Existence of positive solutions for integral boundary value problems of fractional differential equations with p-laplacian

Transcription

1 Zhang et al. Advances in Difference Equations :36 DOI /s R E S E A R C H Open Access Existence of positive solutions for integral boundary value problems of fractional differential equations with p-laplacian Luchao Zhang 1,2, Weiguo Zhang 2*, Xiping Liu 2 and Mei Jia 2 * Correspondence: zwgzwm@126.com 2 College of Science, University of Shanghai for Science and Technology, Shanghai, 293, P.R. China Full list of author information is available at the end of the article Abstract This paper is concerned with the existence of positive solutions for integral boundary value problems of Caputo fractional differential equations with p-laplacian operator. By means of the properties of the Green s function, Avery-Peterson fixed point theorems, we establish conditions ensuring the existence of positive solutions for the problem. As an application, an example is given to demonstrate the main result. Keywords: fractional differential equations; Caputo derivative; p-laplacian operator; integral boundary conditions; positive solutions 1 Introduction Recently, fractional differential equations have been proved to be valuable tools in the modeling of many phenomena in various fields of science and engineering, such as rheology, dynamical processes in self-similar and porous structures, heat conduction, control theory, electroanalytical chemistry, chemical physics, economics, etc. Many researchers have shown their interest in fractional differential equations. The motivation for this work stems from both the intensive development of the theory of fractional calculus itself and the applications. Many papers and books have appeared on fractional calculus and fractional differential equations see [1 1]. It is well known that the p-laplacian operator is also used in analyzing mechanics, physics, and dynamic systems, and the related fields of mathematical modeling. However, there are few studies of the existence of positive solution of fractional differential equations with the p-laplacian operator;see [11 18] and the references therein. In [11], Liu et al. studied the solvability of the Caputo fractional differential equation with boundary value conditions involving the p-laplacian operator. The existence and uniqueness of the problem is found by the Banach fixed point theorem. The problem is given in the following: ϕ p D α + xt = f t, xt, for t, 1, x = r x1, x = r 1 x 1, x j =, The Authors 217. This article is distributed under the terms of the Creative Commons Attribution 4. International License which permits unrestricted use, distribution, and reproduction in any medium, provided you give appropriate credit to the original authors and the source, provide a link to the Creative Commons license, and indicate if changes were made.

2 Zhangetal. Advances in Difference Equations :36 Page 2 of 19 where i =2,3,...,[α] 1.Here,ϕ p is the p-laplacian operator and D α + is the Caputo fractional derivative, 1 < α R, and the nonlinear function f C[, 1] R, Risgiven. In [12], Lu et al. studied the existence of non-negative solutions of a nonlinear fractional boundary value problem with the p-laplacianoperator D β + ϕ pd α + ut = f t, ut, for < t <1, u = u = u 1 =, D α + u = Dα + u1 =, where 2 < α 3, 1 < β 3, and D α +, Dβ + are the standard Riemann-Liouville fractional derivatives. Green s functions, the Guo-Krasnoselskii theorem, and the Leggett-Williams fixed point theorems are used. Boundary value problems with integral boundary conditions for ordinary differential equations represent a very interesting and important class of problems. They include two-point, three-point and multi-point boundary value problems as special cases. For an overview of the literature on integral boundary value problems and symmetric solutions, see [19 26] and the references therein. In [24], Zhi et al. studied the existence of positive solutions for nonlocal boundary value problem of the fractional differential equations with p-laplacian operator. The problem is given in the following: ϕ p D α + ut = f t, ut, D β + ut, for < t <1, u = u =, u1 = gsus ds, ϕ p D α + u = λ 1 ϕ p D α + uξ 1, ϕ p D α + u1 = λ 2ϕ p D α + uξ 2, where < ξ 1 ξ 2 <1,2<α <3,1<β < α 1<2, λ 1, λ 2 <1.D α + is the Caputo fractional derivative of order α. In [26], Mahmudov and Unul studied the existence of the solutions of the fractional differential equation with p-laplacian operator and integral conditions is discussed. The problem is given in the following: D β + φ pd α + ut = f t, ut, Dγ + ut, 1 u + μ 1 u1 = σ 1 gs, us ds, u + μ 2 u 1 1 = σ 2 hs, us ds, D α + u =, D α + u1 = υdα + uη, where D α +, Dβ + are the Caputo fractional derivative operators with 1 < α <2,1<β <2.μ 1, σ i i = 1, 2 are non-negative constants. f, g, h are continuous functions. Motivated by the above work, we investigate the following integral boundary value problems for short, BVP of fractional differential equations with p-laplacian: D β + ϕ pd α + xt = f t, xt, Dβ + xt, t, 1 ϕ p D α + xi = ϕ p D α + x1 =, i =1,2,...,m 1, x1 = g 1sxs ds, x = g 2sxs ds, x j =, j =2,3,...,n 1, 1.1

