Open Journal of Applied Sciences, 2013, 3, 53-61
doi:10.4236/ojapps.2013.31B1011 Published Online April 2013 (http://www.scirp.org/journal/ojapps)
Existence and Uniqueness of Positive Solutions for a
Coupled System of Nonlinear Fractional
Differential Equations
Minjie Li, Yiliang Liu
College of Sciences, Guangxi University for Nationalities, Nanning 530006, Guangxi Province, P. R. China
Received 2013
ABSTRACT
In this paper, we research the existence and uniqueness of po sitive solu tions for a coupled system of fractional differen-
tial equations. By means of some standard fixed point principles, some results on the existence and uniqueness of posi-
tive solutions for coupled systems are obtained.
Keywords: Caputo Fractional Derivative; Fractional Differential Equations; Coupled System; Fixed Point Theorem;
Positive Solutions
1. Introduction
Fractional differential equations can describe many phe-
nomena in various fields of engineering and scientific
disciplines such as control theory, physics, chemistry,
biology, economics, mechanics and electromagnetic. In
recent years, there are a large number of papers dealing
with the existence of positive solutions of boundary
value problems for nonlinear differential equations of
fractional order. We refer readers to the monographs
such as Kilbas etal. [8], Miller and Ross [20], Podlubny
[21], and the papers [1,3-5,12-19,28-33] and references
therein.
In [12], Li, Luo and Zhou considered the existence of
positive solutions of the following boundary value prob-
lem of fractional order differential equations:
0
00
()(()) 001
(0)0(1)( )


 
 
Dut ftutt
uDuaDu




where 0
D

is the standard Riemann-Liouville fractional
derivative of order 12, 01, 01, (0a1),


1   [0 1][0)[0)f and
satisfies Caratheodory type conditions.
2 1, 0a 

In [30], Yang, Wei and Dong investigated the follow-
ing existence of positive solutions of fractional order
differential equations:
0()(() ())01
(0)(0) 0(1)(1) 0

 


cDut ftututt
uu uu


where 0
cD

is the Caputo fractional derivative of order
12

and ([0 1][0))


f
CR
In addition, recently some authors also pay close at-
tention to the existence of solutions for coupled systems
of fractional differential equations (see[2,3,25,26]).
In [26], Su studied the existence of solutions for a
coupled system of fractional differential equations:
00
00
()(()()) 01
()(()()) 01
(0)(1)(0) (1)




 



 




Dut ftvtDvtt
Dvt gtutDutt
uuvv



where 2
1 2 0, 1, 1, [01]  
f
gR  
DR
0
0
c
c
D
D


are given functions and 0 is the standard Riemann-
Liouville fractional derivative.
In [25], Sun, Liu and Liu considered the following
systems of fractional differential equations with antipe-
riodic boundary cond itions:
1
00
2
00
( )(,( ),(),( ),()),
[0, ],
()(,(), (),(),()),
[0, ],
(0)( ),'(0)'( ),
(0)( ),'(0)'( ),
pq
cc
pq
cc
utf tutvtutvt
tJ T
vtftut vtutvt
tJ T
uuTu uT
vvTv vT
DD
DD











 

 

(1.1)
where 0

cD

denotes the Caputo fractional derivative,
4
12
, 01, (1 2).

   pqffCJRR
However, the research on the systems of positive solu-
tions of fractional differential equations hasn’t received
remarkable attention. In this paper, we shall concern with
the existence and uniqueness of positive solutions for a cou-
R.

M. J. LI, Y. L. LIU
54
pled system of nonlinearfractional differential equations.
More precisely, we will consider the following problem:
1
00
2
00
12
0
12
0
()(,(), (),(),()),
[0, ],
()(,(), (),(),()),
[0, ],
(0)'(0) 0(1)(1) 0
(0)'(0) 0(1)(1) 0











 

 

pq
cc
pq
cc
p
c
q
c
utftutvtutvt
tJ T
vtftut vtutvt
tJ T
uu uu
vv vv
DD
DD
D
D




，，
，，
0
0


c
c
D
D
where 0
cD

denotes the Caputo fractional derivative for
22
12
0 12 0 1, ([01])


 pqffCRRR 
Thisis organized as follows: In section 2, we in-
  
paper
tro
2. Preliminaries
space:
duce some preliminary results, including basic defini-
tions of fractional integrals an d derivatives, some proper-
ties and a fixed point theorems. In section 3, by applying
some standard fixed point principles, we prove the exis-
tence and uniqueness of positive solutions for a coupled
system of nonlinear fractional differential equations.
Let us introduce a 1 () ()([01])
X
{u tu tC}
endowed with the norm
[01]
max
Xt
u[01] ( )max( ).
 

