+ All Categories
Home > Documents > Research Article On Coupled Systems of Time-Fractional...

Research Article On Coupled Systems of Time-Fractional...

Date post: 15-Jul-2020
Category:

Documents
Upload: others
View: 4 times
Download: 0 times
Download Report this document
Share this document with a friend
9
Research Article On Coupled Systems of Time-Fractional Differential Problems by Using a New Fractional Derivative Ahmed Alsaedi, 1 Dumitru Baleanu, 2,3 Sina Etemad, 4 and Shahram Rezapour 4 1 Department of Mathematics, Faculty of Science, King Abdulaziz University, Jeddah 21589, Saudi Arabia 2 Department of Mathematics, Cankaya University, Ogretmenler Caddesi 14, Balgat, 06530 Ankara, Turkey 3 Institute of Space Sciences, Magurele, Bucharest, Romania 4 Department of Mathematics, Azarbaijan Shahid Madani University, Tabriz, Iran Correspondence should be addressed to Dumitru Baleanu; [email protected] Received 19 August 2015; Revised 12 October 2015; Accepted 27 October 2015 Academic Editor: Ismat Beg Copyright © 2016 Ahmed Alsaedi et al. is is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited. e existence of solutions for a coupled system of time-fractional differential equations including continuous functions and the Caputo-Fabrizio fractional derivative is examined. Aſter that we investigated a coupled system of time-fractional differential inclusions including compact- and convex-valued 1 -Caratheodory multifunctions and the Caputo-Fabrizio fractional derivative. 1. Introduction e fractional calculus is nowadays an excellent mathemat- ical tool which opens the gates for finding hidden aspects of the dynamics of the complex processes which appear naturally in many branches of science and engineering [1– 6]. e methods and techniques of this type of calculus are continuously generalized and improved especially during the last few decades. We recall that the existence and multiplicity of positive solutions corresponding to singular fractional boundary value problems were discussed in [7]. Also, the existence results for several nonlinear fractional differential equations were reported in [8]. Besides, the existence of positive solutions corresponding to a coupled system of multiterm singular fractional integrodifferential boundary value problems was shown in [9]. Inventing new derivatives and applying them to study the dynamics of complex systems are an important priority for researchers. As a result, very recently, a new fractional derivative without singular kernel has been provided [10, 11]. By using the main results presented in these two new works, we present the next definition. Definition 1 (see [10]). e order Caputo-Fabrizio time- fractional differential derivative of the function is written as ( CF ) (, ) = (2 − ) () 2 (1 − ) ∫ 0 exp [− ( − ) 1− ] , ( ≥ 0) , (1) where () represents a normalization function, 0<<1, and ∈ 1 [(0, 1) × (0, 1)]. Note that ( CF )(, ) = 0 whenever is a constant function and the kernel has no singularity at = [10, 11]. Also, Losada and Nieto defined the new time-fractional integral based on the new definition of Caputo-Fabrizio fractional derivative [11]. By using this idea, we provide the notion of Caputo-Fabrizio time-fractional integral. Hindawi Publishing Corporation Journal of Function Spaces Volume 2016, Article ID 4626940, 8 pages http://dx.doi.org/10.1155/2016/4626940
Transcript
Page 1: Research Article On Coupled Systems of Time-Fractional ...downloads.hindawi.com/journals/jfs/2016/4626940.pdf · Research Article On Coupled Systems of Time-Fractional Differential

Research ArticleOn Coupled Systems of Time-Fractional Differential Problemsby Using a New Fractional Derivative

Ahmed Alsaedi1 Dumitru Baleanu23 Sina Etemad4 and Shahram Rezapour4

1Department of Mathematics Faculty of Science King Abdulaziz University Jeddah 21589 Saudi Arabia2Department of Mathematics Cankaya University Ogretmenler Caddesi 14 Balgat 06530 Ankara Turkey3Institute of Space Sciences Magurele Bucharest Romania4Department of Mathematics Azarbaijan Shahid Madani University Tabriz Iran

Correspondence should be addressed to Dumitru Baleanu dumitrucankayaedutr

Received 19 August 2015 Revised 12 October 2015 Accepted 27 October 2015

Academic Editor Ismat Beg

Copyright copy 2016 Ahmed Alsaedi et alThis is an open access article distributed under the Creative Commons Attribution Licensewhich permits unrestricted use distribution and reproduction in any medium provided the original work is properly cited

The existence of solutions for a coupled system of time-fractional differential equations including continuous functions and theCaputo-Fabrizio fractional derivative is examined After that we investigated a coupled system of time-fractional differentialinclusions including compact- and convex-valued 1198711-Caratheodory multifunctions and the Caputo-Fabrizio fractional derivative

1 Introduction

The fractional calculus is nowadays an excellent mathemat-ical tool which opens the gates for finding hidden aspectsof the dynamics of the complex processes which appearnaturally in many branches of science and engineering [1ndash6] The methods and techniques of this type of calculus arecontinuously generalized and improved especially during thelast few decades We recall that the existence and multiplicityof positive solutions corresponding to singular fractionalboundary value problems were discussed in [7] Also theexistence results for several nonlinear fractional differentialequations were reported in [8] Besides the existence ofpositive solutions corresponding to a coupled system ofmultiterm singular fractional integrodifferential boundaryvalue problems was shown in [9] Inventing new derivativesand applying them to study the dynamics of complex systemsare an important priority for researchers As a result veryrecently a new fractional derivative without singular kernelhas been provided [10 11] By using themain results presentedin these two new works we present the next definition

Definition 1 (see [10]) The 120572 order Caputo-Fabrizio time-fractional differential derivative of the function 119906 is writtenas

(CF119863120572

119905119906) (119909 119905)

=(2 minus 120572)119872 (120572)

2 (1 minus 120572)int

119905

0

exp [minus120572 (119905 minus 119904)1 minus 120572

]120597119906

120597119905119889119904

(119905 ge 0)

(1)

where119872(120572) represents a normalization function 0 lt 120572 lt 1and 119906 isin 1198671[(0 1) times (0 1)]

Note that (CF119863120572119905119906)(119909 119905) = 0 whenever 119906 is a constant

function and the kernel has no singularity at 119905 = 119904 [1011] Also Losada and Nieto defined the new time-fractionalintegral based on the new definition of Caputo-Fabriziofractional derivative [11] By using this idea we provide thenotion of Caputo-Fabrizio time-fractional integral

Hindawi Publishing CorporationJournal of Function SpacesVolume 2016 Article ID 4626940 8 pageshttpdxdoiorg10115520164626940

2 Journal of Function Spaces

Definition 2 The 120572 order time-fractional integral of a func-tion 119906 has the form [11]

(CF119868120572

119905119906) (119909 119905) =

2 (1 minus 120572)

(2 minus 120572)119872 (120572)119906 (119909 119905)

+2120572

(2 minus 120572)119872 (120572)int

119905

0

119906 (119909 119904) 119889119904

(119905 ge 0)

(2)

where119872(120572) represents a normalization function and 0 lt 120572 lt1

Losada and Nieto showed that119872(120572) = 2(2 minus 120572) for all0 le 120572 le 1 [11] By substituting 119872(120572) in (1) we obtain thedefinition of the time-fractional Caputo-Fabrizio derivativeof order 120572 for a function 119906 as follows

(CF119863120572

119905119906) (119909 119905) =

1

1 minus 120572int

119905

0

exp [minus120572 (119905 minus 119904)1 minus 120572

]120597119906

120597119904119889119904

((119909 119905) isin [0 1] times [0 1])

(3)

They proved that solution of (CF119863120572119905V)(119909 119905) = 119892(119909 119905) is given

by

V (119909 119905) =2 (1 minus 120572)

(2 minus 120572)119872 (120572)119892 (119909 119905)

+2120572

(2 minus 120572)119872 (120572)int

119905

0

119892 (119909 119904) 119889119904 + V (0 0) (4)

where 0 lt 120572 lt 1 and (119909 119905) isin [0 1] times [0 1] [11]The next step is to consider (119883 119889) being a metric space

Let us denote by P(119883) and 2119883 the class of all subsets andthe class of all nonempty subsets of 119883 respectively HencePcl(119883)Pbd(119883)Pcv(119883)Pcp(119883) andPcpcv(119883) are the classof all closed subsets the class of all bounded subsets the classof all convex subsets the class of all compact subsets and theclass of all compact and convex subsets of 119883 respectivelyWe claim that 119906 isin 119883 is a fixed point of the multifunction119865 119883 rarr 2

119883 whenever 119906 isin 119865119906 [12] A multifunction 119865

[0 1] times [0 1] rarr Pcl(R) is called measurable whenever thefunction (119909 119905) 997891rarr 119889(119908 119865(119909 119905)) = inf 119908 minus V V isin 119865(119909 119905) ismeasurable for all119908 isin R [12]ThePompeiu-Hausdorffmetric119867119889 2119883times 2119883rarr [0infin) is defined by

119867119889(119860 119861) = maxsup

119886isin119860

119889 (119886 119861) sup119887isin119861

119889 (119860 119887) (5)

such that 119889(119860 119887) = inf119886isin119860

119889(119886 119887) [13] (CB(119883)119867119889) is a

metric space and (CB(119883)119867119889) depicts a generalized metric

space Here CB(119883) denotes the set of closed and boundedsubsets of119883 and119862(119883) represents the set of closed subsets of119883[12 13]We recall that119865 is said to be convex-valued (compact-valued) whenever 119865119906 is convex (compact) set for each 119906 isin 119883[12] We mention that a multifunction 119865 119883 rarr 119862(119883) is acontraction whenever there exists a constant 120574 isin (0 1) suchthat 119867

119889(119865119906 119865V) le 120574119889(119906 V) for all 119906 V isin 119883 [12] In 1970

Covitz andNadler proved that each closed-valued contractive

multifunction on a complete metric space has a fixed point[14]

Below we examine the existence of solutions for two cou-pled systems of nonlinear time-fractional differential equa-tions and inclusions within Caputo-Fabrizio time-fractionalderivative First we discuss the coupled system namely

(CF119863120572

119905119906) (119909 119905) = 119891

1(119909 119905 119906 (119909 119905) V (119909 119905))

(CF119863120573

119905V) (119909 119905) = 119891

2(119909 119905 119906 (119909 119905) V (119909 119905))

(6)

such that

119906 (0 0) = 0

V (0 0) = 0(7)

where 0 lt 120572 lt 1 0 lt 120573 lt 1 (119909 119905) isin [0 1] times [0 1] andthe mappings 119891

1 1198912 [0 1] times [0 1] times R times R rarr R are

continuous functions In addition we discuss the existence ofsolutions for the coupled system of nonlinear time-fractionaldifferential inclusions

(CF119863120572

119905119906) (119909 119905) isin 119865

1(119909 119905 119906 (119909 119905) V (119909 119905))

(CF119863120573

119905V) (119909 119905) isin 119865

2(119909 119905 119906 (119909 119905) V (119909 119905))

(8)

such that

119906 (0 0) = 0

V (0 0) = 0(9)

where 1198651 1198652 [0 1] times [0 1] times R times R rarr P(R) are some

multivalued mapsWe say that 119865 [0 1] times [0 1] times R times R rarr 2

R is aCaratheodory multifunction whenever (119909 119905) 997891rarr 119865(119909 119905 119906

1

1199062) is measurable for all 119906

119894isin R and (119906

1 1199062) 997891rarr 119865(119909 119905 119906

1 1199062)

is upper semicontinuous (usc) for almost all (119909 119905) isin [0 1] times[0 1] and 119906

1 1199062isin 119883 [12] A Caratheodory multifunction

119865 [0 1] times [0 1] times R times R rarr 2R is said to be an 1198711-

Caratheodory whenever for each 120588 gt 0 there exists 120601120588isin

1198711([0 1] times [0 1]R+) such that1003817100381710038171003817119865 (119909 119905 1199061 1199062)

1003817100381710038171003817

= sup(119909119905)isin[01]times[01]

|119904| 119904 isin 119865 (119909 119905 1199061 1199062) le 120601120588 (119909 119905)(10)

for all |119906119894| le 120588 and for almost all (119909 119905) isin [0 1] times [0 1] [12]

The set of selections of 119865119894at 119906119894is defined by

119878119865119894(119906119894)

= 119908119894isin 1198711([0 1] times [0 1] R) 119908119894 (119909 119905)

isin 119865 (119909 119905 119906119894(119909 119905) 119906

1015840

119894(119909 119905)) for almost all (119909 119905)

isin [0 1] times [0 1]

(11)

for all 119906119894 1199061015840

119894isin 119862R([0 1]times [0 1]) for 119894 = 1 2The sets 119878

119865119894(119906119894)are

nonempty for all119906119894isin 119862119870([0 1]times[0 1])whenever dim119870 lt infin

[12 15]The graph of themultifunction119865 119883 rarr 119884 is defined

Journal of Function Spaces 3

by the set Gr(119865) = (119909 119910) isin 119883 times 119884 119910 isin 119865(119909) (see [12 13])We say that the graph Gr(119865) of 119865 119883 rarr Pcl(119884) is a closedsubset of 119883 times 119884 whenever for all sequences 119906

119899119899isinN in 119883 and

119910119899119899isinN in 119884 with 119906

119899rarr 1199060 119910119899rarr 1199100 and 119910

119899isin 119865(119906

119899) for all

119899 we have 1199100isin 119865(119906

0) [12] Below we introduce the following

results which will be required in our proofs

Theorem 3 (see [12]) Suppose that 119883 is a Banach space 119879

119883 rarr 119883 is a completely continuous operator and the set 119870 =

119906 isin 119883 119906 = 120582119879119906 119891119900119903 119904119900119898119890 120582 isin [0 1] is bounded Then 119879has a fixed point

Lemma 4 (see [12 Proposition 12]) If 119865 119883 rarr P119888119897(119884) is

upper semicontinuous then119866119903(119865) is a closed subset of119883times119884 If119865 is completely continuous with a closed graph then it is uppersemicontinuous

Lemma 5 (see [12]) Let 119883 be a separable Banach space and119865 [0 1] times [0 1] times 119883 times 119883 rarr P

119888119901119888V(119883) an 1198711-Caratheodoryfunction Then the operator Θ sdot 119878

119865 119862119883([0 1] times [0 1]) rarr

P119888119901119888V(119862119883([0 1]times[0 1])) defined by 119906 997891rarr (Θsdot119878

119865)(119906) = Θ(119878

119865119906)

is a closed graph operator where Θ is a linear continuousmapping from 119871

1([0 1] times [0 1] 119883) into 119862

119883([0 1] times [0 1])

Theorem 6 (see [12]) Let 119864 be a Banach space 119862 a closedconvex subset of 119864 119880 an open subset of 119862 and 0 isin 119880Let us suppose that 119865 119880 rarr P

119888119901119888V(119862) depicts an uppersemicontinuous compact map such thatP

119888119901119888V(119862) denotes thefamily of nonempty compact convex subsets of 119862 Then either119865 admits a fixed point in119880 or there exist 119906 isin 120597119880 and 120582 isin (0 1)such that 119906 isin 120582119865(119906)

2 Main Results

First we investigate the coupled system

(CF119863120572

119905119906) (119909 119905) = 119891

1(119909 119905 119906 (119909 119905) V (119909 119905))

(CF119863120573

119905V) (119909 119905) = 119891

2(119909 119905 119906 (119909 119905) V (119909 119905))

(12)

equipped with the boundary value conditions 119906(0 0) = 0

and V(0 0) = 0 where 1198911 1198912 [0 1] times [0 1] times 119883

2rarr

119883 are continuous mappings 120572 120573 isin (0 1) 119909 119905 isin [0 1]and CF

119863120572

119905and CF

119863120573

119905are the Caputo-Fabrizio time-fractional

derivatives Now consider the Banach space 119883 = 119906

119906 isin 119862R([0 1] times [0 1]) endowed with the sup-norm 119906119883=

sup(119909119905)isin[01]times[01]

|119906(119909 119905)| Thus the product space (119883 times

119883 sdot 119883times119883

) is also a Banach space via the product norm(119906 V)

119883times119883= 119906119883+V119883 First we prove the next key lemma

Lemma 7 Suppose that119891 isin 1198711119883([0 1]times[0 1]) and 0 lt 120572 lt 1

The function 1199060isin 119862119883([0 1] times [0 1]) is a solution for the time-

fractional integral equation

119906 (119909 119905) =2 (1 minus 120572)

(2 minus 120572)119872 (120572)(119891 (119909 119905) minus 119891 (0 0))

+2120572

(2 minus 120572)119872 (120572)int

119905

0

119891 (119909 119904) 119889119904

(13)

if and only if1199060is a unique solution of the time-fractional differ-

ential equation

(119862119865119863120572

119905119906) (119909 119905) = 119891 (119909 119905) (119909 119905) isin [0 1] times [0 1]

119906 (0 0) = 0

(14)

Proof A solution of initial value problem (14) is denoted by1199060 As a result (CF119863120572

1199051199060)(119909 119905) = 119891(119909 119905) and 119906

0(0 0) = 0 By

integrating both sides we get

1199060(119909 119905) minus 119906

0(0 0)

=2 (1 minus 120572)

(2 minus 120572)119872 (120572)(119891 (119909 119905) minus 119891 (0 0))

+2120572

(2 minus 120572)119872 (120572)int

119905

0

119891 (119909 119904) 119889119904

(15)

and so 1199060(119909 119905) = (2(1 minus 120572)(2 minus 120572)119872(120572))(119891(119909 119905) minus 119891(0 0)) +

(2120572(2 minus 120572)119872(120572)) int119905

0119891(119909 119904)119889119904 This shows that 119906

0represents

the solution of time-fractional integral equation (13) If1199061and

1199062are two distinct solutions for initial value problem (14)

then CF119863120572

1199051199061(119909 119905) minus

CF119863120572

1199051199062(119909 119905) = [

CF119863120572

1199051199061minus 1199062](119909 119905) = 0

and (1199061minus 1199062)(0 0) = 0 By the property of the Caputo-

Fabrizio time-fractional derivative in [11] we get 1199061= 1199062

Hence 1199060is a unique solution of initial value problem (14)

Now suppose that 1199060is a solution of time-fractional integral

equation (13)Then we conclude that 1199060(119909 119905) = (2(1minus120572)(2minus

120572)119872(120572))(119891(119909 119905)minus119891(0 0))+(2120572(2minus120572)119872(120572)) int119905

0119891(119909 119904)119889119904 By

using (4) one can see that this function represents a solutionfor initial value problem (14) Note that 119906

0(0 0) = 0

Now we consider (1)-(2) For each (119909 119905) isin [0 1] times [0 1]define the operators 119879

1 1198792 119883 rarr 119883 by

(1198791V) (119909 119905)

=2 (1 minus 120572)

(2 minus 120572)119872 (120572)1198911(119909 119905 119906 (119909 119905) V (119909 119905))

minus2 (1 minus 120572)

(2 minus 120572)119872 (120572)1198911(0 0 119906 (0 0) V (0 0))

+2120572

(2 minus 120572)119872 (120572)int

119905

0

1198911(119909 119904 119906 (119909 119904) V (119909 119904)) 119889119904

(1198792119906) (119909 119905)

=2 (1 minus 120573)

(2 minus 120573)119872 (120573)1198912(119909 119905 119906 (119909 119905) V (119909 119905))

minus2 (1 minus 120573)

(2 minus 120573)119872 (120573)1198912(0 0 119906 (0 0) V (0 0))

+2120573

(2 minus 120573)119872(120573)int

119905

0

1198912(119909 119904 119906 (119909 119904) V (119909 119904)) 119889119904

(16)

4 Journal of Function Spaces

and put

1198731=

4 minus 2120572

(2 minus 120572)119872 (120572)

1198732=

4 minus 2120573

(2 minus 120573)119872 (120573)

(17)

Theorem 8 Suppose that 1198911 1198912 [0 1]times [0 1]times119883times119883 rarr 119883

are the continuous mappings in system (6)-(7) and there existpositive constants 119871

1and 119871

2fulfilling |119891

1(119909 119905 119906

1 1199062)| le 119871

1

and |1198912(119909 119905 119906

1 1199062)| le 119871

2for all (119909 119905) isin [0 1] times [0 1] and

1199061 1199062isin 119883 Then system (6)-(7) possesses at least one solution

Proof Let the operators 1198791 1198792 119883 rarr 119883 defined by

(16) We define the operator 119879 119883 times 119883 rarr 119883 times 119883 by119879(119906 V)(119909 119905)fl ((119879

1V)(119909 119905) (119879

2119906)(119909 119905)) for all (119909 119905) isin [0 1] times

[0 1] Note that 119879 is continuous because the mappings 1198911

and 1198912are continuous We prove that the operator 119879 maps

bounded sets into the bounded subsets of 119883 times 119883 Let Ω be abounded subset of119883times119883 (119906 V) isin Ω and (119909 119905) isin [0 1]times[0 1]Then we have

1003816100381610038161003816(1198791V) (119909 119905)1003816100381610038161003816

=

10038161003816100381610038161003816100381610038161003816

2 (1 minus 120572)

(2 minus 120572)119872 (120572)1198911(119909 119905 119906 (119909 119905) V (119909 119905))

minus2 (1 minus 120572)

(2 minus 120572)119872 (120572)1198911(0 0 119906 (0 0) V (0 0))

+2120572

(2 minus 120572)119872 (120572)int

119905

0

1198911(119909 119904 119906 (119909 119904) V (119909 119904)) 119889119904

10038161003816100381610038161003816100381610038161003816

le2 (1 minus 120572)

(2 minus 120572)119872 (120572)

10038161003816100381610038161198911 (119909 119905 119906 (119909 119905) V (119909 119905))1003816100381610038161003816

+2 (1 minus 120572)

(2 minus 120572)119872 (120572)

10038161003816100381610038161198911 (0 0 0 0)1003816100381610038161003816 +

2120572

(2 minus 120572)119872 (120572)

sdot int

119905

0

10038161003816100381610038161198911 (119909 119904 119906 (119909 119904) V (119909 119904))1003816100381610038161003816 119889119904

le 1198711

2 (1 minus 120572)

(2 minus 120572)119872 (120572)+

2 (1 minus 120572)

(2 minus 120572)119872 (120572)

+2120572

(2 minus 120572)119872 (120572)119905 le 119871

1

4 (1 minus 120572)

(2 minus 120572)119872 (120572)

+2120572

(2 minus 120572)119872 (120572) le 119871

1

4 minus 2120572

(2 minus 120572)119872 (120572) = 119871

11198731

(18)

and so (1198791V)(119909 119905)

119883le 11987111198731 Also we have

1003816100381610038161003816(1198792119906) (119909 119905)1003816100381610038161003816

=

100381610038161003816100381610038161003816100381610038161003816

2 (1 minus 120573)

(2 minus 120573)119872 (120573)1198912(119909 119905 119906 (119909 119905) V (119909 119905))

minus2 (1 minus 120573)

(2 minus 120573)119872 (120573)1198912(0 0 119906 (0 0) V (0 0))

+2120573

(2 minus 120573)119872(120573)int

119905

0

1198912(119909 119904 119906 (119909 119904) V (119909 119904)) 119889119904

100381610038161003816100381610038161003816100381610038161003816

le2 (1 minus 120573)

(2 minus 120573)119872 (120573)

10038161003816100381610038161198912 (119909 119905 119906 (119909 119905) V (119909 119905))1003816100381610038161003816

+2 (1 minus 120573)

(2 minus 120573)119872 (120573)

10038161003816100381610038161198912 (0 0 0 0)1003816100381610038161003816 +

2120573

(2 minus 120573)119872(120573)

sdot int

119905

0

10038161003816100381610038161198912 (119909 119904 119906 (119909 119904) V (119909 s))1003816100381610038161003816 119889119904

le 1198712

2 (1 minus 120573)

(2 minus 120573)119872 (120573)+

2 (1 minus 120573)

(2 minus 120573)119872 (120573)

+2120573

(2 minus 120573)119872(120573)119905 le 119871

2

4 (1 minus 120573)

(2 minus 120573)119872 (120573)

+2120573

(2 minus 120573)119872(120573) le 119871

2

4 minus 2120573

(2 minus 120573)119872 (120573)

= 11987121198732

(19)

and so (1198792119906)(119909 119905)

119883le 11987121198732 Thus 119879(119906 V)(119909 119905)

119883times119883le

11987111198731+ 11987121198732 This shows that the operator 119879maps bounded

sets into the bounded sets of 119883 times 119883 Now we show that theoperator119879 is equicontinuous Let (119909 119905

1) (119909 119905

2) isin [0 1]times[0 1]

with 1199051lt 1199052 Then we have

1003816100381610038161003816(1198791V) (119909 1199052) minus (1198791V) (119909 1199051)1003816100381610038161003816

=

10038161003816100381610038161003816100381610038161003816

2 (1 minus 120572)

(2 minus 120572)119872 (120572)1198911(119909 1199052 119906 (119909 119905

2) V (119909 119905

2))

minus2 (1 minus 120572)

(2 minus 120572)119872 (120572)1198911(0 0 119906 (0 0) V (0 0))

+2120572

(2 minus 120572)119872 (120572)int

1199052

0

1198911(119909 119904 119906 (119909 119904) V (119909 119904)) 119889119904

minus2 (1 minus 120572)

(2 minus 120572)119872 (120572)1198911(119909 1199051 119906 (119909 119905

1) V (119909 119905

1))

+2 (1 minus 120572)

(2 minus 120572)119872 (120572)1198911(0 0 119906 (0 0) V (0 0))

minus2120572

(2 minus 120572)119872 (120572)int

1199051

0

1198911(119909 119904 119906 (119909 119904) V (119909 119904)) 119889119904

10038161003816100381610038161003816100381610038161003816

le2 (1 minus 120572)

(2 minus 120572)119872 (120572)

10038161003816100381610038161198911 (119909 1199052 119906 (119909 1199052) V (119909 1199052))

minus 1198911(119909 1199051 119906 (119909 119905

1) V (119909 119905

1))1003816100381610038161003816 +

2120572

(2 minus 120572)119872 (120572)

sdot int

1199052

1199051

10038161003816100381610038161198911 (119909 119904 119906 (119909 119904) V (119909 119904))1003816100381610038161003816 119889119904

le2 (1 minus 120572)

(2 minus 120572)119872 (120572)

10038161003816100381610038161198911 (119909 1199052 119906 (119909 1199052) V (119909 1199052))

minus 1198911(119909 1199051 119906 (119909 119905

1) V (119909 119905

1))1003816100381610038161003816 +

21205721198711

(2 minus 120572)119872 (120572)(1199052

minus 1199051)

(20)

Journal of Function Spaces 5

This implies that |(1198791V)(119909 119905

2) minus (119879

1V)(119909 119905

1)| rarr 0 whenever

(119909 1199052) rarr (119909 119905

1) By utilizing the Arzela-Ascoli theorem 119879

1

is completely continuous Similarly we have1003816100381610038161003816(1198792119906) (119909 1199052) minus (1198792119906) (119909 1199051)

1003816100381610038161003816

=

100381610038161003816100381610038161003816100381610038161003816

2 (1 minus 120573)

(2 minus 120573)119872 (120573)1198912(119909 1199052 119906 (119909 119905

2) V (119909 119905

2))

minus2 (1 minus 120573)

(2 minus 120573)119872 (120573)1198912(0 0 119906 (0 0) V (0 0))

+2120573

(2 minus 120573)119872 (120573)int

1199052

0

1198912(119909 119904 119906 (119909 119904) V (119909 119904)) 119889119904

minus2 (1 minus 120573)

(2 minus 120573)119872 (120573)1198912(119909 1199051 119906 (119909 119905

1) V (119909 119905

1))

+2 (1 minus 120573)

(2 minus 120573)119872 (120573)1198912(0 0 119906 (0 0) V (0 0))

minus2120573

(2 minus 120573)119872 (120573)int

1199051

0

1198912(119909 119904 119906 (119909 119904) V (119909 119904)) 119889119904

100381610038161003816100381610038161003816100381610038161003816

le2 (1 minus 120573)

(2 minus 120573)119872 (120573)

10038161003816100381610038161198912 (119909 1199052 119906 (119909 1199052) V (119909 1199052))

minus 1198912(119909 1199051 119906 (119909 119905

1) V (119909 119905

1))1003816100381610038161003816 +

2120573

(2 minus 120573)119872(120573)

sdot int

1199052

1199051

10038161003816100381610038161198912 (119909 119904 119906 (119909 119904) V (119909 119904))1003816100381610038161003816 119889119904

le2 (1 minus 120573)

(2 minus 120573)119872 (120573)

10038161003816100381610038161198912 (119909 1199052 119906 (119909 1199052) V (119909 1199052))

minus 1198912(119909 1199051 119906 (119909 119905

1) V (119909 119905

1))1003816100381610038161003816 +

21205731198712

(2 minus 120573)119872 (120573)(1199052

minus 1199051)

(21)

Again by utilizing the Arzela-Ascoli theorem we ob-serve that 119879

2is completely continuous Therefore we get

119879(119906 V)(119909 1199052) minus 119879(119906 V)(119909 119905

2)119883times119883

rarr 0 whenever (119909 1199052)

tends to (119909 1199051) Thus 119879 is completely continuous In the next

step we prove that

Ω = (119906 V) isin 119883 times 119883 (119906 V) = 120582119879 (119906 V) for some 120582

isin [0 1]

(22)

is bounded Let (119906 V) be an arbitrary element of Ω Choose120582 isin [0 1] fulfilling (119906 V) = 120582119879(119906 V) Hence V(119909 119905) =

120582(1198791V)(119909 119905) and 119906(119909 119905) = 120582(119879

2119906)(119909 119905) for all (119909 119905) isin [0 1] times

[0 1] Since

1

120582|V (119909 119905)| = 1003816100381610038161003816(1198791V) (119909 119905)

1003816100381610038161003816 le 11987111198731(23)

we get |V(119909 119905)| le 12058211987111198731and so V(119909 119905)

119883le 12058211987111198731 Simi-

larly we prove that 119906(119909 119905)119883le 12058211987121198732 Thus (119906 V)

119883times119883le

120582[11987111198731+ 11987121198732] and so Ω is a bounded set Now by using

Theorem 3 we get that 119879 has a fixed point which is a solutionfor the coupled system of the time-fractional differentialequations

Next we study the existence of solution for the coupledsystem of time-fractional differential inclusions

(CF119863120572

119905119906) (119909 119905) isin 119865

1(119909 119905 119906 (119909 119905) V (119909 119905))

(CF119863120573

119905V) (119909 119905) isin 119865

2(119909 119905 119906 (119909 119905) V (119909 119905))

(24)

with the initial value conditions 119906(0 0) = 0 and V(0 0) = 0where 119865

1 1198652 [0 1] times [0 1] times R times R rarr P(R) are some

multivalued maps

Definition 9 One says that (1199061 1199062) isin 119862([0 1] times [0 1] 119883) times

