Infinitely many Solutions to boundary value problems for a coupled
system of fractional differential equations
Peiluan Li
1,*
， Changjin Xu
2
(1.School of Mathematics and Statistics, Henan University of Science and Technology, Luoyang, 471023,China;
2.Guizhou Key Laboratory of Economics System Simulation，Guizhou College of
Finance and
Economics，Guiyang，550004, China)
Abstract
Using variational methods, we investigate the solutions to the boundary value problems for a coupled
system of fractional order differential equations. First, we obtain the existence of at least one weak
solution by the minimization result due to Mawhin and Willem. Then, the existence criteria of infinitely
many solutions are established by a critical point theorem due to Rabinowitz. At last, some examples are
also provided to illustrate the results.
Keywords: Fractional Differential Equations; Coupled system; Variational methods; Infinitely many
solutions
2010 MSC: 34B37, 34K10, 26A33
1
Introduction
Recently, a great attention has been focused on the study of boundary value problems (BVP) for
fractional differential equations. Fractional calculus provide a powerful tool for the description of
hereditary properties of various materials and memory processes [1-2]. Fractional differential equations
have also recently proved to be strong tools in the modeling of medical, physics, economics and
technical sciences. For more details on fractional calculus theory, one can see the monographs of
Kilbas et al. [3], Lakshmikantham et al. [4], Podlubny [5] and Tarasov [6]. Fractional differential
equations involving the Riemann–Liouville fractional derivative or the Caputo fractional derivative
have been paid more and more attentions.
In recent years, some fixed point theorems and monotone iterative methods have been applied
successfully to investigate the existence of solutions for nonlinear fractional boundary-value problems,
see for example, [7-12] and the references therein.
In [11], Bai and Fang studied the following singular coupled system of fractional differential
equations
 DT (u (t ))  f (t , v),
 
 DT (u (t ))  g (t , u ),
where
0<t  1,
0<t  1,
0   ,   1 , D , D  are two standard Riemann-Liouville fractional derivatives. By
applying the Krasnoselskii’s fixed point theorem and the nonlinear alternative of Leray–Schauder
theorem in a cone, the authors have obtained the existence of positive solutions for the coupled system.
*Corresponding author.
Email addresses: lpllpl_lpl@163.com(PeiluanLi); xcj403@126.com (Changjin Xu)
1
By means of the nonlinear alternative of Leray–Schauder theorem, Ahmad and Alsaedi in [12]
established the existence and uniqueness results for the following fractional differential equations
c 
c 
(k )
 DT u (t )  f (t , DT v(t )), u   k , 0  t  1,
c 
c 
(k )
 DT u (t )  g (t , DT u (t )), v   k , 0  t  1,
c
where
D denotes the Caputo fractional derivative,  ,   (m  1, m),  ,   (n  1, n) with
   ,    , k  0,1, 2, …, m  1,  , , ,  N ,
and
k ,  k are suitable real constants.
On the other hand, critical point theory and the variational methods have been very useful in
dealing with the existence and multiplicity of solutions for integer order differential equations with
some boundary conditions. But until now, there are few works that deal with the fractional differential
equations via the variational methods; see [13-21]. It is often very difficult to establish a suitable space
and variational functional for fractional boundary value problem for several reasons. First and
foremost, the composition rule in general fails to be satisfied by fractional integral and fractional
derivative operators. Furthermore, the fractional integral is a singular integral operator and fractional
derivative operator is non-local. Besides, the adjoint of a fractional differential operator is not the
negative of itself. By means of critical point theory, Jiao and Zhou [13] first considered the following
fractional boundary value problems
 t DT (0 Dt u (t )  F (t , u (t )),

 u (0)  u (T )  0,
where
  (0,1) , 0 Dt
t
DT are the left and right Riemann-Loiuville fractional derivatives
F :[0, T ]  R N  R (with N  1 ) is a suitable given function and F (t , x) is the
respectively.
gradient of
and
a.e. t  [0, T ],
F with respect to x .
In [21], the authors investigated the existence of weak solution for the following coupled system of
fractional differential equations
 t DT (a (t ) 0 Dt (u (t ))   Fu (t , u (t ), v(t )),
 

 t DT (b(t ) 0 Dt (u (t ))   Fv (t , u (t ), v(t )),
 u (0)  u (T )  0, v(0)  v(T )  0,

where

is a positive real parameters,
b0 : ess inf b(t )  0 ,  ,   (0,1] ,
[0,T ]
fractional derivatives of order

