# Lower and Upper Solutions Method for Positive Solutions of Fractional Boundary Value Problems

Authors: R. Darzi; B. Mohammadzadeh; A. Neamaty; D. Bǎleanu
Journal: Abstract and Applied Analysis
 (2013)

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## Abstract

We apply the lower and upper solutions method and fixed-point theorems to prove the existence of positive solution to fractional boundary value problemD0+αut+ft,ut=0, 0<t<1, 2<α≤3, u0=u′0=0, D0+α−1u1=βuξ, 0<ξ<1, where D0+α denotes Riemann-Liouville fractional derivative, β is positive real number, βξα−1≥2Γα, and f is continuous on 0,1×0,∞. As an application, one example is given to illustrate the main result.


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## Body

## 1. Introduction

In the recent years, fractional calculus has been one of the most interesting issues that have attracted many scientists, especially in the fields of mathematics and engineering sciences. Many natural phenomena can be presented by boundary value problems of fractional differential equations. Many authors in different fields such as chemical physics, fluid flows, electrical networks, and viscoelasticity try to present a model of these phenomena by boundary value problems of fractional differential equations [1–4]. In order to achieve extra information in fractional calculus, interested readers can refer to more valuable books that are written by other authors [5–20].The existence and multiplicity of solutions or positive solutions of nonlinear fractional differential equation (FDE) by the use of fixed point theorems, Leray-Shauder theory, and so forth are mentioned in some papers [6, 8, 12, 20, 21]. Few papers have considered the boundary value problems of fractional differential equations [12, 14]. By the use of some fixed point theorems on cones, Zhang [15] obtained the existence of positive solution for the equation
(1)D0+αu(t)+f(t,u(t))=0,0<t<1,1<α≤2,
with the boundary conditions
(2)u(0)=u(1)=0.
In [22], Liang and Zhang applied lower and upper solutions method and fixed point theorems to obtain some results on the existence of positive solutions for the following BVPs:
(3)D0+αu(t)=f(t,u(t)),0<t<1,3<α≤4,u(0)=u(1)=u′(0)=u′(1)=0,
where Dα denotes Riemann-Liouville fractional derivative.In this paper, we investigate the existence of positive solution for a nonlocal BVP of FDE,(4)D0+αu(t)+f(t,u(t))=0,0<t<1,2<α≤3,(5)u(0)=u′(0)=0,D0+α-1u(1)=βu(ξ),
using lower and upper solutions method and fixed point theorem, where Dα denotes standard Riemann-Liouville fractional derivative, βξα-1≥2Γ(α), and f∈C([0,1]×[0,∞),[0,∞)).The main result of this paper can be seen in Theorem10. In Theorem 10, we use the following conditions:(H1)
f(t,u(t))∈C([0,1]×[0,∞),ℝ+) is nondecreasing with respect to u,(H2)
f(t,ϱ(t))≠0 for t∈(0,1),(H3)
there exist a positive constantλ<1 such that kλf(t,u)≤f(t,ku), for all 0≤k≤1, and the Schauder fixed-point theorem to show that problem (4)-(5) has a positive solution.


