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(1)

Positive Solution for Higher-order Singular Infinite-point Fractional Differential Equation with p-Laplacian

QIUYAN ZHONG Jining Medical College

Department of Information Engineering Hehua Road No. 16, 272067, Jining

CHINA

zhqy197308@163.com

XINGQIU ZHANG Jining Medical College

Department of Information Engineering Hehua Road No. 16, 272067, Jining

CHINA

zhxq197508@163.com

Abstract: By means of the method of upper and lower solutions together with the Schauder fixed point theorem, the conditions for the existence of at least one positive solution are established for some higher-order singular infinite-point fractional differential equation with p-Laplacian. The nonlinear term may be singular with respect to both the time and space variables.

Key–Words:Fractional differential equations, p-Laplacian, Singularity, Upper and lower solutions, Positive solu- tion

1 Introduction

We investigate the existence of positive solutions for the following fractional differential equations con- taning p-Laplacian operator (PFDE, for short) and infinite-point boundary value conditions

D0+βp(D0+α u(t))) + f (t, u(t)) = 0, 0 < t < 1, u(0) = u0(0) = · · · = u(n−2)(0) = 0, D0+α u(0) = 0, u(i)(1) =

P

j=1

αju(ξj),

(1)

where Dα0+, Dβ0+ is the standard Riemann-Liouville derivative, ϕp(s) = |s|p−2s, p > 1, f ∈ C((0, 1) × J, J ), J = (0, +∞), R+ = [0, +∞), h ∈ L1[0, 1] is nonnegative. f (t, u) may be singular at t = 0, 1 and u = 0, i ∈ [1, n − 2] is a fixed integer, n − 1 < α ≤ n, n ≥ 3, 0 < β ≤ 1, αj ≥ 0, 0 < ξ1 < ξ2 <

· · · < ξj−1 < ξj < · · · < 1 (j = 1, 2, . . .), ∆ −

P

j=1

αjξjα−1> 0, ∆ = (α − 1)(α − 2) · · · (α − i).

In recent years, many excellent results of fraction- al differential equations have been widely reported for their numerous applications such as in electrody- namics of complex medium, control, electromagnetic, polymer rheology, and so on, see [1-3] for an exten- sive collection of such results. In [4-6], by means of fixed point theorem and theory of fixed point index to- gether with eigenvalue with respect to the relevant lin- ear operator, the existence and multiplicity of positive solutions, pseudo-solutions are obtained for m-point

boundary value problem of fractional differential e- quations

Dα0+u(t) + q(t)f (t, u(t)) = 0, 0 < t < 1 (A)

subject to the following boundary conditions

u(0) = · · · = u(n−2)(0) = 0, u(1) =

m

X

i=1

αiu(ξi).

(B1) Similar results are extended to more general bound- ary value problems in [7]. Motivated by [8], by in- troducing height functions of the nonlinear term on some bounded sets, we considered local existence and multiplicity of positive solutions for BVP (A) with infinite-point boundary value conditions in [9]. On the other hand, there have been some papers dealing with the fractional differential equations involving p- Laplacian operator [10-16]. The purpose of this paper is to study the existence of at least one solution for PFDE (1) by means of the upper and lower solutions and the Schauder fixed point theorem.

Compared to [5-7], this paper admits the follow- ing three new features. Firstly, the facts p-Laplacian operator is involved in differential operator and infi- nite points is contained in boundary value problems make the problem considered more general. Second- ly, nonlinear term permits singularities with respect to both the time and space variables.

(2)

2 Preliminaries and several lemmas

Let E be the Banach space of continuous function- s u : [0, 1] → R equipped with the norm kuk = max0≤t≤1|u(t)|. Here, we list some definitions and useful lemmas from fractional calculus theory.

Definition 1 ([3]) The Riemann-Liouville fractional integral of orderα > 0 of a function y : (0, ∞) → R is given by

I0+α y(t) = 1 Γ(α)

Z t 0

(t − s)α−1y(s)ds provided the right-hand side is pointwise defined on (0, ∞).

