Positive solutions to p-laplacian fractional differential equations with infinite-point boundary value conditions

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1 Wang et al. Advances in Difference Equations ( :425 R E S E A R C H Open Access Positive solutions to p-laplacian fractional differential equations with infinite-point boundary value conditions Han Wang 1,SuliLiu 1 and Huilai Li 1* * Correspondence: lihuilai@jlu.edu.cn 1 Department of Mathematics, Jilin University, Changchun, P.R. China Abstract This paper is devoted to an investigation of the multiple positive solutions to a class of infinite-point boundary value problems of nonlinear fractional differential equations coupled with the p-laplacian operators and infinite-point boundary value conditions. By means of the properties of Green s function and fixed point theorems, we establish the suitable criteria to guarantee the existence of positive solutions. Finally, an example is given in order to illustrate the main results. Keywords: Fractional differential equations; Infinite-point boundary conditions; p-laplacian operators; Fixed point theorems 1 Introduction With the advance of technology, researchers are not satisfied with the limitations of the integer order calculus anymore; therefore, fractional calculus the expansion and extension of the integer one is brought into the public. There is no doubt the study on nonlinear fractional differential equations has become an issue of focus and is widely being used in many fields, such asphysics [1, 2], biology [3], energy [4], chemical engineering [5], pharmacokinetics [6]. Based on practical application, researchers established lots of fractional differential equation models, so how to get the solutions to the equations is a big challenge to them. As is well known, it is difficult to give the exact solution, so giving the properties of the solutionsisasignificantpoint. Agreatnumberofresearchersarefocusedontheexistenceofsolutionsforlinearornonlinear FODE with different boundaryconditions. Take Zhang [7] for example, one studied the existence of minimal non-negative solution for a class of nonlinear fractional integrodifferential equations by applying the cone theory and the monotone iterative technique: D α +x(t+f (t, x(t, Tx(t, Sx(t =, t [,, x( =, D α 1 + x( =x, where (Tx(t= t k(t, sx(s ds, (Sx(t = h(t, sx(s ds, The Author(s 218. 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 author(s and the source, provide a link to the Creative Commons license, and indicate if changes were made.

2 Wang et al. Advances in Difference Equations ( :425 Page 2 of 15 and 1 < α 2, D α + is the Riemann Liouville derivative, f, k, h are continuous functions. Wang [8] considered the iterative positive solutions for a class of nonlocal fractional differential equations with nonlocal Hadamard integral and discrete boundary conditions by the monotone iterative method: H D α x(t+δ(tf (t, x(t =, t (1,, x(1 = x (1 =, H D α 1 x( =a H I β x(ξ+b m 2 i=1 c ix(η i, where 2 < α 3, a, b are real constants, c i (i =1,2,...,,m 2 are positive real constants, 1<ξ < η 1 < η 2 < < η m 2 <+,and H D α is the Hadamard fractional derivative. As for Guo [9], they investigated the existence of solutions for the following infinitepoint BVP with Caputo fractional derivative by Avery Peterson s fixed point theorem: c D α x(t+f (t, x(t, x (t =, < t <1, x( = x ( =, x (1 = j=1 ζ jx(ξ j, where 2 < α 3, ζ j, < ξ 1 < ξ 2 < < ξ i < <1,and i=1 ζ iξ i <1. Recently, some scientists have become interested in the study of the fractional differential equations with p-laplacian operators [1 13]. For example, Xiping Liu [12, 13] obtained some new results on the existence of positive solutions for the four-point BVP with mixed fractional derivatives and p-laplacian operator by a new method of lower and upper solutions which is based on the monotone iterative technique: D β +(φ p( C D α +x(t = f (t, x(t,c D α +x(t, t (, 1, c D α +x( = x ( =, x(1 = r 1 x(η, C D α +x(1 = rc 2 Dα +x(ξ, where 1 < α, β 2, r 1, r 2, f C([, 1] [, + (, ], [, +, c D α + is the Caputo derivative, D α + is the Riemann Liouville derivative, φ p is the p-laplacian operator which is defined as φ p (s= s p 2 s(p >1,φ q is the inverse function of φ p and 1 p + 1 q =1. In this paper, we consider a class of nonlinear fractional differential equations coupled with the p-laplacian operator and infinite-point boundary value conditions: D β +(φ p(d α +x(t λx(t = f (t, x(t, t (, 1], lim t + t 1 α x(t= i=1 μ ix(ξ i, (1 lim t + t 2 β (φ p (D α +x(t λx(t = φ p(d α +x(1 λx(1 =, where < α 1, 1 < β 2, λ <,μ i, < ξ i <1,i R +, α 1 <1,f C([, 1] [, +, [, +, φ p is the p-laplacian operator and D α, D β are the Riemann Liouville fractional derivatives. To the best of our knowledge, few researchers studied fractional differential equations with p-laplacian by the Riemann Liouville derivative; this condition coupled with values