## 2. Basic Definitions and Preliminaries

In this section, we present the necessary definitions and lemmas that will be used to prove our new results.Definition 1 (see [5, 6]).
The Riemann-Liouville fractional integral of orderα>0 of a function f:(0,∞)→ℝ is defined by
(6)I0+αf(t)=1Γ(α)∫0t‍(t-s)α-1f(s)ds,n-1<α≤n,
provided that the right-hand side is pointwise defined on (0,∞).Definition 2 (see [5, 6]).
The Riemann-Liouville fractional derivative of orderα>0 of a function f:(0,∞)→ℝ is defined by
(7)D0+αf(t)=1Γ(n-α)dndxn∫0t‍(t-s)α-1f(s)ds,00000000000000000000000000n-1<α≤n,
where n=[α]+1, provided that the right-hand side is pointwise defined on (0,∞).Definition 3 (see [5, 6]).
A functionμ(t)∈C2[0,1] is called a lower solution of problem (4)-(5) if μ(t) satisfies
(8)-D0+αμ(t)≤f(t,μ(t)),0<t<1,2<α≤3,μ(0)≤0,μ′(0)≤0,D0+αμ(t)≤βμ(ξ).Definition 4 (see [7, 8]).
A functionμ(t)∈C2[0,1] is called an upper solution of problem (4)-(5) if μ(t) satisfies
(9)-D0+αμ(t)≥f(t,μ(t)),0<t<1,2<α≤3,μ(0)≥0,μ′(0)≥0,D0+αμ(t)≥βμ(ξ).Lemma 5 (see [7, 8]).
Letu∈C(0,1)∩L1(0,1). Then the fractional differential equation
(10)Dαu(t)=0
has
(11)u(t)=c1tα-1+c2tα-2+⋯+cntα-n,forsomeci∈ℝ,i=1,…,n,
as a unique solution.Lemma 6 (see [7, 8]).
Letu∈C(0,1)∩L1(0,1) with a fractional derivative of order α>0, n-1<α≤n(n∈ℕ), that belongs to C(0,1)∩L(0,1). Then
(12)I0+αD0+αu(t)=u(t)+c1tα-1+c2tα-2+⋯+cntα-n,forsomeci∈ℝ,i=1,…,n.Lemma 7.
Ifβξα-1≥2Γ(α), then for 0≤y(t)∈C[0,1], the problem,
(13)D0+αu(t)+y(t)=0,0<t<1,2<α≤3,(14)u(0)=u′(0)=0,D0+α-1u(1)=βu(ξ),
has a unique positive solution
(15)u(t)=-1Γ(α)∫0t‍(t-s)α-1y(s)ds+1βξα-1-Γ(α)∫01‍tα-1y(s)ds+β[βξα-1-Γ(α)]Γ(α)∫0ξ‍tα-1(ξ-s)α-1y(s)ds.Proof.
We can apply Lemma6 to reduce (13) to an equivalent integral equation
(16)u(t)=-1Γ(α)∫0t‍(t-s)α-1y(s)ds+c1tα-1+c2tα-2+c3tα-3,
for some c1,c2,c3∈ℝ. From u(0)=0 and u′(0)=0 in (14), we have c2=c3=0. On the other hand, Dα-1u(1)=βu(ξ) yields
(17)c1=1βξα-1-Γ(α)∫0t‍y(s)ds+β[βξα-1-Γ(α)]Γ(α)∫0ξ‍(ξ-s)α-1y(s)ds.
Then, the unique solution of problem is given by u(t). Obviously, u(t)≥0 if y(t)≥0 on t∈[0,1]. The proof is complete.