Definition 2 ([3]) The Riemann-Liouville fractional derivative of order α > 0 of a continuous function y : (0, ∞) → R is given by

Dα0+y(t) = 1 Γ(n − α)

d dt

nZ t 0

y(s) (t − s)α−n+1ds wheren = [α] + 1, [α] denotes the integer part of the numberα, provided that the right-hand side is point- wise defined on(0, ∞).

Now, we consider the linear fractional differential equation

Dα0+u(t) + y(t) = 0, 0 < t < 1, u(0) = u0(0) = · · · = u(n−2)(0) = 0, u(i)(1) =

P

j=1

αju(ξj).

(2)

Lemma 3 ([9]) Given y ∈ C[0, 1], then the unique solution of the problem

Dα0+u(t) + y(t) = 0, 0 < t < 1, u(0) = u0(0) = · · · = u(n−2)(0) = 0, u(i)(1) =

P

j=1

αju(ξj),

(3)

can be expressed by u(t) =

Z 1 0

G(t, s)y(s)ds, where

G(t, s) = 1 p(0)Γ(α)

tα−1p(s)(1 − s)α−1−i

−p(0)(t − s)α−1, 0 ≤ s ≤ t ≤ 1,

tα−1p(s)(1 − s)α−1−i, 0 ≤ t ≤ s ≤ 1,

(4)

wherep(s) = ∆ − P

s≤ξj

αjξ1−sj−sα−1(1 − s)i. Obvi- ously,G(t, s) is continuous on [0, 1] × [0, 1].

Lemma 4 [7] The function G(t, s) defined by (4) has the following properties:

(1) p(0)Γ(α)G(t, s) ≥ m1s(1 − s)α−1−itα−1, ∀ t, s ∈ [0, 1];

(2) p(0)Γ(α)G(t, s) ≤ [M1 + p(0)n] (1 − s)α−1−i tα−1, ∀ t, s ∈ [0, 1];

(3) p(0)Γ(α)G(t, s) ≤ [M1+ p(0)n]s(1 − s)α−1−i,

∀ t, s ∈ [0, 1];

(4) G(t, s) > 0, ∀ t, s ∈ (0, 1) where M1 = sup

0<s≤1

p(s)−p(0)

s , m1 = inf

0<s≤1

p(s)−p(0) s

are positive numbers.

Proof. The proof of (1) and (3) is almost as the same as that in [7] and (4) is obvious. To get (2), check the proof of Lemma 2.4 in [7]. For 0 < s ≤ t ≤ 1, we get that

p(0)Γ(α)G(t, s) = p(s)(1 − s)α−1−itα−1

−p(0)(t − s)α−1

= [p(s) − p(0)](1 − s)α−1−itα−1 +p(0)[(1 − s)α−1−itα−1

−(t − s)α−1]

≤ M1s(1 − s)α−1−itα−1 +p(0)(1 − s)α−1−itα−1 [1 − (1 −s

t)][1 + (1 −s t) + (1 −s

t)2+ · · · + (1 − s t)n−1]

≤ M1s(1 − s)α−1−itα−1 +p(0)(1 − s)α−1−itα−2sn

≤ M1s(1 − s)α−1−itα−1 +p(0)(1 − s)α−1−itα−2tn

≤ [M1+ p(0)n](1 − s)α−1−itα−1. For 0 < t ≤ s ≤ 1, we have that

p(0)Γ(α)G(t, s) = p(s)(1 − s)α−1−itα−1

= [p(s) − p(0)](1 − s)α−1−itα−1 +p(0)(1 − s)α−1−itα−1

≤ [M1+ p(0)n](1 − s)α−1−itα−1. Let q > 1 satisfy 1p + 1q = 1. Then, ϕ−1p (s) = ϕq(s). To study the PFDE (1), we first consider the associated linear PFDE

Dβ0+p(Dα0+u(t))) + y(t) = 0, 0 < t < 1, u(0) = u0(0) = · · · = u(n−2)(0) = 0, Dα0+u(0) = 0, u(i)(1) =

P

j=1

ηju(ξj),

(5)

for y ∈ L1[0, 1] and y ≥ 0.

