Journal of Applied Mathematics and Physics, 2014, 2, 108-114
Published Online April 2014 in SciRes. http://www.scirp.org/journal/jamp
http://dx.doi.org/10.4236/jamp.2014.25014
How to cite this paper: Zhou, L.N. and Jiang, W.H. (2014) Positive Solutions for Fractional Differential Equations with
Multi-Point Boundary Value Problems. Journal of Applied Mathematics and Physics, 2, 108-114.
http://dx.doi.org/10.4236/jamp.2014.25014
Positive Solutions for Fractional Differential
Equations with Multi-Point Boundary Value
Problems
Lina Zhou1, Weihua Jiang2
1College of Mathematics and Information Science, Hebei Normal University, Shijiazhuang, China
2College of Science, Hebei University of Science and Technology, Shijiazhuang , China
Received Novemb er 20 13
Abstract
In this paper, a fractional multi-point boundary value problem is considered. By using the fixed
point index theory and Krein-Rutman theorem, some results on existence are obtained.
Keywords
Caputo Fractional Derivative, Fractional Integral, Boundary Value Problem, Fixed Point Index
Theory
1. Introduction
Fractional differential equations have been of great interest recently. This is due to the intensive development of
the theory of fractional calculus itself as well as its applications. Apart from diverse areas of mathematics, frac-
tional differential equations arise in rheology, dynamical processes in self similar and porous structures, elec-
trical networks, visco-elasticity, chemical physics, and many other branches of science. For details, see
[1]-[7].
It should be noted that most of papers and books on fractional calculus are devoted to the solvability of linear
initial fractional differential equations on terms of special functions. Recently, there are some papers dealing
with the existence and multiplicity of solution to the nonlinear fractional differential equations boundary value
problems, see [8]-[14].
Zhao [11] investigated the existence and uniqueness of positive solutions for a local boundary value problem
of fractional differential equation.
0()(, ())0 01
(0)() 0,(1)() 0
Dutf tutt
u uuu
α
βξ γη
+
+= <<

′′
−= +=


where
α
is a real number with
12,0, 1,01 αβγξ η <≤≤≤ ≤≤≤
,
0
D
α
+
is the Caputo’s derivative.
Inspired by above work, we will consider the fractional boundary value problem

L. N. Zhou, W. H. Jiang
109
0
3
1
()(, ())0 01
(0) ()0,(1)()0
m
ii
i
Dutf tutt
uuu u
α
βξγ η
+
−
=
+= <<

′′
−=+ =


∑
(1)
where
α
is a real number with
12, 01, 01,1, 2,,3,
iim
α βγ <≤≤≤≤≤=−
12 3
01
m
ξη ηη
−
≤< << <≤
,
0
D
α
+
is the Caputo’s derivative.
Let
33
11
(1)(1 ).
mm
ii i
ii
βγ ηγβξ
−−
= =
∆= ++−
∑∑
Now we list some conditions for convenience.
(H1)
3
1
(1)(1),1(1) 0
m
ii
i
αβξγ η
−
=
−−>∆ +−>
∑
(H2)
: [0,1]f RR
++
×→
satisfied Carathéodory condition,that is
(, )fu⋅
is measurable for each fixed
uR
+
∈
and
(,)ft⋅
is continuous for a.e.
[0,1]t∈
. For any
0r>
, there existed
1
( )[0,1]tLΦ∈
such that
(, )()f tut≤Φ
, where
[0, ]ur∈
; a.e.
[0,1]t∈
.
(H3)
0,0,(,)lLftuL∀>∃ ><
where
[0, ]ul∈
; a.e.
[0,1]t∈
.
2. Preliminary
For the convenience of readers, we provide some background material in this section. {\bf Definition 2.1[7] The
Riemann-Liouville fractional of order
α
for function y is defined as
1
00
1
()( )()
()
t
Iy ttsys ds
αα
α
−
+= −
Γ∫
Definition 2.2 [7] The Caputo’s derivative for function y is defined as
()
01
0
1 ()
() ()
()
n
t
n
ys
Dytds
nts
α
α
α
++−
=Γ− −
∫
Lemma 2.1 [14] Let
0 α >
, then the fractional differential equation
0()0D ut
α
+=
has solutions
21
12 3
(),,1, 2,,,[]1
n
ni
utcct ctctcRinn α
−
=+ +++∈==+
Lemma 2.2 [14] Let
0 α >
, then for some
,1, 2,,,[]1
i
c Rinn α ∈= =+
Lemma 2.3 (Krein-Rutman) [15] Let
K
be a reproducing cone in a real Banach space
X
and let
:LX X→
be a compact linear operator with
()LKK⊆
.
()rL
is the spectral radius of
L
If
() 0rL>
, then
there exists
1
\ {0}K ϕ ∈
such that
11
()L rL ϕϕ =
.
Lemma 2.4 [16] Let
X
is a Banach space,
P
be a cone in
X
and
()PΩ
be a bounded open subset in
P
.
Suppose that
: ()APPΩ→
is a completely continuous operator. Then the following results hold:
(1) If there exists
0\ {0}uP∈
such that
0
,( ),0u AuuuP λλ ≠+∀∈∂Ω>
then the fixed point index
( ,( ),)0iAPPΩ=
.
(2) If
0 ()P∈Ω
and
,( ),1Auu uP λλ ≠∀∈ ∂Ω≥
, then the fixed point index
( ,( ),)1iAPPΩ=
.
Take
[0,1]XC=
with norm
[0,1]
()max()
t
xt xt
∈
=
,
1
[0,1]YL=
with norm
1
10
() () xtxt dt=
∫
.
{|()0,[0,1]}KuX utt=∈ ≥∈
Obviously
K
is a reproducing cone of
X
.
Lemma 2.5 If
[0,1],1 2,yC α ∈ <≤
then the unique solution of
0
3
1
()() 001
(0) ()0,(1)()0
m
ii
i
D utytt
uuu u
α
βξγ η
+
−
=
+ =<<