3 Zhangetal. Advances in Difference Equations :36 Page 3 of 19 where ϕ p is the p-laplacian operator, 1 < n 1<α < n,1<m 1<β < m, α β >1,andD α + and D β + are the Caputo fractional derivatives. g k C[, 1], [, +, k =1,2,f C[, 1] [,+,], [,+ are given functions. In this paper, a positive solution x = xtofbvp1.1meansasolutionof1.1satisfying xt>,t [, 1]. Throughout this paper, we always assume that the following condition is satisfied. L g 1 t>g 2 t, g 2s ds, g 1s ds <1. The organization of the paper is as follows. In Section 2, we present some necessary definitions and lemmas which will be used to prove our main results. In Section 3,byusing the Avery-Peterson fixed point theorems, the results for the existence of multiple positive solutionsofbvp 1.1 are established. In Section 4, we give anexampletodemonstrate the main result. 2 Preliminaries and lemmas In this section, we give some definitions and basic lemmas that will be used and important to us in the following. Definition 2.1 see [1] The Riemann-Liouville integral of fractional order α >ofafunction g is defined as I α +gt:= 1 Ɣα t s α 1 gs ds. Definition 2.2 see [2] The Caputo derivative of fractional order α >ofafunctiong is defined as D α + gt= 1 Ɣn α t s n α 1 g n s ds, where n is the smallest integer greater than or equal to α. Definition 2.3 see [4] Let E be a real Banach space. A nonempty, closed, and convex set P E is a cone if the following two conditions are satisfied: 1 if x P and μ,thenμx P; 2 if x P and x P,thenx =. Every cone P E induces the ordering in E given by x 1 x 2 if and only if x 2 x 1 P. Definition 2.4 see [4] The map γ is said to be a continuous non-negative convex concavefunctiononaconep of a real Banach space E provided that γ : P [, + is continuous and γ tx +1 ty tγ x+1 tγ y, x, y P, t [, 1]. Lemma 2.1 see [8] Let α >,assume that u, D α +u C, 1 L1, 1, then I α + Dα + ut=ut+c + C 1 t + C 2 t C n 1 t n 1

4 Zhangetal. Advances in Difference Equations :36 Page 4 of 19 holds for some C i R, i =,1,2,...,n 1,where n is the smallest integer greater than or equal to α. Lemma 2.2 see [26] The Caputo fractional derivative of order n 1<α < nfort β is given by D α + tβ = { Ɣβ+1 Ɣβ α+1 tβ α, β > n 1,, β {,1,...,n 1}. Lemma 2.3 Let h C[, 1] and 1<m 1<β < m. Then the BVP { D β +ut=ht, < t <1, u1 = u i =, i =1,2,...,m 1, 2.1 has an unique solution ut Ht, shs ds, 2.2 where { Ht, s= 1 1 s β 1 t s β 1, s t 1, Ɣβ 1 s β 1, t s Proof From 2.1 and Lemma 2.1,wehave ut= 1 Ɣβ t s β 1 hs ds + C + C 1 t + C 2 t C m 1 t m Since u i = i =1,2,...,m 1,wehaveC 1 = C 2 = = C m 1 =and u1 = 1 Ɣβ Combined with u1 =, we know Thus C 1 Ɣβ ut= 1 Ɣβ = 1 Ɣβ 1 Ɣβ 1 s β 1 hs ds + C s β 1 hs ds. t s β 1 hs ds 1 Ɣβ t s β 1 1 s β 1 hs ds t 1 s β 1 hs ds Ht, shs ds, where Ht, sisgivenby s β 1 hs ds

