Research Article Existence and Uniqueness of Positive Solutions for a Fractional Switched System Zhi-Wei Lv 1,2 and Bao-Feng Chen 2 1 Department of Mathematics, Zhengzhou University, Zhengzhou, Henan 450001, China 2 Department of Mathematics and Physics, Anyang Institute of Technology, Anyang, Henan 455000, China Correspondence should be addressed to Zhi-Wei Lv; [email protected] Received 25 January 2014; Accepted 11 March 2014; Published 13 April 2014 Academic Editor: Xinan Hao Copyright © 2014 Z.-W. Lv and B.-F. Chen. is is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited. We discuss the existence and uniqueness of positive solutions for the following fractional switched system: ( 0+ ()+ () (, ())+ () (, ()) = 0, ∈ = [0, 1]); ((0) = (0) = 0, (1) = ∫ 1 0 () ), where 0+ is the Caputo fractional derivative with 2<≤3, () : → {1, 2, . . . , } is a piecewise constant function depending on , and R + = [0, +∞), , ∈ [ × R + , R + ], = 1, 2, . . . , . Our results are based on a fixed point theorem of a sum operator and contraction mapping principle. Furthermore, two examples are also given to illustrate the results. 1. Introduction Fractional differential equations arise in various areas of science and engineering. Due to their applications, fractional differential equations have gained considerable attention (cf., e.g., [1–15] and references therein). Moreover, the theory of boundary value problems with integral boundary conditions has various applications in applied fields. For example, heat conduction, chemical engineering, underground water flow, thermoelasticity, and population dynamics can be reduced to the nonlocal problems with integral boundary conditions. In [2], Cabada and Wang considered the following m-point boundary value problem for fractional differential equation 0+ () + (, ()) = 0, 0 < < 1, (0) = (0) = 0, (1) = ∫ 1 0 () , (1) where 2<≤3, 0+ is the Caputo fractional derivative, and : [0, 1] × [0, ∞) → [0, ∞) is a continuous function. On the other hand, a switched system consists of a family of subsystems described by differential or difference equations, which has many applications in traffic control, switching power converters, network control, multiagent consensus, and so forth (see [16–18]). When we consider a switched system, we always suppose that the solution is unique. So it is important to study the uniqueness of solution for a switched system. In [1], Li and Liu investigated the uniqueness of positive solution for the following switched system: () + () (, ()) = 0, ∈ = [0, 1] , (0) = 0, (1) = ∫ 1 0 () () , (2) where () : → {1, 2, . . . , } is a piecewise constant function depending on , and R + = [0, +∞), ∈ [ × R + , R + ], = 1, 2, . . . , . In this paper, we discuss the existence and uniqueness of positive solutions for the following fractional switched system: 0+ () + () (, ()) + () (, ()) = 0, ∈ = [0, 1] , (0) = (0) = 0, (1) = ∫ 1 0 () , (3) Hindawi Publishing Corporation Abstract and Applied Analysis Volume 2014, Article ID 828721, 7 pages http://dx.doi.org/10.1155/2014/828721
Welcome message from author
This document is posted to help you gain knowledge. Please leave a comment to let me know what you think about it! Share it to your friends and learn new things together.
Transcript
Research ArticleExistence and Uniqueness of Positive Solutions for a FractionalSwitched System
Zhi-Wei Lv12 and Bao-Feng Chen2
1 Department of Mathematics Zhengzhou University Zhengzhou Henan 450001 China2Department of Mathematics and Physics Anyang Institute of Technology Anyang Henan 455000 China
Correspondence should be addressed to Zhi-Wei Lv sdlllzwmailustceducn
Received 25 January 2014 Accepted 11 March 2014 Published 13 April 2014
Academic Editor Xinan Hao
Copyright copy 2014 Z-W Lv and B-F Chen 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
We discuss the existence and uniqueness of positive solutions for the following fractional switched system ( 1198881198631205720+119906(119905)+119891
120590(119905)(119905 119906(119905))+
119892120590(119905)
(119905 119906(119905)) = 0 119905 isin 119869 = [0 1]) (119906(0) = 11990610158401015840
(0) = 0 119906(1) = int1
0
119906(119904) 119889119904) where 1198881198631205720+is theCaputo fractional derivativewith 2 lt 120572 le 3
120590(119905) 119869 rarr 1 2 119873 is a piecewise constant function depending on 119905 andR+ = [0 +infin) 119891119894 119892119894isin 119862[119869timesR+R+] 119894 = 1 2 119873
Our results are based on a fixed point theorem of a sum operator and contraction mapping principle Furthermore two examplesare also given to illustrate the results
1 Introduction
Fractional differential equations arise in various areas ofscience and engineering Due to their applications fractionaldifferential equations have gained considerable attention (cfeg [1ndash15] and references therein) Moreover the theory ofboundary value problems with integral boundary conditionshas various applications in applied fields For example heatconduction chemical engineering underground water flowthermoelasticity and population dynamics can be reducedto the nonlocal problems with integral boundary conditionsIn [2] Cabada and Wang considered the following m-pointboundary value problem for fractional differential equation
119888
119863120572
0+119906 (119905) + 119891 (119905 119906 (119905)) = 0 0 lt 119905 lt 1
119906 (0) = 11990610158401015840
(0) = 0 119906 (1) = 120582int
1
0
119906 (119904) 119889119904
(1)
where 2 lt 120572 le 3 1198881198631205720+
is the Caputo fractional derivativeand 119891 [0 1] times [0infin) rarr [0infin) is a continuous function
On the other hand a switched system consists of afamily of subsystems described by differential or differenceequations which has many applications in traffic controlswitching power converters network control multiagent
consensus and so forth (see [16ndash18]) When we considera switched system we always suppose that the solution isunique So it is important to study the uniqueness of solutionfor a switched system In [1] Li and Liu investigated theuniqueness of positive solution for the following switchedsystem
11990910158401015840
(119905) + 119891120590(119905)
(119905 119909 (119905)) = 0 119905 isin 119869 = [0 1]
119909 (0) = 0 119909 (1) = int
1
0
119886 (119904) 119909 (119904) 119889119904
(2)
where 120590(119905) 119869 rarr 1 2 119873 is a piecewise constantfunction depending on 119905 and R+ = [0 +infin) 119891
119894isin 119862[119869 times
R+R+] 119894 = 1 2 119873In this paper we discuss the existence and uniqueness
of positive solutions for the following fractional switchedsystem
119888
119863120572
0+119906 (119905) + 119891
120590(119905)(119905 119906 (119905)) + 119892
120590(119905)(119905 119906 (119905)) = 0
119905 isin 119869 = [0 1]
119906 (0) = 11990610158401015840
(0) = 0 119906 (1) = int
1
0
119906 (119904) 119889119904
(3)
Hindawi Publishing CorporationAbstract and Applied AnalysisVolume 2014 Article ID 828721 7 pageshttpdxdoiorg1011552014828721
2 Abstract and Applied Analysis
where 1198881198631205720+
is the Caputo fractional derivative with 2 lt 120572 le
3 120590(119905) 119869 rarr 1 2 119873 is a piecewise constant functiondepending on 119905 and R+ = [0 +infin) 119891
119894 119892119894isin 119862[119869 times R+R+]
119894 = 1 2 119873The paper is organized as follows In Section 2 we present
some background materials and preliminaries Section 3deals with some existence results In Section 4 two examplesare given to illustrate the results
2 Background Materials and Preliminaries
Definition 1 (see [3 4]) The fractional integral of order120572withthe lower limit 119905
0for a function 119891 is defined as
119868120572
119891 (119905) =1
Γ (120572)int
119905
1199050
(119905 minus 119904)120572minus1
119891 (119904) 119889119904 119905 gt 1199050 120572 gt 0 (4)
where Γ is the gamma function
Definition 2 (see [3 4]) For a function 119891 [0infin) rarr R theCaputo derivative of fractional order is defined as
119888
119863120572
0+119891 (119905) =
1
Γ (119899 minus 120572)int
119905
0
(119905 minus 119904)119899minus120572minus1
119891(119899)
(119904) 119889119904
120572 gt 0 119899 = [120572] + 1
(5)
In the following let us recall some basic information oncone (see more from [19 20]) Let 119864 be a real Banach spaceand let 119875 be a cone in 119864 which defined a partial ordering in119864 by 119909 le 119910 if and only if 119910 minus 119909 isin 119875 119875 is said to be normal ifthere exists a positive constant119873 such that 120579 le 119909 le 119910 implies119909 le 119873119910 119875 is called solid if its interior
∘
119875 is nonempty If119909 le 119910 and 119909 = 119910 we write 119909 lt 119910 We say that an operator119860 isincreasing if 119909 le 119910 implies 119860119909 le 119860119910
For all 119909 119910 isin 119864 the notation 119909 sim 119910means that there exist120582 gt 0 and 120583 gt 0 such that 120582119909 le 119910 le 120583119909 Clearly sim is anequivalence relation Given ℎ gt 120579 (ie ℎ ge 120579 and ℎ = 120579) wedenote by 119875
ℎthe set 119875
ℎ= 119909 isin 119864 | 119909 sim ℎ It is easy to see
that 119875ℎsub 119875
Definition 3 Let 119863 = 119875 or 119863 =∘
119875 and let 120574 be a real numberwith 0 le 120574 lt 1 An operator 119860 119875 rarr 119875 is said to be120574-concave if it satisfies
119860 (119905119909) ge 119905120574
119860119909 forall119905 isin (0 1) 119909 isin 119863 (6)
Definition 4 An operator 119860 119864 rarr 119864 is said to behomogeneous if it satisfies
119860 (119905119909) = 119905119860119909 forall119905 gt 0 119909 isin 119864 (7)
An operator 119860 119875 rarr 119875 is said to be subhomogeneous if itsatisfies
119860 (119905119909) ge 119905119860119909 forall119905 isin (0 1) 119909 isin 119875
(8)
From [2] we have the following result
Lemma 5 Assume that 2 lt 120572 le 3 and 119891119894 119892119894isin 119862[119869timesR+R+]
119894 = 1 2 119873 Then the problem (3) has a solution if and onlyif 119906 is a solution of the integral equation
119906 (119905) = int
1
0
119866 (119905 119904) (119891120590(119904)
(119904 119906 (119904)) + 119892120590(119904)
(119904 119906 (119904))) 119889119904 (9)
where
119866 (119905 119904)
=
2119905(1 minus 119904)120572minus1
(120572 minus 1 + 119904) minus 120572(119905 minus 119904)120572minus1
Γ (120572 + 1) 0 le 119904 le 119905 le 1
2119905(1 minus 119904)120572minus1
(120572 minus 1 + 119904)
Γ (120572 + 1) 0 le 119905 le 119904 le 1
(10)
Lemma 6 119866(119905 119904) in Lemma 5 has the following property
(i) 119866(119905 119904) gt 0 119891119900119903 119886119897119897 119905 119904 isin (0 1)
(ii) (1Γ(120572 + 1))ℎ(119905)(1 minus 119904)120572minus1
(120572 minus 2 + 2119904) le 119866(119905 119904) le
(2Γ(120572 + 1))ℎ(119905)(1 minus 119904)120572minus1
(120572 minus 1 + 119904) 119905 119904 isin [0 1] 2 lt
120572 le 3 ℎ(119905) = 119905
Proof From [2] we know that (i) is obvious For 0 le 119904 le 119905 le
1 2 lt 120572 le 3 we have
2119905(1 minus 119904)120572minus1
(120572 minus 1 + 119904) minus 120572(119905 minus 119904)120572minus1
= 2119905(1 minus 119904)120572minus1
(120572 minus 1 + 119904) minus 120572119905120572minus1
(1 minus119904
119905)
120572minus1
ge 2119905(1 minus 119904)120572minus1
(120572 minus 1 + 119904) minus 120572119905(1 minus 119904)120572minus1
= 119905(1 minus 119904)120572minus1
(120572 minus 2 + 2119904)
(11)
This means that (ii) holds
Theorem 7 (see [19]) Let 119875 be a normal cone in a real Banachspace 119864 119860 119875 rarr 119875 an increasing 120574-concave operator and119861 119875 rarr 119875 an increasing subhomogeneous operator Assumethat
(i) there is ℎ gt 120579 such that 119860ℎ isin 119875ℎand 119861ℎ isin 119875
ℎ
(ii) there exists a constant 1205750gt 0 such that 119860119909 ge 120575
0119861119909
forall119909 isin 119875
Then the operator equation119860119909+119861119909 = 119909 has a unique solution119909lowast in 119875ℎ Moreover constructing successively the sequence 119910
119899=
119860119910119899minus1
+ 119861119910119899minus1
119899 = 1 2 for any initial value 1199100isin 119875ℎ we
have 119910119899rarr 119909lowast as 119899 rarr infin
3 Main Results
In this section we will deal with the existence and uniquenessof positive solutions for problem (3) Let
1198661(119904 119904) =
2
Γ (120572 + 1)(1 minus 119904)
120572minus1
(120572 minus 1 + 119904) (12)
Abstract and Applied Analysis 3
It is obvious that
119866 (119905 119904) le 1198661(119904 119904) 119905 119904 isin [0 1] (13)
We consider the Banach space119864 = 119862[[0 1]R] endowedwiththe norm defined by 119906 = sup
0le119905le1|119906(119905)| Letting 119875 = 119906 isin
119864 | 119906(119905) ge 0 then 119875 is a cone in 119864 Define an operator ϝ
119864 rarr 119864 as
(ϝ119906) (119905) = int
1
0
119866 (119905 119904) (119891120590(119904)
(119904 119906 (119904)) + 119892120590(119904)
(119904 119906 (119904))) 119889119904
(14)
Then ϝ has a solution if and only if the operator ϝ has a fixedpoint
Theorem 8 Let 119891119894 119892119894isin 119862[119869 times R+R+] 119894 = 1 2 119873
Suppose that the following conditions are satisfied1003816100381610038161003816119891119894 (119905 119906 (119905)) minus 119891
119894(119905 V (119905))1003816100381610038161003816 le 119897
119894(119905) |119906 (119905) minus V (119905)|
1003816100381610038161003816119892119894 (119905 119906 (119905)) minus 119892119894(119905 V (119905))1003816100381610038161003816 le 119897
119894(119905) |119906 (119905) minus V (119905)|
0 lt int
1
0
1198661(119904 119904) (119897
119894(119904) + 119897
119894(119904)) 119889119904 lt 1
(15)
where
119897119894 119897119894isin 119862 [119869R
+
] 119894 = 1 2 119873 (16)
Then the problem (3) has a unique solution on [0 1]
