Positive Solutions for a System of Semipositone Fractional Difference

Hindawi Journal of Function Spaces Volume 2018, Article ID 6835028, 11 pages https://doi.org/10.1155/2018/6835028

Research Article Positive Solutions for a System of Semipositone Fractional Difference Boundary Value Problems Cheng Chen,1 Jiafa Xu

,1 Donal O’Regan,2 and Zhengqing Fu

3

1

School of Mathematical Sciences, Chongqing Normal University, Chongqing 40133, China School of Mathematics, Statistics and Applied Mathematics, National University of Ireland, Galway, Ireland 3 College of Mathematics and System Sciences, Shandong University of Science and Technology, Qingdao 266590, China 2

Correspondence should be addressed to Jiafa Xu; [email protected] Received 16 April 2018; Accepted 8 July 2018; Published 18 July 2018 Academic Editor: Lishan Liu Copyright © 2018 Cheng Chen et al. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited. Using the fixed point index, we establish two existence theorems for positive solutions to a system of semipositone fractional difference boundary value problems. We adopt nonnegative concave functions and nonnegative matrices to characterize the coupling behavior of our nonlinear terms.

𝑓, 𝑔 (𝑡, 𝑥, 𝑦) ≥ −𝑀,

1. Introduction In this paper we study the existence of positive solutions for the system of fractional difference boundary value problems involving semipositone nonlinearities: −Δ]]−3 𝑥 (𝑡) = 𝑓 (𝑡 + ] − 1, 𝑥 (𝑡 + ] − 1) , 𝑦 (𝑡 + ] − 1)) , −Δ]]−3 𝑦 (𝑡)

𝑡 ∈ [0, 𝑏 + 2]N0 , = 𝑔 (𝑡 + ] − 1, 𝑥 (𝑡 + ] − 1) , 𝑦 (𝑡 + ] − 1)) ,

𝑥 (] − 3) = = 𝑦 (] − 3) = =

𝑡 ∈ [0, 𝑏 + 2]N0 , 󵄨 [Δ𝛼]−3 𝑥 (𝑡)]󵄨󵄨󵄨𝑡=]−𝛼−2 (1) 󵄨󵄨 𝛽 [Δ ]−3 𝑥 (𝑡)]󵄨󵄨󵄨 = 0, 󵄨𝑡=]+𝑏+2−𝛽 󵄨 [Δ𝛼]−3 𝑦 (𝑡)]󵄨󵄨󵄨𝑡=]−𝛼−2 𝛽 󵄨󵄨 [Δ ]−3 𝑦 (𝑡)]󵄨󵄨󵄨 = 0, 󵄨𝑡=]+𝑏+2−𝛽

where 2 < ] ≤ 3, 1 < 𝛽 < 2, ] − 𝛽 > 1, 0 < 𝛼 < 1, 𝑏 > 3 (𝑏 ∈ N), and Δ]]−3 is a discrete fractional operator. For the nonlinear terms 𝑓, 𝑔, we assume the following. (H0) 𝑓, 𝑔 : [] − 1, 𝑏 + ] + 1]N]−1 × R+ × R+ 󳨀→ R are two continuous functions; moreover, there exists a positive constant 𝑀 > 0 such that

for all (𝑡, 𝑥, 𝑦) ∈ [] − 1, 𝑏 + ] + 1]N]−1 × R+ × R+ .

(2)

Note that, in this paper, we use [𝑎, 𝑏]N𝑎 to stand for {𝑎, 𝑎 + 1, 𝑎+2, . . . , 𝑏} with 𝑏−𝑎 ∈ N1 , where N𝑎 fl {𝑎, 𝑎+1, 𝑎+2, ⋅ ⋅ ⋅ }. Fractional calculus has been applied in physics, chemistry, aerodynamics, biophysics, and blood flow phenomena. For example, CD4+ T cells’ infections can be depicted by a fractional order model 𝐷𝛼1 (𝑇) = 𝑠 − 𝐾𝑉𝑇 − 𝑑𝑇 + 𝑏𝐼, 𝐷𝛼2 (𝐼) = 𝐾𝑉𝑇 − (𝑏 + 𝛿) 𝐼,

(3)

𝛼3

𝐷 (𝑉) = 𝑁𝛿𝐼 − 𝑐𝑉, where 𝐷𝛼𝑖 (𝑖 = 1, 2, 3) are fractional derivatives (see [1, 2]); we also refer the reader to [1–45] and the references therein. In [3], the authors considered the existence of positive solutions for the semipositone discrete fractional system − Δ]1 𝑦1 (𝑡) = 𝜆 1 𝑓1 (𝑡 + ]1 − 1, 𝑦1 (𝑡 + ]1 − 1) , 𝑦2 (𝑡 + ]2 − 1)) , 𝑡 ∈ [1, 𝑏 + 1]N ,

2

Journal of Function Spaces − Δ]2 𝑦2 (𝑡)

Definition 1. We define 𝑡] fl Γ(𝑡 + 1)/Γ(𝑡 + 1 − ]) for any 𝑡, ] ∈ R for which the right-hand side is well-defined. We use the convention that if 𝑡+1−] is a pole of the Gamma function and 𝑡 + 1 is not a pole, then 𝑡] = 0.

