We establish the existence and uniqueness of a positive solution u for the following
fractional boundary value problem:Dαu(x)=−a(x)uσ(x), x∈(0,1) with the conditions limx→0+x2−αu(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. We also give the global behavior of such a solution.

1. Introduction

Recently, the theory of fractional differential equations has been developed very quickly and the investigation for the existence of solutions of these differential equations has attracted considerable attention of researchers in the last few years (see [1–11] and the references therein).

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 [12–14]). In consequence, the subject of fractional differential equations is gaining much importance. Motivated by the surge in the development of this subject, we consider the following singular Dirichlet problem:
(1)Dαu(x)=-a(x)uσ(x),x∈(0,1),limx→0+x2-αu(x)=0,u(1)=0,
where 1<α≤2, -1<σ<1, and a is a nonnegative continuous function on (0,1) that may be singular at x=0 or x=1. 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)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 integer part of the number β and Γ is the Euler Gamma function.

Moreover, we have the following well-known properties (see [3, 13, 15]):

IβIγv(x)=Iβ+γv(x) for x∈[0,1], v∈L1((0,1]), β+γ≥1;

DβIβv(x)=v(x) for a.e. x∈[0,1], where v∈L1((0,1]), β>0;

if v∈C((0,1))∩L1((0,1)) and Dβv(x)=0, then v(x)=∑j=1ncjtβ-j, where (c1,c2,…,cn)∈ℝn and n is the smallest integer greater than or equal to β.

Several results are obtained for fractional differential equations with different boundary conditions, but none of them deal with the existence of a positive solution to problem (1).

Our aim in this paper is to establish the existence and uniqueness of a positive solution u∈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].

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 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 by x+=max(x,0) for x∈ℝ and by B+((0,1)) the set of all nonnegative measurable functions on (0,1).

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

(H0)a∈C((0,1)) such that for t∈(0,1)(5)a(t)≈t-λL1(t)(1-t)-μL2(1-t),
where λ≤α+(2-α)(1-σ), μ≤α, L1,L2∈𝒦 satisfying
(6)∫0ηL1(t)tλ+(2-α)σ-1dt<∞,∫0ηL2(t)tμ-α+1dt<∞.
In the sequel, we introduce the function θ defined on (0,1) by
(7)θ(x)=xmin(1,(2-λ+(α-2)σ)/(1-σ))(L~1(x))1/(1-σ)×(1-x)min(1,(α-μ)/(1-σ))(L~2(1-x))1/(1-σ),
where
(8)L~1(x)={1,ifλ<α-(α-1)×(1-σ),∫xηL1(s)sds,ifλ=α-(α-1)×(1-σ),L1(x),ifα-(α-1)(1-σ)<λ<α+(2-α)(1-σ),∫0xL1(s)sds,ifλ=α+(2-α)×(1-σ),L~2(x)={1,ifμ<α+σ-1,∫xηL2(s)sds,ifμ=α+σ-1,L2(x),ifα+σ-1<μ<α,∫0xL2(s)sds,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 u∈C2-α([0,1]) satisfying for x∈(0,1),
(9)u(x)≈xα-2θ(x).

Remark 2.

Note that, for x∈(0,1), we have
(10)xα-2θ(x)≈xmin(α-1,(α-λ)/(1-σ))×(L~1(x))1/(1-σ)(1-x)min(1,(α-μ)/(1-σ))×(L~2(1-x))1/(1-σ).
This implies in particular that, for 1<α<2 and α<λ≤α+(2-α)(1-σ), the solution u blows up at x=0 and for λ<α, limx→0+u(x)=0.

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 let the paper be self-contained, we begin this section by recapitulating some properties of Karamata regular variation theory. The following is due to [16, 17].

Lemma 3.

The following hold.

Letting L∈𝒦 and ɛ>0, then one has
(11)limt→0+tɛL(t)=0.

Let L1,L2∈𝒦 and p∈ℝ. Then one has L1+L2∈𝒦, L1L2∈𝒦, and L1p∈𝒦.

Example 4.

Let m be a positive integer. Let c>0, (μ1,μ2,…,μm)∈ℝm, and d be a sufficiently large positive real number such that the function
(12)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 5.

Let μ∈ℝ and 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~t→0+-t1+μL(t)/(μ+1);

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

Lemma 6.

Let L∈𝒦 be defined on (0,η]. Then one has
(13)limt→0+L(t)∫tη(L(s)/s)ds=0.
If further ∫0η(L(s)/s)ds converges, then one has
(14)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
(15)L(t)≤∫0tL(s)sds.
It follows that limt→0+L(t)=0. So we deduce (13).