Proof It follows from Lemma 6 that ϝ 119875 rarr 119875 For 119905 isin
119869 119894 = 1 2 119873 we set max119894=12119873
sup119905isin119869
|119891119894(119905 0)| = 119872
max119894=12119873
sup119905isin119869
|119892119894(119905 0)| = 119872 and 119861
119903= 119906 isin 119862[119869R+]
119906 le 119903 where
119903 ge
(119872 +119872)int1
0
1198661(119904 119904) 119889119904
1 minusmax119894=12119873
int1
0
1198661(119904 119904) (119897
119894(119904) + 119897
119894(119904)) 119889119904
(17)
Step 1We show that ϝ (119861119903) sub 119861119903
For 119906 isin 119861119903and 119905 isin 119869 119894 = 1 2 119873
int
1
0
119866 (119905 119904)1003816100381610038161003816119891119894 (119904 119906 (119904)) + 119892
119894(119904 119906 (119904))
1003816100381610038161003816 119889119904
le int
1
0
1198661(119904 119904) (
1003816100381610038161003816119891119894 (119904 119906 (119904)) minus 119891119894(119904 0)
1003816100381610038161003816 +1003816100381610038161003816119891119894 (119904 0)
1003816100381610038161003816) 119889119904
+ int
1
0
1198661(119904 119904) (
1003816100381610038161003816119892119894 (119904 119906 (119904)) minus 119892119894(119904 0)
1003816100381610038161003816 +1003816100381610038161003816119892119894 (119904 0)
1003816100381610038161003816) 119889119904
le 119903 max119894=12119873
int
1
0
1198661(119904 119904) (119897
119894(119904) + 119897
119894(119904)) 119889119904
+ (119872 +119872)int
1
0
1198661(119904 119904) 119889119904
le 119903
(18)
which implies that |(ϝ119906)(119905)| le 119903 Thus ϝ119906 le 119903 Therefore
ϝ (119861119903) sub 119861119903 (19)
Step 2We show that ϝ is a contraction mappingFor 119906 V isin 119861
119903and for each 119905 isin 119869 119894 = 1 2 119873 we have
int
1
0
119866 (119905 119904)1003816100381610038161003816119891119894 (119904 119906 (119904)) minus 119891
119894(119904 V (119904))1003816100381610038161003816 119889119904
+ int
1
0
119866 (119905 119904)1003816100381610038161003816119892119894 (119904 119906 (119904)) minus 119892
119894(119904 V (119904))1003816100381610038161003816 119889119904
le int
1
0
1198661(119904 119904) 119897
119894(119904) |119906 (119904) minus V (119904)| 119889119904
+ int
1
0
1198661(119904 119904) 119897
119894(119904) |119906 (119904) minus V (119904)| 119889119904
le int
1
0
1198661(119904 119904) (119897
119894(119904) + 119897
119894(119904)) 119889119904 119906 minus V
(20)
This together with 0 lt int1
0
1198661(119904 119904)(119897
119894(119904) + 119897
119894(119904))119889119904 lt 1 119894 =
1 2 119873 yields that1003816100381610038161003816(ϝ119906) (119905) minus (ϝV) (119905)1003816100381610038161003816 le 119896 119906 minus V (21)
where
0 lt 119896 = max119894=12119873
int
1
0
1198661(119904 119904) (119897
119894(119904) + 119897
119894(119904)) 119889119904 lt 1 (22)
Thus1003817100381710038171003817ϝ119906 minus ϝV1003817100381710038171003817 le 119896 119906 minus V (23)
This means that ϝ is a contraction mappingIt follows from Banachrsquos contraction mapping that ϝ has
a unique fixed point in 119861119903 Therefore the problem (3) has a
unique solution
Corollary 9 Let 119891119894isin 119862[119869 timesR+R+] 119894 = 1 2 119873 Suppose
that the following conditions are satisfied1003816100381610038161003816119891119894 (119905 119906 (119905)) minus 119891
119894(119905 V (119905))1003816100381610038161003816 le 119897
119894(119905) |119906 (119905) minus V (119905)|
0 lt int
1
0
1198661(119904 119904) 119897
119894(119904) 119889119904 lt 1
(24)
where
119897119894isin 119862 [119869R
+
] 119894 = 1 2 119873 (25)
Then the following fractional switched system119888
119863120572
0+119906 (119905) + 119891
120590(119905)(119905 119906 (119905)) = 0 119905 isin 119869 = [0 1]
119906 (0) = 11990610158401015840
(0) = 0 119906 (1) = int
1
0
119906 (119904) 119889119904
(26)
has a unique solution on [0 1]
Theorem 10 Assume that
(H1) 119891119894 119892119894
isin 119862[119869 times R+R+] and 119891119894(119905 119909) 119892
119894(119905 119909) are
increasing in 119909 for 119909 isin R+ 119892119894(119905 0) = 0 119894 = 1 2 119873
(H2) 119892119894(119905 120582119909) ge 120582119892
119894(119905 119909) for 120582 isin (0 1) 119905 isin 119869 119909 isin
R+ and there exists a constant 120574 isin (0 1) such that119891119894(119905 120582119909) ge 120582
120574
119891119894(119905 119909) forall119905 isin 119869 120582 isin (0 1) 119909 isin R+
119894 = 1 2 119873
4 Abstract and Applied Analysis
(H3) there exists a constant 120575
0gt 0 such that 119891
119894(119905 119909) ge
1205750119892119894(119905 119909) 119905 isin 119869 119909 isin R+ 119894 = 1 2 119873
Then problem (3) has a unique solution 119906lowast in 119875ℎ where ℎ(119905) =
119905 119905 isin 119869 Moreover for any initial value 1199060isin 119875ℎ constructing
successively the sequence
119906119899+1
(119905) = int
1
0
119866 (119905 119904) (119891120590(119904)
(119904 119906119899(119904)) + 119892
120590(119904)(119904 119906119899(119904))) 119889119904
119899 = 0 1 2
(27)
we have 119906119899(119905) rarr 119906
lowast
(119905) as 119899 rarr infin
Proof Define the two operators
119860119906 (119905) = int
1
0
119866 (119905 119904) 119891120590(119904)
(119904 119906 (119904)) 119889119904
119861119906 (119905) = int
1
0
119866 (119905 119904) 119892120590(119904)
(119904 119906 (119904)) 119889119904
(28)
From Lemma 6 we have 119860 119875 rarr 119875 and 119861 119875 rarr 119875 It isobvious that 119906 is the solution of problem (3) if and only if 119906 =
119860119906+119861119906 It follows from (H1) that119860 and 119861 are two increasing
operators Thus for 119906 V isin 119875 119906 ge V we have 119860119906 ge 119860V and119861119906 ge 119861VStep 1 We show that 119860 is a 120574-concave operator and 119861 is asubhomogeneous operator
In fact for 120582 isin (0 1) 119906 isin 119875 119905 isin 119869 119894 = 1 2 119873 from(H2) we have
int
1
0
119866 (119905 119904) 119891119894(119904 120582119906 (119904)) 119889119904 ge 120582
120574
int
1
0
119866 (119905 119904) 119891119894(119904 119906 (119904)) 119889119904
(29)
which yields that
119860 (120582119906) (119905) ge 120582120574
119860119906 (119905) (30)
Thus 119860 is a 120574-concave operator By a closely similar way wecan see that 119861 is a subhomogeneous operator
Step 2 We show that 119860ℎ isin 119875ℎand 119861ℎ isin 119875
ℎ
From Lemma 6 and (H1) we have for 119905 isin 119869 119894 =
1 2 119873
int
1
0
119866 (119905 119904) 119891119894(119904 ℎ (119904)) 119889119904
le2
Γ (120572 + 1)ℎ (119905)
times int
1
0
(1 minus 119904)120572minus1
(120572 minus 1 + 119904) 119891119894(119904 1) 119889119904
int
1
0
119866 (119905 119904) 119891119894(119904 ℎ (119904)) 119889119904
ge1
Γ (120572 + 1)ℎ (119905)
times int
1
0
(1 minus 119904)120572minus1
(120572 minus 2 + 2119904) 119891119894(119904 0) 119889119904
(31)
For 119894 = 1 2 119873 let
119898119894=
1
Γ (120572 + 1)int
1
0
(1 minus 119904)120572minus1
(120572 minus 2 + 2119904) 119891119894(119904 0) 119889119904
119898119894=
2
Γ (120572 + 1)int
1
0
1 minus 119904120572minus1
(120572 minus 1 + 119904) 119891119894(119904 1) 119889119904
(32)
It follows from 119892(119905 0) = 0 that int10
119891119894(119904 1) 119889119904 ge int
1
0
119891119894(119904 0) 119889119904 ge
1205750int1
0
119892119894(119904 0) 119889119904 gt 0
Thus
119898119894gt 0 119898
119894gt 0 119894 = 1 2 119873 (33)
Letting = min119898119894 119894 = 1 2 119873 and = max119898
119894 119894 =
1 2 119873 then gt 0 and gt 0 Therefore
ℎ (119905) le 119860ℎ (119905) le ℎ (119905) (34)
which implies that
119860ℎ isin 119875ℎ (35)
Similarly we have 119861ℎ isin 119875ℎ
Step 3 There exists a constant 1205750gt 0 such that 119860119906 ge 120575
0119861119906
forall119906 isin 119875For 119906 isin 119875 and 119905 isin 119869 119894 = 1 2 119873 by (H
3) we have
int
1
0
119866 (119905 119904) 119891119894(119904 119906 (119904)) 119889119904 ge 120575
0int
1
0
119866 (119905 119904) 119892119894(119904 119906 (119904)) 119889119904 (36)
This means that
119860119906 ge 1205750119861119906 119906 isin 119875 (37)
Therefore the conditions of Theorem 7 are satisfied Bymeans of Theorem 7 we obtain that the operator equation119860119906+119861119906 = 119906 has a unique solution 119906lowast in119875
ℎ Moreover for any
initial value 1199060isin 119875ℎ constructing successively the sequence
119906119899+1
(119905) = int
1
0
119866 (119905 119904) (119891120590(119904)
(119904 119906119899(119904)) + 119892
120590(119904)(119904 119906119899(119904))) 119889119904
119899 = 0 1 2
(38)
we have 119906119899(119905) rarr 119906
lowast
(119905) as 119899 rarr infin
In Theorem 10 if we let 119861 be a null operator we have thefollowing conclusion
Corollary 11 Assume that
(H4) 119891119894isin 119862[119869 times R+R+] and 119891
119894(119905 119909) is increasing in 119909 for
119909 isin R+ 119891119894(119905 0) = 0 119894 = 1 2 119873
Abstract and Applied Analysis 5
(H5) there exists a constant 120574 isin (0 1) such that 119891
119894(119905 120582119909) ge
120582120574
119891119894(119905 119909) forall119905 isin 119869 120582 isin (0 1) 119909 isin R+ 119894 = 1 2 119873
Then the following fractional switched system
119888
119863120572
0+119906 (119905) + 119891
120590(119905)(119905 119906 (119905)) = 0 119905 isin 119869 = [0 1]
119906 (0) = 11990610158401015840
(0) = 0 119906 (1) = int
1
0
119906 (119904) 119889119904
(39)
has a unique solution119906lowast in119875ℎ where ℎ(119905) = 119905 119905 isin 119869Moreover
for any initial value 1199060
isin 119875ℎ constructing successively the
sequence
119906119899+1
(119905) = int
1
0
119866 (119905 119904) 119891120590(119904)
(119904 119906119899(119904)) 119889119904 119899 = 0 1 2
(40)
we have 119906119899(119905) rarr 119906
lowast
(119905) as 119899 rarr infin
4 Examples
Example 1 Consider the following boundary value problem
119888
119863120572
0+119906 (119905) + 119891
120590(119905)(119905 119906 (119905)) + 119892
120590(119905)(119905 119906 (119905)) = 0
119905 isin 119869 = [0 1]
119906 (0) = 11990610158401015840
(0) = 0 119906 (1) = int
1
0
119906 (119904) 119889119904
(41)
where 120572 = 52 120590(119905) 119869 rarr 119872 = 1 2 is a finite switchingsignal
1198911(119905 119909) =
1
4(119905 + 2)2
119909
1 + 119909+ 1
1198921(119905 119909) =
1
16sin2119909 + 119905
2
1198912(119905 119909) =
1
8(119905 + 2)2
119909
1 + 119909+ 1
1198922(119905 119909) =
1
32sin2119909 + 119905
2
(42)
Thus
119891119894 119892119894isin 119862 [119869 timesR
+
R+
] 119894 = 1 2 (43)
By computation we deduce that
10038161003816100381610038161198911 (119905 1199091) minus 1198911(119905 1199092)1003816100381610038161003816 le
1
16
10038161003816100381610038161199092 minus 1199091
1003816100381610038161003816
10038161003816100381610038161198921 (119905 1199091) minus 1198921(119905 1199092)1003816100381610038161003816 le
1
16
10038161003816100381610038161199092 minus 1199091
1003816100381610038161003816
10038161003816100381610038161198912 (119905 1199091) minus 1198912(119905 1199092)1003816100381610038161003816 le
1
32
10038161003816100381610038161199092 minus 1199091
1003816100381610038161003816
10038161003816100381610038161198922 (119905 1199091) minus 1198922(119905 1199092)1003816100381610038161003816 le
1
32
10038161003816100381610038161199092 minus 1199091
1003816100381610038161003816
(44)
On the other hand
int
1
0
1198661(119904 119904) (119897
1(119904) + 119897
1(119904)) 119889119904
= int
1
0
2(1 minus 119904)(52)minus1
((52) minus 1 + 119904)
Γ ((52) + 1)(1
16+
1
16) 119889119904
=1
4Γ (72)int
1
0
(1 minus 119904)32
(3
2+ 119904) 119889119904
le1
4Γ (72)int
1
0
(1 minus 119904)32
(3
2+ 1) 119889119904
=1
3radic120587times2
5
lt 1
int
1
0
1198661(119904 119904) (119897
1(119904) + 119897
1(119904)) 119889119904
= int
1
0
2(1 minus 119904)(52)minus1
((52) minus 1 + 119904)
Γ ((52) + 1)(1
32+
1
32) 119889119904
=1
3radic120587times1
5
lt 1
(45)
Hence byTheorem 8 BVP (41) has a unique positive solutionon [0 1]
Example 2 Consider the following boundary value problem
119888119863120572
0+119906 (119905) + 119891
120590(119905)(119905 119906 (119905)) + 119892
120590(119905)(119905 119906 (119905)) = 0
119905 isin 119869 = [0 1]
119906 (0) = 11990610158401015840
(0) = 0 119906 (1) = int
1
0
119906 (119904) 119889119904
(46)
where 120572 = 52 120590(119905) 119869 rarr 1 2 3 is a finite switching signal
1198911(119905 119909) = 119909
13
+ 1199052
+ 119888
1198921(119905 119909) =
119909
(1 + 1199052) (1 + 119909)+ 119887 minus 119888
1198912(119905 119909) = 2119909
13
+ 1199052
+ 2119888
1198922(119905 119909) =
2119909
(1 + 1199052) (1 + 119909)+ 2 (119887 minus 119888)
6 Abstract and Applied Analysis
1198913(119905 119909) = 3119909
13
+ 1199052
+ 3119888
1198923(119905 119909) =
3119909
(1 + 1199052) (1 + 119909)+ 3 (119887 minus 119888)
(47)
Let 120574 = 13 and 0 lt 119888 lt 119887 It is obvious that 119891119894 119892119894isin
119862[119869 times R+R+] and are increasing with respect to the secondargument 119892
119894(119905 0) = 119887 minus 119888 gt 0 119894 = 1 2 3 On the other hand
for 120582 isin (0 1) 119905 isin 119869 119909 isin [0 +infin) 119894 = 1 2 3 we have
119892119894(119905 120582119909) =
119894120582119909
(1 + 1199052) (1 + 120582119909)+ 119894 (119887 minus 119888)
ge119894120582119909
(1 + 1199052) (1 + 120582119909)+ 119894120582 (119887 minus 119888)
= 120582119892119894(119905 119909)
119891119894(119905 120582119909) = 119894120582
13
11990913
+ 1199052
+ 119894119888
ge 12058213
(11989411990913
+ 1199052
+ 119894119888)
= 120582120574
119891119894(119905 119909)
(48)
Moreover for 119905 isin 119869 119909 isin R+ 119894 = 1 2 3 we have
119891119894(119905 119909) = 119894119909
13
+ 1199052
