int
119904
0
120596 (120591) 119891 (120591 119909 (120591)) 119889120591) 119889119904
ge120590
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902 (120582int
119904
0
120596 (120591) 119871lowast1120601119901 (119909 (120591)) 119889120591) 119889119904
ge1205902120601119902 (120582119871lowast1) 119909
Γ (120572)
times int
1
0
(1 minus 119904)120572minus1
times 120601119902 (int
119904
0
120596 (120591) 119889120591) 119889119904
=1205902120601119902 (120582119871lowast1) 1205742
Γ (120572)119909
(71)
Abstract and Applied Analysis 13
which implies that
Ψ119909 ge 119909 forall119909 isin 119875 cap 120597Ω1199031
(72)
Next from 119891infin
120601= +infin there exists 1199032 gt 0 such that
119891 (119905 119909) ge 119871lowast2120601119901 (119909) forall119909 ge 1199032 (73)
where 119871lowast2 satisfies
1205902120601119902 (120582119871lowast2) 1205742
Γ (120572)ge 1 (74)
Let 1199032 gt max1199031 1199032120590 then for 119909 isin 119875 cap 120597Ω1199032
we have
119909 (119905) ge 120590 119909 = 1205901199032 gt 1199032 (75)
Similar to the above proof we can get
Ψ119909 ge 119909 forall119909 isin 119875 cap 120597Ω1199032
(76)
Let 1199031 lt 1199033 lt 11990321198721199033
= max119891(119905 119909) | 119905 isin [0 1] 119909 le 1199033+1 gt
0 and 120582lowast0 = (11198721199033
)(Γ(120572+1)1199033(1minus120578 int1
0119892(119904)119889119904)1205742)
119901minus1Thenfor 119909 isin 119875 cap 120597Ω119903
3
we have
119891 (119905 119909) le 1198721199033
forall119905 isin [0 1] (77)
Hence
(Ψ119909) (119905)
=1
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902 (120582int
119904
0
120596 (120591) 119891 (120591 119909 (120591)) 119889120591) 119889119904
minus int
119905
0
(119905 minus 119904)120572minus1
times 120601119902 (120582int
119904
0
120596 (120591) 119891 (120591 119909 (120591)) 119889120591) 119889119904
+120578
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
119892 (119904)
times (int
1
0
(1 minus 120591)120572minus1
times 120601119902 (120582int
120591
0
120596 (120583) 119891 (120583 119909 (120583)) 119889120583)119889120591
minus int
119904
0
(119904 minus 120591)120572minus1
times120601119902 (120582int
120591
0
120596 (120583) 119891 (120583 119909 (120583)) 119889120583)119889120591)119889119904
le120578
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
119892 (119904)
times int
1
0
(1 minus 120591)120572minus1
120601119902
times (120582int
120591
0
120596 (120583) 119891 (120583 119909 (120583)) 119889120583)119889120591 119889119904
+1
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902 (120582int
119904
0
120596 (120591) 119891 (120591 119909 (120591)) 119889120591) 119889119904
le120578
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
119892 (119904)
times int
1
0
(1 minus 120591)120572minus1
120601119902
times (120582int
120591
0
120596 (120583)1198721199033
119889120583)119889120591119889119904
+1
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902 (120582int
119904
0
120596 (120591)1198721199033
119889120591)119889119904
le120578120601119902 (120582119872119903
3
) int1
0119892 (119904) 119889119904
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
(1 minus 120591)120572minus1
times 120601119902 (int
1
0
120596 (120583) 119889120583)119889120591
+120601119902 (120582119872119903
3
)
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902 (int
1
0
120596 (120591) 119889120591) 119889119904
=1205742120601119902 (120582119872119903
3
)
Γ (120572 + 1) (1 minus 120578 int1
0119892 (119904) 119889119904)
lt1205742120601119902 (120582lowast0119872119903
3
)
Γ (120572 + 1) (1 minus 120578 int1
0119892 (119904) 119889119904)
= 1199033
(78)
14 Abstract and Applied Analysis
This implies that
Ψ119909 lt 119909 forall119909 isin 119875 cap 120597Ω1199033
(79)
Applying Lemma 7 to (72) (76) and (79) yields that Ψ hastwo fixed points 1199091 1199092 such that 1199091 isin 119875 cap (Ω119903
3
Ω1199031
) and1199092 isin 119875 cap (Ω119903
2
Ω1199033
)(ii) By the proof of (i) and the continuity of 119891(119905 119909) there
exists 119909 isin [1199031 1199032] such that min119909isin[11990311199032](119891(119905 119909)120601119901(119909)) =
119891(119905 119909)120601119901(119909)Let Υ = min119871lowast1 119871lowast2 119891(119905 119909)120601119901(119909) since (H6) holds
then Υ gt 0 and
119891 (119905 119909) ge Υ120601119901 (119909) forall119909 ge 0 119905 isin [0 1] (80)
Assuming 119909(119905) is a positive solution of problem (1) we willshow that this leads to contradiction for all 120582 gt 120582 where 120582 =
(1Υ)(Γ(120572)1205741120590)119901minus1 It follows form (26) that
119909 = 119909 (0)
=120578
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
119892 (119904)
times (int
1
0
(1 minus 120591)120572minus1
times 120601119902 (120582int
120591
0
120596 (120583) 119891 (120583 119909 (120583)) 119889120583)119889120591
minus int
119904
0
(119904 minus 120591)120572minus1
times120601119902 (120582int
120591
0
120596 (120583) 119891 (120583 119909 (120583)) 119889120583)119889120591)119889119904
+1
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902 (120582int
119904
0
120596 (120591) 119891 (120591 119909 (120591)) 119889120591) 119889119904
ge1
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902 (120582int
119904
0
120596 (120591) 119891 (120591 119909 (120591)) 119889120591) 119889119904
ge1
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902 (120582int
119904
0
120596 (120591) Υ120601119901 (119909 (120591)) 119889120591) 119889119904
ge1
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902 (120582int
119904
0
120596 (120591) Υ120601119901 (120590 119909 119889120591)) 119889119904
ge120601119902 (120582Υ) 120590 119909
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902 (int
119904
0
120596 (120591) 119889120591) 119889119904
=120601119902 (120582Υ) 1205741120590
Γ (120572)119909
(81)
This implies that 1 ge 120601119902(120582Υ)1205741120590Γ(120572) = (120582Υ)119902minus1
1205741120590Γ(120572) gt
(120582Υ)119902minus1
1205741120590Γ(120572) = 1 which is a contradiction This com-pletes the proof of (ii)
Corollary 15 Assume that (H1)ndash(H5) holds
(i) If lim119909rarrinfin
119891(119905 119909) = 119891infin isin (0 +infin] uniformly for 119905 isin
[0 1] and one of 1198910120601
= 0 and 119891infin
120601= 0 is satisfied then
there exists 120582lowast gt 0 such that the problem (1) has at leastone positive solution for any 120582 gt 120582
lowast(ii) If one of 1198910
120601= +infin and 119891
infin
120601= +infin holds then there
exists 120582lowast0 gt 0 such that problem (1) has at least onepositive solutions for any 120582 isin (0 120582lowast0)
Proof (i) The conclusion is a direct consequence of Theo-rem 11(i)
(ii) The conclusion is a direct consequence of Theo-rem 14(i)
Theorem 16 Assume that (H1)ndash(H5) holdThen the followingconclusions hold
(i) If 1198910120601
= 0 and 119891infin
120601= infin then the problem (1) has a
positive solution 119909120582(119905) for all 120582 gt 0(ii) If 1198910
120601= infin and 119891
infin
120601= 0 then the problem (1) has a
positive solution 119909120582(119905) for all 120582 gt 0
Proof Weonly prove (i) the proof of (ii) is similar sowe omitit here
Let 120582 gt 0 since 1198910
120601= 0 there exists 119903 gt 0 119897 gt 0 such that
119891 (119905 119909) le 119897120601119901 (119909) (forall119905 isin [0 1] 119909 isin [0 119903]) (82)
where 119897 satisfied
1205742120582119902minus1
119897119902minus1
Γ (120572 + 1) (1 minus 120578 int1
0119892 (119904) 119889119904)
le 1 (83)
Similar to the proof of Theorem 14 we obtain that
Ψ119909 le 119909 forall119909 isin 119875 cap 120597Ω119903 (84)
Abstract and Applied Analysis 15
From 119891infin
120601= infin there exists 119877 gt 119903 gt 0 119871 gt 0 such that
119891 (119905 119909) ge 119871120601119901 (119909) (forall119905 isin [0 1] 119909 ge 119877) (85)where 119871 satisfied
1205902120582119902minus1
119871119902minus1
1205741
Γ (120572)ge 1 (86)
We get thatΨ119909 ge 119909 forall119909 isin 119875 cap 120597Ω119877 (87)
Applying Lemma 7 to (84) (87) yields thatΨ has a fixed point119909 such that 119909 isin 119875 cap (Ω119877 Ω119903)
Theorem 17 Assume that (H1)ndash(H6) holdThen the followingconclusions hold
If 0 lt 1198910
120601lt infin and 0 lt 119891
infin
120601lt infin then there exists 1 2 gt
0 such that problem (1) has no positive solution for any 120582 gt 1
and 0 lt 120582 lt 2
Proof It follows from 0 lt 1198910
120601lt infin and 0 lt 119891
infin
120601lt infin that
there exist 1 gt 0 2 gt 0 1 gt 0 2 gt 0 and 1199032 gt 1199031 gt 0
such that1120601119901 (119909) le 119891 (119905 119909) le 1120601119901 (119909)
(forall119905 isin [0 1] 0 le 119909 le 1199031)
2120601119901 (119909) le 119891 (119905 119909) le 2120601119901 (119909)
(forall119905 isin [0 1] 119909 ge 1199032)
(88)
Let = max119891(119905 119909)120601119901(119909) 1199031 le 119909 le 1199032 0 le 119905 le 1 =
min119891(119905 119909)120601119901(119909) 1199031 le 119909 le 1199032 0 le 119905 le 1 By the condition(H6) gt 0 Then forall119909 ge 0 we have
120601119901 (119909) le 119891 (119905 119909) le 120601119901 (119909) (89)
where = max1 2 gt 0 and = min1198971 2 gt 0Let 1 = (1)(Γ(120572)1205741120590)
119901minus1 and 2 =
(1)(Γ(120572 + 1)(1 minus 120578 int1
0119892(119904)119889119904)1205742)
119901minus1
If problem (1)has positive solution 119909(119905) in 119875 then
119909 = 119909 (0)
=120578
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
119892 (119904)
times (int
1
0
(1 minus 120591)120572minus1
times 120601119902 (120582int
120591
0
120596 (120583) 119891 (120583 119909 (120583)) 119889120583)119889120591
minus int
119904
0
(119904 minus 120591)120572minus1
times120601119902 (120582int
120591
0
120596 (120583) 119891 (120583 119909 (120583)) 119889120583)119889120591)119889119904
+1
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902 (120582int
119904
0
120596 (120591) 119891 (120591 119909 (120591)) 119889120591) 119889119904
ge1
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902 (120582int
119904
0
120596 (120591) 119891 (120591 119909 (120591)) 119889120591) 119889119904
ge1
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902 (120582int
119904
0
120596 (120591) 120601119901 (119909 (120591)) 119889120591) 119889119904
ge1
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902 (120582int
119904
0
120596 (120591) 120601119901 (120590 119909 119889120591)) 119889119904
ge120601119902 (120582) 120590 119909
Γ (120572)
times int
1
0
(1 minus 119904)120572minus1
120601119902 (int
119904
0
120596 (120591) 119889120591) 119889119904
=120601119902 (120582) 1205741120590
Γ (120572)119909
gt120601119902 (1 ) 1205741120590
Γ (120572)119909 = 119909
119909 = 119909 (0)
=120578
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
119892 (119904)
times (int
1
0
(1 minus 120591)120572minus1
120601119902
times (120582int
120591
0
120596 (120583) 119891 (120583 119909 (120583)) 119889120583)119889120591
minus int
119904
0
(119904 minus 120591)120572minus1
120601119902
times(120582int
120591
0
120596 (120583) 119891 (120583 119909 (120583)) 119889120583)119889120591)119889119904
+1
Γ (120572)int
1
0
(1 minus 119904)120572minus1
120601119902
times (120582int
119904
0
120596 (120591) 119891 (120591 119909 (120591)) 119889120591) 119889119904
16 Abstract and Applied Analysis
le120578
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
119892 (119904)
times int
1
0
(1 minus 120591)120572minus1
times 120601119902 (120582int
120591
0
120596 (120583) 119891 (120583 119909 (120583)) 119889120583)119889120591 119889119904
+1
Γ (120572)int
1
0
(1 minus 119904)120572minus1
120601119902
times (120582int
119904
0
120596 (120591) 119891 (120591 119909 (120591)) 119889120591) 119889119904
=1
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
(1 minus 119904)120572minus1
120601119902
times (120582int
119904
0
120596 (120591) 119891 (120591 119909 (120591)) 119889120591) 119889119904
le1
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
(1 minus 119904)120572minus1
120601119902
times (120582int
1
0
120596 (120591) 120601119901 (119909 (120591)) 119889120591) 119889119904
=120601119902 (120582) 1205742
Γ (120572 + 1) (1 minus 120578 int1
0119892 (119904) 119889119904)
119909
lt120601119902 (2) 1205742
Γ (120572 + 1) (1 minus 120578 int1
0119892 (119904) 119889119904)
119909 = 119909
(90)
which is a contradiction This finishes the proof of Theo-rem 17
4 Examples
In this section we give out same examples to illustrate ourmain results can be used in practice
Example 1 Consider the following fractional differentialequation
(120601119901 (11986352
0+119909 (119905)))
1015840
+ 120582119905(1 + 1199052)12
119909119901minus1 arctan119909 = 0
0 lt 119905 lt 1
1199091015840(0) = 119909
10158401015840(0) = 0
119909 (1) =1
2int
1
0
1
2119909 (119904) 119889119904
(91)
where 120572 = 52 119901 = 32 119902 = 3 120596(119905) = 119905 119891(119905 119909) = (1 +
1199052)12
119909119901minus1 arctan119909 120578 = 12 and 119892(119905) = 12
By simple computation 119897lowast1= 1205872 119897
lowast
2= radic21205872 1205741 = int
1
0(1minus
119904)32
1206013(int119904
0120591 119889120591)119889119904 asymp 00043 1205742 = 1206013(int
1
0120591 119889120591) = 025 120590 asymp
01429 120582lowast
0= (1119897
lowast
1)(Γ(120572)1205901205741)
119901minus1asymp 25149 120582
lowast
0= (1119897
lowast
2)(Γ(120572+
1)(1 minus 120578 int1
0119892(119904)119889119904)1205742)
119901minus1asymp 14222 Then it is easy to see
all the conditions of Theorem 9(ii) are satisfied thus forall120582 isin
[14222 25149]] the problem (91) has at least one positivesolution
Example 2 Consider the boundary value problem
(120601119901 (11986352
0+119909 (119905)))
1015840
+ 120582119905(1 + 1199052)12
119909119901119890minus119909
= 0
0 lt 119905 lt 1
1199091015840(0) = 119909
10158401015840(0) = 0
119909 (1) =1
2int
1
0
1
2119909 (119904) 119889119904
(92)
where 120572 = 52 119901 = 32 119902 = 3 120596(119905) = 119905 119891(119905 119909) = (1 +
1199052)12
119909119901119890minus119909
120578 = 12 and 119892(119905) = 12
1198910
120601= lim119909rarr0
119891 (119905 119909)
120601119901 (119909)= lim119909rarr0
(1 + 1199052)12
119909119901119890minus119909
119909119901minus1= 0
forall119905 isin [0 1]
1198910
120601= lim119909rarrinfin
119891 (119905 119909)
120601119901 (119909)= lim119909rarrinfin
(1 + 1199052)12
119909119901119890minus119909
119909119901minus1= 0
forall119905 isin [0 1]
(93)
Let 1205881= 12 120588
2= 4 then
119891 (119905 119909)
120601119901 (119909)le (1 + 119905
2)12 1
2radic119890le
1
radic2119890
forall119905 isin [0 1] 119909 isin [01
2]
119891 (119905 119909)
120601119901 (119909)le (1 + 119905
2)12 4
1198904le
4radic2
1198904
forall119905 isin [0 1] 119909 isin [4infin)
(94)
so 1205761 = 1radic2119890 1205762 = 4radic21198904 and forall119909 isin
[12 4] max12le119909le4(119891(119905 119909)120601119901(119909)) = radic2119890minus1 We
have Υ = max1radic2119890 4radic21198904 radic2119890minus1
= radic2119890minus1 1205742 = 025
Abstract and Applied Analysis 17
and 120582 = (1Υ)(Γ(120572+1)(1minus120578 int1
0119892(119904)119889119904)1205742)
119901minus1= 3radic5120587
141198904
By Theorem 11(ii) for any 120582 lt 120582 = 3radic512058714
1198904 the problemhas no positive solution
5 Conclusions
Recently differential equations with p-Laplacian operatorwere widely discussed by several authors In this paperby using fixed point techniques combining with partiallyordered structure of Banach space we obtained the existencemultiply and the dependence of parameter These newresults we presented can be used in numerical computationand analyze mathematical models of physical phenomenamechanics nonlinear dynamics and many other relatedfields
Conflict of Interests
The authors declare that there is no conflict of interestsregarding the publication of this paper
Acknowledgments
The authors are very grateful to anonymous referees for theirvaluable and detailed suggestions and comments to improvethe original paper This work is supported by the Programfor New Century Excellent Talents in University (NCET-10-0097)
References
[1] I Podlubny Fractional Differential Equations vol 198 Aca-demic Press San Diego Calif USA 1999
[2] A A Kilbas H M Srivastava and J J Trujillo Theoryand Applications of Fractional Differential Equations vol 204of North-Holland Mathematics Studies Elsevier Science BVAmsterdam The Netherlands 2006
[3] Y A Rossikhin and M V Shitikova ldquoApplication of fractionalderivatives to the analysis of damped vibrations of viscoelasticsingle mass systemsrdquoActa Mechanica vol 120 no 1ndash4 pp 109ndash125 1997
[4] J Sabatier O Agrawal and J T Machado Advances in Frac-tional Calculus Theoretical developments and applications inphysics and engineering Springer Berlin Germany 2007
[5] N Engheta ldquoOn fractional calculus and fractional multipolesin electromagnetismrdquo IEEE Transactions on Antennas andPropagation vol 44 no 4 pp 554ndash566 1996
[6] R Hilfer Applications of Fractional Calculus in Physics WorldScientific Singapore 2000
[7] R L Bagley and R A Calico ldquoFractional order state equationsfor the control of viscoelastically damped structuresrdquo Journalof Guidance Control and Dynamics vol 14 no 2 pp 304ndash3111991
[8] W-X Zhou andY-D Chu ldquoExistence of solutions for fractionaldifferential equations with multi-point boundary conditionsrdquoCommunications in Nonlinear Science and Numerical Simula-tion vol 17 no 3 pp 1142ndash1148 2012
[9] B Ahmad and G Wang ldquoA study of an impulsive four-pointnonlocal boundary value problemof nonlinear fractional differ-ential equationsrdquo Computers amp Mathematics with Applicationsvol 62 no 3 pp 1341ndash1349 2011
[10] B Ahmad and S Sivasundaram ldquoOn four-point nonlocalboundary value problems of nonlinear integro-differentialequations of fractional orderrdquo Applied Mathematics and Com-putation vol 217 no 2 pp 480ndash487 2010
[11] N Nyamoradi ldquoPositive solutions for multi-point boundaryvalue problems for nonlinear fractional differential equationsrdquoJournal of Contemporary Mathematical Analysis vol 48 no 4pp 145ndash157 2013
[12] B Ahmad and J J Nieto ldquoExistence of solutions for nonlocalboundary value problems of higher-order nonlinear fractionaldifferential equationsrdquo Abstract and Applied Analysis vol 2009Article ID 494720 9 pages 2009
[13] X Liu M Jia and X Xiang ldquoOn the solvability of a fractionaldifferential equationmodel involving the119901-Laplacian operatorrdquoComputers ampMathematics with Applications vol 64 no 10 pp3267ndash3275 2012
[14] J Wang H Xiang and Z Liu ldquoExistence of concave positivesolutions for boundary value problem of nonlinear fractionaldifferential equation with 119901-Laplacian operatorrdquo InternationalJournal of Mathematics and Mathematical Sciences vol 2010Article ID 495138 17 pages 2010
