# Positive and Nondecreasing Solutions to anm-Point Boundary Value Problem for Nonlinear Fractional Differential Equation

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

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## Abstract

We are concerned with the existence and uniqueness of a positive and nondecreasing solution for the following nonlinear fractionalm-point boundary value problem: D0+αu(t)+f(t,u(t))=0,0<t<1,2<α≤3,u(0)=u′(0)=0,u′(1)=∑i=1m-2aiu′(ξi), where D0+α denotes the standard Riemann-Liouville fractional derivative, f:[0,1]×[0,∞)→[0,∞) is a continuous function, ai≥0 for i=1,2,…,m-2, and 0<ξ1<ξ2<⋯<ξm-2<1. Our analysis relies on a fixed point theorem in partially ordered sets. Some examples are also presented to illustrate the main results.


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## Body

## 1. Introduction

Many papers and books on fractional differential equations have appeared recently. Most of them are devoted to the solvability of the linear fractional equation in terms of a special function (see, e.g., [1, 2]) and to problems of analyticity in the complex domain [3]. Moreover, Delbosco and Rodino [4] considered the existence of a solution for the nonlinear fractional differential equation(1.1)D0+αu=f(t,u),0<α<1,
where f:[0,a]×ℝ→ℝ, 0<a≤+∞, is a given continuous function in (0,a)×ℝ. They obtained existence results by using the Schauder fixed point theorem and the Banach contraction principle.Recently, El-Shahed [5] considered the following nonlinear fractional boundary value problem:(1.2)D0+αu+λa(t)f(t,u(t))=0,0<t<1,2<α≤3,u(0)=u′(0)=u′(1)=0.
They used the Krasnoselskii’s fixed point theorem on cone expansion and compression to show the existence and nonexistence of positive solutions for the above fractional boundary value problem.In [6], Liang and Zhang considered the following nonlinear fractional boundary value problem:(1.3)D0+αu(t)+f(t,u(t))=0,0<t<1,3<α≤4,u(0)=u′(0)=u′′(0)=u′′(1)=0,
and by means of lower and upper solution method and fixed point theorems, they obtained some results on the existence of positive solutions to the above boundary value problem.The question of uniqueness of the solution is not treated in [6].Recently, in [7] Caballero et al. studied the fractional boundary value problem appearing in [6] by using a fixed point theorem in partially ordered sets, and the authors obtained uniqueness of the solution.In this paper we discuss the existence and uniqueness of a positive and nondecreasing solution for the followingm-point nonlinear boundary value problem of fractional order:(1.4)D0+αu(t)+f(t,u(t))=0,0<t<1,2<α≤3,u(0)=u′(0)=0,u′(1)=∑i=1m-2aiu′(ξi),
where ai≥0 for i=1,2,…,m-2, 0<ξ1<ξ2<⋯<ξm-2<1, and, moreover, 0<∑i=1m-2aiξiα-2<1.Recently, this problem has been studied in [8, 9]. In [8] the author studies the existence and multiplicity of positive solutions for problem (1.4), and he uses Krasnoselskii and Leggett-Williams fixed point theorems. The question of uniqueness and monotonicity of the solution is not treated. In [9], the authors investigate the existence and uniqueness of positive and nondecreasing solutions for problem (1.4), and the main tools in this paper are a fixed point theorem in partially ordered sets and the lower and upper solution method.Our study is based on a different fixed point theorem in partially ordered sets than the one used in [9].Our main interest in this paper is to give an alternative answer to the main results of [8, 9].Existence of fixed points in partially ordered sets has been considered recently in [10–13], among others.For existence theorems for fractional differential equations and applications, we refer to the survey [14]. Concerning the definitions and basic properties, we refer the reader to [15].


