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Research ArticleExistence Results for a Coupled System of NonlinearFractional Hybrid Differential Equations with HomogeneousBoundary Conditions
Josefa Caballero1 Mohamed Abdalla Darwish23
Kishin Sadarangani1 and Wafa M Shammakh2
1 Department of Mathematics University of Las Palmas de Gran Canaria Campus de Tafira Baja35017 Las Palmas de Gran Canaria Spain
2Department of Mathematics Sciences Faculty for Girls King Abdulaziz University Jeddah Saudi Arabia3 Department of Mathematics Faculty of Science Damanhour University Damanhour Egypt
Correspondence should be addressed to Mohamed Abdalla Darwish drmadarwishgmailcom
Received 7 April 2014 Accepted 16 June 2014 Published 14 July 2014
Academic Editor Ljubisa Kocinac
Copyright copy 2014 Josefa Caballero et alThis is an open access article distributed under theCreative CommonsAttribution Licensewhich permits unrestricted use distribution and reproduction in any medium provided the original work is properly cited
We study an existence result for the following coupled system of nonlinear fractional hybrid differential equations withhomogeneous boundary conditions 119863
120572
0+ [119909(119905)119891(119905 119909(119905) 119910(119905))] = 119892(119905 119909(119905) 119910(119905)) 119863
120572
0+ [119910(119905)119891(119905 119910(119905) 119909(119905))] = 119892(119905 119910(119905) 119909(119905)) 0 lt
119905 lt 1 and 119909(0) = 119910(0) = 0 where 120572 isin (0 1) and 119863120572
0+ denotes the Riemann-Liouville fractional derivative The main tools in our
study are the techniques associated to measures of noncompactness in the Banach algebras and a fixed point theorem of Darbotype
1 Introduction
Fractional differential equations arise in many engineeringand scientific disciplines as the mathematical modelling ofa great number of processes which appear in physics chem-istry aerodynamics and so forth and involve also derivativesof fractional order For details see [1ndash5] and the referencestherein
On the other hand about the theory of hybrid differentialequations we refer to the paper [6] where the authors studiedthe hybrid differential equation of first order
119889
119889119905[
119909 (119905)
119891 (119905 119909 (119905))] = 119892 (119905 119909 (119905)) 119905 isin 119869 = [0 119879)
119909 (1199050) = 1199090
isin R
(1)
where 119891 isin 119862(119869 times RR 0) and 119892 isin 119862(119869 times RR)
In [7] the authors studied the fractional version of theabovementioned problem that is
119863120572
0+ [
119909 (119905)
119891 (119905 119909 (119905))] = 119892 (119905 119909 (119905)) 119905 isin 119869 0 lt 120572 lt 1
119909 (0) = 0
(2)
under the same assumptions on 119891 and 119892 in [6]Recently in [8] the authors studied the following frac-
tional hybrid initial value problem with supremum
119863120572
0+ [
119909 (119905)
119891 (119905 119909 (119905) max0le120591le119905
|119909 (120591)|)] = 119892 (119905 119909 (119905))
0 lt 119905 lt 1
119909 (0) = 0
(3)
where 0 lt 120572 lt 1 119891 isin 119862([0 1] times R times RR 0) and 119892 isin
119862([0 1] times RR)
Hindawi Publishing CorporationAbstract and Applied AnalysisVolume 2014 Article ID 672167 10 pageshttpdxdoiorg1011552014672167
2 Abstract and Applied Analysis
The coupled systems involving fractional differentialequations are very important because they occur in numerousproblems of applied nature for instance see [9ndash13]
In this paper we consider the following coupled system
119863120572
0+ [
119909 (119905)
119891 (119905 119909 (119905) 119910 (119905))] = 119892 (119905 119909 (119905) 119910 (119905))
119863120572
0+ [
119910 (119905)
119891 (119905 119910 (119905) 119909 (119905))] = 119892 (119905 119910 (119905) 119909 (119905))
0 lt 119905 lt 1
119909 (0) = 119910 (0) = 0
(4)
where 120572 isin (0 1) and 119863120572
0+ is the standard Riemann-Liouville
fractional derivativeThe main tool in our study is a fixed point theorem of
Darbo type associated to measures of noncompactness
2 Preliminaries
We begin this section with some definitions and results aboutfractional calculus
Let 120572 gt 0 and 119899 = [120572] + 1 where [120572] denotes the integerpart of 120572 For a function 119891 (0 infin) rarr R the Riemann-Liouville fractional integral of order 120572 gt 0 of 119891 is defined as
119868120572
0+119891 (119905) =
1
Γ (120572)int
119905
0
119891 (119904)
(119905 minus 119904)1minus120572
119889119904 (5)
provided that the right side is pointwise defined on (0 infin)The Riemann-Liouville fractional derivative of order 120572 of
a continuous function 119891 is defined by
119863120572
0+119891 (119909) =
1
Γ (119899 minus 120572)
119889119899
119889119909119899int
119909
0
119891 (119904)
(119909 minus 119904)120572minus119899+1
119889119904 (6)
provided that the right side is pointwise defined on (0 infin)The following lemma will be useful for our study [14]
Lemma 1 Let ℎ isin 1198711
(0 1) and 0 lt 120572 lt 1 Then
(a)
119863120572
0+119868120572
0+ℎ (119909) = ℎ (119909) (7)
(b)
119868120572
0+119863120572
0+ℎ (119909) = ℎ (119909) minus
1198681minus120572
0+ ℎ (119909)
10038161003816100381610038161003816119909=0
Γ (120572)119909120572minus1
119886119890 on (0 1)
(8)
Lemma 2 Let 0 lt 120572 lt 1 and suppose that 119891 isin 119862([0 1]R
0) and119910 isin 119862[0 1]Then the unique solution of the fractionalhybrid initial value problem
119863120572
0+ [
119909 (119905)
119891 (119905)] = 119910 (119905) 0 lt 119905 lt 1
119909 (0) = 0
(9)
is given by
119909 (119905) =119891 (119905)
Γ (120572)int
119905
0
119910 (119904)
(119905 minus 119904)1minus120572
119889119904 119905 isin [0 1] (10)
Proof Suppose that 119909(119905) is a solution of problem (9) Usingthe operator 119868
120572
0+ and taking into account Lemma 1 we get
119868120572
0+119863120572
0+ [
119909 (119905)
119891 (119905)] = 119868120572
0+119910 (119905) (11)
or equivalently
119909 (119905)
119891 (119905)minus
1198681minus120572
0+ (119909 (119905) 119891 (119905))
10038161003816100381610038161003816119905=0
Γ (120572)119905120572minus1
= 119868120572
0+119910 (119905) (12)
Since 119909(119905)119891(119905)|119905=0
= 119909(0)119891(0) = 0119891(0) = 0 (because119891(0) = 0) we have
119909 (119905) = 119891 (119905) 119868120572
0+119910 (119905) (13)
This means that
119909 (119905) =119891 (119905)
Γ (120572)int
119905
0
119910 (119904)
(119905 minus 119904)1minus120572
119889119904 (14)
Conversely suppose that 119909(119905) is given by
119909 (119905) =119891 (119905)
Γ (120572)int
119905
0
119910 (119904)
(119905 minus 119904)1minus120572
119889119904 119905 isin [0 1] (15)
This means that
119909 (119905) = 119891 (119905) 119868120572
0+119910 (119905) 119905 isin [0 1] (16)
Applying 119863120572
0+ and taking into account Lemma 1 and that
119891(119905) = 0 for 119905 isin [0 1] we obtain
119863120572
0+ [
119909 (119905)
119891 (119905)] = 119863
120572
0+119868120572
0+119910 (119905) = 119910 (119905) 0 lt 119905 lt 1 (17)
Moreover for 119905 = 0 in (16) we have 119909(0) = 119891(0) sdot 0 = 0 Thiscompletes the proof
In the sequel we recall some definitions and basic factsabout measures of noncompactness
Assume that 119864 is a real Banach space with norm sdot
and zero element 120579 By 119861(119909 119903) we denote the closed ball in 119864
centered at 119909with radius 119903 By 119861119903we denote the ball 119861(120579 119903) If
119883 is a nonempty subset of119864 by the symbols119883 andConv119883wedenote the closure and the convex closure of 119883 respectivelyBy 119883 we denote the quantity 119883 = sup119909 119909 isin 119883Finally by M
119864we will denote the family of all nonempty
and bounded subsets of 119864 and byN119864we denote its subfamily
consisting of all relatively compact subsets of 119864
Definition 3 Amapping120583 M119864
rarr R+
= [0 infin) is said to bea measure of noncompactness in 119864 if it satisfies the followingconditions
(a) The family ker 120583 = 119883 isin M119864
120583(119883) = 0 is nonemptyand ker 120583 sub N
119864
Abstract and Applied Analysis 3
(b) 119883 sub 119884 rArr 120583(119883) le 120583(119884)(c) 120583(119883) = 120583(Conv119883) = 120583(119883)(d) 120583(120582119883 + (1 minus 120582)119884) le 120582120583(119883) + (1 minus 120582)120583(119884) for 120582 isin [0 1](e) If (119883
119899) is a sequence of closed subsets fromM
119864such
that 119883119899+1
sub 119883119899
(119899 ge 1) and lim119899rarrinfin
120583(119883119899) = 0 then
119883infin
= capinfin
119899=1119883119899
= 120601
The family ker 120583 appearing in (a) is called the kernel ofthe measure of noncompactness 120583 Notice that the set 119883
infin
appearing in (e) is an element of ker120583 Indeed since 120583(119883infin
) le
120583(119883119899) for 119899 = 1 2 we infer that 120583(119883
infin) = 0 and this says
that 119883infin
isin ker120583An important theorem about fixed point theorem in
the context of measures of noncompactness is the followingDarborsquos fixed point theorem [15]
Theorem4 Let119862 be a nonempty bounded closed and convexsubset of a Banach space 119864 and let 119879 119862 rarr 119862 be a continuousmapping Suppose that there exists a constant 119896 isin [0 1) suchthat
120583 (119879 (119883)) le 119896120583 (119883) (18)
for any nonempty subset 119883 of 119862Then 119879 has a fixed point
A generalization of Theorem 4 which will be very usefulin our study is the following theorem due to Sadovskiı [16]
Theorem5 Let119862 be a nonempty bounded closed and convexsubset of a Banach space 119864 and let 119879 119862 rarr 119862 be a continuousoperator satisfying
120583 (119879 (119883)) lt 120583 (119883) (19)
for any nonempty subset 119883 of 119862 with 120583(119883) gt 0Then 119879 has a fixed point
Next we will assume that the space 119864 has structure ofBanach algebra By 119909119910 we will denote the product of twoelements 119909 119910 isin 119883 and by 119883119884 we will denote the set definedby 119883119884 = 119909119910 119909 isin 119883 119910 isin 119884
Definition 6 Let 119864 be a Banach algebra We will say thata measure of noncompactness 120583 defined on 119864 satisfiescondition (119898) if
120583 (119883119884) le 119883 120583 (119884) + 119884 120583 (119883) (20)
for any 119883 119884 isin M119864
This definition appears in [17]In this paper we will work in the space 119862[0 1] consisting
of all real functions defined and continuous on [0 1] with thestandard supremum norm
119909 = sup |119909 (119905)| 119905 isin [0 1] (21)
for119909 isin 119862[0 1] It is clear that (119862[0 1] sdot) is a Banach algebrawhere the multiplication is defined as the usual product ofreal functions
Next we present the measure of noncompactness in119862[0 1] which will be used later Let us fix 119883 isin M
119862[01]and
120576 gt 0 For 119909 isin 119883 we denote by 120596(119909 120576) the modulus ofcontinuity of 119909 that is
120596 (119909 120576) = sup |119909 (119905) minus 119909 (119904)| 119905 119904 isin [0 1] |119905 minus 119904| le 120576 (22)
Put
120596 (119883 120576) = sup 120596 (119909 120576) 119909 isin 119883
1205960
(119883) = lim120576rarr0
120596 (119883 120576) (23)
In [15] it is proved that 1205960(119883) is a measure of noncompact-
ness in 119862[0 1]
Proposition 7 Themeasure of noncompactness 1205960on 119862[0 1]
satisfies condition (119898)
Proof Fix 119883 119884 isin M119862[01]
120576 gt 0 and 119905 119904 isin [0 1] with |119905 minus 119904| le
120576 Then we have1003816100381610038161003816119909 (119905) 119910 (119905) minus 119909 (119904) 119910 (119904)
1003816100381610038161003816 le1003816100381610038161003816119909 (119905) 119910 (119905) minus 119909 (119905) 119910 (119904)
1003816100381610038161003816
+1003816100381610038161003816119909 (119905) 119910 (119904) minus 119909 (119904) 119910 (119904)
1003816100381610038161003816
= |119909 (119905)|1003816100381610038161003816119910 (119905) minus 119910 (119904)
1003816100381610038161003816
+1003816100381610038161003816119910 (119904)
1003816100381610038161003816 |119909 (119905) minus 119909 (119904)|
le 119909 120596 (119910 120576) +1003817100381710038171003817119910
1003817100381710038171003817 120596 (119909 120576)
(24)
This means that
120596 (119909119910 120576) le 119909 120596 (119910 120576) +1003817100381710038171003817119910
1003817100381710038171003817 120596 (119909 120576) (25)
and therefore
120596 (119883119884 120576) le 119883 120596 (119884 120576) + 119884 120596 (119883 120576) (26)
Taking 120576 rarr 0 we get
1205960
(119883119884) le 119883 1205960
(119884) + 119884 1205960
(119883) (27)
This completes the proof
Proposition 7 appears in [17] and we have given the prooffor the paper is self-contained
3 Main Results
We begin this section introducing the following class A offunctions
A = 120593 R+
997888rarr R+
120593 is nondecreasing
and lim119899rarrinfin
120593119899
(119905) = 0 for any 119905 gt 0
(28)
where 120593119899 denotes the 119899-iteration of 120593
4 Abstract and Applied Analysis
Remark 8 Notice that if 120593 isin A then 120593(119905) lt 119905 for any 119905 gt 0Indeed in contrary case we can find 119905
0gt 0 and 119905
0le 120593(119905
0)
Since 120593 is nondecreasing we have
0 lt 1199050
le 120593 (1199050) le 1205932
(1199050) le sdot sdot sdot le 120593
119899
(1199050) le sdot sdot sdot (29)
and therefore 0 lt 1199050
le lim119899rarrinfin
120593119899
(1199050) and this contradicts
the fact that 120593 isin AMoreover the fact that 120593(119905) lt 119905 for any 119905 gt 0 proves that
if 120593 isin A then 120593 is continuous at 1199050
= 0
Using Remark 8 and Theorem 5 we have the followingfixed point theorem
Theorem9 Let119862 be a nonempty bounded closed and convexsubset of a Banach space 119864 and let 119879 119862 rarr 119862 be a continuousoperator satisfying
120583 (119879 (119883)) le 120593 (120583 (119883)) (30)
for any nonempty subset 119883 of 119862 where 120593 isin AThen 119879 has a fixed point
Theorem 9 appears in [18] where the authors present aproof without usingTheorem 5
The following result which appears in [19] will be inter-esting in our study
Theorem 10 Let 1205831 1205832 120583
119899be measures of noncompact-
ness in the Banach spaces 1198641 1198642 119864
119899 respectively Suppose
that 119865 [0 infin)119899
rarr [0 infin) is a convex function such that119865(1199091 1199092 119909
119899) = 0 if and only if 119909
119894= 0 for 119894 = 1 2 119899
Then
120583 (119883) = 119865 (1205831
(1198831) 1205832
(1198832) 120583
119899(119883119899)) (31)
defines a measure compactness in 1198641
times 1198642
times sdot sdot sdot times 119864119899 where 119883
119894
denotes the natural projection of 119883 into 119864119894 for 119894 = 1 2 119899
Remark 11 As a consequence ofTheorem 10 we have that if 120583
is a measure of noncompactness on a Banach space 119864 and weconsider the function 119865 [0 infin) times [0 infin) rarr [0 infin) definedby 119865(119909 119910) = max(119909 119910) then since 119865 is convex and 119865(119909 119910) =
0 if and only if 119909 = 119910 = 0 120583(119883) = max120583(1198831) 120583(119883
2) defines
a measure of noncompactness in the space 119864 times 119864
Next we present the definition of a coupled fixed point
Definition 12 An element (119909 119910) isin 119883 times 119883 is said to be acoupled fixed point of a mapping 119866 119883 times 119883 rarr 119883 if119866(119909 119910) = 119909 and 119866(119910 119909) = 119910
The following result is crucial for our study
Theorem 13 Let Ω be a nonempty bounded closed andconvex subset of a Banach space 119864 and let 120583 be a measure ofnoncompactness in 119864 Suppose that 119866 Ω times Ω rarr Ω is acontinuous operator satisfying
120583 (119866 (1198831
times 1198832)) le 120593 (max (120583 (119883
1) 120583 (119883
2))) (32)
for all nonempty subsets 1198831and 119883
2of Ω where 120593 isin A
Then 119866 has at least a coupled fixed point
Proof Notice that by Remark 11 120583(119883) = max120583(1198831) 120583(119883
2)
is a measure of noncompactness in the space 119864 times 119864 where 1198831
and 1198832are the projections of 119883 into 119864
Now we consider the mapping 119866 Ω times Ω rarr Ω times Ω
defined by 119866(119909 119910) = (119866(119909 119910) 119866(119910 119909)) It is easily seen thatΩ times Ω is a nonempty bounded closed and convex subsetof 119864 times 119864 Since 119866 is continuous it is clear that 119866 is alsocontinuous
Next we take a nonempty 119883 of Ω times Ω Then
120583 (119866 (119883)) = 120583 (119866 (1198831
times 1198832))
= 120583 (119866 (1198831
times 1198832) times 119866 (119883
2times 1198831))
= max 120583 (119866 (1198831
times 1198832)) 120583 (119866 (119883
2times 1198831))
le max 120593 (max (120583 (1198831) 120583 (119883
2)))
120593 (max (120583 (1198832) 120583 (119883
1)))
= 120593 (max (120583 (1198831) 120583 (119883
2)))
= 120593 (120583 (1198831
times 1198832))
(33)
Since 120593 isin A by Theorem 9 the mapping 119866 has at least onefixed pointThismeans that there exists (119909
0 1199100) isin ΩtimesΩ such
that 119866(1199090 1199100) = (119909
0 1199100) or equivalently 119866(119909
0 1199100) = 1199090and
119866(1199100 1199090) = 1199100 This proves that 119866 has at least a coupled fixed
point
Now we consider the coupled system of integral equa-tions
119909 (119905) =119891 (119905 119909 (119905) 119910 (119905))
Γ (120572)int
119905
0
119892 (119904 119909 (119904) 119910 (119904))
(119905 minus 119904)1minus120572
119889119904
119910 (119905) =119891 (119905 119910 (119905) 119909 (119905))
Γ (120572)int
119905
0
119892 (119904 119910 (119904) 119909 (119904))
(119905 minus 119904)1minus120572
119889119904 119905 isin [0 1]
(34)
Lemma 14 Assume that 119891 isin 119862([0 1] times R times RR 0) and119892 isin 119862([0 1] times R times RR) Then (119909 119910) isin 119862[0 1] times 119862[0 1] isa solution of (34) if and only if (119909 119910) isin 119862[0 1] times 119862[0 1] is asolution of (4)
Proof The proof is an immediate consequence of Lemma 2so we omit it
Next we will study problem (4) under the followingassumptions
(H1) 119891 isin 119862([0 1] times R times RR 0) and 119892 isin 119862([0 1] times R times
RR)(H2) The functions 119891 and 119892 satisfy
1003816100381610038161003816119891 (119905 1199091 1199101) minus 119891 (119905 119909
2 1199102)1003816100381610038161003816
le 1205931
(max (10038161003816100381610038161199091 minus 119909
2
1003816100381610038161003816 10038161003816100381610038161199101 minus 119910
2
1003816100381610038161003816))
1003816100381610038161003816119892 (119905 1199091 1199102) minus 119892 (119905 119909
2 1199102)1003816100381610038161003816
le 1205932
(max (10038161003816100381610038161199091 minus 119910
1
1003816100381610038161003816 10038161003816100381610038161199092 minus 119910
2
1003816100381610038161003816))
(35)
Abstract and Applied Analysis 5
respectively for any 119905 isin [0 1] and 1199091 1199092 1199101 1199102
isin Rwhere 120593
1 1205932
isin A and 1205931is continuous
Notice that assumption (H1) gives us the existence
of two nonnegative constants 1198961and 119896
2such that
|119891(119905 0 0)| le 1198961and |119892(119905 0 0)| le 119896
2 for any 119905 isin [0 1]
(H3) There exists 119903
0gt 0 satisfying the inequalities
(1205931
(119903) + 1198961) sdot (1205932
(119903) + 1198962) le 119903Γ (120572 + 1)
1205932
(119903) + 1198962
le Γ (120572 + 1)
(36)
Theorem 15 Under assumptions (1198671)ndash(1198673) problem (4) has
at least one solution in 119862[0 1] times 119862[0 1]
Proof In virtue of Lemma 14 a solution (119909 119910) isin 119862[0 1] times
119862[0 1] of problem (4) satisfies
119909 (119905) =119891 (119905 119909 (119905) 119910 (119905))
Γ (120572)int
119905
0
119892 (119904 119909 (119904) 119910 (119904))
(119905 minus 119904)1minus120572
119889119904
119910 (119905) =119891 (119905 119910 (119905) 119909 (119905))
Γ (120572)int
119905
0
119892 (119904 119910 (119904) 119909 (119904))
(119905 minus 119904)1minus120572
119889119904
119905 isin [0 1]
(37)
We consider the space 119862[0 1] times 119862[0 1] equipped with thenorm (119909 119910)
119862[01]times119862[01]= max119909 119910 for any (119909 119910) isin
119862[0 1] times 119862[0 1]In 119862[0 1] times 119862[0 1] we define the operator
119866 (119909 119910) (119905) =119891 (119905 119909 (119905) 119910 (119905))
Γ (120572)int
119905
0
119892 (119904 119909 (119904) 119910 (119904))
(119905 minus 119904)1minus120572
119889119904
119905 isin [0 1]
(38)
LetF andG be the operators given by
F (119909 119910) (119905) = 119891 (119905 119909 (119905) 119910 (119905))
G (119909 119910) (119905) =1
Γ (120572)int
119905
0
119892 (119904 119909 (119904) 119910 (119904))
(119905 minus 119904)1minus120572
119889119904
(39)
for any (119909 119910) isin 119862[0 1] times 119862[0 1] and any 119905 isin [0 1] Then
119866 (119909 119910) = F (119909 119910) sdot G (119909 119910) (40)
Firstly we will prove that 119866 applies 119862[0 1] times 119862[0 1] into119862[0 1] To do this it is sufficient to prove that F(119909 119910)G(119909 119910) isin 119862[0 1] for any (119909 119910) isin 119862[0 1] times 119862[0 1] since theproduct of continuous functions is continuous
In virtue of assumption (H1) it is clear that F(119909 119910) isin
119862[0 1] for (119909 119910) isin 119862[0 1] times 119862[0 1] In order to prove thatG(119909 119910) isin 119862[0 1] for (119909 119910) isin 119862[0 1] times 119862[0 1] we fix 119905
0isin
[0 1] and we consider a sequence (119905119899) isin [0 1] such that 119905
119899rarr
1199050 and we have to prove that G(119909 119910)(119905
119899) rarr G(119909 119910)(119905
0)
Without loss of generality we can suppose that 119905119899
gt 1199050 Then
we have
1003816100381610038161003816G (119909 119910) (119905119899) minus G (119909 119910) (119905
0)1003816100381610038161003816
=1
Γ (120572)
1003816100381610038161003816100381610038161003816100381610038161003816
int
119905119899
0
119892 (119904 119909 (119904) 119910 (119904))
(119905119899
minus 119904)1minus120572
119889119904
minus int
1199050
0
119892 (119904 119909 (119904) 119910 (119904))
(1199050
minus 119904)1minus120572
119889119904
1003816100381610038161003816100381610038161003816100381610038161003816
le1
Γ (120572)
1003816100381610038161003816100381610038161003816100381610038161003816
int
119905119899
0
119892 (119904 119909 (119904) 119910 (119904))
(119905119899
minus 119904)1minus120572
119889119904
minus int
119905119899
0
119892 (119904 119909 (119904) 119910 (119904))
(1199050
minus 119904)1minus120572
119889119904
1003816100381610038161003816100381610038161003816100381610038161003816
+1
Γ (120572)
1003816100381610038161003816100381610038161003816100381610038161003816
int
119905119899
0
119892 (119904 119909 (119904) 119910 (119904))
(1199050
minus 119904)1minus120572
119889119904
minus int
1199050
0
119892 (119904 119909 (119904) 119910 (119904))
(1199050
minus 119904)1minus120572
119889119904
1003816100381610038161003816100381610038161003816100381610038161003816
le1
Γ (120572)int
119905119899
0
10038161003816100381610038161003816(119905119899
minus 119904)120572minus1
minus (1199050
minus 119904)120572minus110038161003816100381610038161003816
times1003816100381610038161003816119892 (119904 119909 (119904) 119910 (119904))
1003816100381610038161003816 119889119904
+1
Γ (120572)int
119905119899
1199050
10038161003816100381610038161003816(1199050
minus 119904)120572minus110038161003816100381610038161003816
1003816100381610038161003816119892 (119904 119909 (119904) 119910 (119904))1003816100381610038161003816 119889119904
(41)
By assumption (H1) since 119892 isin 119862([0 1] times R times RR) 119892 is
bounded on the compact set [0 1]times[minus119909 119909]times[minus119910 119910]Denote by
119872 = sup 1003816100381610038161003816119892 (119904 119909
1 1199101)1003816100381610038161003816 119904 isin [0 1] 119909
1isin [minus 119909 119909]
1199101
isin [minus1003817100381710038171003817119910
1003817100381710038171003817 1003817100381710038171003817119910
1003817100381710038171003817]
(42)
From the last estimate we obtain
1003816100381610038161003816G (119909 119910) (119905119899) minus G (119909 119910) (119905
0)1003816100381610038161003816
le119872
Γ (120572)int
119905119899
0
10038161003816100381610038161003816(119905119899
minus 119904)120572minus1
minus (1199050
minus 119904)120572minus110038161003816100381610038161003816
119889119904
+119872
Γ (120572)int
119905119899
1199050
10038161003816100381610038161003816(1199050
minus 119904)120572minus110038161003816100381610038161003816
119889119904
(43)
6 Abstract and Applied Analysis
As 0 lt 120572 lt 1 and 119905119899
gt 1199050 we infer that
1003816100381610038161003816G (119909 119910) (119905119899) minus G (119909 119910) (119905
0)1003816100381610038161003816
le119872
Γ (120572)[int
1199050
0
10038161003816100381610038161003816(119905119899
minus 119904)120572minus1
minus (1199050
minus 119904)120572minus110038161003816100381610038161003816
119889119904
+ int
119905119899
1199050
10038161003816100381610038161003816(119905119899
minus 119904)120572minus1
minus (1199050
minus 119904)120572minus110038161003816100381610038161003816
119889119904]
+119872
Γ (120572)int
119905119899
1199050
1
(119904 minus 1199050)1minus120572
119889119904
=119872
Γ (120572)[ int
1199050
0
[(1199050
minus 119904)120572minus1
minus (119905119899
minus 119904)120572minus1
] 119889119904
+ int
119905119899
1199050
119889119904
(119905119899
minus 119904)1minus120572
+ int
119905119899
1199050
119889119904
(119904 minus 1199050)1minus120572
]
+119872
Γ (120572)int
119905119899
1199050
1
(119904 minus 1199050)1minus120572
119889119904
le119872
Γ (120572 + 1)[(119905119899
minus 1199050)120572
+ 119905120572
0minus 119905120572
119899
+ (119905119899
minus 1199050)120572
+ (119905119899
minus 1199050)120572
]
+119872
Γ (120572 + 1)(119905119899
minus 1199050)120572
=4119872
Γ (120572 + 1)(119905119899
minus 1199050)120572
+119872
Γ (120572 + 1)(119905120572
0minus 119905120572
119899)
lt4119872
Γ (120572 + 1)(119905119899
minus 1199050)120572
(44)
where the last inequality has been obtained by using the factthat 119905120572
0minus 119905120572
119899lt 0
Therefore since 119905119899
rarr 1199050 from the last estimate we
deduce that G(119909 119910)(119905119899) rarr G(119909 119910)(119905
0) This proves that
G(119909 119910) isin 119862[0 1] Consequently G 119862[0 1] times 119862[0 1] rarr
119862[0 1] On the other hand for (119909 119910) isin 119862[0 1] times 119862[0 1] and119905 isin 119862[0 1] we have
1003816100381610038161003816119866 (119909 119910) (119905)1003816100381610038161003816
=1003816100381610038161003816F (119909 119910) (119905) sdot G (119909 119910) (119905)
1003816100381610038161003816
=1003816100381610038161003816119891 (119905 119909 (119905) 119910 (119905))
1003816100381610038161003816
100381610038161003816100381610038161003816100381610038161003816
1
Γ (120572)int
119905
0
119892 (119904 119909 (119904) 119910 (119904))
(119905 minus 119904)1minus120572
119889119904
100381610038161003816100381610038161003816100381610038161003816
le [1003816100381610038161003816119891 (119905 119909 (119905) 119910 (119905)) minus 119891 (119905 0 0)
1003816100381610038161003816 +1003816100381610038161003816119891 (119905 0 0)
1003816100381610038161003816]
times [1
Γ (120572)
100381610038161003816100381610038161003816100381610038161003816
int
119905
0
119892 (119904 119909 (119904) 119910 (119904)) minus 119892 (119904 0 0)
(119905 minus 119904)1minus120572
119889119904
+ int
119905
0
119892 (119904 0 0)
(119905 minus 119904)1minus120572
119889119904
100381610038161003816100381610038161003816100381610038161003816
]
le1
Γ (120572)[1205931
(max (|119909 (119905)| 1003816100381610038161003816119910 (119905)
1003816100381610038161003816)) + 1198961]
times [int
119905
0
1003816100381610038161003816119892 (119904 119909 (119904) 119910 (119904)) minus 119892 (119904 0 0)1003816100381610038161003816
(119905 minus 119904)1minus120572
119889119904
+ int
119905
0
1003816100381610038161003816119892 (119904 0 0)1003816100381610038161003816
(119905 minus 119904)1minus120572
119889119904]
le1
Γ (120572)[1205931
(max (119909 1003817100381710038171003817119910
1003817100381710038171003817)) + 1198961]
times [int
119905
0
1205932
(max (|119909 (119904)| 1003816100381610038161003816119910 (119904)
1003816100381610038161003816))
(119905 minus 119904)1minus120572
119889119904
+ 1198962
int
119905
0
119889119904
(119905 minus 119904)1minus120572
]
le1
Γ (120572)[1205931
(max (119909 1003817100381710038171003817119910
1003817100381710038171003817)) + 1198961]
sdot [1205932
(max (119909 1003817100381710038171003817119910
1003817100381710038171003817)) + 1198962] int
119905
0
119889119904
(119905 minus 119904)1minus120572
le1
Γ (120572 + 1)(1205931
(1003817100381710038171003817(119909 119910)
1003817100381710038171003817) + 1198961)
sdot (1205932
(1003817100381710038171003817(119909 119910)
1003817100381710038171003817) + 1198962)
(45)
Now taking into account assumption (H3) we infer that the
operator 119866 applies 1198611199030
times 1198611199030
into 1198611199030
Moreover from the lastestimates it follows that
10038171003817100381710038171003817F (1198611199030
times 1198611199030
)10038171003817100381710038171003817
le 1205931
(1199030) + 1198961
10038171003817100381710038171003817G (1198611199030
times 1198611199030
)10038171003817100381710038171003817
le1205932
(1199030) + 1198962
Γ (120572 + 1)
(46)
Next we will prove that the operators F and G arecontinuous on the ball 119861
1199030
times 1198611199030
and consequently 119866 will bealso continuous
In fact we fix 120576 gt 0 and we take (1199090 1199100) (119909 119910) isin 119861
1199030
times 1198611199030
with (119909 119910)minus(1199090 1199100) = (119909minus119909
0 119910minus1199100) =max119909minus119909
0 119910minus
1199100 le 120576 Then for 119905 isin [0 1] we have
1003816100381610038161003816F (119909 119910) (119905) minus F (1199090 1199100) (119905)
1003816100381610038161003816
=1003816100381610038161003816119891 (119905 119909 (119905) 119910 (119905)) minus 119891 (119905 119909
0(119905) 1199100
(119905))1003816100381610038161003816
le 1205931
(max (1003816100381610038161003816119909 (119905) minus 119909
0(119905)
1003816100381610038161003816 1003816100381610038161003816119910 (119905) 119910
0(119905)
1003816100381610038161003816))
le 1205931
(max (1003817100381710038171003817119909 minus 119909
0
1003817100381710038171003817 1003817100381710038171003817119910 minus 119910
0
1003817100381710038171003817))
le 1205931
(120576) lt 120576
(47)
where we have used Remark 8 This proves the continuity ofF on 119861
1199030
times 1198611199030
Abstract and Applied Analysis 7
In order to prove the continuity ofG on 1198611199030
times1198611199030
we have1003816100381610038161003816G (119909 119910) (119905) minus G (119909
0 1199100) (119905)
1003816100381610038161003816
=1
Γ (120572)
100381610038161003816100381610038161003816100381610038161003816
int
119905
0
119892 (119904 119909 (119904) 119910 (119904))
(119905 minus 119904)1minus120572
119889119904
minus int
119905
0
119892 (119904 1199090
(119904) 1199100
(119904))
(119905 minus 119904)1minus120572
119889119904
100381610038161003816100381610038161003816100381610038161003816
le1
Γ (120572)int
119905
0
1003816100381610038161003816119892 (119904 119909 (119904) 119910 (119904)) minus 119892 (119904 1199090
(119904) 1199100
(119904))1003816100381610038161003816
(119905 minus 119904)1minus120572
119889119904
le1
Γ (120572)int
119905
0
1205932
(max (1003816100381610038161003816119909 (119904) minus 119909
0(119904)
1003816100381610038161003816 1003816100381610038161003816119910 (119904) minus 119910
0(119904)
1003816100381610038161003816))
(119905 minus 119904)1minus120572
119889119904
le1
Γ (120572)1205932
(max (1003817100381710038171003817119909 minus 119909
0
1003817100381710038171003817 1003817100381710038171003817119910 minus 119910
0
1003817100381710038171003817)) int
119905
0
119889119904
(119905 minus 119904)1minus120572
le1
Γ (120572 + 1)1205932
(120576)
lt120576
Γ (120572 + 1)
(48)
Therefore1003817100381710038171003817G (119909 119910) minus G (119909
0 1199100)1003817100381710038171003817 lt
120576
Γ (120572 + 1)(49)
and consequentlyG is a continuous operator on 1198611199030
times 1198611199030
In order to prove that G satisfies assumptions of
Theorem 13 only we have to check the condition
120583 (119866 (1198831
times 1198832)) le 120593 (max (120583 (119883
1) 120583 (119883
2))) (50)
for any subsets 1198831and 119883
2of 1198611199030
To do this we fix 120576 gt 0 119905
1 1199052
isin [0 1] with |1199051minus 1199052| le 120576 and
(119909 119910) isin 1198831
times 1198832 then we have
1003816100381610038161003816F (119909 119910) (1199051) minus F (119909 119910) (119905
2)1003816100381610038161003816
=1003816100381610038161003816119891 (1199051 119909 (1199051) 119910 (119905
1)) minus 119891 (119905
2 119909 (1199052) 119910 (119905
2))
1003816100381610038161003816
le1003816100381610038161003816119891 (1199051 119909 (1199051) 119910 (119905
1)) minus 119891 (119905
1 119909 (1199052) 119910 (119905
2))
1003816100381610038161003816
+1003816100381610038161003816119891 (1199051 119909 (1199052) 119910 (119905
2)) minus 119891 (119905
2 119909 (1199052) 119910 (119905
2))
1003816100381610038161003816
le 1205931
(max (1003816100381610038161003816119909 (1199051) minus 119909 (119905
2)1003816100381610038161003816
1003816100381610038161003816119910 (1199051) minus 119910 (119905
2)1003816100381610038161003816))
+ 120596 (119891 120576)
le 1205931
(max (120596 (119909 120576) 120596 (119910 120576))) + 120596 (119891 120576)
(51)
where 120596(119891 120576) denotes the quantity
120596 (119891 120576) = sup 1003816100381610038161003816119891 (119905 119909 119910) minus 119891 (119904 119909 119910)
1003816100381610038161003816 119905 119904 isin [0 1]
|119905 minus 119904| le 120576 119909 119910 isin [minus1199030 1199030]
(52)
From the last estimate we infer that120596 (F (119883
1times 1198832) 120576)
le 1205931
(max (120596 (1198831 120576) 120596 (119883
2 120576))) + 120596 (119891 120576)
(53)
Since 119891(119905 119909 119910) is uniformly continuous on bounded subsetsof [0 1] times R times R we deduce that 120596(119891 120576) rarr 0 as 120576 rarr 0 andtherefore
1205960
(F (1198831
times 1198832)) le lim120576rarr0
1205931
(max (120596 (1198831 120576) 120596 (119883
2 120576)))
(54)
By assumption (H2) since 120593
1is continuous we infer
1205960
(F (1198831
times 1198832)) le 120593
1(max (120596
0(1198831) 1205960
(1198832))) (55)
Now we estimate the quantity 1205960(G(1198831
times 1198832))
Fix 120576 gt 0 1199051 1199052
isin [0 1] with |1199051
minus 1199052| le 120576 and (119909 119910) isin 119883
1times
1198832 Without loss of generality we can suppose that 119905
1lt 1199052
then we have1003816100381610038161003816G (119909 119910) (119905
2) minus G (119909 119910) (119905
1)1003816100381610038161003816
=1
Γ (120572)
1003816100381610038161003816100381610038161003816100381610038161003816
int
1199052
0
119892 (119904 119909 (119904) 119910 (119904))
(1199052
minus 119904)1minus120572
119889119904
minus int
1199051
0
119892 (119904 119909 (119904) 119910 (119904))
(1199051
minus 119904)1minus120572
119889119904
1003816100381610038161003816100381610038161003816100381610038161003816
le1
Γ (120572)[int
1199051
0
10038161003816100381610038161003816(1199052
minus 119904)120572minus1
minus (1199051
minus 119904)120572minus110038161003816100381610038161003816
times1003816100381610038161003816119892 (119904 119909 (119904) 119909 (120590119904))
1003816100381610038161003816 119889119904
+ int
1199052
1199051
(1199052
minus 119904)120572minus1 1003816100381610038161003816119892 (119904 119909 (119904) 119910 (119904))
1003816100381610038161003816 119889119904]
le1
Γ (120572)[int
1199051
0
[(1199051
minus 119904)120572minus1
minus (1199052
minus 119904)120572minus1
]
times1003816100381610038161003816119892 (119904 119909 (119904) 119909 (120590119904))
1003816100381610038161003816 119889119904
+ int
1199052
1199051
(1199052
minus 119904)120572minus1 1003816100381610038161003816119892 (119904 119909 (119904) 119910 (119904))
1003816100381610038161003816 119889119904]
(56)