[15] J Wang and H Xiang ldquoUpper and lower solutions method fora class of singular fractional boundary value problems with 119901-Laplacian operatorrdquo Abstract and Applied Analysis vol 2010Article ID 971824 12 pages 2010
[16] H Lu andZHan ldquoExistence of positive solutions for boundary-value problem of fractional differential equation with 119901-laplacian operatorrdquo The American Jouranl of Engineering andTechnology Research vol 11 pp 3757ndash3764 2011
[17] G Chai ldquoPositive solutions for boundary value problem offractional differential equation with 119901-Laplacian operatorrdquoBoundary Value Problems vol 2012 article 18 2012
[18] T Chen W Liu and Z Hu ldquoA boundary value problem forfractional differential equation with 119901-Laplacian operator atresonancerdquoNonlinear AnalysisTheoryMethodsampApplicationsvol 75 no 6 pp 3210ndash3217 2012
[19] T Chen andW Liu ldquoAn anti-periodic boundary value problemfor the fractional differential equation with a 119901-LaplacianoperatorrdquoAppliedMathematics Letters of Rapid Publication vol25 no 11 pp 1671ndash1675 2012
[20] S J Li X G Zhang Y H Wu and L Caccetta ldquoExtremalsolutions for 119901-Laplacian differential systems via iterative com-putationrdquo Applied Mathematics and Computation vol 26 pp1151ndash1158 2013
[21] Z Liu and L Lu ldquoA class of BVPs for nonlinear fractionaldifferential equations with 119901-Laplacian operatorrdquo ElectronicJournal of Qualitative Theory of Differential Equations vol 70pp 1ndash16 2012
[22] Z Liu L Lu and I Szanto ldquoExistence of solutions for fractionalimpulsive differential equations with 119901-Laplacian operatorrdquoActa Mathematica Hungarica vol 141 no 3 pp 203ndash219 2013
[23] Z Bai ldquoEigenvalue intervals for a class of fractional boundaryvalue problemrdquo Computers amp Mathematics with Applicationsvol 64 no 10 pp 3253ndash3257 2012
[24] X Zhang L Wang and Q Sun ldquoExistence of positive solutionsfor a class of nonlinear fractional differential equations withintegral boundary conditions and a parameterrdquo Applied Mathe-matics and Computation vol 226 pp 708ndash718 2014
18 Abstract and Applied Analysis
[25] D J Guo and V Lakshmikantham Nonlinear Problems inAbstract Cones vol 5 Academic Press New York NY USA1988
[26] M A Krasnoselrsquoskii Topological Methods in the Theory ofNonlinear Integral Equations Translated by A H ArmstrongElmsford Pergamon Turkey 1964
Submit your manuscripts athttpwwwhindawicom
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
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Mathematical Problems in Engineering
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Differential EquationsInternational Journal of
Volume 2014
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International Journal of
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Function Spaces
Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of Mathematics and Mathematical Sciences
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Algebra
Discrete Dynamics in Nature and Society
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
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Decision SciencesAdvances in
Discrete MathematicsJournal of
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Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Stochastic AnalysisInternational Journal of
Abstract and Applied Analysis 11
times 120601119902 (120582int
119904
0
120596 (120591) 119891 (120591 119909 (120591)) 119889120591) 119889119904
gt120590
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902 (120582int
119904
0
120596 (120591) 120577 119889120591) 119889119904
gt120590120601119902 (120582120577)
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902 (int
119904
0
120596 (120591) 119889120591) 119889119904
=120590120601119902 (120582120577) 1205741
Γ (120572)= 1205883
(62)
which implies
Ψ119909 gt 119909 forall119909 isin 119875 cap 120597Ω1205883
(63)
Next for 119891infin
120601= 0 there exist 1205762 gt 0 120588
2gt 1205883gt 0 such
that
119891 (119905 119909) le 1205762120601119901 (119909) (forall119905 isin [0 1] 119909 ge 1205882) (64)
where 1205762 satisfied 120601119902(1205821205762)1205742Γ(120572 + 1)(1 minus 120578 int1
0119892(119904)119889119904) le 1
Similar to the above proof we get
Ψ119909 le 119909 forall119909 isin 119875 cap 120597Ω1205882
(65)
Applying Lemma 7 to (61) (65) and (63) yields that Ψ
has two fixed points 1199091 1199092 such that 1199091 isin 119875cap (Ω1205883
Ω1205881
) and1199092 isin 119875 cap (Ω120588
2
Ω1205883
)(ii) It follows from 119891
0
120601= 0 and 119891
infin
120601= 0 that there exists
1205761 1205762 1205881 1205882 gt 0 such that
119891 (119905 119909) le 1205761120601119901 (119909) (forall119905 isin [0 1] 0 le 119909 le 1205881)
119891 (119905 119909) le 1205762120601119901 (119909) (forall119905 isin [0 1] 119909 ge 1205882)
(66)
then for 119909 isin [1205881 1205882] by the continuous of 119891(119905 119909(119905)) there
exists 119909 isin [1205881 1205882] such that
max1205881le119909le120588
2
119891 (119905 119909)
120601119901 (119909)=
119891 (119905 119909)
120601119901 (119909) (67)
Let Υ = max1205761 1205762 119891(119905 119909)120601119901(119909) gt 0 then
119891 (119905 119909) le Υ120601119901 (119909) (forall0 le 119905 le 1 119909 ge 0) (68)
Assuming 119909(119905) is a positive solution of problem (1) we willshow that this leads to a contradiction for 120582 lt 120582 where
120582 = (1Υ)(Γ(120572 + 1)(1 minus 120578 int1
0119892(119904)119889119904)1205742)
119901minus1 It follows from(26) that
119909 = 119909 (0)
=120578
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
119892 (119904)
times (int
1
0
(1 minus 120591)120572minus1
times 120601119902 (120582int
120591
0
120596 (120583) 119891 (120583 119909 (120583)) 119889120583)119889120591
minus int
119904
0
(119904 minus 120591)120572minus1
times 120601119902(120582int
120591
0
120596 (120583) 119891 (120583 119909 (120583)) 119889120583)119889120591)119889119904
+1
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902 (120582int
119904
0
120596 (120591) 119891 (120591 119909 (120591)) 119889120591) 119889119904
le120578
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
119892 (119904)
times int
1
0
(1 minus 120591)120572minus1
times 120601119902(120582int
120591
0
120596 (120583) 119891 (120583 119909 (120583)) 119889120583)119889120591 119889119904
+1
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902 (120582int
119904
0
120596 (120591) 119891 (120591 119909 (120591)) 119889120591) 119889119904
=1
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
(1 minus 119904)120572minus1
times 120601119902 (120582int
119904
0
120596 (120591) 119891 (120591 119909 (120591)) 119889120591) 119889119904
le1
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
(1 minus 119904)120572minus1
12 Abstract and Applied Analysis
times 120601119902 (120582int
1
0
120596 (120591) Υ120601119901 (119909 (120591)) 119889120591) 119889119904
=120601119902 (120582Υ) 1205742
Γ (120572 + 1) (1 minus 120578 int1
0119892 (119904) 119889119904)
119909
(69)
This implies that 1 le 120601119902(120582Υ)1205742Γ(120572 + 1)(1 minus 120578 int1
0119892(119904)119889119904) =
(120582Υ)119902minus1
1205742Γ(120572+1)(1minus120578 int1
0119892(119904)119889119904) lt (120582 Υ)
119902minus11205742Γ(120572+1)(1minus
120578 int1
0119892(119904)119889119904) = 1 which is a contradiction This finishes the
proof of (ii)
Remark 12 In (i) the condition lim119909rarrinfin
119891(119905 119909) = 119891infin isin
(0 +infin] uniformly for 119905 isin [0 1] can be replaced withcondition (H6) The same methods can be used to prove theresults
Remark 13 From the above proof the condition 119891infin isin
(0 +infin] is important to keep the problem (1) having at leasttwo positive solutions if this condition is not satisfied thenthere exists 120582 gt 0 small enough such that the problem (1) hasno positive solutions for all 120582 lt 120582
Theorem 14 Assume that (H1)ndash(H6) hold(i) If 1198910
120601= +infin and 119891
infin
120601= +infin then there exists 120582lowast0 gt 0
such that problem (1) has at least two solutions for 120582 isin
(0 120582lowast0)(ii) If 1198910
120601= +infin and 119891
infin
120601= +infin then there exists 120582 gt 0
such that problem (1) has no solution for all 120582 gt 120582
Proof (i) It follows from 1198910
120601= +infin that there exists 1199031 gt 0
such that
119891 (119905 119909) ge 119871lowast1120601119901 (119909) (forall119905 isin [0 1] 0 le 119909 le 1199031) (70)
where 119871lowast1 satisfies 1205902120601119902(120582119871lowast1)1205741Γ(120572) ge 1
Then for 119909 isin 119875cap120597Ω1199031
we have 120590119909 le 119909 le 1199031 for 119905 isin [0 1]
and
(Ψ119909) (119905)
=1
Γ (120572)
times int
1
0
(1 minus 119904 )120572minus1
times 120601119902 (120582int
119904
0
120596 (120591) 119891 (120591 119909 (120591)) 119889120591) 119889119904
minus int
119905
0
(119905 minus 119904)120572minus1
times 120601119902 (120582int
119904
0
120596 (120591) 119891 (120591 119909 (120591)) 119889120591) 119889119904
+120578
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
119892 (119904)
times (int
1
0
(1 minus 120591)120572minus1
times 120601119902 (120582int
120591
0
120596 (120583) 119891 (120583 119909 (120583)) 119889120583)119889120591
minus int
119904
0
(119904 minus 120591)120572minus1
times 120601119902 (120582int
120591
0
120596 (120583) 119891 (120583 119909 (120583)) 119889120583)119889120591)119889119904
ge120590120578
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
119892 (119904)
times (int
1
0
(1 minus 120591)120572minus1
times 120601119902 (120582int
120591
0
120596 (120583) 119891 (120583 119909 (120583)) 119889120583)119889120591
minus int
119904
0
(119904 minus 120591)120572minus1
times120601119902 (120582int
120591
0
120596 (120583) 119891 (120583 119909 (120583)) 119889120583)119889120591)119889119904
+120590
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902 (120582int
119904
0
120596 (120591) 119891 (120591 119909 (120591)) 119889120591) 119889119904
ge120590
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902 (120582int
119904
0
120596 (120591) 119891 (120591 119909 (120591)) 119889120591) 119889119904
ge120590
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902 (120582int
119904
0
120596 (120591) 119871lowast1120601119901 (119909 (120591)) 119889120591) 119889119904
ge1205902120601119902 (120582119871lowast1) 119909
Γ (120572)
times int
1
0
(1 minus 119904)120572minus1
times 120601119902 (int
119904
0
120596 (120591) 119889120591) 119889119904
=1205902120601119902 (120582119871lowast1) 1205742
Γ (120572)119909
(71)
Abstract and Applied Analysis 13
which implies that
Ψ119909 ge 119909 forall119909 isin 119875 cap 120597Ω1199031
(72)
Next from 119891infin
120601= +infin there exists 1199032 gt 0 such that
119891 (119905 119909) ge 119871lowast2120601119901 (119909) forall119909 ge 1199032 (73)
where 119871lowast2 satisfies
1205902120601119902 (120582119871lowast2) 1205742
Γ (120572)ge 1 (74)
Let 1199032 gt max1199031 1199032120590 then for 119909 isin 119875 cap 120597Ω1199032
we have
119909 (119905) ge 120590 119909 = 1205901199032 gt 1199032 (75)
Similar to the above proof we can get
Ψ119909 ge 119909 forall119909 isin 119875 cap 120597Ω1199032
(76)
Let 1199031 lt 1199033 lt 11990321198721199033
= max119891(119905 119909) | 119905 isin [0 1] 119909 le 1199033+1 gt
0 and 120582lowast0 = (11198721199033
)(Γ(120572+1)1199033(1minus120578 int1
0119892(119904)119889119904)1205742)
119901minus1Thenfor 119909 isin 119875 cap 120597Ω119903
3
we have
119891 (119905 119909) le 1198721199033
forall119905 isin [0 1] (77)
Hence
(Ψ119909) (119905)
=1
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902 (120582int
119904
0
120596 (120591) 119891 (120591 119909 (120591)) 119889120591) 119889119904
minus int
119905
0
(119905 minus 119904)120572minus1
times 120601119902 (120582int
119904
0
120596 (120591) 119891 (120591 119909 (120591)) 119889120591) 119889119904
+120578
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
119892 (119904)
times (int
1
0
(1 minus 120591)120572minus1
times 120601119902 (120582int
120591
0
120596 (120583) 119891 (120583 119909 (120583)) 119889120583)119889120591
minus int
119904
0
(119904 minus 120591)120572minus1
times120601119902 (120582int
120591
0
120596 (120583) 119891 (120583 119909 (120583)) 119889120583)119889120591)119889119904
le120578
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
119892 (119904)
times int
1
0
(1 minus 120591)120572minus1
120601119902
times (120582int
120591
0
120596 (120583) 119891 (120583 119909 (120583)) 119889120583)119889120591 119889119904
+1
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902 (120582int
119904
0
120596 (120591) 119891 (120591 119909 (120591)) 119889120591) 119889119904
le120578
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
119892 (119904)
times int
1
0
(1 minus 120591)120572minus1
120601119902
times (120582int
120591
0
120596 (120583)1198721199033
119889120583)119889120591119889119904
+1
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902 (120582int
119904
0
120596 (120591)1198721199033
119889120591)119889119904
le120578120601119902 (120582119872119903
3
) int1
0119892 (119904) 119889119904
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
(1 minus 120591)120572minus1
times 120601119902 (int
1
0
120596 (120583) 119889120583)119889120591
+120601119902 (120582119872119903
3
)
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902 (int
1
0
120596 (120591) 119889120591) 119889119904
=1205742120601119902 (120582119872119903
3
)
Γ (120572 + 1) (1 minus 120578 int1
0119892 (119904) 119889119904)
lt1205742120601119902 (120582lowast0119872119903
3
)
Γ (120572 + 1) (1 minus 120578 int1
0119892 (119904) 119889119904)
= 1199033
(78)
14 Abstract and Applied Analysis
This implies that
Ψ119909 lt 119909 forall119909 isin 119875 cap 120597Ω1199033
(79)
Applying Lemma 7 to (72) (76) and (79) yields that Ψ hastwo fixed points 1199091 1199092 such that 1199091 isin 119875 cap (Ω119903
3
Ω1199031
) and1199092 isin 119875 cap (Ω119903
2
Ω1199033
)(ii) By the proof of (i) and the continuity of 119891(119905 119909) there
exists 119909 isin [1199031 1199032] such that min119909isin[11990311199032](119891(119905 119909)120601119901(119909)) =
119891(119905 119909)120601119901(119909)Let Υ = min119871lowast1 119871lowast2 119891(119905 119909)120601119901(119909) since (H6) holds
then Υ gt 0 and
119891 (119905 119909) ge Υ120601119901 (119909) forall119909 ge 0 119905 isin [0 1] (80)
Assuming 119909(119905) is a positive solution of problem (1) we willshow that this leads to contradiction for all 120582 gt 120582 where 120582 =
(1Υ)(Γ(120572)1205741120590)119901minus1 It follows form (26) that
119909 = 119909 (0)
=120578
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
119892 (119904)
times (int
1
0
(1 minus 120591)120572minus1
times 120601119902 (120582int
120591
0
120596 (120583) 119891 (120583 119909 (120583)) 119889120583)119889120591
minus int
119904
0
(119904 minus 120591)120572minus1
times120601119902 (120582int
120591
0
120596 (120583) 119891 (120583 119909 (120583)) 119889120583)119889120591)119889119904
+1
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902 (120582int
119904
0
120596 (120591) 119891 (120591 119909 (120591)) 119889120591) 119889119904
ge1
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902 (120582int
119904
0
120596 (120591) 119891 (120591 119909 (120591)) 119889120591) 119889119904
ge1
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902 (120582int
119904
0
120596 (120591) Υ120601119901 (119909 (120591)) 119889120591) 119889119904
ge1
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902 (120582int
119904
0
120596 (120591) Υ120601119901 (120590 119909 119889120591)) 119889119904
ge120601119902 (120582Υ) 120590 119909
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902 (int
119904
0
120596 (120591) 119889120591) 119889119904
=120601119902 (120582Υ) 1205741120590
Γ (120572)119909
(81)
This implies that 1 ge 120601119902(120582Υ)1205741120590Γ(120572) = (120582Υ)119902minus1
1205741120590Γ(120572) gt
(120582Υ)119902minus1
1205741120590Γ(120572) = 1 which is a contradiction This com-pletes the proof of (ii)
Corollary 15 Assume that (H1)ndash(H5) holds
(i) If lim119909rarrinfin
119891(119905 119909) = 119891infin isin (0 +infin] uniformly for 119905 isin
[0 1] and one of 1198910120601
= 0 and 119891infin
120601= 0 is satisfied then
there exists 120582lowast gt 0 such that the problem (1) has at leastone positive solution for any 120582 gt 120582
lowast(ii) If one of 1198910
120601= +infin and 119891
infin
120601= +infin holds then there
exists 120582lowast0 gt 0 such that problem (1) has at least onepositive solutions for any 120582 isin (0 120582lowast0)
Proof (i) The conclusion is a direct consequence of Theo-rem 11(i)
(ii) The conclusion is a direct consequence of Theo-rem 14(i)
Theorem 16 Assume that (H1)ndash(H5) holdThen the followingconclusions hold
(i) If 1198910120601
= 0 and 119891infin
120601= infin then the problem (1) has a
positive solution 119909120582(119905) for all 120582 gt 0(ii) If 1198910
120601= infin and 119891
infin
120601= 0 then the problem (1) has a
positive solution 119909120582(119905) for all 120582 gt 0
Proof Weonly prove (i) the proof of (ii) is similar sowe omitit here
Let 120582 gt 0 since 1198910
120601= 0 there exists 119903 gt 0 119897 gt 0 such that
119891 (119905 119909) le 119897120601119901 (119909) (forall119905 isin [0 1] 119909 isin [0 119903]) (82)
where 119897 satisfied
1205742120582119902minus1
119897119902minus1
Γ (120572 + 1) (1 minus 120578 int1
0119892 (119904) 119889119904)
le 1 (83)
Similar to the proof of Theorem 14 we obtain that
Ψ119909 le 119909 forall119909 isin 119875 cap 120597Ω119903 (84)
Abstract and Applied Analysis 15
From 119891infin
120601= infin there exists 119877 gt 119903 gt 0 119871 gt 0 such that
119891 (119905 119909) ge 119871120601119901 (119909) (forall119905 isin [0 1] 119909 ge 119877) (85)where 119871 satisfied
1205902120582119902minus1
119871119902minus1
1205741
Γ (120572)ge 1 (86)
We get thatΨ119909 ge 119909 forall119909 isin 119875 cap 120597Ω119877 (87)
Applying Lemma 7 to (84) (87) yields thatΨ has a fixed point119909 such that 119909 isin 119875 cap (Ω119877 Ω119903)
Theorem 17 Assume that (H1)ndash(H6) holdThen the followingconclusions hold
If 0 lt 1198910
120601lt infin and 0 lt 119891
infin
120601lt infin then there exists 1 2 gt
0 such that problem (1) has no positive solution for any 120582 gt 1
and 0 lt 120582 lt 2
Proof It follows from 0 lt 1198910
120601lt infin and 0 lt 119891
infin
120601lt infin that
there exist 1 gt 0 2 gt 0 1 gt 0 2 gt 0 and 1199032 gt 1199031 gt 0
such that1120601119901 (119909) le 119891 (119905 119909) le 1120601119901 (119909)
(forall119905 isin [0 1] 0 le 119909 le 1199031)
