Existence of positive solutions for discrete delta-nabla fractional boundary value problems with p-laplacian

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1 Liu et al. Boundary Value Problems :60 DOI 0.86/s R E S E A R C H Open Access Existence of positive solutions for discrete delta-nabla fractional boundary value problems with p-laplacian Huan Liu, Yuanfeng Jin and Chengmin Hou * * Correspondence: cmhou@foxmail.com Department of Mathematics, Yanbian University, Yanji, 002, P.R. China Abstract In this paper, we consider a discrete delta-nabla boundary value problem for the fractional difference equation with p-laplacian β v 2ϕ p b ν xt + λft ν + β +,xt ν + β +, [ b ε xt ] t ν+β+ε+ =0, xb=0, [ b ν xt ] b ν 2 =0, x = xtat, where t T =[ν β,b + ν β ] Nν. β ν 2, b ν are left and right fractional difference operators, respectively, and ϕ p s= s p 2 s, p >. By using the method of upper and lower solution and the Schauder fixed point theorem, we obtain the existence of positive solutions for the above boundary value problem; and applying a monotone iterative technique, we establish iterative schemes for approximating the solution. MSC: 9A06; 9A22 Keywords: discrete delta-nabla; boundary value problem; positive solutions; upper and lower solution; monotone iteration; fixed point theorem Introduction In this paper, we investigate the existence of positive solutions for the following discrete delta-nabla fractional boundary value problem FBVP with p-laplacian: β ν 2 ϕp b ν xt + λf t ν + β +,xt ν + β +, [ b ε xt ] t ν+β+ε+ =0, t T,. xb=0, [ b ν xt ] b ν 2 =0, x = xtat,.2 where t T =[ν,b+ν ] Nν, β ν 2, b ν are left and right fractional difference operators, respectively. ϕ p s= s p 2 s, p >,ϕ p = ϕ q, p + =,β, ν, ε 0, +, and they q satisfy the following: The Authors 207. This article is distributed under the terms of the Creative Commons Attribution 4.0 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 Liuet al. Boundary Value Problems :60 Page 2 of 2 H ν, 2], β, ε 0, ], ν ε >0; H 2 At is a function defined on [0, b] N0 ; H f :[0,b] N0 R R [0, is continuous for any t [0, b] N0, f t,0,0 0, f t,, 0,andlet σ = max f t,, 0.. t [0,b] N0 The equation with p-laplacianoperator arises in the modeling of different physical and natural phenomena, non-newtonian mechanics [], combustion theory [2], population biology [], nonlinear flow laws [4] and the system of Monge-Kantorovich mass transfer [5]. Integral and derivative operators of fractional order can describe the characteristics exhibited in many complex processes and systems having long-memory in time. Then many classical integer order models for complex systems are substituted by fractional order models. Fractional calculus has recently developed into a relatively vibrant research area. It also provides an excellent tool to describe the hereditary properties of materials and processes. Many successful new applications of fractional calculus in various fields have also been reported recently. For example, Nieto and Pimentel [6]extendedasecondorder thermostat model to the fractional model; Ding and Jiang [7] usedwaveformrelax- ation methods to study some fractional functional differential equation models. For the basic theories of fractional calculus and some recent work in application, the reader is referred to Refs. [8 ]. On the other hand, discrete fractional calculus has attracted slowly but steadily increasing attention in the past seven years or so. In particular, several recent papers by Atici and Eloe as well as other recent papers by the present authors have addressed some basic theory of both discrete fractional initial value problems and discrete FBVPs. More specifically, Atici and Eloe [4] have already analyzed a transform method in discrete fractional calculus. Goodrich [5] considered a discrete right-focal fractional boundary value problem. All of the fundamental background in discrete fractional calculus can be found in [6]which is written by Goodrich and Peterson. Other recent work has considered discrete FBVPs with a variety of boundary conditions, see [7 20]. There are also a few papers for the discrete delta-nabla boundary value problems. For example, Malinowska and Torres [2] propose a more general approach to the calculus of variations on time scales that allows to obtain both delta and nabla results as particular cases. Martins and Torres [22]studythe calculus of variations on time scales with nabla derivatives and so on. What is more, [2] is the first paper to consider a discrete fractional difference equation with a p-laplacian operator. From the above works, we can see the fact that although the discrete delta-nabla boundary value problem has been studied by many authors, to the best of our knowledge, there are very few papers on the discrete delta-nabla FBVPs. For example, Xie, Jin and Hou [24] obtained some results which ensure the existence of a well precise interval of the parameter for which the problem admits multiple solutions. Our aim is to use the method of upper and lower solution and the Schauder fixed point theorem to obtain the existence of positive solutions for the above boundary value problem; and to apply a monotone iterative technique to establish iterative schemes for approximating the solution.

