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Volume 2011, Article ID 475126, 17 pagesdoi:10.1155/2011/475126 Research Article Existence of Positive Solutions to a Boundary Value Problem for a Delayed Nonlinear Fractional Differenti

Volume 2011, Article ID 475126, 17 pages

doi:10.1155/2011/475126

Research Article

Existence of Positive Solutions to a Boundary Value Problem for a Delayed Nonlinear Fractional

Differential System

1 School of Mathematics and Physics, School of Nuclear Science and Technology, University of South China, Hengyang 421001, China

2 Department of Mathematics, Wilfrid Laurier University, Waterloo, Ontario, Canada N2L 3C5

3 School of Nuclear Science and Technology, University of South China, Hengyang 421001, China

Correspondence should be addressed to Zigen Ouyang,zigenouyang@yahoo.com.cn

Received 14 November 2010; Accepted 24 February 2011

Academic Editor: Gary Lieberman

Copyrightq 2011 Zigen Ouyang et al This 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

Though boundary value problems for fractional differential equations have been extensively studied, most of the studies focus on scalar equations and the fractional order between 1 and

2 On the other hand, delay is natural in practical systems However, not much has been done for fractional differential equations with delays Therefore, in this paper, we consider a boundary value problem of a general delayed nonlinear fractional system With the help of some fixed point theorems and the properties of the Green function, we establish several sets of sufficient conditions

on the existence of positive solutions The obtained results extend and include some existing ones and are illustrated with some examples for their feasibility

1 Introduction

In the past decades, fractional differential equations have been intensively studied This is due to the rapid development of the theory of fractional differential equations itself and the applications of such construction in various sciences such as physics, mechanics, chemistry, and engineering1,2 For the basic theory of fractional differential equations, we refer the readers to3 7

Recently, many researchers have devoted their attention to studying the existence of

positive solutions of boundary value problems for differential equations with fractional order8 23 We mention that the fractional order α involved is generally in 1, 2 with the exception that α ∈ 2, 3 in 12,23 and α ∈ 3, 4 in 8,17 Though there have been extensive

study on systems of fractional differential equations, not much has been done for boundary value problems for systems of fractional differential equations 18–20

On the other hand, we know that delay arises naturally in practical systems due to the transmission of signal or the mechanical transmission Though theory of ordinary differential equations with delays is mature, not much has been done for fractional differential equations with delays24–31

As a result, in this paper, we consider the following nonlinear system of fractional order differential equations with delays,

D α i u it  fit, u1τi1 t, , uNτiN t 0, t ∈ 0, 1,

u j i 0 0, j 0, 1, , ni − 2, i 1, 2, , N,

u n i−1

i 1 ηi , i 1, 2, , N,

1.1

where D α i is the standard Riemann-Liouville fractional derivative of order αi ∈ ni − 1, ni for some integer n i > 1, η i ≥ 0 for i 1, , N, 0 ≤ τij t ≤ t for i, j 1, 2, , N, and fi is

a nonlinear function from0, 1 ×Ê

N

 toÊ  0, ∞ The purpose is to establish sufficient

conditions on the existence of positive solutions to1.1 by using some fixed point theorems and some properties of the Green function By a positive solution to1.1 we mean a mapping with positive components on0, 1 such that 1.1 is satisfied Obviously, 1.1 includes the usual system of fractional differential equations when τijt ≡ t for all i and j Therefore, the obtained results generalize and include some existing ones

The remaining part of this paper is organized as follows In Section2, we introduce some basics of fractional derivative and the fixed point theorems which will be used in Section3to establish the existence of positive solutions To conclude the paper, the feasibility

of some of the results is illustrated with concrete examples in Section4

2 Preliminaries

We first introduce some basic definitions of fractional derivative for the readers’ convenience

Definition 2.1see 3,32 The fractional integral of order α> 0 of a function f : 0, ∞ → Ê

is defined as

I α f t 1

Γα

t

0

f s

t − s1−αds 2.1

provided that the integral exists on0, ∞, where Γα 0∞e −t t α−1dt is the Gamma function.