2120601119901 (119909) le 119891 (119905 119909) le 2120601119901 (119909)
(forall119905 isin [0 1] 119909 ge 1199032)
(88)
Let = max119891(119905 119909)120601119901(119909) 1199031 le 119909 le 1199032 0 le 119905 le 1 =
min119891(119905 119909)120601119901(119909) 1199031 le 119909 le 1199032 0 le 119905 le 1 By the condition(H6) gt 0 Then forall119909 ge 0 we have
120601119901 (119909) le 119891 (119905 119909) le 120601119901 (119909) (89)
where = max1 2 gt 0 and = min1198971 2 gt 0Let 1 = (1)(Γ(120572)1205741120590)
119901minus1 and 2 =
(1)(Γ(120572 + 1)(1 minus 120578 int1
0119892(119904)119889119904)1205742)
119901minus1
If problem (1)has positive solution 119909(119905) in 119875 then
119909 = 119909 (0)
=120578
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
119892 (119904)
times (int
1
0
(1 minus 120591)120572minus1
times 120601119902 (120582int
120591
0
120596 (120583) 119891 (120583 119909 (120583)) 119889120583)119889120591
minus int
119904
0
(119904 minus 120591)120572minus1
times120601119902 (120582int
120591
0
120596 (120583) 119891 (120583 119909 (120583)) 119889120583)119889120591)119889119904
+1
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902 (120582int
119904
0
120596 (120591) 119891 (120591 119909 (120591)) 119889120591) 119889119904
ge1
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902 (120582int
119904
0
120596 (120591) 119891 (120591 119909 (120591)) 119889120591) 119889119904
ge1
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902 (120582int
119904
0
120596 (120591) 120601119901 (119909 (120591)) 119889120591) 119889119904
ge1
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902 (120582int
119904
0
120596 (120591) 120601119901 (120590 119909 119889120591)) 119889119904
ge120601119902 (120582) 120590 119909
Γ (120572)
times int
1
0
(1 minus 119904)120572minus1
120601119902 (int
119904
0
120596 (120591) 119889120591) 119889119904
=120601119902 (120582) 1205741120590
Γ (120572)119909
gt120601119902 (1 ) 1205741120590
Γ (120572)119909 = 119909
119909 = 119909 (0)
=120578
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
119892 (119904)
times (int
1
0
(1 minus 120591)120572minus1
120601119902
times (120582int
120591
0
120596 (120583) 119891 (120583 119909 (120583)) 119889120583)119889120591
minus int
119904
0
(119904 minus 120591)120572minus1
120601119902
times(120582int
120591
0
120596 (120583) 119891 (120583 119909 (120583)) 119889120583)119889120591)119889119904
+1
Γ (120572)int
1
0
(1 minus 119904)120572minus1
120601119902
times (120582int
119904
0
120596 (120591) 119891 (120591 119909 (120591)) 119889120591) 119889119904
16 Abstract and Applied Analysis
le120578
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
119892 (119904)
times int
1
0
(1 minus 120591)120572minus1
times 120601119902 (120582int
120591
0
120596 (120583) 119891 (120583 119909 (120583)) 119889120583)119889120591 119889119904
+1
Γ (120572)int
1
0
(1 minus 119904)120572minus1
120601119902
times (120582int
119904
0
120596 (120591) 119891 (120591 119909 (120591)) 119889120591) 119889119904
=1
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
(1 minus 119904)120572minus1
120601119902
times (120582int
119904
0
120596 (120591) 119891 (120591 119909 (120591)) 119889120591) 119889119904
le1
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
(1 minus 119904)120572minus1
120601119902
times (120582int
1
0
120596 (120591) 120601119901 (119909 (120591)) 119889120591) 119889119904
=120601119902 (120582) 1205742
Γ (120572 + 1) (1 minus 120578 int1
0119892 (119904) 119889119904)
119909
lt120601119902 (2) 1205742
Γ (120572 + 1) (1 minus 120578 int1
0119892 (119904) 119889119904)
119909 = 119909
(90)
which is a contradiction This finishes the proof of Theo-rem 17
4 Examples
In this section we give out same examples to illustrate ourmain results can be used in practice
Example 1 Consider the following fractional differentialequation
(120601119901 (11986352
0+119909 (119905)))
1015840
+ 120582119905(1 + 1199052)12
119909119901minus1 arctan119909 = 0
0 lt 119905 lt 1
1199091015840(0) = 119909
10158401015840(0) = 0
119909 (1) =1
2int
1
0
1
2119909 (119904) 119889119904
(91)
where 120572 = 52 119901 = 32 119902 = 3 120596(119905) = 119905 119891(119905 119909) = (1 +
1199052)12
119909119901minus1 arctan119909 120578 = 12 and 119892(119905) = 12
By simple computation 119897lowast1= 1205872 119897
lowast
2= radic21205872 1205741 = int
1
0(1minus
119904)32
1206013(int119904
0120591 119889120591)119889119904 asymp 00043 1205742 = 1206013(int
1
0120591 119889120591) = 025 120590 asymp
01429 120582lowast
0= (1119897
lowast
1)(Γ(120572)1205901205741)
119901minus1asymp 25149 120582
lowast
0= (1119897
lowast
2)(Γ(120572+
1)(1 minus 120578 int1
0119892(119904)119889119904)1205742)
119901minus1asymp 14222 Then it is easy to see
all the conditions of Theorem 9(ii) are satisfied thus forall120582 isin
[14222 25149]] the problem (91) has at least one positivesolution
Example 2 Consider the boundary value problem
(120601119901 (11986352
0+119909 (119905)))
1015840
+ 120582119905(1 + 1199052)12
119909119901119890minus119909
= 0
0 lt 119905 lt 1
1199091015840(0) = 119909
10158401015840(0) = 0
119909 (1) =1
2int
1
0
1
2119909 (119904) 119889119904
(92)
where 120572 = 52 119901 = 32 119902 = 3 120596(119905) = 119905 119891(119905 119909) = (1 +
1199052)12
119909119901119890minus119909
120578 = 12 and 119892(119905) = 12
1198910
120601= lim119909rarr0
119891 (119905 119909)
120601119901 (119909)= lim119909rarr0
(1 + 1199052)12
119909119901119890minus119909
119909119901minus1= 0
forall119905 isin [0 1]
1198910
120601= lim119909rarrinfin
119891 (119905 119909)
120601119901 (119909)= lim119909rarrinfin
(1 + 1199052)12
119909119901119890minus119909
119909119901minus1= 0
forall119905 isin [0 1]
(93)
Let 1205881= 12 120588
2= 4 then
119891 (119905 119909)
120601119901 (119909)le (1 + 119905
2)12 1
2radic119890le
1
radic2119890
forall119905 isin [0 1] 119909 isin [01
2]
119891 (119905 119909)
120601119901 (119909)le (1 + 119905
2)12 4
1198904le
4radic2
1198904
forall119905 isin [0 1] 119909 isin [4infin)
(94)
so 1205761 = 1radic2119890 1205762 = 4radic21198904 and forall119909 isin
[12 4] max12le119909le4(119891(119905 119909)120601119901(119909)) = radic2119890minus1 We
have Υ = max1radic2119890 4radic21198904 radic2119890minus1
= radic2119890minus1 1205742 = 025
Abstract and Applied Analysis 17
and 120582 = (1Υ)(Γ(120572+1)(1minus120578 int1
0119892(119904)119889119904)1205742)
119901minus1= 3radic5120587
141198904
By Theorem 11(ii) for any 120582 lt 120582 = 3radic512058714
1198904 the problemhas no positive solution
5 Conclusions
Recently differential equations with p-Laplacian operatorwere widely discussed by several authors In this paperby using fixed point techniques combining with partiallyordered structure of Banach space we obtained the existencemultiply and the dependence of parameter These newresults we presented can be used in numerical computationand analyze mathematical models of physical phenomenamechanics nonlinear dynamics and many other relatedfields
Conflict of Interests
The authors declare that there is no conflict of interestsregarding the publication of this paper
Acknowledgments
The authors are very grateful to anonymous referees for theirvaluable and detailed suggestions and comments to improvethe original paper This work is supported by the Programfor New Century Excellent Talents in University (NCET-10-0097)
References
[1] I Podlubny Fractional Differential Equations vol 198 Aca-demic Press San Diego Calif USA 1999
[2] A A Kilbas H M Srivastava and J J Trujillo Theoryand Applications of Fractional Differential Equations vol 204of North-Holland Mathematics Studies Elsevier Science BVAmsterdam The Netherlands 2006
[3] Y A Rossikhin and M V Shitikova ldquoApplication of fractionalderivatives to the analysis of damped vibrations of viscoelasticsingle mass systemsrdquoActa Mechanica vol 120 no 1ndash4 pp 109ndash125 1997
[4] J Sabatier O Agrawal and J T Machado Advances in Frac-tional Calculus Theoretical developments and applications inphysics and engineering Springer Berlin Germany 2007
[5] N Engheta ldquoOn fractional calculus and fractional multipolesin electromagnetismrdquo IEEE Transactions on Antennas andPropagation vol 44 no 4 pp 554ndash566 1996
[6] R Hilfer Applications of Fractional Calculus in Physics WorldScientific Singapore 2000
[7] R L Bagley and R A Calico ldquoFractional order state equationsfor the control of viscoelastically damped structuresrdquo Journalof Guidance Control and Dynamics vol 14 no 2 pp 304ndash3111991
[8] W-X Zhou andY-D Chu ldquoExistence of solutions for fractionaldifferential equations with multi-point boundary conditionsrdquoCommunications in Nonlinear Science and Numerical Simula-tion vol 17 no 3 pp 1142ndash1148 2012
[9] B Ahmad and G Wang ldquoA study of an impulsive four-pointnonlocal boundary value problemof nonlinear fractional differ-ential equationsrdquo Computers amp Mathematics with Applicationsvol 62 no 3 pp 1341ndash1349 2011
[10] B Ahmad and S Sivasundaram ldquoOn four-point nonlocalboundary value problems of nonlinear integro-differentialequations of fractional orderrdquo Applied Mathematics and Com-putation vol 217 no 2 pp 480ndash487 2010
[11] N Nyamoradi ldquoPositive solutions for multi-point boundaryvalue problems for nonlinear fractional differential equationsrdquoJournal of Contemporary Mathematical Analysis vol 48 no 4pp 145ndash157 2013
[12] B Ahmad and J J Nieto ldquoExistence of solutions for nonlocalboundary value problems of higher-order nonlinear fractionaldifferential equationsrdquo Abstract and Applied Analysis vol 2009Article ID 494720 9 pages 2009
[13] X Liu M Jia and X Xiang ldquoOn the solvability of a fractionaldifferential equationmodel involving the119901-Laplacian operatorrdquoComputers ampMathematics with Applications vol 64 no 10 pp3267ndash3275 2012
[14] J Wang H Xiang and Z Liu ldquoExistence of concave positivesolutions for boundary value problem of nonlinear fractionaldifferential equation with 119901-Laplacian operatorrdquo InternationalJournal of Mathematics and Mathematical Sciences vol 2010Article ID 495138 17 pages 2010
[15] J Wang and H Xiang ldquoUpper and lower solutions method fora class of singular fractional boundary value problems with 119901-Laplacian operatorrdquo Abstract and Applied Analysis vol 2010Article ID 971824 12 pages 2010
[16] H Lu andZHan ldquoExistence of positive solutions for boundary-value problem of fractional differential equation with 119901-laplacian operatorrdquo The American Jouranl of Engineering andTechnology Research vol 11 pp 3757ndash3764 2011
[17] G Chai ldquoPositive solutions for boundary value problem offractional differential equation with 119901-Laplacian operatorrdquoBoundary Value Problems vol 2012 article 18 2012
[18] T Chen W Liu and Z Hu ldquoA boundary value problem forfractional differential equation with 119901-Laplacian operator atresonancerdquoNonlinear AnalysisTheoryMethodsampApplicationsvol 75 no 6 pp 3210ndash3217 2012
[19] T Chen andW Liu ldquoAn anti-periodic boundary value problemfor the fractional differential equation with a 119901-LaplacianoperatorrdquoAppliedMathematics Letters of Rapid Publication vol25 no 11 pp 1671ndash1675 2012
[20] S J Li X G Zhang Y H Wu and L Caccetta ldquoExtremalsolutions for 119901-Laplacian differential systems via iterative com-putationrdquo Applied Mathematics and Computation vol 26 pp1151ndash1158 2013
[21] Z Liu and L Lu ldquoA class of BVPs for nonlinear fractionaldifferential equations with 119901-Laplacian operatorrdquo ElectronicJournal of Qualitative Theory of Differential Equations vol 70pp 1ndash16 2012
[22] Z Liu L Lu and I Szanto ldquoExistence of solutions for fractionalimpulsive differential equations with 119901-Laplacian operatorrdquoActa Mathematica Hungarica vol 141 no 3 pp 203ndash219 2013
[23] Z Bai ldquoEigenvalue intervals for a class of fractional boundaryvalue problemrdquo Computers amp Mathematics with Applicationsvol 64 no 10 pp 3253ndash3257 2012
[24] X Zhang L Wang and Q Sun ldquoExistence of positive solutionsfor a class of nonlinear fractional differential equations withintegral boundary conditions and a parameterrdquo Applied Mathe-matics and Computation vol 226 pp 708ndash718 2014
18 Abstract and Applied Analysis
[25] D J Guo and V Lakshmikantham Nonlinear Problems inAbstract Cones vol 5 Academic Press New York NY USA1988
[26] M A Krasnoselrsquoskii Topological Methods in the Theory ofNonlinear Integral Equations Translated by A H ArmstrongElmsford Pergamon Turkey 1964
Submit your manuscripts athttpwwwhindawicom
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
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Differential EquationsInternational Journal of
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Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Stochastic AnalysisInternational Journal of
12 Abstract and Applied Analysis
times 120601119902 (120582int
1
0
120596 (120591) Υ120601119901 (119909 (120591)) 119889120591) 119889119904
=120601119902 (120582Υ) 1205742
Γ (120572 + 1) (1 minus 120578 int1
0119892 (119904) 119889119904)
119909
(69)
This implies that 1 le 120601119902(120582Υ)1205742Γ(120572 + 1)(1 minus 120578 int1
0119892(119904)119889119904) =
(120582Υ)119902minus1
1205742Γ(120572+1)(1minus120578 int1
0119892(119904)119889119904) lt (120582 Υ)
119902minus11205742Γ(120572+1)(1minus
120578 int1
0119892(119904)119889119904) = 1 which is a contradiction This finishes the
proof of (ii)
Remark 12 In (i) the condition lim119909rarrinfin
119891(119905 119909) = 119891infin isin
(0 +infin] uniformly for 119905 isin [0 1] can be replaced withcondition (H6) The same methods can be used to prove theresults
Remark 13 From the above proof the condition 119891infin isin
(0 +infin] is important to keep the problem (1) having at leasttwo positive solutions if this condition is not satisfied thenthere exists 120582 gt 0 small enough such that the problem (1) hasno positive solutions for all 120582 lt 120582
Theorem 14 Assume that (H1)ndash(H6) hold(i) If 1198910
120601= +infin and 119891
infin
120601= +infin then there exists 120582lowast0 gt 0
such that problem (1) has at least two solutions for 120582 isin
(0 120582lowast0)(ii) If 1198910
120601= +infin and 119891
infin
120601= +infin then there exists 120582 gt 0
such that problem (1) has no solution for all 120582 gt 120582
Proof (i) It follows from 1198910
120601= +infin that there exists 1199031 gt 0
such that
119891 (119905 119909) ge 119871lowast1120601119901 (119909) (forall119905 isin [0 1] 0 le 119909 le 1199031) (70)
where 119871lowast1 satisfies 1205902120601119902(120582119871lowast1)1205741Γ(120572) ge 1
Then for 119909 isin 119875cap120597Ω1199031
we have 120590119909 le 119909 le 1199031 for 119905 isin [0 1]
and
(Ψ119909) (119905)
=1
Γ (120572)
times int
1
0
(1 minus 119904 )120572minus1
times 120601119902 (120582int
119904
0
120596 (120591) 119891 (120591 119909 (120591)) 119889120591) 119889119904
minus int
119905
0
(119905 minus 119904)120572minus1
times 120601119902 (120582int
119904
0
120596 (120591) 119891 (120591 119909 (120591)) 119889120591) 119889119904
+120578
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
119892 (119904)
times (int
1
0
(1 minus 120591)120572minus1
times 120601119902 (120582int
120591
0
120596 (120583) 119891 (120583 119909 (120583)) 119889120583)119889120591
minus int
119904
0
(119904 minus 120591)120572minus1
times 120601119902 (120582int
120591
0
120596 (120583) 119891 (120583 119909 (120583)) 119889120583)119889120591)119889119904
ge120590120578
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
119892 (119904)
times (int
1
0
(1 minus 120591)120572minus1
times 120601119902 (120582int
120591
0
120596 (120583) 119891 (120583 119909 (120583)) 119889120583)119889120591
minus int
119904
0
(119904 minus 120591)120572minus1
times120601119902 (120582int
120591
0
120596 (120583) 119891 (120583 119909 (120583)) 119889120583)119889120591)119889119904
+120590
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902 (120582int
119904
0
120596 (120591) 119891 (120591 119909 (120591)) 119889120591) 119889119904
ge120590
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902 (120582int
119904
0
120596 (120591) 119891 (120591 119909 (120591)) 119889120591) 119889119904
ge120590
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902 (120582int
119904
0
120596 (120591) 119871lowast1120601119901 (119909 (120591)) 119889120591) 119889119904
ge1205902120601119902 (120582119871lowast1) 119909
Γ (120572)
times int
1
0
(1 minus 119904)120572minus1
times 120601119902 (int
119904
0
120596 (120591) 119889120591) 119889119904
=1205902120601119902 (120582119871lowast1) 1205742
Γ (120572)119909
(71)
Abstract and Applied Analysis 13
which implies that
Ψ119909 ge 119909 forall119909 isin 119875 cap 120597Ω1199031
(72)
Next from 119891infin
120601= +infin there exists 1199032 gt 0 such that
119891 (119905 119909) ge 119871lowast2120601119901 (119909) forall119909 ge 1199032 (73)
where 119871lowast2 satisfies
1205902120601119902 (120582119871lowast2) 1205742
Γ (120572)ge 1 (74)
Let 1199032 gt max1199031 1199032120590 then for 119909 isin 119875 cap 120597Ω1199032
we have
119909 (119905) ge 120590 119909 = 1205901199032 gt 1199032 (75)
Similar to the above proof we can get
Ψ119909 ge 119909 forall119909 isin 119875 cap 120597Ω1199032
(76)
Let 1199031 lt 1199033 lt 11990321198721199033
= max119891(119905 119909) | 119905 isin [0 1] 119909 le 1199033+1 gt
0 and 120582lowast0 = (11198721199033
)(Γ(120572+1)1199033(1minus120578 int1
0119892(119904)119889119904)1205742)
119901minus1Thenfor 119909 isin 119875 cap 120597Ω119903
3
we have
119891 (119905 119909) le 1198721199033
forall119905 isin [0 1] (77)
Hence
(Ψ119909) (119905)
=1
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902 (120582int
119904
0
120596 (120591) 119891 (120591 119909 (120591)) 119889120591) 119889119904
minus int
119905
0
(119905 minus 119904)120572minus1
times 120601119902 (120582int
119904
0
120596 (120591) 119891 (120591 119909 (120591)) 119889120591) 119889119904
+120578
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
119892 (119904)
times (int
1
0
(1 minus 120591)120572minus1
times 120601119902 (120582int
120591
0
120596 (120583) 119891 (120583 119909 (120583)) 119889120583)119889120591
minus int
119904
0
(119904 minus 120591)120572minus1
times120601119902 (120582int
120591
0
120596 (120583) 119891 (120583 119909 (120583)) 119889120583)119889120591)119889119904
le120578
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
119892 (119904)
times int
1
0
(1 minus 120591)120572minus1
120601119902
times (120582int
120591
0
120596 (120583) 119891 (120583 119909 (120583)) 119889120583)119889120591 119889119904
+1
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902 (120582int
119904
0
120596 (120591) 119891 (120591 119909 (120591)) 119889120591) 119889119904
le120578