= 𝜆 2 𝑓2 (𝑡 + ]2 − 1, 𝑦1 (𝑡 + ]1 − 1) , 𝑦2 (𝑡 + ]2 − 1)) , 𝑡 ∈ [1, 𝑏 + 1]N ,

Definition 2. For ] > 0, the ]−th fractional sum of a function 𝑓 is

𝑦1 (]1 − 2) = 𝑦1 (]1 + 𝑏 + 1) = 0, 𝑦2 (]2 − 2) = 𝑦2 (]2 + 𝑏 + 1) = 0, (4) where ]1 , ]2 ∈ (1, 2]. Using the Guo-Krasnosel’ski˘ı fixed point theorem, the authors showed that the problem has positive solutions for sufficiently small values of 𝜆 1 , 𝜆 2 > 0. The growth conditions on 𝑓𝑖 (𝑖 = 1, 2) are superlinear; i.e., lim

𝑦1 +𝑦2 󳨀→+∞

𝑓𝑖 (𝑡, 𝑥, 𝑦) = +∞, 𝑦1 + 𝑦2

𝑓 (𝑡, 𝑥, 𝑦) lim + 𝑖 = 0, 𝑦1 +𝑦2 󳨀→0 𝑦1 + 𝑦2

Δ−] 𝑎 𝑓 (𝑡) =

for 𝑡 ∈ N𝑎+] . (6)

We also define the ]−th fractional difference for ] > 0 by Δ]𝑎 𝑓 (𝑡) = Δ𝑁Δ]−𝑁 𝑎 𝑓 (𝑡) , for 𝑡 ∈ N𝑎+𝑁−] ,

−Δ]]−3 𝑦 (𝑡) = ℎ (𝑡 + ] − 1) , 𝑡 ∈ [0, 𝑏 + 2]N0 , 󵄨 𝑦 (] − 3) = [Δ𝛼]−3 𝑦 (𝑡)]󵄨󵄨󵄨𝑡=]−𝛼−2

(8)

󵄨󵄨 𝛽 = [Δ ]−3 𝑦 (𝑡)]󵄨󵄨󵄨 = 0, 󵄨𝑡=]+𝑏+2−𝛽 where ], 𝛼, 𝛽, 𝑏 are as in (1). The following two lemmas are in [9], so we omit their proofs. Lemma 3 (see [9], Lemma 4). Problem (8) has a unique solution 𝑏+2

𝑦 (𝑡) = ∑ 𝐺 (𝑡, 𝑠) ℎ (𝑠 + ] − 1) , 𝑠=0

(9)

2. Preliminaries We first recall some background materials from discrete fractional calculus; for more details we refer the reader to [10].

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where 𝑁 ∈ N with 0 ≤ 𝑁 − 1 < ] ≤ 𝑁. Let ℎ : []−1, 𝑏+]+1]N]−1 󳨀→ R be a continuous function. Then we consider the fractional difference boundary value problems

(5)

uniformly for 𝑡 ∈ []𝑖 , ]𝑖 + 𝑏]N] −2 . Using conditions of (5) type 𝑖 the existence of solutions for various fractional boundary value problems was considered in [1, 4–6, 9, 11–13]. In this paper, we use the fixed point index to obtain two existence theorems for positive solutions to (1) with semipositone nonlinearities. We adopt some appropriate nonnegative concave functions and nonnegative matrices to characterize the coupling behavior of our nonlinear terms. Moreover, the growth conditions on +∞ of our nonlinearities 𝑓, 𝑔 are an improvement of (5); see conditions (H1) and (H3) in Section 3.

1 𝑡−] ∑ (𝑡 − 𝑠 − 1)]−1 𝑓 (𝑠) , Γ (]) 𝑠=𝑎

𝑡 ∈ [] − 1, 𝑏 + ] + 1]N]−1 , where

]−𝛽−1

𝑡]−1 (] + 𝑏 − 𝛽 − 𝑠 + 1) { { − (𝑡 − 𝑠 − 1)]−1 , { { ]−𝛽−1 { (] + 𝑏 − 𝛽 + 2) 1 { 𝐺 (𝑡, 𝑠) = ]−𝛽−1 ]−1 { Γ (]) { { 𝑡 (] + 𝑏 − 𝛽 − 𝑠 + 1) { { , { ]−𝛽−1 (] + 𝑏 − 𝛽 + 2) {