0<t  T ,
0<t  T ,
a, b  L [0, T ] with a0 : ess inf a(t )  0 and
[0,T ]

0
Dt and t DT are the left and right Riemann-Liouville
respectively, and
F :[0, T ]  R 2  R is a function such that
F (, x, y ) is continuous in [0, T ] for every ( x, y )  R 2 and F (t , , ) is a C 1 function in R 2
for any
t  [0, T ] , and Fs denotes the partial derivative of F with respect to s . By means of the
2
variational methods and a critical point theorem due to Bonanno and Marano, the authors get the
existence of three distinct weak solutions.
Motivated by the works above, in this article, we consider the following coupled system of
fractional differential equations
 t DT (a(t ) 0 Dt u (t ))   v(t )  Fu (t , u (t ), v(t )),
 

 t DT (b(t ) 0 Dt (v(t ))  u (t )  Fv (t , u (t ), v(t )),
 u (0)  u (T )  0, v(0)  v(T )  0,

0<t  T ,
0<t  T ,
(1.1).
First, we obtain the existence of at least one weak solution by the minimization result due to
Mawhin and Willem. Then, the existence criteria of infinitely many solutions are established by a
critical point theorem due to Rabinowitz. At last, some examples are given to illustrate the results.
The rest of this paper is organized as follows. In Section 2, some definitions and lemmas which are
essential to prove our main results are stated. In Section 3, we give the main results. And, two examples
are offered to demonstrate the application of our main results.
2
Preliminaries
At first, we present the necessary definitions for the fractional calculus theory and several lemmas
which are used further in this paper.
f be a function defined by [a, b] . The left and right Riemann-Liouville
Definition 2.1 ([3]).Let
fractional derivatives of order

for function
f denoted by
a
Dt f (t ) and t Db f (t ) respectively,
are defined by

a Dt f (t ) 
dn
1 dn
 n
D
f
(
t
)

 t
dt n
( ) dt n
t
 (t  s)
dn
(1) n d n
 n
f (t ) 
t Db f (t )  ( 1)
t Db
dt n
( ) dt n

n  1
0
n

b
t
f ( s )ds,
( s  t ) n  1 f ( s )ds, t  [a, b],   0,
provide that the right-hand side integral is pointwise defined on
[a, b].
Lemma 2.2 ([3]). The left and right Riemann-Liouville fractional integral operators have the property of
a semigroup; that is,
b
[
a
provided that
q  1,
b
a
Dt f (t )]g (t )dt   [ t Db g (t )] f (t )dt ,   0,
a
f  Lp ([a, b], R), g  Lq ([a, b], R) and p  q, q  1,
1 1
  1 .
p q
3
1 1
  1   or p  1,
p q
Lemma 2.3 ([3]). The left and right Riemann-Liouville fractional integral operators have the property of
a semigroup; that is,
b
[
a
provided that
b
a
Dt f (t )]g (t )dt   [ t Db g (t )] f (t )dt ,   0,
a
f (a )  f (b)  0, f   L ([a, b], R N ) and g  L1 ([a, b], R N ) or g (a ) 
g (b)  0, f   L ([a, b], R N ) and f  L1 ([a, b], R N ) .
In order to establish a variational structure which enables us to reduce the existence of solutions of
problem (1.1) to one of finding critical points of corresponding functional, it is necessary to construct
appropriate function spaces.
Let us recall that for any fixed
u

t  [0, T ] and 1  p  ,
 max u (t ) ,
u
t[0,T ]
T
p
1
p
 (  u ( s ) ds ) .
Lp
0
0    1, We define the fractional derivative spaces E0 by the closure of
Let
C0 ([0, T ], R ) with respect to the weighted norm
u
T