119862([0 1] times [0 1] 119883) is a solution for the system of thetime-fractional differential inclusions whenever it satisfiesthe initial value conditions and there exists (119908

1 1199082) isin

1198711([0 1] times [0 1]) times 119871

1([0 1] times [0 1]) such that 119908

119894(119909 119905) isin

119865119894(119909 119905 119906(119909 119905) V(119909 119905)) for almost all (119909 119905) isin [0 1] times [0 1] and

119894 = 1 2 and also

119906119894(119909 119905)

=2 (1 minus 120572)

(2 minus 120572)119872 (120572)119908119894(119909 119905 119906 (119909 119905) V (119909 119905))

minus2 (1 minus 120572)

(2 minus 120572)119872 (120572)119908119894(0 0 119906 (0 0) V (0 0))

+2120572

(2 minus 120572)119872 (120572)int

119905

0

119908119894(119909 119904 119906 (119909 119904) V (119909 119904)) 119889119904

(25)

for all (119909 119905) isin [0 1] times [0 1] and 119894 = 1 2

Theorem 10 Let 1198651 1198652 [0 1] times [0 1] timesR timesR rarr P

119888119901119888V(R)

be 1198711-Caratheodory multifunctions Suppose that there exist anondecreasing bounded continuousmap120595 [0infin) rarr (0infin)

and a continuous function 119901 [0 1] times [0 1] rarr (0infin) suchthat 119865

119894(119909 119905 119906

119894(119909 119905) 119906

1015840

119894(119909 119905)) le 119901(119909 119905)120595(119906

119894) for all (119909 119905) isin

[0 1] times [0 1] 119906119894 1199061015840

119894isin 119883 for 119894 = 1 2 Then coupled system

of time-fractional differential inclusions (8)-(9) has at least onesolution

Proof Define the operator119873 119883times119883 rarr 2119883times119883 by119873(119906

1 1199062) =

(1198731(1199061 1199062)

1198732(1199061 1199062)) where

1198731(1199061 1199062) = ℎ

1isin 119883 times 119883 there exists V

1

isin 1198781198651 1199061

such that ℎ1(119909 119905) = V

1(119909 119905) forall (119909 119905)

isin [0 1] times [0 1]

1198732(1199061 1199062) = ℎ

2isin 119883 times 119883 there exists V

2

isin 1198781198652 1199062

such that ℎ2(119909 119905) = V

2(119909 119905) forall (119909 119905)

isin [0 1] times [0 1]

6 Journal of Function Spaces

ℎ1(119909 119905) =

2 (1 minus 120572)

(2 minus 120572)119872 (120572)V1(119909 119905) minus

2 (1 minus 120572)

(2 minus 120572)119872 (120572)

sdot V1(0 0) +

2120572

(2 minus 120572)119872 (120572)int

119905

0

V1(119909 119904) 119889119904

ℎ2(119909 119905) =

2 (1 minus 120573)

(2 minus 120573)119872 (120573)V2(119909 119905) minus

2 (1 minus 120573)

(2 minus 120573)119872 (120573)

sdot V2(0 0) +

2120573

(2 minus 120573)119872(120573)int

119905

0

V2(119909 119904) 119889119904

(26)

By Lemma 7 it is clear that each fixed point of the operator119873 is a solution for system of time-fractional differentialinclusions (8) First we prove that the multifunction 119873 isconvex-valued Let (119906

1 1199062) isin 119883 times 119883 (ℎ

1 ℎ2) (ℎ1015840

1 ℎ1015840

2) isin

119873(1199061 1199062) Choose V

119894 V1015840119894isin 119878119865119894(1199061 1199062)

such that

ℎ119894(119909 119905) =

2 (1 minus 120572)

(2 minus 120572)119872 (120572)V119894(119909 119905)

minus2 (1 minus 120572)

(2 minus 120572)119872 (120572)V119894(0 0)

+2120572

(2 minus 120572)119872 (120572)int

119905

0

V119894(119909 119904) 119889119904

ℎ1015840

119894(119909 119905) =

2 (1 minus 120572)

(2 minus 120572)119872 (120572)V1015840119894(119909 119905)

minus2 (1 minus 120572)

(2 minus 120572)119872 (120572)V1015840119894(0 0)

+2120572

(2 minus 120572)119872 (120572)int

119905

0

V1015840119894(119909 119904) 119889119904

(27)

for almost all (119909 119905) isin [0 1] times [0 1] and 119894 = 1 2 Let 0 le 120582 le 1be given Then we have

[120582ℎ119894+ (1 minus 120582) ℎ

1015840

119894] (119909 119905)

=2 (1 minus 120572)

(2 minus 120572)119872 (120572)[120582V119894(119909 119905) + (1 minus 120582) V1015840

119894(119909 119905)]

minus2 (1 minus 120572)

(2 minus 120572)119872 (120572)[120582V119894(0 0) + (1 minus 120582) V1015840

119894(0 0)]

+2120572

(2 minus 120572)119872 (120572)

sdot int

119905

0

[120582V119894(119909 119904) + (1 minus 120582) V1015840

119894(119909 119904)] 119889119904

(28)

for 119894 = 1 2 Since the operator 119865119894has convex values 119878

119865119894(119906119894)is a

convex set and [120582ℎ119894+ (1 minus 120582)ℎ

1015840

119894] isin 119873119894(1199061 1199062) for 119894 = 1 2 This

implies that the operator119873 has convex values Now we provethat119873maps bounded sets of119883 into bounded sets Let 119903 gt 0119861119903= (1199061 1199062) isin 119883 times 119883 (119906

1 1199062) le 119903 be a bounded subset

of 119883 times 119883 (ℎ1 ℎ2) isin 119873(119906

1 1199062) and (119906

1 1199062) isin 119861119903 Then there

exists (V1 V2) isin 1198781198651(1199061)

times 1198781198652(1199062)

such that

ℎ1(119909 119905) =

2 (1 minus 120572)

(2 minus 120572)119872 (120572)V1(119909 119905)

minus2 (1 minus 120572)

(2 minus 120572)119872 (120572)V1(0 0)

+2120572

(2 minus 120572)119872 (120572)int

119905

0

V1(119909 119904) 119889119904

(29)

and ℎ2(119909 119905) = (2(1 minus120573)(2 minus120573)119872(120573))V

2(119909 119905) minus (2(1 minus120573)(2 minus

120573)119872(120573))V2(0 0) + (2120573(2 minus 120573)119872(120573)) int

119905

0V2(119909 119904)119889119904 for almost

all (119909 119905) isin [0 1] times [0 1] If 119901infin= sup

(119909119905)isin[01]times[01]|119901(119909 119905)|

then we obtain

1003816100381610038161003816(ℎ1) (119909 119905)1003816100381610038161003816 =

10038161003816100381610038161003816100381610038161003816

2 (1 minus 120572)

(2 minus 120572)119872 (120572)V1(119909 119905)

minus2 (1 minus 120572)

(2 minus 120572)119872 (120572)V1(0 0)

+2120572

(2 minus 120572)119872 (120572)int

119905

0

V1(119909 119904) 119889119904

10038161003816100381610038161003816100381610038161003816

le2 (1 minus 120572)

(2 minus 120572)119872 (120572)

1003816100381610038161003816V1 (119909 119905)1003816100381610038161003816

+2 (1 minus 120572)

(2 minus 120572)119872 (120572)

1003816100381610038161003816V1 (0 0)1003816100381610038161003816 +

2120572

(2 minus 120572)119872 (120572)

sdot int

119905

0

1003816100381610038161003816V1 (119909 119904)1003816100381610038161003816 119889119904 le 119901 (119909 119905) 120595 (

100381710038171003817100381711990611003817100381710038171003817)

sdot 2 (1 minus 120572)

(2 minus 120572)119872 (120572)

+2 (1 minus 120572)

(2 minus 120572)119872 (120572)+

2120572

(2 minus 120572)119872 (120572)119905 le

10038171003817100381710038171199011003817100381710038171003817infin

sdot 120595 (10038171003817100381710038171199061

1003817100381710038171003817) 4 (1 minus 120572)

(2 minus 120572)119872 (120572)+

2120572

(2 minus 120572)119872 (120572)

le10038171003817100381710038171199011003817100381710038171003817infin

120595 (10038171003817100381710038171199061

1003817100381710038171003817) 4 minus 2120572

(2 minus 120572)119872 (120572) =

10038171003817100381710038171199011003817100381710038171003817infin

sdot 120595 (10038171003817100381710038171199061

1003817100381710038171003817)1198731

(30)

where the constant 1198731is defined by (17) This implies

that ℎ1 le 119901

infin120595(1199061)1198731 Similarly we get ℎ

2 le

119901infin120595(1199062)1198732 where the constant 119873

2is defined by (17)

Thus (ℎ1 ℎ2) le 119901

infin120595((119906

1 1199062))(1198731+ 1198732) Now we

prove that119873maps bounded sets into equicontinuous subsets

Journal of Function Spaces 7

of 119883 times 119883 Let (1199061 1199062) isin 119861119903and (119909 119905

1) (119909 119905

2) isin [0 1] times [0 1]

with 1199051lt 1199052 Then we have

1003816100381610038161003816(ℎ1) (119909 1199052) minus (ℎ1) (119909 1199051)1003816100381610038161003816 =

10038161003816100381610038161003816100381610038161003816

2 (1 minus 120572)

(2 minus 120572)119872 (120572)V1(119909 1199052)

minus2 (1 minus 120572)

(2 minus 120572)119872 (120572)V1(0 0) +

2120572

(2 minus 120572)119872 (120572)

sdot int

1199052

0

V1(119909 119904) 119889119904 minus

2 (1 minus 120572)

(2 minus 120572)119872 (120572)V1(119909 1199051)

+2 (1 minus 120572)

(2 minus 120572)119872 (120572)V1(0 0)

minus2120572

(2 minus 120572)119872 (120572)int

1199051

0

V1(119909 119904) 119889119904

10038161003816100381610038161003816100381610038161003816

le2 (1 minus 120572)

(2 minus 120572)119872 (120572)

1003816100381610038161003816V1 (119909 1199052) minus V1(119909 1199051)1003816100381610038161003816

+2120572

(2 minus 120572)119872 (120572)int

1199052

1199051

1003816100381610038161003816V1 (119909 119904)1003816100381610038161003816 119889119904

le2 (1 minus 120572)

(2 minus 120572)119872 (120572)

1003816100381610038161003816V1 (119909 1199052) minus V1(119909 1199051)1003816100381610038161003816

+212057210038171003817100381710038171199011003817100381710038171003817infin

120595 (10038171003817100381710038171199061

1003817100381710038171003817)

(2 minus 120572)119872 (120572)(1199052minus 1199051)

(31)

By using a similar method we obtain1003816100381610038161003816(ℎ2) (119909 1199052) minus (ℎ2) (119909 1199051)

1003816100381610038161003816

le2 (1 minus 120573)

(2 minus 120573)119872 (120573)

1003816100381610038161003816V2 (119909 1199052) minus V2(119909 1199051)1003816100381610038161003816

+212057310038171003817100381710038171199011003817100381710038171003817infin

120595 (10038171003817100381710038171199062

1003817100381710038171003817)

(2 minus 120573)119872 (120573)(1199052minus 1199051)

(32)

Hence |ℎ119894(119909 1199052) minus ℎ119894(119909 1199051)| rarr 0 as (119909 119905

2) rarr (119909 119905

1) By

using the Arzela-Ascoli theorem we get that119873 is completelycontinuous Here we prove that119873 is upper semicontinuousBy using Lemma 4 119873 is upper semicontinuous whenever ithas a closed graph Since119873 is completely continuouswemustshow that119873 has a closed graph

Let (1199061198991 119906119899

2) be a sequence in 119883 times 119883 with (119906119899

1 119906119899

2) rarr

(1199060

1 1199060

2) and (ℎ119899

1 ℎ119899

2) isin 119873(119906

119899

1 119906119899

2)with (ℎ119899

1 ℎ119899

2) rarr (ℎ

0

1 ℎ0

2)We

show that (ℎ01 ℎ0

2) isin 119873(119906

0

1 1199060

2) For each (ℎ119899

1 ℎ119899

2) isin 119873(119906

119899

1 119906119899

2)

we can choose (V1198991 V1198992) isin 1198781198651(119906119899

1)times 1198781198652(119906119899

2)such that

ℎ119899

1(119909 119905) =

2 (1 minus 120572)

(2 minus 120572)119872 (120572)V1198991(119909 119905)

minus2 (1 minus 120572)

(2 minus 120572)119872 (120572)V1198991(0 0)

+2120572

(2 minus 120572)119872 (120572)int

119905

0

V1198991(119909 119904) 119889119904

(33)

and ℎ119899

2(119909 119905) = (2(1 minus 120573)(2 minus 120573)119872(120573))V119899

2(119909 119905) minus (2(1 minus

120573)(2 minus 120573)119872(120573))V1198992(0 0) + (2120573(2 minus 120573)119872(120573)) int

119905

0V1198992(119909 119904)119889119904 for

all (119909 119905) isin [0 1]times[0 1] It is sufficient to show that there exists(V01 V02) isin 1198781198651(1199060

1)times 1198781198652(1199060

2)such that

ℎ0

1(119909 119905) =

2 (1 minus 120572)

(2 minus 120572)119872 (120572)V01(119909 119905)

minus2 (1 minus 120572)

(2 minus 120572)119872 (120572)V01(0 0)

+2120572

(2 minus 120572)119872 (120572)int

119905

0

V01(119909 119904) 119889119904

(34)

and ℎ0

2(119909 119905) = (2(1 minus 120573)(2 minus 120573)119872(120573))V0

2(119909 119905) minus (2(1 minus

120573)(2 minus 120573)119872(120573))V02(0 0) + (2120573(2 minus 120573)119872(120573)) int

119905

0V02(119909 119904)119889119904 for

all (119909 119905) isin [0 1] times [0 1] Now consider the linear operatorsΘ1 Θ2 1198711([0 1] times [0 1] 119883) rarr 119862([0 1] times [0 1] 119883) defined

by

Θ1(V) (119909 119905) =

2 (1 minus 120572)

(2 minus 120572)119872 (120572)V (119909 119905)

minus2 (1 minus 120572)

(2 minus 120572)119872 (120572)V (0 0)

+2120572

(2 minus 120572)119872 (120572)int

119905

0

V (119909 119904) 119889119904

(35)

andΘ2(V)(119909 119905) = (2(1minus120573)(2minus120573)119872(120573))V(119909 119905)minus(2(1minus120573)(2minus

120573)119872(120573))V(0 0) + (2120573(2 minus 120573)119872(120573)) int119905

0V(119909 119904)119889119904 Note that

10038171003817100381710038171003817ℎ119899

1(119909 119905) minus ℎ

0

1(119909 119905)

10038171003817100381710038171003817

=

10038171003817100381710038171003817100381710038171003817

2 (1 minus 120572)

(2 minus 120572)119872 (120572)[V1198991(119909 119905) minus V0

1(119909 119905)]

minus2 (1 minus 120572)

(2 minus 120572)119872 (120572)[V1198991(0 0) minus V0

1(0 0)]

+2120572

(2 minus 120572)119872 (120572)int

119905

0

[V1198991(119909 119904) minus V0

1(119909 119904)] 119889119904

10038171003817100381710038171003817100381710038171003817

997888rarr 0

10038171003817100381710038171003817ℎ119899

2(119909 119905) minus ℎ

0

2(119909 119905)

10038171003817100381710038171003817

=

100381710038171003817100381710038171003817100381710038171003817

2 (1 minus 120573)

(2 minus 120573)119872 (120573)[V1198992(119909 119905) minus V0

2(119909 119905)]

minus2 (1 minus 120573)

(2 minus 120573)119872 (120573)[V1198992(0 0) minus V0

2(0 0)]

+2120573

(2 minus 120573)119872(120573)int

119905

0

[V1198992(119909 119904) minus V0

2(119909 119904)] 119889119904

100381710038171003817100381710038171003817100381710038171003817

997888rarr 0

(36)

8 Journal of Function Spaces

By using Lemma 5 we get that Θ119894sdot 119878119865119894

is a closed graphoperator for 119894 = 1 2 Also we get ℎ119899

119894(119909 119905) isin Θ

119894(119878119865119894(119906119899

119894)) for

all 119899 Since 119906119899119894rarr 1199060

119894 we get

ℎ0

1(119909 119905) =

2 (1 minus 120572)

(2 minus 120572)119872 (120572)V01(119909 119905)

minus2 (1 minus 120572)

(2 minus 120572)119872 (120572)V01(0 0)

+2120572

(2 minus 120572)119872 (120572)int

119905

0

V01(119909 119904) 119889119904

(37)

and ℎ02(119909 119905) = (2(1minus120573)(2minus120573)119872(120573))V0

2(119909 119905) minus (2(1minus120573)(2minus

120573)119872(120573))V02(0 0) + (2120573(2 minus 120573)119872(120573)) int

119905

0V02(119909 119904)119889119904 for some

V0119894isin 119878119865119894(1199060

119894)(119894 = 1 2) Thus119873 has a closed graph

Now we prove that there is an open set 119880 sube 119883 with(1199061 1199062) notin 119873(119906

1 1199062) for all 120582 isin (0 1) and (119906

1 1199062) isin 120597119880

Let 120582 isin (0 1) and (1199061 1199062) isin 120582119873(119906

1 1199062) Then there exists

V119894isin 1198711([0 1] times [0 1]R) with V

119894isin 119878119865119894(119906119894)

(119894 = 1 2) such that

1199061(119909 119905) =

2 (1 minus 120572)

(2 minus 120572)119872 (120572)V1(119909 119905)

minus2 (1 minus 120572)

(2 minus 120572)119872 (120572)V1(0 0)

+2120572

(2 minus 120572)119872 (120572)int

119905

0

V1(119909 119904) 119889119904

(38)

and 1199062(119909 119905) = (2(1 minus 120573)(2 minus 120573)119872(120573))V

2(119909 119905) minus (2(1 minus

120573)(2 minus 120573)119872(120573))V2(0 0) + (2120573(2 minus 120573)119872(120573)) int

119905

0V2(119909 119904)119889119904 for

all (119909 119905) isin [0 1] times [0 1] By using the above computedvalues we obtain 119906

119894 le 119901

infin120595(119906119894)sum119899

119894=1119873119894for 119894 = 1 2

This follows that 119906119894119901infin120595(119906119894)sum119899

119894=1119873119894le 1 for 119894 =

1 2 Choose 119872119894gt 0 with 119906

119894 = 119872

119894in such a way that

119872119894119901infin120595(119906119894)sum119899

119894=1119873119894gt 1 for 119894 = 1 2 Put 119880 = (119906

1 1199062) isin

119883 times 119883 (1199061 1199062) lt min 119872

11198722 We note that the

operator 119873 119880 rarr P(119883) is upper semicontinuous andcompletely continuous Also we showed that there is no(1199061 1199062) isin 120597119880 such that (119906

1 1199062) isin 120582119873(119906

1 1199062) for some

120582 isin (0 1) Hence with the help of Theorem 6 we get that119873 has a fixed point (119906

1 1199062) isin 119880 which is a solution for time-

fractional differential inclusion (8)-(9)

Conflict of Interests

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

Acknowledgment

Research of the last two authors was supported by AzarbaijanShahid Madani University

References

[1] R P Agarwal M Benchohra and S Hamani ldquoA survey onexistence results for boundary value problems of nonlinear frac-tional differential equations and inclusionsrdquo Acta ApplicandaeMathematicae vol 109 no 3 pp 973ndash1033 2010

[2] D Baleanu H Mohammadi and S Rezapour ldquoOn a nonlinearfractional differential equation on partially ordered metricspacesrdquo Advances in Difference Equations vol 2013 article 832013

[3] D Baleanu S Rezapour S Etemad and A Alsaedi ldquoOn a time-fractional integrodifferential equation via three-point boundaryvalue conditionsrdquo Mathematical Problems in Engineering vol2015 Article ID 785738 12 pages 2015

[4] K Diethelm The Analysis of Fractional Differential EquationsLecture Notes inMathematics Springer Berlin Germany 2010

[5] R Gorenflo and F Mainardi ldquoFractional calculus integraland differential equations of fractional orderrdquo in Fractals andFractional Calculus in ContinuumMechanics A Carpinteri andF Mainardi Eds Springer New York NY USA 1997

[6] G Samko A Kilbas and O Marichev Fractional Integrals andDerivatives Theory and Applications Gordon and Breach 1993

[7] Z Bai and W Sun ldquoExistence and multiplicity of positivesolutions for singular fractional boundary value problemsrdquoComputers amp Mathematics with Applications vol 63 no 9 pp1369ndash1381 2012

[8] D Baleanu R P Agarwal H Mohammadi and S RezapourldquoSome existence results for a nonlinear fractional differentialequation on partially ordered Banach spacesrdquo Boundary ValueProblems 2013112 8 pages 2013

[9] D Baleanu S Zahra Nazemi and S Rezapour ldquoThe existenceof positive solutions for a new coupled system of multiterm sin-gular fractional integrodifferential boundary value problemsrdquoAbstract and Applied Analysis vol 2013 Article ID 368659 15pages 2013

[10] M Caputo and M Fabrizio ldquoA new definition of fractionalderivative without singular kernelrdquo Progress in Fractional Dif-ferentiation and Applications vol 1 no 2 pp 73ndash85 2015

[11] J Losada and J J Nieto ldquoProperties of a new fractionalderivative without singular kernelrdquo Progress in Fractional Dif-ferentiation and Applications vol 1 no 2 pp 87ndash92 2015

[12] D R Smart Fixed PointTheorems Cambridge University PressCambridge UK 1980

[13] S M Aleomraninejad S Rezapour and N Shahzad ldquoOn fixedpoint generalizations of Suzukirsquos methodrdquo Applied MathematicsLetters vol 24 no 7 pp 1037ndash1040 2011

[14] H Covitz and S B Nadler ldquoMulti-valued contractionmappingsin generalizedmetric spacesrdquo Israel Journal of Mathematics vol8 pp 5ndash11 1970

[15] K S Miller and B Ross An Introduction to the FractionalCalculus and Fractional Differential Eqautions John Wiley ampSons 1993

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

Page 2: Research Article On Coupled Systems of Time-Fractional ...downloads.hindawi.com/journals/jfs/2016/4626940.pdf · Research Article On Coupled Systems of Time-Fractional Differential

2 Journal of Function Spaces

Definition 2 The 120572 order time-fractional integral of a func-tion 119906 has the form [11]

(CF119868120572

119905119906) (119909 119905) =

2 (1 minus 120572)

(2 minus 120572)119872 (120572)119906 (119909 119905)

+2120572

(2 minus 120572)119872 (120572)int

119905

0

119906 (119909 119904) 119889119904

(119905 ge 0)

(2)

where119872(120572) represents a normalization function and 0 lt 120572 lt1

Losada and Nieto showed that119872(120572) = 2(2 minus 120572) for all0 le 120572 le 1 [11] By substituting 119872(120572) in (1) we obtain thedefinition of the time-fractional Caputo-Fabrizio derivativeof order 120572 for a function 119906 as follows

(CF119863120572

119905119906) (119909 119905) =

1

1 minus 120572int

119905

0

exp [minus120572 (119905 minus 119904)1 minus 120572

]120597119906

120597119904119889119904

((119909 119905) isin [0 1] times [0 1])

(3)

They proved that solution of (CF119863120572119905V)(119909 119905) = 119892(119909 119905) is given

by

V (119909 119905) =2 (1 minus 120572)

(2 minus 120572)119872 (120572)119892 (119909 119905)

+2120572

(2 minus 120572)119872 (120572)int

119905

0

119892 (119909 119904) 119889119904 + V (0 0) (4)

where 0 lt 120572 lt 1 and (119909 119905) isin [0 1] times [0 1] [11]The next step is to consider (119883 119889) being a metric space

Let us denote by P(119883) and 2119883 the class of all subsets andthe class of all nonempty subsets of 119883 respectively HencePcl(119883)Pbd(119883)Pcv(119883)Pcp(119883) andPcpcv(119883) are the classof all closed subsets the class of all bounded subsets the classof all convex subsets the class of all compact subsets and theclass of all compact and convex subsets of 119883 respectivelyWe claim that 119906 isin 119883 is a fixed point of the multifunction119865 119883 rarr 2

119883 whenever 119906 isin 119865119906 [12] A multifunction 119865

[0 1] times [0 1] rarr Pcl(R) is called measurable whenever thefunction (119909 119905) 997891rarr 119889(119908 119865(119909 119905)) = inf 119908 minus V V isin 119865(119909 119905) ismeasurable for all119908 isin R [12]ThePompeiu-Hausdorffmetric119867119889 2119883times 2119883rarr [0infin) is defined by

119867119889(119860 119861) = maxsup

119886isin119860

119889 (119886 119861) sup119887isin119861

119889 (119860 119887) (5)

such that 119889(119860 119887) = inf119886isin119860

119889(119886 119887) [13] (CB(119883)119867119889) is a

metric space and (CB(119883)119867119889) depicts a generalized metric

space Here CB(119883) denotes the set of closed and boundedsubsets of119883 and119862(119883) represents the set of closed subsets of119883[12 13]We recall that119865 is said to be convex-valued (compact-valued) whenever 119865119906 is convex (compact) set for each 119906 isin 119883[12] We mention that a multifunction 119865 119883 rarr 119862(119883) is acontraction whenever there exists a constant 120574 isin (0 1) suchthat 119867

119889(119865119906 119865V) le 120574119889(119906 V) for all 119906 V isin 119883 [12] In 1970

Covitz andNadler proved that each closed-valued contractive

multifunction on a complete metric space has a fixed point[14]

Below we examine the existence of solutions for two cou-pled systems of nonlinear time-fractional differential equa-tions and inclusions within Caputo-Fabrizio time-fractionalderivative First we discuss the coupled system namely

(CF119863120572

119905119906) (119909 119905) = 119891

1(119909 119905 119906 (119909 119905) V (119909 119905))

(CF119863120573

119905V) (119909 119905) = 119891

2(119909 119905 119906 (119909 119905) V (119909 119905))

(6)

such that

119906 (0 0) = 0

V (0 0) = 0(7)

where 0 lt 120572 lt 1 0 lt 120573 lt 1 (119909 119905) isin [0 1] times [0 1] andthe mappings 119891

1 1198912 [0 1] times [0 1] times R times R rarr R are

continuous functions In addition we discuss the existence ofsolutions for the coupled system of nonlinear time-fractionaldifferential inclusions

(CF119863120572

119905119906) (119909 119905) isin 119865

1(119909 119905 119906 (119909 119905) V (119909 119905))

(CF119863120573

119905V) (119909 119905) isin 119865

2(119909 119905 119906 (119909 119905) V (119909 119905))

(8)

such that

119906 (0 0) = 0

V (0 0) = 0(9)

where 1198651 1198652 [0 1] times [0 1] times R times R rarr P(R) are some

multivalued mapsWe say that 119865 [0 1] times [0 1] times R times R rarr 2

R is aCaratheodory multifunction whenever (119909 119905) 997891rarr 119865(119909 119905 119906

1

1199062) is measurable for all 119906

119894isin R and (119906

1 1199062) 997891rarr 119865(119909 119905 119906

1 1199062)

is upper semicontinuous (usc) for almost all (119909 119905) isin [0 1] times[0 1] and 119906

1 1199062isin 119883 [12] A Caratheodory multifunction

119865 [0 1] times [0 1] times R times R rarr 2R is said to be an 1198711-

Caratheodory whenever for each 120588 gt 0 there exists 120601120588isin

1198711([0 1] times [0 1]R+) such that1003817100381710038171003817119865 (119909 119905 1199061 1199062)

1003817100381710038171003817

= sup(119909119905)isin[01]times[01]

|119904| 119904 isin 119865 (119909 119905 1199061 1199062) le 120601120588 (119909 119905)(10)

for all |119906119894| le 120588 and for almost all (119909 119905) isin [0 1] times [0 1] [12]

The set of selections of 119865119894at 119906119894is defined by

119878119865119894(119906119894)

= 119908119894isin 1198711([0 1] times [0 1] R) 119908119894 (119909 119905)

isin 119865 (119909 119905 119906119894(119909 119905) 119906

1015840

119894(119909 119905)) for almost all (119909 119905)

isin [0 1] times [0 1]

(11)

for all 119906119894 1199061015840

119894isin 119862R([0 1]times [0 1]) for 119894 = 1 2The sets 119878

119865119894(119906119894)are

nonempty for all119906119894isin 119862119870([0 1]times[0 1])whenever dim119870 lt infin

[12 15]The graph of themultifunction119865 119883 rarr 119884 is defined

Journal of Function Spaces 3

by the set Gr(119865) = (119909 119910) isin 119883 times 119884 119910 isin 119865(119909) (see [12 13])We say that the graph Gr(119865) of 119865 119883 rarr Pcl(119884) is a closedsubset of 119883 times 119884 whenever for all sequences 119906

119899119899isinN in 119883 and

119910119899119899isinN in 119884 with 119906

119899rarr 1199060 119910119899rarr 1199100 and 119910

119899isin 119865(119906

119899) for all

119899 we have 1199100isin 119865(119906

0) [12] Below we introduce the following

results which will be required in our proofs

Theorem 3 (see [12]) Suppose that 119883 is a Banach space 119879

119883 rarr 119883 is a completely continuous operator and the set 119870 =

119906 isin 119883 119906 = 120582119879119906 119891119900119903 119904119900119898119890 120582 isin [0 1] is bounded Then 119879has a fixed point

Lemma 4 (see [12 Proposition 12]) If 119865 119883 rarr P119888119897(119884) is

upper semicontinuous then119866119903(119865) is a closed subset of119883times119884 If119865 is completely continuous with a closed graph then it is uppersemicontinuous

Lemma 5 (see [12]) Let 119883 be a separable Banach space and119865 [0 1] times [0 1] times 119883 times 119883 rarr P

119888119901119888V(119883) an 1198711-Caratheodoryfunction Then the operator Θ sdot 119878

119865 119862119883([0 1] times [0 1]) rarr

P119888119901119888V(119862119883([0 1]times[0 1])) defined by 119906 997891rarr (Θsdot119878

119865)(119906) = Θ(119878

119865119906)

is a closed graph operator where Θ is a linear continuousmapping from 119871

1([0 1] times [0 1] 119883) into 119862

119883([0 1] times [0 1])

Theorem 6 (see [12]) Let 119864 be a Banach space 119862 a closedconvex subset of 119864 119880 an open subset of 119862 and 0 isin 119880Let us suppose that 119865 119880 rarr P

119888119901119888V(119862) depicts an uppersemicontinuous compact map such thatP

119888119901119888V(119862) denotes thefamily of nonempty compact convex subsets of 119862 Then either119865 admits a fixed point in119880 or there exist 119906 isin 120597119880 and 120582 isin (0 1)such that 119906 isin 120582119865(119906)