Lemma 4 (see [9], Lemma 5). Green’s function (10) has the following properties. (i) 𝐺(𝑡, 𝑠) > 0, (𝑡, 𝑠) ∈ [] − 1, 𝑏 + ] + 1]N]−1 × [0, 𝑏 + 2]N0 , (ii) 𝑞∗ (𝑡)𝐺(𝑏 + ] + 1, 𝑠) ≤ 𝐺(𝑡, 𝑠) ≤ 𝐺(𝑏 + ] + 1, 𝑠), (𝑡, 𝑠) ∈ []−1, 𝑏+]+1]N]−1 ×[0, 𝑏+2]N0 , where 𝑞∗ (𝑡) = 𝑡]−1 /(𝑏+]+1)]−1 . Let 𝜑(𝑠 + ] − 1) = 𝐺(𝑏 + ] + 1, 𝑠) for 𝑠 ∈ [0, 𝑏 + 2]N0 . Then 𝜑(𝑡) = 𝐺(𝑏 + ] + 1, 𝑡 − ] + 1) for 𝑡 ∈ [] − 1, 𝑏 + ] + 1]N]−1 . From Lemma ??, the following inequalities are satisfied:

0 ≤ 𝑠 < 𝑡 − ] + 1 ≤ 𝑏 + 2, (10) 0 ≤ 𝑡 − ] + 1 ≤ 𝑠 ≤ 𝑏 + 2.

𝑏+]+1

𝑏+]+1

𝑡=]−1

𝑡=]−1

∑ 𝑞∗ (𝑡) 𝜑 (𝑡) ⋅ 𝜑 (𝑠 + ] − 1) ≤ ∑ 𝐺 (𝑡, 𝑠) 𝜑 (𝑡) (11) 𝑏+]+1

≤ ∑ 𝜑 (𝑡) ⋅ 𝜑 (𝑠 + ] − 1) , for 𝑠 ∈ [0, 𝑏 + 2]N0 . 𝑡=]−1

For convenience, we let

Journal of Function Spaces

3

𝑏+]+1

where ], 𝛼, 𝛽, 𝑏 are as in (1). Define 𝑧 = 𝑀𝜔, and then, from Lemmas 3 and ??, we have

𝜅1 = ∑ 𝑞∗ (𝑡) 𝜑 (𝑡) and 𝑡=]−1

(12)

𝑏+]+1

𝜅2 = ∑ 𝜑 (𝑡) .

𝑏+2

𝑏+2

𝑠=0

𝑠=0

𝑧 (𝑡) = 𝑀𝜔 (𝑡) = 𝑀 ∑ 𝐺 (𝑡, 𝑠) ≤ 𝑀 ∑ 𝜑 (𝑠 + ] − 1)

𝑡=]−1

Let 𝐸 be the collection of all maps from [] − 3, 𝑏 + ] + 1]N]−3 to R equipped with the max norm, ‖ ⋅ ‖. Then 𝐸 is a Banach space. Define a set 𝑃 ⊂ 𝐸 by 𝑃 = {𝑦 ∈ 𝐸 : 𝑦(𝑡) ≥ 0, 𝑡 ∈ [] − 1, 𝑏 + ] + 1]N]−1 }. Then 𝑃 is a cone in 𝐸. Note that 𝐸 × 𝐸 is a Banach space with the norm ‖(𝑥, 𝑦)‖ fl max{‖𝑥‖, ‖𝑦‖}, and 𝑃 × 𝑃 is a cone in 𝐸 × 𝐸. From Lemma 3, for all 𝑡 ∈ [] − 1, 𝑏 + ] + 1]N]−1 , we have that (1) is equivalent to

𝑏+]+1

= 𝑀 ∑ 𝜑 (𝑠) = 𝑀𝜅2 . 𝑠=]−1

We note that (1) has a positive solution (𝑥, 𝑦) ∈ (𝑃 × 𝑃) \ {0} if ̃ 𝑦) ̃ = (𝑥 + 𝑧, 𝑦 + 𝑧) is a solution of the fractional and only if (𝑥, difference boundary value problems − Δ]]−3 𝑥 (𝑡) = 𝑓̃ (𝑡 + ] − 1, 𝑥 (𝑡 + ] − 1)

𝑏+2

− 𝑧 (𝑡 + ] − 1) , 𝑦 (𝑡 + ] − 1) − 𝑧 (𝑡 + ] − 1)) ,

𝑥 (𝑡) = ∑ 𝐺 (𝑡, 𝑠) 𝑠=0

⋅ 𝑓 (𝑠 + ] − 1, 𝑥 (𝑠 + ] − 1) , 𝑦 (𝑠 + ] − 1)) , 𝑏+2

𝑡 ∈ [0, 𝑏 + 2]N0 , (13)

− Δ]]−3 𝑦 (𝑡) = 𝑔̃ (𝑡 + ] − 1, 𝑥 (𝑡 + ] − 1)

𝑦 (𝑡) = ∑ 𝐺 (𝑡, 𝑠)