(3)

Lemma 5 The unique solution for the associated lin- ear PFDE (5) can be written by

u(t) =  1 Γ(β)

q−1Z 1 0

G(t, s)

·ϕ−1p  Z s

0

(s − τ )β−1y(τ )dτds.

Proof. Let w = Dα0+u, v = ϕp(w). Then, the initial value problem

( Dβ0+v(t) + y(t) = 0, t ∈ (0, 1),

v(0) = 0 (6)

has the solution v(t) = c1tβ−1− Iβy(t), t ∈ [0, 1].

Noticing that v(0) = 0, 0 < β ≤ 1, we have that c1 = 0. As a consequence,

v(t) = −Iβy(t), t ∈ [0, 1]. (7) Considering that D0+α u = w, w = ϕ−1p (v), we have from (7) that

Dα0+u(t) = ϕ−1p (−Iβ(y(t))), 0 < t < 1, u(0) = u0(0) = · · · = u(n−2)(0) = 0, u(i)(1) =

P

j=1

ηju(ξj).

(8)

By Lemma 3, the solution of (8) can be expressed by u(t) =  1

Γ(β)

q−1Z 1 0

G(t, s)

·ϕ−1p  Z s

0

(s − τ )β−1y(τ )dτds.

Definition 6 A continuous function Ψ(t) is called a lower solution of the PFDE (1) if it satisfies

−Dβ0+p(D0+α Ψ(t))) ≤ f (t, Ψ(t)), 0 < t < 1, Ψ(0) ≥ 0, Ψ0(0) ≥ 0, · · · , Ψ(n−2)(0) ≥ 0, D0+α Ψ(0) ≥ 0, Ψ(i)(1) ≥

P

j=1

αjΨ(ξj).

Definition 7 A continuous function Φ(t) is called a upper solution of the PFDE (1) if it satisfies

−Dβ0+p(D0+α Φ(t))) ≥ f (t, Φ(t)), 0 < t < 1, Φ(0) ≤ 0, Φ0(0) ≤ 0, · · · , Φ(n−2)(0) ≤ 0, D0+α Φ(0) ≤ 0, Φ(i)(1) ≤

P

j=1

αjΦ(ξj).

Let

F = {u ∈ C([0, 1], R), u(0) = u0(0) = · · ·

= u(n−2)(0) = 0, u(i)(1) =

X

j=1

αju(ξj)}.

Lemma 8 Let u ∈ F such that −D0+α u(t) ≥ 0, t ∈ [0, 1]. Then u(t) ≥ 0, t ∈ [0, 1].

Proof. Let −Dα0+u(t) = h(t). Noticing that u ∈ F , by Lemma 3, we know that

u(t) = Z 1

0

G(t, s)h(s)ds.

It follows from Lemma 4 and h(t) ≥ 0 that u(t) ≥ 0, t ∈ [0, 1].

Lemma 9 (Leray-Schauder fixed point theorem) Let T be a continuous and compact mapping of a Banach spaceE into itself, such that the set

{x ∈ E : x = σT x, for some 0 ≤ σ ≤ 1} (9) is bounded. ThenT has a fixed point.

3 Main results

Denote e(t) = tα−1, m = p(0)Γ(α)m1 , M = Mp(0)Γ(α)1+p(0)n. We list below some assumptions used in this paper.

(H0) 0 <

Z 1 0

ϕ−1p  Z s

0

(s − τ )β−1f (τ, e(τ )) dτds < +∞.