t
ut
ut
Indeed is a Banach space. Obviously, the pro-
readers, we first present
, ()
X
X
ce X


X
u
duct spa )


XX
X is also a Banach space with
()


XX
uv For the convenience of the
some useful definitions and
(


X
v.
fundamental facts of fractional calculus theory, which
can be found in [8,21].
Definition 2.1. For 0

, the integral
01
0
1()
tfs
() () ()




I
ft ds
ts


(2.1)
is called the Riemann-Liouville fractional integral.

Definition 2.2. For a function ()
f
t given in the in-
terval [0 ) , the expression
01
0
1

()
()( )
() ()
[] 1


 

t
Ln
n
dfs
Dft ds
ndt
ts
n



(2.2)
is called the Riemann-Liouville fractional derivative of
order 0

, where []

denotes the integer part of real
number

.
Definionti 2.3 [6]. The Caputo’s derivative of order

for a function ([0) )
n
f
CR can be written as
()
(0)
kk
cL tf

1
00
() [ ()]
1


 

 

n
k
Df
t Dftk
nn

(2.3)
Lemma 2.4 [8,21]. Let 1[01]



n
uC and (1]

 qn n

nN. Then for [0 1]

t,
1()
00 
() ()(0)



k
n
qc qk
t
ID ututu (2.4)
Lemma 2.5. Let

kk
[0 1].

C

If 2
1(2 )
11

 
p
H 

20

, then tion of the following frac-
fferent equat
()ut
ial is a solu ions: tional di
0
120
() ()2
)(0(1)(1)0 01


 


[01]1
(0 0)
 




c
u

 
cp
Dutt t
u uDup


)
if and only if ()ut is a solution
(2.5
of the fractional integral
equations
1
0
()() ()



utG tssds

 (2.6)
where
11
1
1
21
11
121
() ( )(1)
()
()(1)
()
() 01
()(1) ()(1)
()( )
01



 



 
 


 

 






p
p
tst s
ts
p
Gtss t
tsts
p
ts











(2.7)
21
1(2 )p



 

Furthermore, if the assumption H holds, then
()([0 1)[0 1))

Gts C

(0 1) and , for any ()0Gts


ts .
Proof. Assume ()ut satisfies (2.4), (2.5),
we .5). By (2 have
1
0)
() ()(0) (0)
(


()

 
tts
usds uut




Hence,
t
2
0
()
()() (0)
(1)







tts
utsdsu



By definition 2.3 together with the facts that
1
0(2 )





p
Lp t
Dt p 0() ()


Lp p
DIutI ut

and the linearity of fractional differential, we get
11
()
 

00
() () (0)
() (2
)


 
p
cp
Du
tsds u


p
tts t

pp
Applying the boundary conditions
120
(0)(0)0 (1)(1)0



 
cp
uu uDu  ,
we obtain that

M. J. LI, Y. L. LIU 55
11
1
0
12
112
0
11
0
(0)(1)( )
()( )
1(1)()
()( )
(1)()





  
 


p
ussds
p
sds
p
ssds










Consequently,
11
0
(1)()

 
p
ss ds


(0)

us
111
1
00
11
21
0
1
0
(
Gt
s

()
()()(1 )()
() ()
()
(1)()
()
)( )








 



t
p
t
ts
utsdss sds
tssds
p
sds





 


Conversely, assumeis a solution of fractional in-
tegral equations (2.6e definition of Caputo’s
derivative (2.3) and the that
()ut
). Using th
fact2
0(1 )



p
Lp Ct
p
DC and 0

LD

is the left inverse of 0
I

we
ession of
get (2.5).
Observing the expr in (2
easily obtain ()Gts

[01)) Let .6), we () ([01)Gts C

1 1
1
21
)(1) () 1
1
(
()
1
t

(2.8
() ()
()(1)
0
 

 
 



p
ts ts
gts
ts s
 



)
()  p 
11
121
2()(1))(1) (
() ()( )
01

 
 