− 𝑧 (𝑡 + ] − 1) , 𝑦 (𝑡 + ] − 1) − 𝑧 (𝑡 + ] − 1)) ,

𝑠=0

𝑡 ∈ [0, 𝑏 + 2]N0 , (18)

⋅ 𝑔 (𝑠 + ] − 1, 𝑥 (𝑠 + ] − 1) , 𝑦 (𝑠 + ] − 1)) ,

󵄨 𝑥 (] − 3) = [Δ𝛼]−3 𝑥 (𝑡)]󵄨󵄨󵄨𝑡=]−𝛼−2

where 𝐺 is defined in (10). Lemma 5 (see [46]). Let 𝐸 be a real Banach space and 𝑃 a cone on 𝐸. Suppose that Ω ⊂ 𝐸 is a bounded open set and that 𝐴 : Ω ∩ 𝑃 󳨀→ 𝑃 is a continuous compact operator. If there exists 𝜔0 ∈ 𝑃 \ {0} such that 𝜔 − 𝐴𝜔 ≠ 𝜆𝜔0 ,

∀𝜆 ≥ 0, 𝜔 ∈ 𝜕Ω ∩ 𝑃,

(14)

then 𝑖(𝐴, Ω ∩ 𝑃, 𝑃) = 0, where 𝑖 denotes the fixed point index on 𝑃. Lemma 6 (see [46]). Let 𝐸 be a real Banach space and 𝑃 a cone on 𝐸. Suppose that Ω ⊂ 𝐸 is a bounded open set with 0 ∈ Ω and that 𝐴 : Ω ∩ 𝑃 󳨀→ 𝑃 is a continuous compact operator. If 𝜔 − 𝜆𝐴𝜔 ≠ 0,

∀𝜆 ∈ [0, 1] , 𝜔 ∈ 𝜕Ω ∩ 𝑃,

(15)

then 𝑖(𝐴, Ω ∩ 𝑃, 𝑃) = 1.

󵄨󵄨 𝛽 = [Δ ]−3 𝑥 (𝑡)]󵄨󵄨󵄨 = 0, 󵄨𝑡=]+𝑏+2−𝛽 󵄨 𝑦 (] − 3) = [Δ𝛼]−3 𝑦 (𝑡)]󵄨󵄨󵄨𝑡=]−𝛼−2 󵄨󵄨 𝛽 = [Δ ]−3 𝑦 (𝑡)]󵄨󵄨󵄨 = 0, 󵄨𝑡=]+𝑏+2−𝛽

̃ 𝑦)(𝑡) ̃ and (𝑥, ≥ (𝑧, 𝑧)(𝑡) for 𝑡 ∈ [] − 1, 𝑏 + ] + 1]N]−1 , where ], 𝛼, 𝛽, 𝑏 are as in (1) and 𝑓̃ (𝑡, 𝑥, 𝑦) 𝑓 (𝑡, 𝑥, 𝑦) + 𝑀, { { { { { {𝑓 (𝑡, 𝑥, 0) + 𝑀, { ={ { { 𝑓 (𝑡, 0, 𝑦) + 𝑀, { { { { {𝑓 (𝑡, 0, 0) + 𝑀,

𝑡 ∈ [] − 1, 𝑏 + ] + 1]N]−1 , 𝑥, 𝑦 ≥ 0, 𝑡 ∈ [] − 1, 𝑏 + ] + 1]N]−1 , 𝑥 > 0, 𝑦 < 0, (19) 𝑡 ∈ [] − 1, 𝑏 + ] + 1]N]−1 , 𝑥 < 0, 𝑦 > 0, 𝑡 ∈ [] − 1, 𝑏 + ] + 1]N]−1 , 𝑥, 𝑦 < 0,

and 𝑔̃ (𝑡, 𝑥, 𝑦)

3. Main Results Let 𝜔 be a solution of −Δ]]−3 𝑦 (𝑡)

(17)

= 1, 𝑡 ∈ [0, 𝑏 + 2]N0 ,

󵄨 𝑦 (] − 3) = [Δ𝛼]−3 𝑦 (𝑡)]󵄨󵄨󵄨𝑡=]−𝛼−2

󵄨󵄨 𝛽 = [Δ ]−3 𝑦 (𝑡)]󵄨󵄨󵄨 = 0, 󵄨𝑡=]+𝑏+2−𝛽

(16)

𝑔 (𝑡, 𝑥, 𝑦) + 𝑀, { { { { { { {𝑔 (𝑡, 𝑥, 0) + 𝑀, ={ { { 𝑔 (𝑡, 0, 𝑦) + 𝑀, { { { { {𝑔 (𝑡, 0, 0) + 𝑀,

𝑡 ∈ [] − 1, 𝑏 + ] + 1]N]−1 , 𝑥, 𝑦 ≥ 0, 𝑡 ∈ [] − 1, 𝑏 + ] + 1]N]−1 , 𝑥 > 0, 𝑦 < 0, (20) 𝑡 ∈ [] − 1, 𝑏 + ] + 1]N]−1 , 𝑥 < 0, 𝑦 > 0, 𝑡 ∈ [] − 1, 𝑏 + ] + 1]N]−1 , 𝑥, 𝑦 < 0.

