# Existence and Uniqueness of Positive Solutions for a Singular Fractional Three-Point Boundary Value Problem

Authors: I. J. Cabrera; J. Harjani; K. B. Sadarangani
Journal: Abstract and Applied Analysis
 (2012)

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## Abstract

We investigate the existence and uniqueness of positive solutions for the following singular fractional three-point boundary value problemD0+αu(t)+f(t,u(t))=0, 0<t<1,u(0)=u′(0)=u′′(0)=0,u′′(1)=βu′′(η), where 3<α≤4, D0+α is the standard Riemann-Liouville derivative and f:(0,1]×[0,∞)→[0,∞) with limt→0+f(t,·)=∞ (i.e., f is singular at t=0). Our analysis relies on a fixed point theorem in partially ordered metric spaces.


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## Body

## 1. Introduction

Fractional differential equations have been of great interest recently. It is caused both by the intensive development of the theory of fractional calculus itself and by the applications (see, e.g., [1–5]).Recently, many papers have appeared dealing with the existence of solutions of nonlinear fractional boundary value problems.In [6], the authors studied the existence and multiplicity of positive solutions for the boundary value problem:
(1.1)D0+αu(t)+f(t,u(t))=0,0<t<1,u(0)=u(1)=0,
where 1<α≤2 and f:[0,1]×[0,∞)→[0,∞) is continuous, by using some fixed point theorem on cones.In [7], the authors considered the following nonlinear fractional boundary value problem:
(1.2)D0+αu(t)+f(t,u(t))=0,0<t<1,u(0)=u′(0)=u′′(0)=u′′(1)=0,
where 3<α≤4 and f:[0,1]×[0,∞)→[0,∞) is continuous. They obtained their results by using lower and upper solution method and fixed point theorems.In [8] the authors investigated the existence and uniqueness of positive and nondecreasing solutions for a class of singular fractional boundary value problems by using a fixed point theorem in partially ordered metric spaces.Very recently, in [9] the authors studied the existence of solutions of the following three-point boundary value problem:
(1.3)D0+αu(t)+f(t,u(t))=0,0<t<1,u(0)=u′(0)=u′′(0)=0,u′′(1)=βu′′(η),
where 3<α≤4, 0<η<1, 0<βηα-3<1 and f:[0,1]×[0,∞)→[0,∞) is continuous.Motivated by [8, 9], in this paper we discuss the existence and uniqueness of positive solutions for Problem (1.3) assuming that f:(0,1]×[0,∞)→[0,∞) is such that limt→0+f(t,·)=∞ (i.e., f is singular at t=0). Our main tool is a fixed point theorem in partially ordered metric spaces which appears in [10].


## 2. Preliminaries and Basic Facts

For the convenience of the reader, we present some notations and lemmas which will be used in the proof of our results.Definition 2.1 (see [5]).
The Riemann-Liouville fractional integral of orderα>0 of a function f:(0,∞)→ℝ is given by
(2.1)I0+αf(t)=1Γ(α)∫0t‍(t-s)α-1f(s)ds,
provided that the right-hand side is pointwise defined on (0,∞) and where Γ(α) denotes the classical gamma function.Definition 2.2 (see [2]).
The Riemann-Liouville fractional derivative of orderα>0 of a function f:(0,∞)→ℝ is defined as
(2.2)D0+αf(t)=1Γ(n-α)(ddt)n∫0t‍f(s)(t-s)α-n+1ds,
where n=[α]+1 and [α] denotes the integer part of α.The following two lemmas can be found in [2] and they are crucial in finding an integral representation of the boundary value problem (1.3).Lemma 2.3 (see [2]).
Assume thatu∈C(0,1)∩L1(0,1) and α>0.
Then the fractional differential equation(2.3)D0+αu(t)=0
has
(2.4)u(t)=c1tα-1+c2tα-2+⋯+cntα-n,
where ci∈ℝ,i=1,2,…,n and n=[α]+1, as unique solution.Lemma 2.4 (see [2]).
