# Positive Solution to a Fractional Boundary Value Problem

Authors: A. Guezane-Lakoud; R. Khaldi
Journal: International Journal of Differential Equations
 (2011)

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## Abstract

A fractional boundary value problem is considered. By
means of Banach contraction principle, Leray-Schauder nonlinear alternative,
properties of the Green function, and Guo-Krasnosel'skii fixed point
theorem on cone, some results on the existence, uniqueness, and positivity
of solutions are obtained.


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## Body

## 1. Introduction

Fractional differential equations are a natural generalization of ordinary differential equations. In the last few decades many authors pointed out that differential equations of fractional order are suitable for the metallization of various physical phenomena and that they have numerous applications in viscoelasticity, electrochemistry, control and electromagnetic, and so forth, see [1–4].This work is devoted to the study of the following fractional boundary value problem (P1):(1.1)cD0+qu(t)=f(t,u(t),cD0+σu(t)),0<t<1,(1.2)u(0)=u′′(0)=0,u′(1)=αu′′(1),
where f:[0,1]×ℝ×ℝ→ℝ is a given function, 2<q<3, 1<σ<2 and cD0+q denotes the Caputo's fractional derivative. Our results allow the function f to depend on the fractional derivative cD0+σu(t) which leads to extra difficulties. No contributions exist, as far as we know, concerning the existence of positive solutions of the fractional differential equation (1.1) jointly with the nonlocal condition (1.2).Our mean objective is to investigate the existence, uniqueness, and existence of positive solutions for the fractional boundary value problem (P1), by using Banach contraction principle, Leray-Schauder nonlinear alternative, properties of the Green function and Guo-Krasnosel'skii fixed point theorem on cone.The research in this area has grown significantly and many papers appeared on this subject, using techniques of nonlinear analysis, see [5–14].In [6], El-Shahed considered the following nonlinear fractional boundary value problem(1.3)D0+qu(t)+λa(t)f(u(t))=0,0<t<1,u(0)=u′(0)=u′(1)=0,
where 2<q≤3 and D0+q is the Riemann-Liouville fractional derivative. Using the Krasnoselskii's fixed-point theorem on cone, he proved the existence and nonexistence of positive solutions for the above fractional boundary value problem.Liang and Zhang in [9] studied the existence and uniqueness of positive solutions by the properties of the Green function, the lower and upper solution method and fixed point theorem for the fractional boundary value problem (1.4)D0+qu(t)+f(t,u(t))=0,0<t<1,u(0)=u′(0)=0,u′(1)=∑i=1m-2βiu′(ζi),
where 2<q≤3 and D0+q is the Riemann-Liouville fractional derivative.In [5] Bai and Lü investigated the existence and multiplicity of positive solutions for nonlinear fractional differential equation boundary value problem:(1.5)D0+qu(t)+f(t,u(t))=0,0<t<1,u(0)=u(1)=0,
where 1<q≤1 and D0+q is the Riemann-Liouville fractional derivative. Applying fixed-point theorems on cone, they prove some existence and multiplicity results of positive solutions.This paper is organized as follows, in the Section2 we cite some definitions and lemmas needed in our proofs. Section 3 treats the existence and uniqueness of solution by using Banach contraction principle, Leray Schauder nonlinear alternative. Section 4 is devoted to prove the existence of positive solutions with the help of Guo-Krasnoselskii Theorem, then we give some examples illustrating the previous results.


## 2. Preliminaries and Lemmas

In this section, we present some lemmas and definitions from fractional calculus theory which will be needed later.Definition 2.1.
Ifg∈C([a,b]) and α>0, then the Riemann-Liouville fractional integral is defined by
(2.1)Ia+αg(t)=1Γ(α)∫atg(s)(t-s)1-αds.Definition 2.2.
Letα≥0, n=[α]+1. If f∈ACn[a,b] then the Caputo fractional derivative of order α of f defined by
(2.2)cDa+αg(t)=1Γ(n-α)∫atg(n)(s)(t-s)α-n+1ds,
exists almost everywhere on [a,b] ([α] is the entire part of α).Lemma 2.3 (see [15]).
Letα,β>0 and n=[α]+1, then the following relations hold: cD0+αtβ-1=(Γ(β)/Γ(β-α))tβ-1, β>n and cD0+αtk=0,  k=0,1,2,…,n-1.Lemma 2.4 (see [15]).