2 Main Results

First we investigate the coupled system

(CF119863120572

119905119906) (119909 119905) = 119891

1(119909 119905 119906 (119909 119905) V (119909 119905))

(CF119863120573

119905V) (119909 119905) = 119891

2(119909 119905 119906 (119909 119905) V (119909 119905))

(12)

equipped with the boundary value conditions 119906(0 0) = 0

and V(0 0) = 0 where 1198911 1198912 [0 1] times [0 1] times 119883

2rarr

119883 are continuous mappings 120572 120573 isin (0 1) 119909 119905 isin [0 1]and CF

119863120572

119905and CF

119863120573

119905are the Caputo-Fabrizio time-fractional

derivatives Now consider the Banach space 119883 = 119906

119906 isin 119862R([0 1] times [0 1]) endowed with the sup-norm 119906119883=

sup(119909119905)isin[01]times[01]

|119906(119909 119905)| Thus the product space (119883 times

119883 sdot 119883times119883

) is also a Banach space via the product norm(119906 V)

119883times119883= 119906119883+V119883 First we prove the next key lemma

Lemma 7 Suppose that119891 isin 1198711119883([0 1]times[0 1]) and 0 lt 120572 lt 1

The function 1199060isin 119862119883([0 1] times [0 1]) is a solution for the time-

fractional integral equation

119906 (119909 119905) =2 (1 minus 120572)

(2 minus 120572)119872 (120572)(119891 (119909 119905) minus 119891 (0 0))

+2120572

(2 minus 120572)119872 (120572)int

119905

0

119891 (119909 119904) 119889119904

(13)

if and only if1199060is a unique solution of the time-fractional differ-

ential equation

(119862119865119863120572

119905119906) (119909 119905) = 119891 (119909 119905) (119909 119905) isin [0 1] times [0 1]

119906 (0 0) = 0

(14)

Proof A solution of initial value problem (14) is denoted by1199060 As a result (CF119863120572

1199051199060)(119909 119905) = 119891(119909 119905) and 119906

0(0 0) = 0 By

integrating both sides we get

1199060(119909 119905) minus 119906

0(0 0)

=2 (1 minus 120572)

(2 minus 120572)119872 (120572)(119891 (119909 119905) minus 119891 (0 0))

+2120572

(2 minus 120572)119872 (120572)int

119905

0

119891 (119909 119904) 119889119904

(15)

and so 1199060(119909 119905) = (2(1 minus 120572)(2 minus 120572)119872(120572))(119891(119909 119905) minus 119891(0 0)) +

(2120572(2 minus 120572)119872(120572)) int119905

0119891(119909 119904)119889119904 This shows that 119906

0represents

the solution of time-fractional integral equation (13) If1199061and

1199062are two distinct solutions for initial value problem (14)

then CF119863120572

1199051199061(119909 119905) minus

CF119863120572

1199051199062(119909 119905) = [

CF119863120572

1199051199061minus 1199062](119909 119905) = 0

and (1199061minus 1199062)(0 0) = 0 By the property of the Caputo-

Fabrizio time-fractional derivative in [11] we get 1199061= 1199062

Hence 1199060is a unique solution of initial value problem (14)

Now suppose that 1199060is a solution of time-fractional integral

equation (13)Then we conclude that 1199060(119909 119905) = (2(1minus120572)(2minus

120572)119872(120572))(119891(119909 119905)minus119891(0 0))+(2120572(2minus120572)119872(120572)) int119905

0119891(119909 119904)119889119904 By

using (4) one can see that this function represents a solutionfor initial value problem (14) Note that 119906

0(0 0) = 0

Now we consider (1)-(2) For each (119909 119905) isin [0 1] times [0 1]define the operators 119879

1 1198792 119883 rarr 119883 by

(1198791V) (119909 119905)

=2 (1 minus 120572)

(2 minus 120572)119872 (120572)1198911(119909 119905 119906 (119909 119905) V (119909 119905))

minus2 (1 minus 120572)

(2 minus 120572)119872 (120572)1198911(0 0 119906 (0 0) V (0 0))

+2120572

(2 minus 120572)119872 (120572)int

119905

0

1198911(119909 119904 119906 (119909 119904) V (119909 119904)) 119889119904

(1198792119906) (119909 119905)

=2 (1 minus 120573)

(2 minus 120573)119872 (120573)1198912(119909 119905 119906 (119909 119905) V (119909 119905))

minus2 (1 minus 120573)

(2 minus 120573)119872 (120573)1198912(0 0 119906 (0 0) V (0 0))

+2120573

(2 minus 120573)119872(120573)int

119905

0

1198912(119909 119904 119906 (119909 119904) V (119909 119904)) 119889119904

(16)

4 Journal of Function Spaces

and put

1198731=

4 minus 2120572

(2 minus 120572)119872 (120572)

1198732=

4 minus 2120573

(2 minus 120573)119872 (120573)

(17)

Theorem 8 Suppose that 1198911 1198912 [0 1]times [0 1]times119883times119883 rarr 119883

are the continuous mappings in system (6)-(7) and there existpositive constants 119871

1and 119871

2fulfilling |119891

1(119909 119905 119906

1 1199062)| le 119871

1

and |1198912(119909 119905 119906

1 1199062)| le 119871

2for all (119909 119905) isin [0 1] times [0 1] and

1199061 1199062isin 119883 Then system (6)-(7) possesses at least one solution

Proof Let the operators 1198791 1198792 119883 rarr 119883 defined by

(16) We define the operator 119879 119883 times 119883 rarr 119883 times 119883 by119879(119906 V)(119909 119905)fl ((119879

1V)(119909 119905) (119879

2119906)(119909 119905)) for all (119909 119905) isin [0 1] times

[0 1] Note that 119879 is continuous because the mappings 1198911

and 1198912are continuous We prove that the operator 119879 maps

bounded sets into the bounded subsets of 119883 times 119883 Let Ω be abounded subset of119883times119883 (119906 V) isin Ω and (119909 119905) isin [0 1]times[0 1]Then we have

1003816100381610038161003816(1198791V) (119909 119905)1003816100381610038161003816

=

10038161003816100381610038161003816100381610038161003816

2 (1 minus 120572)

(2 minus 120572)119872 (120572)1198911(119909 119905 119906 (119909 119905) V (119909 119905))

minus2 (1 minus 120572)

(2 minus 120572)119872 (120572)1198911(0 0 119906 (0 0) V (0 0))

+2120572

(2 minus 120572)119872 (120572)int

119905

0

1198911(119909 119904 119906 (119909 119904) V (119909 119904)) 119889119904

10038161003816100381610038161003816100381610038161003816

le2 (1 minus 120572)

(2 minus 120572)119872 (120572)

10038161003816100381610038161198911 (119909 119905 119906 (119909 119905) V (119909 119905))1003816100381610038161003816

+2 (1 minus 120572)

(2 minus 120572)119872 (120572)

10038161003816100381610038161198911 (0 0 0 0)1003816100381610038161003816 +

2120572

(2 minus 120572)119872 (120572)

sdot int

119905

0

10038161003816100381610038161198911 (119909 119904 119906 (119909 119904) V (119909 119904))1003816100381610038161003816 119889119904

le 1198711

2 (1 minus 120572)

(2 minus 120572)119872 (120572)+

2 (1 minus 120572)

(2 minus 120572)119872 (120572)

+2120572

(2 minus 120572)119872 (120572)119905 le 119871

1

4 (1 minus 120572)

(2 minus 120572)119872 (120572)

+2120572

(2 minus 120572)119872 (120572) le 119871

1

4 minus 2120572

(2 minus 120572)119872 (120572) = 119871

11198731

(18)

and so (1198791V)(119909 119905)

119883le 11987111198731 Also we have

1003816100381610038161003816(1198792119906) (119909 119905)1003816100381610038161003816

=

100381610038161003816100381610038161003816100381610038161003816

2 (1 minus 120573)

(2 minus 120573)119872 (120573)1198912(119909 119905 119906 (119909 119905) V (119909 119905))

minus2 (1 minus 120573)

(2 minus 120573)119872 (120573)1198912(0 0 119906 (0 0) V (0 0))

+2120573

(2 minus 120573)119872(120573)int

119905

0

1198912(119909 119904 119906 (119909 119904) V (119909 119904)) 119889119904

100381610038161003816100381610038161003816100381610038161003816

le2 (1 minus 120573)

(2 minus 120573)119872 (120573)

10038161003816100381610038161198912 (119909 119905 119906 (119909 119905) V (119909 119905))1003816100381610038161003816

+2 (1 minus 120573)

(2 minus 120573)119872 (120573)

10038161003816100381610038161198912 (0 0 0 0)1003816100381610038161003816 +

2120573

(2 minus 120573)119872(120573)

sdot int

119905

0

10038161003816100381610038161198912 (119909 119904 119906 (119909 119904) V (119909 s))1003816100381610038161003816 119889119904

le 1198712

2 (1 minus 120573)

(2 minus 120573)119872 (120573)+

2 (1 minus 120573)

(2 minus 120573)119872 (120573)

+2120573

(2 minus 120573)119872(120573)119905 le 119871

2

4 (1 minus 120573)

(2 minus 120573)119872 (120573)

+2120573

(2 minus 120573)119872(120573) le 119871

2

4 minus 2120573

(2 minus 120573)119872 (120573)

= 11987121198732

(19)

and so (1198792119906)(119909 119905)

119883le 11987121198732 Thus 119879(119906 V)(119909 119905)

119883times119883le

11987111198731+ 11987121198732 This shows that the operator 119879maps bounded

sets into the bounded sets of 119883 times 119883 Now we show that theoperator119879 is equicontinuous Let (119909 119905

1) (119909 119905

2) isin [0 1]times[0 1]

with 1199051lt 1199052 Then we have

1003816100381610038161003816(1198791V) (119909 1199052) minus (1198791V) (119909 1199051)1003816100381610038161003816

=

10038161003816100381610038161003816100381610038161003816

2 (1 minus 120572)

(2 minus 120572)119872 (120572)1198911(119909 1199052 119906 (119909 119905

2) V (119909 119905

2))

minus2 (1 minus 120572)

(2 minus 120572)119872 (120572)1198911(0 0 119906 (0 0) V (0 0))

+2120572

(2 minus 120572)119872 (120572)int

1199052

0

1198911(119909 119904 119906 (119909 119904) V (119909 119904)) 119889119904

minus2 (1 minus 120572)

(2 minus 120572)119872 (120572)1198911(119909 1199051 119906 (119909 119905

1) V (119909 119905

1))

+2 (1 minus 120572)

(2 minus 120572)119872 (120572)1198911(0 0 119906 (0 0) V (0 0))

minus2120572

(2 minus 120572)119872 (120572)int

1199051

0

1198911(119909 119904 119906 (119909 119904) V (119909 119904)) 119889119904

10038161003816100381610038161003816100381610038161003816

le2 (1 minus 120572)

(2 minus 120572)119872 (120572)

10038161003816100381610038161198911 (119909 1199052 119906 (119909 1199052) V (119909 1199052))

minus 1198911(119909 1199051 119906 (119909 119905

1) V (119909 119905

1))1003816100381610038161003816 +

2120572

(2 minus 120572)119872 (120572)

sdot int

1199052

1199051

10038161003816100381610038161198911 (119909 119904 119906 (119909 119904) V (119909 119904))1003816100381610038161003816 119889119904

le2 (1 minus 120572)

(2 minus 120572)119872 (120572)

10038161003816100381610038161198911 (119909 1199052 119906 (119909 1199052) V (119909 1199052))

minus 1198911(119909 1199051 119906 (119909 119905

1) V (119909 119905

1))1003816100381610038161003816 +

21205721198711

(2 minus 120572)119872 (120572)(1199052

minus 1199051)

(20)

Journal of Function Spaces 5

This implies that |(1198791V)(119909 119905

2) minus (119879

1V)(119909 119905

1)| rarr 0 whenever

(119909 1199052) rarr (119909 119905

1) By utilizing the Arzela-Ascoli theorem 119879

1

is completely continuous Similarly we have1003816100381610038161003816(1198792119906) (119909 1199052) minus (1198792119906) (119909 1199051)

1003816100381610038161003816

=

100381610038161003816100381610038161003816100381610038161003816

2 (1 minus 120573)

(2 minus 120573)119872 (120573)1198912(119909 1199052 119906 (119909 119905

2) V (119909 119905

2))

minus2 (1 minus 120573)

(2 minus 120573)119872 (120573)1198912(0 0 119906 (0 0) V (0 0))

+2120573

(2 minus 120573)119872 (120573)int

1199052

0

1198912(119909 119904 119906 (119909 119904) V (119909 119904)) 119889119904

minus2 (1 minus 120573)

(2 minus 120573)119872 (120573)1198912(119909 1199051 119906 (119909 119905

1) V (119909 119905

1))

+2 (1 minus 120573)

(2 minus 120573)119872 (120573)1198912(0 0 119906 (0 0) V (0 0))

minus2120573

(2 minus 120573)119872 (120573)int

1199051

0

1198912(119909 119904 119906 (119909 119904) V (119909 119904)) 119889119904

100381610038161003816100381610038161003816100381610038161003816

le2 (1 minus 120573)

(2 minus 120573)119872 (120573)

10038161003816100381610038161198912 (119909 1199052 119906 (119909 1199052) V (119909 1199052))

minus 1198912(119909 1199051 119906 (119909 119905

1) V (119909 119905

1))1003816100381610038161003816 +

2120573

(2 minus 120573)119872(120573)

sdot int

1199052

1199051

10038161003816100381610038161198912 (119909 119904 119906 (119909 119904) V (119909 119904))1003816100381610038161003816 119889119904

le2 (1 minus 120573)

(2 minus 120573)119872 (120573)

10038161003816100381610038161198912 (119909 1199052 119906 (119909 1199052) V (119909 1199052))

minus 1198912(119909 1199051 119906 (119909 119905

1) V (119909 119905

1))1003816100381610038161003816 +

21205731198712

(2 minus 120573)119872 (120573)(1199052

minus 1199051)

(21)

Again by utilizing the Arzela-Ascoli theorem we ob-serve that 119879

2is completely continuous Therefore we get

119879(119906 V)(119909 1199052) minus 119879(119906 V)(119909 119905

2)119883times119883

rarr 0 whenever (119909 1199052)

tends to (119909 1199051) Thus 119879 is completely continuous In the next

step we prove that

Ω = (119906 V) isin 119883 times 119883 (119906 V) = 120582119879 (119906 V) for some 120582

isin [0 1]

(22)

is bounded Let (119906 V) be an arbitrary element of Ω Choose120582 isin [0 1] fulfilling (119906 V) = 120582119879(119906 V) Hence V(119909 119905) =

120582(1198791V)(119909 119905) and 119906(119909 119905) = 120582(119879

2119906)(119909 119905) for all (119909 119905) isin [0 1] times

[0 1] Since

1

120582|V (119909 119905)| = 1003816100381610038161003816(1198791V) (119909 119905)

1003816100381610038161003816 le 11987111198731(23)

we get |V(119909 119905)| le 12058211987111198731and so V(119909 119905)

119883le 12058211987111198731 Simi-

larly we prove that 119906(119909 119905)119883le 12058211987121198732 Thus (119906 V)

119883times119883le

120582[11987111198731+ 11987121198732] and so Ω is a bounded set Now by using

Theorem 3 we get that 119879 has a fixed point which is a solutionfor the coupled system of the time-fractional differentialequations

Next we study the existence of solution for the coupledsystem of time-fractional differential inclusions

(CF119863120572

119905119906) (119909 119905) isin 119865

1(119909 119905 119906 (119909 119905) V (119909 119905))

(CF119863120573

119905V) (119909 119905) isin 119865

2(119909 119905 119906 (119909 119905) V (119909 119905))

(24)

with the initial value conditions 119906(0 0) = 0 and V(0 0) = 0where 119865

1 1198652 [0 1] times [0 1] times R times R rarr P(R) are some

multivalued maps

Definition 9 One says that (1199061 1199062) isin 119862([0 1] times [0 1] 119883) times

119862([0 1] times [0 1] 119883) is a solution for the system of thetime-fractional differential inclusions whenever it satisfiesthe initial value conditions and there exists (119908

1 1199082) isin

1198711([0 1] times [0 1]) times 119871

1([0 1] times [0 1]) such that 119908

119894(119909 119905) isin

119865119894(119909 119905 119906(119909 119905) V(119909 119905)) for almost all (119909 119905) isin [0 1] times [0 1] and

119894 = 1 2 and also

119906119894(119909 119905)

=2 (1 minus 120572)

(2 minus 120572)119872 (120572)119908119894(119909 119905 119906 (119909 119905) V (119909 119905))

minus2 (1 minus 120572)

(2 minus 120572)119872 (120572)119908119894(0 0 119906 (0 0) V (0 0))

+2120572

(2 minus 120572)119872 (120572)int

119905

0

119908119894(119909 119904 119906 (119909 119904) V (119909 119904)) 119889119904

(25)

for all (119909 119905) isin [0 1] times [0 1] and 119894 = 1 2

Theorem 10 Let 1198651 1198652 [0 1] times [0 1] timesR timesR rarr P

119888119901119888V(R)

be 1198711-Caratheodory multifunctions Suppose that there exist anondecreasing bounded continuousmap120595 [0infin) rarr (0infin)

and a continuous function 119901 [0 1] times [0 1] rarr (0infin) suchthat 119865

119894(119909 119905 119906

119894(119909 119905) 119906

1015840

119894(119909 119905)) le 119901(119909 119905)120595(119906

119894) for all (119909 119905) isin

[0 1] times [0 1] 119906119894 1199061015840

119894isin 119883 for 119894 = 1 2 Then coupled system

of time-fractional differential inclusions (8)-(9) has at least onesolution

Proof Define the operator119873 119883times119883 rarr 2119883times119883 by119873(119906

1 1199062) =

(1198731(1199061 1199062)

1198732(1199061 1199062)) where

1198731(1199061 1199062) = ℎ

1isin 119883 times 119883 there exists V

1

isin 1198781198651 1199061

such that ℎ1(119909 119905) = V

1(119909 119905) forall (119909 119905)

isin [0 1] times [0 1]

1198732(1199061 1199062) = ℎ

2isin 119883 times 119883 there exists V

2

isin 1198781198652 1199062

such that ℎ2(119909 119905) = V

2(119909 119905) forall (119909 119905)

isin [0 1] times [0 1]

6 Journal of Function Spaces

ℎ1(119909 119905) =

2 (1 minus 120572)

(2 minus 120572)119872 (120572)V1(119909 119905) minus

2 (1 minus 120572)

(2 minus 120572)119872 (120572)

sdot V1(0 0) +

2120572

(2 minus 120572)119872 (120572)int

119905

0

V1(119909 119904) 119889119904

ℎ2(119909 119905) =

2 (1 minus 120573)

(2 minus 120573)119872 (120573)V2(119909 119905) minus

2 (1 minus 120573)

(2 minus 120573)119872 (120573)

sdot V2(0 0) +

2120573

(2 minus 120573)119872(120573)int

119905

0

V2(119909 119904) 119889119904

(26)

By Lemma 7 it is clear that each fixed point of the operator119873 is a solution for system of time-fractional differentialinclusions (8) First we prove that the multifunction 119873 isconvex-valued Let (119906

1 1199062) isin 119883 times 119883 (ℎ

1 ℎ2) (ℎ1015840

1 ℎ1015840

2) isin

119873(1199061 1199062) Choose V

119894 V1015840119894isin 119878119865119894(1199061 1199062)

such that

ℎ119894(119909 119905) =

2 (1 minus 120572)

(2 minus 120572)119872 (120572)V119894(119909 119905)

minus2 (1 minus 120572)

(2 minus 120572)119872 (120572)V119894(0 0)

+2120572

(2 minus 120572)119872 (120572)int

119905

0

V119894(119909 119904) 119889119904

ℎ1015840

119894(119909 119905) =

2 (1 minus 120572)

(2 minus 120572)119872 (120572)V1015840119894(119909 119905)

minus2 (1 minus 120572)

(2 minus 120572)119872 (120572)V1015840119894(0 0)

+2120572

(2 minus 120572)119872 (120572)int

119905

0

V1015840119894(119909 119904) 119889119904

(27)

for almost all (119909 119905) isin [0 1] times [0 1] and 119894 = 1 2 Let 0 le 120582 le 1be given Then we have

[120582ℎ119894+ (1 minus 120582) ℎ

1015840

119894] (119909 119905)

=2 (1 minus 120572)

(2 minus 120572)119872 (120572)[120582V119894(119909 119905) + (1 minus 120582) V1015840

119894(119909 119905)]

minus2 (1 minus 120572)

(2 minus 120572)119872 (120572)[120582V119894(0 0) + (1 minus 120582) V1015840

119894(0 0)]

+2120572

(2 minus 120572)119872 (120572)

sdot int

119905

0

[120582V119894(119909 119904) + (1 minus 120582) V1015840

119894(119909 119904)] 119889119904

(28)

for 119894 = 1 2 Since the operator 119865119894has convex values 119878

119865119894(119906119894)is a

convex set and [120582ℎ119894+ (1 minus 120582)ℎ

1015840

119894] isin 119873119894(1199061 1199062) for 119894 = 1 2 This

implies that the operator119873 has convex values Now we provethat119873maps bounded sets of119883 into bounded sets Let 119903 gt 0119861119903= (1199061 1199062) isin 119883 times 119883 (119906

1 1199062) le 119903 be a bounded subset

of 119883 times 119883 (ℎ1 ℎ2) isin 119873(119906

1 1199062) and (119906

1 1199062) isin 119861119903 Then there

exists (V1 V2) isin 1198781198651(1199061)

times 1198781198652(1199062)

such that

ℎ1(119909 119905) =

2 (1 minus 120572)

(2 minus 120572)119872 (120572)V1(119909 119905)

minus2 (1 minus 120572)

(2 minus 120572)119872 (120572)V1(0 0)

+2120572

(2 minus 120572)119872 (120572)int

119905

0

V1(119909 119904) 119889119904

(29)

and ℎ2(119909 119905) = (2(1 minus120573)(2 minus120573)119872(120573))V

2(119909 119905) minus (2(1 minus120573)(2 minus

120573)119872(120573))V2(0 0) + (2120573(2 minus 120573)119872(120573)) int

119905

0V2(119909 119904)119889119904 for almost

all (119909 119905) isin [0 1] times [0 1] If 119901infin= sup

(119909119905)isin[01]times[01]|119901(119909 119905)|

then we obtain

1003816100381610038161003816(ℎ1) (119909 119905)1003816100381610038161003816 =

10038161003816100381610038161003816100381610038161003816

2 (1 minus 120572)

(2 minus 120572)119872 (120572)V1(119909 119905)

minus2 (1 minus 120572)

(2 minus 120572)119872 (120572)V1(0 0)

+2120572

(2 minus 120572)119872 (120572)int

119905

0

V1(119909 119904) 119889119904

10038161003816100381610038161003816100381610038161003816

le2 (1 minus 120572)

(2 minus 120572)119872 (120572)

1003816100381610038161003816V1 (119909 119905)1003816100381610038161003816

+2 (1 minus 120572)

(2 minus 120572)119872 (120572)

1003816100381610038161003816V1 (0 0)1003816100381610038161003816 +

2120572

(2 minus 120572)119872 (120572)

sdot int

119905

0

1003816100381610038161003816V1 (119909 119904)1003816100381610038161003816 119889119904 le 119901 (119909 119905) 120595 (

100381710038171003817100381711990611003817100381710038171003817)

sdot 2 (1 minus 120572)

(2 minus 120572)119872 (120572)

+2 (1 minus 120572)

(2 minus 120572)119872 (120572)+

2120572

(2 minus 120572)119872 (120572)119905 le

10038171003817100381710038171199011003817100381710038171003817infin

sdot 120595 (10038171003817100381710038171199061

1003817100381710038171003817) 4 (1 minus 120572)

(2 minus 120572)119872 (120572)+

2120572

(2 minus 120572)119872 (120572)

le10038171003817100381710038171199011003817100381710038171003817infin

120595 (10038171003817100381710038171199061

1003817100381710038171003817) 4 minus 2120572

(2 minus 120572)119872 (120572) =

10038171003817100381710038171199011003817100381710038171003817infin

sdot 120595 (10038171003817100381710038171199061

1003817100381710038171003817)1198731

(30)

where the constant 1198731is defined by (17) This implies

that ℎ1 le 119901

infin120595(1199061)1198731 Similarly we get ℎ

2 le

119901infin120595(1199062)1198732 where the constant 119873

2is defined by (17)

Thus (ℎ1 ℎ2) le 119901

infin120595((119906

1 1199062))(1198731+ 1198732) Now we

prove that119873maps bounded sets into equicontinuous subsets

Journal of Function Spaces 7

of 119883 times 119883 Let (1199061 1199062) isin 119861119903and (119909 119905

1) (119909 119905

2) isin [0 1] times [0 1]

with 1199051lt 1199052 Then we have

1003816100381610038161003816(ℎ1) (119909 1199052) minus (ℎ1) (119909 1199051)1003816100381610038161003816 =

10038161003816100381610038161003816100381610038161003816

2 (1 minus 120572)

(2 minus 120572)119872 (120572)V1(119909 1199052)

minus2 (1 minus 120572)

(2 minus 120572)119872 (120572)V1(0 0) +

2120572

(2 minus 120572)119872 (120572)

sdot int

1199052

0

V1(119909 119904) 119889119904 minus

2 (1 minus 120572)

(2 minus 120572)119872 (120572)V1(119909 1199051)

+2 (1 minus 120572)

(2 minus 120572)119872 (120572)V1(0 0)

minus2120572

(2 minus 120572)119872 (120572)int

1199051

0

V1(119909 119904) 119889119904

10038161003816100381610038161003816100381610038161003816

le2 (1 minus 120572)

(2 minus 120572)119872 (120572)

1003816100381610038161003816V1 (119909 1199052) minus V1(119909 1199051)1003816100381610038161003816

+2120572

(2 minus 120572)119872 (120572)int

1199052

1199051

1003816100381610038161003816V1 (119909 119904)1003816100381610038161003816 119889119904

le2 (1 minus 120572)

(2 minus 120572)119872 (120572)

1003816100381610038161003816V1 (119909 1199052) minus V1(119909 1199051)1003816100381610038161003816

+212057210038171003817100381710038171199011003817100381710038171003817infin

120595 (10038171003817100381710038171199061

1003817100381710038171003817)

(2 minus 120572)119872 (120572)(1199052minus 1199051)

(31)

By using a similar method we obtain1003816100381610038161003816(ℎ2) (119909 1199052) minus (ℎ2) (119909 1199051)

1003816100381610038161003816

le2 (1 minus 120573)

(2 minus 120573)119872 (120573)

1003816100381610038161003816V2 (119909 1199052) minus V2(119909 1199051)1003816100381610038161003816

+212057310038171003817100381710038171199011003817100381710038171003817infin

120595 (10038171003817100381710038171199062

1003817100381710038171003817)

(2 minus 120573)119872 (120573)(1199052minus 1199051)

(32)

Hence |ℎ119894(119909 1199052) minus ℎ119894(119909 1199051)| rarr 0 as (119909 119905

2) rarr (119909 119905

1) By

using the Arzela-Ascoli theorem we get that119873 is completelycontinuous Here we prove that119873 is upper semicontinuousBy using Lemma 4 119873 is upper semicontinuous whenever ithas a closed graph Since119873 is completely continuouswemustshow that119873 has a closed graph

Let (1199061198991 119906119899

2) be a sequence in 119883 times 119883 with (119906119899

1 119906119899

2) rarr

(1199060

1 1199060

2) and (ℎ119899

1 ℎ119899

2) isin 119873(119906

119899

1 119906119899

2)with (ℎ119899

1 ℎ119899

2) rarr (ℎ

0

1 ℎ0

2)We

show that (ℎ01 ℎ0

2) isin 119873(119906

0

1 1199060

2) For each (ℎ119899

1 ℎ119899

2) isin 119873(119906

119899

1 119906119899

2)

we can choose (V1198991 V1198992) isin 1198781198651(119906119899

1)times 1198781198652(119906119899

2)such that

ℎ119899

1(119909 119905) =

2 (1 minus 120572)

(2 minus 120572)119872 (120572)V1198991(119909 119905)

minus2 (1 minus 120572)

(2 minus 120572)119872 (120572)V1198991(0 0)

+2120572

(2 minus 120572)119872 (120572)int

119905

0

V1198991(119909 119904) 119889119904

(33)

and ℎ119899

2(119909 119905) = (2(1 minus 120573)(2 minus 120573)119872(120573))V119899

2(119909 119905) minus (2(1 minus

120573)(2 minus 120573)119872(120573))V1198992(0 0) + (2120573(2 minus 120573)119872(120573)) int

119905

0V1198992(119909 119904)119889119904 for

all (119909 119905) isin [0 1]times[0 1] It is sufficient to show that there exists(V01 V02) isin 1198781198651(1199060

1)times 1198781198652(1199060

2)such that

ℎ0

1(119909 119905) =

2 (1 minus 120572)

(2 minus 120572)119872 (120572)V01(119909 119905)

minus2 (1 minus 120572)

(2 minus 120572)119872 (120572)V01(0 0)

+2120572

(2 minus 120572)119872 (120572)int

119905

0

V01(119909 119904) 119889119904

(34)

and ℎ0

2(119909 119905) = (2(1 minus 120573)(2 minus 120573)119872(120573))V0

2(119909 119905) minus (2(1 minus

120573)(2 minus 120573)119872(120573))V02(0 0) + (2120573(2 minus 120573)119872(120573)) int

119905

0V02(119909 119904)119889119904 for

all (119909 119905) isin [0 1] times [0 1] Now consider the linear operatorsΘ1 Θ2 1198711([0 1] times [0 1] 119883) rarr 119862([0 1] times [0 1] 119883) defined

by

Θ1(V) (119909 119905) =

2 (1 minus 120572)

(2 minus 120572)119872 (120572)V (119909 119905)

minus2 (1 minus 120572)

(2 minus 120572)119872 (120572)V (0 0)

+2120572

(2 minus 120572)119872 (120572)int

119905

0

V (119909 119904) 119889119904

(35)

andΘ2(V)(119909 119905) = (2(1minus120573)(2minus120573)119872(120573))V(119909 119905)minus(2(1minus120573)(2minus

120573)119872(120573))V(0 0) + (2120573(2 minus 120573)119872(120573)) int119905

0V(119909 119904)119889119904 Note that

10038171003817100381710038171003817ℎ119899

1(119909 119905) minus ℎ

0

1(119909 119905)

10038171003817100381710038171003817

=

10038171003817100381710038171003817100381710038171003817

2 (1 minus 120572)

(2 minus 120572)119872 (120572)[V1198991(119909 119905) minus V0

1(119909 119905)]

minus2 (1 minus 120572)

(2 minus 120572)119872 (120572)[V1198991(0 0) minus V0

1(0 0)]