Assume thatu∈C(0,1)∩L1(0,1) with fractional derivative of order α>0 that belongs to C(0,1)∩L1(0,1). Then
(2.5)I0+αD0+αu(t)=u(t)+c1tα-1+c2tα-2+⋯+cntα-n,
for some ci∈ℝ,i=1,2,…,n and n=[α]+1.By using Lemma2.4, in [9] the authors proved the following result.Lemma 2.5 (see [9]).
Let0<η<1 and β≠1/ηα-3 and h∈C[0,1].
Then the boundary value problem(2.6)D0+αu(t)+h(t)=0,0<t<1,u(0)=u′(0)=u′′(0)=0,u′′(1)=βu′′(η),
where 3<α≤4, has as unique solution
(2.7)u(t)=∫01‍G(t,s)h(s)ds+βtα-1(α-1)(α-2)(1-βηα-3)∫01‍H(η,s)h(s)ds,
where
(2.8)G(t,s)={tα-1(1-s)α-3-(t-s)α-1Γ(α),0≤s≤t≤1,tα-1(1-s)α-3Γ(α),0≤t≤s≤1,H(t,s)=∂2G(t,s)∂t2={(α-1)(α-2)Γ(α)[tα-3(1-s)α-3-(t-s)α-3],0≤s≤t≤1,(α-1)(α-2)Γ(α)tα-3(1-s)α-3,0≤t≤s≤1.Remark 2.6.
In [9] it is proved that G is a continuous function on [0,1]×[0,1], G(t,s)≥0, G(t,1)=0 and
(2.9)sup⁡0≤t≤1∫01‍G(t,s)ds=2(α-2)Γ(α+1),∫01‍H(η,s)ds=ηα-3(α-1)(1-η)Γ(α).In the sequel, we present the fixed point theorem which we will use later. Previously, we present the following class of functions.By𝒮 we denote the class of functions β:[0,∞)→[0,1) satisfying the following condition:
(2.10)β(tn)⟶1impliestn⟶0.
Examples of functions belonging to 𝒮 are β(t)=kt with 0≤k<1 and β(t)=1/(1+t).The fixed point theorem which we will use later appears in [10].Theorem 2.7 (see [10]).
Let(X,≤) be a partially ordered set and suppose that there exists a metric d in X such that (X,d) is a complete metric space. Let T:X→X be a nondecreasing mapping such that there exists an element x0∈X with x0≤Tx0. Suppose that there exists β∈𝒮 such that
(2.11)d(Tx,Ty)≤β(d(x,y))d(x,y)forx,y∈Xwithx≥y.
Assume that either T is continuous or X is such that
(2.12)if(xn)isanondecreasingsequenceinXsuchthatxn⟶xthenxn≤x∀n∈N.
Besides, if
(2.13)forallx,y∈Xthereexistsz∈Xwhichiscomparabletox,y,
then T has a unique fixed point.In our considerations, we will work in the Banach spaceC[0,1]={x:[0,1]→ℝ,xiscontinuous} with the classical metric given by d(x,y)=sup0≤t≤1|x(t)-y(t)|.Notice that this space can be equipped with a partial order given by(2.14)x,y∈C[0,1],x≤y⟺x(t)≤y(t)foranyt∈[0,1].
In [11] it is proved that (C[0,1],≤) satisfies condition (2.12) of Theorem 2.7. Moreover, for x,y∈C[0,1], as the function max{x,y}∈C[0,1], (C[0,1],≤) satisfies condition (2.13).


## 3. Main Result

Our starting point of this section is the following lemma.Lemma 3.1.
Suppose that0<σ<1, 3<α≤4, and F:(0,1]→ℝ  is a continuous function with limt→0+F(t)=∞. If tσF(t) is a continuous function on [0,1] then the function defined by
(3.1)L(t)=∫01‍G(t,s)F(s)ds
is continuous on [0,1], where G(t,s) is the Green’s function appearing in Lemma 2.5.Proof.
We divide the proof into three cases.
Case1 (t0=0). It is clear that L(0)=0.