Forα>0, g(t)∈C(0,1), the homogenous fractional differential equation
(2.3)cDa+αg(t)=0
has a solution
(2.4)g(t)=c1+c2t+c3t2+⋯+cntn-1,
where, ci∈R, i=0,…,n, and n=[α]+1.Denote byL1([0,1],ℝ) the Banach space of Lebesgue integrable functions from [0,1] into ℝ with the norm ||y||L1=∫01|y(t)|dt.The following Lemmas gives some properties of Riemann-Liouville fractional integrals and Caputo fractional derivative.Lemma 2.5 (see [16]).
Letp,q≥0, f∈L1[a,b]. Then I0+pI0+qf(t)=I0+p+qf(t)=I0+qI0+pf(t) and cD0+qI0+qf(t)=f(t), for all t∈a,b].Lemma 2.6 (see [15]).
Letβ>α>0. Then the formula cD0+αI0+βf(t)=I0+β-αf(t), holds almost everywhere on t∈a,b], for f∈L1[a,b] and it is valid at any point x∈[a,b] if f∈C[a,b].Now we start by solving an auxiliary problem.Lemma 2.7.
Let2<q<3, 1<σ<2 and y∈C[0,1]. The unique solution of the fractional boundary value problem
(2.5)Dc0+qu(t)=y(t),0<t<1,u(0)=u′′(0)=0,u′(1)=αu′′(1)
is given by
(2.6)u(t)=1Γ(q-2)∫01G(t,s)y(s)ds,
where
(2.7)G(t,s)={(t-s)q-1q(q-1)+αt(1-s)3-q-t(1-s)q-2q-1,s≤tαt(1-s)3-q-t(1-s)q-2q-1,t≤s.Proof.
Applying Lemmas2.4 and 2.5 to (2.5) we get
(2.8)u(t)=I0+qy(t)+c1+c2t+c3t2.
Differentiating both sides of (2.8) and using Lemma 2.6 it yields
(2.9)u′(t)=I0+q-1y(t)+c2+c3t,u′′(t)=I0+q-2y(t)+c3.
The first condition in (2.5) implies c1=c3=0, the second one gives c2=αI0+q-2y(1)-I0+q-1y(1). Substituting c2 by its value in (2.8), we obtain
(2.10)u(t)=I0+qy(t)+t(αI0+q-2y(1)-I0+q-1y(1))
that can be written as
(2.11)u(t)=1Γ(q-2)∫0t[(t-s)q-1q(q-1)+αt(1-s)3-q-t(1-s)q-2q-1]y(s)ds+1Γ(q-2)∫t1[αt(1-s)3-q-t(1-s)q-2q-1]y(s)ds
that is equivalent to
(2.12)u(t)=1Γ(q-2)∫01G(t,s)y(s)ds,
where G is defined by (2.7). The proof is complete.


## 3. Existence and Uniqueness Results

In this section we prove the existence and uniqueness of solutions in the Banach spaceE of all functions u∈C2[0,1] into ℝ, with the norm ||u||=maxt∈[0,1]|u(t)|+maxt∈[0,1]|cD0+σu(t)|. We know that cD0+σu∈C[0,1], 1<σ<2, see [15]. Denote by E+={u∈E,u(t)≥0,t∈[0,1]}. Throughout this section, we suppose that f∈C([0,1]×ℝ×ℝ,ℝ). Define the integral operator T:E→E by(3.1)Tu(t)=1Γ(q-2)∫01G(t,s)f(s,u(s),cD0+σu(s))ds.Lemma 3.1.
The functionu∈E is solution of the fractional boundary value problem (P1) if and only if Tu(t)=u(t), for all t∈[0,1].Proof.
Letu be solution of (P1) and v(t)=∫01G(t,s)f(s,u(s),cD0+σu(s))ds. In view of (2.10) we have
(3.2)v(t)=I0+qf(t,u(t),cD0+σu(t))+αtI0+q-2f(1,u(1),cD0+σu(1))-tI0+q-1f(1,u(1),cD0+σu(1)).
