doi:10.1155/2010/186928
Research Article
Positive Solution to Nonzero Boundary
Values Problem for a Coupled System of Nonlinear Fractional Differential Equations
Jinhua Wang, Hongjun Xiang, and Zhigang Liu
Department of Mathematics, Xiangnan University, Chenzhou 423000, China
Correspondence should be addressed to Hongjun Xiang,[email protected] Received 13 April 2009; Accepted 9 June 2009
Academic Editor: Shaher Momani
Copyrightq2010 Jinhua Wang 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.
We consider the existence and uniqueness of positive solution to nonzero boundary values problem for a coupled system of fractional differential equations. The differential operator is taken in the standard Riemann-Liouville sense. By using Banach fixed point theorem and nonlinear differentiation of Leray-Schauder type, the existence and uniqueness of positive solution are obtained. Two examples are given to demonstrate the feasibility of the obtained results.
1. Introduction
Fractional differential equation can describe many phenomena in various fields of science and engineering such as control, porous media, electrochemistry, viscoelasticity, and electromagnetic. There are many papers dealing with the existence and uniqueness of solution for nonlinear fractional differential equation; see, for example, 1–5. In 1, the authors investigated a singular coupled system with initial value problems of fractional order. In2, Su discussed a boundary value problem of coupled system with zero boundary values. By means of Schauder fixed point theorem, the existence of the solution is obtained.
The nonzero boundary values problem of nonlinear fractional differential equations is more difficult and complicated. No contributions exist, as far as we know, concerning the existence of positive solution for coupled system of nonlinear fractional differential equations with nonzero boundary values.
In this paper, we consider the existence and uniqueness of positive solution to nonzero boundary values problem for a coupled system of nonlinear fractional differential equations:
Dαut ft, vt 0, 0< t <1, Dβvt gt, ut 0, 0< t <1,
u0 0, u1 auξ,
v0 0, v1 bvξ,
1.1
where 1 < α < 2,1 < β < 2,0 a, b 1,0 < ξ < 1,f, g : 0,1×0,∞ → 0,∞are given functions, andDis the standard Riemann-Liouville differentiation. By using Banach fixed point theorem and nonlinear differentiation of Leray-Schauder type, some sufficient conditions for the existence and uniqueness of positive solution to the above coupled boundary values problem are obtained.
The rest of the paper is organized as follows. InSection 2, we introduce some basic definitions and preliminaries used in later. In Section 3, the existence and uniqueness of positive solution for the coupled boundary values problem 1.1 will be discussed, and examples are given to demonstrate the feasibility of the obtained results.
2. Basic Definitions and Preliminaries
In this section, we introduce some basic definitions and lemmas which are used throughout this paper.
Definition 2.1see6,7. The fractional integral of orderαα >0of a functiony:0,∞ → Ris given by
Iαyt 1 Γα
t
0
t−sα−1ysds, 2.1
provided that the right side is pointwise defined on0,∞.
Definition 2.2see6,7. The fractional derivative of orderα > 0 of a continuous function y:0,∞ → Ris given by
Dαyt 1
Γn−α d
dt nt
0
t−sn−α−1ysds, 2.2
wheren α 1 provided that the right side is pointwise defined on0,∞.
Remark 2.3see3. The following properties are useful for our discussion:
1IαDαut ut−N
k 1Cktα−k, Dαut∈C0,1
L0,1,Ck∈R,N α 1, 2DαIαut ut,
3Dαtγ Γγ1/Γγ1−αtγ−α, α >0, γ >−1, γ > α−1, t >0.
Lemma 2.4the nonlinear alternative of Leray and Schauder type8. LetEbe a Banach space withC⊆Eclosed and convex. LetUbe a relatively open subset ofCwith 0∈Uand letT :U → C be a continuous and compact mapping. Then either
athe mappingT has a fixed point inU, or bthere existu∈∂Uandλ∈0,1withu λTu.