3 Wang et al. Advances in Difference Equations ( :425 Page 3 of 15 at infinite number of points is not covered in the previous situations, and it is more general for the fractional differential equation models. By means of the properties of Green s function, Krasnosel skii fixed point theorem, the corollary of Krasnosel skii fixed point theorem and Avery Peterson fixed point theorem, we establish the suitable criteria to warrant the existence of the positive solutions for the BVP (1. The rest of the paper is organized as follows. In Sect. 2, we list several basic definitions and lemmas to prepare for the further study in this paper. On top of that, the main result is showed in Sect. 3.InSect.4, we give an example to illustrate our main results. 2 Preliminaries As is well known, C[a, b] is the Banach space of all continuous real functions defined on [a, b] withthenorm x = sup t [a,b] x(t and L 1 [a, b] isforthelebesgueintegrablereal functions defined on [a, b]. In this paper, we consider the following space: C 1 α [, 1] = { x(t:t 1 α x(t C[, 1], < α 1 }, endowed with its norm x 1 α = sup t [,1] t 1 α x(t. Obviously,C 1 α [, 1] is also a Banach space; moreover, C 1 α [, 1] L 1 [, 1]. Definition 2.1 ([14] Let α >, the Riemann Liouville fractional integral of the function f is defined by I α 1 +f (t= Γ (α t (t s α 1 f (s ds. Definition 2.2 ([14] Let α >, the Riemann Liouville fractional derivative of the function f is defined by ( D α +f (t= 1 d n t (t s n α 1 f (s ds, Γ (n α dt where n is the smallest integer greater than or equal to α. Definition 2.3 ([15] The two-parameter Mittag Leffler function is defined by E α,β (ζ = k= ζ k, α, β >,ζ R. Γ (αk + β Definition 2.4 ([16] If the map T satisfies T ( ax +(1 ay ( at(x+(1 at(y, x, y P, a [, 1], then T is called a non-negative convex (concave operator on a cone P. Lemma 2.1 ([17] Assume α >,u C(, 1 L(, 1, then I α +Dα +f (t=u(t+c 1t α 1 + c 2 t α c n t α n, where c i R, i =1,2,...,n with n is the smallest integer greater than or equal to α.

