Hao et al. Boundary Value Problems (2017) 2017:182 DOI 10.1186/s13661-017-0915-5

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Positive solutions for a system of nonlinear fractional nonlocal boundary value problems with parameters and p-Laplacian operator Xinan Hao1* , Huaqing Wang1 , Lishan Liu1,2 and Yujun Cui3 *

Correspondence: [email protected] 1 School of Mathematical Sciences, Qufu Normal University, Qufu, Shandong 273165, P.R. China Full list of author information is available at the end of the article

Abstract In this paper, we investigate the existence of positive solutions for a system of nonlinear fractional differential equations nonlocal boundary value problems with parameters and p-Laplacian operator. Under different combinations of superlinearity and sublinearity of the nonlinearities, various existence results for positive solutions are derived in terms of different values of parameters via the Guo-Krasnosel’skii fixed point theorem. Keywords: positive solution; p-Laplacian operator; fractional differential system; nonlocal boundary value problem

1 Introduction In this paper, we investigate the following system of nonlinear fractional differential equations nonlocal boundary value problems with parameters and p-Laplacian operator: ⎧ β ⎪ –Dα0+1 (ϕp1 (D0+1 u(t))) = λf (t, u(t), v(t)), ⎪ ⎪ ⎪ ⎪ ⎨–Dα+2 (ϕ (Dβ+2 v(t))) = μg(t, u(t), v(t)), 0

p2

0

⎪ ⎪ u(0) = u(1) = u (0) = u (1) = 0, ⎪ ⎪ ⎪ ⎩ v(0) = v(1) = v (0) = v (1) = 0, 



α

0 < t < 1, 0 < t < 1,

β D0+1 u(0) = 0, β D0+2 v(0) = 0,

β

β

D0+1 u(1) = b1 D0+1 u(η1 ), β

(1.1)

β

D0+2 v(1) = b2 D0+2 v(η2 ),

β

where αi ∈ (1, 2], βi ∈ (3, 4], D0+i and D0+i are the standard Riemann-Liouville derivatives, 1–αi p –1

ϕpi (s) = |s|pi –2 s, pi > 1, ϕp–1i = ϕqi , p1i + q1i = 1, ηi ∈ (0, 1), bi ∈ (0, ηi i ), i = 1, 2.f , g ∈ C([0, 1] × [0, +∞)2 , [0, +∞)), λ and μ are positive parameters. Fractional differential equation models are proved to be more adequate than integer order models for some problems in science and engineering. Fractional differential equations play a very important role in various fields due to their deep real world background. For an introduction of fractional calculus and fractional differential equations, we refer the reader to [1–3] and the references therein. Turbulent flow in a porous medium is a fundamental mechanics problem. For studying this type of problem, Leibenson [4] introduced differential equations with p-Laplacian © The Author(s) 2017. 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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operator       ϕp u (t) = f t, u(t) .

(1.2)

The study of differential equation with p-Laplacian operator is of significance theoretically and practically. It is quite natural to study fractional differential equation relative to equation (1.2). Recently, many scholars have paid more attention to the fractional order differential equation boundary value problems with p-Laplacian operator, see [5–23]. In [15], Lu et al. investigated a class of boundary value problems for fractional differential equations with p-Laplacian ⎧ ⎨Dα+ (ϕ (Dβ+ u(t))) = f (t, u(t)), 0 < t < 1, p 0 0 β β ⎩u(0) = u (0) = u (1) = 0, D + u(0) = D + u(1) = 0, 0

0

β

where α ∈ (1, 2], β ∈ (2, 3], Dα0+ and D0+ are the standard Riemann-Liouville derivatives, f ∈ C([0, 1] × [0, +∞), [0, +∞)). The existence and multiplicity results of positive solutions were obtained by using the Guo-Krasnosel’skii fixed point theorem, the Leggett-Williams fixed point theorem and the upper and lower solutions method. Xu and Dong [17] considered the following three point boundary value problem of fractional differential equation with p-Laplacian operator: ⎧ ⎨–Dα+ (ϕ (Dβ+ u(t))) = f (t, u(t)), p 0 0 ⎩u(0) = u(1) = u (0) = u (1) = 0,

0 < t < 1, β

β

D0+ u(0) = 0,

β

D0+ u(1) = bD0+ u(η),

β

where α ∈ (1, 2], β ∈ (3, 4], Dα0+ and D0+ are the standard Riemann-Liouville derivatives, 1–α p–1

η ∈ (0, 1), b ∈ (0, η ), f ∈ C([0, 1] × [0, +∞), [0, +∞)). The existence and uniqueness of positive solutions were obtained by using the upper and lower solutions method and Schauder’s fixed point theorem, the iterative sequences for the unique solution were also given. In [18], Zhang et al. considered the eigenvalue problems of fractional differential equations with integral boundary conditions and p-Laplacian operator ⎧ ⎨–Dα+ (ϕ (Dβ+ u(t))) = λf (t, u(t), 0 < t < 1, p 0 0 1 β ⎩u(0) = 0, D0+ u(0) = 0, u(1) = 0 u(s) dA(s),

(1.3)

β

where α ∈ (0, 1], β ∈ (1, 2], Dα0+ and D0+ are the standard Riemann-Liouville derivatives, 1 A is a function of bounded variation, 0 u(s) dA(s) is the Riemann-Stieltjes integral, f (t, u) : (0, 1) × (0, +∞) → [0, +∞) is a continuous function that may be singular at t = 0, 1 and u = 0. The existence of positive solutions of problem (1.3) was established by using the upper and lower solutions method and Schauder’s fixed point theorem. Lv [20] discussed an m-point boundary value problem of fractional differential equation with p-Laplacian operator ⎧ ⎨Dα+ (ϕ (Dβ+ u(t))) + ϕ (λ)f (t, u(t)) = 0, 0 < t < 1, p p 0 0 γ γ β ⎩u(0) = 0, D + u(1) = m–2 ξi D + u(ηi ), D + u(0) = 0, 0

i=1

0

0

(1.4)

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where α, γ ∈ (0, 1], β ∈ (1, 2], β – α – 1 ≥ 0, β – γ – 1 ≥ 0, λ is a positive parameter, 0 < β–α–1 ξi , ηi < 1, m–2 < 1, f ∈ C([0, 1] × [0, +∞), [0, +∞)). The existence and multiplicity i=1 ξi ηi of positive solutions for system (1.4) were established via the monotone iterative method and the fixed point index theory. The system of fractional differential equations boundary value problems with pLaplacian operator have also received much attention and have developed very rapidly, see [24–32]. In [24], Li et al. studied the following fractional differential system involving the p-Laplacian operator and nonlocal boundary conditions: ⎧ β ⎪ Dα0+1 (ϕp1 (D0+1 u(t))) = f (t, v(t)), 0 < t < 1, ⎪ ⎪ ⎪ ⎪ ⎨Dα+2 (ϕ (Dβ+2 v(t))) = g(t, u(t)), 0 < t < 1, p2 0 0 γ γ1 ⎪u(0) = Dβ+1 u(0) = 0, ⎪ D01+ u(1) = m–2 ⎪ j=1 a1j D0+ u(ηj ), 0 ⎪ ⎪ ⎩ γ γ2 β v(0) = D0+2 v(0) = 0, D02+ v(1) = m–2 j=1 a2j D0+ v(ηj ), α