## 2. Preliminaries and Previous Results

For the convenience of the reader, we present here some definitions, lemmas, and results that will be used in the proofs of our main results.Definition 2.1.
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 Euler gamma function given by
(2.2)Γ(α)=∫0+∞tα-1e-tdt,α>0.Definition 2.2.
The Riemann-Liouville fractional derivative of orderα>0 of a function f:(0,∞)→ℝ is defined by
(2.3)D0+αf(t)=1Γ(n-α)(ddt)n∫0tf(t)(t-s)α-n+1ds,
where n=[α]+1 and [α] denote the integer part of α.The following two lemmas can be found in [16, 17].Lemma 2.3.
Letα>0 and u∈C(0,1)∩L1(0,1). Then the fractional differential equation
(2.4)D0+αu(t)=0
has
(2.5)u(t)=c1tα-1+c2tα-2+⋯+cntα-n,ci∈R(i=1,2,…,n),n=[α]+1,
as unique solutions.Lemma 2.4.
Assume thatu∈C(0,1)∩L1(0,1) with a fractional derivative of order α>0 that belongs to C(0,1)∩L1(0,1). Then
(2.6)I0+αD0+αu(t)=u(t)+c1tα-1+c2tα-2+⋯+cntα-n,
for some ci∈ℝ(i=1,2,…,n) and n=[α]+1.Using Lemma2.4, in [8] the following result is proved.Lemma 2.5.
Givenf∈C[0,1], then the unique solution of
(2.7)D0+αu(t)+f(t)=0,0<t<1,2<α≤3,u(0)=u′(0)=0,u′(1)=∑i=1m-2aiu′(ξi)
is
(2.8)u(t)=∫01G(t,s)f(s)ds,
where Green’s function G(t,s) is given by
(2.9)G(t,s)=G1(t,s)+G2(t,s),taking(2.10)G1(t,s)={(1-s)α-2tα-1-(t-s)α-1Γ(α),0≤s≤t≤1,(1-s)α-2tα-1Γ(α),0≤t≤s≤1,G2(t,s)=1(1-ϱ)Γ(α)(ϱ(1-s)α-2-∑i=1m-2ai(ξi-s)α-2χEi(s))tα-1,
where ϱ=∑i=1m-2aiξiα-2 and χEi denotes the characteristic function of the set Ei=[0,ξi] for i=1,2,…,m-2.Remark 2.6.
Notice that Lemma 3 appears in [8] under assumption 2<α<3. In [9] the authors prove this result in the same way for 2<α≤3.The following result is proved in [8, 9].Lemma 2.7.
Under the assumptionai≥0 for i=1,2,…,m-2,0<ξ1<ξ2<⋯<ξm-2<1 and 1-ϱ>0 where ϱ=∑i=1m-2aiξiα-2, the Green’s function appearing in Lemma 2.5 satisfies G(t,s)≥0.Remark 2.8.
It is easily checked thatG(t,s) is a continuous function on [0,1]×[0,1].The following lemmas appear in [9].Lemma 2.9.
The functionG1(t,s) appearing in Lemma 2.5 is strictly increasing in the first variable.Lemma 2.10.
The functionG(t,s) appearing in Lemma 2.5 satisfies
(2.11)max⁡0≤t≤1∫01G(t,s)ds=1(α-1)Γ(α)[1α+11-ϱ(∑i=1m-2aiξiα-2(1-ξi))].For convenience, we will denote byL the constant(2.12)L=1(α-1)Γ(α)[1α+11-ϱ(∑i=1m-2aiξiα-2(1-ξi))].
In the sequel we present the fixed point theorem we will use later and which appears in [18].Firstly, we need to introduce the following class of functions. By𝒮 we denote the class of those functions β:[0,∞)→[0,1) satisfying β(tn)→1 implies tn→0.Theorem 2.11 (see [18]).
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 existsβ∈𝒮 such that(2.13)d(Tx,Ty)≤β(d(x,y))d(x,y),
for x,y∈X with y≥x.
Assume that eitherT is continuous or X is such that(2.14)if(xn)isanondecreasingsequencewithxn⟶xinX,thenxn≤x,∀n∈N.
Besides, if
(2.15)forx,y∈Xthereexistsz∈Xwhichiscomparabletoxandy,
then T has a unique fixed point.In our considerations, we will work in the spaceC[0,1]={x:[0,1]→ℝ,continuous}, with the standard distance given by d(x,y)=max0≤t≤1{|x(t)-y(t)|} for x,y∈C[0,1].Notice that this space can be equipped with a partial order given by(2.16)x≤y⟺x(t)≤y(t),
for x,y∈C[0,1] and t∈[0,1].In [12] it is proved that (C[0,1],≤) with the above-mentioned distance satisfies condition (2.14) of Theorem 2.11. Moreover, for x,y∈C[0,1], as the function max{x,y} is continuous in [0,1], (C[0,1],≤) satisfies condition (2.15) of Theorem 2.11.