Consider
Dαut yt 0, 0< t <1,
u0 0, u1 auξ, 2.3
then one has the following lemma.
Lemma 2.5. Lety∈C0,1and 1< α <2, thenutis a solution of BVP2.3if and only ifutis a solution of the integral equation:
ut 1
0
G1t, sysds, 2.4
where
G1t, s
⎧⎪
⎪⎪
⎪⎪
⎪⎪
⎪⎪
⎪⎪
⎪⎪
⎪⎪
⎪⎪
⎨
⎪⎪
⎪⎪
⎪⎪
⎪⎪
⎪⎪
⎪⎪
⎪⎪
⎪⎪
⎪⎩
t1−sα−1−atα−1ξ−sα−1−t−sα−1
1−aξα−1 1−aξα−1
Γα , 0st1, sξ, t1−sα−1−t−sα−1
1−aξα−1 1−aξα−1
Γα , 0< ξst1,
t1−s α−1−atα−1ξ−sα−1 1−aξα−1
Γα , 0tsξ1, t1−sα−1
1−aξα−1
Γα, 0ts1, ξs.
2.5
Proof. Assume thatutis a solution of BVP2.3, then byRemark 2.3, we have ut −Iαyt C1tα−1C2tα−2
− t
0
t−sα−1
Γα ysdsC1tα−1C2tα−2.
2.6
By2.3, we have
C2 0, C1 1
0
1−sα−1 Γα
1−aξα−1ysds−a ξ
0
ξ−sα−1 Γα
1−aξα−1ysds. 2.7
Therefore, we obtain
ut − t
0
t−sα−1
Γα ysds 1
0
tα−11−sα−1 Γα
1−aξα−1ysds−a ξ
0
tα−1ξ−sα−1 Γα
1−aξα−1ysds 1
0
G1t, sysds.
2.8
Conversely, ifutis a solution of integral equation2.4, using the relationDαtα−m 0, m 1,2, . . . , N,whereNis the smallest integer greater than or equal toα3, Remark 2.1, we have
Dαut −Dα t
0
t−sα−1 Γα ysds
Dαtα−1 1
0
1−sα−1 Γα
1−aξα−1ysds−a ξ
0
ξ−sα−1 Γα
1−aξα−1ysds
−DαIαyt −yt.
2.9
A simple computation showedu0 0, u1 auξ. The proof is complete.
Let
G2t, s
⎧⎪
⎪⎪
⎪⎪
⎪⎪
⎪⎪
⎪⎪
⎪⎪
⎪⎪
⎪⎪
⎪⎨
⎪⎪
⎪⎪
⎪⎪
⎪⎪
⎪⎪
⎪⎪
⎪⎪
⎪⎪
⎪⎪
⎩
t1−sβ−1−btβ−1ξ−sβ−1−t−sβ−1
1−bξβ−1 1−bξβ−1
Γ
β , 0st1, sξ,
t1−sβ−1−t−sβ−1
1−bξβ−1 1−bξβ−1
Γ
β , 0< ξst1,
t1−s β−1−btβ−1ξ−sβ−1 1−bξβ−1
Γ
β , 0tsξ1,
t1−sβ−1 1−bξβ−1
Γ
β, 0ts1, ξs,
2.10
we callGt, s G1t, s, G2t, sGreen’s function of the boundary value problem1.1.
Lemma 2.6. Let 0a, b1, then the functionGt, sis continuous and satisfies 1Gt, s>0, fort, s∈0,1,
2Gt, sGs, s, fort, s∈0,1.
Proof. It is easy to prove thatGt, sis continuous on0,1×0,1, here we omit it. Now we proveG1t, s>0. Let
g1t, s t1−sα−1−atα−1ξ−sα−1−t−sα−1
1−aξα−1 1−aξα−1
Γα , 0< st1, sξ,
g2t, s t1−sα−1−t−sα−1
1−aξα−1 1−aξα−1
Γα , 0< ξst1,
g3t, s t1−s α−1−atα−1ξ−sα−1 1−aξα−1
Γα , 0< tsξ1, g4t, s t1−sα−1
1−aξα−1
Γα, 0< ts1, ξs.