β

γ

where αi , γi ∈ (0, 1], βi ∈ (1, 2], D0+i , D0+i and D0i+ are the standard Riemann-Liouville derivatives, i = 1, 2. The conditions for the existence of the maximal and minimal solutions to the system were established. Ren et al. [25] considered the following nonlocal fractional differential system: ⎧ β1 α1 ⎪ ⎪ ⎪–D0+ (ϕp1 (–D0+ u(t))) = f (u(t), v(t)), 0 < t < 1, ⎪ ⎪ ⎨–Dα+2 (ϕ (–Dβ+2 v(t))) = g(u(t), v(t)), 0 < t < 1, p2 0 0 1 β1 β ⎪ ⎪ u(0) = D u(0) = D0+1 u(1) = 0, u(1) = 0 u(s) dA(s), + ⎪ 0 ⎪ ⎪ 1 ⎩ β β v(0) = D0+2 v(0) = D0+2 v(1) = 0, v(1) = 0 v(s) dB(s), α

β

where αi , βi ∈ (1, 2], D0+i and D0+i are the standard Riemann-Liouville derivatives. A and B 1 1 are functions of bounded variations, 0 u(s) dA(s) and 0 v(s) dB(s) are Riemann-Stieltjes integrals. By introducing a new type of growth conditions and using the monotone iterative technique, some new results about the existence of maximal and minimal solutions were established, and the estimation of the lower and upper bounds of the maximum and minimum solutions was also derived. By means of the Avery-Henderson fixed point theorem and six functionals fixed point theorem, Rao [26] investigated the existence of multiple positive solutions for a coupled system of p-Laplacian fractional order two point boundary value problems ⎧ β1 α1 ⎪ ⎪–Da+ (ϕp (Da+ u(t))) = f (t, u(t), v(t)), ⎪ ⎪ ⎪ ⎨–Dα+2 (ϕ (Dβ+2 v(t))) = g(t, u(t), v(t)), a

p

a

⎪ ⎪ ξ u(a) – ηu (a) = 0, ⎪ ⎪ ⎪ ⎩ ξ v(a) – ηv (a) = 0,

a < t < b, a < t < b,

γ u(b) + δu (b) = 0, γ v(b) + δv (b) = 0, α

β

β

Da+1 u(a) = 0, β

Da+2 v(a) = 0,

where αi ∈ (0, 1], βi ∈ (1, 2], Da+i and Da+i are the standard Riemann-Liouville derivatives.

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He and Song [29] discussed the following fractional order differential system with pLaplacian operator: ⎧ β ⎪ Dα0+1 (ϕp1 (D0+1 u(t))) = λf (t, v(t)), 0 < t < 1, ⎪ ⎪ ⎪ ⎪ ⎨Dα+2 (ϕ (Dβ+2 v(t))) = μg(t, u(t)), 0 < t < 1, p2 0 0 β ⎪u(0) = 0, u(1) = a1 u(ξ1 ), ⎪ D0+1 u(0) = 0, ⎪ ⎪ ⎪ ⎩ β v(0) = 0, v(1) = a2 v(ξ2 ), D0+2 v(0) = 0, α

β

β

D0+1 u(1) = b1 D0+1 u(η1 ), β

β

D0+2 v(1) = b2 D0+2 v(η2 ),

β

where αi , βi ∈ (1, 2], D0+i and D0+i are the standard Riemann-Liouville derivatives, ξi , ηi ∈ (0, 1), ai , bi ∈ [0, 1], i = 1, 2. λ and μ are positive parameters. The uniqueness of solution was established by using the Banach contraction mapping principle. Khan et al. [31] considered the existence and uniqueness of solutions to a coupled system of fractional differential equations with p-Laplacian operator. The functions involved in the proposed coupled system were continuous and satisfied certain growth conditions. By using topological degree theory, some conditions were established which ensured the existence and uniqueness of solution to the proposed problem. Motivated by the papers mentioned above, in this paper, we study the existence of positive solutions for a system of nonlinear fractional differential equations nonlocal boundary value problems with parameters and p-Laplacian operator. Under different combinations of superlinearity and sublinearity of the functions f and g, various existence results for positive solutions are derived in terms of different values of λ and μ via the Guo-Krasnosel’skii fixed point theorem. Moreover, in this paper it is possible to replace the four point boundary conditions by multi-point boundary conditions or integral boundary conditions with minor modifications.

2 Preliminaries and lemmas We present here the definitions, some lemmas from the theory of fractional calculus and some auxiliary results that will be used to prove our main theorems. Definition 2.1 ([1–3]) The Riemann-Liouville fractional integral of order α > 0 of a function f : (0, +∞) → (–∞, +∞) is given by I0α+ f (t) =

1

(α)



t

(t – s)α–1 f (s) ds 0

provided the right-hand side is pointwise defined on (0, +∞). Definition 2.2 ([1–3]) The Riemann-Liouville fractional derivative of order α > 0 of a continuous function f : (0, +∞) → (–∞, +∞) is given by Dα0+ f (t) =

 n t d 1 (t – s)n–α–1 f (s) ds,

(n – α) dt 0

where n is the smallest integer not less than α, provided the right-hand side is pointwise defined on (0, +∞).

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Lemma 2.1 ([1–3]) Let α > 0. Then the following equality holds for u ∈ L(0, 1) and Dα0+ u(t) ∈ L(0, 1): I0α+ Dα0+ u(t) = u(t) + c1 t α–1 + c2 t α–2 + · · · + cn t α–n , where c1 , c2 , . . . , cn ∈ (–∞, +∞), n – 1 < α ≤ n. β

We transform problem (1.1) to its equivalent integral equations. Denote ϕp1 (D0+1 u(t)) = z(t), then p –1

z(1) = b11 z(η1 ).

z(0) = 0,

We now consider the following fractional differential equation: ⎧ ⎨–Dα+1 z(t) = y(t),

0 < t < 1,

0

⎩z(0) = 0,

(2.1)

p –1

z(1) = b11 z(η1 ).