## 3. Main Result

In this section, we present and prove our main result.Lemma 8.
Suppose thatβξα-1≥2Γ(α). Given that y∈C[0,1], the Green function for the problem (13)-(14) is given by
(18)G(t,s)={(-(t-s)α-1[βξα-1-Γ(α)]+Γ(α)tα-1+βtα-1(ξ-s)α-1)×([βξα-1-Γ(α)]Γ(α))-1,0≤s≤t≤1,s≤ξ,(-(t-s)α-1[βξα-1-Γ(α)]+Γ(α)tα-1)×([βξα-1-Γ(α)]Γ(α))-1,0≤ξ≤s≤t≤1,Γ(α)tα-1+βtα-1(ξ-s)α-1[βξα-1-Γ(α)]Γ(α),0≤t≤s≤ξ≤1,tα-1βξα-1-Γ(α),t≤s,s≥ξ.Proof.
By Lemma7, for t≤ξ, we have
(19)u(t)=-1Γ(α)∫0t‍(t-s)α-1y(s)ds+1βξα-1-Γ(α)×[(∫0t‍+∫tξ‍+∫ξ1‍)tα-1y(s)ds]+β[βξα-1-Γ(α)]Γ(α)×[(∫0t‍+∫tξ‍)tα-1(ξ-s)α-1y(s)ds]=∫0t‍(([βξα-1-Γ(α)]Γ(α))-1(-(t-s)α-1[βξα-1-Γ(α)]+Γ(α)tα-1+βtα-1(ξ-s)α-1)×([βξα-1-Γ(α)]Γ(α))-1)y(s)ds+∫tξ‍Γ(α)tα-1+βtα-1(ξ-s)α-1[βξα-1-Γ(α)]Γ(α)y(s)ds+∫ξ1‍Γ(α)tα-1[βξα-1-Γ(α)]Γ(α)y(s)ds=∫01‍G(t,s)y(s)ds.
For t≥ξ, we have
(20)u(t)=-1Γ(α)[(∫0ξ‍∫ξt‍)](t-s)α-1y(s)ds+1βξα-1-Γ(α)×[(∫0ξ‍+∫ξt‍+∫t1‍)tα-1y(s)ds]+β[βξα-1-Γ(α)]Γ(α)∫0ξ‍tα-1(ξ-s)α-1y(s)ds=∫0ξ‍(([βξα-1-Γ(α)]Γ(α))-1(-(t-s)α-1[βξα-1-Γ(α)]+Γ(α)tα-1+βtα-1(ξ-s)α-1)×([βξα-1-Γ(α)]Γ(α))-1)y(s)ds+∫ξt‍(([βξα-1-Γ(α)]Γ(α))-1(-(t-s)α-1[βξα-1-Γ(α)]+Γ(α)tα-1)l×([βξα-1-Γ(α)]Γ(α))-1)y(s)ds+∫t1‍Γ(α)tα-1[βξα-1-Γ(α)]Γ(α)y(s)ds=∫01‍G(t,s)y(s)ds.
The proof is complete.Lemma 9.
Suppose thatu(t)∈C2[0,1] and is a positive solution of (4)-(5). Then
(21)mϱ(t)≤u(t)≤Mϱ(t),
where
(22)ϱ(t)=1cΓ(α+1)[βξα-Γ(α+1)]tα-1-ctα,
where c=βξα-1-Γ(α), βξα-1≠Γ(α), and m and M are two constants.Proof.
Sinceu(t)∈C2[0,1], there exists T>0 so that |u(t)|≤T for t∈[0,1]. We define
(23)m:=min(t,u)∈[0,1]×[0,T]f(t,u(t)),M:=max(t,u)∈[0,1]×[0,T]f(t,u(t)).
Therefore, we have
(24)m∫01‍G(t,s)ds≤∫01‍G(t,s)f(s,u(s))ds≤M∫01‍G(t,s)ds.
On the other hand, by direct computation, we get
(25)∫01‍G(t,s)ds=1[βξα-1-Γ(α)]Γ(α+1)×([βξα-1-Γ(α+1)]tα-1-[βξα-1-Γ(α)]tα).
This completes the proof of the lemma.Theorem 10.
The fractional boundary value problem (4)-(5) has a positive solution u(t) if the conditions (H1)–(H3) are satisfied.Proof.
Suppose that(26)α1=min{1,inft∈[0,1]f(t,ϱ(t))},α2=max{1,supt∈[0,1]f(t,ϱ(t))},0<k1≤min{1α2,(α1)λ/(1-λ)},k2≥max{1α2,(α1)λ/(1-λ)},
and h(t)=∫01‍G(t,s)f(s,ϱ(s))ds. We show that μ(t)=k1h(t) and ν(t)=k2h(t) are lower and upper solutions of (4)-(5), respectively. From Lemma 7, h(t) is a positive solution of the following problem:
(27)-D0+αu(t)≤f(t,ϱ(t)),0<t<1,2<α≤3,u(0)=u′(0)=0,D0+α-1u(1)=βu(ξ).
We know that α1ϱ(t)≤h(t)≤α2ϱ(t). Now, using the assumption of the theorem, we get
(28)k1α1≤h(t)ϱ(t)≤k2α2,1k1α2≤ϱ(t)h(t)≤1k2α1≤1,(k1α1)λ≥k1,(k2α2)λ≥k2.
Therefore, from (H3) and since (k1α1)λ≥k1, the following relations satisfy
(29)f(t,μ(t))=f(t,μ(t)ϱ(t)ϱ(t))≥(μ(t)ϱ(t))λf(t,ϱ(t))≥(k1α1)λf(t,ϱ(t))>k1f(t,ϱ(t)),k2f(t,ϱ(t))=f(t,ϱ(t)ν(t)ν(t))≥k2(ϱ(t)ν(t))λf(t,ν(t))≥k2(k1α1)-λf(t,ν(t))>f(t,ν(t)).
Consequently
(30)-D0+αμ(t)=k1f(t,ϱ(t))≤f(t,μ(t)),0<t<1,2<α≤3,-D0+αν(t)=k2f(t,ϱ(t))≥f(t,ν(t)),0<t<1,2<α≤3.
Since μ(t)=k1h(t) and ν(t)=k2h(t) satisfy the boundary conditions, μ(t) and ν(t) are lower and upper solutions of (4)-(5), respectively. Now, we suppose that
(31)g(t,u(t))={f(t,μ(t)),u(t)≤μ(t),f(t,u(t)),μ(t)≤u(t)≤ν(t),f(t,ν(t)),u(t)≤ν(t),
and prove that FBVP,
(32)-D0+αu(t)=g(t,u(t)),0<t<1,2<α≤3,(33)u(0)=u′(0)=0,D0+α-1u(1)=βu(ξ),
has a solution. Consider operator T:C2[0,1]→C2[0,1], with Tu(t)=∫01‍G(t,s)g(s,u(s))ds, where G(t,s) is defined as in Lemma 8. It is easy to see that T is continuous in C2[0,1]. Since f is nondecreasing in u (from(H1)), for u∈C2[0,1], we have
(34)f(t,μ(t))≤g(t,u(t))≤f(t,ν(t)),t∈[0,1].
So, there exists a positive constant M, such that |g(t,u(t))|≤M. We will show that the operator T is equicontinuous.
Case  1. If s≤ξ,
(35)|Tu(t1)-Tu(t2)|≤|∫01‍[G(t2,s)-G(t1,s)]g(s,u(s))ds|=|∫0t2‍-(t2-s)α-1Γ(α)g(s,u(s))ds000+Γ(α)t2α-1[βξα-1-Γ(α)]Γ(α)∫0t2‍g(s,u(s))ds000+βt2α-1[βξα-1-Γ(α)]Γ(α)∫0t2‍(ξ-s)α-1g(s,u(s))ds000+Γ(α)t2α-1[βξα-1-Γ(α)]Γ(α)×∫t21‍g(s,u(s))ds000+βt2α-1[βξα-1-Γ(α)]Γ(α)∫t21‍(ξ-s)α-1g(s,u(s))ds000-∫0t1‍-(t2-s)α-1Γ(α)g(s,u(s))ds000-Γ(α)t1α-1[βξα-1-Γ(α)]Γ(α)∫0t1‍g(s,u(s))ds000-βt1α-1[βξα-1-Γ(α)]Γ(α)∫0t1‍(ξ-s)α-1g(s,u(s))ds000-Γ(α)t2α-1[βξα-1-Γ(α)]Γ(α)×∫t11‍g(s,u(s))ds000-βt1α-1[βξα-1-Γ(α)]Γ(α)∫t11‍(ξ-s)α-1g(s,u(s))ds|=|∫0t1‍[-(t2-s)α-1Γ(α)+(t1-s)α-1Γ(α)]g(s,u(s))ds000+∫t1t2‍-(t2-s)α-1Γ(α)g(s,u(s))ds000+Γ(α)(t2α-1-t1α-1)[βξα-1-Γ(α)]Γ(α)∫0t1‍g(s,u(s))ds000+Γ(α)t2α-1[βξα-1-Γ(α)]Γ(α)∫t1t2‍g(s,u(s))ds000+β(t2α-1-t1α-1)[βξα-1-Γ(α)]Γ(α)∫0t1‍(ξ-s)α-1g(s,u(s))ds000+βt2α-1[βξα-1-Γ(α)]Γ(α)×∫t1t2‍(ξ-s)α-1g(s,u(s))ds000+Γ(α)(t2α-1-t1α-1)[βξα-1-Γ(α)]Γ(α)∫t21‍g(s,u(s))ds