 

p
ts ts
gtsp
ts



(2.9)
By H, we have and 0
11
121
2()(1) ()(1)
() 0
()( )
(0 1)

 
 
 
 
p
ts ts
gtsp
st




which also implies by (2.8). Hence
lproof is completed.
6. If w
1
()0ts

for al(0 1). Th
()0gts
e
ake use of
G
Remark 2. st e m
12()
1 ()

 

p
H

 
instead of H, we may simder the problem
(1.1). We omit it here.
2.7 [22]. a Banach space. Assume
ilarly consi
th ntinuous operator and
the set s bounded. Then
T has a
For the sake of convenience, we set
Lemma Let E be
at TE E is a co mpletely co

01VuEuTu  i
fixed point in E.
3. Main Result
121
1(1)
1)( 1)
 
1
2
1
(
1(1
(1)( 1)


1
2
 

)

 
 

Mp
 


 p


M

(3.1)
3112113
max [( 23
122 2142 4
1
) ]
(2)
1
[()]
(2 )


 





 

d
p
Mc MdMc Md
(3.2)
We denote
  q
MM
cMdMcM
22
11
11
(2)(2 )



 
 pq



and give the following assumption
2
2
12
(2 )
12
(2 )
11 0
11 0.





 
 
p
q
H




Define the operat or as  TX XXX
12
(()( )()( ))
1
10 0
0
1
200
0
()()
()(()()()())
()(()()()())




 




cp cq
cp cq
Tuv t
Gtsfsusvs Dus Dvsds
GtsfsusvsDus Dvsds


（
）
TuvtTuvt

(3.3)
which implies
12
2
100
0
11100
0
11
2
10
0
(())()((())()(())())
() (()()())
(1)
1
(1)(()()())
()
(1 )(()()
()















 



tcpcq
cp cq
pcpc
Tuvt TuvtTuvt
ts fsus Dus Dvsds

s
fsus Dus Dvsds
sfsusDus
p






（

0
2
200
0
11200
0
11
220
0
())
()
(()()())
(1)
1
(1)(()()())
()
(1)(()()
()










 




 



q
tcp cq
cp cq
pcpc
Dvsds
ts fsus DusDvsds
s
fsus DusDvsds
sfsusDus
p







0
())

q
Dvsds）
(3.4)
Lemma 3.1. The operator is
completely continuous.
Proof. Firstly, we show that the operator
TX XX X
TX X

X
X is continuous.

M. J. LI, Y. L. LIU
56
For such that )
in 01 
nn
p{uv}XX
 00
()(
nn
uv uv
X
X we have
000
[01]
0
[01]
0
0
0
0
[01]
0
[01]
max()()
11
=max()( )
(1 )(1 )
()()
1
=max()[( )( )]
(1 )
1
=max()()
(2 )
1
=(2 )











 
 














cp cp
n
t
tpn
t
tp
tpn
t
n
t
n
Dut Dut
ts sds
u
pp
ts usds
tssusds
u
p
tut
u
p
u
p0X
u
By , we get the sequence
conve with
0
we get the sequence
with
Since

Combining (3.3),(3.4) with the continuity of
00 nX
uu
rges uniformly on 0()

cp
n
Dut
[0 1]
00
lim( )( )
 

cp cp
nn
Dut Dut.
Similarly, by 0nX
vv
0() 
q
n
Dvt converges uniformly o
limcq cq
0,
n
c[0 1]
000
( )( )
 

nn
Dvt Dvt.
00
100
[01]
1100
[01]
2200
[01]
2200
[01]
()(
)( )()( )
max(())()(())()
max()()()()
max(())()(() )()



 
 

 
 

 

nn
n
t
nn
t
nn
t
nn
t
Tu vTu v
tTuvt
Tu vtTuvt
Tuv tTuv t
Tuvt Tuvt
1
)
max (

 
XX
n
Tu v
1
f
, 2
f
,
we can get
)
Thus T is continuous in
00 00
()() 0(()()

 
nnXX nn
TuvTuvuvuv

X
X.
Let  
X
X
stants 
i
Lbe bounded. Then there exist posi- tive con such that 0
00
()()) (
()

(()()12)

