# Existence and Exact Asymptotic Behavior of Positive Solutions for a Fractional Boundary Value Problem

Authors: Habib Mâagli; Noureddine Mhadhebi; Noureddine Zeddini
Journal: Abstract and Applied Analysis
 (2013)

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## Abstract

We establish the existence and uniqueness of a positive solutionu for the fractional boundary value problem Dαu(x)=-a(x)uσ(x), x∈(0,1) with the condition limx→0Dα-1u(x)=0,u(1)=0, where 1<α≤2, σ∈(-1,1), and a is a nonnegative continuous function on (0,1) that may be singular at x=0 or x=1.


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## Body

## 1. Introduction

Fractional differential equations arise in various fields of science and engineering such as control, porous media, electrochemistry, viscoelasticity, and electromagnetism. They also serve as an excellent tool for the description of hereditary properties of various materials and processes (see[1–3]). In consequence, the subject of fractional differential equations is gaining much importance and attention. Motivated by the surge in the development of this subject, we consider the following problem:
(1)Dαu(x)=-a(x)uσ(x),x∈(0,1),limx→0+Dα-1u(x)=0,u(1)=0,
where 1<α≤2, -1<σ<1, a is a nonnegative continuous function on (0,1) that may be singular at x=0 or x=1 and Dα is the Riemann-Liouville fractional derivative. Then we study the existence and exact asymptotic behavior of positive solutions for this problem.We recall that for a measurable functionv, the Riemann-Liouville fractional integral Iβv and the Riemann-Liouville derivative Dβv of order β>0 are, respectively, defined by
(2)Iβv(x)=1Γ(β)∫0x(x-t)β-1v(t)dt,Dβv(x)=1Γ(n-β)(ddx)n∫0x(x-t)n-β-1v(t)dt=(ddx)nIn-βv(x),
provided that the right-hand sides are pointwise defined on (0,1]. Here n=[β]+1 and [β] means the integral part of the number β and Γ is the Euler Gamma function.Moreover, we have the following well-known properties (see [2, 4]):(i)
IβIγv(x)=Iβ+γv(x) for x∈[0,1], v∈L1((0,1]), β+γ≥1,(ii)
DβIβv(x)=v(x) for a.e. x∈[0,1], where v∈L1((0,1]), β>0,(iii)
Dβv(x)=0 if and only if v(x)=∑j=1n‍cjtβ-j, where n=[β]+1 and (c1,c2,…,cn)∈ℝn.Several results are obtained for fractional differential equation with different boundary conditions (see [5–15] and the references therein), but none of them deals with the existence of a positive solution for problem (1).Our aim in this paper is to establish the existence and uniqueness of a positive solutionu∈C2-α([0,1]) for problem (1) with a precise asymptotic behavior, where C2-α([0,1]) is the set of all functions f such that t→t2-αf(t) is continuous on [0,1]. Note that for 0<α<2, the solution u for problem (1) blows up at x=0.To state our result, we need some notations. We will use𝒦 to denote the set of Karamata functions L defined on (0,η] by
(3)L(t)∶=cexp(∫tηz(s)sds),
for some η>1, where c>0 and z∈C([0,η]) such that z(0)=0. It is clear that a function L is in 𝒦 if and only if L is a positive function in C1((0,η]) such that
(4)limt→0+tL′(t)L(t)=0.
For two nonnegative functions f and g defined on a set S, the notation f(x)≈g(x), x∈S, means that there exists c>0 such that (1/c)f(x)≤g(x)≤cf(x), for all x∈S. We denote also x+=max(x,0) for x∈ℝ.Throughout this paper we assume thata is nonnegative on (0,1) and satisfies the following condition.(
H
0
)  a∈C((0,1)) such that
(5)a(t)≈t-λL1(t)(1-t)-μL2(1-t),t∈(0,1),
where λ+(2-α)σ≤1, μ≤α, L1,L2∈𝒦 satisfying
(6)∫0ηL1(t)tλ+(2-α)σdt<∞,∫0ηL2(t)tμ-α+1dt<∞.
In the sequel, we introduce the function θ defined on (0,1) by
(7)θ(x)={1-xifμ<σ+α-1,(1-x)(∫1-xηL2(s)sds)1/(1-σ)ifμ=σ+α-1,(1-x)(α-μ)/(1-σ)(L2(1-x))1/(1-σ)ifσ+α-1<μ<α,(∫01-xL2(s)sds)1/(1-σ)ifμ=α.
Our main result is the following.Theorem 1.
Letσ∈(-1,1) and assume that a satisfies (H0). Then problem (1) has a unique positive solution u∈C2-α([0,1]) satisfying for x∈(0,1)(8)u(x)≈xα-2θ(x).This paper is organized as follows. Some preliminary lemmas are stated and proved in the next section, involving some already known results on Karamata functions. In Section3, we give the proof of Theorem 1.