## 3. Main Result

Our starting point in this section is to present the class of functions𝒜 which we use later. By 𝒜 we will denote the class of those functions ϕ:[0,∞)→[0,∞) satisfying the following conditions:(C1)
ϕ is nondecreasing;(C2)
For anyx>0, ϕ(x)<x;(C3)
ϕ(x)/x∈𝒮, where 𝒮  is the class of functions appearing in Section 2.Examples of functions in𝒜 are ϕ(x)=μx with 0≤μ<1,ϕ(x)=x/1+x and ϕ(x)=ln(1+x).In what follows, we formulate our main result.Theorem 3.1.
Suppose that the following assumptions are satisfied:(a)
f:[0,1]×[0,∞)→[0,∞) is continuous;(b)
f(t,x) is nondecreasing with respect to the second variable for each t∈[0,1];(c)
there exists0<λ≤1/L and ϕ∈𝒜 such that(3.1)f(t,y)-f(t,x)≤λϕ(y-x),
for x,y∈[0,∞) with y≥x and t∈[0,1].
Then problem (1.4) has a unique nonnegative and nondecreasing solution.Proof.
Consider the coneP={u∈C[0,1]:u≥0}.
Notice that, asP is a closed set of C[0,1], P is a complete metric space with the distance given by d(x,y)=max0≤t≤1{|x(t)-y(t)|} satisfying conditions (2.14) and (2.15) of Theorem 2.11.
Now, foru∈P we define the operator T by(3.2)(Tu)(t)=∫01G(t,s)f(s,u(s))ds,
where G(t,s) is Green’s function defined in Section 2.
By Lemma2.7, Remark 2.8, and (a), it is clear that T applies the cone P into itself.
Now, we will check that assumptions in Theorem2.11 are satisfied.
Firstly, the operatorT is nondecreasing.
In fact, by assumption (b), foru≥v we have(3.3)(Tu)(t)=∫01G(t,s)f(s,u(s))ds≥∫01G(t,s)f(s,v(s))ds=(Tv)(t).
On the other hand, for u≥v and u≠v and taking into account our hypotheses, we can obtain
(3.4)d(Tu,Tv)=max⁡0≤t≤1{|(Tu)(t)-(Tv)(t)|}=max⁡0≤t≤1[(Tu)(t)-(Tv)(t)]=max⁡0≤t≤1[∫01G(t,s)(f(s,u(s))-f(s,v(s)))ds]≤max⁡0≤t≤1[∫01G(t,s)λϕ(u(s)-v(s))ds].
Since ϕ is not nondecreasing and taking into account Lemma 2.10 and assumption (c), we get
(3.5)d(Tu,Tv)≤λϕ(‖u-v‖)max⁡0≤t≤1∫01G(t,s)ds=λϕ(d(u,v))L≤ϕ(d(u,v))=ϕ(d(u,v))d(u,v)d(u,v).
Thus, for u≥v and u≠v,
(3.6)d(Tu,Tv)≤β(d(u,v))d(u,v),
where β(x)=ϕ(x)/x∈𝒮.
Obviously, the last inequality is satisfied foru=v.
Thus condition (2.13) in Theorem 2.11 holds with β(x)=ϕ(x)/x. Moreover, since f and G are nonnegative functions,(3.7)(T0)(t)=∫01G(t,s)f(s,0)ds≥0.
Finally, Theorem 2.11 tells us that problem (1.4) has a unique nonnegative solution u(t).In what follows, we will prove that the unique nonnegative solutionu(t) for problem (1.4) is nondecreasing.In fact, by Lemmas2.5 and 2.9 it is easily seen that G(t,s) is strictly increasing and, since u(t) is a fixed point of the operator T, we have(3.8)u(t)=∫01G(t,s)f(s,u(s))ds.
These facts and the nonnegative character of f give us that u(t) is nondecreasing.Now, we present a sufficient condition for the existence and uniqueness of a positive and strictly increasing solution for problem (1.4) (positive solution means a solution satisfying x(t)>0 for t∈(0,1)). The proof of this fact is similar to the proof of Theorem 3.6 of [18]. We present it for completeness.Theorem 3.2.
Under assumptions of Theorem3.1 and adding the following assumption (d)
f(t0,0)≠0 for certain t0∈[0,1], one obtains existence and uniqueness of a positive and strictly increasing solution for problem (1.4).Proof.
Consider the nonnegative solutionx(t) for Problem (1.4) whose existence is guaranteed by Theorem 3.1.
Notice thatx(t) satisfies(3.9)x(t)=∫01G(t,s)f(s,x(s))ds.
Firstly, we will prove that x(t)>0 for t∈(0,1).
In fact, in contrary case we can find0<t*<1 such that x(t*)=0. Consequently,(3.10)x(t*)=∫01G(t*,s)f(s,x(s))ds=0.