SincetσF(t) is a continuous function on [0,1], we can find a constant M>0 such that |tσF(t)|≤M for any t∈[0,1].
Then, we get(3.2)|L(t)-L(0)|=|L(t)|=|∫01‍G(t,s)F(s)ds|=|∫01‍G(t,s)s-σsσF(s)ds|=|∫0t‍tα-1(1-s)α-3-(t-s)α-1Γ(α)s-σsσF(s)ds+∫t1‍tα-1(1-s)α-3Γ(α)s-σsσF(s)ds|=|∫01‍tα-1(1-s)α-3Γ(α)s-σsσF(s)ds-∫0t‍(t-s)α-1Γ(α)s-σsσF(s)ds|≤|∫01‍tα-1(1-s)α-3Γ(α)s-σsσF(s)ds|+|∫0t‍(t-s)α-1Γ(α)s-σsσF(s)ds|≤Mtα-1Γ(α)∫01‍(1-s)α-3s-αds+MΓ(α)∫0t‍(t-s)α-1s-αds=Mtα-1Γ(α)∫01‍(1-s)α-3s-αds+Mtα-1Γ(α)∫0t‍(1-st)α-1s-αds=Mtα-1Γ(α)β(1-σ,α-2)+Mtα-1Γ(α)∫0t‍(1-st)α-1s-αds.
If in the integral ∫0t‍(1-(s/t))α-1s-σds we use the change of variables u=s/t then we have
(3.3)∫0t‍(1-st)α-1s-σds=t1-σ∫01‍(1-u)α-1u-σdu=t1-σβ(1-σ,α).
This and (3.2) give us
(3.4)|L(t)|≤Mtα-1Γ(α)β(1-σ,α-2)+Mtα-σΓ(α)β(1-σ,α)
and letting t→0, we see that |L(t)|→0.
This proves the continuity ofL at t0=0.
Case2(t0∈(0,1)). We take tn→t0 and we have to prove that L(tn)→L(t0).
Without loss of generality, we can taketn>t0 (the same argument works for tn<t0).
In fact,(3.5)|L(tn)-L(t0)|=|∫0tn‍tnα-1(1-s)α-3-(tn-s)α-1Γ(α)s-σsσF(s)ds+∫tn1‍tnα-1(1-s)α-3Γ(α)s-σsσF(s)ds-∫0t0‍t0α-1(1-s)α-3-(t0-s)α-1Γ(α)s-σsσF(s)ds-∫t01‍t0α-1(1-s)α-3Γ(α)s-σsσF(s)ds|=|∫01‍tnα-1(1-s)α-3Γ(α)s-σsσF(s)ds-∫0tn‍(tn-s)α-1Γ(α)s-σsσF(s)ds-∫01‍t0α-1(1-s)α-3Γ(α)s-σsσF(s)ds+∫0t0‍(t0-s)α-1Γ(α)s-σsσF(s)ds|=|∫01‍(tnα-1-t0α-1)(1-s)α-3Γ(α)s-σsσF(s)ds-∫0t0‍(tn-s)α-1-(t0-s)α-1Γ(α)s-σsσF(s)ds-∫t0tn‍(tn-s)α-1Γ(α)s-σsσF(s)ds|≤M(tnα-1-t0α-1)Γ(α)∫01‍(1-s)α-3s-σds+MΓ(α)∫0t0‍((tn-s)α-1-(t0-s)α-1)s-σds+MΓ(α)∫t0tn‍(tn-s)α-1s-σds≤M(tnα-1-t0α-1)Γ(α)β(1-σ,α-2)+MΓ(α)In1+MΓ(α)In2,
where
(3.6)In1=∫0t0‍((tn-s)α-1-(t0-s)α-1)s-σds,In2=∫t0tn‍(tn-s)α-1s-σds.
In the sequel, we will prove that In1→0 when n→∞.
In fact, astn→t0, then
(3.7)((tn-s)α-1-(t0-s)α-1)s-σ⟶0whenn⟶∞.