T
2
1
2
 (  a(t ) 0 Dt u (t ) dt   u (t ) dt ) 2 , u  E0 .
0
(2.1)
0
where
C0 ([0, T ], R)  {u  C  ([0, T ], R) : u (0)  u (T )} . Clearly, the fractional derivative
space
E0 is the space of functions u  L2 [0, T ] having an  -order fractional derivative
0
Dt u (t )  L2 [0, T ] and u (0)  u (T ) .
Lemma 2.4 ([13]).Let
(i)
u
Let
2
L

T
(  1)
1
   1; for all u  E0 , one has
2

0 Dt u (t )
u
(ii)
L2


T  1/2
( ) 2  1
0
Dt u (t )
L2
a0  min a (t ) , from Lemma 2.4, one has
tJ
1
u
L2
T
2
T

(  a(t ) 0 Dt u (t ) dt ) 2 ,
0
(  1) a0
(2.2)
1
u

T
2
T  1/2

(  a(t ) 0 Dt u (t ) dt ) 2 .
( ) a0 (2  1) 0
(2.3.)
By (2.2),(2.3), we can also defined
u
T

2

1
2
 (  a(t ) 0 Dt u (t ) dt ) , u  E0 .
0
4
(2.4)
u
Then we can conclude that
In the sequel, we will consider

defined in (2.1) is equivalent to the norm
E0 with the norm u

u

defined in (2.4).
defined in (2.4). Obviously,
E0 is a
u .
reflexive and separable Banach space with the norm
It follows (2.2)-(2.4) that
u
L2

T
u ,
(  1) a0 
u


T  1/2
u .
( ) a0 (2  1) 
(2.5)
0    1, we define the fractional derivative spaces E0 by the closure of
Similarly, let
C0 ([0, T ], R ) with respect to the weighted norm
T
v
Let

T
2

1
2
2
 (  b(t ) 0 Dt v(t ) dt   v(t ) dt ) .
0
(2.6)
0
b0  min b(t ) . According to Lemma 2.4, one has
t J
v

2
L
1
T
2
T
(  b(t ) 0 Dt v(t ) dt ) 2 ,
(   1) b0 0
(2.7)
1
v

T
2
T  1/ 2

(  b(t ) 0 Dt v(t ) dt ) 2 .
(  ) b0 (2   1) 0
From (2.7), (2.8), it is easy to see that
v


It is easy to see that the norm
In the sequel, we will consider
E0 is a Hilbert space with the norm
  b(t)
T
0
(2.8)
2

0
Dt u (t ) dt

1
2
,
 E0 .
(2.9)
v  defined in (2.9) is equivalent to the norm v  defined in (2.6).
E0 with the norm v  defined in (2.9) .
From (2.7),(2.8), we have
v
We denote
v

L2

T
v ,
(   1) b0 
v


T  1/ 2
v .
(  ) b0 (2  1) 
X  E0  E0 equipped with the norm (u , v)
X
are defined in (2.4) and (2.9).
Similar to some properties in [13], we have the following results.
5
 u

 v  , where u
(2.10)

and
 ,   (0,1] .
Lemma 2.5 Let
The fractional derivative space
X  E0  E0 is a reflexive and
separable Banach space.
 ,   (0,1]
Lemma 2.6 Let
and the sequence
{uk } converges weakly to u in E0 , the sequence
{vk } converges weakly to v in E0 , then uk  u , vk  v in C ([0, T ], R ); that is,
un  u
 0 , vn  v


 0 , as k  .
Definition 2.7 By the solution of the coupled problem (1.1) , we mean any
(i)
t
(u , v)  X such that
DT 1 (a (t )0 Dt u (t )) , 0 Dt 1u (t ) , t DT 1 (b(t ) 0 Dt v(t )) , 0 Dt 1v(t ) are derivatives for every
t  [0, T ] , and
(ii)
(u , v)  X satisfies (1.1).
Definition 2.8
(u , v)  X is called a weak solution of problem (1.1) if