4 Wang et al. Advances in Difference Equations ( :425 Page 4 of 15 Lemma 2.2 ([18] Assume α (, 1], β >,λ is a negative constant, then the following conclusions hold: (i lim t + E α,β (λt α = 1 Γ (β, (ii E α,α (λt α >for t [, +, (iii E α,α (λt α is decreasing in t for t (, +, (iv E α,α (λt α E α,α+1 (λt α for t (, +. Lemma 2.3 Assume σ (t C(, 1], < α 1, then the problem D α + x(t λx(t=σ (t, t (, 1], lim t + t 1 α x(t= i=1 μ ix(ξ i, (2 has an unique solution x(t= G(t, sσ (s ds, (3 where G(t, s= Γ (αe α,α (λt α s ξ μ i i (ξ i s α 1 E α,α (λ(ξ i s α t 1 α (1 Γ (α + Eα,α(λ(t αα, i=1 μ i ξi α 1 E α,α (λξi α (t s 1 α s < t 1, Γ (αe α,α (λt α s ξ i μ i (ξ i s α 1 E α,α (λ(ξ i s α t 1 α (1 Γ (α i=1 μ i ξ α 1 i E α,α (λξ α i, <t s 1. (4 Proof According to ([19], if lim t + t (1 α x(t=c,wecaneasilyget x(t=c Γ (αt α 1 E α,α ( λt α + t (t s α 1 E α,α ( λ(t α α σ (s ds. In particular, if we choose c = i=1 μ ix(ξ i, by simple calculation, Hence, i=1 c = μ ξi i (ξ i s α 1 E α,α (λ(ξ i s α σ (s ds 1 Γ (α α 1 E α,α (λξi α. Γ (αt α 1 x(t= 1 Γ (α α 1 E α,α (λξi α ( μ i E α,α λt α ξ i (ξ i s α 1 ( E α,α λ(ξi s α σ (s ds = + i=1 t t + (t s α 1 E α,α ( λ(t α α σ (s ds Γ (αe α,α (λt α s ξ i μ i (ξ i s α 1 E α,α (λ(ξ i s α t 1 α (1 Γ (α α 1 E α,α (λξi α σ (s ds E α,α (λ(t α α (t s 1 α σ (s ds

5 Wang et al. Advances in Difference Equations ( :425 Page 5 of 15 = G(t, sσ (s ds, (5 where G(t, sisgivenby(4. The proof is finished. Lemma 2.4 Assume α (, 1], λ is a negative constant, then (i G(t, s>for t (, 1], s [, 1, (ii t 1 α G(t, s Γ (α(e α,α(λ 2 s ξ μ i i ξi α 1 1 Γ (α m for t, s [, 1], i=1 μ i ξi α 1 E α,α (λ (iii min t [,1] t 1 α G(t, s ρ max t [,1] t 1 α G(t, s for ρ (, 1, (iv t α 1 G(t, s ds (Γ (α(1 i= μ iξi α 1 1 n for t, s [, 1]. Proof (i By simple calculation, 1 Γ (α i=1 μ i ξi α 1 ( E α,α λξ α i 1 i=1 μ i ξ α 1 i >. In addition, according to Lemma 2.2, we can easilysee that (iholds. (ii Since (t s α 1 E α,α (λ(t s α for all s < t 1, t 1 α G(t, s Γ (αe α,α(λt α s ξ i μ i (ξ i s α 1 E α,α (λ(ξ i s α 1 Γ (α α 1 E α,α (λξi α Γ (α(e α,α(λ 2 s ξ i μ i ξi α 1 1 Γ (αe α,α (λ α 1 m >. (iii As t 1 α G(t, sisincreasingins and decreasing in t max t [,1] t1 α G(t, s= According to Lemma 2.4(i, we have s ξ i μ i (ξ i s α 1 E α,α (λ(ξ i s α 1 Γ (α α 1 E α,α (λξi α. min t [,1] t1 α G(t, s Γ (αe α,α (λ max t [,1] t1 α G(t, s ρ max t [,1] t1 α G(t, s, where ρ = Γ (αe α,α (λ, so < ρ <1. (iv We divide this proof into two parts: Part 1: For s t 1, t 1 α G(t, s ds = t + Γ (αe α,α (λt α s ξ i μ i (ξ i s α 1 E α,α (λ(ξ i s α 1 Γ (α α 1 E α,α (λξi α ds t 1 α (t s α 1 E α,α ( λ(t α α ds Γ (αe α,α(λt α α 1 1 Γ (α α 1 E α,α (λξi α + te ( α,α+1 λt α