 


 
ut DvtLi
uv

cp cq
ii
ftutvt D
Thus, for any we have ()uv ,
1
110
0
1
()()() ()
() 
 

cp c
uvtvs Dus

0
q
11
110
0
( )(()
())
(1)(()()()
()



 






t
cpc
Tt sfsus
Dvsds
tsfsusvsDus




111 121
11 121
(1) (1) (1
()
(1) (1)
 )




  
LL L
p
LL
p




2
11
0
00
1
())
1

Du s Dv
11
0
1
(( ))()()(()()
(1)
()
(1)(( )( )
()



 





tc
pcq
c
Tuvtt sfsusvs
s ds
sfsusvs




0
11
21 10
0
0
())
(1)(()()()
()
())







 


q
p
cpc
q
Dvsds
t
s
fsusvs Dus
p
Dvsd



s
00
00
( )( ))
(
()()



11
21
0(1)(( )( )
)


 

pcq
p
c
pcq
s DvsdsDu
s
fsusvs
p




Du s Dvs
11 21
121
)
)(1)
(1)
(1) (1)
(1) (


 

 
  
ds
LLL
p
p

 


Hence


LL

11 211
111
(1)(1)
() (1)( 1)
 
 
 

X
LL
TuvML
p
 

(3.5)
where 1
M
In the sam
is given by (3.1).
e way ,we can verify that
21
2
22 122
(1
() (1)
(1)
(1)




 
 


 

XL
Tuv
L
)
M
L
q




(3.6)
where 2
M
is given by (3.1). Thus,
112 2
()


 
XX
TuvML MLM
which implies that the operator T is uniformly bounded.
Next we show that T is equicontinuous.
For any 12 01

tt,
1
2
1
2
12211 1
211 0
0
0
211 0
0
1
(()())(()())
[( )()](()()()
())
[()()](()()()
())
[(






 








tcp c
q
tcp c
t
q
t
TutvtTut vt
GtsGtsfsusvsDus
Dvs ds
GtsGtsfsusvsDus
Dvs ds
G




211 0
0
212 1221
1
)()](()() ()
())
()
[]
(1)(1)(1)



 

 

   
cp c
q
tsGtsfsusvs Dus
Dvs ds
t ttt
p



 

tt
L

M. J. LI, Y. L. LIU 57
2
1
1211
2
21 0
0
0
())

t
Dv
s ds
2
11 0
0
0
1
(())()(())()
1()(()() ()
(1)
()(()()()
())






 


 




tcp c
cp c
q
TuvtTuv t
ts fsusvsDus
ts fsusvsDus
Dvs ds
L



q
21
22
21
00
121
() ()
(1)
()

  



tt
ts dsts ds
L



11
()

 tt

Analogously, we can obtain the following inequalities:
2 2
t
2211
(()())(()())TutvtTut vt
21 21221
2()
[]
(1
)
(1)( 1)


 
 

  
ttttt
Lq




222
11
121
(())()(())()
()
()


 


Tuv tTuv t
Ltt


11
1

Since the functions


 tttt
 
[0 1], we can are uniformly con- tinuous on the interval conclude that ()

Tuv
is equicontinuous on
Thus, the operator is completely
continuous. The proo f is completed.
Theorem 3.2. Assume that there exist positive con-
stants 

[0 1].
 TXXXX
00
(01234) 0 0 0 (1
iii i
cdicdc di
) such that 234
22
1234
()[0 1][01]

 tx xxxRRt
1123401122
33 44
2123401122
33 44
()
()




f
tx xxxccxcx
cx cx
f
tx xxxddxdx
dx dx
(3.7)
In addition, assume that
112 1132 3
122 2142 4
1
()(2 )
1
1()1
 
 
Mc MdMcMdp
Mc MdMc Md
where
(2 )q
1
M
and 2
M
lem (1
are define d by (3.1).
Thenprob.1) has at least one positive solu-
tion.
Proof. Let us verify that the set
is bounded. Let , then.
the

()() ()01 VuvXXuvTuv 
()uv V ()() uv Tuv

For any
, we ha [0 1]tve
1
1
0
()( )()
1( )(
tcp
utTuvt
ts fs
10
10
0
()()()
(1)(( )( )()
suvs Dus
ds
0
()
())
cq
Dv s
11
1
()
cp
t
s
fsusvsDus 