Moreover,
(3.8)|((tn-s)α-1-(t0-s)α-1)s-σ|≤(|tn-s|α-1+|t0-s|α-1)s-σ≤2s-σ
and, as
(3.9)∫01‍2s-σds=2[s-σ+1-σ+1]01=21-σ<∞,
we have that the sequence ((tn-s)α-1-(t0-s)α-1)s-σ converges pointwise to the zero function and |((tn-s)α-1-(t0-s)α-1)s-σ| is bounded by a function belonging to L1[0,1], then by Lebesgue’s dominated convergence theorem
(3.10)In1⟶0whenn⟶∞.
Now, we will prove that In2→0 when n→∞.
In fact, as(3.11)In2=∫t0tn‍(tn-s)α-1s-σds≤∫t0tn‍s-σds=11-σ(tn1-σ-t01-σ)
and letting n→∞ and, taking into account that tn→t0, from the last expression we get
(3.12)In2⟶0whenn⟶∞.
Finally, from (3.5), (3.10), and (3.12) we get
(3.13)|L(tn)-L(t0)|⟶0whenn⟶∞.
This proves the continuity of L at t0.
Case3 (t0=1). It is easily checked that L(1)=0.
Now, following the same lines that in the proof of Case 1, we can demonstrate the continuity ofL at t0=1.
This finishes the proof.Lemma 3.2.
Suppose that0<σ<1, 3<α≤4, 0<βηα-3<1, and F:(0,1]→ℝ  is  a continuous function with limt→0+F(t)=∞.
IftσF(t) is a continuous function on [0,1] then the function defined by
(3.14)N(t)=βtα-1(α-1)(α-2)(1-βηα-3)∫01‍H(η,s)F(s)ds
is continuous on [0,1], where H(t,s) is the function appearing in Lemma 2.5.Proof.
SincetσF(t) is continuous on [0,1], there exists a constant M>0 such that |tσF(t)|≤M for any t∈[0,1].
Taking into account that(3.15)|H(t,s)|≤2(α-1)(α-2)Γ(α)
we have
(3.16)|∫01‍H(η,s)F(s)ds|=|∫01‍H(η,s)s-σsσF(s)ds|≤2M(α-1)(α-2)Γ(α)∫01‍s-σds=2M(α-1)(α-2)Γ(α)(1-α)<∞,
and, consequently, the function N is continuous at any point t∈[0,1].Remark 3.3.
Notice that the functionH(t,s) appearing in Lemma 2.5 which is defined as
(3.17)H(t,s)={(α-1)(α-2)Γ(α)[tα-3(1-s)α-3-(t-s)α-3],0≤s≤t≤1,(α-1)(α-2)Γ(α)tα-3(1-s)α-3,0≤t≤s≤1,
is continuous function on [0,1]×[0,1] and, moreover, H(t,s)≥0.
In fact, for0≤t≤s≤1 it is clear that H(t,s)≥0.
In the case,0≤s≤t≤1, we have
(3.18)H(t,s)=(α-1)(α-2)Γ(α)[tα-3(1-s)α-3-(t-s)α-3]=(α-1)(α-2)Γ(α)[(t-ts)α-3-(t-s)α-3]≥0.
This proves the nonnegative character of the function H on [0,1]×[0,1].Lemma 3.4.
Suppose that0<σ<1. Then
(3.19)sup⁡0≤t≤1∫01‍G(t,s)s-σds=1Γ(α)(β(1-σ,α-2)-β(1-σ,α)),
where G(t,s) is the function appearing in Lemma 2.5.Proof.
By definition ofG(t,s), we have
(3.20)∫01‍G(t,s)s-σds=∫0t‍G(t,s)s-σds+∫t1‍G(t,s)s-σds=∫0t‍tα-1(1-s)α-3-(t-s)α-1Γ(α)s-σds+∫t1‍tα-1(1-s)α-3Γ(α)s-σds=∫01‍tα-1(1-s)α-3Γ(α)s-σds-∫0t‍(t-s)α-1Γ(α)s-αds=tα-1Γ(α)∫01‍(1-s)α-3s-σds-1Γ(α)∫0t‍(t-s)α-1s-σds.