′′
−=+ =


∑
is
1
0
()(, ) (),utGtsys ds=
∫
where

L. N. Zhou, W. H. Jiang ,
110
3
11 1
3
21
1
3
12 1
3
1
[1( )]
1()( )
() ()
1(1 )
(1 )(),
(1)( )
[1( )]1
()(1 )
( )(1)
(1)(
()
(,)
m
ii
i
m
ii
i
m
ii
i
m
i
i
t
ts s
tt
sss st
tt
ss
t
Gts
αα
αα
αα
β γη
ξ
αα
βξ ββξ βγη ξ
αα
β γηβξ β
ξ
αα
βξ βγη
α
−
−− =
−
−−
=
−
−− =
−
=
+−
− −+−
Γ ∆Γ
−+ −+
+−+−≤≤
∆Γ −∆Γ
+− −+
−+ −
∆Γ∆Γ −
−+
+∆Γ
=
∑
∑
∑
∑1
12
311
1
3
21
1
1
12
),
11
()(1 )
( )(1)
(1)() ,
()
1(1 )
(1 )(),
(1)( )
11
()(1 )
( )(1)
(1
i
m
ii
i
m
ii
i
ss ts
t
ts s
tss st
tt
sss ts
t
ts s
α
αα
α
αα
αα
ξ
βξ β
αα
βξ βγηξ η
α
βξ ββξ βγηξ η
αα
βξ β
αα
−
−−
−−
=
−
−−
=
−−
− ≤≤
−+
− −+−
Γ∆Γ −
−+
+−≤≤ ≤
∆Γ
−+ −+
−+−≤≤ ≤
∆Γ −∆Γ
−+
− −+−
Γ∆Γ −
−
+
∑
∑
311
3
21
1
)() ,
()
1(1 )
(1 )(),
(1)( )
m
jji i
ji
m
jji i
ji
tss st
tt
sss ts
α
αα
βξ βγηη η
α
βξ ββξ βγηηη
αα
−−
−
=
−
−−
−
=




























+
−≤≤ ≤
∆Γ


−+ −+
−+−≤≤≤
∆Γ −∆Γ


∑
∑
Proof: The equation
0
()() 0D utyt
α
+
+=
has a unique solution
112
0
1
()( )()
()
t
uttsysdscc t
α
α
−
=− −++
Γ∫
where
12
,.cc R∈
By
3
1
(0)()0,(1)()0
m
ii
i
uuu u βξγ η
−
=
′′ −=+ =
∑
, we have
33
1
1 000
11
1[(1)() (1)(1) (1)()]
mm
iii i
ii
cIyI yIy
αα α
βγ ηξβξβξγη
−−
−
++ +
= =
= ++−+−
∆
∑∑
33
1
20 00
11
1[(1)()( )]
mm
ii i
ii
cI yIyIy
ααα
ββγηβγξ
−−
−
+++
= =
=+−
∆
∑∑
3
1
1 12 1
0 00
31
0
1
[1( )]
11
()() ()() ()(1)()
()()( 1)
(1) () ()
()
i
m
ii
tt
i
m
ii
i
tt
uttsy sdssy s dssy s ds
tsy s ds
α αα
ηα
β γηβξ β
ξ
αα α
βξ βγη
α
−
− −− =
−−
=
+− −+
=−−+−+ −
Γ∆Γ∆Γ −
−+
+−
∆Γ
∑
∫ ∫∫
∑∫
The proof is complete.
Lemma 2.6 If (H1) hold, then there exist a constant
M
such that
2
0( ,)(1),,[0,1]GtsMsts
α −
≤ ≤−∈
Proof: Obviously
( ,)0,Gts ≥