6 Wang et al. Advances in Difference Equations ( :425 Page 6 of 15 Part 2: For t s 1, t 1 α G(t, s ds = i=1 μ iξi α 1 Γ (α(1 α Γ (α 1 = Γ (α(1 α 1 n. Γ (αe α,α (λt α s ξ i μ i (ξ i s α 1 E α,α (λ(ξ i s α 1 Γ (α α 1 E α,α (λξi α ds Γ (αe α,α(λt α α 1 1 Γ (α α 1 E α,α (λξi α i=1 μ iξi α 1 Γ (α(1 α 1 1 < Γ (α(1 α 1 n. The proof is completed. Lemma 2.5 Assume h(t C(, 1], 1 < β 2, then the problem D β + u(t=h(t, t (, 1], lim t + t 2 β u(t=u(1 =, (6 has an unique solution u(t= H(t, sh(s ds, (7 where H(t, s= 1 Γ (β t (1 s β 1 (t s β 1, s < t 1, t (1 s β 1, <t s 1. (8 Proof According to Lemma 2.1,we can easily get u(t= I β +h(t+c 1t β 1 + c 2 t β 2, by the condition (6, we obtain c 1 = I β +h(1, c 2 =. Hence, u(t= tβ 1 Γ (β (1 s β 1 h(s ds 1 Γ (β t (t s β 1 h(s ds = H(t, sh(s ds, where H(t, sisgivenby(8. The proof is finished.

7 Wang et al. Advances in Difference Equations ( :425 Page 7 of 15 Lemma 2.6 ([2] Assume β (1, 2], then (i H(t, s>for t (, 1], s [, 1, (ii H(t, s =H(1 s, 1 t for t, s (, 1, (iii t β 1 (1 ts(1 s β 1 Γ (βh(t, s (β 1s(1 s β 1 for t (, 1], s [, 1, (iv t β 1 (1 ts(1 s β 1 Γ (βh(t, s (β 1(1 tt β 1 for t (, 1], s [, 1. Lemma 2.7 Suppose h(t=f (t, x(t, σ (t=φ q (u(t hold, then we obtain the unique solution of BVP (1: ( x(t= G(t, sφ q H(s, τf ( τ, x(τ. (9 3 Main results In this section, we establish the related condition to warrant the existence of the multiple positive solutions for BVP (1 by means of the properties of Green s function we proved in the last section and the well-known Krasnosel skii fixed point theorem, the corollary of Krasnosel skii fixed point theorem and the Avery Peterson fixed point theorem. Let E = {x(t:x(t C 1 α [, 1]}.DefineP E, { } P = x(t E : x(t, min t [,1] t1 α x(t ρ max t [,1] t1 α x(t, where ρ is defined in Lemma 2.4. Suppose T : P E is the operator defined by ( Tx(t:= G(t, sφ q H(s, τf ( τ, x(τ. Throughout this section, suppose the following assumption holds: (H f (t, x M. Lemma 3.1 T : P P is a completely continuous operator. Proof Step 1: We show that T(P P. For x P, since the continuity of G(t, s, H(t, s, f (t, x(t,itiseasytogett 1 α Tx(t C[, 1]. According to Lemma 2.4 and Lemma 2.6, weobtaing(t, s, H(t, s, thanks to f (t, x(t being non-negative, therefore, Tx(t. Furthermore, min t [,1] t1 α Tx(t= min t [,1] t1 α Hence, T(P P. = G(t, sφ q ( min t [,1] t1 α G(t, sφ q ( ρ max t [,1] t1 α G(t, sφ q = ρ max t [,1] Tx(t. ( H(s, τf ( τ, x(τ H(s, τf ( τ, x(τ H(s, τf ( τ, x(τ

8 Wang et al. Advances in Difference Equations ( :425 Page 8 of 15 Step 2: We will prove that T is uniformly bounded. We have Tx(t ( 1 α = t1 α G(t, sφ q H(s, τf ( τ, x(τ t 1 α G(t, s ( ( φq H(s, τ f τ, x(τ ( (β 1M < nφ q Γ (β ( M = nφ q M 1. Γ (β 1 Therefore, Tx(t 1 α < M 1,whichmeansT is uniformly bounded. Step 3: We will give the proof of the equicontinuity of T. Obviously, t 1 α G(t, s is uniformly continuous in the interval [, 1] [, 1]. Therefore, for all ε >,thereexistsδ >, without loss of generality, we choose t 1, t 2 [, 1], when t 1 t 2 < δ,wehave Thus, G(t1, s G(t 2, s < Tx(t 1 Tx(t 2 1 α = < t1 α 1 t 1 α 2 ε φ q (M φ q( H(s, τ dτ ds. ( G(t 1, sφ q H(s, τf ( τ, x(τ ( G(t 2, sφ q H(s, τf ( τ, x(τ 2 G(t 2, s ( ( φq H(s, τ f τ, x(τ t 1 α 1 G(t 1, s t 1 α ε φ q (M φ q( H(s, τ dτ dsφ q(m = ε. φ q ( H(s, τ So, T is equicontinuous. AccordingtotheArzelà Ascolitheorem,T is a completely continuous operator. The proof is finished. Define P k = {x P : x 1 α < r k } andthenwedefinetherelatedboundary P k = {x P : x 1 α = r k } where k is the positive integer. Theorem 3.1 ([21]Krasnosel skiifixedpointtheorem Suppose E is a Banach space, and P Eisacone. Assume P 1, P 2 are two open and bounded subsets of E coupled with P 1, P 1 P 2 and suppose T : P (P 2 \ P 1 P is a completely continuous operator such that one of the following conditions holds: (L 1 Tx x, if x P P 1, and Tx x, if x P P 2,