Since 119892 isin 119862([0 1] times R times R timesR) is bounded on the compactsubsets of [0 1] times R times R particularly on [0 1] times [minus119903
0 1199030] times
[minus1199030 1199030] Put 119871 = sup|119892(119905 119909 119910)| 119905 isin [0 1] 119909 119910 isin [minus119903
0 1199030]
Then from the last inequality we infer that1003816100381610038161003816G (119909 119910) (119905
2) minus G (119909 119910) (119905
1)1003816100381610038161003816
le119871
Γ (120572)[int
1199051
0
[(1199051
minus 119904)120572minus1
minus (1199052
minus 119904)120572minus1
] 119889119904
+ int
1199052
1199051
(1199052
minus 119904)120572minus1
119889119904]
le119871
Γ (120572 + 1)[(1199052
minus 1199051)120572
+ 119905120572
1minus 119905120572
2+ (1199052
minus 1199051)120572
]
le2119871
Γ (120572 + 1)(1199052
minus 1199051)120572
le2119871
Γ (120572 + 1)120576120572
(57)
8 Abstract and Applied Analysis
where we have used the fact that 119905120572
1minus 119905120572
2le 0 Therefore
120596 (G (1198831
times 1198832) 120576) le
2119871
Γ (120572 + 1)120576120572
(58)
From this it follows that
1205960
(G (1198831
times 1198832)) = 0 (59)
Next by Proposition 7 (46) (55) and (59) we have
1205960
(119866 (1198831
times 1198832))
= 1205960
(F (1198831
times 1198832) sdot G (119883
1times 1198832))
le1003817100381710038171003817F (119883
1times 1198832)1003817100381710038171003817 1205960
(G (1198831
times 1198832))
+1003817100381710038171003817G (119883
1times 1198832)1003817100381710038171003817 1205960
(F (1198831
times 1198832))
le10038171003817100381710038171003817F (1198611199030
times 1198611199030
)10038171003817100381710038171003817
1205960
(G (1198831
times 1198832))
+10038171003817100381710038171003817G (1198611199030
times 1198611199030
)10038171003817100381710038171003817
1205960
(F (1198831
times 1198832))
le1205932
(1199030) + 1198962
Γ (120572 + 1)1205931
(max (1205960
(1198831) 1205960
(1198832)))
(60)
By assumption (H3) since 120593
2(1199030) + 1198962
le Γ(120572 + 1) and since itis easily proved that if 120572 isin [0 1] and 120593 isin A then 120572120593 isin A wededuce that
1205960
(119866 (1198831
times 1198832)) le 120593 (max (120596
0(1198831) 1205960
(1198832))) (61)
where 120593 isin AFinally by Theorem 13 the operator G has at least a
coupled fixed point and this is the desired result Thiscompletes the proof
The nonoscillary character of the solutions of problem(4) seems to be an interesting question from the practicalpoint of view This means that the solutions of problem (4)have a constant sign on the interval (0 1) In connection withthis question we notice that if 119891(119905 119909 119910) and 119892(119905 119909 119910) haveconstant sign and are equal (this means that 119891(119905 119909 119910) gt 0
and 119892(119905 119909 119910) ge 0 or 119891(119905 119909 119910) lt 0 and 119892(119905 119909 119910) le 0
for any 119905 isin [0 1] and 119909 119910 isin R) and under assumptionsof Theorem 15 then the solution (119909 119910) isin 119862[0 1] times 119862[0 1]
of problem (4) given by Theorem 15 satisfies 119909(119905) ge 0 and119910(119905) ge 0 for 119905 isin [0 1] since the solution (119909 119910) satisfies thesystem of integral equations
119909 (119905) =119891 (119905 119909 (119905) 119910 (119905))
Γ (120572)int
119905
0
119892 (119904 119909 (119904) 119910 (119904))
(119905 minus 119904)1minus120572
119889119904
119910 (119905) =119891 (119905 119910 (119905) 119909 (119905))
Γ (120572)int
119905
0
119892 (119904 119910 (119904) 119909 (119904))
(119905 minus 119904)1minus120572
119889119904 0 le 119905 le 1
(62)
On the other hand if we perturb the data function in problem(4) of the following manner
119863120572
0+ [
119909 (119905)
119891 (119905 119909 (119905) 119910 (119905))] = 119892 (119905 119909 (119905) 119910 (119905)) + 120578 (119905)
119863120572
0+ [
119910 (119905)
119891 (119905 119910 (119905) 119909 (119905))] = 119892 (119905 119910 (119905) 119909 (119905)) + 120578 (119905)
0 lt 119905 lt 1
(63)
where 0 lt 120572 lt 1 119891 isin 119862([0 1] times R times RR 0)119892 isin 119862([0 1] times R times RR) and 120578 isin 119862[0 1] then underassumptions of Theorem 15 problem (63) can be studiedby using Theorem 15 where assumptions (H
1) and (H
2) are
automatically satisfied and we only have to check assumption(H3) This fact gives a great applicability to Theorem 15Before presenting an example illustrating our results we
need some facts about the functions involving this exampleThe following lemma appears in [18]
Lemma 16 Let 120593 R+
rarr R+be a nondecreasing and upper
semicontinuous function Then the following conditions areequivalent
(i) lim119899rarrinfin
120593119899
(119905) = 0 for any 119905 ge 0
(ii) 120593(119905) lt 119905 for any 119905 gt 0
Particularly the functions 1205721 1205722
R+
rarr R+given by
1205721(119905) = arctan 119905 and 120572
2(119905) = 119905(1 + 119905) belong to the class
A since they are nondecreasing and continuous and as itis easily seen they satisfy (ii) of Lemma 16
On the other hand since the function 1205721(119905) = arctan 119905 is
concave (because 12057210158401015840
(119905) le 0) and 1205721(0) = 0 we infer that 120572
1
is subadditive and therefore for any 119905 1199051015840
isin R+ we have
100381610038161003816100381610038161205721
(119905) minus 1205721
(1199051015840
)10038161003816100381610038161003816
=10038161003816100381610038161003816arctan 119905 minus arctan 119905
101584010038161003816100381610038161003816
le arctan 10038161003816100381610038161003816119905 minus 119905101584010038161003816100381610038161003816
(64)
Moreover it is easily seen that max(1205721 1205722) is a nondecreasing
and continuous function because 1205721and 120572
2are nondecreas-
ing and continuous andmax(1205721 1205722) satisfies (ii) of Lemma 16
Therefore max(1205721 1205722) isin A
Nowwe are ready to present an examplewhere our resultscan be applied
Example 17 Consider the following coupled system of frac-tional hybrid differential equations
11986312
0+ [119909 (119905) times (
1
4+ (
1
10) arctan |119909 (119905)|
+(1
20) (
1003816100381610038161003816119910 (119905)1003816100381610038161003816
(1 +1003816100381610038161003816119910 (119905)
1003816100381610038161003816)))
minus1
]
=1
7+
1
9119909 (119905) +
1
10119910 (119905)
Abstract and Applied Analysis 9
11986312
0+ [119910 (119905) times (
1
4+ (
1
10) arctan 1003816100381610038161003816119910 (119905)
1003816100381610038161003816
+ (1
20) (
|119909 (119905)|
(1 + |119909 (119905)|)))
minus1
]
=1
7+
1
9119910 (119905) +
1
10119909 (119905)
0 lt 119905 lt 1
119909 (0) = 119910 (0) = 0
(65)
Notice that problem (17) is a particular case of problem (4)where 120572 = 12 119891(119905 119909 119910) = 14 + (110) arctan |119909| +
(120)(|119910|(1 + |119910|)) and 119892(119905 119909 119910) = 17 + (19)119909 + (110)119910It is clear that 119891 isin 119862([0 1] times R times RR 0) and 119892 isin
119862([0 1] times R times RR) and moreover 1198961
= sup|119891(119905 0 0)| 119905 isin
[0 1] = 14 and 1198962
= sup|119892(119905 0 0)| 119905 isin [0 1] = 17Therefore assumption (H
1) of Theorem 15 is satisfied
Moreover for 119905 isin [0 1] and 1199091 1199092 1199101 1199102
isin R we have1003816100381610038161003816119891 (119905 119909
1 1199101) minus 119891 (119905 119909
2 1199102)1003816100381610038161003816
le1
10
1003816100381610038161003816arctan10038161003816100381610038161199091
1003816100381610038161003816 minus arctan 100381610038161003816100381611990921003816100381610038161003816
1003816100381610038161003816
+1
20
100381610038161003816100381610038161003816100381610038161003816
100381610038161003816100381611991011003816100381610038161003816
1 +10038161003816100381610038161199101
1003816100381610038161003816
minus
100381610038161003816100381611991021003816100381610038161003816
1 +10038161003816100381610038161199102
1003816100381610038161003816
100381610038161003816100381610038161003816100381610038161003816
le1
10arctan 1003816100381610038161003816
100381610038161003816100381611990911003816100381610038161003816 minus
100381610038161003816100381611990921003816100381610038161003816
1003816100381610038161003816
+1
20
100381610038161003816100381610038161003816100381610038161003816
100381610038161003816100381611991011003816100381610038161003816 minus
100381610038161003816100381611991021003816100381610038161003816
(1 +10038161003816100381610038161199101
1003816100381610038161003816) (1 +10038161003816100381610038161199102
1003816100381610038161003816)
100381610038161003816100381610038161003816100381610038161003816
le1
10arctan (
10038161003816100381610038161199091 minus 1199092
1003816100381610038161003816)
+1
20
10038161003816100381610038161199101 minus 1199102
1003816100381610038161003816
1 +10038161003816100381610038161199101 minus 119910
2
1003816100381610038161003816
=1
101205721
(10038161003816100381610038161199091 minus 119909
2
1003816100381610038161003816)
+1
201205722
(10038161003816100381610038161199101 minus 119910
2
1003816100381610038161003816)
le1
10max (120572
1 1205722) (
10038161003816100381610038161199091 minus 1199092
1003816100381610038161003816)
+1
10max (120572
1 1205722) (
10038161003816100381610038161199101 minus 1199102
1003816100381610038161003816)
le1
10[2max (120572
1 1205722)
times max (10038161003816100381610038161199091 minus 119909
2
1003816100381610038161003816 10038161003816100381610038161199101 minus 119910
2
1003816100381610038161003816)]
=1
5max (120572
1 1205722) (max (
10038161003816100381610038161199091 minus 1199092
1003816100381610038161003816 10038161003816100381610038161199101 minus 119910
2
1003816100381610038161003816))
(66)
Therefore 1205931(119905) = (15)max(120572
1(119905) 1205722(119905)) and 120593
1isin A
On the other hand1003816100381610038161003816119892 (119905 119909
1 1199101) minus 119892 (119905 119909
2 1199102)1003816100381610038161003816
le1
9
10038161003816100381610038161199091 minus 1199092
1003816100381610038161003816 +1
10
10038161003816100381610038161199092 minus 1199102
1003816100381610038161003816
le1
9(10038161003816100381610038161199091 minus 119910
1
1003816100381610038161003816 +10038161003816100381610038161199102 minus 119910
2
1003816100381610038161003816)
le1
9(2max (
10038161003816100381610038161199091 minus 1199101
1003816100381610038161003816 10038161003816100381610038161199102 minus 119910
2
1003816100381610038161003816))
=2
9max (
10038161003816100381610038161199091 minus 1199101
1003816100381610038161003816 10038161003816100381610038161199102 minus 119910
2
1003816100381610038161003816)
(67)
and 1205932(119905) = (29)119905 It is clear that 120593
2isin A Therefore
assumption (H2) of Theorem 15 is satisfied
In our case the inequality appearing in assumption (H3)
of Theorem 15 has the expression
[1
5max(arctan 119903
119903
1 + 119903) +
1
4] [
2
9119903 +
1
7] le 119903Γ (
3
2) (68)
It is easily seen that 1199030
= 1 satisfies the last inequalityMoreover
2
91199030
+1
7=
2
9+
1
7le Γ (
3
2) cong 088623 (69)
Finally Theorem 15 says that problem (17) has at least onesolution (119909 119910) isin 119862[0 1] such that max(119909 119910) le 1
Conflict of Interests
The authors declare that there is no conflict of interests in thesubmitted paper
References
[1] W Deng ldquoNumerical algorithm for the time fractional Fokker-Planck equationrdquo Journal of Computational Physics vol 227 no2 pp 1510ndash1522 2007
[2] S Z Rida H M El-Sherbiny and A A M Arafa ldquoOn thesolution of the fractional nonlinear Schrodinger equationrdquoPhysics Letters A vol 372 no 5 pp 553ndash558 2008
[3] K Diethelm and A D Freed ldquoOn the solution of nonlinearfractional order differential equations used in the modeling ofviscoplasticityrdquo in Scientific Computing in Chemical EngineeringII pp 217ndash224 Springer Berlin Germany 1999
[4] F Mainardi ldquoFractional calculus some basic proble ms incontinuum and statistical mechanicsrdquo in Fractals and FractionalCalculus in Continuum Mechanics (Udine 1996) vol 378 ofCISM Courses and Lectures pp 291ndash348 Springer ViennaAustria 1997
[5] R Metzler W Schick H Kilian and T F NonnenmacherldquoRelaxation in filled polymers a fractional calculus approachrdquoThe Journal of Chemical Physics vol 103 no 16 pp 7180ndash71861995
[6] B C Dhage and V Lakshmikantham ldquoBasic results on hybriddifferential equationsrdquo Nonlinear Analysis Hybrid Systems vol4 no 3 pp 414ndash424 2010
10 Abstract and Applied Analysis
[7] Y Zhao S Sun Z Han and Q Li ldquoTheory of fractionalhybrid differential equationsrdquo Computers amp Mathematics withApplications vol 62 no 3 pp 1312ndash1324 2011
[8] J Caballero M A Darwish and K Sadarangani ldquoSolvabilityof a fractional hybrid initial value problem with supremum byusingmeasures of noncompactness in Banach algebrasrdquoAppliedMathematics and Computation vol 224 pp 553ndash563 2013
[9] C Bai and J Fang ldquoThe existence of a positive solution fora singular coupled system of nonlinear fractional differentialequationsrdquo Applied Mathematics and Computation vol 150 no3 pp 611ndash621 2004
[10] Y Chen and H An ldquoNumerical solutions of coupled Burgersequations with time- and space-fractional derivativesrdquo AppliedMathematics and Computation vol 200 no 1 pp 87ndash95 2008
[11] M P Lazarevic ldquoFinite time stability analysis of PD120572 fractionalcontrol of robotic time-delay systemsrdquo Mechanics ResearchCommunications vol 33 no 2 pp 269ndash279 2006
[12] V Gafiychuk B Datsko V Meleshko and D Blackmore ldquoAnal-ysis of the solutions of coupled nonlinear fractional reaction-diffusion equationsrdquo Chaos Solitons and Fractals vol 41 no 3pp 1095ndash1104 2009
[13] B Ahmad and J J Nieto ldquoExistence results for a coupledsystem of nonlinear fractional differential equations with three-point boundary conditionsrdquo Computers amp Mathematics withApplications vol 58 no 9 pp 1838ndash1843 2009
[14] A A Kilbas H M Srivastava and J J Trujillo Theory andApplications of Fractional Differential Equations vol 204 ofNorth-Holland Mathematics Studies Elsevier Science Amster-dam The Netherlands 2006
[15] J Banas and K GoebelMeasures of Noncompactness in BanachSpaces Lecture Notes in Pure andAppliedMathematicsMarcelDekker New York NY USA 1980
[16] B N Sadovskii ldquoOn a fixed point principlerdquo Functional Analysisand Its Applications vol 1 pp 151ndash153 1967
[17] J Banas and L Olszowy ldquoOn a class of measures of noncom-pactness in Banach algebras and their application to nonlinearintegral equationsrdquo Journal of Analysis and its Applications vol28 no 4 pp 475ndash498 2009
[18] A Aghajani J Banas and N Sabzali ldquoSome generalizationsof Darbo fixed point theorem and applicationsrdquo Bulletin of theBelgian Mathematical Society vol 20 no 2 pp 345ndash358 2013
[19] R R Akhmerov M I Kamenskii A S Potapov A E Rod-kina and B N Sadovskii Measures of Noncompactness andCondensing Operators vol 55 ofOperatorTheory Advances andApplications Birkhaauser Basel Switzerland 1992
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Function Spaces
Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of Mathematics and Mathematical Sciences
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Algebra
Discrete Dynamics in Nature and Society
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Decision SciencesAdvances in
Discrete MathematicsJournal of
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Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Stochastic AnalysisInternational Journal of
2 Abstract and Applied Analysis
The coupled systems involving fractional differentialequations are very important because they occur in numerousproblems of applied nature for instance see [9ndash13]
In this paper we consider the following coupled system
119863120572
0+ [
119909 (119905)
119891 (119905 119909 (119905) 119910 (119905))] = 119892 (119905 119909 (119905) 119910 (119905))
119863120572
0+ [
119910 (119905)
119891 (119905 119910 (119905) 119909 (119905))] = 119892 (119905 119910 (119905) 119909 (119905))
0 lt 119905 lt 1
119909 (0) = 119910 (0) = 0
(4)
where 120572 isin (0 1) and 119863120572
0+ is the standard Riemann-Liouville
fractional derivativeThe main tool in our study is a fixed point theorem of
Darbo type associated to measures of noncompactness
2 Preliminaries
We begin this section with some definitions and results aboutfractional calculus
Let 120572 gt 0 and 119899 = [120572] + 1 where [120572] denotes the integerpart of 120572 For a function 119891 (0 infin) rarr R the Riemann-Liouville fractional integral of order 120572 gt 0 of 119891 is defined as
119868120572
0+119891 (119905) =
1
Γ (120572)int
119905
0
119891 (119904)
(119905 minus 119904)1minus120572
119889119904 (5)
provided that the right side is pointwise defined on (0 infin)The Riemann-Liouville fractional derivative of order 120572 of
a continuous function 119891 is defined by
119863120572
0+119891 (119909) =
1
Γ (119899 minus 120572)
119889119899
119889119909119899int
119909
0
119891 (119904)
(119909 minus 119904)120572minus119899+1
119889119904 (6)
provided that the right side is pointwise defined on (0 infin)The following lemma will be useful for our study [14]
Lemma 1 Let ℎ isin 1198711
(0 1) and 0 lt 120572 lt 1 Then
(a)
119863120572
0+119868120572
0+ℎ (119909) = ℎ (119909) (7)
(b)
119868120572
0+119863120572
0+ℎ (119909) = ℎ (119909) minus
1198681minus120572
0+ ℎ (119909)
10038161003816100381610038161003816119909=0
Γ (120572)119909120572minus1
119886119890 on (0 1)
(8)
Lemma 2 Let 0 lt 120572 lt 1 and suppose that 119891 isin 119862([0 1]R
0) and119910 isin 119862[0 1]Then the unique solution of the fractionalhybrid initial value problem
119863120572
0+ [
119909 (119905)
119891 (119905)] = 119910 (119905) 0 lt 119905 lt 1
119909 (0) = 0
(9)
is given by
119909 (119905) =119891 (119905)
Γ (120572)int
119905
0
119910 (119904)
(119905 minus 119904)1minus120572
119889119904 119905 isin [0 1] (10)
Proof Suppose that 119909(119905) is a solution of problem (9) Usingthe operator 119868
120572
0+ and taking into account Lemma 1 we get
119868120572
0+119863120572
0+ [
119909 (119905)
119891 (119905)] = 119868120572
0+119910 (119905) (11)
or equivalently
119909 (119905)
119891 (119905)minus
1198681minus120572
0+ (119909 (119905) 119891 (119905))
10038161003816100381610038161003816119905=0
Γ (120572)119905120572minus1
= 119868120572
0+119910 (119905) (12)
Since 119909(119905)119891(119905)|119905=0
= 119909(0)119891(0) = 0119891(0) = 0 (because119891(0) = 0) we have
119909 (119905) = 119891 (119905) 119868120572
0+119910 (119905) (13)
This means that
119909 (119905) =119891 (119905)
Γ (120572)int
119905
0
119910 (119904)
(119905 minus 119904)1minus120572
119889119904 (14)
Conversely suppose that 119909(119905) is given by
119909 (119905) =119891 (119905)
Γ (120572)int
119905
0
119910 (119904)
(119905 minus 119904)1minus120572
119889119904 119905 isin [0 1] (15)
This means that
119909 (119905) = 119891 (119905) 119868120572
0+119910 (119905) 119905 isin [0 1] (16)
Applying 119863120572
0+ and taking into account Lemma 1 and that
119891(119905) = 0 for 119905 isin [0 1] we obtain
119863120572
0+ [
119909 (119905)
119891 (119905)] = 119863
120572
0+119868120572
0+119910 (119905) = 119910 (119905) 0 lt 119905 lt 1 (17)
Moreover for 119905 = 0 in (16) we have 119909(0) = 119891(0) sdot 0 = 0 Thiscompletes the proof
In the sequel we recall some definitions and basic factsabout measures of noncompactness
Assume that 119864 is a real Banach space with norm sdot
and zero element 120579 By 119861(119909 119903) we denote the closed ball in 119864
centered at 119909with radius 119903 By 119861119903we denote the ball 119861(120579 119903) If
119883 is a nonempty subset of119864 by the symbols119883 andConv119883wedenote the closure and the convex closure of 119883 respectivelyBy 119883 we denote the quantity 119883 = sup119909 119909 isin 119883Finally by M
119864we will denote the family of all nonempty
and bounded subsets of 119864 and byN119864we denote its subfamily
consisting of all relatively compact subsets of 119864
Definition 3 Amapping120583 M119864
rarr R+
= [0 infin) is said to bea measure of noncompactness in 119864 if it satisfies the followingconditions
(a) The family ker 120583 = 119883 isin M119864
120583(119883) = 0 is nonemptyand ker 120583 sub N
119864
Abstract and Applied Analysis 3
(b) 119883 sub 119884 rArr 120583(119883) le 120583(119884)(c) 120583(119883) = 120583(Conv119883) = 120583(119883)(d) 120583(120582119883 + (1 minus 120582)119884) le 120582120583(119883) + (1 minus 120582)120583(119884) for 120582 isin [0 1](e) If (119883
119899) is a sequence of closed subsets fromM
119864such
that 119883119899+1
sub 119883119899
(119899 ge 1) and lim119899rarrinfin
120583(119883119899) = 0 then
119883infin
= capinfin
119899=1119883119899
= 120601
The family ker 120583 appearing in (a) is called the kernel ofthe measure of noncompactness 120583 Notice that the set 119883
infin
appearing in (e) is an element of ker120583 Indeed since 120583(119883infin
) le
120583(119883119899) for 119899 = 1 2 we infer that 120583(119883
infin) = 0 and this says
that 119883infin
isin ker120583An important theorem about fixed point theorem in
the context of measures of noncompactness is the followingDarborsquos fixed point theorem [15]
Theorem4 Let119862 be a nonempty bounded closed and convexsubset of a Banach space 119864 and let 119879 119862 rarr 119862 be a continuousmapping Suppose that there exists a constant 119896 isin [0 1) suchthat
120583 (119879 (119883)) le 119896120583 (119883) (18)
for any nonempty subset 119883 of 119862Then 119879 has a fixed point
A generalization of Theorem 4 which will be very usefulin our study is the following theorem due to Sadovskiı [16]
Theorem5 Let119862 be a nonempty bounded closed and convexsubset of a Banach space 119864 and let 119879 119862 rarr 119862 be a continuousoperator satisfying
120583 (119879 (119883)) lt 120583 (119883) (19)
for any nonempty subset 119883 of 119862 with 120583(119883) gt 0Then 119879 has a fixed point
Next we will assume that the space 119864 has structure ofBanach algebra By 119909119910 we will denote the product of twoelements 119909 119910 isin 119883 and by 119883119884 we will denote the set definedby 119883119884 = 119909119910 119909 isin 119883 119910 isin 119884
Definition 6 Let 119864 be a Banach algebra We will say thata measure of noncompactness 120583 defined on 119864 satisfiescondition (119898) if
120583 (119883119884) le 119883 120583 (119884) + 119884 120583 (119883) (20)
for any 119883 119884 isin M119864
This definition appears in [17]In this paper we will work in the space 119862[0 1] consisting
of all real functions defined and continuous on [0 1] with thestandard supremum norm
119909 = sup |119909 (119905)| 119905 isin [0 1] (21)
for119909 isin 119862[0 1] It is clear that (119862[0 1] sdot) is a Banach algebrawhere the multiplication is defined as the usual product ofreal functions
Next we present the measure of noncompactness in119862[0 1] which will be used later Let us fix 119883 isin M
119862[01]and
120576 gt 0 For 119909 isin 119883 we denote by 120596(119909 120576) the modulus ofcontinuity of 119909 that is
120596 (119909 120576) = sup |119909 (119905) minus 119909 (119904)| 119905 119904 isin [0 1] |119905 minus 119904| le 120576 (22)
Put
120596 (119883 120576) = sup 120596 (119909 120576) 119909 isin 119883
1205960
(119883) = lim120576rarr0
120596 (119883 120576) (23)
In [15] it is proved that 1205960(119883) is a measure of noncompact-
ness in 119862[0 1]
Proposition 7 Themeasure of noncompactness 1205960on 119862[0 1]
satisfies condition (119898)
Proof Fix 119883 119884 isin M119862[01]
120576 gt 0 and 119905 119904 isin [0 1] with |119905 minus 119904| le
120576 Then we have1003816100381610038161003816119909 (119905) 119910 (119905) minus 119909 (119904) 119910 (119904)
1003816100381610038161003816 le1003816100381610038161003816119909 (119905) 119910 (119905) minus 119909 (119905) 119910 (119904)
1003816100381610038161003816
+1003816100381610038161003816119909 (119905) 119910 (119904) minus 119909 (119904) 119910 (119904)
1003816100381610038161003816
= |119909 (119905)|1003816100381610038161003816119910 (119905) minus 119910 (119904)
1003816100381610038161003816
+1003816100381610038161003816119910 (119904)
1003816100381610038161003816 |119909 (119905) minus 119909 (119904)|
le 119909 120596 (119910 120576) +1003817100381710038171003817119910
1003817100381710038171003817 120596 (119909 120576)
(24)
This means that
120596 (119909119910 120576) le 119909 120596 (119910 120576) +1003817100381710038171003817119910
1003817100381710038171003817 120596 (119909 120576) (25)
and therefore
120596 (119883119884 120576) le 119883 120596 (119884 120576) + 119884 120596 (119883 120576) (26)
Taking 120576 rarr 0 we get
1205960
(119883119884) le 119883 1205960
(119884) + 119884 1205960
(119883) (27)
This completes the proof
Proposition 7 appears in [17] and we have given the prooffor the paper is self-contained
3 Main Results
We begin this section introducing the following class A offunctions
A = 120593 R+
997888rarr R+
120593 is nondecreasing
and lim119899rarrinfin
120593119899
(119905) = 0 for any 119905 gt 0
(28)
where 120593119899 denotes the 119899-iteration of 120593
4 Abstract and Applied Analysis
Remark 8 Notice that if 120593 isin A then 120593(119905) lt 119905 for any 119905 gt 0Indeed in contrary case we can find 119905
0gt 0 and 119905
0le 120593(119905
0)
Since 120593 is nondecreasing we have
0 lt 1199050
le 120593 (1199050) le 1205932
(1199050) le sdot sdot sdot le 120593
119899
(1199050) le sdot sdot sdot (29)
and therefore 0 lt 1199050
le lim119899rarrinfin
120593119899
(1199050) and this contradicts
the fact that 120593 isin AMoreover the fact that 120593(119905) lt 119905 for any 119905 gt 0 proves that
if 120593 isin A then 120593 is continuous at 1199050
= 0
Using Remark 8 and Theorem 5 we have the followingfixed point theorem
Theorem9 Let119862 be a nonempty bounded closed and convexsubset of a Banach space 119864 and let 119879 119862 rarr 119862 be a continuousoperator satisfying
120583 (119879 (119883)) le 120593 (120583 (119883)) (30)
for any nonempty subset 119883 of 119862 where 120593 isin AThen 119879 has a fixed point
Theorem 9 appears in [18] where the authors present aproof without usingTheorem 5
The following result which appears in [19] will be inter-esting in our study
Theorem 10 Let 1205831 1205832 120583
119899be measures of noncompact-
ness in the Banach spaces 1198641 1198642 119864
119899 respectively Suppose
that 119865 [0 infin)119899
rarr [0 infin) is a convex function such that119865(1199091 1199092 119909
119899) = 0 if and only if 119909
119894= 0 for 119894 = 1 2 119899
Then
120583 (119883) = 119865 (1205831
(1198831) 1205832
(1198832) 120583
119899(119883119899)) (31)
defines a measure compactness in 1198641
times 1198642
times sdot sdot sdot times 119864119899 where 119883
119894
denotes the natural projection of 119883 into 119864119894 for 119894 = 1 2 119899
Remark 11 As a consequence ofTheorem 10 we have that if 120583
is a measure of noncompactness on a Banach space 119864 and weconsider the function 119865 [0 infin) times [0 infin) rarr [0 infin) definedby 119865(119909 119910) = max(119909 119910) then since 119865 is convex and 119865(119909 119910) =
0 if and only if 119909 = 119910 = 0 120583(119883) = max120583(1198831) 120583(119883
2) defines
a measure of noncompactness in the space 119864 times 119864
Next we present the definition of a coupled fixed point
Definition 12 An element (119909 119910) isin 119883 times 119883 is said to be acoupled fixed point of a mapping 119866 119883 times 119883 rarr 119883 if119866(119909 119910) = 119909 and 119866(119910 119909) = 119910
The following result is crucial for our study
Theorem 13 Let Ω be a nonempty bounded closed andconvex subset of a Banach space 119864 and let 120583 be a measure ofnoncompactness in 119864 Suppose that 119866 Ω times Ω rarr Ω is acontinuous operator satisfying
120583 (119866 (1198831
times 1198832)) le 120593 (max (120583 (119883
1) 120583 (119883
2))) (32)
for all nonempty subsets 1198831and 119883
2of Ω where 120593 isin A
Then 119866 has at least a coupled fixed point
Proof Notice that by Remark 11 120583(119883) = max120583(1198831) 120583(119883
2)
is a measure of noncompactness in the space 119864 times 119864 where 1198831
and 1198832are the projections of 119883 into 119864
Now we consider the mapping 119866 Ω times Ω rarr Ω times Ω
defined by 119866(119909 119910) = (119866(119909 119910) 119866(119910 119909)) It is easily seen thatΩ times Ω is a nonempty bounded closed and convex subsetof 119864 times 119864 Since 119866 is continuous it is clear that 119866 is alsocontinuous
Next we take a nonempty 119883 of Ω times Ω Then
120583 (119866 (119883)) = 120583 (119866 (1198831
times 1198832))
= 120583 (119866 (1198831
times 1198832) times 119866 (119883
2times 1198831))
= max 120583 (119866 (1198831
times 1198832)) 120583 (119866 (119883
2times 1198831))
le max 120593 (max (120583 (1198831) 120583 (119883
2)))
120593 (max (120583 (1198832) 120583 (119883
1)))
= 120593 (max (120583 (1198831) 120583 (119883
2)))
= 120593 (120583 (1198831
times 1198832))
(33)
Since 120593 isin A by Theorem 9 the mapping 119866 has at least onefixed pointThismeans that there exists (119909
0 1199100) isin ΩtimesΩ such
that 119866(1199090 1199100) = (119909
0 1199100) or equivalently 119866(119909
0 1199100) = 1199090and
119866(1199100 1199090) = 1199100 This proves that 119866 has at least a coupled fixed
point
Now we consider the coupled system of integral equa-tions
119909 (119905) =119891 (119905 119909 (119905) 119910 (119905))
Γ (120572)int
119905
0
119892 (119904 119909 (119904) 119910 (119904))
(119905 minus 119904)1minus120572
119889119904
119910 (119905) =119891 (119905 119910 (119905) 119909 (119905))
Γ (120572)int
119905
0
119892 (119904 119910 (119904) 119909 (119904))
(119905 minus 119904)1minus120572
119889119904 119905 isin [0 1]
(34)
Lemma 14 Assume that 119891 isin 119862([0 1] times R times RR 0) and119892 isin 119862([0 1] times R times RR) Then (119909 119910) isin 119862[0 1] times 119862[0 1] isa solution of (34) if and only if (119909 119910) isin 119862[0 1] times 119862[0 1] is asolution of (4)
Proof The proof is an immediate consequence of Lemma 2so we omit it
Next we will study problem (4) under the followingassumptions
(H1) 119891 isin 119862([0 1] times R times RR 0) and 119892 isin 119862([0 1] times R times
RR)(H2) The functions 119891 and 119892 satisfy
1003816100381610038161003816119891 (119905 1199091 1199101) minus 119891 (119905 119909
2 1199102)1003816100381610038161003816
le 1205931
(max (10038161003816100381610038161199091 minus 119909
2
1003816100381610038161003816 10038161003816100381610038161199101 minus 119910
2
1003816100381610038161003816))
1003816100381610038161003816119892 (119905 1199091 1199102) minus 119892 (119905 119909
2 1199102)1003816100381610038161003816
le 1205932
(max (10038161003816100381610038161199091 minus 119910
1
1003816100381610038161003816 10038161003816100381610038161199092 minus 119910
2
1003816100381610038161003816))
(35)
Abstract and Applied Analysis 5
respectively for any 119905 isin [0 1] and 1199091 1199092 1199101 1199102
isin Rwhere 120593
1 1205932
isin A and 1205931is continuous
Notice that assumption (H1) gives us the existence
of two nonnegative constants 1198961and 119896
2such that
|119891(119905 0 0)| le 1198961and |119892(119905 0 0)| le 119896
2 for any 119905 isin [0 1]
(H3) There exists 119903
0gt 0 satisfying the inequalities
(1205931
(119903) + 1198961) sdot (1205932
(119903) + 1198962) le 119903Γ (120572 + 1)
1205932
(119903) + 1198962
le Γ (120572 + 1)
(36)
Theorem 15 Under assumptions (1198671)ndash(1198673) problem (4) has
at least one solution in 119862[0 1] times 119862[0 1]
Proof In virtue of Lemma 14 a solution (119909 119910) isin 119862[0 1] times
119862[0 1] of problem (4) satisfies
119909 (119905) =119891 (119905 119909 (119905) 119910 (119905))
Γ (120572)int
119905
0
119892 (119904 119909 (119904) 119910 (119904))
(119905 minus 119904)1minus120572
119889119904
119910 (119905) =119891 (119905 119910 (119905) 119909 (119905))
Γ (120572)int
119905
0
119892 (119904 119910 (119904) 119909 (119904))
(119905 minus 119904)1minus120572
119889119904
119905 isin [0 1]
(37)
We consider the space 119862[0 1] times 119862[0 1] equipped with thenorm (119909 119910)
119862[01]times119862[01]= max119909 119910 for any (119909 119910) isin
119862[0 1] times 119862[0 1]In 119862[0 1] times 119862[0 1] we define the operator
119866 (119909 119910) (119905) =119891 (119905 119909 (119905) 119910 (119905))
Γ (120572)int
119905
0
119892 (119904 119909 (119904) 119910 (119904))
(119905 minus 119904)1minus120572
119889119904
119905 isin [0 1]
(38)
LetF andG be the operators given by
F (119909 119910) (119905) = 119891 (119905 119909 (119905) 119910 (119905))
G (119909 119910) (119905) =1
Γ (120572)int
119905
0
119892 (119904 119909 (119904) 119910 (119904))
(119905 minus 119904)1minus120572
119889119904
(39)
for any (119909 119910) isin 119862[0 1] times 119862[0 1] and any 119905 isin [0 1] Then
119866 (119909 119910) = F (119909 119910) sdot G (119909 119910) (40)
Firstly we will prove that 119866 applies 119862[0 1] times 119862[0 1] into119862[0 1] To do this it is sufficient to prove that F(119909 119910)G(119909 119910) isin 119862[0 1] for any (119909 119910) isin 119862[0 1] times 119862[0 1] since theproduct of continuous functions is continuous
In virtue of assumption (H1) it is clear that F(119909 119910) isin
119862[0 1] for (119909 119910) isin 119862[0 1] times 119862[0 1] In order to prove thatG(119909 119910) isin 119862[0 1] for (119909 119910) isin 119862[0 1] times 119862[0 1] we fix 119905
0isin
[0 1] and we consider a sequence (119905119899) isin [0 1] such that 119905
119899rarr
1199050 and we have to prove that G(119909 119910)(119905
119899) rarr G(119909 119910)(119905
0)
Without loss of generality we can suppose that 119905119899
gt 1199050 Then
we have
1003816100381610038161003816G (119909 119910) (119905119899) minus G (119909 119910) (119905
0)1003816100381610038161003816
=1
Γ (120572)
1003816100381610038161003816100381610038161003816100381610038161003816
int
119905119899
0
119892 (119904 119909 (119904) 119910 (119904))
(119905119899
minus 119904)1minus120572
119889119904
minus int
1199050
0
119892 (119904 119909 (119904) 119910 (119904))
(1199050
minus 119904)1minus120572
119889119904
1003816100381610038161003816100381610038161003816100381610038161003816
le1
Γ (120572)
1003816100381610038161003816100381610038161003816100381610038161003816
int
119905119899
0
119892 (119904 119909 (119904) 119910 (119904))
(119905119899
minus 119904)1minus120572
119889119904
minus int
119905119899
0
119892 (119904 119909 (119904) 119910 (119904))
(1199050
minus 119904)1minus120572
119889119904
1003816100381610038161003816100381610038161003816100381610038161003816
+1
Γ (120572)
1003816100381610038161003816100381610038161003816100381610038161003816
int
119905119899
0
119892 (119904 119909 (119904) 119910 (119904))
(1199050
minus 119904)1minus120572
119889119904
minus int
1199050
0
119892 (119904 119909 (119904) 119910 (119904))
(1199050
minus 119904)1minus120572
119889119904
1003816100381610038161003816100381610038161003816100381610038161003816
le1
Γ (120572)int
119905119899
0
10038161003816100381610038161003816(119905119899
minus 119904)120572minus1
minus (1199050
minus 119904)120572minus110038161003816100381610038161003816
times1003816100381610038161003816119892 (119904 119909 (119904) 119910 (119904))
1003816100381610038161003816 119889119904
+1
Γ (120572)int
119905119899
1199050
10038161003816100381610038161003816(1199050
minus 119904)120572minus110038161003816100381610038161003816
1003816100381610038161003816119892 (119904 119909 (119904) 119910 (119904))1003816100381610038161003816 119889119904
(41)
By assumption (H1) since 119892 isin 119862([0 1] times R times RR) 119892 is
bounded on the compact set [0 1]times[minus119909 119909]times[minus119910 119910]Denote by
119872 = sup 1003816100381610038161003816119892 (119904 119909
1 1199101)1003816100381610038161003816 119904 isin [0 1] 119909
1isin [minus 119909 119909]
1199101
isin [minus1003817100381710038171003817119910
1003817100381710038171003817 1003817100381710038171003817119910
1003817100381710038171003817]
(42)
From the last estimate we obtain
1003816100381610038161003816G (119909 119910) (119905119899) minus G (119909 119910) (119905