5 Zhangetal. Advances in Difference Equations :36 Page 5 of 19 The proof is completed. Lemma 2.4 Assume L holds, let y C[, 1] and 1<n 1<α < n,1<m 1<β < m. Then the following boundary value problems: D α +xt=yt, < t <1, x j =, j =2,3,...,n 1, x1 = g 1sxs ds, x = g 2sxs ds, 2.6 has an unique solution and where xt D β + xt= 1 Ɣα β Gt, sys ds 2.7 t s α β 1 ys ds, 2.8 here Gt, s=g 1 t, s+g 2 t, s, 2.9 { G 1 t, s= 1 1 s α 1 t s α 1, s t 1, 2.1 Ɣα 1 s α 1, t s 1, G 2 t, s=δ Pt g 1 τg 1 τ, s dτ + Qt g 2 τg 1 τ, s dτ, 2.11 δ 1 =1 M 1 1 N 2 +N 1 1 M 2, 2.12 Pt=1 N 2 + N 1 t, Qt=M M 1 t, 2.13 M 1 = N 1 = g 1 s ds, M 2 = g 2 s ds, N 2 = Proof FromLemma 2.1, considering BVP 2.6, we have xt= 1 Ɣα sg 1 s ds, 2.14 sg 2 s ds t s α 1 ys ds + C + C 1 t + C 2 t C n 1 t n 1. Because x j = j =1,2,...,n 1,wehaveC 2 = C 3 = = C n 1 =,and x1 = 1 Ɣα 1 s α 1 ys ds + C + C 1.

6 Zhangetal. Advances in Difference Equations :36 Page 6 of 19 From the second condition of BVP 2.6, we have g 1 sxs ds = 1 Ɣα From the last condition of BVP 2.6, we have C 1 = x = By 2.16and2.17, we obtain So C = xt= 1 Ɣα = 1 Ɣα g 1 sxs ds 1 s α 1 ys ds + C + C g 2 sxs ds g 2 sxs ds 1 Ɣα t s α 1 ys ds + C + C 1 t t s α 1 ys ds + g 1 sxs ds 1 1 s α 1 ys ds + t g 2 sxs ds Ɣα = 1 t s α 1 1 s α 1 ys ds 1 Ɣα Ɣα + g 1 sxs ds g 2 sxs ds + t 1 s α 1 ys ds g 2 sxs ds t g 2 sxs ds 1 s α 1 ys ds G 1 t, sys ds + A 1 A 2 + ta 2, 2.19 where G 1 t, s is given in 2.1, and A 1 = A 2 = g 1 sxs ds, g 2 sxs ds. In view of 2.19, we get g 1 txtg 1 t Integrating 2.2from to 1,weobtain A 1 = g 1 sxs ds G 1 t, sys ds + A 1 g 1 t A 2 g 1 t+ta 2 g 1 t. 2.2 g 1 s G 1 s, τyτ dτ ds

7 Zhangetal. Advances in Difference Equations :36 Page 7 of 19 + A 1 g 1 s ds A 2 g 1 s ds + A 2 sg 1 s ds = I 1 + A 1 M 1 A 2 M 1 + A 2 M 2, 2.21 where M 1 and M 2 are given in 2.14, and I 1 Similarly, we obtain g 1 s G 1 s, τyτ dτ ds g 1 sg 1 s, τ ds yτ dτ. A 2 = g 2 sxs ds g 2 s G 1 s, τyτ dτ ds + A 1 g 2 s ds A 2 g 2 s ds + A 2 sg 2 s ds = I 2 + A 1 N 1 A 2 N 1 + A 2 N 2, 2.22 where N 1 and N 2 are given in 2.15, and I 2 g 2 s G 1 s, τyτ dτ ds From 2.21and2.22, we get g 2 sg 1 s, τ ds yτ dτ. A 1 = I 2 M 2 M 1 +I N 1 N 2 δ, A 2 = I 2 1 M 1 +I 1 N 1 δ, where δ 1 is given in Hence, xt + Qt G 1 t, sys ds + A 1 A 2 + ta 2 G 1 t, sys ds + δ I 2 M 2 1+t M 1 t+i 1 1 N 2 + N 1 t G 1 t, sys ds + δi 1 Pt+δI 2 Qt G 1 t, sys ds δqt g 2 τg 1 τ, s dτ G 1 t, sys ds g 2 τg 1 τ, s dτ δpt g 1 τg 1 τ, s dτ ys ds δ Pt ys ds ys ds g 1 τg 1 τ, s dτ

