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Uniqueness and existence of positive solutions for a multi-point boundary value problem of singular fractional differential equations

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Abstract

In this paper, we study the uniqueness of a positive solution for the singular nonlinear fractional differential equation boundary value problem D 0 + α u(t)+f(t,u(t))=0, 0<t<1, u(0)=0, u(1)=a D α 1 2 u(t) | t = ξ , where 1<α2 is a real number, D 0 + α is the standard Riemann-Liouville differentiation and f:(0,1]×[0,+)[0,+), with lim t 0 + f(t,)=+. Our analysis relies on a fixed-point theorem in partially ordered set. As an application, an example is presented to illustrate the main result.

MSC:26A33, 34B15, 34K37.

1 Introduction

Fractional differential equations have gained considerable importance due to their application in various sciences, such as physics, mechanics, chemistry, engineering, etc. In recent years, there has been a significant development in ordinary and partial differential equations involving fractional derivatives; see the monographs of Kilbas et al. [1], Miller and Ross [2], Oldham and Spanier [3], Podlubny [4], Samko [5], and the papers [616] and the references therein.

However, there are few papers, which have considered the singular boundary value problems of fractional differential equations; see [1723]. In particular, Delbosco and Rodino [17] considered the existence of a solution for the nonlinear fractional differential equation D 0 + α u=f(t,u), where 0<α<1 and f:[0,a]×RR, 0<a+ is a given continuous function in (0,a)×R. They obtained some results for solutions by using the Schauder fixed-point theorem and the Banach contraction principle.

Qiu and Bai [18] considered the existence of a positive solution to boundary value problems of the nonlinear fractional differential equation

D 0 + α c u ( t ) + f ( t , u ( t ) ) = 0 , 0 < t < 1 , 2 < α 3 , u ( 0 ) = u ( 1 ) = u ( 0 ) = 0 ,

where D 0 + α c is the Caputo fractional derivative, and f:(0,1]×[0,+)[0,+), with lim t 0 + f(t,)=+ (i.e., f is singular at t=0). They obtained the existence of positive solutions by means of the Guo-Krasnosel’skii fixed-point theorem and nonlinear alternative of Leray-Schauder type in a cone. In [18], the uniqueness of the solution is not treated.

From the above works, we can see a fact, although the fractional boundary value problems have been investigated by some authors, to the best of our knowledge, there have been few papers which deal with the problem (1.1)-(1.2) for nonlinear singular fractional differential equation. Motivated by all the works above, this paper is mainly concerned with the uniqueness of a positive solution for the singular nonlinear fractional differential equation boundary value problem

D 0 + α u(t)+f ( t , u ( t ) ) =0,0<t<1,
(1.1)
u(0)=0,u(1)=a D α 1 2 u(t) | t = ξ ,
(1.2)

where 1<α2 is a real number, ξ(0, 1 2 ], a(0,+) satisfy aΓ(α) ξ α 1 2 <Γ( α + 1 2 ), and D 0 + α is the standard Riemann-Liouville differentiation, and f:(0,1]×[0,+)[0,+), with lim t 0 + f(t,)=+. In this article, by using a fixed- point theorem in partially ordered set, existence and uniqueness results of a positive solution for the problem (1.1)-(1.2) are given.

The paper is organized as follows. In Section 2, we give some preliminary results that will be used in the proof of the main results. In Section 3, we establish the uniqueness of a positive solution for the singular nonlinear fractional differential equation boundary value problem (1.1)-(1.2). In the end, we illustrate a simple use of the main result.

2 Preliminaries and lemmas

For the convenience of the reader, we present here the necessary definitions from fractional calculus theory. These definitions can be found in the recent literature such as [1] and [4].

Definition 2.1 [1, 4]

The Riemann-Liouville fractional integral of order α>0 of a function f:(0,+)R is given by

I 0 + α f(t)= 0 t ( t s ) α 1 Γ ( α ) f(s)ds,

provided that the right side is pointwise defined on (0,+), where Γ is the gamma function.