000-Γ(α)t1α-1[βξα-1-Γ(α)]Γ(α)∫t1t2‍g(s,u(s))ds000+β(t2α-1-t1α-1)[βξα-1-Γ(α)]Γ(α)∫t21‍(ξ-s)α-1g(s,u(s))ds000-βt1α-1[βξα-1-Γ(α)]Γ(α)×∫t1t2‍(ξ-s)α-1g(s,u(s))ds|≤2MΓ(α+1)(t2-t1)α+MΓ(α)(t2α-1-t1α-1)t1[βξα-1-Γ(α)]Γ(α)+MΓ(α)t2α-1(t2-t1)[βξα-1-Γ(α)]Γ(α)+Mβ(t2α-1-t1α-1)t1[βξα-1-Γ(α)]Γ(α)+Mβt2α-1(t2-t1)[βξα-1-Γ(α)]Γ(α)+MΓ(α)(t2α-1-t1α-1)(1-t2)[βξα-1-Γ(α)]Γ(α)+MΓ(α)t1α-1(t2-t1)[βξα-1-Γ(α)]Γ(α)+Mβ(t2α-1-t1α-1)(1-t2)[βξα-1-Γ(α)]Γ(α)+Mβt1α-1(t2-t1)[βξα-1-Γ(α)]Γ(α).
Case  2. If s≥ξ,
(36)|Tu(t1)-Tu(t2)|≤|∫01‍[G(t2,s)-G(t1,s)]g(s,u(s))ds|=|∫0t2‍-(t2-s)α-1Γ(α)g(s,u(s))ds000+1[βξα-1-Γ(α)]∫0t2‍t2α-1g(s,u(s))ds000+1βξα-1-Γ(α)∫t21‍t2α-1g(s,u(s))ds000+∫0t1‍-(t1-s)α-1Γ(α)g(s,u(s))ds000-1[βξα-1-Γ(α)]∫0t1‍t1α-1g(s,u(s))ds000-1βξα-1-Γ(α)∫t11‍t1α-1g(s,u(s))ds|=|∫0t1‍[-(t2-s)α-1Γ(α)+(t1-s)α-1Γ(α)]g(s,u(s))ds000+∫t1t2‍-(t2-s)α-1Γ(α)g(s,u(s))ds000+(t2α-1-t1α-1)βξα-1-Γ(α)∫0t1‍g(s,u(s))ds000+t2α-1βξα-1-Γ(α)∫t1t2‍g(s,u(s))ds000+(t2α-1-t1α-1)βξα-1-Γ(α)∫t21‍g(s,u(s))ds000+t1α-1βξα-1-Γ(α)∫t1t2‍g(s,u(s))ds|≤2MΓ(α+1)(t2-t1)α+M(t2α-1-t1α-1)t1βξα-1-Γ(α)00+Mt2α-1(t2-t1)βξα-1-Γ(α)+M(t2α-1-t1α-1)(1-t2)βξα-1-Γ(α)00+Mt1α-1(t2-t1)βξα-1-Γ(α).
Therefore, the operator T is equicontinuous, and by Arzela-Ascoli theorem, T is a compact operator. Now, the Schauder fixed-point theorem [23] shows that the operator T has a fixed-point theorem and so FBVP (32)-(33) has a solution. Finally, we will prove that FBVP (4)-(5) has a positive solution. Suppose that u1(t) is a solution of FBVP (32)-(33). Since the function f is nondecreasing in u, we have
(37)f(t,μ(t))≤g(t,u1(t))≤f(t,ν(t)),t∈[0,1].
Assuming X(t)=ν(t)-u1(t),
(38)DαX(t)≥f(t,ν(t))-g(t,u1(t))≥0,X(0)=X′(0)=0,Dα-1X(1)=βX(ξ).
By Lemma 7, X(t)≥0; that is, u1(t)≤ν(t) for t∈[0,1]. Similarly, μ(t)≤u1(t) for t∈[0,1]. Therefore u1(t) is a positive solution of FBVP (4)-(5). The proof is complete.Example 11.
Consider the following fractional boundary value problem:(39)-D0+5/2u(t)=f(t,u(t)),0<t<1,u(0)=u′(0)=0,D0+α-1u(1)=βu(ξ),
where
(40)f(t,u(t))=sin(πt2)+u.
For 0≤k≤1, we have k≤1. Therefore
(41)k1/2f(t,u(t))=k1/2(sinπ2t+u)≤sinπ2t+ku=f(t,ku(t)).
Now, by Theorem 10, we obtain that the FBVP (39) has a positive solution.



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