Note that I αhas the semigroup property, that is,

I α I β I α β I β I α for α > 0, β > 0. 2.2

Definition 2.2see 3,32 The Riemann-Liouville derivative of order α> 0 of a function

f : 0, ∞ → Êis given by

D α f t 1

Γn − α

d n

dt n

t

0

f s

t − s α 1−n ds 2.3

provided that the right-hand side is pointwise defined on0, ∞, where n α  1.

It is well known that if n − 1 < α ≤ n then D α t α −k 0, k 1, 2, , n Furthermore, if

y t ∈ L10, T and α > 0 then D α I α y t yt for t ∈ 0, T.

The following results on fractional integral and fractional derivative will be needed in establishing our main results

Lemma 2.3 see 10 Let α > 0 Then solutions to the fractional equation D α h t 0 can be

written as

h t c1t α−1 c2t α−2 · · ·  cn t α −n , 2.4

where c i∈Ê, i 1, 2, , n α  1.

Lemma 2.4 see 10 Let α > 0 Then

I α D α h t ht  c1t α−1 c2t α−2 · · ·  cn t α −n 2.5

for some c i∈Ê, i 1, 2, , n α  1.

Now, we cite the fixed point theorems to be used in Section 3

Lemma 2.5 the Banach contraction mapping theorem 33 Let M be a complete metric space

and let T : M → M be a contraction mapping Then T has a unique fixed point.

Lemma 2.6 see 16,34 Let C be a closed and convex subset of a Banach space X Assume that U

is a relatively open subset of C with 0 ∈ U and T : U → C is completely continuous Then at least

one of the following two properties holds:

i T has a fixed point in U;

ii there exists u ∈ ∂U and λ ∈ 0, 1 with u λTu.

Lemma 2.7 the Krasnosel’skii fixed point theorem 33,35 Let P be a cone in a Banach space

X Assume thatΩ1 and Ω2 are open subsets of X with 0 ∈ Ω1 and Ω1 ⊆ Ω2 Suppose that T :

P

Ω2\ Ω1 → P is a completely continuous operator such that either

i Tu ≤ u for u ∈ P∂Ω1and Tu ≥ u for u ∈ P∂Ω2

or

ii Tu ≥ u for u ∈ P∂Ω1and Tu ≤ u for u ∈ P∂Ω2.

Then T has a fixed point inΩ2\ Ω1.

3 Existence of Positive Solutions

Throughout this paper, we let E C0, 1,Ê

N  Then E, · E is a Banach space, where

u E max

1≤i≤Nmax

0≤t≤1|uit| for u u1, , u N T ∈ E. 3.1

In this section, we always assume that f f1, , f N T ∈ C0, 1 ×Ê

N

 ,Ê

N

 

Lemma 3.1 System 1.1 is equivalent to the following system of integral equations:

u it

1

0

G it, sfis, u1τi1s, , uNτiN sds

 η i t α i−1

αi − 1 · · · αi − ni 1, i 1, 2, , N,

3.2

where

G it, s

⎧

⎪

⎪

t α i−11 − s α i −n i − t − s α i−1

Γαi , 0 ≤ s ≤ t ≤ 1,

t α i−11 − s α i −n i

Γαi , 0≤ t ≤ s ≤ 1.

3.3

Proof It is easy to see that if u1, u2, , u N T satisfies3.2 then it also satisfies 3.2 So, assume thatu1, u2, , u N Tis a solution to1.1 Integrating both sides of the first equation

of1.1 of order αi with respect to t gives us

u it − 1

Γαi

t

0

t − s α i−1f is, u1τi1s, , uNτiN sds

 c 1i t α i−1 c 2i t α i−2 · · ·  cn,i t α i −n i

3.4

for 0≤ t ≤ 1, i 1, 2, , N It follows that

ui t − α i− 1

Γαi

t

0

t − s α i−2f is, u1τi1s, , uNτiNsds

 αi − 1c 1i t α i−2 αi − 2c 2i t α i−3 · · ·  αi − ni  1cn −1,i t α i −n i

3.5

for 0≤ t ≤ 1, i 1, 2, , N This, combined with the boundary conditions in 1.1, yields