Note that for (𝑥1 , 𝑦1 )(𝑡) ≥ (𝑥2 , 𝑦2 )(𝑡), we mean 𝑥1 (𝑡) ≥ 𝑥2 (𝑡), 𝑦1 (𝑡) ≥ 𝑦2 (𝑡) for all 𝑡 ∈ [] − 1, 𝑏 + ] + 1]N]−1 .

4


For (𝑥, 𝑦) ∈ 𝑃 × 𝑃, and 𝑡 ∈ [] − 1, 𝑏 + ] + 1]N]−1 , we define the operators 𝑏+2

𝐵1 (𝑥, 𝑦) (𝑡) = ∑ 𝐺 (𝑡, 𝑠) 𝑓̃ (𝑠 + ] − 1, 𝑥 (𝑠 + ] − 1)

𝑒42

(H3) There exist 𝑒𝑖 ≥ 0 (𝑖 = 1, 2, 3, 4) with 𝑒12 + 𝑒22 ≠ 0, 𝑒32 + ≠ 0 such that (i) 𝜅 = (1 − 𝑒1 𝜅2 )(1 − 𝑒4 𝜅2 ) − 𝑒2 𝑒3 𝜅22 > 0, 𝑒1 , 𝑒4 < 𝜅2−1 , (ii) there exist 𝑐2 > 0 such that

𝑠=0

− 𝑧 (𝑠 + ] − 1) , 𝑦 (𝑠 + ] − 1) − 𝑧 (𝑠 + ] − 1)) , 𝑏+2

(21)

(

𝑒1 𝑥 + 𝑒2 𝑦 + 𝑐2 𝑓̃ (𝑡, 𝑥, 𝑦) )≤( ), 𝑒3 𝑥 + 𝑒4 𝑦 + 𝑐2 𝑔̃ (𝑡, 𝑥, 𝑦)

𝐵2 (𝑥, 𝑦) (𝑡) = ∑ 𝐺 (𝑡, 𝑠) 𝑔̃ (𝑠 + ] − 1, 𝑥 (𝑠 + ] − 1) − 𝑧 (𝑠 + ] − 1) , 𝑦 (𝑠 + ] − 1) − 𝑧 (𝑠 + ] − 1)) ,

(H4) For any (𝑡, 𝑥, 𝑦) ∈ []−1, 𝑏+]+1]N]−1 ×[0, 𝑞0−1 𝑀𝜅2 ]× [0, 𝑞0−1 𝑀𝜅2 ], assume

and (22)

Then (H0) and using the Arzelà-Ascoli theorem in a standard way establish that 𝐵 : 𝑃 × 𝑃 󳨀→ 𝑃 × 𝑃 is a completely continuous operator. It is clear that (𝑥, 𝑦) ∈ (𝑃 × 𝑃) \ {0} is a positive solution for (18) if and only if (𝑥, 𝑦) ∈ (𝑃 × 𝑃) \ {0} is a fixed point of 𝐵. Let 𝑃0 = {𝑦 ∈ 𝑃 : 𝑦(𝑡) ≥ 𝑞∗ (𝑡)‖𝑦‖, ∀𝑡 ∈ []−1, 𝑏+]+1]N]−1 }. Then from Lemma ?? we have 𝐵𝑖 (𝑃 × 𝑃) ⊂ 𝑃0 , 𝑖 = 1, 2.

(23)

̃ 𝑦) ̃ of 𝐵 then 𝑥, ̃ 𝑦̃ ∈ 𝑃0 and If we seek a fixed point (𝑥, 𝑤 (𝑡) − 𝑧 (𝑡) ≥ 𝑞∗ (𝑡) ‖𝑤‖ − 𝑀𝜅2 ≥ 𝑞0 ‖𝑤‖ − 𝑀𝜅2 , for 𝑡 ∈ [] − 1, 𝑏 + ] + 1]N]−1 ,

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̃ 𝑦, ̃ and 𝑞0 = min𝑡∈[]−1,𝑏+]+1]N 𝑞∗ (𝑡) > 0, so where 𝑤 = 𝑥,

̃ 𝑔̃ (𝑡, 𝑥, 𝑦) > 𝑞−2 𝑀. 𝑓, 0

𝑓̃ (𝑡, 𝑥, 𝑦) ≥ 𝑥 + 𝑝 (𝑦) − 𝑐1 ,

𝑝 (𝜔𝑞 (𝑥)) 𝜔19/20 𝑥19/10 = lim inf = +∞. 𝑥󳨀→+∞ 𝑥󳨀→+∞ 𝑥 𝑥

lim inf

𝑓 (𝑡, 𝑥, 𝑦) + 𝑀 =
0 such that 𝜅1 (1 + 𝛾1 𝜅1 ) > 1 and ≥ 𝜅2 𝛾1 (𝑥 (𝑡 + ] − 1) − 𝑧 (𝑡 + ] − 1)) − 𝑐1