L. N. Zhou, W. H. Jiang
111
3
3
12 1 1
01 1
3
3
22 2
1
1
3
1
[1( )]1(1 )
max( ,)()(1)()
()( 1)()
[1 ]1(1 )
(1 )(1 )(1 )
()(1)()
(1
m
ii m
iii
ti
m
ii m
ii
i
m
i
i
ttt
Gtss ss
ss s
αα α
αα α
β γηβξ ββξ β
ξ γη
αα α
β γηβξββξβγ
αα α
βγ
−
−
−− − =
≤≤ =
−
−
−− −
=
=
−
=
+− −+ −+
≤−+− +−
∆Γ∆Γ −∆Γ
+−+ −+
≤−+− +−
∆Γ∆Γ −∆Γ
+
=
∑∑
∑∑
3
2 1
) (1)(1)(1)
(1 )
()
m
ii
i
s
α
ηαβξ βγβξ β
α
−
− =
+−−++ −+
−
∆Γ
∑∑
Let
33
11
(1 )(1)(1 )(1 )
()
mm
ii i
ii
M
βγ ηαβξβγβξβ
α
−−
= =
++−−++ −+
=∆Γ
∑∑
.The proof is completed.
Define an operator
:AK K→
and a linear operator
:TX X→
as follows:
1
0
( )( ,)(,())Au tG tsfsusds=
∫
1
0
()(, ) ()TutGts usds=
∫
Then the fixed point
\ {0}uK∈
of
A
is the positive solutions of (1)
3. Main Results
In order to obtain our main results, we firstly present and prove some lemmas.
Lemma 3.1 If (H1)-(H3) hold, then
:AK K→
and
:TX X→
are completely continuous.
Proof: According to the Lebesgue Dominated Convergence Theorem and Lemma 2.6, we have
:AK K→
is uniformly bounded and equicontinuous. It follows from Ascoli-Arzela theorem that
: AK K →
is com-
pletely continuous.
By the same method, we can get that
:TX X→
is completely continuous also.
Lemma3.2 If (H1)-(H2) hold, then
() 0rT>
(
r
is the spectral radius of
T
)
Proof: Take
() 1ut≡
1
0
3
1
1 12
1
0 00
31
0
1
3
1
()(, ) ()
[1( )]
11
()()(1 )
()()(1)
(1) ()
()
[1( )]
11
()()(1
i
m
ii
tt
i
m
ii
i
m
ii
i
Tu tG tsu sds
tt
ts dss dssds
ts ds
t
tt
α αα
ηα
αα
β γηβξ β
ξ
αα α
βξ βγη
α
βγη ξβξ β
αααα α
−
− −−
=
−−
=
−
=
=
+− −+
=− −+−+−
Γ∆Γ∆Γ −
−+
+−
∆Γ
+− −+
=−++
Γ∆Γ∆Γ −
∫
∑
∫ ∫∫
∑∫
∑
3
1
3
1
1 (1)
) 1()
11(1):
(1) () (1)
m
ii
i
m
ii
i
t
l
α
α
γη
βξ β
α αα
βξβξ γη
α αα
−
=
−
=
−+
+
− ∆Γ
−−
≥−+ +=
Γ +∆Γ∆Γ +
∑
∑
22
()(())(),( ())
nn
TutTTutTll Tutl= ≥≥≥
11
, ()lim0
nn
nn
n
TlrT Tl
→∞
≥ =≥>
The proof is completed.
By Lemma 2.3, we can get there exists
0\{0}K
ϕ ∈
such that
00
()T rT ϕϕ =