Lemma 2.2 ([17]) If y ∈ C[0, 1], then problem (2.1) has a unique solution

1

H1 (t, s)y(s) ds,

z(t) = 0

where p –1

H1 (t, s) = h1 (t, s) +

b11 t α1 –1 p –1

1 – b11 η1α1 –1

h1 (η1 , s),

⎧ 0 ≤ t ≤ s ≤ 1, 1 ⎨[t(1 – s)]α1 –1 , h1 (t, s) =

(α1 ) ⎩[t(1 – s)]α1 –1 – (t – s)α1 –1 , 0 ≤ s ≤ t ≤ 1. From the above analysis, the boundary value problem ⎧ ⎨–Dα+1 (ϕ (Dβ+1 u(t))) = y(t), 0 < t < 1, p1 0 0 β ⎩u(0) = u(1) = u (0) = u (1) = 0, D +1 u(0) = 0, 0

β

β

D0+1 u(1) = b1 D0+1 u(η1 ),

is equal to ⎧ ⎨Dβ+1 u(t) = ϕ ( 1 H (t, s)y(s) ds), q1 0 1 0 ⎩u(0) = u(1) = u (0) = u (1) = 0. Lemma 2.3 ([17]) If y ∈ C[0, 1], then problem (2.2) has a unique solution

u(t) = 0



1

G1 (t, s)ϕq1

0

1

H1 (s, τ )y(τ ) dτ ds,

(2.2)

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where

G1 (t, s) =

⎧ ⎪ t β1 –2 (1 – s)β1 –2 [(s – t) + (β1 – 2)(1 – t)s], ⎪ ⎪ ⎪ ⎪ ⎨ 0 ≤ t ≤ s ≤ 1,

1

(β1 ) ⎪ ⎪ t β1 –2 (1 – s)β1 –2 [(s – t) + (β1 – 2)(1 – t)s] + (t – s)β1 –1 , ⎪ ⎪ ⎪ ⎩ 0 ≤ s ≤ t ≤ 1.

(2.3)

Lemma 2.4 ([17, 33]) The function G1 (t, s) defined by (2.3) is continuous on [0, 1] × [0, 1] and has the following properties: (a) G1 (t, s) > 0 for all (t, s) ∈ (0, 1) × (0, 1); (b) (β1 – 2)k1 (t)l1 (s) ≤ (β1 )G1 (t, s) ≤ M1 l1 (s), (t, s) ∈ (0, 1) × (0, 1); (c) (β1 – 2)k1 (t)l1 (s) ≤ (β1 )G1 (t, s) ≤ M1 k1 (t), (t, s) ∈ (0, 1) × (0, 1); where k1 (t) = t β1 –2 (1 – t)2 , l1 (s) = s2 (1 – s)β1 –2 , 

M1 = max β1 – 1, (β1 – 2)2 . Similarly, we can obtain the following Lemmas 2.5 and 2.6 for the following boundary value problem: ⎧ ⎨–Dα+2 (ϕ (Dβ+2 v(t))) = y(t), 0 < t < 1, p2 0 0 β ⎩v(0) = v(1) = v (0) = v (1) = 0, D +2 v(0) = 0, 0

β

β

D0+2 v(1) = b2 D0+2 u(η2 ).

(2.4)

Lemma 2.5 If y ∈ C[0, 1], then problem (2.4) has a unique solution



1

v(t) = 0

G2 (t, s)ϕq2

1

H2 (s, τ )y(τ ) dτ ds,

0

where ⎧ ⎪ t β2 –2 (1 – s)β2 –2 [(s – t) + (β2 – 2)(1 – t)s], ⎪ ⎪ ⎪ ⎪ 1 ⎨ 0 ≤ t ≤ s ≤ 1, G2 (t, s) =

(β2 ) ⎪ ⎪ t β2 –2 (1 – s)β2 –2 [(s – t) + (β2 – 2)(1 – t)s] + (t – s)β2 –1 , ⎪ ⎪ ⎪ ⎩ 0 ≤ s ≤ t ≤ 1, p –1

H2 (t, s) = h2 (t, s) +

b22 t α2 –1 p –1

1 – b22 η2α2 –1

h1 (η2 , s),

⎧ 0 ≤ t ≤ s ≤ 1, 1 ⎨[t(1 – s)]α2 –1 , h2 (t, s) =

(α2 ) ⎩[t(1 – s)]α2 –1 – (t – s)α2 –1 , 0 ≤ s ≤ t ≤ 1. Lemma 2.6 The function G2 (t, s) given by (2.5) has the properties: (a) G2 (t, s) > 0, (t, s) ∈ (0, 1) × (0, 1); (b) (β2 – 2)k2 (t)l2 (s) ≤ (β2 )G2 (t, s) ≤ M2 l2 (s), (t, s) ∈ (0, 1) × (0, 1); (c) (β2 – 2)k2 (t)l2 (s) ≤ (β2 )G2 (t, s) ≤ M2 k2 (t), (t, s) ∈ (0, 1) × (0, 1);

(2.5)

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where k2 (t) = t β2 –2 (1 – t)2 ,

 M2 = max β2 – 1, (β2 – 2)2 .

l2 (s) = s2 (1 – s)β2 –2 ,

Let X = C[0, 1], then X is a Banach space with the norm u = supt∈[0,1] |u(t)|. Let Y = X × X, then Y is a Banach space with the norm (u, v)Y = u + v. For θ1 , θ2 ∈ (0, 1) and θ1 < θ2 , denote       P = (u, v) ∈ Y : u(t) ≥ 0, v(t) ≥ 0, ∀t ∈ [0, 1], min u(t) + v(t) ≥ γ (u, v)Y , t∈[θ1 ,θ2 ]

i –2) where γ = min{γ1 , γ2 } and γi = (βM mint∈[θ1 ,θ2 ] ki (t), i = 1, 2, then P is a cone of Y . Define i operators T1 , T2 : Y → X and Q : Y → Y as follows:

T1 (u, v)(t) = ϕq1 (λ)

G1 (t, s)ϕq1

0

T2 (u, v)(t) = ϕq2 (μ)



1

0



1

G2 (t, s)ϕq2

0

  Q(u, v) = T1 (u, v), T2 (u, v) ,



1

1





H1 (s, τ )f τ , u(τ ), v(τ ) dτ ds,

t ∈ [0, 1],

  H2 (s, τ )g τ , u(τ ), v(τ ) dτ ds,

t ∈ [0, 1],

0

(u, v) ∈ Y .

It is well known that if (u, v) is a fixed point of the operator Q in P, then (u, v) is a positive solution of system (1.1). Lemma 2.7 Q : P → P is a completely continuous operator. Proof For (u, v) ∈ P and t ∈ [0, 1], obviously, T1 (u, v)(t) ≥ 0, T2 (u, v)(t) ≥ 0. It follows from Lemmas 2.4 and 2.6 that

T1 (u, v)(t) = ϕq1 (λ)

G1 (t, s)ϕq1

0

≤ ϕq1 (λ)



1

0

1

1

  H1 (s, τ )f τ , u(τ ), v(τ ) dτ ds

0

M1 l1 (s) ϕq

(β1 ) 1



1

  H1 (s, τ )f τ , u(τ ), v(τ ) dτ ds,

0

then   T1 (u, v) ≤ M1 ϕq (λ)

(β1 ) 1





1

l1 (s)ϕq1

0

1

  H1 (s, τ )f τ , u(τ ), v(τ ) dτ ds.

1

  H2 (s, τ )g τ , u(τ ), v(τ ) dτ ds.