(30)

Let 𝑓(𝑡, 𝑥, 𝑦) = (1/2𝜅2 )𝑥 + (1/2𝜅2 𝑒𝛽1 +cos(𝑡𝑥) )𝑦 − 𝑀 and −𝛽 𝑔(𝑡, 𝑥, 𝑦) = (1/𝑒𝛽2 +sin(𝑡𝑥) )(𝑞0−1 𝑀)1−𝛽3 𝜅2 3 𝑥𝛽3 − 𝑀, where 𝛽1 , 𝛽2 > 1, 𝛽3 > 2, for (𝑡, 𝑥, 𝑦) ∈ []−1, 𝑏+]+1]N]−1 ×R+ ×R+ . Then, for any (𝑡, 𝑥, 𝑦) ∈ [] − 1, 𝑏 + ] + 1]N]−1 × [0, 𝑞0−1 𝑀𝜅2 ] × [0, 𝑞0−1 𝑀𝜅2 ], we have

∀ (𝑡, 𝑥, 𝑦) ∈ [] − 1, 𝑏 + ] + 1]N]−1 × R+ × R+ , 𝑝 (𝜅2 𝑞 (𝑥 (𝑡 + ] − 1) − 𝑧 (𝑡 + ] − 1)))

(29)

Example 7. Let 𝑝(𝑦) = 𝑦19/20 and 𝑞(𝑥) = 𝑥2 for 𝑥, 𝑦 ∈ R+ . Then, for any 𝜔 > 0, we have

]−1

̃ 𝑦)(𝑡) ̃ ̃ ‖𝑦‖ ̃ ≥ 𝑞0−1 𝑀𝜅2 then (𝑥, ≥ (𝑧, 𝑧)(𝑡) for as a result if ‖𝑥‖, 𝑡 ∈ [] − 1, 𝑏 + ] + 1]N]−1 (i.e., (𝑥̃ − 𝑧, 𝑦̃ − 𝑧)(𝑡) is a positive solution for (1)). For convenience, we use 𝑐1 , 𝑐2 , . . . to stand for different positive constants. Let 𝐵󰜚 fl {𝑥 ∈ 𝐸 : ‖𝑥‖ < 󰜚} for 󰜚 > 0. Now, ̃ 𝑔̃ (the first two are needed for we list our assumptions on 𝑓, Theorem 10 and the last two are needed for Theorem 11): (H1) There exist 𝑝, 𝑞 ∈ 𝐶(R+ , R+ ) such that (i) 𝑝 is concave and strictly increasing on R+ , (ii) there exists 𝑐1 > 0 such that

𝑔̃ (𝑡, 𝑥, 𝑦) ≥ 𝑞 (𝑥) − 𝑐1 ,

+

∀ (𝑡, 𝑥, 𝑦) ∈ [] − 1, 𝑏 + ] + 1]N]−1 × R × R .

𝑠=0

𝐵 (𝑥, 𝑦) (𝑡) = (𝐵1 , 𝐵2 ) (𝑥, 𝑦) (𝑡) .

(28) +

(27)

𝑥󳨀→+∞,𝑦󳨀→+∞

=

𝑓 (𝑡, 𝑥, 𝑦) + 𝑀 𝑥 + 𝑝 (𝑦)

lim inf

𝑥󳨀→+∞,𝑦󳨀→+∞

(1/2𝜅2 ) 𝑥 + (1/2𝜅2 𝑒𝛽1 +cos(𝑡𝑥) ) 𝑦 𝑥+

𝑦19/20

= +∞, for 𝑡 ∈ [] − 1, 𝑏 + ] + 1]N]−1 ,

(33)


5

and lim inf

𝑥󳨀→+∞

Proof. We first claim that there exists a sufficiently large positive number 𝑅 > 𝑞0−1 𝑀𝜅2 such that

𝑔 (𝑡, 𝑥, 𝑦) + 𝑀 𝑞 (𝑥)

(𝑥, 𝑦) ≠ 𝐵 (𝑥, 𝑦) + 𝜆 (𝑥0 , 𝑦0 ) , 1−𝛽3

= lim inf

(1/𝑒𝛽2 +sin(𝑡𝑥) ) (𝑞0−1 𝑀)

−𝛽

𝜅2 3 𝑥𝛽3

𝑥2

𝑥󳨀→+∞

(34)

uniformly on (𝑡, 𝑦) ∈ [] − 1, 𝑏 + ] + 1]N]−1 × R+ . Thus, (H1)-(H2) are satisfied.