3 Liuet al. Boundary Value Problems :60 Page of 2 The rest of this paper has the following structure. In Section 2, we recall some basic definitions of fractional calculus, establish some lemmas and use symbols to replace with some formula which plays a pivotal role in the text. Section contains an existence result for problem.and.2 which is established by applying the method of upper and lower solution andthe Schauder fixedpoint theorem. In Section 4, we show the iterative schemes for approximating the solution by using a monotone iterative technique. 2 Preliminaries In this section, we collect some basic definitions and lemmas for manipulating discrete fractional operators. For any real number β,letn β = {β, β +,β +2,...}, β N = {...,β 2,β,β}. We define t ν = Ɣt+ for any t, ν R, for which the right-hand side is well defined. We Ɣt+ ν also appeal to the convention that if t + ν is a pole of the gamma function and t +is not a pole, then t ν =0. Definition 2. [7] Let f : N a R and ν >0begiven.Theνth left fractional sum of f is given by ν a f t= t ν t s ν f s fort N a+ν. Ɣν s=a Also, let N N be chosen such that N <ν N. Then the νth left fractional difference of f is given by ν a f t:=n ν N a f t fort N a+n ν. Definition 2.2 [8] The νth right fractional sum of f tfor ν >0 is defined by b ν f t= Ɣν b s t ν f s fort b ν N. s=t+ν We also define the νth right fractional difference for ν >0 by b ν f t:= N N b ν N f t fort b N+ν N, where N N is chosen so that 0 N <ν N. Lemma 2. [8] Let b R and μ >0be given. Then b t μ = μb t μ for any t, for which both sides are well defined. Furthermore, for ν >0with N <ν N, N N, b μ ν b t μ = μ ν b t μ+ν, t b μ ν N,

4 Liuet al. Boundary Value Problems :60 Page 4 of 2 and b μ ν b t μ = μ ν b t μ ν, t b μ N+ν N. Lemma 2.2 [7] Let f : N a R be given, and suppose ν, μ >0with N <ν N. Then ν a+μ μ a f t=ν μ a f t, t N a+μ+n ν. Lemma 2. [8] Let f : b N R be given, and suppose ν, μ >0with N <ν N. Then b μ ν b μ f t= b ν μ f t, t b μ N+ν N. Lemma 2.4 [7] Let f : N a R and ν >0be given with N <ν N. The following two definitions for the left fractional difference ν a f : N a+n ν R are equivalent: ν a f t= N N ν a f t, t+ν ν a f t= Ɣ ν s=a t s ν f s, N <ν N, N f t, ν = N. Lemma 2.5 [8] Let f : b N R and ν >0be given with N <ν N. The following two definitions for the right fractional difference b ν f : b N+ν N R are equivalent: b ν f t= N b N N ν f t, b b ν Ɣ ν s=t ν f t= s t ν f s, N <ν N, N N f t, ν = N. Lemma 2.6 [7] Let f : N a R be given and suppose k N 0 and ν >0.Then for t N a+m μ+ν, k a k f t= k ν j f a a f t Ɣν k + j + t aν k+j. ν j=0 Moreover, if μ >0with M <μ M, then for t N a+ν, a+m μ μ a f M t=μ ν a f t ν j=0 j M+μ a f a + M μ Ɣν M + j + t a M + μ ν M+j. Lemma 2.7 [8] Let f : b N R be given, and suppose k N 0 and ν >0.Then for t b ν N, k b ν b k f t= b k ν b j f b f t Ɣν k + j + b tν k+j. j=0

5 Liuet al. Boundary Value Problems :60 Page 5 of 2 Moreover, if μ >0with M <μ M, then for t b M+μ ν N, M b M+μ ν b μ f t= b μ ν f t j=0 b j M+μ f b M + μ b M + μ t ν M+j. Ɣν M + j + Remark 2. When we choose t T =[ν β,b + ν β ] Nν in., problem..2 is significative. In fact, by Definitions 2., 2.2, Lemmas2.4and 2.5, wehave and β ν 2 ϕ p b ν xt = [b ε xt ] t ν+β+ε+ = Ɣ ε t+β t s ϕ p Ɣ Ɣ ν s=ν 2 b s=t ν+β+ b u s ν xu, u=s ν s + ν β ε t 2 ε xs. We can see that the domain of the function x is { 2,,0,...,b}. Inthefollowingparagraphs,wedefine j t=i yt=0forj < i. Next, we denote Gt, s= Ɣν b G A s= s ν t ν b+ν ν s t ν, ν t + ν <s b + ν, s ν t ν b+ν ν, ν s t + ν b + ν. Gt, sat b + ν, C = b ν b + ν 2 t ν b + ν ν At, 2. Jt, s= b + ν 2 tν G A s+gt, s. 2.2 CƔν For variable t,wedenote t = t ν + β +, t = b +2ν t β. By Lemmas 2. and 2.5,forν s b + ν,wehave b t ε s t ν = Ɣ ε = Ɣ ε b u t ε s u ν u=t ε s ν u=t ε = s ν t ε s t ν =ν ε s t ν ε = Ɣν Ɣν ε s t ν ε. u t ε s u ν

6 Liuet al. Boundary Value Problems :60 Page 6 of 2 Thus b ε t Gt, s= Ɣν ε s ν t ν ε b+ν ν s t ν ε, ν t + ν ε <s b + ν, s ν t ν ε b+ν ν, ν s t + ν ε b + ν. 2. We denote Gt, s:= b t ε Gt, s, + ν 2 tν ε Jt, s:=gt, s+b G A s. 2.4 CƔν ε Lemma 2.8 Let 0 C <and h :[0,b] N0 R be given, the problem b ν xt+ht ν +=0, t [ν,b + ν ] Nν, xb=0, x = b xtat, 2.5 has the unique solution xt= Jt, shs ν +, s=ν where Jt, s is given by 2.2. Proof Denote hs= hs ν + b + ν ν. By Lemma 2.7,wehave xt= ν ht ν ++k b + ν 2 t ν + k 2 b + ν 2 t ν 2. From 2.5, we have k 2 =0and Then k = s ν hs CƔν s=ν b xt= s t ν hs ν + Ɣν + s=t+ν b + ν 2 tν CƔν s=ν Gt, shs ν + Ɣν = s=ν s t ν hs. At s=t+ν b s ν hs s=ν s t ν hs At s=t+ν s ν b + ν 2 t ν hs