As we saw in Case 1 of Lemma 3.1.
(3.21)∫0t‍(t-s)α-1s-σds=tα-σΓ(α)β(1-σ,α)
and, therefore,
(3.22)∫0t‍G(t,s)s-σds=tα-1Γ(α)β(1-σ,α-2)-tα-σΓ(α)β(1-σ,α).
Now, using elemental calculus it is easily seen that the function
(3.23)φ(t)=β(1-σ,α-2)Γ(α)tα-1-β(1-σ,α)Γ(α)tα-σ
is increasing on the interval [0,1] and, therefore,
(3.24)sup⁡0≤t≤1∫01‍G(t,s)s-σds=sup⁡0≤t≤1φ(t)=φ(1)=1Γ(α)(β(1-σ,α-2)-β(1-σ,α)).Lemma 3.5.
Suppose that0<σ<1 then
(3.25)∫01‍H(η,s)s-σds=(α-1)(α-2)Γ(α)(ηα-3-ηα-σ-2)β(1-σ,α-2),
where H(t,s) is the function appearing in Lemma 2.5.Proof.
By definition ofH(t,s), we have
(3.26)∫01‍H(η,s)s-σds=∫0η‍H(η,s)s-σds+∫η1‍H(η,s)s-σds=∫0η‍(α-1)(α-2)Γ(α)[ηα-3(1-s)α-3-(η-s)α-3]s-σds+∫η1‍(α-1)(α-2)Γ(α)ηα-3(1-s)α-3s-σds=∫01‍(α-1)(α-2)Γ(α)ηα-3(1-s)α-3s-σds-∫0η‍(α-1)(α-2)Γ(α)(η-s)α-3s-σds=(α-1)(α-2)Γ(α)ηα-3∫01‍(1-s)α-3s-σds-(α-1)(α-2)Γ(α)∫0η‍(η-s)α-3s-σds=(α-1)(α-2)Γ(α)ηα-3β(1-σ,α-2)-(α-1)(α-2)Γ(α)∫0η‍(η-s)α-3s-σds.
By a similar argument that the one used in the Case 1 of Lemma 3.1, we have
(3.27)∫01‍H(η,s)s-σds=(α-1)(α-2)Γ(α)ηα-3β(1-σ,α-2)-(α-1)(α-2)Γ(α)ηα-σ-2β(1-σ,α-2)=(α-1)(α-2)Γ(α)(ηα-3-ηα-σ-2)β(1-σ,α-2).
This finishes the proof.By commodity, we denote byK the constant given by
(3.28)K=1Γ(α)[(1+β(ηα-3-ηα-σ-2)1-βηα-3)β(1-σ,α-2)-β(1-σ,α)].
Moreover, we introduce the following class of functions which will be used in the main result of the paper. By 𝒜 we denote the class of functions ϕ:[0,∞)→[0,∞) satisfying the following:  (i)
ϕis nondecreasing.(ii)
ϕ(x)<x for any x>0.(iii)
β(x)=ϕ(x)/x∈𝒮, where 𝒮 is the class of functions introduced in Remark 2.6.Theorem 3.6.
Let0<σ<1, 3<α≤4, 0<η<1, 0<βηα-3<1, and f:(0,1]×[0,∞)→[0,∞)  is  a continuous function with limt→0+f(t,·)=∞ and such that tσf(t,y) is a continuous function on [0,1]×[0,∞). Assume that there exists 0<λ≤1/K such that for x,y∈[0,∞) with y≥x and t∈[0,1](3.29)0≤tσ(f(t,y)-f(t,x))≤λϕ(y-x),
where ϕ∈𝒜.
Then Problem (1.3) has a unique positive solution (this means that x(t)>0 for t∈(0,1)).Proof.
Consider the cone:(3.30)P={u∈C[0,1]:u(t)≥0}.
Since P is a closed set of C[0,1], P is a complete metric space with the distance given by d(u,v)=sup0≤t≤1|u(t)-v(t)|, for u,v∈P.