0)1003816100381610038161003816
le119872
Γ (120572)int
119905119899
0
10038161003816100381610038161003816(119905119899
minus 119904)120572minus1
minus (1199050
minus 119904)120572minus110038161003816100381610038161003816
119889119904
+119872
Γ (120572)int
119905119899
1199050
10038161003816100381610038161003816(1199050
minus 119904)120572minus110038161003816100381610038161003816
119889119904
(43)
6 Abstract and Applied Analysis
As 0 lt 120572 lt 1 and 119905119899
gt 1199050 we infer that
1003816100381610038161003816G (119909 119910) (119905119899) minus G (119909 119910) (119905
0)1003816100381610038161003816
le119872
Γ (120572)[int
1199050
0
10038161003816100381610038161003816(119905119899
minus 119904)120572minus1
minus (1199050
minus 119904)120572minus110038161003816100381610038161003816
119889119904
+ int
119905119899
1199050
10038161003816100381610038161003816(119905119899
minus 119904)120572minus1
minus (1199050
minus 119904)120572minus110038161003816100381610038161003816
119889119904]
+119872
Γ (120572)int
119905119899
1199050
1
(119904 minus 1199050)1minus120572
119889119904
=119872
Γ (120572)[ int
1199050
0
[(1199050
minus 119904)120572minus1
minus (119905119899
minus 119904)120572minus1
] 119889119904
+ int
119905119899
1199050
119889119904
(119905119899
minus 119904)1minus120572
+ int
119905119899
1199050
119889119904
(119904 minus 1199050)1minus120572
]
+119872
Γ (120572)int
119905119899
1199050
1
(119904 minus 1199050)1minus120572
119889119904
le119872
Γ (120572 + 1)[(119905119899
minus 1199050)120572
+ 119905120572
0minus 119905120572
119899
+ (119905119899
minus 1199050)120572
+ (119905119899
minus 1199050)120572
]
+119872
Γ (120572 + 1)(119905119899
minus 1199050)120572
=4119872
Γ (120572 + 1)(119905119899
minus 1199050)120572
+119872
Γ (120572 + 1)(119905120572
0minus 119905120572
119899)
lt4119872
Γ (120572 + 1)(119905119899
minus 1199050)120572
(44)
where the last inequality has been obtained by using the factthat 119905120572
0minus 119905120572
119899lt 0
Therefore since 119905119899
rarr 1199050 from the last estimate we
deduce that G(119909 119910)(119905119899) rarr G(119909 119910)(119905
0) This proves that
G(119909 119910) isin 119862[0 1] Consequently G 119862[0 1] times 119862[0 1] rarr
119862[0 1] On the other hand for (119909 119910) isin 119862[0 1] times 119862[0 1] and119905 isin 119862[0 1] we have
1003816100381610038161003816119866 (119909 119910) (119905)1003816100381610038161003816
=1003816100381610038161003816F (119909 119910) (119905) sdot G (119909 119910) (119905)
1003816100381610038161003816
=1003816100381610038161003816119891 (119905 119909 (119905) 119910 (119905))
1003816100381610038161003816
100381610038161003816100381610038161003816100381610038161003816
1
Γ (120572)int
119905
0
119892 (119904 119909 (119904) 119910 (119904))
(119905 minus 119904)1minus120572
119889119904
100381610038161003816100381610038161003816100381610038161003816
le [1003816100381610038161003816119891 (119905 119909 (119905) 119910 (119905)) minus 119891 (119905 0 0)
1003816100381610038161003816 +1003816100381610038161003816119891 (119905 0 0)
1003816100381610038161003816]
times [1
Γ (120572)
100381610038161003816100381610038161003816100381610038161003816
int
119905
0
119892 (119904 119909 (119904) 119910 (119904)) minus 119892 (119904 0 0)
(119905 minus 119904)1minus120572
119889119904
+ int
119905
0
119892 (119904 0 0)
(119905 minus 119904)1minus120572
119889119904
100381610038161003816100381610038161003816100381610038161003816
]
le1
Γ (120572)[1205931
(max (|119909 (119905)| 1003816100381610038161003816119910 (119905)
1003816100381610038161003816)) + 1198961]
times [int
119905
0
1003816100381610038161003816119892 (119904 119909 (119904) 119910 (119904)) minus 119892 (119904 0 0)1003816100381610038161003816
(119905 minus 119904)1minus120572
119889119904
+ int
119905
0
1003816100381610038161003816119892 (119904 0 0)1003816100381610038161003816
(119905 minus 119904)1minus120572
119889119904]
le1
Γ (120572)[1205931
(max (119909 1003817100381710038171003817119910
1003817100381710038171003817)) + 1198961]
times [int
119905
0
1205932
(max (|119909 (119904)| 1003816100381610038161003816119910 (119904)
1003816100381610038161003816))
(119905 minus 119904)1minus120572
119889119904
+ 1198962
int
119905
0
119889119904
(119905 minus 119904)1minus120572
]
le1
Γ (120572)[1205931
(max (119909 1003817100381710038171003817119910
1003817100381710038171003817)) + 1198961]
sdot [1205932
(max (119909 1003817100381710038171003817119910
1003817100381710038171003817)) + 1198962] int
119905
0
119889119904
(119905 minus 119904)1minus120572
le1
Γ (120572 + 1)(1205931
(1003817100381710038171003817(119909 119910)
1003817100381710038171003817) + 1198961)
sdot (1205932
(1003817100381710038171003817(119909 119910)
1003817100381710038171003817) + 1198962)
(45)
Now taking into account assumption (H3) we infer that the
operator 119866 applies 1198611199030
times 1198611199030
into 1198611199030
Moreover from the lastestimates it follows that
10038171003817100381710038171003817F (1198611199030
times 1198611199030
)10038171003817100381710038171003817
le 1205931
(1199030) + 1198961
10038171003817100381710038171003817G (1198611199030
times 1198611199030
)10038171003817100381710038171003817
le1205932
(1199030) + 1198962
Γ (120572 + 1)
(46)
Next we will prove that the operators F and G arecontinuous on the ball 119861
1199030
times 1198611199030
and consequently 119866 will bealso continuous
In fact we fix 120576 gt 0 and we take (1199090 1199100) (119909 119910) isin 119861
1199030
times 1198611199030
with (119909 119910)minus(1199090 1199100) = (119909minus119909
0 119910minus1199100) =max119909minus119909
0 119910minus
1199100 le 120576 Then for 119905 isin [0 1] we have
1003816100381610038161003816F (119909 119910) (119905) minus F (1199090 1199100) (119905)
1003816100381610038161003816
=1003816100381610038161003816119891 (119905 119909 (119905) 119910 (119905)) minus 119891 (119905 119909
0(119905) 1199100
(119905))1003816100381610038161003816
le 1205931
(max (1003816100381610038161003816119909 (119905) minus 119909
0(119905)
1003816100381610038161003816 1003816100381610038161003816119910 (119905) 119910
0(119905)
1003816100381610038161003816))
le 1205931
(max (1003817100381710038171003817119909 minus 119909
0
1003817100381710038171003817 1003817100381710038171003817119910 minus 119910
0
1003817100381710038171003817))
le 1205931
(120576) lt 120576
(47)
where we have used Remark 8 This proves the continuity ofF on 119861
1199030
times 1198611199030
Abstract and Applied Analysis 7
In order to prove the continuity ofG on 1198611199030
times1198611199030
we have1003816100381610038161003816G (119909 119910) (119905) minus G (119909
0 1199100) (119905)
1003816100381610038161003816
=1
Γ (120572)
100381610038161003816100381610038161003816100381610038161003816
int
119905
0
119892 (119904 119909 (119904) 119910 (119904))
(119905 minus 119904)1minus120572
119889119904
minus int
119905
0
119892 (119904 1199090
(119904) 1199100
(119904))
(119905 minus 119904)1minus120572
119889119904
100381610038161003816100381610038161003816100381610038161003816
le1
Γ (120572)int
119905
0
1003816100381610038161003816119892 (119904 119909 (119904) 119910 (119904)) minus 119892 (119904 1199090
(119904) 1199100
(119904))1003816100381610038161003816
(119905 minus 119904)1minus120572
119889119904
le1
Γ (120572)int
119905
0
1205932
(max (1003816100381610038161003816119909 (119904) minus 119909
0(119904)
1003816100381610038161003816 1003816100381610038161003816119910 (119904) minus 119910
0(119904)
1003816100381610038161003816))
(119905 minus 119904)1minus120572
119889119904
le1
Γ (120572)1205932
(max (1003817100381710038171003817119909 minus 119909
0
1003817100381710038171003817 1003817100381710038171003817119910 minus 119910
0
1003817100381710038171003817)) int
119905
0
119889119904
(119905 minus 119904)1minus120572
le1
Γ (120572 + 1)1205932
(120576)
lt120576
Γ (120572 + 1)
(48)
Therefore1003817100381710038171003817G (119909 119910) minus G (119909
0 1199100)1003817100381710038171003817 lt
120576
Γ (120572 + 1)(49)
and consequentlyG is a continuous operator on 1198611199030
times 1198611199030
In order to prove that G satisfies assumptions of
Theorem 13 only we have to check the condition
120583 (119866 (1198831
times 1198832)) le 120593 (max (120583 (119883
1) 120583 (119883
2))) (50)
for any subsets 1198831and 119883
2of 1198611199030
To do this we fix 120576 gt 0 119905
1 1199052
isin [0 1] with |1199051minus 1199052| le 120576 and
(119909 119910) isin 1198831
times 1198832 then we have
1003816100381610038161003816F (119909 119910) (1199051) minus F (119909 119910) (119905
2)1003816100381610038161003816
=1003816100381610038161003816119891 (1199051 119909 (1199051) 119910 (119905
1)) minus 119891 (119905
2 119909 (1199052) 119910 (119905
2))
1003816100381610038161003816
le1003816100381610038161003816119891 (1199051 119909 (1199051) 119910 (119905
1)) minus 119891 (119905
1 119909 (1199052) 119910 (119905
2))
1003816100381610038161003816
+1003816100381610038161003816119891 (1199051 119909 (1199052) 119910 (119905
2)) minus 119891 (119905
2 119909 (1199052) 119910 (119905
2))
1003816100381610038161003816
le 1205931
(max (1003816100381610038161003816119909 (1199051) minus 119909 (119905
2)1003816100381610038161003816
1003816100381610038161003816119910 (1199051) minus 119910 (119905
2)1003816100381610038161003816))
+ 120596 (119891 120576)
le 1205931
(max (120596 (119909 120576) 120596 (119910 120576))) + 120596 (119891 120576)
(51)
where 120596(119891 120576) denotes the quantity
120596 (119891 120576) = sup 1003816100381610038161003816119891 (119905 119909 119910) minus 119891 (119904 119909 119910)
1003816100381610038161003816 119905 119904 isin [0 1]
|119905 minus 119904| le 120576 119909 119910 isin [minus1199030 1199030]
(52)
From the last estimate we infer that120596 (F (119883
1times 1198832) 120576)
le 1205931
(max (120596 (1198831 120576) 120596 (119883
2 120576))) + 120596 (119891 120576)
(53)
Since 119891(119905 119909 119910) is uniformly continuous on bounded subsetsof [0 1] times R times R we deduce that 120596(119891 120576) rarr 0 as 120576 rarr 0 andtherefore
1205960
(F (1198831
times 1198832)) le lim120576rarr0
1205931
(max (120596 (1198831 120576) 120596 (119883
2 120576)))
(54)
By assumption (H2) since 120593
1is continuous we infer
1205960
(F (1198831
times 1198832)) le 120593
1(max (120596
0(1198831) 1205960
(1198832))) (55)
Now we estimate the quantity 1205960(G(1198831
times 1198832))
Fix 120576 gt 0 1199051 1199052
isin [0 1] with |1199051
minus 1199052| le 120576 and (119909 119910) isin 119883
1times
1198832 Without loss of generality we can suppose that 119905
1lt 1199052
then we have1003816100381610038161003816G (119909 119910) (119905
2) minus G (119909 119910) (119905
1)1003816100381610038161003816
=1
Γ (120572)
1003816100381610038161003816100381610038161003816100381610038161003816
int
1199052
0
119892 (119904 119909 (119904) 119910 (119904))
(1199052
minus 119904)1minus120572
119889119904
minus int
1199051
0
119892 (119904 119909 (119904) 119910 (119904))
(1199051
minus 119904)1minus120572
119889119904
1003816100381610038161003816100381610038161003816100381610038161003816
le1
Γ (120572)[int
1199051
0
10038161003816100381610038161003816(1199052
minus 119904)120572minus1
minus (1199051
minus 119904)120572minus110038161003816100381610038161003816
times1003816100381610038161003816119892 (119904 119909 (119904) 119909 (120590119904))
1003816100381610038161003816 119889119904
+ int
1199052
1199051
(1199052
minus 119904)120572minus1 1003816100381610038161003816119892 (119904 119909 (119904) 119910 (119904))
1003816100381610038161003816 119889119904]
le1
Γ (120572)[int
1199051
0
[(1199051
minus 119904)120572minus1
minus (1199052
minus 119904)120572minus1
]
times1003816100381610038161003816119892 (119904 119909 (119904) 119909 (120590119904))
1003816100381610038161003816 119889119904
+ int
1199052
1199051
(1199052
minus 119904)120572minus1 1003816100381610038161003816119892 (119904 119909 (119904) 119910 (119904))
1003816100381610038161003816 119889119904]
(56)
Since 119892 isin 119862([0 1] times R times R timesR) is bounded on the compactsubsets of [0 1] times R times R particularly on [0 1] times [minus119903
0 1199030] times
[minus1199030 1199030] Put 119871 = sup|119892(119905 119909 119910)| 119905 isin [0 1] 119909 119910 isin [minus119903
0 1199030]
Then from the last inequality we infer that1003816100381610038161003816G (119909 119910) (119905
2) minus G (119909 119910) (119905
1)1003816100381610038161003816
le119871
Γ (120572)[int
1199051
0
[(1199051
minus 119904)120572minus1
minus (1199052
minus 119904)120572minus1
] 119889119904
+ int
1199052
1199051
(1199052
minus 119904)120572minus1
119889119904]
le119871
Γ (120572 + 1)[(1199052
minus 1199051)120572
+ 119905120572
1minus 119905120572
2+ (1199052
minus 1199051)120572
]
le2119871
Γ (120572 + 1)(1199052
minus 1199051)120572
le2119871
Γ (120572 + 1)120576120572
(57)
8 Abstract and Applied Analysis
where we have used the fact that 119905120572
1minus 119905120572
2le 0 Therefore
120596 (G (1198831
times 1198832) 120576) le
2119871
Γ (120572 + 1)120576120572
(58)
From this it follows that
1205960
(G (1198831
times 1198832)) = 0 (59)
Next by Proposition 7 (46) (55) and (59) we have
1205960
(119866 (1198831
times 1198832))
= 1205960
(F (1198831
times 1198832) sdot G (119883
1times 1198832))
le1003817100381710038171003817F (119883
1times 1198832)1003817100381710038171003817 1205960
(G (1198831
times 1198832))
+1003817100381710038171003817G (119883
1times 1198832)1003817100381710038171003817 1205960
(F (1198831
times 1198832))
le10038171003817100381710038171003817F (1198611199030
times 1198611199030
)10038171003817100381710038171003817
1205960
(G (1198831
times 1198832))
+10038171003817100381710038171003817G (1198611199030
times 1198611199030
)10038171003817100381710038171003817
1205960
(F (1198831
times 1198832))
le1205932
(1199030) + 1198962
Γ (120572 + 1)1205931
(max (1205960
(1198831) 1205960
(1198832)))
(60)
By assumption (H3) since 120593
2(1199030) + 1198962
le Γ(120572 + 1) and since itis easily proved that if 120572 isin [0 1] and 120593 isin A then 120572120593 isin A wededuce that
1205960
(119866 (1198831
times 1198832)) le 120593 (max (120596
0(1198831) 1205960
(1198832))) (61)
where 120593 isin AFinally by Theorem 13 the operator G has at least a
coupled fixed point and this is the desired result Thiscompletes the proof
The nonoscillary character of the solutions of problem(4) seems to be an interesting question from the practicalpoint of view This means that the solutions of problem (4)have a constant sign on the interval (0 1) In connection withthis question we notice that if 119891(119905 119909 119910) and 119892(119905 119909 119910) haveconstant sign and are equal (this means that 119891(119905 119909 119910) gt 0
and 119892(119905 119909 119910) ge 0 or 119891(119905 119909 119910) lt 0 and 119892(119905 119909 119910) le 0
for any 119905 isin [0 1] and 119909 119910 isin R) and under assumptionsof Theorem 15 then the solution (119909 119910) isin 119862[0 1] times 119862[0 1]
of problem (4) given by Theorem 15 satisfies 119909(119905) ge 0 and119910(119905) ge 0 for 119905 isin [0 1] since the solution (119909 119910) satisfies thesystem of integral equations
119909 (119905) =119891 (119905 119909 (119905) 119910 (119905))
Γ (120572)int
119905
0
119892 (119904 119909 (119904) 119910 (119904))
(119905 minus 119904)1minus120572
119889119904
119910 (119905) =119891 (119905 119910 (119905) 119909 (119905))
Γ (120572)int
119905
0
119892 (119904 119910 (119904) 119909 (119904))
(119905 minus 119904)1minus120572
119889119904 0 le 119905 le 1
(62)
On the other hand if we perturb the data function in problem(4) of the following manner
119863120572
0+ [
119909 (119905)
119891 (119905 119909 (119905) 119910 (119905))] = 119892 (119905 119909 (119905) 119910 (119905)) + 120578 (119905)
119863120572
0+ [
119910 (119905)
119891 (119905 119910 (119905) 119909 (119905))] = 119892 (119905 119910 (119905) 119909 (119905)) + 120578 (119905)
0 lt 119905 lt 1
(63)
where 0 lt 120572 lt 1 119891 isin 119862([0 1] times R times RR 0)119892 isin 119862([0 1] times R times RR) and 120578 isin 119862[0 1] then underassumptions of Theorem 15 problem (63) can be studiedby using Theorem 15 where assumptions (H
1) and (H
2) are
automatically satisfied and we only have to check assumption(H3) This fact gives a great applicability to Theorem 15Before presenting an example illustrating our results we
need some facts about the functions involving this exampleThe following lemma appears in [18]
Lemma 16 Let 120593 R+
rarr R+be a nondecreasing and upper
semicontinuous function Then the following conditions areequivalent
(i) lim119899rarrinfin
120593119899
(119905) = 0 for any 119905 ge 0
(ii) 120593(119905) lt 119905 for any 119905 gt 0
Particularly the functions 1205721 1205722
R+
rarr R+given by
1205721(119905) = arctan 119905 and 120572
2(119905) = 119905(1 + 119905) belong to the class
A since they are nondecreasing and continuous and as itis easily seen they satisfy (ii) of Lemma 16
On the other hand since the function 1205721(119905) = arctan 119905 is
concave (because 12057210158401015840
(119905) le 0) and 1205721(0) = 0 we infer that 120572
1
is subadditive and therefore for any 119905 1199051015840
isin R+ we have
100381610038161003816100381610038161205721
(119905) minus 1205721
(1199051015840
)10038161003816100381610038161003816
=10038161003816100381610038161003816arctan 119905 minus arctan 119905
101584010038161003816100381610038161003816
le arctan 10038161003816100381610038161003816119905 minus 119905101584010038161003816100381610038161003816
(64)
Moreover it is easily seen that max(1205721 1205722) is a nondecreasing
and continuous function because 1205721and 120572
2are nondecreas-
ing and continuous andmax(1205721 1205722) satisfies (ii) of Lemma 16
Therefore max(1205721 1205722) isin A
Nowwe are ready to present an examplewhere our resultscan be applied
Example 17 Consider the following coupled system of frac-tional hybrid differential equations
11986312
0+ [119909 (119905) times (
1
4+ (
1
10) arctan |119909 (119905)|
+(1
20) (
1003816100381610038161003816119910 (119905)1003816100381610038161003816
(1 +1003816100381610038161003816119910 (119905)
1003816100381610038161003816)))
minus1
]
=1
7+
1
9119909 (119905) +
1
10119910 (119905)
Abstract and Applied Analysis 9
11986312
0+ [119910 (119905) times (
1
4+ (
1
10) arctan 1003816100381610038161003816119910 (119905)
1003816100381610038161003816
+ (1
20) (
|119909 (119905)|
(1 + |119909 (119905)|)))
minus1
]
=1
7+
1
9119910 (119905) +
1
10119909 (119905)
0 lt 119905 lt 1
119909 (0) = 119910 (0) = 0
(65)
Notice that problem (17) is a particular case of problem (4)where 120572 = 12 119891(119905 119909 119910) = 14 + (110) arctan |119909| +
(120)(|119910|(1 + |119910|)) and 119892(119905 119909 119910) = 17 + (19)119909 + (110)119910It is clear that 119891 isin 119862([0 1] times R times RR 0) and 119892 isin
119862([0 1] times R times RR) and moreover 1198961
= sup|119891(119905 0 0)| 119905 isin
[0 1] = 14 and 1198962
= sup|119892(119905 0 0)| 119905 isin [0 1] = 17Therefore assumption (H
1) of Theorem 15 is satisfied
Moreover for 119905 isin [0 1] and 1199091 1199092 1199101 1199102
isin R we have1003816100381610038161003816119891 (119905 119909
1 1199101) minus 119891 (119905 119909
2 1199102)1003816100381610038161003816
le1
10
1003816100381610038161003816arctan10038161003816100381610038161199091
1003816100381610038161003816 minus arctan 100381610038161003816100381611990921003816100381610038161003816
1003816100381610038161003816
+1
20
100381610038161003816100381610038161003816100381610038161003816
100381610038161003816100381611991011003816100381610038161003816
1 +10038161003816100381610038161199101
1003816100381610038161003816
minus
100381610038161003816100381611991021003816100381610038161003816
1 +10038161003816100381610038161199102
1003816100381610038161003816
100381610038161003816100381610038161003816100381610038161003816
le1
10arctan 1003816100381610038161003816
100381610038161003816100381611990911003816100381610038161003816 minus
100381610038161003816100381611990921003816100381610038161003816
1003816100381610038161003816
+1
20
100381610038161003816100381610038161003816100381610038161003816
100381610038161003816100381611991011003816100381610038161003816 minus
100381610038161003816100381611991021003816100381610038161003816
(1 +10038161003816100381610038161199101
1003816100381610038161003816) (1 +10038161003816100381610038161199102
1003816100381610038161003816)
100381610038161003816100381610038161003816100381610038161003816
le1
10arctan (
10038161003816100381610038161199091 minus 1199092
1003816100381610038161003816)
+1
20
10038161003816100381610038161199101 minus 1199102
1003816100381610038161003816
1 +10038161003816100381610038161199101 minus 119910
2
1003816100381610038161003816
=1
101205721
(10038161003816100381610038161199091 minus 119909
2
1003816100381610038161003816)
+1
201205722
(10038161003816100381610038161199101 minus 119910
2
1003816100381610038161003816)
le1
10max (120572
1 1205722) (
10038161003816100381610038161199091 minus 1199092
1003816100381610038161003816)
+1
10max (120572
1 1205722) (
10038161003816100381610038161199101 minus 1199102
1003816100381610038161003816)
le1
10[2max (120572
1 1205722)
times max (10038161003816100381610038161199091 minus 119909
2
1003816100381610038161003816 10038161003816100381610038161199101 minus 119910
2
1003816100381610038161003816)]
=1
5max (120572
1 1205722) (max (
10038161003816100381610038161199091 minus 1199092
1003816100381610038161003816 10038161003816100381610038161199101 minus 119910
2
1003816100381610038161003816))
(66)
Therefore 1205931(119905) = (15)max(120572
1(119905) 1205722(119905)) and 120593
1isin A
On the other hand1003816100381610038161003816119892 (119905 119909
1 1199101) minus 119892 (119905 119909
2 1199102)1003816100381610038161003816
le1
9
10038161003816100381610038161199091 minus 1199092
1003816100381610038161003816 +1
10
10038161003816100381610038161199092 minus 1199102
1003816100381610038161003816
le1
9(10038161003816100381610038161199091 minus 119910
1
1003816100381610038161003816 +10038161003816100381610038161199102 minus 119910
2
1003816100381610038161003816)
le1
9(2max (
10038161003816100381610038161199091 minus 1199101
1003816100381610038161003816 10038161003816100381610038161199102 minus 119910
2
1003816100381610038161003816))
=2
9max (
10038161003816100381610038161199091 minus 1199101
1003816100381610038161003816 10038161003816100381610038161199102 minus 119910
2
1003816100381610038161003816)
(67)
and 1205932(119905) = (29)119905 It is clear that 120593
2isin A Therefore
assumption (H2) of Theorem 15 is satisfied
In our case the inequality appearing in assumption (H3)
of Theorem 15 has the expression
[1
5max(arctan 119903
119903
1 + 119903) +
1
4] [
2
9119903 +
1
7] le 119903Γ (
3
2) (68)
It is easily seen that 1199030
= 1 satisfies the last inequalityMoreover
2
91199030
+1
7=
2
9+
1
7le Γ (
3
2) cong 088623 (69)
Finally Theorem 15 says that problem (17) has at least onesolution (119909 119910) isin 119862[0 1] such that max(119909 119910) le 1
Conflict of Interests
The authors declare that there is no conflict of interests in thesubmitted paper
References
[1] W Deng ldquoNumerical algorithm for the time fractional Fokker-Planck equationrdquo Journal of Computational Physics vol 227 no2 pp 1510ndash1522 2007
[2] S Z Rida H M El-Sherbiny and A A M Arafa ldquoOn thesolution of the fractional nonlinear Schrodinger equationrdquoPhysics Letters A vol 372 no 5 pp 553ndash558 2008
[3] K Diethelm and A D Freed ldquoOn the solution of nonlinearfractional order differential equations used in the modeling ofviscoplasticityrdquo in Scientific Computing in Chemical EngineeringII pp 217ndash224 Springer Berlin Germany 1999
[4] F Mainardi ldquoFractional calculus some basic proble ms incontinuum and statistical mechanicsrdquo in Fractals and FractionalCalculus in Continuum Mechanics (Udine 1996) vol 378 ofCISM Courses and Lectures pp 291ndash348 Springer ViennaAustria 1997
[5] R Metzler W Schick H Kilian and T F NonnenmacherldquoRelaxation in filled polymers a fractional calculus approachrdquoThe Journal of Chemical Physics vol 103 no 16 pp 7180ndash71861995
[6] B C Dhage and V Lakshmikantham ldquoBasic results on hybriddifferential equationsrdquo Nonlinear Analysis Hybrid Systems vol4 no 3 pp 414ndash424 2010
10 Abstract and Applied Analysis
[7] Y Zhao S Sun Z Han and Q Li ldquoTheory of fractionalhybrid differential equationsrdquo Computers amp Mathematics withApplications vol 62 no 3 pp 1312ndash1324 2011
[8] J Caballero M A Darwish and K Sadarangani ldquoSolvabilityof a fractional hybrid initial value problem with supremum byusingmeasures of noncompactness in Banach algebrasrdquoAppliedMathematics and Computation vol 224 pp 553ndash563 2013
[9] C Bai and J Fang ldquoThe existence of a positive solution fora singular coupled system of nonlinear fractional differentialequationsrdquo Applied Mathematics and Computation vol 150 no3 pp 611ndash621 2004
[10] Y Chen and H An ldquoNumerical solutions of coupled Burgersequations with time- and space-fractional derivativesrdquo AppliedMathematics and Computation vol 200 no 1 pp 87ndash95 2008
[11] M P Lazarevic ldquoFinite time stability analysis of PD120572 fractionalcontrol of robotic time-delay systemsrdquo Mechanics ResearchCommunications vol 33 no 2 pp 269ndash279 2006
[12] V Gafiychuk B Datsko V Meleshko and D Blackmore ldquoAnal-ysis of the solutions of coupled nonlinear fractional reaction-diffusion equationsrdquo Chaos Solitons and Fractals vol 41 no 3pp 1095ndash1104 2009
[13] B Ahmad and J J Nieto ldquoExistence results for a coupledsystem of nonlinear fractional differential equations with three-point boundary conditionsrdquo Computers amp Mathematics withApplications vol 58 no 9 pp 1838ndash1843 2009
[14] A A Kilbas H M Srivastava and J J Trujillo Theory andApplications of Fractional Differential Equations vol 204 ofNorth-Holland Mathematics Studies Elsevier Science Amster-dam The Netherlands 2006
[15] J Banas and K GoebelMeasures of Noncompactness in BanachSpaces Lecture Notes in Pure andAppliedMathematicsMarcelDekker New York NY USA 1980
[16] B N Sadovskii ldquoOn a fixed point principlerdquo Functional Analysisand Its Applications vol 1 pp 151ndash153 1967
[17] J Banas and L Olszowy ldquoOn a class of measures of noncom-pactness in Banach algebras and their application to nonlinearintegral equationsrdquo Journal of Analysis and its Applications vol28 no 4 pp 475ndash498 2009
[18] A Aghajani J Banas and N Sabzali ldquoSome generalizationsof Darbo fixed point theorem and applicationsrdquo Bulletin of theBelgian Mathematical Society vol 20 no 2 pp 345ndash358 2013
[19] R R Akhmerov M I Kamenskii A S Potapov A E Rod-kina and B N Sadovskii Measures of Noncompactness andCondensing Operators vol 55 ofOperatorTheory Advances andApplications Birkhaauser Basel Switzerland 1992
Submit your manuscripts athttpwwwhindawicom
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Decision SciencesAdvances in
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Volume 2014
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Stochastic AnalysisInternational Journal of
Abstract and Applied Analysis 3
(b) 119883 sub 119884 rArr 120583(119883) le 120583(119884)(c) 120583(119883) = 120583(Conv119883) = 120583(119883)(d) 120583(120582119883 + (1 minus 120582)119884) le 120582120583(119883) + (1 minus 120582)120583(119884) for 120582 isin [0 1](e) If (119883
119899) is a sequence of closed subsets fromM
119864such
that 119883119899+1
sub 119883119899
(119899 ge 1) and lim119899rarrinfin
120583(119883119899) = 0 then
119883infin
= capinfin
119899=1119883119899
= 120601
The family ker 120583 appearing in (a) is called the kernel ofthe measure of noncompactness 120583 Notice that the set 119883
infin
appearing in (e) is an element of ker120583 Indeed since 120583(119883infin
) le
120583(119883119899) for 119899 = 1 2 we infer that 120583(119883
infin) = 0 and this says
that 119883infin
isin ker120583An important theorem about fixed point theorem in
the context of measures of noncompactness is the followingDarborsquos fixed point theorem [15]
Theorem4 Let119862 be a nonempty bounded closed and convexsubset of a Banach space 119864 and let 119879 119862 rarr 119862 be a continuousmapping Suppose that there exists a constant 119896 isin [0 1) suchthat
120583 (119879 (119883)) le 119896120583 (119883) (18)
for any nonempty subset 119883 of 119862Then 119879 has a fixed point
A generalization of Theorem 4 which will be very usefulin our study is the following theorem due to Sadovskiı [16]
Theorem5 Let119862 be a nonempty bounded closed and convexsubset of a Banach space 119864 and let 119879 119862 rarr 119862 be a continuousoperator satisfying
120583 (119879 (119883)) lt 120583 (119883) (19)
for any nonempty subset 119883 of 119862 with 120583(119883) gt 0Then 119879 has a fixed point
Next we will assume that the space 119864 has structure ofBanach algebra By 119909119910 we will denote the product of twoelements 119909 119910 isin 119883 and by 119883119884 we will denote the set definedby 119883119884 = 119909119910 119909 isin 119883 119910 isin 119884
Definition 6 Let 119864 be a Banach algebra We will say thata measure of noncompactness 120583 defined on 119864 satisfiescondition (119898) if
120583 (119883119884) le 119883 120583 (119884) + 119884 120583 (119883) (20)
for any 119883 119884 isin M119864
This definition appears in [17]In this paper we will work in the space 119862[0 1] consisting
of all real functions defined and continuous on [0 1] with thestandard supremum norm
119909 = sup |119909 (119905)| 119905 isin [0 1] (21)
for119909 isin 119862[0 1] It is clear that (119862[0 1] sdot) is a Banach algebrawhere the multiplication is defined as the usual product ofreal functions
Next we present the measure of noncompactness in119862[0 1] which will be used later Let us fix 119883 isin M
119862[01]and
120576 gt 0 For 119909 isin 119883 we denote by 120596(119909 120576) the modulus ofcontinuity of 119909 that is
120596 (119909 120576) = sup |119909 (119905) minus 119909 (119904)| 119905 119904 isin [0 1] |119905 minus 119904| le 120576 (22)
Put
120596 (119883 120576) = sup 120596 (119909 120576) 119909 isin 119883
1205960
(119883) = lim120576rarr0
120596 (119883 120576) (23)
In [15] it is proved that 1205960(119883) is a measure of noncompact-
ness in 119862[0 1]
Proposition 7 Themeasure of noncompactness 1205960on 119862[0 1]
satisfies condition (119898)
Proof Fix 119883 119884 isin M119862[01]
120576 gt 0 and 119905 119904 isin [0 1] with |119905 minus 119904| le
120576 Then we have1003816100381610038161003816119909 (119905) 119910 (119905) minus 119909 (119904) 119910 (119904)
1003816100381610038161003816 le1003816100381610038161003816119909 (119905) 119910 (119905) minus 119909 (119905) 119910 (119904)
1003816100381610038161003816
+1003816100381610038161003816119909 (119905) 119910 (119904) minus 119909 (119904) 119910 (119904)
1003816100381610038161003816
= |119909 (119905)|1003816100381610038161003816119910 (119905) minus 119910 (119904)
1003816100381610038161003816
+1003816100381610038161003816119910 (119904)
1003816100381610038161003816 |119909 (119905) minus 119909 (119904)|
le 119909 120596 (119910 120576) +1003817100381710038171003817119910
1003817100381710038171003817 120596 (119909 120576)
(24)
This means that
120596 (119909119910 120576) le 119909 120596 (119910 120576) +1003817100381710038171003817119910
1003817100381710038171003817 120596 (119909 120576) (25)
and therefore
120596 (119883119884 120576) le 119883 120596 (119884 120576) + 119884 120596 (119883 120576) (26)
Taking 120576 rarr 0 we get
1205960
(119883119884) le 119883 1205960
(119884) + 119884 1205960
(119883) (27)
This completes the proof
Proposition 7 appears in [17] and we have given the prooffor the paper is self-contained
3 Main Results
We begin this section introducing the following class A offunctions
A = 120593 R+
997888rarr R+
120593 is nondecreasing
and lim119899rarrinfin
120593119899
(119905) = 0 for any 119905 gt 0
(28)
where 120593119899 denotes the 119899-iteration of 120593
4 Abstract and Applied Analysis
Remark 8 Notice that if 120593 isin A then 120593(119905) lt 119905 for any 119905 gt 0Indeed in contrary case we can find 119905
0gt 0 and 119905
0le 120593(119905
0)
Since 120593 is nondecreasing we have
0 lt 1199050
le 120593 (1199050) le 1205932
(1199050) le sdot sdot sdot le 120593
119899
(1199050) le sdot sdot sdot (29)
and therefore 0 lt 1199050
le lim119899rarrinfin
120593119899
(1199050) and this contradicts
the fact that 120593 isin AMoreover the fact that 120593(119905) lt 119905 for any 119905 gt 0 proves that
if 120593 isin A then 120593 is continuous at 1199050
= 0
Using Remark 8 and Theorem 5 we have the followingfixed point theorem
Theorem9 Let119862 be a nonempty bounded closed and convexsubset of a Banach space 119864 and let 119879 119862 rarr 119862 be a continuousoperator satisfying
120583 (119879 (119883)) le 120593 (120583 (119883)) (30)
for any nonempty subset 119883 of 119862 where 120593 isin AThen 119879 has a fixed point
Theorem 9 appears in [18] where the authors present aproof without usingTheorem 5
The following result which appears in [19] will be inter-esting in our study
Theorem 10 Let 1205831 1205832 120583
119899be measures of noncompact-
ness in the Banach spaces 1198641 1198642 119864
119899 respectively Suppose
that 119865 [0 infin)119899
rarr [0 infin) is a convex function such that119865(1199091 1199092 119909
119899) = 0 if and only if 119909
119894= 0 for 119894 = 1 2 119899
Then
120583 (119883) = 119865 (1205831
(1198831) 1205832
(1198832) 120583
119899(119883119899)) (31)
defines a measure compactness in 1198641
times 1198642
times sdot sdot sdot times 119864119899 where 119883
119894
denotes the natural projection of 119883 into 119864119894 for 119894 = 1 2 119899
Remark 11 As a consequence ofTheorem 10 we have that if 120583
is a measure of noncompactness on a Banach space 119864 and weconsider the function 119865 [0 infin) times [0 infin) rarr [0 infin) definedby 119865(119909 119910) = max(119909 119910) then since 119865 is convex and 119865(119909 119910) =
0 if and only if 119909 = 119910 = 0 120583(119883) = max120583(1198831) 120583(119883
2) defines
a measure of noncompactness in the space 119864 times 119864
Next we present the definition of a coupled fixed point
Definition 12 An element (119909 119910) isin 119883 times 119883 is said to be acoupled fixed point of a mapping 119866 119883 times 119883 rarr 119883 if119866(119909 119910) = 119909 and 119866(119910 119909) = 119910
The following result is crucial for our study
Theorem 13 Let Ω be a nonempty bounded closed andconvex subset of a Banach space 119864 and let 120583 be a measure ofnoncompactness in 119864 Suppose that 119866 Ω times Ω rarr Ω is acontinuous operator satisfying
120583 (119866 (1198831
times 1198832)) le 120593 (max (120583 (119883
1) 120583 (119883
2))) (32)
for all nonempty subsets 1198831and 119883
2of Ω where 120593 isin A
Then 119866 has at least a coupled fixed point
Proof Notice that by Remark 11 120583(119883) = max120583(1198831) 120583(119883
2)
is a measure of noncompactness in the space 119864 times 119864 where 1198831
and 1198832are the projections of 119883 into 119864
Now we consider the mapping 119866 Ω times Ω rarr Ω times Ω
defined by 119866(119909 119910) = (119866(119909 119910) 119866(119910 119909)) It is easily seen thatΩ times Ω is a nonempty bounded closed and convex subsetof 119864 times 119864 Since 119866 is continuous it is clear that 119866 is alsocontinuous
Next we take a nonempty 119883 of Ω times Ω Then
120583 (119866 (119883)) = 120583 (119866 (1198831
times 1198832))
= 120583 (119866 (1198831
times 1198832) times 119866 (119883
2times 1198831))
= max 120583 (119866 (1198831
times 1198832)) 120583 (119866 (119883
2times 1198831))
le max 120593 (max (120583 (1198831) 120583 (119883
2)))
120593 (max (120583 (1198832) 120583 (119883
1)))
= 120593 (max (120583 (1198831) 120583 (119883
2)))
= 120593 (120583 (1198831
times 1198832))
(33)
Since 120593 isin A by Theorem 9 the mapping 119866 has at least onefixed pointThismeans that there exists (119909
0 1199100) isin ΩtimesΩ such
that 119866(1199090 1199100) = (119909