With the help of Lemma 2.6 we obtain
(3.3)cD0+qv(t)=D0+qcI0+qf(t,u(t),cD0+σu(t))+tαI0+q-2f(1,u(1),cD0+σu(1))-tI0+q-1f(1,u(1),cD0+σu(1))=f(t,u(t),cD0+σu(t)).
It is clear that v satisfies conditions (1.2), then it is a solution for the problem (P1). The proof is complete.Theorem 3.2.
Assume that there exist nonnegative functionsg,h∈L1([0,1],ℝ+) such that for all x,y∈ℝ and t∈[0,1], one has
(3.4)|f(t,x,x¯)-f(t,y,y¯)|≤g(t)|x-y|+h(t)|x¯-y¯|,(3.5)Cg+Ch<1,Ag+Ah<(2-σ)Γ(2-σ),
where
(3.6)Cg=‖I0+q-1g‖L1+|α|I0+q-2g(1)+I0+q-1g(1),Ag=2I0+q-1g(1)+|α|I0+q-2g(1),Ch=‖I0+q-1h‖L1+|α|I0+q-2h(1)+I0+q-1h(1),Ah=2I0+q-1h(1)+|α|I0+q-2h(1).
Then the fractional boundary value problem (P1) has a unique solution u in E.To prove Theorem3.2, we use the following property of Riemann-Liouville fractional integrals.Lemma 3.3.
Letq>0, f∈L1([a,b],ℝ+). Then, for all t∈a,b] we have
(3.7)I0+q+1f(t)≤‖I0+qf‖L1.Proof.
Letf∈L1([a,b],ℝ+), then
(3.8)‖I0+qf‖L1=∫01I0+qf(r)dr≥1Γ(q)∫at∫arf(s)(r-s)1-qdsdr=1Γ(q)∫at(∫stf(s)(r-s)1-qdr)ds=I0+q+1f(t).Now we prove Theorem3.2.Proof.
We transform the fractional boundary value problem to a fixed point problem. By Lemma3.1, the fractional boundary value problem (P1) has a solution if and only if the operator T has a fixed point in E. Now we will prove that T is a contraction. Let u,v∈E, in view of (2.10) we get
(3.9)Tu(t)-Tv(t)=1Γ(q-2)∫01G(t,s)(f(s,u(s),cD0+σu(s))-f(s,v(s),cD0+σv(s)))ds=I0+q(f(t,u(t),cD0+σu(t))-f(s,v(t),cD0+σv(t)))+tα[I0+q-2(f(1,u(1),cD0+σu(1))-f(1,v(1),cD0+σv(1)))-tI0+q-1(f(1,u(1),cD0+σu(1))-f(1,v(1),cD0+σv(1)))],
with the help of (3.4) we obtain
(3.10)|Tu(t)-Tv(t)|≤max⁡⁡|u(t)-v(t)|(I0+qg(t)+|α|I0+q-2g(1)+I0+q-1g(1))+max⁡⁡|D0+σcu(t)-cD0+σv(t)|(I0+qh(t)+|α|I0+q-2h(1)+I0+q-1h(1)).
Lemma 3.3 implies
(3.11)|Tu(t)-Tv(t)|≤‖u-v‖[‖I0+q-1g‖L1+|α|I0+q-2g(1)+I0+q-1g(1)+‖I0+q-1h‖L1+|α|I0+q-2h(1)+I0+q-1h(1)]=‖u-v‖(Cg+Ch).
In view of (3.5) it yields
(3.12)|Tu-Tv|<‖u-v‖.
On the other hand we have
(3.13)cD0+σTu-cD0+σTv=1Γ(2-σ)∫0t(Tu)′(s)-(Tv)′(s)(t-s)σ-1ds,
where
(3.14)(Tu)′(t)=1Γ(q-2)∫01G1(t,s)f(s,u(s),cD0+σu(s))ds,G1(t,s)=∂G(t,s)∂t={(t-s)q-2q+α(1-s)3-q-(1-s)q-2q-1,0≤s≤t≤1α(1-s)3-q-(1-s)q-2q-1,0≤t≤s≤1.
Therefore(3.15)cD0+σTu-cD0+σTv=1Γ(q-2)Γ(2-σ)∫0t∫01(t-s)-σ+1G1(s,r)×(f(r,u(r),cD0+σu(r))-f(r,v(r),cD0+σv(r)))drds.