+ 119894119888
ge 119894119888 ge119888
3 + (119887 minus 119888)(119894 + 119894 (119887 minus 119888))
ge119888
3 + (119887 minus 119888)(
119894119909
(1 + 1199052) (1 + 119909)+ 119894 (119887 minus 119888))
= 1205750119892119894(119905 119909)
(49)
where
1205750=
119888
3 + (119887 minus 119888) (50)
Hence all the conditions of Theorem 10 are satisfied ThusBVP (46) has a unique positive solution in 119875
ℎ where ℎ(119905) = 119905
119905 isin [0 1]
Conflict of Interests
The authors declare that there is no conflict of interestsregarding the publication of this paper
Acknowledgments
This research was supported by Henan Province CollegeYouth Backbone Teacher Funds (2011GGJS-213) and theNational Natural Science Foundation of China (11271336)
References
[1] H T Li and Y S Liu ldquoOn the uniqueness of the positivesolution for a second-order integral boundary value problemwith switched nonlinearityrdquo Applied Mathematics Letters vol24 no 12 pp 2201ndash2205 2011
[2] A Cabada and G Wang ldquoPositive solutions of nonlinearfractional differential equations with integral boundary valueconditionsrdquo Journal of Mathematical Analysis and Applicationsvol 389 no 1 pp 403ndash411 2012
[3] A A Kilbas H M Srivastava and J J Trujjllo Theory andApplications of Fractional Differential Equations vol 204 ofNorth-Holland Mathematics Studies Elsevier Amsterdam TheNetherlands 2006
[4] I Podlubny Fractional Differential Equations Academic PressNew York NY USA 1993
[5] X Zhang L Liu and Y Wu ldquoThe eigenvalue problem for asingular higher order fractional differential equation involvingfractional derivativesrdquo Applied Mathematics and Computationvol 218 no 17 pp 8526ndash8536 2012
[6] Z-W Lv J Liang and T-J Xiao ldquoSolutions to the Cauchy prob-lem for differential equations in Banach spaces with fractionalorderrdquo Computers ampMathematics with Applications vol 62 no3 pp 1303ndash1311 2011
[7] R P Agarwal V Lakshmikantham and J J Nieto ldquoOn theconcept of solution for fractional differential equations withuncertaintyrdquo Nonlinear Analysis Theory Methods amp Applica-tions vol 72 no 6 pp 2859ndash2862 2010
[8] R-N Wang T-J Xiao and J Liang ldquoA note on the fractionalCauchy problems with nonlocal initial conditionsrdquo AppliedMathematics Letters vol 24 no 8 pp 1435ndash1442 2011
[9] R-N Wang D-H Chen and T-J Xiao ldquoAbstract fractionalCauchy problems with almost sectorial operatorsrdquo Journal ofDifferential Equations vol 252 no 1 pp 202ndash235 2012
[10] J Henderson and A Ouahab ldquoFractional functional differentialinclusions with finite delayrdquo Nonlinear Analysis Theory Meth-ods amp Applications vol 70 no 5 pp 2091ndash2105 2009
[11] V Lakshmikantham ldquoTheory of fractional functional differen-tial equationsrdquo Nonlinear Analysis Theory Methods amp Applica-tions vol 69 no 10 pp 3337ndash3343 2008
[12] V Lakshmikantham and A S Vatsala ldquoBasic theory of frac-tional differential equationsrdquo Nonlinear Analysis Theory Meth-ods amp Applications vol 69 no 8 pp 2677ndash2682 2008
[13] F Li ldquoMild solutions for fractional differential equations withnonlocal conditionsrdquo Advances in Difference Equations ArticleID 287861 9 pages 2010
[14] X Q Zhang ldquoPositive solution for a class of singular semiposi-tone fractional differential equations with integral boundaryconditionsrdquo Boundary Value Problems vol 2012 article 1232012
[15] C Yang and C Zhai ldquoUniqueness of positive solutions for afractional differential equation via a fixed point theorem of asum operatorrdquo Electronic Journal of Differential Equations vol2012 no 70 pp 1ndash8 2012
[16] A A Agrachev and D Liberzon ldquoLie-algebraic stability criteriafor switched systemsrdquo SIAM Journal on Control and Optimiza-tion vol 40 no 1 pp 253ndash269 2001
[17] J Daafouz P Riedinger and C Iung ldquoStability analysis andcontrol synthesis for switched systems a switched Lyapunovfunction approachrdquo IEEE Transactions on Automatic Controlvol 47 no 11 pp 1883ndash1887 2002
[18] L Gurvits R Shorten and O Mason ldquoOn the stability ofswitched positive linear systemsrdquo IEEE Transactions on Auto-matic Control vol 52 no 6 pp 1099ndash1103 2007
Abstract and Applied Analysis 7
[19] C Zhai and D R Anderson ldquoA sum operator equation andapplications to nonlinear elastic beam equations and Lane-Emden-Fowler equationsrdquo Journal of Mathematical Analysisand Applications vol 375 no 2 pp 388ndash400 2011
[20] D J Guo and V Lakshmikantham Nonlinear Problems inAbstract Cone vol 5 Academic Press San Diego Calif USA1988
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
2 Abstract and Applied Analysis
where 1198881198631205720+
is the Caputo fractional derivative with 2 lt 120572 le
3 120590(119905) 119869 rarr 1 2 119873 is a piecewise constant functiondepending on 119905 and R+ = [0 +infin) 119891
119894 119892119894isin 119862[119869 times R+R+]
119894 = 1 2 119873The paper is organized as follows In Section 2 we present
some background materials and preliminaries Section 3deals with some existence results In Section 4 two examplesare given to illustrate the results
2 Background Materials and Preliminaries
Definition 1 (see [3 4]) The fractional integral of order120572withthe lower limit 119905
0for a function 119891 is defined as
119868120572
119891 (119905) =1
Γ (120572)int
119905
1199050
(119905 minus 119904)120572minus1
119891 (119904) 119889119904 119905 gt 1199050 120572 gt 0 (4)
where Γ is the gamma function
Definition 2 (see [3 4]) For a function 119891 [0infin) rarr R theCaputo derivative of fractional order is defined as
119888
119863120572
0+119891 (119905) =
1
Γ (119899 minus 120572)int
119905
0
(119905 minus 119904)119899minus120572minus1
119891(119899)
(119904) 119889119904
120572 gt 0 119899 = [120572] + 1
(5)
In the following let us recall some basic information oncone (see more from [19 20]) Let 119864 be a real Banach spaceand let 119875 be a cone in 119864 which defined a partial ordering in119864 by 119909 le 119910 if and only if 119910 minus 119909 isin 119875 119875 is said to be normal ifthere exists a positive constant119873 such that 120579 le 119909 le 119910 implies119909 le 119873119910 119875 is called solid if its interior
∘
119875 is nonempty If119909 le 119910 and 119909 = 119910 we write 119909 lt 119910 We say that an operator119860 isincreasing if 119909 le 119910 implies 119860119909 le 119860119910
For all 119909 119910 isin 119864 the notation 119909 sim 119910means that there exist120582 gt 0 and 120583 gt 0 such that 120582119909 le 119910 le 120583119909 Clearly sim is anequivalence relation Given ℎ gt 120579 (ie ℎ ge 120579 and ℎ = 120579) wedenote by 119875
ℎthe set 119875
ℎ= 119909 isin 119864 | 119909 sim ℎ It is easy to see
that 119875ℎsub 119875
Definition 3 Let 119863 = 119875 or 119863 =∘
119875 and let 120574 be a real numberwith 0 le 120574 lt 1 An operator 119860 119875 rarr 119875 is said to be120574-concave if it satisfies
119860 (119905119909) ge 119905120574
119860119909 forall119905 isin (0 1) 119909 isin 119863 (6)
Definition 4 An operator 119860 119864 rarr 119864 is said to behomogeneous if it satisfies
119860 (119905119909) = 119905119860119909 forall119905 gt 0 119909 isin 119864 (7)
An operator 119860 119875 rarr 119875 is said to be subhomogeneous if itsatisfies
119860 (119905119909) ge 119905119860119909 forall119905 isin (0 1) 119909 isin 119875
(8)
From [2] we have the following result
Lemma 5 Assume that 2 lt 120572 le 3 and 119891119894 119892119894isin 119862[119869timesR+R+]
119894 = 1 2 119873 Then the problem (3) has a solution if and onlyif 119906 is a solution of the integral equation
119906 (119905) = int
1
0
119866 (119905 119904) (119891120590(119904)
(119904 119906 (119904)) + 119892120590(119904)
(119904 119906 (119904))) 119889119904 (9)
where
119866 (119905 119904)
=
2119905(1 minus 119904)120572minus1
(120572 minus 1 + 119904) minus 120572(119905 minus 119904)120572minus1
Γ (120572 + 1) 0 le 119904 le 119905 le 1
2119905(1 minus 119904)120572minus1
(120572 minus 1 + 119904)
Γ (120572 + 1) 0 le 119905 le 119904 le 1
(10)
Lemma 6 119866(119905 119904) in Lemma 5 has the following property
(i) 119866(119905 119904) gt 0 119891119900119903 119886119897119897 119905 119904 isin (0 1)
(ii) (1Γ(120572 + 1))ℎ(119905)(1 minus 119904)120572minus1
(120572 minus 2 + 2119904) le 119866(119905 119904) le
(2Γ(120572 + 1))ℎ(119905)(1 minus 119904)120572minus1
(120572 minus 1 + 119904) 119905 119904 isin [0 1] 2 lt
120572 le 3 ℎ(119905) = 119905
Proof From [2] we know that (i) is obvious For 0 le 119904 le 119905 le
1 2 lt 120572 le 3 we have
2119905(1 minus 119904)120572minus1
(120572 minus 1 + 119904) minus 120572(119905 minus 119904)120572minus1
= 2119905(1 minus 119904)120572minus1
(120572 minus 1 + 119904) minus 120572119905120572minus1
(1 minus119904
119905)
120572minus1
ge 2119905(1 minus 119904)120572minus1
(120572 minus 1 + 119904) minus 120572119905(1 minus 119904)120572minus1
= 119905(1 minus 119904)120572minus1
(120572 minus 2 + 2119904)
(11)
This means that (ii) holds
Theorem 7 (see [19]) Let 119875 be a normal cone in a real Banachspace 119864 119860 119875 rarr 119875 an increasing 120574-concave operator and119861 119875 rarr 119875 an increasing subhomogeneous operator Assumethat
(i) there is ℎ gt 120579 such that 119860ℎ isin 119875ℎand 119861ℎ isin 119875
ℎ
(ii) there exists a constant 1205750gt 0 such that 119860119909 ge 120575
0119861119909
forall119909 isin 119875
Then the operator equation119860119909+119861119909 = 119909 has a unique solution119909lowast in 119875ℎ Moreover constructing successively the sequence 119910
119899=
119860119910119899minus1
+ 119861119910119899minus1
119899 = 1 2 for any initial value 1199100isin 119875ℎ we
have 119910119899rarr 119909lowast as 119899 rarr infin
3 Main Results
In this section we will deal with the existence and uniquenessof positive solutions for problem (3) Let
1198661(119904 119904) =
2
Γ (120572 + 1)(1 minus 119904)
120572minus1
(120572 minus 1 + 119904) (12)
Abstract and Applied Analysis 3
It is obvious that
119866 (119905 119904) le 1198661(119904 119904) 119905 119904 isin [0 1] (13)
We consider the Banach space119864 = 119862[[0 1]R] endowedwiththe norm defined by 119906 = sup
0le119905le1|119906(119905)| Letting 119875 = 119906 isin
119864 | 119906(119905) ge 0 then 119875 is a cone in 119864 Define an operator ϝ
119864 rarr 119864 as
(ϝ119906) (119905) = int
1
0
119866 (119905 119904) (119891120590(119904)
(119904 119906 (119904)) + 119892120590(119904)
(119904 119906 (119904))) 119889119904
(14)
Then ϝ has a solution if and only if the operator ϝ has a fixedpoint
Theorem 8 Let 119891119894 119892119894isin 119862[119869 times R+R+] 119894 = 1 2 119873
Suppose that the following conditions are satisfied1003816100381610038161003816119891119894 (119905 119906 (119905)) minus 119891
119894(119905 V (119905))1003816100381610038161003816 le 119897
119894(119905) |119906 (119905) minus V (119905)|
1003816100381610038161003816119892119894 (119905 119906 (119905)) minus 119892119894(119905 V (119905))1003816100381610038161003816 le 119897
119894(119905) |119906 (119905) minus V (119905)|
0 lt int
1
0
1198661(119904 119904) (119897
119894(119904) + 119897
119894(119904)) 119889119904 lt 1
(15)
where
119897119894 119897119894isin 119862 [119869R
+
] 119894 = 1 2 119873 (16)
Then the problem (3) has a unique solution on [0 1]
Proof It follows from Lemma 6 that ϝ 119875 rarr 119875 For 119905 isin
119869 119894 = 1 2 119873 we set max119894=12119873
sup119905isin119869
|119891119894(119905 0)| = 119872
max119894=12119873
sup119905isin119869
|119892119894(119905 0)| = 119872 and 119861
119903= 119906 isin 119862[119869R+]
119906 le 119903 where
119903 ge
(119872 +119872)int1
0
1198661(119904 119904) 119889119904
1 minusmax119894=12119873
int1
0
1198661(119904 119904) (119897
119894(119904) + 119897
119894(119904)) 119889119904
(17)
Step 1We show that ϝ (119861119903) sub 119861119903
For 119906 isin 119861119903and 119905 isin 119869 119894 = 1 2 119873
int
1
0
119866 (119905 119904)1003816100381610038161003816119891119894 (119904 119906 (119904)) + 119892
119894(119904 119906 (119904))
1003816100381610038161003816 119889119904
le int
1
0
1198661(119904 119904) (
1003816100381610038161003816119891119894 (119904 119906 (119904)) minus 119891119894(119904 0)
1003816100381610038161003816 +1003816100381610038161003816119891119894 (119904 0)
1003816100381610038161003816) 119889119904
+ int
1
0
1198661(119904 119904) (
1003816100381610038161003816119892119894 (119904 119906 (119904)) minus 119892119894(119904 0)
1003816100381610038161003816 +1003816100381610038161003816119892119894 (119904 0)
1003816100381610038161003816) 119889119904
le 119903 max119894=12119873
int
1
0
1198661(119904 119904) (119897
119894(119904) + 119897
119894(119904)) 119889119904
+ (119872 +119872)int
1
0
1198661(119904 119904) 119889119904
le 119903
(18)
which implies that |(ϝ119906)(119905)| le 119903 Thus ϝ119906 le 119903 Therefore
ϝ (119861119903) sub 119861119903 (19)
Step 2We show that ϝ is a contraction mappingFor 119906 V isin 119861
119903and for each 119905 isin 119869 119894 = 1 2 119873 we have
int
1
0
119866 (119905 119904)1003816100381610038161003816119891119894 (119904 119906 (119904)) minus 119891
119894(119904 V (119904))1003816100381610038161003816 119889119904
+ int
1
0
119866 (119905 119904)1003816100381610038161003816119892119894 (119904 119906 (119904)) minus 119892
119894(119904 V (119904))1003816100381610038161003816 119889119904
le int
1
0
1198661(119904 119904) 119897
119894(119904) |119906 (119904) minus V (119904)| 119889119904
+ int
1
0
1198661(119904 119904) 119897
119894(119904) |119906 (119904) minus V (119904)| 119889119904
le int
1
0
1198661(119904 119904) (119897
119894(119904) + 119897
