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. In [2], 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:
Dαu(t)+f(t,v(t))=0,0<t<1,Dβv(t)+g(t,u(t))=0,0<t<1,u(0)=0,u(1)=au(ξ),v(0)=0,v(1)=bv(ξ),
where 1<α<2,1<β<2,0⩽a,b⩽1,0<ξ<1, f,g:[0,1]×[0,+∞)→[0,+∞) are given functions, and D is 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. In Section 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 2.1 (see [<xref ref-type="bibr" rid="B6">6</xref>, <xref ref-type="bibr" rid="B7">7</xref>]).

The fractional integral of order α(α>0) of a function y:(0,∞)→R is given by
Iαy(t)=1Γ(α)∫0t(t-s)α-1y(s)ds,
provided that the right side is pointwise defined on (0,∞).

Definition 2.2 (see [<xref ref-type="bibr" rid="B6">6</xref>, <xref ref-type="bibr" rid="B7">7</xref>]).

The fractional derivative of order α>0 of a continuous function y:(0,∞)→R is given by
Dαy(t)=1Γ(n-α)(ddt)n∫0t(t-s)n-α-1y(s)ds,
where n=[α]+1 provided that the right side is pointwise defined on (0,∞).

Remark 2.3 (see [<xref ref-type="bibr" rid="B2">3</xref>]).

The following properties are useful for our discussion:

Lemma 2.4 (the nonlinear alternative of Leray and Schauder type [<xref ref-type="bibr" rid="B8">8</xref>]).

Let E be a Banach space with C⊆E closed and convex. Let U be a relatively open subset of C with 0∈U and let T:U¯→C be a continuous and compact mapping. Then either

the mapping T has a fixed point in U̅, or

there exist u∈∂U and λ∈(0,1) with u=λTu.

Consider
Dαu(t)+y(t)=0,0<t<1,u(0)=0,u(1)=au(ξ),
then one has the following lemma.Lemma 2.5.

Let y∈C[0,1] and 1<α<2, then u(t) is a solution of BVP (2.3) if and only if u(t) is a solution of the integral equation:
u(t)=∫01G1(t,s)y(s)ds,
where
G1(t,s)={[t(1-s)]α-1-atα-1(ξ-s)α-1-(t-s)α-1(1-aξα-1)(1-aξα-1)Γ(α),0⩽s⩽t⩽1,s⩽ξ,[t(1-s)]α-1-(t-s)α-1(1-aξα-1)(1-aξα-1)Γ(α),0<ξ⩽s⩽t⩽1,[t(1-s)]α-1-atα-1(ξ-s)α-1(1-aξα-1)Γ(α),0⩽t⩽s⩽ξ⩽1,[t(1-s)]α-1(1-aξα-1)Γ(α),0⩽t⩽s⩽1,ξ⩽s.

Proof.

Assume that u(t) is a solution of BVP (2.3), then by Remark 2.3, we have
u(t)=-Iαy(t)+C1tα-1+C2tα-2=-∫0t(t-s)α-1Γ(α)y(s)ds+C1tα-1+C2tα-2.

By (2.3), we have
C2=0,C1=∫01(1-s)α-1Γ(α)(1-aξα-1)y(s)ds-a∫0ξ(ξ-s)α-1Γ(α)(1-aξα-1)y(s)ds.

Therefore, we obtain
u(t)=-∫0t(t-s)α-1Γ(α)y(s)ds+∫01tα-1(1-s)α-1Γ(α)(1-aξα-1)y(s)ds-a∫0ξtα-1(ξ-s)α-1Γ(α)(1-aξα-1)y(s)ds=∫01G1(t,s)y(s)ds.

Conversely, if u(t) is a solution of integral equation (2.4), using the relation Dαtα-m=0,m=1,2,…,N, where N is the smallest integer greater than or equal to α [3, Remark 2.1], we have
Dαu(t)=-Dα(∫0t(t-s)α-1Γ(α)y(s)ds)+Dαtα-1[∫01(1-s)α-1Γ(α)(1-aξα-1)y(s)ds-a∫0ξ(ξ-s)α-1Γ(α)(1-aξα-1)y(s)ds]=-DαIαy(t)=-y(t).
A simple computation showed u(0)=0,u(1)=au(ξ). The proof is complete.

Let
G2(t,s)={[t(1-s)]β-1-btβ-1(ξ-s)β-1-(t-s)β-1(1-bξβ-1)(1-bξβ-1)Γ(β),0⩽s⩽t⩽1,s⩽ξ,[t(1-s)]β-1-(t-s)β-1(1-bξβ-1)(1-bξβ-1)Γ(β),0<ξ⩽s⩽t⩽1,[t(1-s)]β-1-btβ-1(ξ-s)β-1(1-bξβ-1)Γ(β),0⩽t⩽s⩽ξ⩽1,[t(1-s)]β-1(1-bξβ-1)Γ(β),0⩽t⩽s⩽1,ξ⩽s,
we call G(t,s)=(G1(t,s),G2(t,s)) Green's function of the boundary value problem (1.1).