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
119892 (119904)
times int
1
0
(1 minus 120591)120572minus1
120601119902
times (120582int
120591
0
120596 (120583)1198721199033
119889120583)119889120591119889119904
+1
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902 (120582int
119904
0
120596 (120591)1198721199033
119889120591)119889119904
le120578120601119902 (120582119872119903
3
) int1
0119892 (119904) 119889119904
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
(1 minus 120591)120572minus1
times 120601119902 (int
1
0
120596 (120583) 119889120583)119889120591
+120601119902 (120582119872119903
3
)
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902 (int
1
0
120596 (120591) 119889120591) 119889119904
=1205742120601119902 (120582119872119903
3
)
Γ (120572 + 1) (1 minus 120578 int1
0119892 (119904) 119889119904)
lt1205742120601119902 (120582lowast0119872119903
3
)
Γ (120572 + 1) (1 minus 120578 int1
0119892 (119904) 119889119904)
= 1199033
(78)
14 Abstract and Applied Analysis
This implies that
Ψ119909 lt 119909 forall119909 isin 119875 cap 120597Ω1199033
(79)
Applying Lemma 7 to (72) (76) and (79) yields that Ψ hastwo fixed points 1199091 1199092 such that 1199091 isin 119875 cap (Ω119903
3
Ω1199031
) and1199092 isin 119875 cap (Ω119903
2
Ω1199033
)(ii) By the proof of (i) and the continuity of 119891(119905 119909) there
exists 119909 isin [1199031 1199032] such that min119909isin[11990311199032](119891(119905 119909)120601119901(119909)) =
119891(119905 119909)120601119901(119909)Let Υ = min119871lowast1 119871lowast2 119891(119905 119909)120601119901(119909) since (H6) holds
then Υ gt 0 and
119891 (119905 119909) ge Υ120601119901 (119909) forall119909 ge 0 119905 isin [0 1] (80)
Assuming 119909(119905) is a positive solution of problem (1) we willshow that this leads to contradiction for all 120582 gt 120582 where 120582 =
(1Υ)(Γ(120572)1205741120590)119901minus1 It follows form (26) that
119909 = 119909 (0)
=120578
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
119892 (119904)
times (int
1
0
(1 minus 120591)120572minus1
times 120601119902 (120582int
120591
0
120596 (120583) 119891 (120583 119909 (120583)) 119889120583)119889120591
minus int
119904
0
(119904 minus 120591)120572minus1
times120601119902 (120582int
120591
0
120596 (120583) 119891 (120583 119909 (120583)) 119889120583)119889120591)119889119904
+1
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902 (120582int
119904
0
120596 (120591) 119891 (120591 119909 (120591)) 119889120591) 119889119904
ge1
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902 (120582int
119904
0
120596 (120591) 119891 (120591 119909 (120591)) 119889120591) 119889119904
ge1
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902 (120582int
119904
0
120596 (120591) Υ120601119901 (119909 (120591)) 119889120591) 119889119904
ge1
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902 (120582int
119904
0
120596 (120591) Υ120601119901 (120590 119909 119889120591)) 119889119904
ge120601119902 (120582Υ) 120590 119909
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902 (int
119904
0
120596 (120591) 119889120591) 119889119904
=120601119902 (120582Υ) 1205741120590
Γ (120572)119909
(81)
This implies that 1 ge 120601119902(120582Υ)1205741120590Γ(120572) = (120582Υ)119902minus1
1205741120590Γ(120572) gt
(120582Υ)119902minus1
1205741120590Γ(120572) = 1 which is a contradiction This com-pletes the proof of (ii)
Corollary 15 Assume that (H1)ndash(H5) holds
(i) If lim119909rarrinfin
119891(119905 119909) = 119891infin isin (0 +infin] uniformly for 119905 isin
[0 1] and one of 1198910120601
= 0 and 119891infin
120601= 0 is satisfied then
there exists 120582lowast gt 0 such that the problem (1) has at leastone positive solution for any 120582 gt 120582
lowast(ii) If one of 1198910
120601= +infin and 119891
infin
120601= +infin holds then there
exists 120582lowast0 gt 0 such that problem (1) has at least onepositive solutions for any 120582 isin (0 120582lowast0)
Proof (i) The conclusion is a direct consequence of Theo-rem 11(i)
(ii) The conclusion is a direct consequence of Theo-rem 14(i)
Theorem 16 Assume that (H1)ndash(H5) holdThen the followingconclusions hold
(i) If 1198910120601
= 0 and 119891infin
120601= infin then the problem (1) has a
positive solution 119909120582(119905) for all 120582 gt 0(ii) If 1198910
120601= infin and 119891
infin
120601= 0 then the problem (1) has a
positive solution 119909120582(119905) for all 120582 gt 0
Proof Weonly prove (i) the proof of (ii) is similar sowe omitit here
Let 120582 gt 0 since 1198910
120601= 0 there exists 119903 gt 0 119897 gt 0 such that
119891 (119905 119909) le 119897120601119901 (119909) (forall119905 isin [0 1] 119909 isin [0 119903]) (82)
where 119897 satisfied
1205742120582119902minus1
119897119902minus1
Γ (120572 + 1) (1 minus 120578 int1
0119892 (119904) 119889119904)
le 1 (83)
Similar to the proof of Theorem 14 we obtain that
Ψ119909 le 119909 forall119909 isin 119875 cap 120597Ω119903 (84)
Abstract and Applied Analysis 15
From 119891infin
120601= infin there exists 119877 gt 119903 gt 0 119871 gt 0 such that
119891 (119905 119909) ge 119871120601119901 (119909) (forall119905 isin [0 1] 119909 ge 119877) (85)where 119871 satisfied
1205902120582119902minus1
119871119902minus1
1205741
Γ (120572)ge 1 (86)
We get thatΨ119909 ge 119909 forall119909 isin 119875 cap 120597Ω119877 (87)
Applying Lemma 7 to (84) (87) yields thatΨ has a fixed point119909 such that 119909 isin 119875 cap (Ω119877 Ω119903)
Theorem 17 Assume that (H1)ndash(H6) holdThen the followingconclusions hold
If 0 lt 1198910
120601lt infin and 0 lt 119891
infin
120601lt infin then there exists 1 2 gt
0 such that problem (1) has no positive solution for any 120582 gt 1
and 0 lt 120582 lt 2
Proof It follows from 0 lt 1198910
120601lt infin and 0 lt 119891
infin
120601lt infin that
there exist 1 gt 0 2 gt 0 1 gt 0 2 gt 0 and 1199032 gt 1199031 gt 0
such that1120601119901 (119909) le 119891 (119905 119909) le 1120601119901 (119909)
(forall119905 isin [0 1] 0 le 119909 le 1199031)
2120601119901 (119909) le 119891 (119905 119909) le 2120601119901 (119909)
(forall119905 isin [0 1] 119909 ge 1199032)
(88)
Let = max119891(119905 119909)120601119901(119909) 1199031 le 119909 le 1199032 0 le 119905 le 1 =
min119891(119905 119909)120601119901(119909) 1199031 le 119909 le 1199032 0 le 119905 le 1 By the condition(H6) gt 0 Then forall119909 ge 0 we have
120601119901 (119909) le 119891 (119905 119909) le 120601119901 (119909) (89)
where = max1 2 gt 0 and = min1198971 2 gt 0Let 1 = (1)(Γ(120572)1205741120590)
119901minus1 and 2 =
(1)(Γ(120572 + 1)(1 minus 120578 int1
0119892(119904)119889119904)1205742)
119901minus1
If problem (1)has positive solution 119909(119905) in 119875 then
119909 = 119909 (0)
=120578
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
119892 (119904)
times (int
1
0
(1 minus 120591)120572minus1
times 120601119902 (120582int
120591
0
120596 (120583) 119891 (120583 119909 (120583)) 119889120583)119889120591
minus int
119904
0
(119904 minus 120591)120572minus1
times120601119902 (120582int
120591
0
120596 (120583) 119891 (120583 119909 (120583)) 119889120583)119889120591)119889119904
+1
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902 (120582int
119904
0
120596 (120591) 119891 (120591 119909 (120591)) 119889120591) 119889119904
ge1
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902 (120582int
119904
0
120596 (120591) 119891 (120591 119909 (120591)) 119889120591) 119889119904
ge1
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902 (120582int
119904
0
120596 (120591) 120601119901 (119909 (120591)) 119889120591) 119889119904
ge1
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902 (120582int
119904
0
120596 (120591) 120601119901 (120590 119909 119889120591)) 119889119904
ge120601119902 (120582) 120590 119909
Γ (120572)
times int
1
0
(1 minus 119904)120572minus1
120601119902 (int
119904
0
120596 (120591) 119889120591) 119889119904
=120601119902 (120582) 1205741120590
Γ (120572)119909
gt120601119902 (1 ) 1205741120590
Γ (120572)119909 = 119909
119909 = 119909 (0)
=120578
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
119892 (119904)
times (int
1
0
(1 minus 120591)120572minus1
120601119902
times (120582int
120591
0
120596 (120583) 119891 (120583 119909 (120583)) 119889120583)119889120591
minus int
119904
0
(119904 minus 120591)120572minus1
120601119902
times(120582int
120591
0
120596 (120583) 119891 (120583 119909 (120583)) 119889120583)119889120591)119889119904
+1
Γ (120572)int
1
0
(1 minus 119904)120572minus1
120601119902
times (120582int
119904
0
120596 (120591) 119891 (120591 119909 (120591)) 119889120591) 119889119904
16 Abstract and Applied Analysis
le120578
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
119892 (119904)
times int
1
0
(1 minus 120591)120572minus1
times 120601119902 (120582int
120591
0
120596 (120583) 119891 (120583 119909 (120583)) 119889120583)119889120591 119889119904
+1
Γ (120572)int
1
0
(1 minus 119904)120572minus1
120601119902
times (120582int
119904
0
120596 (120591) 119891 (120591 119909 (120591)) 119889120591) 119889119904
=1
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
(1 minus 119904)120572minus1
120601119902
times (120582int
119904
0
120596 (120591) 119891 (120591 119909 (120591)) 119889120591) 119889119904
le1
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
(1 minus 119904)120572minus1
120601119902
times (120582int
1
0
120596 (120591) 120601119901 (119909 (120591)) 119889120591) 119889119904
=120601119902 (120582) 1205742
Γ (120572 + 1) (1 minus 120578 int1
0119892 (119904) 119889119904)
119909
lt120601119902 (2) 1205742
Γ (120572 + 1) (1 minus 120578 int1
0119892 (119904) 119889119904)
119909 = 119909
(90)
which is a contradiction This finishes the proof of Theo-rem 17
4 Examples
In this section we give out same examples to illustrate ourmain results can be used in practice
Example 1 Consider the following fractional differentialequation
(120601119901 (11986352
0+119909 (119905)))
1015840
+ 120582119905(1 + 1199052)12
119909119901minus1 arctan119909 = 0
0 lt 119905 lt 1
1199091015840(0) = 119909
10158401015840(0) = 0
119909 (1) =1
2int
1
0
1
2119909 (119904) 119889119904
(91)
where 120572 = 52 119901 = 32 119902 = 3 120596(119905) = 119905 119891(119905 119909) = (1 +
1199052)12
119909119901minus1 arctan119909 120578 = 12 and 119892(119905) = 12
By simple computation 119897lowast1= 1205872 119897
lowast
2= radic21205872 1205741 = int
1
0(1minus
119904)32
1206013(int119904
0120591 119889120591)119889119904 asymp 00043 1205742 = 1206013(int
1
0120591 119889120591) = 025 120590 asymp
01429 120582lowast
0= (1119897
lowast
1)(Γ(120572)1205901205741)
119901minus1asymp 25149 120582
lowast
0= (1119897
lowast
2)(Γ(120572+
1)(1 minus 120578 int1
0119892(119904)119889119904)1205742)
119901minus1asymp 14222 Then it is easy to see
all the conditions of Theorem 9(ii) are satisfied thus forall120582 isin
[14222 25149]] the problem (91) has at least one positivesolution
Example 2 Consider the boundary value problem
(120601119901 (11986352
0+119909 (119905)))
1015840
+ 120582119905(1 + 1199052)12
119909119901119890minus119909
= 0
0 lt 119905 lt 1
1199091015840(0) = 119909
10158401015840(0) = 0
119909 (1) =1
2int
1
0
1
2119909 (119904) 119889119904
(92)
where 120572 = 52 119901 = 32 119902 = 3 120596(119905) = 119905 119891(119905 119909) = (1 +
1199052)12
119909119901119890minus119909
120578 = 12 and 119892(119905) = 12
1198910
120601= lim119909rarr0
119891 (119905 119909)
120601119901 (119909)= lim119909rarr0
(1 + 1199052)12
119909119901119890minus119909
119909119901minus1= 0
forall119905 isin [0 1]
1198910
120601= lim119909rarrinfin
119891 (119905 119909)
120601119901 (119909)= lim119909rarrinfin
(1 + 1199052)12
119909119901119890minus119909
119909119901minus1= 0
forall119905 isin [0 1]
(93)
Let 1205881= 12 120588
2= 4 then
119891 (119905 119909)
120601119901 (119909)le (1 + 119905
2)12 1
2radic119890le
1
radic2119890
forall119905 isin [0 1] 119909 isin [01
2]
119891 (119905 119909)
120601119901 (119909)le (1 + 119905
2)12 4
1198904le
4radic2
1198904
forall119905 isin [0 1] 119909 isin [4infin)
(94)
so 1205761 = 1radic2119890 1205762 = 4radic21198904 and forall119909 isin
[12 4] max12le119909le4(119891(119905 119909)120601119901(119909)) = radic2119890minus1 We
have Υ = max1radic2119890 4radic21198904 radic2119890minus1
= radic2119890minus1 1205742 = 025
Abstract and Applied Analysis 17
and 120582 = (1Υ)(Γ(120572+1)(1minus120578 int1
0119892(119904)119889119904)1205742)
119901minus1= 3radic5120587
141198904
By Theorem 11(ii) for any 120582 lt 120582 = 3radic512058714
1198904 the problemhas no positive solution
5 Conclusions
Recently differential equations with p-Laplacian operatorwere widely discussed by several authors In this paperby using fixed point techniques combining with partiallyordered structure of Banach space we obtained the existencemultiply and the dependence of parameter These newresults we presented can be used in numerical computationand analyze mathematical models of physical phenomenamechanics nonlinear dynamics and many other relatedfields
Conflict of Interests
The authors declare that there is no conflict of interestsregarding the publication of this paper
Acknowledgments
The authors are very grateful to anonymous referees for theirvaluable and detailed suggestions and comments to improvethe original paper This work is supported by the Programfor New Century Excellent Talents in University (NCET-10-0097)
References
[1] I Podlubny Fractional Differential Equations vol 198 Aca-demic Press San Diego Calif USA 1999
[2] A A Kilbas H M Srivastava and J J Trujillo Theoryand Applications of Fractional Differential Equations vol 204of North-Holland Mathematics Studies Elsevier Science BVAmsterdam The Netherlands 2006
[3] Y A Rossikhin and M V Shitikova ldquoApplication of fractionalderivatives to the analysis of damped vibrations of viscoelasticsingle mass systemsrdquoActa Mechanica vol 120 no 1ndash4 pp 109ndash125 1997
[4] J Sabatier O Agrawal and J T Machado Advances in Frac-tional Calculus Theoretical developments and applications inphysics and engineering Springer Berlin Germany 2007
[5] N Engheta ldquoOn fractional calculus and fractional multipolesin electromagnetismrdquo IEEE Transactions on Antennas andPropagation vol 44 no 4 pp 554ndash566 1996
[6] R Hilfer Applications of Fractional Calculus in Physics WorldScientific Singapore 2000
[7] R L Bagley and R A Calico ldquoFractional order state equationsfor the control of viscoelastically damped structuresrdquo Journalof Guidance Control and Dynamics vol 14 no 2 pp 304ndash3111991
[8] W-X Zhou andY-D Chu ldquoExistence of solutions for fractionaldifferential equations with multi-point boundary conditionsrdquoCommunications in Nonlinear Science and Numerical Simula-tion vol 17 no 3 pp 1142ndash1148 2012
[9] B Ahmad and G Wang ldquoA study of an impulsive four-pointnonlocal boundary value problemof nonlinear fractional differ-ential equationsrdquo Computers amp Mathematics with Applicationsvol 62 no 3 pp 1341ndash1349 2011
[10] B Ahmad and S Sivasundaram ldquoOn four-point nonlocalboundary value problems of nonlinear integro-differentialequations of fractional orderrdquo Applied Mathematics and Com-putation vol 217 no 2 pp 480ndash487 2010
[11] N Nyamoradi ldquoPositive solutions for multi-point boundaryvalue problems for nonlinear fractional differential equationsrdquoJournal of Contemporary Mathematical Analysis vol 48 no 4pp 145ndash157 2013
[12] B Ahmad and J J Nieto ldquoExistence of solutions for nonlocalboundary value problems of higher-order nonlinear fractionaldifferential equationsrdquo Abstract and Applied Analysis vol 2009Article ID 494720 9 pages 2009
[13] X Liu M Jia and X Xiang ldquoOn the solvability of a fractionaldifferential equationmodel involving the119901-Laplacian operatorrdquoComputers ampMathematics with Applications vol 64 no 10 pp3267ndash3275 2012
[14] J Wang H Xiang and Z Liu ldquoExistence of concave positivesolutions for boundary value problem of nonlinear fractionaldifferential equation with 119901-Laplacian operatorrdquo InternationalJournal of Mathematics and Mathematical Sciences vol 2010Article ID 495138 17 pages 2010
[15] J Wang and H Xiang ldquoUpper and lower solutions method fora class of singular fractional boundary value problems with 119901-Laplacian operatorrdquo Abstract and Applied Analysis vol 2010Article ID 971824 12 pages 2010
[16] H Lu andZHan ldquoExistence of positive solutions for boundary-value problem of fractional differential equation with 119901-laplacian operatorrdquo The American Jouranl of Engineering andTechnology Research vol 11 pp 3757ndash3764 2011
[17] G Chai ldquoPositive solutions for boundary value problem offractional differential equation with 119901-Laplacian operatorrdquoBoundary Value Problems vol 2012 article 18 2012
[18] T Chen W Liu and Z Hu ldquoA boundary value problem forfractional differential equation with 119901-Laplacian operator atresonancerdquoNonlinear AnalysisTheoryMethodsampApplicationsvol 75 no 6 pp 3210ndash3217 2012
[19] T Chen andW Liu ldquoAn anti-periodic boundary value problemfor the fractional differential equation with a 119901-LaplacianoperatorrdquoAppliedMathematics Letters of Rapid Publication vol25 no 11 pp 1671ndash1675 2012
[20] S J Li X G Zhang Y H Wu and L Caccetta ldquoExtremalsolutions for 119901-Laplacian differential systems via iterative com-putationrdquo Applied Mathematics and Computation vol 26 pp1151ndash1158 2013
[21] Z Liu and L Lu ldquoA class of BVPs for nonlinear fractionaldifferential equations with 119901-Laplacian operatorrdquo ElectronicJournal of Qualitative Theory of Differential Equations vol 70pp 1ndash16 2012
[22] Z Liu L Lu and I Szanto ldquoExistence of solutions for fractionalimpulsive differential equations with 119901-Laplacian operatorrdquoActa Mathematica Hungarica vol 141 no 3 pp 203ndash219 2013
[23] Z Bai ldquoEigenvalue intervals for a class of fractional boundaryvalue problemrdquo Computers amp Mathematics with Applicationsvol 64 no 10 pp 3253ndash3257 2012
[24] X Zhang L Wang and Q Sun ldquoExistence of positive solutionsfor a class of nonlinear fractional differential equations withintegral boundary conditions and a parameterrdquo Applied Mathe-matics and Computation vol 226 pp 708ndash718 2014
18 Abstract and Applied Analysis
[25] D J Guo and V Lakshmikantham Nonlinear Problems inAbstract Cones vol 5 Academic Press New York NY USA1988
[26] M A Krasnoselrsquoskii Topological Methods in the Theory ofNonlinear Integral Equations Translated by A H ArmstrongElmsford Pergamon Turkey 1964
Submit your manuscripts athttpwwwhindawicom
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
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Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
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Hindawi Publishing Corporationhttpwwwhindawicom