Definition 2.2 [1, 4]

The Riemann-Liouville fractional derivative of order α>0 of a continuous function f:(0,+)R is given by

( D 0 + α f ) (t)= 1 Γ ( n α ) ( d d t ) n 0 t ( t s ) n α + 1 f(s)ds,

provided that the right side is pointwise defined on (0,+). Here, n=[α]+1 and [α] denotes the integer part of α.

Lemma 2.1 [1]

Let α>0. If we assume uC(0,1)L(0,1), then the fractional differential equation

D 0 + α u(t)=0

has

u(t)= C 1 t α 1 + C 2 t α 2 ++ C N t α N , C i R,i=1,2,,N,

as unique solutions, where N is the smallest integer greater than or equal to α.

Lemma 2.2 [1]

Assume that hC(0,1)L(0,1) with a fractional derivative of order α>0 that belongs to C(0,1)L(0,1). Then

I 0 + α D 0 + α h(t)=h(t)+ C 1 t α 1 + C 2 t α 2 ++ C N t α N ,

for some C i R, i=1,2,,N, where N is the smallest integer greater than or equal to α.

Lemma 2.3 [21]

Let hC(0,1)L(0,1) and 1<α2, ξ(0,1), aR satisfy that aΓ(α) ξ α 1 2 Γ( α + 1 2 ), then the unique solution of

D 0 + α u(t)+h(t)=0,0<t<1,
(2.1)
u(0)=0,u(1)=a D α 1 2 u(t) | t = ξ
(2.2)

is given by

u ( t ) = 0 t ( t s ) α 1 Γ ( α ) h ( s ) d s + t α 1 Γ ( α + 1 2 ) Γ ( α + 1 2 ) a Γ ( α ) ξ α 1 2 { 0 1 ( 1 s ) α 1 Γ ( α ) h ( s ) d s a 0 ξ ( ξ s ) α 1 2 Γ ( α + 1 2 ) h ( s ) d s } .
(2.3)

Lemma 2.4 [21]

Let hC((0,1),[0,+))L(0,1) and 1<α2, ξ(0, 1 2 ], a(0,+) satisfy that aΓ(α) ξ α 1 2 <Γ( α + 1 2 ), then the unique solution of the problem (2.1)-(2.2)

u ( t ) = 0 t ( t s ) α 1 Γ ( α ) h ( s ) d s + t α 1 Γ ( α + 1 2 ) Γ ( α + 1 2 ) a Γ ( α ) ξ α 1 2 { 0 1 ( 1 s ) α 1 Γ ( α ) h ( s ) d s a 0 ξ ( ξ s ) α 1 2 Γ ( α + 1 2 ) h ( s ) d s } = 0 1 G ( t , s ) h ( s ) d s + a t α 1 Γ ( α + 1 2 ) a Γ ( α ) ξ α 1 2 × { 0 ξ [ ( 1 s ) α 1 ξ α 1 2 ( ξ s ) α 1 2 ] h ( s ) d s + ξ 1 ( 1 s ) α 1 ξ α 1 2 h ( s ) d s }
(2.4)

is nonnegative on [0,1], where

G(t,s)={ [ t ( 1 s ) ] q 1 ( t s ) q 1 Γ ( q ) , if 0 s t 1 , [ t ( 1 s ) ] q 1 Γ ( q ) , if 0 t s 1 .

The following two lemmas are fundamental in the proofs of our main result.

Lemma 2.5 [24]

Let (E,) be a partially ordered set and suppose that there exists a metric d in E such that (E,d) is a complete metric space. Assume that E satisfies:

If x n is a nondecreasing sequence in  E such that x n x then x n x,nN.
(2.5)

Let f:EE be a nondecreasing mapping such that

d ( f ( x ) , f ( y ) ) d(x,y)φ ( d ( x , y ) ) ,for xy,
(2.6)

where φ:[0,+)[0,+) is continuous and nondecreasing function such that φ is positive in (0,+), φ(0)=0 and lim t + φ(t)=+. If there exists x 0 E with x 0 f( x 0 ), then f has a fixed point.