7 Liuet al. Boundary Value Problems :60 Page 7 of 2 + = b + ν 2 tν s ν hs CƔν s=ν Gt, shs ν ++ s=ν C = s=ν b s ν hs Gt, shs ν ++ s=ν b = s=ν Gt, shs ν ++ s=ν Jt, shs ν +. = s=ν b b + ν 2 tν CƔν At s t ν hs At s=t+ν s t ν hs s=t+ν b + ν 2 tν CƔν s ν b + ν 2 t ν Aths b b + ν ν b 2+ν tν CƔν b s t ν hs At s=t+ν Gt, saths s=ν The proof is complete. Lemma 2.9 Let 0 C <and h :[0,b] N0 R. Problem 2.5 is equivalent to the following problem: b+ε ν ε yt+ht + ν =0, t [ν,b + ν ] Nν, yb + ε=0, [ b+ε ε yt] = b 2.6 b+ε ε ytat. Proof Suppose that xt is a solution of 2.5. Let yt= b ε xt. Then by Lemma 2.7 and xb=0,wehavext= b+ε ε yt. By Lemma 2.,weget b ν xt= b ν xt= b ν b+ε ε yt= b+ε ν ε yt. Therefore and b+ε ν ε yt+ht + ν =0, xb=yb + ε=0, x = [ b+ε ε yt ] b = b+ε ε yt At. Same if vice versa. Using Lemmas 2.8 and 2.9, we may easily obtain the following Lemmas 2.0 and 2..

8 Liuet al. Boundary Value Problems :60 Page 8 of 2 Lemma 2.0 Let 0 C <, problem. and.2 is equivalent to the following problem: β ν 2 ϕ p b+ε ν ε yt = λf t, b+ε ε yt, yt + ε, yb + ε=0, [ b+ε ν ε yt] ν 2 =0, [ b+ε ε yt] = b b+ε ε ytat. 2.7 Lemma 2. FBVP 2.7 has the unique solution yt= s=ν Jt, sϕ q λ ν f s, b+ε ε y s, y s + ε. 2.8 Conversely, if yt satisfies 2.8, then yt is a solution of 2.7, where Jt, s is given by 2.4. Lemma 2.2 The function Jt, s has the following properties: i Jt, s 0, t, s [ε, b + ε] Nε [ν,b + ν ] Nν, ii b + ν 2 t ν ε ms Jt, s Mb + ν 2 t ν ε, t [ε, b + ε] Nε, where Jt, s is given by 2.4, and ms= M =+ G A s CƔν ε, G A CƔν ε, G A = max s [ν,b+ν ] Nν GA s. The proof of i is similar to Theorem.2 of [8],henceitisomitted.Forii,wenote that Jt, s sν b + ν 2 t ν ε + G A b + ν 2 t ν ε b + ν ν CƔν ε G A + b + ν 2 t ν ε. CƔν ε Then it is easy to get properties ii. Definition 2. Afunctionφt is called a lower solution of 2.7ifitsatisfies β ν 2 ϕ p b+ε ν ε φt λf t, b+ε ε φt, φt + ε, φb + ε 0, [ b+ε ν ε φt] ν 2 0, [ b+ε ε φt] b b+ε ε φtat. 2.9 Definition 2.4 Afunctionψt is called an upper solution of 2.7ifitsatisfies β ν 2 ϕ p b+ε ν ε ψt λf t, b+ε ε ψt, ψt + ε, ψb + ε 0, [ b+ε ν ε ψt] ν 2 0, [ b+ε ε ψt] b b+ε ε ψtat. 2.0

9 Liuet al. Boundary Value Problems :60 Page 9 of 2 Remark 2.2 Assume 0 C <,G A s 0 for s [ν,b + ν ] Nν, and y :[ε, b + ε] Nε R with yb + ε=0, [ b+ε ε yt ] b = b+ε ε yt At, b+ε ν ε yt 0, t [ν, b + ν] Nν. Then yt 0, t [ε, b + ε] Nε. In fact, let b+ε ν ε yt=ηt. Then yt= s=ν Jt, sηs. From ηt 0, we can get the conclusion yt 0, t [ε, b + ε] Nε. Lemma 2. Schauder fixed point theorem Let T be a continuous and compact mapping of a Banach space E into itself such that the set {x E : x = σ Tx} for some 0 σ is bounded. Then T has a fixed point. The method of upper and lower solutions To establish the existence of a solution for the boundary value problem, we need to make the following assumptions. H 4 A is defined on [0, b] N0, satisfying G A s 0 for s [ν,b + ν ] Nν,and0 C <. H 5 f, u, s:[0,b] N0 [0, + [0, + [0, + is continuous and is nonincreasing on u and s. For all λ 0,, thereexisttwoconstantsμ, μ 2 >0such that, for any t, u, s [0, b] N0 [0, + [0, +, f t, λu, s λ μ f t, u, s,. f t, u, λs λ μ 2 f t, u, s..2 Remark. Inequalities.,.2 are equivalent to the following inequalities.,.4, respectively: f t, λu, s λ μ f t, u, s, λ >,. f t, u, λs λ μ 2 f t, u, s, λ >..4 Now we denote ξ = l y μ +μ 2, ζ = l y μ +μ 2 Ɣβ Ɣβ and gt=f t, b+ε ε t ν ε,t ε ν ε, t T.Thengt CT, R, for m 0,, we define g m := b+ν 2 m g m s. s=ν