8 Zhangetal. Advances in Difference Equations :36 Page 8 of 19 G 1 t, sys ds Gt, sys ds, G 2 t, sys ds where G 2 t, s is given in 2.11andPt, Qtaregivenin2.13. On the other hand, in view of 2.19, because m 1<β < α 1<n 1, by Lemma 2.2,we have D β 1 t + xt=dβ + t s α 1 ys ds + C + C 1 t Ɣα = D β + I α + yt+c + C 1 t = D β + Iα + yt+dβ + C +D β + C 1t = D β + Iα + yt = I α β + yt = 1 Ɣα β t s α β 1 ys ds The proof is completed. Lemma 2.5 BVP 1.1 equivalent to the following integral equation: xt= Gt, sϕ q Hs, τf τ, xτ, D β + xτ dτ ds 2.24 and D β + xt 1 Ɣα β t s α β 1 ϕ q Hs, τf τ, xτ, D β + xτ dτ ds, 2.25 where Ht, s and Gt, s are given in 2.3 and 2.9. Proof From Lemma 2.4 and Lemma 2.5,letyt=ϕ q ut, ht=f t, xt, D β + xt, we have 1 yt=ϕ q ut = ϕq Ht, sf s, xs, D β + xs ds ϕ q Ht, sf s, xs, D β + xs ds. Immediately we obtain xt Gt, sys ds = Gt, sϕ q Hs, τf τ, xτ, D β + xτ dτ ds. From 2.23, we have The proof is completed. Lemma 2.6 Assume L holds, then the function Ht, s defined by 2.3 and the function Gt, s defined by 2.9 satisfies

9 Zhangetal. Advances in Difference Equations :36 Page 9 of 19 1 Ht, s is continuous for all t, s [, 1]; 2 Ht, s Hs, s for all t, s [, 1]; 3 Gt, s is continuous for all t, s [, 1]. Proof 1 From 2.3, it is easy to show that Ht, s is continuous on [, 1] [, 1] and obviously, Ht, s, for s t. For s t 1, we have 1 s β 1 t s β 1 =1 s β 1 1 t s β 1. 1 s By 2.3, we know Ht, s, t, s [, 1], and Ht, s>,t, s, 1. 2 For s t 1, we have Ht, s=1 s β 1 t s β 1 1 s β 1 = Hs, s and for s t, Ht, s=hs, s, so that Ht, s Hs, s, for all t, s [, 1]. 3 From 2.9, we know Gt, s=g 1 t, s+g 2 t, s. Firstly, from 2.1, similarly, we can obtain G 1 t, s, t, s [, 1], and G 1 t, s>,t, s, 1. On the other hand, from L, we know, for t, 1, g 1 t>tg 1 t>tg 2 t and 1> 1> g 1 s ds > g 1 s ds > sg 1 s ds > g 2 s ds > sg 2 s ds >, sg 2 s ds >. That implies 1>M 1 > M 2 > N 2 >, 1>M 1 > N 1 > N 2 > Therefore δ 1 =1 M 1 1 N 2 +N 1 1 M 2 >. From L andg 1 t, s, we know G 2 t, s t = δ N 1 g 1 τg 1 τ, s dτ +1 M 1 g 2 τg 1 τ, s dτ >, 2.27 which implies that G 2 t, s is a monotone increasing function with respect to t [, 1]. So, from 2.11, we have G 2 t, s G 2, s=δ P = δ 1 N 2 g 1 τg 1 τ, s dτ + Q g 2 τg 1 τ, s dτ g 1 τg 1 τ, s dτ +M 2 1 g 2 τg 1 τ, s dτ

