AAA Abstract and Applied Analysis 1687-0409 1085-3375 Hindawi Publishing Corporation 420514 10.1155/2013/420514 420514 Research Article Existence and Exact Asymptotic Behavior of Positive Solutions for a Fractional Boundary Value Problem Mâagli Habib Mhadhebi Noureddine 0000-0001-5763-0538 Zeddini Noureddine Bai Chuanzhi King Abdulaziz University, Rabigh Campus College of Sciences and Arts Department of Mathematics P.O. Box 344, Rabigh 21911 Saudi Arabia kau.edu.sa 2013 31 12 2013 2013 04 11 2012 25 12 2012 2013 Copyright © 2013 Habib Mâagli et al. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

We establish the existence and uniqueness of a positive solution u for the fractional boundary value problem Dαu(x)=-a(x)uσ(x), x(0,1) with the condition limx0Dα-1u(x)=0,  u(1)=0, where 1<α2, σ(-1,1), and a is a nonnegative continuous function on (0,1) that may be singular at x=0 or x=1.

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). 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)Dα  u(x)=-a(x)uσ(x),x(0,1),limx0+Dα-1u(x)=0,u(1)=0, where 1<α2, -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 function v, 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)n0x(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]):

IβIγv(x)=Iβ+γv(x) for x[0,1], vL1((0,1]), β+γ1,

DβIβv(x)=v(x) for a.e. x[0,1], where vL1((0,1]), β>0,

Dβv(x)=0 if and only if v(x)=j=1ncjtβ-j, where n=[β]+1 and (c1,c2,,cn)n.

Several results are obtained for fractional differential equation with different boundary conditions (see  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 solution uC2-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 tt2-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 use 𝒦 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 zC([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)limt0+tL(t)L(t)=0. For two nonnegative functions f and g defined on a set S, the notation f(x)g(x), xS, means that there exists c>0 such that (1/c)f(x)g(x)cf(x), for all xS. We denote also x+=max(x,0) for x.

Throughout this paper we assume that a is nonnegative on (0,1) and satisfies the following condition.

( H 0 )   aC((0,1)) such that (5)a(t)t-λL1(t)(1-t)-μL2(1-t),  t(0,1), where λ+(2-α)σ1, μα, L1,L2𝒦 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.

Let σ(-1,1) and assume that a satisfies (H0). Then problem (1) has a unique positive solution uC2-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 Section 3, 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.

Let L𝒦 and ε>0, then one has (9)limt0+tεL(t)=0.

Let L1, L2𝒦 and p. Then one has L1+L2𝒦, L1L2𝒦, and L1p𝒦.

Example 3.

Let m*. Let c>0, (μ1,μ2,,μm)m and let d be a sufficiently large positive real number such that the function (10)L(t)=ck=1m(logk(dt))-μk is defined and positive on (0,η], for some η>1, where logkx=logloglogx (k times). Then L𝒦.

Applying Karamata’s theorem (see [16, 17]), we get the following.

Lemma 4.

Let μ and let L be a function in 𝒦 defined on (0,η]. One has the following.

If μ<-1, then 0ηsμL(s)ds diverges and tηsμL(s)ds~t0+-(t1+μL(t)/(μ+1)).

If μ>-1, then 0ηsμL(s)ds converges and 0tsμL(s)ds~t0+(t1+μL(t)/(μ+1)).

Lemma 5.

Let L𝒦 be defined on (0,η]. Then one has (11)limt0+L(t)tη(L(s)/s)ds=0. If further 0η(L(s)/s)ds converges, then one has (12)limt0+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 tL(t)/t is nonincreasing in (0,ω], for some ω<η, it follows that, for each tω, we have (13)L(t)0tL(s)sds. It follows that limt0+L(t)=0. So we deduce (11).

Now put (14)φ(t)=L(t)t,fort(0,η). Using that limt0+(tφ(t)/φ(t))=-1, we obtain (15)0tφ(s)ds~t0+-0tsφ(s)ds=-tφ(t)+0tφ(s)ds. This implies that (16)0tL(s)sds~t0+-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~t0+tφ(t)-ωφ(ω)+tωφ(s)ds. This implies that (18)tωL(s)sds~t0+L(t)-ωφ(ω)+tωL(s)sds. This proves (11) and completes the proof.