(H1) f ∈ C((0, 1) × J, R+), for any fixed t ∈ (0, 1), f (t, u) is non-increasing in u, for any c ∈ (0, 1), there exist λ > 0 such that for all (t, u) ∈ (0, 1] × J ,

f (t, cu) ≤ c−λf (t, u). (10) From (10), it is easy to see that if c ∈ [1, +∞), then

f (t, cu) ≥ c−λf (t, u). (11) Let

P = {x ∈ C[0, 1] : x(t) ≥ 0, t ∈ [0, 1]}.

Obviously, P is a normal cone in the Banach space E.

Now, define a subset D in E as follows D = {u ∈ P : there exist two positive

numbers lu < 1 < Lu such that lue(t) ≤ u(t)

≤ Lue(t), t ∈ [0, 1]}.

(12)

Obviously, D is nonempty since e(t) ∈ P. Now define an operator A as follows:

(Au)(t) =  1 Γ(β)

q−1Z 1 0

G(t, s) (13)

·ϕ−1p  Z s

0

(s − τ )β−1

f (τ, u(τ ))dτds, t ∈ [0, 1].

(4)

Theorem 10 Assume that (H0) and (H1)hold. Then the PFDE (1) has at least one positive solutionw ∈ D, and there exist constants 0 < k < 1 and K > 1 such thatke(t) ≤ w(t) ≤ Ke(t), t ∈ [0, 1].

Proof. Firstly, we show that A : D → D is well defined.

In fact, for any u ∈ D, there exist two positive numbers Lu> 1 > lu such that

lue(t) ≤ u(t) ≤ Lue(t), t ∈ [0, 1]. (14) We have from (H0), (H1), Lemma 4, (10), (11), (14) that

 1 Γ(β)

q−1Z 1 0

G(t, s) ϕ−1p 

Z s 0

(s − τ )β−1f (τ, u(τ ))dτds

 1

Γ(β)

q−1

M tα−1 Z 1

0

ϕ−1p  Z s

0

(s − τ )β−1f (τ, lue(τ ))dτds

 1

Γ(β)

q−1

l−λ(q−1)u M

· Z 1

0

ϕ−1p  Z s

0

(s − τ )β−1f (τ, e(τ ))dτds

·e(t)

< +∞, and

 1 Γ(β)

q−1Z 1 0

G(t, s) ϕ−1p 

Z s 0

(s − τ )β−1f (τ, u(τ ))dτds

 1 Γ(β)

q−1

L−λ(q−1)u m · Z 1

0

s(1 − s)α−1−i ϕ−1p 

Z s 0

(s − τ )β−1f (τ, e(τ ))dτds · e(t).

By (H0), it is clear that Z 1

0

s(1 − s)α−1−i ϕ−1p 

Z s 0

(s − τ )β−1f (τ, e(τ ))dτds

Z 1

0

ϕ−1p  Z s

0

(s − τ )β−1f (τ, e(τ ))dτds

< +∞.

Thus, we have proved that A : D → D is well de- fined.

By Lemma 5, we know that Au(t) satisfy the fol- lowing equation

−D0+βp(D0+α (Au)(t))) = f (t, u(t)), 0 < t < 1, (Au)(0) = (Au)0(0) = · · ·

= (Au)(n−2)(0) = 0, Dα0+(Au)(0) = 0, (Au)(i)(1) =

P

j=1

ηj(Au)(ξj).

(15)

Now, we are in position to find a pair of upper and lower solutions for PFDE (1). Let

u0(t) =  1 Γ(β)

q−1Z 1 0

G(t, s) ϕ−1p 

Z s 0

(s − τ )β−1f (τ, e(τ ))dτds, t ∈ [0, 1].

By Lemma 4, we get that u0(t) ≥  1

Γ(β)

q−1

m Z 1

0

ϕ−1p  Z s

0

(s − τ )β−1 f (τ, e(τ ))dτds · e(t), t ∈ [0, 1].