1

  H1 (s, τ )f τ , u(τ ), v(τ ) dτ ds

0

Similarly,   T2 (u, v) ≤ M2 ϕq (μ)

(β2 ) 2





1

l2 (s)ϕq2

0

0

Therefore   Q(u, v) ≤ M1 ϕq (λ) Y

(β1 ) 1 +

0

M2 ϕq (μ)

(β2 ) 2



1

l1 (s)ϕq1

0

0



1

l2 (s)ϕq2

1 0

  H2 (s, τ )g τ , u(τ ), v(τ ) dτ ds.

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On the other hand, for any t ∈ [θ1 , θ2 ], by Lemma 2.4, we have  1   β1 – 2 T1 (u, v)(t) ≥ ϕq1 (λ) k1 (t)l1 (s)ϕq1 H1 (s, τ )f τ , u(τ ), v(τ ) dτ ds 0 (β1 ) 0  1

1   β1 – 2 k1 (t)ϕq1 (λ) l1 (s)ϕq1 H1 (s, τ )f τ , u(τ ), v(τ ) dτ ds =

(β1 ) 0 0   β1 – 2 k1 (t)T1 (u, v), ≥ M1

1

then   min T1 (u, v)(t) ≥ γ1 T1 (u, v).

t∈[θ1 ,θ2 ]

Similarly,   min T2 (u, v)(t) ≥ γ2 T2 (u, v).

t∈[θ1 ,θ2 ]

Hence   min T1 (u, v)(t) + T2 (u, v)(t)

t∈[θ1 ,θ2 ]

≥ min T1 (u, v)(t) + min T2 (u, v)(t) t∈[θ1 ,θ2 ]

t∈[θ1 ,θ2 ]

    ≥ γ1 T1 (u, v) + γ2 T2 (u, v)       ≥ γ T1 (u, v) + T2 (u, v) = γ Q(u, v)Y , that is, Q(P) ⊂ P. By the Ascoli-Arzela theorem and the continuity of f , g, Gi and Hi , we deduce that T1 and T2 are completely continuous operators, then Q is a completely continuous operator. This completes the proof.  Lemma 2.8 ([34]) Let P be a positive cone in a Banach space E, 1 and 2 are bounded open sets in E, θ ∈ 1 , 1 ⊂ 2 , A : P ∩ ( 2 \ 1 ) → P is a completely continuous operator. If the following conditions are satisfied: (i) Ax ≤ x, ∀x ∈ P ∩ ∂ 1 , Ax ≥ x, ∀x ∈ P ∩ ∂ 2 , or (ii) Ax ≥ x, ∀x ∈ P ∩ ∂ 1 , Ax ≤ x, ∀x ∈ P ∩ ∂ 2 , then A has at least one fixed point in P ∩ ( 2 \ 1 ).

3 Main results Denote f0s = lim sup max

f (t, u, v) , ϕp1 (u + v)

g0s = lim sup max

f0i = lim inf min

f (t, u, v) , ϕp1 (u + v)

g0i = lim inf min

u+v→0 t∈[0,1]

u+v→0 t∈[θ1 ,θ2 ]

f (t, u, v) , u+v→∞ t∈[0,1] ϕp1 (u + v)

s = lim sup max f∞

u+v→0 t∈[0,1]

g(t, u, v) , ϕp2 (u + v)

u+v→0 t∈[θ1 ,θ2 ]

g(t, u, v) , ϕp2 (u + v)

g(t, u, v) , u+v→∞ t∈[0,1] ϕp2 (u + v)

s g∞ = lim sup max

Hao et al. Boundary Value Problems (2017) 2017:182

i f∞ = lim inf min

u+v→∞ t∈[θ1 ,θ2 ]





1

A=

l1 (s)ϕq1

0

C=

f (t, u, v) , ϕp1 (u + v) 1

u+v→∞ t∈[θ1 ,θ2 ]



0



θ2

l1 (s)ϕq1 θ1

g(t, u, v) . ϕp2 (u + v)  1

1 B= l2 (s)ϕq2 H2 (s, τ ) dτ ds,

i = lim inf min g∞

H1 (s, τ ) dτ ds, 0



θ2

Page 9 of 18



H1 (s, τ ) dτ ds,

0

D=

θ1



θ2

l2 (s)ϕq2 θ1



θ2

H2 (s, τ ) dτ ds. θ1

i i For f0s , g0s , f∞ , g∞ ∈ (0, ∞), we define the symbols L1 , L2 , L3 and L4 as follows:

 L1 = ϕp1  L3 = ϕp2

(β1 ) 2(β1 – 2)Cγ1 γ

(β2 ) 2(β2 – 2)Dγ2 γ



1 , i f∞ 1 , i g∞





(β1 ) 1 , 2AM1 f0s 

(β2 ) 1 L4 = ϕp2 . 2BM2 g0s

L2 = ϕp1

i i , g∞ ∈ (0, ∞), L1 < L2 , L3 < L4 , then for each λ ∈ (L1 , L2 ) and Theorem 3.1 (1) If f0s , g0s , f∞ μ ∈ (L3 , L4 ), system (1.1) has at least one positive solution (u(t), v(t)), t ∈ (0, 1). i i (2) If f0s = 0, g0s , f∞ , g∞ ∈ (0, ∞), L3 < L4 , then for each λ ∈ (L1 , ∞) and μ ∈ (L3 , L4 ), system (1.1) has at least one positive solution (u(t), v(t)), t ∈ (0, 1). i i (3) If g0s = 0, f0s , f∞ , g∞ ∈ (0, ∞), L1 < L2 , then for each λ ∈ (L1 , L2 ) and μ ∈ (L3 , ∞), system (1.1) has at least one positive solution (u(t), v(t)), t ∈ (0, 1). i i (4) If f0s = g0s = 0, f∞ , g∞ ∈ (0, ∞), then for each λ ∈ (L1 , ∞) and μ ∈ (L3 , ∞), system (1.1) has at least one positive solution (u(t), v(t)), t ∈ (0, 1). i i (5) If {f0s , g0s ∈ (0, ∞), f∞ = ∞} or {f0s , g0s ∈ (0, ∞), g∞ = ∞}, then for each λ ∈ (0, L2 ) and μ ∈ (0, L4 ), system (1.1) has at least one positive solution (u(t), v(t)), t ∈ (0, 1). i i (6) If {f0s = 0, g0s ∈ (0, ∞), g∞ = ∞} or {f0s = 0, g0s ∈ (0, ∞), f∞ = ∞}, then for each λ ∈ (0, ∞) and μ ∈ (0, L4 ), system (1.1) has at least one positive solution (u(t), v(t)), t ∈ (0, 1). i i (7) If {f0s ∈ (0, ∞), g0s = 0, g∞ = ∞} or {f0s ∈ (0, ∞), g0s = 0, f∞ = ∞}, then for each λ ∈ (0, L2 ) and μ ∈ (0, ∞), system (1.1) has at least one positive solution (u(t), v(t)), t ∈ (0, 1). i i (8) If {f0s = g0s = 0, g∞ = ∞} or {f0s = g0s = 0, f∞ = ∞}, then for each λ ∈ (0, ∞) and μ ∈ (0, ∞), system (1.1) has at least one positive solution (u(t), v(t)), t ∈ (0, 1).