If we consider that (E,) satisfies the following condition:

For x,yE there exists zE which is comparable to x and y,
(2.7)

then we have the following result.

Lemma 2.6 [24]

Adding condition (2.7) to the hypotheses of Lemma  2.5, we obtain uniqueness of the fixed point of f.

3 Main results

Theorem 3.1 Let 1<α2, ξ(0, 1 2 ], a(0,+) satisfy that aΓ(α) ξ α 1 2 <Γ( α + 1 2 ), F:(0,1][0,+) is continuous, and lim t 0 + F(t)=+. Suppose that there exists a constant σ: 0<σ<1 such that t σ F(t) is a continuous function on [0,1]. Then the unique solution of the problem (2.1)-(2.2) is given by

H ( t ) = 0 t ( t s ) α 1 Γ ( α ) F ( s ) d s + t α 1 Γ ( α + 1 2 ) Γ ( α + 1 2 ) a Γ ( α ) ξ α 1 2 { 0 1 ( 1 s ) α 1 Γ ( α ) F ( s ) d s a 0 ξ ( ξ s ) α 1 2 Γ ( α + 1 2 ) F ( s ) d s } = 0 1 G ( t , s ) F ( s ) d s + a t α 1 Γ ( α + 1 2 ) a Γ ( α ) ξ α 1 2 × { 0 ξ [ ( 1 s ) α 1 ξ α 1 2 ( ξ s ) α 1 2 ] F ( s ) d s + ξ 1 ( 1 s ) α 1 ξ α 1 2 F ( s ) d s }
(3.1)

and is continuous on [0,1].

Proof By the continuity of t σ F(t), it is easy to check that H(0)=0. The proof is divided into three cases.

Case 1. t 0 =0, t(0,1].

Since t σ F(t) is continuous in [0,1], there exists a constant M>0, such that | t σ F(t)|M, t[0,1]. Hence,

| H ( t ) H ( 0 ) | = | 0 1 G ( t , s ) F ( s ) d s + a t α 1 Γ ( α + 1 2 ) a Γ ( α ) ξ α 1 2 × { 0 ξ [ ( 1 s ) α 1 ξ α 1 2 ( ξ s ) α 1 2 ] F ( s ) d s + ξ 1 ( 1 s ) α 1 ξ α 1 2 F ( s ) d s } | = | 0 1 G ( t , s ) s σ s σ F ( s ) d s + a t α 1 Γ ( α + 1 2 ) a Γ ( α ) ξ α 1 2 × { 0 ξ [ ( 1 s ) α 1 ξ α 1 2 ( ξ s ) α 1 2 ] s σ s σ F ( s ) d s + ξ 1 ( 1 s ) α 1 ξ α 1 2 s σ s σ F ( s ) d s } | M | 0 t [ [ t ( 1 s ) ] α 1 ( t s ) α 1 Γ ( α ) ] s σ d s + t 1 [ t ( 1 s ) ] α 1 Γ ( α ) s σ d s + a t α 1 Γ ( α + 1 2 ) a Γ ( α ) ξ α 1 2 × { 0 ξ [ ( 1 s ) α 1 ξ α 1 2 ( ξ s ) α 1 2 ] s σ d s + ξ 1 ( 1 s ) α 1 ξ α 1 2 s σ d s } | M t α 1 Γ ( α ) B ( 1 σ , α ) + M t α σ Γ ( α ) B ( 1 σ , α ) + a M t α 1 ξ α 1 2 Γ ( α + 1 2 ) a Γ ( α ) ξ α 1 2 B ( 1 σ , α ) + a M t α 1 ξ 1 σ + α 1 2 Γ ( α + 1 2 ) a Γ ( α ) ξ α 1 2 B ( 1 σ , α + 1 2 ) M Γ ( 1 σ ) Γ ( 1 σ + α ) [ t α 1 + t α σ ] + a M t α 1 ξ α 1 2 Γ ( α + 1 2 ) a Γ ( α ) ξ α 1 2 Γ ( 1 σ ) Γ ( α ) Γ ( 1 σ + α ) + a M t α 1 ξ 1 σ + α 1 2 Γ ( α + 1 2 ) a Γ ( α ) ξ α 1 2 Γ ( 1 σ ) Γ ( α + 1 2 ) Γ ( 1 σ + α + 1 2 ) 0 ( t 0 ) ,