Remark 6.

Let L𝒦 defined on (0,η]; then, using (4) and (11), we deduce that (19)ttηL(s)sds𝒦. If further 0η(L(s)/s)ds converges, we have by (11) that (20)t0tL(s)sds𝒦.

Lemma 7.

Given 1<α2 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),limx0Dα-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.

Since u0=-Iαf is a solution of the equation Dαu=-f, then Dα(u+Iαf)=0. Consequently there exist two constants c1, c2 such that u(x)+Iαf(x)=c1xα-1+c2xα-2. Using the fact that limx0Dα-1u(x)=0 and u(1)=0, we obtain c1=0 and c2=Iαf(1). So (24)u(x)=1Γ(α)xα-201(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.

For λ,μ(0,), and a,t[0,1], one has (25)min(1,μλ)(1-atλ)1-atμmax(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.

For x,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,L4𝒦 and we aim to give some estimates on x2-αGαb(x).

Proposition 10.

Assume that L3,L4𝒦,β1,γα 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 Proposition 9, 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<x1/2, we use Lemma 4 and hypotheses (30) to deduce that (33)I(x){x1-βL3(x)ifβ<1,0xL3(t)t  dtifβ=1,J(x)1/21(1-t)α-1-γL4(1-t)dt+x1/2t-βL3(t)dt1+x1/2t-βL3(t)dt1. Hence, it follows from Lemma 2 and hypothesis (30) that, for 0<x1/2, we have (34)x2-αGαb(x)1. Now, for 1/2x<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)dt1+1-x1/2tα-2-γL4(t)dt{1ifγ<α-1,1-xηL4(t)t  dtifγ=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 Theorem <xref ref-type="statement" rid="thm1.1">1</xref>

We begin this section by giving a preliminary result that will play a crucial role in the proof of Theorem 1.

Proposition 11.

Assume that the function a satisfies (H0) and put ω(t)=a(t)t(α-2)σ  θσ(t) for t(0,1). Then one has, for x(0,1), (37)x2-α  Gαω(x)θ(x).

Proof.

For t(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 β1, γα and, according to Lemma 2, the functions tL~1(t) and tL~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 Theorem <xref ref-type="statement" rid="thm1.1">1</xref>.

From Proposition 11, 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).

Put c0=M1/(1-|σ|) and let (41)Λ={vC([0,1]):1c0θvc0θ}. 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 vΛ, we have Tvc0θ 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 vΛ such that Tv=v. Put u(x)=xα-2v(x). Then uC2-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 that u is the unique positive continuous solution satisfying (8). To this aim, we assume that (1) has two positive solutions u,vC2-α([0,1]) satisfying (8) and consider the nonempty set J={m1:1/mu/vm} and put c=infJ. Then c1 and we have (1/c)vucv. It follows that uσc|σ|vσ and consequently (45)-Dα(c|σ|v-u)=a(c|σ|  vσ-uσ)0,limt0+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 vc|σ|u. Hence c|σ|J and cc|σ|. Since |σ|<1, then c=1 and consequently u=v.

Example 12.

Let σ(-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 or μ=α and β>1. Then, using Theorem 1, problem (1) has a unique positive continuous solution u satisfying the following estimates.

If μ<σ+α-1 or μ=σ+α-1 and β>1, then for x(0,1), (47)u(x)xα-2(1-x).

If μ=σ+α-1  and  β=1, then for x(0,1), (48)u(x)xα-2(1-x)[log(log(31-x))]1/(1-σ).

If μ=σ+α-1  and  β<1, then for x(0,1), (49)u(x)xα-2(1-x)[log(31-x)](1-β)/(1-σ).

If σ+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).

If μ=α  and  β>1, then for x(0,1), (51)u(x)xα-2[log(31-x)](1-β)/(1-σ).

Acknowledgment

The authors thank the anonymous referees for a careful reading of the paper and for their helpful suggestions.

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