Sheng et al. Boundary Value Problems (2018) 2018:70 https://doi.org/10.1186/s13661-018-0990-2

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Positive solutions to fractional boundary-value problems with p-Laplacian on time scales Kai Sheng1 , Wei Zhang1 and Zhanbing Bai1* *

Correspondence: [email protected] 1 College of Mathematics and System Science, Shandong University of Science and Technology, Qingdao, P.R. China

Abstract In this article, we consider the following boundary-value problem of nonlinear fractional differential equation with p-Laplacian operator: Dα (φp (Dα u(t))) = f (t, u(t)),

t ∈ [0, 1]T ,

u(0) = u(σ (1)) = Dα u(0) = Dα u(σ (1)) = 0, where 1 < α ≤ 2 is a real number, the time scale T is a nonempty closed subset of R. Dα is the conformable fractional derivative on time scales, φp (s) = |s|p–2 s, p > 1, φp–1 = φq , 1/p + 1/q = 1, and f : [0, σ (1)] × [0, +∞) → [0, +∞) is continuous. By the use of the approach method and fixed-point theorems on cone, some existence and multiplicity results of positive solutions are acquired. Some examples are presented to illustrate the main results. MSC: 34B15 Keywords: Conformable fractional derivative; Time scales; Fixed-point theorems on cone; p-Laplacian operator

1 Introduction In this paper, the existence and multiplicity of positive solutions for the following fractional differential boundary-value problem on time scales is studied:      Dα φp Dα u(t) = f t, u(t) , t ∈ [0, 1]T ,     u(0) = u σ (1) = Dα u(0) = Dα u σ (1) = 0,

(1.1) (1.2)

where 1 < α ≤ 2, Dα is the conformable fractional derivative on time scales, φp (s) = |s|p–2 s, p > 1, φp–1 = φq , 1/p + 1/q = 1, and f : [0, σ (1)] × [0, +∞) → [0, +∞) is continuous. The existence of positive solutions for boundary-value problem on time scales has become the focus in recent years; for details, see [1–6]. Due to the wide applications, many researchers studied the existence of positive solutions for fractional derivatives boundaryvalue problem [7–21] and the references therein. Meanwhile, the boundary-value problem with p-Laplacian operator have also been discussed extensively in the literature; for example, see [4, 11, 22–27]. © The Author(s) 2018. This article is distributed under the terms of the Creative Commons Attribution 4.0 International License (http://creativecommons.org/licenses/by/4.0/), which permits unrestricted use, distribution, and reproduction in any medium, provided you give appropriate credit to the original author(s) and the source, provide a link to the Creative Commons license, and indicate if changes were made.

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For α = 2, problem (1.1), (1.2) is called a fourth order p-Laplacian boundary-value problem which has been studied in [4]. Dong et al. [22] investigated the boundary-value problem for a fractional differential equation with the p-Laplacian operator      Dα φp Dα u(t) = f t, u(t) ,

0 < t < 1,

u(0) = u(1) = Dα u(0) = Dα u(1) = 0,

(1.3) (1.4)

where 1 < α ≤ 2 is a real number, Dα is the conformable fractional derivative, φp (s) = |s|p–2 s, p > 1, φp–1 = φq , 1/p + 1/q = 1,f : [0, 1] × [0, +∞) → [0, +∞) is continuous. By the use of the fixed-point theorems on cone, some existence and multiplicity results of positive solutions are obtained. Motivated by the work mentioned above, we investigate the existence and multiplicity of positive solutions for (1.3), (1.4) on time scales. The rest of this paper is organized as follows. In Sect. 2, we recall some concepts relative to the new conformable fractional calculus and give some lemmas with respect to the corresponding Green’s function. In Sect. 3, we investigate the existence and multiplicity of positive solution for boundaryvalue problem (1.1), (1.2). In Sect. 4, we present some examples to illustrate our main results, respectively.

2 Preliminaries and lemmas In this section, we introduced notations and definitions of conformable fractional derivative on time scales and some lemmas. Let T be a time scale and denote [a, b]T =: [a, b] ∩ T. These results can be found in the recent literature; see [2, 3, 6]. Definition 2.1 A time scale T is a nonempty closed subset of R; assume that T has the topology that it inherits from the standard topology on R. Define the forward and backward jump operators σ , ρ: T → T by σ (t) = inf{τ > t | τ ∈ T},

ρ(t) = sup{τ < t | τ ∈ T}.