c n −1,i 0, i 1, 2, , N. 3.6

Similarly, one can obtain

c n i −2,i cn i −3,i · · · c 2,i 0, 3.7

u n i−1

i t − αi − 1 · · · αi − ni 1

Γαi

t

0

t − s α i −n i f is, u1τi1s, , uNτiNsds

 αi − 1 · · · αi − ni  1c 1i t α i −n i ,

3.8

i 1, 2, , N Then it follows from 3.8 and the boundary condition u n i−1

i 1 ηithat

c 1,i η i

αi − 1 · · · αi − ni 1

1

Γαi

1

0

1 − s α i −n i f is, u1τi1s, , uNτiN sds 3.9 Therefore, for i 1, 2, , N,

u it − 1

Γαi

t

0

t − s α i−1f is, u1τi1s, , uNτiNsds  η i t α i−1

αi − 1 · · · αi − ni 1

 t α i−1

Γαi

1

0

1 − s α i −n i f is, u1τi1s, , uNτiNsds

1

Γαi

t

0

t α i−11 − s α i −n i − t − s α i−1 f is, u1τi1s, , uNτiNsds

 1

Γαi

1

t

t α i−11 − s α i −n i f is, u1τi1s, , uNτiN sds  η i t α i−1

αi − 1 · · · αi − ni 1

1

0

G it, sfis, u1τi1s, , uNτiNsds  η i t α i−1

αi − 1 · · · αi − ni 1.

3.10 This completes the proof

The following two results give some properties of the Green functions Git, s.

Lemma 3.2 For i 1, 2, , N, G it, s is continuous on 0, 1 × 0, 1 and Git, s > 0 for t, s ∈

0, 1 × 0, 1.

Proof Obviously, G it, s is continuous on 0, 1×0, 1 It remains to show that Git, s > 0 for

t, s ∈ 0, 1 × 0, 1 It is easy to see that Git, s > 0 for 0 < t ≤ s < 1 We only need to show that G it, s > 0 for 0 < s ≤ t < 1 For 0 < s ≤ t ≤ 1, let

g it, s t α i−11 − s α i −n i − t − s α i−1, 3.11

h it, s 1 − s α i −n i− 1−s

t

α i−1

g it, s t α i−1h it, s, 0 < s ≤ t < 1. 3.13

Note that h is, s > 0 and ∂hi /∂t t, s −αi − 11 − s/t α i−2st−2 < 0 for 0 < s ≤ t < 1 It follows that hit, s > 0 and hence git, s > 0 for 0 < s ≤ t < 1.

Therefore, G it, s > 0 for 0 < s ≤ t < 1 and the proof is complete.

Lemma 3.3 (i) If n i 2, then Git, s ≤ Gis, s for t, s ∈ 0, 1 × 0, 1.

(ii) If n i > 2, then G it, s < Gi1, s for t, s ∈ 0, 1 × 0, 1.

Proof.

i Obviously, Git, s ≤ Gis, s for 0 < t ≤ s < 1 Now, for 0 < s ≤ t < 1, we have

∂g it, s

∂t αi − 1t α i−2

1 − s α i−2− 1−s

t

α i−2

≤ 0, 3.14

where giis the function defined by3.11 It follows that Git, s ≤ Gis, s for 0 < s ≤ t < 1 In

summary, we have provedi

ii Again, one can easily see that Git, s < Gi1, s for 0 < t ≤ s < 1 When 0 < s ≤ t ≤ 1,

we have in this case that

∂g it, s

∂t αi − 1t α i−2

1 − s α i −n i − 1−s

t

α i−2

≥ αi − 1t α i−2 1 − s α i −n i − 1 − s α i−2

> 0,

3.15

which implies that G it, s ≤ Gi1, s for 0 < s ≤ t < 1 To summarize, we have proved ii and