Differential EquationsInternational Journal of
Volume 2014
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Journal of
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OptimizationJournal of
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International Journal of
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Function Spaces
Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of Mathematics and Mathematical Sciences
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Algebra
Discrete Dynamics in Nature and Society
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
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Decision SciencesAdvances in
Discrete MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom
Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Stochastic AnalysisInternational Journal of
Abstract and Applied Analysis 13
which implies that
Ψ119909 ge 119909 forall119909 isin 119875 cap 120597Ω1199031
(72)
Next from 119891infin
120601= +infin there exists 1199032 gt 0 such that
119891 (119905 119909) ge 119871lowast2120601119901 (119909) forall119909 ge 1199032 (73)
where 119871lowast2 satisfies
1205902120601119902 (120582119871lowast2) 1205742
Γ (120572)ge 1 (74)
Let 1199032 gt max1199031 1199032120590 then for 119909 isin 119875 cap 120597Ω1199032
we have
119909 (119905) ge 120590 119909 = 1205901199032 gt 1199032 (75)
Similar to the above proof we can get
Ψ119909 ge 119909 forall119909 isin 119875 cap 120597Ω1199032
(76)
Let 1199031 lt 1199033 lt 11990321198721199033
= max119891(119905 119909) | 119905 isin [0 1] 119909 le 1199033+1 gt
0 and 120582lowast0 = (11198721199033
)(Γ(120572+1)1199033(1minus120578 int1
0119892(119904)119889119904)1205742)
119901minus1Thenfor 119909 isin 119875 cap 120597Ω119903
3
we have
119891 (119905 119909) le 1198721199033
forall119905 isin [0 1] (77)
Hence
(Ψ119909) (119905)
=1
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902 (120582int
119904
0
120596 (120591) 119891 (120591 119909 (120591)) 119889120591) 119889119904
minus int
119905
0
(119905 minus 119904)120572minus1
times 120601119902 (120582int
119904
0
120596 (120591) 119891 (120591 119909 (120591)) 119889120591) 119889119904
+120578
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
119892 (119904)
times (int
1
0
(1 minus 120591)120572minus1
times 120601119902 (120582int
120591
0
120596 (120583) 119891 (120583 119909 (120583)) 119889120583)119889120591
minus int
119904
0
(119904 minus 120591)120572minus1
times120601119902 (120582int
120591
0
120596 (120583) 119891 (120583 119909 (120583)) 119889120583)119889120591)119889119904
le120578
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
119892 (119904)
times int
1
0
(1 minus 120591)120572minus1
120601119902
times (120582int
120591
0
120596 (120583) 119891 (120583 119909 (120583)) 119889120583)119889120591 119889119904
+1
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902 (120582int
119904
0
120596 (120591) 119891 (120591 119909 (120591)) 119889120591) 119889119904
le120578
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
119892 (119904)
times int
1
0
(1 minus 120591)120572minus1
120601119902
times (120582int
120591
0
120596 (120583)1198721199033
119889120583)119889120591119889119904
+1
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902 (120582int
119904
0
120596 (120591)1198721199033
119889120591)119889119904
le120578120601119902 (120582119872119903
3
) int1
0119892 (119904) 119889119904
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
(1 minus 120591)120572minus1
times 120601119902 (int
1
0
120596 (120583) 119889120583)119889120591
+120601119902 (120582119872119903
3
)
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902 (int
1
0
120596 (120591) 119889120591) 119889119904
=1205742120601119902 (120582119872119903
3
)
Γ (120572 + 1) (1 minus 120578 int1
0119892 (119904) 119889119904)
lt1205742120601119902 (120582lowast0119872119903
3
)
Γ (120572 + 1) (1 minus 120578 int1
0119892 (119904) 119889119904)
= 1199033
(78)
14 Abstract and Applied Analysis
This implies that
Ψ119909 lt 119909 forall119909 isin 119875 cap 120597Ω1199033
(79)
Applying Lemma 7 to (72) (76) and (79) yields that Ψ hastwo fixed points 1199091 1199092 such that 1199091 isin 119875 cap (Ω119903
3
Ω1199031
) and1199092 isin 119875 cap (Ω119903
2
Ω1199033
)(ii) By the proof of (i) and the continuity of 119891(119905 119909) there
exists 119909 isin [1199031 1199032] such that min119909isin[11990311199032](119891(119905 119909)120601119901(119909)) =
119891(119905 119909)120601119901(119909)Let Υ = min119871lowast1 119871lowast2 119891(119905 119909)120601119901(119909) since (H6) holds
then Υ gt 0 and
119891 (119905 119909) ge Υ120601119901 (119909) forall119909 ge 0 119905 isin [0 1] (80)
Assuming 119909(119905) is a positive solution of problem (1) we willshow that this leads to contradiction for all 120582 gt 120582 where 120582 =
(1Υ)(Γ(120572)1205741120590)119901minus1 It follows form (26) that
119909 = 119909 (0)
=120578
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
119892 (119904)
times (int
1
0
(1 minus 120591)120572minus1
times 120601119902 (120582int
120591
0
120596 (120583) 119891 (120583 119909 (120583)) 119889120583)119889120591
minus int
119904
0
(119904 minus 120591)120572minus1
times120601119902 (120582int
120591
0
120596 (120583) 119891 (120583 119909 (120583)) 119889120583)119889120591)119889119904
+1
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902 (120582int
119904
0
120596 (120591) 119891 (120591 119909 (120591)) 119889120591) 119889119904
ge1
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902 (120582int
119904
0
120596 (120591) 119891 (120591 119909 (120591)) 119889120591) 119889119904
ge1
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902 (120582int
119904
0
120596 (120591) Υ120601119901 (119909 (120591)) 119889120591) 119889119904
ge1
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902 (120582int
119904
0
120596 (120591) Υ120601119901 (120590 119909 119889120591)) 119889119904
ge120601119902 (120582Υ) 120590 119909
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902 (int
119904
0
120596 (120591) 119889120591) 119889119904
=120601119902 (120582Υ) 1205741120590
Γ (120572)119909
(81)
This implies that 1 ge 120601119902(120582Υ)1205741120590Γ(120572) = (120582Υ)119902minus1
1205741120590Γ(120572) gt
(120582Υ)119902minus1
1205741120590Γ(120572) = 1 which is a contradiction This com-pletes the proof of (ii)
Corollary 15 Assume that (H1)ndash(H5) holds
(i) If lim119909rarrinfin
119891(119905 119909) = 119891infin isin (0 +infin] uniformly for 119905 isin
[0 1] and one of 1198910120601
= 0 and 119891infin
120601= 0 is satisfied then
there exists 120582lowast gt 0 such that the problem (1) has at leastone positive solution for any 120582 gt 120582
lowast(ii) If one of 1198910
120601= +infin and 119891
infin
120601= +infin holds then there
exists 120582lowast0 gt 0 such that problem (1) has at least onepositive solutions for any 120582 isin (0 120582lowast0)
Proof (i) The conclusion is a direct consequence of Theo-rem 11(i)
(ii) The conclusion is a direct consequence of Theo-rem 14(i)
Theorem 16 Assume that (H1)ndash(H5) holdThen the followingconclusions hold
(i) If 1198910120601
= 0 and 119891infin
120601= infin then the problem (1) has a
positive solution 119909120582(119905) for all 120582 gt 0(ii) If 1198910
120601= infin and 119891
infin
120601= 0 then the problem (1) has a
positive solution 119909120582(119905) for all 120582 gt 0
Proof Weonly prove (i) the proof of (ii) is similar sowe omitit here
Let 120582 gt 0 since 1198910
120601= 0 there exists 119903 gt 0 119897 gt 0 such that
119891 (119905 119909) le 119897120601119901 (119909) (forall119905 isin [0 1] 119909 isin [0 119903]) (82)
where 119897 satisfied
1205742120582119902minus1
119897119902minus1
Γ (120572 + 1) (1 minus 120578 int1
0119892 (119904) 119889119904)
le 1 (83)
Similar to the proof of Theorem 14 we obtain that
Ψ119909 le 119909 forall119909 isin 119875 cap 120597Ω119903 (84)
Abstract and Applied Analysis 15
From 119891infin
120601= infin there exists 119877 gt 119903 gt 0 119871 gt 0 such that
119891 (119905 119909) ge 119871120601119901 (119909) (forall119905 isin [0 1] 119909 ge 119877) (85)where 119871 satisfied
1205902120582119902minus1
119871119902minus1
1205741
Γ (120572)ge 1 (86)
We get thatΨ119909 ge 119909 forall119909 isin 119875 cap 120597Ω119877 (87)
Applying Lemma 7 to (84) (87) yields thatΨ has a fixed point119909 such that 119909 isin 119875 cap (Ω119877 Ω119903)
Theorem 17 Assume that (H1)ndash(H6) holdThen the followingconclusions hold
If 0 lt 1198910
120601lt infin and 0 lt 119891
infin
120601lt infin then there exists 1 2 gt
0 such that problem (1) has no positive solution for any 120582 gt 1
and 0 lt 120582 lt 2
Proof It follows from 0 lt 1198910
120601lt infin and 0 lt 119891
infin
120601lt infin that
there exist 1 gt 0 2 gt 0 1 gt 0 2 gt 0 and 1199032 gt 1199031 gt 0
such that1120601119901 (119909) le 119891 (119905 119909) le 1120601119901 (119909)
(forall119905 isin [0 1] 0 le 119909 le 1199031)
2120601119901 (119909) le 119891 (119905 119909) le 2120601119901 (119909)
(forall119905 isin [0 1] 119909 ge 1199032)
(88)
Let = max119891(119905 119909)120601119901(119909) 1199031 le 119909 le 1199032 0 le 119905 le 1 =
min119891(119905 119909)120601119901(119909) 1199031 le 119909 le 1199032 0 le 119905 le 1 By the condition(H6) gt 0 Then forall119909 ge 0 we have
120601119901 (119909) le 119891 (119905 119909) le 120601119901 (119909) (89)
where = max1 2 gt 0 and = min1198971 2 gt 0Let 1 = (1)(Γ(120572)1205741120590)
119901minus1 and 2 =
(1)(Γ(120572 + 1)(1 minus 120578 int1
0119892(119904)119889119904)1205742)
119901minus1
If problem (1)has positive solution 119909(119905) in 119875 then
119909 = 119909 (0)
=120578
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
119892 (119904)
times (int
1
0
(1 minus 120591)120572minus1
times 120601119902 (120582int
120591
0
120596 (120583) 119891 (120583 119909 (120583)) 119889120583)119889120591
minus int
119904
0
(119904 minus 120591)120572minus1
times120601119902 (120582int
120591
0
120596 (120583) 119891 (120583 119909 (120583)) 119889120583)119889120591)119889119904
+1
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902 (120582int
119904
0
120596 (120591) 119891 (120591 119909 (120591)) 119889120591) 119889119904
ge1
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902 (120582int
119904
0
120596 (120591) 119891 (120591 119909 (120591)) 119889120591) 119889119904
ge1
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902 (120582int
119904
0
120596 (120591) 120601119901 (119909 (120591)) 119889120591) 119889119904
ge1
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902 (120582int
119904
0
120596 (120591) 120601119901 (120590 119909 119889120591)) 119889119904
ge120601119902 (120582) 120590 119909
Γ (120572)
times int
1
0
(1 minus 119904)120572minus1
120601119902 (int
119904
0
120596 (120591) 119889120591) 119889119904
=120601119902 (120582) 1205741120590
Γ (120572)119909
gt120601119902 (1 ) 1205741120590
Γ (120572)119909 = 119909
119909 = 119909 (0)
=120578
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
119892 (119904)
times (int
1
0
(1 minus 120591)120572minus1
120601119902
times (120582int
120591
0
120596 (120583) 119891 (120583 119909 (120583)) 119889120583)119889120591
minus int
119904
0
(119904 minus 120591)120572minus1
120601119902
times(120582int
120591
0
120596 (120583) 119891 (120583 119909 (120583)) 119889120583)119889120591)119889119904
+1
Γ (120572)int
1
0
(1 minus 119904)120572minus1
120601119902
times (120582int
119904
0
120596 (120591) 119891 (120591 119909 (120591)) 119889120591) 119889119904
16 Abstract and Applied Analysis
le120578
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
119892 (119904)
times int
1
0
(1 minus 120591)120572minus1
times 120601119902 (120582int
120591
0
120596 (120583) 119891 (120583 119909 (120583)) 119889120583)119889120591 119889119904
+1
Γ (120572)int
1
0
(1 minus 119904)120572minus1
120601119902
times (120582int
119904
0
120596 (120591) 119891 (120591 119909 (120591)) 119889120591) 119889119904
=1
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
(1 minus 119904)120572minus1
120601119902
times (120582int
119904
0
120596 (120591) 119891 (120591 119909 (120591)) 119889120591) 119889119904
le1
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
(1 minus 119904)120572minus1
120601119902
times (120582int
1
0
120596 (120591) 120601119901 (119909 (120591)) 119889120591) 119889119904
=120601119902 (120582) 1205742
Γ (120572 + 1) (1 minus 120578 int1
0119892 (119904) 119889119904)
119909
lt120601119902 (2) 1205742
Γ (120572 + 1) (1 minus 120578 int1
0119892 (119904) 119889119904)
119909 = 119909
(90)
which is a contradiction This finishes the proof of Theo-rem 17
4 Examples
In this section we give out same examples to illustrate ourmain results can be used in practice
Example 1 Consider the following fractional differentialequation
(120601119901 (11986352
0+119909 (119905)))
1015840
+ 120582119905(1 + 1199052)12
119909119901minus1 arctan119909 = 0
0 lt 119905 lt 1
1199091015840(0) = 119909
10158401015840(0) = 0
119909 (1) =1
2int
1
0
1
2119909 (119904) 119889119904
(91)
where 120572 = 52 119901 = 32 119902 = 3 120596(119905) = 119905 119891(119905 119909) = (1 +
1199052)12
119909119901minus1 arctan119909 120578 = 12 and 119892(119905) = 12
By simple computation 119897lowast1= 1205872 119897
lowast
2= radic21205872 1205741 = int
1
0(1minus
119904)32
1206013(int119904
0120591 119889120591)119889119904 asymp 00043 1205742 = 1206013(int
1
0120591 119889120591) = 025 120590 asymp
01429 120582lowast
0= (1119897
lowast
1)(Γ(120572)1205901205741)
119901minus1asymp 25149 120582
lowast
0= (1119897
lowast
2)(Γ(120572+
1)(1 minus 120578 int1
0119892(119904)119889119904)1205742)
119901minus1asymp 14222 Then it is easy to see
all the conditions of Theorem 9(ii) are satisfied thus forall120582 isin
[14222 25149]] the problem (91) has at least one positivesolution
Example 2 Consider the boundary value problem
(120601119901 (11986352
0+119909 (119905)))
1015840
+ 120582119905(1 + 1199052)12
119909119901119890minus119909
= 0
0 lt 119905 lt 1
1199091015840(0) = 119909
10158401015840(0) = 0
119909 (1) =1
2int
1
0
1
2119909 (119904) 119889119904
(92)
where 120572 = 52 119901 = 32 119902 = 3 120596(119905) = 119905 119891(119905 119909) = (1 +
1199052)12
119909119901119890minus119909
120578 = 12 and 119892(119905) = 12
1198910
120601= lim119909rarr0
119891 (119905 119909)
120601119901 (119909)= lim119909rarr0
(1 + 1199052)12
119909119901119890minus119909
119909119901minus1= 0
forall119905 isin [0 1]
1198910
120601= lim119909rarrinfin
119891 (119905 119909)
120601119901 (119909)= lim119909rarrinfin
(1 + 1199052)12
119909119901119890minus119909
119909119901minus1= 0
forall119905 isin [0 1]
(93)
Let 1205881= 12 120588
2= 4 then
119891 (119905 119909)
120601119901 (119909)le (1 + 119905
2)12 1
2radic119890le
1
radic2119890
forall119905 isin [0 1] 119909 isin [01
2]
119891 (119905 119909)
120601119901 (119909)le (1 + 119905
2)12 4
1198904le
4radic2
1198904
forall119905 isin [0 1] 119909 isin [4infin)
(94)
so 1205761 = 1radic2119890 1205762 = 4radic21198904 and forall119909 isin
[12 4] max12le119909le4(119891(119905 119909)120601119901(119909)) = radic2119890minus1 We
have Υ = max1radic2119890 4radic21198904 radic2119890minus1
= radic2119890minus1 1205742 = 025
Abstract and Applied Analysis 17
and 120582 = (1Υ)(Γ(120572+1)(1minus120578 int1
0119892(119904)119889119904)1205742)
119901minus1= 3radic5120587
141198904
By Theorem 11(ii) for any 120582 lt 120582 = 3radic512058714
1198904 the problemhas no positive solution
5 Conclusions
Recently differential equations with p-Laplacian operatorwere widely discussed by several authors In this paperby using fixed point techniques combining with partiallyordered structure of Banach space we obtained the existencemultiply and the dependence of parameter These newresults we presented can be used in numerical computationand analyze mathematical models of physical phenomenamechanics nonlinear dynamics and many other relatedfields
Conflict of Interests
The authors declare that there is no conflict of interestsregarding the publication of this paper
Acknowledgments
The authors are very grateful to anonymous referees for theirvaluable and detailed suggestions and comments to improvethe original paper This work is supported by the Programfor New Century Excellent Talents in University (NCET-10-0097)
References
[1] I Podlubny Fractional Differential Equations vol 198 Aca-demic Press San Diego Calif USA 1999
[2] A A Kilbas H M Srivastava and J J Trujillo Theoryand Applications of Fractional Differential Equations vol 204of North-Holland Mathematics Studies Elsevier Science BVAmsterdam The Netherlands 2006
[3] Y A Rossikhin and M V Shitikova ldquoApplication of fractionalderivatives to the analysis of damped vibrations of viscoelasticsingle mass systemsrdquoActa Mechanica vol 120 no 1ndash4 pp 109ndash125 1997
[4] J Sabatier O Agrawal and J T Machado Advances in Frac-tional Calculus Theoretical developments and applications inphysics and engineering Springer Berlin Germany 2007
[5] N Engheta ldquoOn fractional calculus and fractional multipolesin electromagnetismrdquo IEEE Transactions on Antennas andPropagation vol 44 no 4 pp 554ndash566 1996
[6] R Hilfer Applications of Fractional Calculus in Physics WorldScientific Singapore 2000
[7] R L Bagley and R A Calico ldquoFractional order state equationsfor the control of viscoelastically damped structuresrdquo Journalof Guidance Control and Dynamics vol 14 no 2 pp 304ndash3111991
[8] W-X Zhou andY-D Chu ldquoExistence of solutions for fractionaldifferential equations with multi-point boundary conditionsrdquoCommunications in Nonlinear Science and Numerical Simula-tion vol 17 no 3 pp 1142ndash1148 2012
[9] B Ahmad and G Wang ldquoA study of an impulsive four-pointnonlocal boundary value problemof nonlinear fractional differ-ential equationsrdquo Computers amp Mathematics with Applicationsvol 62 no 3 pp 1341ndash1349 2011
[10] B Ahmad and S Sivasundaram ldquoOn four-point nonlocalboundary value problems of nonlinear integro-differentialequations of fractional orderrdquo Applied Mathematics and Com-putation vol 217 no 2 pp 480ndash487 2010
[11] N Nyamoradi ldquoPositive solutions for multi-point boundaryvalue problems for nonlinear fractional differential equationsrdquoJournal of Contemporary Mathematical Analysis vol 48 no 4pp 145ndash157 2013
[12] B Ahmad and J J Nieto ldquoExistence of solutions for nonlocalboundary value problems of higher-order nonlinear fractionaldifferential equationsrdquo Abstract and Applied Analysis vol 2009Article ID 494720 9 pages 2009
[13] X Liu M Jia and X Xiang ldquoOn the solvability of a fractionaldifferential equationmodel involving the119901-Laplacian operatorrdquoComputers ampMathematics with Applications vol 64 no 10 pp3267ndash3275 2012