## 2. Technical Lemmas

To keep the paper self-contained, we begin this section by recapitulating some properties of Karamata regular variation theory. The following is due to [16, 17].Lemma 2.
The following hold.(i)
LetL∈𝒦 and ε>0, then one has
(9)limt→0+tεL(t)=0.(ii)
LetL1, L2∈𝒦 and p∈ℝ. Then one has L1+L2∈𝒦, L1L2∈𝒦, and L1p∈𝒦.Example 3.
Letm∈ℕ*. Let c>0, (μ1,μ2,…,μm)∈ℝm and let d be a sufficiently large positive real number such that the function
(10)L(t)=c∏k=1m(logk(dt))-μk
is defined and positive on (0,η], for some η>1, where logkx=log∘log∘⋯∘logx (k times). Then L∈𝒦.Applying Karamata’s theorem (see [16, 17]), we get the following.Lemma 4.
Letμ∈ℝ and let L be a function in 𝒦 defined on (0,η]. One has the following.(i)
Ifμ<-1, then ∫0η‍sμL(s)ds diverges and ∫tη‍sμL(s)ds~t→0+-(t1+μL(t)/(μ+1)).(ii)
Ifμ>-1, then ∫0η‍sμL(s)ds converges and ∫0t‍sμL(s)ds~t→0+(t1+μL(t)/(μ+1)).Lemma 5.
LetL∈𝒦 be defined on (0,η]. Then one has
(11)limt→0+L(t)∫tη(L(s)/s)ds=0.
If further ∫0η‍(L(s)/s)ds converges, then one has
(12)limt→0+L(t)∫0t‍(L(s)/s)ds=0.Proof.
We distinguish two cases.
Case  1. We suppose that ∫0η‍(L(s)/s)ds converges. Since the function t→L(t)/t is nonincreasing in (0,ω], for some ω<η, it follows that, for each t≤ω, we have
(13)L(t)≤∫0tL(s)sds.
It follows that limt→0+L(t)=0. So we deduce (11).
Now put(14)φ(t)=L(t)t,fort∈(0,η).
Using that limt→0+(tφ′(t)/φ(t))=-1, we obtain
(15)∫0tφ(s)ds~t→0+-∫0tsφ′(s)ds=-tφ(t)+∫0tφ(s)ds.
This implies that
(16)∫0t‍L(s)sds~t→0+-L(t)+∫0tL(s)sds.
So (12) holds.
Case  2. We suppose that ∫0η‍(L(s)/s)ds diverges. We have, for some ω<η,
(17)∫tωφ(s)ds~t→0+tφ(t)-ωφ(ω)+∫tωφ(s)ds.
This implies that
(18)∫tωL(s)sds~t→0+L(t)-ωφ(ω)+∫tωL(s)sds.
This proves (11) and completes the proof.Remark 6.
LetL∈𝒦 defined on (0,η]; then, using (4) and (11), we deduce that
(19)t→∫tη‍L(s)sds∈𝒦.
If further ∫0η‍(L(s)/s)ds converges, we have by (11) that
(20)t→∫0tL(s)sds∈𝒦.Lemma 7.
Given1<α≤2 and f is such that the function t→(1-t)α-1f(t) is continuous and integrable on (0,1), then the boundary value problem
(21)Dαu(t)=-f(t),t∈(0,1),limx→0Dα-1u(x)=0,u(1)=0,
has a unique solution given by
(22)u(x)=Gαf(x)∶=∫01Gα(x,t)f(t)dt,
where