In this definition we put inf ∅ = sup T, sup ∅ = inf T. Set σ 2 (t) = σ (σ (t)), ρ 2 (t) = ρ(ρ(t)). The sets T k and Tk which are derived from the time scale T are as follows:   T k := t ∈ T : t is not maximal or ρ(t) = t ,   Tk := t ∈ T : t is not minimal or σ (t) = t . Denote interval I on T by IT = I ∩ T. Definition 2.2 If f : T → R is a function and t ∈ Tk , then the delta derivative of f at the point t is defined to be the number f  (t) (provided it exists) with the property that, for each  > 0, there is a neighborhood U of t such that        f σ (t) – f (s) – f  (t) σ (t) – s  ≤  σ (t) – s for all s ∈ U. The function f is called -differentiable on T k if f  (t) exists for all t ∈ T k .

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Definition 2.3 ([3]) Let α ∈ (1, 2] and f : T → R, t ∈ T k . For t > 0, we define Dα f (t) to be the number (provided it exists) with the property that, given any ε > 0, there is a δneighborhood Vt ⊂ T of t, δ > 0, such that          f σ (t) – f (s) t 2–α – Dα f (t) σ (t) – s  ≤  σ (t) – s. We call Dα f (t) the conformable fractional derivative of f of order α at t, and we define the conformable fractional derivative at 0 as Dα f (0) = limt→0+ Dα f (t). Lemma 2.1 ([3]) Let α ∈ (1, 2] and f be two times delta differentiable at t ∈ T k . The following relation holds: Dα f (t) = t 2–α f (t). Definition 2.4 ([2]) A function f : T → R is called regulated provided its right-sided limits exist (and are finite) at all right-dense points in T and its left-sided limits exist (and are finite) at all left-dense points in T. Definition 2.5 ([2]) A function f : T → R is called rd-continuous provided it is continuous at right-dense points in T and its left-sided limit exist (finite) at all left-dense points in T. The set of rd-continuous functions f : T → R will be denoted by Crd (T, R). Lemma 2.2 ([2]) Assume f : T → R. (i) If f is continuous, then f is rd-continuous. (ii) If f is rd-continuous, then f is regulated. Definition 2.6 ([3]) Let f : T → R be a regulated function and 1 < α ≤ 2. Then the αfractional integral of f is defined by   I α f (t) = I 2 t α–2 f (t) =



t

(t – s)sα–2 f (s)s.

(2.1)

0

Lemma 2.3 Let t > 0, α ∈ (1, 2], and the function f : [0, ∞)T → R be rd-continuous, then Dα I α f (t) = f (t). Proof Since f (t) is rd-continuous, then f (t) is regulated, and I α f (t) is twice times differentiable. In view of Lemma 2.1, one has   Dα I α f (t) = t 2–α





t

(t – s)sα–2 f (s)s 0

=t

2–α

f (t)t α–2

= f (t). The proof is complete.



Lemma 2.4 (Mean value theorem [6]) Let a ≥ 0 and f : T → R be a function continuous on [a, b]T which is conformable fractional differentiable of order with α on [a, b]T . Then there exist ξ , τ ∈ [a, b]T such that ξ α–1 Dα f (ξ ) ≤

f (b) – f (a) ≤ τ α–1 Dα f (τ ). b–a

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Lemma 2.5 Let α ∈ (1, 2], f be a α-differentiable function at t > 0, then Dα f (t) = 0 for t ∈ [0, 1]T if and only if f (t) = a0 + a1 t, where ak ∈ R, for k = 0, 1. Proof The sufficiency follows by the definition of the delta derivative on time scales. Next, given t1 , t2 ∈ [0, 1]T with t1 < t2 , by Lemma 2.4, there exists ξ , τ ∈ (t1 , t2 )T such that ξ α–1 Dα f (ξ ) ≤

f  (t2 ) – f  (t1 ) ≤ τ α–1 Dα f (τ ). t2 – t1

By means of Dα f (ξ ) = Dα f (τ ) = 0, we have f  (t2 ) = f  (t1 ), with the arbitrariness of t1 , t2 , one has f  (t) is a constant, so f (t) = a0 + a1 t, for t ∈ [0, 1]T .  With Lemma 2.3 and Lemma 2.5, the following lemma is immediate. Lemma 2.6 Assume that u ∈ C(0, +∞)T with a fractional derivative of order α ∈ (1, 2]. Then I α Dα u(t) = u(t) + c0 + c1 t,