+2120572

(2 minus 120572)119872 (120572)int

119905

0

[V1198991(119909 119904) minus V0

1(119909 119904)] 119889119904

10038171003817100381710038171003817100381710038171003817

997888rarr 0

10038171003817100381710038171003817ℎ119899

2(119909 119905) minus ℎ

0

2(119909 119905)

10038171003817100381710038171003817

=

100381710038171003817100381710038171003817100381710038171003817

2 (1 minus 120573)

(2 minus 120573)119872 (120573)[V1198992(119909 119905) minus V0

2(119909 119905)]

minus2 (1 minus 120573)

(2 minus 120573)119872 (120573)[V1198992(0 0) minus V0

2(0 0)]

+2120573

(2 minus 120573)119872(120573)int

119905

0

[V1198992(119909 119904) minus V0

2(119909 119904)] 119889119904

100381710038171003817100381710038171003817100381710038171003817

997888rarr 0

(36)

8 Journal of Function Spaces

By using Lemma 5 we get that Θ119894sdot 119878119865119894

is a closed graphoperator for 119894 = 1 2 Also we get ℎ119899

119894(119909 119905) isin Θ

119894(119878119865119894(119906119899

119894)) for

all 119899 Since 119906119899119894rarr 1199060

119894 we get

ℎ0

1(119909 119905) =

2 (1 minus 120572)

(2 minus 120572)119872 (120572)V01(119909 119905)

minus2 (1 minus 120572)

(2 minus 120572)119872 (120572)V01(0 0)

+2120572

(2 minus 120572)119872 (120572)int

119905

0

V01(119909 119904) 119889119904

(37)

and ℎ02(119909 119905) = (2(1minus120573)(2minus120573)119872(120573))V0

2(119909 119905) minus (2(1minus120573)(2minus

120573)119872(120573))V02(0 0) + (2120573(2 minus 120573)119872(120573)) int

119905

0V02(119909 119904)119889119904 for some

V0119894isin 119878119865119894(1199060

119894)(119894 = 1 2) Thus119873 has a closed graph

Now we prove that there is an open set 119880 sube 119883 with(1199061 1199062) notin 119873(119906

1 1199062) for all 120582 isin (0 1) and (119906

1 1199062) isin 120597119880

Let 120582 isin (0 1) and (1199061 1199062) isin 120582119873(119906

1 1199062) Then there exists

V119894isin 1198711([0 1] times [0 1]R) with V

119894isin 119878119865119894(119906119894)

(119894 = 1 2) such that

1199061(119909 119905) =

2 (1 minus 120572)

(2 minus 120572)119872 (120572)V1(119909 119905)

minus2 (1 minus 120572)

(2 minus 120572)119872 (120572)V1(0 0)

+2120572

(2 minus 120572)119872 (120572)int

119905

0

V1(119909 119904) 119889119904

(38)

and 1199062(119909 119905) = (2(1 minus 120573)(2 minus 120573)119872(120573))V

2(119909 119905) minus (2(1 minus

120573)(2 minus 120573)119872(120573))V2(0 0) + (2120573(2 minus 120573)119872(120573)) int

119905

0V2(119909 119904)119889119904 for

all (119909 119905) isin [0 1] times [0 1] By using the above computedvalues we obtain 119906

119894 le 119901

infin120595(119906119894)sum119899

119894=1119873119894for 119894 = 1 2

This follows that 119906119894119901infin120595(119906119894)sum119899

119894=1119873119894le 1 for 119894 =

1 2 Choose 119872119894gt 0 with 119906

119894 = 119872

119894in such a way that

119872119894119901infin120595(119906119894)sum119899

119894=1119873119894gt 1 for 119894 = 1 2 Put 119880 = (119906

1 1199062) isin

119883 times 119883 (1199061 1199062) lt min 119872

11198722 We note that the

operator 119873 119880 rarr P(119883) is upper semicontinuous andcompletely continuous Also we showed that there is no(1199061 1199062) isin 120597119880 such that (119906

1 1199062) isin 120582119873(119906

1 1199062) for some

120582 isin (0 1) Hence with the help of Theorem 6 we get that119873 has a fixed point (119906

1 1199062) isin 119880 which is a solution for time-

fractional differential inclusion (8)-(9)

Conflict of Interests

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

Acknowledgment

Research of the last two authors was supported by AzarbaijanShahid Madani University

References

[1] R P Agarwal M Benchohra and S Hamani ldquoA survey onexistence results for boundary value problems of nonlinear frac-tional differential equations and inclusionsrdquo Acta ApplicandaeMathematicae vol 109 no 3 pp 973ndash1033 2010

[2] D Baleanu H Mohammadi and S Rezapour ldquoOn a nonlinearfractional differential equation on partially ordered metricspacesrdquo Advances in Difference Equations vol 2013 article 832013

[3] D Baleanu S Rezapour S Etemad and A Alsaedi ldquoOn a time-fractional integrodifferential equation via three-point boundaryvalue conditionsrdquo Mathematical Problems in Engineering vol2015 Article ID 785738 12 pages 2015

[4] K Diethelm The Analysis of Fractional Differential EquationsLecture Notes inMathematics Springer Berlin Germany 2010

[5] R Gorenflo and F Mainardi ldquoFractional calculus integraland differential equations of fractional orderrdquo in Fractals andFractional Calculus in ContinuumMechanics A Carpinteri andF Mainardi Eds Springer New York NY USA 1997

[6] G Samko A Kilbas and O Marichev Fractional Integrals andDerivatives Theory and Applications Gordon and Breach 1993

[7] Z Bai and W Sun ldquoExistence and multiplicity of positivesolutions for singular fractional boundary value problemsrdquoComputers amp Mathematics with Applications vol 63 no 9 pp1369ndash1381 2012

[8] D Baleanu R P Agarwal H Mohammadi and S RezapourldquoSome existence results for a nonlinear fractional differentialequation on partially ordered Banach spacesrdquo Boundary ValueProblems 2013112 8 pages 2013

[9] D Baleanu S Zahra Nazemi and S Rezapour ldquoThe existenceof positive solutions for a new coupled system of multiterm sin-gular fractional integrodifferential boundary value problemsrdquoAbstract and Applied Analysis vol 2013 Article ID 368659 15pages 2013

[10] M Caputo and M Fabrizio ldquoA new definition of fractionalderivative without singular kernelrdquo Progress in Fractional Dif-ferentiation and Applications vol 1 no 2 pp 73ndash85 2015

[11] J Losada and J J Nieto ldquoProperties of a new fractionalderivative without singular kernelrdquo Progress in Fractional Dif-ferentiation and Applications vol 1 no 2 pp 87ndash92 2015

[12] D R Smart Fixed PointTheorems Cambridge University PressCambridge UK 1980

[13] S M Aleomraninejad S Rezapour and N Shahzad ldquoOn fixedpoint generalizations of Suzukirsquos methodrdquo Applied MathematicsLetters vol 24 no 7 pp 1037ndash1040 2011

[14] H Covitz and S B Nadler ldquoMulti-valued contractionmappingsin generalizedmetric spacesrdquo Israel Journal of Mathematics vol8 pp 5ndash11 1970

[15] K S Miller and B Ross An Introduction to the FractionalCalculus and Fractional Differential Eqautions John Wiley ampSons 1993

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

Page 3: Research Article On Coupled Systems of Time-Fractional ...downloads.hindawi.com/journals/jfs/2016/4626940.pdf · Research Article On Coupled Systems of Time-Fractional Differential

Journal of Function Spaces 3

by the set Gr(119865) = (119909 119910) isin 119883 times 119884 119910 isin 119865(119909) (see [12 13])We say that the graph Gr(119865) of 119865 119883 rarr Pcl(119884) is a closedsubset of 119883 times 119884 whenever for all sequences 119906

119899119899isinN in 119883 and

119910119899119899isinN in 119884 with 119906

119899rarr 1199060 119910119899rarr 1199100 and 119910

119899isin 119865(119906

119899) for all

119899 we have 1199100isin 119865(119906

0) [12] Below we introduce the following

results which will be required in our proofs

Theorem 3 (see [12]) Suppose that 119883 is a Banach space 119879

119883 rarr 119883 is a completely continuous operator and the set 119870 =

119906 isin 119883 119906 = 120582119879119906 119891119900119903 119904119900119898119890 120582 isin [0 1] is bounded Then 119879has a fixed point

Lemma 4 (see [12 Proposition 12]) If 119865 119883 rarr P119888119897(119884) is

upper semicontinuous then119866119903(119865) is a closed subset of119883times119884 If119865 is completely continuous with a closed graph then it is uppersemicontinuous

Lemma 5 (see [12]) Let 119883 be a separable Banach space and119865 [0 1] times [0 1] times 119883 times 119883 rarr P

119888119901119888V(119883) an 1198711-Caratheodoryfunction Then the operator Θ sdot 119878

119865 119862119883([0 1] times [0 1]) rarr

P119888119901119888V(119862119883([0 1]times[0 1])) defined by 119906 997891rarr (Θsdot119878

119865)(119906) = Θ(119878

119865119906)

is a closed graph operator where Θ is a linear continuousmapping from 119871

1([0 1] times [0 1] 119883) into 119862

119883([0 1] times [0 1])

Theorem 6 (see [12]) Let 119864 be a Banach space 119862 a closedconvex subset of 119864 119880 an open subset of 119862 and 0 isin 119880Let us suppose that 119865 119880 rarr P

119888119901119888V(119862) depicts an uppersemicontinuous compact map such thatP

119888119901119888V(119862) denotes thefamily of nonempty compact convex subsets of 119862 Then either119865 admits a fixed point in119880 or there exist 119906 isin 120597119880 and 120582 isin (0 1)such that 119906 isin 120582119865(119906)

2 Main Results

First we investigate the coupled system

(CF119863120572

119905119906) (119909 119905) = 119891

1(119909 119905 119906 (119909 119905) V (119909 119905))

(CF119863120573

119905V) (119909 119905) = 119891

2(119909 119905 119906 (119909 119905) V (119909 119905))

(12)

equipped with the boundary value conditions 119906(0 0) = 0

and V(0 0) = 0 where 1198911 1198912 [0 1] times [0 1] times 119883

2rarr

119883 are continuous mappings 120572 120573 isin (0 1) 119909 119905 isin [0 1]and CF

119863120572

119905and CF

119863120573

119905are the Caputo-Fabrizio time-fractional

derivatives Now consider the Banach space 119883 = 119906

119906 isin 119862R([0 1] times [0 1]) endowed with the sup-norm 119906119883=

sup(119909119905)isin[01]times[01]

|119906(119909 119905)| Thus the product space (119883 times

119883 sdot 119883times119883

) is also a Banach space via the product norm(119906 V)

119883times119883= 119906119883+V119883 First we prove the next key lemma

Lemma 7 Suppose that119891 isin 1198711119883([0 1]times[0 1]) and 0 lt 120572 lt 1

The function 1199060isin 119862119883([0 1] times [0 1]) is a solution for the time-

fractional integral equation

119906 (119909 119905) =2 (1 minus 120572)

(2 minus 120572)119872 (120572)(119891 (119909 119905) minus 119891 (0 0))

+2120572

(2 minus 120572)119872 (120572)int

119905

0

119891 (119909 119904) 119889119904

(13)

if and only if1199060is a unique solution of the time-fractional differ-

ential equation

(119862119865119863120572

119905119906) (119909 119905) = 119891 (119909 119905) (119909 119905) isin [0 1] times [0 1]

119906 (0 0) = 0

(14)

Proof A solution of initial value problem (14) is denoted by1199060 As a result (CF119863120572

1199051199060)(119909 119905) = 119891(119909 119905) and 119906

0(0 0) = 0 By

integrating both sides we get

1199060(119909 119905) minus 119906

0(0 0)

=2 (1 minus 120572)

(2 minus 120572)119872 (120572)(119891 (119909 119905) minus 119891 (0 0))

+2120572

(2 minus 120572)119872 (120572)int

119905

0

119891 (119909 119904) 119889119904

(15)

and so 1199060(119909 119905) = (2(1 minus 120572)(2 minus 120572)119872(120572))(119891(119909 119905) minus 119891(0 0)) +

(2120572(2 minus 120572)119872(120572)) int119905

0119891(119909 119904)119889119904 This shows that 119906

0represents

the solution of time-fractional integral equation (13) If1199061and

1199062are two distinct solutions for initial value problem (14)

then CF119863120572

1199051199061(119909 119905) minus

CF119863120572

1199051199062(119909 119905) = [

CF119863120572

1199051199061minus 1199062](119909 119905) = 0

and (1199061minus 1199062)(0 0) = 0 By the property of the Caputo-

Fabrizio time-fractional derivative in [11] we get 1199061= 1199062

Hence 1199060is a unique solution of initial value problem (14)

Now suppose that 1199060is a solution of time-fractional integral

equation (13)Then we conclude that 1199060(119909 119905) = (2(1minus120572)(2minus

120572)119872(120572))(119891(119909 119905)minus119891(0 0))+(2120572(2minus120572)119872(120572)) int119905

0119891(119909 119904)119889119904 By

using (4) one can see that this function represents a solutionfor initial value problem (14) Note that 119906

0(0 0) = 0

Now we consider (1)-(2) For each (119909 119905) isin [0 1] times [0 1]define the operators 119879

1 1198792 119883 rarr 119883 by

(1198791V) (119909 119905)

=2 (1 minus 120572)

(2 minus 120572)119872 (120572)1198911(119909 119905 119906 (119909 119905) V (119909 119905))

minus2 (1 minus 120572)

(2 minus 120572)119872 (120572)1198911(0 0 119906 (0 0) V (0 0))

+2120572

(2 minus 120572)119872 (120572)int

119905

0

1198911(119909 119904 119906 (119909 119904) V (119909 119904)) 119889119904

(1198792119906) (119909 119905)

=2 (1 minus 120573)

(2 minus 120573)119872 (120573)1198912(119909 119905 119906 (119909 119905) V (119909 119905))

minus2 (1 minus 120573)

(2 minus 120573)119872 (120573)1198912(0 0 119906 (0 0) V (0 0))

+2120573

(2 minus 120573)119872(120573)int

119905

0

1198912(119909 119904 119906 (119909 119904) V (119909 119904)) 119889119904

(16)

4 Journal of Function Spaces

and put

1198731=

4 minus 2120572

(2 minus 120572)119872 (120572)

1198732=

4 minus 2120573

(2 minus 120573)119872 (120573)

(17)

Theorem 8 Suppose that 1198911 1198912 [0 1]times [0 1]times119883times119883 rarr 119883

are the continuous mappings in system (6)-(7) and there existpositive constants 119871

1and 119871

2fulfilling |119891

1(119909 119905 119906

1 1199062)| le 119871

1

and |1198912(119909 119905 119906

1 1199062)| le 119871

2for all (119909 119905) isin [0 1] times [0 1] and

1199061 1199062isin 119883 Then system (6)-(7) possesses at least one solution

Proof Let the operators 1198791 1198792 119883 rarr 119883 defined by

(16) We define the operator 119879 119883 times 119883 rarr 119883 times 119883 by119879(119906 V)(119909 119905)fl ((119879

1V)(119909 119905) (119879

2119906)(119909 119905)) for all (119909 119905) isin [0 1] times

[0 1] Note that 119879 is continuous because the mappings 1198911

and 1198912are continuous We prove that the operator 119879 maps

bounded sets into the bounded subsets of 119883 times 119883 Let Ω be abounded subset of119883times119883 (119906 V) isin Ω and (119909 119905) isin [0 1]times[0 1]Then we have

1003816100381610038161003816(1198791V) (119909 119905)1003816100381610038161003816

=

10038161003816100381610038161003816100381610038161003816

2 (1 minus 120572)

(2 minus 120572)119872 (120572)1198911(119909 119905 119906 (119909 119905) V (119909 119905))

minus2 (1 minus 120572)

(2 minus 120572)119872 (120572)1198911(0 0 119906 (0 0) V (0 0))

+2120572

(2 minus 120572)119872 (120572)int

119905

0

1198911(119909 119904 119906 (119909 119904) V (119909 119904)) 119889119904

10038161003816100381610038161003816100381610038161003816

le2 (1 minus 120572)

(2 minus 120572)119872 (120572)

10038161003816100381610038161198911 (119909 119905 119906 (119909 119905) V (119909 119905))1003816100381610038161003816

+2 (1 minus 120572)

(2 minus 120572)119872 (120572)

10038161003816100381610038161198911 (0 0 0 0)1003816100381610038161003816 +

2120572

(2 minus 120572)119872 (120572)

sdot int

119905

0

10038161003816100381610038161198911 (119909 119904 119906 (119909 119904) V (119909 119904))1003816100381610038161003816 119889119904

le 1198711

2 (1 minus 120572)

(2 minus 120572)119872 (120572)+

2 (1 minus 120572)

(2 minus 120572)119872 (120572)

+2120572

(2 minus 120572)119872 (120572)119905 le 119871

1

4 (1 minus 120572)

(2 minus 120572)119872 (120572)

+2120572

(2 minus 120572)119872 (120572) le 119871

1

4 minus 2120572

(2 minus 120572)119872 (120572) = 119871

11198731

(18)

and so (1198791V)(119909 119905)

119883le 11987111198731 Also we have

1003816100381610038161003816(1198792119906) (119909 119905)1003816100381610038161003816

=

100381610038161003816100381610038161003816100381610038161003816

2 (1 minus 120573)

(2 minus 120573)119872 (120573)1198912(119909 119905 119906 (119909 119905) V (119909 119905))

minus2 (1 minus 120573)

(2 minus 120573)119872 (120573)1198912(0 0 119906 (0 0) V (0 0))

+2120573

(2 minus 120573)119872(120573)int

119905

0

1198912(119909 119904 119906 (119909 119904) V (119909 119904)) 119889119904

100381610038161003816100381610038161003816100381610038161003816

le2 (1 minus 120573)

(2 minus 120573)119872 (120573)

10038161003816100381610038161198912 (119909 119905 119906 (119909 119905) V (119909 119905))1003816100381610038161003816

+2 (1 minus 120573)

(2 minus 120573)119872 (120573)

10038161003816100381610038161198912 (0 0 0 0)1003816100381610038161003816 +

2120573

(2 minus 120573)119872(120573)

sdot int

119905

0

10038161003816100381610038161198912 (119909 119904 119906 (119909 119904) V (119909 s))1003816100381610038161003816 119889119904

le 1198712

2 (1 minus 120573)

(2 minus 120573)119872 (120573)+

2 (1 minus 120573)

(2 minus 120573)119872 (120573)

+2120573

(2 minus 120573)119872(120573)119905 le 119871

2

4 (1 minus 120573)

(2 minus 120573)119872 (120573)

+2120573

(2 minus 120573)119872(120573) le 119871

2

4 minus 2120573

(2 minus 120573)119872 (120573)

= 11987121198732

(19)

and so (1198792119906)(119909 119905)

119883le 11987121198732 Thus 119879(119906 V)(119909 119905)

119883times119883le

11987111198731+ 11987121198732 This shows that the operator 119879maps bounded

sets into the bounded sets of 119883 times 119883 Now we show that theoperator119879 is equicontinuous Let (119909 119905

1) (119909 119905

2) isin [0 1]times[0 1]

with 1199051lt 1199052 Then we have

1003816100381610038161003816(1198791V) (119909 1199052) minus (1198791V) (119909 1199051)1003816100381610038161003816

=

10038161003816100381610038161003816100381610038161003816

2 (1 minus 120572)

(2 minus 120572)119872 (120572)1198911(119909 1199052 119906 (119909 119905

2) V (119909 119905

2))

minus2 (1 minus 120572)

(2 minus 120572)119872 (120572)1198911(0 0 119906 (0 0) V (0 0))

+2120572

(2 minus 120572)119872 (120572)int

1199052

0

1198911(119909 119904 119906 (119909 119904) V (119909 119904)) 119889119904

minus2 (1 minus 120572)

(2 minus 120572)119872 (120572)1198911(119909 1199051 119906 (119909 119905

1) V (119909 119905

1))

+2 (1 minus 120572)

(2 minus 120572)119872 (120572)1198911(0 0 119906 (0 0) V (0 0))

minus2120572

(2 minus 120572)119872 (120572)int

1199051

0

1198911(119909 119904 119906 (119909 119904) V (119909 119904)) 119889119904

10038161003816100381610038161003816100381610038161003816

le2 (1 minus 120572)

(2 minus 120572)119872 (120572)

10038161003816100381610038161198911 (119909 1199052 119906 (119909 1199052) V (119909 1199052))

minus 1198911(119909 1199051 119906 (119909 119905

1) V (119909 119905

1))1003816100381610038161003816 +

2120572

(2 minus 120572)119872 (120572)

sdot int

1199052

1199051

10038161003816100381610038161198911 (119909 119904 119906 (119909 119904) V (119909 119904))1003816100381610038161003816 119889119904

le2 (1 minus 120572)

(2 minus 120572)119872 (120572)

10038161003816100381610038161198911 (119909 1199052 119906 (119909 1199052) V (119909 1199052))

minus 1198911(119909 1199051 119906 (119909 119905

1) V (119909 119905

1))1003816100381610038161003816 +

21205721198711

(2 minus 120572)119872 (120572)(1199052

minus 1199051)

(20)

Journal of Function Spaces 5

This implies that |(1198791V)(119909 119905

2) minus (119879

1V)(119909 119905

1)| rarr 0 whenever

(119909 1199052) rarr (119909 119905

1) By utilizing the Arzela-Ascoli theorem 119879

1

is completely continuous Similarly we have1003816100381610038161003816(1198792119906) (119909 1199052) minus (1198792119906) (119909 1199051)

1003816100381610038161003816

=

100381610038161003816100381610038161003816100381610038161003816

2 (1 minus 120573)

(2 minus 120573)119872 (120573)1198912(119909 1199052 119906 (119909 119905

2) V (119909 119905

2))

minus2 (1 minus 120573)

(2 minus 120573)119872 (120573)1198912(0 0 119906 (0 0) V (0 0))

+2120573

(2 minus 120573)119872 (120573)int

1199052

0

1198912(119909 119904 119906 (119909 119904) V (119909 119904)) 119889119904

minus2 (1 minus 120573)

(2 minus 120573)119872 (120573)1198912(119909 1199051 119906 (119909 119905

1) V (119909 119905

1))

+2 (1 minus 120573)

(2 minus 120573)119872 (120573)1198912(0 0 119906 (0 0) V (0 0))

minus2120573

(2 minus 120573)119872 (120573)int

1199051

0

1198912(119909 119904 119906 (119909 119904) V (119909 119904)) 119889119904

100381610038161003816100381610038161003816100381610038161003816

le2 (1 minus 120573)

(2 minus 120573)119872 (120573)

10038161003816100381610038161198912 (119909 1199052 119906 (119909 1199052) V (119909 1199052))

minus 1198912(119909 1199051 119906 (119909 119905

1) V (119909 119905

1))1003816100381610038161003816 +

2120573

(2 minus 120573)119872(120573)

sdot int

1199052

1199051

10038161003816100381610038161198912 (119909 119904 119906 (119909 119904) V (119909 119904))1003816100381610038161003816 119889119904

le2 (1 minus 120573)

(2 minus 120573)119872 (120573)

10038161003816100381610038161198912 (119909 1199052 119906 (119909 1199052) V (119909 1199052))

minus 1198912(119909 1199051 119906 (119909 119905

1) V (119909 119905

1))1003816100381610038161003816 +

21205731198712

(2 minus 120573)119872 (120573)(1199052

minus 1199051)

(21)

Again by utilizing the Arzela-Ascoli theorem we ob-serve that 119879

2is completely continuous Therefore we get

119879(119906 V)(119909 1199052) minus 119879(119906 V)(119909 119905

2)119883times119883

rarr 0 whenever (119909 1199052)

tends to (119909 1199051) Thus 119879 is completely continuous In the next

step we prove that

Ω = (119906 V) isin 119883 times 119883 (119906 V) = 120582119879 (119906 V) for some 120582

isin [0 1]

(22)

is bounded Let (119906 V) be an arbitrary element of Ω Choose120582 isin [0 1] fulfilling (119906 V) = 120582119879(119906 V) Hence V(119909 119905) =

120582(1198791V)(119909 119905) and 119906(119909 119905) = 120582(119879

2119906)(119909 119905) for all (119909 119905) isin [0 1] times

[0 1] Since

1

120582|V (119909 119905)| = 1003816100381610038161003816(1198791V) (119909 119905)

1003816100381610038161003816 le 11987111198731(23)

we get |V(119909 119905)| le 12058211987111198731and so V(119909 119905)

119883le 12058211987111198731 Simi-

larly we prove that 119906(119909 119905)119883le 12058211987121198732 Thus (119906 V)

119883times119883le

120582[11987111198731+ 11987121198732] and so Ω is a bounded set Now by using

Theorem 3 we get that 119879 has a fixed point which is a solutionfor the coupled system of the time-fractional differentialequations

Next we study the existence of solution for the coupledsystem of time-fractional differential inclusions

(CF119863120572

119905119906) (119909 119905) isin 119865

1(119909 119905 119906 (119909 119905) V (119909 119905))

(CF119863120573

119905V) (119909 119905) isin 119865

2(119909 119905 119906 (119909 119905) V (119909 119905))

(24)

with the initial value conditions 119906(0 0) = 0 and V(0 0) = 0where 119865

1 1198652 [0 1] times [0 1] times R times R rarr P(R) are some

multivalued maps

Definition 9 One says that (1199061 1199062) isin 119862([0 1] times [0 1] 119883) times

119862([0 1] times [0 1] 119883) is a solution for the system of thetime-fractional differential inclusions whenever it satisfiesthe initial value conditions and there exists (119908

1 1199082) isin

1198711([0 1] times [0 1]) times 119871

1([0 1] times [0 1]) such that 119908

119894(119909 119905) isin

119865119894(119909 119905 119906(119909 119905) V(119909 119905)) for almost all (119909 119905) isin [0 1] times [0 1] and

119894 = 1 2 and also

119906119894(119909 119905)

=2 (1 minus 120572)

(2 minus 120572)119872 (120572)119908119894(119909 119905 119906 (119909 119905) V (119909 119905))

minus2 (1 minus 120572)

(2 minus 120572)119872 (120572)119908119894(0 0 119906 (0 0) V (0 0))

+2120572

(2 minus 120572)119872 (120572)int

119905

0

119908119894(119909 119904 119906 (119909 119904) V (119909 119904)) 119889119904

(25)

for all (119909 119905) isin [0 1] times [0 1] and 119894 = 1 2

Theorem 10 Let 1198651 1198652 [0 1] times [0 1] timesR timesR rarr P

119888119901119888V(R)

be 1198711-Caratheodory multifunctions Suppose that there exist anondecreasing bounded continuousmap120595 [0infin) rarr (0infin)

and a continuous function 119901 [0 1] times [0 1] rarr (0infin) suchthat 119865

119894(119909 119905 119906

119894(119909 119905) 119906

1015840

119894(119909 119905)) le 119901(119909 119905)120595(119906

119894) for all (119909 119905) isin

[0 1] times [0 1] 119906119894 1199061015840

119894isin 119883 for 119894 = 1 2 Then coupled system

of time-fractional differential inclusions (8)-(9) has at least onesolution

Proof Define the operator119873 119883times119883 rarr 2119883times119883 by119873(119906

1 1199062) =

(1198731(1199061 1199062)

1198732(1199061 1199062)) where

1198731(1199061 1199062) = ℎ

1isin 119883 times 119883 there exists V

1

isin 1198781198651 1199061

such that ℎ1(119909 119905) = V

1(119909 119905) forall (119909 119905)

isin [0 1] times [0 1]

1198732(1199061 1199062) = ℎ

2isin 119883 times 119883 there exists V

2

isin 1198781198652 1199062

such that ℎ2(119909 119905) = V

2(119909 119905) forall (119909 119905)

isin [0 1] times [0 1]

6 Journal of Function Spaces

ℎ1(119909 119905) =

2 (1 minus 120572)

(2 minus 120572)119872 (120572)V1(119909 119905) minus

2 (1 minus 120572)

(2 minus 120572)119872 (120572)

sdot V1(0 0) +

2120572

(2 minus 120572)119872 (120572)int

119905

0

V1(119909 119904) 119889119904

ℎ2(119909 119905) =

2 (1 minus 120573)

(2 minus 120573)119872 (120573)V2(119909 119905) minus

2 (1 minus 120573)

(2 minus 120573)119872 (120573)

sdot V2(0 0) +

2120573

(2 minus 120573)119872(120573)int

119905

0

V2(119909 119904) 119889119904

(26)

By Lemma 7 it is clear that each fixed point of the operator119873 is a solution for system of time-fractional differentialinclusions (8) First we prove that the multifunction 119873 isconvex-valued Let (119906

1 1199062) isin 119883 times 119883 (ℎ

1 ℎ2) (ℎ1015840

1 ℎ1015840

2) isin

119873(1199061 1199062) Choose V

119894 V1015840119894isin 119878119865119894(1199061 1199062)

such that

ℎ119894(119909 119905) =

2 (1 minus 120572)

(2 minus 120572)119872 (120572)V119894(119909 119905)

minus2 (1 minus 120572)

(2 minus 120572)119872 (120572)V119894(0 0)

+2120572

(2 minus 120572)119872 (120572)int

119905

0

V119894(119909 119904) 119889119904

ℎ1015840

119894(119909 119905) =

2 (1 minus 120572)

(2 minus 120572)119872 (120572)V1015840119894(119909 119905)

minus2 (1 minus 120572)

(2 minus 120572)119872 (120572)V1015840119894(0 0)

+2120572

(2 minus 120572)119872 (120572)int

119905

0

V1015840119894(119909 119904) 119889119904

(27)

for almost all (119909 119905) isin [0 1] times [0 1] and 119894 = 1 2 Let 0 le 120582 le 1be given Then we have

[120582ℎ119894+ (1 minus 120582) ℎ

1015840

119894] (119909 119905)

=2 (1 minus 120572)

(2 minus 120572)119872 (120572)[120582V119894(119909 119905) + (1 minus 120582) V1015840

119894(119909 119905)]

minus2 (1 minus 120572)

(2 minus 120572)119872 (120572)[120582V119894(0 0) + (1 minus 120582) V1015840

119894(0 0)]

+2120572

(2 minus 120572)119872 (120572)

sdot int

119905

0

[120582V119894(119909 119904) + (1 minus 120582) V1015840

119894(119909 119904)] 119889119904

(28)

for 119894 = 1 2 Since the operator 119865119894has convex values 119878

119865119894(119906119894)is a

convex set and [120582ℎ119894+ (1 minus 120582)ℎ

1015840

119894] isin 119873119894(1199061 1199062) for 119894 = 1 2 This

implies that the operator119873 has convex values Now we provethat119873maps bounded sets of119883 into bounded sets Let 119903 gt 0119861119903= (1199061 1199062) isin 119883 times 119883 (119906