It is easily checked thatP satisfies conditions (2.12) and (2.13) of Theorem 2.7.
Now, foru∈P we define the operator T by
(3.31)(Tu)(t)=∫01‍G(t,s)f(s,u(s))ds+βtα-1(α-1)(α-2)(1-βηα-3)∫01‍H(η,s)f(s,u(s))ds.
By Lemmas 3.1 and 3.2, for u∈P we have Tu∈C[0,1].
Moreover, in view of the nonnegative character ofG(t,s), H(η,s), and f(s,x), we have that Tu∈P for u∈P.
In what follows, we check that assumptions in Theorem2.7 are satisfied.
Firstly, we will prove thatT is nondecreasing.
In fact, by (3.29), for u≥v we have
(3.32)(Tu)(t)=∫01‍G(t,s)f(s,u(s))ds+βtα-1(α-1)(α-2)(1-βηα-3)∫01‍H(η,s)f(s,u(s))ds=∫01‍G(t,s)s-σsσf(s,u(s))ds+βtα-1(α-1)(α-2)(1-βηα-3)∫01‍H(η,s)s-σsσf(s,u(s))ds≥∫01‍G(t,s)s-σsσf(s,v(s))ds+βtα-1(α-1)(α-2)(1-βηα-3)∫01‍H(η,s)s-σsσf(s,v(s))ds=∫01‍G(t,s)f(s,v(s))ds+βtα-1(α-1)(α-2)(1-βηα-3)∫01‍H(η,s)f(s,v(s))ds=(Tv)(t).
This proves that T is a nondecreasing operator.
On the other hand, foru≥v and u≠v, we have
(3.33)d(Tu,Tv)=sup⁡0≤t≤1|(Tu)(t)-(Tv)(t)|=sup⁡0≤t≤1((Tu)(t)-(Tv)(t))=sup⁡0≤t≤1[∫01‍G(t,s)(f(s,u(s))-f(s,v(s)))ds+βtα-1(α-1)(α-2)(1-βηα-3)∫01‍H(η,s)(f(s,u(s))-f(s,v(s)))ds]≤sup⁡0≤t≤1∫01‍G(t,s)s-σsσ(f(s,u(s))-f(s,v(s)))ds+β(α-1)(α-2)(1-βηα-3)∫01‍H(η,s)s-σsσ(f(s,u(s))-f(s,v(s)))ds≤sup⁡0≤t≤1∫01‍G(t,s)s-σλ(ϕ(u(s)-v(s)))ds+β(α-1)(α-2)(1-βηα-3)∫01‍H(η,s)s-σλ(ϕ(u(s)-v(s)))ds.
Since ϕ is nondecreasing, the last inequality implies
(3.34)d(Tu,Tv)≤λϕ(d(u,v))sup⁡0≤t≤0∫01‍G(t,s)s-σds+β(α-1)(α-2)(1-βηα-3)λϕ(d(u,v))∫01‍H(η,s)s-σds=λϕ(d(u,v))[sup⁡0≤t≤0∫01‍G(t,s)s-σds+β(α-1)(α-2)(1-βηα-3)∫01‍H(η,s)s-σds].
Now, from Lemmas 3.4 and 3.5 it follows:
(3.35)d(Tu,Tv)≤λϕ(d(u,v))[1Γ(α)(β(1-σ,α-2)-β(1-σ,α))+β(α-1)(α-2)(1-βηα-3)×(α-1)(α-2)Γ(α)(ηα-3-ηα-σ-2)β(1-σ,α-2)]=λϕ(d(u,v))[1Γ(α)(β(1-σ,α-2)-β(1-σ,α))+β(ηα-3-ηα-σ-2)(1-βηα-3)Γ(α)β(1-σ,α-2)]=λϕ(d(u,v))[1Γ(α)[(1+β(ηα-3-ηα-σ-2)1-βηα-3)β(1-σ,α-2)-β(1-σ,α)]]=λϕ(d(u,v))K.