As a consequence, there exists a constant k0 ≥ 1 such that

k0u0(t) ≥ e(t), ∀ t ∈ [0, 1]. (16) It following from (H0), (H1) and (16) that A is de- creasing on u, thus for k > k0, we have

 1 Γ(β)

q−1Z 1 0

G(t, s)

ϕ−1p R0s(s − τ )β−1f (τ, ku0(τ ))dτds

 1 Γ(β)

q−1Z 1 0

G(t, s) ϕ−1p 

Z s 0

(s − τ )β−1f (τ, k0u0(τ ))dτds

 1 Γ(β)

q−1Z 1 0

G(t, s)

ϕ−1p R0s(s − τ )β−1f (τ, e(τ ))dτds < +∞, and

u0(t) ≤  1 Γ(β)

q−1

M Z 1

0

ϕ−1p  Z s

0

(s − τ )β−1 f (τ, e(τ ))dτds < +∞.

Let ρ =  1 Γ(β)

q−1

M Z 1

0

ϕ−1p  Z s

0

(s − τ )β−1 f (τ, e(τ ))dτds + 1. Take

k = maxnk0,h 1 Γ(β)

q−1

m Z 1

0

ϕ−1p  Z s

0

(s − τ )β−1f (τ, ρ)dτdsi

1 λ(q−1)o

(5)

Then, we have

+∞ > 1 Γ(β)

q−1Z 1 0

G(t, s) ϕ−1p 

Z s 0

(s − τ )β−1 f (τ, ku0(τ ))dτds

≥ (k)−λ(q−1) 1 Γ(β)

q−1

mtα−1 Z 1

0

ϕ−1p  Z s

0

(s − τ )β−1 f (τ, u0(τ ))dτds

≥ (k)−λ(q−1) 1 Γ(β)

q−1

mtα−1 Z 1

0

ϕ−1p  Z s

0

(s − τ )β−1f (τ, ρ)dτds

≥ tα−1, ∀ t ∈ [0, 1].

(17)

Let

Φ(t) = ku0(t), Ψ(t) = (AΦ)(t). (18) Then, it follows from (16) and (17) that

Φ(t) = k 1 Γ(β)

q−1Z 1 0

G(t, s) (19)

ϕ−1p  Z s

0

(s − τ )β−1f (τ, e(τ ))dτds

≥ tα−1,

Ψ(t) =  1 Γ(β)

q−1Z 1 0

G(t, s) (20)

ϕ−1p  Z s

0

(s − τ )β−1f (τ, ku0(τ ))dτds

≥ tα−1.

In addition, by (15) and (18), we know that Φ(0) = Φ0(0) = · · · = Φ(n−2)(0) = 0, Dα0+Φ(0) = 0, Φ(i)(1) =

X

j=1

ηjΦ(ξj),

Ψ(0) = Φ0(0) = · · · = Ψ(n−2)(0) = 0, D0+α Ψ(0) = 0, Ψ(i)(1) =

X

j=1

ηjΨ(ξj).

By (18),

Ψ(t) = (AΦ)(t) = 1 Γ(β)

q−1Z 1 0

G(t, s)ϕ−1p  Z s

0

(s − τ )β−1 f (τ, ku0(τ ))dτds

≤ k 1 Γ(β)

q−1Z 1 0

G(t, s) ϕ−1p 

Z s 0

(s − τ )β−1 f (τ, u0(τ ))dτds

= Φ(t), ∀ t ∈ [0, 1].

(21)

Considering the fact that f is non-increasing in u, we get from (18)-(21) that

D0+βp(D0+α Ψ(t))) + f (t, Ψ(t))

= D0+βp(D0+α (AΦ)(t))) + f (t, Ψ(t))

≥ D0+βp(D0+α (AΦ)(t))) + f (t, Φ(t))

= −f (t, Φ(t)) + f (t, Φ(t)) = 0,

(22)

D0+βp(D0+α Φ(t))) + f (t, Φ(t))

= D0+βp(D0+α A(tα−1))) + f (t, Φ(t))

= −f (t, tα−1) + f (t, Φ(t))

≤ −f (t, tα−1) + f (t, tα−1) = 0.

(23)

By (22) and (23), we know that Φ, Ψ ∈ P are desired upper and lower solutions of the PFDE (1), respective- ly.