1 1199062) le 119903 be a bounded subset

of 119883 times 119883 (ℎ1 ℎ2) isin 119873(119906

1 1199062) and (119906

1 1199062) isin 119861119903 Then there

exists (V1 V2) isin 1198781198651(1199061)

times 1198781198652(1199062)

such that

ℎ1(119909 119905) =

2 (1 minus 120572)

(2 minus 120572)119872 (120572)V1(119909 119905)

minus2 (1 minus 120572)

(2 minus 120572)119872 (120572)V1(0 0)

+2120572

(2 minus 120572)119872 (120572)int

119905

0

V1(119909 119904) 119889119904

(29)

and ℎ2(119909 119905) = (2(1 minus120573)(2 minus120573)119872(120573))V

2(119909 119905) minus (2(1 minus120573)(2 minus

120573)119872(120573))V2(0 0) + (2120573(2 minus 120573)119872(120573)) int

119905

0V2(119909 119904)119889119904 for almost

all (119909 119905) isin [0 1] times [0 1] If 119901infin= sup

(119909119905)isin[01]times[01]|119901(119909 119905)|

then we obtain

1003816100381610038161003816(ℎ1) (119909 119905)1003816100381610038161003816 =

10038161003816100381610038161003816100381610038161003816

2 (1 minus 120572)

(2 minus 120572)119872 (120572)V1(119909 119905)

minus2 (1 minus 120572)

(2 minus 120572)119872 (120572)V1(0 0)

+2120572

(2 minus 120572)119872 (120572)int

119905

0

V1(119909 119904) 119889119904

10038161003816100381610038161003816100381610038161003816

le2 (1 minus 120572)

(2 minus 120572)119872 (120572)

1003816100381610038161003816V1 (119909 119905)1003816100381610038161003816

+2 (1 minus 120572)

(2 minus 120572)119872 (120572)

1003816100381610038161003816V1 (0 0)1003816100381610038161003816 +

2120572

(2 minus 120572)119872 (120572)

sdot int

119905

0

1003816100381610038161003816V1 (119909 119904)1003816100381610038161003816 119889119904 le 119901 (119909 119905) 120595 (

100381710038171003817100381711990611003817100381710038171003817)

sdot 2 (1 minus 120572)

(2 minus 120572)119872 (120572)

+2 (1 minus 120572)

(2 minus 120572)119872 (120572)+

2120572

(2 minus 120572)119872 (120572)119905 le

10038171003817100381710038171199011003817100381710038171003817infin

sdot 120595 (10038171003817100381710038171199061

1003817100381710038171003817) 4 (1 minus 120572)

(2 minus 120572)119872 (120572)+

2120572

(2 minus 120572)119872 (120572)

le10038171003817100381710038171199011003817100381710038171003817infin

120595 (10038171003817100381710038171199061

1003817100381710038171003817) 4 minus 2120572

(2 minus 120572)119872 (120572) =

10038171003817100381710038171199011003817100381710038171003817infin

sdot 120595 (10038171003817100381710038171199061

1003817100381710038171003817)1198731

(30)

where the constant 1198731is defined by (17) This implies

that ℎ1 le 119901

infin120595(1199061)1198731 Similarly we get ℎ

2 le

119901infin120595(1199062)1198732 where the constant 119873

2is defined by (17)

Thus (ℎ1 ℎ2) le 119901

infin120595((119906

1 1199062))(1198731+ 1198732) Now we

prove that119873maps bounded sets into equicontinuous subsets

Journal of Function Spaces 7

of 119883 times 119883 Let (1199061 1199062) isin 119861119903and (119909 119905

1) (119909 119905

2) isin [0 1] times [0 1]

with 1199051lt 1199052 Then we have

1003816100381610038161003816(ℎ1) (119909 1199052) minus (ℎ1) (119909 1199051)1003816100381610038161003816 =

10038161003816100381610038161003816100381610038161003816

2 (1 minus 120572)

(2 minus 120572)119872 (120572)V1(119909 1199052)

minus2 (1 minus 120572)

(2 minus 120572)119872 (120572)V1(0 0) +

2120572

(2 minus 120572)119872 (120572)

sdot int

1199052

0

V1(119909 119904) 119889119904 minus

2 (1 minus 120572)

(2 minus 120572)119872 (120572)V1(119909 1199051)

+2 (1 minus 120572)

(2 minus 120572)119872 (120572)V1(0 0)

minus2120572

(2 minus 120572)119872 (120572)int

1199051

0

V1(119909 119904) 119889119904

10038161003816100381610038161003816100381610038161003816

le2 (1 minus 120572)

(2 minus 120572)119872 (120572)

1003816100381610038161003816V1 (119909 1199052) minus V1(119909 1199051)1003816100381610038161003816

+2120572

(2 minus 120572)119872 (120572)int

1199052

1199051

1003816100381610038161003816V1 (119909 119904)1003816100381610038161003816 119889119904

le2 (1 minus 120572)

(2 minus 120572)119872 (120572)

1003816100381610038161003816V1 (119909 1199052) minus V1(119909 1199051)1003816100381610038161003816

+212057210038171003817100381710038171199011003817100381710038171003817infin

120595 (10038171003817100381710038171199061

1003817100381710038171003817)

(2 minus 120572)119872 (120572)(1199052minus 1199051)

(31)

By using a similar method we obtain1003816100381610038161003816(ℎ2) (119909 1199052) minus (ℎ2) (119909 1199051)

1003816100381610038161003816

le2 (1 minus 120573)

(2 minus 120573)119872 (120573)

1003816100381610038161003816V2 (119909 1199052) minus V2(119909 1199051)1003816100381610038161003816

+212057310038171003817100381710038171199011003817100381710038171003817infin

120595 (10038171003817100381710038171199062

1003817100381710038171003817)

(2 minus 120573)119872 (120573)(1199052minus 1199051)

(32)

Hence |ℎ119894(119909 1199052) minus ℎ119894(119909 1199051)| rarr 0 as (119909 119905

2) rarr (119909 119905

1) By

using the Arzela-Ascoli theorem we get that119873 is completelycontinuous Here we prove that119873 is upper semicontinuousBy using Lemma 4 119873 is upper semicontinuous whenever ithas a closed graph Since119873 is completely continuouswemustshow that119873 has a closed graph

Let (1199061198991 119906119899

2) be a sequence in 119883 times 119883 with (119906119899

1 119906119899

2) rarr

(1199060

1 1199060

2) and (ℎ119899

1 ℎ119899

2) isin 119873(119906

119899

1 119906119899

2)with (ℎ119899

1 ℎ119899

2) rarr (ℎ

0

1 ℎ0

2)We

show that (ℎ01 ℎ0

2) isin 119873(119906

0

1 1199060

2) For each (ℎ119899

1 ℎ119899

2) isin 119873(119906

119899

1 119906119899

2)

we can choose (V1198991 V1198992) isin 1198781198651(119906119899

1)times 1198781198652(119906119899

2)such that

ℎ119899

1(119909 119905) =

2 (1 minus 120572)

(2 minus 120572)119872 (120572)V1198991(119909 119905)

minus2 (1 minus 120572)

(2 minus 120572)119872 (120572)V1198991(0 0)

+2120572

(2 minus 120572)119872 (120572)int

119905

0

V1198991(119909 119904) 119889119904

(33)

and ℎ119899

2(119909 119905) = (2(1 minus 120573)(2 minus 120573)119872(120573))V119899

2(119909 119905) minus (2(1 minus

120573)(2 minus 120573)119872(120573))V1198992(0 0) + (2120573(2 minus 120573)119872(120573)) int

119905

0V1198992(119909 119904)119889119904 for

all (119909 119905) isin [0 1]times[0 1] It is sufficient to show that there exists(V01 V02) isin 1198781198651(1199060

1)times 1198781198652(1199060

2)such that

ℎ0

1(119909 119905) =

2 (1 minus 120572)

(2 minus 120572)119872 (120572)V01(119909 119905)

minus2 (1 minus 120572)

(2 minus 120572)119872 (120572)V01(0 0)

+2120572

(2 minus 120572)119872 (120572)int

119905

0

V01(119909 119904) 119889119904

(34)

and ℎ0

2(119909 119905) = (2(1 minus 120573)(2 minus 120573)119872(120573))V0

2(119909 119905) minus (2(1 minus

120573)(2 minus 120573)119872(120573))V02(0 0) + (2120573(2 minus 120573)119872(120573)) int

119905

0V02(119909 119904)119889119904 for

all (119909 119905) isin [0 1] times [0 1] Now consider the linear operatorsΘ1 Θ2 1198711([0 1] times [0 1] 119883) rarr 119862([0 1] times [0 1] 119883) defined

by

Θ1(V) (119909 119905) =

2 (1 minus 120572)

(2 minus 120572)119872 (120572)V (119909 119905)

minus2 (1 minus 120572)

(2 minus 120572)119872 (120572)V (0 0)

+2120572

(2 minus 120572)119872 (120572)int

119905

0

V (119909 119904) 119889119904

(35)

andΘ2(V)(119909 119905) = (2(1minus120573)(2minus120573)119872(120573))V(119909 119905)minus(2(1minus120573)(2minus

120573)119872(120573))V(0 0) + (2120573(2 minus 120573)119872(120573)) int119905

0V(119909 119904)119889119904 Note that

10038171003817100381710038171003817ℎ119899

1(119909 119905) minus ℎ

0

1(119909 119905)

10038171003817100381710038171003817

=

10038171003817100381710038171003817100381710038171003817

2 (1 minus 120572)

(2 minus 120572)119872 (120572)[V1198991(119909 119905) minus V0

1(119909 119905)]

minus2 (1 minus 120572)

(2 minus 120572)119872 (120572)[V1198991(0 0) minus V0

1(0 0)]

+2120572

(2 minus 120572)119872 (120572)int

119905

0

[V1198991(119909 119904) minus V0

1(119909 119904)] 119889119904

10038171003817100381710038171003817100381710038171003817

997888rarr 0

10038171003817100381710038171003817ℎ119899

2(119909 119905) minus ℎ

0

2(119909 119905)

10038171003817100381710038171003817

=

100381710038171003817100381710038171003817100381710038171003817

2 (1 minus 120573)

(2 minus 120573)119872 (120573)[V1198992(119909 119905) minus V0

2(119909 119905)]

minus2 (1 minus 120573)

(2 minus 120573)119872 (120573)[V1198992(0 0) minus V0

2(0 0)]

+2120573

(2 minus 120573)119872(120573)int

119905

0

[V1198992(119909 119904) minus V0

2(119909 119904)] 119889119904

100381710038171003817100381710038171003817100381710038171003817

997888rarr 0

(36)

8 Journal of Function Spaces

By using Lemma 5 we get that Θ119894sdot 119878119865119894

is a closed graphoperator for 119894 = 1 2 Also we get ℎ119899

119894(119909 119905) isin Θ

119894(119878119865119894(119906119899

119894)) for

all 119899 Since 119906119899119894rarr 1199060

119894 we get

ℎ0

1(119909 119905) =

2 (1 minus 120572)

(2 minus 120572)119872 (120572)V01(119909 119905)

minus2 (1 minus 120572)

(2 minus 120572)119872 (120572)V01(0 0)

+2120572

(2 minus 120572)119872 (120572)int

119905

0

V01(119909 119904) 119889119904

(37)

and ℎ02(119909 119905) = (2(1minus120573)(2minus120573)119872(120573))V0

2(119909 119905) minus (2(1minus120573)(2minus

120573)119872(120573))V02(0 0) + (2120573(2 minus 120573)119872(120573)) int

119905

0V02(119909 119904)119889119904 for some

V0119894isin 119878119865119894(1199060

119894)(119894 = 1 2) Thus119873 has a closed graph

Now we prove that there is an open set 119880 sube 119883 with(1199061 1199062) notin 119873(119906

1 1199062) for all 120582 isin (0 1) and (119906

1 1199062) isin 120597119880

Let 120582 isin (0 1) and (1199061 1199062) isin 120582119873(119906

1 1199062) Then there exists

V119894isin 1198711([0 1] times [0 1]R) with V

119894isin 119878119865119894(119906119894)

(119894 = 1 2) such that

1199061(119909 119905) =

2 (1 minus 120572)

(2 minus 120572)119872 (120572)V1(119909 119905)

minus2 (1 minus 120572)

(2 minus 120572)119872 (120572)V1(0 0)

+2120572

(2 minus 120572)119872 (120572)int

119905

0

V1(119909 119904) 119889119904

(38)

and 1199062(119909 119905) = (2(1 minus 120573)(2 minus 120573)119872(120573))V

2(119909 119905) minus (2(1 minus

120573)(2 minus 120573)119872(120573))V2(0 0) + (2120573(2 minus 120573)119872(120573)) int

119905

0V2(119909 119904)119889119904 for

all (119909 119905) isin [0 1] times [0 1] By using the above computedvalues we obtain 119906

119894 le 119901

infin120595(119906119894)sum119899

119894=1119873119894for 119894 = 1 2

This follows that 119906119894119901infin120595(119906119894)sum119899

119894=1119873119894le 1 for 119894 =

1 2 Choose 119872119894gt 0 with 119906

119894 = 119872

119894in such a way that

119872119894119901infin120595(119906119894)sum119899

119894=1119873119894gt 1 for 119894 = 1 2 Put 119880 = (119906

1 1199062) isin

119883 times 119883 (1199061 1199062) lt min 119872

11198722 We note that the

operator 119873 119880 rarr P(119883) is upper semicontinuous andcompletely continuous Also we showed that there is no(1199061 1199062) isin 120597119880 such that (119906

1 1199062) isin 120582119873(119906

1 1199062) for some

120582 isin (0 1) Hence with the help of Theorem 6 we get that119873 has a fixed point (119906

1 1199062) isin 119880 which is a solution for time-

fractional differential inclusion (8)-(9)

Conflict of Interests

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

Acknowledgment

Research of the last two authors was supported by AzarbaijanShahid Madani University

References

[1] R P Agarwal M Benchohra and S Hamani ldquoA survey onexistence results for boundary value problems of nonlinear frac-tional differential equations and inclusionsrdquo Acta ApplicandaeMathematicae vol 109 no 3 pp 973ndash1033 2010

[2] D Baleanu H Mohammadi and S Rezapour ldquoOn a nonlinearfractional differential equation on partially ordered metricspacesrdquo Advances in Difference Equations vol 2013 article 832013

[3] D Baleanu S Rezapour S Etemad and A Alsaedi ldquoOn a time-fractional integrodifferential equation via three-point boundaryvalue conditionsrdquo Mathematical Problems in Engineering vol2015 Article ID 785738 12 pages 2015

[4] K Diethelm The Analysis of Fractional Differential EquationsLecture Notes inMathematics Springer Berlin Germany 2010

[5] R Gorenflo and F Mainardi ldquoFractional calculus integraland differential equations of fractional orderrdquo in Fractals andFractional Calculus in ContinuumMechanics A Carpinteri andF Mainardi Eds Springer New York NY USA 1997

[6] G Samko A Kilbas and O Marichev Fractional Integrals andDerivatives Theory and Applications Gordon and Breach 1993

[7] Z Bai and W Sun ldquoExistence and multiplicity of positivesolutions for singular fractional boundary value problemsrdquoComputers amp Mathematics with Applications vol 63 no 9 pp1369ndash1381 2012

[8] D Baleanu R P Agarwal H Mohammadi and S RezapourldquoSome existence results for a nonlinear fractional differentialequation on partially ordered Banach spacesrdquo Boundary ValueProblems 2013112 8 pages 2013

[9] D Baleanu S Zahra Nazemi and S Rezapour ldquoThe existenceof positive solutions for a new coupled system of multiterm sin-gular fractional integrodifferential boundary value problemsrdquoAbstract and Applied Analysis vol 2013 Article ID 368659 15pages 2013

[10] M Caputo and M Fabrizio ldquoA new definition of fractionalderivative without singular kernelrdquo Progress in Fractional Dif-ferentiation and Applications vol 1 no 2 pp 73ndash85 2015

[11] J Losada and J J Nieto ldquoProperties of a new fractionalderivative without singular kernelrdquo Progress in Fractional Dif-ferentiation and Applications vol 1 no 2 pp 87ndash92 2015

[12] D R Smart Fixed PointTheorems Cambridge University PressCambridge UK 1980

[13] S M Aleomraninejad S Rezapour and N Shahzad ldquoOn fixedpoint generalizations of Suzukirsquos methodrdquo Applied MathematicsLetters vol 24 no 7 pp 1037ndash1040 2011

[14] H Covitz and S B Nadler ldquoMulti-valued contractionmappingsin generalizedmetric spacesrdquo Israel Journal of Mathematics vol8 pp 5ndash11 1970

[15] K S Miller and B Ross An Introduction to the FractionalCalculus and Fractional Differential Eqautions John Wiley ampSons 1993

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

Page 4: Research Article On Coupled Systems of Time-Fractional ...downloads.hindawi.com/journals/jfs/2016/4626940.pdf · Research Article On Coupled Systems of Time-Fractional Differential

4 Journal of Function Spaces

and put

1198731=

4 minus 2120572

(2 minus 120572)119872 (120572)

1198732=

4 minus 2120573

(2 minus 120573)119872 (120573)

(17)

Theorem 8 Suppose that 1198911 1198912 [0 1]times [0 1]times119883times119883 rarr 119883

are the continuous mappings in system (6)-(7) and there existpositive constants 119871

1and 119871

2fulfilling |119891

1(119909 119905 119906

1 1199062)| le 119871

1

and |1198912(119909 119905 119906

1 1199062)| le 119871

2for all (119909 119905) isin [0 1] times [0 1] and

1199061 1199062isin 119883 Then system (6)-(7) possesses at least one solution

Proof Let the operators 1198791 1198792 119883 rarr 119883 defined by

(16) We define the operator 119879 119883 times 119883 rarr 119883 times 119883 by119879(119906 V)(119909 119905)fl ((119879

1V)(119909 119905) (119879

2119906)(119909 119905)) for all (119909 119905) isin [0 1] times

[0 1] Note that 119879 is continuous because the mappings 1198911

and 1198912are continuous We prove that the operator 119879 maps

bounded sets into the bounded subsets of 119883 times 119883 Let Ω be abounded subset of119883times119883 (119906 V) isin Ω and (119909 119905) isin [0 1]times[0 1]Then we have

1003816100381610038161003816(1198791V) (119909 119905)1003816100381610038161003816

=

10038161003816100381610038161003816100381610038161003816

2 (1 minus 120572)

(2 minus 120572)119872 (120572)1198911(119909 119905 119906 (119909 119905) V (119909 119905))

minus2 (1 minus 120572)

(2 minus 120572)119872 (120572)1198911(0 0 119906 (0 0) V (0 0))

+2120572

(2 minus 120572)119872 (120572)int

119905

0

1198911(119909 119904 119906 (119909 119904) V (119909 119904)) 119889119904

10038161003816100381610038161003816100381610038161003816

le2 (1 minus 120572)

(2 minus 120572)119872 (120572)

10038161003816100381610038161198911 (119909 119905 119906 (119909 119905) V (119909 119905))1003816100381610038161003816

+2 (1 minus 120572)

(2 minus 120572)119872 (120572)

10038161003816100381610038161198911 (0 0 0 0)1003816100381610038161003816 +

2120572

(2 minus 120572)119872 (120572)

sdot int

119905

0

10038161003816100381610038161198911 (119909 119904 119906 (119909 119904) V (119909 119904))1003816100381610038161003816 119889119904

le 1198711

2 (1 minus 120572)

(2 minus 120572)119872 (120572)+

2 (1 minus 120572)

(2 minus 120572)119872 (120572)

+2120572

(2 minus 120572)119872 (120572)119905 le 119871

1

4 (1 minus 120572)

(2 minus 120572)119872 (120572)

+2120572

(2 minus 120572)119872 (120572) le 119871

1

4 minus 2120572

(2 minus 120572)119872 (120572) = 119871

11198731

(18)

and so (1198791V)(119909 119905)

119883le 11987111198731 Also we have

1003816100381610038161003816(1198792119906) (119909 119905)1003816100381610038161003816

=

100381610038161003816100381610038161003816100381610038161003816

2 (1 minus 120573)

(2 minus 120573)119872 (120573)1198912(119909 119905 119906 (119909 119905) V (119909 119905))

minus2 (1 minus 120573)

(2 minus 120573)119872 (120573)1198912(0 0 119906 (0 0) V (0 0))

+2120573

(2 minus 120573)119872(120573)int

119905

0

1198912(119909 119904 119906 (119909 119904) V (119909 119904)) 119889119904

100381610038161003816100381610038161003816100381610038161003816

le2 (1 minus 120573)

(2 minus 120573)119872 (120573)

10038161003816100381610038161198912 (119909 119905 119906 (119909 119905) V (119909 119905))1003816100381610038161003816

+2 (1 minus 120573)

(2 minus 120573)119872 (120573)

10038161003816100381610038161198912 (0 0 0 0)1003816100381610038161003816 +

2120573

(2 minus 120573)119872(120573)

sdot int

119905

0

10038161003816100381610038161198912 (119909 119904 119906 (119909 119904) V (119909 s))1003816100381610038161003816 119889119904

le 1198712

2 (1 minus 120573)

(2 minus 120573)119872 (120573)+

2 (1 minus 120573)

(2 minus 120573)119872 (120573)

+2120573

(2 minus 120573)119872(120573)119905 le 119871

2

4 (1 minus 120573)

(2 minus 120573)119872 (120573)

+2120573

(2 minus 120573)119872(120573) le 119871

2

4 minus 2120573

(2 minus 120573)119872 (120573)

= 11987121198732

(19)

and so (1198792119906)(119909 119905)

119883le 11987121198732 Thus 119879(119906 V)(119909 119905)

119883times119883le

11987111198731+ 11987121198732 This shows that the operator 119879maps bounded

sets into the bounded sets of 119883 times 119883 Now we show that theoperator119879 is equicontinuous Let (119909 119905

1) (119909 119905

2) isin [0 1]times[0 1]

with 1199051lt 1199052 Then we have

1003816100381610038161003816(1198791V) (119909 1199052) minus (1198791V) (119909 1199051)1003816100381610038161003816

=

10038161003816100381610038161003816100381610038161003816

2 (1 minus 120572)

(2 minus 120572)119872 (120572)1198911(119909 1199052 119906 (119909 119905

2) V (119909 119905

2))

minus2 (1 minus 120572)

(2 minus 120572)119872 (120572)1198911(0 0 119906 (0 0) V (0 0))

+2120572

(2 minus 120572)119872 (120572)int

1199052

0

1198911(119909 119904 119906 (119909 119904) V (119909 119904)) 119889119904

minus2 (1 minus 120572)

(2 minus 120572)119872 (120572)1198911(119909 1199051 119906 (119909 119905

1) V (119909 119905

1))

+2 (1 minus 120572)

(2 minus 120572)119872 (120572)1198911(0 0 119906 (0 0) V (0 0))

minus2120572

(2 minus 120572)119872 (120572)int

1199051

0

1198911(119909 119904 119906 (119909 119904) V (119909 119904)) 119889119904

10038161003816100381610038161003816100381610038161003816

le2 (1 minus 120572)

(2 minus 120572)119872 (120572)

10038161003816100381610038161198911 (119909 1199052 119906 (119909 1199052) V (119909 1199052))

minus 1198911(119909 1199051 119906 (119909 119905

1) V (119909 119905

1))1003816100381610038161003816 +

2120572

(2 minus 120572)119872 (120572)

sdot int

1199052

1199051

10038161003816100381610038161198911 (119909 119904 119906 (119909 119904) V (119909 119904))1003816100381610038161003816 119889119904

le2 (1 minus 120572)

(2 minus 120572)119872 (120572)

10038161003816100381610038161198911 (119909 1199052 119906 (119909 1199052) V (119909 1199052))

minus 1198911(119909 1199051 119906 (119909 119905

1) V (119909 119905

1))1003816100381610038161003816 +

21205721198711

(2 minus 120572)119872 (120572)(1199052

minus 1199051)

(20)

Journal of Function Spaces 5

This implies that |(1198791V)(119909 119905

2) minus (119879

1V)(119909 119905

1)| rarr 0 whenever

(119909 1199052) rarr (119909 119905

1) By utilizing the Arzela-Ascoli theorem 119879

1

is completely continuous Similarly we have1003816100381610038161003816(1198792119906) (119909 1199052) minus (1198792119906) (119909 1199051)

1003816100381610038161003816

=

100381610038161003816100381610038161003816100381610038161003816

2 (1 minus 120573)

(2 minus 120573)119872 (120573)1198912(119909 1199052 119906 (119909 119905

2) V (119909 119905

2))

minus2 (1 minus 120573)

(2 minus 120573)119872 (120573)1198912(0 0 119906 (0 0) V (0 0))

+2120573

(2 minus 120573)119872 (120573)int

1199052

0

1198912(119909 119904 119906 (119909 119904) V (119909 119904)) 119889119904

minus2 (1 minus 120573)

(2 minus 120573)119872 (120573)1198912(119909 1199051 119906 (119909 119905

1) V (119909 119905

1))

+2 (1 minus 120573)

(2 minus 120573)119872 (120573)1198912(0 0 119906 (0 0) V (0 0))

minus2120573

(2 minus 120573)119872 (120573)int

1199051

0

1198912(119909 119904 119906 (119909 119904) V (119909 119904)) 119889119904

100381610038161003816100381610038161003816100381610038161003816

le2 (1 minus 120573)

(2 minus 120573)119872 (120573)

10038161003816100381610038161198912 (119909 1199052 119906 (119909 1199052) V (119909 1199052))

minus 1198912(119909 1199051 119906 (119909 119905

1) V (119909 119905

1))1003816100381610038161003816 +

2120573

(2 minus 120573)119872(120573)

sdot int

1199052

1199051

10038161003816100381610038161198912 (119909 119904 119906 (119909 119904) V (119909 119904))1003816100381610038161003816 119889119904

le2 (1 minus 120573)

(2 minus 120573)119872 (120573)

10038161003816100381610038161198912 (119909 1199052 119906 (119909 1199052) V (119909 1199052))

minus 1198912(119909 1199051 119906 (119909 119905

1) V (119909 119905

1))1003816100381610038161003816 +

21205731198712

(2 minus 120573)119872 (120573)(1199052

minus 1199051)

(21)

Again by utilizing the Arzela-Ascoli theorem we ob-serve that 119879

2is completely continuous Therefore we get

119879(119906 V)(119909 1199052) minus 119879(119906 V)(119909 119905

2)119883times119883

rarr 0 whenever (119909 1199052)

tends to (119909 1199051) Thus 119879 is completely continuous In the next

step we prove that

Ω = (119906 V) isin 119883 times 119883 (119906 V) = 120582119879 (119906 V) for some 120582

isin [0 1]

(22)

is bounded Let (119906 V) be an arbitrary element of Ω Choose120582 isin [0 1] fulfilling (119906 V) = 120582119879(119906 V) Hence V(119909 119905) =

120582(1198791V)(119909 119905) and 119906(119909 119905) = 120582(119879

2119906)(119909 119905) for all (119909 119905) isin [0 1] times

[0 1] Since

1

120582|V (119909 119905)| = 1003816100381610038161003816(1198791V) (119909 119905)

1003816100381610038161003816 le 11987111198731(23)

we get |V(119909 119905)| le 12058211987111198731and so V(119909 119905)

119883le 12058211987111198731 Simi-

larly we prove that 119906(119909 119905)119883le 12058211987121198732 Thus (119906 V)

119883times119883le

120582[11987111198731+ 11987121198732] and so Ω is a bounded set Now by using

Theorem 3 we get that 119879 has a fixed point which is a solutionfor the coupled system of the time-fractional differentialequations

Next we study the existence of solution for the coupledsystem of time-fractional differential inclusions

(CF119863120572

119905119906) (119909 119905) isin 119865

1(119909 119905 119906 (119909 119905) V (119909 119905))

(CF119863120573

119905V) (119909 119905) isin 119865

2(119909 119905 119906 (119909 119905) V (119909 119905))

(24)

with the initial value conditions 119906(0 0) = 0 and V(0 0) = 0where 119865

1 1198652 [0 1] times [0 1] times R times R rarr P(R) are some

multivalued maps

Definition 9 One says that (1199061 1199062) isin 119862([0 1] times [0 1] 119883) times

119862([0 1] times [0 1] 119883) is a solution for the system of thetime-fractional differential inclusions whenever it satisfiesthe initial value conditions and there exists (119908

1 1199082) isin

1198711([0 1] times [0 1]) times 119871

1([0 1] times [0 1]) such that 119908

119894(119909 119905) isin

119865119894(119909 119905 119906(119909 119905) V(119909 119905)) for almost all (119909 119905) isin [0 1] times [0 1] and

119894 = 1 2 and also

119906119894(119909 119905)

=2 (1 minus 120572)

(2 minus 120572)119872 (120572)119908119894(119909 119905 119906 (119909 119905) V (119909 119905))

minus2 (1 minus 120572)

(2 minus 120572)119872 (120572)119908119894(0 0 119906 (0 0) V (0 0))

+2120572

(2 minus 120572)119872 (120572)int

119905

0

119908119894(119909 119904 119906 (119909 119904) V (119909 119904)) 119889119904

(25)

for all (119909 119905) isin [0 1] times [0 1] and 119894 = 1 2

Theorem 10 Let 1198651 1198652 [0 1] times [0 1] timesR timesR rarr P

119888119901119888V(R)

be 1198711-Caratheodory multifunctions Suppose that there exist anondecreasing bounded continuousmap120595 [0infin) rarr (0infin)

and a continuous function 119901 [0 1] times [0 1] rarr (0infin) suchthat 119865

119894(119909 119905 119906

119894(119909 119905) 119906

1015840

119894(119909 119905)) le 119901(119909 119905)120595(119906

119894) for all (119909 119905) isin

[0 1] times [0 1] 119906119894 1199061015840

119894isin 119883 for 119894 = 1 2 Then coupled system

of time-fractional differential inclusions (8)-(9) has at least onesolution

Proof Define the operator119873 119883times119883 rarr 2119883times119883 by119873(119906

1 1199062) =

(1198731(1199061 1199062)

1198732(1199061 1199062)) where

1198731(1199061 1199062) = ℎ

1isin 119883 times 119883 there exists V

1

isin 1198781198651 1199061

such that ℎ1(119909 119905) = V

1(119909 119905) forall (119909 119905)

isin [0 1] times [0 1]

1198732(1199061 1199062) = ℎ

2isin 119883 times 119883 there exists V

2

isin 1198781198652 1199062

such that ℎ2(119909 119905) = V

2(119909 119905) forall (119909 119905)

isin [0 1] times [0 1]

6 Journal of Function Spaces

ℎ1(119909 119905) =

2 (1 minus 120572)

(2 minus 120572)119872 (120572)V1(119909 119905) minus

2 (1 minus 120572)