(2.2)

for some ck ∈ R, k = 0, 1. We present below the Green’s function and its properties. Lemma 2.7 Given y ∈ C[0, σ (1)]T , the unique solution of Dα u(t) + y(t) = 0,

t ∈ [0, 1]T ,

  u(0) = u σ (1) = 0,

(2.3)

(2.4)

is

σ (1)

u(t) =

G(t, s)y(s)s, 0

where ⎧ 1 ⎨(σ (1) – t)sα–1 , for 0 ≤ s ≤ t; G(t, s) = σ (1) ⎩tsα–2 (σ (1) – s), for t ≤ s ≤ 1.

(2.5)

Proof By the use of the Lemma 2.6, we can deduce from equation (2.3) an equivalent integral equation, u(t) = –I α y(t) + c0 + c1 t t = – (t – s)sα–2 y(s)s + c0 + c1 t, 0

for some c0 , c1 ∈ R. By (2.4), there are c0 = 0,

c1 =

1 σ (1)

0

σ (1) 

 σ (1) – s sα–2 y(s)s.

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Therefore, the unique solution of Problem (2.3), (2.4) is

t

(t – s)sα–2 y(s)s +

u(t) = – 0

=

1 σ (1)



t

t σ (1)



σ (1) 

 σ (1) – s sα–2 y(s)s

0

 –σ (1)tsα–2 + σ (1)sα–1 y(s)s

0

σ (1)   1 tσ (1)sα–2 – tsα–2 y(s)s + σ (1) 0 t   1 = σ (1) – t (s)α–1 y(s)s σ (1) 0 σ (1)   1 tsα–2 σ (1) – s y(s)s + σ (1) t σ (1) G(t, s)y(s)s. = 0



The proof is complete.

We point out here that (2.5) becomes the usual Green’s function when α = 2 on time scales. Lemma 2.8 Let y ∈ C[0, σ (1)] and 1 < α ≤ 2. Then the problem    Dα φp Dα u(t) = y(t), t ∈ [0, 1]T ,     u(0) = u σ (1) = Dα u(0) = Dα u σ (1) = 0,

(2.6) (2.7)

has a unique solution



σ (1)

u(t) =

σ (1)

G(t, s)φq 0

 G(s, τ )y(τ )τ s.

(2.8)

0

Proof Applying operator I α on both sides of (2.6), with Lemma 2.6,   φp Dα u(t) + C0 + C1 t = I α y(t). So,   φp Dα u(t) = I α y(t) – C0 – C1 t t (t – τ )τ α–2 y(τ )τ – C0 – C1 t = 0

for some C0 , C1 ∈ R. By the boundary conditions Dα u(0) = Dα u(σ (1)) = 0, as a consequence we have

C0 = 0,

C1 =

1 σ (1)

0

σ (1) 

 σ (1) – τ τ α–2 y(τ )τ .

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Therefore, the solution u(t) of fractional differential equation boundary-value problem (2.6) and (2.7) satisfies   φp Dα u(t) =



t

(t – τ )τ α–2 y(τ )τ 0

σ (1)   1 σ (1) – τ τ α–2 y(τ )τ t σ (1) 0 σ (1) G(t, τ )y(τ )τ . =– –

0

Thus, the fractional differential equation boundary-value problem (2.6) and (2.7) is equivalent to the problem





σ (1)

Dα u(t) + φq

G(t, τ )y(τ )τ

= 0,

0 < t < 1,

0

  u(0) = u σ (1) = 0. Lemma 2.7 implies that fractional differential equation boundary-value problem (2.6), (2.7) has a unique solution u(t) =



σ (1)

G(t, s)φq 0

σ (1)

 G(s, τ )y(τ )τ s.

0



The proof is complete. Lemma 2.9 The function G(t, s) defined by (2.5) satisfies: (i) G(t, s) ≥ 0, for t ∈ [0, σ (1)], s ∈ [0, 1], and G(t, s) > 0, for t ∈ (0, σ (1)), s ∈ (0, 1); (ii) G(t, s) ≤ G(s, s), for t ∈ [0, σ (1)], s ∈ [0, 1]; (iii) G(t, s) ≥ σ 4(1) G(s, s), for t ∈ [ σ 4(1) , 3σ4(1) ], s ∈ [0, 1].