[14] J Wang H Xiang and Z Liu ldquoExistence of concave positivesolutions for boundary value problem of nonlinear fractionaldifferential equation with 119901-Laplacian operatorrdquo InternationalJournal of Mathematics and Mathematical Sciences vol 2010Article ID 495138 17 pages 2010
[15] J Wang and H Xiang ldquoUpper and lower solutions method fora class of singular fractional boundary value problems with 119901-Laplacian operatorrdquo Abstract and Applied Analysis vol 2010Article ID 971824 12 pages 2010
[16] H Lu andZHan ldquoExistence of positive solutions for boundary-value problem of fractional differential equation with 119901-laplacian operatorrdquo The American Jouranl of Engineering andTechnology Research vol 11 pp 3757ndash3764 2011
[17] G Chai ldquoPositive solutions for boundary value problem offractional differential equation with 119901-Laplacian operatorrdquoBoundary Value Problems vol 2012 article 18 2012
[18] T Chen W Liu and Z Hu ldquoA boundary value problem forfractional differential equation with 119901-Laplacian operator atresonancerdquoNonlinear AnalysisTheoryMethodsampApplicationsvol 75 no 6 pp 3210ndash3217 2012
[19] T Chen andW Liu ldquoAn anti-periodic boundary value problemfor the fractional differential equation with a 119901-LaplacianoperatorrdquoAppliedMathematics Letters of Rapid Publication vol25 no 11 pp 1671ndash1675 2012
[20] S J Li X G Zhang Y H Wu and L Caccetta ldquoExtremalsolutions for 119901-Laplacian differential systems via iterative com-putationrdquo Applied Mathematics and Computation vol 26 pp1151ndash1158 2013
[21] Z Liu and L Lu ldquoA class of BVPs for nonlinear fractionaldifferential equations with 119901-Laplacian operatorrdquo ElectronicJournal of Qualitative Theory of Differential Equations vol 70pp 1ndash16 2012
[22] Z Liu L Lu and I Szanto ldquoExistence of solutions for fractionalimpulsive differential equations with 119901-Laplacian operatorrdquoActa Mathematica Hungarica vol 141 no 3 pp 203ndash219 2013
[23] Z Bai ldquoEigenvalue intervals for a class of fractional boundaryvalue problemrdquo Computers amp Mathematics with Applicationsvol 64 no 10 pp 3253ndash3257 2012
[24] X Zhang L Wang and Q Sun ldquoExistence of positive solutionsfor a class of nonlinear fractional differential equations withintegral boundary conditions and a parameterrdquo Applied Mathe-matics and Computation vol 226 pp 708ndash718 2014
18 Abstract and Applied Analysis
[25] D J Guo and V Lakshmikantham Nonlinear Problems inAbstract Cones vol 5 Academic Press New York NY USA1988
[26] M A Krasnoselrsquoskii Topological Methods in the Theory ofNonlinear Integral Equations Translated by A H ArmstrongElmsford Pergamon Turkey 1964
Submit your manuscripts athttpwwwhindawicom
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
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Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
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Differential EquationsInternational Journal of
Volume 2014
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Function Spaces
Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
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Algebra
Discrete Dynamics in Nature and Society
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
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Decision SciencesAdvances in
Discrete MathematicsJournal of
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Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Stochastic AnalysisInternational Journal of
14 Abstract and Applied Analysis
This implies that
Ψ119909 lt 119909 forall119909 isin 119875 cap 120597Ω1199033
(79)
Applying Lemma 7 to (72) (76) and (79) yields that Ψ hastwo fixed points 1199091 1199092 such that 1199091 isin 119875 cap (Ω119903
3
Ω1199031
) and1199092 isin 119875 cap (Ω119903
2
Ω1199033
)(ii) By the proof of (i) and the continuity of 119891(119905 119909) there
exists 119909 isin [1199031 1199032] such that min119909isin[11990311199032](119891(119905 119909)120601119901(119909)) =
119891(119905 119909)120601119901(119909)Let Υ = min119871lowast1 119871lowast2 119891(119905 119909)120601119901(119909) since (H6) holds
then Υ gt 0 and
119891 (119905 119909) ge Υ120601119901 (119909) forall119909 ge 0 119905 isin [0 1] (80)
Assuming 119909(119905) is a positive solution of problem (1) we willshow that this leads to contradiction for all 120582 gt 120582 where 120582 =
(1Υ)(Γ(120572)1205741120590)119901minus1 It follows form (26) that
119909 = 119909 (0)
=120578
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
119892 (119904)
times (int
1
0
(1 minus 120591)120572minus1
times 120601119902 (120582int
120591
0
120596 (120583) 119891 (120583 119909 (120583)) 119889120583)119889120591
minus int
119904
0
(119904 minus 120591)120572minus1
times120601119902 (120582int
120591
0
120596 (120583) 119891 (120583 119909 (120583)) 119889120583)119889120591)119889119904
+1
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902 (120582int
119904
0
120596 (120591) 119891 (120591 119909 (120591)) 119889120591) 119889119904
ge1
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902 (120582int
119904
0
120596 (120591) 119891 (120591 119909 (120591)) 119889120591) 119889119904
ge1
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902 (120582int
119904
0
120596 (120591) Υ120601119901 (119909 (120591)) 119889120591) 119889119904
ge1
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902 (120582int
119904
0
120596 (120591) Υ120601119901 (120590 119909 119889120591)) 119889119904
ge120601119902 (120582Υ) 120590 119909
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902 (int
119904
0
120596 (120591) 119889120591) 119889119904
=120601119902 (120582Υ) 1205741120590
Γ (120572)119909
(81)
This implies that 1 ge 120601119902(120582Υ)1205741120590Γ(120572) = (120582Υ)119902minus1
1205741120590Γ(120572) gt
(120582Υ)119902minus1
1205741120590Γ(120572) = 1 which is a contradiction This com-pletes the proof of (ii)
Corollary 15 Assume that (H1)ndash(H5) holds
(i) If lim119909rarrinfin
119891(119905 119909) = 119891infin isin (0 +infin] uniformly for 119905 isin
[0 1] and one of 1198910120601
= 0 and 119891infin
120601= 0 is satisfied then
there exists 120582lowast gt 0 such that the problem (1) has at leastone positive solution for any 120582 gt 120582
lowast(ii) If one of 1198910
120601= +infin and 119891
infin
120601= +infin holds then there
exists 120582lowast0 gt 0 such that problem (1) has at least onepositive solutions for any 120582 isin (0 120582lowast0)
Proof (i) The conclusion is a direct consequence of Theo-rem 11(i)
(ii) The conclusion is a direct consequence of Theo-rem 14(i)
Theorem 16 Assume that (H1)ndash(H5) holdThen the followingconclusions hold
(i) If 1198910120601
= 0 and 119891infin
120601= infin then the problem (1) has a
positive solution 119909120582(119905) for all 120582 gt 0(ii) If 1198910
120601= infin and 119891
infin
120601= 0 then the problem (1) has a
positive solution 119909120582(119905) for all 120582 gt 0
Proof Weonly prove (i) the proof of (ii) is similar sowe omitit here
Let 120582 gt 0 since 1198910
120601= 0 there exists 119903 gt 0 119897 gt 0 such that
119891 (119905 119909) le 119897120601119901 (119909) (forall119905 isin [0 1] 119909 isin [0 119903]) (82)
where 119897 satisfied
1205742120582119902minus1
119897119902minus1
Γ (120572 + 1) (1 minus 120578 int1
0119892 (119904) 119889119904)
le 1 (83)
Similar to the proof of Theorem 14 we obtain that
Ψ119909 le 119909 forall119909 isin 119875 cap 120597Ω119903 (84)
Abstract and Applied Analysis 15
From 119891infin
120601= infin there exists 119877 gt 119903 gt 0 119871 gt 0 such that
119891 (119905 119909) ge 119871120601119901 (119909) (forall119905 isin [0 1] 119909 ge 119877) (85)where 119871 satisfied
1205902120582119902minus1
119871119902minus1
1205741
Γ (120572)ge 1 (86)
We get thatΨ119909 ge 119909 forall119909 isin 119875 cap 120597Ω119877 (87)
Applying Lemma 7 to (84) (87) yields thatΨ has a fixed point119909 such that 119909 isin 119875 cap (Ω119877 Ω119903)
Theorem 17 Assume that (H1)ndash(H6) holdThen the followingconclusions hold
If 0 lt 1198910
120601lt infin and 0 lt 119891
infin
120601lt infin then there exists 1 2 gt
0 such that problem (1) has no positive solution for any 120582 gt 1
and 0 lt 120582 lt 2
Proof It follows from 0 lt 1198910
120601lt infin and 0 lt 119891
infin
120601lt infin that
there exist 1 gt 0 2 gt 0 1 gt 0 2 gt 0 and 1199032 gt 1199031 gt 0
such that1120601119901 (119909) le 119891 (119905 119909) le 1120601119901 (119909)
(forall119905 isin [0 1] 0 le 119909 le 1199031)
2120601119901 (119909) le 119891 (119905 119909) le 2120601119901 (119909)
(forall119905 isin [0 1] 119909 ge 1199032)
(88)
Let = max119891(119905 119909)120601119901(119909) 1199031 le 119909 le 1199032 0 le 119905 le 1 =
min119891(119905 119909)120601119901(119909) 1199031 le 119909 le 1199032 0 le 119905 le 1 By the condition(H6) gt 0 Then forall119909 ge 0 we have
120601119901 (119909) le 119891 (119905 119909) le 120601119901 (119909) (89)
where = max1 2 gt 0 and = min1198971 2 gt 0Let 1 = (1)(Γ(120572)1205741120590)
119901minus1 and 2 =
(1)(Γ(120572 + 1)(1 minus 120578 int1
0119892(119904)119889119904)1205742)
119901minus1
If problem (1)has positive solution 119909(119905) in 119875 then
119909 = 119909 (0)
=120578
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
119892 (119904)
times (int
1
0
(1 minus 120591)120572minus1
times 120601119902 (120582int
120591
0
120596 (120583) 119891 (120583 119909 (120583)) 119889120583)119889120591
minus int
119904
0
(119904 minus 120591)120572minus1
times120601119902 (120582int
120591
0
120596 (120583) 119891 (120583 119909 (120583)) 119889120583)119889120591)119889119904
+1
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902 (120582int
119904
0
120596 (120591) 119891 (120591 119909 (120591)) 119889120591) 119889119904
ge1
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902 (120582int
119904
0
120596 (120591) 119891 (120591 119909 (120591)) 119889120591) 119889119904
ge1
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902 (120582int
119904
0
120596 (120591) 120601119901 (119909 (120591)) 119889120591) 119889119904
ge1
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902 (120582int
119904
0
120596 (120591) 120601119901 (120590 119909 119889120591)) 119889119904
ge120601119902 (120582) 120590 119909
Γ (120572)
times int
1
0
(1 minus 119904)120572minus1
120601119902 (int
119904
0
120596 (120591) 119889120591) 119889119904
=120601119902 (120582) 1205741120590
Γ (120572)119909
gt120601119902 (1 ) 1205741120590
Γ (120572)119909 = 119909
119909 = 119909 (0)
=120578
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
119892 (119904)
times (int
1
0
(1 minus 120591)120572minus1
120601119902
times (120582int
120591
0
120596 (120583) 119891 (120583 119909 (120583)) 119889120583)119889120591
minus int
119904
0
(119904 minus 120591)120572minus1
120601119902
times(120582int
120591
0
120596 (120583) 119891 (120583 119909 (120583)) 119889120583)119889120591)119889119904
+1
Γ (120572)int
1
0
(1 minus 119904)120572minus1
120601119902
times (120582int
119904
0
120596 (120591) 119891 (120591 119909 (120591)) 119889120591) 119889119904
16 Abstract and Applied Analysis
le120578
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
119892 (119904)
times int
1
0
(1 minus 120591)120572minus1
times 120601119902 (120582int
120591
0
120596 (120583) 119891 (120583 119909 (120583)) 119889120583)119889120591 119889119904
+1
Γ (120572)int
1
0
(1 minus 119904)120572minus1
120601119902
times (120582int
119904
0
120596 (120591) 119891 (120591 119909 (120591)) 119889120591) 119889119904
=1
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
(1 minus 119904)120572minus1
120601119902
times (120582int
119904
0
120596 (120591) 119891 (120591 119909 (120591)) 119889120591) 119889119904
le1
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
(1 minus 119904)120572minus1
120601119902
times (120582int
1
0
120596 (120591) 120601119901 (119909 (120591)) 119889120591) 119889119904
=120601119902 (120582) 1205742
Γ (120572 + 1) (1 minus 120578 int1
0119892 (119904) 119889119904)
119909
lt120601119902 (2) 1205742
Γ (120572 + 1) (1 minus 120578 int1
0119892 (119904) 119889119904)
119909 = 119909
(90)
which is a contradiction This finishes the proof of Theo-rem 17
4 Examples
In this section we give out same examples to illustrate ourmain results can be used in practice
Example 1 Consider the following fractional differentialequation
(120601119901 (11986352
0+119909 (119905)))
1015840
+ 120582119905(1 + 1199052)12
119909119901minus1 arctan119909 = 0
0 lt 119905 lt 1
1199091015840(0) = 119909
10158401015840(0) = 0
119909 (1) =1
2int
1
0
1
2119909 (119904) 119889119904
(91)
where 120572 = 52 119901 = 32 119902 = 3 120596(119905) = 119905 119891(119905 119909) = (1 +
1199052)12
119909119901minus1 arctan119909 120578 = 12 and 119892(119905) = 12
By simple computation 119897lowast1= 1205872 119897
lowast
2= radic21205872 1205741 = int
1
0(1minus
119904)32
1206013(int119904
0120591 119889120591)119889119904 asymp 00043 1205742 = 1206013(int
1
0120591 119889120591) = 025 120590 asymp
01429 120582lowast
0= (1119897
lowast
1)(Γ(120572)1205901205741)
119901minus1asymp 25149 120582
lowast
0= (1119897
lowast
2)(Γ(120572+
1)(1 minus 120578 int1
0119892(119904)119889119904)1205742)
119901minus1asymp 14222 Then it is easy to see
all the conditions of Theorem 9(ii) are satisfied thus forall120582 isin
[14222 25149]] the problem (91) has at least one positivesolution
Example 2 Consider the boundary value problem
(120601119901 (11986352
0+119909 (119905)))
1015840
+ 120582119905(1 + 1199052)12
119909119901119890minus119909
= 0
0 lt 119905 lt 1
1199091015840(0) = 119909
10158401015840(0) = 0
119909 (1) =1
2int
1
0
1
2119909 (119904) 119889119904
(92)
where 120572 = 52 119901 = 32 119902 = 3 120596(119905) = 119905 119891(119905 119909) = (1 +
1199052)12
119909119901119890minus119909
120578 = 12 and 119892(119905) = 12
1198910
120601= lim119909rarr0
119891 (119905 119909)
120601119901 (119909)= lim119909rarr0
(1 + 1199052)12
119909119901119890minus119909
119909119901minus1= 0
forall119905 isin [0 1]
1198910
120601= lim119909rarrinfin
119891 (119905 119909)
120601119901 (119909)= lim119909rarrinfin
(1 + 1199052)12
119909119901119890minus119909
119909119901minus1= 0
forall119905 isin [0 1]
(93)
Let 1205881= 12 120588
2= 4 then
119891 (119905 119909)
120601119901 (119909)le (1 + 119905
2)12 1
2radic119890le
1
radic2119890
forall119905 isin [0 1] 119909 isin [01
2]
119891 (119905 119909)
120601119901 (119909)le (1 + 119905
2)12 4
1198904le
4radic2
1198904
forall119905 isin [0 1] 119909 isin [4infin)
(94)
so 1205761 = 1radic2119890 1205762 = 4radic21198904 and forall119909 isin
[12 4] max12le119909le4(119891(119905 119909)120601119901(119909)) = radic2119890minus1 We
have Υ = max1radic2119890 4radic21198904 radic2119890minus1
= radic2119890minus1 1205742 = 025
Abstract and Applied Analysis 17
and 120582 = (1Υ)(Γ(120572+1)(1minus120578 int1
0119892(119904)119889119904)1205742)
119901minus1= 3radic5120587
141198904
By Theorem 11(ii) for any 120582 lt 120582 = 3radic512058714
1198904 the problemhas no positive solution
5 Conclusions
Recently differential equations with p-Laplacian operatorwere widely discussed by several authors In this paperby using fixed point techniques combining with partiallyordered structure of Banach space we obtained the existencemultiply and the dependence of parameter These newresults we presented can be used in numerical computationand analyze mathematical models of physical phenomenamechanics nonlinear dynamics and many other relatedfields
Conflict of Interests
The authors declare that there is no conflict of interestsregarding the publication of this paper
Acknowledgments
The authors are very grateful to anonymous referees for theirvaluable and detailed suggestions and comments to improvethe original paper This work is supported by the Programfor New Century Excellent Talents in University (NCET-10-0097)
References
[1] I Podlubny Fractional Differential Equations vol 198 Aca-demic Press San Diego Calif USA 1999
[2] A A Kilbas H M Srivastava and J J Trujillo Theoryand Applications of Fractional Differential Equations vol 204of North-Holland Mathematics Studies Elsevier Science BVAmsterdam The Netherlands 2006
[3] Y A Rossikhin and M V Shitikova ldquoApplication of fractionalderivatives to the analysis of damped vibrations of viscoelasticsingle mass systemsrdquoActa Mechanica vol 120 no 1ndash4 pp 109ndash125 1997
[4] J Sabatier O Agrawal and J T Machado Advances in Frac-tional Calculus Theoretical developments and applications inphysics and engineering Springer Berlin Germany 2007
[5] N Engheta ldquoOn fractional calculus and fractional multipolesin electromagnetismrdquo IEEE Transactions on Antennas andPropagation vol 44 no 4 pp 554ndash566 1996
[6] R Hilfer Applications of Fractional Calculus in Physics WorldScientific Singapore 2000
[7] R L Bagley and R A Calico ldquoFractional order state equationsfor the control of viscoelastically damped structuresrdquo Journalof Guidance Control and Dynamics vol 14 no 2 pp 304ndash3111991
[8] W-X Zhou andY-D Chu ldquoExistence of solutions for fractionaldifferential equations with multi-point boundary conditionsrdquoCommunications in Nonlinear Science and Numerical Simula-tion vol 17 no 3 pp 1142ndash1148 2012
[9] B Ahmad and G Wang ldquoA study of an impulsive four-pointnonlocal boundary value problemof nonlinear fractional differ-ential equationsrdquo Computers amp Mathematics with Applicationsvol 62 no 3 pp 1341ndash1349 2011
[10] B Ahmad and S Sivasundaram ldquoOn four-point nonlocalboundary value problems of nonlinear integro-differentialequations of fractional orderrdquo Applied Mathematics and Com-putation vol 217 no 2 pp 480ndash487 2010
[11] N Nyamoradi ldquoPositive solutions for multi-point boundaryvalue problems for nonlinear fractional differential equationsrdquoJournal of Contemporary Mathematical Analysis vol 48 no 4pp 145ndash157 2013
[12] B Ahmad and J J Nieto ldquoExistence of solutions for nonlocalboundary value problems of higher-order nonlinear fractionaldifferential equationsrdquo Abstract and Applied Analysis vol 2009Article ID 494720 9 pages 2009
[13] X Liu M Jia and X Xiang ldquoOn the solvability of a fractionaldifferential equationmodel involving the119901-Laplacian operatorrdquoComputers ampMathematics with Applications vol 64 no 10 pp3267ndash3275 2012
[14] J Wang H Xiang and Z Liu ldquoExistence of concave positivesolutions for boundary value problem of nonlinear fractionaldifferential equation with 119901-Laplacian operatorrdquo InternationalJournal of Mathematics and Mathematical Sciences vol 2010Article ID 495138 17 pages 2010