0 1199100) or equivalently 119866(119909
0 1199100) = 1199090and
119866(1199100 1199090) = 1199100 This proves that 119866 has at least a coupled fixed
point
Now we consider the coupled system of integral equa-tions
119909 (119905) =119891 (119905 119909 (119905) 119910 (119905))
Γ (120572)int
119905
0
119892 (119904 119909 (119904) 119910 (119904))
(119905 minus 119904)1minus120572
119889119904
119910 (119905) =119891 (119905 119910 (119905) 119909 (119905))
Γ (120572)int
119905
0
119892 (119904 119910 (119904) 119909 (119904))
(119905 minus 119904)1minus120572
119889119904 119905 isin [0 1]
(34)
Lemma 14 Assume that 119891 isin 119862([0 1] times R times RR 0) and119892 isin 119862([0 1] times R times RR) Then (119909 119910) isin 119862[0 1] times 119862[0 1] isa solution of (34) if and only if (119909 119910) isin 119862[0 1] times 119862[0 1] is asolution of (4)
Proof The proof is an immediate consequence of Lemma 2so we omit it
Next we will study problem (4) under the followingassumptions
(H1) 119891 isin 119862([0 1] times R times RR 0) and 119892 isin 119862([0 1] times R times
RR)(H2) The functions 119891 and 119892 satisfy
1003816100381610038161003816119891 (119905 1199091 1199101) minus 119891 (119905 119909
2 1199102)1003816100381610038161003816
le 1205931
(max (10038161003816100381610038161199091 minus 119909
2
1003816100381610038161003816 10038161003816100381610038161199101 minus 119910
2
1003816100381610038161003816))
1003816100381610038161003816119892 (119905 1199091 1199102) minus 119892 (119905 119909
2 1199102)1003816100381610038161003816
le 1205932
(max (10038161003816100381610038161199091 minus 119910
1
1003816100381610038161003816 10038161003816100381610038161199092 minus 119910
2
1003816100381610038161003816))
(35)
Abstract and Applied Analysis 5
respectively for any 119905 isin [0 1] and 1199091 1199092 1199101 1199102
isin Rwhere 120593
1 1205932
isin A and 1205931is continuous
Notice that assumption (H1) gives us the existence
of two nonnegative constants 1198961and 119896
2such that
|119891(119905 0 0)| le 1198961and |119892(119905 0 0)| le 119896
2 for any 119905 isin [0 1]
(H3) There exists 119903
0gt 0 satisfying the inequalities
(1205931
(119903) + 1198961) sdot (1205932
(119903) + 1198962) le 119903Γ (120572 + 1)
1205932
(119903) + 1198962
le Γ (120572 + 1)
(36)
Theorem 15 Under assumptions (1198671)ndash(1198673) problem (4) has
at least one solution in 119862[0 1] times 119862[0 1]
Proof In virtue of Lemma 14 a solution (119909 119910) isin 119862[0 1] times
119862[0 1] of problem (4) satisfies
119909 (119905) =119891 (119905 119909 (119905) 119910 (119905))
Γ (120572)int
119905
0
119892 (119904 119909 (119904) 119910 (119904))
(119905 minus 119904)1minus120572
119889119904
119910 (119905) =119891 (119905 119910 (119905) 119909 (119905))
Γ (120572)int
119905
0
119892 (119904 119910 (119904) 119909 (119904))
(119905 minus 119904)1minus120572
119889119904
119905 isin [0 1]
(37)
We consider the space 119862[0 1] times 119862[0 1] equipped with thenorm (119909 119910)
119862[01]times119862[01]= max119909 119910 for any (119909 119910) isin
119862[0 1] times 119862[0 1]In 119862[0 1] times 119862[0 1] we define the operator
119866 (119909 119910) (119905) =119891 (119905 119909 (119905) 119910 (119905))
Γ (120572)int
119905
0
119892 (119904 119909 (119904) 119910 (119904))
(119905 minus 119904)1minus120572
119889119904
119905 isin [0 1]
(38)
LetF andG be the operators given by
F (119909 119910) (119905) = 119891 (119905 119909 (119905) 119910 (119905))
G (119909 119910) (119905) =1
Γ (120572)int
119905
0
119892 (119904 119909 (119904) 119910 (119904))
(119905 minus 119904)1minus120572
119889119904
(39)
for any (119909 119910) isin 119862[0 1] times 119862[0 1] and any 119905 isin [0 1] Then
119866 (119909 119910) = F (119909 119910) sdot G (119909 119910) (40)
Firstly we will prove that 119866 applies 119862[0 1] times 119862[0 1] into119862[0 1] To do this it is sufficient to prove that F(119909 119910)G(119909 119910) isin 119862[0 1] for any (119909 119910) isin 119862[0 1] times 119862[0 1] since theproduct of continuous functions is continuous
In virtue of assumption (H1) it is clear that F(119909 119910) isin
119862[0 1] for (119909 119910) isin 119862[0 1] times 119862[0 1] In order to prove thatG(119909 119910) isin 119862[0 1] for (119909 119910) isin 119862[0 1] times 119862[0 1] we fix 119905
0isin
[0 1] and we consider a sequence (119905119899) isin [0 1] such that 119905
119899rarr
1199050 and we have to prove that G(119909 119910)(119905
119899) rarr G(119909 119910)(119905
0)
Without loss of generality we can suppose that 119905119899
gt 1199050 Then
we have
1003816100381610038161003816G (119909 119910) (119905119899) minus G (119909 119910) (119905
0)1003816100381610038161003816
=1
Γ (120572)
1003816100381610038161003816100381610038161003816100381610038161003816
int
119905119899
0
119892 (119904 119909 (119904) 119910 (119904))
(119905119899
minus 119904)1minus120572
119889119904
minus int
1199050
0
119892 (119904 119909 (119904) 119910 (119904))
(1199050
minus 119904)1minus120572
119889119904
1003816100381610038161003816100381610038161003816100381610038161003816
le1
Γ (120572)
1003816100381610038161003816100381610038161003816100381610038161003816
int
119905119899
0
119892 (119904 119909 (119904) 119910 (119904))
(119905119899
minus 119904)1minus120572
119889119904
minus int
119905119899
0
119892 (119904 119909 (119904) 119910 (119904))
(1199050
minus 119904)1minus120572
119889119904
1003816100381610038161003816100381610038161003816100381610038161003816
+1
Γ (120572)
1003816100381610038161003816100381610038161003816100381610038161003816
int
119905119899
0
119892 (119904 119909 (119904) 119910 (119904))
(1199050
minus 119904)1minus120572
119889119904
minus int
1199050
0
119892 (119904 119909 (119904) 119910 (119904))
(1199050
minus 119904)1minus120572
119889119904
1003816100381610038161003816100381610038161003816100381610038161003816
le1
Γ (120572)int
119905119899
0
10038161003816100381610038161003816(119905119899
minus 119904)120572minus1
minus (1199050
minus 119904)120572minus110038161003816100381610038161003816
times1003816100381610038161003816119892 (119904 119909 (119904) 119910 (119904))
1003816100381610038161003816 119889119904
+1
Γ (120572)int
119905119899
1199050
10038161003816100381610038161003816(1199050
minus 119904)120572minus110038161003816100381610038161003816
1003816100381610038161003816119892 (119904 119909 (119904) 119910 (119904))1003816100381610038161003816 119889119904
(41)
By assumption (H1) since 119892 isin 119862([0 1] times R times RR) 119892 is
bounded on the compact set [0 1]times[minus119909 119909]times[minus119910 119910]Denote by
119872 = sup 1003816100381610038161003816119892 (119904 119909
1 1199101)1003816100381610038161003816 119904 isin [0 1] 119909
1isin [minus 119909 119909]
1199101
isin [minus1003817100381710038171003817119910
1003817100381710038171003817 1003817100381710038171003817119910
1003817100381710038171003817]
(42)
From the last estimate we obtain
1003816100381610038161003816G (119909 119910) (119905119899) minus G (119909 119910) (119905
0)1003816100381610038161003816
le119872
Γ (120572)int
119905119899
0
10038161003816100381610038161003816(119905119899
minus 119904)120572minus1
minus (1199050
minus 119904)120572minus110038161003816100381610038161003816
119889119904
+119872
Γ (120572)int
119905119899
1199050
10038161003816100381610038161003816(1199050
minus 119904)120572minus110038161003816100381610038161003816
119889119904
(43)
6 Abstract and Applied Analysis
As 0 lt 120572 lt 1 and 119905119899
gt 1199050 we infer that
1003816100381610038161003816G (119909 119910) (119905119899) minus G (119909 119910) (119905
0)1003816100381610038161003816
le119872
Γ (120572)[int
1199050
0
10038161003816100381610038161003816(119905119899
minus 119904)120572minus1
minus (1199050
minus 119904)120572minus110038161003816100381610038161003816
119889119904
+ int
119905119899
1199050
10038161003816100381610038161003816(119905119899
minus 119904)120572minus1
minus (1199050
minus 119904)120572minus110038161003816100381610038161003816
119889119904]
+119872
Γ (120572)int
119905119899
1199050
1
(119904 minus 1199050)1minus120572
119889119904
=119872
Γ (120572)[ int
1199050
0
[(1199050
minus 119904)120572minus1
minus (119905119899
minus 119904)120572minus1
] 119889119904
+ int
119905119899
1199050
119889119904
(119905119899
minus 119904)1minus120572
+ int
119905119899
1199050
119889119904
(119904 minus 1199050)1minus120572
]
+119872
Γ (120572)int
119905119899
1199050
1
(119904 minus 1199050)1minus120572
119889119904
le119872
Γ (120572 + 1)[(119905119899
minus 1199050)120572
+ 119905120572
0minus 119905120572
119899
+ (119905119899
minus 1199050)120572
+ (119905119899
minus 1199050)120572
]
+119872
Γ (120572 + 1)(119905119899
minus 1199050)120572
=4119872
Γ (120572 + 1)(119905119899
minus 1199050)120572
+119872
Γ (120572 + 1)(119905120572
0minus 119905120572
119899)
lt4119872
Γ (120572 + 1)(119905119899
minus 1199050)120572
(44)
where the last inequality has been obtained by using the factthat 119905120572
0minus 119905120572
119899lt 0
Therefore since 119905119899
rarr 1199050 from the last estimate we
deduce that G(119909 119910)(119905119899) rarr G(119909 119910)(119905
0) This proves that
G(119909 119910) isin 119862[0 1] Consequently G 119862[0 1] times 119862[0 1] rarr
119862[0 1] On the other hand for (119909 119910) isin 119862[0 1] times 119862[0 1] and119905 isin 119862[0 1] we have
1003816100381610038161003816119866 (119909 119910) (119905)1003816100381610038161003816
=1003816100381610038161003816F (119909 119910) (119905) sdot G (119909 119910) (119905)
1003816100381610038161003816
=1003816100381610038161003816119891 (119905 119909 (119905) 119910 (119905))
1003816100381610038161003816
100381610038161003816100381610038161003816100381610038161003816
1
Γ (120572)int
119905
0
119892 (119904 119909 (119904) 119910 (119904))
(119905 minus 119904)1minus120572
119889119904
100381610038161003816100381610038161003816100381610038161003816
le [1003816100381610038161003816119891 (119905 119909 (119905) 119910 (119905)) minus 119891 (119905 0 0)
1003816100381610038161003816 +1003816100381610038161003816119891 (119905 0 0)
1003816100381610038161003816]
times [1
Γ (120572)
100381610038161003816100381610038161003816100381610038161003816
int
119905
0
119892 (119904 119909 (119904) 119910 (119904)) minus 119892 (119904 0 0)
(119905 minus 119904)1minus120572
119889119904
+ int
119905
0
119892 (119904 0 0)
(119905 minus 119904)1minus120572
119889119904
100381610038161003816100381610038161003816100381610038161003816
]
le1
Γ (120572)[1205931
(max (|119909 (119905)| 1003816100381610038161003816119910 (119905)
1003816100381610038161003816)) + 1198961]
times [int
119905
0
1003816100381610038161003816119892 (119904 119909 (119904) 119910 (119904)) minus 119892 (119904 0 0)1003816100381610038161003816
(119905 minus 119904)1minus120572
119889119904
+ int
119905
0
1003816100381610038161003816119892 (119904 0 0)1003816100381610038161003816
(119905 minus 119904)1minus120572
119889119904]
le1
Γ (120572)[1205931
(max (119909 1003817100381710038171003817119910
1003817100381710038171003817)) + 1198961]
times [int
119905
0
1205932
(max (|119909 (119904)| 1003816100381610038161003816119910 (119904)
1003816100381610038161003816))
(119905 minus 119904)1minus120572
119889119904
+ 1198962
int
119905
0
119889119904
(119905 minus 119904)1minus120572
]
le1
Γ (120572)[1205931
(max (119909 1003817100381710038171003817119910
1003817100381710038171003817)) + 1198961]
sdot [1205932
(max (119909 1003817100381710038171003817119910
1003817100381710038171003817)) + 1198962] int
119905
0
119889119904
(119905 minus 119904)1minus120572
le1
Γ (120572 + 1)(1205931
(1003817100381710038171003817(119909 119910)
1003817100381710038171003817) + 1198961)
sdot (1205932
(1003817100381710038171003817(119909 119910)
1003817100381710038171003817) + 1198962)
(45)
Now taking into account assumption (H3) we infer that the
operator 119866 applies 1198611199030
times 1198611199030
into 1198611199030
Moreover from the lastestimates it follows that
10038171003817100381710038171003817F (1198611199030
times 1198611199030
)10038171003817100381710038171003817
le 1205931
(1199030) + 1198961
10038171003817100381710038171003817G (1198611199030
times 1198611199030
)10038171003817100381710038171003817
le1205932
(1199030) + 1198962
Γ (120572 + 1)
(46)
Next we will prove that the operators F and G arecontinuous on the ball 119861
1199030
times 1198611199030
and consequently 119866 will bealso continuous
In fact we fix 120576 gt 0 and we take (1199090 1199100) (119909 119910) isin 119861
1199030
times 1198611199030
with (119909 119910)minus(1199090 1199100) = (119909minus119909
0 119910minus1199100) =max119909minus119909
0 119910minus
1199100 le 120576 Then for 119905 isin [0 1] we have
1003816100381610038161003816F (119909 119910) (119905) minus F (1199090 1199100) (119905)
1003816100381610038161003816
=1003816100381610038161003816119891 (119905 119909 (119905) 119910 (119905)) minus 119891 (119905 119909
0(119905) 1199100
(119905))1003816100381610038161003816
le 1205931
(max (1003816100381610038161003816119909 (119905) minus 119909
0(119905)
1003816100381610038161003816 1003816100381610038161003816119910 (119905) 119910
0(119905)
1003816100381610038161003816))
le 1205931
(max (1003817100381710038171003817119909 minus 119909
0
1003817100381710038171003817 1003817100381710038171003817119910 minus 119910
0
1003817100381710038171003817))
le 1205931
(120576) lt 120576
(47)
where we have used Remark 8 This proves the continuity ofF on 119861
1199030
times 1198611199030
Abstract and Applied Analysis 7
In order to prove the continuity ofG on 1198611199030
times1198611199030
we have1003816100381610038161003816G (119909 119910) (119905) minus G (119909
0 1199100) (119905)
1003816100381610038161003816
=1
Γ (120572)
100381610038161003816100381610038161003816100381610038161003816
int
119905
0
119892 (119904 119909 (119904) 119910 (119904))
(119905 minus 119904)1minus120572
119889119904
minus int
119905
0
119892 (119904 1199090
(119904) 1199100
(119904))
(119905 minus 119904)1minus120572
119889119904
100381610038161003816100381610038161003816100381610038161003816
le1
Γ (120572)int
119905
0
1003816100381610038161003816119892 (119904 119909 (119904) 119910 (119904)) minus 119892 (119904 1199090
(119904) 1199100
(119904))1003816100381610038161003816
(119905 minus 119904)1minus120572
119889119904
le1
Γ (120572)int
119905
0
1205932
(max (1003816100381610038161003816119909 (119904) minus 119909
0(119904)
1003816100381610038161003816 1003816100381610038161003816119910 (119904) minus 119910
0(119904)
1003816100381610038161003816))
(119905 minus 119904)1minus120572
119889119904
le1
Γ (120572)1205932
(max (1003817100381710038171003817119909 minus 119909
0
1003817100381710038171003817 1003817100381710038171003817119910 minus 119910
0
1003817100381710038171003817)) int
119905
0
119889119904
(119905 minus 119904)1minus120572
le1
Γ (120572 + 1)1205932
(120576)
lt120576
Γ (120572 + 1)
(48)
Therefore1003817100381710038171003817G (119909 119910) minus G (119909
0 1199100)1003817100381710038171003817 lt
120576
Γ (120572 + 1)(49)
and consequentlyG is a continuous operator on 1198611199030
times 1198611199030
In order to prove that G satisfies assumptions of
Theorem 13 only we have to check the condition
120583 (119866 (1198831
times 1198832)) le 120593 (max (120583 (119883
1) 120583 (119883
2))) (50)
for any subsets 1198831and 119883
2of 1198611199030
To do this we fix 120576 gt 0 119905
1 1199052
isin [0 1] with |1199051minus 1199052| le 120576 and
(119909 119910) isin 1198831
times 1198832 then we have
1003816100381610038161003816F (119909 119910) (1199051) minus F (119909 119910) (119905
2)1003816100381610038161003816
=1003816100381610038161003816119891 (1199051 119909 (1199051) 119910 (119905
1)) minus 119891 (119905
2 119909 (1199052) 119910 (119905
2))
1003816100381610038161003816
le1003816100381610038161003816119891 (1199051 119909 (1199051) 119910 (119905
1)) minus 119891 (119905
1 119909 (1199052) 119910 (119905
2))
1003816100381610038161003816
+1003816100381610038161003816119891 (1199051 119909 (1199052) 119910 (119905
2)) minus 119891 (119905
2 119909 (1199052) 119910 (119905
2))
1003816100381610038161003816
le 1205931
(max (1003816100381610038161003816119909 (1199051) minus 119909 (119905
2)1003816100381610038161003816
1003816100381610038161003816119910 (1199051) minus 119910 (119905
2)1003816100381610038161003816))
+ 120596 (119891 120576)
le 1205931
(max (120596 (119909 120576) 120596 (119910 120576))) + 120596 (119891 120576)
(51)
where 120596(119891 120576) denotes the quantity
120596 (119891 120576) = sup 1003816100381610038161003816119891 (119905 119909 119910) minus 119891 (119904 119909 119910)
1003816100381610038161003816 119905 119904 isin [0 1]
|119905 minus 119904| le 120576 119909 119910 isin [minus1199030 1199030]
(52)
From the last estimate we infer that120596 (F (119883
1times 1198832) 120576)
le 1205931
(max (120596 (1198831 120576) 120596 (119883
2 120576))) + 120596 (119891 120576)
(53)
Since 119891(119905 119909 119910) is uniformly continuous on bounded subsetsof [0 1] times R times R we deduce that 120596(119891 120576) rarr 0 as 120576 rarr 0 andtherefore
1205960
(F (1198831
times 1198832)) le lim120576rarr0
1205931
(max (120596 (1198831 120576) 120596 (119883
2 120576)))
(54)
By assumption (H2) since 120593
1is continuous we infer
1205960
(F (1198831
times 1198832)) le 120593
1(max (120596
0(1198831) 1205960
(1198832))) (55)
Now we estimate the quantity 1205960(G(1198831
times 1198832))
Fix 120576 gt 0 1199051 1199052
isin [0 1] with |1199051
minus 1199052| le 120576 and (119909 119910) isin 119883
1times
1198832 Without loss of generality we can suppose that 119905
1lt 1199052
then we have1003816100381610038161003816G (119909 119910) (119905
2) minus G (119909 119910) (119905
1)1003816100381610038161003816
=1
Γ (120572)
1003816100381610038161003816100381610038161003816100381610038161003816
int
1199052
0
119892 (119904 119909 (119904) 119910 (119904))
(1199052
minus 119904)1minus120572
119889119904
minus int
1199051
0
119892 (119904 119909 (119904) 119910 (119904))
(1199051
minus 119904)1minus120572
119889119904
1003816100381610038161003816100381610038161003816100381610038161003816
le1
Γ (120572)[int
1199051
0
10038161003816100381610038161003816(1199052
minus 119904)120572minus1
minus (1199051
minus 119904)120572minus110038161003816100381610038161003816
times1003816100381610038161003816119892 (119904 119909 (119904) 119909 (120590119904))
1003816100381610038161003816 119889119904
+ int
1199052
1199051
(1199052
minus 119904)120572minus1 1003816100381610038161003816119892 (119904 119909 (119904) 119910 (119904))
1003816100381610038161003816 119889119904]
le1
Γ (120572)[int
1199051
0
[(1199051
minus 119904)120572minus1
minus (1199052
minus 119904)120572minus1
]
times1003816100381610038161003816119892 (119904 119909 (119904) 119909 (120590119904))
1003816100381610038161003816 119889119904
+ int
1199052
1199051
(1199052
minus 119904)120572minus1 1003816100381610038161003816119892 (119904 119909 (119904) 119910 (119904))
1003816100381610038161003816 119889119904]
(56)
Since 119892 isin 119862([0 1] times R times R timesR) is bounded on the compactsubsets of [0 1] times R times R particularly on [0 1] times [minus119903
0 1199030] times
[minus1199030 1199030] Put 119871 = sup|119892(119905 119909 119910)| 119905 isin [0 1] 119909 119910 isin [minus119903
0 1199030]
Then from the last inequality we infer that1003816100381610038161003816G (119909 119910) (119905
2) minus G (119909 119910) (119905
1)1003816100381610038161003816
le119871
Γ (120572)[int
1199051
0
[(1199051
minus 119904)120572minus1
minus (1199052
minus 119904)120572minus1
] 119889119904
+ int
1199052
1199051
(1199052
minus 119904)120572minus1
119889119904]
le119871
Γ (120572 + 1)[(1199052
minus 1199051)120572
+ 119905120572
1minus 119905120572
2+ (1199052
minus 1199051)120572
]
le2119871
Γ (120572 + 1)(1199052
minus 1199051)120572
le2119871
Γ (120572 + 1)120576120572
(57)
8 Abstract and Applied Analysis
where we have used the fact that 119905120572
1minus 119905120572
2le 0 Therefore
120596 (G (1198831
times 1198832) 120576) le
2119871
Γ (120572 + 1)120576120572
(58)
From this it follows that
1205960
(G (1198831
times 1198832)) = 0 (59)
Next by Proposition 7 (46) (55) and (59) we have
1205960
(119866 (1198831
times 1198832))
= 1205960
(F (1198831
times 1198832) sdot G (119883
1times 1198832))
le1003817100381710038171003817F (119883
1times 1198832)1003817100381710038171003817 1205960
(G (1198831
times 1198832))
+1003817100381710038171003817G (119883
1times 1198832)1003817100381710038171003817 1205960
(F (1198831
times 1198832))
le10038171003817100381710038171003817F (1198611199030
times 1198611199030
)10038171003817100381710038171003817
1205960
(G (1198831
times 1198832))
+10038171003817100381710038171003817G (1198611199030
times 1198611199030
)10038171003817100381710038171003817
1205960
(F (1198831
times 1198832))
le1205932
(1199030) + 1198962
Γ (120572 + 1)1205931
(max (1205960
(1198831) 1205960
(1198832)))
(60)
By assumption (H3) since 120593
2(1199030) + 1198962
le Γ(120572 + 1) and since itis easily proved that if 120572 isin [0 1] and 120593 isin A then 120572120593 isin A wededuce that
1205960
(119866 (1198831
times 1198832)) le 120593 (max (120596
0(1198831) 1205960
(1198832))) (61)
where 120593 isin AFinally by Theorem 13 the operator G has at least a
coupled fixed point and this is the desired result Thiscompletes the proof
The nonoscillary character of the solutions of problem(4) seems to be an interesting question from the practicalpoint of view This means that the solutions of problem (4)have a constant sign on the interval (0 1) In connection withthis question we notice that if 119891(119905 119909 119910) and 119892(119905 119909 119910) haveconstant sign and are equal (this means that 119891(119905 119909 119910) gt 0
and 119892(119905 119909 119910) ge 0 or 119891(119905 119909 119910) lt 0 and 119892(119905 119909 119910) le 0
for any 119905 isin [0 1] and 119909 119910 isin R) and under assumptionsof Theorem 15 then the solution (119909 119910) isin 119862[0 1] times 119862[0 1]
of problem (4) given by Theorem 15 satisfies 119909(119905) ge 0 and119910(119905) ge 0 for 119905 isin [0 1] since the solution (119909 119910) satisfies thesystem of integral equations
119909 (119905) =119891 (119905 119909 (119905) 119910 (119905))
Γ (120572)int
119905
0
119892 (119904 119909 (119904) 119910 (119904))
(119905 minus 119904)1minus120572
119889119904
119910 (119905) =119891 (119905 119910 (119905) 119909 (119905))
Γ (120572)int
119905
0
119892 (119904 119910 (119904) 119909 (119904))
(119905 minus 119904)1minus120572
119889119904 0 le 119905 le 1
(62)
On the other hand if we perturb the data function in problem(4) of the following manner
119863120572
0+ [
119909 (119905)
119891 (119905 119909 (119905) 119910 (119905))] = 119892 (119905 119909 (119905) 119910 (119905)) + 120578 (119905)
119863120572
0+ [
119910 (119905)
119891 (119905 119910 (119905) 119909 (119905))] = 119892 (119905 119910 (119905) 119909 (119905)) + 120578 (119905)
0 lt 119905 lt 1
(63)
where 0 lt 120572 lt 1 119891 isin 119862([0 1] times R times RR 0)119892 isin 119862([0 1] times R times RR) and 120578 isin 119862[0 1] then underassumptions of Theorem 15 problem (63) can be studiedby using Theorem 15 where assumptions (H
1) and (H
2) are
automatically satisfied and we only have to check assumption(H3) This fact gives a great applicability to Theorem 15Before presenting an example illustrating our results we
need some facts about the functions involving this exampleThe following lemma appears in [18]
Lemma 16 Let 120593 R+
rarr R+be a nondecreasing and upper
semicontinuous function Then the following conditions areequivalent
(i) lim119899rarrinfin
120593119899
(119905) = 0 for any 119905 ge 0
(ii) 120593(119905) lt 119905 for any 119905 gt 0
Particularly the functions 1205721 1205722
R+
rarr R+given by
1205721(119905) = arctan 119905 and 120572
2(119905) = 119905(1 + 119905) belong to the class
A since they are nondecreasing and continuous and as itis easily seen they satisfy (ii) of Lemma 16
On the other hand since the function 1205721(119905) = arctan 119905 is
concave (because 12057210158401015840
(119905) le 0) and 1205721(0) = 0 we infer that 120572
1
is subadditive and therefore for any 119905 1199051015840
isin R+ we have
100381610038161003816100381610038161205721
(119905) minus 1205721
(1199051015840
)10038161003816100381610038161003816
=10038161003816100381610038161003816arctan 119905 minus arctan 119905
101584010038161003816100381610038161003816
le arctan 10038161003816100381610038161003816119905 minus 119905101584010038161003816100381610038161003816
(64)
Moreover it is easily seen that max(1205721 1205722) is a nondecreasing
and continuous function because 1205721and 120572
2are nondecreas-
ing and continuous andmax(1205721 1205722) satisfies (ii) of Lemma 16
Therefore max(1205721 1205722) isin A
Nowwe are ready to present an examplewhere our resultscan be applied
Example 17 Consider the following coupled system of frac-tional hybrid differential equations
11986312
0+ [119909 (119905) times (
1
4+ (
1
10) arctan |119909 (119905)|
+(1
20) (
1003816100381610038161003816119910 (119905)1003816100381610038161003816
(1 +1003816100381610038161003816119910 (119905)
1003816100381610038161003816)))
minus1
]
=1
7+
1
9119909 (119905) +
1
10119910 (119905)
Abstract and Applied Analysis 9
11986312
0+ [119910 (119905) times (
1
4+ (
1
10) arctan 1003816100381610038161003816119910 (119905)
1003816100381610038161003816
+ (1
20) (
|119909 (119905)|
(1 + |119909 (119905)|)))
minus1
]
=1
7+
1
9119910 (119905) +
1
10119909 (119905)
0 lt 119905 lt 1
119909 (0) = 119910 (0) = 0
(65)
Notice that problem (17) is a particular case of problem (4)where 120572 = 12 119891(119905 119909 119910) = 14 + (110) arctan |119909| +
(120)(|119910|(1 + |119910|)) and 119892(119905 119909 119910) = 17 + (19)119909 + (110)119910It is clear that 119891 isin 119862([0 1] times R times RR 0) and 119892 isin
119862([0 1] times R times RR) and moreover 1198961
= sup|119891(119905 0 0)| 119905 isin
[0 1] = 14 and 1198962
= sup|119892(119905 0 0)| 119905 isin [0 1] = 17Therefore assumption (H
1) of Theorem 15 is satisfied
Moreover for 119905 isin [0 1] and 1199091 1199092 1199101 1199102
isin R we have1003816100381610038161003816119891 (119905 119909
1 1199101) minus 119891 (119905 119909
2 1199102)1003816100381610038161003816
le1
10
1003816100381610038161003816arctan10038161003816100381610038161199091
1003816100381610038161003816 minus arctan 100381610038161003816100381611990921003816100381610038161003816
1003816100381610038161003816
+1
20
100381610038161003816100381610038161003816100381610038161003816
100381610038161003816100381611991011003816100381610038161003816
1 +10038161003816100381610038161199101
1003816100381610038161003816
minus
100381610038161003816100381611991021003816100381610038161003816
1 +10038161003816100381610038161199102
1003816100381610038161003816
100381610038161003816100381610038161003816100381610038161003816
le1
10arctan 1003816100381610038161003816
100381610038161003816100381611990911003816100381610038161003816 minus
100381610038161003816100381611990921003816100381610038161003816
1003816100381610038161003816
+1
20
100381610038161003816100381610038161003816100381610038161003816
100381610038161003816100381611991011003816100381610038161003816 minus
100381610038161003816100381611991021003816100381610038161003816
(1 +10038161003816100381610038161199101
1003816100381610038161003816) (1 +10038161003816100381610038161199102
1003816100381610038161003816)
100381610038161003816100381610038161003816100381610038161003816
le1
10arctan (
10038161003816100381610038161199091 minus 1199092
1003816100381610038161003816)
+1
20
10038161003816100381610038161199101 minus 1199102
1003816100381610038161003816
1 +10038161003816100381610038161199101 minus 119910
2
1003816100381610038161003816
=1
101205721
(10038161003816100381610038161199091 minus 119909
2
1003816100381610038161003816)
+1
201205722
(10038161003816100381610038161199101 minus 119910
2
1003816100381610038161003816)
le1
10max (120572
1 1205722) (
10038161003816100381610038161199091 minus 1199092
1003816100381610038161003816)
+1
10max (120572
1 1205722) (
10038161003816100381610038161199101 minus 1199102
1003816100381610038161003816)
le1
10[2max (120572
1 1205722)
times max (10038161003816100381610038161199091 minus 119909
2
1003816100381610038161003816 10038161003816100381610038161199101 minus 119910
2
1003816100381610038161003816)]
=1
5max (120572
1 1205722) (max (
10038161003816100381610038161199091 minus 1199092
1003816100381610038161003816 10038161003816100381610038161199101 minus 119910
2
1003816100381610038161003816))
(66)
Therefore 1205931(119905) = (15)max(120572
1(119905) 1205722(119905)) and 120593
1isin A
On the other hand1003816100381610038161003816119892 (119905 119909
1 1199101) minus 119892 (119905 119909
2 1199102)1003816100381610038161003816
le1
9
10038161003816100381610038161199091 minus 1199092
1003816100381610038161003816 +1
10
10038161003816100381610038161199092 minus 1199102
1003816100381610038161003816
le1
9(10038161003816100381610038161199091 minus 119910
1
1003816100381610038161003816 +10038161003816100381610038161199102 minus 119910
2
1003816100381610038161003816)
le1
9(2max (
10038161003816100381610038161199091 minus 1199101
1003816100381610038161003816 10038161003816100381610038161199102 minus 119910
2
1003816100381610038161003816))
=2
9max (
10038161003816100381610038161199091 minus 1199101
1003816100381610038161003816 10038161003816100381610038161199102 minus 119910
2
1003816100381610038161003816)
(67)
and 1205932(119905) = (29)119905 It is clear that 120593
2isin A Therefore
assumption (H2) of Theorem 15 is satisfied
In our case the inequality appearing in assumption (H3)
of Theorem 15 has the expression
[1
5max(arctan 119903
119903
1 + 119903) +
1
4] [
2
9119903 +
1
7] le 119903Γ (
3
2) (68)
It is easily seen that 1199030
= 1 satisfies the last inequalityMoreover
2
91199030
+1
7=
2
9+
1
7le Γ (
3
2) cong 088623 (69)
Finally Theorem 15 says that problem (17) has at least onesolution (119909 119910) isin 119862[0 1] such that max(119909 119910) le 1
Conflict of Interests
The authors declare that there is no conflict of interests in thesubmitted paper
References
[1] W Deng ldquoNumerical algorithm for the time fractional Fokker-Planck equationrdquo Journal of Computational Physics vol 227 no2 pp 1510ndash1522 2007
[2] S Z Rida H M El-Sherbiny and A A M Arafa ldquoOn thesolution of the fractional nonlinear Schrodinger equationrdquoPhysics Letters A vol 372 no 5 pp 553ndash558 2008
[3] K Diethelm and A D Freed ldquoOn the solution of nonlinearfractional order differential equations used in the modeling ofviscoplasticityrdquo in Scientific Computing in Chemical EngineeringII pp 217ndash224 Springer Berlin Germany 1999
[4] F Mainardi ldquoFractional calculus some basic proble ms incontinuum and statistical mechanicsrdquo in Fractals and FractionalCalculus in Continuum Mechanics (Udine 1996) vol 378 ofCISM Courses and Lectures pp 291ndash348 Springer ViennaAustria 1997
[5] R Metzler W Schick H Kilian and T F NonnenmacherldquoRelaxation in filled polymers a fractional calculus approachrdquoThe Journal of Chemical Physics vol 103 no 16 pp 7180ndash71861995
[6] B C Dhage and V Lakshmikantham ldquoBasic results on hybriddifferential equationsrdquo Nonlinear Analysis Hybrid Systems vol4 no 3 pp 414ndash424 2010
10 Abstract and Applied Analysis
[7] Y Zhao S Sun Z Han and Q Li ldquoTheory of fractionalhybrid differential equationsrdquo Computers amp Mathematics withApplications vol 62 no 3 pp 1312ndash1324 2011
[8] J Caballero M A Darwish and K Sadarangani ldquoSolvabilityof a fractional hybrid initial value problem with supremum byusingmeasures of noncompactness in Banach algebrasrdquoAppliedMathematics and Computation vol 224 pp 553ndash563 2013
[9] C Bai and J Fang ldquoThe existence of a positive solution fora singular coupled system of nonlinear fractional differentialequationsrdquo Applied Mathematics and Computation vol 150 no3 pp 611ndash621 2004
[10] Y Chen and H An ldquoNumerical solutions of coupled Burgersequations with time- and space-fractional derivativesrdquo AppliedMathematics and Computation vol 200 no 1 pp 87ndash95 2008
[11] M P Lazarevic ldquoFinite time stability analysis of PD120572 fractionalcontrol of robotic time-delay systemsrdquo Mechanics ResearchCommunications vol 33 no 2 pp 269ndash279 2006
[12] V Gafiychuk B Datsko V Meleshko and D Blackmore ldquoAnal-ysis of the solutions of coupled nonlinear fractional reaction-diffusion equationsrdquo Chaos Solitons and Fractals vol 41 no 3pp 1095ndash1104 2009
[13] B Ahmad and J J Nieto ldquoExistence results for a coupledsystem of nonlinear fractional differential equations with three-point boundary conditionsrdquo Computers amp Mathematics withApplications vol 58 no 9 pp 1838ndash1843 2009
[14] A A Kilbas H M Srivastava and J J Trujillo Theory andApplications of Fractional Differential Equations vol 204 ofNorth-Holland Mathematics Studies Elsevier Science Amster-dam The Netherlands 2006
[15] J Banas and K GoebelMeasures of Noncompactness in BanachSpaces Lecture Notes in Pure andAppliedMathematicsMarcelDekker New York NY USA 1980
[16] B N Sadovskii ldquoOn a fixed point principlerdquo Functional Analysisand Its Applications vol 1 pp 151ndash153 1967
[17] J Banas and L Olszowy ldquoOn a class of measures of noncom-pactness in Banach algebras and their application to nonlinearintegral equationsrdquo Journal of Analysis and its Applications vol28 no 4 pp 475ndash498 2009
[18] A Aghajani J Banas and N Sabzali ldquoSome generalizationsof Darbo fixed point theorem and applicationsrdquo Bulletin of theBelgian Mathematical Society vol 20 no 2 pp 345ndash358 2013
[19] R R Akhmerov M I Kamenskii A S Potapov A E Rod-kina and B N Sadovskii Measures of Noncompactness andCondensing Operators vol 55 ofOperatorTheory Advances andApplications Birkhaauser Basel Switzerland 1992
Submit your manuscripts athttpwwwhindawicom
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
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Mathematical Problems in Engineering
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Differential EquationsInternational Journal of
Volume 2014
Applied MathematicsJournal of
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Journal of
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Mathematical PhysicsAdvances in
Complex AnalysisJournal of
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OptimizationJournal of
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CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of
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Operations ResearchAdvances in
Journal of
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
Remark 8 Notice that if 120593 isin A then 120593(119905) lt 119905 for any 119905 gt 0Indeed in contrary case we can find 119905
0gt 0 and 119905
0le 120593(119905
0)
Since 120593 is nondecreasing we have
0 lt 1199050
le 120593 (1199050) le 1205932
(1199050) le sdot sdot sdot le 120593
119899
(1199050) le sdot sdot sdot (29)
and therefore 0 lt 1199050
le lim119899rarrinfin
120593119899
(1199050) and this contradicts
the fact that 120593 isin AMoreover the fact that 120593(119905) lt 119905 for any 119905 gt 0 proves that
if 120593 isin A then 120593 is continuous at 1199050
= 0
Using Remark 8 and Theorem 5 we have the followingfixed point theorem
Theorem9 Let119862 be a nonempty bounded closed and convexsubset of a Banach space 119864 and let 119879 119862 rarr 119862 be a continuousoperator satisfying
120583 (119879 (119883)) le 120593 (120583 (119883)) (30)
for any nonempty subset 119883 of 119862 where 120593 isin AThen 119879 has a fixed point
Theorem 9 appears in [18] where the authors present aproof without usingTheorem 5
The following result which appears in [19] will be inter-esting in our study
Theorem 10 Let 1205831 1205832 120583
119899be measures of noncompact-
ness in the Banach spaces 1198641 1198642 119864
119899 respectively Suppose
that 119865 [0 infin)119899
rarr [0 infin) is a convex function such that119865(1199091 1199092 119909
119899) = 0 if and only if 119909
119894= 0 for 119894 = 1 2 119899
Then
120583 (119883) = 119865 (1205831