119894(119904)) 119889119904 119906 minus V
(20)
This together with 0 lt int1
0
1198661(119904 119904)(119897
119894(119904) + 119897
119894(119904))119889119904 lt 1 119894 =
1 2 119873 yields that1003816100381610038161003816(ϝ119906) (119905) minus (ϝV) (119905)1003816100381610038161003816 le 119896 119906 minus V (21)
where
0 lt 119896 = max119894=12119873
int
1
0
1198661(119904 119904) (119897
119894(119904) + 119897
119894(119904)) 119889119904 lt 1 (22)
Thus1003817100381710038171003817ϝ119906 minus ϝV1003817100381710038171003817 le 119896 119906 minus V (23)
This means that ϝ is a contraction mappingIt follows from Banachrsquos contraction mapping that ϝ has
a unique fixed point in 119861119903 Therefore the problem (3) has a
unique solution
Corollary 9 Let 119891119894isin 119862[119869 timesR+R+] 119894 = 1 2 119873 Suppose
that the following conditions are satisfied1003816100381610038161003816119891119894 (119905 119906 (119905)) minus 119891
119894(119905 V (119905))1003816100381610038161003816 le 119897
119894(119905) |119906 (119905) minus V (119905)|
0 lt int
1
0
1198661(119904 119904) 119897
119894(119904) 119889119904 lt 1
(24)
where
119897119894isin 119862 [119869R
+
] 119894 = 1 2 119873 (25)
Then the following fractional switched system119888
119863120572
0+119906 (119905) + 119891
120590(119905)(119905 119906 (119905)) = 0 119905 isin 119869 = [0 1]
119906 (0) = 11990610158401015840
(0) = 0 119906 (1) = int
1
0
119906 (119904) 119889119904
(26)
has a unique solution on [0 1]
Theorem 10 Assume that
(H1) 119891119894 119892119894
isin 119862[119869 times R+R+] and 119891119894(119905 119909) 119892
119894(119905 119909) are
increasing in 119909 for 119909 isin R+ 119892119894(119905 0) = 0 119894 = 1 2 119873
(H2) 119892119894(119905 120582119909) ge 120582119892
119894(119905 119909) for 120582 isin (0 1) 119905 isin 119869 119909 isin
R+ and there exists a constant 120574 isin (0 1) such that119891119894(119905 120582119909) ge 120582
120574
119891119894(119905 119909) forall119905 isin 119869 120582 isin (0 1) 119909 isin R+
119894 = 1 2 119873
4 Abstract and Applied Analysis
(H3) there exists a constant 120575
0gt 0 such that 119891
119894(119905 119909) ge
1205750119892119894(119905 119909) 119905 isin 119869 119909 isin R+ 119894 = 1 2 119873
Then problem (3) has a unique solution 119906lowast in 119875ℎ where ℎ(119905) =
119905 119905 isin 119869 Moreover for any initial value 1199060isin 119875ℎ constructing
successively the sequence
119906119899+1
(119905) = int
1
0
119866 (119905 119904) (119891120590(119904)
(119904 119906119899(119904)) + 119892
120590(119904)(119904 119906119899(119904))) 119889119904
119899 = 0 1 2
(27)
we have 119906119899(119905) rarr 119906
lowast
(119905) as 119899 rarr infin
Proof Define the two operators
119860119906 (119905) = int
1
0
119866 (119905 119904) 119891120590(119904)
(119904 119906 (119904)) 119889119904
119861119906 (119905) = int
1
0
119866 (119905 119904) 119892120590(119904)
(119904 119906 (119904)) 119889119904
(28)
From Lemma 6 we have 119860 119875 rarr 119875 and 119861 119875 rarr 119875 It isobvious that 119906 is the solution of problem (3) if and only if 119906 =
119860119906+119861119906 It follows from (H1) that119860 and 119861 are two increasing
operators Thus for 119906 V isin 119875 119906 ge V we have 119860119906 ge 119860V and119861119906 ge 119861VStep 1 We show that 119860 is a 120574-concave operator and 119861 is asubhomogeneous operator
In fact for 120582 isin (0 1) 119906 isin 119875 119905 isin 119869 119894 = 1 2 119873 from(H2) we have
int
1
0
119866 (119905 119904) 119891119894(119904 120582119906 (119904)) 119889119904 ge 120582
120574
int
1
0
119866 (119905 119904) 119891119894(119904 119906 (119904)) 119889119904
(29)
which yields that
119860 (120582119906) (119905) ge 120582120574
119860119906 (119905) (30)
Thus 119860 is a 120574-concave operator By a closely similar way wecan see that 119861 is a subhomogeneous operator
Step 2 We show that 119860ℎ isin 119875ℎand 119861ℎ isin 119875
ℎ
From Lemma 6 and (H1) we have for 119905 isin 119869 119894 =
1 2 119873
int
1
0
119866 (119905 119904) 119891119894(119904 ℎ (119904)) 119889119904
le2
Γ (120572 + 1)ℎ (119905)
times int
1
0
(1 minus 119904)120572minus1
(120572 minus 1 + 119904) 119891119894(119904 1) 119889119904
int
1
0
119866 (119905 119904) 119891119894(119904 ℎ (119904)) 119889119904
ge1
Γ (120572 + 1)ℎ (119905)
times int
1
0
(1 minus 119904)120572minus1
(120572 minus 2 + 2119904) 119891119894(119904 0) 119889119904
(31)
For 119894 = 1 2 119873 let
119898119894=
1
Γ (120572 + 1)int
1
0
(1 minus 119904)120572minus1
(120572 minus 2 + 2119904) 119891119894(119904 0) 119889119904
119898119894=
2
Γ (120572 + 1)int
1
0
1 minus 119904120572minus1
(120572 minus 1 + 119904) 119891119894(119904 1) 119889119904
(32)
It follows from 119892(119905 0) = 0 that int10
119891119894(119904 1) 119889119904 ge int
1
0
119891119894(119904 0) 119889119904 ge
1205750int1
0
119892119894(119904 0) 119889119904 gt 0
Thus
119898119894gt 0 119898
119894gt 0 119894 = 1 2 119873 (33)
Letting = min119898119894 119894 = 1 2 119873 and = max119898
119894 119894 =
1 2 119873 then gt 0 and gt 0 Therefore
ℎ (119905) le 119860ℎ (119905) le ℎ (119905) (34)
which implies that
119860ℎ isin 119875ℎ (35)
Similarly we have 119861ℎ isin 119875ℎ
Step 3 There exists a constant 1205750gt 0 such that 119860119906 ge 120575
0119861119906
forall119906 isin 119875For 119906 isin 119875 and 119905 isin 119869 119894 = 1 2 119873 by (H
3) we have
int
1
0
119866 (119905 119904) 119891119894(119904 119906 (119904)) 119889119904 ge 120575
0int
1
0
119866 (119905 119904) 119892119894(119904 119906 (119904)) 119889119904 (36)
This means that
119860119906 ge 1205750119861119906 119906 isin 119875 (37)
Therefore the conditions of Theorem 7 are satisfied Bymeans of Theorem 7 we obtain that the operator equation119860119906+119861119906 = 119906 has a unique solution 119906lowast in119875
ℎ Moreover for any
initial value 1199060isin 119875ℎ constructing successively the sequence
119906119899+1
(119905) = int
1
0
119866 (119905 119904) (119891120590(119904)
(119904 119906119899(119904)) + 119892
120590(119904)(119904 119906119899(119904))) 119889119904
119899 = 0 1 2
(38)
we have 119906119899(119905) rarr 119906
lowast
(119905) as 119899 rarr infin
In Theorem 10 if we let 119861 be a null operator we have thefollowing conclusion
Corollary 11 Assume that
(H4) 119891119894isin 119862[119869 times R+R+] and 119891
119894(119905 119909) is increasing in 119909 for
119909 isin R+ 119891119894(119905 0) = 0 119894 = 1 2 119873
Abstract and Applied Analysis 5
(H5) there exists a constant 120574 isin (0 1) such that 119891
119894(119905 120582119909) ge
120582120574
119891119894(119905 119909) forall119905 isin 119869 120582 isin (0 1) 119909 isin R+ 119894 = 1 2 119873
Then the following fractional switched system
119888
119863120572
0+119906 (119905) + 119891
120590(119905)(119905 119906 (119905)) = 0 119905 isin 119869 = [0 1]
119906 (0) = 11990610158401015840
(0) = 0 119906 (1) = int
1
0
119906 (119904) 119889119904
(39)
has a unique solution119906lowast in119875ℎ where ℎ(119905) = 119905 119905 isin 119869Moreover
for any initial value 1199060
isin 119875ℎ constructing successively the
sequence
119906119899+1
(119905) = int
1
0
119866 (119905 119904) 119891120590(119904)
(119904 119906119899(119904)) 119889119904 119899 = 0 1 2
(40)
we have 119906119899(119905) rarr 119906
lowast
(119905) as 119899 rarr infin
4 Examples
Example 1 Consider the following boundary value problem
119888
119863120572
0+119906 (119905) + 119891
120590(119905)(119905 119906 (119905)) + 119892
120590(119905)(119905 119906 (119905)) = 0
119905 isin 119869 = [0 1]
119906 (0) = 11990610158401015840
(0) = 0 119906 (1) = int
1
0
119906 (119904) 119889119904
(41)
where 120572 = 52 120590(119905) 119869 rarr 119872 = 1 2 is a finite switchingsignal
1198911(119905 119909) =
1
4(119905 + 2)2
119909
1 + 119909+ 1
1198921(119905 119909) =
1
16sin2119909 + 119905
2
1198912(119905 119909) =
1
8(119905 + 2)2
119909
1 + 119909+ 1
1198922(119905 119909) =
1
32sin2119909 + 119905
2
(42)
Thus
119891119894 119892119894isin 119862 [119869 timesR
+
R+
] 119894 = 1 2 (43)
By computation we deduce that
10038161003816100381610038161198911 (119905 1199091) minus 1198911(119905 1199092)1003816100381610038161003816 le
1
16
10038161003816100381610038161199092 minus 1199091
1003816100381610038161003816
10038161003816100381610038161198921 (119905 1199091) minus 1198921(119905 1199092)1003816100381610038161003816 le
1
16
10038161003816100381610038161199092 minus 1199091
1003816100381610038161003816
10038161003816100381610038161198912 (119905 1199091) minus 1198912(119905 1199092)1003816100381610038161003816 le
1
32
10038161003816100381610038161199092 minus 1199091
1003816100381610038161003816
10038161003816100381610038161198922 (119905 1199091) minus 1198922(119905 1199092)1003816100381610038161003816 le
1
32
10038161003816100381610038161199092 minus 1199091
1003816100381610038161003816
(44)
On the other hand
int
1
0
1198661(119904 119904) (119897
1(119904) + 119897
1(119904)) 119889119904
= int
1
0
2(1 minus 119904)(52)minus1
((52) minus 1 + 119904)
Γ ((52) + 1)(1
16+
1
16) 119889119904
=1
4Γ (72)int
1
0
(1 minus 119904)32
(3
2+ 119904) 119889119904
le1
4Γ (72)int
1
0
(1 minus 119904)32
(3
2+ 1) 119889119904
=1
3radic120587times2
5
lt 1
int
1
0
1198661(119904 119904) (119897
1(119904) + 119897
1(119904)) 119889119904
= int
1
0
2(1 minus 119904)(52)minus1
((52) minus 1 + 119904)
Γ ((52) + 1)(1
32+
1
32) 119889119904
=1
3radic120587times1
5
lt 1
(45)
Hence byTheorem 8 BVP (41) has a unique positive solutionon [0 1]
Example 2 Consider the following boundary value problem
119888119863120572
0+119906 (119905) + 119891
120590(119905)(119905 119906 (119905)) + 119892
120590(119905)(119905 119906 (119905)) = 0
119905 isin 119869 = [0 1]
119906 (0) = 11990610158401015840
(0) = 0 119906 (1) = int
1
0
119906 (119904) 119889119904
(46)
where 120572 = 52 120590(119905) 119869 rarr 1 2 3 is a finite switching signal
1198911(119905 119909) = 119909
13
+ 1199052
+ 119888
1198921(119905 119909) =
119909
(1 + 1199052) (1 + 119909)+ 119887 minus 119888
1198912(119905 119909) = 2119909
13
+ 1199052
+ 2119888
1198922(119905 119909) =
2119909
(1 + 1199052) (1 + 119909)+ 2 (119887 minus 119888)
6 Abstract and Applied Analysis
1198913(119905 119909) = 3119909
13
+ 1199052
+ 3119888
1198923(119905 119909) =
3119909
(1 + 1199052) (1 + 119909)+ 3 (119887 minus 119888)
(47)
Let 120574 = 13 and 0 lt 119888 lt 119887 It is obvious that 119891119894 119892119894isin
119862[119869 times R+R+] and are increasing with respect to the secondargument 119892
119894(119905 0) = 119887 minus 119888 gt 0 119894 = 1 2 3 On the other hand
for 120582 isin (0 1) 119905 isin 119869 119909 isin [0 +infin) 119894 = 1 2 3 we have
119892119894(119905 120582119909) =
119894120582119909
(1 + 1199052) (1 + 120582119909)+ 119894 (119887 minus 119888)
ge119894120582119909
(1 + 1199052) (1 + 120582119909)+ 119894120582 (119887 minus 119888)
= 120582119892119894(119905 119909)
119891119894(119905 120582119909) = 119894120582
13
11990913
+ 1199052
+ 119894119888
ge 12058213
(11989411990913
+ 1199052
+ 119894119888)
= 120582120574
119891119894(119905 119909)
(48)
Moreover for 119905 isin 119869 119909 isin R+ 119894 = 1 2 3 we have
119891119894(119905 119909) = 119894119909
13
+ 1199052
+ 119894119888
ge 119894119888 ge119888
3 + (119887 minus 119888)(119894 + 119894 (119887 minus 119888))
ge119888
3 + (119887 minus 119888)(
119894119909
(1 + 1199052) (1 + 119909)+ 119894 (119887 minus 119888))
= 1205750119892119894(119905 119909)
(49)
where
1205750=
119888
3 + (119887 minus 119888) (50)
Hence all the conditions of Theorem 10 are satisfied ThusBVP (46) has a unique positive solution in 119875
ℎ where ℎ(119905) = 119905
119905 isin [0 1]
Conflict of Interests
The authors declare that there is no conflict of interestsregarding the publication of this paper
Acknowledgments
This research was supported by Henan Province CollegeYouth Backbone Teacher Funds (2011GGJS-213) and theNational Natural Science Foundation of China (11271336)
References
[1] H T Li and Y S Liu ldquoOn the uniqueness of the positivesolution for a second-order integral boundary value problemwith switched nonlinearityrdquo Applied Mathematics Letters vol24 no 12 pp 2201ndash2205 2011
[2] A Cabada and G Wang ldquoPositive solutions of nonlinearfractional differential equations with integral boundary valueconditionsrdquo Journal of Mathematical Analysis and Applicationsvol 389 no 1 pp 403ndash411 2012
[3] A A Kilbas H M Srivastava and J J Trujjllo Theory andApplications of Fractional Differential Equations vol 204 ofNorth-Holland Mathematics Studies Elsevier Amsterdam TheNetherlands 2006
[4] I Podlubny Fractional Differential Equations Academic PressNew York NY USA 1993
[5] X Zhang L Liu and Y Wu ldquoThe eigenvalue problem for asingular higher order fractional differential equation involvingfractional derivativesrdquo Applied Mathematics and Computationvol 218 no 17 pp 8526ndash8536 2012
[6] Z-W Lv J Liang and T-J Xiao ldquoSolutions to the Cauchy prob-lem for differential equations in Banach spaces with fractionalorderrdquo Computers ampMathematics with Applications vol 62 no3 pp 1303ndash1311 2011
[7] R P Agarwal V Lakshmikantham and J J Nieto ldquoOn theconcept of solution for fractional differential equations withuncertaintyrdquo Nonlinear Analysis Theory Methods amp Applica-tions vol 72 no 6 pp 2859ndash2862 2010
[8] R-N Wang T-J Xiao and J Liang ldquoA note on the fractionalCauchy problems with nonlocal initial conditionsrdquo AppliedMathematics Letters vol 24 no 8 pp 1435ndash1442 2011