[15] J Wang and H Xiang ldquoUpper and lower solutions method fora class of singular fractional boundary value problems with 119901-Laplacian operatorrdquo Abstract and Applied Analysis vol 2010Article ID 971824 12 pages 2010
[16] H Lu andZHan ldquoExistence of positive solutions for boundary-value problem of fractional differential equation with 119901-laplacian operatorrdquo The American Jouranl of Engineering andTechnology Research vol 11 pp 3757ndash3764 2011
[17] G Chai ldquoPositive solutions for boundary value problem offractional differential equation with 119901-Laplacian operatorrdquoBoundary Value Problems vol 2012 article 18 2012
[18] T Chen W Liu and Z Hu ldquoA boundary value problem forfractional differential equation with 119901-Laplacian operator atresonancerdquoNonlinear AnalysisTheoryMethodsampApplicationsvol 75 no 6 pp 3210ndash3217 2012
[19] T Chen andW Liu ldquoAn anti-periodic boundary value problemfor the fractional differential equation with a 119901-LaplacianoperatorrdquoAppliedMathematics Letters of Rapid Publication vol25 no 11 pp 1671ndash1675 2012
[20] S J Li X G Zhang Y H Wu and L Caccetta ldquoExtremalsolutions for 119901-Laplacian differential systems via iterative com-putationrdquo Applied Mathematics and Computation vol 26 pp1151ndash1158 2013
[21] Z Liu and L Lu ldquoA class of BVPs for nonlinear fractionaldifferential equations with 119901-Laplacian operatorrdquo ElectronicJournal of Qualitative Theory of Differential Equations vol 70pp 1ndash16 2012
[22] Z Liu L Lu and I Szanto ldquoExistence of solutions for fractionalimpulsive differential equations with 119901-Laplacian operatorrdquoActa Mathematica Hungarica vol 141 no 3 pp 203ndash219 2013
[23] Z Bai ldquoEigenvalue intervals for a class of fractional boundaryvalue problemrdquo Computers amp Mathematics with Applicationsvol 64 no 10 pp 3253ndash3257 2012
[24] X Zhang L Wang and Q Sun ldquoExistence of positive solutionsfor a class of nonlinear fractional differential equations withintegral boundary conditions and a parameterrdquo Applied Mathe-matics and Computation vol 226 pp 708ndash718 2014
18 Abstract and Applied Analysis
[25] D J Guo and V Lakshmikantham Nonlinear Problems inAbstract Cones vol 5 Academic Press New York NY USA1988
[26] M A Krasnoselrsquoskii Topological Methods in the Theory ofNonlinear Integral Equations Translated by A H ArmstrongElmsford Pergamon Turkey 1964
Submit your manuscripts athttpwwwhindawicom
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
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Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical Problems in Engineering
Hindawi Publishing Corporationhttpwwwhindawicom
Differential EquationsInternational Journal of
Volume 2014
Applied MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical PhysicsAdvances in
Complex AnalysisJournal of
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OptimizationJournal of
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CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of
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Operations ResearchAdvances in
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Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Function Spaces
Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of Mathematics and Mathematical Sciences
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Algebra
Discrete Dynamics in Nature and Society
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Decision SciencesAdvances in
Discrete MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom
Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Stochastic AnalysisInternational Journal of
Abstract and Applied Analysis 15
From 119891infin
120601= infin there exists 119877 gt 119903 gt 0 119871 gt 0 such that
119891 (119905 119909) ge 119871120601119901 (119909) (forall119905 isin [0 1] 119909 ge 119877) (85)where 119871 satisfied
1205902120582119902minus1
119871119902minus1
1205741
Γ (120572)ge 1 (86)
We get thatΨ119909 ge 119909 forall119909 isin 119875 cap 120597Ω119877 (87)
Applying Lemma 7 to (84) (87) yields thatΨ has a fixed point119909 such that 119909 isin 119875 cap (Ω119877 Ω119903)
Theorem 17 Assume that (H1)ndash(H6) holdThen the followingconclusions hold
If 0 lt 1198910
120601lt infin and 0 lt 119891
infin
120601lt infin then there exists 1 2 gt
0 such that problem (1) has no positive solution for any 120582 gt 1
and 0 lt 120582 lt 2
Proof It follows from 0 lt 1198910
120601lt infin and 0 lt 119891
infin
120601lt infin that
there exist 1 gt 0 2 gt 0 1 gt 0 2 gt 0 and 1199032 gt 1199031 gt 0
such that1120601119901 (119909) le 119891 (119905 119909) le 1120601119901 (119909)
(forall119905 isin [0 1] 0 le 119909 le 1199031)
2120601119901 (119909) le 119891 (119905 119909) le 2120601119901 (119909)
(forall119905 isin [0 1] 119909 ge 1199032)
(88)
Let = max119891(119905 119909)120601119901(119909) 1199031 le 119909 le 1199032 0 le 119905 le 1 =
min119891(119905 119909)120601119901(119909) 1199031 le 119909 le 1199032 0 le 119905 le 1 By the condition(H6) gt 0 Then forall119909 ge 0 we have
120601119901 (119909) le 119891 (119905 119909) le 120601119901 (119909) (89)
where = max1 2 gt 0 and = min1198971 2 gt 0Let 1 = (1)(Γ(120572)1205741120590)
119901minus1 and 2 =
(1)(Γ(120572 + 1)(1 minus 120578 int1
0119892(119904)119889119904)1205742)
119901minus1
If problem (1)has positive solution 119909(119905) in 119875 then
119909 = 119909 (0)
=120578
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
119892 (119904)
times (int
1
0
(1 minus 120591)120572minus1
times 120601119902 (120582int
120591
0
120596 (120583) 119891 (120583 119909 (120583)) 119889120583)119889120591
minus int
119904
0
(119904 minus 120591)120572minus1
times120601119902 (120582int
120591
0
120596 (120583) 119891 (120583 119909 (120583)) 119889120583)119889120591)119889119904
+1
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902 (120582int
119904
0
120596 (120591) 119891 (120591 119909 (120591)) 119889120591) 119889119904
ge1
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902 (120582int
119904
0
120596 (120591) 119891 (120591 119909 (120591)) 119889120591) 119889119904
ge1
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902 (120582int
119904
0
120596 (120591) 120601119901 (119909 (120591)) 119889120591) 119889119904
ge1
Γ (120572)int
1
0
(1 minus 119904)120572minus1
times 120601119902 (120582int
119904
0
120596 (120591) 120601119901 (120590 119909 119889120591)) 119889119904
ge120601119902 (120582) 120590 119909
Γ (120572)
times int
1
0
(1 minus 119904)120572minus1
120601119902 (int
119904
0
120596 (120591) 119889120591) 119889119904
=120601119902 (120582) 1205741120590
Γ (120572)119909
gt120601119902 (1 ) 1205741120590
Γ (120572)119909 = 119909
119909 = 119909 (0)
=120578
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
119892 (119904)
times (int
1
0
(1 minus 120591)120572minus1
120601119902
times (120582int
120591
0
120596 (120583) 119891 (120583 119909 (120583)) 119889120583)119889120591
minus int
119904
0
(119904 minus 120591)120572minus1
120601119902
times(120582int
120591
0
120596 (120583) 119891 (120583 119909 (120583)) 119889120583)119889120591)119889119904
+1
Γ (120572)int
1
0
(1 minus 119904)120572minus1
120601119902
times (120582int
119904
0
120596 (120591) 119891 (120591 119909 (120591)) 119889120591) 119889119904
16 Abstract and Applied Analysis
le120578
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
119892 (119904)
times int
1
0
(1 minus 120591)120572minus1
times 120601119902 (120582int
120591
0
120596 (120583) 119891 (120583 119909 (120583)) 119889120583)119889120591 119889119904
+1
Γ (120572)int
1
0
(1 minus 119904)120572minus1
120601119902
times (120582int
119904
0
120596 (120591) 119891 (120591 119909 (120591)) 119889120591) 119889119904
=1
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
(1 minus 119904)120572minus1
120601119902
times (120582int
119904
0
120596 (120591) 119891 (120591 119909 (120591)) 119889120591) 119889119904
le1
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
(1 minus 119904)120572minus1
120601119902
times (120582int
1
0
120596 (120591) 120601119901 (119909 (120591)) 119889120591) 119889119904
=120601119902 (120582) 1205742
Γ (120572 + 1) (1 minus 120578 int1
0119892 (119904) 119889119904)
119909
lt120601119902 (2) 1205742
Γ (120572 + 1) (1 minus 120578 int1
0119892 (119904) 119889119904)
119909 = 119909
(90)
which is a contradiction This finishes the proof of Theo-rem 17
4 Examples
In this section we give out same examples to illustrate ourmain results can be used in practice
Example 1 Consider the following fractional differentialequation
(120601119901 (11986352
0+119909 (119905)))
1015840
+ 120582119905(1 + 1199052)12
119909119901minus1 arctan119909 = 0
0 lt 119905 lt 1
1199091015840(0) = 119909
10158401015840(0) = 0
119909 (1) =1
2int
1
0
1
2119909 (119904) 119889119904
(91)
where 120572 = 52 119901 = 32 119902 = 3 120596(119905) = 119905 119891(119905 119909) = (1 +
1199052)12
119909119901minus1 arctan119909 120578 = 12 and 119892(119905) = 12
By simple computation 119897lowast1= 1205872 119897
lowast
2= radic21205872 1205741 = int
1
0(1minus
119904)32
1206013(int119904
0120591 119889120591)119889119904 asymp 00043 1205742 = 1206013(int
1
0120591 119889120591) = 025 120590 asymp
01429 120582lowast
0= (1119897
lowast
1)(Γ(120572)1205901205741)
119901minus1asymp 25149 120582
lowast
0= (1119897
lowast
2)(Γ(120572+
1)(1 minus 120578 int1
0119892(119904)119889119904)1205742)
119901minus1asymp 14222 Then it is easy to see
all the conditions of Theorem 9(ii) are satisfied thus forall120582 isin
[14222 25149]] the problem (91) has at least one positivesolution
Example 2 Consider the boundary value problem
(120601119901 (11986352
0+119909 (119905)))
1015840
+ 120582119905(1 + 1199052)12
119909119901119890minus119909
= 0
0 lt 119905 lt 1
1199091015840(0) = 119909
10158401015840(0) = 0
119909 (1) =1
2int
1
0
1
2119909 (119904) 119889119904
(92)
where 120572 = 52 119901 = 32 119902 = 3 120596(119905) = 119905 119891(119905 119909) = (1 +
1199052)12
119909119901119890minus119909
120578 = 12 and 119892(119905) = 12
1198910
120601= lim119909rarr0
119891 (119905 119909)
120601119901 (119909)= lim119909rarr0
(1 + 1199052)12
119909119901119890minus119909
119909119901minus1= 0
forall119905 isin [0 1]
1198910
120601= lim119909rarrinfin
119891 (119905 119909)
120601119901 (119909)= lim119909rarrinfin
(1 + 1199052)12
119909119901119890minus119909
119909119901minus1= 0
forall119905 isin [0 1]
(93)
Let 1205881= 12 120588
2= 4 then
119891 (119905 119909)
120601119901 (119909)le (1 + 119905
2)12 1
2radic119890le
1
radic2119890
forall119905 isin [0 1] 119909 isin [01
2]
119891 (119905 119909)
120601119901 (119909)le (1 + 119905
2)12 4
1198904le
4radic2
1198904
forall119905 isin [0 1] 119909 isin [4infin)
(94)
so 1205761 = 1radic2119890 1205762 = 4radic21198904 and forall119909 isin
[12 4] max12le119909le4(119891(119905 119909)120601119901(119909)) = radic2119890minus1 We
have Υ = max1radic2119890 4radic21198904 radic2119890minus1
= radic2119890minus1 1205742 = 025
Abstract and Applied Analysis 17
and 120582 = (1Υ)(Γ(120572+1)(1minus120578 int1
0119892(119904)119889119904)1205742)
119901minus1= 3radic5120587
141198904
By Theorem 11(ii) for any 120582 lt 120582 = 3radic512058714
1198904 the problemhas no positive solution
5 Conclusions
Recently differential equations with p-Laplacian operatorwere widely discussed by several authors In this paperby using fixed point techniques combining with partiallyordered structure of Banach space we obtained the existencemultiply and the dependence of parameter These newresults we presented can be used in numerical computationand analyze mathematical models of physical phenomenamechanics nonlinear dynamics and many other relatedfields
Conflict of Interests
The authors declare that there is no conflict of interestsregarding the publication of this paper
Acknowledgments
The authors are very grateful to anonymous referees for theirvaluable and detailed suggestions and comments to improvethe original paper This work is supported by the Programfor New Century Excellent Talents in University (NCET-10-0097)
References
[1] I Podlubny Fractional Differential Equations vol 198 Aca-demic Press San Diego Calif USA 1999
[2] A A Kilbas H M Srivastava and J J Trujillo Theoryand Applications of Fractional Differential Equations vol 204of North-Holland Mathematics Studies Elsevier Science BVAmsterdam The Netherlands 2006
[3] Y A Rossikhin and M V Shitikova ldquoApplication of fractionalderivatives to the analysis of damped vibrations of viscoelasticsingle mass systemsrdquoActa Mechanica vol 120 no 1ndash4 pp 109ndash125 1997
[4] J Sabatier O Agrawal and J T Machado Advances in Frac-tional Calculus Theoretical developments and applications inphysics and engineering Springer Berlin Germany 2007
[5] N Engheta ldquoOn fractional calculus and fractional multipolesin electromagnetismrdquo IEEE Transactions on Antennas andPropagation vol 44 no 4 pp 554ndash566 1996
[6] R Hilfer Applications of Fractional Calculus in Physics WorldScientific Singapore 2000
[7] R L Bagley and R A Calico ldquoFractional order state equationsfor the control of viscoelastically damped structuresrdquo Journalof Guidance Control and Dynamics vol 14 no 2 pp 304ndash3111991
[8] W-X Zhou andY-D Chu ldquoExistence of solutions for fractionaldifferential equations with multi-point boundary conditionsrdquoCommunications in Nonlinear Science and Numerical Simula-tion vol 17 no 3 pp 1142ndash1148 2012
[9] B Ahmad and G Wang ldquoA study of an impulsive four-pointnonlocal boundary value problemof nonlinear fractional differ-ential equationsrdquo Computers amp Mathematics with Applicationsvol 62 no 3 pp 1341ndash1349 2011
[10] B Ahmad and S Sivasundaram ldquoOn four-point nonlocalboundary value problems of nonlinear integro-differentialequations of fractional orderrdquo Applied Mathematics and Com-putation vol 217 no 2 pp 480ndash487 2010
[11] N Nyamoradi ldquoPositive solutions for multi-point boundaryvalue problems for nonlinear fractional differential equationsrdquoJournal of Contemporary Mathematical Analysis vol 48 no 4pp 145ndash157 2013
[12] B Ahmad and J J Nieto ldquoExistence of solutions for nonlocalboundary value problems of higher-order nonlinear fractionaldifferential equationsrdquo Abstract and Applied Analysis vol 2009Article ID 494720 9 pages 2009
[13] X Liu M Jia and X Xiang ldquoOn the solvability of a fractionaldifferential equationmodel involving the119901-Laplacian operatorrdquoComputers ampMathematics with Applications vol 64 no 10 pp3267ndash3275 2012
[14] J Wang H Xiang and Z Liu ldquoExistence of concave positivesolutions for boundary value problem of nonlinear fractionaldifferential equation with 119901-Laplacian operatorrdquo InternationalJournal of Mathematics and Mathematical Sciences vol 2010Article ID 495138 17 pages 2010
[15] J Wang and H Xiang ldquoUpper and lower solutions method fora class of singular fractional boundary value problems with 119901-Laplacian operatorrdquo Abstract and Applied Analysis vol 2010Article ID 971824 12 pages 2010
[16] H Lu andZHan ldquoExistence of positive solutions for boundary-value problem of fractional differential equation with 119901-laplacian operatorrdquo The American Jouranl of Engineering andTechnology Research vol 11 pp 3757ndash3764 2011
[17] G Chai ldquoPositive solutions for boundary value problem offractional differential equation with 119901-Laplacian operatorrdquoBoundary Value Problems vol 2012 article 18 2012
[18] T Chen W Liu and Z Hu ldquoA boundary value problem forfractional differential equation with 119901-Laplacian operator atresonancerdquoNonlinear AnalysisTheoryMethodsampApplicationsvol 75 no 6 pp 3210ndash3217 2012
[19] T Chen andW Liu ldquoAn anti-periodic boundary value problemfor the fractional differential equation with a 119901-LaplacianoperatorrdquoAppliedMathematics Letters of Rapid Publication vol25 no 11 pp 1671ndash1675 2012
[20] S J Li X G Zhang Y H Wu and L Caccetta ldquoExtremalsolutions for 119901-Laplacian differential systems via iterative com-putationrdquo Applied Mathematics and Computation vol 26 pp1151ndash1158 2013
[21] Z Liu and L Lu ldquoA class of BVPs for nonlinear fractionaldifferential equations with 119901-Laplacian operatorrdquo ElectronicJournal of Qualitative Theory of Differential Equations vol 70pp 1ndash16 2012
[22] Z Liu L Lu and I Szanto ldquoExistence of solutions for fractionalimpulsive differential equations with 119901-Laplacian operatorrdquoActa Mathematica Hungarica vol 141 no 3 pp 203ndash219 2013
[23] Z Bai ldquoEigenvalue intervals for a class of fractional boundaryvalue problemrdquo Computers amp Mathematics with Applicationsvol 64 no 10 pp 3253ndash3257 2012
[24] X Zhang L Wang and Q Sun ldquoExistence of positive solutionsfor a class of nonlinear fractional differential equations withintegral boundary conditions and a parameterrdquo Applied Mathe-matics and Computation vol 226 pp 708ndash718 2014
18 Abstract and Applied Analysis
[25] D J Guo and V Lakshmikantham Nonlinear Problems inAbstract Cones vol 5 Academic Press New York NY USA1988
[26] M A Krasnoselrsquoskii Topological Methods in the Theory ofNonlinear Integral Equations Translated by A H ArmstrongElmsford Pergamon Turkey 1964
Submit your manuscripts athttpwwwhindawicom
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical Problems in Engineering
Hindawi Publishing Corporationhttpwwwhindawicom
Differential EquationsInternational Journal of
Volume 2014
Applied MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical PhysicsAdvances in
Complex AnalysisJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
OptimizationJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Operations ResearchAdvances in
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Function Spaces
Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of Mathematics and Mathematical Sciences
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Algebra
Discrete Dynamics in Nature and Society
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Decision SciencesAdvances in
Discrete MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom
Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Stochastic AnalysisInternational Journal of
16 Abstract and Applied Analysis
le120578
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
119892 (119904)
times int
1
0
(1 minus 120591)120572minus1
times 120601119902 (120582int
120591
0
120596 (120583) 119891 (120583 119909 (120583)) 119889120583)119889120591 119889119904
+1
Γ (120572)int
1
0
(1 minus 119904)120572minus1
120601119902
times (120582int
119904
0
120596 (120591) 119891 (120591 119909 (120591)) 119889120591) 119889119904
=1
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
(1 minus 119904)120572minus1
120601119902
times (120582int
119904
0
120596 (120591) 119891 (120591 119909 (120591)) 119889120591) 119889119904
le1
Γ (120572) (1 minus 120578 int1
0119892 (119904) 119889119904)