Applying hypothesis (3.4) we get
(3.16)|cD0+σTu-cD0+σTv|≤max⁡|u-v|Γ(q-2)Γ(2-σ)∫0t∫01(t-s)-σ+1|G1(s,r)|g(r)drds+max⁡|cD0+σu-cD0+σv|Γ(q-2)Γ(2-σ)∫0t∫01(t-s)-σ+1|G1(s,r)|h(r)drds.
Let us estimate the term ∫01(∂G(s,r)/∂s)g(r)dr. We have
(3.17)∫01|G1(s,r)|g(r)dr≤Γ(q-2)(2(q-2)I0+q-1g(1)q-1+|α|I0+q-2g(1))=Γ(q-2)(2I0+q-1g(1)+|α|I0+q-2g(1))=Γ(q-2)Ag.
Consequently (3.16) becomes
(3.18)|cD0+σTu-cD0+σTv|≤‖u-v‖1(2-σ)Γ(2-σ)(Ag+Ah).
With the help of hypothesis (3.5) it yields
(3.19)|cD0+σTu-cD0+σTv|≤‖u-v‖.
Taking into account (3.12)–(3.19) we obtain
(3.20)‖Tu-Tv‖<‖u-v‖,
from here, the contraction principle ensures the uniqueness of solution for the fractional boundary value problem (P1). This finishes the proof.Now we give an existence result for the fractional boundary value problem (P1).Theorem 3.4.
Assume thatf(t,0,0)≠0 and there exist nonnegative functions k,h,g∈L1([0,1],ℝ+), ϕ,ψ∈C(ℝ+,ℝ+*) nondecreasing on ℝ+ and r>0, such that
(3.21)|f(t,x,x¯)|≤k(t)ψ(|x|)+h(t)ϕ(|x¯|)+g(t),a.e.(t,x)∈[0,1]×R.(3.22)(ψ(r)+ϕ(r)+1)(C1Γ(q-2)+C2(2-σ)Γ(2-σ))<r,
where C1=max{Ck,Ch,Cg}, C2=max{Ak,Ah,Ag}, Ch and Cg are defined as in Theorem 3.2 and
(3.23)Ck=‖I0+q-1k‖L1+|α|I0+q-2k(1)+I0+q-1k(1),Ak=2I0+q-1k(1)+|α|I0+q-2k(1).
Then the fractional boundary value problem (P1) has at least one nontrivial solution u*∈E.To prove this Theorem, we apply Leray-Schauder nonlinear alternative.Lemma 3.5 (see [17]).
LetF be a Banach space and Ω a bounded open subset of F, 0∈Ω. T:Ω¯→F be a completely continuous operator. Then, either there exists x∈∂Ω, λ>1 such that T(x)=λx, or there exists a fixed point x*∈Ω¯.Proof.
First let us prove thatT is completely continuous. It is clear that T is continuous since f and G are continuous. Let Br={u∈E,∥u∥≤r} be a bounded subset in E. We shall prove that T(Br) is relatively compact.
(i) Foru∈Br and using (3.21) we get
(3.24)|Tu(t)|≤1Γ(q-2)∫01|G(t,s)|[k(s)ψ(|u(s)|)+h(s)ϕ(|cD0+σu(s)|)+g(s)]ds.
Since ψ and ϕ are nondecreasing then (3.24) implies
(3.25)|Tu(t)|≤1Γ(q-2)∫01|G(t,s)|[k(s)ψ(‖u‖)+h(s)ϕ(‖u‖)+g(s)]ds≤1Γ(q-2)∫01|G(t,s)|[k(s)ψ(r)+h(s)ϕ(r)+g(s)]ds,
using similar techniques as to get (3.12) it yields
(3.26)|Tu(t)|≤1Γ(q-2)[ψ(r)(‖I0+q-1k‖L1+|α|I0+q-2k(1)+I0+q-1k(1))+ϕ(r)(‖I0+q-1h‖L1+|α|I0+q-2h(1)+I0+q-1h(1))+(‖I0+q-1g‖L1+|α|I0+q-2g(1)+I0+q-1g(1))]=1Γ(q-2)(Ckψ(r)+Chϕ(r)+Cg).
Hence
(3.27)|Tu(t)|≤C1Γ(q-2)[ψ(r)+ϕ(r)+1].