10 Liuet al.boundary Value Problems :60 Page 0 of 2 Theorem. Suppose that H 4 and H 5 hold. Then there exists a constant λ >0such that FBVP 2.7 has at least one positive solution wt for any λ λ,+. Moreover, there exists a constant 0<l <such that lb + ν 2 t ν ε wt l b + ν 2 t ν ε. Proof Let Q = C[ε, b + ε] Nε, R, and define a subset P of Q as follows: P = { y Q : l 0,, such that lb + ν 2 t ν ε yt l b + ν 2 t ν ε }. Clearly, P is a nonempty set since b + ν 2 t ν ε P. Now define the operator T λ in P. T λ yt= Jt, sϕ q ν λf s, b+ε ε y s, y s + ε,.5 s=ν where Jt, sisgivenby2.4. We assert that T λ is well defined and T λ P P. In fact, for any y P, there exists a positive number 0 < l y <suchthat l y b + ν 2 t ν ε yt l y b + ν 2 tν ε, t [ε, ε + b] Nε. Thus, by Lemma 2.2, condition H 5, Hölder s inequality and noticing m 0,, we get T λ yt = s=ν Jt, sϕ q ν λf s, b+ε ε y s, y s + ε λ q Jt, sϕ q ν f s, b+ε ε l y s ν ε, ly s ε ν ε s=ν λ q Jt, sϕ q ξ s=ν s u=ν s u β gu λ q Mb + ν 2 t ν ε ξ q s=ν ξλ q Mb + ν 2 ε ν ε g q m <+, s s=ν u=ν s u β m s u=ν mq s u β gu q i.e., T λ yt<+..6

11 Liuet al.boundary Value Problems :60 Page of 2 On the other hand, using Lemma 2.2 and Remark.,wehave T λ yt = s=ν Jt, sϕ q ν λf s, b+ε ε y s, y s + ε λ q Jt, sϕ q ν f s, b+ε ε l y s ν ε, l y s ε ν ε s=ν λ q Jt, sϕ q ζ s=ν s u=ν s u β gu λζ q msb + ν 2 t ν ε ϕ q s=ν s u=ν s u β gu. Therefore T λ yt λζ q b + ν 2 t ν ε ms s s=ν u=ν s u β gu q..7 Choose { [ I y = min 2, λξ q M g q m λζ q ms s=ν s u=ν s s=ν u=ν s u β m mq ], q } s u β gu..8 Then it follows from.6,.7and.8that I y b + ν 2 t ν ε T λ yt I y b + ν 2 tν ε. Next we shall devote our attention to finding the upper and lower solutions of FBVP 2.7. Let et= s=ν Jt, sϕ q ν gs. By Lemma 2.2,wehave et b + ν 2 t ν ε msϕ q ν gs, t [ε, b + ε] Nε, s=ν and consequently there exists a constant λ suchthat λ et b + ν 2 t ν ε..9

12 Liuet al.boundary Value Problems :60 Page 2 of 2 Thus, for any λ > λ,byh 5 andsimilarto.6, we have s=ν Jt, sϕ q ν f s, b+ε ε λe s, λe s + ε s=ν Jt, sϕ q ν f s, b+ε ε λ e s, λ e s + ε [ Jt, sϕ q s=ν [ Jt, sϕ q = s=ν <+, s u=ν s u=ν s u β f u, Ɣβ b+ε ε u ν ε, u ε ν ε ] s u β Ɣβ ] gu and et Mb + ν 2 t ν ε ϕ q s=ν s u=ν Ɣβ s u gu <+. Now let Take q ρ = Mb + ν 2 ε ν ε g q Ɣβ m +. { λ = max λ [ q, ρ μ +μ 2 q s=ν s s=ν u=ν s u β mq m msϕ q ν f s, b+ε ε, ] } [μ +μ 2 q ]q. Then by Lemma 2.2,.and.4, for t [ε, b + ε] Nε,wecanget + > s=ν Jt, sϕ q ν λ f s, b+ε ε λ q e s, λ q e s + ε b + ν 2 t ν ε λ [ μ +μ 2 q ]q s=ν msϕ q ν f s, b+ε ε e s, e s + ε b + ν 2 t ν ε λ [ μ +μ 2 q ]q msϕ q ν f s, b+ε ε ρ, ρ s=ν b + ν 2 t ν ε λ [ μ +μ 2 q ]q ρ μ +μ 2 q