9 Wang et al. Advances in Difference Equations ( :425 Page 9 of 15 (L 2 Tx x, if x P P 1, and Tx x, if x P P 2. Then T has at least one fixed point in P (P 2 \ P 1. Theorem 3.2 If there exist <r 1 < r 2, N 1 (ρm φ q( H(s, τ dτ ds 1, N 2 (n φ q( H(s, τ dτ ds 1 and N 1 r 1 < N 2 r 2 such that the following conditions hold: (L 3 f (t, x N 1 r 1,(t, x [, 1] [, r 1 ], (L 4 f (t, x N 2 r 2,(t, x [, 1] [, r 2 ], then T has at least one fixed point in P (P 2 \ P 1. In addition, r 1 x 1 α r 2. Proof Step 1: Let P 1 = {x P : x 1 α < r 1 }, by assumption (L 3, for all t [, 1], ( Tx 1 α = t 1 α G(t, sφ q H(s, τf ( τ, x(τ ( N 1 r 1 ρmφ q H(s, τ r 1 = x 1 α. Then we can get Tx 1 α x 1 α ( x P P 1. Step 2: Let P 2 = {x P : x 1 α < r 2 },byassumption(l 4, for all t [, 1], ( Tx 1 α = t 1 α G(t, sφ q H(s, τf ( τ, x(τ ( N 2 r 2 nφ q H(s, τ r 2 = x 1 α. It is easy to obtain Tx 1 α x 1 α ( x P P 2. To conclude, the above proof satisfies the condition (L 1 oftheorem3.1, thatistosay, T has at least one fixed point in P (P 2 \ P 1. In addition, r 1 x 1 α r 2. The proof is finished. Theorem 3.3 ([22] Suppose E is a Banach space, and P Eisacone. Assume P 3, P 4 are two open and bounded subsets of E and T : P 3 P is a completely continuous operator. If Tx 1 α < r 3, for x P 3 and Tx 1 α r 4, for x P 4 hold, then T has at least two fixed points in P 3, and x 1 1 α < r 3 < x 2 1 α r 4. Theorem 3.4 If there exist <r 3 < r 4, N 3 <(ρm φ q( H(s, τ dτ ds 1, N 4 (n φ q( H(s, τ dτ ds 1 such that the following conditions hold: (L 5 f (t, x<n 3 r 3, x P 3, (L 6 f (t, x N 4 r 4, x P 4, then T has at least two fixed points in P 3, and x 1 1 α < r 3 < x 2 1 α r 4.