As x≥0,G(t,s)≥0, and f is nondecreasing with respect to the second variable (assumption (b)), from the last expression we can get
(3.11)0=x(t*)=∫01G(t*,s)f(s,x(s))ds≥∫01G(t*,s)f(s,0)ds≥0
and, thus,
(3.12)∫01G(t*,s)f(s,0)ds=0.
This fact and the nonnegative character of the functions G(t*,s) and f(t,0) imply
(3.13)G(t*,s)f(s,0)=0a.e.(s).
By Lemma 2.9, and since G(0,s)=0 and G(t*,s)>0 for any s∈[0,1], we have
(3.14)f(s,0)=0a.e.(s).
On the other hand, assumption (d) gives us that f(t0,0)≠0 for certain t0∈[0,1] and, thus, f(t0,0)>0.
This fact and the continuity off imply the existence of a set A⊂[0,1] with t0∈A and μ(A)>0, where μ is the Lebesgue measure, such that f(t,0)>0 for any t∈A.
This contradicts (3.14).
Therefore,x(t)>0 for t∈(0,1).
In the sequel, we will show thatx(t) is strictly increasing.
In fact, sinceG(0,s)=0, we have(3.15)x(0)=∫01G(0,s)f(s,x(s))ds=0.
Now, we take t1,t2∈[0,1] with t1<t2.
We can consider two cases.
Case 1 (t=0).
Suppose thatx(t2)=0.
Using a similar argument similar o the one used in the proof of the positive character ofx(t), we obtain a contradiction.Case 2 (t1>0).
Suppose thatx(t1)=x(t2).
In this case, we have(3.16)∫01(G(t1,s)-G(t2,s))f(s,x(s))ds=0.
SinceG(t1,s)-G(t2,s)>0 (Lemma 2.3), we get(3.17)f(s,x(s))=0a.e.(s).
Again, the same reasoning that we use earlier gives us a contradiction.
Therefore,x(t1)<x(t2).
This finished the proof.Remark 3.3.
In Theorem3.2, the condition f(t0,0)≠0 for certain t0∈[0,1] seems to be a strong condition in order to obtain a positive solution for problem (1.4), but when the solution is unique, we will see that this condition is very adjusted one. In fact, suppose that problem (1.4) has a unique nonnegative solution x(t), then we have
(3.18)f(t,0)=0foranyt∈[0,1]iffx(t)≡0.
Indeed, if f(t,0)=0 for any t∈[0,1], then it is easily seen that the zero function is a solution for problem (1.4) and the uniqueness of solution gives us x(t)≡0.
The reverse implication is obvious.Remark 3.4.
Notice that assumptions in Theorem3.1 are invariant by nonnegative and continuous perturbations. More precisely, if f(t,0)=0 for any t∈[0,1] and f satisfies conditions (a), (b), and (c) of Theorem 3.1, then g(t,x)=a(t)+f(t,x), where a:[0,1]→[0,∞)  continuously  and a≠0, satisfies assumptions of Theorem 3.2, and this means that the boundary value problem
(3.19)D0+αu(t)+g(t,u(t))=0,0<t<1,2<α≤3,u(0)=u′(0)=0,u′(1)=∑i=1m-2aiu′(ξi),
with ai≥0 for i=1,2,…,m-2,0<ξ1<ξ2<⋯<ξm-2<1, and 0<∑i=1m-2aiξiα-2<,1 has a unique positive and strictly increasing solution.In the sequel, we present an example which illustrates our results.Example 3.5.
Consider the following boundary value problem:(3.20)D0+5/2u(t)+t+λu(t)1+u(t)=0,0<t<1,λ>0,u(0)=u′(0)=0,u′(1)=(12)u′(14)+(14)u′(12).
In this case, α=5/2, f(t,u)=t+(λu/(1+u)), 0<ξ1=1/4<ξ2=1/2<1, a1=1/2, and a2=1/4. Besides, ϱ=(1/2)(1/2)+(1/4)(1/2)≃0.4267767, and
(3.21)L=1(3/2)Γ(5/2)[25+11-ϱ(12(14)1/234+14(12)1/212)]≃0.44197,
and 1/L=2.2626.
It is easily seen thatf(t,u) satisfies condition (a) of Theorem 3.1.
Since∂f/∂u=λ/(1+u)2>0 for u∈[0,∞), f(t,u) satisfies (b) of Theorem 3.1.
Moreover, foru≥v and t∈[0,1], we have(3.22)f(t,u)-f(t,v)=t+λu1+u-λ-tv1+v=λ(u1+u-v1+v)=λ(u-v(1+u)(1+v))≤λu-v1+u-v=λϕ(u-v),
where ϕ(x)=x/1+x. It is easily proved that ϕ belongs to the class 𝒜. Since f(t,0)=t≠0 for t≠0, Theorem 3.2 states that problem (3.20) has a unique positive and strictly increasing solution for 0<λ≤1/L≃2.2626.