T
0
T
a(t )0 Dt u (t ) 0 Dt x(t )dt   b(t )0 Dt v(t ) 0 Dt y (t )dt
0
T
T
0
0
  (v(t ) x(t )  u (t ) y (t ))dt   ( Fu (t , u (t ), v(t )) x(t )  Fv (t , u (t ), v(t )) y (t ))dt  0
for all
( x, y )  X .
Similarly to the proof of Theorem 5.1 in [13], we have the following Lemma 2.12.
Lemma 2.9 Let
0   ,   1 and u  E0 . If (u , v)  X is a non-trivial weak solution of the
problem (1.1), then
(u , v)  X is also a non-trivial solution of the problem (1.1).
Throughout this paper, we assume that the following condition
( H1 ) is satisfied.
( H1 ) .   0 is a real parameters, a, b  L [0, T ] ,  ,   (0,1] , and F :[0, T ]  R 2  R is
a function such that
F (, x, y ) is continuous in [0, T ] for every ( x, y )  R 2 with F (t , 0, 0)  0 ,
F (t , , ) is a C 1 function in R 2 for any t  [0, T ] , and Fs denotes the partial derivative of F
with respect to s .
We consider the functional
 (u, v) 
 : X  R , defined by
T
2
2
1 T
[a (t ) 0 Dt u (t )  b(t ) 0 Dt v(t )   (u 2 (t )  v 2 (t )]dt   F (t , u (t ), v(t ))dt .

0
2 0
(2.11)
6
Then

is continuously differentiable under the assumption
T
T
0
0
( H1 ) , and we have
 (u , v), ( x, y )   a(t )0 Dt u (t ) 0 Dt x(t )dt   b(t )0 Dt v(t ) 0 Dt y (t )dt
T
T
0
0
  (v(t ) x(t )  u (t ) y (t ))dt   ( Fu (t , u (t ), v(t )) x(t )  Fv (t , u (t ), v(t )) y (t ))dt ,
(2.12)
for all
( x, y )  X . Hence the critical point of  is the weak solution of problem (1.1). Next, we
consider the critical point of
.
Finally, we need the following results in critical point theory.
Definition 2.10 Let
E be a real Banach space, and   C 1 ( E , R) . We say that  satisfies the
Palais–Smale condition if any
{um }  E for which  (um ) is bounded and  (um )  0, as
m  , posses a convergent subsequence.
The proofs of the main results in this paper are based on the following critical point theorems.
 is weakly lower semi-continuous (w.l.s.c) on a reflexive
Banach space X and has a bounded minimizing sequence, then  has a minimum on X.
Lemma 2.11 ([22, Theorem 1.1]). If
X be a real Banach space and let   C 1 ( X , R) be an even
Lemma 2.12 ([23, Theorem 9.12]) Let
functional, which satisfies the Palais–Smale condition and
where dimV   , and
(1) there exist

 (0)  0 .
Suppose that X  V  E ,
satisfies that:
a,   0 such that  B
 E
 a, where B  {u  X : u  },
(2) for any finite dimensional subspace W  X , there is
R  R (W ) such that  (u )  0 on
W \ BR (W ) .
Then,

possesses an unbounded sequence of critical values.
3. Main results
Lemma 3.1
(u , v)  X is bounded if and only if u  E0 , v  E0 are all bounded.
Proof. From
(u , v)
 u
X
(u , v)
X

 v  , for M  0 is a constant, it is easy to proof
M  u

 v

M  u

 M, v

M ,
which shows the conclusion of Lemma 3.1.
Next we give the first result which is based on the minimization theorem due to Mawhin and
Willem.
Theorem 3.2 Let
( H1 ) hold and   1 ,
7
1  2 (  1)a0
1  2 (   1)b0
(