2.11
We only need to proveg1t, s>0, 0< st1, sξ.Since t1−sα−1−atα−1ξ−sα−1−t−sα−1
1−aξα−1 tα−11−sα−1−aξ−sα−1−
1−s t
α−1
1−aξα−1 ,
2.12
setgt 1−sα−1−aξ−sα−1−1−s/tα−11−aξα−1, we have
gt −α−1 1− s
t α−2s
t2
1−aξα−1
0, for 0< s < t1, sξ. 2.13
Thengtis decreasing on0,1. Meanwhile,
g1 1−sα−1−aξ−sα−1−1−sα−1
1−aξα−1 aξα−1
1−sα−1−
1−s ξ
α−1
>0, 0< s < t1, sξ.
2.14
Therefore,g1t, s> 0,for 0< s < t1, s ξ.Clearlyg1t, s >0, t s,sog1t, s >
0, s, t ∈ 0,1.It is easy to show thatg2t, s > 0, g3t, s > 0,g4t, s > 0.Hence,G1t, s >
0, s, t∈0,1.
Similarly,G2t, s>0, s, t∈0,1.The proof of1is completed.
Let
g2t t1−sα−1−t−sα−1
1−aξα−1 1−aξα−1
Γα , 0< ξst1, 2.15
then,
g2t α−1tα−21−sα−1−1−s/tα−2
1−aξα−1 1−aξα−1
Γα , 0< ξs < t1,
1−sα−1−1−s/tα−2
1−aξα−1 1−sα−1−1−sα−2
1−aξα−1 1−sα−2
aξα−1−s
0, 0< ξs < t1,
2.16
therefore,
g2t0, 0< ξs < t1. 2.17
So,g2t, s is decreasing with respect tot. Similarly, g1t, sis decreasing with respect to t.
Alsog3t, sandg4t, sare increasing with respect tot. We obtain thatG1t, sis decreasing with respect totforstand increasing with respect totforts.
With the use of the monotonicity ofG1t, s, we have
0t1maxG1t, s G1s, s
⎧⎪
⎪⎪
⎪⎪
⎨
⎪⎪
⎪⎪
⎪⎩
s1−sα−1−asξ−sα−1 Γα
1−aξα−1 , s∈0, ξ, s1−sα−1
Γα
1−aξα−1, s∈ξ,1.
2.18
Similarly,
0t1maxG2t, s G2s, s
⎧⎪
⎪⎪
⎪⎪
⎨
⎪⎪
⎪⎪
⎪⎩
s1−sβ−1−bsξ−sβ−1 Γ
β
1−bξβ−1 , s∈0, ξ, s1−sβ−1
Γ β
1−bξβ−1, s∈ξ,1.
2.19
The proof of2is completed.
3. Main Result
In this section, we will discuss the existence and uniqueness of positive solution for boundary value problem1.1.
We define the spaceX {ut|ut∈C0,1}endowed withuX max0t1|ut|, Y {vt|vt∈C0,1}endowed withuY max0t1|vt|.
Foru, v∈X×Y, letu, vX×Y max{uX,vY}.
DefineP {u, v∈X×Y |ut0, vt0}, then the coneP ⊂X×Y.
FromLemma 2.5inSection 2, we can obtain the following lemma.
Lemma 3.1. Suppose thatft, vandgt, uare continuous, thenu, v ∈ X×Y is a solution of BVP1.1if and only ifu, v∈X×Y is a solution of the integral equations
ut 1
0
G1t, sfs, vsds,
vt 1
0
G2t, sgs, usds.
3.1
LetT :X×Y → X×Y be the operator defined as Tu, vt
1
0
G1t, sfs, vsds, 1
0
G2t, sgs, usds
:T1vt, T2ut,
3.2
then byLemma 3.1, the fixed point of operatorTcoincides with the solution of system1.1.