Moreover, we have(3.28)|(Tu(t))′|≤1Γ(q-2)[ψ(r)∫01|∂G∂t(t,s)|k(s)ds+ϕ(r)∫01|∂G∂t(t,s)|h(s)+∫01|∂G∂t(t,s)|g(s)ds].
Using (3.17) we obtain
(3.29)|cD0+σTu|≤C2(2-σ)Γ(2-σ)(ψ(r)+ϕ(r)+1).
From (3.27) and (3.29) we get
(3.30)‖Tu‖=(ψ(r)+ϕ(r)+1)(C1Γ(q-2)+C2(2-σ)Γ(2-σ)),
then T(Br) is uniformly bounded.
(ii)T(Br) is equicontinuous. Indeed for all t1, t2∈[0,1], t1<t2, u∈Br, let C=max(|f(t,u(t),cD0+σu(t))|, 0≤t≤1,∥u∥<r), therefore
(3.31)|Tu(t1)-Tu(t2)|≤CΓ(q-2)(∫0t1|G(t1,s)-G(t2,s)|ds+∫t1t2|G(t1,s)-G(t2,s)|ds+∫t21|G(t1,s)-G(t2,s)|ds),
that implies
(3.32)|Tu(t1)-Tu(t2)|≤CΓ(q-2)∫0t1(t2-s)q-1-(t1-s)q-1q(q-1)+(t2-t1)(|α|(1-s)3-q-(1-s)q-2q-1)ds+∫t1t2(t2-s)q-1q(q-1)+(t2-t1)(|α|(1-s)3-q-(1-s)q-2q-1)ds+∫t21(t1-t2)(|α|(1-s)3-q-(1-s)q-2q-1)ds.
Let us consider the functionΦ(x)=xq-1-(q-1)x, we see that Φ is decreasing on [0,1], consequently (t2-s)q-1-(t1-s)q-1≤(q-1)(t2-t1), from which we deduce
(3.33)|Tu(t1)-Tu(t2)|≤CΓ(q-2)[(t2-t1)∫0t11q(q-1)+|α(1-s)3-q-(1-s)q-2q-1|ds+∫t1t2(t2-s)q-1q(q-1)+(t2-t1)||α|(1-s)3-q-(1-s)q-2q-1|ds+(t1-t2)∫t21||α|(1-s)3-q-(1-s)q-2q-1|ds].
Some computations give
(3.34)|Tu(t1)-Tu(t2)|≤C(t2-t1)Γ(q-2)(1q(q-1)+3|α|(q-2)+3q-1)+CΓ(q-2)∫t1t2(t2-s)q-1q(q-1)ds.
On the other hand we have
(3.35)|cD0+σTu(t1)-cD0+σTu(t2)|≤+1Γ(2-σ)∫0t1((t1-s)-σ+1-(t2-s)-σ+1)|(Tu(s))′|ds+1Γ(2-σ)∫t1t2(t2-s)-σ+1|(Tu(s))′|ds.
Using (3.17) and (3.28) it yields
(3.36)|(Tu(t))′|≤[ψ(r)+ϕ(r)+1]C2,
then
(3.37)|cD0+σTu(t1)-cD0+σTu(t2)|≤[ψ(r)+ϕ(r)+1]C2(2-σ)Γ(2-σ)[2(t2-t1)2-σ+t22-σ-t12-σ],
when t1→t2, in (3.34) and (3.37) then |Tu(t1)-Tu(t2)| and |cD0+σTu(t1)-cD0+σTu(t2)| tend to 0. Consequently T(Br) is equicontinuous. From Arzelá-Ascoli Theorem we deduce that T is completely continuous operator.
Now we apply Leray Schauder nonlinear alternative to prove thatT has at least one nontrivial solution in E.
LettingΩ={u∈E:∥u∥<r}, for any u∈∂Ω, such that u=λTu, 0<λ<1, we get, with the help of (3.27),
(3.38)|u(t)|=λ|Tu(t)|≤|Tu(t)|≤C1Γ(q-2)[ψ(r)+ϕ(r)+1].
Taking into account (3.29) we obtain
(3.39)|cD0+σu(t)|≤C2(2-σ)Γ(2-σ)(ψ(r)+ϕ(r)+1).