),
(
  )}
2
T 2
2
2T 2 
( H 2 ) . There exist a positive constant a1  min{
such that
lim sup
x  , y 
uniformly for
F (t , x, y )
2
x  y
2
 a1 ,
(x, y )  R 2 , t  [0, T ] .
Then (1.1) possesses at least one weak solution.
Proof. First, we prove that

is weakly lower semi-continuous. Since X is a separable and reflexive
real Banach space, we assume that
2.6, we can obtain that
{uk , vk }  X converges weakly to (u , v) in  X . By Lemma
uk  u , vk  v uniformly in C ([0, T ], R ) , as k   , that is,
uk  u

vk  v
 0,

 0 , as k   .
and
lim inf (uk , vk )
k 
Then it follows
X
 lim inf ( uk
k 

 vk

) u

 v

 (u, vk ) X .
( H1 ) that
lim inf  (uk , vk )  lim inf{
k 
k 
2
2
1 T
[a (t ) 0 Dt uk (t )  b(t ) 0 Dt vk (t )

2 0
T
 (uk 2 (t )  vk 2 (t ))]dt   F (t , uk (t ), vk (t ))dt}
0

2
1


[
a
(
t
)
D
u
(
t
)

b
(
t
)
D
v
(
t
)
0
t
0
t
2 0
T
2
T
 (u 2 (t )  v 2 (t ))]dt   F (t , u (t ), v(t ))dt   (u, v) .
0
which implies that

is weakly lower semi-continuous.
Now, we are in the position to show that the functional
From

is coercive.
( H 2 ) , we know there exist two positive constants a2 , a3 large enough such that
2
2
F (t , u , v)  a1 ( u  v ), for u  a2 , v  a3 , t  [0, T ]
On the other hand, from the continuity of
for
F (t , u , v) , we con clued that F (t , u , v) is bounded
u  a2 , v  a3 , t  [0, T ] . Then there exists a constant b1  0 such that
F (t , u, v)  b1 ,
for
u  a2 , v  a3 , t  [0, T ] .
8
(t , u , v)  [0, T ]  R 2 , we can get
Hence, for all
2
2
F (t , u , v)  a1 ( u  v )  b1.
Together with (2.5), (2.10), (2.11) , one has
T
2
2
1 T


2
2
[
a
(
t
)
D
u
(
t
)

b
(
t
)
D
v
(
t
)


(
u
(
t
)

v
(
t
)]
dt

0 t
0 t
0 F (t , u (t ), v(t ))dt
2 0
2
2
1 T
  [a (t ) 0 Dt u (t )  b(t ) 0 Dt v(t )   (u 2 (t )  v 2 (t )]dt
2 0
 (u, v) 
T
2
2
  [a1 ( u  v )  b1 ]dt
0

1
(u
2
2

2

T
 v  )  (  a1 )  [(u 2 (t )  v 2 (t )]dt  b1T
0
2
1 
T 2
 [  (  a1 ) 2
)] u
2 2
 (  1)a0
In view of
a1  min{


u
as

 , v


 b1T ,
 .
is coercive. Thus, by virtue of Lemma 2.11, the functional
a critical point of
2
1  2 (  1)a0
1  2 (   1)b0
(


),
(
  )} , we can conclude
2
T 2
2
2T 2 
 (u, v)  ,
Then
1 
T 2
 [  (  a1 ) 2
]v
2 2
 (   1)b0
2

has a minimum, which is
 . It follows that the boundary value problem (1.1) has one weak solution.
( H 2 ) becomes the subquadratic case, that is
Remark 3.3 If the asymptotically quadratic case in
e
e
lim sup F (t , x, y )  e x 1  e0 y 2 , 1  e1 , e2  2, e, e0  0 ,
x  , y 
we can get the similar result.

Our task is now to use Lemma 2.12 to find infinitely many critical points of functional
Theorem 3.4 Let
on
X.
( H1 ) hold. If the following assumptions ( H 3 ) - ( H 5 ) are satisfied.
1  2 (  1)a0
1  2 (   1)b0
( H 3 ) . There exists a positive constant c0  min{ (