where B(,) denotes the beta function.

Case 2. t 0 (0,1), t( t 0 ,1].

| H ( t ) H ( t 0 ) | = | 0 1 G ( t , s ) s σ s σ F ( s ) d s 0 1 G ( t 0 , s ) s σ s σ F ( s ) d s + a [ t α 1 t 0 α 1 ] Γ ( α + 1 2 ) a Γ ( α ) ξ α 1 2 × { 0 ξ [ ( 1 s ) α 1 ξ α 1 2 ( ξ s ) α 1 2 ] s σ s σ F ( s ) d s + ξ 1 ( 1 s ) α 1 ξ α 1 2 s σ s σ F ( s ) d s } | M | 0 t [ t ( 1 s ) ] α 1 ( t s ) α 1 Γ ( α ) s σ d s + t 1 [ t ( 1 s ) ] α 1 Γ ( α ) s σ d s t 0 1 [ t 0 ( 1 s ) ] α 1 Γ ( α ) s σ d s 0 t 0 [ t 0 ( 1 s ) ] α 1 ( t 0 s ) α 1 Γ ( α ) s σ d s + a [ t α 1 t 0 α 1 ] Γ ( α + 1 2 ) a Γ ( α ) ξ α 1 2 × { 0 ξ [ ( 1 s ) α 1 ξ α 1 2 ( ξ s ) α 1 2 ] s σ d s + ξ 1 ( 1 s ) α 1 ξ α 1 2 s σ d s } | = M | t α 1 t 0 α 1 Γ ( α ) 0 1 ( 1 s ) α 1 s σ d s + t 0 t ( t s ) α 1 Γ ( α ) s σ d s + 0 t 0 ( t s ) α 1 ( t 0 s ) α 1 Γ ( α ) s σ d s + a [ t α 1 t 0 α 1 ] Γ ( α + 1 2 ) a Γ ( α ) ξ α 1 2 × { 0 ξ [ ( 1 s ) α 1 ξ α 1 2 ( ξ s ) α 1 2 ] s σ d s + ξ 1 ( 1 s ) α 1 ξ α 1 2 s σ d s } | M { t α 1 t 0 α 1 Γ ( α ) B ( 1 σ , α ) + t α σ t 0 α σ Γ ( α ) B ( 1 σ , α ) + a [ t α 1 t 0 α 1 ] Γ ( α + 1 2 ) a Γ ( α ) ξ α 1 2 × [ ξ α 1 2 B ( 1 σ , α ) + ξ 1 σ + α 1 2 B ( 1 σ , α 1 2 ) ] } 0 ( t t 0 ) .

Case 3. t 0 (0,1], t[0, t 0 ). The proof is similar to that of Case 2, so we omit it. □

Let Banach space E=C[0,1] be endowed with the norm u= max t [ 0 , 1 ] |u(t)|. Note that this space can be equipped with a partial order given by

x,yE,xyx(t)y(t),t[0,1].
(3.2)

It is easy to check that (E,) with the classic metric given by

d(x,y)= max t [ 0 , 1 ] { | x ( t ) y ( t ) | }
(3.3)

satisfies condition (2.6) of Lemma 2.5. Moreover, for x,yE, as the function max{x,y} is continuous in [0,1], (E,) satisfies condition (2.7).