this completes the proof

Now, we are ready to present the main results

Theorem 3.4 Suppose that there exist functions λ ijt ∈ C0, 1,Ê , i, j 1, 2, , N, such that

f it, u1, , u N − fit, v1, , v N ≤N

j 1

λ ijtu j − vj 3.16

for t ∈ 0, 1, i 1,2, , N If

max

1≤i≤N, ni >2

1

0

G i1, s

⎛

⎝N

j 1

λ ijs

⎞

⎠ds < 1, 3.17

1≤i≤N, ni 2

1

0

G is, s

⎛

⎝N

j 1

λ ijs

⎞

⎠ds < 1, 3.18

then1.1 has a unique positive solution.

Proof Let

Ω {u ∈ E | uit ≥ 0 for t ∈ 0, 1, i 1, 2, , N}. 3.19

It is easy to see thatΩ is a complete metric space Define an operator T on Ω by

Tu t

1

0

G t, sgsds  diag



, η i t

α i−1

αi − 1 · · · αi − ni 1, ,



, 3.20

where Gt, s diagG1t, s, G2t, s, , GNt, s and

g t

⎛

⎜

⎜

⎜

⎝

f1t, u1τ11t, u2τ12t, , uNτ 1N t

f2t, u1τ21t, u2τ22t, , uNτ 2N t

f Nt, u1τN1 t, u2τ N2 t, , uNτNN t

⎞

⎟

⎟

⎟

⎠

. 3.21

Because of the continuity of G and f, it follows easily from Lemma3.2that T mapsΩ into

itself To finish the proof, we only need to show that T is a contraction Indeed, for u, v∈ Ω,

by3.16 we have

|Tut i − Tvt i|







1

0

G it, sf is, u1τi1s, , uNτiNs − fis, v1τi1s, , vNτiNsds







≤

1

0

G it, sf is, u1τi1s, , uNτiNs − fis, v1τi1 s, , vNτiN sds

≤

1

0

G it, s

⎛

⎝N

j 1

λ ijsu j

τ ij s− vjτ ijs⎞⎠ds.

3.22

This, combined with Lemma3.3and3.17 and 3.18, immediately implies that T : Ω → Ω

is a contraction Therefore, the proof is complete with the help of Lemmas3.1and2.5

The following result can be proved in the same spirit as that for Theorem3.4

Theorem 3.5 For i 1, 2, , N, suppose that there exist nonnegative function λ it and

nonnegative constants q i1, qi2, , qi N such thatN

j 1q ij 1 and

f it, u1, , u N − fit, v1, , v N ≤ λitN

j 1

u j − vjq ij 3.23

for t ∈ 0, 1, u1, u2, , u N T , v1, v2, , v NT ∈Ê

N

 If

max

1≤i≤N, ni >2

1

0

G i1, sλisds < 1, max

1≤i≤N, ni 2

1

0

G is, sλisds < 1, 3.24

then1.1 has a unique positive solution.

Theorem 3.6 For i 1, 2, , N, suppose that there exist nonnegative real-valued functions

m i , n i1 , , n iN ∈ L0, 1 such that

f it, u1, , u N ≤ mit N

j 1

n ijtuj 3.25

for almost every t ∈ 0, 1 and all u1, u2, , u N T ∈Ê

N

 If

max

1≤i≤N, ni >2

⎧

⎨

⎩

1

0

G i1, s

⎛

⎝N

j 1

n ijs

⎞

⎠ds

⎫

⎬

⎭< 1,

max

1≤i≤N, ni 2

⎧

⎨

⎩

1

0

G is, s

⎛

⎝N

j 1

n ijs

⎞

⎠ds

⎫

⎬

⎭< 1,

3.26

then1.1 has at least one positive solution.

Proof Let Ω and T : Ω → Ω be defined by 3.19 and 3.20, respectively We first show that

T is completely continuous through the following three steps.