10 Zhangetal.Advances in Difference Equations :36 Page 1 of 19 = δ 1 N2 g 1 τ+m 2 1g 2 τ G 1 τ, s dτ δ 1 N2 g 2 τ+m 2 1g 2 τ G 1 τ, s dτ = δm 2 N 2 g 2 τg 1 τ, s dτ. So G 2 t, s. Hence, Gt, s. The proof is completed. Lemma 2.7 Let η, 1 2, denote I η =[,η] and ρ 1 =1 η 1 η α 1, then min G 1 t, s ρ 1 G 1 s, s=ρ 1 max G 1t, s t [,1] Proof For s < t 1andt I η, G 1 t, s=1 s α 1 t s α 1 1 s α 1 = G 1 s, s and G 1 t, s G 1 s, s = 1 sα 1 t s α 1 1 s α 1 t s α 1 =1 1 s η 1 1 s η 1 1 η α 1 α 1 = ρ 1 > For s t and t I η, η α 1 G 1 t, s=g 1 s, s> 1 G 1 s, s=ρ 1 G 1 s, s η Therefore, min G 1 t, s ρ 1 G 1 s, s. The proof is completed. Lemma 2.8 Assume L holds, then the function G 2 t, s satisfies the following properties: 1 G 2 t, s G 2 1, s=max t [,1] G 2 t, s; 2 min t Iη G 2 t, s ρ 2 max t [,1] G 2 t, s, where <ρ 2 = M 2 N 2 1 N 2 +N 1 <1.

11 Zhangetal.Advances in Difference Equations :36 Page 11 of 19 Proof From Lemma 2.7 and 2.27, we obtain G 2 t, s>and and max t [,1] G 2t, s=g 2 1, s = δ 1 N 2 + N 1 g 1 τg 1 τ, s dτ +M 2 M 1 g 2 τg 1 τ, s dτ min G 2 t, s =G 2, s = δ 1 N 2. g 1 τg 1 τ, s dτ +M 2 1 g 2 τg 1 τ, s dτ Furthermore, G 2, s G 2 1, s = 1 N 2 g 1τG 1 τ, s dτ +M 2 1 g 2τG 1 τ, s dτ 1 N 2 + N 1 g 1τG 1 τ, s dτ +M 2 M 1 g 2τG 1 τ, s dτ 1 N 2 > g 1τG 1 τ, s dτ +M 2 1 g 1τG 1 τ, s dτ 1 N 2 + N 1 g 1τG 1 τ, s dτ +M 2 M 1 g 2τG 1 τ, s dτ > 1 N 2 + M 2 1 g 1τG 1 τ, s dτ 1 N 2 + N 1 g 1τG 1 τ, s dτ = M 2 N 2 1 N 2 + N 1 = ρ 2 and Hence, <ρ 2 = M 2 N 2 1 N 2 + N 1 < M 2 N 2 1 N 2 <1. min G 2 t, s ρ 2 G 2 1, s. The proof is completed. From Lemma 2.7 and Lemma 2.8, we can easily show that the following result holds. Lemma 2.9 min Gt, s ρ G 1 s, s+g 2 1, s and Gt, s G 1 s, s+g 2 1, s, where ρ = min{ρ 1, ρ 2 }.

12 Zhangetal.Advances in Difference Equations :36 Page 12 of 19 3 Main results In this section we deduce the existence of positive solutions to BVP 1.1 by using the wellknown Avery-Peterson fixed point theorem; see [4]. Let γ and θ be non-negative continuous convex functionals on P, ϕ be a non-negative continuous concave functional on P,andψ be a non-negative continuous functional on P. For a, b, c, d >, we define the following convex set: Pγ ; d= { x P : γ x<d }, Pγ, ϕ; b, d= { x P : b ϕx, γ x d }, Pγ, θ, ϕ; b, c, d= { x P : b ϕx, θx c, γ x d }, and a closed set Pγ, ψ; a, d= { x P : a ψx, γ x d }. Lemma 3.1 see [4] Let P be a cone in a real Banach space E. Let γ and θ be non-negative continuous convex functionals on P, ϕ be a non-negative continuous concave functional on P, and ψ be a non-negative continuous functional on P satisfying ψλx λψx for λ 1, such that, for some positive numbers M and d, ϕx ψx, x Mγ x 3.1 for all x Pγ ; d. Suppose T : Pγ ; d Pγ ; d is completely continuous and there exist positive numbers a, b, and c with a < bsuchthat H1 {x Pγ, θ, ϕ; b, c, d:ϕx>b} Ø, and ϕx>b for x Pγ, θ, ϕ; b, c, d; H2 ϕtx>b for x Pγ, ϕ; b, d with θtx>c; H3 / Pγ, ψ; a, d and ψtx<a for x Pγ, ψ; a, d with ψx=a. Then T has at least three fixed point x 1, x 2, x 3 Pγ ; d such that γ x i d, i =1,2,3; ϕx 1 >b, a < ϕx 2, ψx 2 <b; ψx 3 <a. Let E = {x C[, 1] : D β + x C[, 1], x = g 2sxs ds, x j =, j =2,3,...,m 1} be endowed with the norm { x = max max xt, max D β + xt }, t [,1] then E is a Banach space. We define a set P E by t [,1] { } P = x E : xt, D β + xt, min xt ρ max xt. t [,1]