Theorem 3.2 Let 0<σ<1, 1<α2, ξ(0, 1 2 ], a(0,+) satisfy that aΓ(α) ξ α 1 2 <Γ( α + 1 2 ), f:(0,1]×[0,+)[0,+) is continuous, with lim t 0 + f(t,)=+, and t σ f(t,u) is continuous function on [0,1]×[0,+). Assume that there exists λ satisfying

0 < λ [ Γ ( 1 σ ) Γ ( 1 σ + α ) + Γ ( α + 1 2 ) Γ ( 1 σ ) [ Γ ( α + 1 2 ) a Γ ( α ) ξ α 1 2 ] Γ ( 1 σ + α ) + a ξ ( 1 σ + α 1 2 ) Γ ( 1 σ ) Γ ( α + 1 2 ) [ Γ ( α + 1 2 ) a Γ ( α ) ξ α 1 2 ] Γ ( 1 σ + α + 1 2 ) ] 1 ,

such that for u,v[0,+) with uv and t[0,1],

0 t σ [ f ( t , u ) f ( t , v ) ] λϕ(uv),
(3.4)

where ϕ:[0,+)[0,+) is continuous and nondecreasing, φ(u)=uϕ(u) satisfies

  1. (a)

    φ:[0,+)[0,+) and nondecreasing;

  2. (b)

    φ(0)=0;

  3. (c)

    φ is positive in (0,+).

Then the problem (1.1)-(1.2) has an unique positive solution.

Proof Define the cone KE by

K= { u E : u ( t ) 0 , t [ 0 , 1 ] } .

Note that, as K is a closed subset of E, K is a complete metric space.

Suppose that u is a solution of boundary value problem (1.1) and (1.2). Then

u ( t ) = 0 1 G ( t , s ) f ( s , u ( s ) ) d s + a t α 1 Γ ( α + 1 2 ) a Γ ( α ) ξ α 1 2 × { 0 ξ [ ( 1 s ) α 1 ξ α 1 2 ( ξ s ) α 1 2 ] f ( s , u ( s ) ) d s + ξ 1 ( 1 s ) α 1 ξ α 1 2 f ( s , u ( s ) ) d s } , t [ 0 , 1 ] .

Define an operator A:KE as follows:

( A u ) ( t ) = 0 1 G ( t , s ) f ( s , u ( s ) ) d s + a t α 1 Γ ( α + 1 2 ) a Γ ( α ) ξ α 1 2 × { 0 ξ [ ( 1 s ) α 1 ξ α 1 2 ( ξ s ) α 1 2 ] f ( s , u ( s ) ) d s + ξ 1 ( 1 s ) α 1 ξ α 1 2 f ( s , u ( s ) ) d s } , t [ 0 , 1 ] .
(3.5)

By Theorem 3.1, AuE. Moreover, in view of Lemma 2.4 and t σ f(t,u)0 for (t,u)[0,1]×[0,+), by hypothesis, we get

(Au)(t)0,t[0,1],

so, A(K)K.

In what follows, we check that hypotheses in Lemmas 2.5 and 2.6 are satisfied. Firstly, the operator A is nondecreasing. By hypothesis, for uv, we get

( A u ) ( t ) = 0 1 G ( t , s ) f ( s , u ( s ) ) d s + a t α 1 Γ ( α + 1 2 ) a Γ ( α ) ξ α 1 2 × { 0 ξ [ ( 1 s ) α 1 ξ α 1 2 ( ξ s ) α 1 2 ] f ( s , u ( s ) ) d s + ξ 1 ( 1 s ) α 1 ξ α 1 2 f ( s , u ( s ) ) d s } 0 1 G ( t , s ) f ( s , v ( s ) ) d s + a t α 1 Γ ( α + 1 2 ) a Γ ( α ) ξ α 1 2 × { 0 ξ [ ( 1 s ) α 1 ξ α 1 2 ( ξ s ) α 1 2 ] f ( s , v ( s ) ) d s + ξ 1 ( 1 s ) α 1 ξ α 1 2 f ( s , v ( s ) ) d s } = ( A v ) ( t ) .