),
(
  )}
2
T 2
2
T 2
such that
lim sup
x  0, y  0
uniformly for
F (t , x, y )
2
x  y
2
 c0 ,
(x, y )  R 2 , t  [0, T ] .
( H 4 ) . There are constants   2 , M  0 such that
2
2
0   F (t , x, y )  xFx (t , x, y )  yFy (t , x, y ) , for all t  [0, T ] and x  y  M .
9
here,
Fs denotes the partial derivative of F with respect to s .
( H 5 ) . F (t , u , v)  F (t , u , v) .
  (0, min{
Then for every
 2 (  1)a0  2 (   1)b0
,
) , the problem (1.1) has infinitely
T 2
T 2
many solutions.
( H 4 ) implies there exist d 0  0, d1 , d 2 , d3  0 such that
Proof. We note that
1
F (t , x, y ) 
[ xFx (t , x, y )  yFy (t , x, y )]  d 0 , for t  [0, T ], (x, y )  R 2 ,



F (t , x, y )  d1 x  d 2 y  d 3 , for t  [0, T ], (x, y )  R 2 .
The assumption
(3.1)
(3.2)
( H1 ) implies that  is continuous and continuously differentiable. In view of
the expression (2.11) and
( H 5 ) , it is obvious that  is even and  (0)  0 .
we divide our proof into three steps.
Step 1. We proof that  satisfies the Palais-Smale condition.
Let
we show
{uk , vk }  X such that  (uk , vk ) is bounded and  (uk , vk )  0, as k   . First,
{uk , vk }  X is bounded. It follows (2.10), (2.12) and (3.1) that
 (uk , vk ) 
2
2
1 T
[a (t ) 0 Dt uk (t )  b(t ) 0 Dt vk (t )   (uk 2 (t )  vk 2 (t ))]dt

0
2
T
  F (t , uk (t ), vk (t ))dt
0
1
 ( uk
2
2
2
 vk

)


2
T
0
[(uk 2 (t )  vk 2 (t )]dt
T 1
  ( [uk Fuk (t , uk , vk )  vk Fv (t , uk , vk )]  d 0 )dt
0

1
 ( uk
2

1

2

 vk
T
2

)

2
T
0
2
[(uk 2 (t )  vk 2 (t )]dt
T
2
{ a (t ) 0 Dt uk (t ) dt   b(t ) 0 Dt vk (t ) dt   (uk , vk ), (uk , vk )
0
0
T
  (uk 2 (t )  vk 2 (t ))dt}  d 0T
0
1 1
 (  )( uk
2 
2

1 1
2
 vk  )   (  )( uk
2 
10
2
L2
 vk
2
L2
)

1

 (uk , vk ) (uk , vk )  d 0T
1 1
 (  )[ uk
2 

1

2
(1  

T 2
T 2
2
)

(1


) vk  )
2
2
 (  1)a0
 (   1)b0
 (uk , vk ) (uk , vk )  d 0T ,
 2 (  1)a0  2 (   1)b0
For   (0, min{
,
) , according to the condition  (uk , vk ) is
T 2
T 2
bounded with
 (uk , vk )  0,
Then Lemma 3.1 shows
as
k  , it easy to proof that u  E0 , v  E0 are all bounded.
{uk , vk }  X is bounded. From the reflexivity of X  E   E  , we know
{uk , vk }  X has a weakly convergent subsequence. Without loss of generality, we assume that
{uk , vk }  X converges weakly to (u , v) in  X . By Lemma 2.6, we can obtain that uk  u ,
vk  v in C ([0, T ], R) , as k  , that is,
uk  u

vk  v
 0,

 0 , as k   .
(3.3)
From (2.12), we have
T
T
2
2
 (uk , vk ), (uk , vk )  { a(t ) 0 Dt uk (t ) dt   b(t ) 0 Dt vk (t ) dt
0
0
T
T
0
0
  (uk 2 (t )  vk 2 (t ))dt   [uk Fuk (t , uk , vk )  vk Fv (t , uk , vk )]dt
 ( uk
2
 vk