[9] R-N Wang D-H Chen and T-J Xiao ldquoAbstract fractionalCauchy problems with almost sectorial operatorsrdquo Journal ofDifferential Equations vol 252 no 1 pp 202ndash235 2012
[10] J Henderson and A Ouahab ldquoFractional functional differentialinclusions with finite delayrdquo Nonlinear Analysis Theory Meth-ods amp Applications vol 70 no 5 pp 2091ndash2105 2009
[11] V Lakshmikantham ldquoTheory of fractional functional differen-tial equationsrdquo Nonlinear Analysis Theory Methods amp Applica-tions vol 69 no 10 pp 3337ndash3343 2008
[12] V Lakshmikantham and A S Vatsala ldquoBasic theory of frac-tional differential equationsrdquo Nonlinear Analysis Theory Meth-ods amp Applications vol 69 no 8 pp 2677ndash2682 2008
[13] F Li ldquoMild solutions for fractional differential equations withnonlocal conditionsrdquo Advances in Difference Equations ArticleID 287861 9 pages 2010
[14] X Q Zhang ldquoPositive solution for a class of singular semiposi-tone fractional differential equations with integral boundaryconditionsrdquo Boundary Value Problems vol 2012 article 1232012
[15] C Yang and C Zhai ldquoUniqueness of positive solutions for afractional differential equation via a fixed point theorem of asum operatorrdquo Electronic Journal of Differential Equations vol2012 no 70 pp 1ndash8 2012
[16] A A Agrachev and D Liberzon ldquoLie-algebraic stability criteriafor switched systemsrdquo SIAM Journal on Control and Optimiza-tion vol 40 no 1 pp 253ndash269 2001
[17] J Daafouz P Riedinger and C Iung ldquoStability analysis andcontrol synthesis for switched systems a switched Lyapunovfunction approachrdquo IEEE Transactions on Automatic Controlvol 47 no 11 pp 1883ndash1887 2002
[18] L Gurvits R Shorten and O Mason ldquoOn the stability ofswitched positive linear systemsrdquo IEEE Transactions on Auto-matic Control vol 52 no 6 pp 1099ndash1103 2007
Abstract and Applied Analysis 7
[19] C Zhai and D R Anderson ldquoA sum operator equation andapplications to nonlinear elastic beam equations and Lane-Emden-Fowler equationsrdquo Journal of Mathematical Analysisand Applications vol 375 no 2 pp 388ndash400 2011
[20] D J Guo and V Lakshmikantham Nonlinear Problems inAbstract Cone vol 5 Academic Press San Diego Calif USA1988
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 3
It is obvious that
119866 (119905 119904) le 1198661(119904 119904) 119905 119904 isin [0 1] (13)
We consider the Banach space119864 = 119862[[0 1]R] endowedwiththe norm defined by 119906 = sup
0le119905le1|119906(119905)| Letting 119875 = 119906 isin
119864 | 119906(119905) ge 0 then 119875 is a cone in 119864 Define an operator ϝ
119864 rarr 119864 as
(ϝ119906) (119905) = int
1
0
119866 (119905 119904) (119891120590(119904)
(119904 119906 (119904)) + 119892120590(119904)
(119904 119906 (119904))) 119889119904
(14)
Then ϝ has a solution if and only if the operator ϝ has a fixedpoint
Theorem 8 Let 119891119894 119892119894isin 119862[119869 times R+R+] 119894 = 1 2 119873
Suppose that the following conditions are satisfied1003816100381610038161003816119891119894 (119905 119906 (119905)) minus 119891
119894(119905 V (119905))1003816100381610038161003816 le 119897
119894(119905) |119906 (119905) minus V (119905)|
1003816100381610038161003816119892119894 (119905 119906 (119905)) minus 119892119894(119905 V (119905))1003816100381610038161003816 le 119897
119894(119905) |119906 (119905) minus V (119905)|
0 lt int
1
0
1198661(119904 119904) (119897
119894(119904) + 119897
119894(119904)) 119889119904 lt 1
(15)
where
119897119894 119897119894isin 119862 [119869R
+
] 119894 = 1 2 119873 (16)
Then the problem (3) has a unique solution on [0 1]
Proof It follows from Lemma 6 that ϝ 119875 rarr 119875 For 119905 isin
119869 119894 = 1 2 119873 we set max119894=12119873
sup119905isin119869
|119891119894(119905 0)| = 119872
max119894=12119873
sup119905isin119869
|119892119894(119905 0)| = 119872 and 119861
119903= 119906 isin 119862[119869R+]
119906 le 119903 where
119903 ge
(119872 +119872)int1
0
1198661(119904 119904) 119889119904
1 minusmax119894=12119873
int1
0
1198661(119904 119904) (119897
119894(119904) + 119897
119894(119904)) 119889119904
(17)
Step 1We show that ϝ (119861119903) sub 119861119903
For 119906 isin 119861119903and 119905 isin 119869 119894 = 1 2 119873
int
1
0
119866 (119905 119904)1003816100381610038161003816119891119894 (119904 119906 (119904)) + 119892
119894(119904 119906 (119904))
1003816100381610038161003816 119889119904
le int
1
0
1198661(119904 119904) (
1003816100381610038161003816119891119894 (119904 119906 (119904)) minus 119891119894(119904 0)
1003816100381610038161003816 +1003816100381610038161003816119891119894 (119904 0)
1003816100381610038161003816) 119889119904
+ int
1
0
1198661(119904 119904) (
1003816100381610038161003816119892119894 (119904 119906 (119904)) minus 119892119894(119904 0)
1003816100381610038161003816 +1003816100381610038161003816119892119894 (119904 0)
1003816100381610038161003816) 119889119904
le 119903 max119894=12119873
int
1
0
1198661(119904 119904) (119897
119894(119904) + 119897
119894(119904)) 119889119904
+ (119872 +119872)int
1
0
1198661(119904 119904) 119889119904
le 119903
(18)
which implies that |(ϝ119906)(119905)| le 119903 Thus ϝ119906 le 119903 Therefore
ϝ (119861119903) sub 119861119903 (19)
Step 2We show that ϝ is a contraction mappingFor 119906 V isin 119861
119903and for each 119905 isin 119869 119894 = 1 2 119873 we have
int
1
0
119866 (119905 119904)1003816100381610038161003816119891119894 (119904 119906 (119904)) minus 119891
119894(119904 V (119904))1003816100381610038161003816 119889119904
+ int
1
0
119866 (119905 119904)1003816100381610038161003816119892119894 (119904 119906 (119904)) minus 119892
119894(119904 V (119904))1003816100381610038161003816 119889119904
le int
1
0
1198661(119904 119904) 119897
119894(119904) |119906 (119904) minus V (119904)| 119889119904
+ int
1
0
1198661(119904 119904) 119897
119894(119904) |119906 (119904) minus V (119904)| 119889119904
le int
1
0
1198661(119904 119904) (119897
119894(119904) + 119897
119894(119904)) 119889119904 119906 minus V
(20)
This together with 0 lt int1
0
1198661(119904 119904)(119897
119894(119904) + 119897
119894(119904))119889119904 lt 1 119894 =
1 2 119873 yields that1003816100381610038161003816(ϝ119906) (119905) minus (ϝV) (119905)1003816100381610038161003816 le 119896 119906 minus V (21)
where
0 lt 119896 = max119894=12119873
int
1
0
1198661(119904 119904) (119897
119894(119904) + 119897
119894(119904)) 119889119904 lt 1 (22)
Thus1003817100381710038171003817ϝ119906 minus ϝV1003817100381710038171003817 le 119896 119906 minus V (23)
This means that ϝ is a contraction mappingIt follows from Banachrsquos contraction mapping that ϝ has
a unique fixed point in 119861119903 Therefore the problem (3) has a
unique solution
Corollary 9 Let 119891119894isin 119862[119869 timesR+R+] 119894 = 1 2 119873 Suppose
that the following conditions are satisfied1003816100381610038161003816119891119894 (119905 119906 (119905)) minus 119891
119894(119905 V (119905))1003816100381610038161003816 le 119897
119894(119905) |119906 (119905) minus V (119905)|
0 lt int
1
0
1198661(119904 119904) 119897
119894(119904) 119889119904 lt 1
(24)
where
119897119894isin 119862 [119869R
+
] 119894 = 1 2 119873 (25)
Then the following fractional switched system119888
119863120572
0+119906 (119905) + 119891
120590(119905)(119905 119906 (119905)) = 0 119905 isin 119869 = [0 1]
119906 (0) = 11990610158401015840
(0) = 0 119906 (1) = int
1
0
119906 (119904) 119889119904
(26)
has a unique solution on [0 1]
Theorem 10 Assume that
(H1) 119891119894 119892119894
isin 119862[119869 times R+R+] and 119891119894(119905 119909) 119892
119894(119905 119909) are
increasing in 119909 for 119909 isin R+ 119892119894(119905 0) = 0 119894 = 1 2 119873
(H2) 119892119894(119905 120582119909) ge 120582119892
119894(119905 119909) for 120582 isin (0 1) 119905 isin 119869 119909 isin
R+ and there exists a constant 120574 isin (0 1) such that119891119894(119905 120582119909) ge 120582
120574
119891119894(119905 119909) forall119905 isin 119869 120582 isin (0 1) 119909 isin R+
119894 = 1 2 119873
4 Abstract and Applied Analysis
(H3) there exists a constant 120575
0gt 0 such that 119891
119894(119905 119909) ge
1205750119892119894(119905 119909) 119905 isin 119869 119909 isin R+ 119894 = 1 2 119873
Then problem (3) has a unique solution 119906lowast in 119875ℎ where ℎ(119905) =
119905 119905 isin 119869 Moreover for any initial value 1199060isin 119875ℎ constructing
successively the sequence
119906119899+1
(119905) = int
1
0
119866 (119905 119904) (119891120590(119904)
(119904 119906119899(119904)) + 119892
120590(119904)(119904 119906119899(119904))) 119889119904
119899 = 0 1 2
(27)
we have 119906119899(119905) rarr 119906
lowast
(119905) as 119899 rarr infin
Proof Define the two operators
119860119906 (119905) = int
1
0
119866 (119905 119904) 119891120590(119904)
(119904 119906 (119904)) 119889119904
119861119906 (119905) = int
1
0
119866 (119905 119904) 119892120590(119904)
(119904 119906 (119904)) 119889119904
(28)
From Lemma 6 we have 119860 119875 rarr 119875 and 119861 119875 rarr 119875 It isobvious that 119906 is the solution of problem (3) if and only if 119906 =
119860119906+119861119906 It follows from (H1) that119860 and 119861 are two increasing
operators Thus for 119906 V isin 119875 119906 ge V we have 119860119906 ge 119860V and119861119906 ge 119861VStep 1 We show that 119860 is a 120574-concave operator and 119861 is asubhomogeneous operator
In fact for 120582 isin (0 1) 119906 isin 119875 119905 isin 119869 119894 = 1 2 119873 from(H2) we have
int
1
0
119866 (119905 119904) 119891119894(119904 120582119906 (119904)) 119889119904 ge 120582
120574
int
1
0
119866 (119905 119904) 119891119894(119904 119906 (119904)) 119889119904
(29)
which yields that
119860 (120582119906) (119905) ge 120582120574
119860119906 (119905) (30)
Thus 119860 is a 120574-concave operator By a closely similar way wecan see that 119861 is a subhomogeneous operator
Step 2 We show that 119860ℎ isin 119875ℎand 119861ℎ isin 119875
ℎ
From Lemma 6 and (H1) we have for 119905 isin 119869 119894 =
1 2 119873
int
1
0
119866 (119905 119904) 119891119894(119904 ℎ (119904)) 119889119904
le2
Γ (120572 + 1)ℎ (119905)
times int
1
0
(1 minus 119904)120572minus1
(120572 minus 1 + 119904) 119891119894(119904 1) 119889119904
int
1
0
119866 (119905 119904) 119891119894(119904 ℎ (119904)) 119889119904
ge1
Γ (120572 + 1)ℎ (119905)
times int
1
0
(1 minus 119904)120572minus1
(120572 minus 2 + 2119904) 119891119894(119904 0) 119889119904
(31)
For 119894 = 1 2 119873 let
119898119894=
1
Γ (120572 + 1)int
1
0
(1 minus 119904)120572minus1
(120572 minus 2 + 2119904) 119891119894(119904 0) 119889119904
119898119894=
2
Γ (120572 + 1)int
1
0
1 minus 119904120572minus1
(120572 minus 1 + 119904) 119891119894(119904 1) 119889119904
(32)
It follows from 119892(119905 0) = 0 that int10
119891119894(119904 1) 119889119904 ge int
1
0
119891119894(119904 0) 119889119904 ge
1205750int1
0
119892119894(119904 0) 119889119904 gt 0
Thus
119898119894gt 0 119898
119894gt 0 119894 = 1 2 119873 (33)
Letting = min119898119894 119894 = 1 2 119873 and = max119898
119894 119894 =
1 2 119873 then gt 0 and gt 0 Therefore
ℎ (119905) le 119860ℎ (119905) le ℎ (119905) (34)
which implies that
119860ℎ isin 119875ℎ (35)
Similarly we have 119861ℎ isin 119875ℎ
Step 3 There exists a constant 1205750gt 0 such that 119860119906 ge 120575
0119861119906
forall119906 isin 119875For 119906 isin 119875 and 119905 isin 119869 119894 = 1 2 119873 by (H
3) we have
int
1
0
119866 (119905 119904) 119891119894(119904 119906 (119904)) 119889119904 ge 120575
0int
1
0
119866 (119905 119904) 119892119894(119904 119906 (119904)) 119889119904 (36)
This means that
119860119906 ge 1205750119861119906 119906 isin 119875 (37)
Therefore the conditions of Theorem 7 are satisfied Bymeans of Theorem 7 we obtain that the operator equation119860119906+119861119906 = 119906 has a unique solution 119906lowast in119875
ℎ Moreover for any
initial value 1199060isin 119875ℎ constructing successively the sequence
119906119899+1
(119905) = int
1
0
119866 (119905 119904) (119891120590(119904)
(119904 119906119899(119904)) + 119892
120590(119904)(119904 119906119899(119904))) 119889119904
119899 = 0 1 2
(38)
we have 119906119899(119905) rarr 119906
lowast
(119905) as 119899 rarr infin
In Theorem 10 if we let 119861 be a null operator we have thefollowing conclusion
Corollary 11 Assume that
(H4) 119891119894isin 119862[119869 times R+R+] and 119891
119894(119905 119909) is increasing in 119909 for
119909 isin R+ 119891119894(119905 0) = 0 119894 = 1 2 119873
Abstract and Applied Analysis 5
(H5) there exists a constant 120574 isin (0 1) such that 119891
119894(119905 120582119909) ge
120582120574
119891119894(119905 119909) forall119905 isin 119869 120582 isin (0 1) 119909 isin R+ 119894 = 1 2 119873
Then the following fractional switched system
119888
119863120572
0+119906 (119905) + 119891
120590(119905)(119905 119906 (119905)) = 0 119905 isin 119869 = [0 1]
119906 (0) = 11990610158401015840
(0) = 0 119906 (1) = int
1
0
119906 (119904) 119889119904
(39)
has a unique solution119906lowast in119875ℎ where ℎ(119905) = 119905 119905 isin 119869Moreover
for any initial value 1199060
isin 119875ℎ constructing successively the
sequence
119906119899+1
(119905) = int
1
0
119866 (119905 119904) 119891120590(119904)
(119904 119906119899(119904)) 119889119904 119899 = 0 1 2
(40)
we have 119906119899(119905) rarr 119906
lowast
(119905) as 119899 rarr infin
4 Examples
Example 1 Consider the following boundary value problem
119888
119863120572
0+119906 (119905) + 119891
120590(119905)(119905 119906 (119905)) + 119892
120590(119905)(119905 119906 (119905)) = 0
119905 isin 119869 = [0 1]
119906 (0) = 11990610158401015840
(0) = 0 119906 (1) = int
1
0
119906 (119904) 119889119904
(41)