Now put
(16)φ(t)=L(t)t,fort∈(0,η).
Using that limt→0+(tφ′(t)/φ(t))=-1, we obtain
(17)∫0tφ(s)ds~t→0+-∫0tsφ′(s)ds=-tφ(t)+∫0tφ(s)ds.
This implies that
(18)∫0tL(s)sds~t→0+-L(t)+∫0tL(s)sds.
So (14) holds.

Case 2. We suppose that ∫0η(L(s)/s)ds diverges. We have, for some ω<η,
(19)∫tωφ(s)ds~t→0+tφ(t)-ωφ(ω)+∫tωφ(s)ds.
This implies that
(20)∫tωL(s)sds~t→0+L(t)-ωφ(ω)+∫tωL(s)sds.
This proves (13) and completes the proof.

Remark 7.

Let L∈𝒦 be defined on (0,η]; then using (4) and (13), we deduce that
(21)t⟶∫tηL(s)sds∈𝒦.
If further ∫0η(L(s)/s)ds converges, we have by (13) that
(22)t⟶∫0tL(s)sds∈𝒦.

Lemma 8.

Given 1<α≤2 and φ∈C([0,1]), then the unique continuous solution of
(23)Dαu(x)=-φ(x),x∈(0,1),limx→0x2-αu(x)=0,u(1)=0
is given by
(24)u(x)=Gαφ(x)∶=∫01Gα(x,t)φ(t)dt,
where
(25)Gα(x,t)=1Γ(α)[xα-1(1-t)α-1-((x-t)+)α-1]
is Green's function for the boundary value problem (23).

Proof.

Since φ∈C([0,1]), then u0=-Iαφ is a solution of the equation Dαu=-φ. Hence Dα(u+Iαφ)=0. Consequently there exist two constants c1,c2∈ℝ such that u(x)+Iαφ(x)=c1xα-1+c2xα-2. Using the fact that limx→0x2-αu(x)=0 and u(1)=0, we obtain c2=0 and c1=Iαφ(1). So
(26)u(x)=1Γ(α)xα-1∫01(1-t)α-1φ(t)dt-1Γ(α)∫0x(x-t)α-1φ(t)dt=∫01Gα(x,t)φ(t)dt.

In the following, we give some estimates on the Green function Gα(x,y). So, we need the following lemma.

Lemma 9.

For λ,μ∈(0,∞) and t∈[0,1] one has
(27)min(1,μλ)(1-tλ)≤1-tμ≤max(1,μλ)(1-tλ).

Proposition 10.

On (0,1)×(0,1), one has

Gα(x,t)≈xα-2(1-t)α-2min(x,t)(1-max(x,t));

there exist two constantsc1,c2>0suchthat(28)c1xα-1t(1-t)α-1(1-x)≤Gα(x,t)≤c2xα-2t(1-t)α-1.

Proof.

(i) For x,t∈(0,1)×(0,1) we have
(29)Gα(x,t)=(1-t)α-1xα-1Γ(α)[1-((x-t)+x(1-t))α-1].
Since (x-t)+/x(1-t)∈(0,1) for x,t∈(0,1), then by applying Lemma 9 with μ=α-1 and λ=1, we obtain
(30)Gα(x,t)≈xα-1(1-t)α-1(1-(x-t)+x(1-t))=xα-2(1-t)α-2min(x,t)(1-max(x,t)).

(ii) Using the following inequalities for x,t∈[0,1],
(31)x(1-x)t(1-t)≤min(x,t)(1-max(x,t))≤t(1-t),
we deduce the result from (i).

As a consequence of Proposition 10, we obtain the following.

Corollary 11.

Let f∈B+((0,1)) and put Gαf(x):=∫01Gα(x,t)f(t)dt for x∈(0,1]. Then
(32)Gαf(x)<∞forx∈(0,1)iff∫01t(1-t)α-1f(t)dt<∞.

Proposition 12.

Given 1<α<2 and f such that the function t→t(1-t)α-1f(t) is continuous and integrable on (0,1), then Gαf is the unique solution in C2-α([0,1]) of the following boundary value problem:
(33)Dαu(x)=-f(x),x∈(0,1),limx→0+x2-αu(x)=0,u(1)=0.

Proof.