Lemma 3.2. Letft, vandgt, ube continuous on0,1×0,∞ → 0,∞, thenT : P → P defined by3.2is completely continuous.
Proof. Letu, v∈P, in view of nonnegativeness and continuity of functionsGt, s,f, andg, we conclude thatT:P → Pis continuous.
LetΩ∈Pbe bounded, that is, there exists a positive constanth >0 such thatu, v hfor allu, v∈Ω.
Let
M maxft, vt1 : 0t1,0vh , N maxgt, ut1 : 0t1,0uh
, 3.3
then we have
|T1vt|
1
0
G1t, sfs, vsds M
1
0
G1s, sds,
|T2ut|
1
0
G2t, sgs, usds N
1
0
G2s, sds.
3.4
Hence,Tu, vmax{M1
0G1s, sds, N1
0G2s, sds}.TΩis uniformly bounded.
SinceG1t, sis continuous on0,1×0,1, it is uniformly continuous on0,1×0,1.
Thus, for fixed s ∈ 0,1and for any ε > 0, there exists a constant δ > 0, such that any t1, t2∈0,1and|t1−t2|< δ,
|G1t1, s−G1t2, s|< ε/M. 3.5
Then
|T1vt2−T1vt1|M 1
0
|G1t2, s−G1t1, s|ds < ε. 3.6
Similarly,
|T2ut2−T2ut1|N 1
0
|G2t2, s−G2t1, s|ds < ε. 3.7
For the Euclidean distancedonR2, we have that ift1, t2∈0,1are such that|t2−t1|< δ, then dTu, vt2, Tu, vt1
T1vt2−T1vt12 T2ut2−T2ut12<√
2ε. 3.8
That is to say,TPis equicontinuous. By the means of the Arzela-Ascoli theorem, we have T :P → P is completely continuous. The proof is completed.
Theorem 3.3. Assume thatft, vandgt, uare continuous on0,1×0,∞ → 0,∞, and there exist two positive functionsmt, ntthat satisfy
H1|ft, v2−ft, v1|mt|v2−v1|, fort∈0,1, v1, v2∈0,∞, H2|gt, u2−gt, u1|nt|u2−u1|, fort∈0,1, u1, u2∈0,∞.
Then system1.1has a unique positive solution if
ρ 1
0
G1s, smsds <1, θ 1
0
G2s, snsds <1. 3.9
Proof. For all u, v ∈ P, by the nonegativeness of Gt, s and ft, v, gt, u, we have Tu, vt0. Hence,TP⊆P.
T1v2−T1v1 max
t∈0,1|T1v2−T1v1|
t∈0,1max
1
0
G1t, s
fs, v2s−fs, v1s ds
1
0
G1s, smsdsv2−v1 ρv2−v1.
3.10
Similarly,
T2u2−T2u1θu2−u1. 3.11
We have,
Tu2, v2−Tu1, v1max ρ, θ
u2, v2−u1, v1. 3.12
FromLemma 3.2,T is completely continuous, by Banach fixed point theorem, the operator T has a unique fixed point in P, which is the unique positive solution of system1.1. This completes the proof.
Theorem 3.4. Assume thatft, vand gt, uare continuous on 0,1×0,∞ → 0,∞and satisfy
H3|ft, vt|a1t a2t|vt|, H4|gt, ut|b1t b2t|ut|, H5A1 1
0G1s, sa2sds <1, 0< B1 1
0G1s, sa1sds <∞, H6A2
1
0G2s, sb2sds <1, 0< B2
1
0G2s, sb1sds <∞.
Then the system1.1has at least one positive solutionu, vin
C
u, v∈P | u, v<min B1
1−A1
, B2 1−A2
. 3.13
Proof. LetC {u, v∈X×Y :u, v< r}withr minB1/1−A1, B2/1−A2, define the operatorT :C → Pas3.2.