(2 minus 120572)119872 (120572)

sdot V1(0 0) +

2120572

(2 minus 120572)119872 (120572)int

119905

0

V1(119909 119904) 119889119904

ℎ2(119909 119905) =

2 (1 minus 120573)

(2 minus 120573)119872 (120573)V2(119909 119905) minus

2 (1 minus 120573)

(2 minus 120573)119872 (120573)

sdot V2(0 0) +

2120573

(2 minus 120573)119872(120573)int

119905

0

V2(119909 119904) 119889119904

(26)

By Lemma 7 it is clear that each fixed point of the operator119873 is a solution for system of time-fractional differentialinclusions (8) First we prove that the multifunction 119873 isconvex-valued Let (119906

1 1199062) isin 119883 times 119883 (ℎ

1 ℎ2) (ℎ1015840

1 ℎ1015840

2) isin

119873(1199061 1199062) Choose V

119894 V1015840119894isin 119878119865119894(1199061 1199062)

such that

ℎ119894(119909 119905) =

2 (1 minus 120572)

(2 minus 120572)119872 (120572)V119894(119909 119905)

minus2 (1 minus 120572)

(2 minus 120572)119872 (120572)V119894(0 0)

+2120572

(2 minus 120572)119872 (120572)int

119905

0

V119894(119909 119904) 119889119904

ℎ1015840

119894(119909 119905) =

2 (1 minus 120572)

(2 minus 120572)119872 (120572)V1015840119894(119909 119905)

minus2 (1 minus 120572)

(2 minus 120572)119872 (120572)V1015840119894(0 0)

+2120572

(2 minus 120572)119872 (120572)int

119905

0

V1015840119894(119909 119904) 119889119904

(27)

for almost all (119909 119905) isin [0 1] times [0 1] and 119894 = 1 2 Let 0 le 120582 le 1be given Then we have

[120582ℎ119894+ (1 minus 120582) ℎ

1015840

119894] (119909 119905)

=2 (1 minus 120572)

(2 minus 120572)119872 (120572)[120582V119894(119909 119905) + (1 minus 120582) V1015840

119894(119909 119905)]

minus2 (1 minus 120572)

(2 minus 120572)119872 (120572)[120582V119894(0 0) + (1 minus 120582) V1015840

119894(0 0)]

+2120572

(2 minus 120572)119872 (120572)

sdot int

119905

0

[120582V119894(119909 119904) + (1 minus 120582) V1015840

119894(119909 119904)] 119889119904

(28)

for 119894 = 1 2 Since the operator 119865119894has convex values 119878

119865119894(119906119894)is a

convex set and [120582ℎ119894+ (1 minus 120582)ℎ

1015840

119894] isin 119873119894(1199061 1199062) for 119894 = 1 2 This

implies that the operator119873 has convex values Now we provethat119873maps bounded sets of119883 into bounded sets Let 119903 gt 0119861119903= (1199061 1199062) isin 119883 times 119883 (119906

1 1199062) le 119903 be a bounded subset

of 119883 times 119883 (ℎ1 ℎ2) isin 119873(119906

1 1199062) and (119906

1 1199062) isin 119861119903 Then there

exists (V1 V2) isin 1198781198651(1199061)

times 1198781198652(1199062)

such that

ℎ1(119909 119905) =

2 (1 minus 120572)

(2 minus 120572)119872 (120572)V1(119909 119905)

minus2 (1 minus 120572)

(2 minus 120572)119872 (120572)V1(0 0)

+2120572

(2 minus 120572)119872 (120572)int

119905

0

V1(119909 119904) 119889119904

(29)

and ℎ2(119909 119905) = (2(1 minus120573)(2 minus120573)119872(120573))V

2(119909 119905) minus (2(1 minus120573)(2 minus

120573)119872(120573))V2(0 0) + (2120573(2 minus 120573)119872(120573)) int

119905

0V2(119909 119904)119889119904 for almost

all (119909 119905) isin [0 1] times [0 1] If 119901infin= sup

(119909119905)isin[01]times[01]|119901(119909 119905)|

then we obtain

1003816100381610038161003816(ℎ1) (119909 119905)1003816100381610038161003816 =

10038161003816100381610038161003816100381610038161003816

2 (1 minus 120572)

(2 minus 120572)119872 (120572)V1(119909 119905)

minus2 (1 minus 120572)

(2 minus 120572)119872 (120572)V1(0 0)

+2120572

(2 minus 120572)119872 (120572)int

119905

0

V1(119909 119904) 119889119904

10038161003816100381610038161003816100381610038161003816

le2 (1 minus 120572)

(2 minus 120572)119872 (120572)

1003816100381610038161003816V1 (119909 119905)1003816100381610038161003816

+2 (1 minus 120572)

(2 minus 120572)119872 (120572)

1003816100381610038161003816V1 (0 0)1003816100381610038161003816 +

2120572

(2 minus 120572)119872 (120572)

sdot int

119905

0

1003816100381610038161003816V1 (119909 119904)1003816100381610038161003816 119889119904 le 119901 (119909 119905) 120595 (

100381710038171003817100381711990611003817100381710038171003817)

sdot 2 (1 minus 120572)

(2 minus 120572)119872 (120572)

+2 (1 minus 120572)

(2 minus 120572)119872 (120572)+

2120572

(2 minus 120572)119872 (120572)119905 le

10038171003817100381710038171199011003817100381710038171003817infin

sdot 120595 (10038171003817100381710038171199061

1003817100381710038171003817) 4 (1 minus 120572)

(2 minus 120572)119872 (120572)+

2120572

(2 minus 120572)119872 (120572)

le10038171003817100381710038171199011003817100381710038171003817infin

120595 (10038171003817100381710038171199061

1003817100381710038171003817) 4 minus 2120572

(2 minus 120572)119872 (120572) =

10038171003817100381710038171199011003817100381710038171003817infin

sdot 120595 (10038171003817100381710038171199061

1003817100381710038171003817)1198731

(30)

where the constant 1198731is defined by (17) This implies

that ℎ1 le 119901

infin120595(1199061)1198731 Similarly we get ℎ

2 le

119901infin120595(1199062)1198732 where the constant 119873

2is defined by (17)

Thus (ℎ1 ℎ2) le 119901

infin120595((119906

1 1199062))(1198731+ 1198732) Now we

prove that119873maps bounded sets into equicontinuous subsets

Journal of Function Spaces 7

of 119883 times 119883 Let (1199061 1199062) isin 119861119903and (119909 119905

1) (119909 119905

2) isin [0 1] times [0 1]

with 1199051lt 1199052 Then we have

1003816100381610038161003816(ℎ1) (119909 1199052) minus (ℎ1) (119909 1199051)1003816100381610038161003816 =

10038161003816100381610038161003816100381610038161003816

2 (1 minus 120572)

(2 minus 120572)119872 (120572)V1(119909 1199052)

minus2 (1 minus 120572)

(2 minus 120572)119872 (120572)V1(0 0) +

2120572

(2 minus 120572)119872 (120572)

sdot int

1199052

0

V1(119909 119904) 119889119904 minus

2 (1 minus 120572)

(2 minus 120572)119872 (120572)V1(119909 1199051)

+2 (1 minus 120572)

(2 minus 120572)119872 (120572)V1(0 0)

minus2120572

(2 minus 120572)119872 (120572)int

1199051

0

V1(119909 119904) 119889119904

10038161003816100381610038161003816100381610038161003816

le2 (1 minus 120572)

(2 minus 120572)119872 (120572)

1003816100381610038161003816V1 (119909 1199052) minus V1(119909 1199051)1003816100381610038161003816

+2120572

(2 minus 120572)119872 (120572)int

1199052

1199051

1003816100381610038161003816V1 (119909 119904)1003816100381610038161003816 119889119904

le2 (1 minus 120572)

(2 minus 120572)119872 (120572)

1003816100381610038161003816V1 (119909 1199052) minus V1(119909 1199051)1003816100381610038161003816

+212057210038171003817100381710038171199011003817100381710038171003817infin

120595 (10038171003817100381710038171199061

1003817100381710038171003817)

(2 minus 120572)119872 (120572)(1199052minus 1199051)

(31)

By using a similar method we obtain1003816100381610038161003816(ℎ2) (119909 1199052) minus (ℎ2) (119909 1199051)

1003816100381610038161003816

le2 (1 minus 120573)

(2 minus 120573)119872 (120573)

1003816100381610038161003816V2 (119909 1199052) minus V2(119909 1199051)1003816100381610038161003816

+212057310038171003817100381710038171199011003817100381710038171003817infin

120595 (10038171003817100381710038171199062

1003817100381710038171003817)

(2 minus 120573)119872 (120573)(1199052minus 1199051)

(32)

Hence |ℎ119894(119909 1199052) minus ℎ119894(119909 1199051)| rarr 0 as (119909 119905

2) rarr (119909 119905

1) By

using the Arzela-Ascoli theorem we get that119873 is completelycontinuous Here we prove that119873 is upper semicontinuousBy using Lemma 4 119873 is upper semicontinuous whenever ithas a closed graph Since119873 is completely continuouswemustshow that119873 has a closed graph

Let (1199061198991 119906119899

2) be a sequence in 119883 times 119883 with (119906119899

1 119906119899

2) rarr

(1199060

1 1199060

2) and (ℎ119899

1 ℎ119899

2) isin 119873(119906

119899

1 119906119899

2)with (ℎ119899

1 ℎ119899

2) rarr (ℎ

0

1 ℎ0

2)We

show that (ℎ01 ℎ0

2) isin 119873(119906

0

1 1199060

2) For each (ℎ119899

1 ℎ119899

2) isin 119873(119906

119899

1 119906119899

2)

we can choose (V1198991 V1198992) isin 1198781198651(119906119899

1)times 1198781198652(119906119899

2)such that

ℎ119899

1(119909 119905) =

2 (1 minus 120572)

(2 minus 120572)119872 (120572)V1198991(119909 119905)

minus2 (1 minus 120572)

(2 minus 120572)119872 (120572)V1198991(0 0)

+2120572

(2 minus 120572)119872 (120572)int

119905

0

V1198991(119909 119904) 119889119904

(33)

and ℎ119899

2(119909 119905) = (2(1 minus 120573)(2 minus 120573)119872(120573))V119899

2(119909 119905) minus (2(1 minus

120573)(2 minus 120573)119872(120573))V1198992(0 0) + (2120573(2 minus 120573)119872(120573)) int

119905

0V1198992(119909 119904)119889119904 for

all (119909 119905) isin [0 1]times[0 1] It is sufficient to show that there exists(V01 V02) isin 1198781198651(1199060

1)times 1198781198652(1199060

2)such that

ℎ0

1(119909 119905) =

2 (1 minus 120572)

(2 minus 120572)119872 (120572)V01(119909 119905)

minus2 (1 minus 120572)

(2 minus 120572)119872 (120572)V01(0 0)

+2120572

(2 minus 120572)119872 (120572)int

119905

0

V01(119909 119904) 119889119904

(34)

and ℎ0

2(119909 119905) = (2(1 minus 120573)(2 minus 120573)119872(120573))V0

2(119909 119905) minus (2(1 minus

120573)(2 minus 120573)119872(120573))V02(0 0) + (2120573(2 minus 120573)119872(120573)) int

119905

0V02(119909 119904)119889119904 for

all (119909 119905) isin [0 1] times [0 1] Now consider the linear operatorsΘ1 Θ2 1198711([0 1] times [0 1] 119883) rarr 119862([0 1] times [0 1] 119883) defined

by

Θ1(V) (119909 119905) =

2 (1 minus 120572)

(2 minus 120572)119872 (120572)V (119909 119905)

minus2 (1 minus 120572)

(2 minus 120572)119872 (120572)V (0 0)

+2120572

(2 minus 120572)119872 (120572)int

119905

0

V (119909 119904) 119889119904

(35)

andΘ2(V)(119909 119905) = (2(1minus120573)(2minus120573)119872(120573))V(119909 119905)minus(2(1minus120573)(2minus

120573)119872(120573))V(0 0) + (2120573(2 minus 120573)119872(120573)) int119905

0V(119909 119904)119889119904 Note that

10038171003817100381710038171003817ℎ119899

1(119909 119905) minus ℎ

0

1(119909 119905)

10038171003817100381710038171003817

=

10038171003817100381710038171003817100381710038171003817

2 (1 minus 120572)

(2 minus 120572)119872 (120572)[V1198991(119909 119905) minus V0

1(119909 119905)]

minus2 (1 minus 120572)

(2 minus 120572)119872 (120572)[V1198991(0 0) minus V0

1(0 0)]

+2120572

(2 minus 120572)119872 (120572)int

119905

0

[V1198991(119909 119904) minus V0

1(119909 119904)] 119889119904

10038171003817100381710038171003817100381710038171003817

997888rarr 0

10038171003817100381710038171003817ℎ119899

2(119909 119905) minus ℎ

0

2(119909 119905)

10038171003817100381710038171003817

=

100381710038171003817100381710038171003817100381710038171003817

2 (1 minus 120573)

(2 minus 120573)119872 (120573)[V1198992(119909 119905) minus V0

2(119909 119905)]

minus2 (1 minus 120573)

(2 minus 120573)119872 (120573)[V1198992(0 0) minus V0

2(0 0)]

+2120573

(2 minus 120573)119872(120573)int

119905

0

[V1198992(119909 119904) minus V0

2(119909 119904)] 119889119904

100381710038171003817100381710038171003817100381710038171003817

997888rarr 0

(36)

8 Journal of Function Spaces

By using Lemma 5 we get that Θ119894sdot 119878119865119894

is a closed graphoperator for 119894 = 1 2 Also we get ℎ119899

119894(119909 119905) isin Θ

119894(119878119865119894(119906119899

119894)) for

all 119899 Since 119906119899119894rarr 1199060

119894 we get

ℎ0

1(119909 119905) =

2 (1 minus 120572)

(2 minus 120572)119872 (120572)V01(119909 119905)

minus2 (1 minus 120572)

(2 minus 120572)119872 (120572)V01(0 0)

+2120572

(2 minus 120572)119872 (120572)int

119905

0

V01(119909 119904) 119889119904

(37)

and ℎ02(119909 119905) = (2(1minus120573)(2minus120573)119872(120573))V0

2(119909 119905) minus (2(1minus120573)(2minus

120573)119872(120573))V02(0 0) + (2120573(2 minus 120573)119872(120573)) int

119905

0V02(119909 119904)119889119904 for some

V0119894isin 119878119865119894(1199060

119894)(119894 = 1 2) Thus119873 has a closed graph

Now we prove that there is an open set 119880 sube 119883 with(1199061 1199062) notin 119873(119906

1 1199062) for all 120582 isin (0 1) and (119906

1 1199062) isin 120597119880

Let 120582 isin (0 1) and (1199061 1199062) isin 120582119873(119906

1 1199062) Then there exists

V119894isin 1198711([0 1] times [0 1]R) with V

119894isin 119878119865119894(119906119894)

(119894 = 1 2) such that

1199061(119909 119905) =

2 (1 minus 120572)

(2 minus 120572)119872 (120572)V1(119909 119905)

minus2 (1 minus 120572)

(2 minus 120572)119872 (120572)V1(0 0)

+2120572

(2 minus 120572)119872 (120572)int

119905

0

V1(119909 119904) 119889119904

(38)

and 1199062(119909 119905) = (2(1 minus 120573)(2 minus 120573)119872(120573))V

2(119909 119905) minus (2(1 minus

120573)(2 minus 120573)119872(120573))V2(0 0) + (2120573(2 minus 120573)119872(120573)) int

119905

0V2(119909 119904)119889119904 for

all (119909 119905) isin [0 1] times [0 1] By using the above computedvalues we obtain 119906

119894 le 119901

infin120595(119906119894)sum119899

119894=1119873119894for 119894 = 1 2

This follows that 119906119894119901infin120595(119906119894)sum119899

119894=1119873119894le 1 for 119894 =

1 2 Choose 119872119894gt 0 with 119906

119894 = 119872

119894in such a way that

119872119894119901infin120595(119906119894)sum119899

119894=1119873119894gt 1 for 119894 = 1 2 Put 119880 = (119906

1 1199062) isin

119883 times 119883 (1199061 1199062) lt min 119872

11198722 We note that the

operator 119873 119880 rarr P(119883) is upper semicontinuous andcompletely continuous Also we showed that there is no(1199061 1199062) isin 120597119880 such that (119906

1 1199062) isin 120582119873(119906

1 1199062) for some

120582 isin (0 1) Hence with the help of Theorem 6 we get that119873 has a fixed point (119906

1 1199062) isin 119880 which is a solution for time-

fractional differential inclusion (8)-(9)

Conflict of Interests

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

Acknowledgment

Research of the last two authors was supported by AzarbaijanShahid Madani University

References

[1] R P Agarwal M Benchohra and S Hamani ldquoA survey onexistence results for boundary value problems of nonlinear frac-tional differential equations and inclusionsrdquo Acta ApplicandaeMathematicae vol 109 no 3 pp 973ndash1033 2010

[2] D Baleanu H Mohammadi and S Rezapour ldquoOn a nonlinearfractional differential equation on partially ordered metricspacesrdquo Advances in Difference Equations vol 2013 article 832013

[3] D Baleanu S Rezapour S Etemad and A Alsaedi ldquoOn a time-fractional integrodifferential equation via three-point boundaryvalue conditionsrdquo Mathematical Problems in Engineering vol2015 Article ID 785738 12 pages 2015

[4] K Diethelm The Analysis of Fractional Differential EquationsLecture Notes inMathematics Springer Berlin Germany 2010

[5] R Gorenflo and F Mainardi ldquoFractional calculus integraland differential equations of fractional orderrdquo in Fractals andFractional Calculus in ContinuumMechanics A Carpinteri andF Mainardi Eds Springer New York NY USA 1997

[6] G Samko A Kilbas and O Marichev Fractional Integrals andDerivatives Theory and Applications Gordon and Breach 1993

[7] Z Bai and W Sun ldquoExistence and multiplicity of positivesolutions for singular fractional boundary value problemsrdquoComputers amp Mathematics with Applications vol 63 no 9 pp1369ndash1381 2012

[8] D Baleanu R P Agarwal H Mohammadi and S RezapourldquoSome existence results for a nonlinear fractional differentialequation on partially ordered Banach spacesrdquo Boundary ValueProblems 2013112 8 pages 2013

[9] D Baleanu S Zahra Nazemi and S Rezapour ldquoThe existenceof positive solutions for a new coupled system of multiterm sin-gular fractional integrodifferential boundary value problemsrdquoAbstract and Applied Analysis vol 2013 Article ID 368659 15pages 2013

[10] M Caputo and M Fabrizio ldquoA new definition of fractionalderivative without singular kernelrdquo Progress in Fractional Dif-ferentiation and Applications vol 1 no 2 pp 73ndash85 2015

[11] J Losada and J J Nieto ldquoProperties of a new fractionalderivative without singular kernelrdquo Progress in Fractional Dif-ferentiation and Applications vol 1 no 2 pp 87ndash92 2015

[12] D R Smart Fixed PointTheorems Cambridge University PressCambridge UK 1980

[13] S M Aleomraninejad S Rezapour and N Shahzad ldquoOn fixedpoint generalizations of Suzukirsquos methodrdquo Applied MathematicsLetters vol 24 no 7 pp 1037ndash1040 2011

[14] H Covitz and S B Nadler ldquoMulti-valued contractionmappingsin generalizedmetric spacesrdquo Israel Journal of Mathematics vol8 pp 5ndash11 1970

[15] K S Miller and B Ross An Introduction to the FractionalCalculus and Fractional Differential Eqautions John Wiley ampSons 1993

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

Page 5: Research Article On Coupled Systems of Time-Fractional ...downloads.hindawi.com/journals/jfs/2016/4626940.pdf · Research Article On Coupled Systems of Time-Fractional Differential

Journal of Function Spaces 5

This implies that |(1198791V)(119909 119905

2) minus (119879

1V)(119909 119905

1)| rarr 0 whenever

(119909 1199052) rarr (119909 119905

1) By utilizing the Arzela-Ascoli theorem 119879

1

is completely continuous Similarly we have1003816100381610038161003816(1198792119906) (119909 1199052) minus (1198792119906) (119909 1199051)

1003816100381610038161003816

=

100381610038161003816100381610038161003816100381610038161003816

2 (1 minus 120573)

(2 minus 120573)119872 (120573)1198912(119909 1199052 119906 (119909 119905

2) V (119909 119905

2))

minus2 (1 minus 120573)

(2 minus 120573)119872 (120573)1198912(0 0 119906 (0 0) V (0 0))

+2120573

(2 minus 120573)119872 (120573)int

1199052

0

1198912(119909 119904 119906 (119909 119904) V (119909 119904)) 119889119904

minus2 (1 minus 120573)

(2 minus 120573)119872 (120573)1198912(119909 1199051 119906 (119909 119905

1) V (119909 119905

1))

+2 (1 minus 120573)

(2 minus 120573)119872 (120573)1198912(0 0 119906 (0 0) V (0 0))

minus2120573

(2 minus 120573)119872 (120573)int

1199051

0

1198912(119909 119904 119906 (119909 119904) V (119909 119904)) 119889119904

100381610038161003816100381610038161003816100381610038161003816

le2 (1 minus 120573)

(2 minus 120573)119872 (120573)

10038161003816100381610038161198912 (119909 1199052 119906 (119909 1199052) V (119909 1199052))

minus 1198912(119909 1199051 119906 (119909 119905

1) V (119909 119905

1))1003816100381610038161003816 +

2120573

(2 minus 120573)119872(120573)

sdot int

1199052

1199051

10038161003816100381610038161198912 (119909 119904 119906 (119909 119904) V (119909 119904))1003816100381610038161003816 119889119904

le2 (1 minus 120573)

(2 minus 120573)119872 (120573)

10038161003816100381610038161198912 (119909 1199052 119906 (119909 1199052) V (119909 1199052))

minus 1198912(119909 1199051 119906 (119909 119905

1) V (119909 119905

1))1003816100381610038161003816 +

21205731198712

(2 minus 120573)119872 (120573)(1199052

minus 1199051)

(21)

Again by utilizing the Arzela-Ascoli theorem we ob-serve that 119879

2is completely continuous Therefore we get

119879(119906 V)(119909 1199052) minus 119879(119906 V)(119909 119905

2)119883times119883

rarr 0 whenever (119909 1199052)

tends to (119909 1199051) Thus 119879 is completely continuous In the next

step we prove that

Ω = (119906 V) isin 119883 times 119883 (119906 V) = 120582119879 (119906 V) for some 120582

isin [0 1]

(22)

is bounded Let (119906 V) be an arbitrary element of Ω Choose120582 isin [0 1] fulfilling (119906 V) = 120582119879(119906 V) Hence V(119909 119905) =

120582(1198791V)(119909 119905) and 119906(119909 119905) = 120582(119879

2119906)(119909 119905) for all (119909 119905) isin [0 1] times

[0 1] Since

1

120582|V (119909 119905)| = 1003816100381610038161003816(1198791V) (119909 119905)

1003816100381610038161003816 le 11987111198731(23)

we get |V(119909 119905)| le 12058211987111198731and so V(119909 119905)

119883le 12058211987111198731 Simi-

larly we prove that 119906(119909 119905)119883le 12058211987121198732 Thus (119906 V)

119883times119883le

120582[11987111198731+ 11987121198732] and so Ω is a bounded set Now by using

Theorem 3 we get that 119879 has a fixed point which is a solutionfor the coupled system of the time-fractional differentialequations

Next we study the existence of solution for the coupledsystem of time-fractional differential inclusions

(CF119863120572

119905119906) (119909 119905) isin 119865

1(119909 119905 119906 (119909 119905) V (119909 119905))

(CF119863120573

119905V) (119909 119905) isin 119865

2(119909 119905 119906 (119909 119905) V (119909 119905))

(24)

with the initial value conditions 119906(0 0) = 0 and V(0 0) = 0where 119865

1 1198652 [0 1] times [0 1] times R times R rarr P(R) are some

multivalued maps

Definition 9 One says that (1199061 1199062) isin 119862([0 1] times [0 1] 119883) times

119862([0 1] times [0 1] 119883) is a solution for the system of thetime-fractional differential inclusions whenever it satisfiesthe initial value conditions and there exists (119908

1 1199082) isin

1198711([0 1] times [0 1]) times 119871

1([0 1] times [0 1]) such that 119908

119894(119909 119905) isin

119865119894(119909 119905 119906(119909 119905) V(119909 119905)) for almost all (119909 119905) isin [0 1] times [0 1] and

119894 = 1 2 and also

119906119894(119909 119905)

=2 (1 minus 120572)

(2 minus 120572)119872 (120572)119908119894(119909 119905 119906 (119909 119905) V (119909 119905))

minus2 (1 minus 120572)

(2 minus 120572)119872 (120572)119908119894(0 0 119906 (0 0) V (0 0))

+2120572

(2 minus 120572)119872 (120572)int

119905

0

119908119894(119909 119904 119906 (119909 119904) V (119909 119904)) 119889119904

(25)

for all (119909 119905) isin [0 1] times [0 1] and 119894 = 1 2

Theorem 10 Let 1198651 1198652 [0 1] times [0 1] timesR timesR rarr P

119888119901119888V(R)

be 1198711-Caratheodory multifunctions Suppose that there exist anondecreasing bounded continuousmap120595 [0infin) rarr (0infin)

and a continuous function 119901 [0 1] times [0 1] rarr (0infin) suchthat 119865

119894(119909 119905 119906

119894(119909 119905) 119906

1015840

119894(119909 119905)) le 119901(119909 119905)120595(119906

119894) for all (119909 119905) isin

[0 1] times [0 1] 119906119894 1199061015840

119894isin 119883 for 119894 = 1 2 Then coupled system

of time-fractional differential inclusions (8)-(9) has at least onesolution

Proof Define the operator119873 119883times119883 rarr 2119883times119883 by119873(119906

1 1199062) =

(1198731(1199061 1199062)

1198732(1199061 1199062)) where

1198731(1199061 1199062) = ℎ

1isin 119883 times 119883 there exists V

1

isin 1198781198651 1199061

such that ℎ1(119909 119905) = V

1(119909 119905) forall (119909 119905)

isin [0 1] times [0 1]

1198732(1199061 1199062) = ℎ

2isin 119883 times 119883 there exists V

2

isin 1198781198652 1199062

such that ℎ2(119909 119905) = V

2(119909 119905) forall (119909 119905)

isin [0 1] times [0 1]

6 Journal of Function Spaces

ℎ1(119909 119905) =

2 (1 minus 120572)

(2 minus 120572)119872 (120572)V1(119909 119905) minus

2 (1 minus 120572)

(2 minus 120572)119872 (120572)

sdot V1(0 0) +

2120572

(2 minus 120572)119872 (120572)int

119905

0

V1(119909 119904) 119889119904

ℎ2(119909 119905) =

2 (1 minus 120573)

(2 minus 120573)119872 (120573)V2(119909 119905) minus

2 (1 minus 120573)

(2 minus 120573)119872 (120573)

sdot V2(0 0) +

2120573

(2 minus 120573)119872(120573)int

119905

0

V2(119909 119904) 119889119904

(26)

By Lemma 7 it is clear that each fixed point of the operator119873 is a solution for system of time-fractional differentialinclusions (8) First we prove that the multifunction 119873 isconvex-valued Let (119906

1 1199062) isin 119883 times 119883 (ℎ

1 ℎ2) (ℎ1015840

1 ℎ1015840

2) isin

119873(1199061 1199062) Choose V

119894 V1015840119894isin 119878119865119894(1199061 1199062)

such that

ℎ119894(119909 119905) =

2 (1 minus 120572)

(2 minus 120572)119872 (120572)V119894(119909 119905)

minus2 (1 minus 120572)

(2 minus 120572)119872 (120572)V119894(0 0)

+2120572

(2 minus 120572)119872 (120572)int

119905

0

V119894(119909 119904) 119889119904

ℎ1015840

119894(119909 119905) =

2 (1 minus 120572)

(2 minus 120572)119872 (120572)V1015840119894(119909 119905)

minus2 (1 minus 120572)

(2 minus 120572)119872 (120572)V1015840119894(0 0)

+2120572

(2 minus 120572)119872 (120572)int

119905

0

V1015840119894(119909 119904) 119889119904

(27)

for almost all (119909 119905) isin [0 1] times [0 1] and 119894 = 1 2 Let 0 le 120582 le 1be given Then we have

[120582ℎ119894+ (1 minus 120582) ℎ

1015840

119894] (119909 119905)

=2 (1 minus 120572)

(2 minus 120572)119872 (120572)[120582V119894(119909 119905) + (1 minus 120582) V1015840

119894(119909 119905)]

minus2 (1 minus 120572)

(2 minus 120572)119872 (120572)[120582V119894(0 0) + (1 minus 120582) V1015840

119894(0 0)]

+2120572

(2 minus 120572)119872 (120572)

sdot int

119905

0

[120582V119894(119909 119904) + (1 minus 120582) V1015840

119894(119909 119904)] 119889119904

(28)

for 119894 = 1 2 Since the operator 119865119894has convex values 119878

119865119894(119906119894)is a

convex set and [120582ℎ119894+ (1 minus 120582)ℎ

1015840

119894] isin 119873119894(1199061 1199062) for 119894 = 1 2 This

implies that the operator119873 has convex values Now we provethat119873maps bounded sets of119883 into bounded sets Let 119903 gt 0119861119903= (1199061 1199062) isin 119883 times 119883 (119906

1 1199062) le 119903 be a bounded subset

of 119883 times 119883 (ℎ1 ℎ2) isin 119873(119906

1 1199062) and (119906

1 1199062) isin 119861119903 Then there

exists (V1 V2) isin 1198781198651(1199061)

times 1198781198652(1199062)

such that

ℎ1(119909 119905) =

2 (1 minus 120572)

(2 minus 120572)119872 (120572)V1(119909 119905)

minus2 (1 minus 120572)

(2 minus 120572)119872 (120572)V1(0 0)

+2120572

(2 minus 120572)119872 (120572)int

119905

0

V1(119909 119904) 119889119904

(29)

and ℎ2(119909 119905) = (2(1 minus120573)(2 minus120573)119872(120573))V

2(119909 119905) minus (2(1 minus120573)(2 minus

120573)119872(120573))V2(0 0) + (2120573(2 minus 120573)119872(120573)) int

119905

0V2(119909 119904)119889119904 for almost

all (119909 119905) isin [0 1] times [0 1] If 119901infin= sup

(119909119905)isin[01]times[01]|119901(119909 119905)|

then we obtain

1003816100381610038161003816(ℎ1) (119909 119905)1003816100381610038161003816 =

10038161003816100381610038161003816100381610038161003816

2 (1 minus 120572)

(2 minus 120572)119872 (120572)V1(119909 119905)

minus2 (1 minus 120572)

(2 minus 120572)119872 (120572)V1(0 0)