(23)Gα(x,t)=1Γ(α)[xα-2(1-t)α-1-((x-t)+)α-1],
is the Green function for the boundary value problem (21).Proof.
Sinceu0=-Iαf is a solution of the equation Dαu=-f, then Dα(u+Iαf)=0. Consequently there exist two constants c1, c2∈ℝ such that u(x)+Iαf(x)=c1xα-1+c2xα-2. Using the fact that limx→0Dα-1u(x)=0 and u(1)=0, we obtain c1=0 and c2=Iαf(1). So
(24)u(x)=1Γ(α)xα-2∫01(1-t)α-1f(t)dt-1Γ(α)∫0x(x-t)α-1f(t)dt=∫01Gα(x,t)f(t)dt.
In the following, we give some estimates on the function Gα. So, we need the following lemma.Lemma 8.
Forλ,μ∈(0,∞), and a,t∈[0,1], one has
(25)min(1,μλ)(1-atλ)≤1-atμ≤max(1,μλ)(1-atλ).Proposition 9.
On(0,1)×(0,1), one has
(26)Gα(x,t)≈xα-2(1-t)α-2(1-max(x,t)).Proof.
Forx,t∈(0,1)×(0,1), we have
(27)Gα(x,t)=(1-t)α-1xα-2Γ(α)[1-x((x-t)+x(1-t))α-1].
Since (x-t)+/x(1-t)∈(0,1) for t∈(0,1), then, by applying Lemma 8 with μ=α-1 and λ=1, we obtain
(28)Gα(x,t)≈xα-2(1-t)α-1(1-(x-t)+1-t)=xα-2(1-t)α-2(1-max(x,t)),
which completes the proof.In the sequel we put(29)b(t)=t-βL3(t)(1-t)-γL4(1-t),
where L3,L4∈𝒦 and we aim to give some estimates on x2-αGαb(x).Proposition 10.
Assume thatL3,L4∈𝒦,β≤1,γ≤α with
(30)∫0ηt-βL3(t)dt<∞,∫0ηtα-1-γL4(t)dt<∞.
Then for x∈(0,1),
(31)x2-αGαb(x)≈{1-xifγ<α-1,(1-x)∫1-xηL4(t)tdtifγ=α-1,(1-x)α-γL4(1-x)ifα-1<γ<α,∫01-xL4(t)tdtifγ=α.Proof.
Using Proposition9, we have
(32)x2-αGαb(x)≈(1-x)∫0x(1-t)α-2-γt-βL3(t)L4(1-t)dt+∫x1(1-t)α-1-γt-βL3(t)L4(1-t)dt=(1-x)I(x)+J(x).
For 0<x≤1/2, we use Lemma 4 and hypotheses (30) to deduce that
(33)I(x)≈{x1-βL3(x)ifβ<1,∫0x‍L3(t)tdtifβ=1,J(x)≈∫1/21(1-t)α-1-γL4(1-t)dt+∫x1/2t-βL3(t)dt≈1+∫x1/2t-βL3(t)dt≈1.
Hence, it follows from Lemma 2 and hypothesis (30) that, for 0<x≤1/2, we have
(34)x2-αGαb(x)≈1.
Now, for 1/2≤x<1, we use again Lemma 4 and hypothesis (30) to deduce that
(35)I(x)≈∫01/2t-βL3(t)dt+∫1/2x(1-t)α-2-γL4(1-t)dt≈1+∫1-x1/2tα-2-γL4(t)dt≈{1ifγ<α-1,∫1-xη‍L4(t)tdtifγ=α-1,(1-x)α-1-γL4(1-x)ifγ>α-1,J(x)≈∫01-xtα-1-γL4(t)dt≈{(1-x)α-γL4(1-x)ifγ<α,∫01-xL4(t)tdtifγ=α.
Hence, it follows from Lemmas 2 and 5 that, for x∈[1/2,1), we have
(36)x2-αGαb(x)≈{1-xifγ<α-1,(1-x)∫1-xηL4(t)tdtifγ=α-1,(1-x)α-γL4(1-x)ifα-1<γ<α,∫01-xL4(t)tdtifγ=α.
This together with (34) implies that (36) holds on (0,1).