10 Wang et al. Advances in Difference Equations ( :425 Page 1 of 15 Proof On the one hand, P 3 = {x P : x 1 α < r 3 },byassumption(l 5, for all t [, 1], ( Tx 1 α = t 1 α G(t, sφ q H(s, τf ( τ, x(τ ( N 3 r 3 nφ q H(s, τ < r 3. Then we can get Tx 1 α < r 3 ( x P 3. On the other hand, P 4 = {x P : x 1 α < r 4 }, by assumption (L 6, for all t [, 1], ( Tx 1 α = t 1 α G(t, sφ q H(s, τf ( τ, x(τ ( N 4 r 4 ρmφ q H(s, τ r 4. Obviously, Tx 1 α r 4 ( x P 4 holds. To conclude, the above proof satisfies the condition of Corollary 3.3,thatistosay,T has at least two fixed points in P 3,whichsatisfy x 1 1 α < r 3 < x 2 1 α r 4. The proof is finished. Suppose ϕ, ψ are non-negative continuous convex functionals on P, θ is a non-negative continuous concave functional on P,and η is a non-negative continuous functional on P. Define the following convex set: P(ϕ; d= { x P : ϕ(x<d }, P(ϕ, θ; b, d= { x P : b θ(x, ϕ(x d }, P(ϕ, ψ, θ; b, c, d= { x P : b θ(x, ψ(x c, ϕ(x d }, and a closed set P(ϕ, η; a, d= { x P : a θ(x, ϕ(x d }, where a, b, c, d are positive constants. Theorem 3.5 ([23]Avery Petersonfixedpointtheorem Suppose P is a cone on the real Banach space E. Suppose ϕ, ψ are non-negative continuous convex functionals on P, θ is a non-negative continuous concave functional on P, and η is a non-negative continuous functional on P satisfying η(ρx ρη(x for ρ 1, for positive constants k and d, we have θ(x η(x, x kϕ(x, x P(ϕ; d. (1

11 Wang et al. Advances in Difference Equations ( :425 Page 11 of 15 Suppose T : P(ϕ; d P(ϕ; d is a completely continuous operator and there exist a, b, c > with a< b such that the following conditions hold: (H 1 {x P(ϕ, ψ, θ; b, c, d:θ(x>b} and θ(x>b for x P(ϕ, ψ, θ; b, c, d, (H 2 θ(tx>b for x P(ϕ, θ; b, d with η(tx>c, (H 3 / P(ϕ, η; a, d and η(tx<a, for P(ϕ, η; a, d with η(x=a. Then T has at least three fixed points x 1, x 2, x 3 P(ϕ; d such that ϕ(x i d, θ(x 1 >b, θ(x 2 >a, η(x 2 <b, η(x 3 <a, where i =1,2,3. Define the following functionals: ϕ(x= x 1 α, N = ρm ψ(x=η(x= max t 1 α x(t, t [,1] ( φ q H(s, τf ( τ, x(τ, θ(x= min t 1 α x(t, t [,1] therefore, ϕ, ψ, η, θ satisfy the condition as mentioned above, and ρη(x θ(x ψ(x=η(x, x 1 α ϕ(x. Theorem 3.6 Suppose (H holds, there exist positive constants a, b, c, dwitha< b < c = bρ 1, (H 4 f (t, x Γ (β 1φ p (dn 1 for (t, x [, 1] [, d], (H 5 f (t, x>φ p (bn 1 for (t, x [, 1] [b, c], (H 6 f (t, x<γ (β 1φ p (an 1 for (t, x [, 1] [, a], Then T has at least three fixed points x 1, x 2, x 3 P(ϕ; d such that ϕ(x i d, θ(x 1 >b, θ(x 2 >a, η(x 2 <b, η(x 3 <a, where i =1,2,3. Proof For x(t P(ϕ; d, we have ϕ(x= x 1 α d. According to (H 4, we have Tx(t ( = t1 α G(t, sφ q H(s, τf ( τ, x(τ t 1 α G(t, s ( ( φq H(s, τ f τ, x(τ < nφ q ( Γ (β 1φq ( d n Γ (β 1 = d.