Proof Because the proofs of the above cases are similar, in what follows we will prove two of them, namely cases (1) and (6). i i (1) For any λ ∈ (L1 , L2 ) and μ ∈ (L3 , L4 ), there exists 0 < ε < min{f∞ , g∞ } such that 



(β1 ) 1 1 ≤ λ ≤ ϕ , p1 i –ε f∞ 2AM1 f0s + ε  

(β2 )

(β2 ) 1 1 ≤ μ ≤ ϕ . ϕp2 p 2 s i 2(β2 – 2)Dγ2 γ g∞ – ε 2BM2 g0 + ε

ϕp1

(β1 ) 2(β1 – 2)Cγ1 γ



By the definitions of f0s and g0s , there exists R1 > 0 such that   f (t, u, v) < f0s + ε ϕp1 (u + v),   g(t, u, v) < g0s + ε ϕp2 (u + v),

t ∈ [0, 1], 0 ≤ u + v ≤ R1 , t ∈ [0, 1], 0 ≤ u + v ≤ R1 .

Hao et al. Boundary Value Problems (2017) 2017:182

Page 10 of 18

Denote 1 = {(u, v) ∈ Y : (u, v)Y < R1 }, for any (u, v) ∈ P ∩ ∂ 1 and t ∈ [0, 1], we have 0 ≤ u(t) + v(t) ≤ u + v = (u, v)Y = R1 , then

T1 (u, v)(t) = ϕq1 (λ)

M1

(β1 )

≤

G1 (t, s)ϕq1

0



1





H1 (s, τ )f τ , u(τ ), v(τ ) dτ ds 0

 1     M1 l1 (s)ϕq1 H1 (s, τ ) f0s + ε ϕp1 u(τ ) + v(τ ) dτ ds 0 (β1 ) 0  1

 s    1 ϕq1 (λ)ϕq1 f0 + ε l1 (s)ϕq1 H1 (s, τ )ϕp1 u + v dτ ds

≤ ϕq1 (λ)



1

1

0

0

     ϕq1 λ f0s + ε A(u, v)Y

M1

(β1 )  1 ≤ (u, v)Y , 2 =

so T1 (u, v) ≤ 12 (u, v)Y , (u, v) ∈ P ∩ ∂ 1 . In a similar manner, we deduce

T2 (u, v)(t) = ϕq2 (μ)



1

G2 (t, s)ϕq2

0

1

  H2 (s, τ )g τ , u(τ ), v(τ ) dτ ds

0

 1     M2 H2 (s, τ ) g0s + ε ϕp2 u(τ ) + v(τ ) dτ ds ≤ ϕq2 (μ) l2 (s)ϕq2 0 (β2 ) 0  1

1      M2 ≤ ϕq2 μ g0s + ε l2 (s)ϕq2 H2 (s, τ ) dτ ds(u, v)Y

(β2 ) 0 0       M2 1 ϕq2 μ g0s + ε B(u, v)Y ≤ (u, v)Y , = (3.1)

(β2 ) 2

1

then T2 (u, v) ≤ 12 (u, v)Y , (u, v) ∈ P ∩ ∂ 1 . Hence         Q(u, v) = T1 (u, v) + T2 (u, v) ≤ (u, v) , Y Y

(u, v) ∈ P ∩ ∂ 1 .

(3.2)

i i On the other hand, by the definitions of f∞ and g∞ , there exists R2 > 0 such that

 i  f (t, u, v) ≥ f∞ – ε ϕp1 (u + v), t ∈ [θ1 , θ2 ], u, v ≥ 0, u + v ≥ R2 ,   i – ε ϕp2 (u + v), t ∈ [θ1 , θ2 ], u, v ≥ 0, u + v ≥ R2 . g(t, u, v) ≥ g∞ Denote R2 = max{2R1 , Rγ2 } and 2 = {(u, v) ∈ Y : (u, v)Y < R2 }. For any (u, v) ∈ P ∩ ∂ 2 , we have mint∈[θ1 ,θ2 ] (u(t) + v(t)) ≥ γ (u, v)Y = γ R2 ≥ R2 , then T1 (u, v)(θ1 )

= ϕq1 (λ)



1

G1 (θ1 , s)ϕq1

0

0

1

  H1 (s, τ )f τ , u(τ ), v(τ ) dτ ds

 θ2   β1 – 2 γ1 l1 (s)ϕq1 H1 (s, τ )f τ , u(τ ), v(τ ) dτ ds 0 (β1 ) θ1  θ2

θ2   i   β1 – 2 ≥ γ1 ϕq1 (λ) l1 (s)ϕq1 H1 (s, τ ) f∞ – ε ϕp1 u(τ ) + v(τ ) dτ ds

(β1 ) θ1 θ1

≥ ϕq1 (λ)

1

Hao et al. Boundary Value Problems (2017) 2017:182

≥

Page 11 of 18

  i  β1 – 2 γ1 ϕq1 λ f∞ –ε

(β1 )





θ2

θ2

l1 (s)ϕq1 θ1

θ1

     i β1 – 2 γ1 γ ϕq1 λ f∞ – ε C (u, v)Y ≥ =

(β1 )

    H1 (s, τ )ϕp1 γ (u, v)Y dτ ds

 1 (u, v) , Y 2

and T1 (u, v) ≥ 12 (u, v)Y , (u, v) ∈ P ∩ ∂ 2 . Similarly, we have T2 (u, v)(θ2 )

= ϕq2 (μ)



1

G2 (θ2 , s)ϕq2

0

1

  H2 (s, τ )g τ , u(τ ), v(τ ) dτ ds

0

 θ2   β2 – 2 H2 (s, τ )g τ , u(τ ), v(τ ) dτ ds γ2 l2 (s)ϕq2 θ1 0 (β2 )  θ2

1   i   β2 – 2 ≥ γ2 ϕq2 (μ) l2 (s)ϕq2 H2 (s, τ ) g∞ – ε ϕp2 u(τ ) + v(τ ) dτ ds

(β2 ) 0 θ1  θ2

θ2    i    β2 – 2 l2 (s)ϕq2 H2 (s, τ ) g∞ – ε ϕp2 γ (u, v)Y dτ ds ≥ γ2 ϕq2 (μ)

(β2 ) θ1 θ1

≥ ϕq2 (μ)

=

1

    i   1 β2 – 2 γ2 γ ϕq2 μ g∞ – ε D(u, v)Y ≥ (u, v)Y ,

(β2 ) 2

then T2 (u, v) ≥ 12 (u, v)Y , (u, v) ∈ P ∩ ∂ 2 . So,     Q(u, v) ≥ (u, v) , Y Y

(u, v) ∈ P ∩ ∂ 2 .