+2120572

(2 minus 120572)119872 (120572)int

119905

0

V1(119909 119904) 119889119904

10038161003816100381610038161003816100381610038161003816

le2 (1 minus 120572)

(2 minus 120572)119872 (120572)

1003816100381610038161003816V1 (119909 119905)1003816100381610038161003816

+2 (1 minus 120572)

(2 minus 120572)119872 (120572)

1003816100381610038161003816V1 (0 0)1003816100381610038161003816 +

2120572

(2 minus 120572)119872 (120572)

sdot int

119905

0

1003816100381610038161003816V1 (119909 119904)1003816100381610038161003816 119889119904 le 119901 (119909 119905) 120595 (

100381710038171003817100381711990611003817100381710038171003817)

sdot 2 (1 minus 120572)

(2 minus 120572)119872 (120572)

+2 (1 minus 120572)

(2 minus 120572)119872 (120572)+

2120572

(2 minus 120572)119872 (120572)119905 le

10038171003817100381710038171199011003817100381710038171003817infin

sdot 120595 (10038171003817100381710038171199061

1003817100381710038171003817) 4 (1 minus 120572)

(2 minus 120572)119872 (120572)+

2120572

(2 minus 120572)119872 (120572)

le10038171003817100381710038171199011003817100381710038171003817infin

120595 (10038171003817100381710038171199061

1003817100381710038171003817) 4 minus 2120572

(2 minus 120572)119872 (120572) =

10038171003817100381710038171199011003817100381710038171003817infin

sdot 120595 (10038171003817100381710038171199061

1003817100381710038171003817)1198731

(30)

where the constant 1198731is defined by (17) This implies

that ℎ1 le 119901

infin120595(1199061)1198731 Similarly we get ℎ

2 le

119901infin120595(1199062)1198732 where the constant 119873

2is defined by (17)

Thus (ℎ1 ℎ2) le 119901

infin120595((119906

1 1199062))(1198731+ 1198732) Now we

prove that119873maps bounded sets into equicontinuous subsets

Journal of Function Spaces 7

of 119883 times 119883 Let (1199061 1199062) isin 119861119903and (119909 119905

1) (119909 119905

2) isin [0 1] times [0 1]

with 1199051lt 1199052 Then we have

1003816100381610038161003816(ℎ1) (119909 1199052) minus (ℎ1) (119909 1199051)1003816100381610038161003816 =

10038161003816100381610038161003816100381610038161003816

2 (1 minus 120572)

(2 minus 120572)119872 (120572)V1(119909 1199052)

minus2 (1 minus 120572)

(2 minus 120572)119872 (120572)V1(0 0) +

2120572

(2 minus 120572)119872 (120572)

sdot int

1199052

0

V1(119909 119904) 119889119904 minus

2 (1 minus 120572)

(2 minus 120572)119872 (120572)V1(119909 1199051)

+2 (1 minus 120572)

(2 minus 120572)119872 (120572)V1(0 0)

minus2120572

(2 minus 120572)119872 (120572)int

1199051

0

V1(119909 119904) 119889119904

10038161003816100381610038161003816100381610038161003816

le2 (1 minus 120572)

(2 minus 120572)119872 (120572)

1003816100381610038161003816V1 (119909 1199052) minus V1(119909 1199051)1003816100381610038161003816

+2120572

(2 minus 120572)119872 (120572)int

1199052

1199051

1003816100381610038161003816V1 (119909 119904)1003816100381610038161003816 119889119904

le2 (1 minus 120572)

(2 minus 120572)119872 (120572)

1003816100381610038161003816V1 (119909 1199052) minus V1(119909 1199051)1003816100381610038161003816

+212057210038171003817100381710038171199011003817100381710038171003817infin

120595 (10038171003817100381710038171199061

1003817100381710038171003817)

(2 minus 120572)119872 (120572)(1199052minus 1199051)

(31)

By using a similar method we obtain1003816100381610038161003816(ℎ2) (119909 1199052) minus (ℎ2) (119909 1199051)

1003816100381610038161003816

le2 (1 minus 120573)

(2 minus 120573)119872 (120573)

1003816100381610038161003816V2 (119909 1199052) minus V2(119909 1199051)1003816100381610038161003816

+212057310038171003817100381710038171199011003817100381710038171003817infin

120595 (10038171003817100381710038171199062

1003817100381710038171003817)

(2 minus 120573)119872 (120573)(1199052minus 1199051)

(32)

Hence |ℎ119894(119909 1199052) minus ℎ119894(119909 1199051)| rarr 0 as (119909 119905

2) rarr (119909 119905

1) By

using the Arzela-Ascoli theorem we get that119873 is completelycontinuous Here we prove that119873 is upper semicontinuousBy using Lemma 4 119873 is upper semicontinuous whenever ithas a closed graph Since119873 is completely continuouswemustshow that119873 has a closed graph

Let (1199061198991 119906119899

2) be a sequence in 119883 times 119883 with (119906119899

1 119906119899

2) rarr

(1199060

1 1199060

2) and (ℎ119899

1 ℎ119899

2) isin 119873(119906

119899

1 119906119899

2)with (ℎ119899

1 ℎ119899

2) rarr (ℎ

0

1 ℎ0

2)We

show that (ℎ01 ℎ0

2) isin 119873(119906

0

1 1199060

2) For each (ℎ119899

1 ℎ119899

2) isin 119873(119906

119899

1 119906119899

2)

we can choose (V1198991 V1198992) isin 1198781198651(119906119899

1)times 1198781198652(119906119899

2)such that

ℎ119899

1(119909 119905) =

2 (1 minus 120572)

(2 minus 120572)119872 (120572)V1198991(119909 119905)

minus2 (1 minus 120572)

(2 minus 120572)119872 (120572)V1198991(0 0)

+2120572

(2 minus 120572)119872 (120572)int

119905

0

V1198991(119909 119904) 119889119904

(33)

and ℎ119899

2(119909 119905) = (2(1 minus 120573)(2 minus 120573)119872(120573))V119899

2(119909 119905) minus (2(1 minus

120573)(2 minus 120573)119872(120573))V1198992(0 0) + (2120573(2 minus 120573)119872(120573)) int

119905

0V1198992(119909 119904)119889119904 for

all (119909 119905) isin [0 1]times[0 1] It is sufficient to show that there exists(V01 V02) isin 1198781198651(1199060

1)times 1198781198652(1199060

2)such that

ℎ0

1(119909 119905) =

2 (1 minus 120572)

(2 minus 120572)119872 (120572)V01(119909 119905)

minus2 (1 minus 120572)

(2 minus 120572)119872 (120572)V01(0 0)

+2120572

(2 minus 120572)119872 (120572)int

119905

0

V01(119909 119904) 119889119904

(34)

and ℎ0

2(119909 119905) = (2(1 minus 120573)(2 minus 120573)119872(120573))V0

2(119909 119905) minus (2(1 minus

120573)(2 minus 120573)119872(120573))V02(0 0) + (2120573(2 minus 120573)119872(120573)) int

119905

0V02(119909 119904)119889119904 for

all (119909 119905) isin [0 1] times [0 1] Now consider the linear operatorsΘ1 Θ2 1198711([0 1] times [0 1] 119883) rarr 119862([0 1] times [0 1] 119883) defined

by

Θ1(V) (119909 119905) =

2 (1 minus 120572)

(2 minus 120572)119872 (120572)V (119909 119905)

minus2 (1 minus 120572)

(2 minus 120572)119872 (120572)V (0 0)

+2120572

(2 minus 120572)119872 (120572)int

119905

0

V (119909 119904) 119889119904

(35)

andΘ2(V)(119909 119905) = (2(1minus120573)(2minus120573)119872(120573))V(119909 119905)minus(2(1minus120573)(2minus

120573)119872(120573))V(0 0) + (2120573(2 minus 120573)119872(120573)) int119905

0V(119909 119904)119889119904 Note that

10038171003817100381710038171003817ℎ119899

1(119909 119905) minus ℎ

0

1(119909 119905)

10038171003817100381710038171003817

=

10038171003817100381710038171003817100381710038171003817

2 (1 minus 120572)

(2 minus 120572)119872 (120572)[V1198991(119909 119905) minus V0

1(119909 119905)]

minus2 (1 minus 120572)

(2 minus 120572)119872 (120572)[V1198991(0 0) minus V0

1(0 0)]

+2120572

(2 minus 120572)119872 (120572)int

119905

0

[V1198991(119909 119904) minus V0

1(119909 119904)] 119889119904

10038171003817100381710038171003817100381710038171003817

997888rarr 0

10038171003817100381710038171003817ℎ119899

2(119909 119905) minus ℎ

0

2(119909 119905)

10038171003817100381710038171003817

=

100381710038171003817100381710038171003817100381710038171003817

2 (1 minus 120573)

(2 minus 120573)119872 (120573)[V1198992(119909 119905) minus V0

2(119909 119905)]

minus2 (1 minus 120573)

(2 minus 120573)119872 (120573)[V1198992(0 0) minus V0

2(0 0)]

+2120573

(2 minus 120573)119872(120573)int

119905

0

[V1198992(119909 119904) minus V0

2(119909 119904)] 119889119904

100381710038171003817100381710038171003817100381710038171003817

997888rarr 0

(36)

8 Journal of Function Spaces

By using Lemma 5 we get that Θ119894sdot 119878119865119894

is a closed graphoperator for 119894 = 1 2 Also we get ℎ119899

119894(119909 119905) isin Θ

119894(119878119865119894(119906119899

119894)) for

all 119899 Since 119906119899119894rarr 1199060

119894 we get

ℎ0

1(119909 119905) =

2 (1 minus 120572)

(2 minus 120572)119872 (120572)V01(119909 119905)

minus2 (1 minus 120572)

(2 minus 120572)119872 (120572)V01(0 0)

+2120572

(2 minus 120572)119872 (120572)int

119905

0

V01(119909 119904) 119889119904

(37)

and ℎ02(119909 119905) = (2(1minus120573)(2minus120573)119872(120573))V0

2(119909 119905) minus (2(1minus120573)(2minus

120573)119872(120573))V02(0 0) + (2120573(2 minus 120573)119872(120573)) int

119905

0V02(119909 119904)119889119904 for some

V0119894isin 119878119865119894(1199060

119894)(119894 = 1 2) Thus119873 has a closed graph

Now we prove that there is an open set 119880 sube 119883 with(1199061 1199062) notin 119873(119906

1 1199062) for all 120582 isin (0 1) and (119906

1 1199062) isin 120597119880

Let 120582 isin (0 1) and (1199061 1199062) isin 120582119873(119906

1 1199062) Then there exists

V119894isin 1198711([0 1] times [0 1]R) with V

119894isin 119878119865119894(119906119894)

(119894 = 1 2) such that

1199061(119909 119905) =

2 (1 minus 120572)

(2 minus 120572)119872 (120572)V1(119909 119905)

minus2 (1 minus 120572)

(2 minus 120572)119872 (120572)V1(0 0)

+2120572

(2 minus 120572)119872 (120572)int

119905

0

V1(119909 119904) 119889119904

(38)

and 1199062(119909 119905) = (2(1 minus 120573)(2 minus 120573)119872(120573))V

2(119909 119905) minus (2(1 minus

120573)(2 minus 120573)119872(120573))V2(0 0) + (2120573(2 minus 120573)119872(120573)) int

119905

0V2(119909 119904)119889119904 for

all (119909 119905) isin [0 1] times [0 1] By using the above computedvalues we obtain 119906

119894 le 119901

infin120595(119906119894)sum119899

119894=1119873119894for 119894 = 1 2

This follows that 119906119894119901infin120595(119906119894)sum119899

119894=1119873119894le 1 for 119894 =

1 2 Choose 119872119894gt 0 with 119906

119894 = 119872

119894in such a way that

119872119894119901infin120595(119906119894)sum119899

119894=1119873119894gt 1 for 119894 = 1 2 Put 119880 = (119906

1 1199062) isin

119883 times 119883 (1199061 1199062) lt min 119872

11198722 We note that the

operator 119873 119880 rarr P(119883) is upper semicontinuous andcompletely continuous Also we showed that there is no(1199061 1199062) isin 120597119880 such that (119906

1 1199062) isin 120582119873(119906

1 1199062) for some

120582 isin (0 1) Hence with the help of Theorem 6 we get that119873 has a fixed point (119906

1 1199062) isin 119880 which is a solution for time-

fractional differential inclusion (8)-(9)

Conflict of Interests

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

Acknowledgment

Research of the last two authors was supported by AzarbaijanShahid Madani University

References

[1] R P Agarwal M Benchohra and S Hamani ldquoA survey onexistence results for boundary value problems of nonlinear frac-tional differential equations and inclusionsrdquo Acta ApplicandaeMathematicae vol 109 no 3 pp 973ndash1033 2010

[2] D Baleanu H Mohammadi and S Rezapour ldquoOn a nonlinearfractional differential equation on partially ordered metricspacesrdquo Advances in Difference Equations vol 2013 article 832013

[3] D Baleanu S Rezapour S Etemad and A Alsaedi ldquoOn a time-fractional integrodifferential equation via three-point boundaryvalue conditionsrdquo Mathematical Problems in Engineering vol2015 Article ID 785738 12 pages 2015

[4] K Diethelm The Analysis of Fractional Differential EquationsLecture Notes inMathematics Springer Berlin Germany 2010

[5] R Gorenflo and F Mainardi ldquoFractional calculus integraland differential equations of fractional orderrdquo in Fractals andFractional Calculus in ContinuumMechanics A Carpinteri andF Mainardi Eds Springer New York NY USA 1997

[6] G Samko A Kilbas and O Marichev Fractional Integrals andDerivatives Theory and Applications Gordon and Breach 1993

[7] Z Bai and W Sun ldquoExistence and multiplicity of positivesolutions for singular fractional boundary value problemsrdquoComputers amp Mathematics with Applications vol 63 no 9 pp1369ndash1381 2012

[8] D Baleanu R P Agarwal H Mohammadi and S RezapourldquoSome existence results for a nonlinear fractional differentialequation on partially ordered Banach spacesrdquo Boundary ValueProblems 2013112 8 pages 2013

[9] D Baleanu S Zahra Nazemi and S Rezapour ldquoThe existenceof positive solutions for a new coupled system of multiterm sin-gular fractional integrodifferential boundary value problemsrdquoAbstract and Applied Analysis vol 2013 Article ID 368659 15pages 2013

[10] M Caputo and M Fabrizio ldquoA new definition of fractionalderivative without singular kernelrdquo Progress in Fractional Dif-ferentiation and Applications vol 1 no 2 pp 73ndash85 2015

[11] J Losada and J J Nieto ldquoProperties of a new fractionalderivative without singular kernelrdquo Progress in Fractional Dif-ferentiation and Applications vol 1 no 2 pp 87ndash92 2015

[12] D R Smart Fixed PointTheorems Cambridge University PressCambridge UK 1980

[13] S M Aleomraninejad S Rezapour and N Shahzad ldquoOn fixedpoint generalizations of Suzukirsquos methodrdquo Applied MathematicsLetters vol 24 no 7 pp 1037ndash1040 2011

[14] H Covitz and S B Nadler ldquoMulti-valued contractionmappingsin generalizedmetric spacesrdquo Israel Journal of Mathematics vol8 pp 5ndash11 1970

[15] K S Miller and B Ross An Introduction to the FractionalCalculus and Fractional Differential Eqautions John Wiley ampSons 1993

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

Page 6: Research Article On Coupled Systems of Time-Fractional ...downloads.hindawi.com/journals/jfs/2016/4626940.pdf · Research Article On Coupled Systems of Time-Fractional Differential

6 Journal of Function Spaces

ℎ1(119909 119905) =

2 (1 minus 120572)

(2 minus 120572)119872 (120572)V1(119909 119905) minus

2 (1 minus 120572)

(2 minus 120572)119872 (120572)

sdot V1(0 0) +

2120572

(2 minus 120572)119872 (120572)int

119905

0

V1(119909 119904) 119889119904

ℎ2(119909 119905) =

2 (1 minus 120573)

(2 minus 120573)119872 (120573)V2(119909 119905) minus

2 (1 minus 120573)

(2 minus 120573)119872 (120573)

sdot V2(0 0) +

2120573

(2 minus 120573)119872(120573)int

119905

0

V2(119909 119904) 119889119904

(26)

By Lemma 7 it is clear that each fixed point of the operator119873 is a solution for system of time-fractional differentialinclusions (8) First we prove that the multifunction 119873 isconvex-valued Let (119906

1 1199062) isin 119883 times 119883 (ℎ

1 ℎ2) (ℎ1015840

1 ℎ1015840

2) isin

119873(1199061 1199062) Choose V

119894 V1015840119894isin 119878119865119894(1199061 1199062)

such that

ℎ119894(119909 119905) =

2 (1 minus 120572)

(2 minus 120572)119872 (120572)V119894(119909 119905)

minus2 (1 minus 120572)

(2 minus 120572)119872 (120572)V119894(0 0)

+2120572

(2 minus 120572)119872 (120572)int

119905

0

V119894(119909 119904) 119889119904

ℎ1015840

119894(119909 119905) =

2 (1 minus 120572)

(2 minus 120572)119872 (120572)V1015840119894(119909 119905)

minus2 (1 minus 120572)

(2 minus 120572)119872 (120572)V1015840119894(0 0)

+2120572

(2 minus 120572)119872 (120572)int

119905

0

V1015840119894(119909 119904) 119889119904

(27)

for almost all (119909 119905) isin [0 1] times [0 1] and 119894 = 1 2 Let 0 le 120582 le 1be given Then we have

[120582ℎ119894+ (1 minus 120582) ℎ

1015840

119894] (119909 119905)

=2 (1 minus 120572)

(2 minus 120572)119872 (120572)[120582V119894(119909 119905) + (1 minus 120582) V1015840

119894(119909 119905)]

minus2 (1 minus 120572)

(2 minus 120572)119872 (120572)[120582V119894(0 0) + (1 minus 120582) V1015840

119894(0 0)]

+2120572

(2 minus 120572)119872 (120572)

sdot int

119905

0

[120582V119894(119909 119904) + (1 minus 120582) V1015840

119894(119909 119904)] 119889119904

(28)

for 119894 = 1 2 Since the operator 119865119894has convex values 119878

119865119894(119906119894)is a

convex set and [120582ℎ119894+ (1 minus 120582)ℎ

1015840

119894] isin 119873119894(1199061 1199062) for 119894 = 1 2 This

implies that the operator119873 has convex values Now we provethat119873maps bounded sets of119883 into bounded sets Let 119903 gt 0119861119903= (1199061 1199062) isin 119883 times 119883 (119906

1 1199062) le 119903 be a bounded subset

of 119883 times 119883 (ℎ1 ℎ2) isin 119873(119906

1 1199062) and (119906

1 1199062) isin 119861119903 Then there

exists (V1 V2) isin 1198781198651(1199061)

times 1198781198652(1199062)

such that

ℎ1(119909 119905) =

2 (1 minus 120572)

(2 minus 120572)119872 (120572)V1(119909 119905)

minus2 (1 minus 120572)

(2 minus 120572)119872 (120572)V1(0 0)

+2120572

(2 minus 120572)119872 (120572)int

119905

0

V1(119909 119904) 119889119904

(29)

and ℎ2(119909 119905) = (2(1 minus120573)(2 minus120573)119872(120573))V

2(119909 119905) minus (2(1 minus120573)(2 minus

120573)119872(120573))V2(0 0) + (2120573(2 minus 120573)119872(120573)) int

119905

0V2(119909 119904)119889119904 for almost

all (119909 119905) isin [0 1] times [0 1] If 119901infin= sup

(119909119905)isin[01]times[01]|119901(119909 119905)|

then we obtain

1003816100381610038161003816(ℎ1) (119909 119905)1003816100381610038161003816 =

10038161003816100381610038161003816100381610038161003816

2 (1 minus 120572)

(2 minus 120572)119872 (120572)V1(119909 119905)

minus2 (1 minus 120572)

(2 minus 120572)119872 (120572)V1(0 0)

+2120572

(2 minus 120572)119872 (120572)int

119905

0

V1(119909 119904) 119889119904

10038161003816100381610038161003816100381610038161003816

le2 (1 minus 120572)

(2 minus 120572)119872 (120572)

1003816100381610038161003816V1 (119909 119905)1003816100381610038161003816

+2 (1 minus 120572)

(2 minus 120572)119872 (120572)

1003816100381610038161003816V1 (0 0)1003816100381610038161003816 +

2120572

(2 minus 120572)119872 (120572)

sdot int

119905

0

1003816100381610038161003816V1 (119909 119904)1003816100381610038161003816 119889119904 le 119901 (119909 119905) 120595 (

100381710038171003817100381711990611003817100381710038171003817)

sdot 2 (1 minus 120572)

(2 minus 120572)119872 (120572)

+2 (1 minus 120572)

(2 minus 120572)119872 (120572)+

2120572

(2 minus 120572)119872 (120572)119905 le

10038171003817100381710038171199011003817100381710038171003817infin

sdot 120595 (10038171003817100381710038171199061

1003817100381710038171003817) 4 (1 minus 120572)

(2 minus 120572)119872 (120572)+

2120572

(2 minus 120572)119872 (120572)

le10038171003817100381710038171199011003817100381710038171003817infin

120595 (10038171003817100381710038171199061

1003817100381710038171003817) 4 minus 2120572

(2 minus 120572)119872 (120572) =

10038171003817100381710038171199011003817100381710038171003817infin

sdot 120595 (10038171003817100381710038171199061

1003817100381710038171003817)1198731

(30)

where the constant 1198731is defined by (17) This implies

that ℎ1 le 119901

infin120595(1199061)1198731 Similarly we get ℎ

2 le

119901infin120595(1199062)1198732 where the constant 119873

2is defined by (17)

Thus (ℎ1 ℎ2) le 119901

infin120595((119906

1 1199062))(1198731+ 1198732) Now we

prove that119873maps bounded sets into equicontinuous subsets

Journal of Function Spaces 7

of 119883 times 119883 Let (1199061 1199062) isin 119861119903and (119909 119905

1) (119909 119905

2) isin [0 1] times [0 1]

with 1199051lt 1199052 Then we have

1003816100381610038161003816(ℎ1) (119909 1199052) minus (ℎ1) (119909 1199051)1003816100381610038161003816 =

10038161003816100381610038161003816100381610038161003816

2 (1 minus 120572)

(2 minus 120572)119872 (120572)V1(119909 1199052)

minus2 (1 minus 120572)

(2 minus 120572)119872 (120572)V1(0 0) +

2120572

(2 minus 120572)119872 (120572)

sdot int

1199052

0

V1(119909 119904) 119889119904 minus

2 (1 minus 120572)

(2 minus 120572)119872 (120572)V1(119909 1199051)

+2 (1 minus 120572)

(2 minus 120572)119872 (120572)V1(0 0)

minus2120572

(2 minus 120572)119872 (120572)int

1199051

0

V1(119909 119904) 119889119904

10038161003816100381610038161003816100381610038161003816

le2 (1 minus 120572)

(2 minus 120572)119872 (120572)

1003816100381610038161003816V1 (119909 1199052) minus V1(119909 1199051)1003816100381610038161003816

+2120572

(2 minus 120572)119872 (120572)int

1199052

1199051

1003816100381610038161003816V1 (119909 119904)1003816100381610038161003816 119889119904

le2 (1 minus 120572)

(2 minus 120572)119872 (120572)

1003816100381610038161003816V1 (119909 1199052) minus V1(119909 1199051)1003816100381610038161003816

+212057210038171003817100381710038171199011003817100381710038171003817infin

120595 (10038171003817100381710038171199061

1003817100381710038171003817)

(2 minus 120572)119872 (120572)(1199052minus 1199051)

(31)

By using a similar method we obtain1003816100381610038161003816(ℎ2) (119909 1199052) minus (ℎ2) (119909 1199051)

1003816100381610038161003816

le2 (1 minus 120573)

(2 minus 120573)119872 (120573)

1003816100381610038161003816V2 (119909 1199052) minus V2(119909 1199051)1003816100381610038161003816

+212057310038171003817100381710038171199011003817100381710038171003817infin

120595 (10038171003817100381710038171199062

1003817100381710038171003817)

(2 minus 120573)119872 (120573)(1199052minus 1199051)

(32)

Hence |ℎ119894(119909 1199052) minus ℎ119894(119909 1199051)| rarr 0 as (119909 119905

2) rarr (119909 119905

1) By

using the Arzela-Ascoli theorem we get that119873 is completelycontinuous Here we prove that119873 is upper semicontinuousBy using Lemma 4 119873 is upper semicontinuous whenever ithas a closed graph Since119873 is completely continuouswemustshow that119873 has a closed graph

Let (1199061198991 119906119899

2) be a sequence in 119883 times 119883 with (119906119899

1 119906119899

2) rarr

(1199060

1 1199060

2) and (ℎ119899

1 ℎ119899

2) isin 119873(119906

119899

1 119906119899

2)with (ℎ119899

1 ℎ119899

2) rarr (ℎ

0

1 ℎ0

2)We

show that (ℎ01 ℎ0

2) isin 119873(119906

0

1 1199060

2) For each (ℎ119899

1 ℎ119899

2) isin 119873(119906

119899

1 119906119899

2)

we can choose (V1198991 V1198992) isin 1198781198651(119906119899

1)times 1198781198652(119906119899

2)such that

ℎ119899

1(119909 119905) =

2 (1 minus 120572)

(2 minus 120572)119872 (120572)V1198991(119909 119905)

minus2 (1 minus 120572)

(2 minus 120572)119872 (120572)V1198991(0 0)

+2120572

(2 minus 120572)119872 (120572)int

119905

0

V1198991(119909 119904) 119889119904

(33)

and ℎ119899

2(119909 119905) = (2(1 minus 120573)(2 minus 120573)119872(120573))V119899

2(119909 119905) minus (2(1 minus

120573)(2 minus 120573)119872(120573))V1198992(0 0) + (2120573(2 minus 120573)119872(120573)) int

119905

0V1198992(119909 119904)119889119904 for

all (119909 119905) isin [0 1]times[0 1] It is sufficient to show that there exists(V01 V02) isin 1198781198651(1199060

1)times 1198781198652(1199060

2)such that

ℎ0

1(119909 119905) =

2 (1 minus 120572)

(2 minus 120572)119872 (120572)V01(119909 119905)

minus2 (1 minus 120572)

(2 minus 120572)119872 (120572)V01(0 0)

+2120572

(2 minus 120572)119872 (120572)int

119905

0

V01(119909 119904) 119889119904

(34)

and ℎ0

2(119909 119905) = (2(1 minus 120573)(2 minus 120573)119872(120573))V0

2(119909 119905) minus (2(1 minus

120573)(2 minus 120573)119872(120573))V02(0 0) + (2120573(2 minus 120573)119872(120573)) int

119905

0V02(119909 119904)119889119904 for

all (119909 119905) isin [0 1] times [0 1] Now consider the linear operatorsΘ1 Θ2 1198711([0 1] times [0 1] 119883) rarr 119862([0 1] times [0 1] 119883) defined

by

Θ1(V) (119909 119905) =

2 (1 minus 120572)

(2 minus 120572)119872 (120572)V (119909 119905)

minus2 (1 minus 120572)

(2 minus 120572)119872 (120572)V (0 0)

+2120572

(2 minus 120572)119872 (120572)int

119905

0

V (119909 119904) 119889119904

(35)

andΘ2(V)(119909 119905) = (2(1minus120573)(2minus120573)119872(120573))V(119909 119905)minus(2(1minus120573)(2minus

120573)119872(120573))V(0 0) + (2120573(2 minus 120573)119872(120573)) int119905

0V(119909 119904)119889119904 Note that

10038171003817100381710038171003817ℎ119899

1(119909 119905) minus ℎ

0

1(119909 119905)

10038171003817100381710038171003817

=

10038171003817100381710038171003817100381710038171003817

2 (1 minus 120572)

(2 minus 120572)119872 (120572)[V1198991(119909 119905) minus V0

1(119909 119905)]

minus2 (1 minus 120572)

(2 minus 120572)119872 (120572)[V1198991(0 0) minus V0

1(0 0)]

+2120572

(2 minus 120572)119872 (120572)int

119905

0

[V1198991(119909 119904) minus V0

1(119909 119904)] 119889119904

10038171003817100381710038171003817100381710038171003817

997888rarr 0

10038171003817100381710038171003817ℎ119899

2(119909 119905) minus ℎ

0

2(119909 119905)

10038171003817100381710038171003817

=

100381710038171003817100381710038171003817100381710038171003817

2 (1 minus 120573)

(2 minus 120573)119872 (120573)[V1198992(119909 119905) minus V0

2(119909 119905)]

minus2 (1 minus 120573)

(2 minus 120573)119872 (120573)[V1198992(0 0) minus V0

2(0 0)]

+2120573

(2 minus 120573)119872(120573)int

119905

0

[V1198992(119909 119904) minus V0

2(119909 119904)] 119889119904

100381710038171003817100381710038171003817100381710038171003817

997888rarr 0

(36)

8 Journal of Function Spaces

By using Lemma 5 we get that Θ119894sdot 119878119865119894

is a closed graphoperator for 119894 = 1 2 Also we get ℎ119899

119894(119909 119905) isin Θ

119894(119878119865119894(119906119899

119894)) for

all 119899 Since 119906119899119894rarr 1199060

119894 we get

ℎ0

1(119909 119905) =

2 (1 minus 120572)

(2 minus 120572)119872 (120572)V01(119909 119905)

minus2 (1 minus 120572)

(2 minus 120572)119872 (120572)V01(0 0)

+2120572

(2 minus 120572)119872 (120572)int

119905

0

V01(119909 119904) 119889119904

(37)

and ℎ02(119909 119905) = (2(1minus120573)(2minus120573)119872(120573))V0

2(119909 119905) minus (2(1minus120573)(2minus

120573)119872(120573))V02(0 0) + (2120573(2 minus 120573)119872(120573)) int

119905

0V02(119909 119904)119889119904 for some

V0119894isin 119878119865119894(1199060

119894)(119894 = 1 2) Thus119873 has a closed graph

Now we prove that there is an open set 119880 sube 119883 with(1199061 1199062) notin 119873(119906

1 1199062) for all 120582 isin (0 1) and (119906

1 1199062) isin 120597119880

Let 120582 isin (0 1) and (1199061 1199062) isin 120582119873(119906

1 1199062) Then there exists

V119894isin 1198711([0 1] times [0 1]R) with V

119894isin 119878119865119894(119906119894)

(119894 = 1 2) such that

1199061(119909 119905) =

2 (1 minus 120572)

(2 minus 120572)119872 (120572)V1(119909 119905)

minus2 (1 minus 120572)

(2 minus 120572)119872 (120572)V1(0 0)