where 120572 = 52 120590(119905) 119869 rarr 119872 = 1 2 is a finite switchingsignal
1198911(119905 119909) =
1
4(119905 + 2)2
119909
1 + 119909+ 1
1198921(119905 119909) =
1
16sin2119909 + 119905
2
1198912(119905 119909) =
1
8(119905 + 2)2
119909
1 + 119909+ 1
1198922(119905 119909) =
1
32sin2119909 + 119905
2
(42)
Thus
119891119894 119892119894isin 119862 [119869 timesR
+
R+
] 119894 = 1 2 (43)
By computation we deduce that
10038161003816100381610038161198911 (119905 1199091) minus 1198911(119905 1199092)1003816100381610038161003816 le
1
16
10038161003816100381610038161199092 minus 1199091
1003816100381610038161003816
10038161003816100381610038161198921 (119905 1199091) minus 1198921(119905 1199092)1003816100381610038161003816 le
1
16
10038161003816100381610038161199092 minus 1199091
1003816100381610038161003816
10038161003816100381610038161198912 (119905 1199091) minus 1198912(119905 1199092)1003816100381610038161003816 le
1
32
10038161003816100381610038161199092 minus 1199091
1003816100381610038161003816
10038161003816100381610038161198922 (119905 1199091) minus 1198922(119905 1199092)1003816100381610038161003816 le
1
32
10038161003816100381610038161199092 minus 1199091
1003816100381610038161003816
(44)
On the other hand
int
1
0
1198661(119904 119904) (119897
1(119904) + 119897
1(119904)) 119889119904
= int
1
0
2(1 minus 119904)(52)minus1
((52) minus 1 + 119904)
Γ ((52) + 1)(1
16+
1
16) 119889119904
=1
4Γ (72)int
1
0
(1 minus 119904)32
(3
2+ 119904) 119889119904
le1
4Γ (72)int
1
0
(1 minus 119904)32
(3
2+ 1) 119889119904
=1
3radic120587times2
5
lt 1
int
1
0
1198661(119904 119904) (119897
1(119904) + 119897
1(119904)) 119889119904
= int
1
0
2(1 minus 119904)(52)minus1
((52) minus 1 + 119904)
Γ ((52) + 1)(1
32+
1
32) 119889119904
=1
3radic120587times1
5
lt 1
(45)
Hence byTheorem 8 BVP (41) has a unique positive solutionon [0 1]
Example 2 Consider the following boundary value problem
119888119863120572
0+119906 (119905) + 119891
120590(119905)(119905 119906 (119905)) + 119892
120590(119905)(119905 119906 (119905)) = 0
119905 isin 119869 = [0 1]
119906 (0) = 11990610158401015840
(0) = 0 119906 (1) = int
1
0
119906 (119904) 119889119904
(46)
where 120572 = 52 120590(119905) 119869 rarr 1 2 3 is a finite switching signal
1198911(119905 119909) = 119909
13
+ 1199052
+ 119888
1198921(119905 119909) =
119909
(1 + 1199052) (1 + 119909)+ 119887 minus 119888
1198912(119905 119909) = 2119909
13
+ 1199052
+ 2119888
1198922(119905 119909) =
2119909
(1 + 1199052) (1 + 119909)+ 2 (119887 minus 119888)
6 Abstract and Applied Analysis
1198913(119905 119909) = 3119909
13
+ 1199052
+ 3119888
1198923(119905 119909) =
3119909
(1 + 1199052) (1 + 119909)+ 3 (119887 minus 119888)
(47)
Let 120574 = 13 and 0 lt 119888 lt 119887 It is obvious that 119891119894 119892119894isin
119862[119869 times R+R+] and are increasing with respect to the secondargument 119892
119894(119905 0) = 119887 minus 119888 gt 0 119894 = 1 2 3 On the other hand
for 120582 isin (0 1) 119905 isin 119869 119909 isin [0 +infin) 119894 = 1 2 3 we have
119892119894(119905 120582119909) =
119894120582119909
(1 + 1199052) (1 + 120582119909)+ 119894 (119887 minus 119888)
ge119894120582119909
(1 + 1199052) (1 + 120582119909)+ 119894120582 (119887 minus 119888)
= 120582119892119894(119905 119909)
119891119894(119905 120582119909) = 119894120582
13
11990913
+ 1199052
+ 119894119888
ge 12058213
(11989411990913
+ 1199052
+ 119894119888)
= 120582120574
119891119894(119905 119909)
(48)
Moreover for 119905 isin 119869 119909 isin R+ 119894 = 1 2 3 we have
119891119894(119905 119909) = 119894119909
13
+ 1199052
+ 119894119888
ge 119894119888 ge119888
3 + (119887 minus 119888)(119894 + 119894 (119887 minus 119888))
ge119888
3 + (119887 minus 119888)(
119894119909
(1 + 1199052) (1 + 119909)+ 119894 (119887 minus 119888))
= 1205750119892119894(119905 119909)
(49)
where
1205750=
119888
3 + (119887 minus 119888) (50)
Hence all the conditions of Theorem 10 are satisfied ThusBVP (46) has a unique positive solution in 119875
ℎ where ℎ(119905) = 119905
119905 isin [0 1]
Conflict of Interests
The authors declare that there is no conflict of interestsregarding the publication of this paper
Acknowledgments
This research was supported by Henan Province CollegeYouth Backbone Teacher Funds (2011GGJS-213) and theNational Natural Science Foundation of China (11271336)
References
[1] H T Li and Y S Liu ldquoOn the uniqueness of the positivesolution for a second-order integral boundary value problemwith switched nonlinearityrdquo Applied Mathematics Letters vol24 no 12 pp 2201ndash2205 2011
[2] A Cabada and G Wang ldquoPositive solutions of nonlinearfractional differential equations with integral boundary valueconditionsrdquo Journal of Mathematical Analysis and Applicationsvol 389 no 1 pp 403ndash411 2012
[3] A A Kilbas H M Srivastava and J J Trujjllo Theory andApplications of Fractional Differential Equations vol 204 ofNorth-Holland Mathematics Studies Elsevier Amsterdam TheNetherlands 2006
[4] I Podlubny Fractional Differential Equations Academic PressNew York NY USA 1993
[5] X Zhang L Liu and Y Wu ldquoThe eigenvalue problem for asingular higher order fractional differential equation involvingfractional derivativesrdquo Applied Mathematics and Computationvol 218 no 17 pp 8526ndash8536 2012
[6] Z-W Lv J Liang and T-J Xiao ldquoSolutions to the Cauchy prob-lem for differential equations in Banach spaces with fractionalorderrdquo Computers ampMathematics with Applications vol 62 no3 pp 1303ndash1311 2011
[7] R P Agarwal V Lakshmikantham and J J Nieto ldquoOn theconcept of solution for fractional differential equations withuncertaintyrdquo Nonlinear Analysis Theory Methods amp Applica-tions vol 72 no 6 pp 2859ndash2862 2010
[8] R-N Wang T-J Xiao and J Liang ldquoA note on the fractionalCauchy problems with nonlocal initial conditionsrdquo AppliedMathematics Letters vol 24 no 8 pp 1435ndash1442 2011
[9] R-N Wang D-H Chen and T-J Xiao ldquoAbstract fractionalCauchy problems with almost sectorial operatorsrdquo Journal ofDifferential Equations vol 252 no 1 pp 202ndash235 2012
[10] J Henderson and A Ouahab ldquoFractional functional differentialinclusions with finite delayrdquo Nonlinear Analysis Theory Meth-ods amp Applications vol 70 no 5 pp 2091ndash2105 2009
[11] V Lakshmikantham ldquoTheory of fractional functional differen-tial equationsrdquo Nonlinear Analysis Theory Methods amp Applica-tions vol 69 no 10 pp 3337ndash3343 2008
[12] V Lakshmikantham and A S Vatsala ldquoBasic theory of frac-tional differential equationsrdquo Nonlinear Analysis Theory Meth-ods amp Applications vol 69 no 8 pp 2677ndash2682 2008
[13] F Li ldquoMild solutions for fractional differential equations withnonlocal conditionsrdquo Advances in Difference Equations ArticleID 287861 9 pages 2010
[14] X Q Zhang ldquoPositive solution for a class of singular semiposi-tone fractional differential equations with integral boundaryconditionsrdquo Boundary Value Problems vol 2012 article 1232012
[15] C Yang and C Zhai ldquoUniqueness of positive solutions for afractional differential equation via a fixed point theorem of asum operatorrdquo Electronic Journal of Differential Equations vol2012 no 70 pp 1ndash8 2012
[16] A A Agrachev and D Liberzon ldquoLie-algebraic stability criteriafor switched systemsrdquo SIAM Journal on Control and Optimiza-tion vol 40 no 1 pp 253ndash269 2001
[17] J Daafouz P Riedinger and C Iung ldquoStability analysis andcontrol synthesis for switched systems a switched Lyapunovfunction approachrdquo IEEE Transactions on Automatic Controlvol 47 no 11 pp 1883ndash1887 2002
[18] L Gurvits R Shorten and O Mason ldquoOn the stability ofswitched positive linear systemsrdquo IEEE Transactions on Auto-matic Control vol 52 no 6 pp 1099ndash1103 2007
Abstract and Applied Analysis 7
[19] C Zhai and D R Anderson ldquoA sum operator equation andapplications to nonlinear elastic beam equations and Lane-Emden-Fowler equationsrdquo Journal of Mathematical Analysisand Applications vol 375 no 2 pp 388ndash400 2011
[20] D J Guo and V Lakshmikantham Nonlinear Problems inAbstract Cone vol 5 Academic Press San Diego Calif USA1988
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
4 Abstract and Applied Analysis
(H3) there exists a constant 120575
0gt 0 such that 119891
119894(119905 119909) ge
1205750119892119894(119905 119909) 119905 isin 119869 119909 isin R+ 119894 = 1 2 119873
Then problem (3) has a unique solution 119906lowast in 119875ℎ where ℎ(119905) =
119905 119905 isin 119869 Moreover for any initial value 1199060isin 119875ℎ constructing
successively the sequence
119906119899+1
(119905) = int
1
0
119866 (119905 119904) (119891120590(119904)
(119904 119906119899(119904)) + 119892
120590(119904)(119904 119906119899(119904))) 119889119904
119899 = 0 1 2
(27)
we have 119906119899(119905) rarr 119906
lowast
(119905) as 119899 rarr infin
Proof Define the two operators
119860119906 (119905) = int
1
0
119866 (119905 119904) 119891120590(119904)
(119904 119906 (119904)) 119889119904
119861119906 (119905) = int
1
0
119866 (119905 119904) 119892120590(119904)
(119904 119906 (119904)) 119889119904
(28)
From Lemma 6 we have 119860 119875 rarr 119875 and 119861 119875 rarr 119875 It isobvious that 119906 is the solution of problem (3) if and only if 119906 =
119860119906+119861119906 It follows from (H1) that119860 and 119861 are two increasing
operators Thus for 119906 V isin 119875 119906 ge V we have 119860119906 ge 119860V and119861119906 ge 119861VStep 1 We show that 119860 is a 120574-concave operator and 119861 is asubhomogeneous operator
In fact for 120582 isin (0 1) 119906 isin 119875 119905 isin 119869 119894 = 1 2 119873 from(H2) we have
int
1
0
119866 (119905 119904) 119891119894(119904 120582119906 (119904)) 119889119904 ge 120582
120574
int
1
0
119866 (119905 119904) 119891119894(119904 119906 (119904)) 119889119904
(29)
which yields that
119860 (120582119906) (119905) ge 120582120574
119860119906 (119905) (30)
Thus 119860 is a 120574-concave operator By a closely similar way wecan see that 119861 is a subhomogeneous operator
Step 2 We show that 119860ℎ isin 119875ℎand 119861ℎ isin 119875
ℎ
From Lemma 6 and (H1) we have for 119905 isin 119869 119894 =
1 2 119873
int
1
0
119866 (119905 119904) 119891119894(119904 ℎ (119904)) 119889119904
le2
Γ (120572 + 1)ℎ (119905)
times int
1
0
(1 minus 119904)120572minus1
(120572 minus 1 + 119904) 119891119894(119904 1) 119889119904
int
1
0
119866 (119905 119904) 119891119894(119904 ℎ (119904)) 119889119904
ge1
Γ (120572 + 1)ℎ (119905)
times int
1
0
(1 minus 119904)120572minus1
(120572 minus 2 + 2119904) 119891119894(119904 0) 119889119904
(31)
For 119894 = 1 2 119873 let
119898119894=
1
Γ (120572 + 1)int
1
0
(1 minus 119904)120572minus1
(120572 minus 2 + 2119904) 119891119894(119904 0) 119889119904
119898119894=
2
Γ (120572 + 1)int
1
0
1 minus 119904120572minus1
(120572 minus 1 + 119904) 119891119894(119904 1) 119889119904
(32)
It follows from 119892(119905 0) = 0 that int10
119891119894(119904 1) 119889119904 ge int
1
0
119891119894(119904 0) 119889119904 ge
1205750int1
0
119892119894(119904 0) 119889119904 gt 0
Thus
119898119894gt 0 119898
119894gt 0 119894 = 1 2 119873 (33)
Letting = min119898119894 119894 = 1 2 119873 and = max119898
119894 119894 =
1 2 119873 then gt 0 and gt 0 Therefore
ℎ (119905) le 119860ℎ (119905) le ℎ (119905) (34)
which implies that
119860ℎ isin 119875ℎ (35)
Similarly we have 119861ℎ isin 119875ℎ
Step 3 There exists a constant 1205750gt 0 such that 119860119906 ge 120575
0119861119906
forall119906 isin 119875For 119906 isin 119875 and 119905 isin 119869 119894 = 1 2 119873 by (H
3) we have
int
1
0
119866 (119905 119904) 119891119894(119904 119906 (119904)) 119889119904 ge 120575
0int
1
0
119866 (119905 119904) 119892119894(119904 119906 (119904)) 119889119904 (36)
This means that
119860119906 ge 1205750119861119906 119906 isin 119875 (37)
Therefore the conditions of Theorem 7 are satisfied Bymeans of Theorem 7 we obtain that the operator equation119860119906+119861119906 = 119906 has a unique solution 119906lowast in119875
ℎ Moreover for any
initial value 1199060isin 119875ℎ constructing successively the sequence
119906119899+1
(119905) = int
1
0
119866 (119905 119904) (119891120590(119904)
(119904 119906119899(119904)) + 119892
120590(119904)(119904 119906119899(119904))) 119889119904
119899 = 0 1 2
(38)
we have 119906119899(119905) rarr 119906
lowast
(119905) as 119899 rarr infin
In Theorem 10 if we let 119861 be a null operator we have thefollowing conclusion
Corollary 11 Assume that
(H4) 119891119894isin 119862[119869 times R+R+] and 119891
119894(119905 119909) is increasing in 119909 for
119909 isin R+ 119891119894(119905 0) = 0 119894 = 1 2 119873
Abstract and Applied Analysis 5
(H5) there exists a constant 120574 isin (0 1) such that 119891
119894(119905 120582119909) ge
120582120574
119891119894(119905 119909) forall119905 isin 119869 120582 isin (0 1) 119909 isin R+ 119894 = 1 2 119873
Then the following fractional switched system
119888
119863120572
0+119906 (119905) + 119891
120590(119905)(119905 119906 (119905)) = 0 119905 isin 119869 = [0 1]
119906 (0) = 11990610158401015840
(0) = 0 119906 (1) = int
1
0
119906 (119904) 119889119904
(39)
has a unique solution119906lowast in119875ℎ where ℎ(119905) = 119905 119905 isin 119869Moreover
for any initial value 1199060
isin 119875ℎ constructing successively the
sequence
119906119899+1
(119905) = int
1
0
119866 (119905 119904) 119891120590(119904)
(119904 119906119899(119904)) 119889119904 119899 = 0 1 2
(40)
we have 119906119899(119905) rarr 119906
lowast
(119905) as 119899 rarr infin
4 Examples
Example 1 Consider the following boundary value problem
119888
119863120572
0+119906 (119905) + 119891
120590(119905)(119905 119906 (119905)) + 119892
120590(119905)(119905 119906 (119905)) = 0