13 Liuet al.boundary Value Problems :60 Page of 2 s=ν msϕ q ν f s, b+ε ε, b + ν 2 t ν ε. That is to say, s=ν Jt, sϕ q ν λ f s, b+ε ε λ q e s, λ q e s + ε b + ν 2 t ν ε..0 Let φt= λ q et=tλ b + ν 2 t ν ε, ψt=t λ φt.. It follows from.9and.0 that for any t [ε, b + ε] Nε, φt= s=ν Jt, sϕ q ν λ gs λ et b + ν 2 t ν ε, ψt= s=ν Jt, sϕ q ν λ f s, b+ε ε λ q es, λ q es + ε..2 Moreover, by. and.2, we know φb + ε=0, [ b+ε ν ε φt] ν 2 =0, [ b+ε ε φt] = b b+ε ε φtat, ψb + ε=0, [ b+ε ν ε ψt] ν 2 =0, [ b+ε ε ψt] = b b+ε ε ψtat.. Proceeding as in.6-.8, we get that φt, ψt P.By.0, we have ψt=t λ φt b + ν 2 t ν ε,.4 which implies ψt=t λ φt = s=ν s=ν Jt, sϕ q ν λ f s, b+ε ε λ q e s, λ q e s + ε Jt, sϕ q ν λ gs = φt, t [ε, b + ε] Nε..5 Thus, taking account of f being nonincreasing, and by.,.4and.5, we have β ν 2 ϕp b+ε ν ε ψ t+λ f t, b+ε ε ψ t, ψ t + ε = β ν 2 ϕp b+ε ν ε T λ φ t+λ f t, b+ε ε ψ t, ψ t + ε ϕp b+ε ν ε T λ φ t+λ f t, b+ε ε φ t, φ t + ε β ν 2

14 Liuet al.boundary Value Problems :60 Page 4 of 2 = λ f t, b+ε ε φ t, φ t + ε + λ f t, b+ε ε φ t, φ t + ε =0,.6 ϕp b+ε ν ε φ t+λ f t, b+ε ε φ t, φ t + ε β ν 2 = β ν 2 ϕp b+ε ν ε T λ b + ν 2 t ν ε + λ f t, b+ε ε φ t, φ t + ε = λ f t, b+ε ν ε t ν ε, t ε ν ε + λ f t, b+ε ε φ t, φ t + ε λ f t, b+ε ν ε t ν ε, t ε ν ε + λ f t, b+ε ν ε t ν ε, t ε ν ε =0..7 It follows from. and.5-.7 thatψt, φt are upper and lower solutions of FBVP 2.7 andψt,φt P. Now we define a function F t, b+ε ε y t, y f t, b+ε ε ψt, ψt + ε, t = f t, b+ε ε yt, yt + ε, f t, b+ε ε φt, φt + ε, y < ψt, ψt y φt, y > φt..8 It then follows from H 5 and.8thatft, u, s:[0,b] N0 [0, + [0, + [0, + is continuous. We now show that the FBVP β ν 2 ϕ p b+ε ν ε yt = λ Ft, b+ε ε yt, yt + ε, t [ε, b + ε] Nε, yb + ε=0, [ b+ε ν ε yt] ν 2 =0, [ b+ε ε yt] = b b+ε ε ytat.9 has a positive solution. Define the operator D λ by D λ yt= Jt, sϕ q ν λ F s, b+ε ε y s, y s + ε, s=ν t [ε, b + ε] Nε..20 Then D λ : C[ε, b + ε] Nε, R C[ε, b + ε] Nε, R, and a fixed point of the operator D λ is asolutionoffbvp.9. On the other hand, from the definition of F and the fact that the function f is nonincreasing on the second and third variable, we obtain f t, b+ε ε φ t, φ t + ε F t, b+ε ε y t, y t + ε f t, b+ε ε ψ t, ψ t + ε,

15 Liuet al.boundary Value Problems :60 Page 5 of 2 provided that ψt yt φt; F t, b+ε ε y t, y t + ε = f t, b+ε ε ψ t, ψ t + ε provided that yt<ψt; F t, b+ε ε y t, y t + ε = f t, b+ε ε φ t, φ t + ε provided that yt>φt. So we have f t, b+ε ε φ t, φ t + ε F t, b+ε ε y t, y t + ε f t, b+ε ε ψ t, ψ t + ε..2 Furthermore, by.,.4 and.2, we have f t, b+ε ε φ t, φ t + ε f t, b+ε ε ψ t, ψ t + ε f t, b+ε ε t ν ε, t ε ν ε = gt..22 It follows from Lemma 2.2 and.22 that for any y P, D λ yt = s=ν Jt, sϕ q ν λ F s, b+ε ε y s, y s + ε λ q Mb + ν 2 t ν ε Ɣβ λ Ɣβ s s=ν s q Mb + ν 2 ν ε g q m s=ν u=ν s u β mq m u=ν s u β gu q <+,.2 namely the operator D λ is uniformly bounded. Next, let P be bounded. Since the right side of.20 is finite sum, we can prove that D is equicontinuous. By the Arzela-Ascoli theorem, we have D λ : P P is completely continuous. Moreover,.2 impliesthatd λ satisfies the conditions of Lemma 2.. Thus, by using the Schauder fixed point theorem, D λ has at least one fixed point w such that w = D λ w. Now we prove ψt wt φt, t [ε, b + ε] Nε.