13 Zhangetal.Advances in Difference Equations :36 Page 13 of 19 For x, y P and k 1, k 2, it is easy to obtain k 1 xt+k 2 yt, D β + k1 xt+k 2 yt = k 1 D β + xt+k 2D β + yt and { min k1 xt+k 2 yt } { min k1 xt } { + min k2 yt } { ρ max k1 xt } { + ρ max k2 yt } t [,1] t [,1] = ρ max t [,1] ρ max t [,1] { k1 xt } { + max k2 yt } t [,1] { k1 xt+k 2 yt }. Thus, for x, y P and k 1, k 2, k 1 xt+k 2 yt P. And if x P, x,itiseasytoprove that x / P. Therefore, P isaconeine. Let T : P E be the operator defined by Txt:= Gt, sϕ q Hs, τf τ, xτ, D β + xτ dτ ds. Then we have the following conclusion. Lemma 3.2 Assume L holds, then T : P P is a completely continuous operator. Proof For x P, from the non-negativity and continuity of Gt, s, Ht, s, f t, xt, D β + xt, we know that T is a continuous operator and Txt. By 2.25, we have D β + Txt 1 Ɣα β Furthermore, min Txt =min = t s α β 1 ϕ q Gt, sϕ q min Gt, sϕ q ρ max t [,1] Gt, sϕ q Hs, τf τ, xτ, D β + xτ dτ ds. Hs, τf τ, xτ, D β + xτ dτ ds Hs, τf τ, xτ, D β + xτ dτ ds Hs, τf τ, xτ, D β + xτ dτ ds = ρ max Txt. t [,1] Thus, TP P. Next, we show that T is uniformly bounded. Let D P be bounded, i.e., there exists a positive constant r such that x r, for all x D.LetM = max t [,1],x D f t, xt, D β + xt +1>,forx D,wehave 1 Txt = Gt, sϕ q Hs, τf τ, xτ, D β + xτ dτ ds Gt, s ϕ q Hs, τ f τ, xτ, D β + xτ dτ ds

14 Zhangetal.Advances in Difference Equations :36 Page 14 of 19 ϕ q M ϕ q M Gt, sϕ q G1 s, s+g 2 1, s ϕ q Furthermore, for any x D and t [, 1], we have Hs, τ ds Hs, τ ds := M 1. β D + Txt 1 t = t s α β 1 ϕ q Hs, τf τ, xτ, D β + Ɣα β xτ dτ ds 1 t t s α β 1 ϕ q Hs, τf τ, xτ, D β + Ɣα β xτ dτ ds 1 1 ϕ q M 1 s α β 1 ϕ q Hs, τ dτ ds := M 2. Ɣα β So Tx max{m 1, M 2 }, which implies T is uniformly bounded. Inthefollowingoftheproof,wewillprovethatT is equicontinuous. Since Gt, s is continuous on [, 1] [, 1], it is uniformly continuous on [, 1] [, 1]. Thus, for any ɛ >, there exists a constant δ 1 >,suchthat Gt1, s Gt 2, s < ɛ ϕ q M ϕ q Hs, τ dτ ds for t 1, t 2 [, 1] with t 1, t 2 < δ 1. Therefore, Txt1 Txt 2 1 [ = Gt 1, sϕ q Hs, τf ] τ, xτ, D β + xτ dτ ds Gt 2, sϕ q Hs, τf τ, xτ, D β + xτ dτ ds Gt1, s Gt 2, s ϕq Hs, τf τ, xτ, D β + xτ dτ ds ɛ < ϕ q M ϕ q Hs, τ dτ dsϕ qm = ɛ. ϕ q Hs, τ dτ ds On the other hand, for 1 < m 1<β < α 1<n 1,t α β is uniformly continuous on [, 1]. ϕ We denote M 3 = q M α βɣα β. Then there exists a constant < δ ɛ 2 < 4M 3 α β 1,suchthat, for any < t 1 < t 2 <1and t 2 t 1 < δ 2,wehave t α β 2 t α β ɛ 1 <. 2M 3 Thus, from Lemma 2.7,wehave β D + Txt 2 D β + Txt 1 = 1 t2 Ɣα β 1 Ɣα β 1 t 2 s α β 1 ϕ q t 1 s α β 1 ϕ q Hs, τf τ, xτ, D β + xτ dτ ds Hs, τf τ, xτ, D β + xτ dτ ds