Proof Observing the expression of G(t, s), it is clear that G(t, s) ≥ 0 for t ∈ [0, σ (1)], s ∈ [0, 1], and G(t, s) > 0, for t ∈ (0, σ (1)), s ∈ (0, 1). Moreover, G(t, s) is decreasing with respect to t for s ≤ t, and increasing for t ≤ s. By the fact ⎧ ⎨t,

G(t, s) = s G(s, s) ⎩ σ (1)–t , σ (1)–s

t ≤ s; s ≤ t.

We have G(t, s) ≤ G(s, s), for t ∈ [0, σ (1)], s ∈ [0, 1]. Furthermore, if t ∈ [ σ 4(1) , 3σ4(1) ], s ∈ [0, 1], one has G(t, s) σ (1) ≥ , G(s, s) 4 which implies the desired results.



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Lemma 2.10 ([27]) The following relations hold: (1) If 1 < q ≤ 2, then |φq (u + v) – φq (u)| ≤ 22–q |v|q–1 for u, v ∈ R. (2) If q > 2, then |φq (u + v) – φq (u)| ≤ (q – 1)(|u| + |v|)q–2 |v| for u, v ∈ R. Lemma 2.11 ([28]) Suppose E is a Banach space and Tn : E → E, n = 3, 4, . . . are completely continuous operators, T : E → E. If Tn u – Tu uniformly to zero when n → ∞ for all bounded set ⊆ E, then T : E → E is completely continuous. Definition 2.7 The map θ is said to be a nonnegative continuous concave functional on a cone P of a Banach space E provided that θ : P → [0, ∞) is continuous and   θ tx + (1 – t)y ≥ tθ (x) + (1 – t)θ (y) for all x, y ∈ P and 0 < t < 1. The following fixed-point theorems are useful in our proofs. Lemma 2.12 ([29]) Let E be a Banach space, P ⊆ E be a cone, and 1 , 2 be two bounded open balls of E centered at the origin with 1 ⊂ 2 . Suppose that A : P ∩ ( 2 \ 1 ) → P is a completely continuous operator such that either (i) Ax ≤ x, x ∈ P ∩ ∂ 1 , and Ax ≥ x, x ∈ P ∩ ∂ 2 , or (ii) Ax ≥ x, x ∈ P ∩ ∂ 1 , and Ax ≤ x, x ∈ P ∩ ∂ 2 , holds. Then A has a fixed point in P ∩ ( 2 \ 1 ). Lemma 2.13 ([30]) Let P be a cone in a real Banach space E, Pc = {x ∈ P | x ≤ c}, θ be a nonnegative continuous concave functional on P such that θ (x) ≤ x, for all x ∈ Pc , and P(θ , b, d) = {x ∈ P | b ≤ θ (x), x ≤ d}. Suppose A : Pc → Pc is a completely continuous and there exist constants 0 < a < b < d ≤ c such that (C1) {x ∈ P(θ , b, d) | θ (x) > b} is nonempty, and θ (Ax) > b, for x ∈ P(θ , b, d); (C2) Ax < a, for x ≤ a; (C3) θ (Ax) > b, for x ∈ P(θ , b, c) with Ax > d. Then A has at least three fixed points x1 , x2 , x3 with x1  < a,

b < θ (x2 ),

a < x3 ,

θ (x3 ) < b.

Remark 2.1 ([30]) If we have d = c, then condition (C1) of Lemma 2.13 implies condition (C3) of Lemma 2.13.

3 Existence results Let E = {u : [0, σ (1)] → R} be endowed with the ordering u ≤ v if u(t) ≤ v(t) for all t ∈ [0, σ (1)], and the norm u = max0≤t≤σ (1) |u(t)|. Define      σ (1) 3σ (1) σ (1) u for t ∈ , . P = u ∈ E | u(t) ≥ 0 on 0, σ (1) , u(t) ≥ 4 4 4

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Given a function f ∈ C([0, σ (1)] × [0, ∞), [0, ∞)), define T, Tn : P → E as



σ (1)

(Tu)(t) :=

σ (1)

G(t, s)φq 0

0

(Tn u)(t) :=

   G(s, τ )f τ , u(τ ) τ s,



σ (1)

G(t, s)φq

1 n

   G(s, τ )f τ , u(τ ) τ s,

σ (1) 1 n

n = 3, 4, . . . .