16 Liuet al.boundary Value Problems :60 Page 6 of 2 Since w is a fixed point of D λ,wehave [ wb + ε=0, b+ε ν ε wt ] ν 2 =0, [ b+ε ε wt ] b = b+ε ε wt At..24 From.,.22 and noticing that w is a fixed point of D λ,wealsohave Let Then β ν 2 ϕp b+ε ν ε φt β ν 2 ϕp b+ε ν ε wt = λ f t, b+ε ε t ν ε, t ε ν ε + λ F t, b+ε ε w t, w t + ε = λ f t, b+ε ε t ν ε, t ε ν ε + λ f t, b+ε ε ψ t, ψ t + ε λ f t, b+ε ε t ν ε, t ε ν ε + λ f t, b+ε ε t ν ε, t ε ν ε =0. zt=ϕ p b+ε ν ε φt ϕ p b+ε ν ε wt. β ν 2 zt=β ν 2 ϕp b+ε ν ε φt β ν 2 ϕp b+ε ν ε wt 0, t [ε, b + ε] Nε, zν 2=ϕ p b+ε ν ε φν 2 ϕ p b+ε ν ε wν 2 =0. Moreover, we have zt 0, i.e., ϕ p b+ε ν ε φt ϕ p b+ε ν ε wt 0. In fact, if we denote that β ν 2zt= ηt 0, according to Lemma 2.6,we have zt= ν ηt+k t ν + β + β, zν 2=0. In view of K =0,hencezt 0. Noticing that ϕ p is monotone increasing and b+ε ν ε is a linear operator, we have b+ε ν ε φ wt 0. It follows from Remark 2.2 and.24that φt wt 0.

17 Liuet al.boundary Value Problems :60 Page 7 of 2 Thus we have wt φtfort [ε, b + ε] Nε. In the same way, we also have wt ψtfor t [ε, b + ε] Nε,so ψt wt φt, t [ε, b + ε] Nε..25 Consequently, F t, b+ε ε w t, w t + ε = f t, b+ε ε w t, w t + ε, t [ε, b + ε] Nε. Hence wt is a positive solution of FBVP.9, i.e., yt= b+ε ε wt isapositivesolution of problem.and.2. Finally, by.25andφ, ψ P,wehave l ψ b + ν 2 t ν ε ψt wt φt l φ b + ν 2 tν ε. Let l y = min{l ψ, l φ },then l y b + ν 2 t ν ε ψt wt φt l y b + ν 2 tν ε. 4 Iteration of positive solutions To study the iteration of positive solutions to FBVP. and.2, we need the following assumption. H 6 f, u, s:[0,b] N0 [0, + [0, + [0, + is continuous, and there exist two constants r, r 2 >0such that for any t [0, b] N0, u, s [0, +. f t, λu, s λ r f t, u, s, λ 0,, 4. f t, u, λs λ r 2 f t, u, s, λ 0,. 4.2 Remark 4. Inequalities 4., 4.2 are equivalent to the following inequalities, respectively: f t, λu, s λ r f t, u, s, λ >, 4. f t, u, λs λ r 2 f t, u, s, λ >. 4.4 Definition 4. [7] Let E be a real Banach space. Let P be a nonempty, convex closed set in E.WesaythatP is a cone if it satisfies the following properties: i λu P for u P, λ 0; ii u, u P implies u = θ θ denotes the null element of E. Let the Banach space E = C[ε, ε + b] Nε be endowed with the norm { y = max max yt, max b+ε ν ε yt }. t [ε,ε+b] Nε t [ν,ν+b] Nν In addition, E + = {u E ut 0, t [ε, ε + b] Nε }.

18 Liuet al.boundary Value Problems :60 Page 8 of 2 Define the cone P E by P = { y E + b+ε ν ε yt 0, t [ν, ν + b] Nν } for any yt E +, λ > 0. Define an operator Tyt= Jt, sϕ q ν λf s, b+ε ε y s, y s + ε, 4.5 s=ν where Jt, sisgivenby2.4. Lemma 4. Assume that H, H 2 and H 6 hold, then the operator T : P Piscompletely continuous. Proof From H, H 2, H 6 and the definition of T,wededucethatforanyy P,thereis Tyt 0. b+ε ν ε Tyt= ϕ q ν λf t, b+ε ε y t, y t + ε 0, t [ν, ν + b] N0, which implies TP P. For convenience, we use the following notations. Let B = max { max t [ε,ε+b] Nε s=ν Jt, s, }, N = Bϕ q β + b. Ɣβ + We now give our results for the iteration of a positive solution for 2.7. Theorem 4. Suppose that H -H and H 6 hold. If there exists a positive constant a > such that H 7 f t, u, s f t, u 2, s 2 for any t [0, b] N0, 0 u < u 2 a, 0 s < s 2 a, H 8 ϕ p N ϕ pa λσ a r +r 2 k r b+ε ε, where k = Ɣε+. r, r 2, σ are defined by 4., 4.2 and., respectively. Then FBVP 2.7 has two positive solutions u and w such that 0 u a, 0 w a. Moreover, lim u n = lim T n u 0 = u, n n lim w n = lim T n w 0 = w, n n lim n b+ε ν ε u n = b+ε ν ε u, lim n b+ε ν ε w n = b+ε ν ε w, where u 0 t= a B s=ν Jt, s, w 0t=0,t [ε, ε + b] Nε. Tisdefinedby4.5. The iterative schemes in the theorem are and u 0 t, u n+ = Tu n = T n u 0, n =0,,2,..., w 0 t, w n+ = Tw n = T n w 0, n =0,,2,...