15 Zhangetal.Advances in Difference Equations :36 Page 15 of 19 t1 t2 s α β 1 t 1 s α β 1 Ɣα β ϕ q Hs, τf τ, xτ, D β + xτ dτ ds 2 + t 2 s α β 1 ϕ q Hs, τf τ, xτ, D β + xτ dτ ds t 1 ϕ q M Hτ, τ dτ t1 t2 s α β 1 t 1 s α β 1 ds Ɣα β 2 + t 2 s α β 1 ds t 1 t1 = M 3 d [ t 1 s α β t 2 s α β] t2 + dt 2 s α β t 1 = M t2 3 t 1 α β t α β 1 + t α β 2 + t2 t 1 α β M 2t2 3 t 1 α β + t α β 2 t α β 1 ɛ α β 1 α β ɛ < M M 3 2M 3 = ɛ. Hence, T is equicontinuous. AccordingtotheArzelà-Ascolitheorem,T is a completely continuous operator. The proof is completed. Denote the positive constants J 1 = J 2 = Hs, τ dτ ds, 1 s α β 1 ϕ q Hs, τ dτ ds, G1 s, s+g 2 1, s ϕ q 1 Ɣα β and J 3 = G1 s, s+g 2 1, s ϕ q η Hs, τ dτ ds. Define the functionals as follows: γ x= x, θx=ψx= max xt, t [,1] ϕx=min xt, then γ and θ are continuous non-negative convex functionals, ϕ is a continuous nonnegative concave functional, ψ is a continuous non-negative functional, and ρθx ϕx θx=ψx, x Mγ x, where M = 1. Therefore, condition 3.1 in Lemma 3.1 is satisfied.

16 Zhangetal.Advances in Difference Equations :36 Page 16 of 19 Theorem 3.1 Suppose L holds, and there exist constants <a, b, dwitha< b < ρd min{ J 3 J1, J 3 J2 } and c = b, such that ρ L 1 f t, x, y min{ϕ p d J 1, ϕ p d J 2 }, t, x, y [, 1] [, d] [ d,]; L 2 f t, x, y>ϕ p b ρj 3, t, x, y [, η] [b, b ρ ] [ d,]; L 3 f t, x, y<ϕ p a J 1, t, x, y [, 1] [, a] [ d,]. Then BVP 1.1 has at least three positive solutions x 1, x 2, x 3, satisfying x i d i =1,2,3, 3.2 min x1 t > b, a < minx2 t, max x2 t < b, max x3 t < a. 3.3 t [,1] t [,1] Proof It is clear that the fixed points of operator T are equivalent the solutions of BVP 1.1. For x Pγ ; d, we have γ x= x d, this implies max xt d, t [,1] max D β + xt d, t [,1] then xt d, d D β + xt. By L 1, we have max Txt = max t [,1] t [,1] Gt, sϕ q G1 s, s+g 2 1, s ϕ q G1 s, s+g 2 1, s ϕ q ϕ p d J 1 Hs, τf τ, xτ, D β + xτ dτ ds Hs, τf τ, xτ, D β + xτ dτ ds Hs, τ dτ ds = d G1 s, s+g 2 1, s ϕ q Hs, τ dτ ds J 1 = d and max β D + Txt = max 1 t t [,1] t [,1] t s α β 1 Ɣα β ϕ q Hs, τf τ, xτ, D β + xτ dτ ds 1 1 d 1 1 s α β 1 ϕ q ϕ p Hs, τ dτ ds Ɣα β J 2