Since 0<λ≤1/K, from the last inequality we obtain
(3.36)d(Tu,Tv)≤λϕ(d(u,v))K≤ϕ(d(u,v)),
and, since u≠v,
(3.37)d(Tu,Tv)≤ϕ(d(u,v))d(u,v)d(u,v)=β(d(u,v))d(u,v).
Since this inequality is obviously satisfied for u=v, we have
(3.38)d(Tu,Tv)≤β(d(u,v))d(u,v)foranyu,v∈Pwithu≥v.
Finally, since the zero function satisfies 0≤T0, Theorem 2.7 says us that the operator T has a unique fixed point in P, or, equivalently, Problem (1.3) has a unique nonnegative solution x in C[0,1].
Now, we will prove thatx is a positive solution.
In contrary case, we can find0<t*<1 such that x(t*)=0.
Taking into account that the nonnegative solutionx of Problem (1.3) is a fixed point of the operator, we have
(3.39)x(t)=∫01‍G(t,s)f(s,x(s))ds+βtα-1(α-1)(α-2)(1-βηα-3)∫01‍H(η,s)f(s,x(s))ds
and, particularly,
(3.40)x(t*)=∫01‍G(t*,s)f(s,x(s))ds+βt*α-1(α-1)(α-2)(1-βηα-3)∫01‍H(η,s)f(s,x(s))ds=0.
Since both summands in the right hand are nonnegative (see Remarks 2.6 and 3.3) we have
(3.41)∫01‍G(t*,s)f(s,x(s))ds=0,∫01‍H(η,s)f(s,x(s))ds=0.
Given the nonnegative character of G(t,s), H(η,s), and f(s,u), we have
(3.42)G(t*,s)f(s,x(s))=0a.e.(s),H(η,s)f(s,x(s))=0a.e.(s).
Taking into account that limt→0+f(t,0)=∞, this means that for M>0 we can find δ>0 such that for s∈[0,1]∩(0,δ) we have f(s,0)>M. Notice that [0,1]∩(0,δ)⊂{s∈[0,1]:f(s,x(s))>M} and μ([0,1]∩(0,δ))>0, where μ is the Lebesgue measure on [0,1].
This and (3.42) give us that
(3.43)G(t*,s)=0a.e.(s),H(η,s)=0a.e.(s),
and this is a contradiction since G(t*,s) and H(η,s) are rational functions in the variable s.
Therefore,x(t)>0 for t∈(0,1).
This finishes the proof.In order to present an example which illustrates our results, we need to prove some properties about the hyperbolic tangent function.Previously, we recalled some definitions.Definition 3.7.
A functionf:[0,∞)→[0,∞) is said to be subadditive if it satisfies
(3.44)f(x+y)≤f(x)+f(y)foranyx,y∈[0,∞).
An example of subadditive function is the square root function, that is, f(x)=x.Remark 3.8.
Suppose thatf:[0,∞)→[0,∞) is subadditive and y≤x then
(3.45)f(x)-f(y)≤f(x-y).
In fact, since(3.46)f(x)=f(x-y+y)≤f(x-y)+f(y)this inequality implies that(3.47)f(x)-f(y)≤f(x-y).
Recall that a functionf:[0,∞)→[0,∞) is concave if for any x,y∈[0,∞) and λ∈[0,1].
(3.48)f(λx+(1-λ)y)≥λf(x)+(1-λ)f(y).Lemma 3.9.
Letf:[0,∞)→[0,∞) be a concave function with f(0)=0. Then f is subadditive.Proof.
We takex,y∈[0,∞).
Sincef is concave and f(0)=0, we get
(3.49)f(x)=f(yx+y0+xx+y(x+y))≥yx+yf(0)+xx+yf(x+y)=xx+yf(x+y),f(y)=f(xx+y0+yx+y(x+y))≥xx+yf(0)+yx+yf(x+y)=yx+yf(x+y).
Adding these inequalities, we have
(3.50)f(x)+f(y)≥xx+yf(x+y)+yx+yf(x+y)=f(x+y).