times int
1
0
(1 minus 119904)120572minus1
120601119902
times (120582int
1
0
120596 (120591) 120601119901 (119909 (120591)) 119889120591) 119889119904
=120601119902 (120582) 1205742
Γ (120572 + 1) (1 minus 120578 int1
0119892 (119904) 119889119904)
119909
lt120601119902 (2) 1205742
Γ (120572 + 1) (1 minus 120578 int1
0119892 (119904) 119889119904)
119909 = 119909
(90)
which is a contradiction This finishes the proof of Theo-rem 17
4 Examples
In this section we give out same examples to illustrate ourmain results can be used in practice
Example 1 Consider the following fractional differentialequation
(120601119901 (11986352
0+119909 (119905)))
1015840
+ 120582119905(1 + 1199052)12
119909119901minus1 arctan119909 = 0
0 lt 119905 lt 1
1199091015840(0) = 119909
10158401015840(0) = 0
119909 (1) =1
2int
1
0
1
2119909 (119904) 119889119904
(91)
where 120572 = 52 119901 = 32 119902 = 3 120596(119905) = 119905 119891(119905 119909) = (1 +
1199052)12
119909119901minus1 arctan119909 120578 = 12 and 119892(119905) = 12
By simple computation 119897lowast1= 1205872 119897
lowast
2= radic21205872 1205741 = int
1
0(1minus
119904)32
1206013(int119904
0120591 119889120591)119889119904 asymp 00043 1205742 = 1206013(int
1
0120591 119889120591) = 025 120590 asymp
01429 120582lowast
0= (1119897
lowast
1)(Γ(120572)1205901205741)
119901minus1asymp 25149 120582
lowast
0= (1119897
lowast
2)(Γ(120572+
1)(1 minus 120578 int1
0119892(119904)119889119904)1205742)
119901minus1asymp 14222 Then it is easy to see
all the conditions of Theorem 9(ii) are satisfied thus forall120582 isin
[14222 25149]] the problem (91) has at least one positivesolution
Example 2 Consider the boundary value problem
(120601119901 (11986352
0+119909 (119905)))
1015840
+ 120582119905(1 + 1199052)12
119909119901119890minus119909
= 0
0 lt 119905 lt 1
1199091015840(0) = 119909
10158401015840(0) = 0
119909 (1) =1
2int
1
0
1
2119909 (119904) 119889119904
(92)
where 120572 = 52 119901 = 32 119902 = 3 120596(119905) = 119905 119891(119905 119909) = (1 +
1199052)12
119909119901119890minus119909
120578 = 12 and 119892(119905) = 12
1198910
120601= lim119909rarr0
119891 (119905 119909)
120601119901 (119909)= lim119909rarr0
(1 + 1199052)12
119909119901119890minus119909
119909119901minus1= 0
forall119905 isin [0 1]
1198910
120601= lim119909rarrinfin
119891 (119905 119909)
120601119901 (119909)= lim119909rarrinfin
(1 + 1199052)12
119909119901119890minus119909
119909119901minus1= 0
forall119905 isin [0 1]
(93)
Let 1205881= 12 120588
2= 4 then
119891 (119905 119909)
120601119901 (119909)le (1 + 119905
2)12 1
2radic119890le
1
radic2119890
forall119905 isin [0 1] 119909 isin [01
2]
119891 (119905 119909)
120601119901 (119909)le (1 + 119905
2)12 4
1198904le
4radic2
1198904
forall119905 isin [0 1] 119909 isin [4infin)
(94)
so 1205761 = 1radic2119890 1205762 = 4radic21198904 and forall119909 isin
[12 4] max12le119909le4(119891(119905 119909)120601119901(119909)) = radic2119890minus1 We
have Υ = max1radic2119890 4radic21198904 radic2119890minus1
= radic2119890minus1 1205742 = 025
Abstract and Applied Analysis 17
and 120582 = (1Υ)(Γ(120572+1)(1minus120578 int1
0119892(119904)119889119904)1205742)
119901minus1= 3radic5120587
141198904
By Theorem 11(ii) for any 120582 lt 120582 = 3radic512058714
1198904 the problemhas no positive solution
5 Conclusions
Recently differential equations with p-Laplacian operatorwere widely discussed by several authors In this paperby using fixed point techniques combining with partiallyordered structure of Banach space we obtained the existencemultiply and the dependence of parameter These newresults we presented can be used in numerical computationand analyze mathematical models of physical phenomenamechanics nonlinear dynamics and many other relatedfields
Conflict of Interests
The authors declare that there is no conflict of interestsregarding the publication of this paper
Acknowledgments
The authors are very grateful to anonymous referees for theirvaluable and detailed suggestions and comments to improvethe original paper This work is supported by the Programfor New Century Excellent Talents in University (NCET-10-0097)
References
[1] I Podlubny Fractional Differential Equations vol 198 Aca-demic Press San Diego Calif USA 1999
[2] A A Kilbas H M Srivastava and J J Trujillo Theoryand Applications of Fractional Differential Equations vol 204of North-Holland Mathematics Studies Elsevier Science BVAmsterdam The Netherlands 2006
[3] Y A Rossikhin and M V Shitikova ldquoApplication of fractionalderivatives to the analysis of damped vibrations of viscoelasticsingle mass systemsrdquoActa Mechanica vol 120 no 1ndash4 pp 109ndash125 1997
[4] J Sabatier O Agrawal and J T Machado Advances in Frac-tional Calculus Theoretical developments and applications inphysics and engineering Springer Berlin Germany 2007
[5] N Engheta ldquoOn fractional calculus and fractional multipolesin electromagnetismrdquo IEEE Transactions on Antennas andPropagation vol 44 no 4 pp 554ndash566 1996
[6] R Hilfer Applications of Fractional Calculus in Physics WorldScientific Singapore 2000
[7] R L Bagley and R A Calico ldquoFractional order state equationsfor the control of viscoelastically damped structuresrdquo Journalof Guidance Control and Dynamics vol 14 no 2 pp 304ndash3111991
[8] W-X Zhou andY-D Chu ldquoExistence of solutions for fractionaldifferential equations with multi-point boundary conditionsrdquoCommunications in Nonlinear Science and Numerical Simula-tion vol 17 no 3 pp 1142ndash1148 2012
[9] B Ahmad and G Wang ldquoA study of an impulsive four-pointnonlocal boundary value problemof nonlinear fractional differ-ential equationsrdquo Computers amp Mathematics with Applicationsvol 62 no 3 pp 1341ndash1349 2011
[10] B Ahmad and S Sivasundaram ldquoOn four-point nonlocalboundary value problems of nonlinear integro-differentialequations of fractional orderrdquo Applied Mathematics and Com-putation vol 217 no 2 pp 480ndash487 2010
[11] N Nyamoradi ldquoPositive solutions for multi-point boundaryvalue problems for nonlinear fractional differential equationsrdquoJournal of Contemporary Mathematical Analysis vol 48 no 4pp 145ndash157 2013
[12] B Ahmad and J J Nieto ldquoExistence of solutions for nonlocalboundary value problems of higher-order nonlinear fractionaldifferential equationsrdquo Abstract and Applied Analysis vol 2009Article ID 494720 9 pages 2009
[13] X Liu M Jia and X Xiang ldquoOn the solvability of a fractionaldifferential equationmodel involving the119901-Laplacian operatorrdquoComputers ampMathematics with Applications vol 64 no 10 pp3267ndash3275 2012
[14] J Wang H Xiang and Z Liu ldquoExistence of concave positivesolutions for boundary value problem of nonlinear fractionaldifferential equation with 119901-Laplacian operatorrdquo InternationalJournal of Mathematics and Mathematical Sciences vol 2010Article ID 495138 17 pages 2010
[15] J Wang and H Xiang ldquoUpper and lower solutions method fora class of singular fractional boundary value problems with 119901-Laplacian operatorrdquo Abstract and Applied Analysis vol 2010Article ID 971824 12 pages 2010
[16] H Lu andZHan ldquoExistence of positive solutions for boundary-value problem of fractional differential equation with 119901-laplacian operatorrdquo The American Jouranl of Engineering andTechnology Research vol 11 pp 3757ndash3764 2011
[17] G Chai ldquoPositive solutions for boundary value problem offractional differential equation with 119901-Laplacian operatorrdquoBoundary Value Problems vol 2012 article 18 2012
[18] T Chen W Liu and Z Hu ldquoA boundary value problem forfractional differential equation with 119901-Laplacian operator atresonancerdquoNonlinear AnalysisTheoryMethodsampApplicationsvol 75 no 6 pp 3210ndash3217 2012
[19] T Chen andW Liu ldquoAn anti-periodic boundary value problemfor the fractional differential equation with a 119901-LaplacianoperatorrdquoAppliedMathematics Letters of Rapid Publication vol25 no 11 pp 1671ndash1675 2012
[20] S J Li X G Zhang Y H Wu and L Caccetta ldquoExtremalsolutions for 119901-Laplacian differential systems via iterative com-putationrdquo Applied Mathematics and Computation vol 26 pp1151ndash1158 2013
[21] Z Liu and L Lu ldquoA class of BVPs for nonlinear fractionaldifferential equations with 119901-Laplacian operatorrdquo ElectronicJournal of Qualitative Theory of Differential Equations vol 70pp 1ndash16 2012
[22] Z Liu L Lu and I Szanto ldquoExistence of solutions for fractionalimpulsive differential equations with 119901-Laplacian operatorrdquoActa Mathematica Hungarica vol 141 no 3 pp 203ndash219 2013
[23] Z Bai ldquoEigenvalue intervals for a class of fractional boundaryvalue problemrdquo Computers amp Mathematics with Applicationsvol 64 no 10 pp 3253ndash3257 2012
[24] X Zhang L Wang and Q Sun ldquoExistence of positive solutionsfor a class of nonlinear fractional differential equations withintegral boundary conditions and a parameterrdquo Applied Mathe-matics and Computation vol 226 pp 708ndash718 2014
18 Abstract and Applied Analysis
[25] D J Guo and V Lakshmikantham Nonlinear Problems inAbstract Cones vol 5 Academic Press New York NY USA1988
[26] M A Krasnoselrsquoskii Topological Methods in the Theory ofNonlinear Integral Equations Translated by A H ArmstrongElmsford Pergamon Turkey 1964
Submit your manuscripts athttpwwwhindawicom
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical Problems in Engineering
Hindawi Publishing Corporationhttpwwwhindawicom
Differential EquationsInternational Journal of
Volume 2014
Applied MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical PhysicsAdvances in
Complex AnalysisJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
OptimizationJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Operations ResearchAdvances in
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Function Spaces
Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of Mathematics and Mathematical Sciences
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Algebra
Discrete Dynamics in Nature and Society
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Decision SciencesAdvances in
Discrete MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom
Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Stochastic AnalysisInternational Journal of
Abstract and Applied Analysis 17
and 120582 = (1Υ)(Γ(120572+1)(1minus120578 int1
0119892(119904)119889119904)1205742)
119901minus1= 3radic5120587
141198904
By Theorem 11(ii) for any 120582 lt 120582 = 3radic512058714
1198904 the problemhas no positive solution
5 Conclusions
Recently differential equations with p-Laplacian operatorwere widely discussed by several authors In this paperby using fixed point techniques combining with partiallyordered structure of Banach space we obtained the existencemultiply and the dependence of parameter These newresults we presented can be used in numerical computationand analyze mathematical models of physical phenomenamechanics nonlinear dynamics and many other relatedfields
Conflict of Interests
The authors declare that there is no conflict of interestsregarding the publication of this paper
Acknowledgments
The authors are very grateful to anonymous referees for theirvaluable and detailed suggestions and comments to improvethe original paper This work is supported by the Programfor New Century Excellent Talents in University (NCET-10-0097)
References
[1] I Podlubny Fractional Differential Equations vol 198 Aca-demic Press San Diego Calif USA 1999
[2] A A Kilbas H M Srivastava and J J Trujillo Theoryand Applications of Fractional Differential Equations vol 204of North-Holland Mathematics Studies Elsevier Science BVAmsterdam The Netherlands 2006
[3] Y A Rossikhin and M V Shitikova ldquoApplication of fractionalderivatives to the analysis of damped vibrations of viscoelasticsingle mass systemsrdquoActa Mechanica vol 120 no 1ndash4 pp 109ndash125 1997
[4] J Sabatier O Agrawal and J T Machado Advances in Frac-tional Calculus Theoretical developments and applications inphysics and engineering Springer Berlin Germany 2007
[5] N Engheta ldquoOn fractional calculus and fractional multipolesin electromagnetismrdquo IEEE Transactions on Antennas andPropagation vol 44 no 4 pp 554ndash566 1996
[6] R Hilfer Applications of Fractional Calculus in Physics WorldScientific Singapore 2000
[7] R L Bagley and R A Calico ldquoFractional order state equationsfor the control of viscoelastically damped structuresrdquo Journalof Guidance Control and Dynamics vol 14 no 2 pp 304ndash3111991
[8] W-X Zhou andY-D Chu ldquoExistence of solutions for fractionaldifferential equations with multi-point boundary conditionsrdquoCommunications in Nonlinear Science and Numerical Simula-tion vol 17 no 3 pp 1142ndash1148 2012
[9] B Ahmad and G Wang ldquoA study of an impulsive four-pointnonlocal boundary value problemof nonlinear fractional differ-ential equationsrdquo Computers amp Mathematics with Applicationsvol 62 no 3 pp 1341ndash1349 2011
[10] B Ahmad and S Sivasundaram ldquoOn four-point nonlocalboundary value problems of nonlinear integro-differentialequations of fractional orderrdquo Applied Mathematics and Com-putation vol 217 no 2 pp 480ndash487 2010
[11] N Nyamoradi ldquoPositive solutions for multi-point boundaryvalue problems for nonlinear fractional differential equationsrdquoJournal of Contemporary Mathematical Analysis vol 48 no 4pp 145ndash157 2013
[12] B Ahmad and J J Nieto ldquoExistence of solutions for nonlocalboundary value problems of higher-order nonlinear fractionaldifferential equationsrdquo Abstract and Applied Analysis vol 2009Article ID 494720 9 pages 2009
[13] X Liu M Jia and X Xiang ldquoOn the solvability of a fractionaldifferential equationmodel involving the119901-Laplacian operatorrdquoComputers ampMathematics with Applications vol 64 no 10 pp3267ndash3275 2012
[14] J Wang H Xiang and Z Liu ldquoExistence of concave positivesolutions for boundary value problem of nonlinear fractionaldifferential equation with 119901-Laplacian operatorrdquo InternationalJournal of Mathematics and Mathematical Sciences vol 2010Article ID 495138 17 pages 2010
[15] J Wang and H Xiang ldquoUpper and lower solutions method fora class of singular fractional boundary value problems with 119901-Laplacian operatorrdquo Abstract and Applied Analysis vol 2010Article ID 971824 12 pages 2010
[16] H Lu andZHan ldquoExistence of positive solutions for boundary-value problem of fractional differential equation with 119901-laplacian operatorrdquo The American Jouranl of Engineering andTechnology Research vol 11 pp 3757ndash3764 2011
[17] G Chai ldquoPositive solutions for boundary value problem offractional differential equation with 119901-Laplacian operatorrdquoBoundary Value Problems vol 2012 article 18 2012
[18] T Chen W Liu and Z Hu ldquoA boundary value problem forfractional differential equation with 119901-Laplacian operator atresonancerdquoNonlinear AnalysisTheoryMethodsampApplicationsvol 75 no 6 pp 3210ndash3217 2012
[19] T Chen andW Liu ldquoAn anti-periodic boundary value problemfor the fractional differential equation with a 119901-LaplacianoperatorrdquoAppliedMathematics Letters of Rapid Publication vol25 no 11 pp 1671ndash1675 2012
[20] S J Li X G Zhang Y H Wu and L Caccetta ldquoExtremalsolutions for 119901-Laplacian differential systems via iterative com-putationrdquo Applied Mathematics and Computation vol 26 pp1151ndash1158 2013
[21] Z Liu and L Lu ldquoA class of BVPs for nonlinear fractionaldifferential equations with 119901-Laplacian operatorrdquo ElectronicJournal of Qualitative Theory of Differential Equations vol 70pp 1ndash16 2012
[22] Z Liu L Lu and I Szanto ldquoExistence of solutions for fractionalimpulsive differential equations with 119901-Laplacian operatorrdquoActa Mathematica Hungarica vol 141 no 3 pp 203ndash219 2013
[23] Z Bai ldquoEigenvalue intervals for a class of fractional boundaryvalue problemrdquo Computers amp Mathematics with Applicationsvol 64 no 10 pp 3253ndash3257 2012
[24] X Zhang L Wang and Q Sun ldquoExistence of positive solutionsfor a class of nonlinear fractional differential equations withintegral boundary conditions and a parameterrdquo Applied Mathe-matics and Computation vol 226 pp 708ndash718 2014
18 Abstract and Applied Analysis
[25] D J Guo and V Lakshmikantham Nonlinear Problems inAbstract Cones vol 5 Academic Press New York NY USA1988
[26] M A Krasnoselrsquoskii Topological Methods in the Theory ofNonlinear Integral Equations Translated by A H ArmstrongElmsford Pergamon Turkey 1964
Submit your manuscripts athttpwwwhindawicom
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical Problems in Engineering
Hindawi Publishing Corporationhttpwwwhindawicom
Differential EquationsInternational Journal of
Volume 2014
Applied MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical PhysicsAdvances in
Complex AnalysisJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
OptimizationJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Operations ResearchAdvances in
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Function Spaces
Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of Mathematics and Mathematical Sciences
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Algebra
Discrete Dynamics in Nature and Society
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Decision SciencesAdvances in
Discrete MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom
Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Stochastic AnalysisInternational Journal of
18 Abstract and Applied Analysis
[25] D J Guo and V Lakshmikantham Nonlinear Problems inAbstract Cones vol 5 Academic Press New York NY USA1988
[26] M A Krasnoselrsquoskii Topological Methods in the Theory ofNonlinear Integral Equations Translated by A H ArmstrongElmsford Pergamon Turkey 1964
Submit your manuscripts athttpwwwhindawicom
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical Problems in Engineering
Hindawi Publishing Corporationhttpwwwhindawicom
Differential EquationsInternational Journal of
Volume 2014
Applied MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical PhysicsAdvances in
Complex AnalysisJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
OptimizationJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Operations ResearchAdvances in
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Function Spaces
Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of Mathematics and Mathematical Sciences
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Algebra
Discrete Dynamics in Nature and Society
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Decision SciencesAdvances in
Discrete MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom
Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Stochastic AnalysisInternational Journal of
Submit your manuscripts athttpwwwhindawicom
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
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