12 Wang et al. Advances in Difference Equations ( :425 Page 12 of 15 Hence, ϕ(tx= Tx 1 α d,thatistosay,t : P(ϕ; d P(ϕ; d is a completely continuous operator. Let x(t= b ρ, x(t P(ϕ, ψ, θ; b, c, d andθ( b ρ >b, andso{x(t P(ϕ, ψ, θ; b, c, d:θ( b ρ > b}. For x(t P(ϕ, ψ, θ; b, c, d, we obtain b x(t c = b ρ,then θ(tx= min t 1 α Tx(t t [,1] = min t [,1] t1 α > ρm = b. G(t, sφ q ( ρ max t [,1] t1 α G(t, sφ q ( ( φ q H(s, τf ( τ, x(τ dτ H(s, τf ( τ, x(τ H(s, τf ( τ, x(τ ds b N Hence, θ(tx>b for x P(ϕ, ψ, θ; b, c, d. That is to say, (H 1 holds. Let x P(ϕ, θ; b, dwithψ(tx>c = b ρ,weobtain θ(tx ρψ(tx>ρc = ρ b ρ = b. Therefore, (H 2 issatisfied. By θ( = < a,wehave / P(ϕ, η; a, d. Let x P(ϕ, η; a, dandη(x=a,weobtainϕ(x d, according to (H 6, η(tx= max t 1 α Tx(t t [,1] = max t [,1] t1 α n G(t, sφ q ( ( φ q H(s, τf ( τ, x(τ < nφ q ( Γ (β 1φq ( a n Γ (β 1 = a. H(s, τf ( τ, x(τ Hence, the condition (H 3 holds. The proof is finished. 4 Illustrative example Consider the following nonlinear fractional differential equations: D (φ 3.5(D.5 + x(t 3x(t = f (t, x(t, t (, 1], lim t + t.5 x(t= 1 k=1 2 x( 1, 2k (k 1! 2 2k lim t + t.3 (φ 3.5 (D.5 + x(t 3x(t = φ 3.5(D.5 + x(1 3x(1 =. (11

13 Wang et al. Advances in Difference Equations ( :425 Page 13 of 15 1 Obviously, we choose α =.5,β =1.7,p =3.5,λ = 3,μ k =, ξ 2 2k (k 1! k = 1, 2 2k k=1 μ kξk α 1 = < 1. By simple calculation, we obtain ρ.4819, m.11246, n e.5 2 Part 1: In order to satisfy Theorem 3.1, bymatlab,wegetn 1 = , N 2 =.5,r 1 =.2, r 2 =4 1 4, so we know that the BVP (11 has at least one positive solution x 1 satisfy.2 x 1 1 α 4 1 4,where f (t, x=7 1 3 e t + sin x, (t, x [, 1] [, 4 1 4], where f (t, xsatisfies (i f (t, x N 1 r 1 = for (t, x [, 1] [,.2], (ii f (t, x>n 2 r 2 =2 1 4 for (t, x [, 1] [, ]. Part 2: In order to make Theorem 3.3 hold, we choose N 3 = 5, N 4 =1,r 3 =2,r 4 =5, then the BVP (11 has at least two positive solutions x 1, x 2,and x 1 1 α <2< x 2 1 α 5, where f (t, x=(t x2, (t, x [, 1] [, 1], where f (t, xsatisfies (i f (t, x<n 3 r 3 = 1 for (t, x [, 1] [, 2], (ii f (t, x>n 4 r 4 =5for (t, x [, 1] [, 5]. Part 3: In order to make Theorem 3.5 hold, we choose a =.75, b =2,c = bρ 1 = 41.51, d =.5, then we know the BVP (11 has at least three positive solutions x 1, x 2, x 3 satisfying x.5 (x i.5 (i =1,2,3, min t.5 x 1 >2, max t.5 x 2 <2, t [,1] t [,1] max t.5 x 3 <.75, t [,1] min t.5 x 2 >.75, t [,1] where f (t, x= t + x 13, (t, x [, 1] [, 2], where f (t, xsatisfies (i f (t, x<γ (β 1φ p (dn 1 =.617 for (t, x [, 1] [,.5], (ii f (t, x>φ p (bn 1 = for (t, x [, 1] [2, 41.51], (iii f (t, x<γ (β 1φ p (an 1 =.342 for (t, x [, 1] [,.75]. 5 Conclusions In this paper, we consider a class of nonlinear fractional differential equations coupled with the p-laplacian operator and infinite-point boundary value conditions by means of the properties of Green s function, Krasnosel skii fixed point theorem, the corollary of Krasnosel skii fixed point theorem and the Avery Peterson fixed point theorem. On top of that, we establish the suitable criteria to warrant the existence of the positive solutions

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