L. N. Zhou, W. H. Jiang ,
112
Define
00 [0,1]\[0,1]\
(, )(, )
liminfinf,limsup sup
utE u tE
f tuf tu
ff
uu
∞
→∈ →∞ ∈
= =
, where
[0,1]E⊂
, with
() 0mE=
(
()mE
is
the lebesgue measure of
E
and the same as follows).
Set
1
{ |},.
()
K uKurT
ρ
ρµ =∈ <=
Lemma 3.3 Suppose
0
f µ < ≤∞
, then there exists
0
0 ρ >
such that for
0
(0, ] ρρ ∈
,
if
,uAu uK
ρ
≠∀ ∈∂
,then
( ,,)0.iAKK
ρ
=
Proof: It follows from
0
f µ <
that there exists
0 ε >
and
0
0 ρ >
such that for a.e.
0
[0,1], 0tu ρ ∈ ≤≤
(, )()f tuu µε ≥+
(2)
For
0
0 ρρ <≤
, assume
, uAu uK
ρ
≠∀ ∈∂
By Lemma 2.4,we need only to prove that
0, ,0u AuuK ρ
λϕ λ ≠+∀ ∈∂>
where
0 00
\ {0},()KTrT ϕ ϕϕ ∈=
.
Otherwise, there exists
00
,0 uK
ρ
λ ∈∂>
such that
0000
u Au λϕ = +
(3)
Then
00000
,, uAu u λϕ ≥≥
by (2), we can get that
1
00 0
0( ,)(,())()
AuG tsfsusdsTu µε
= ≥+
∫
(4)
Considering
0 00
,u λϕ ≥
we get
00000
()Au T µε λϕλϕ ≥+ >
.
This together with (3) means that
0 00
2,u λϕ ≥
by (4) we get
0 00
2Au λϕ ≥
.So
0 00
3u λϕ ≥
.
Repeating this process, we get that
0 00
un λϕ ≥
, so we have
0 00
,.un n λϕ ≥→∞ →∞
This is a contradic-
tion.
It follows from Lemma 2.4 that
0
( ,,)0,(0,]iAK K
ρ
ρρ = ∈
. The proof is completed.
Lemma 3.4 Suppose
0f µ
∞
≤<
, then there exists
0
0r>
such that
( ,,)1,
r
iAKK =
for each
0
rr>
Proof: Let
0 ε >
satisfy
f µε
∞<−
, then there exists
10r>
such that
1
(,)(),for ,a.e.[0,1]f tuuurt µε ≤−> ∈
By (H3),there existed
1
( )[0,1]tLΦ∈
such that
(, )() f tut≤Φ
, where
1
[0, ]ur∈
; a.e.
[0,1]t∈
.
Thus, for all
uR
+
∈
a.e.
[0,1]t∈
(, )()()f tuut µε ≤ −+Φ
(5)
Since
1
()rT
µ =
,
1
()
IT
µε
−
−
−
exists. Let
11
0
0
(, )(),()
IC
CGtss dsrT
µε µε
−
= Φ=−
−−
∫
Take
0
rr>
, we will show
,Au u λ ≠
for each
,1
r
uK λ ∈∂ ≥
.
Otherwise, there exist
00
,1
r
uK λ ∈∂ ≥
, such that
0 00
,Au u λ =
This together with (5), implies
0 0000
()uuAuTu C λ µε ≤= ≤−+
Then
0
()()
IC
Tut
µε µε
−≤
−−
. So we get
0
()().
CI
Tu tK
µε µε
−− ∈
−−
It
follows from
11
0
()()
nn
n
ITT
µε
µε
∞
−+
=
−= −
−
∑
and
()TK K⊂
, we get
1
0
() ()
IC
ut T
µε µε
−
≤−
−−
. Therefore,
we have
00
u rr≤≤
. This is a contradiction.
By Lemma 2.4 (2), we get
( ,,)1,
r
iAKK =
for each
0
rr>
. The proof is completed.
Theorem 3.1 Suppose
0
f µ < ≤∞
and
0f µ
∞
≤<
then (1) has at least one positive solution.
Proof: It follows from
0f µ
∞
≤<
and Lemma 3.4 there exists
0r>
such that
(,, ) 1
r
iAKK =
.
By
0
f µ < <∞
and Lemma 3.3, we can get there exists
0r ρ <<
such that either there exists
uK∈∂
with
or (,,)0uAuiA KK
ρ
= =
. In the second case,
A
has a fixed point
uK∈
with
ur ρ ≤<
by the
properties of index. The proof is completed.

L. N. Zhou, W. H. Jiang
113
4. Example
Let’s consider the following boundary value problem
3
2
0
()(, ())001
11 13
(0)() 0,(1)() 0
5454
Dutf tutt
uu uu
+
+= <<


′′
−=+=

(6)
where
,[0,1] is a irrational number
(, )0,[0,1] is a rational number
u tt
f tut
+∈

=∈


.
Corresponding to the problem (1), we have that
31311
,,,,
24455
αξηβγ = = ===
. Let
{|[0,1] is a rational number}E tt= ∈
, then
() 0mE=
. Obviously, (H1)-(H3) are satisfied. By simple calcula-
tion, we get
0,0ff ∞
=∞=
. By Theorem 3.1, we get that (6) has at least one solution. This problem can be not
solved by the theorem in [11].
Acknowledgements
This work is supported by the Natural Science Foundation of China (11171088), the Natural Youth science
Foundation of China (11101118) and the Natural Science Foundation of Hebei Province (A2012205074).
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