119905 isin 119869 = [0 1]
119906 (0) = 11990610158401015840
(0) = 0 119906 (1) = int
1
0
119906 (119904) 119889119904
(41)
where 120572 = 52 120590(119905) 119869 rarr 119872 = 1 2 is a finite switchingsignal
1198911(119905 119909) =
1
4(119905 + 2)2
119909
1 + 119909+ 1
1198921(119905 119909) =
1
16sin2119909 + 119905
2
1198912(119905 119909) =
1
8(119905 + 2)2
119909
1 + 119909+ 1
1198922(119905 119909) =
1
32sin2119909 + 119905
2
(42)
Thus
119891119894 119892119894isin 119862 [119869 timesR
+
R+
] 119894 = 1 2 (43)
By computation we deduce that
10038161003816100381610038161198911 (119905 1199091) minus 1198911(119905 1199092)1003816100381610038161003816 le
1
16
10038161003816100381610038161199092 minus 1199091
1003816100381610038161003816
10038161003816100381610038161198921 (119905 1199091) minus 1198921(119905 1199092)1003816100381610038161003816 le
1
16
10038161003816100381610038161199092 minus 1199091
1003816100381610038161003816
10038161003816100381610038161198912 (119905 1199091) minus 1198912(119905 1199092)1003816100381610038161003816 le
1
32
10038161003816100381610038161199092 minus 1199091
1003816100381610038161003816
10038161003816100381610038161198922 (119905 1199091) minus 1198922(119905 1199092)1003816100381610038161003816 le
1
32
10038161003816100381610038161199092 minus 1199091
1003816100381610038161003816
(44)
On the other hand
int
1
0
1198661(119904 119904) (119897
1(119904) + 119897
1(119904)) 119889119904
= int
1
0
2(1 minus 119904)(52)minus1
((52) minus 1 + 119904)
Γ ((52) + 1)(1
16+
1
16) 119889119904
=1
4Γ (72)int
1
0
(1 minus 119904)32
(3
2+ 119904) 119889119904
le1
4Γ (72)int
1
0
(1 minus 119904)32
(3
2+ 1) 119889119904
=1
3radic120587times2
5
lt 1
int
1
0
1198661(119904 119904) (119897
1(119904) + 119897
1(119904)) 119889119904
= int
1
0
2(1 minus 119904)(52)minus1
((52) minus 1 + 119904)
Γ ((52) + 1)(1
32+
1
32) 119889119904
=1
3radic120587times1
5
lt 1
(45)
Hence byTheorem 8 BVP (41) has a unique positive solutionon [0 1]
Example 2 Consider the following boundary value problem
119888119863120572
0+119906 (119905) + 119891
120590(119905)(119905 119906 (119905)) + 119892
120590(119905)(119905 119906 (119905)) = 0
119905 isin 119869 = [0 1]
119906 (0) = 11990610158401015840
(0) = 0 119906 (1) = int
1
0
119906 (119904) 119889119904
(46)
where 120572 = 52 120590(119905) 119869 rarr 1 2 3 is a finite switching signal
1198911(119905 119909) = 119909
13
+ 1199052
+ 119888
1198921(119905 119909) =
119909
(1 + 1199052) (1 + 119909)+ 119887 minus 119888
1198912(119905 119909) = 2119909
13
+ 1199052
+ 2119888
1198922(119905 119909) =
2119909
(1 + 1199052) (1 + 119909)+ 2 (119887 minus 119888)
6 Abstract and Applied Analysis
1198913(119905 119909) = 3119909
13
+ 1199052
+ 3119888
1198923(119905 119909) =
3119909
(1 + 1199052) (1 + 119909)+ 3 (119887 minus 119888)
(47)
Let 120574 = 13 and 0 lt 119888 lt 119887 It is obvious that 119891119894 119892119894isin
119862[119869 times R+R+] and are increasing with respect to the secondargument 119892
119894(119905 0) = 119887 minus 119888 gt 0 119894 = 1 2 3 On the other hand
for 120582 isin (0 1) 119905 isin 119869 119909 isin [0 +infin) 119894 = 1 2 3 we have
119892119894(119905 120582119909) =
119894120582119909
(1 + 1199052) (1 + 120582119909)+ 119894 (119887 minus 119888)
ge119894120582119909
(1 + 1199052) (1 + 120582119909)+ 119894120582 (119887 minus 119888)
= 120582119892119894(119905 119909)
119891119894(119905 120582119909) = 119894120582
13
11990913
+ 1199052
+ 119894119888
ge 12058213
(11989411990913
+ 1199052
+ 119894119888)
= 120582120574
119891119894(119905 119909)
(48)
Moreover for 119905 isin 119869 119909 isin R+ 119894 = 1 2 3 we have
119891119894(119905 119909) = 119894119909
13
+ 1199052
+ 119894119888
ge 119894119888 ge119888
3 + (119887 minus 119888)(119894 + 119894 (119887 minus 119888))
ge119888
3 + (119887 minus 119888)(
119894119909
(1 + 1199052) (1 + 119909)+ 119894 (119887 minus 119888))
= 1205750119892119894(119905 119909)
(49)
where
1205750=
119888
3 + (119887 minus 119888) (50)
Hence all the conditions of Theorem 10 are satisfied ThusBVP (46) has a unique positive solution in 119875
ℎ where ℎ(119905) = 119905
119905 isin [0 1]
Conflict of Interests
The authors declare that there is no conflict of interestsregarding the publication of this paper
Acknowledgments
This research was supported by Henan Province CollegeYouth Backbone Teacher Funds (2011GGJS-213) and theNational Natural Science Foundation of China (11271336)
References
[1] H T Li and Y S Liu ldquoOn the uniqueness of the positivesolution for a second-order integral boundary value problemwith switched nonlinearityrdquo Applied Mathematics Letters vol24 no 12 pp 2201ndash2205 2011
[2] A Cabada and G Wang ldquoPositive solutions of nonlinearfractional differential equations with integral boundary valueconditionsrdquo Journal of Mathematical Analysis and Applicationsvol 389 no 1 pp 403ndash411 2012
[3] A A Kilbas H M Srivastava and J J Trujjllo Theory andApplications of Fractional Differential Equations vol 204 ofNorth-Holland Mathematics Studies Elsevier Amsterdam TheNetherlands 2006
[4] I Podlubny Fractional Differential Equations Academic PressNew York NY USA 1993
[5] X Zhang L Liu and Y Wu ldquoThe eigenvalue problem for asingular higher order fractional differential equation involvingfractional derivativesrdquo Applied Mathematics and Computationvol 218 no 17 pp 8526ndash8536 2012
[6] Z-W Lv J Liang and T-J Xiao ldquoSolutions to the Cauchy prob-lem for differential equations in Banach spaces with fractionalorderrdquo Computers ampMathematics with Applications vol 62 no3 pp 1303ndash1311 2011
[7] R P Agarwal V Lakshmikantham and J J Nieto ldquoOn theconcept of solution for fractional differential equations withuncertaintyrdquo Nonlinear Analysis Theory Methods amp Applica-tions vol 72 no 6 pp 2859ndash2862 2010
[8] R-N Wang T-J Xiao and J Liang ldquoA note on the fractionalCauchy problems with nonlocal initial conditionsrdquo AppliedMathematics Letters vol 24 no 8 pp 1435ndash1442 2011
[9] R-N Wang D-H Chen and T-J Xiao ldquoAbstract fractionalCauchy problems with almost sectorial operatorsrdquo Journal ofDifferential Equations vol 252 no 1 pp 202ndash235 2012
[10] J Henderson and A Ouahab ldquoFractional functional differentialinclusions with finite delayrdquo Nonlinear Analysis Theory Meth-ods amp Applications vol 70 no 5 pp 2091ndash2105 2009
[11] V Lakshmikantham ldquoTheory of fractional functional differen-tial equationsrdquo Nonlinear Analysis Theory Methods amp Applica-tions vol 69 no 10 pp 3337ndash3343 2008
[12] V Lakshmikantham and A S Vatsala ldquoBasic theory of frac-tional differential equationsrdquo Nonlinear Analysis Theory Meth-ods amp Applications vol 69 no 8 pp 2677ndash2682 2008
[13] F Li ldquoMild solutions for fractional differential equations withnonlocal conditionsrdquo Advances in Difference Equations ArticleID 287861 9 pages 2010
[14] X Q Zhang ldquoPositive solution for a class of singular semiposi-tone fractional differential equations with integral boundaryconditionsrdquo Boundary Value Problems vol 2012 article 1232012
[15] C Yang and C Zhai ldquoUniqueness of positive solutions for afractional differential equation via a fixed point theorem of asum operatorrdquo Electronic Journal of Differential Equations vol2012 no 70 pp 1ndash8 2012
[16] A A Agrachev and D Liberzon ldquoLie-algebraic stability criteriafor switched systemsrdquo SIAM Journal on Control and Optimiza-tion vol 40 no 1 pp 253ndash269 2001
[17] J Daafouz P Riedinger and C Iung ldquoStability analysis andcontrol synthesis for switched systems a switched Lyapunovfunction approachrdquo IEEE Transactions on Automatic Controlvol 47 no 11 pp 1883ndash1887 2002
[18] L Gurvits R Shorten and O Mason ldquoOn the stability ofswitched positive linear systemsrdquo IEEE Transactions on Auto-matic Control vol 52 no 6 pp 1099ndash1103 2007
Abstract and Applied Analysis 7
[19] C Zhai and D R Anderson ldquoA sum operator equation andapplications to nonlinear elastic beam equations and Lane-Emden-Fowler equationsrdquo Journal of Mathematical Analysisand Applications vol 375 no 2 pp 388ndash400 2011
[20] D J Guo and V Lakshmikantham Nonlinear Problems inAbstract Cone vol 5 Academic Press San Diego Calif USA1988
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 5
(H5) there exists a constant 120574 isin (0 1) such that 119891
119894(119905 120582119909) ge
120582120574
119891119894(119905 119909) forall119905 isin 119869 120582 isin (0 1) 119909 isin R+ 119894 = 1 2 119873
Then the following fractional switched system
119888
119863120572
0+119906 (119905) + 119891
120590(119905)(119905 119906 (119905)) = 0 119905 isin 119869 = [0 1]
119906 (0) = 11990610158401015840
(0) = 0 119906 (1) = int
1
0
119906 (119904) 119889119904
(39)
has a unique solution119906lowast in119875ℎ where ℎ(119905) = 119905 119905 isin 119869Moreover
for any initial value 1199060
isin 119875ℎ constructing successively the
sequence
119906119899+1
(119905) = int
1
0
119866 (119905 119904) 119891120590(119904)
(119904 119906119899(119904)) 119889119904 119899 = 0 1 2
(40)
we have 119906119899(119905) rarr 119906
lowast
(119905) as 119899 rarr infin
4 Examples
Example 1 Consider the following boundary value problem
119888
119863120572
0+119906 (119905) + 119891
120590(119905)(119905 119906 (119905)) + 119892
120590(119905)(119905 119906 (119905)) = 0
119905 isin 119869 = [0 1]
119906 (0) = 11990610158401015840
(0) = 0 119906 (1) = int
1
0
119906 (119904) 119889119904
(41)
where 120572 = 52 120590(119905) 119869 rarr 119872 = 1 2 is a finite switchingsignal
1198911(119905 119909) =
1
4(119905 + 2)2
119909
1 + 119909+ 1
1198921(119905 119909) =
1
16sin2119909 + 119905
2
1198912(119905 119909) =
1
8(119905 + 2)2
119909
1 + 119909+ 1
1198922(119905 119909) =
1
32sin2119909 + 119905
2
(42)
Thus
119891119894 119892119894isin 119862 [119869 timesR
+
R+
] 119894 = 1 2 (43)
By computation we deduce that
10038161003816100381610038161198911 (119905 1199091) minus 1198911(119905 1199092)1003816100381610038161003816 le
1
16
10038161003816100381610038161199092 minus 1199091
1003816100381610038161003816
10038161003816100381610038161198921 (119905 1199091) minus 1198921(119905 1199092)1003816100381610038161003816 le
1
16
10038161003816100381610038161199092 minus 1199091
1003816100381610038161003816
10038161003816100381610038161198912 (119905 1199091) minus 1198912(119905 1199092)1003816100381610038161003816 le
1
32
10038161003816100381610038161199092 minus 1199091
1003816100381610038161003816
10038161003816100381610038161198922 (119905 1199091) minus 1198922(119905 1199092)1003816100381610038161003816 le
1
32
10038161003816100381610038161199092 minus 1199091
1003816100381610038161003816
(44)
On the other hand
int
1
0
1198661(119904 119904) (119897
1(119904) + 119897
1(119904)) 119889119904
= int
1
0
2(1 minus 119904)(52)minus1
((52) minus 1 + 119904)
Γ ((52) + 1)(1
16+
1
16) 119889119904
=1
4Γ (72)int
1
0
(1 minus 119904)32
(3
2+ 119904) 119889119904
le1
4Γ (72)int
1
0
(1 minus 119904)32
(3
2+ 1) 119889119904
=1
3radic120587times2
5
lt 1
int
1
0
1198661(119904 119904) (119897
1(119904) + 119897
1(119904)) 119889119904
= int
1
0
2(1 minus 119904)(52)minus1
((52) minus 1 + 119904)
Γ ((52) + 1)(1
32+
1
32) 119889119904
=1
3radic120587times1
5
lt 1
(45)
Hence byTheorem 8 BVP (41) has a unique positive solutionon [0 1]
Example 2 Consider the following boundary value problem
119888119863120572
0+119906 (119905) + 119891
120590(119905)(119905 119906 (119905)) + 119892
120590(119905)(119905 119906 (119905)) = 0
119905 isin 119869 = [0 1]
119906 (0) = 11990610158401015840
(0) = 0 119906 (1) = int
1
0
119906 (119904) 119889119904
(46)
where 120572 = 52 120590(119905) 119869 rarr 1 2 3 is a finite switching signal
1198911(119905 119909) = 119909
13
+ 1199052
+ 119888
1198921(119905 119909) =
119909
(1 + 1199052) (1 + 119909)+ 119887 minus 119888
1198912(119905 119909) = 2119909
13
+ 1199052
+ 2119888
1198922(119905 119909) =
2119909
(1 + 1199052) (1 + 119909)+ 2 (119887 minus 119888)
6 Abstract and Applied Analysis
1198913(119905 119909) = 3119909
13
+ 1199052
+ 3119888
1198923(119905 119909) =
3119909
(1 + 1199052) (1 + 119909)+ 3 (119887 minus 119888)
(47)
Let 120574 = 13 and 0 lt 119888 lt 119887 It is obvious that 119891119894 119892119894isin
119862[119869 times R+R+] and are increasing with respect to the secondargument 119892
119894(119905 0) = 119887 minus 119888 gt 0 119894 = 1 2 3 On the other hand
for 120582 isin (0 1) 119905 isin 119869 119909 isin [0 +infin) 119894 = 1 2 3 we have
119892119894(119905 120582119909) =
119894120582119909
(1 + 1199052) (1 + 120582119909)+ 119894 (119887 minus 119888)
ge119894120582119909
(1 + 1199052) (1 + 120582119909)+ 119894120582 (119887 minus 119888)
= 120582119892119894(119905 119909)
119891119894(119905 120582119909) = 119894120582
13
11990913
+ 1199052
+ 119894119888
ge 12058213
(11989411990913
+ 1199052
+ 119894119888)
= 120582120574
119891119894(119905 119909)
(48)
Moreover for 119905 isin 119869 119909 isin R+ 119894 = 1 2 3 we have
119891119894(119905 119909) = 119894119909
13
+ 1199052
+ 119894119888
ge 119894119888 ge119888
3 + (119887 minus 119888)(119894 + 119894 (119887 minus 119888))
ge119888
3 + (119887 minus 119888)(
119894119909
(1 + 1199052) (1 + 119909)+ 119894 (119887 minus 119888))
= 1205750119892119894(119905 119909)
(49)
where
1205750=
119888
3 + (119887 minus 119888) (50)
Hence all the conditions of Theorem 10 are satisfied ThusBVP (46) has a unique positive solution in 119875
ℎ where ℎ(119905) = 119905
119905 isin [0 1]
Conflict of Interests
The authors declare that there is no conflict of interestsregarding the publication of this paper
Acknowledgments
This research was supported by Henan Province CollegeYouth Backbone Teacher Funds (2011GGJS-213) and theNational Natural Science Foundation of China (11271336)