Lemma 3.1 T : P → P is completely continuous. Proof Firstly, take the constant in the second member to be independent on n, Hence, we show that Tn : P → P are completely continuous for n = 3, 4, . . . . Given u ∈ P, with Lemma 2.9 and the nonnegativity of f (t, u), one has (Tn u)(t) =

G(t, s)φq

1 n

≤



σ (1)

  G(s, τ )f τ , u(τ ) τ s

σ (1)

   G(s, τ )f τ , u(τ ) τ s,

1 n



σ (1)

G(s, s)φq

1 n

σ (1)

1 n



so Tn u ≤



σ (1)

G(s, s)φq

1 n

σ (1) 1 n

   G(s, τ )f τ , u(τ ) τ s.

For u ∈ P, min

σ (1) 3σ (1) 4 ≤t≤ 4

(Tn u)(t) = ≥

min

σ (1) 3σ (1) 4 ≤t≤ 4



σ (1) 4



σ (1)

G(t, s)φq

1 n



σ (1)

G(s, s)φq

1 n

σ (1) 1 n

σ (1) 1 n

   G(s, τ )f τ , u(τ ) τ s

   G(s, τ )f τ , u(τ ) τ s.

It follows that min

σ (1) 3σ (1) 4 ≤t≤ 4

(Tn u)(t) ≥

σ (1) Tn u. 4

Hence, Tn u ∈ P, and so Tn : P → P. Let ⊂ P be bounded, i.e., there exists a positive constant M > 0 such that u ≤ M for all u ∈ . Let L=

max

0≤t≤σ (1),0≤u≤M



  f (t, u) + 1,

H=

σ (1)

G(s, s)s + 1, 0

then, for u ∈ , we have   (Tn u)(t) =





σ (1) 1 n

G(t, s)φq

σ (1) 1 n

   G(s, τ )f τ , u(τ ) τ s

≤ Lq–1 H q < +∞. Hence, Tn ( ) is bounded for n = 3, 4, . . . .

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On the other hand, given  > 0, let δ=

 , 2σ (1)Lq–1 H q–1

then, for each u ∈ , t1 , t2 ∈ [0, σ (1)], t1 ≤ t2 , and t2 – t1 < δ, one has   (Tn u)(t2 ) – (Tn u)(t1 ) < . That is to say Tn ( ) has equicontinuity. In fact, we consider three situations. (1) 0 < t1 ≤ t2 < n1 .   (Tn u)(t2 ) – (Tn u)(t1 )  σ (1)

  G(t2 , s)φq = 1 n



–

σ (1) 1 n



σ (1)

G(t1 , s)φq

1 n



≤ Lq–1 H q–1

1 n

σ (1) 

 G(t2 , s) – G(t1 , s)s

1 n



σ (1)

  (t2 – t1 )sα–2 σ (1) – s s

1 q–1 q–1 L H σ (1)

≤

1 q–1 q–1 L H (t2 – t1 ) σ (1)

≤L

H

q–1

     G(s, τ )f τ , u(τ ) τ s

σ (1)

=

q–1

  G(s, τ )f τ , u(τ ) τ s 

1 n



σ (1) 

 σ (1) – s s

0

(t2 – t1 )σ (1)

< . (2) 0 < t1 ≤

1 n

≤ t2 < 1.

  (Tn u)(t2 ) – (Tn u)(t1 )

t2   G(t2 , s) – G(t1 , s)s + ≤ Lq–1 H q–1 1 n

≤ Lq–1 H q–1

1 σ (1)

(t2 – t1 )s

t2 

 (t1 – s)sα–2 σ (1) + (t2 – t1 )sα–1 s

1 n

α–2



 σ (1) – s s

t2

1 q–1 q–1 ≤ L H (t2 – t1 ) σ (1) ≤

1 q–1 q–1 L H (t2 – t1 ) σ (1)

≤2



1 q–1 q–1 L H (t2 – t1 ) σ (1)

≤ 2Lq–1 H q–1 σ (1)(t2 – t1 ) < .

 G(t2 , s) – G(t1 , s)s

t2



σ (1)

+

σ (1) 

σ (1) 



 sα–2 σ (1) + sα–1 s

0 σ (1) 

 σ (1) + s s

0



σ (1)

σ (1)s 0



Sheng et al. Boundary Value Problems (2018) 2018:70

(3)

1 n

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< t1 ≤ t2 < 1.