T
2
)    [(uk 2 (t )  vk 2 (t )]dt

0
T
  [uk Fuk (t , uk , vk )  vk Fv (t , uk , vk )]dt ,
0
Then, it follows that
uk  u
2

 vk  v
2

  (uk , vk )   (u, v), (uk  u, vk  v)
  ( uk  u


T
0
uk (t )  u (t ) dt  vk  v


T
0
vk (t )  v(t ) dt )
T
  Fuk (t , (uk  u )(t ), (vk  v)(t ))dt uk  u
0

T
  Fvk (t , (uk  u )(t ), (vk  v)(t ))dt vk  v
0
11

,
(3.4)
From (3.3) and
 (uk , vk )  0,
as k   , we have
 (uk , vk )   (u, v), (uk  u, vk  v)
  (uk , vk ), (uk  u, vk  v)   (u , v ), (uk  u, vk  v)
  (uk , vk ) uk  u, vk  v
  (u , v ), (uk  u , vk  v)
X
 0, as k   .
In view of (3.3)-(3.5), we know that
uk  u
 0, vk  v


(3.5)
uk  u
2

2
 vk  v

 0 , as k   , which shows
 0 , as k   . So we known (uk  u, vk  v)
X
 0, as
k   . Then {uk , vk } converges strongly to (u , v) in X . Therefore  satisfies Palais-Smale
condition.
Step 2. We show that the condition (1) in Lemma 2.12 holds.
( H 3 ) , there exists a constant c  0 small enough and a constant c0  0 such that
From
2
2
F (t , x, y )  c0 ( x  y ),
for
2
2
1
( u  v ) 2  c , t  [0, T ]
Let
1
T 2
1
T 2
c1  min{ [1  (  2c0 ) 2
],
[1  (  2c0 ) 2
]} .
2
 (  1)a0
2
 (   1)b0
( H 3 ) implies c1  0 .
The assumption
Then for
2
2
1
( u  v ) 2  c , t  [0, T ] , it follows from (2.5), (2.10) and (2.11) that
T
2
2
1 T
[a (t ) 0 Dt u (t )  b(t ) 0 Dt v(t )   (u 2 (t )  v 2 (t ))]dt   F (t , u (t ), v(t ))dt

0
2 0
T
T
1

2
2
 ( u   v  )   [(u 2 (t )  v 2 (t )]dt   c0 [(u 2 (t )  v 2 (t )]3 dt
0
2
2 0
 (u, v) 
1
 (u
2
2


T 2
2
 v  )  (  c0 )[ 2
)u
2
 (  1)a0
1
T 2
 [1  (  2c0 ) 2
]u
2
 (  1)a0
2

2


T 2
2
v ]
2
 (   1)b0
1
T 2
 [1  (  2c0 ) 2
]v
2
 (   1)b0
12
2

 c1 ( u
For
2
2

 v ),
0    c, let
2
1
2
2
2
2
1
2
(u , v) 2  ( u  v ) , B  {u  X : (u , v) 2  ( u  v )   } .
Then, we can easily choose a constant
a1  0 such that  B
 E
Step3. For any finite dimensional subspace W  X , we prove
(u , v) W \ (0, 0) with u
For any r  0 and
  (0, min{
 (ru, rv) 

 a1 .
 (u )  0
 v

on
W \ BR (W ) .
 1 , by the conditions   2 ,
 2 (  1)a0  2 (   1)b0
,
) and (3.2) , we have
T 2
T 2
r2
2

T
0
2
2
[a (t ) 0 Dt u (t )  b(t ) 0 Dt v(t )   (u 2 (t )  v 2 (t ))]dt
T
  F (t , ru (t ), rv(t ))dt
0
T
 r 2   F (t , ru (t ), rv(t ))dt
0
T

T

 r 2  d1r   u dt  d 2 r   v dt  d3T
0
0
  , as r   .
Hence, there exists a constant
r0  0 such that (ru, rv)   and  (ru , rv)  0 for any
r  r0 . Since W is a finite dimensional subspace of X , we know all the norms in W are
equivalent. For all
(u , v) W \ (0, 0) with u
can choose the same

 v

 1 , similarly to the procedure in [24], we
r0  0 such that there exists R(W )  0 and  (u , v)  0 on W \ BR (W ) .
All the conditions in Lemma 2.12 hold. Then it follows Lemma 2.12 that the function

has
infinitely many critical points. That is, the boundary value problem (1.1) has infinitely many weak
solutions. As a consequence of Lemma 2.9, we deduce that the boundary value problem (1.1) has
infinitely many solutions.
Finally, we give two examples to illustrate the usefulness of our main result. Consider the following
coupled system of fractional differential equations
Example 3.5
13
1
 12
v (t )
2
D
(
D
 Fu (t , u (t ), v(t )), 0<t  1,
 t T 0 t u (t )) 
2