Step 1 Show that T : Ω → Ω is continuous Let {u k t} be a sequence in Ω such that u k t →

u t ∈ Ω Then Ω0 0, 1 × {ut | u k t, ut ∈ Ω, t ∈ 0, 1, k ≥ 1} is bounded in 0, 1 ×Ê

N



Since f is continuous, it is uniformly continuous on any compact set In particular, for any

ε > 0, there exists a positive integer K0such that



fi t, u k1τi1t, , u k

N τiNt − fit, u1τi1t, , uNτiNt

max1≤i≤Nmaxt ∈0,11

0G it, sds

3.27

for t ∈ 0, 1 and k ≥ K0, i 1, 2, , N Then, for k ≥ K0, we have



 Tu k t

i − Tuti 





1

0

G it, sfi s, u k1τi1s, , u k

N τiN s

−fis, u1τi1s, , uNτiNs



ds

≤

1

0

G it, sfi

s, u k1τi1s, , u k

N τiN s

−fi s, u1τi1s, , uNτiNsds

max1≤i≤Nmaxt∈0,11

0G it, sds

1

0

G it, sds ≤ ε

3.28

for k ≥ K0and t ∈ 0, 1, i 1, 2, , N Therefore,

Tu k t − Tut < ε for k ≥ K

which implies that T is continuous.

Step 2 Show that T maps bounded sets of Ω into bounded sets Let A be a bounded subset of

Ω Then 0, 1 × {ut | t ∈ 0, 1, u ∈ A} ⊆ 0, 1 ×Ê

N

 is bounded Since f is continuous, there exists an M > 0 such that

f it, u1τi1t, , uNτiNt ≤ M for u ∈ A, t ∈ 0, 1, 1 ≤ i ≤ N. 3.30

It follows that, for u ∈ A, t ∈ 0, 1 and 1 ≤ i ≤ N,

Tut i

1

0

G it, sfis, u1τi1s, , uNτiNsds  η i

αi − 1 · · · αi − ni 1

≤ M

1

0

G it, sds  αi η i

− 1 · · · αi − ni 1

≤ max

1≤i≤N

!

M max

t ∈0,1

1

0

G it, sds  η i

αi − 1 · · · αi − ni 1

"

.

3.31

Immediately, we can easily see that TA is a bounded subset ofΩ

Step 3 Show that T maps bounded sets of Ω into equicontinuous sets Let B be a bounded

subset ofΩ Similarly as in Step2, there exists L > 0 such that

f it, u1τi1t, , uNτiNt ≤ L for u ∈ B, t ∈ 0, 1, 1 ≤ i ≤ N. 3.32

Then, for any u ∈ B and t1, t2∈ 0, 1 and 1 ≤ i ≤ N,

|Tut2i − Tut1i|









η i

t α i−1

2 − t α i−1 1

αi − 1 · · · αi − ni 1



1

0

Git2, s  − Git1, s fis, u1τi1s, , uNτiNsds









≤ η i



t α i−1

2 − t α i−1

1 

αi − 1 · · · αi − ni 1

1

0

|Git2, s  − Git1, s |Lds

≤ η i



t α i−1

2 − t α i−1

1 

αi − 1 · · · αi − ni 1 max0≤s≤1|Git2, s  − Git1, s L.

3.33

Now the equicontituity of T on B follows easily from the fact that G iis continuous and hence uniformly continuous on0, 1 × 0, 1.

Now we have shown that T is completely continuous To apply Lemma2.6, let

μ max1≤i≤N,ni >2

#1

0G i1, smisds  ηi / αi − 1 · · · αi − ni 1$

1− max1≤i≤N, ni >2

#1

0G i1, s N

j 1n ij s ds$ ,

ν max1≤i≤N,ni 2

#1

0G is, smisds  ηi / αi − 1 · · · αi − ni 1$

1− max1≤i≤N, ni 2#1

0G is, s N

j 1n ijs ds$ .

3.34

Fix r > max{μ, ν} and define

U {u ∈ Ωu E < r }. 3.35

g it, s t α i−1h it,...

0G it, sds

3.27