# Positive Solution to Nonzero Boundary Values Problem for a Coupled System of Nonlinear Fractional Differential Equations

Authors: Jinhua Wang; Hongjun Xiang; Zhigang Liu
Journal: International Journal of Differential Equations
 (2010)

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## Abstract

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.


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## Body

## 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:(1.1)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 Section2, 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 [6, 7]).
The fractional integral of orderα(α>0) of a function y:(0,∞)→R is given by
(2.1)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 [6, 7]).
The fractional derivative of orderα>0 of a continuous function y:(0,∞)→R is given by
(2.2)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 [3]).
The following properties are useful for our discussion:(1)
IαDαu(t)=u(t)-∑k=1NCktα-k,Dαu(t)∈C(0,1)⋂L(0,1), Ck∈R, N=[α]+1,(2)
DαIαu(t)=u(t),(3)
Dαtγ=Γ(γ+1)/Γ(γ+1-α)tγ-α,α>0,γ>-1,γ>α-1,t>0.Lemma 2.4 (the nonlinear alternative of Leray and Schauder type [8]).
LetE 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(a)
the mappingT has a fixed point in U̅,   or(b)
there existu∈∂U and λ∈(0,1) with u=λTu.Consider
(2.3)Dαu(t)+y(t)=0,0<t<1,u(0)=0,u(1)=au(ξ),
then one has the following lemma.Lemma 2.5.
Lety∈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:
(2.4)u(t)=∫01G1(t,s)y(s)ds,
where
(2.5)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 thatu(t) is a solution of BVP (2.3), then by Remark 2.3, we have
(2.6)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
(2.7)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(2.8)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, ifu(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
(2.9)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(2.10)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.
Let0⩽a,b⩽1, then the function G(t,s) is continuous and satisfies(1)
G(t,s)>0,fort,s∈(0,1),(2)
G(t,s)⩽G(s,s),fort,s∈(0,1).Proof.
It is easy to prove thatG(t,s) is continuous on [0,1]×[0,1], here we omit it. Now we prove G1(t,s)>0. Let
(2.11)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
(2.12)[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
(2.13)g′(t)=-(α-1)(1-st)α-2st2(1-aξα-1)⩽0,for0<s<t⩽1,s⩽ξ.
Theng(t) is decreasing on (0,1). Meanwhile,
(2.14)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(2.15)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,
(2.16)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,
(2.17)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 ofG1(t,s), we have
(2.18)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,
(2.19)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 spaceX={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}.DefineP={(u,v)∈X×Y∣u(t)⩾0,v(t)⩾0}, then the cone P⊂X×Y.From Lemma2.5 in Section 2, we can obtain the following lemma.Lemma 3.1.
Suppose thatf(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
(3.1)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
(3.2)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.
Letf(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.
LetΩ∈P be bounded, that is, there exists a positive constant h>0 such that ∥(u,v)∥⩽h for all (u,v)∈Ω.
Let(3.3)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
(3.4)|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.
SinceG1(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,(3.5)|G1(t1,s)-G1(t2,s)|<ε/M.
Then(3.6)|T1(v)(t2)-T1(v)(t1)|⩽M∫01|G1(t2,s)-G1(t1,s)|ds<ε.
Similarly,(3.7)|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
(3.8)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 thatf(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(H1)
|f(t,v2)-f(t,v1)|⩽m(t)|v2-v1|,fort∈[0,1],v1,v2∈[0,∞),(H2)
|g(t,u2)-g(t,u1)|⩽n(t)|u2-u1|,fort∈[0,1],u1,u2∈[0,∞).
Then system (1.1) has a unique positive solution if
(3.9)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.(3.10)∥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,
(3.11)∥T2u2-T2u1∥⩽θ∥u2-u1∥.
We have,
(3.12)∥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 thatf(t,v) and g(t,u) are continuous on [0,1]×[0,∞)→[0,∞) and satisfy(H3)
|f(t,v(t))|⩽a1(t)+a2(t)|v(t)|,(H4)
|g(t,u(t))|⩽b1(t)+b2(t)|u(t)|,(H5)
A1=∫01G1(s,s)a2(s)ds<1,0<B1=∫01G1(s,s)a1(s)ds<∞,(H6)
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
(3.13)C={(u,v)∈P∣∥(u,v)∥<min(B11-A1,B21-A2)}.Proof.
LetC={(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
(3.14)∥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 3.2   T:C¯→C¯ is completely continuous.
Consider the eigenvalue problem(3.15)(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
(3.16)∥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(3.17)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
(3.18)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
(3.19)|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,
(3.20)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(3.21)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
(3.22)f(t,v(t))=t2+t1+tln(1+v(t)),g(t,u(t))=10+t220+u(t).
We have
(3.23)|f(t,v(t))|t2+t1+t·|v(t)|,|g(t,u(t))|⩽(10+t220)+|u(t)|.
Hence,
(3.24)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
(3.25)C={(u,v)∈P∣∥u,v∥<min(B11-A1,B21-A2)}.



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