(1198831) 1205832
(1198832) 120583
119899(119883119899)) (31)
defines a measure compactness in 1198641
times 1198642
times sdot sdot sdot times 119864119899 where 119883
119894
denotes the natural projection of 119883 into 119864119894 for 119894 = 1 2 119899
Remark 11 As a consequence ofTheorem 10 we have that if 120583
is a measure of noncompactness on a Banach space 119864 and weconsider the function 119865 [0 infin) times [0 infin) rarr [0 infin) definedby 119865(119909 119910) = max(119909 119910) then since 119865 is convex and 119865(119909 119910) =
0 if and only if 119909 = 119910 = 0 120583(119883) = max120583(1198831) 120583(119883
2) defines
a measure of noncompactness in the space 119864 times 119864
Next we present the definition of a coupled fixed point
Definition 12 An element (119909 119910) isin 119883 times 119883 is said to be acoupled fixed point of a mapping 119866 119883 times 119883 rarr 119883 if119866(119909 119910) = 119909 and 119866(119910 119909) = 119910
The following result is crucial for our study
Theorem 13 Let Ω be a nonempty bounded closed andconvex subset of a Banach space 119864 and let 120583 be a measure ofnoncompactness in 119864 Suppose that 119866 Ω times Ω rarr Ω is acontinuous operator satisfying
120583 (119866 (1198831
times 1198832)) le 120593 (max (120583 (119883
1) 120583 (119883
2))) (32)
for all nonempty subsets 1198831and 119883
2of Ω where 120593 isin A
Then 119866 has at least a coupled fixed point
Proof Notice that by Remark 11 120583(119883) = max120583(1198831) 120583(119883
2)
is a measure of noncompactness in the space 119864 times 119864 where 1198831
and 1198832are the projections of 119883 into 119864
Now we consider the mapping 119866 Ω times Ω rarr Ω times Ω
defined by 119866(119909 119910) = (119866(119909 119910) 119866(119910 119909)) It is easily seen thatΩ times Ω is a nonempty bounded closed and convex subsetof 119864 times 119864 Since 119866 is continuous it is clear that 119866 is alsocontinuous
Next we take a nonempty 119883 of Ω times Ω Then
120583 (119866 (119883)) = 120583 (119866 (1198831
times 1198832))
= 120583 (119866 (1198831
times 1198832) times 119866 (119883
2times 1198831))
= max 120583 (119866 (1198831
times 1198832)) 120583 (119866 (119883
2times 1198831))
le max 120593 (max (120583 (1198831) 120583 (119883
2)))
120593 (max (120583 (1198832) 120583 (119883
1)))
= 120593 (max (120583 (1198831) 120583 (119883
2)))
= 120593 (120583 (1198831
times 1198832))
(33)
Since 120593 isin A by Theorem 9 the mapping 119866 has at least onefixed pointThismeans that there exists (119909
0 1199100) isin ΩtimesΩ such
that 119866(1199090 1199100) = (119909
0 1199100) or equivalently 119866(119909
0 1199100) = 1199090and
119866(1199100 1199090) = 1199100 This proves that 119866 has at least a coupled fixed
point
Now we consider the coupled system of integral equa-tions
119909 (119905) =119891 (119905 119909 (119905) 119910 (119905))
Γ (120572)int
119905
0
119892 (119904 119909 (119904) 119910 (119904))
(119905 minus 119904)1minus120572
119889119904
119910 (119905) =119891 (119905 119910 (119905) 119909 (119905))
Γ (120572)int
119905
0
119892 (119904 119910 (119904) 119909 (119904))
(119905 minus 119904)1minus120572
119889119904 119905 isin [0 1]
(34)
Lemma 14 Assume that 119891 isin 119862([0 1] times R times RR 0) and119892 isin 119862([0 1] times R times RR) Then (119909 119910) isin 119862[0 1] times 119862[0 1] isa solution of (34) if and only if (119909 119910) isin 119862[0 1] times 119862[0 1] is asolution of (4)
Proof The proof is an immediate consequence of Lemma 2so we omit it
Next we will study problem (4) under the followingassumptions
(H1) 119891 isin 119862([0 1] times R times RR 0) and 119892 isin 119862([0 1] times R times
RR)(H2) The functions 119891 and 119892 satisfy
1003816100381610038161003816119891 (119905 1199091 1199101) minus 119891 (119905 119909
2 1199102)1003816100381610038161003816
le 1205931
(max (10038161003816100381610038161199091 minus 119909
2
1003816100381610038161003816 10038161003816100381610038161199101 minus 119910
2
1003816100381610038161003816))
1003816100381610038161003816119892 (119905 1199091 1199102) minus 119892 (119905 119909
2 1199102)1003816100381610038161003816
le 1205932
(max (10038161003816100381610038161199091 minus 119910
1
1003816100381610038161003816 10038161003816100381610038161199092 minus 119910
2
1003816100381610038161003816))
(35)
Abstract and Applied Analysis 5
respectively for any 119905 isin [0 1] and 1199091 1199092 1199101 1199102
isin Rwhere 120593
1 1205932
isin A and 1205931is continuous
Notice that assumption (H1) gives us the existence
of two nonnegative constants 1198961and 119896
2such that
|119891(119905 0 0)| le 1198961and |119892(119905 0 0)| le 119896
2 for any 119905 isin [0 1]
(H3) There exists 119903
0gt 0 satisfying the inequalities
(1205931
(119903) + 1198961) sdot (1205932
(119903) + 1198962) le 119903Γ (120572 + 1)
1205932
(119903) + 1198962
le Γ (120572 + 1)
(36)
Theorem 15 Under assumptions (1198671)ndash(1198673) problem (4) has
at least one solution in 119862[0 1] times 119862[0 1]
Proof In virtue of Lemma 14 a solution (119909 119910) isin 119862[0 1] times
119862[0 1] of problem (4) satisfies
119909 (119905) =119891 (119905 119909 (119905) 119910 (119905))
Γ (120572)int
119905
0
119892 (119904 119909 (119904) 119910 (119904))
(119905 minus 119904)1minus120572
119889119904
119910 (119905) =119891 (119905 119910 (119905) 119909 (119905))
Γ (120572)int
119905
0
119892 (119904 119910 (119904) 119909 (119904))
(119905 minus 119904)1minus120572
119889119904
119905 isin [0 1]
(37)
We consider the space 119862[0 1] times 119862[0 1] equipped with thenorm (119909 119910)
119862[01]times119862[01]= max119909 119910 for any (119909 119910) isin
119862[0 1] times 119862[0 1]In 119862[0 1] times 119862[0 1] we define the operator
119866 (119909 119910) (119905) =119891 (119905 119909 (119905) 119910 (119905))
Γ (120572)int
119905
0
119892 (119904 119909 (119904) 119910 (119904))
(119905 minus 119904)1minus120572
119889119904
119905 isin [0 1]
(38)
LetF andG be the operators given by
F (119909 119910) (119905) = 119891 (119905 119909 (119905) 119910 (119905))
G (119909 119910) (119905) =1
Γ (120572)int
119905
0
119892 (119904 119909 (119904) 119910 (119904))
(119905 minus 119904)1minus120572
119889119904
(39)
for any (119909 119910) isin 119862[0 1] times 119862[0 1] and any 119905 isin [0 1] Then
119866 (119909 119910) = F (119909 119910) sdot G (119909 119910) (40)
Firstly we will prove that 119866 applies 119862[0 1] times 119862[0 1] into119862[0 1] To do this it is sufficient to prove that F(119909 119910)G(119909 119910) isin 119862[0 1] for any (119909 119910) isin 119862[0 1] times 119862[0 1] since theproduct of continuous functions is continuous
In virtue of assumption (H1) it is clear that F(119909 119910) isin
119862[0 1] for (119909 119910) isin 119862[0 1] times 119862[0 1] In order to prove thatG(119909 119910) isin 119862[0 1] for (119909 119910) isin 119862[0 1] times 119862[0 1] we fix 119905
0isin
[0 1] and we consider a sequence (119905119899) isin [0 1] such that 119905
119899rarr
1199050 and we have to prove that G(119909 119910)(119905
119899) rarr G(119909 119910)(119905
0)
Without loss of generality we can suppose that 119905119899
gt 1199050 Then
we have
1003816100381610038161003816G (119909 119910) (119905119899) minus G (119909 119910) (119905
0)1003816100381610038161003816
=1
Γ (120572)
1003816100381610038161003816100381610038161003816100381610038161003816
int
119905119899
0
119892 (119904 119909 (119904) 119910 (119904))
(119905119899
minus 119904)1minus120572
119889119904
minus int
1199050
0
119892 (119904 119909 (119904) 119910 (119904))
(1199050
minus 119904)1minus120572
119889119904
1003816100381610038161003816100381610038161003816100381610038161003816
le1
Γ (120572)
1003816100381610038161003816100381610038161003816100381610038161003816
int
119905119899
0
119892 (119904 119909 (119904) 119910 (119904))
(119905119899
minus 119904)1minus120572
119889119904
minus int
119905119899
0
119892 (119904 119909 (119904) 119910 (119904))
(1199050
minus 119904)1minus120572
119889119904
1003816100381610038161003816100381610038161003816100381610038161003816
+1
Γ (120572)
1003816100381610038161003816100381610038161003816100381610038161003816
int
119905119899
0
119892 (119904 119909 (119904) 119910 (119904))
(1199050
minus 119904)1minus120572
119889119904
minus int
1199050
0
119892 (119904 119909 (119904) 119910 (119904))
(1199050
minus 119904)1minus120572
119889119904
1003816100381610038161003816100381610038161003816100381610038161003816
le1
Γ (120572)int
119905119899
0
10038161003816100381610038161003816(119905119899
minus 119904)120572minus1
minus (1199050
minus 119904)120572minus110038161003816100381610038161003816
times1003816100381610038161003816119892 (119904 119909 (119904) 119910 (119904))
1003816100381610038161003816 119889119904
+1
Γ (120572)int
119905119899
1199050
10038161003816100381610038161003816(1199050
minus 119904)120572minus110038161003816100381610038161003816
1003816100381610038161003816119892 (119904 119909 (119904) 119910 (119904))1003816100381610038161003816 119889119904
(41)
By assumption (H1) since 119892 isin 119862([0 1] times R times RR) 119892 is
bounded on the compact set [0 1]times[minus119909 119909]times[minus119910 119910]Denote by
119872 = sup 1003816100381610038161003816119892 (119904 119909
1 1199101)1003816100381610038161003816 119904 isin [0 1] 119909
1isin [minus 119909 119909]
1199101
isin [minus1003817100381710038171003817119910
1003817100381710038171003817 1003817100381710038171003817119910
1003817100381710038171003817]
(42)
From the last estimate we obtain
1003816100381610038161003816G (119909 119910) (119905119899) minus G (119909 119910) (119905
0)1003816100381610038161003816
le119872
Γ (120572)int
119905119899
0
10038161003816100381610038161003816(119905119899
minus 119904)120572minus1
minus (1199050
minus 119904)120572minus110038161003816100381610038161003816
119889119904
+119872
Γ (120572)int
119905119899
1199050
10038161003816100381610038161003816(1199050
minus 119904)120572minus110038161003816100381610038161003816
119889119904
(43)
6 Abstract and Applied Analysis
As 0 lt 120572 lt 1 and 119905119899
gt 1199050 we infer that
1003816100381610038161003816G (119909 119910) (119905119899) minus G (119909 119910) (119905
0)1003816100381610038161003816
le119872
Γ (120572)[int
1199050
0
10038161003816100381610038161003816(119905119899
minus 119904)120572minus1
minus (1199050
minus 119904)120572minus110038161003816100381610038161003816
119889119904
+ int
119905119899
1199050
10038161003816100381610038161003816(119905119899
minus 119904)120572minus1
minus (1199050
minus 119904)120572minus110038161003816100381610038161003816
119889119904]
+119872
Γ (120572)int
119905119899
1199050
1
(119904 minus 1199050)1minus120572
119889119904
=119872
Γ (120572)[ int
1199050
0
[(1199050
minus 119904)120572minus1
minus (119905119899
minus 119904)120572minus1
] 119889119904
+ int
119905119899
1199050
119889119904
(119905119899
minus 119904)1minus120572
+ int
119905119899
1199050
119889119904
(119904 minus 1199050)1minus120572
]
+119872
Γ (120572)int
119905119899
1199050
1
(119904 minus 1199050)1minus120572
119889119904
le119872
Γ (120572 + 1)[(119905119899
minus 1199050)120572
+ 119905120572
0minus 119905120572
119899
+ (119905119899
minus 1199050)120572
+ (119905119899
minus 1199050)120572
]
+119872
Γ (120572 + 1)(119905119899
minus 1199050)120572
=4119872
Γ (120572 + 1)(119905119899
minus 1199050)120572
+119872
Γ (120572 + 1)(119905120572
0minus 119905120572
119899)
lt4119872
Γ (120572 + 1)(119905119899
minus 1199050)120572
(44)
where the last inequality has been obtained by using the factthat 119905120572
0minus 119905120572
119899lt 0
Therefore since 119905119899
rarr 1199050 from the last estimate we
deduce that G(119909 119910)(119905119899) rarr G(119909 119910)(119905
0) This proves that
G(119909 119910) isin 119862[0 1] Consequently G 119862[0 1] times 119862[0 1] rarr
119862[0 1] On the other hand for (119909 119910) isin 119862[0 1] times 119862[0 1] and119905 isin 119862[0 1] we have
1003816100381610038161003816119866 (119909 119910) (119905)1003816100381610038161003816
=1003816100381610038161003816F (119909 119910) (119905) sdot G (119909 119910) (119905)
1003816100381610038161003816
=1003816100381610038161003816119891 (119905 119909 (119905) 119910 (119905))
1003816100381610038161003816
100381610038161003816100381610038161003816100381610038161003816
1
Γ (120572)int
119905
0
119892 (119904 119909 (119904) 119910 (119904))
(119905 minus 119904)1minus120572
119889119904
100381610038161003816100381610038161003816100381610038161003816
le [1003816100381610038161003816119891 (119905 119909 (119905) 119910 (119905)) minus 119891 (119905 0 0)
1003816100381610038161003816 +1003816100381610038161003816119891 (119905 0 0)
1003816100381610038161003816]
times [1
Γ (120572)
100381610038161003816100381610038161003816100381610038161003816
int
119905
0
119892 (119904 119909 (119904) 119910 (119904)) minus 119892 (119904 0 0)
(119905 minus 119904)1minus120572
119889119904
+ int
119905
0
119892 (119904 0 0)
(119905 minus 119904)1minus120572
119889119904
100381610038161003816100381610038161003816100381610038161003816
]
le1
Γ (120572)[1205931
(max (|119909 (119905)| 1003816100381610038161003816119910 (119905)
1003816100381610038161003816)) + 1198961]
times [int
119905
0
1003816100381610038161003816119892 (119904 119909 (119904) 119910 (119904)) minus 119892 (119904 0 0)1003816100381610038161003816
(119905 minus 119904)1minus120572
119889119904
+ int
119905
0
1003816100381610038161003816119892 (119904 0 0)1003816100381610038161003816
(119905 minus 119904)1minus120572
119889119904]
le1
Γ (120572)[1205931
(max (119909 1003817100381710038171003817119910
1003817100381710038171003817)) + 1198961]
times [int
119905
0
1205932
(max (|119909 (119904)| 1003816100381610038161003816119910 (119904)
1003816100381610038161003816))
(119905 minus 119904)1minus120572
119889119904
+ 1198962
int
119905
0
119889119904
(119905 minus 119904)1minus120572
]
le1
Γ (120572)[1205931
(max (119909 1003817100381710038171003817119910
1003817100381710038171003817)) + 1198961]
sdot [1205932
(max (119909 1003817100381710038171003817119910
1003817100381710038171003817)) + 1198962] int
119905
0
119889119904
(119905 minus 119904)1minus120572
le1
Γ (120572 + 1)(1205931
(1003817100381710038171003817(119909 119910)
1003817100381710038171003817) + 1198961)
sdot (1205932
(1003817100381710038171003817(119909 119910)
1003817100381710038171003817) + 1198962)
(45)
Now taking into account assumption (H3) we infer that the
operator 119866 applies 1198611199030
times 1198611199030
into 1198611199030
Moreover from the lastestimates it follows that
10038171003817100381710038171003817F (1198611199030
times 1198611199030
)10038171003817100381710038171003817
le 1205931
(1199030) + 1198961
10038171003817100381710038171003817G (1198611199030
times 1198611199030
)10038171003817100381710038171003817
le1205932
(1199030) + 1198962
Γ (120572 + 1)
(46)
Next we will prove that the operators F and G arecontinuous on the ball 119861
1199030
times 1198611199030
and consequently 119866 will bealso continuous
In fact we fix 120576 gt 0 and we take (1199090 1199100) (119909 119910) isin 119861
1199030
times 1198611199030
with (119909 119910)minus(1199090 1199100) = (119909minus119909
0 119910minus1199100) =max119909minus119909
0 119910minus
1199100 le 120576 Then for 119905 isin [0 1] we have
1003816100381610038161003816F (119909 119910) (119905) minus F (1199090 1199100) (119905)
1003816100381610038161003816
=1003816100381610038161003816119891 (119905 119909 (119905) 119910 (119905)) minus 119891 (119905 119909
0(119905) 1199100
(119905))1003816100381610038161003816
le 1205931
(max (1003816100381610038161003816119909 (119905) minus 119909
0(119905)
1003816100381610038161003816 1003816100381610038161003816119910 (119905) 119910
0(119905)
1003816100381610038161003816))
le 1205931
(max (1003817100381710038171003817119909 minus 119909
0
1003817100381710038171003817 1003817100381710038171003817119910 minus 119910
0
1003817100381710038171003817))
le 1205931
(120576) lt 120576
(47)
where we have used Remark 8 This proves the continuity ofF on 119861
1199030
times 1198611199030
Abstract and Applied Analysis 7
In order to prove the continuity ofG on 1198611199030
times1198611199030
we have1003816100381610038161003816G (119909 119910) (119905) minus G (119909
0 1199100) (119905)
1003816100381610038161003816
=1
Γ (120572)
100381610038161003816100381610038161003816100381610038161003816
int
119905
0
119892 (119904 119909 (119904) 119910 (119904))
(119905 minus 119904)1minus120572
119889119904
minus int
119905
0
119892 (119904 1199090
(119904) 1199100
(119904))
(119905 minus 119904)1minus120572
119889119904
100381610038161003816100381610038161003816100381610038161003816
le1
Γ (120572)int
119905
0
1003816100381610038161003816119892 (119904 119909 (119904) 119910 (119904)) minus 119892 (119904 1199090
(119904) 1199100
(119904))1003816100381610038161003816
(119905 minus 119904)1minus120572
119889119904
le1
Γ (120572)int
119905
0
1205932
(max (1003816100381610038161003816119909 (119904) minus 119909
0(119904)
1003816100381610038161003816 1003816100381610038161003816119910 (119904) minus 119910
0(119904)
1003816100381610038161003816))
(119905 minus 119904)1minus120572
119889119904
le1
Γ (120572)1205932
(max (1003817100381710038171003817119909 minus 119909
0
1003817100381710038171003817 1003817100381710038171003817119910 minus 119910
0
1003817100381710038171003817)) int
119905
0
119889119904
(119905 minus 119904)1minus120572
le1
Γ (120572 + 1)1205932
(120576)
lt120576
Γ (120572 + 1)
(48)
Therefore1003817100381710038171003817G (119909 119910) minus G (119909
0 1199100)1003817100381710038171003817 lt
120576
Γ (120572 + 1)(49)
and consequentlyG is a continuous operator on 1198611199030
times 1198611199030
In order to prove that G satisfies assumptions of
Theorem 13 only we have to check the condition
120583 (119866 (1198831
times 1198832)) le 120593 (max (120583 (119883
1) 120583 (119883
2))) (50)
for any subsets 1198831and 119883
2of 1198611199030
To do this we fix 120576 gt 0 119905
1 1199052
isin [0 1] with |1199051minus 1199052| le 120576 and
(119909 119910) isin 1198831
times 1198832 then we have
1003816100381610038161003816F (119909 119910) (1199051) minus F (119909 119910) (119905
2)1003816100381610038161003816
=1003816100381610038161003816119891 (1199051 119909 (1199051) 119910 (119905
1)) minus 119891 (119905
2 119909 (1199052) 119910 (119905
2))
1003816100381610038161003816
le1003816100381610038161003816119891 (1199051 119909 (1199051) 119910 (119905
1)) minus 119891 (119905
1 119909 (1199052) 119910 (119905
2))
1003816100381610038161003816
+1003816100381610038161003816119891 (1199051 119909 (1199052) 119910 (119905
2)) minus 119891 (119905
2 119909 (1199052) 119910 (119905
2))
1003816100381610038161003816
le 1205931
(max (1003816100381610038161003816119909 (1199051) minus 119909 (119905
2)1003816100381610038161003816
1003816100381610038161003816119910 (1199051) minus 119910 (119905
2)1003816100381610038161003816))
+ 120596 (119891 120576)
le 1205931
(max (120596 (119909 120576) 120596 (119910 120576))) + 120596 (119891 120576)
(51)
where 120596(119891 120576) denotes the quantity
120596 (119891 120576) = sup 1003816100381610038161003816119891 (119905 119909 119910) minus 119891 (119904 119909 119910)
1003816100381610038161003816 119905 119904 isin [0 1]
|119905 minus 119904| le 120576 119909 119910 isin [minus1199030 1199030]
(52)
From the last estimate we infer that120596 (F (119883
1times 1198832) 120576)
le 1205931
(max (120596 (1198831 120576) 120596 (119883
2 120576))) + 120596 (119891 120576)
(53)
Since 119891(119905 119909 119910) is uniformly continuous on bounded subsetsof [0 1] times R times R we deduce that 120596(119891 120576) rarr 0 as 120576 rarr 0 andtherefore
1205960
(F (1198831
times 1198832)) le lim120576rarr0
1205931
(max (120596 (1198831 120576) 120596 (119883
2 120576)))
(54)
By assumption (H2) since 120593
1is continuous we infer
1205960
(F (1198831
times 1198832)) le 120593
1(max (120596
0(1198831) 1205960
(1198832))) (55)
Now we estimate the quantity 1205960(G(1198831
times 1198832))
Fix 120576 gt 0 1199051 1199052
isin [0 1] with |1199051
minus 1199052| le 120576 and (119909 119910) isin 119883
1times
1198832 Without loss of generality we can suppose that 119905
1lt 1199052
then we have1003816100381610038161003816G (119909 119910) (119905
2) minus G (119909 119910) (119905
1)1003816100381610038161003816
=1
Γ (120572)
1003816100381610038161003816100381610038161003816100381610038161003816
int
1199052
0
119892 (119904 119909 (119904) 119910 (119904))
(1199052
minus 119904)1minus120572
119889119904
minus int
1199051
0
119892 (119904 119909 (119904) 119910 (119904))
(1199051
minus 119904)1minus120572
119889119904
1003816100381610038161003816100381610038161003816100381610038161003816
le1
Γ (120572)[int
1199051
0
10038161003816100381610038161003816(1199052
minus 119904)120572minus1
minus (1199051
minus 119904)120572minus110038161003816100381610038161003816
times1003816100381610038161003816119892 (119904 119909 (119904) 119909 (120590119904))
1003816100381610038161003816 119889119904
+ int
1199052
1199051
(1199052
minus 119904)120572minus1 1003816100381610038161003816119892 (119904 119909 (119904) 119910 (119904))
1003816100381610038161003816 119889119904]
le1
Γ (120572)[int
1199051
0
[(1199051
minus 119904)120572minus1
minus (1199052
minus 119904)120572minus1
]
times1003816100381610038161003816119892 (119904 119909 (119904) 119909 (120590119904))
1003816100381610038161003816 119889119904
+ int
1199052
1199051
(1199052
minus 119904)120572minus1 1003816100381610038161003816119892 (119904 119909 (119904) 119910 (119904))
1003816100381610038161003816 119889119904]
(56)
Since 119892 isin 119862([0 1] times R times R timesR) is bounded on the compactsubsets of [0 1] times R times R particularly on [0 1] times [minus119903
0 1199030] times
[minus1199030 1199030] Put 119871 = sup|119892(119905 119909 119910)| 119905 isin [0 1] 119909 119910 isin [minus119903
0 1199030]
Then from the last inequality we infer that1003816100381610038161003816G (119909 119910) (119905
2) minus G (119909 119910) (119905
1)1003816100381610038161003816
le119871
Γ (120572)[int
1199051
0
[(1199051
minus 119904)120572minus1
minus (1199052
minus 119904)120572minus1
] 119889119904
+ int
1199052
1199051
(1199052
minus 119904)120572minus1
119889119904]
le119871
Γ (120572 + 1)[(1199052
minus 1199051)120572
+ 119905120572
1minus 119905120572
2+ (1199052
minus 1199051)120572
]
le2119871
Γ (120572 + 1)(1199052
minus 1199051)120572
le2119871
Γ (120572 + 1)120576120572
(57)
8 Abstract and Applied Analysis
where we have used the fact that 119905120572
1minus 119905120572
2le 0 Therefore
120596 (G (1198831
times 1198832) 120576) le
2119871
Γ (120572 + 1)120576120572
(58)
From this it follows that
1205960
(G (1198831
times 1198832)) = 0 (59)
Next by Proposition 7 (46) (55) and (59) we have
1205960
(119866 (1198831
times 1198832))
= 1205960
(F (1198831
times 1198832) sdot G (119883
1times 1198832))
le1003817100381710038171003817F (119883
1times 1198832)1003817100381710038171003817 1205960
(G (1198831
times 1198832))
+1003817100381710038171003817G (119883
1times 1198832)1003817100381710038171003817 1205960
(F (1198831
times 1198832))
le10038171003817100381710038171003817F (1198611199030
times 1198611199030
)10038171003817100381710038171003817
1205960
(G (1198831
times 1198832))
+10038171003817100381710038171003817G (1198611199030
times 1198611199030
)10038171003817100381710038171003817
1205960
(F (1198831
times 1198832))
le1205932
(1199030) + 1198962
Γ (120572 + 1)1205931
(max (1205960
(1198831) 1205960
(1198832)))
(60)
By assumption (H3) since 120593
2(1199030) + 1198962
le Γ(120572 + 1) and since itis easily proved that if 120572 isin [0 1] and 120593 isin A then 120572120593 isin A wededuce that
1205960
(119866 (1198831
times 1198832)) le 120593 (max (120596
0(1198831) 1205960
(1198832))) (61)
where 120593 isin AFinally by Theorem 13 the operator G has at least a
coupled fixed point and this is the desired result Thiscompletes the proof
The nonoscillary character of the solutions of problem(4) seems to be an interesting question from the practicalpoint of view This means that the solutions of problem (4)have a constant sign on the interval (0 1) In connection withthis question we notice that if 119891(119905 119909 119910) and 119892(119905 119909 119910) haveconstant sign and are equal (this means that 119891(119905 119909 119910) gt 0
and 119892(119905 119909 119910) ge 0 or 119891(119905 119909 119910) lt 0 and 119892(119905 119909 119910) le 0
for any 119905 isin [0 1] and 119909 119910 isin R) and under assumptionsof Theorem 15 then the solution (119909 119910) isin 119862[0 1] times 119862[0 1]
of problem (4) given by Theorem 15 satisfies 119909(119905) ge 0 and119910(119905) ge 0 for 119905 isin [0 1] since the solution (119909 119910) satisfies thesystem of integral equations
119909 (119905) =119891 (119905 119909 (119905) 119910 (119905))
Γ (120572)int
119905
0
119892 (119904 119909 (119904) 119910 (119904))
(119905 minus 119904)1minus120572
119889119904
119910 (119905) =119891 (119905 119910 (119905) 119909 (119905))
Γ (120572)int
119905
0
119892 (119904 119910 (119904) 119909 (119904))
(119905 minus 119904)1minus120572
119889119904 0 le 119905 le 1
(62)
On the other hand if we perturb the data function in problem(4) of the following manner
119863120572
0+ [
119909 (119905)
119891 (119905 119909 (119905) 119910 (119905))] = 119892 (119905 119909 (119905) 119910 (119905)) + 120578 (119905)
119863120572
0+ [
119910 (119905)
119891 (119905 119910 (119905) 119909 (119905))] = 119892 (119905 119910 (119905) 119909 (119905)) + 120578 (119905)
0 lt 119905 lt 1
(63)
where 0 lt 120572 lt 1 119891 isin 119862([0 1] times R times RR 0)119892 isin 119862([0 1] times R times RR) and 120578 isin 119862[0 1] then underassumptions of Theorem 15 problem (63) can be studiedby using Theorem 15 where assumptions (H
1) and (H
2) are
automatically satisfied and we only have to check assumption(H3) This fact gives a great applicability to Theorem 15Before presenting an example illustrating our results we
need some facts about the functions involving this exampleThe following lemma appears in [18]
Lemma 16 Let 120593 R+
rarr R+be a nondecreasing and upper
semicontinuous function Then the following conditions areequivalent
(i) lim119899rarrinfin
120593119899
(119905) = 0 for any 119905 ge 0
(ii) 120593(119905) lt 119905 for any 119905 gt 0
Particularly the functions 1205721 1205722
R+
rarr R+given by
1205721(119905) = arctan 119905 and 120572
2(119905) = 119905(1 + 119905) belong to the class
A since they are nondecreasing and continuous and as itis easily seen they satisfy (ii) of Lemma 16
On the other hand since the function 1205721(119905) = arctan 119905 is
concave (because 12057210158401015840
(119905) le 0) and 1205721(0) = 0 we infer that 120572
1
is subadditive and therefore for any 119905 1199051015840
isin R+ we have
100381610038161003816100381610038161205721
(119905) minus 1205721
(1199051015840
)10038161003816100381610038161003816
=10038161003816100381610038161003816arctan 119905 minus arctan 119905
101584010038161003816100381610038161003816
le arctan 10038161003816100381610038161003816119905 minus 119905101584010038161003816100381610038161003816
(64)
Moreover it is easily seen that max(1205721 1205722) is a nondecreasing
and continuous function because 1205721and 120572
2are nondecreas-
ing and continuous andmax(1205721 1205722) satisfies (ii) of Lemma 16
Therefore max(1205721 1205722) isin A
Nowwe are ready to present an examplewhere our resultscan be applied
Example 17 Consider the following coupled system of frac-tional hybrid differential equations
11986312
0+ [119909 (119905) times (
1
4+ (
1
10) arctan |119909 (119905)|
+(1
20) (
1003816100381610038161003816119910 (119905)1003816100381610038161003816
(1 +1003816100381610038161003816119910 (119905)
1003816100381610038161003816)))
minus1
]
=1
7+
1
9119909 (119905) +
1
10119910 (119905)
Abstract and Applied Analysis 9
11986312
0+ [119910 (119905) times (
1
4+ (
1
10) arctan 1003816100381610038161003816119910 (119905)
1003816100381610038161003816
+ (1
20) (
|119909 (119905)|
(1 + |119909 (119905)|)))
minus1
]
=1
7+
1
9119910 (119905) +
1
10119909 (119905)
0 lt 119905 lt 1
119909 (0) = 119910 (0) = 0
(65)
Notice that problem (17) is a particular case of problem (4)where 120572 = 12 119891(119905 119909 119910) = 14 + (110) arctan |119909| +
(120)(|119910|(1 + |119910|)) and 119892(119905 119909 119910) = 17 + (19)119909 + (110)119910It is clear that 119891 isin 119862([0 1] times R times RR 0) and 119892 isin
119862([0 1] times R times RR) and moreover 1198961
= sup|119891(119905 0 0)| 119905 isin
[0 1] = 14 and 1198962
= sup|119892(119905 0 0)| 119905 isin [0 1] = 17Therefore assumption (H
1) of Theorem 15 is satisfied
Moreover for 119905 isin [0 1] and 1199091 1199092 1199101 1199102
isin R we have1003816100381610038161003816119891 (119905 119909
1 1199101) minus 119891 (119905 119909
2 1199102)1003816100381610038161003816
le1
10
1003816100381610038161003816arctan10038161003816100381610038161199091
1003816100381610038161003816 minus arctan 100381610038161003816100381611990921003816100381610038161003816
1003816100381610038161003816
+1
20
100381610038161003816100381610038161003816100381610038161003816
100381610038161003816100381611991011003816100381610038161003816
1 +10038161003816100381610038161199101
1003816100381610038161003816
minus
100381610038161003816100381611991021003816100381610038161003816
1 +10038161003816100381610038161199102
1003816100381610038161003816
100381610038161003816100381610038161003816100381610038161003816
le1
10arctan 1003816100381610038161003816
100381610038161003816100381611990911003816100381610038161003816 minus
100381610038161003816100381611990921003816100381610038161003816
1003816100381610038161003816
+1
20
100381610038161003816100381610038161003816100381610038161003816
100381610038161003816100381611991011003816100381610038161003816 minus
100381610038161003816100381611991021003816100381610038161003816
(1 +10038161003816100381610038161199101
1003816100381610038161003816) (1 +10038161003816100381610038161199102
1003816100381610038161003816)
100381610038161003816100381610038161003816100381610038161003816
le1
10arctan (
10038161003816100381610038161199091 minus 1199092
1003816100381610038161003816)
+1
20
10038161003816100381610038161199101 minus 1199102
1003816100381610038161003816
1 +10038161003816100381610038161199101 minus 119910
2
1003816100381610038161003816
=1
101205721
(10038161003816100381610038161199091 minus 119909
2
1003816100381610038161003816)
+1
201205722
(10038161003816100381610038161199101 minus 119910
2
1003816100381610038161003816)
le1
10max (120572
1 1205722) (
10038161003816100381610038161199091 minus 1199092
1003816100381610038161003816)
+1
10max (120572
1 1205722) (
10038161003816100381610038161199101 minus 1199102
1003816100381610038161003816)
le1
10[2max (120572
1 1205722)
times max (10038161003816100381610038161199091 minus 119909
2
1003816100381610038161003816 10038161003816100381610038161199101 minus 119910
2
1003816100381610038161003816)]
=1
5max (120572
1 1205722) (max (
10038161003816100381610038161199091 minus 1199092
1003816100381610038161003816 10038161003816100381610038161199101 minus 119910
2
1003816100381610038161003816))
(66)
Therefore 1205931(119905) = (15)max(120572
1(119905) 1205722(119905)) and 120593
1isin A
On the other hand1003816100381610038161003816119892 (119905 119909
1 1199101) minus 119892 (119905 119909
2 1199102)1003816100381610038161003816
le1
9
10038161003816100381610038161199091 minus 1199092
1003816100381610038161003816 +1
10
10038161003816100381610038161199092 minus 1199102
1003816100381610038161003816
le1
9(10038161003816100381610038161199091 minus 119910
1
1003816100381610038161003816 +10038161003816100381610038161199102 minus 119910
2
1003816100381610038161003816)
le1
9(2max (
10038161003816100381610038161199091 minus 1199101
1003816100381610038161003816 10038161003816100381610038161199102 minus 119910
2
1003816100381610038161003816))
=2
9max (
10038161003816100381610038161199091 minus 1199101
1003816100381610038161003816 10038161003816100381610038161199102 minus 119910
2
1003816100381610038161003816)
(67)
and 1205932(119905) = (29)119905 It is clear that 120593
2isin A Therefore
assumption (H2) of Theorem 15 is satisfied
In our case the inequality appearing in assumption (H3)
of Theorem 15 has the expression
[1
5max(arctan 119903
119903
1 + 119903) +
1
4] [
2
9119903 +
1
7] le 119903Γ (
3
2) (68)
It is easily seen that 1199030
= 1 satisfies the last inequalityMoreover
2
91199030
+1
7=
2
9+
1
7le Γ (
3
2) cong 088623 (69)
Finally Theorem 15 says that problem (17) has at least onesolution (119909 119910) isin 119862[0 1] such that max(119909 119910) le 1
Conflict of Interests
The authors declare that there is no conflict of interests in thesubmitted paper
References
[1] W Deng ldquoNumerical algorithm for the time fractional Fokker-Planck equationrdquo Journal of Computational Physics vol 227 no2 pp 1510ndash1522 2007
[2] S Z Rida H M El-Sherbiny and A A M Arafa ldquoOn thesolution of the fractional nonlinear Schrodinger equationrdquoPhysics Letters A vol 372 no 5 pp 553ndash558 2008
[3] K Diethelm and A D Freed ldquoOn the solution of nonlinearfractional order differential equations used in the modeling ofviscoplasticityrdquo in Scientific Computing in Chemical EngineeringII pp 217ndash224 Springer Berlin Germany 1999
[4] F Mainardi ldquoFractional calculus some basic proble ms incontinuum and statistical mechanicsrdquo in Fractals and FractionalCalculus in Continuum Mechanics (Udine 1996) vol 378 ofCISM Courses and Lectures pp 291ndash348 Springer ViennaAustria 1997
[5] R Metzler W Schick H Kilian and T F NonnenmacherldquoRelaxation in filled polymers a fractional calculus approachrdquoThe Journal of Chemical Physics vol 103 no 16 pp 7180ndash71861995
[6] B C Dhage and V Lakshmikantham ldquoBasic results on hybriddifferential equationsrdquo Nonlinear Analysis Hybrid Systems vol4 no 3 pp 414ndash424 2010
10 Abstract and Applied Analysis
[7] Y Zhao S Sun Z Han and Q Li ldquoTheory of fractionalhybrid differential equationsrdquo Computers amp Mathematics withApplications vol 62 no 3 pp 1312ndash1324 2011
[8] J Caballero M A Darwish and K Sadarangani ldquoSolvabilityof a fractional hybrid initial value problem with supremum byusingmeasures of noncompactness in Banach algebrasrdquoAppliedMathematics and Computation vol 224 pp 553ndash563 2013
[9] C Bai and J Fang ldquoThe existence of a positive solution fora singular coupled system of nonlinear fractional differentialequationsrdquo Applied Mathematics and Computation vol 150 no3 pp 611ndash621 2004
[10] Y Chen and H An ldquoNumerical solutions of coupled Burgersequations with time- and space-fractional derivativesrdquo AppliedMathematics and Computation vol 200 no 1 pp 87ndash95 2008
[11] M P Lazarevic ldquoFinite time stability analysis of PD120572 fractionalcontrol of robotic time-delay systemsrdquo Mechanics ResearchCommunications vol 33 no 2 pp 269ndash279 2006
[12] V Gafiychuk B Datsko V Meleshko and D Blackmore ldquoAnal-ysis of the solutions of coupled nonlinear fractional reaction-diffusion equationsrdquo Chaos Solitons and Fractals vol 41 no 3pp 1095ndash1104 2009
[13] B Ahmad and J J Nieto ldquoExistence results for a coupledsystem of nonlinear fractional differential equations with three-point boundary conditionsrdquo Computers amp Mathematics withApplications vol 58 no 9 pp 1838ndash1843 2009
[14] A A Kilbas H M Srivastava and J J Trujillo Theory andApplications of Fractional Differential Equations vol 204 ofNorth-Holland Mathematics Studies Elsevier Science Amster-dam The Netherlands 2006
[15] J Banas and K GoebelMeasures of Noncompactness in BanachSpaces Lecture Notes in Pure andAppliedMathematicsMarcelDekker New York NY USA 1980
[16] B N Sadovskii ldquoOn a fixed point principlerdquo Functional Analysisand Its Applications vol 1 pp 151ndash153 1967
[17] J Banas and L Olszowy ldquoOn a class of measures of noncom-pactness in Banach algebras and their application to nonlinearintegral equationsrdquo Journal of Analysis and its Applications vol28 no 4 pp 475ndash498 2009