(3.3)

Therefore, by (3.2), (3.3) and Lemma 2.8, we conclude that Q has at least one fixed point (u, v) ∈ P ∩ ( 2 \ 1 ) with R1 ≤ (u, v)Y ≤ R2 . i (6) Suppose f0s = 0, g0s ∈ (0, ∞), g∞ = ∞, then for any λ ∈ (0, ∞) and μ ∈ (0, L4 ), there exists ε > 0 such that  0 < λ < ϕp1



(β1 ) 1 , 2AM1 ε

 ϕp2



(β2 )

(β2 ) 1 ε < μ < ϕp2 . (β2 – 2)Dγ2 γ 2BM2 g0s + ε

By the definitions of f0s and g0s , there exists R3 > 0 such that f (t, u, v) < εϕp1 (u + v), t ∈ [0, 1], 0 ≤ u + v ≤ R3 ,   g(t, u, v) < g0s + ε ϕp2 (u + v), t ∈ [0, 1], 0 ≤ u + v ≤ R3 . Denote 3 = {(u, v) ∈ Y : (u, v)Y < R3 }. For any (u, v) ∈ P ∩ ∂ 3 and t ∈ [0, 1], we have  1   M1 l1 (s)ϕq1 H1 (s, τ )εϕp1 u(τ ) + v(τ ) dτ ds T1 (u, v)(t) ≤ ϕq1 (λ) 0 (β1 ) 0  1

1   M1 ≤ ϕq1 (λε) l1 (s)ϕq1 H1 (s, τ ) dτ ds(u, v)Y

(β1 ) 0 0     M1 1 ϕq1 (λε)A(u, v)Y < (u, v)Y , =

(β1 ) 2

1

then T1 (u, v) ≤ 12 (u, v)Y , (u, v) ∈ P ∩ ∂ 3 .

Hao et al. Boundary Value Problems (2017) 2017:182

Page 12 of 18

Similar to (3.1) of (1), we get T2 (u, v) ≤ 12 (u, v)Y , (u, v) ∈ P ∩ ∂ 3 , then     Q(u, v) ≤ (u, v) , Y Y

(u, v) ∈ P ∩ ∂ 3 .

(3.4)

i = ∞, there exists R4 > 0 such that On the other hand, by g∞

1 g(t, u, v) ≥ ϕp2 (u + v), ε

t ∈ [θ1 , θ2 ], u, v ≥ 0, u + v ≥ R4 .

Let R4 = max{2R3 , Rγ4 } and 4 = {(u, v) ∈ Y : (u, v)Y < R4 }. For any (u, v) ∈ P ∩ ∂ 4 , we have mint∈[θ1 ,θ2 ] (u(t) + v(t)) ≥ γ (u, v)Y = γ R4 ≥ R4 , then

T2 (u, v)(θ2 ) = ϕq2 (μ)



1

G2 (θ2 , s)ϕq2

0

1

  H2 (s, τ )g τ , u(τ ), v(τ ) dτ ds

0

 θ2   β2 – 2 γ2 l2 (s)ϕq2 H2 (s, τ )g τ , u(τ ), v(τ ) dτ ds 0 (β2 ) θ1  θ2

1  β2 – 2 1  γ2 ϕq2 (μ) ≥ l2 (s)ϕq2 H2 (s, τ ) ϕp2 u(τ ) + v(τ ) dτ ds

(β2 ) ε θ1 0  θ2

θ2   1   β2 – 2 γ2 ϕq2 (μ) l2 (s)ϕq2 H2 (s, τ ) ϕp2 γ (u, v)Y dτ ds ≥

(β2 ) ε θ1 θ1      β2 – 2 μ γ2 γ ϕq2 D(u, v)Y > (u, v)Y . =

(β2 ) ε

≥ ϕq2 (μ)

1

Therefore       Q(u, v) ≥ T2 (u, v) ≥ (u, v) , Y Y

(u, v) ∈ P ∩ ∂ 4 .

(3.5)

By (3.4), (3.5) and Lemma 2.8, we conclude that Q has at least one fixed point (u, v) ∈ P ∩ ( 4 \ 3 ) with R3 ≤ (u, v)Y ≤ R4 . This completes the proof.  s s , g∞ ∈ (0, ∞), we define the symbols L1 , L2 , L3 , L4 as follows: For f0i , g0i , f∞

L1 = ϕp1 L3 = ϕp2

 

(β1 ) 2(β1 – 2)Cγ1 γ

(β2 ) 2(β2 – 2)Dγ2 γ



1 , f0i 1 , g0i



(β1 ) 1 , s 2AM1 f∞ 

(β2 ) 1 L4 = ϕp2 . s 2BM2 g∞

L2 = ϕp1



s s , g∞ , f0i , g0i ∈ (0, ∞), and L1 < L2 , L3 < L4 , then for each λ ∈ (L1 , L2 ) Theorem 3.2 (1) If f∞ and μ ∈ (L3 , L4 ), system (1.1) has at least one positive solution (u(t), v(t)), t ∈ (0, 1). s s (2) If f∞ , f0i , g0i ∈ (0, ∞), g∞ = 0, and L1 < L2 , then for each λ ∈ (L1 , L2 ) and μ ∈ (L3 , ∞), system (1.1) has at least one positive solution (u(t), v(t)), t ∈ (0, 1). s s , f0i , g0i ∈ (0, ∞), f∞ = 0, L3 < L4 , then for each λ ∈ (L1 , ∞) and μ ∈ (L3 , L4 ), system (3) If g∞ (1.1) has at least one positive solution (u(t), v(t)), t ∈ (0, 1). s s (4) If f0i , g0i ∈ (0, ∞), f∞ = g∞ = 0, then for each λ ∈ (L1 , ∞) and μ ∈ (L3 , ∞), system (1.1) has at least one positive solution (u(t), v(t)), t ∈ (0, 1). s s s s (5) If {f∞ , g∞ ∈ (0, ∞), f0i = ∞} or {f∞ , g∞ ∈ (0, ∞), g0i = ∞}, then for each λ ∈ (0, L2 ) and μ ∈ (0, L4 ), system (1.1) has at least one positive solution (u(t), v(t)), t ∈ (0, 1).

Hao et al. Boundary Value Problems (2017) 2017:182

Page 13 of 18

s s s s (6) If {f0i = ∞, f∞ ∈ (0, ∞), g∞ = 0} or {f∞ ∈ (0, ∞), g∞ = 0, g0i = ∞}, then for each λ ∈ (0, L2 ) and μ ∈ (0, ∞), system (1.1) has at least one positive solution (u(t), v(t)), t ∈ (0, 1). s s s s (7) If {f0i = ∞, g∞ ∈ (0, ∞), f∞ = 0} or {g∞ ∈ (0, ∞), g0i = ∞, f∞ = 0}, then for each λ ∈ (0, ∞) and μ ∈ (0, L4 ), system (1.1) has at least one positive solution (u(t), v(t)), t ∈ (0, 1). s s s s (8) If {f∞ = g∞ = 0, f0i = ∞} or {f∞ = g∞ = 0, g0i = ∞}, then for each λ ∈ (0, ∞) and μ ∈ (0, ∞), system (1.1) has at least one positive solution (u(t), v(t)), t ∈ (0, 1).