From (3.38), (3.39), and (3.22) we deduce that
(3.40)‖u‖≤(ψ(r)+ϕ(r)+1)(C1Γ(q-2)+C2(2-σ)Γ(2-σ))<r,
this contradicts the fact that u∈∂Ω. Lemma 3.5 allows us to conclude that the operator T has a fixed point u*∈Ω¯ and then the fractional boundary value problem (P1) has a nontrivial solution u*∈E. The proof is complete.


## 4. Existence of Positive Solutions

In this section we investigate the positivity of solution for the fractional boundary value problem (P1), for this we make the following hypotheses.(H1)
f(t,u,v)=a(t)f1(u,v) where a∈C((0,1),(0,∞)) and f1∈C(ℝ+×ℝ,ℝ+).(H2)
0<∫01G(s,s)a(s)ds<∞.Now we give the properties of the Green function.Lemma 4.1.
LetG(t,s) be the function defined by (2.7). If α≥1 then G(t,s) has the following properties:(i)
G(t,s)∈C([0,1]×[0,1]), G(t,s)>0, for all t, s∈]0,1[.(ii)
Ift,s∈(τ,1), τ>0, then(4.1)0<τG(s,s)≤G(t,s)≤2τG(s,s).Proof.
(i) It is obvious thatG(t,s)∈C([0,1]×[0,1]), moreover, we have
(4.2)αt(1-s)3-q-t(1-s)q-2q-1=tq-1(1-s)3-q[(q-1)α-(1-s)]≥tq-1(1-s)3-q[α-1+s],
which is positive if α≥1. Hence G(t,s) is nonnegative for all t,s∈]0,1[.
(ii) Lett,s∈(τ,1), it is easy to see that G(s,s)≠0, then we have
(4.3)G(t,s)G(s,s)=(t-s)q-1(1-s)3-qqs[(q-1)α-(1-s)]+ts≤1+(1-s)2s≤2τ,0<τ≤s≤t<1,G(t,s)G(s,s)=ts≤2τ,0<τ≤t≤s≤1.
Now we look for lower bounds of G(t,s)(4.4)G(t,s)G(s,s)≥ts≥τs≥τ,0<τ≤s≤t<1,0<τ≤t≤s≤1.
Finally, since G(s,s) is nonnegative we obtain 0<τG(s,s)≤G(t,s)≤(2/τ)G(s,s).We recall the definition of positive of solution.Definition 4.2.
A functionu(t) is called positive solution of the fractional boundary value problem (P1) if u(t)≥0, for all t∈[0,1].Lemma 4.3.
Ifu∈E+ and α≥1, then the solution of the fractional boundary value problem (P1) is positive and satisfies
(4.5)min⁡t∈(τ,1)(u(t)+cD0+σu(t))≥τ22‖u‖.Proof.
First let us remark that under the assumptions onu and f, the function cD0+σu is nonnegative. From Lemma 3.1 we have
(4.6)u(t)=1Γ(q-2)∫01G(t,s)a(s)f1(u(s),cD0+σu(s))ds.
Applying the right-hand side of inequality (4.1) we get
(4.7)u(t)≤2τΓ(q-2)∫01G(s,s)a(s)f1(u(s),cD0+σu(s))ds.
Moreover, (4.1) gives
(4.8)cD0+σu(t)=1Γ(q-2)Γ(2-σ)∫0t∫01(t-s)-σ+1G1(s,r)×a(r)f1(u(r),cD0+σu(r))dsdr≤1τ(2-σ)Γ(q-2)Γ(2-σ)∫01G1(r,r)a(r)f1(u(r),cD0+σu(r))dr.
Combining (4.7) and (4.8) yields
(4.9)‖u‖≤2τΓ(q-2)∫01[G(s,s)+G1(s,s)(2-σ)Γ(2-σ)]a(s)f1(u(s),cD0+σu(s))ds,
hence
(4.10)∫01[G(s,s)+G1(s,s)(2-σ)Γ(2-σ)]a(s)f1(u(s),cD0+σu(s))ds≥τΓ(q-2)2‖u‖.
In view of the-left hand side of (4.1), we obtain for all t∈(τ,1)(4.11)u(t)≥τΓ(q-2)∫01G(s,s)a(s)f1(u(s),cD0+σu(s))ds,
on the other hand we have
(4.12)cD0+σu(t)≥τ2-σ(2-σ)Γ(q-2)Γ(2-σ)∫01G1(r,r)a(r)f1(u(r),cD0+σu(r))dr.