17 Zhangetal.Advances in Difference Equations :36 Page 17 of 19 = d 1 J 2 Ɣα β = d, 1 s α β 1 ϕ q Hs, τ dτ ds so { γ Tx= Tx = max max Txt, max D β + Txt } d. t [,1] t [,1] Therefore T : Pγ ; d Pγ ; d. Let xt= b ρ, xt Pγ, θ, ϕ; b, c, dandϕ b >b, which implies that ρ { } x Pγ, θ, ϕ; b, c, d:ϕx>b. For x Pγ, θ, ϕ; b, c, d, we know that b xt c = b ρ for t I η and d D β + xt. In view of L 2, ϕtx=min Txt = min > = ρ b ρj 3 Gt, sϕ q Hs, τf τ, xτ, D β + xτ dτ ds ρ G 1 s, s+g 2 1, s η b ϕ q Hs, τϕ p ρj 3 G1 s, s+g 2 1, s η ϕ q Hs, τ dτ dτ ds ds = b. 3.4 So ϕtx>b for all x Pγ, θ, ϕ; b, c, d. Hence, the condition H1 of Lemma 3.1 is satisfied. By 3.2, for all x Pγ, ϕ; b, dwithθtx>c = b ρ,wehave ϕtx ρθtx>ρc = ρ b ρ = b. Thus, the condition H2 of Lemma 3.1 holds. Because of ψ = < a, then/ Pγ, ψ; a, d. For x Pγ, ψ; a, d withψx =a, we know γ x d.itmeansthatmax t [,1] xt=a and d D β + xt. From L 3, we can obtain ψtx= max Txt t [,1] = max t [,1] < Gt, sϕ q G1 s, s+g 2 1, s ϕ q = a G1 s, s+g 2 1, s ϕ q J 1 = a. Therefore, the condition H3 of Lemma 3.1 holds. Hs, τf τ, xτ, D β + xτ dτ ds a Hs, τϕ p dτ ds J 1 Hs, τ dτ ds

18 Zhangetal.Advances in Difference Equations :36 Page 18 of 19 To sum up, the conditions of Lemma 3.1 are all verified. Hence, BVP 1.1 has at least three positive solutions x 1, x 2, x 3 satisfying 3.2and3.3. The proof is completed. 4 Example In this section, we present an example to illustrate the main result. Consider the following boundary value problems: 4 D ϕ 3 D xt = f t, xt, D 5 + xt, t, 1, 2 ϕ 3 D x 2 = ϕ 3 D x1 =, x1 = sxs ds, x = s2 xs ds, x =, 4.1 where α = 5 2, β = 5 4, p = 3 2, g 1t=t, g 2 t=t 2,and f t, x, y= { tan.6t+.1e x/5 x 2 + siny, x 18, tan.6t+siny+92, 18<x 5,. 4.2 Choose a =8,b =18,d = 5,, η = 1. By a simple computation, we have 3 ρ 1 = 1 2, ρ 2 = 1 13, ρ = 1 13, r =1, M 1 = 1 2, M 2 = 1 3, N 1 = 1 3, N 2 = 1 4, δ = 72 43, J 1 =.26227, J 2 = , J 3 = We can check that the nonlinear term f t, x, ysatisfies L 1 f t, x, y min{ϕ p d J 1, ϕ p d J 2 } , t, x, y [, 1] [, 5,] [ 5,,]; L 2 f t, x, y>ϕ p b ρj , t, x, y [, 1 ] [18, 234] [ 5,,]; 3 L 3 f t, x, y<ϕ p a J , t, x, y [, 1] [, 8] [ 5,,]. Then all assumptions of Theorem 3.1 are satisfied. Thus, BVP 4.1 hasat leastthree positive solutions x 1, x 2, x 3,satisfying x i 5, i =1,2,3, 4.3 min x 1 >18, 8<min x 2, max x 2 <18, t [,1] max x 3 < t [,1] Competing interests The authors declare that they have no competing interests. Authors contributions All authors jointly worked on the results and they read and approved the final manuscript. Author details 1 Business School, University of Shanghai for Science and Technology, Shanghai, 293, P.R. China. 2 College of Science, University of Shanghai for Science and Technology, Shanghai, 293, P.R. China.

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