1
 12
u (t )
 Fv (t , u (t ), v(t )), 0<t  1,
 t D1 ( 0 Dt2 v(t )) 
2

u (0)  u (1)  0, v(0)  v(1)  0,



Let
F (t , u (t ), v(t )) 
u 2 (t )  v 2 (t )
, we can easily verify that all the conditions of ( H1 ) is
9
satisfied. From (3.6), we know
We Choose
(3.6)
1
2
1
2
    , a(t )  b(t )  1, T  1 ,   .
1
a1  , it follows that
8
1  2 (  1)a0
1  2 (   1)b0
1
3 1  1
1
(


)

(
  )}  ( 2 ( )  )    a1  .
2
2
2
T
2
T
2
2 2
8 4
8
It is also see to see
1 2
2
(u  v )
F (t , u , v)
1
1
lim sup 2
 lim sup 9 2
  a1  ,
2
2
9
8
u  , v  u  v
u  , v 
u v
which implies condition
( H 2 ) holds.
Then the problem (3.6) satisfies all the conditions in Theorem 3.2. In view of Theorem 3.2, the
problem (3.11) has at east weak solution.
Example 3.6
1
 12
2
D
(
D
 t T 0 t u (t ))   v(t )  Fu (t , u (t ), v(t )),
 1
1
 2
D
(
D
 t 1 0 t2 v(t ))   u (t )  Fv (t , u (t ), v(t )),

u (0)  u (1)  0, v(0)  v(1)  0,


Let
0<t  1,
0<t  1,
(3.7)
F (t , u (t ), v(t ))  u 4 (t )  v 4 (t ) , it is easy to check the hypothesis ( H1 ) and ( H 5 ) hold.
(3.7) shows that 
1
   , a (t )  b(t )  1, T  1 . A direct calculation shows
2
 2 (  1)a0  2 (   1)b0
3 
min{
,
}  2 ( )  .
2
2
T
T
2
4
Then for each
c0 

1 
  (0, ) , we choose c0  (   ) , which implies
4
4 4
1 
1 
1  2 (  1)a0
1  2 (   1)b0
(   )  (   )  min{ (


),
(
  )} .
4 4
2 4
2
T 2
2
T 2
14
It is also easy to see
lim sup
x  0, y  0
F (t , x, y )
2
x  y
which shows
Let
2
 lim sup
x  0, y 0
4
4
2
2
x  y
x  y
 0  c0 , uniformly for (x, y )  R 2 , t  [0,1] ,
( H 4 ) is satisfied.
  3 , for (x, y )  R 2 , t  [0,1] , we can get
0   F (t , x, y )  3(u 4  v 4 )  xFx (t , x, y )  yFy (t , x, y )  4(u 4  v 4 ) .
Hence
each
( H 4 ) holds. Then all the conditions in Theorem 3.3 are satisfied. Owing to Theorem 3.3, for

  (0, ) the coupled system (3.7) possesses infinitely many solutions.
4
Acknowledgments
The authors thank the referees for their careful reading of the manuscript and insightful comments,
which help to improve the quality of the paper. We would also like to acknowledge the valuable
comments and suggestions from the editors, which vastly contribute to the perfection of the paper.
This work is supported by National Natural Science Foundation of China (No.11001274，11101126,
11261010), China Postdoctoral Science Foundation (No.20110491249), Key Scientific and
Technological Research Project of Department of Education of Henan Province (NO. 12B110006),
Youth Science Foundation of Henan University of Science and Technology(NO. 2012QN010), The
Natural Science Foundation to cultivating innovation ability of Henan University of Science and
Technology(NO. 2013ZCX020).
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