Lemma 2.6.

Let 0⩽a,b⩽1, then the function G(t,s) is continuous and satisfies

G(t,s)>0,fort,s∈(0,1),

G(t,s)⩽G(s,s),fort,s∈(0,1).

Proof.

It is easy to prove that G(t,s) is continuous on [0,1]×[0,1], here we omit it. Now we prove G1(t,s)>0. Let
g1(t,s)=[t(1-s)]α-1-atα-1(ξ-s)α-1-(t-s)α-1(1-aξα-1)(1-aξα-1)Γ(α),0<s⩽t⩽1,s⩽ξ,g2(t,s)=[t(1-s)]α-1-(t-s)α-1(1-aξα-1)(1-aξα-1)Γ(α),0<ξ⩽s⩽t⩽1,g3(t,s)=[t(1-s)]α-1-atα-1(ξ-s)α-1(1-aξα-1)Γ(α),0<t⩽s⩽ξ⩽1,g4(t,s)=[t(1-s)]α-1(1-aξα-1)Γ(α),0<t⩽s⩽1,ξ⩽s.
We only need to prove g1(t,s)>0,0<s⩽t⩽1,s⩽ξ. Since
[t(1-s)]α-1-atα-1(ξ-s)α-1-(t-s)α-1(1-aξα-1)=tα-1[(1-s)]α-1-a(ξ-s)α-1-(1-st)α-1(1-aξα-1),
set g(t)=(1-s)α-1-a(ξ-s)α-1-(1-s/t)α-1(1-aξα-1), we have
g′(t)=-(α-1)(1-st)α-2st2(1-aξα-1)⩽0,for0<s<t⩽1,s⩽ξ.

Then g(t) is decreasing on (0,1). Meanwhile,
g(1)=(1-s)α-1-a(ξ-s)α-1-(1-s)α-1(1-aξα-1)=aξα-1[(1-s)α-1-(1-sξ)α-1]>0,0<s<t⩽1,s⩽ξ.

Therefore, g1(t,s)>0, for 0<s<t⩽1,s⩽ξ. Clearly g1(t,s)>0,t=s, so g1(t,s)>0,s,t∈(0,1). It is easy to show that g2(t,s)>0,g3(t,s)>0, g4(t,s)>0. Hence, G1(t,s)>0,s,t∈(0,1).

Similarly, G2(t,s)>0,s,t∈(0,1). The proof of (1) is completed.

Let
g2(t)=[t(1-s)]α-1-(t-s)α-1(1-aξα-1)(1-aξα-1)Γ(α),0<ξ⩽s⩽t⩽1,
then,
g2′(t)=(α-1)tα-2[(1-s)]α-1-(1-s/t)α-2(1-aξα-1)(1-aξα-1)Γ(α),0<ξ⩽s<t⩽1,[(1-s)]α-1-(1-s/t)α-2(1-aξα-1)⩽[(1-s)]α-1-(1-s)α-2(1-aξα-1)=[(1-s)]α-2(aξα-1-s)⩽0,0<ξ⩽s<t⩽1,
therefore,
g2′(t)⩽0,0<ξ⩽s<t⩽1.
So, g2(t,s) is decreasing with respect to t. Similarly, g1(t,s) is decreasing with respect to t. Also g3(t,s) and g4(t,s) are increasing with respect to t. We obtain that G1(t,s) is decreasing with respect to t for s⩽t and increasing with respect to t for t⩽s.

With the use of the monotonicity of G1(t,s), we have
max0⩽t⩽1G1(t,s)=G1(s,s)={[s(1-s)]α-1-a[s(ξ-s)]α-1Γ(α)(1-aξα-1),s∈(0,ξ],[s(1-s)]α-1Γ(α)(1-aξα-1),s∈[ξ,1).
Similarly,
max0⩽t⩽1G2(t,s)=G2(s,s)={[s(1-s)]β-1-b[s(ξ-s)]β-1Γ(β)(1-bξβ-1),s∈(0,ξ],[s(1-s)]β-1Γ(β)(1-bξβ-1),s∈[ξ,1).
The proof of (2) is completed.

3. Main Result

In this section, we will discuss the existence and uniqueness of positive solution for boundary value problem (1.1).