Define a function F as follows F (t, u) =

f (t, Ψ(t)), u < Ψ(t), f (t, u(t)), Ψ(t) ≤ u ≤ Φ(t), f (t, Φ(t)), Φ(t) < u.

(24)

This together with (H1) shows that F : (0, 1)×R+→ R+is continuous.

In the following, we shall show that the fractional boundary value problem

D0+βp(D0+α u(t))) + F (t, u(t)) = 0, 0 < t < 1, u(0) = u0(0) = · · · = u(n−2)(0) = 0, D0+α u(0) = 0, u(i)(1) =

P

j=1

αju(ξj)

(25)

has a positive solution.

Let

(T u)(t) =  1 Γ(β)

q−1Z 1 0

G(t, s) (26) ϕ−1p 

Z s 0

(s − τ )β−1

F (τ, u(τ ))dτds, t ∈ [0, 1].

(6)

Then T : E → E and a fixed point of the operator T is a solution of the PFDE (25). By (20), (21), the def- inition of F and the fact that f (t, u) is non-increasing in u, we have that

f (t, Φ(t)) ≤ F (t, u(t)) ≤ f (t, Ψ(t)), ∀ x ∈ E, (27) and

f (t, Φ(t)) ≤ F (t, u(t)) ≤ f (t, tα−1), ∀ x ∈ E. (28) By Lemma 4 and (28), for u ∈ E, we have

(T u)(t) =  1 Γ(β)

q−1Z 1 0

G(t, s) ϕ−1p 

Z s 0

(s − τ )β−1 F (τ, u(τ ))dτds

 1 Γ(β)

q−1

M tα−1 Z 1

0

ϕ−1p  Z s

0

(s − τ )β−1 f (τ, e(τ ))dτds < +∞,

(29)

which means that T is uniformly bounded. Consider- ing the uniform continuity of G(t, s) on [0, 1] × [0, 1], it can be easily seen that T : E → E is completely continuous. In addition, we have from (29) that (9) holds. Thus, Schauder fixed point theorem guarantees that T has at least one fixed point w.

Now, we are in position to show that

Ψ(t) ≤ w(t) ≤ Φ(t), t ∈ [0, 1]. (30) Since w is a fixed point of T , we have by (25) that w(0) = w0(0) = · · · = w(n−2)(0) = 0, (31) D0+α w(0) = 0, w(i)(1) =

X

j=1

αjw(ξj),

Φ(0) = Φ0(0) = · · · = Φ(n−2)(0) = 0, (32) Dα0+Φ(0) = 0, Φ(i)(1) =

X

j=1

ηjΦ(ξj).

Let z(t) = ϕp(D0+α Φ(t)) − ϕp(Dα0+w(t)). Then Dβ0+z(t) = Dβ0+p(D0+α Φ(t)))

−Dβ0+p(Dα0+w(t)))

= −f (t, tα−1) + F (t, w(t))

≤ 0, t ∈ [0, 1], Dβ0+z(0) = Dβ0+p(D0+α Φ(0)))

−Dβ0+p(Dα0+w(0))) = 0.

By (6) and (8), we know that z(t) ≤ 0, which means that

ϕp(Dα0+Φ(t)) − ϕp(D0+α w(t)) ≤ 0.

We get from the fact that ϕpis monotone increasing ϕp(Dα0+Φ(t)) ≤ ϕp(Dα0+w(t)),

i.e.,

−ϕp(D0+α (Φ − w))(t) ≥ 0.

It follows from Lemma 8, (32) and (33) that Φ(t) − w(t) ≥ 0.

Thus, we have proved that w(t) ≤ Φ(t) on [0, 1].

Similarly, we can get that w(t) ≥ Ψ(t) on [0, 1].