[18] A Aghajani J Banas and N Sabzali ldquoSome generalizationsof Darbo fixed point theorem and applicationsrdquo Bulletin of theBelgian Mathematical Society vol 20 no 2 pp 345ndash358 2013
[19] R R Akhmerov M I Kamenskii A S Potapov A E Rod-kina and B N Sadovskii Measures of Noncompactness andCondensing Operators vol 55 ofOperatorTheory Advances andApplications Birkhaauser Basel Switzerland 1992
Submit your manuscripts athttpwwwhindawicom
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
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Differential EquationsInternational Journal of
Volume 2014
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OptimizationJournal of
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CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of
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Operations ResearchAdvances in
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Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Function Spaces
Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of Mathematics and Mathematical Sciences
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Algebra
Discrete Dynamics in Nature and Society
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Decision SciencesAdvances in
Discrete MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom
Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Stochastic AnalysisInternational Journal of
Abstract and Applied Analysis 5
respectively for any 119905 isin [0 1] and 1199091 1199092 1199101 1199102
isin Rwhere 120593
1 1205932
isin A and 1205931is continuous
Notice that assumption (H1) gives us the existence
of two nonnegative constants 1198961and 119896
2such that
|119891(119905 0 0)| le 1198961and |119892(119905 0 0)| le 119896
2 for any 119905 isin [0 1]
(H3) There exists 119903
0gt 0 satisfying the inequalities
(1205931
(119903) + 1198961) sdot (1205932
(119903) + 1198962) le 119903Γ (120572 + 1)
1205932
(119903) + 1198962
le Γ (120572 + 1)
(36)
Theorem 15 Under assumptions (1198671)ndash(1198673) problem (4) has
at least one solution in 119862[0 1] times 119862[0 1]
Proof In virtue of Lemma 14 a solution (119909 119910) isin 119862[0 1] times
119862[0 1] of problem (4) satisfies
119909 (119905) =119891 (119905 119909 (119905) 119910 (119905))
Γ (120572)int
119905
0
119892 (119904 119909 (119904) 119910 (119904))
(119905 minus 119904)1minus120572
119889119904
119910 (119905) =119891 (119905 119910 (119905) 119909 (119905))
Γ (120572)int
119905
0
119892 (119904 119910 (119904) 119909 (119904))
(119905 minus 119904)1minus120572
119889119904
119905 isin [0 1]
(37)
We consider the space 119862[0 1] times 119862[0 1] equipped with thenorm (119909 119910)
119862[01]times119862[01]= max119909 119910 for any (119909 119910) isin
119862[0 1] times 119862[0 1]In 119862[0 1] times 119862[0 1] we define the operator
119866 (119909 119910) (119905) =119891 (119905 119909 (119905) 119910 (119905))
Γ (120572)int
119905
0
119892 (119904 119909 (119904) 119910 (119904))
(119905 minus 119904)1minus120572
119889119904
119905 isin [0 1]
(38)
LetF andG be the operators given by
F (119909 119910) (119905) = 119891 (119905 119909 (119905) 119910 (119905))
G (119909 119910) (119905) =1
Γ (120572)int
119905
0
119892 (119904 119909 (119904) 119910 (119904))
(119905 minus 119904)1minus120572
119889119904
(39)
for any (119909 119910) isin 119862[0 1] times 119862[0 1] and any 119905 isin [0 1] Then
119866 (119909 119910) = F (119909 119910) sdot G (119909 119910) (40)
Firstly we will prove that 119866 applies 119862[0 1] times 119862[0 1] into119862[0 1] To do this it is sufficient to prove that F(119909 119910)G(119909 119910) isin 119862[0 1] for any (119909 119910) isin 119862[0 1] times 119862[0 1] since theproduct of continuous functions is continuous
In virtue of assumption (H1) it is clear that F(119909 119910) isin
119862[0 1] for (119909 119910) isin 119862[0 1] times 119862[0 1] In order to prove thatG(119909 119910) isin 119862[0 1] for (119909 119910) isin 119862[0 1] times 119862[0 1] we fix 119905
0isin
[0 1] and we consider a sequence (119905119899) isin [0 1] such that 119905
119899rarr
1199050 and we have to prove that G(119909 119910)(119905
119899) rarr G(119909 119910)(119905
0)
Without loss of generality we can suppose that 119905119899
gt 1199050 Then
we have
1003816100381610038161003816G (119909 119910) (119905119899) minus G (119909 119910) (119905
0)1003816100381610038161003816
=1
Γ (120572)
1003816100381610038161003816100381610038161003816100381610038161003816
int
119905119899
0
119892 (119904 119909 (119904) 119910 (119904))
(119905119899
minus 119904)1minus120572
119889119904
minus int
1199050
0
119892 (119904 119909 (119904) 119910 (119904))
(1199050
minus 119904)1minus120572
119889119904
1003816100381610038161003816100381610038161003816100381610038161003816
le1
Γ (120572)
1003816100381610038161003816100381610038161003816100381610038161003816
int
119905119899
0
119892 (119904 119909 (119904) 119910 (119904))
(119905119899
minus 119904)1minus120572
119889119904
minus int
119905119899
0
119892 (119904 119909 (119904) 119910 (119904))
(1199050
minus 119904)1minus120572
119889119904
1003816100381610038161003816100381610038161003816100381610038161003816
+1
Γ (120572)
1003816100381610038161003816100381610038161003816100381610038161003816
int
119905119899
0
119892 (119904 119909 (119904) 119910 (119904))
(1199050
minus 119904)1minus120572
119889119904
minus int
1199050
0
119892 (119904 119909 (119904) 119910 (119904))
(1199050
minus 119904)1minus120572
119889119904
1003816100381610038161003816100381610038161003816100381610038161003816
le1
Γ (120572)int
119905119899
0
10038161003816100381610038161003816(119905119899
minus 119904)120572minus1
minus (1199050
minus 119904)120572minus110038161003816100381610038161003816
times1003816100381610038161003816119892 (119904 119909 (119904) 119910 (119904))
1003816100381610038161003816 119889119904
+1
Γ (120572)int
119905119899
1199050
10038161003816100381610038161003816(1199050
minus 119904)120572minus110038161003816100381610038161003816
1003816100381610038161003816119892 (119904 119909 (119904) 119910 (119904))1003816100381610038161003816 119889119904
(41)
By assumption (H1) since 119892 isin 119862([0 1] times R times RR) 119892 is
bounded on the compact set [0 1]times[minus119909 119909]times[minus119910 119910]Denote by
119872 = sup 1003816100381610038161003816119892 (119904 119909
1 1199101)1003816100381610038161003816 119904 isin [0 1] 119909
1isin [minus 119909 119909]
1199101
isin [minus1003817100381710038171003817119910
1003817100381710038171003817 1003817100381710038171003817119910
1003817100381710038171003817]
(42)
From the last estimate we obtain
1003816100381610038161003816G (119909 119910) (119905119899) minus G (119909 119910) (119905
0)1003816100381610038161003816
le119872
Γ (120572)int
119905119899
0
10038161003816100381610038161003816(119905119899
minus 119904)120572minus1
minus (1199050
minus 119904)120572minus110038161003816100381610038161003816
119889119904
+119872
Γ (120572)int
119905119899
1199050
10038161003816100381610038161003816(1199050
minus 119904)120572minus110038161003816100381610038161003816
119889119904
(43)
6 Abstract and Applied Analysis
As 0 lt 120572 lt 1 and 119905119899
gt 1199050 we infer that
1003816100381610038161003816G (119909 119910) (119905119899) minus G (119909 119910) (119905
0)1003816100381610038161003816
le119872
Γ (120572)[int
1199050
0
10038161003816100381610038161003816(119905119899
minus 119904)120572minus1
minus (1199050
minus 119904)120572minus110038161003816100381610038161003816
119889119904
+ int
119905119899
1199050
10038161003816100381610038161003816(119905119899
minus 119904)120572minus1
minus (1199050
minus 119904)120572minus110038161003816100381610038161003816
119889119904]
+119872
Γ (120572)int
119905119899
1199050
1
(119904 minus 1199050)1minus120572
119889119904
=119872
Γ (120572)[ int
1199050
0
[(1199050
minus 119904)120572minus1
minus (119905119899
minus 119904)120572minus1
] 119889119904
+ int
119905119899
1199050
119889119904
(119905119899
minus 119904)1minus120572
+ int
119905119899
1199050
119889119904
(119904 minus 1199050)1minus120572
]
+119872
Γ (120572)int
119905119899
1199050
1
(119904 minus 1199050)1minus120572
119889119904
le119872
Γ (120572 + 1)[(119905119899
minus 1199050)120572
+ 119905120572
0minus 119905120572
119899
+ (119905119899
minus 1199050)120572
+ (119905119899
minus 1199050)120572
]
+119872
Γ (120572 + 1)(119905119899
minus 1199050)120572
=4119872
Γ (120572 + 1)(119905119899
minus 1199050)120572
+119872
Γ (120572 + 1)(119905120572
0minus 119905120572
119899)
lt4119872
Γ (120572 + 1)(119905119899
minus 1199050)120572
(44)
where the last inequality has been obtained by using the factthat 119905120572
0minus 119905120572
119899lt 0
Therefore since 119905119899
rarr 1199050 from the last estimate we
deduce that G(119909 119910)(119905119899) rarr G(119909 119910)(119905
0) This proves that
G(119909 119910) isin 119862[0 1] Consequently G 119862[0 1] times 119862[0 1] rarr
119862[0 1] On the other hand for (119909 119910) isin 119862[0 1] times 119862[0 1] and119905 isin 119862[0 1] we have
1003816100381610038161003816119866 (119909 119910) (119905)1003816100381610038161003816
=1003816100381610038161003816F (119909 119910) (119905) sdot G (119909 119910) (119905)
1003816100381610038161003816
=1003816100381610038161003816119891 (119905 119909 (119905) 119910 (119905))
1003816100381610038161003816
100381610038161003816100381610038161003816100381610038161003816
1
Γ (120572)int
119905
0
119892 (119904 119909 (119904) 119910 (119904))
(119905 minus 119904)1minus120572
119889119904
100381610038161003816100381610038161003816100381610038161003816
le [1003816100381610038161003816119891 (119905 119909 (119905) 119910 (119905)) minus 119891 (119905 0 0)
1003816100381610038161003816 +1003816100381610038161003816119891 (119905 0 0)
1003816100381610038161003816]
times [1
Γ (120572)
100381610038161003816100381610038161003816100381610038161003816
int
119905
0
119892 (119904 119909 (119904) 119910 (119904)) minus 119892 (119904 0 0)
(119905 minus 119904)1minus120572
119889119904
+ int
119905
0
119892 (119904 0 0)
(119905 minus 119904)1minus120572
119889119904
100381610038161003816100381610038161003816100381610038161003816
]
le1
Γ (120572)[1205931
(max (|119909 (119905)| 1003816100381610038161003816119910 (119905)
1003816100381610038161003816)) + 1198961]
times [int
119905
0
1003816100381610038161003816119892 (119904 119909 (119904) 119910 (119904)) minus 119892 (119904 0 0)1003816100381610038161003816
(119905 minus 119904)1minus120572
119889119904
+ int
119905
0
1003816100381610038161003816119892 (119904 0 0)1003816100381610038161003816
(119905 minus 119904)1minus120572
119889119904]
le1
Γ (120572)[1205931
(max (119909 1003817100381710038171003817119910
1003817100381710038171003817)) + 1198961]
times [int
119905
0
1205932
(max (|119909 (119904)| 1003816100381610038161003816119910 (119904)
1003816100381610038161003816))
(119905 minus 119904)1minus120572
119889119904
+ 1198962
int
119905
0
119889119904
(119905 minus 119904)1minus120572
]
le1
Γ (120572)[1205931
(max (119909 1003817100381710038171003817119910
1003817100381710038171003817)) + 1198961]
sdot [1205932
(max (119909 1003817100381710038171003817119910
1003817100381710038171003817)) + 1198962] int
119905
0
119889119904
(119905 minus 119904)1minus120572
le1
Γ (120572 + 1)(1205931
(1003817100381710038171003817(119909 119910)
1003817100381710038171003817) + 1198961)
sdot (1205932
(1003817100381710038171003817(119909 119910)
1003817100381710038171003817) + 1198962)
(45)
Now taking into account assumption (H3) we infer that the
operator 119866 applies 1198611199030
times 1198611199030
into 1198611199030
Moreover from the lastestimates it follows that
10038171003817100381710038171003817F (1198611199030
times 1198611199030
)10038171003817100381710038171003817
le 1205931
(1199030) + 1198961
10038171003817100381710038171003817G (1198611199030
times 1198611199030
)10038171003817100381710038171003817
le1205932
(1199030) + 1198962
Γ (120572 + 1)
(46)
Next we will prove that the operators F and G arecontinuous on the ball 119861
1199030
times 1198611199030
and consequently 119866 will bealso continuous
In fact we fix 120576 gt 0 and we take (1199090 1199100) (119909 119910) isin 119861
1199030
times 1198611199030
with (119909 119910)minus(1199090 1199100) = (119909minus119909
0 119910minus1199100) =max119909minus119909
0 119910minus
1199100 le 120576 Then for 119905 isin [0 1] we have
1003816100381610038161003816F (119909 119910) (119905) minus F (1199090 1199100) (119905)
1003816100381610038161003816
=1003816100381610038161003816119891 (119905 119909 (119905) 119910 (119905)) minus 119891 (119905 119909
0(119905) 1199100
(119905))1003816100381610038161003816
le 1205931
(max (1003816100381610038161003816119909 (119905) minus 119909
0(119905)
1003816100381610038161003816 1003816100381610038161003816119910 (119905) 119910
0(119905)
1003816100381610038161003816))
le 1205931
(max (1003817100381710038171003817119909 minus 119909
0
1003817100381710038171003817 1003817100381710038171003817119910 minus 119910
0
1003817100381710038171003817))
le 1205931
(120576) lt 120576
(47)
where we have used Remark 8 This proves the continuity ofF on 119861
1199030
times 1198611199030
Abstract and Applied Analysis 7
In order to prove the continuity ofG on 1198611199030
times1198611199030
we have1003816100381610038161003816G (119909 119910) (119905) minus G (119909
0 1199100) (119905)
1003816100381610038161003816
=1
Γ (120572)
100381610038161003816100381610038161003816100381610038161003816
int
119905
0
119892 (119904 119909 (119904) 119910 (119904))
(119905 minus 119904)1minus120572
119889119904
minus int
119905
0
119892 (119904 1199090
(119904) 1199100
(119904))
(119905 minus 119904)1minus120572
119889119904
100381610038161003816100381610038161003816100381610038161003816
le1
Γ (120572)int
119905
0
1003816100381610038161003816119892 (119904 119909 (119904) 119910 (119904)) minus 119892 (119904 1199090
(119904) 1199100
(119904))1003816100381610038161003816
(119905 minus 119904)1minus120572
119889119904
le1
Γ (120572)int
119905
0
1205932
(max (1003816100381610038161003816119909 (119904) minus 119909
0(119904)
1003816100381610038161003816 1003816100381610038161003816119910 (119904) minus 119910
0(119904)
1003816100381610038161003816))
(119905 minus 119904)1minus120572
119889119904
le1
Γ (120572)1205932
(max (1003817100381710038171003817119909 minus 119909
0
1003817100381710038171003817 1003817100381710038171003817119910 minus 119910
0
1003817100381710038171003817)) int
119905
0
119889119904
(119905 minus 119904)1minus120572
le1
Γ (120572 + 1)1205932
(120576)
lt120576
Γ (120572 + 1)
(48)
Therefore1003817100381710038171003817G (119909 119910) minus G (119909
0 1199100)1003817100381710038171003817 lt
120576
Γ (120572 + 1)(49)
and consequentlyG is a continuous operator on 1198611199030
times 1198611199030
In order to prove that G satisfies assumptions of
Theorem 13 only we have to check the condition
120583 (119866 (1198831
times 1198832)) le 120593 (max (120583 (119883
1) 120583 (119883
2))) (50)
for any subsets 1198831and 119883
2of 1198611199030
To do this we fix 120576 gt 0 119905
1 1199052
isin [0 1] with |1199051minus 1199052| le 120576 and
(119909 119910) isin 1198831
times 1198832 then we have
1003816100381610038161003816F (119909 119910) (1199051) minus F (119909 119910) (119905
2)1003816100381610038161003816
=1003816100381610038161003816119891 (1199051 119909 (1199051) 119910 (119905
1)) minus 119891 (119905
2 119909 (1199052) 119910 (119905
2))
1003816100381610038161003816
le1003816100381610038161003816119891 (1199051 119909 (1199051) 119910 (119905
1)) minus 119891 (119905
1 119909 (1199052) 119910 (119905
2))
1003816100381610038161003816
+1003816100381610038161003816119891 (1199051 119909 (1199052) 119910 (119905
2)) minus 119891 (119905
2 119909 (1199052) 119910 (119905
2))
1003816100381610038161003816
le 1205931
(max (1003816100381610038161003816119909 (1199051) minus 119909 (119905
2)1003816100381610038161003816
1003816100381610038161003816119910 (1199051) minus 119910 (119905
2)1003816100381610038161003816))
+ 120596 (119891 120576)
le 1205931
(max (120596 (119909 120576) 120596 (119910 120576))) + 120596 (119891 120576)
(51)
where 120596(119891 120576) denotes the quantity
120596 (119891 120576) = sup 1003816100381610038161003816119891 (119905 119909 119910) minus 119891 (119904 119909 119910)
1003816100381610038161003816 119905 119904 isin [0 1]
|119905 minus 119904| le 120576 119909 119910 isin [minus1199030 1199030]
(52)
From the last estimate we infer that120596 (F (119883
1times 1198832) 120576)
le 1205931
(max (120596 (1198831 120576) 120596 (119883
2 120576))) + 120596 (119891 120576)
(53)
Since 119891(119905 119909 119910) is uniformly continuous on bounded subsetsof [0 1] times R times R we deduce that 120596(119891 120576) rarr 0 as 120576 rarr 0 andtherefore
1205960
(F (1198831
times 1198832)) le lim120576rarr0
1205931
(max (120596 (1198831 120576) 120596 (119883
2 120576)))
(54)
By assumption (H2) since 120593
1is continuous we infer
1205960
(F (1198831
times 1198832)) le 120593
1(max (120596
0(1198831) 1205960
(1198832))) (55)
Now we estimate the quantity 1205960(G(1198831
times 1198832))
Fix 120576 gt 0 1199051 1199052
isin [0 1] with |1199051
minus 1199052| le 120576 and (119909 119910) isin 119883
1times
1198832 Without loss of generality we can suppose that 119905
1lt 1199052
then we have1003816100381610038161003816G (119909 119910) (119905
2) minus G (119909 119910) (119905
1)1003816100381610038161003816
=1
Γ (120572)
1003816100381610038161003816100381610038161003816100381610038161003816
int
1199052
0
119892 (119904 119909 (119904) 119910 (119904))
(1199052
minus 119904)1minus120572
119889119904
minus int
1199051
0
119892 (119904 119909 (119904) 119910 (119904))
(1199051
minus 119904)1minus120572
119889119904
1003816100381610038161003816100381610038161003816100381610038161003816
le1
Γ (120572)[int
1199051
0
10038161003816100381610038161003816(1199052
minus 119904)120572minus1
minus (1199051
minus 119904)120572minus110038161003816100381610038161003816
times1003816100381610038161003816119892 (119904 119909 (119904) 119909 (120590119904))
1003816100381610038161003816 119889119904
+ int
1199052
1199051
(1199052
minus 119904)120572minus1 1003816100381610038161003816119892 (119904 119909 (119904) 119910 (119904))
1003816100381610038161003816 119889119904]
le1
Γ (120572)[int
1199051
0
[(1199051
minus 119904)120572minus1
minus (1199052
minus 119904)120572minus1
]
times1003816100381610038161003816119892 (119904 119909 (119904) 119909 (120590119904))
1003816100381610038161003816 119889119904
+ int
1199052
1199051
(1199052
minus 119904)120572minus1 1003816100381610038161003816119892 (119904 119909 (119904) 119910 (119904))
1003816100381610038161003816 119889119904]
(56)
Since 119892 isin 119862([0 1] times R times R timesR) is bounded on the compactsubsets of [0 1] times R times R particularly on [0 1] times [minus119903
0 1199030] times
[minus1199030 1199030] Put 119871 = sup|119892(119905 119909 119910)| 119905 isin [0 1] 119909 119910 isin [minus119903
0 1199030]
Then from the last inequality we infer that1003816100381610038161003816G (119909 119910) (119905
2) minus G (119909 119910) (119905
1)1003816100381610038161003816
le119871
Γ (120572)[int
1199051
0
[(1199051
minus 119904)120572minus1
minus (1199052
minus 119904)120572minus1
] 119889119904
+ int
1199052
1199051
(1199052
minus 119904)120572minus1
119889119904]
le119871
Γ (120572 + 1)[(1199052
minus 1199051)120572
+ 119905120572
1minus 119905120572
2+ (1199052
minus 1199051)120572
]
le2119871
Γ (120572 + 1)(1199052
minus 1199051)120572
le2119871
Γ (120572 + 1)120576120572
(57)
8 Abstract and Applied Analysis
where we have used the fact that 119905120572
1minus 119905120572
2le 0 Therefore
120596 (G (1198831
times 1198832) 120576) le
2119871
Γ (120572 + 1)120576120572
(58)
From this it follows that
1205960
(G (1198831
times 1198832)) = 0 (59)
Next by Proposition 7 (46) (55) and (59) we have
1205960
(119866 (1198831
times 1198832))
= 1205960
(F (1198831
times 1198832) sdot G (119883
1times 1198832))
le1003817100381710038171003817F (119883
1times 1198832)1003817100381710038171003817 1205960
(G (1198831
times 1198832))
+1003817100381710038171003817G (119883
1times 1198832)1003817100381710038171003817 1205960
(F (1198831
times 1198832))
le10038171003817100381710038171003817F (1198611199030
times 1198611199030
)10038171003817100381710038171003817
1205960
(G (1198831
times 1198832))
+10038171003817100381710038171003817G (1198611199030
times 1198611199030
)10038171003817100381710038171003817
1205960
(F (1198831
times 1198832))
le1205932
(1199030) + 1198962
Γ (120572 + 1)1205931
(max (1205960
(1198831) 1205960
(1198832)))
(60)
By assumption (H3) since 120593
2(1199030) + 1198962
le Γ(120572 + 1) and since itis easily proved that if 120572 isin [0 1] and 120593 isin A then 120572120593 isin A wededuce that
1205960
(119866 (1198831
times 1198832)) le 120593 (max (120596
0(1198831) 1205960
(1198832))) (61)
where 120593 isin AFinally by Theorem 13 the operator G has at least a
coupled fixed point and this is the desired result Thiscompletes the proof
The nonoscillary character of the solutions of problem(4) seems to be an interesting question from the practicalpoint of view This means that the solutions of problem (4)have a constant sign on the interval (0 1) In connection withthis question we notice that if 119891(119905 119909 119910) and 119892(119905 119909 119910) haveconstant sign and are equal (this means that 119891(119905 119909 119910) gt 0
and 119892(119905 119909 119910) ge 0 or 119891(119905 119909 119910) lt 0 and 119892(119905 119909 119910) le 0
for any 119905 isin [0 1] and 119909 119910 isin R) and under assumptionsof Theorem 15 then the solution (119909 119910) isin 119862[0 1] times 119862[0 1]
of problem (4) given by Theorem 15 satisfies 119909(119905) ge 0 and119910(119905) ge 0 for 119905 isin [0 1] since the solution (119909 119910) satisfies thesystem of integral equations
119909 (119905) =119891 (119905 119909 (119905) 119910 (119905))
Γ (120572)int
119905
0
119892 (119904 119909 (119904) 119910 (119904))
(119905 minus 119904)1minus120572
119889119904
119910 (119905) =119891 (119905 119910 (119905) 119909 (119905))
Γ (120572)int
119905
0
119892 (119904 119910 (119904) 119909 (119904))
(119905 minus 119904)1minus120572
119889119904 0 le 119905 le 1
(62)
On the other hand if we perturb the data function in problem(4) of the following manner
119863120572
0+ [
119909 (119905)
119891 (119905 119909 (119905) 119910 (119905))] = 119892 (119905 119909 (119905) 119910 (119905)) + 120578 (119905)
119863120572
0+ [
119910 (119905)
119891 (119905 119910 (119905) 119909 (119905))] = 119892 (119905 119910 (119905) 119909 (119905)) + 120578 (119905)
0 lt 119905 lt 1
(63)
where 0 lt 120572 lt 1 119891 isin 119862([0 1] times R times RR 0)119892 isin 119862([0 1] times R times RR) and 120578 isin 119862[0 1] then underassumptions of Theorem 15 problem (63) can be studiedby using Theorem 15 where assumptions (H
1) and (H
2) are
automatically satisfied and we only have to check assumption(H3) This fact gives a great applicability to Theorem 15Before presenting an example illustrating our results we
need some facts about the functions involving this exampleThe following lemma appears in [18]
Lemma 16 Let 120593 R+
rarr R+be a nondecreasing and upper
semicontinuous function Then the following conditions areequivalent
(i) lim119899rarrinfin
120593119899
(119905) = 0 for any 119905 ge 0
(ii) 120593(119905) lt 119905 for any 119905 gt 0
Particularly the functions 1205721 1205722
R+
rarr R+given by
1205721(119905) = arctan 119905 and 120572
2(119905) = 119905(1 + 119905) belong to the class
A since they are nondecreasing and continuous and as itis easily seen they satisfy (ii) of Lemma 16
On the other hand since the function 1205721(119905) = arctan 119905 is
concave (because 12057210158401015840
(119905) le 0) and 1205721(0) = 0 we infer that 120572
1
is subadditive and therefore for any 119905 1199051015840
isin R+ we have
100381610038161003816100381610038161205721
(119905) minus 1205721
(1199051015840
)10038161003816100381610038161003816
=10038161003816100381610038161003816arctan 119905 minus arctan 119905
101584010038161003816100381610038161003816
le arctan 10038161003816100381610038161003816119905 minus 119905101584010038161003816100381610038161003816
(64)
Moreover it is easily seen that max(1205721 1205722) is a nondecreasing
and continuous function because 1205721and 120572
2are nondecreas-
ing and continuous andmax(1205721 1205722) satisfies (ii) of Lemma 16
Therefore max(1205721 1205722) isin A
Nowwe are ready to present an examplewhere our resultscan be applied
Example 17 Consider the following coupled system of frac-tional hybrid differential equations
11986312
0+ [119909 (119905) times (
1
4+ (
1
10) arctan |119909 (119905)|
+(1
20) (
1003816100381610038161003816119910 (119905)1003816100381610038161003816
(1 +1003816100381610038161003816119910 (119905)
1003816100381610038161003816)))
minus1
]
=1
7+
1
9119909 (119905) +
1
10119910 (119905)
Abstract and Applied Analysis 9
11986312
0+ [119910 (119905) times (
1
4+ (
1
10) arctan 1003816100381610038161003816119910 (119905)
1003816100381610038161003816
+ (1
20) (
|119909 (119905)|
(1 + |119909 (119905)|)))
minus1
]
=1
7+
1
9119910 (119905) +
1
10119909 (119905)
0 lt 119905 lt 1
119909 (0) = 119910 (0) = 0
(65)
Notice that problem (17) is a particular case of problem (4)where 120572 = 12 119891(119905 119909 119910) = 14 + (110) arctan |119909| +
(120)(|119910|(1 + |119910|)) and 119892(119905 119909 119910) = 17 + (19)119909 + (110)119910It is clear that 119891 isin 119862([0 1] times R times RR 0) and 119892 isin
119862([0 1] times R times RR) and moreover 1198961
= sup|119891(119905 0 0)| 119905 isin
[0 1] = 14 and 1198962
= sup|119892(119905 0 0)| 119905 isin [0 1] = 17Therefore assumption (H
1) of Theorem 15 is satisfied
Moreover for 119905 isin [0 1] and 1199091 1199092 1199101 1199102
isin R we have1003816100381610038161003816119891 (119905 119909
1 1199101) minus 119891 (119905 119909
2 1199102)1003816100381610038161003816
le1
10
1003816100381610038161003816arctan10038161003816100381610038161199091
1003816100381610038161003816 minus arctan 100381610038161003816100381611990921003816100381610038161003816
1003816100381610038161003816
+1
20
100381610038161003816100381610038161003816100381610038161003816
100381610038161003816100381611991011003816100381610038161003816
1 +10038161003816100381610038161199101
1003816100381610038161003816
minus
100381610038161003816100381611991021003816100381610038161003816
1 +10038161003816100381610038161199102
1003816100381610038161003816
100381610038161003816100381610038161003816100381610038161003816
le1
10arctan 1003816100381610038161003816
100381610038161003816100381611990911003816100381610038161003816 minus
100381610038161003816100381611990921003816100381610038161003816
1003816100381610038161003816
+1
20
100381610038161003816100381610038161003816100381610038161003816
100381610038161003816100381611991011003816100381610038161003816 minus
100381610038161003816100381611991021003816100381610038161003816
(1 +10038161003816100381610038161199101
1003816100381610038161003816) (1 +10038161003816100381610038161199102
1003816100381610038161003816)
100381610038161003816100381610038161003816100381610038161003816
le1
10arctan (
10038161003816100381610038161199091 minus 1199092
1003816100381610038161003816)
+1
20
10038161003816100381610038161199101 minus 1199102
1003816100381610038161003816
1 +10038161003816100381610038161199101 minus 119910
2
1003816100381610038161003816
=1
101205721
(10038161003816100381610038161199091 minus 119909
2
1003816100381610038161003816)
+1
201205722
(10038161003816100381610038161199101 minus 119910
2
1003816100381610038161003816)
le1
10max (120572
1 1205722) (
10038161003816100381610038161199091 minus 1199092
1003816100381610038161003816)
+1
10max (120572
1 1205722) (
10038161003816100381610038161199101 minus 1199102
1003816100381610038161003816)
le1
10[2max (120572
1 1205722)
times max (10038161003816100381610038161199091 minus 119909
2
1003816100381610038161003816 10038161003816100381610038161199101 minus 119910
2
1003816100381610038161003816)]
=1
5max (120572
1 1205722) (max (
10038161003816100381610038161199091 minus 1199092
1003816100381610038161003816 10038161003816100381610038161199101 minus 119910
2
1003816100381610038161003816))
(66)
Therefore 1205931(119905) = (15)max(120572
1(119905) 1205722(119905)) and 120593
1isin A
On the other hand1003816100381610038161003816119892 (119905 119909
1 1199101) minus 119892 (119905 119909
2 1199102)1003816100381610038161003816
le1
9
10038161003816100381610038161199091 minus 1199092
1003816100381610038161003816 +1
10
10038161003816100381610038161199092 minus 1199102
1003816100381610038161003816
le1
9(10038161003816100381610038161199091 minus 119910
1
1003816100381610038161003816 +10038161003816100381610038161199102 minus 119910
2
1003816100381610038161003816)
le1
9(2max (
10038161003816100381610038161199091 minus 1199101
1003816100381610038161003816 10038161003816100381610038161199102 minus 119910
2
1003816100381610038161003816))
=2
9max (
10038161003816100381610038161199091 minus 1199101
1003816100381610038161003816 10038161003816100381610038161199102 minus 119910
2
1003816100381610038161003816)
(67)
and 1205932(119905) = (29)119905 It is clear that 120593
2isin A Therefore
assumption (H2) of Theorem 15 is satisfied
In our case the inequality appearing in assumption (H3)
of Theorem 15 has the expression
[1
5max(arctan 119903
119903
1 + 119903) +
1
4] [
2
9119903 +
1
7] le 119903Γ (
3
2) (68)
It is easily seen that 1199030
= 1 satisfies the last inequalityMoreover
2
91199030
+1
7=
2
9+
1
7le Γ (
3
2) cong 088623 (69)
Finally Theorem 15 says that problem (17) has at least onesolution (119909 119910) isin 119862[0 1] such that max(119909 119910) le 1
Conflict of Interests
The authors declare that there is no conflict of interests in thesubmitted paper
References
[1] W Deng ldquoNumerical algorithm for the time fractional Fokker-Planck equationrdquo Journal of Computational Physics vol 227 no2 pp 1510ndash1522 2007
[2] S Z Rida H M El-Sherbiny and A A M Arafa ldquoOn thesolution of the fractional nonlinear Schrodinger equationrdquoPhysics Letters A vol 372 no 5 pp 553ndash558 2008
[3] K Diethelm and A D Freed ldquoOn the solution of nonlinearfractional order differential equations used in the modeling ofviscoplasticityrdquo in Scientific Computing in Chemical EngineeringII pp 217ndash224 Springer Berlin Germany 1999
[4] F Mainardi ldquoFractional calculus some basic proble ms incontinuum and statistical mechanicsrdquo in Fractals and FractionalCalculus in Continuum Mechanics (Udine 1996) vol 378 ofCISM Courses and Lectures pp 291ndash348 Springer ViennaAustria 1997
[5] R Metzler W Schick H Kilian and T F NonnenmacherldquoRelaxation in filled polymers a fractional calculus approachrdquoThe Journal of Chemical Physics vol 103 no 16 pp 7180ndash71861995
[6] B C Dhage and V Lakshmikantham ldquoBasic results on hybriddifferential equationsrdquo Nonlinear Analysis Hybrid Systems vol4 no 3 pp 414ndash424 2010
10 Abstract and Applied Analysis
[7] Y Zhao S Sun Z Han and Q Li ldquoTheory of fractionalhybrid differential equationsrdquo Computers amp Mathematics withApplications vol 62 no 3 pp 1312ndash1324 2011
[8] J Caballero M A Darwish and K Sadarangani ldquoSolvabilityof a fractional hybrid initial value problem with supremum byusingmeasures of noncompactness in Banach algebrasrdquoAppliedMathematics and Computation vol 224 pp 553ndash563 2013
[9] C Bai and J Fang ldquoThe existence of a positive solution fora singular coupled system of nonlinear fractional differentialequationsrdquo Applied Mathematics and Computation vol 150 no3 pp 611ndash621 2004
[10] Y Chen and H An ldquoNumerical solutions of coupled Burgersequations with time- and space-fractional derivativesrdquo AppliedMathematics and Computation vol 200 no 1 pp 87ndash95 2008
[11] M P Lazarevic ldquoFinite time stability analysis of PD120572 fractionalcontrol of robotic time-delay systemsrdquo Mechanics ResearchCommunications vol 33 no 2 pp 269ndash279 2006
[12] V Gafiychuk B Datsko V Meleshko and D Blackmore ldquoAnal-ysis of the solutions of coupled nonlinear fractional reaction-diffusion equationsrdquo Chaos Solitons and Fractals vol 41 no 3pp 1095ndash1104 2009
[13] B Ahmad and J J Nieto ldquoExistence results for a coupledsystem of nonlinear fractional differential equations with three-point boundary conditionsrdquo Computers amp Mathematics withApplications vol 58 no 9 pp 1838ndash1843 2009
[14] A A Kilbas H M Srivastava and J J Trujillo Theory andApplications of Fractional Differential Equations vol 204 ofNorth-Holland Mathematics Studies Elsevier Science Amster-dam The Netherlands 2006
[15] J Banas and K GoebelMeasures of Noncompactness in BanachSpaces Lecture Notes in Pure andAppliedMathematicsMarcelDekker New York NY USA 1980
[16] B N Sadovskii ldquoOn a fixed point principlerdquo Functional Analysisand Its Applications vol 1 pp 151ndash153 1967
[17] J Banas and L Olszowy ldquoOn a class of measures of noncom-pactness in Banach algebras and their application to nonlinearintegral equationsrdquo Journal of Analysis and its Applications vol28 no 4 pp 475ndash498 2009
[18] A Aghajani J Banas and N Sabzali ldquoSome generalizationsof Darbo fixed point theorem and applicationsrdquo Bulletin of theBelgian Mathematical Society vol 20 no 2 pp 345ndash358 2013
[19] R R Akhmerov M I Kamenskii A S Potapov A E Rod-kina and B N Sadovskii Measures of Noncompactness andCondensing Operators vol 55 ofOperatorTheory Advances andApplications Birkhaauser Basel Switzerland 1992
Submit your manuscripts athttpwwwhindawicom
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
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Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
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Differential EquationsInternational Journal of
Volume 2014
Applied MathematicsJournal of
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Journal of
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Mathematical PhysicsAdvances in
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OptimizationJournal of
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CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of
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Operations ResearchAdvances in
Journal of
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
As 0 lt 120572 lt 1 and 119905119899
gt 1199050 we infer that
1003816100381610038161003816G (119909 119910) (119905119899) minus G (119909 119910) (119905
0)1003816100381610038161003816
le119872
Γ (120572)[int
1199050
0
10038161003816100381610038161003816(119905119899
minus 119904)120572minus1
minus (1199050
minus 119904)120572minus110038161003816100381610038161003816
119889119904
+ int
119905119899
1199050
10038161003816100381610038161003816(119905119899
minus 119904)120572minus1
minus (1199050
minus 119904)120572minus110038161003816100381610038161003816
119889119904]
+119872
Γ (120572)int
119905119899
1199050
1
(119904 minus 1199050)1minus120572
119889119904
=119872
Γ (120572)[ int
1199050
0
[(1199050
minus 119904)120572minus1
minus (119905119899
minus 119904)120572minus1
] 119889119904
+ int
119905119899
1199050
119889119904
(119905119899
minus 119904)1minus120572
+ int
119905119899
1199050
119889119904
(119904 minus 1199050)1minus120572
]
+119872
Γ (120572)int
119905119899
1199050
1
(119904 minus 1199050)1minus120572
119889119904
le119872
Γ (120572 + 1)[(119905119899
minus 1199050)120572
+ 119905120572
0minus 119905120572
119899
+ (119905119899
minus 1199050)120572
+ (119905119899
minus 1199050)120572
]
+119872
Γ (120572 + 1)(119905119899
minus 1199050)120572
=4119872
Γ (120572 + 1)(119905119899
minus 1199050)120572
+119872
Γ (120572 + 1)(119905120572
0minus 119905120572
119899)
lt4119872
Γ (120572 + 1)(119905119899
minus 1199050)120572
(44)
where the last inequality has been obtained by using the factthat 119905120572
0minus 119905120572
119899lt 0
Therefore since 119905119899
rarr 1199050 from the last estimate we
deduce that G(119909 119910)(119905119899) rarr G(119909 119910)(119905
0) This proves that
G(119909 119910) isin 119862[0 1] Consequently G 119862[0 1] times 119862[0 1] rarr