From Corollary 11, the function Gαf is defined on (0,1) and by Proposition 10, we have
(34)Gα|f|(x)≤c2xα-2∫01t(1-t)α-1|f(t)|dt,
which implies that I2-α(Gα|f|) is bounded on (0,1). Now, using Fubini's theorem, we have
(35)I2-α(Gαf)(x)=1Γ(2-α)∫0x(x-t)1-αGαf(t)dt=1Γ(2-α)∫01(∫0x(x-t)1-αGα(t,s)dt)f(s)ds.
On the other hand, we have
(36)1Γ(2-α)∫0x(x-t)1-αGα(t,s)dt=1Γ(2-α)Γ(α)[(1-s)α-1∫0x(x-t)1-αtα-1dt-∫0x(x-t)1-α((t-s)+)α-1dt]=x(1-s)α-1-1Γ(2-α)Γ(α)∫0x(x-t)1-α×((t-s)+)α-1dt.
Now, suppose that s≤x; then we have
(37)∫0x(x-t)1-α((t-s)+)α-1dt=∫sx(x-t)1-α(t-s)α-1dt.
By considering the substitution t=s+θ(x-s), we obtain
(38)∫sx(x-t)1-α(t-s)α-1dt=Γ(α)Γ(2-α)(x-s).
Moreover if x≤s and t∈(0,x), we have ∫0x(x-t)1-α((t-s)+)α-1dt=0.

So, it follows that
(39)1Γ(2-α)∫0x(x-t)1-αGα(t,s)dt=x(1-s)α-1-(x-s)+.
This implies that
(40)I2-α(Gαf)(x)=∫01[x(1-s)α-1-(x-s)+]f(s)ds=x∫0x((1-s)α-1-1)f(s)ds+∫0xsf(s)ds+x∫x1(1-s)α-1f(s)ds,Dα(Gαf)(x)=d2dx2(I2-α(Gαf))(x)=-f(x),forx∈(0,1).
Moreover, using part (i) of Proposition 10 and the dominated convergence theorem, we conclude that limx→0+x2-αGαf(x)=0 and Gαf(1)=0.

Finally, we prove the uniqueness. Let u,v∈C2-α([0,1]) be two solutions of (33) and put w=v-u. Then w∈C2-α([0,1])⊂L1((0,1))∩C((0,1)) and Dαw=0. Hence, it follows that w(x)=c1xα-1+c2xα-2. Using the fact that limx→0+x2-αw(x)=w(1)=0, we conclude that w=0 and so u=v.

In the sequel, we assume that β≤2 and γ≤α and we put
(41)b(t)=t-βL3(t)(1-t)-γL4(1-t),
where L3,L4∈𝒦 satisfy
(42)∫0ηL3(t)tβ-1dt<∞,∫0ηtα-1-γL4(t)dt<∞.
So, we aim to give some estimates on the potential function Gαb(x).

We define the Karamata functions ψβ, ϕγ by
(43)ψβ(x)={∫0xL3(t)tdt,ifβ=2,L3(x),if1<β<2,∫xηL3(t)tdt,ifβ=1,1,ifβ<1,(44)ϕγ(x)={∫0xL4(t)tdt,ifγ=α,L4(x),ifα-1<γ<α,∫xηL4(t)tdt,ifγ=α-1,1,ifγ<α-1.

Then, we have the following.

Proposition 13.

For x∈(0,1),
(45)Gαb(x)≈xmin(α-1,α-β)(1-x)min(1,α-γ)ψβ(x)ϕγ(1-x).

Proof.

Using Proposition 10, we have
(46)x2-αGαb(x)≈∫01(1-t)α-2-γt-βmin(x,t)×(1-max(x,t))L3(t)L4(1-t)dt≈(1-x)∫0x(1-t)α-2-γt1-β×L3(t)L4(1-t)dt+x∫x1(1-t)α-1-γt-β×L3(t)L4(1-t)dt=(1-x)I(x)+xJ(x).
For 0<x≤1/2, we have I(x)≈∫0xt1-βL3(t)dt. So, using Lemma 5 and hypothesis (42), we deduce that
(47)I(x)≈{x2-βL3(x),ifβ<2,∫0xL3(t)tdt,ifβ=2.
Now, we have
(48)J(x)≈∫x1/2t-βL3(t)dt+∫1/21(1-t)α-1-γL4(1-t)dt≈1+∫x1/2t-βL3(t)dt,
which implies by Lemma 5 that
(49)J(x)≈{x1-βL3(x),if1<β≤2,∫xηL3(t)tdt,ifβ=1,1,ifβ<1.
Hence, it follows by Lemma 6 and hypothesis (42) that, for 0<x≤1/2, we get
(50)x2-αGαb(x)≈{∫0xL3(t)tdt,ifβ=2,x2-βL3(x),if1<β<2,x∫xηL3(t)tdt,ifβ=1,x,ifβ<1.
That is
(51)Gαb(x)≈xmin(α-1,α-β)ψβ(x).
Now, for 1/2≤x<1, we use again Lemma 5 and hypothesis (42) to deduce that
(52)I(x)≈∫01/2t1-βL3(t)dt+∫1/2x(1-t)α-2-γL4(1-t)dt≈1+∫1/2x(1-t)α-2-γL4(1-t)dt≈{(1-x)α-1-γL4(1-x),ifα-1<γ≤α,∫1-xηL4(t)tdt,ifγ=α-1,1,ifγ<α-1,J(x)≈∫01-xtα-1-γL4(t)dt≈{(1-x)α-γL4(1-x),ifγ<α,∫01-xL4(t)tdt,ifγ=α.
Hence, it follows from Lemma 3 that, for x∈[1/2,1), we get
(53)x2-αGαb(x)≈{∫01-xL4(t)tdt,ifγ=α,(1-x)α-γL4(1-x),ifα-1<γ<α,(1-x)∫1-xηL4(t)tdt,ifγ=α-1,1-x,ifγ<α-1.
That is
(54)x2-αGαb(x)≈(1-x)min(1,α-γ)ϕγ(1-x).
This together with (51) implies that, for x∈(0,1), we have
(55)Gαb(x)≈xmin(α-1,α-β)(1-x)min(1,α-γ)ψβ(x)ϕγ(1-x).