## 4. Some Remark

In [9] the authors consider our problem (1.4) and they prove the following result.Theorem 4.1.
Problem (1.4) has a unique positive and strictly increasing with solution u(t) if the following conditions are satisfied:(1)
f:[0,1]×[0,∞)→[0,∞) is continuous and nondecreasing with respect to the second variable and f(t,u(t))≠0 for t∈Z⊂[0,1] with μ(Z)>0 (μ denotes the Lebesgue measure),(2)
there exists0<λ<1/L such that for u,v∈[0,∞) with u≥v and t∈[0,1](4.1)f(t,u)-f(t,v)≤λln⁡(u-v+1).The main tool used by the authors in [9] for the proof of Theorem 4.1 is a fixed point theorem in partially ordered sets which appears in [19]. This fixed point theorem uses the following class of functions ℳ.Byℳ we denote the class of functions ϕ:[0,∞)→[0,∞) continuously such that if φ(x)=x-ϕ(x), then the following conditions are satisfied:(1)
φ:[0,∞)→[0,∞), and it is nondecreasing;(2)
φ(0)=0;(3)
φ is positive in (0,∞).The same proof used by the authors in [9] gives us that the conclusion of Theorem 4.1 is true if we replace condition 2 of Theorem 4.1 by the following:(2′) there exists 0<λ<1/L such that for u,v∈[0,∞) with u≥v and t∈[0,1](4.2)f(t,u)-f(t,v)≤λϕ(u-v),
where ϕ∈ℳ.In what follows we will prove that the classes of functions𝒜 (see Section 3) and ℳ are not comparable.Example 4.2.
This example appears in [10].
Consider the functionϕ:[0,∞)→[0,∞) defined by(4.3)ϕ(t)={0,0≤t≤2,2t-4,2<t≤3,(23)t,3<t.
It is easily proved that ϕ∈𝒜.
On the other hand, it is easily seen thatφ(t)=t-ϕ(t) is not increasing, and, consequently, ϕ∉ℳ.Example 4.3.
Consider the functionϕ:[0,∞)→[0,∞) given by ϕ(x)=x-arctanx.
It is easily proved thatϕ∈ℳ.
On the other hand, sinceβ(x)=ϕ(x)/x=1-(arctanx)/x and β(tn)→1 when tn→∞, we have that ϕ∉𝒜.Examples4.2 and 4.3 tell us that the results of this paper cover cases which cannot be treated by Theorem 4 which appears in [9] and vice versa.In [9] the following result is also proved ([9, Theorem 5.1]).Theorem 4.4.
Problem (1.4) has a positive solution u(t) if the following conditions are satisfied:(Hf)
f(t,u)∈C([0,1]×[0,∞),[0,∞)) is nondecreasing relative to u, f(t,γ(t))≢0 for t∈(0,1), and there exists a positive constant μ<1 such that(4.4)Kμf(t,u)≤f(t,Ku),
for K∈[0,1], where
(4.5)γ(t)=∫01G(t,s)ds+1(1-ϱ)Γ(α)[∫01ϱ(1-s)α-2ds-∑i=1m-2ai∫01(ξi-s)α-2χEi(s)ds]tα-1.In the sequel, we present an example which can be treated by Theorem3.2, and it cannot be studied by Theorem 4.4.Example 4.5.
Consider the fractional boundary value problem(4.6)D0+5/2u(t)+(t2+1)(λu(t)+c)=0,0<t<1,c>0,0<λ<1,u(0)=u′(0)=0,u′(1)=(12)u′(14)+(14)u′(12).
In this case, α=5/2,f(t,u)=(t2+1)(λu+c),0<ξ1=1/4<ξ2=1/2<1, a1=1/2 and a2=1/4. Besides, ϱ=(1/2)(1/4)1/2+(1/4)(1/2)1/2≃0.4267767.It is easily seen thatf(t,u) satisfies condition (a) of Theorem 3.1. Since ∂f/∂u=λ(t2+1)>0, f(t,u) satisfies (b) of Theorem 3.1.Moreover, foru≥v and t∈[0,1], we have(4.7)f(t,u)-f(t,v)=λ(t2+1)(u-v)≤2λ(u-v)=2ϕ(u-v),