We define the space X={u(t)∣u(t)∈C[0,1]} endowed with ∥u∥X=max0⩽t⩽1|u(t)|, Y={v(t)∣v(t)∈C[0,1]} endowed with ∥u∥Y=max0⩽t⩽1|v(t)|.

For (u,v)∈X×Y, let ∥(u,v)∥X×Y=max{∥u∥X,∥v∥Y}.

Define P={(u,v)∈X×Y∣u(t)⩾0,v(t)⩾0}, then the cone P⊂X×Y.

From Lemma 2.5 in Section 2, we can obtain the following lemma.

Lemma 3.1.

Suppose that f(t,v) and g(t,u) are continuous, then (u,v)∈X×Y is a solution of BVP (1.1) if and only if (u,v)∈X×Y is a solution of the integral equations
u(t)=∫01G1(t,s)f(s,v(s))ds,v(t)=∫01G2(t,s)g(s,u(s))ds.
Let T:X×Y→X×Y be the operator defined as
T(u,v)(t)=(∫01G1(t,s)f(s,v(s))ds,∫01G2(t,s)g(s,u(s))ds)=:(T1v(t),T2u(t)),
then by Lemma 3.1, the fixed point of operator T coincides with the solution of system (1.1).

Lemma 3.2.

Let f(t,v) and g(t,u) be continuous on [0,1]×[0,∞)→[0,∞), then T:P→P defined by (3.2) is completely continuous.

Proof.

Let (u,v)∈P, in view of nonnegativeness and continuity of functions G(t,s), f, and g, we conclude that T:P→P is continuous.

Let Ω∈P be bounded, that is, there exists a positive constant h>0 such that ∥(u,v)∥⩽h for all (u,v)∈Ω.

Let
M=max{|f(t,v(t))|+1:0⩽t⩽1,0⩽v⩽h},N=max{|g(t,u(t))|+1:0⩽t⩽1,0⩽u⩽h},
then we have
|T1v(t)|=|∫01G1(t,s)f(s,v(s))ds|⩽M∫01G1(s,s)ds,|T2u(t)|=|∫01G2(t,s)g(s,u(s))ds|⩽N∫01G2(s,s)ds.
Hence, ∥T(u,v)∥⩽max{M∫01G1(s,s)ds,N∫01G2(s,s)ds}. T(Ω) is uniformly bounded.

Since G1(t,s) is continuous on [0,1]×[0,1], it is uniformly continuous on [0,1]×[0,1]. Thus, for fixed s∈[0,1] and for any ε>0, there exists a constant δ>0, such that any t1,t2∈[0,1] and |t1-t2|<δ,|G1(t1,s)-G1(t2,s)|<ε/M.

Then
|T1(v)(t2)-T1(v)(t1)|⩽M∫01|G1(t2,s)-G1(t1,s)|ds<ε.

Similarly,
|T2(u)(t2)-T2(u)(t1)|⩽N∫01|G2(t2,s)-G2(t1,s)|ds<ε.
For the Euclidean distance d on R2, we have that if t1,t2∈[0,1] are such that |t2-t1|<δ, then
d(T(u,v)(t2),T(u,v)(t1))=(T1v(t2)-T1v(t1))2+(T2u(t2)-T2u(t1))2<2ε.
That is to say, T(P) is 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 that f(t,v) and g(t,u) are continuous on [0,1]×[0,∞)→[0,∞), and there exist two positive functions m(t),n(t) that satisfy

Then system (1.1) has a unique positive solution if
ρ=∫01G1(s,s)m(s)ds<1,θ=∫01G2(s,s)n(s)ds<1.

Proof.

For all (u,v)∈P, by the nonnegativeness of G(t,s) and f(t,v),g(t,u), we have T(u,v)(t)⩾0. Hence, T(P)⊆P.∥T1v2-T1v1∥=maxt∈[0,1]|T1v2-T1v1|=maxt∈[0,1]|∫01G1(t,s)[f(s,v2(s))-f(s,v1(s))]ds|⩽∫01G1(s,s)m(s)ds∥v2-v1∥⩽ρ∥v2-v1∥.
Similarly,
∥T2u2-T2u1∥⩽θ∥u2-u1∥.
We have,
∥T(u2,v2)-T(u1,v1)∥⩽max(ρ,θ)∥(u2,v2)-(u1,v1)∥.
From Lemma 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 system (1.1). This completes the proof.

Theorem 3.4.

Assume that f(t,v) and g(t,u) are continuous on [0,1]×[0,∞)→[0,∞) and satisfy

|f(t,v(t))|⩽a1(t)+a2(t)|v(t)|,

|g(t,u(t))|⩽b1(t)+b2(t)|u(t)|,

A1=∫01G1(s,s)a2(s)ds<1,0<B1=∫01G1(s,s)a1(s)ds<∞,

A2=∫01G2(s,s)b2(s)ds<1,0<B2=∫01G2(s,s)b1(s)ds<∞.