3. Proof of Theorem <xref ref-type="statement" rid="thm1.1">1</xref>

In order to prove Theorem 1, we need the following Lemma.

Lemma 14.

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

Proof.

Put r=min(α-1,(α-λ)/(1-σ)) and s=min(1,(α-μ)/(1-σ)). Then for t∈(0,1), we have
(57)ω(t)=t-λ+rσL1(t)(L~1(t))σ/(1-σ)×(1-t)-μ+sσL2(1-t)(L~2(1-t))σ/(1-σ).
Let β=λ-rσ, γ=μ-sσ, L3(t)=L1(t)(L~1(t))σ/(1-σ), and L4(t)=L2(t)(L~2(t))σ/(1-σ). Then, using Proposition 13, we obtain by a simple computation that
(58)x2-αGα(ω)(x)≈θ(x).

Proof of Theorem <xref ref-type="statement" rid="thm1.1">1</xref>.

From Lemma 14, there exists M>1 such that, for each x∈(0,1),
(59)1Mθ(x)≤x2-αGαω(x)≤Mθ(x),
where ω(t)=a(t)t(α-2)σθσ(t).

Put c0=M1/(1-|σ|) and let
(60)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
(61)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
(62)Tv(x)=x2-αΓ(α)∫01Gα(x,t)a~(t)vσ(t)dt=x2-αΓ(α)∫01[-((x-t)+)α-1xα-1(1-t)α-1-((x-t)+)α-1]a~(t)vσ(t)dt.
Since the function (x,t)→xα-1(1-t)α-1-((x-t)+)α-1 is continuous on [0,1]×[0,1] and by Proposition 10, Corollary 11, and Lemma 14, the function t→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
(63)u(x)=Gα(auσ)(x).
Since the function t→t(1-t)α-1a(t)uσ(t) is continuous and integrable on (0,1), then by Proposition 12, the function u is a positive continuous solution of problem (1).

Finally, let us prove that u is the unique positive continuous solution satisfying (9). To this aim, we assume that (1) has two positive solutions u,v∈C2-α([0,1]) satisfying (9) 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
(64)-Dα(c|σ|v-u)=a(c|σ|vσ-uσ)≥0,limt→0+x2-α(c|σ|v-u)(t)=0,(c|σ|v-u)(1)=0,
which implies by Proposition 12 that c|σ|v-u=Gα(a(c|σ|vσ-uσ))≥0. By symmetry, we also obtain that v≤c|σ|u. Hence, c|σ|∈J and c≤c|σ|. Since |σ|<1, then c=1 and consequently u=v.

Example 15.

Let σ∈(-1,1) and a be a positive continuous function on (0,1) such that
(65)a(t)≈t-λ(1-t)-μlog(21-t),
where λ<α+(2-α)(1-σ) and μ<α. Then, using Theorem 1, problem (1) has a unique positive continuous solution u satisfying the following estimates:
(66)u(x)≈xmin(α-1,(α-λ)/(1-σ))(L~1(x))1/(1-σ)×(1-x)min(1,(α-μ)/(1-σ))(L~2(1-x))1/(1-σ),
where
(67)L~1(x)={1,ifλ≠α-(α-1)(1-σ),log(2x),ifλ=α-(α-1)(1-σ),L~2(x)={1,ifμ<α+σ-1,(log(2x))2,ifμ=α+σ-1,log(2x),ifα+σ-1<μ<α.

Conflict of Interests

The authors declare that there is no conflict of interests regarding the publication of this paper.

Acknowledgment

This paper was funded by the Deanship of Scientific Research (DSR), King Abdulaziz University, Jeddah. The authors, therefore, acknowledge with thanks DSR technical and financial support.

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