 




 



0
11
21 1
0
00
01230
())
()
(1)(()()
()
()())
[()()
cq
p
cpc q
c
Dvs ds
tsfsusvs
p
Dus Dvs
ccut cvt cD

 







 



ds
112
40
012 3
4
()
()]()
(1) (1)
1
[(2)
1
]
(2)
()
p
cq
XXX
X
ut
cDvt p
ccu cv cu
p
cv
q







 
 
 





 

11
2
)
(1 (1)
p


 

1
2
1
()(
tts fsu
10
0
110
0
( )( )( )
(1)
())
(1)(()()()
cp
cq
cp
svs Dus
s ds
0
1
1
()
Dv
()(())()utTuvt
s
fsusvs Dus
















 





 
0
11
21
0
00
0
())
(1)(()()
()
()())
[
cq
p
cpc q
Dvs ds
sfsusvs
p
Dus Dvsds
c








 



123
040
2
012 3
4
2
() ()
()()]
1
()
(1)(1)
1
[(2 )
1
]
(2)
1
((1) (
c
pcq
XXX
X
cut cvt c
Dutc Dvt
p
ccucvcu
p
cv
q
p








 
 
  
 




  
 

)
1) 
Hence,
10 12
34
[
11
]
(2)(2)



 
 
 
XXX
XX
uMccucv
cucv
pq

M. J. LI, Y. L. LIU
58
1
M
where is defined by (3.1). Similarly, we can get
20 12
34
[
11
]
(2)(2 )
 

 
 
 
XXX
XX
vMddudv
dud
pq
v
(3.9)
where 2
M
), we obis defined by (3.1). Combining (3.8) with (3.9tain
112 11323
12 2214 24
102 0
31020
()
1
[()]
(2 )
1
[()
(2 )
()




 



 


 
XX
XX
X
X
XX
uv
uv
Mc MdMcMdu
p
]
M
cMd McMdv
q
Mc Md
Muv McMd
As a result
102 0
3
() 1





XX 
M
cMd
uv M
for any where [0 1]t3
M
is given by (3.2). So the
set V is
Thus, by Lemma 2.7, the operator T has at least one
fixed point. Hence the problem (1.1) has at least one
positive solution.The proof is completed.
Theorem 3.3. Assume that both and
bounded.
1
f
22
2[0 1]
 
f
RR R
are continuous functions and there exist constants
, such that
R
)
)


0 (1234)

 
ii
nn i
222 2
1234 1234
[01]() ()

 tuuuuRsRvvvvR
112341 1234
1112223 33
444
2123421234
11 1222333
44 4
()(
()(
 





 

f
tu uuuftv vvv
nu vnuvnuv
nuv
f
tu uuuftv vvv
nu vnuvnuv
nuv
In addition, assume that
123 41
123 42
11
(2)(2) 4
11
(2)(2) 4
 

 

 
nnn n
pq
nnn n
pq
1
1

M
M
where 12 
M
M
nique
are defined by (3.1). Then the problem
(1 solution.
Proof. Define
.1) has a u
[01] 11
sup( 0000)
   tftN
[01] 22
sup( 0000)





tftN

hat such t


112 2
4maxrMNMN We show that

rr
TBB where () ()


r
B{uvXX
()  uv r}
For

B, we have
r
uv
1
1100
1
)(()()()())
()


0
1
110
0
0
1
21
0
()()
1(
(1)(()()()
()
())
(1)
()



 

 


cp cq
ts fsusvs Dus Dvsds
t



 




 



t
cp
cq
Tuvt
sfsusvsDus
Dvs ds
ts
p




11
00
1100
0
11
(()()
()())
1()[(() ()()())
()
(0000)(0000)]
(








  


p
cpc q
tcp cq
fsusvs
Dus Dvsds
tsfsusvs Dus Dvs
fsfs ds




1
 t

1110
0
011
1
1
(1)[(( )( )()
)
())(0000)()]
[



 
  
 1
210
0
(1)(()()()
()



01
())(0000)

0000

 
 
pc p
sf susvsDus
p



 
cq
Dv s fs


cp
cq
sfsusvsDus
Dvs fsfsds
t


1
123 41
( 0000)]
11
[() ]
(2)(2 )
 
 
 
11
2
()
(1) (1)
 