  (Tn u)(t2 ) – (Tn u)(t1 )

t1   G(t2 , s) – G(t1 , s)s + ≤ Lq–1 H q–1 1 n



 G(t2 , s) – G(t1 , s)s

σ (1) 

+

t2 

 G(t2 , s) – G(t1 , s)s

t1



t2

 t2     1 q–1 q–1 σ (1) – t2 sα–1 – t1 sα–2 σ (1) – s s = L H σ (1) t1  t1 σ (1)   α–1 α–2 α–1 + (t2 – t1 )s σ (1) – s s (t2 – t1 )s s + 1 n

≤

1 q–1 q–1 L H (t2 – t1 ) σ (1)

1 q–1 q–1 L H (t2 – t1 ) ≤ σ (1)

t2



σ (1) 

 sα–2 σ (1) + sα–1 s

0 σ (1) 

 σ (1) + s s

0

< . By the means of the Arzela–Ascoli theorem, we see that Tn : P → P are completely continuous operators. Secondly, it is clear that T : P → P. We prove that Tn : P → P have uniform convergence to T and T : P → P is completely continuous too. With the use of Lemma 2.10, φq (A + B) < φq (A) + 2φq (B) + (q – 1)(A + B)q–2 B. Given  > 0, let

K=

 σ (1) · (2Lq–1 H + qLq–1 H q–1 ) , 

then Tn u – Tu < , for all n > N . In fact, Tn u – Tu

  = max (Tn u)(t) – (Tu)(t) 0≤t≤σ (1)

  = max  0≤t≤σ (1)

G(t, s)φq

σ (1)

< max

0≤t≤σ (1) 0

+ 2φq 0

1 n

   G(s, τ )f τ , u(τ ) τ s

0



σ (1) 1 n

σ (1)

G(t, s)φq

0

–



σ (1)

σ (1) 1 n

    G(s, τ )f τ , u(τ ) τ s

 G(t, s) φq



σ (1) 1 n

  G(s, τ )f τ , u(τ ) τ

  G(s, τ )f τ , u(τ ) τ





Sheng et al. Boundary Value Problems (2018) 2018:70



σ (1)

+ (q – 1) –

  G(s, τ )f τ , u(τ ) τ

0



G(t, s)φq

1 n

≤ max

0≤t≤σ (1) 0



1 n

   G(s, τ )f τ , u(τ ) τ s

  G(s, τ )f s, u(s) τ s 

σ (1) 1 n



G(t, s)φq

σ (1) 1 n

σ (1)

+ 2Lq–1

q–2 0



σ (1) 1 n

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   G(s, τ )f τ , u(τ ) τ s



1 n

G(s, s)sφq 0

 G(τ , τ )τ

0





σ (1)

+ (q – 1)Lq–1

q–2

σ (1)

1 n

G(τ , τ )τ

G(s, s)s 0

0

  ≤ σ (1) · Lq–1 H q–1 + 2Lq–1 H + (q – 1)Lq–1 H q–1 ·

 1 n

G(τ , τ )τ

0

< . By the use of Lemma 2.11, T : P → P is completely continuous.



We take into account that the Green’s function satisfy G(t, s) ≥ 0 for t ∈ [0, σ (1)], s ∈ [0, 1], and G(t, s) > 0, for t ∈ (0, σ (1)), s ∈ (0, 1). The following constants are well defined:





σ (1)

M=

N=

G(τ , τ )τ

G(s, s)sφq 0

–1

σ (1)

,

0 3σ (1) 4

σ (1) 4

σ (1) G(s, s)sφq 4



3σ (1) 4 σ (1) 4

σ (1) G(τ , τ )τ 4

–1 .

Theorem 3.1 Let f ∈ C([0, σ (1)] × [0, ∞), [0, ∞)). Assume that there exist two different positive constants r2 , r1 , and r2 = r1 such that (H1) f (t, u) ≤ φp (Mr1 ), for (t, u) ∈ [0, σ (1)] × [0, r1 ]; (H2) f (t, u) ≥ φp (Nr2 ), for (t, u) ∈ [ σ 4(1) , 3σ4(1) ] × [ σ 4(1) r2 , r2 ]. Then Problem (1.1), (1.2) has at least one positive solution u such that min{r2 , r1 } ≤ u ≤ max{r2 , r1 }. Proof By Lemma 3.1, T : P → P is completely continuous. Without loss of generality, suppose 0 < r1 < r2 , and let   1 := u ∈ P | u < r1 ,

  2 := u ∈ P | u < r2 .