+2120572

(2 minus 120572)119872 (120572)int

119905

0

V1(119909 119904) 119889119904

(38)

and 1199062(119909 119905) = (2(1 minus 120573)(2 minus 120573)119872(120573))V

2(119909 119905) minus (2(1 minus

120573)(2 minus 120573)119872(120573))V2(0 0) + (2120573(2 minus 120573)119872(120573)) int

119905

0V2(119909 119904)119889119904 for

all (119909 119905) isin [0 1] times [0 1] By using the above computedvalues we obtain 119906

119894 le 119901

infin120595(119906119894)sum119899

119894=1119873119894for 119894 = 1 2

This follows that 119906119894119901infin120595(119906119894)sum119899

119894=1119873119894le 1 for 119894 =

1 2 Choose 119872119894gt 0 with 119906

119894 = 119872

119894in such a way that

119872119894119901infin120595(119906119894)sum119899

119894=1119873119894gt 1 for 119894 = 1 2 Put 119880 = (119906

1 1199062) isin

119883 times 119883 (1199061 1199062) lt min 119872

11198722 We note that the

operator 119873 119880 rarr P(119883) is upper semicontinuous andcompletely continuous Also we showed that there is no(1199061 1199062) isin 120597119880 such that (119906

1 1199062) isin 120582119873(119906

1 1199062) for some

120582 isin (0 1) Hence with the help of Theorem 6 we get that119873 has a fixed point (119906

1 1199062) isin 119880 which is a solution for time-

fractional differential inclusion (8)-(9)

Conflict of Interests

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

Acknowledgment

Research of the last two authors was supported by AzarbaijanShahid Madani University

References

[1] R P Agarwal M Benchohra and S Hamani ldquoA survey onexistence results for boundary value problems of nonlinear frac-tional differential equations and inclusionsrdquo Acta ApplicandaeMathematicae vol 109 no 3 pp 973ndash1033 2010

[2] D Baleanu H Mohammadi and S Rezapour ldquoOn a nonlinearfractional differential equation on partially ordered metricspacesrdquo Advances in Difference Equations vol 2013 article 832013

[3] D Baleanu S Rezapour S Etemad and A Alsaedi ldquoOn a time-fractional integrodifferential equation via three-point boundaryvalue conditionsrdquo Mathematical Problems in Engineering vol2015 Article ID 785738 12 pages 2015

[4] K Diethelm The Analysis of Fractional Differential EquationsLecture Notes inMathematics Springer Berlin Germany 2010

[5] R Gorenflo and F Mainardi ldquoFractional calculus integraland differential equations of fractional orderrdquo in Fractals andFractional Calculus in ContinuumMechanics A Carpinteri andF Mainardi Eds Springer New York NY USA 1997

[6] G Samko A Kilbas and O Marichev Fractional Integrals andDerivatives Theory and Applications Gordon and Breach 1993

[7] Z Bai and W Sun ldquoExistence and multiplicity of positivesolutions for singular fractional boundary value problemsrdquoComputers amp Mathematics with Applications vol 63 no 9 pp1369ndash1381 2012

[8] D Baleanu R P Agarwal H Mohammadi and S RezapourldquoSome existence results for a nonlinear fractional differentialequation on partially ordered Banach spacesrdquo Boundary ValueProblems 2013112 8 pages 2013

[9] D Baleanu S Zahra Nazemi and S Rezapour ldquoThe existenceof positive solutions for a new coupled system of multiterm sin-gular fractional integrodifferential boundary value problemsrdquoAbstract and Applied Analysis vol 2013 Article ID 368659 15pages 2013

[10] M Caputo and M Fabrizio ldquoA new definition of fractionalderivative without singular kernelrdquo Progress in Fractional Dif-ferentiation and Applications vol 1 no 2 pp 73ndash85 2015

[11] J Losada and J J Nieto ldquoProperties of a new fractionalderivative without singular kernelrdquo Progress in Fractional Dif-ferentiation and Applications vol 1 no 2 pp 87ndash92 2015

[12] D R Smart Fixed PointTheorems Cambridge University PressCambridge UK 1980

[13] S M Aleomraninejad S Rezapour and N Shahzad ldquoOn fixedpoint generalizations of Suzukirsquos methodrdquo Applied MathematicsLetters vol 24 no 7 pp 1037ndash1040 2011

[14] H Covitz and S B Nadler ldquoMulti-valued contractionmappingsin generalizedmetric spacesrdquo Israel Journal of Mathematics vol8 pp 5ndash11 1970

[15] K S Miller and B Ross An Introduction to the FractionalCalculus and Fractional Differential Eqautions John Wiley ampSons 1993

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

Page 7: Research Article On Coupled Systems of Time-Fractional ...downloads.hindawi.com/journals/jfs/2016/4626940.pdf · Research Article On Coupled Systems of Time-Fractional Differential

Journal of Function Spaces 7

of 119883 times 119883 Let (1199061 1199062) isin 119861119903and (119909 119905

1) (119909 119905

2) isin [0 1] times [0 1]

with 1199051lt 1199052 Then we have

1003816100381610038161003816(ℎ1) (119909 1199052) minus (ℎ1) (119909 1199051)1003816100381610038161003816 =

10038161003816100381610038161003816100381610038161003816

2 (1 minus 120572)

(2 minus 120572)119872 (120572)V1(119909 1199052)

minus2 (1 minus 120572)

(2 minus 120572)119872 (120572)V1(0 0) +

2120572

(2 minus 120572)119872 (120572)

sdot int

1199052

0

V1(119909 119904) 119889119904 minus

2 (1 minus 120572)

(2 minus 120572)119872 (120572)V1(119909 1199051)

+2 (1 minus 120572)

(2 minus 120572)119872 (120572)V1(0 0)

minus2120572

(2 minus 120572)119872 (120572)int

1199051

0

V1(119909 119904) 119889119904

10038161003816100381610038161003816100381610038161003816

le2 (1 minus 120572)

(2 minus 120572)119872 (120572)

1003816100381610038161003816V1 (119909 1199052) minus V1(119909 1199051)1003816100381610038161003816

+2120572

(2 minus 120572)119872 (120572)int

1199052

1199051

1003816100381610038161003816V1 (119909 119904)1003816100381610038161003816 119889119904

le2 (1 minus 120572)

(2 minus 120572)119872 (120572)

1003816100381610038161003816V1 (119909 1199052) minus V1(119909 1199051)1003816100381610038161003816

+212057210038171003817100381710038171199011003817100381710038171003817infin

120595 (10038171003817100381710038171199061

1003817100381710038171003817)

(2 minus 120572)119872 (120572)(1199052minus 1199051)

(31)

By using a similar method we obtain1003816100381610038161003816(ℎ2) (119909 1199052) minus (ℎ2) (119909 1199051)

1003816100381610038161003816

le2 (1 minus 120573)

(2 minus 120573)119872 (120573)

1003816100381610038161003816V2 (119909 1199052) minus V2(119909 1199051)1003816100381610038161003816

+212057310038171003817100381710038171199011003817100381710038171003817infin

120595 (10038171003817100381710038171199062

1003817100381710038171003817)

(2 minus 120573)119872 (120573)(1199052minus 1199051)

(32)

Hence |ℎ119894(119909 1199052) minus ℎ119894(119909 1199051)| rarr 0 as (119909 119905

2) rarr (119909 119905

1) By

using the Arzela-Ascoli theorem we get that119873 is completelycontinuous Here we prove that119873 is upper semicontinuousBy using Lemma 4 119873 is upper semicontinuous whenever ithas a closed graph Since119873 is completely continuouswemustshow that119873 has a closed graph

Let (1199061198991 119906119899

2) be a sequence in 119883 times 119883 with (119906119899

1 119906119899

2) rarr

(1199060

1 1199060

2) and (ℎ119899

1 ℎ119899

2) isin 119873(119906

119899

1 119906119899

2)with (ℎ119899

1 ℎ119899

2) rarr (ℎ

0

1 ℎ0

2)We

show that (ℎ01 ℎ0

2) isin 119873(119906

0

1 1199060

2) For each (ℎ119899

1 ℎ119899

2) isin 119873(119906

119899

1 119906119899

2)

we can choose (V1198991 V1198992) isin 1198781198651(119906119899

1)times 1198781198652(119906119899

2)such that

ℎ119899

1(119909 119905) =

2 (1 minus 120572)

(2 minus 120572)119872 (120572)V1198991(119909 119905)

minus2 (1 minus 120572)

(2 minus 120572)119872 (120572)V1198991(0 0)

+2120572

(2 minus 120572)119872 (120572)int

119905

0

V1198991(119909 119904) 119889119904

(33)

and ℎ119899

2(119909 119905) = (2(1 minus 120573)(2 minus 120573)119872(120573))V119899

2(119909 119905) minus (2(1 minus

120573)(2 minus 120573)119872(120573))V1198992(0 0) + (2120573(2 minus 120573)119872(120573)) int

119905

0V1198992(119909 119904)119889119904 for

all (119909 119905) isin [0 1]times[0 1] It is sufficient to show that there exists(V01 V02) isin 1198781198651(1199060

1)times 1198781198652(1199060

2)such that

ℎ0

1(119909 119905) =

2 (1 minus 120572)

(2 minus 120572)119872 (120572)V01(119909 119905)

minus2 (1 minus 120572)

(2 minus 120572)119872 (120572)V01(0 0)

+2120572

(2 minus 120572)119872 (120572)int

119905

0

V01(119909 119904) 119889119904

(34)

and ℎ0

2(119909 119905) = (2(1 minus 120573)(2 minus 120573)119872(120573))V0

2(119909 119905) minus (2(1 minus

120573)(2 minus 120573)119872(120573))V02(0 0) + (2120573(2 minus 120573)119872(120573)) int

119905

0V02(119909 119904)119889119904 for

all (119909 119905) isin [0 1] times [0 1] Now consider the linear operatorsΘ1 Θ2 1198711([0 1] times [0 1] 119883) rarr 119862([0 1] times [0 1] 119883) defined

by

Θ1(V) (119909 119905) =

2 (1 minus 120572)

(2 minus 120572)119872 (120572)V (119909 119905)

minus2 (1 minus 120572)

(2 minus 120572)119872 (120572)V (0 0)

+2120572

(2 minus 120572)119872 (120572)int

119905

0

V (119909 119904) 119889119904

(35)

andΘ2(V)(119909 119905) = (2(1minus120573)(2minus120573)119872(120573))V(119909 119905)minus(2(1minus120573)(2minus

120573)119872(120573))V(0 0) + (2120573(2 minus 120573)119872(120573)) int119905

0V(119909 119904)119889119904 Note that

10038171003817100381710038171003817ℎ119899

1(119909 119905) minus ℎ

0

1(119909 119905)

10038171003817100381710038171003817

=

10038171003817100381710038171003817100381710038171003817

2 (1 minus 120572)

(2 minus 120572)119872 (120572)[V1198991(119909 119905) minus V0

1(119909 119905)]

minus2 (1 minus 120572)

(2 minus 120572)119872 (120572)[V1198991(0 0) minus V0

1(0 0)]

+2120572

(2 minus 120572)119872 (120572)int

119905

0

[V1198991(119909 119904) minus V0

1(119909 119904)] 119889119904

10038171003817100381710038171003817100381710038171003817

997888rarr 0

10038171003817100381710038171003817ℎ119899

2(119909 119905) minus ℎ

0

2(119909 119905)

10038171003817100381710038171003817

=

100381710038171003817100381710038171003817100381710038171003817

2 (1 minus 120573)

(2 minus 120573)119872 (120573)[V1198992(119909 119905) minus V0

2(119909 119905)]

minus2 (1 minus 120573)

(2 minus 120573)119872 (120573)[V1198992(0 0) minus V0

2(0 0)]

+2120573

(2 minus 120573)119872(120573)int

119905

0

[V1198992(119909 119904) minus V0

2(119909 119904)] 119889119904

100381710038171003817100381710038171003817100381710038171003817

997888rarr 0

(36)

8 Journal of Function Spaces

By using Lemma 5 we get that Θ119894sdot 119878119865119894

is a closed graphoperator for 119894 = 1 2 Also we get ℎ119899

119894(119909 119905) isin Θ

119894(119878119865119894(119906119899

119894)) for

all 119899 Since 119906119899119894rarr 1199060

119894 we get

ℎ0

1(119909 119905) =

2 (1 minus 120572)

(2 minus 120572)119872 (120572)V01(119909 119905)

minus2 (1 minus 120572)

(2 minus 120572)119872 (120572)V01(0 0)

+2120572

(2 minus 120572)119872 (120572)int

119905

0

V01(119909 119904) 119889119904

(37)

and ℎ02(119909 119905) = (2(1minus120573)(2minus120573)119872(120573))V0

2(119909 119905) minus (2(1minus120573)(2minus

120573)119872(120573))V02(0 0) + (2120573(2 minus 120573)119872(120573)) int

119905

0V02(119909 119904)119889119904 for some

V0119894isin 119878119865119894(1199060

119894)(119894 = 1 2) Thus119873 has a closed graph

Now we prove that there is an open set 119880 sube 119883 with(1199061 1199062) notin 119873(119906

1 1199062) for all 120582 isin (0 1) and (119906

1 1199062) isin 120597119880

Let 120582 isin (0 1) and (1199061 1199062) isin 120582119873(119906

1 1199062) Then there exists

V119894isin 1198711([0 1] times [0 1]R) with V

119894isin 119878119865119894(119906119894)

(119894 = 1 2) such that

1199061(119909 119905) =

2 (1 minus 120572)

(2 minus 120572)119872 (120572)V1(119909 119905)

minus2 (1 minus 120572)

(2 minus 120572)119872 (120572)V1(0 0)

+2120572

(2 minus 120572)119872 (120572)int

119905

0

V1(119909 119904) 119889119904

(38)

and 1199062(119909 119905) = (2(1 minus 120573)(2 minus 120573)119872(120573))V

2(119909 119905) minus (2(1 minus

120573)(2 minus 120573)119872(120573))V2(0 0) + (2120573(2 minus 120573)119872(120573)) int

119905

0V2(119909 119904)119889119904 for

all (119909 119905) isin [0 1] times [0 1] By using the above computedvalues we obtain 119906

119894 le 119901

infin120595(119906119894)sum119899

119894=1119873119894for 119894 = 1 2

This follows that 119906119894119901infin120595(119906119894)sum119899

119894=1119873119894le 1 for 119894 =

1 2 Choose 119872119894gt 0 with 119906

119894 = 119872

119894in such a way that

119872119894119901infin120595(119906119894)sum119899

119894=1119873119894gt 1 for 119894 = 1 2 Put 119880 = (119906

1 1199062) isin

119883 times 119883 (1199061 1199062) lt min 119872

11198722 We note that the

operator 119873 119880 rarr P(119883) is upper semicontinuous andcompletely continuous Also we showed that there is no(1199061 1199062) isin 120597119880 such that (119906

1 1199062) isin 120582119873(119906

1 1199062) for some

120582 isin (0 1) Hence with the help of Theorem 6 we get that119873 has a fixed point (119906

1 1199062) isin 119880 which is a solution for time-

fractional differential inclusion (8)-(9)

Conflict of Interests

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

Acknowledgment

Research of the last two authors was supported by AzarbaijanShahid Madani University

References

[1] R P Agarwal M Benchohra and S Hamani ldquoA survey onexistence results for boundary value problems of nonlinear frac-tional differential equations and inclusionsrdquo Acta ApplicandaeMathematicae vol 109 no 3 pp 973ndash1033 2010

[2] D Baleanu H Mohammadi and S Rezapour ldquoOn a nonlinearfractional differential equation on partially ordered metricspacesrdquo Advances in Difference Equations vol 2013 article 832013

[3] D Baleanu S Rezapour S Etemad and A Alsaedi ldquoOn a time-fractional integrodifferential equation via three-point boundaryvalue conditionsrdquo Mathematical Problems in Engineering vol2015 Article ID 785738 12 pages 2015

[4] K Diethelm The Analysis of Fractional Differential EquationsLecture Notes inMathematics Springer Berlin Germany 2010

[5] R Gorenflo and F Mainardi ldquoFractional calculus integraland differential equations of fractional orderrdquo in Fractals andFractional Calculus in ContinuumMechanics A Carpinteri andF Mainardi Eds Springer New York NY USA 1997

[6] G Samko A Kilbas and O Marichev Fractional Integrals andDerivatives Theory and Applications Gordon and Breach 1993

[7] Z Bai and W Sun ldquoExistence and multiplicity of positivesolutions for singular fractional boundary value problemsrdquoComputers amp Mathematics with Applications vol 63 no 9 pp1369ndash1381 2012

[8] D Baleanu R P Agarwal H Mohammadi and S RezapourldquoSome existence results for a nonlinear fractional differentialequation on partially ordered Banach spacesrdquo Boundary ValueProblems 2013112 8 pages 2013

[9] D Baleanu S Zahra Nazemi and S Rezapour ldquoThe existenceof positive solutions for a new coupled system of multiterm sin-gular fractional integrodifferential boundary value problemsrdquoAbstract and Applied Analysis vol 2013 Article ID 368659 15pages 2013

[10] M Caputo and M Fabrizio ldquoA new definition of fractionalderivative without singular kernelrdquo Progress in Fractional Dif-ferentiation and Applications vol 1 no 2 pp 73ndash85 2015

[11] J Losada and J J Nieto ldquoProperties of a new fractionalderivative without singular kernelrdquo Progress in Fractional Dif-ferentiation and Applications vol 1 no 2 pp 87ndash92 2015

[12] D R Smart Fixed PointTheorems Cambridge University PressCambridge UK 1980

[13] S M Aleomraninejad S Rezapour and N Shahzad ldquoOn fixedpoint generalizations of Suzukirsquos methodrdquo Applied MathematicsLetters vol 24 no 7 pp 1037ndash1040 2011

[14] H Covitz and S B Nadler ldquoMulti-valued contractionmappingsin generalizedmetric spacesrdquo Israel Journal of Mathematics vol8 pp 5ndash11 1970

[15] K S Miller and B Ross An Introduction to the FractionalCalculus and Fractional Differential Eqautions John Wiley ampSons 1993

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

Page 8: Research Article On Coupled Systems of Time-Fractional ...downloads.hindawi.com/journals/jfs/2016/4626940.pdf · Research Article On Coupled Systems of Time-Fractional Differential

8 Journal of Function Spaces

By using Lemma 5 we get that Θ119894sdot 119878119865119894

is a closed graphoperator for 119894 = 1 2 Also we get ℎ119899

119894(119909 119905) isin Θ

119894(119878119865119894(119906119899

119894)) for

all 119899 Since 119906119899119894rarr 1199060

119894 we get

ℎ0

1(119909 119905) =

2 (1 minus 120572)

(2 minus 120572)119872 (120572)V01(119909 119905)

minus2 (1 minus 120572)

(2 minus 120572)119872 (120572)V01(0 0)

+2120572

(2 minus 120572)119872 (120572)int

119905

0

V01(119909 119904) 119889119904

(37)

and ℎ02(119909 119905) = (2(1minus120573)(2minus120573)119872(120573))V0

2(119909 119905) minus (2(1minus120573)(2minus

120573)119872(120573))V02(0 0) + (2120573(2 minus 120573)119872(120573)) int

119905

0V02(119909 119904)119889119904 for some

V0119894isin 119878119865119894(1199060

119894)(119894 = 1 2) Thus119873 has a closed graph

Now we prove that there is an open set 119880 sube 119883 with(1199061 1199062) notin 119873(119906

1 1199062) for all 120582 isin (0 1) and (119906

1 1199062) isin 120597119880

Let 120582 isin (0 1) and (1199061 1199062) isin 120582119873(119906

1 1199062) Then there exists

V119894isin 1198711([0 1] times [0 1]R) with V

119894isin 119878119865119894(119906119894)

(119894 = 1 2) such that

1199061(119909 119905) =

2 (1 minus 120572)

(2 minus 120572)119872 (120572)V1(119909 119905)

minus2 (1 minus 120572)

(2 minus 120572)119872 (120572)V1(0 0)

+2120572

(2 minus 120572)119872 (120572)int

119905

0

V1(119909 119904) 119889119904

(38)

and 1199062(119909 119905) = (2(1 minus 120573)(2 minus 120573)119872(120573))V

2(119909 119905) minus (2(1 minus

120573)(2 minus 120573)119872(120573))V2(0 0) + (2120573(2 minus 120573)119872(120573)) int

119905

0V2(119909 119904)119889119904 for

all (119909 119905) isin [0 1] times [0 1] By using the above computedvalues we obtain 119906

119894 le 119901

infin120595(119906119894)sum119899

119894=1119873119894for 119894 = 1 2

This follows that 119906119894119901infin120595(119906119894)sum119899

119894=1119873119894le 1 for 119894 =

1 2 Choose 119872119894gt 0 with 119906

119894 = 119872

119894in such a way that

119872119894119901infin120595(119906119894)sum119899

119894=1119873119894gt 1 for 119894 = 1 2 Put 119880 = (119906

1 1199062) isin

119883 times 119883 (1199061 1199062) lt min 119872

11198722 We note that the

operator 119873 119880 rarr P(119883) is upper semicontinuous andcompletely continuous Also we showed that there is no(1199061 1199062) isin 120597119880 such that (119906

1 1199062) isin 120582119873(119906

1 1199062) for some

120582 isin (0 1) Hence with the help of Theorem 6 we get that119873 has a fixed point (119906

1 1199062) isin 119880 which is a solution for time-

fractional differential inclusion (8)-(9)

Conflict of Interests

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

Acknowledgment

Research of the last two authors was supported by AzarbaijanShahid Madani University

References

[1] R P Agarwal M Benchohra and S Hamani ldquoA survey onexistence results for boundary value problems of nonlinear frac-tional differential equations and inclusionsrdquo Acta ApplicandaeMathematicae vol 109 no 3 pp 973ndash1033 2010

[2] D Baleanu H Mohammadi and S Rezapour ldquoOn a nonlinearfractional differential equation on partially ordered metricspacesrdquo Advances in Difference Equations vol 2013 article 832013

[3] D Baleanu S Rezapour S Etemad and A Alsaedi ldquoOn a time-fractional integrodifferential equation via three-point boundaryvalue conditionsrdquo Mathematical Problems in Engineering vol2015 Article ID 785738 12 pages 2015

[4] K Diethelm The Analysis of Fractional Differential EquationsLecture Notes inMathematics Springer Berlin Germany 2010

[5] R Gorenflo and F Mainardi ldquoFractional calculus integraland differential equations of fractional orderrdquo in Fractals andFractional Calculus in ContinuumMechanics A Carpinteri andF Mainardi Eds Springer New York NY USA 1997

[6] G Samko A Kilbas and O Marichev Fractional Integrals andDerivatives Theory and Applications Gordon and Breach 1993

[7] Z Bai and W Sun ldquoExistence and multiplicity of positivesolutions for singular fractional boundary value problemsrdquoComputers amp Mathematics with Applications vol 63 no 9 pp1369ndash1381 2012

[8] D Baleanu R P Agarwal H Mohammadi and S RezapourldquoSome existence results for a nonlinear fractional differentialequation on partially ordered Banach spacesrdquo Boundary ValueProblems 2013112 8 pages 2013

[9] D Baleanu S Zahra Nazemi and S Rezapour ldquoThe existenceof positive solutions for a new coupled system of multiterm sin-gular fractional integrodifferential boundary value problemsrdquoAbstract and Applied Analysis vol 2013 Article ID 368659 15pages 2013

[10] M Caputo and M Fabrizio ldquoA new definition of fractionalderivative without singular kernelrdquo Progress in Fractional Dif-ferentiation and Applications vol 1 no 2 pp 73ndash85 2015

[11] J Losada and J J Nieto ldquoProperties of a new fractionalderivative without singular kernelrdquo Progress in Fractional Dif-ferentiation and Applications vol 1 no 2 pp 87ndash92 2015

[12] D R Smart Fixed PointTheorems Cambridge University PressCambridge UK 1980

[13] S M Aleomraninejad S Rezapour and N Shahzad ldquoOn fixedpoint generalizations of Suzukirsquos methodrdquo Applied MathematicsLetters vol 24 no 7 pp 1037ndash1040 2011

[14] H Covitz and S B Nadler ldquoMulti-valued contractionmappingsin generalizedmetric spacesrdquo Israel Journal of Mathematics vol8 pp 5ndash11 1970

[15] K S Miller and B Ross An Introduction to the FractionalCalculus and Fractional Differential Eqautions John Wiley ampSons 1993

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

Page 9: Research Article On Coupled Systems of Time-Fractional ...downloads.hindawi.com/journals/jfs/2016/4626940.pdf · Research Article On Coupled Systems of Time-Fractional Differential

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


Recommended
A Coupled Wavelet-Based Method for Wave and … and fractional partial differential equations ... the wave and fractional wave equations play ... equations. In wavelet based approximation

A Coupled Wavelet-Based Method for Wave and … and fractional partial differential equations ... the wave and fractional wave equations play ... equations. In wavelet based approximation

Documents

Research Article Optimization of Fractional-N-PLL …downloads.hindawi.com/journals/vlsi/2014/406416.pdfResearch Article Optimization of Fractional-N-PLL Frequency Synthesizer for

Research Article Optimization of Fractional-N-PLL …downloads.hindawi.com/journals/vlsi/2014/406416.pdfResearch Article Optimization of Fractional-N-PLL Frequency Synthesizer for

Documents

Research Article Fractional Calculus of Fractal ...downloads.hindawi.com/journals/aaa/2014/640628.pdfResearch Article Fractional Calculus of Fractal Interpolation Function on [0,](

Research Article Fractional Calculus of Fractal ...downloads.hindawi.com/journals/aaa/2014/640628.pdfResearch Article Fractional Calculus of Fractal Interpolation Function on [0,](

Documents

Research Article Exact Solutions of the Time Fractional ...downloads.hindawi.com/journals/amp/2014/181594.pdf · Research Article Exact Solutions of the Time Fractional BBM-Burger

Research Article Exact Solutions of the Time Fractional ...downloads.hindawi.com/journals/amp/2014/181594.pdf · Research Article Exact Solutions of the Time Fractional BBM-Burger

Documents

Research Article Numerical Solution of Fractional Integro ...downloads.hindawi.com/journals/mpe/2014/431965.pdf · Numerical Solution of Fractional Integro-Differential Equations

Research Article Numerical Solution of Fractional Integro ...downloads.hindawi.com/journals/mpe/2014/431965.pdf · Numerical Solution of Fractional Integro-Differential Equations

Documents

Research Article A 3.9 s Settling-Time Fractional …downloads.hindawi.com/journals/jece/2015/765485.pdfResearch Article A 3.9 s Settling-Time Fractional Spread-Spectrum Clock Generator

Research Article A 3.9 s Settling-Time Fractional …downloads.hindawi.com/journals/jece/2015/765485.pdfResearch Article A 3.9 s Settling-Time Fractional Spread-Spectrum Clock Generator

Documents

Research Article The Use of Fractional Order Derivative to ...

Research Article The Use of Fractional Order Derivative to ...

Documents

HIGH DIMENSIONAL FRACTIONAL COUPLED SYSTEMS: NEW … · 2016-12-06 · KragujevacJournalofMathematics Volume40(2)(2016),Pages224–236. HIGH DIMENSIONAL FRACTIONAL COUPLED SYSTEMS:

HIGH DIMENSIONAL FRACTIONAL COUPLED SYSTEMS: NEW … · 2016-12-06 · KragujevacJournalofMathematics Volume40(2)(2016),Pages224–236. HIGH DIMENSIONAL FRACTIONAL COUPLED SYSTEMS:

Documents

Research Article Numerical Fractional-Calculus Model for ...downloads.hindawi.com/journals/amp/2013/429835.pdf · Research Article Numerical Fractional-Calculus Model for Two-Phase

Research Article Numerical Fractional-Calculus Model for ...downloads.hindawi.com/journals/amp/2013/429835.pdf · Research Article Numerical Fractional-Calculus Model for Two-Phase

Documents

Research Article On Generalized Fractional Kinetic ...downloads.hindawi.com/journals/mpe/2015/289387.pdf · Research Article On Generalized Fractional Kinetic Equations Involving

Research Article On Generalized Fractional Kinetic ...downloads.hindawi.com/journals/mpe/2015/289387.pdf · Research Article On Generalized Fractional Kinetic Equations Involving

Documents

Research Article A Fractional Factorial Design to …downloads.hindawi.com/journals/er/2014/162962.pdfResearch Article A Fractional Factorial Design to Study the Effect of Process

Research Article A Fractional Factorial Design to …downloads.hindawi.com/journals/er/2014/162962.pdfResearch Article A Fractional Factorial Design to Study the Effect of Process

Documents

Research Article A Study of Caputo-Hadamard-Type Fractional Differential Equations ... · 2019. 7. 30. · Research Article A Study of Caputo-Hadamard-Type Fractional Differential

Research Article A Study of Caputo-Hadamard-Type Fractional Differential Equations ... · 2019. 7. 30. · Research Article A Study of Caputo-Hadamard-Type Fractional Differential

Documents

Research Article Stability Analysis for a Fractional HIV ...downloads.hindawi.com/journals/ddns/2015/563127.pdf · Research Article Stability Analysis for a Fractional HIV Infection

Research Article Stability Analysis for a Fractional HIV ...downloads.hindawi.com/journals/ddns/2015/563127.pdf · Research Article Stability Analysis for a Fractional HIV Infection

Documents

Multiple positive solutions to a coupled systems of nonlinear fractional differential ... · 2017. 8. 27. · Shah and Khan SpringerPlus Page 3 of 20 systems of fractional order differential

Multiple positive solutions to a coupled systems of nonlinear fractional differential ... · 2017. 8. 27. · Shah and Khan SpringerPlus Page 3 of 20 systems of fractional order differential

Documents

Research Article An Enhanced Fractional Order Model of ...downloads.hindawi.com/journals/amp/2013/717659.pdf · Research Article An Enhanced Fractional Order Model of Ionic Polymer-Metal

Research Article An Enhanced Fractional Order Model of ...downloads.hindawi.com/journals/amp/2013/717659.pdf · Research Article An Enhanced Fractional Order Model of Ionic Polymer-Metal

Documents

Research Article Local Fractional Laplace Variational ...downloads.hindawi.com/journals/mpe/2014/914725.pdf · Research Article Local Fractional Laplace Variational Iteration Method

Research Article Local Fractional Laplace Variational ...downloads.hindawi.com/journals/mpe/2014/914725.pdf · Research Article Local Fractional Laplace Variational Iteration Method

Documents

Research Article Coupled Fixed Point Theorems with Rational …downloads.hindawi.com/journals/jmath/2014/785357.pdf · Research Article Coupled Fixed Point Theorems with Rational

Research Article Coupled Fixed Point Theorems with Rational …downloads.hindawi.com/journals/jmath/2014/785357.pdf · Research Article Coupled Fixed Point Theorems with Rational

Documents

Review Article A Review of Definitions for Fractional ...

Review Article A Review of Definitions for Fractional ...

Documents