119862[0 1] On the other hand for (119909 119910) isin 119862[0 1] times 119862[0 1] and119905 isin 119862[0 1] we have
1003816100381610038161003816119866 (119909 119910) (119905)1003816100381610038161003816
=1003816100381610038161003816F (119909 119910) (119905) sdot G (119909 119910) (119905)
1003816100381610038161003816
=1003816100381610038161003816119891 (119905 119909 (119905) 119910 (119905))
1003816100381610038161003816
100381610038161003816100381610038161003816100381610038161003816
1
Γ (120572)int
119905
0
119892 (119904 119909 (119904) 119910 (119904))
(119905 minus 119904)1minus120572
119889119904
100381610038161003816100381610038161003816100381610038161003816
le [1003816100381610038161003816119891 (119905 119909 (119905) 119910 (119905)) minus 119891 (119905 0 0)
1003816100381610038161003816 +1003816100381610038161003816119891 (119905 0 0)
1003816100381610038161003816]
times [1
Γ (120572)
100381610038161003816100381610038161003816100381610038161003816
int
119905
0
119892 (119904 119909 (119904) 119910 (119904)) minus 119892 (119904 0 0)
(119905 minus 119904)1minus120572
119889119904
+ int
119905
0
119892 (119904 0 0)
(119905 minus 119904)1minus120572
119889119904
100381610038161003816100381610038161003816100381610038161003816
]
le1
Γ (120572)[1205931
(max (|119909 (119905)| 1003816100381610038161003816119910 (119905)
1003816100381610038161003816)) + 1198961]
times [int
119905
0
1003816100381610038161003816119892 (119904 119909 (119904) 119910 (119904)) minus 119892 (119904 0 0)1003816100381610038161003816
(119905 minus 119904)1minus120572
119889119904
+ int
119905
0
1003816100381610038161003816119892 (119904 0 0)1003816100381610038161003816
(119905 minus 119904)1minus120572
119889119904]
le1
Γ (120572)[1205931
(max (119909 1003817100381710038171003817119910
1003817100381710038171003817)) + 1198961]
times [int
119905
0
1205932
(max (|119909 (119904)| 1003816100381610038161003816119910 (119904)
1003816100381610038161003816))
(119905 minus 119904)1minus120572
119889119904
+ 1198962
int
119905
0
119889119904
(119905 minus 119904)1minus120572
]
le1
Γ (120572)[1205931
(max (119909 1003817100381710038171003817119910
1003817100381710038171003817)) + 1198961]
sdot [1205932
(max (119909 1003817100381710038171003817119910
1003817100381710038171003817)) + 1198962] int
119905
0
119889119904
(119905 minus 119904)1minus120572
le1
Γ (120572 + 1)(1205931
(1003817100381710038171003817(119909 119910)
1003817100381710038171003817) + 1198961)
sdot (1205932
(1003817100381710038171003817(119909 119910)
1003817100381710038171003817) + 1198962)
(45)
Now taking into account assumption (H3) we infer that the
operator 119866 applies 1198611199030
times 1198611199030
into 1198611199030
Moreover from the lastestimates it follows that
10038171003817100381710038171003817F (1198611199030
times 1198611199030
)10038171003817100381710038171003817
le 1205931
(1199030) + 1198961
10038171003817100381710038171003817G (1198611199030
times 1198611199030
)10038171003817100381710038171003817
le1205932
(1199030) + 1198962
Γ (120572 + 1)
(46)
Next we will prove that the operators F and G arecontinuous on the ball 119861
1199030
times 1198611199030
and consequently 119866 will bealso continuous
In fact we fix 120576 gt 0 and we take (1199090 1199100) (119909 119910) isin 119861
1199030
times 1198611199030
with (119909 119910)minus(1199090 1199100) = (119909minus119909
0 119910minus1199100) =max119909minus119909
0 119910minus
1199100 le 120576 Then for 119905 isin [0 1] we have
1003816100381610038161003816F (119909 119910) (119905) minus F (1199090 1199100) (119905)
1003816100381610038161003816
=1003816100381610038161003816119891 (119905 119909 (119905) 119910 (119905)) minus 119891 (119905 119909
0(119905) 1199100
(119905))1003816100381610038161003816
le 1205931
(max (1003816100381610038161003816119909 (119905) minus 119909
0(119905)
1003816100381610038161003816 1003816100381610038161003816119910 (119905) 119910
0(119905)
1003816100381610038161003816))
le 1205931
(max (1003817100381710038171003817119909 minus 119909
0
1003817100381710038171003817 1003817100381710038171003817119910 minus 119910
0
1003817100381710038171003817))
le 1205931
(120576) lt 120576
(47)
where we have used Remark 8 This proves the continuity ofF on 119861
1199030
times 1198611199030
Abstract and Applied Analysis 7
In order to prove the continuity ofG on 1198611199030
times1198611199030
we have1003816100381610038161003816G (119909 119910) (119905) minus G (119909
0 1199100) (119905)
1003816100381610038161003816
=1
Γ (120572)
100381610038161003816100381610038161003816100381610038161003816
int
119905
0
119892 (119904 119909 (119904) 119910 (119904))
(119905 minus 119904)1minus120572
119889119904
minus int
119905
0
119892 (119904 1199090
(119904) 1199100
(119904))
(119905 minus 119904)1minus120572
119889119904
100381610038161003816100381610038161003816100381610038161003816
le1
Γ (120572)int
119905
0
1003816100381610038161003816119892 (119904 119909 (119904) 119910 (119904)) minus 119892 (119904 1199090
(119904) 1199100
(119904))1003816100381610038161003816
(119905 minus 119904)1minus120572
119889119904
le1
Γ (120572)int
119905
0
1205932
(max (1003816100381610038161003816119909 (119904) minus 119909
0(119904)
1003816100381610038161003816 1003816100381610038161003816119910 (119904) minus 119910
0(119904)
1003816100381610038161003816))
(119905 minus 119904)1minus120572
119889119904
le1
Γ (120572)1205932
(max (1003817100381710038171003817119909 minus 119909
0
1003817100381710038171003817 1003817100381710038171003817119910 minus 119910
0
1003817100381710038171003817)) int
119905
0
119889119904
(119905 minus 119904)1minus120572
le1
Γ (120572 + 1)1205932
(120576)
lt120576
Γ (120572 + 1)
(48)
Therefore1003817100381710038171003817G (119909 119910) minus G (119909
0 1199100)1003817100381710038171003817 lt
120576
Γ (120572 + 1)(49)
and consequentlyG is a continuous operator on 1198611199030
times 1198611199030
In order to prove that G satisfies assumptions of
Theorem 13 only we have to check the condition
120583 (119866 (1198831
times 1198832)) le 120593 (max (120583 (119883
1) 120583 (119883
2))) (50)
for any subsets 1198831and 119883
2of 1198611199030
To do this we fix 120576 gt 0 119905
1 1199052
isin [0 1] with |1199051minus 1199052| le 120576 and
(119909 119910) isin 1198831
times 1198832 then we have
1003816100381610038161003816F (119909 119910) (1199051) minus F (119909 119910) (119905
2)1003816100381610038161003816
=1003816100381610038161003816119891 (1199051 119909 (1199051) 119910 (119905
1)) minus 119891 (119905
2 119909 (1199052) 119910 (119905
2))
1003816100381610038161003816
le1003816100381610038161003816119891 (1199051 119909 (1199051) 119910 (119905
1)) minus 119891 (119905
1 119909 (1199052) 119910 (119905
2))
1003816100381610038161003816
+1003816100381610038161003816119891 (1199051 119909 (1199052) 119910 (119905
2)) minus 119891 (119905
2 119909 (1199052) 119910 (119905
2))
1003816100381610038161003816
le 1205931
(max (1003816100381610038161003816119909 (1199051) minus 119909 (119905
2)1003816100381610038161003816
1003816100381610038161003816119910 (1199051) minus 119910 (119905
2)1003816100381610038161003816))
+ 120596 (119891 120576)
le 1205931
(max (120596 (119909 120576) 120596 (119910 120576))) + 120596 (119891 120576)
(51)
where 120596(119891 120576) denotes the quantity
120596 (119891 120576) = sup 1003816100381610038161003816119891 (119905 119909 119910) minus 119891 (119904 119909 119910)
1003816100381610038161003816 119905 119904 isin [0 1]
|119905 minus 119904| le 120576 119909 119910 isin [minus1199030 1199030]
(52)
From the last estimate we infer that120596 (F (119883
1times 1198832) 120576)
le 1205931
(max (120596 (1198831 120576) 120596 (119883
2 120576))) + 120596 (119891 120576)
(53)
Since 119891(119905 119909 119910) is uniformly continuous on bounded subsetsof [0 1] times R times R we deduce that 120596(119891 120576) rarr 0 as 120576 rarr 0 andtherefore
1205960
(F (1198831
times 1198832)) le lim120576rarr0
1205931
(max (120596 (1198831 120576) 120596 (119883
2 120576)))
(54)
By assumption (H2) since 120593
1is continuous we infer
1205960
(F (1198831
times 1198832)) le 120593
1(max (120596
0(1198831) 1205960
(1198832))) (55)
Now we estimate the quantity 1205960(G(1198831
times 1198832))
Fix 120576 gt 0 1199051 1199052
isin [0 1] with |1199051
minus 1199052| le 120576 and (119909 119910) isin 119883
1times
1198832 Without loss of generality we can suppose that 119905
1lt 1199052
then we have1003816100381610038161003816G (119909 119910) (119905
2) minus G (119909 119910) (119905
1)1003816100381610038161003816
=1
Γ (120572)
1003816100381610038161003816100381610038161003816100381610038161003816
int
1199052
0
119892 (119904 119909 (119904) 119910 (119904))
(1199052
minus 119904)1minus120572
119889119904
minus int
1199051
0
119892 (119904 119909 (119904) 119910 (119904))
(1199051
minus 119904)1minus120572
119889119904
1003816100381610038161003816100381610038161003816100381610038161003816
le1
Γ (120572)[int
1199051
0
10038161003816100381610038161003816(1199052
minus 119904)120572minus1
minus (1199051
minus 119904)120572minus110038161003816100381610038161003816
times1003816100381610038161003816119892 (119904 119909 (119904) 119909 (120590119904))
1003816100381610038161003816 119889119904
+ int
1199052
1199051
(1199052
minus 119904)120572minus1 1003816100381610038161003816119892 (119904 119909 (119904) 119910 (119904))
1003816100381610038161003816 119889119904]
le1
Γ (120572)[int
1199051
0
[(1199051
minus 119904)120572minus1
minus (1199052
minus 119904)120572minus1
]
times1003816100381610038161003816119892 (119904 119909 (119904) 119909 (120590119904))
1003816100381610038161003816 119889119904
+ int
1199052
1199051
(1199052
minus 119904)120572minus1 1003816100381610038161003816119892 (119904 119909 (119904) 119910 (119904))
1003816100381610038161003816 119889119904]
(56)
Since 119892 isin 119862([0 1] times R times R timesR) is bounded on the compactsubsets of [0 1] times R times R particularly on [0 1] times [minus119903
0 1199030] times
[minus1199030 1199030] Put 119871 = sup|119892(119905 119909 119910)| 119905 isin [0 1] 119909 119910 isin [minus119903
0 1199030]
Then from the last inequality we infer that1003816100381610038161003816G (119909 119910) (119905
2) minus G (119909 119910) (119905
1)1003816100381610038161003816
le119871
Γ (120572)[int
1199051
0
[(1199051
minus 119904)120572minus1
minus (1199052
minus 119904)120572minus1
] 119889119904
+ int
1199052
1199051
(1199052
minus 119904)120572minus1
119889119904]
le119871
Γ (120572 + 1)[(1199052
minus 1199051)120572
+ 119905120572
1minus 119905120572
2+ (1199052
minus 1199051)120572
]
le2119871
Γ (120572 + 1)(1199052
minus 1199051)120572
le2119871
Γ (120572 + 1)120576120572
(57)
8 Abstract and Applied Analysis
where we have used the fact that 119905120572
1minus 119905120572
2le 0 Therefore
120596 (G (1198831
times 1198832) 120576) le
2119871
Γ (120572 + 1)120576120572
(58)
From this it follows that
1205960
(G (1198831
times 1198832)) = 0 (59)
Next by Proposition 7 (46) (55) and (59) we have
1205960
(119866 (1198831
times 1198832))
= 1205960
(F (1198831
times 1198832) sdot G (119883
1times 1198832))
le1003817100381710038171003817F (119883
1times 1198832)1003817100381710038171003817 1205960
(G (1198831
times 1198832))
+1003817100381710038171003817G (119883
1times 1198832)1003817100381710038171003817 1205960
(F (1198831
times 1198832))
le10038171003817100381710038171003817F (1198611199030
times 1198611199030
)10038171003817100381710038171003817
1205960
(G (1198831
times 1198832))
+10038171003817100381710038171003817G (1198611199030
times 1198611199030
)10038171003817100381710038171003817
1205960
(F (1198831
times 1198832))
le1205932
(1199030) + 1198962
Γ (120572 + 1)1205931
(max (1205960
(1198831) 1205960
(1198832)))
(60)
By assumption (H3) since 120593
2(1199030) + 1198962
le Γ(120572 + 1) and since itis easily proved that if 120572 isin [0 1] and 120593 isin A then 120572120593 isin A wededuce that
1205960
(119866 (1198831
times 1198832)) le 120593 (max (120596
0(1198831) 1205960
(1198832))) (61)
where 120593 isin AFinally by Theorem 13 the operator G has at least a
coupled fixed point and this is the desired result Thiscompletes the proof
The nonoscillary character of the solutions of problem(4) seems to be an interesting question from the practicalpoint of view This means that the solutions of problem (4)have a constant sign on the interval (0 1) In connection withthis question we notice that if 119891(119905 119909 119910) and 119892(119905 119909 119910) haveconstant sign and are equal (this means that 119891(119905 119909 119910) gt 0
and 119892(119905 119909 119910) ge 0 or 119891(119905 119909 119910) lt 0 and 119892(119905 119909 119910) le 0
for any 119905 isin [0 1] and 119909 119910 isin R) and under assumptionsof Theorem 15 then the solution (119909 119910) isin 119862[0 1] times 119862[0 1]
of problem (4) given by Theorem 15 satisfies 119909(119905) ge 0 and119910(119905) ge 0 for 119905 isin [0 1] since the solution (119909 119910) satisfies thesystem of integral equations
119909 (119905) =119891 (119905 119909 (119905) 119910 (119905))
Γ (120572)int
119905
0
119892 (119904 119909 (119904) 119910 (119904))
(119905 minus 119904)1minus120572
119889119904
119910 (119905) =119891 (119905 119910 (119905) 119909 (119905))
Γ (120572)int
119905
0
119892 (119904 119910 (119904) 119909 (119904))
(119905 minus 119904)1minus120572
119889119904 0 le 119905 le 1
(62)
On the other hand if we perturb the data function in problem(4) of the following manner
119863120572
0+ [
119909 (119905)
119891 (119905 119909 (119905) 119910 (119905))] = 119892 (119905 119909 (119905) 119910 (119905)) + 120578 (119905)
119863120572
0+ [
119910 (119905)
119891 (119905 119910 (119905) 119909 (119905))] = 119892 (119905 119910 (119905) 119909 (119905)) + 120578 (119905)
0 lt 119905 lt 1
(63)
where 0 lt 120572 lt 1 119891 isin 119862([0 1] times R times RR 0)119892 isin 119862([0 1] times R times RR) and 120578 isin 119862[0 1] then underassumptions of Theorem 15 problem (63) can be studiedby using Theorem 15 where assumptions (H
1) and (H
2) are
automatically satisfied and we only have to check assumption(H3) This fact gives a great applicability to Theorem 15Before presenting an example illustrating our results we
need some facts about the functions involving this exampleThe following lemma appears in [18]
Lemma 16 Let 120593 R+
rarr R+be a nondecreasing and upper
semicontinuous function Then the following conditions areequivalent
(i) lim119899rarrinfin
120593119899
(119905) = 0 for any 119905 ge 0
(ii) 120593(119905) lt 119905 for any 119905 gt 0
Particularly the functions 1205721 1205722
R+
rarr R+given by
1205721(119905) = arctan 119905 and 120572
2(119905) = 119905(1 + 119905) belong to the class
A since they are nondecreasing and continuous and as itis easily seen they satisfy (ii) of Lemma 16
On the other hand since the function 1205721(119905) = arctan 119905 is
concave (because 12057210158401015840
(119905) le 0) and 1205721(0) = 0 we infer that 120572
1
is subadditive and therefore for any 119905 1199051015840
isin R+ we have
100381610038161003816100381610038161205721
(119905) minus 1205721
(1199051015840
)10038161003816100381610038161003816
=10038161003816100381610038161003816arctan 119905 minus arctan 119905
101584010038161003816100381610038161003816
le arctan 10038161003816100381610038161003816119905 minus 119905101584010038161003816100381610038161003816
(64)
Moreover it is easily seen that max(1205721 1205722) is a nondecreasing
and continuous function because 1205721and 120572
2are nondecreas-
ing and continuous andmax(1205721 1205722) satisfies (ii) of Lemma 16
Therefore max(1205721 1205722) isin A
Nowwe are ready to present an examplewhere our resultscan be applied
Example 17 Consider the following coupled system of frac-tional hybrid differential equations
11986312
0+ [119909 (119905) times (
1
4+ (
1
10) arctan |119909 (119905)|
+(1
20) (
1003816100381610038161003816119910 (119905)1003816100381610038161003816
(1 +1003816100381610038161003816119910 (119905)
1003816100381610038161003816)))
minus1
]
=1
7+
1
9119909 (119905) +
1
10119910 (119905)
Abstract and Applied Analysis 9
11986312
0+ [119910 (119905) times (
1
4+ (
1
10) arctan 1003816100381610038161003816119910 (119905)
1003816100381610038161003816
+ (1
20) (
|119909 (119905)|
(1 + |119909 (119905)|)))
minus1
]
=1
7+
1
9119910 (119905) +
1
10119909 (119905)
0 lt 119905 lt 1
119909 (0) = 119910 (0) = 0
(65)
Notice that problem (17) is a particular case of problem (4)where 120572 = 12 119891(119905 119909 119910) = 14 + (110) arctan |119909| +
(120)(|119910|(1 + |119910|)) and 119892(119905 119909 119910) = 17 + (19)119909 + (110)119910It is clear that 119891 isin 119862([0 1] times R times RR 0) and 119892 isin
119862([0 1] times R times RR) and moreover 1198961
= sup|119891(119905 0 0)| 119905 isin
[0 1] = 14 and 1198962
= sup|119892(119905 0 0)| 119905 isin [0 1] = 17Therefore assumption (H
1) of Theorem 15 is satisfied
Moreover for 119905 isin [0 1] and 1199091 1199092 1199101 1199102
isin R we have1003816100381610038161003816119891 (119905 119909
1 1199101) minus 119891 (119905 119909
2 1199102)1003816100381610038161003816
le1
10
1003816100381610038161003816arctan10038161003816100381610038161199091
1003816100381610038161003816 minus arctan 100381610038161003816100381611990921003816100381610038161003816
1003816100381610038161003816
+1
20
100381610038161003816100381610038161003816100381610038161003816
100381610038161003816100381611991011003816100381610038161003816
1 +10038161003816100381610038161199101
1003816100381610038161003816
minus
100381610038161003816100381611991021003816100381610038161003816
1 +10038161003816100381610038161199102
1003816100381610038161003816
100381610038161003816100381610038161003816100381610038161003816
le1
10arctan 1003816100381610038161003816
100381610038161003816100381611990911003816100381610038161003816 minus
100381610038161003816100381611990921003816100381610038161003816
1003816100381610038161003816
+1
20
100381610038161003816100381610038161003816100381610038161003816
100381610038161003816100381611991011003816100381610038161003816 minus
100381610038161003816100381611991021003816100381610038161003816
(1 +10038161003816100381610038161199101
1003816100381610038161003816) (1 +10038161003816100381610038161199102
1003816100381610038161003816)
100381610038161003816100381610038161003816100381610038161003816
le1
10arctan (
10038161003816100381610038161199091 minus 1199092
1003816100381610038161003816)
+1
20
10038161003816100381610038161199101 minus 1199102
1003816100381610038161003816
1 +10038161003816100381610038161199101 minus 119910
2
1003816100381610038161003816
=1
101205721
(10038161003816100381610038161199091 minus 119909
2
1003816100381610038161003816)
+1
201205722
(10038161003816100381610038161199101 minus 119910
2
1003816100381610038161003816)
le1
10max (120572
1 1205722) (
10038161003816100381610038161199091 minus 1199092
1003816100381610038161003816)
+1
10max (120572
1 1205722) (
10038161003816100381610038161199101 minus 1199102
1003816100381610038161003816)
le1
10[2max (120572
1 1205722)
times max (10038161003816100381610038161199091 minus 119909
2
1003816100381610038161003816 10038161003816100381610038161199101 minus 119910
2
1003816100381610038161003816)]
=1
5max (120572
1 1205722) (max (
10038161003816100381610038161199091 minus 1199092
1003816100381610038161003816 10038161003816100381610038161199101 minus 119910
2
1003816100381610038161003816))
(66)
Therefore 1205931(119905) = (15)max(120572
1(119905) 1205722(119905)) and 120593
1isin A
On the other hand1003816100381610038161003816119892 (119905 119909
1 1199101) minus 119892 (119905 119909
2 1199102)1003816100381610038161003816
le1
9
10038161003816100381610038161199091 minus 1199092
1003816100381610038161003816 +1
10
10038161003816100381610038161199092 minus 1199102
1003816100381610038161003816
le1
9(10038161003816100381610038161199091 minus 119910
1
1003816100381610038161003816 +10038161003816100381610038161199102 minus 119910
2
1003816100381610038161003816)
le1
9(2max (
10038161003816100381610038161199091 minus 1199101
1003816100381610038161003816 10038161003816100381610038161199102 minus 119910
2
1003816100381610038161003816))
=2
9max (
10038161003816100381610038161199091 minus 1199101
1003816100381610038161003816 10038161003816100381610038161199102 minus 119910
2
1003816100381610038161003816)
(67)
and 1205932(119905) = (29)119905 It is clear that 120593
2isin A Therefore
assumption (H2) of Theorem 15 is satisfied
In our case the inequality appearing in assumption (H3)
of Theorem 15 has the expression
[1
5max(arctan 119903
119903
1 + 119903) +
1
4] [
2
9119903 +
1
7] le 119903Γ (
3
2) (68)
It is easily seen that 1199030
= 1 satisfies the last inequalityMoreover
2
91199030
+1
7=
2
9+
1
7le Γ (
3
2) cong 088623 (69)
Finally Theorem 15 says that problem (17) has at least onesolution (119909 119910) isin 119862[0 1] such that max(119909 119910) le 1
Conflict of Interests
The authors declare that there is no conflict of interests in thesubmitted paper
References
[1] W Deng ldquoNumerical algorithm for the time fractional Fokker-Planck equationrdquo Journal of Computational Physics vol 227 no2 pp 1510ndash1522 2007
[2] S Z Rida H M El-Sherbiny and A A M Arafa ldquoOn thesolution of the fractional nonlinear Schrodinger equationrdquoPhysics Letters A vol 372 no 5 pp 553ndash558 2008
[3] K Diethelm and A D Freed ldquoOn the solution of nonlinearfractional order differential equations used in the modeling ofviscoplasticityrdquo in Scientific Computing in Chemical EngineeringII pp 217ndash224 Springer Berlin Germany 1999
[4] F Mainardi ldquoFractional calculus some basic proble ms incontinuum and statistical mechanicsrdquo in Fractals and FractionalCalculus in Continuum Mechanics (Udine 1996) vol 378 ofCISM Courses and Lectures pp 291ndash348 Springer ViennaAustria 1997
[5] R Metzler W Schick H Kilian and T F NonnenmacherldquoRelaxation in filled polymers a fractional calculus approachrdquoThe Journal of Chemical Physics vol 103 no 16 pp 7180ndash71861995
[6] B C Dhage and V Lakshmikantham ldquoBasic results on hybriddifferential equationsrdquo Nonlinear Analysis Hybrid Systems vol4 no 3 pp 414ndash424 2010
10 Abstract and Applied Analysis
[7] Y Zhao S Sun Z Han and Q Li ldquoTheory of fractionalhybrid differential equationsrdquo Computers amp Mathematics withApplications vol 62 no 3 pp 1312ndash1324 2011
[8] J Caballero M A Darwish and K Sadarangani ldquoSolvabilityof a fractional hybrid initial value problem with supremum byusingmeasures of noncompactness in Banach algebrasrdquoAppliedMathematics and Computation vol 224 pp 553ndash563 2013
[9] C Bai and J Fang ldquoThe existence of a positive solution fora singular coupled system of nonlinear fractional differentialequationsrdquo Applied Mathematics and Computation vol 150 no3 pp 611ndash621 2004
[10] Y Chen and H An ldquoNumerical solutions of coupled Burgersequations with time- and space-fractional derivativesrdquo AppliedMathematics and Computation vol 200 no 1 pp 87ndash95 2008
[11] M P Lazarevic ldquoFinite time stability analysis of PD120572 fractionalcontrol of robotic time-delay systemsrdquo Mechanics ResearchCommunications vol 33 no 2 pp 269ndash279 2006
[12] V Gafiychuk B Datsko V Meleshko and D Blackmore ldquoAnal-ysis of the solutions of coupled nonlinear fractional reaction-diffusion equationsrdquo Chaos Solitons and Fractals vol 41 no 3pp 1095ndash1104 2009
[13] B Ahmad and J J Nieto ldquoExistence results for a coupledsystem of nonlinear fractional differential equations with three-point boundary conditionsrdquo Computers amp Mathematics withApplications vol 58 no 9 pp 1838ndash1843 2009
[14] A A Kilbas H M Srivastava and J J Trujillo Theory andApplications of Fractional Differential Equations vol 204 ofNorth-Holland Mathematics Studies Elsevier Science Amster-dam The Netherlands 2006
[15] J Banas and K GoebelMeasures of Noncompactness in BanachSpaces Lecture Notes in Pure andAppliedMathematicsMarcelDekker New York NY USA 1980
[16] B N Sadovskii ldquoOn a fixed point principlerdquo Functional Analysisand Its Applications vol 1 pp 151ndash153 1967
[17] J Banas and L Olszowy ldquoOn a class of measures of noncom-pactness in Banach algebras and their application to nonlinearintegral equationsrdquo Journal of Analysis and its Applications vol28 no 4 pp 475ndash498 2009
[18] A Aghajani J Banas and N Sabzali ldquoSome generalizationsof Darbo fixed point theorem and applicationsrdquo Bulletin of theBelgian Mathematical Society vol 20 no 2 pp 345ndash358 2013
[19] R R Akhmerov M I Kamenskii A S Potapov A E Rod-kina and B N Sadovskii Measures of Noncompactness andCondensing Operators vol 55 ofOperatorTheory Advances andApplications Birkhaauser Basel Switzerland 1992
Submit your manuscripts athttpwwwhindawicom
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
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Differential EquationsInternational Journal of
Volume 2014
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OptimizationJournal of
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International Journal of
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Function Spaces
Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of Mathematics and Mathematical Sciences
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Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Algebra
Discrete Dynamics in Nature and Society
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
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Decision SciencesAdvances in
Discrete MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom
Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Stochastic AnalysisInternational Journal of
Abstract and Applied Analysis 7
In order to prove the continuity ofG on 1198611199030
times1198611199030
we have1003816100381610038161003816G (119909 119910) (119905) minus G (119909
0 1199100) (119905)
1003816100381610038161003816
=1
Γ (120572)
100381610038161003816100381610038161003816100381610038161003816
int
119905
0
119892 (119904 119909 (119904) 119910 (119904))
(119905 minus 119904)1minus120572
119889119904
minus int
119905
0
119892 (119904 1199090
(119904) 1199100
(119904))
(119905 minus 119904)1minus120572
119889119904
100381610038161003816100381610038161003816100381610038161003816
le1
Γ (120572)int
119905
0
1003816100381610038161003816119892 (119904 119909 (119904) 119910 (119904)) minus 119892 (119904 1199090
(119904) 1199100
(119904))1003816100381610038161003816
(119905 minus 119904)1minus120572
119889119904
le1
Γ (120572)int
119905
0
1205932
(max (1003816100381610038161003816119909 (119904) minus 119909
0(119904)
1003816100381610038161003816 1003816100381610038161003816119910 (119904) minus 119910
0(119904)
1003816100381610038161003816))
(119905 minus 119904)1minus120572
119889119904
le1
Γ (120572)1205932
(max (1003817100381710038171003817119909 minus 119909
0
1003817100381710038171003817 1003817100381710038171003817119910 minus 119910
0
1003817100381710038171003817)) int
119905
0
119889119904
(119905 minus 119904)1minus120572
le1
Γ (120572 + 1)1205932
(120576)
lt120576
Γ (120572 + 1)
(48)
Therefore1003817100381710038171003817G (119909 119910) minus G (119909
0 1199100)1003817100381710038171003817 lt
120576
Γ (120572 + 1)(49)
and consequentlyG is a continuous operator on 1198611199030
times 1198611199030
In order to prove that G satisfies assumptions of
Theorem 13 only we have to check the condition
120583 (119866 (1198831
times 1198832)) le 120593 (max (120583 (119883
1) 120583 (119883
2))) (50)
for any subsets 1198831and 119883
2of 1198611199030
To do this we fix 120576 gt 0 119905
1 1199052
isin [0 1] with |1199051minus 1199052| le 120576 and
(119909 119910) isin 1198831
times 1198832 then we have
1003816100381610038161003816F (119909 119910) (1199051) minus F (119909 119910) (119905
2)1003816100381610038161003816
=1003816100381610038161003816119891 (1199051 119909 (1199051) 119910 (119905
1)) minus 119891 (119905
2 119909 (1199052) 119910 (119905
2))
1003816100381610038161003816
le1003816100381610038161003816119891 (1199051 119909 (1199051) 119910 (119905
1)) minus 119891 (119905
1 119909 (1199052) 119910 (119905
2))
1003816100381610038161003816
+1003816100381610038161003816119891 (1199051 119909 (1199052) 119910 (119905
2)) minus 119891 (119905
2 119909 (1199052) 119910 (119905
2))
1003816100381610038161003816
le 1205931
(max (1003816100381610038161003816119909 (1199051) minus 119909 (119905
2)1003816100381610038161003816
1003816100381610038161003816119910 (1199051) minus 119910 (119905
2)1003816100381610038161003816))
+ 120596 (119891 120576)
le 1205931
(max (120596 (119909 120576) 120596 (119910 120576))) + 120596 (119891 120576)
(51)
where 120596(119891 120576) denotes the quantity
120596 (119891 120576) = sup 1003816100381610038161003816119891 (119905 119909 119910) minus 119891 (119904 119909 119910)
1003816100381610038161003816 119905 119904 isin [0 1]
|119905 minus 119904| le 120576 119909 119910 isin [minus1199030 1199030]
(52)
From the last estimate we infer that120596 (F (119883
1times 1198832) 120576)
le 1205931
(max (120596 (1198831 120576) 120596 (119883
2 120576))) + 120596 (119891 120576)
(53)
Since 119891(119905 119909 119910) is uniformly continuous on bounded subsetsof [0 1] times R times R we deduce that 120596(119891 120576) rarr 0 as 120576 rarr 0 andtherefore
1205960
(F (1198831
times 1198832)) le lim120576rarr0
1205931
(max (120596 (1198831 120576) 120596 (119883
2 120576)))
(54)
By assumption (H2) since 120593
1is continuous we infer
1205960
(F (1198831
times 1198832)) le 120593
1(max (120596
0(1198831) 1205960
(1198832))) (55)
Now we estimate the quantity 1205960(G(1198831
times 1198832))
Fix 120576 gt 0 1199051 1199052
isin [0 1] with |1199051
minus 1199052| le 120576 and (119909 119910) isin 119883
1times
1198832 Without loss of generality we can suppose that 119905
1lt 1199052
then we have1003816100381610038161003816G (119909 119910) (119905
2) minus G (119909 119910) (119905
1)1003816100381610038161003816
=1
Γ (120572)
1003816100381610038161003816100381610038161003816100381610038161003816
int
1199052
0
119892 (119904 119909 (119904) 119910 (119904))
(1199052
minus 119904)1minus120572
119889119904
minus int
1199051
0
119892 (119904 119909 (119904) 119910 (119904))
(1199051
minus 119904)1minus120572
119889119904
1003816100381610038161003816100381610038161003816100381610038161003816
le1
Γ (120572)[int
1199051
0
10038161003816100381610038161003816(1199052
minus 119904)120572minus1
minus (1199051
minus 119904)120572minus110038161003816100381610038161003816
times1003816100381610038161003816119892 (119904 119909 (119904) 119909 (120590119904))
1003816100381610038161003816 119889119904
+ int
1199052
1199051
(1199052
minus 119904)120572minus1 1003816100381610038161003816119892 (119904 119909 (119904) 119910 (119904))
1003816100381610038161003816 119889119904]
le1
Γ (120572)[int
1199051
0
[(1199051
minus 119904)120572minus1
minus (1199052
minus 119904)120572minus1
]
times1003816100381610038161003816119892 (119904 119909 (119904) 119909 (120590119904))
1003816100381610038161003816 119889119904
+ int
1199052
1199051
(1199052
minus 119904)120572minus1 1003816100381610038161003816119892 (119904 119909 (119904) 119910 (119904))
1003816100381610038161003816 119889119904]
(56)
Since 119892 isin 119862([0 1] times R times R timesR) is bounded on the compactsubsets of [0 1] times R times R particularly on [0 1] times [minus119903
0 1199030] times
[minus1199030 1199030] Put 119871 = sup|119892(119905 119909 119910)| 119905 isin [0 1] 119909 119910 isin [minus119903
0 1199030]
Then from the last inequality we infer that1003816100381610038161003816G (119909 119910) (119905
2) minus G (119909 119910) (119905
1)1003816100381610038161003816
le119871
Γ (120572)[int
1199051
0
[(1199051
minus 119904)120572minus1
minus (1199052
minus 119904)120572minus1
] 119889119904
+ int
1199052
1199051
(1199052
minus 119904)120572minus1
119889119904]
le119871
Γ (120572 + 1)[(1199052
minus 1199051)120572
+ 119905120572
1minus 119905120572
2+ (1199052
minus 1199051)120572
]
le2119871
Γ (120572 + 1)(1199052
minus 1199051)120572
le2119871
Γ (120572 + 1)120576120572
(57)
8 Abstract and Applied Analysis
where we have used the fact that 119905120572
1minus 119905120572
2le 0 Therefore
120596 (G (1198831
times 1198832) 120576) le
2119871
Γ (120572 + 1)120576120572
(58)
From this it follows that
1205960
(G (1198831
times 1198832)) = 0 (59)
Next by Proposition 7 (46) (55) and (59) we have
1205960
(119866 (1198831
times 1198832))
= 1205960
(F (1198831
times 1198832) sdot G (119883
1times 1198832))
le1003817100381710038171003817F (119883
1times 1198832)1003817100381710038171003817 1205960
(G (1198831
times 1198832))
+1003817100381710038171003817G (119883
1times 1198832)1003817100381710038171003817 1205960
(F (1198831
times 1198832))
le10038171003817100381710038171003817F (1198611199030
times 1198611199030
)10038171003817100381710038171003817
1205960
(G (1198831
times 1198832))
+10038171003817100381710038171003817G (1198611199030
times 1198611199030
)10038171003817100381710038171003817
1205960
(F (1198831
times 1198832))
le1205932
(1199030) + 1198962
Γ (120572 + 1)1205931
(max (1205960
(1198831) 1205960
(1198832)))
(60)
By assumption (H3) since 120593
2(1199030) + 1198962
le Γ(120572 + 1) and since itis easily proved that if 120572 isin [0 1] and 120593 isin A then 120572120593 isin A wededuce that
1205960
(119866 (1198831
times 1198832)) le 120593 (max (120596
0(1198831) 1205960
(1198832))) (61)
where 120593 isin AFinally by Theorem 13 the operator G has at least a
coupled fixed point and this is the desired result Thiscompletes the proof
The nonoscillary character of the solutions of problem(4) seems to be an interesting question from the practicalpoint of view This means that the solutions of problem (4)have a constant sign on the interval (0 1) In connection withthis question we notice that if 119891(119905 119909 119910) and 119892(119905 119909 119910) haveconstant sign and are equal (this means that 119891(119905 119909 119910) gt 0
and 119892(119905 119909 119910) ge 0 or 119891(119905 119909 119910) lt 0 and 119892(119905 119909 119910) le 0
for any 119905 isin [0 1] and 119909 119910 isin R) and under assumptionsof Theorem 15 then the solution (119909 119910) isin 119862[0 1] times 119862[0 1]
of problem (4) given by Theorem 15 satisfies 119909(119905) ge 0 and119910(119905) ge 0 for 119905 isin [0 1] since the solution (119909 119910) satisfies thesystem of integral equations
119909 (119905) =119891 (119905 119909 (119905) 119910 (119905))
Γ (120572)int
119905
0
119892 (119904 119909 (119904) 119910 (119904))
(119905 minus 119904)1minus120572
119889119904
119910 (119905) =119891 (119905 119910 (119905) 119909 (119905))
Γ (120572)int
119905
0
119892 (119904 119910 (119904) 119909 (119904))
(119905 minus 119904)1minus120572
119889119904 0 le 119905 le 1
(62)
On the other hand if we perturb the data function in problem(4) of the following manner
119863120572
0+ [
119909 (119905)
119891 (119905 119909 (119905) 119910 (119905))] = 119892 (119905 119909 (119905) 119910 (119905)) + 120578 (119905)
119863120572
0+ [
119910 (119905)
119891 (119905 119910 (119905) 119909 (119905))] = 119892 (119905 119910 (119905) 119909 (119905)) + 120578 (119905)
0 lt 119905 lt 1
(63)
where 0 lt 120572 lt 1 119891 isin 119862([0 1] times R times RR 0)119892 isin 119862([0 1] times R times RR) and 120578 isin 119862[0 1] then underassumptions of Theorem 15 problem (63) can be studiedby using Theorem 15 where assumptions (H
1) and (H
2) are
automatically satisfied and we only have to check assumption(H3) This fact gives a great applicability to Theorem 15Before presenting an example illustrating our results we
need some facts about the functions involving this exampleThe following lemma appears in [18]
Lemma 16 Let 120593 R+
rarr R+be a nondecreasing and upper
semicontinuous function Then the following conditions areequivalent
(i) lim119899rarrinfin
120593119899
(119905) = 0 for any 119905 ge 0
(ii) 120593(119905) lt 119905 for any 119905 gt 0
Particularly the functions 1205721 1205722
R+
rarr R+given by
1205721(119905) = arctan 119905 and 120572
2(119905) = 119905(1 + 119905) belong to the class
A since they are nondecreasing and continuous and as itis easily seen they satisfy (ii) of Lemma 16
On the other hand since the function 1205721(119905) = arctan 119905 is
concave (because 12057210158401015840