For u ∈ ∂ 1 , we have 0 ≤ u(t) ≤ r1 for all t ∈ [0, σ (1)]. It follows from (H1) that      G(s, τ )f τ , u(τ ) τ s 0 0

σ (1)  σ (1) G(s, s)sq G(τ , τ )τ = r1 = u. ≤ Mr1

  Tu = max  0≤t≤σ (1)



σ (1)

G(t, s)q

0

0

So, Tu ≤ u,

for u ∈ ∂ 1 .

σ (1)

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For u ∈ ∂ 2 , by the definition of P, we have σ (1) σ (1) u = r2 . 4 4

u(t) ≥

min

σ (1) 3σ (1) 4 ≤t≤ 4

By assumption (H2), for t ∈ [ σ 4(1) , 3σ4(1) ], we have



σ (1)

(Tu)(t) =

σ (1)

G(t0 , s)φq 0

   G(s, τ )f τ , u(τ ) τ s

0



σ (1)

≥ 0

≥ Nr2

σ (1) G(s, s)φq 4 3σ (1) 4 σ (1) 4



3σ (1) 4 σ (1) 4

σ (1) G(s, s)φq 4



   σ (1) G(τ , τ )f τ , u(τ ) τ s 4 3σ (1) 4 σ (1) 4

 σ (1) G(τ , τ )τ s 4

= r2 = u. So, Tu ≥ u,

for u ∈ ∂ 2 . 

Therefore, by Lemma 2.12, we complete the proof.

Theorem 3.2 Suppose f ∈ C([0, σ (1)] × [0, ∞), [0, ∞)) and there exist constants 0 < a < b < c such that the following assumptions hold: (A1) f (t, u) ≤ φp (Ma), for (t, u) ∈ [0, σ (1)] × [0, a]; (A2) f (t, u) ≥ φp (Nb), for (t, u) ∈ [ σ 4(1) , 3σ4(1) ] × [b, c]; (A3) f (t, u) ≤ φp (Mc), for (t, u) ∈ [0, σ (1)] × [0, c]. Then the boundary-value problem (1.1), (1.2) has at least three positive solutions u1 , u2 , u3 with   max u1 (t) < a,

b<

0≤t≤σ (1)

  a < max u3 (t) ≤ c,

min

    u2 (t) < max u2 (t) ≤ c,

min

  u3 (t) < b.

σ (1) 3σ (1) 4 ≤t≤ 4

σ (1) 3σ (1) 4 ≤t≤ 4

0≤t≤σ (1)

0≤t≤σ (1)

Proof We show that all the conditions of Lemma 2.13 are satisfied. If u ∈ Pc , then u ≤ c. Assumption (A3) implies f (t, u(t)) ≤ φp (Mc) for 0 ≤ t ≤ σ (1), consequently,   Tu = max  0≤t≤σ (1)

     G(s, τ )f τ , u(τ ) τ s 0 0

σ (1)  σ (1)   G(s, s)φq G(τ , τ )f τ , u(τ ) τ s ≤ 0



σ (1)

0



σ (1)

≤ Mc





σ (1)

G(τ , τ )τ

G(s, s)sφq 0

σ (1)

G(t, s)φq

≤ c.

0

Hence, T : Pc → Pc . Similarly, if u ∈ Pa , then assumption (A1) yields f (t, u(t)) ≤ φp (Ma), 0 ≤ t ≤ σ (1). Therefore, condition (C2) of Lemma 2.13 is satisfied.

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Choose u(t) =

b+c , 2

0 ≤ t ≤ σ (1).

) > b, consequently, Then u(t) ∈ P(θ , b, c), θ (u) = θ ( b+c 2   u ∈ P(θ , b, c) | θ (u) > b = ∅. Hence, if u ∈ P(θ , b, c), then b ≤ u(t) ≤ c for σ 4(1) ≤ t ≤ have f (t, u(t)) ≥ φp (Nb) for σ 4(1) ≤ t ≤ 3σ4(1) . So θ (Tu) =

σ (1) 3σ (1) 4 ≤t≤ 4

> Nb

3σ (1) 4 σ (1) 4



σ (1)

σ (1)

G(t, s)q

min

0

3σ (1) . 4

From assumption (A3), we

   G(s, τ )f τ , u(τ ) τ s

0

σ (1) G(s, s)φq 4



3σ (1) 4 σ (1) 4

 σ (1) G(τ , τ )τ s 4

= b, i.e., θ (Tu) > b,

for all u ∈ P(θ , b, c).