Letu, v∈C, that is,u, v< r. Then
T1v max
t∈0,1
1
0
G1t, sfs, vsds
1
0
G1s, sa1s a2s|vs|ds
≤ 1
0
G1s, sa1sds 1
0
G1s, sa2sdsv B1A1vr.
3.14
Similarly,T2u r, soTu, v r,Tu, v ⊆ C. FromLemma 3.2 T : C → Cis completely continuous.
Consider the eigenvalue problem
u, v λTu, v, λ∈0,1. 3.15
Under the assumption thatu, vis a solution of3.15for aλ∈0,1, one obtains
u λT1v
λmax
t∈0,1
1
0
G1t, sfs, vsds
<
1
0
G1s, sa1s a2s|vs|ds 1
0
G1s, sa1sds 1
0
G1s, sa2sdsv B1A1vr.
3.16
Similarly,v λT2u < r, sou, v < r, which shows thatu, v/∈∂C. ByLemma 2.4,T has a fixed point inC. We complete the proof ofTheorem 3.4.
Example 3.5. Consider the problem
D7/4ut ft, vt 0, 0< t <1, D3/2vt gt, ut 0, 0< t <1,
u0 0, u1 1
2u 1
2
,
v0 0, v1 3
4v 1
2
,
3.17
where
ft, vt tvt
1t1vt, gt, ut arctan t
1t|sinut|. 3.18
Setv1t, v2t, u1t, u2t∈0,∞andt∈0,1, then we have ft, v2t−ft, v1t t
1t|v2t−v1t|, gt, u2t−gt, u1tarctan t
1t|u2t−u1t|.
3.19
Therefore,
ρ 1
0
G1s, smsds 1
0
G1s, sds 1
Γ7/4
1−1/27/4 1/2
0
s1−s3/4ds−
1/2
0
1 2
s 1
2−s 3/4
ds 1
1/2
s1−s3/4ds
2
1 1/27/4
5 ·Γ3/4
Γ1/2< 4 5 <1,
θ 1
0
G2s, snsdsπ 4
1
0
G2s, sds π
4
1 Γ3/2
1−3/41/21/2 1/2
0
s1−s1/2ds−
1/2
0
3 4
s 1
2−s1/2
ds 1
1/2
s1−s1/2ds
π 4
1−3/41/22!
1−3/41/21/2! ·Γ3/2 Γ3
≈0.6018<1.
3.20
With the use ofTheorem 3.3, BVP3.17has a unique positive solution.
Example 3.6. Consider the problem
D7/4ut ft, vt 0, 0< t <1, D3/2vt gt, ut 0, 0< t <1,
u0 0, u1 1
2u 1
2
,
v0 0, v1 3
4v 1
2
,
3.21
where
ft, vt t2 t
1tln1vt, gt, ut 10 t2
20ut. 3.22
We have
ft, vtt2 t
1t· |vt|, gt, ut
10 t2 20
|ut|. 3.23
Hence,
A1 1
0
G1s, sa2sds 1
0
G1s, sds 2
1 1/27/4
5 ·Γ3/4
Γ1/2 <1, B1
1
0
G1s, sa1sds 1
0
G1s, s·s2ds <∞,
A2 1
0
G2s, sb2sds 1
0
G2s, sds≈0.7666<1,
B2 1
0
G2s, sb1sds 1
0
G2s, s
10 s2 20
ds <∞.
3.24
ByTheorem 3.4, BVP3.21has at least one positive solution in
C
u, v∈P | u, v<min B1
1−A1, B2
1−A2
. 3.25
Acknowledgments
This work was jointly supported by the Natural Science Foundation of Hunan Provincial Education Department under Grants 07A066 and 07C700, the Construct Program of the Key Discipline in Hunan Province, Aid Program for Science and Technology Innovative Research Team in Higher Educational Institutions of Hunan Province, and the Foundation of Xiangnan University.
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