where ϕ(x)=λx. It is easily checked that ϕ∈𝒜.Sincef(t,0)=c(t2+1)≠0 for t≠0 and, as 2≤(1/L)=2.2626 (see Example 3.5), Theorem 3.2 gives us existence and uniqueness of a positive and strictly increasing solution for Problem (4.6).On the other hand, we will show that condition(Hf) appearing in Theorem 4.4 is not satisfied.In fact, suppose that there exists0<μ<1 such that(4.8)Kμf(t,u)≤f(t,Ku),
for any K∈[0,1].Sincef(t,u)=(t2+1)(λu+c) with 0<λ<1 and c>0, we have(4.9)Kμ≤f(t,Ku)f(t,u)=(t2+1)(ϱKu+c)(t2+1)(ϱu+c)=ϱKu+cϱu+c.
Taking limit when u→∞ in the last expression, we get Kμ≤K, which is false, since 0<μ<1 and the function h(α)=Kα is decreasing when 0<K<1.Therefore, Problem (4.6) can be covered by Theorem 3.2, and it cannot be treated using Theorem 4.4.In [8], the author proves the following result.Theorem 4.6.
Assume that(H1)
f:[0,1]×[0,∞)→[0,∞) is continuous,(H2)
ai≥0  for i=1,2,…,m-2, 0<ξ1<ξ2<⋯<ξm-2<1 and ϱ=∑i=1m-2aiξiα-2 with ϱ<1.In addition, suppose that one of the following two conditions holds:(H3)
limu→0mint∈[0,1]f(t,u)/u=∞,limu→∞maxt∈[0,1]f(t,u)/u=0,(H4)
limu→0mint∈[0,1]f(t,u)/u=0,limu→∞maxt∈[0,1]f(t,u)/u=∞.Then Problem (1.4) has at least one positive solution.In what follows we present an example which can be treated by our results and it cannot be studied by Theorem4.6.Example 4.7.
Consider Problem (3.20) which appears in Example 3.5.
We prove that this example can be treated by Theorem3.2 and we obtained a unique positive and strictly increasing solution for this problem when 0<λ<1/L≃2.2626.
On the other hand, in this case, sincef(t,u)=t+(λu/(1+u)), we have(4.10)min⁡t∈[0,1](t+(λu/(1+u))u)=λu/(1+u)u=λ1+u,lim⁡u→0min⁡t∈[0,1](t+(λu/(1+u))u)=lim⁡u→0λ1+u=λ.
Since 0<λ<∞, Theorem 4.6 cannot be used in this case.Example 4.8.
Consider the following boundary value problem(4.11)D0+5/2u(t)+c+λarctan⁡u(t)=0,0<t<1,c>0,λ>0,u(0)=u′(0)=0,u′(1)=(12)u′(14)+(14)u′(12).
In this case, α=5/2,f(t,u)=c+λarctanu,0<ξ1=1/4<ξ2=1/2<1, a1=1/2, and a2=1/4. Besides, ϱ=(1/2)(1/2)+(1/4)(1/2)≃0.4267767.It is easily seen thatf(t,u) satisfies conditions (a) and (b) of Theorem 3.1.Moreover, in [20] it is proved that if u≥v≥0,(4.12)arctan⁡u-arctan⁡v≤arctan(u-v).
Using this fact, for u≥v and t∈[0,1], we have(4.13)f(t,u)-f(t,v)=λ(arctan⁡u-arctan⁡v)≤λarctan(u-v)=λϕ(u-v),
where ϕ(x)=arctanx.In [20] it is proved that ϕ∈𝒜.Sincef(t,0)=c≠0, Theorem 3.2 gives us the existence and uniqueness of a positive and strictly increasing solution for Problem (4.11) when 0<λ≤1/L≃2.2626.On the other hand, since(4.14)lim⁡u→0min⁡t∈[0,1]f(t,u)u=lim⁡u→0min⁡t∈[0,1]c+λarctan⁡uu=lim⁡u→0c+λarctan⁡uu=∞,lim⁡u→∞max⁡t∈[0,1]f(t,u)u=lim⁡u→∞c+λarctan⁡uu=0,
Problem (4.11) can be treated by Theorem 4.6, and we obtain the existence of at least one positive solution.Our main contribution is that for0<λ≤1/L≃2.2626 we obtain uniqueness and strictly increasing character for the solution of Problem (4.11).



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*Source: 826580-2012-01-03.xml*