As a consequence, (30) holds. So, F (t, w(t)) = f (t, w(t)), t ∈ [0, 1]. Hence, w(t) is a positive so- lution of the PFDE (1). Noticing that Φ, Ψ ∈ D, by (30), we can easily know that there exist constants 0 < k < 1 and K > 1 such that

ke(t) ≤ w(t) ≤ Ke(t), t ∈ [0, 1].

4 An example

Example Consider the following singular PFDE

D

1 2

0+3(D

7 2

0+u))(t) + 12t121 u16 = 0, 0 < t < 1, u(0) = u0(0) = u00(0) = 0,

D

7 2

0+u(0) = 0, u0(1) =

P

j=1 2 j2u(1j).

(33)

In this situation, f (t, u) = 12t121 u16, α = 72, β =

1

2, p = 3, e(t) = t52, ∆ = 52, αj = j22, ξj = 1j, P

j=1αjξjα−1 ≈ 2.109 < ∆. By simple computa- tion, we have

0 <

Z 1 0

ϕ−1p  Z s

0

(s − τ )β−1f (τ, e(τ ))dτds

= Z 1

0

1 2

Z s 0

(s − τ )12τ121τ125

1 2ds

= Z 1

0

1 2

Z 1 0

(1 − τ )12τ12

1 2ds

=

√2π

2 < +∞.

Therefore, (H1) holds. It is easy to see that (H2) is satisfied for λ = 16. By Theorem 10, PFDE (33) has at least one positive solution w such that there ex- ist constants 0 < k < 1 and K > 1 with ke(t) ≤ w(t) ≤ Ke(t), t ∈ [0, 1].

(7)

Acknowledgements: The project is supported fi- nancially by the Foundations for Jining Medical Col- lege Natural Science (No.JYQ14KJ06), Outstand- ing Middle-Aged and Young Scientists of Shandong Province (BS2010SF004), the National Natural Sci- ence Foundation of China (11371221, 11071141), a Project of Shandong Province Higher Educational Science and Technology Program (No.J10LA53). The corresponding author of this paper is: Dr. Xingqiu Zhang.

References:

[1] S. G. Samko, A. A. Kilbas, O. I. Marichev, Frac- tional Integral and Derivative, in: Theory and Applications, Gordon and Breach, Switzerland, 1993.

[2] I. Podlubny, Fractional Differential Equations, in: Mathematics in science and Engineering, vol. 198, Academic Press, New York, London, Toronto, 1999.

[3] A. A. Kilbas, H. M. Srivastava, J.J. Trujillo, The- ory and Applications of Fractional Differential Equations, in: North-Holland Mathematics S- tudies, vol. 204, Elsevier Science B.V., Amster- dam, 2006.

[4] H. A. H. Salen, On the fractional order m-point boundary value problem in reflexive Banach s- paces and weak topologies, J. Comput. Appl.

Math.224 ,2009, pp. 565-572.

[5] Y. Wang, L. Liu, Positive solutions for fraction- al m-point boundary value problem in Banach spaces, Acta Math. Sci. 32A, 2012, pp. 246-256.

[6] L. Wang, X. Zhang, Positive solutions of m- point boundary value problems for a class of nonlinear fractional differential equations, J. Ap- pl. Math. Comput.42 , 2013, pp. 387-399.

[7] X. Lu, X. Zhang, L. Wang, Existence of posi- tive solutions for a class of fractional differential equations with m-point boundary value condi- tions, J. Sys. Sci. & Math. Scis. 34 (2), 2014, pp.

1-13.

[8] H. Gao, X. Han, Existence of positive solution- s for fractional differential equation with nonlo- cal boundary condition, Int. J. Differ. Equ. 2011, 2011, pp. 1-10.

[9] X. Zhang, Positive solutions for a class of singu- lar fractional differential equation with infinite- point boundary value conditions, Appl. Math.

Lett.39, 2015, pp. 22-27.

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Wu, The eigenvalue for a class of singular p- Laplacian fractional differential equation, Appl.

Math. Comput.235, 2014, pp. 412-422.

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Math. Comput.2014, DOI 10.1007/s12190-014- 0772-7.

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