𝑥 (𝑡) = 𝐵1 (𝑥, 𝑦) (𝑡) + 𝜆𝑥0 (𝑡) , 𝑞0−1 𝑀𝜅2

Example 8. Let 𝑓(𝑡, 𝑥, 𝑦) = (𝑞0−2 𝑀𝑒 +𝑒|sin(𝑡𝑥𝑦)| )𝑒−(𝑥+𝑦)/2 − −2 𝑞0−1 𝑀𝜅2 |cos(𝑡𝑥𝑦)| −(𝑥+𝑦)/2 +𝑒 )𝑒 − 𝑀, 𝑀 and 𝑔(𝑡, 𝑥, 𝑦) = (𝑞0 𝑀𝑒 for (𝑡, 𝑥, 𝑦) ∈ [] − 1, 𝑏 + ] + 1]N]−1 × R+ × R+ . Then, for any (𝑡, 𝑥, 𝑦) ∈ [] − 1, 𝑏 + ] + 1]N]−1 × [0, 𝑞0−1 𝑀𝜅2 ] × [0, 𝑞0−1 𝑀𝜅2 ], we have 𝑓 (𝑡, 𝑥, 𝑦) + 𝑀, 𝑔 (𝑡, 𝑥, 𝑦) + 𝑀 −1

(38)

where 𝑥0 , 𝑦0 ∈ 𝑃0 are two given functions. Suppose not. Then there exist (𝑥, 𝑦) ∈ 𝜕(𝐵𝑅 × 𝐵𝑅 ) ∩ (𝑃 × 𝑃) and 𝜆 ≥ 0 such that (𝑥, 𝑦) = 𝐵(𝑥, 𝑦) + 𝜆(𝑥0 , 𝑦0 ), and so

= +∞,

> 𝑞0−2 𝑀𝑒𝑞0

∀ (𝑥, 𝑦) ∈ 𝜕 (𝐵𝑅 × 𝐵𝑅 ) ∩ (𝑃 × 𝑃) , 𝜆 ≥ 0,

𝑀𝜅2 −𝑞0−1 𝑀𝜅2

𝑒

= 𝑞0−2 𝑀.

𝑦 (𝑡) = 𝐵2 (𝑥, 𝑦) (𝑡) + 𝜆𝑦0 (𝑡) ,

(39)

for 𝑡 ∈ [] − 1, 𝑏 + ] + 1]N]−1 . This implies 𝑥(𝑡) ≥ 𝐵1 (𝑥, 𝑦)(𝑡), and 𝑦(𝑡) ≥ 𝐵2 (𝑥, 𝑦)(𝑡) for 𝑡 ∈ [] − 1, 𝑏 + ] + 1]N]−1 . From (H1) we have

(35)

𝑏+2

𝑥 (𝑡) ≥ 𝐵1 (𝑥, 𝑦) (𝑡) ≥ ∑ 𝐺 (𝑡, 𝑠) [𝑥 (𝑠 + ] − 1) 𝑠=0

Also,

− 𝑧 (𝑠 + ] − 1) + 𝑝 (𝑦 (𝑠 + ] − 1) − 𝑧 (𝑠 + ] − 1))

lim sup

𝑥󳨀→+∞,𝑦󳨀→+∞

=

𝑓 (𝑡, 𝑥, 𝑦) + 𝑀 𝑒1 𝑥 + 𝑒2 𝑦

lim sup

−1 (𝑞0−2 𝑀𝑒𝑞0 𝑀𝜅2

|sin(𝑡𝑥𝑦)|

+𝑒

−(𝑥+𝑦)/2

)𝑒

(36)

𝑏+2

𝑠=0

𝑠=0

𝑏+2

⋅ 𝑝 (𝑦 (𝑠 + ] − 1) − 𝑧 (𝑠 + ] − 1)) − 𝑐3 ≥ ∑ 𝐺 (𝑡, 𝑠)

𝑒1 𝑥 + 𝑒2 𝑦

𝑥󳨀→+∞,𝑦󳨀→+∞

𝑏+2

− 𝑐1 ] ≥ ∑ 𝐺 (𝑡, 𝑠) 𝑥 (𝑠 + ] − 1) + ∑ 𝐺 (𝑡, 𝑠)

𝑠=0

= 0,

𝑏+2

(40)

⋅ 𝑥 (𝑠 + ] − 1) + ∑ 𝐺 (𝑡, 𝑠) [𝑝 (𝑦 (𝑠 + ] − 1)) 𝑠=0

and

𝑏+2

𝑔 (𝑡, 𝑥, 𝑦) + 𝑀 lim sup 𝑒3 𝑥 + 𝑒4 𝑦 𝑥󳨀→+∞,𝑦󳨀→+∞ −1

=

lim sup

𝑥󳨀→+∞,𝑦󳨀→+∞

(𝑞0−2 𝑀𝑒𝑞0

− 𝑝 (𝑧 (𝑠 + ] − 1))] − 𝑐3 ≥ ∑ 𝐺 (𝑡, 𝑠) 𝑥 (𝑠 + ] − 1) 𝑠=0

𝑀𝜅2

𝑏+2

+ 𝑒|cos(𝑡𝑥𝑦)| ) 𝑒−(𝑥+𝑦)/2 (37)