## 3. Proof of Theorem1

We begin this section by giving a preliminary result that will play a crucial role in the proof of Theorem1.Proposition 11.
Assume that the functiona satisfies (H0) and put ω(t)=a(t)t(α-2)σθσ(t) for t∈(0,1). Then one has, for x∈(0,1),
(37)x2-αGαω(x)≈θ(x).Proof.
Fort∈(0,1), we have
(38)ω(t)=a(t)t(α-2)σθσ(t)={t-λ-(2-α)σ(1-t)-μ+σL1(t)L2(1-t)ifμ<σ+α-1,t-λ-(2-α)σ(1-t)-μ+σL1(t)L2(1-t)×(∫1-tη‍L2(s)sds)σ/(1-σ)ifμ=σ+α-1,t-λ-(2-α)σ(1-t)-(μ-σα)/(1-σ)L1(t)×(L2(1-t))σ/(1-σ)ifσ+α-1<μ<α,t-λ-(2-α)σ(1-t)-μL1(t)L2(1-t)×(∫01-tL2(s)sds)σ/(1-σ)ifμ=α.
So, we can see that
(39)ω(t)=t-β(1-t)-γL~1(t)L~2(1-t),
where β≤1, γ≤α and, according to Lemma 2, the functions t→L~1(t) and t→L~2(t) are in 𝒦. Moreover, using Lemma 4, we have ∫0η‍t-βL~1(t)dt<∞ and ∫0η‍t-γL~2(t)dt<∞. So the result follows from Proposition 10.Proof of Theorem1.
From Proposition11, there exists M>1 such that for each x∈(0,1)(40)1Mθ(x)≤x2-αGαω(x)≤Mθ(x),
where ω(t)=a(t)t(α-2)σθσ(t).
Putc0=M1/(1-|σ|) and let
(41)Λ={v∈C([0,1]):1c0θ≤v≤c0θ}.
In order to use a fixed point theorem, we denote a~(t)=a(t)t(α-2)σ and we define the operator T on Λ by
(42)Tv(x)=x2-αGα(a~vσ)(x).For this choice of c0, we can easily prove that for v∈Λ, we have Tv≤c0θ and Tv≥(1/c0)θ.
Now, we have(43)Tv(x)=x2-αΓ(α)∫01Gα(x,t)a~(t)vσ(t)dt=1Γ(α)∫01[(1-t)α-1-x2-α((x-t)+)α-1]×a~(t)vσ(t)dt.
Since the function (x,t)→(1-t)α-1-x2-α((x-t)+)α-1 is continuous on [0,1]×[0,1] and the function t→(1-t)α-1a~(t)θσ(t) is integrable on (0,1), we deduce that the operator T is compact from Λ to itself. It follows by the Schauder fixed point theorem that there exists v∈Λ such that Tv=v. Put u(x)=xα-2v(x). Then u∈C2-α([0,1]) and u satisfies the equation
(44)u(x)=Gα(auσ)(x).
Since the function t→(1-t)α-1a(t)uσ(t) is continuous and integrable on (0,1), then by Lemma 7 the function u is a positive continuous solution of problem (1).
Finally, let us prove thatu is the unique positive continuous solution satisfying (8). To this aim, we assume that (1) has two positive solutions u,v∈C2-α([0,1]) satisfying (8) and consider the nonempty set J={m≥1:1/m≤u/v≤m} and put c=infJ. Then c≥1 and we have (1/c)v≤u≤cv. It follows that uσ≤c|σ|vσ and consequently
(45)-Dα(c|σ|v-u)=a(c|σ|vσ-uσ)≥0,limt→0+Dα-1(c|σ|v-u)(t)=0,(c|σ|v-u)(1)=0.
Which implies by Lemma 7 that c|σ|v-u=Gα(a(c|σ|vσ-uσ))≥0. By symmetry, we obtain also that v≤c|σ|u. Hence c|σ|∈J and c≤c|σ|. Since |σ|<1, then c=1 and consequently u=v.Example 12.
Letσ∈(-1,1) and a be a positive continuous function on (0,1) such that
(46)a(t)≈(1-t)-μlog(31-t)-β,
where μ<α and β∈ℝ or μ=α and β>1. Then, using Theorem 1, problem (1) has a unique positive continuous solution u satisfying the following estimates.(i)
Ifμ<σ+α-1 or μ=σ+α-1 and β>1, then for x∈(0,1),
(47)u(x)≈xα-2(1-x).(ii)
Ifμ=σ+α-1andβ=1, then for x∈(0,1),
(48)u(x)≈xα-2(1-x)[log(log(31-x))]1/(1-σ).(iii)
Ifμ=σ+α-1andβ<1, then for x∈(0,1),
(49)u(x)≈xα-2(1-x)[log(31-x)](1-β)/(1-σ).(iv)
Ifσ+α-1<μ<α, then for x∈(0,1),
(50)u(x)≈xα-2(1-x)(α-μ)/(1-σ)[log(31-x)]-β/(1-σ).(v)
Ifμ=αandβ>1, then for x∈(0,1),
(51)u(x)≈xα-2[log(31-x)](1-β)/(1-σ).



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