Proof Because the proofs of the above cases are similar, in what follows we will prove two of them, namely cases (1) and (6). (1) For any λ ∈ (L1 , L2 ) and μ ∈ (L3 , L4 ), there exists 0 < ε < min{f0i , g0i } such that 



 1

(β1 ) 1 , ≤ λ ≤ ϕ p 1 i s 2AM1 f∞ + ε f0 – ε  

(β2 )

(β2 ) 1 1 ϕp2 . ≤ μ ≤ ϕp2 i s 2(β2 – 2)Dγ2 γ g0 – ε 2BM2 g∞ + ε ϕp1

(β1 ) 2(β1 – 2)Cγ1 γ

By the definitions of f0i and g0i , there exists R1 > 0 such that   f (t, u, v) ≥ f0i – ε ϕp1 (u + v),   g(t, u, v) ≥ g0i – ε ϕp2 (u + v),

t ∈ [θ1 , θ2 ], u, v ≥ 0, u + v ≤ R1 , t ∈ [θ1 , θ2 ], u, v ≥ 0, u + v ≤ R1 .

Denote 1 = {(u, v) ∈ Y : (u, v) < R1 }, for any (u, v) ∈ P ∩ ∂ 1 , we have T1 (u, v)(θ1 )

= ϕq1 (λ)



1

G1 (θ1 , s)ϕq1

0

1

  H1 (s, τ )f τ , u(τ ), v(τ ) dτ ds

0

 θ2   β1 – 2 ≥ ϕq1 (λ) γ1 l1 (s)ϕq1 H1 (s, τ )f τ , u(τ ), v(τ ) dτ ds θ1 (β1 ) θ1  θ2

θ2   i   β1 – 2 γ1 ϕq1 (λ) l1 (s)ϕq1 H1 (s, τ ) f0 – ε ϕp1 u(τ ) + v(τ ) dτ ds ≥

(β1 ) θ1 θ1

≥

θ2

      1 β1 – 2 γ1 γ ϕq1 λ f0i – ε C (u, v)Y ≥ (u, v)Y ,

(β1 ) 2

then T1 (u, v) ≥ 12 (u, v)Y , (u, v) ∈ P ∩ ∂ 1 . Similarly, we have T2 (u, v)(θ2 )

= ϕq2 (μ)



1 0

G2 (θ2 , s)ϕq2

1

  H2 (s, τ )g τ , u(τ ), v(τ ) dτ ds

0

 θ2   β2 – 2 γ2 l2 (s)ϕq2 H2 (s, τ )g τ , u(τ ), v(τ ) dτ ds θ1 (β2 ) θ1  θ2

θ2     β2 – 2 l2 (s)ϕq2 H2 (s, τ ) g0i – ε ϕp2 u(τ ) + v(τ ) dτ ds γ2 ϕq2 (μ) ≥

(β2 ) θ1 θ1

≥ ϕq2 (μ)

≥

θ2

      1 β2 – 2 γ2 γ ϕq2 μ g0i – ε D(u, v)Y ≥ (u, v)Y ,

(β2 ) 2

Hao et al. Boundary Value Problems (2017) 2017:182

Page 14 of 18

then T2 (u, v) ≥ 12 (u, v)Y , (u, v) ∈ P ∩ ∂ 1 . Therefore     Q(u, v) ≥ (u, v) , Y Y

(u, v) ∈ P ∩ ∂ 1 .

(3.6)

On the other hand, we define f ∗ , g ∗ : [0, 1] × [0, +∞) → [0, +∞) as follows: f ∗ (t, x) = max f (t, u, v), 0≤u+v≤x

g ∗ (t, x) = max g(t, u, v), 0≤u+v≤x

then f (t, u, v) ≤ f ∗ (t, x),

t ∈ [0, 1], u, v ≥ 0, u + v ≤ x,

g(t, u, v) ≤ g ∗ (t, x),

t ∈ [0, 1], u, v ≥ 0, u + v ≤ x.

Clearly, f ∗ (t, x) and g ∗ (t, x) are nondecreasing on x, by the proof of [35], we have f ∗ (t, x) s ≤ f∞ , t∈[0,1] ϕp1 (x)

lim sup max x→+∞

g ∗ (t, x) s ≤ g∞ . t∈[0,1] ϕp2 (x)

lim sup max x→+∞

From the above inequalities, there exists R2 > 0 such that f ∗ (t, x) f ∗ (t, x) s ≤ lim sup max + ε ≤ f∞ + ε, ϕp1 (x) x→+∞ t∈[0,1] ϕp1 (x)

t ∈ [0, 1], x ≥ R2 ,

g ∗ (t, x) g ∗ (t, x) s + ε, ≤ lim sup max + ε ≤ g∞ ϕp2 (x) x→+∞ t∈[0,1] ϕp2 (x)

t ∈ [0, 1], x ≥ R2 .

s s + ε)ϕp1 (x), g ∗ (t, x) ≤ (g∞ + ε)ϕp2 (x), t ∈ [0, 1], x ≥ R2 . Then f ∗ (t, x) ≤ (f∞ Denote R2 = max{2R1 , R2 }, 2 = {(u, v) ∈ Y : (u, v)Y < R2 }. For any (u, v) ∈ P ∩ ∂ 2 , by the definitions of f ∗ and g ∗ , we have

      f t, u(t), v(t) ≤ f ∗ t, (u, v)Y ,

      g t, u(t), v(t) ≤ g ∗ t, (u, v)Y ,

t ∈ [0, 1],

so  1     M1 H1 (s, τ )f ∗ τ , (u, v)Y dτ ds l1 (s)ϕq1 0 (β1 ) 0  1

1    s   M1 ≤ ϕq1 (λ) l1 (s)ϕq1 H1 (s, τ ) f∞ + ε ϕp1 (u, v)Y dτ ds

(β1 ) 0 0       M1 1 s = ϕq1 λ f∞ + ε A(u, v)Y ≤ (u, v)Y ,

(β1 ) 2

T1 (u, v)(t) ≤ ϕq1 (λ)

1

and T1 (u, v) ≤ 12 (u, v)Y , (u, v) ∈ P ∩ ∂ 2 . Similarly, we have  1     M2 l2 (s)ϕq2 H2 (s, τ )g ∗ τ , (u, v)Y dτ ds 0 (β2 ) 0  1

1    s   M2 ≤ ϕq2 (μ) l2 (s)ϕq2 H2 (s, τ ) g∞ + ε ϕp2 (u, v)Y dτ ds

(β2 ) 0 0       1 M2 s + ε B(u, v)Y ≤ (u, v)Y , = ϕq2 μ g∞

(β2 ) 2

T2 (u, v)(t) ≤ ϕq2 (μ)

1

Hao et al. Boundary Value Problems (2017) 2017:182

Page 15 of 18

so T2 (u, v) ≤ 12 (u, v)Y , (u, v) ∈ P ∩ ∂ 2 . Therefore         Q(u, v) = T1 (u, v) + T2 (u, v) ≤ (u, v) , Y Y

(u, v) ∈ P ∩ ∂ 2 .