This shows that condition (C1) of Lemma 2.13 is satisfied. By Lemma 2.13 and Remark 2.1, Problem (1.1), (1.2) has at least three positive solutions u1 , u2 , u3 , satisfying   max u1 (t) < a,

0≤t≤σ (1)

  a < max u3 (t), 0≤t≤σ (1)

b<

min

σ (1) 3σ (1) 4 ≤t≤ 4

min

σ (1) 3σ (1) 4 ≤t≤ 4

  u2 (t),

  u3 (t) < b. 

The proof is complete.

4 Examples Example 4.1 Let T = R, α = 32 , p = 3, consider the following fractional differential equation boundary-value problem:  3  3 D 2 φ3 D 2 u(t) = 1 + t + sin u, 3

t ∈ [0, 1]T ,

3

u(0) = u(1) = D 2 u(0) = D 2 u(1) = 0. By a simple computation, we obtain M = 3.75, N ≈ 5.987. Choose r1 = 1, r2 = 15 , then for (t, u) ∈ [0, 1] × [0, 1],     1 1 1 3 , × , . f (t, u) = 1 + t + sin u ≥ 2 > φp (Nr2 ) ≈ 1.1974, for (t, u) ∈ 4 4 20 5 f (t, u) = 1 + t + sin u ≤ 3.5 < φp (Mr1 ) = 3.75,

(4.1) (4.2)

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With the use of Theorem 3.1, the fractional differential equation boundary-value problem (4.1) and (4.2) has at least one positive solution u such that 15 ≤ u ≤ 1. Example 4.2 Let T = R, consider the following fractional differential equation boundaryvalue problem:  3  3 D 2 φ3 D 2 u(t) = f (t, u), 3 2

t ∈ [0, 1]T ,

(4.3)

3 2

u(0) = u(1) = D u(0) = D u(1) = 0,

(4.4)

where ⎧ ⎨2u + 1 t, 10 f (t, u) = ⎩6 + 7(u – 1)2 +

u ≤ 1; 1 t, 10

u ≥ 1.

We obtain M = 3.75, N ≈ 5.987. Choose a = 0.1, b = 1, c = 4, then 1 t < 0.3 < φp (Ma) = 0.375, for (t, u) ∈ [0, 1] × [0, 0.1], 10

   t Nb 1 3 f (t, u) = 6 + 7(u – 1)2 + ≥ 2 > φp ≈ 1.497, (t, u) ∈ , × [1, 4], 10 4 4 4 f (t, u) = 2u +

f (t, u) = 6 + 7(u – 1)2 +

t ≤ 9.1 < φp (Mc) = 15, 10

(t, u) ∈ [0, 1] × [0, 4].

With the use of Theorem 3.2, the fractional differential equation boundary-value problem (4.3) and (4.4) has at least three positive solutions u1 , u2 and u3 with   max u1 (t) < 0.1,

0≤t≤1

  0.1 < max u3 (t) ≤ 4, 0≤t≤1

    1 < min u2 (t) < max u2 (t) ≤ 4, 1 ≤t≤ 3 4 4

0≤t≤1

  min u3 (t) < 1.

1 ≤t≤ 3 4 4

Acknowledgements Not applicable. Funding This work is supported by NSFC (11571207), the Taishan Scholar project and SDUST graduate innovation project SDKDYC170343. Availability of data and materials Data sharing not applicable to this article as no datasets were generated or analyzed during the current study. Competing interests The authors declare that they have no competing interests. Authors’ contributions All authors contributed equally to the writing of this paper. All authors read and approved the final manuscript. Authors’ information The corresponding author is a professor. He has worked on nonlinear functional analysis and fractional boundary-value problems for many years. The first author and the second author are doctorate candidates. Their research field is the solvability of fractional boundary-value problems.

Publisher’s Note Springer Nature remains neutral with regard to jurisdictional claims in published maps and institutional affiliations. Received: 13 March 2018 Accepted: 25 April 2018

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