From (4.11) and (4.12) we get
(4.13)min⁡t∈(τ,1)(u(t)+cD0+σu(t))≥τΓ(q-2)∫01[G(s,s)+G1(s,s)(2-σ)Γ(2-σ)]×a(s)f1(u(s),cD0+σu(s))ds,
with the help of (4.10) we deduce
(4.14)min⁡t∈(τ,1)(u(t)+cD0+σu(t))≥τ22‖u‖.
The proof is complete.Define the quantitiesA0 and A∞ by (4.15)A0=lim⁡(|u|+|v|)→0f1(u,v)|u|+|v|,A∞=lim⁡(|u|+|v|)→∞f1(u,v)|u|+|v|.
The case A0=0 and A∞=∞ is called superlinear case and the case A0=∞ and A∞=0 is called sublinear case.The main result of this section is as follows.Theorem 4.4.
Under the assumption of Lemma4.3, the fractional boundary value problem (P1) has at least one positive solution in the both cases superlinear as well as sublinear.To prove Theorem4.4 we apply the well-known Guo-Krasnosel'skii fixed point theorem on cone.Theorem 4.5 (see [18]).
LetE be a Banach space, and let K⊂E, be a cone. Assume Ω1 and Ω2 are open subsets of E with 0∈Ω1, Ω1¯⊂Ω2 and let
(4.16)A:K∩(Ω2¯∖Ω1)⟶K
be a completely continuous operator such that(i)
||𝒜u||≤||u||, u∈K∩∂Ω1, and ||𝒜u||≥||u||, u∈K∩∂Ω2, or(ii)
||𝒜u||≥||u||, u∈K∩∂Ω1, and ||𝒜u||≤||u||, u∈K∩∂Ω2.Then 𝒜 has a fixed point in K∩(Ω2¯∖Ω1).Proof.
To prove Theorem4.4 we define the cone K by
(4.17)K={u∈E+,min⁡t∈(τ,1)(u(t)+cD0+σu(t))≥τ22‖u‖}.
It is easy to check that K is a nonempty closed and convex subset of E, hence it is a cone. Using Lemma 4.3 we see that TK⊂K. From the prove of Theorem 3.4, we know that T is completely continuous in E.
Let us prove the superlinear case. First, sinceA0=0, for any ε>0, there exists R1>0, such that
(4.18)f1(u,v)≤ε(|u|+|v|)
for 0<|u|+|v|≤R1. Letting Ω1={u∈E,∥u∥<R1}, for any u∈K∩∂Ω1, it yields
(4.19)Tu(t)=1Γ(q-2)∫01G(t,s)a(s)f1(u(s),cD0+σu(s))ds≤2ε‖u‖τΓ(q-2)∫01G(s,s)a(s)ds.
Moreover, we have
(4.20)cD0+σTu(t)=1Γ(q-2)Γ(2-σ)∫0t∫01(t-s)-σ+1G1(s,r)×a(r)f1(u(r),cD0+σu(r))dsdr≤1τ(2-σ)Γ(q-2)Γ(2-σ)∫01G1(r,r)a(r)(|u(r)|+|cD0+σu(r)|)dr≤ε‖u‖τ(2-σ)Γ(q-2)Γ(2-σ)∫01G1(s,s)a(s)ds.
From (4.19) and (4.20) we conclude
(4.21)‖Tu‖≤2ε‖u‖τΓ(q-2)∫01[G(s,s)+G1(s,s)(2-σ)Γ(2-σ)]a(s)ds.
In view of hypothesis (H2), one can chooseε such that
(4.22)ε≤τΓ(q-2)2∫01[G(s,s)+(G1(s,s)/(2-σ)Γ(2-σ))]a(s)ds.
The inequalities (4.21) and (4.22) imply that ||Tu||≤||u||, for all u∈K∩∂Ω1. Second, in view of A∞=∞, then for any M>0, there exists R2>0, such that f1(u,v)≥M(|u|+|v|) for |u|+|v|≥R2. Let R=max{2R1,(2R2/τ2)} and denote by Ω2 the open set {u∈E/||u||<R}. If u∈K∩∂Ω2 then
(4.23)min⁡t∈(τ,1)(u(t)+cD0+σu(t))≥τ22‖u‖=τ22R≥R2.