Besides, for uv, by (3.4), we get

d ( A u , A v ) = max t [ 0 , 1 ] { | A u ( t ) A v ( t ) | } = max t [ 0 , 1 ] [ A u ( t ) A v ( t ) ] max t [ 0 , 1 ] [ 0 1 G ( t , s ) [ f ( s , u ( s ) ) f ( s , v ( s ) ) ] d s + a t α 1 Γ ( α + 1 2 ) a Γ ( α ) ξ α 1 2 × { 0 ξ [ ( 1 s ) α 1 ξ α 1 2 ( ξ s ) α 1 2 ] [ f ( s , u ( s ) ) f ( s , v ( s ) ) ] d s + ξ 1 ( 1 s ) α 1 ξ α 1 2 [ f ( s , u ( s ) ) f ( s , v ( s ) ) ] d s } ] max t [ 0 , 1 ] [ 0 1 G ( t , s ) s σ λ ϕ [ u ( s ) v ( s ) ] d s + a t α 1 Γ ( α + 1 2 ) a Γ ( α ) ξ α 1 2 × { 0 ξ [ ( 1 s ) α 1 ξ α 1 2 ( ξ s ) α 1 2 ] s σ λ ϕ [ u ( s ) v ( s ) ] d s + ξ 1 ( 1 s ) α 1 ξ α 1 2 s σ λ ϕ [ u ( s ) v ( s ) ] d s } ] .

As the function ϕ(u) is nondecreasing, for uv, we get

ϕ [ u ( s ) v ( s ) ] ϕ ( u v ) .
(3.6)

By the last inequality, we get

d ( A u , A v ) max t [ 0 , 1 ] [ 0 1 G ( t , s ) s σ λ ϕ [ u ( s ) v ( s ) ] d s + a t α 1 Γ ( α + 1 2 ) a Γ ( α ) ξ α 1 2 × { 0 ξ [ ( 1 s ) α 1 ξ α 1 2 ( ξ s ) α 1 2 ] s σ λ ϕ [ u ( s ) v ( s ) ] d s + ξ 1 ( 1 s ) α 1 ξ α 1 2 s σ λ ϕ [ u ( s ) v ( s ) ] d s } ] λ ϕ ( u v ) max t [ 0 , 1 ] [ 0 1 G ( t , s ) s σ d s + a t α 1 Γ ( α + 1 2 ) a Γ ( α ) ξ α 1 2 × { 0 ξ [ ( 1 s ) α 1 ξ α 1 2 ( ξ s ) α 1 2 ] s σ d s + ξ 1 ( 1 s ) α 1 ξ α 1 2 s σ d s } ] = λ ϕ ( u v ) max t [ 0 , 1 ] [ t α σ Γ ( α ) B ( 1 σ , α ) + t α 1 Γ ( α + 1 2 ) B ( 1 σ , α ) [ Γ ( α + 1 2 ) a Γ ( α ) ξ α 1 2 ] Γ ( α ) + a t α 1 ξ ( 1 σ + α 1 2 ) [ Γ ( α + 1 2 ) a Γ ( α ) ξ α 1 2 ] B ( 1 σ , α + 1 2 ) ] = λ ϕ ( u v ) [ B ( 1 σ , α ) Γ ( α ) + Γ ( α + 1 2 ) B ( 1 σ , α ) [ Γ ( α + 1 2 ) a Γ ( α ) ξ α 1 2 ] Γ ( α ) + a ξ ( 1 σ + α 1 2 ) [ Γ ( α + 1 2 ) a Γ ( α ) ξ α 1 2 ] B ( 1 σ , α + 1 2 ) ] = λ ϕ ( u v ) [ Γ ( 1 σ ) Γ ( 1 σ + α ) + Γ ( α + 1 2 ) Γ ( 1 σ ) [ Γ ( α + 1 2 ) a Γ ( α ) ξ α 1 2 ] Γ ( 1 σ + α ) + a ξ ( 1 σ + α 1 2 ) Γ ( 1 σ ) Γ ( α + 1 2 ) [ Γ ( α + 1 2 ) a Γ ( α ) ξ α 1 2 ] Γ ( 1 σ + α + 1 2 ) ] ϕ ( u v ) = u v [ u v ϕ ( u v ) ] .