This proves the lemma.In what follows, we will prove that the function(3.51)f(x)=tanhx=e2x-1e2x+1
belongs to the class 𝒜 previously defined.Lemma 3.10.
The functionf:[0,∞)→[0,∞) defined as
(3.52)f(x)=tanhx=e2x-1e2x+1
satisfies:  (a)
f∈𝒜.(b)
f is subadditive.Proof.
(a) Sincef′(x)=4e2x/(e2x+1)2>0 for x>0, f is nondecreasing.
Moreover, the function(3.53)g(x)=x-tanhx=x-e2x-1e2x+1
has as derivative
(3.54)g′(x)=(e2x-1)2(e2x+1)2>0forx>0
and, consequently, g is strictly nondecreasing on (0,∞).
Sinceg(0)=0, we have 0=g(0)<g(x) for x>0 or, equivalently, f(x)=tanhx<x for x>0.
In order to prove thatβ(x)=tanhx/x∈𝒮, notice that if β(tn)→1 then the sequence (tn) is a bounded sequence.
In fact, in contrary casetn→∞ and we have
(3.55)β(tn)=tanhtntn⟶0
which contradicts the fact that β(tn)→1.
Now, we suppose thatβ(tn)→1 and tn↛0.
Then, we can findε>0 such that for each n∈ℕ there exists ϱn≥n with tϱn≥ε.
Since(tn) is a bounded sequence (because β(tn)→1) we can find a subsequence of (tϱn), which we will denote of the same way, such that tϱn→a.
Asβ(tn)→1, it follows that
(3.56)β(tϱn)=tanhtϱntϱn⟶tanhaa=1
and, as the unique solution of the equation tanhx=x on [0,∞) is x0=0, we deduce that a=0.
Therefore,tϱn→0 and this implies that there exists n0∈ℕ such that tϱn<ε for n≥n0.
This contradicts the fact thattϱn≥ε for any n∈ℕ.
Therefore,tn→0.
This proves thatβ(x)=tanhx/x∈𝒮.
Therefore,f∈𝒜.
(b) Sincetanh0=0 and
(3.57)(tanhx)′′=8e2x(1-e2x)(e2x+1)3<0forx>0,
this means that f(x)=tanhx is a concave function with tanh0=0 and, by Lemma 3.9, f(x)=tanhx is subadditive.Remark 3.11.
By Remark3.8 and by (b) of Lemma 3.9, for x,y∈[0,∞) with y≤x(3.58)tanhx-tanhy≤tanh(x-y).Now, we present an example which illustrates our result.Example 3.12.
Consider the following singular fractional boundary value problem(3.59)D0+7/2u(t)+λ(t2+1)tanhu(t)t1/2=0,0<t<1,u(0)=u′(0)=u′′(0)=0,u′′(1)=u′′(14).
In this case, σ=1/2, η=1/4, β=1 and α=7/2.
Moreover, in this casef(t,u)=λ(t2+1)tanhu/t1/2 for (t,u)∈(0,1]×[0,∞).
Notice thatf is continuous in (0,1]×[0,∞) and limt→0+f(t,·)=∞.
Now, we check thatf(t,u) satisfies assumptions appearing in Theorem 3.6.
It is clear thatt1/2f(t,u)=λ(t2+1)tanhu is a continuous function on [0,1]×[0,∞).
Moreover, by Lemma3.10 and Remark 3.11, for u≥v and t∈[0,1] we have
(3.60)0≤t1/2(f(t,u)-f(t,v))=λ(t2+1)(tanhu-tanhv)≤λ(t2+1)tanh(u-v)≤2λtanh(u-v),
where f(x)=tanhx is a function belonging to 𝒜 (see, Lemma 3.10).
Finally, Theorem3.6 says us that Problem (3.59) has a unique positive solution for
(3.61)2λ≤1K=Γ(7/2)(1+((1/4)1/2-1/4)/(1-(1/4)1/2))β(1/2,3/2)-β(1/2,7/2)=307π.
Or, equivalently, for λ≤15/7π.



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*Source: 803417-2012-08-21.xml*