19 Liuet al.boundary Value Problems :60 Page 9 of 2 Proof Let P a = {u P :0 u a},wefirstlyprovetp a P a. If u P a, Tu P,byH 7 wehave 0 ut max ut = u a, 0 f t, b+ε ε u, u f t, ka, a k r a r +r 2 f t,, = σ k r a r +r 2, where k = b+ε ε Ɣε+ b+ε ε =+ b+ε Ɣε s=t+ε+ s t ε >. Since Tu = max Tut = t [ε,ε+b] Nε max t [ε,ε+b] Nε Jt, sϕ q ν λf s, b+ε ε u, u s=ν = max Jt, sϕ q λσ k r a r +r 2 ϕq ν t [ε,ε+b] Nε s=ν ϕ q σλk r a r +r 2 a, max t [ε,ε+b] Nε s=ν β + b β Jt, sϕ q Ɣβ + we get Tu a.sowehaveshownthattp a P a. Let u 0 t= a B s=ν Jt, s, t [ε, ε + b] N ε. Then b+ε ν ε u 0 t= a B 0. It is easy to get u0 t a s=ν Jt, s s=ν Jt, s = a, b+ε ν ε u 0 t a. So u 0 P a. Let u = Tu 0,wehaveu P a. We define u n+ = Tu n = T n+ u 0, n =0,,2,... It follows from TP a P a that u n P a, n =0,,2,...SinceT is completely continuous, we can assert that {u n } is a sequentially compact set. Since u t=tu 0 t λ Jt, sϕ q Ɣβ = s=ν s τ=ν Jt, sϕ q Ɣβ λσ kr a r +r 2 s=ν s τ β λf τ, b+ε ε u 0 τ, u 0 τ + ε s τ=ν u 0tB ϕ q λσ k r a r +r 2 b + β β ϕq a Ɣβ + s τ β

20 Liuet al.boundary Value Problems :60 Page 20 of 2 = u 0tN ϕ q λσ k r a r +r 2 a u 0 t and b+ε ν ε u t = ϕ q ν λf t, b+ε ε u 0 t, u 0 t + ε ϕ q λσ k r a r +r 2 b + β β ϕq Ɣβ + a b + β β N ϕ q = a Ɣβ + B a, we obtain u t u 0 t, b+ε ν ε u t b+ε ν ε u 0 t and u 2 t=tu t Tu 0 t=u t, b+ε ν ε u 2 t = b+ε ν ε Tu t b+ε ν ε Tu 0 t = b+ε ν ε u t, t [ε, ε + b] Nε. By induction we get u n+ t u n t, b+ε ν ε u n+ t b+ε ν ε u n t, t [ε, ε + b]nε, n =0,,2,... Thus, there exists u P a such that u n u. Applying the continuity of T and u n+ t= Tu n t, we get Tu t=u t, which implies that u is a nonnegative solution of FBVP 2.7. On the other hand, let w 0 =0,t [ε, ε + b] Nε,thenw 0 P a.letw = Tw 0,thenw P a. Denote w n+ = Tw n = T n+ w 0, n =0,,2,... It follows from TP a P a that w n P a, n =0,,2,... Since T is completely continuous, we can assert that {w n } is a sequentially compact set. Since w = Tw 0 P a,wehave w t=tw 0 t w 0, t [ε, ε + b] Nε. So w 2 t=tw t w, t [ε, ε + b] Nε, b+ε ν ε w 2 t = b+ε ν ε Tw t b+ε ν ε w t, t [ν, ν + b] Nν. By induction we get w n+ t w n t, b+ε ν ε w n+ t b+ε ν ε w n t, n =0,,2,...

21 Liuet al.boundary Value Problems :60 Page 2 of 2 Hence, there exists w P a such that w n w. Applying the continuity of T and w n+ t= Tw n t, we obtain Tw t=w t, which implies that w is a nonnegative solution of FBVP 2.7. Thus FBVP 2.7 has two positive solutions u, w such that 0 u a,0 w a, and from the above proof, we know that the iterative sequences hold. In order to illustrate the main result, we give the following example. Example 4. Consider the following FBVP: 2 2 ϕ yt = [e t 4 yt 5 + y t + ] +, 4.6 y =0, 4 76 y =0, 4 y = 7 2 where p =,ν = 2, β = 2, ε =, λ =,b =2, Let 0, t =0, At=, t =, 2, t =2. 4 ytat, 4.7 f t, u, s=e t u 5 + s +, r = 5, r 2 =, then for any λ 0, and u, s [0, +, t [0, 2] N0,wehave f t, λu, s=e t[ λu 5 + s ] + λ 5 [ e t u 5 + s + ] λ 5 f t, u, s, f t, u, λs=e t[ u 5 +λs ] + λ [ e t u 5 + s + ] λ f t, u, s, which implies that H 6 holds. On the other hand, it is clear that H, H 2 andh are satisfied, and σ = max f t,,= max 2e t + =2e 2 +. t [0,2] N0 t [0,2] N0 Next we compute k, C, B and N.Wehave C = 8 5, k = 4, 2 s= 2 J, s = 4 Ɣ 6 9 Ɣ 2 Ɣ Ɣ 2, 2 s= 2 4 J, s = 4 Ɣ 6 9 Ɣ 2 Ɣ Ɣ 2, 2 s= 2 7 J, s =0,