Using the left-hand side of (4.1) and Lemma 4.3, we obtain
(4.24)Tu(t)≥τMΓ(q-2)∫01G(s,s)a(s)(|u(s)|+|cD0+σu(s)|)ds≥τ3M||u||2Γ(q-2)∫01G(s,s)a(s)ds.
Moreover, we get with the help of (4.12)
(4.25)cD0+σTu(t)≥τ4-σM||u||2(2-σ)Γ(q-2)Γ(2-σ)∫01G1(s,s)a(s)ds.
In view of (4.26) and (4.24) we can write
(4.26)Tu(t)+cD0+σTu(t)≥τ3M||u||2Γ(q-2)∫01[G(s,s)+τ1-σG1(s,s)(2-σ)Γ(2-σ)]a(s)ds≥τ3M||u||2Γ(q-2)∫01[G(s,s)+G1(s,s)(2-σ)Γ(2-σ)]a(s)ds.
Let us choose M such that
(4.27)M≥2Γ(q-2)τ3∫01[G(s,s)+(G1(s,s)/(2-σ)Γ(2-σ))]a(s)ds,
then we get Tu(t)+cD0+σTu(t)≥||u||. Hence,
(4.28)||Tu||≥||u||,∀u∈K∩∂Ω2.
The first part of Theorem 4.5 implies that T has a fixed point in K∩(Ω̅2∖Ω1) such that R2≤||u||≤R. To prove the sublinear case we apply similar techniques. The proof is complete.In order to illustrate our results, we give the following examples.Example 4.6.
The fractional boundary value problem(4.29)cD0+5/2u=(t-110)3u+t2D0+5/4u+ln⁡t,0<t<1,u(0)=u′′(0)=0,u′(1)=-12u′′(1)
has a unique solution in E.Proof.
In this case we havef(t,x,y)=((t-1)/10)3x+t2y+lnt, 2<q=5/2<3, σ=(5/4)<2, α=-1/2 and
(4.30)|f(t,x,x¯)-f(t,y,y¯)|≤(1-t10)3|x-x¯|+t2|y-y¯|,
then g(t)=((1-t)/10)3 and h(t)=t2. Some calculus give
(4.31)‖I0+q-1g‖L1=0.17846×10-3,I0+q-1g(1)=0.48001×10-3,I0+q-2g(1)=016120×10-3,Cg=7.3907×10-4,Ag=1.0406×10-3,‖I0+q-1h‖L1=0.038210,I0+q-1h(1)=0.17194,I0+q-2h(1)=0.6018,Ch=0.49747,Ah=0.42448,Cg+Ch=0.49821<1,Ag+Ah=0.42552<(2-σ)Γ(2-σ)=0.91906.
Thus Theorem 3.2 implies that fractional boundary value problem (4.29) has a unique in E.Example 4.7.
The fractional boundary value problem(4.32)Dc0+7/3u=(1-t)2(u4100(1+u2)+ln⁡(1+(cD0+6/5u)2)+19)=0,0<t<1,u(0)=u′′(0)=0,u′(1)=32u′′(1)
has at least one nontrivial solution in E.Proof.
We apply Theorem3.4 to prove that the fractional boundary value problem (4.32) has at least one nontrivial solution. We have q=7/3, σ=6/5, α=3/2, and
(4.33)|f(t,x,x¯)|=(1-t)2x4100(1+x2)+(1-t)2ln⁡⁡(1+x2¯)9+(1-t)2≤|x|2100(1-t)2+(1-t)2ln⁡(1+x2¯)9+(1-t)2≤k(t)ψ(|x|)+h(t)ϕ(|x¯|)+g(t),
where k(t)=h(t)=g(t)=(1-t)2, ψ(x)=(x/10)2, ϕ(x¯)=ln(1+x2¯)/9, f(t,0,0)≠0. Let us find r such that (3.22) holds, for this we have
(4.34)‖I0+q-1g‖L1=0.19382,I0+q-2g(1)=0.15998,I0+q-1g(1)=0.33595,Cg=C1=0.76974,Ag=C2=0.91187.
We see that (3.22) is equivalent to 1.2664(((r2)/100)+(ln(1+r2)/9)+1)-r which is negative for r=6.



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*Source: 763456-2011-08-11.xml*