(3.7)

From (3.6), (3.7) and Lemma 2.8, we get that Q has at least one fixed point (u, v) ∈ P ∩ ( 2 \ 1 ) with R1 ≤ (u, v)Y ≤ R2 . s s (6) Suppose f0i = ∞, f∞ ∈ (0, ∞), g∞ = 0, for any λ ∈ (0, L2 ) and μ ∈ (0, ∞), there exists ε > 0 such that  ϕp1



(β1 )

(β1 ) 1 ε < λ < ϕp1 , 0 (β1 – 2)Cγ1 γ 2AM1 f∞ + ε

 0 < μ < ϕp2



(β2 ) 1 . 2BM2 ε

By f0i = ∞, there exists R3 > 0 such that 1 f (t, u, v) ≥ ϕp1 (u + v), ε

t ∈ [θ1 , θ2 ], u, v ≥ 0, 0 ≤ u + v ≤ R3 .

Choose 3 = {(u, v) ∈ Y : (u, v)Y < R3 }, then for any (u, v) ∈ P ∩ ∂ 3 , we have  θ2   β1 – 2 γ1 l1 (s)ϕq1 H1 (s, τ )f τ , u(τ ), v(τ ) dτ ds θ1 (β1 ) θ1  θ2

θ2  β1 – 2 1  γ1 ϕq1 (λ) ≥ l1 (s)ϕq1 H1 (s, τ ) ϕp1 u(τ ) + v(τ ) dτ ds

(β1 ) ε θ1 θ1  θ2  θ2   λ β1 – 2 γ1 γ ϕq1 l1 (s)ϕq1 H1 (s, τ ) dτ ds(u, v)Y ≥

(β1 ) ε θ1 θ1      λ β1 – 2 γ1 γ ϕq1 C (u, v)Y ≥ (u, v)Y . =

(β1 ) ε

T1 (u, v)(θ1 ) ≥ ϕq1 (λ)

θ2

Thus,       Q(u, v) ≥ T1 (u, v) ≥ (u, v) , Y Y

(u, v) ∈ P ∩ ∂ 3 .

On the other hand, we define f ∗ , g ∗ : [0, 1] × [0, +∞) → [0, +∞) as follows: f ∗ (t, x) = max f (t, u, v), 0≤u+v≤x

g ∗ (t, x) = max g(t, u, v). 0≤u+v≤x

By the proof of [35], we have f ∗ (t, x) s ≤ f∞ , t∈[0,1] ϕp1 (x)

lim sup max x→+∞

g ∗ (t, x) = 0. x→+∞ t∈[0,1] ϕp (x) 2 lim max

For above ε > 0, there exists R4 > 0 such that, for any t ∈ [0, 1], x ≥ R4 , we have f ∗ (t, x) f ∗ (t, x) s + ε, ≤ lim sup max + ε ≤ f∞ ϕp1 (x) x→+∞ t∈[0,1] ϕp1 (x) g ∗ (t, x) g ∗ (t, x) ≤ lim max + ε = ε, ϕp2 (x) x→+∞ t∈[0,1] ϕp2 (x) s hence f ∗ (t, x) ≤ (f∞ + ε)ϕp1 (x), g ∗ (t, x) ≤ εϕp2 (x).

(3.8)

Hao et al. Boundary Value Problems (2017) 2017:182

Page 16 of 18

Let R4 = max{2R3 , R4 } and 4 = {(u, v) ∈ Y : (u, v)Y < R4 }. For any (u, v) ∈ P ∩ ∂ 4 and t ∈ [0, 1], we have       f t, u(t), v(t) ≤ f ∗ t, (u, v)Y ,

      g t, u(t), v(t) ≤ g ∗ t, (u, v)Y ,

then  1     M1 H1 (s, τ )f ∗ τ , (u, v)Y dτ ds l1 (s)ϕq1 0 (β1 ) 0  1

1    s   M1 ≤ ϕq1 (λ) l1 (s)ϕq1 H1 (s, τ ) f∞ + ε ϕp1 (u, v)Y dτ ds

(β1 ) 0 0      s  1 M1 ϕq1 λ f∞ + ε A(u, v)Y ≤ (u, v)Y , =

(β1 ) 2

T1 (u, v)(t) ≤ ϕq1 (λ)

1

so T1 (u, v) ≤ 12 (u, v)Y , (u, v) ∈ P ∩ ∂ 4 . In a similar manner, we deduce  1     M2 l2 (s)ϕq2 H2 (s, τ )g ∗ τ , (u, v)Y dτ ds 0 (β2 ) 0  1

1    M2 ≤ ϕq2 (μ) l2 (s)ϕq2 H2 (s, τ )εϕp2 (u, v)Y dτ ds

(β2 ) 0 0    1 M2 ϕq2 (με)B(u, v)Y ≤ (u, v)Y , =

(β2 ) 2

T2 (u, v)(t) ≤ ϕq2 (μ)

1

so T2 (u, v) ≤ 12 (u, v)Y , (u, v) ∈ P ∩ ∂ 4 . Therefore         Q(u, v) = T1 (u, v) + T2 (u, v) ≤ (u, v) , Y Y

(u, v) ∈ P ∩ ∂ 4 .

(3.9)

From (3.8), (3.9) and Lemma 2.8, we conclude that Q has at least one fixed point (u, v) ∈ P ∩ ( 4 \ 3 ) with R3 ≤ (u, v)Y ≤ R4 . This completes the proof. 

4 Conclusion In this paper, we study the existence of positive solutions for a system of nonlinear fractional differential equations nonlocal boundary value problems with parameters and pLaplacian operator. Under different combinations of superlinearity and sublinearity of the functions f and g, various existence results for positive solutions are derived in terms of different values of λ and μ via the Guo-Krasnosel’skii fixed point theorem. Acknowledgements The authors would like to thank the referees for their pertinent comments and valuable suggestions. Funding This work is supported financially by the National Natural Science Foundation of China (11501318, 11371221), the Natural Science Foundation of Shandong Province of China (ZR2015AM022, ZR2017MA036) and the China Postdoctoral Science Foundation (2017M612230). Abbreviations Not applicable. Availability of data and materials Not applicable.

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Competing interests The authors declare that they have no competing interests. Authors’ contributions All authors contributed equally to the writing of this paper. The authors read and approved the final manuscript. Author details 1 School of Mathematical Sciences, Qufu Normal University, Qufu, Shandong 273165, P.R. China. 2 Department of Mathematics and Statistics, Curtin University, Perth, WA 6845, Australia. 3 Department of Applied Mathematics, Shandong University of Science and Technology, Qingdao, 266590, P.R. China.

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