22 Liuet al.boundary Value Problems :60 Page 22 of 2 so B = max { N = Bϕ q 2 max t [ε,ε+b] Nε s=ν } Jt, s, = 4 9 Ɣ 6 Ɣ 2 Ɣ Ɣ 2,, ϕ p N= 2 ϕ pb= 8 Ɣ 6 27 Ɣ 2 Ɣ Ɣ 2. Take a =[ 2 Ɣ 6 Ɣ 2 Ɣ Ɣ 2 2 2e 2 +] 5 6,then ϕ p a Ɣ σ a r +r 2 k r Ɣ 2 Ɣ Ɣ 2 > 8 Ɣ 6 27 Ɣ 2 Ɣ Ɣ 2 = ϕ p N, which implies that H 8 holds. For a =[ 2 Ɣ 6 Ɣ 2 Ɣ Ɣ 2 2 2e 2 +] 5 6,itisclearthatH 7 holds. So by Theorem 4.,FBVP4.6and4.7 has two solutions u and w such that where and 0<u < 0< 4 u < [ Ɣ 6 2 Ɣ 2 Ɣ e 2 ] 6 Ɣ 2 +, lim u n = lim T n u 0 = u, n n u 0 t= 9 4 0<w < 2e < 4 w < [ Ɣ 6 2 Ɣ 2 Ɣ lim n Ɣ [ Ɣ 6 Ɣ 2 Ɣ ] 7 8 Ɣ t 2 2 Ɣ 2 2 t Ɣ 2 2 t 6 7Ɣ 2 Ɣ t Ɣ 2 2 t 6 7Ɣ 2 Ɣ e 2 ] 6 +, 4 u n = lim n 4 T n u 0 = 4 u,, t [0,,, t [, 2], [ Ɣ 6 2 Ɣ 2 Ɣ e 2 ] 6 Ɣ 2 +, [ Ɣ 6 2 Ɣ 2 Ɣ lim n w n = lim n T n w 0 = w, where w 0 t=0,t [0, 2] N0. Ɣ 2 lim n 2 5 2e 2 ] 6 +, 4 w n = lim n 4 T n w 0 = 4 w, Competing interests The authors declare that they have no competing interests. Authors contributions HL and CH worked together in the derivation of the mathematical results. All authors read and approved the final manuscript.

23 Liuet al.boundary Value Problems :60 Page 2 of 2 Acknowledgements The authors would like to thank the referees for invaluable comments and insightful suggestions. Project supported by the National Natural Science Foundation of China Chengmin Hou ; Yuanfeng Jin Publisher s Note Springer Nature remains neutral with regard to jurisdictional claims in published maps and institutional affiliations. Received: 6 December 206 Accepted: 5 March 207 References. Mastorakis, NE, Fathabadi, H: On the solution of p-laplacian for non-newtonian fluid flow. WSEAS Trans. Math. 86, Chhetri, M, Oruganti, S, Shivaji, R: Positive solutions for classes of p-laplacian equations. Differ. Integral Equ. 7-2, Oruganti, S, Shi, J, Shivaji, R: Logistic equation with the p-laplacian and constant yield harvesting. Abstr. Appl. Anal , Liu, C: Weak solutions for a viscous p-laplacian equation. Electron. J. Differ. Equ. 200, Evans, LC, Gangbo, W: Differential equations methods for the Monge-Kantorovich mass transfer problem. Mem. Am. Math. Soc. 7, Nieto, JJ, Pimentel, J: Positive solutions of a fractional thermostat model. Bound. Value Probl. 20,20 7. Ding, XL, Jiang, YL: Waveform relaxation methods for fractional functional differential equations. Fract. Calc. Appl. Anal. 6, Miller, KS, Ross, B: An Introduction to the Fractional Calculus and Fractional Differential Equations. Wiley, New York Nisar, KS, Baleanu, D, Qurashi, MMA: Fractional calculus and application of generalized Struve function. SpringerPlus 5, Oldham, KB, Spanier, J: The Fractional Calculus: Theory and Applications of Differentiation and Integration to Arbitrary Order. Dover, New York Virginia, K: The special functions of fractional calculus as generalized fractional calculus operators of some basic functions. Comput. Math. Appl. 59, Guy, J: On the derivative chain-rules in fractional calculus via fractional difference and their application to systems modelling. Open Phys. 6, Pilipović, S: Fractional calculus with applications in mechanics: vibrations and diffusion processes. Drug Dev. Ind. Pharm Atici, FM, Eloe, PW: A transform method in discrete fractional calculus. Int. J. Difference Equ. 2, Goodrich, SC: Solutions to a discrete right-focal fractional boundary value problem. Int. J. Difference Equ. 52, Goodrich, C, Peterson, AC: Discrete Fractional Calculus Springer, Berlin 205. ISBN: Ibrahim, RW, Jalab, HA: Existence of a class of fractional difference equations with two point boundary value problem. Adv. Differ. Equ. 205, Xie, Z, Hou, C: Properties of right fractional sum and right fractional difference operators and application. Adv. Differ. Equ. 205, Goodrich, CS: Systems of discrete fractional boundary value problems with nonlinearities satisfying no growth conditions. J. Differ. Equ. Appl. 25, Lin, X, Zhao, Z, Guan, Y: Iterative technology in a singular fractional boundary value problem with q-difference. Appl. Math. Irvine 7, Malinowska, AB, Torres, DFM: The delta-nabla calculus of variations. Fasc. Math. 4444, Natália, M, Torres, DFM: Calculus of variations on time scales with nabla derivatives. Nonlinear Anal. 72, e76-e Lv, W: Existence of solutions for discrete fractional boundary value problems with a p-laplacian operator. Adv. Differ. Equ. 202, Xie, Z, Jin, Y, Hou, C: Multiple solutions for a fractional difference boundary value problem via variational approach. Abstr. Appl. Anal. 202,7 202