Put φ(u)=uϕ(u). Obviously, φ:[0,+)[0,+) is continuous, nondecreasing, positive in (0,+), φ(0)=0.

Thus, for uv, we get

d(Au,Av)d(u,v)φ ( d ( u , v ) ) .
(3.7)

Finally, take into account that for the zero function, 0A0, by Lemma 2.5, our problem (1.1)-(1.2) has at least one nonnegative solution. Moreover, this solution is unique, since (K,) satisfies condition (2.7) and Lemma 2.6. This completes the proof. □

In the sequel, we present an example which illustrates Theorem 3.2.

4 An example

Example 4.1

Consider the following fractional boundary value problem:

D 3 2 u(t)= ( t 1 2 ) 2 ln ( 2 + u ( t ) ) t ,0<t<1,
(4.1)
u(0)=0,u(1)= 1 2 D 1 4 u(t) | t = 1 2 ,
(4.2)

where α= 3 2 , a=ξ= 1 2 . In this case, f(t,u)= ( t 1 2 ) 2 ln ( 2 + u ) t for (t,u)(0,1]×[0,+), σ= 1 2 . Note that f is continuous in (0,1]×[0,+) and lim t 0 + f(t,)=. Moreover, for uv and t[0,1], we have

0 t σ [ f ( t , u ) f ( t , v ) ] = [ ( t 1 2 ) 2 ln ( 2 + u ) ( t 1 2 ) 2 ln ( 2 + v ) ] .
(4.3)

Because g(x)=ln(x+2) is nondecreasing on [0,+), and

[ ( t 1 2 ) 2 ln ( 2 + u ) ( t 1 2 ) 2 ln ( 2 + v ) ] = ( t 1 2 ) 2 ln ( 2 + u ) ( 2 + v ) = ( t 1 2 ) 2 ln ( 2 + v + u v ) ( 2 + v ) ( 1 2 ) 2 ln ( 1 + u v ) .

With the aid of a computer, we obtain that

[ Γ ( 1 σ ) Γ ( 1 σ + α ) + Γ ( α + 1 2 ) Γ ( 1 σ ) [ Γ ( α + 1 2 ) a Γ ( α ) ξ α 1 2 ] Γ ( 1 σ + α ) + a ξ ( 1 σ + α 1 2 ) Γ ( 1 σ ) Γ ( α + 1 2 ) [ Γ ( α + 1 2 ) a Γ ( α ) ξ α 1 2 ] Γ ( 1 σ + α + 1 2 ) ] 1 0.282 > 1 4 .

So, by Theorem 3.2, the problem (4.1)-(4.2) has an unique positive solution.

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Acknowledgements

Dedicated to Professor Hari M Srivastava.

The authors are thankful to the referees for their careful reading of the manuscript and insightful comments. The research is supported by the National Natural Science Foundation of China (11161027, 11262009).

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Correspondence to Wen-Xue Zhou.

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The authors declare that they have no competing interests.

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All authors contributed equally to the manuscript. All authors read and approved the final manuscript.

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Keywords

  • boundary value problem
  • singular fractional differential equations
  • Riemann-Liouville fractional derivative
  • uniqueness
  • partially ordered set