Then the system (1.1) has at least one positive solution (u,v) in
C={(u,v)∈P∣∥(u,v)∥<min(B11-A1,B21-A2)}.

Proof.

Let C={(u,v)∈X×Y:∥(u,v)∥<r} with r=min(B1/(1-A1),B2/(1-A2)), define the operator T:C→P as (3.2).

Let (u,v)∈C, that is, ∥(u,v)∥<r. Then
∥T1v∥=maxt∈[0,1]|∫01G1(t,s)f(s,v(s))ds|⩽∫01G1(s,s)(a1(s)+a2(s)|v(s)|)ds≤∫01G1(s,s)a1(s)ds+∫01G1(s,s)a2(s)ds∥v∥=B1+A1∥v∥⩽r.

Similarly, ∥T2u∥⩽r, so ∥T(u,v)∥⩽r, T(u,v)⊆C¯. From Lemma 3.2T:C¯→C¯ is completely continuous.

Consider the eigenvalue problem
(u,v)=λT(u,v),λ∈(0,1).

Under the assumption that (u,v) is a solution of (3.15) for a λ∈(0,1), one obtains
∥u∥=∥λT1v∥=λmaxt∈[0,1]|∫01G1(t,s)f(s,v(s))ds|<∫01G1(s,s)(a1(s)+a2(s)|v(s)|)ds=∫01G1(s,s)a1(s)ds+∫01G1(s,s)a2(s)ds∥v∥=B1+A1∥v∥⩽r.
Similarly, ∥v∥=∥λT2u∥<r, so ∥(u,v)∥<r, which shows that (u,v)∉∂C. By Lemma 2.4, T has a fixed point in C¯. We complete the proof of Theorem 3.4.

Example 3.5.

Consider the problem
D7/4u(t)+f(t,v(t))=0,0<t<1,D3/2v(t)+g(t,u(t))=0,0<t<1,u(0)=0,u(1)=12u(12),v(0)=0,v(1)=34v(12),
where
f(t,v(t))=tv(t)(1+t)(1+v(t)),g(t,u(t))=arctant1+t|sinu(t)|.
Set v1(t),v2(t),u1(t),u2(t)∈[0,∞) and t∈[0,1], then we have
|f(t,v2(t))-f(t,v1(t))|⩽t1+t|v2(t)-v1(t)|,|g(t,u2(t))-g(t,u1(t))|⩽arctant1+t|u2(t)-u1(t)|.
Therefore,
ρ=∫01G1(s,s)m(s)ds⩽∫01G1(s,s)ds=1Γ(7/4)(1-(1/2)7/4){∫01/2[s(1-s)]3/4ds-∫01/212[s(12-s)]3/4ds+∫1/21[s(1-s)]3/4ds}=2(1+(1/2)7/4)5·Γ(3/4)Γ(1/2)<45<1,θ=∫01G2(s,s)n(s)ds⩽π4∫01G2(s,s)ds=π41Γ(3/2)(1-(3/4)(1/2)1/2){∫01/2[s(1-s)]1/2ds-∫01/234[s(12-s)]1/2ds+∫1/21[s(1-s)]1/2ds}=π4[1-(3/4)(1/2)2][1-(3/4)(1/2)1/2]·Γ(3/2)Γ(3)≈0.6018<1.
With the use of Theorem 3.3, BVP (3.17) has a unique positive solution.

Example 3.6.

Consider the problem
D7/4u(t)+f(t,v(t))=0,0<t<1,D3/2v(t)+g(t,u(t))=0,0<t<1,u(0)=0,u(1)=12u(12),v(0)=0,v(1)=34v(12),
where
f(t,v(t))=t2+t1+tln(1+v(t)),g(t,u(t))=10+t220+u(t).
We have
|f(t,v(t))|t2+t1+t·|v(t)|,|g(t,u(t))|⩽(10+t220)+|u(t)|.
Hence,
A1=∫01G1(s,s)a2(s)ds⩽∫01G1(s,s)ds=2(1+(1/2)7/4)5·Γ(3/4)Γ(1/2)<1,B1=∫01G1(s,s)a1(s)ds=∫01G1(s,s)·s2ds<∞,A2=∫01G2(s,s)b2(s)ds=∫01G2(s,s)ds≈0.7666<1,B2=∫01G2(s,s)b1(s)ds=∫01G2(s,s)(10+s220)ds<∞.
By Theorem 3.4, BVP (3.21) has at least one positive solution in
C={(u,v)∈P∣∥u,v∥<min(B11-A1,B21-A2)}.

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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