References
[1] H T Li and Y S Liu ldquoOn the uniqueness of the positivesolution for a second-order integral boundary value problemwith switched nonlinearityrdquo Applied Mathematics Letters vol24 no 12 pp 2201ndash2205 2011
[2] A Cabada and G Wang ldquoPositive solutions of nonlinearfractional differential equations with integral boundary valueconditionsrdquo Journal of Mathematical Analysis and Applicationsvol 389 no 1 pp 403ndash411 2012
[3] A A Kilbas H M Srivastava and J J Trujjllo Theory andApplications of Fractional Differential Equations vol 204 ofNorth-Holland Mathematics Studies Elsevier Amsterdam TheNetherlands 2006
[4] I Podlubny Fractional Differential Equations Academic PressNew York NY USA 1993
[5] X Zhang L Liu and Y Wu ldquoThe eigenvalue problem for asingular higher order fractional differential equation involvingfractional derivativesrdquo Applied Mathematics and Computationvol 218 no 17 pp 8526ndash8536 2012
[6] Z-W Lv J Liang and T-J Xiao ldquoSolutions to the Cauchy prob-lem for differential equations in Banach spaces with fractionalorderrdquo Computers ampMathematics with Applications vol 62 no3 pp 1303ndash1311 2011
[7] R P Agarwal V Lakshmikantham and J J Nieto ldquoOn theconcept of solution for fractional differential equations withuncertaintyrdquo Nonlinear Analysis Theory Methods amp Applica-tions vol 72 no 6 pp 2859ndash2862 2010
[8] R-N Wang T-J Xiao and J Liang ldquoA note on the fractionalCauchy problems with nonlocal initial conditionsrdquo AppliedMathematics Letters vol 24 no 8 pp 1435ndash1442 2011
[9] R-N Wang D-H Chen and T-J Xiao ldquoAbstract fractionalCauchy problems with almost sectorial operatorsrdquo Journal ofDifferential Equations vol 252 no 1 pp 202ndash235 2012
[10] J Henderson and A Ouahab ldquoFractional functional differentialinclusions with finite delayrdquo Nonlinear Analysis Theory Meth-ods amp Applications vol 70 no 5 pp 2091ndash2105 2009
[11] V Lakshmikantham ldquoTheory of fractional functional differen-tial equationsrdquo Nonlinear Analysis Theory Methods amp Applica-tions vol 69 no 10 pp 3337ndash3343 2008
[12] V Lakshmikantham and A S Vatsala ldquoBasic theory of frac-tional differential equationsrdquo Nonlinear Analysis Theory Meth-ods amp Applications vol 69 no 8 pp 2677ndash2682 2008
[13] F Li ldquoMild solutions for fractional differential equations withnonlocal conditionsrdquo Advances in Difference Equations ArticleID 287861 9 pages 2010
[14] X Q Zhang ldquoPositive solution for a class of singular semiposi-tone fractional differential equations with integral boundaryconditionsrdquo Boundary Value Problems vol 2012 article 1232012
[15] C Yang and C Zhai ldquoUniqueness of positive solutions for afractional differential equation via a fixed point theorem of asum operatorrdquo Electronic Journal of Differential Equations vol2012 no 70 pp 1ndash8 2012
[16] A A Agrachev and D Liberzon ldquoLie-algebraic stability criteriafor switched systemsrdquo SIAM Journal on Control and Optimiza-tion vol 40 no 1 pp 253ndash269 2001
[17] J Daafouz P Riedinger and C Iung ldquoStability analysis andcontrol synthesis for switched systems a switched Lyapunovfunction approachrdquo IEEE Transactions on Automatic Controlvol 47 no 11 pp 1883ndash1887 2002
[18] L Gurvits R Shorten and O Mason ldquoOn the stability ofswitched positive linear systemsrdquo IEEE Transactions on Auto-matic Control vol 52 no 6 pp 1099ndash1103 2007
Abstract and Applied Analysis 7
[19] C Zhai and D R Anderson ldquoA sum operator equation andapplications to nonlinear elastic beam equations and Lane-Emden-Fowler equationsrdquo Journal of Mathematical Analysisand Applications vol 375 no 2 pp 388ndash400 2011
[20] D J Guo and V Lakshmikantham Nonlinear Problems inAbstract Cone vol 5 Academic Press San Diego Calif USA1988
Submit your manuscripts athttpwwwhindawicom
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical Problems in Engineering
Hindawi Publishing Corporationhttpwwwhindawicom
Differential EquationsInternational Journal of
Volume 2014
Applied MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical PhysicsAdvances in
Complex AnalysisJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
OptimizationJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Operations ResearchAdvances in
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Function Spaces
Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of Mathematics and Mathematical Sciences
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Algebra
Discrete Dynamics in Nature and Society
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Decision SciencesAdvances in
Discrete MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom
Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Stochastic AnalysisInternational Journal of
6 Abstract and Applied Analysis
1198913(119905 119909) = 3119909
13
+ 1199052
+ 3119888
1198923(119905 119909) =
3119909
(1 + 1199052) (1 + 119909)+ 3 (119887 minus 119888)
(47)
Let 120574 = 13 and 0 lt 119888 lt 119887 It is obvious that 119891119894 119892119894isin
119862[119869 times R+R+] and are increasing with respect to the secondargument 119892
119894(119905 0) = 119887 minus 119888 gt 0 119894 = 1 2 3 On the other hand
for 120582 isin (0 1) 119905 isin 119869 119909 isin [0 +infin) 119894 = 1 2 3 we have
119892119894(119905 120582119909) =
119894120582119909
(1 + 1199052) (1 + 120582119909)+ 119894 (119887 minus 119888)
ge119894120582119909
(1 + 1199052) (1 + 120582119909)+ 119894120582 (119887 minus 119888)
= 120582119892119894(119905 119909)
119891119894(119905 120582119909) = 119894120582
13
11990913
+ 1199052
+ 119894119888
ge 12058213
(11989411990913
+ 1199052
+ 119894119888)
= 120582120574
119891119894(119905 119909)
(48)
Moreover for 119905 isin 119869 119909 isin R+ 119894 = 1 2 3 we have
119891119894(119905 119909) = 119894119909
13
+ 1199052
+ 119894119888
ge 119894119888 ge119888
3 + (119887 minus 119888)(119894 + 119894 (119887 minus 119888))
ge119888
3 + (119887 minus 119888)(
119894119909
(1 + 1199052) (1 + 119909)+ 119894 (119887 minus 119888))
= 1205750119892119894(119905 119909)
(49)
where
1205750=
119888
3 + (119887 minus 119888) (50)
Hence all the conditions of Theorem 10 are satisfied ThusBVP (46) has a unique positive solution in 119875
ℎ where ℎ(119905) = 119905
119905 isin [0 1]
Conflict of Interests
The authors declare that there is no conflict of interestsregarding the publication of this paper
Acknowledgments
This research was supported by Henan Province CollegeYouth Backbone Teacher Funds (2011GGJS-213) and theNational Natural Science Foundation of China (11271336)
References
[1] H T Li and Y S Liu ldquoOn the uniqueness of the positivesolution for a second-order integral boundary value problemwith switched nonlinearityrdquo Applied Mathematics Letters vol24 no 12 pp 2201ndash2205 2011
[2] A Cabada and G Wang ldquoPositive solutions of nonlinearfractional differential equations with integral boundary valueconditionsrdquo Journal of Mathematical Analysis and Applicationsvol 389 no 1 pp 403ndash411 2012
[3] A A Kilbas H M Srivastava and J J Trujjllo Theory andApplications of Fractional Differential Equations vol 204 ofNorth-Holland Mathematics Studies Elsevier Amsterdam TheNetherlands 2006
[4] I Podlubny Fractional Differential Equations Academic PressNew York NY USA 1993
[5] X Zhang L Liu and Y Wu ldquoThe eigenvalue problem for asingular higher order fractional differential equation involvingfractional derivativesrdquo Applied Mathematics and Computationvol 218 no 17 pp 8526ndash8536 2012
[6] Z-W Lv J Liang and T-J Xiao ldquoSolutions to the Cauchy prob-lem for differential equations in Banach spaces with fractionalorderrdquo Computers ampMathematics with Applications vol 62 no3 pp 1303ndash1311 2011
[7] R P Agarwal V Lakshmikantham and J J Nieto ldquoOn theconcept of solution for fractional differential equations withuncertaintyrdquo Nonlinear Analysis Theory Methods amp Applica-tions vol 72 no 6 pp 2859ndash2862 2010
[8] R-N Wang T-J Xiao and J Liang ldquoA note on the fractionalCauchy problems with nonlocal initial conditionsrdquo AppliedMathematics Letters vol 24 no 8 pp 1435ndash1442 2011
[9] R-N Wang D-H Chen and T-J Xiao ldquoAbstract fractionalCauchy problems with almost sectorial operatorsrdquo Journal ofDifferential Equations vol 252 no 1 pp 202ndash235 2012
[10] J Henderson and A Ouahab ldquoFractional functional differentialinclusions with finite delayrdquo Nonlinear Analysis Theory Meth-ods amp Applications vol 70 no 5 pp 2091ndash2105 2009
[11] V Lakshmikantham ldquoTheory of fractional functional differen-tial equationsrdquo Nonlinear Analysis Theory Methods amp Applica-tions vol 69 no 10 pp 3337ndash3343 2008
[12] V Lakshmikantham and A S Vatsala ldquoBasic theory of frac-tional differential equationsrdquo Nonlinear Analysis Theory Meth-ods amp Applications vol 69 no 8 pp 2677ndash2682 2008
[13] F Li ldquoMild solutions for fractional differential equations withnonlocal conditionsrdquo Advances in Difference Equations ArticleID 287861 9 pages 2010
[14] X Q Zhang ldquoPositive solution for a class of singular semiposi-tone fractional differential equations with integral boundaryconditionsrdquo Boundary Value Problems vol 2012 article 1232012
[15] C Yang and C Zhai ldquoUniqueness of positive solutions for afractional differential equation via a fixed point theorem of asum operatorrdquo Electronic Journal of Differential Equations vol2012 no 70 pp 1ndash8 2012
[16] A A Agrachev and D Liberzon ldquoLie-algebraic stability criteriafor switched systemsrdquo SIAM Journal on Control and Optimiza-tion vol 40 no 1 pp 253ndash269 2001
[17] J Daafouz P Riedinger and C Iung ldquoStability analysis andcontrol synthesis for switched systems a switched Lyapunovfunction approachrdquo IEEE Transactions on Automatic Controlvol 47 no 11 pp 1883ndash1887 2002
[18] L Gurvits R Shorten and O Mason ldquoOn the stability ofswitched positive linear systemsrdquo IEEE Transactions on Auto-matic Control vol 52 no 6 pp 1099ndash1103 2007
Abstract and Applied Analysis 7
[19] C Zhai and D R Anderson ldquoA sum operator equation andapplications to nonlinear elastic beam equations and Lane-Emden-Fowler equationsrdquo Journal of Mathematical Analysisand Applications vol 375 no 2 pp 388ndash400 2011
[20] D J Guo and V Lakshmikantham Nonlinear Problems inAbstract Cone vol 5 Academic Press San Diego Calif USA1988
Submit your manuscripts athttpwwwhindawicom
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical Problems in Engineering
Hindawi Publishing Corporationhttpwwwhindawicom
Differential EquationsInternational Journal of
Volume 2014
Applied MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical PhysicsAdvances in
Complex AnalysisJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
OptimizationJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Operations ResearchAdvances in
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Function Spaces
Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of Mathematics and Mathematical Sciences
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Algebra
Discrete Dynamics in Nature and Society
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Decision SciencesAdvances in
Discrete MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom
Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Stochastic AnalysisInternational Journal of
Abstract and Applied Analysis 7
[19] C Zhai and D R Anderson ldquoA sum operator equation andapplications to nonlinear elastic beam equations and Lane-Emden-Fowler equationsrdquo Journal of Mathematical Analysisand Applications vol 375 no 2 pp 388ndash400 2011
[20] D J Guo and V Lakshmikantham Nonlinear Problems inAbstract Cone vol 5 Academic Press San Diego Calif USA1988
Submit your manuscripts athttpwwwhindawicom
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical Problems in Engineering
Hindawi Publishing Corporationhttpwwwhindawicom
Differential EquationsInternational Journal of
Volume 2014
Applied MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical PhysicsAdvances in
Complex AnalysisJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
OptimizationJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Operations ResearchAdvances in
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Function Spaces
Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of Mathematics and Mathematical Sciences
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Algebra
Discrete Dynamics in Nature and Society
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Decision SciencesAdvances in
Discrete MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom
Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Stochastic AnalysisInternational Journal of
Submit your manuscripts athttpwwwhindawicom
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical Problems in Engineering
Hindawi Publishing Corporationhttpwwwhindawicom
Differential EquationsInternational Journal of
Volume 2014
Applied MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical PhysicsAdvances in
Complex AnalysisJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
OptimizationJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Operations ResearchAdvances in
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Function Spaces
Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of Mathematics and Mathematical Sciences
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Algebra
Discrete Dynamics in Nature and Society
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Decision SciencesAdvances in
Discrete MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom
Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Stochastic AnalysisInternational Journal of
Related Documents