(119905) le 0) and 1205721(0) = 0 we infer that 120572
1
is subadditive and therefore for any 119905 1199051015840
isin R+ we have
100381610038161003816100381610038161205721
(119905) minus 1205721
(1199051015840
)10038161003816100381610038161003816
=10038161003816100381610038161003816arctan 119905 minus arctan 119905
101584010038161003816100381610038161003816
le arctan 10038161003816100381610038161003816119905 minus 119905101584010038161003816100381610038161003816
(64)
Moreover it is easily seen that max(1205721 1205722) is a nondecreasing
and continuous function because 1205721and 120572
2are nondecreas-
ing and continuous andmax(1205721 1205722) satisfies (ii) of Lemma 16
Therefore max(1205721 1205722) isin A
Nowwe are ready to present an examplewhere our resultscan be applied
Example 17 Consider the following coupled system of frac-tional hybrid differential equations
11986312
0+ [119909 (119905) times (
1
4+ (
1
10) arctan |119909 (119905)|
+(1
20) (
1003816100381610038161003816119910 (119905)1003816100381610038161003816
(1 +1003816100381610038161003816119910 (119905)
1003816100381610038161003816)))
minus1
]
=1
7+
1
9119909 (119905) +
1
10119910 (119905)
Abstract and Applied Analysis 9
11986312
0+ [119910 (119905) times (
1
4+ (
1
10) arctan 1003816100381610038161003816119910 (119905)
1003816100381610038161003816
+ (1
20) (
|119909 (119905)|
(1 + |119909 (119905)|)))
minus1
]
=1
7+
1
9119910 (119905) +
1
10119909 (119905)
0 lt 119905 lt 1
119909 (0) = 119910 (0) = 0
(65)
Notice that problem (17) is a particular case of problem (4)where 120572 = 12 119891(119905 119909 119910) = 14 + (110) arctan |119909| +
(120)(|119910|(1 + |119910|)) and 119892(119905 119909 119910) = 17 + (19)119909 + (110)119910It is clear that 119891 isin 119862([0 1] times R times RR 0) and 119892 isin
119862([0 1] times R times RR) and moreover 1198961
= sup|119891(119905 0 0)| 119905 isin
[0 1] = 14 and 1198962
= sup|119892(119905 0 0)| 119905 isin [0 1] = 17Therefore assumption (H
1) of Theorem 15 is satisfied
Moreover for 119905 isin [0 1] and 1199091 1199092 1199101 1199102
isin R we have1003816100381610038161003816119891 (119905 119909
1 1199101) minus 119891 (119905 119909
2 1199102)1003816100381610038161003816
le1
10
1003816100381610038161003816arctan10038161003816100381610038161199091
1003816100381610038161003816 minus arctan 100381610038161003816100381611990921003816100381610038161003816
1003816100381610038161003816
+1
20
100381610038161003816100381610038161003816100381610038161003816
100381610038161003816100381611991011003816100381610038161003816
1 +10038161003816100381610038161199101
1003816100381610038161003816
minus
100381610038161003816100381611991021003816100381610038161003816
1 +10038161003816100381610038161199102
1003816100381610038161003816
100381610038161003816100381610038161003816100381610038161003816
le1
10arctan 1003816100381610038161003816
100381610038161003816100381611990911003816100381610038161003816 minus
100381610038161003816100381611990921003816100381610038161003816
1003816100381610038161003816
+1
20
100381610038161003816100381610038161003816100381610038161003816
100381610038161003816100381611991011003816100381610038161003816 minus
100381610038161003816100381611991021003816100381610038161003816
(1 +10038161003816100381610038161199101
1003816100381610038161003816) (1 +10038161003816100381610038161199102
1003816100381610038161003816)
100381610038161003816100381610038161003816100381610038161003816
le1
10arctan (
10038161003816100381610038161199091 minus 1199092
1003816100381610038161003816)
+1
20
10038161003816100381610038161199101 minus 1199102
1003816100381610038161003816
1 +10038161003816100381610038161199101 minus 119910
2
1003816100381610038161003816
=1
101205721
(10038161003816100381610038161199091 minus 119909
2
1003816100381610038161003816)
+1
201205722
(10038161003816100381610038161199101 minus 119910
2
1003816100381610038161003816)
le1
10max (120572
1 1205722) (
10038161003816100381610038161199091 minus 1199092
1003816100381610038161003816)
+1
10max (120572
1 1205722) (
10038161003816100381610038161199101 minus 1199102
1003816100381610038161003816)
le1
10[2max (120572
1 1205722)
times max (10038161003816100381610038161199091 minus 119909
2
1003816100381610038161003816 10038161003816100381610038161199101 minus 119910
2
1003816100381610038161003816)]
=1
5max (120572
1 1205722) (max (
10038161003816100381610038161199091 minus 1199092
1003816100381610038161003816 10038161003816100381610038161199101 minus 119910
2
1003816100381610038161003816))
(66)
Therefore 1205931(119905) = (15)max(120572
1(119905) 1205722(119905)) and 120593
1isin A
On the other hand1003816100381610038161003816119892 (119905 119909
1 1199101) minus 119892 (119905 119909
2 1199102)1003816100381610038161003816
le1
9
10038161003816100381610038161199091 minus 1199092
1003816100381610038161003816 +1
10
10038161003816100381610038161199092 minus 1199102
1003816100381610038161003816
le1
9(10038161003816100381610038161199091 minus 119910
1
1003816100381610038161003816 +10038161003816100381610038161199102 minus 119910
2
1003816100381610038161003816)
le1
9(2max (
10038161003816100381610038161199091 minus 1199101
1003816100381610038161003816 10038161003816100381610038161199102 minus 119910
2
1003816100381610038161003816))
=2
9max (
10038161003816100381610038161199091 minus 1199101
1003816100381610038161003816 10038161003816100381610038161199102 minus 119910
2
1003816100381610038161003816)
(67)
and 1205932(119905) = (29)119905 It is clear that 120593
2isin A Therefore
assumption (H2) of Theorem 15 is satisfied
In our case the inequality appearing in assumption (H3)
of Theorem 15 has the expression
[1
5max(arctan 119903
119903
1 + 119903) +
1
4] [
2
9119903 +
1
7] le 119903Γ (
3
2) (68)
It is easily seen that 1199030
= 1 satisfies the last inequalityMoreover
2
91199030
+1
7=
2
9+
1
7le Γ (
3
2) cong 088623 (69)
Finally Theorem 15 says that problem (17) has at least onesolution (119909 119910) isin 119862[0 1] such that max(119909 119910) le 1
Conflict of Interests
The authors declare that there is no conflict of interests in thesubmitted paper
References
[1] W Deng ldquoNumerical algorithm for the time fractional Fokker-Planck equationrdquo Journal of Computational Physics vol 227 no2 pp 1510ndash1522 2007
[2] S Z Rida H M El-Sherbiny and A A M Arafa ldquoOn thesolution of the fractional nonlinear Schrodinger equationrdquoPhysics Letters A vol 372 no 5 pp 553ndash558 2008
[3] K Diethelm and A D Freed ldquoOn the solution of nonlinearfractional order differential equations used in the modeling ofviscoplasticityrdquo in Scientific Computing in Chemical EngineeringII pp 217ndash224 Springer Berlin Germany 1999
[4] F Mainardi ldquoFractional calculus some basic proble ms incontinuum and statistical mechanicsrdquo in Fractals and FractionalCalculus in Continuum Mechanics (Udine 1996) vol 378 ofCISM Courses and Lectures pp 291ndash348 Springer ViennaAustria 1997
[5] R Metzler W Schick H Kilian and T F NonnenmacherldquoRelaxation in filled polymers a fractional calculus approachrdquoThe Journal of Chemical Physics vol 103 no 16 pp 7180ndash71861995
[6] B C Dhage and V Lakshmikantham ldquoBasic results on hybriddifferential equationsrdquo Nonlinear Analysis Hybrid Systems vol4 no 3 pp 414ndash424 2010
10 Abstract and Applied Analysis
[7] Y Zhao S Sun Z Han and Q Li ldquoTheory of fractionalhybrid differential equationsrdquo Computers amp Mathematics withApplications vol 62 no 3 pp 1312ndash1324 2011
[8] J Caballero M A Darwish and K Sadarangani ldquoSolvabilityof a fractional hybrid initial value problem with supremum byusingmeasures of noncompactness in Banach algebrasrdquoAppliedMathematics and Computation vol 224 pp 553ndash563 2013
[9] C Bai and J Fang ldquoThe existence of a positive solution fora singular coupled system of nonlinear fractional differentialequationsrdquo Applied Mathematics and Computation vol 150 no3 pp 611ndash621 2004
[10] Y Chen and H An ldquoNumerical solutions of coupled Burgersequations with time- and space-fractional derivativesrdquo AppliedMathematics and Computation vol 200 no 1 pp 87ndash95 2008
[11] M P Lazarevic ldquoFinite time stability analysis of PD120572 fractionalcontrol of robotic time-delay systemsrdquo Mechanics ResearchCommunications vol 33 no 2 pp 269ndash279 2006
[12] V Gafiychuk B Datsko V Meleshko and D Blackmore ldquoAnal-ysis of the solutions of coupled nonlinear fractional reaction-diffusion equationsrdquo Chaos Solitons and Fractals vol 41 no 3pp 1095ndash1104 2009
[13] B Ahmad and J J Nieto ldquoExistence results for a coupledsystem of nonlinear fractional differential equations with three-point boundary conditionsrdquo Computers amp Mathematics withApplications vol 58 no 9 pp 1838ndash1843 2009
[14] A A Kilbas H M Srivastava and J J Trujillo Theory andApplications of Fractional Differential Equations vol 204 ofNorth-Holland Mathematics Studies Elsevier Science Amster-dam The Netherlands 2006
[15] J Banas and K GoebelMeasures of Noncompactness in BanachSpaces Lecture Notes in Pure andAppliedMathematicsMarcelDekker New York NY USA 1980
[16] B N Sadovskii ldquoOn a fixed point principlerdquo Functional Analysisand Its Applications vol 1 pp 151ndash153 1967
[17] J Banas and L Olszowy ldquoOn a class of measures of noncom-pactness in Banach algebras and their application to nonlinearintegral equationsrdquo Journal of Analysis and its Applications vol28 no 4 pp 475ndash498 2009
[18] A Aghajani J Banas and N Sabzali ldquoSome generalizationsof Darbo fixed point theorem and applicationsrdquo Bulletin of theBelgian Mathematical Society vol 20 no 2 pp 345ndash358 2013
[19] R R Akhmerov M I Kamenskii A S Potapov A E Rod-kina and B N Sadovskii Measures of Noncompactness andCondensing Operators vol 55 ofOperatorTheory Advances andApplications Birkhaauser Basel Switzerland 1992
Submit your manuscripts athttpwwwhindawicom
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
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Differential EquationsInternational Journal of
Volume 2014
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Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
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Journal of
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Mathematical PhysicsAdvances in
Complex AnalysisJournal of
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OptimizationJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of
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Operations ResearchAdvances in
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Function Spaces
Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of Mathematics and Mathematical Sciences
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Algebra
Discrete Dynamics in Nature and Society
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
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Decision SciencesAdvances in
Discrete MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom
Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Stochastic AnalysisInternational Journal of
8 Abstract and Applied Analysis
where we have used the fact that 119905120572
1minus 119905120572
2le 0 Therefore
120596 (G (1198831
times 1198832) 120576) le
2119871
Γ (120572 + 1)120576120572
(58)
From this it follows that
1205960
(G (1198831
times 1198832)) = 0 (59)
Next by Proposition 7 (46) (55) and (59) we have
1205960
(119866 (1198831
times 1198832))
= 1205960
(F (1198831
times 1198832) sdot G (119883
1times 1198832))
le1003817100381710038171003817F (119883
1times 1198832)1003817100381710038171003817 1205960
(G (1198831
times 1198832))
+1003817100381710038171003817G (119883
1times 1198832)1003817100381710038171003817 1205960
(F (1198831
times 1198832))
le10038171003817100381710038171003817F (1198611199030
times 1198611199030
)10038171003817100381710038171003817
1205960
(G (1198831
times 1198832))
+10038171003817100381710038171003817G (1198611199030
times 1198611199030
)10038171003817100381710038171003817
1205960
(F (1198831
times 1198832))
le1205932
(1199030) + 1198962
Γ (120572 + 1)1205931
(max (1205960
(1198831) 1205960
(1198832)))
(60)
By assumption (H3) since 120593
2(1199030) + 1198962
le Γ(120572 + 1) and since itis easily proved that if 120572 isin [0 1] and 120593 isin A then 120572120593 isin A wededuce that
1205960
(119866 (1198831
times 1198832)) le 120593 (max (120596
0(1198831) 1205960
(1198832))) (61)
where 120593 isin AFinally by Theorem 13 the operator G has at least a
coupled fixed point and this is the desired result Thiscompletes the proof
The nonoscillary character of the solutions of problem(4) seems to be an interesting question from the practicalpoint of view This means that the solutions of problem (4)have a constant sign on the interval (0 1) In connection withthis question we notice that if 119891(119905 119909 119910) and 119892(119905 119909 119910) haveconstant sign and are equal (this means that 119891(119905 119909 119910) gt 0
and 119892(119905 119909 119910) ge 0 or 119891(119905 119909 119910) lt 0 and 119892(119905 119909 119910) le 0
for any 119905 isin [0 1] and 119909 119910 isin R) and under assumptionsof Theorem 15 then the solution (119909 119910) isin 119862[0 1] times 119862[0 1]
of problem (4) given by Theorem 15 satisfies 119909(119905) ge 0 and119910(119905) ge 0 for 119905 isin [0 1] since the solution (119909 119910) satisfies thesystem of integral equations
119909 (119905) =119891 (119905 119909 (119905) 119910 (119905))
Γ (120572)int
119905
0
119892 (119904 119909 (119904) 119910 (119904))
(119905 minus 119904)1minus120572
119889119904
119910 (119905) =119891 (119905 119910 (119905) 119909 (119905))
Γ (120572)int
119905
0
119892 (119904 119910 (119904) 119909 (119904))
(119905 minus 119904)1minus120572
119889119904 0 le 119905 le 1
(62)
On the other hand if we perturb the data function in problem(4) of the following manner
119863120572
0+ [
119909 (119905)
119891 (119905 119909 (119905) 119910 (119905))] = 119892 (119905 119909 (119905) 119910 (119905)) + 120578 (119905)
119863120572
0+ [
119910 (119905)
119891 (119905 119910 (119905) 119909 (119905))] = 119892 (119905 119910 (119905) 119909 (119905)) + 120578 (119905)
0 lt 119905 lt 1
(63)
where 0 lt 120572 lt 1 119891 isin 119862([0 1] times R times RR 0)119892 isin 119862([0 1] times R times RR) and 120578 isin 119862[0 1] then underassumptions of Theorem 15 problem (63) can be studiedby using Theorem 15 where assumptions (H
1) and (H
2) are
automatically satisfied and we only have to check assumption(H3) This fact gives a great applicability to Theorem 15Before presenting an example illustrating our results we
need some facts about the functions involving this exampleThe following lemma appears in [18]
Lemma 16 Let 120593 R+
rarr R+be a nondecreasing and upper
semicontinuous function Then the following conditions areequivalent
(i) lim119899rarrinfin
120593119899
(119905) = 0 for any 119905 ge 0
(ii) 120593(119905) lt 119905 for any 119905 gt 0
Particularly the functions 1205721 1205722
R+
rarr R+given by
1205721(119905) = arctan 119905 and 120572
2(119905) = 119905(1 + 119905) belong to the class
A since they are nondecreasing and continuous and as itis easily seen they satisfy (ii) of Lemma 16
On the other hand since the function 1205721(119905) = arctan 119905 is
concave (because 12057210158401015840
(119905) le 0) and 1205721(0) = 0 we infer that 120572
1
is subadditive and therefore for any 119905 1199051015840
isin R+ we have
100381610038161003816100381610038161205721
(119905) minus 1205721
(1199051015840
)10038161003816100381610038161003816
=10038161003816100381610038161003816arctan 119905 minus arctan 119905
101584010038161003816100381610038161003816
le arctan 10038161003816100381610038161003816119905 minus 119905101584010038161003816100381610038161003816
(64)
Moreover it is easily seen that max(1205721 1205722) is a nondecreasing
and continuous function because 1205721and 120572
2are nondecreas-
ing and continuous andmax(1205721 1205722) satisfies (ii) of Lemma 16
Therefore max(1205721 1205722) isin A
Nowwe are ready to present an examplewhere our resultscan be applied
Example 17 Consider the following coupled system of frac-tional hybrid differential equations
11986312
0+ [119909 (119905) times (
1
4+ (
1
10) arctan |119909 (119905)|
+(1
20) (
1003816100381610038161003816119910 (119905)1003816100381610038161003816
(1 +1003816100381610038161003816119910 (119905)
1003816100381610038161003816)))
minus1
]
=1
7+
1
9119909 (119905) +
1
10119910 (119905)
Abstract and Applied Analysis 9
11986312
0+ [119910 (119905) times (
1
4+ (
1
10) arctan 1003816100381610038161003816119910 (119905)
1003816100381610038161003816
+ (1
20) (
|119909 (119905)|
(1 + |119909 (119905)|)))
minus1
]
=1
7+
1
9119910 (119905) +
1
10119909 (119905)
0 lt 119905 lt 1
119909 (0) = 119910 (0) = 0
(65)
Notice that problem (17) is a particular case of problem (4)where 120572 = 12 119891(119905 119909 119910) = 14 + (110) arctan |119909| +
(120)(|119910|(1 + |119910|)) and 119892(119905 119909 119910) = 17 + (19)119909 + (110)119910It is clear that 119891 isin 119862([0 1] times R times RR 0) and 119892 isin
119862([0 1] times R times RR) and moreover 1198961
= sup|119891(119905 0 0)| 119905 isin
[0 1] = 14 and 1198962
= sup|119892(119905 0 0)| 119905 isin [0 1] = 17Therefore assumption (H
1) of Theorem 15 is satisfied
Moreover for 119905 isin [0 1] and 1199091 1199092 1199101 1199102
isin R we have1003816100381610038161003816119891 (119905 119909
1 1199101) minus 119891 (119905 119909
2 1199102)1003816100381610038161003816
le1
10
1003816100381610038161003816arctan10038161003816100381610038161199091
1003816100381610038161003816 minus arctan 100381610038161003816100381611990921003816100381610038161003816
1003816100381610038161003816
+1
20
100381610038161003816100381610038161003816100381610038161003816
100381610038161003816100381611991011003816100381610038161003816
1 +10038161003816100381610038161199101
1003816100381610038161003816
minus
100381610038161003816100381611991021003816100381610038161003816
1 +10038161003816100381610038161199102
1003816100381610038161003816
100381610038161003816100381610038161003816100381610038161003816
le1
10arctan 1003816100381610038161003816
100381610038161003816100381611990911003816100381610038161003816 minus
100381610038161003816100381611990921003816100381610038161003816
1003816100381610038161003816
+1
20
100381610038161003816100381610038161003816100381610038161003816
100381610038161003816100381611991011003816100381610038161003816 minus
100381610038161003816100381611991021003816100381610038161003816
(1 +10038161003816100381610038161199101
1003816100381610038161003816) (1 +10038161003816100381610038161199102
1003816100381610038161003816)
100381610038161003816100381610038161003816100381610038161003816
le1
10arctan (
10038161003816100381610038161199091 minus 1199092
1003816100381610038161003816)
+1
20
10038161003816100381610038161199101 minus 1199102
1003816100381610038161003816
1 +10038161003816100381610038161199101 minus 119910
2
1003816100381610038161003816
=1
101205721
(10038161003816100381610038161199091 minus 119909
2
1003816100381610038161003816)
+1
201205722
(10038161003816100381610038161199101 minus 119910
2
1003816100381610038161003816)
le1
10max (120572
1 1205722) (
10038161003816100381610038161199091 minus 1199092
1003816100381610038161003816)
+1
10max (120572
1 1205722) (
10038161003816100381610038161199101 minus 1199102
1003816100381610038161003816)
le1
10[2max (120572
1 1205722)
times max (10038161003816100381610038161199091 minus 119909
2
1003816100381610038161003816 10038161003816100381610038161199101 minus 119910
2
1003816100381610038161003816)]
=1
5max (120572
1 1205722) (max (
10038161003816100381610038161199091 minus 1199092
1003816100381610038161003816 10038161003816100381610038161199101 minus 119910
2
1003816100381610038161003816))
(66)
Therefore 1205931(119905) = (15)max(120572
1(119905) 1205722(119905)) and 120593
1isin A
On the other hand1003816100381610038161003816119892 (119905 119909
1 1199101) minus 119892 (119905 119909
2 1199102)1003816100381610038161003816
le1
9
10038161003816100381610038161199091 minus 1199092
1003816100381610038161003816 +1
10
10038161003816100381610038161199092 minus 1199102
1003816100381610038161003816
le1
9(10038161003816100381610038161199091 minus 119910
1
1003816100381610038161003816 +10038161003816100381610038161199102 minus 119910
2
1003816100381610038161003816)
le1
9(2max (
10038161003816100381610038161199091 minus 1199101
1003816100381610038161003816 10038161003816100381610038161199102 minus 119910
2
1003816100381610038161003816))
=2
9max (
10038161003816100381610038161199091 minus 1199101
1003816100381610038161003816 10038161003816100381610038161199102 minus 119910
2
1003816100381610038161003816)
(67)
and 1205932(119905) = (29)119905 It is clear that 120593
2isin A Therefore
assumption (H2) of Theorem 15 is satisfied
In our case the inequality appearing in assumption (H3)
of Theorem 15 has the expression
[1
5max(arctan 119903
119903
1 + 119903) +
1
4] [
2
9119903 +
1
7] le 119903Γ (
3
2) (68)
It is easily seen that 1199030
= 1 satisfies the last inequalityMoreover
2
91199030
+1
7=
2
9+
1
7le Γ (
3
2) cong 088623 (69)
Finally Theorem 15 says that problem (17) has at least onesolution (119909 119910) isin 119862[0 1] such that max(119909 119910) le 1
Conflict of Interests
The authors declare that there is no conflict of interests in thesubmitted paper
References
[1] W Deng ldquoNumerical algorithm for the time fractional Fokker-Planck equationrdquo Journal of Computational Physics vol 227 no2 pp 1510ndash1522 2007
[2] S Z Rida H M El-Sherbiny and A A M Arafa ldquoOn thesolution of the fractional nonlinear Schrodinger equationrdquoPhysics Letters A vol 372 no 5 pp 553ndash558 2008
[3] K Diethelm and A D Freed ldquoOn the solution of nonlinearfractional order differential equations used in the modeling ofviscoplasticityrdquo in Scientific Computing in Chemical EngineeringII pp 217ndash224 Springer Berlin Germany 1999
[4] F Mainardi ldquoFractional calculus some basic proble ms incontinuum and statistical mechanicsrdquo in Fractals and FractionalCalculus in Continuum Mechanics (Udine 1996) vol 378 ofCISM Courses and Lectures pp 291ndash348 Springer ViennaAustria 1997
[5] R Metzler W Schick H Kilian and T F NonnenmacherldquoRelaxation in filled polymers a fractional calculus approachrdquoThe Journal of Chemical Physics vol 103 no 16 pp 7180ndash71861995
[6] B C Dhage and V Lakshmikantham ldquoBasic results on hybriddifferential equationsrdquo Nonlinear Analysis Hybrid Systems vol4 no 3 pp 414ndash424 2010
10 Abstract and Applied Analysis
[7] Y Zhao S Sun Z Han and Q Li ldquoTheory of fractionalhybrid differential equationsrdquo Computers amp Mathematics withApplications vol 62 no 3 pp 1312ndash1324 2011
[8] J Caballero M A Darwish and K Sadarangani ldquoSolvabilityof a fractional hybrid initial value problem with supremum byusingmeasures of noncompactness in Banach algebrasrdquoAppliedMathematics and Computation vol 224 pp 553ndash563 2013
[9] C Bai and J Fang ldquoThe existence of a positive solution fora singular coupled system of nonlinear fractional differentialequationsrdquo Applied Mathematics and Computation vol 150 no3 pp 611ndash621 2004
[10] Y Chen and H An ldquoNumerical solutions of coupled Burgersequations with time- and space-fractional derivativesrdquo AppliedMathematics and Computation vol 200 no 1 pp 87ndash95 2008
[11] M P Lazarevic ldquoFinite time stability analysis of PD120572 fractionalcontrol of robotic time-delay systemsrdquo Mechanics ResearchCommunications vol 33 no 2 pp 269ndash279 2006
[12] V Gafiychuk B Datsko V Meleshko and D Blackmore ldquoAnal-ysis of the solutions of coupled nonlinear fractional reaction-diffusion equationsrdquo Chaos Solitons and Fractals vol 41 no 3pp 1095ndash1104 2009
[13] B Ahmad and J J Nieto ldquoExistence results for a coupledsystem of nonlinear fractional differential equations with three-point boundary conditionsrdquo Computers amp Mathematics withApplications vol 58 no 9 pp 1838ndash1843 2009
[14] A A Kilbas H M Srivastava and J J Trujillo Theory andApplications of Fractional Differential Equations vol 204 ofNorth-Holland Mathematics Studies Elsevier Science Amster-dam The Netherlands 2006
[15] J Banas and K GoebelMeasures of Noncompactness in BanachSpaces Lecture Notes in Pure andAppliedMathematicsMarcelDekker New York NY USA 1980
[16] B N Sadovskii ldquoOn a fixed point principlerdquo Functional Analysisand Its Applications vol 1 pp 151ndash153 1967
[17] J Banas and L Olszowy ldquoOn a class of measures of noncom-pactness in Banach algebras and their application to nonlinearintegral equationsrdquo Journal of Analysis and its Applications vol28 no 4 pp 475ndash498 2009
[18] A Aghajani J Banas and N Sabzali ldquoSome generalizationsof Darbo fixed point theorem and applicationsrdquo Bulletin of theBelgian Mathematical Society vol 20 no 2 pp 345ndash358 2013
[19] R R Akhmerov M I Kamenskii A S Potapov A E Rod-kina and B N Sadovskii Measures of Noncompactness andCondensing Operators vol 55 ofOperatorTheory Advances andApplications Birkhaauser Basel Switzerland 1992
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 9
11986312
0+ [119910 (119905) times (
1
4+ (
1
10) arctan 1003816100381610038161003816119910 (119905)
1003816100381610038161003816
+ (1
20) (
|119909 (119905)|
(1 + |119909 (119905)|)))
minus1
]
=1
7+
1
9119910 (119905) +
1
10119909 (119905)
0 lt 119905 lt 1
119909 (0) = 119910 (0) = 0
(65)
Notice that problem (17) is a particular case of problem (4)where 120572 = 12 119891(119905 119909 119910) = 14 + (110) arctan |119909| +
(120)(|119910|(1 + |119910|)) and 119892(119905 119909 119910) = 17 + (19)119909 + (110)119910It is clear that 119891 isin 119862([0 1] times R times RR 0) and 119892 isin
119862([0 1] times R times RR) and moreover 1198961
= sup|119891(119905 0 0)| 119905 isin
[0 1] = 14 and 1198962
= sup|119892(119905 0 0)| 119905 isin [0 1] = 17Therefore assumption (H
1) of Theorem 15 is satisfied
Moreover for 119905 isin [0 1] and 1199091 1199092 1199101 1199102
isin R we have1003816100381610038161003816119891 (119905 119909
1 1199101) minus 119891 (119905 119909
2 1199102)1003816100381610038161003816
le1
10
1003816100381610038161003816arctan10038161003816100381610038161199091
1003816100381610038161003816 minus arctan 100381610038161003816100381611990921003816100381610038161003816
1003816100381610038161003816
+1
20
100381610038161003816100381610038161003816100381610038161003816
100381610038161003816100381611991011003816100381610038161003816
1 +10038161003816100381610038161199101
1003816100381610038161003816
minus
100381610038161003816100381611991021003816100381610038161003816
1 +10038161003816100381610038161199102
1003816100381610038161003816
100381610038161003816100381610038161003816100381610038161003816
le1
10arctan 1003816100381610038161003816
100381610038161003816100381611990911003816100381610038161003816 minus
100381610038161003816100381611990921003816100381610038161003816
1003816100381610038161003816
+1
20
100381610038161003816100381610038161003816100381610038161003816
100381610038161003816100381611991011003816100381610038161003816 minus
100381610038161003816100381611991021003816100381610038161003816
(1 +10038161003816100381610038161199101
1003816100381610038161003816) (1 +10038161003816100381610038161199102
1003816100381610038161003816)
100381610038161003816100381610038161003816100381610038161003816
le1
10arctan (
10038161003816100381610038161199091 minus 1199092
1003816100381610038161003816)
+1
20
10038161003816100381610038161199101 minus 1199102
1003816100381610038161003816
1 +10038161003816100381610038161199101 minus 119910
2
1003816100381610038161003816
=1
101205721
(10038161003816100381610038161199091 minus 119909
2
1003816100381610038161003816)
+1
201205722
(10038161003816100381610038161199101 minus 119910
2
1003816100381610038161003816)
le1
10max (120572
1 1205722) (
10038161003816100381610038161199091 minus 1199092
1003816100381610038161003816)
+1
10max (120572
1 1205722) (
10038161003816100381610038161199101 minus 1199102
1003816100381610038161003816)
le1
10[2max (120572
1 1205722)
times max (10038161003816100381610038161199091 minus 119909
2
1003816100381610038161003816 10038161003816100381610038161199101 minus 119910
2
1003816100381610038161003816)]
=1
5max (120572
1 1205722) (max (
10038161003816100381610038161199091 minus 1199092
1003816100381610038161003816 10038161003816100381610038161199101 minus 119910
2
1003816100381610038161003816))
(66)
Therefore 1205931(119905) = (15)max(120572
1(119905) 1205722(119905)) and 120593
1isin A
On the other hand1003816100381610038161003816119892 (119905 119909
1 1199101) minus 119892 (119905 119909
2 1199102)1003816100381610038161003816
le1
9
10038161003816100381610038161199091 minus 1199092
1003816100381610038161003816 +1
10
10038161003816100381610038161199092 minus 1199102
1003816100381610038161003816
le1
9(10038161003816100381610038161199091 minus 119910
1
1003816100381610038161003816 +10038161003816100381610038161199102 minus 119910
2
1003816100381610038161003816)
le1
9(2max (
10038161003816100381610038161199091 minus 1199101
1003816100381610038161003816 10038161003816100381610038161199102 minus 119910
2
1003816100381610038161003816))
=2
9max (
10038161003816100381610038161199091 minus 1199101
1003816100381610038161003816 10038161003816100381610038161199102 minus 119910
2
1003816100381610038161003816)
(67)
and 1205932(119905) = (29)119905 It is clear that 120593
2isin A Therefore
assumption (H2) of Theorem 15 is satisfied
In our case the inequality appearing in assumption (H3)
of Theorem 15 has the expression
[1
5max(arctan 119903
119903
1 + 119903) +
1
4] [
2
9119903 +
1
7] le 119903Γ (
3
2) (68)
It is easily seen that 1199030
= 1 satisfies the last inequalityMoreover
2
91199030
+1
7=
2
9+
1
7le Γ (
3
2) cong 088623 (69)
Finally Theorem 15 says that problem (17) has at least onesolution (119909 119910) isin 119862[0 1] such that max(119909 119910) le 1
Conflict of Interests
The authors declare that there is no conflict of interests in thesubmitted paper
References
[1] W Deng ldquoNumerical algorithm for the time fractional Fokker-Planck equationrdquo Journal of Computational Physics vol 227 no2 pp 1510ndash1522 2007
[2] S Z Rida H M El-Sherbiny and A A M Arafa ldquoOn thesolution of the fractional nonlinear Schrodinger equationrdquoPhysics Letters A vol 372 no 5 pp 553ndash558 2008
[3] K Diethelm and A D Freed ldquoOn the solution of nonlinearfractional order differential equations used in the modeling ofviscoplasticityrdquo in Scientific Computing in Chemical EngineeringII pp 217ndash224 Springer Berlin Germany 1999
[4] F Mainardi ldquoFractional calculus some basic proble ms incontinuum and statistical mechanicsrdquo in Fractals and FractionalCalculus in Continuum Mechanics (Udine 1996) vol 378 ofCISM Courses and Lectures pp 291ndash348 Springer ViennaAustria 1997
[5] R Metzler W Schick H Kilian and T F NonnenmacherldquoRelaxation in filled polymers a fractional calculus approachrdquoThe Journal of Chemical Physics vol 103 no 16 pp 7180ndash71861995
[6] B C Dhage and V Lakshmikantham ldquoBasic results on hybriddifferential equationsrdquo Nonlinear Analysis Hybrid Systems vol4 no 3 pp 414ndash424 2010
10 Abstract and Applied Analysis
[7] Y Zhao S Sun Z Han and Q Li ldquoTheory of fractionalhybrid differential equationsrdquo Computers amp Mathematics withApplications vol 62 no 3 pp 1312ndash1324 2011
[8] J Caballero M A Darwish and K Sadarangani ldquoSolvabilityof a fractional hybrid initial value problem with supremum byusingmeasures of noncompactness in Banach algebrasrdquoAppliedMathematics and Computation vol 224 pp 553ndash563 2013
[9] C Bai and J Fang ldquoThe existence of a positive solution fora singular coupled system of nonlinear fractional differentialequationsrdquo Applied Mathematics and Computation vol 150 no3 pp 611ndash621 2004
[10] Y Chen and H An ldquoNumerical solutions of coupled Burgersequations with time- and space-fractional derivativesrdquo AppliedMathematics and Computation vol 200 no 1 pp 87ndash95 2008
[11] M P Lazarevic ldquoFinite time stability analysis of PD120572 fractionalcontrol of robotic time-delay systemsrdquo Mechanics ResearchCommunications vol 33 no 2 pp 269ndash279 2006
[12] V Gafiychuk B Datsko V Meleshko and D Blackmore ldquoAnal-ysis of the solutions of coupled nonlinear fractional reaction-diffusion equationsrdquo Chaos Solitons and Fractals vol 41 no 3pp 1095ndash1104 2009
[13] B Ahmad and J J Nieto ldquoExistence results for a coupledsystem of nonlinear fractional differential equations with three-point boundary conditionsrdquo Computers amp Mathematics withApplications vol 58 no 9 pp 1838ndash1843 2009
[14] A A Kilbas H M Srivastava and J J Trujillo Theory andApplications of Fractional Differential Equations vol 204 ofNorth-Holland Mathematics Studies Elsevier Science Amster-dam The Netherlands 2006
[15] J Banas and K GoebelMeasures of Noncompactness in BanachSpaces Lecture Notes in Pure andAppliedMathematicsMarcelDekker New York NY USA 1980
[16] B N Sadovskii ldquoOn a fixed point principlerdquo Functional Analysisand Its Applications vol 1 pp 151ndash153 1967
[17] J Banas and L Olszowy ldquoOn a class of measures of noncom-pactness in Banach algebras and their application to nonlinearintegral equationsrdquo Journal of Analysis and its Applications vol28 no 4 pp 475ndash498 2009
[18] A Aghajani J Banas and N Sabzali ldquoSome generalizationsof Darbo fixed point theorem and applicationsrdquo Bulletin of theBelgian Mathematical Society vol 20 no 2 pp 345ndash358 2013
[19] R R Akhmerov M I Kamenskii A S Potapov A E Rod-kina and B N Sadovskii Measures of Noncompactness andCondensing Operators vol 55 ofOperatorTheory Advances andApplications Birkhaauser Basel Switzerland 1992
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
10 Abstract and Applied Analysis
[7] Y Zhao S Sun Z Han and Q Li ldquoTheory of fractionalhybrid differential equationsrdquo Computers amp Mathematics withApplications vol 62 no 3 pp 1312ndash1324 2011
[8] J Caballero M A Darwish and K Sadarangani ldquoSolvabilityof a fractional hybrid initial value problem with supremum byusingmeasures of noncompactness in Banach algebrasrdquoAppliedMathematics and Computation vol 224 pp 553ndash563 2013
[9] C Bai and J Fang ldquoThe existence of a positive solution fora singular coupled system of nonlinear fractional differentialequationsrdquo Applied Mathematics and Computation vol 150 no3 pp 611ndash621 2004
[10] Y Chen and H An ldquoNumerical solutions of coupled Burgersequations with time- and space-fractional derivativesrdquo AppliedMathematics and Computation vol 200 no 1 pp 87ndash95 2008
[11] M P Lazarevic ldquoFinite time stability analysis of PD120572 fractionalcontrol of robotic time-delay systemsrdquo Mechanics ResearchCommunications vol 33 no 2 pp 269ndash279 2006
[12] V Gafiychuk B Datsko V Meleshko and D Blackmore ldquoAnal-ysis of the solutions of coupled nonlinear fractional reaction-diffusion equationsrdquo Chaos Solitons and Fractals vol 41 no 3pp 1095ndash1104 2009
[13] B Ahmad and J J Nieto ldquoExistence results for a coupledsystem of nonlinear fractional differential equations with three-point boundary conditionsrdquo Computers amp Mathematics withApplications vol 58 no 9 pp 1838ndash1843 2009
[14] A A Kilbas H M Srivastava and J J Trujillo Theory andApplications of Fractional Differential Equations vol 204 ofNorth-Holland Mathematics Studies Elsevier Science Amster-dam The Netherlands 2006
[15] J Banas and K GoebelMeasures of Noncompactness in BanachSpaces Lecture Notes in Pure andAppliedMathematicsMarcelDekker New York NY USA 1980
[16] B N Sadovskii ldquoOn a fixed point principlerdquo Functional Analysisand Its Applications vol 1 pp 151ndash153 1967
[17] J Banas and L Olszowy ldquoOn a class of measures of noncom-pactness in Banach algebras and their application to nonlinearintegral equationsrdquo Journal of Analysis and its Applications vol28 no 4 pp 475ndash498 2009
[18] A Aghajani J Banas and N Sabzali ldquoSome generalizationsof Darbo fixed point theorem and applicationsrdquo Bulletin of theBelgian Mathematical Society vol 20 no 2 pp 345ndash358 2013
[19] R R Akhmerov M I Kamenskii A S Potapov A E Rod-kina and B N Sadovskii Measures of Noncompactness andCondensing Operators vol 55 ofOperatorTheory Advances andApplications Birkhaauser Basel Switzerland 1992
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