 
()
f
sds
nnnnrN
pq
p 

1
210
0
0
1110
0
0
(())()

1()(()()()
(1)
())
1
(1)(()()()
()






 

 



 



tcp
cq
cp
cq
ts fsusvs Dus
Dvs ds
Tuvt
s
fsusvs Dus
Dv




11
21
0
00
10
0
01 1
())
(1)(()()
()
()())
()[(() ()()
(1)
())(0000)(0000)






 

2
1
 

 


p
cpc q
cp
cq
sds
sfsusvs
p
Dus Dvsds
tsfsusvs Dus
Dvs fsfs




t


1110
0
11
0
0 1
1
(1)[(( )( )()
()
(
(1)[(( )()
)
)())(0000)




01 1
1
))(0000)(0000)]

2
(
0
(

]


 



cq

 



cp
p
cc q
s
p
d
s
fsusvs Dus
Dvs
sfsusvs
p
s Dvsfs



fsfs ds

Du


M. J. LI, Y. L. LIU 59
1
fs
12 4 1
2
( 0000)]
[() ]
)(2)
(1)
()
(1) (1)

 


 
  
ds
nnn rN
q
p



Hence
311
(2
 np
1
123 411
()()
11
[() ]
(2)(2)2


 

X
Tuvt
r
nnnn rNM
pq

In the same way, we can obtain that
2
123 422
()()
11
[() ]
(2)(2)2
 

 
 

X
Tuvt
r
nnnnrNM
pq

Consequently,  Now for ()()


XX
Tuv tr22 ()

uv
11
()uvX X and for any [0 1]t


we get
122 111
112 202
0
021 11
01 01
1
()()()()
1() (()()()
()
())(()()
() ())
tcp
cq
cp cq
Tu vtTuvt
tsfsusvs Dus
Dvs fsusvs
DusDvs ds










 




11122
0
02 02
1110101
11
21 12 2
0
(1)(( )( )
()
()())
(()()()())
(1)(()()
()
cpc q
cp cq
p
tsfsusvs
DusDvs
fsusvs Dus Dvsds
tsfsusvs
p









 


 


 


02 02
1110101
112
123
42121
()())
(()()()())
() 1
()(
(1) (1)(2
1
)()
(2)
cpc q
cp cq
Dus Dvs
fsusvs Dus Dvsds
nnn
pp
nuuvv
q





 

 
  

 
)
122 111
212 202
0
021 11
01 01
(())()(())()
1() (()()()
(1)
())(()()
()





 


 

tcp
cq
cpc q
Tu vtTuvt
tsfsusvs Dus
Dvs fsusvs
Dus Dv



11122
0
02 02
11 101
())
1
(1)(()()
()
()())
(()()()







 
11
212 2
0
02 02
11 10101
2
123
(1)(()()
()
()())
(()()()())
(1)
()
(1) (1)
1
(





 

 


 

p
cpc q
cp cq
sfsusvs
p
DusDvs
fsusvs Dus Dvsds
p
nnn



 

4
21 21
1)
(2)(2 )
()

 

n
pq
uu vv
Hence
122 111
11 2 34
2
( u
he above discu we can ob
121
()()()()
11
()
(2)(2 )
)

 
 
 


X
XX
Tu vtTuvt
Mnn nn
pq
u vv
Similarly to tssion,tain
222 211
212 34
21 21
()()()()
11
()
(2)(2 )
()
 

 
 
 


X
XX
TuvtTuv t
Mnnnn
pq
uuvv
As a result
22 11
11 2 34
212 34
21 21
()()()()
11
[()
(2)(2 )
11
()
(2)(2 )
()



 
 
 
 
  


XX
XX
TuvtTu vt
Mnn nn
pq
Mn nnn
pq
uu vv
]
Since
11
11 2 34
(2)(2 )
11
212 34
(2)(2)
[( )
()
 
 
 
 
]1

 
pq
pq
Mn nnn
Mn nnn
therefore T is a contraction operator.
Thus the conclusion of the theorem holds by using the
fixed point theorem of contraction mapping principle.
The proof is completed.
4. Acknowledgements
This work was financially supported by NNSF of China
Grant No.11271087, and No.61263006, Guangxi Scien-
tific Experimental ( China- ASEAN Research ) Centre
No.20120116.
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DusDvs
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