+ ∑ 𝐺 (𝑡, 𝑠) 𝑝 (𝑦 (𝑠 + ] − 1)) − 𝑐4 , 𝑠=0

𝑒3 𝑥 + 𝑒4 𝑦

for 𝑡 ∈ [] − 1, 𝑏 + ] + 1]N]−1 ,

= 0, for 𝑡 ∈ [] − 1, 𝑏 + ] + 1]N]−1 . Thus, (H3)-(H4) hold. Remark 9. (i) In (H1), the growth condition for nonlinear term 𝑓 depends on two variables 𝑥, 𝑦; however, in [7], this corresponding condition only involves one variable. (ii) When nonlinear terms 𝑓, 𝑔 grow sublinearly at +∞, nonnegative matrices are used to depict the coupling behavior of our nonlinearities. This is different from condition (H4) in [7]. Theorem 10. Suppose that (H0)-(H2) hold. Then (1) has at least one positive solution.

and 𝑦 (𝑡) ≥ 𝐵2 (𝑥, 𝑦) (𝑡) 𝑏+2

≥ ∑ 𝐺 (𝑡, 𝑠) [𝑞 (𝑥 (𝑠 + ] − 1) − 𝑧 (𝑠 + ] − 1)) − 𝑐1 ] 𝑠=0

(41)

𝑏+2

≥ ∑ 𝐺 (𝑡, 𝑠) 𝑞 (𝑥 (𝑠 + ] − 1) − 𝑧 (𝑠 + ] − 1)) − 𝑐3 , 𝑠=0

for 𝑡 ∈ [] − 1, 𝑏 + ] + 1]N]−1 .

6


As a result, for 𝑡 ∈ [0, 𝑏 + 2]N0 , we have

𝑏+2

𝑏+2

𝑠=0

𝑠=0

≥ ∑ 𝐺 (𝑡, 𝑠) 𝑥 (𝑠 + ] − 1) + 𝛾1 ∑ 𝐺 (𝑡, 𝑠)

𝑝 (𝑦 (𝑡 + ] − 1)) + 𝑝 (𝑐3 ) ≥ 𝑝 (𝑦 (𝑡 + ] − 1) + 𝑐3 )

𝑏+2

⋅ ∑ 𝐺 (𝑠 + ] − 1, 𝜏) 𝑥 (𝜏 + ] − 1) − 𝑐8 ,

𝑏+2

𝜏=0

≥ 𝑝 [ ∑ 𝐺 (𝑡 + ] − 1, 𝑠)

for 𝑡 ∈ [] − 1, 𝑏 + ] + 1]N]−1 .

𝑠=0

(44)

⋅ 𝑞 (𝑥 (𝑠 + ] − 1) − 𝑧 (𝑠 + ] − 1))]

Multiply both sides of the above inequality by 𝜑(𝑡) and sum from ] − 1 to 𝑏 + ] + 1 and together with (11) we obtain

𝑏+2

= 𝑝 [ ∑ 𝐺 (𝑡 + ] − 1, 𝑠) 𝑠=0 𝑏+2

⋅ 𝜅2−1 𝜅2 𝑞 (𝑥 (𝑠 + ] − 1) − 𝑧 (𝑠 + ] − 1))] ≥ ∑ 𝐺 (𝑡 +]−

+ ] − 1) − 𝑧 (𝑠 + ] − 1)))

𝑏+2

𝑡=]−1

𝑡=0

∑ 𝑥 (𝑡) 𝜑 (𝑡) = ∑ 𝑥 (𝑡 + ] − 1) 𝜑 (𝑡 + ] − 1)

𝑠=0

1, 𝑠) 𝜅2−1 𝑝 (𝜅2 𝑞 (𝑥 (𝑠

𝑏+]+1

(42)

𝑏+2

𝑏+2

𝑡=0

𝑠=0

≥ ∑ 𝜑 (𝑡 + ] − 1) [ ∑ 𝐺 (𝑡 + ] − 1, 𝑠) 𝑥 (𝑠 + ] − 1)

𝑏+2

≥ ∑ 𝐺 (𝑡 + ] − 1, 𝑠) 𝑠=0

𝑏+2

⋅ 𝜅2−1 [𝜅2 𝛾1 (𝑥 (𝑠 + ] − 1) − 𝑧 (𝑠 + ] − 1)) − 𝑐1 ]

+ 𝛾1 ∑ 𝐺 (𝑡 + ] − 1, 𝑠) 𝑠=0

𝑏+2

(45)

𝑏+2

≥ 𝛾1 ∑ 𝐺 (𝑡 + ] − 1, 𝑠) (𝑥 (𝑠 + ] − 1)

⋅ ∑ 𝐺 (𝑠 + ] − 1, 𝜏) 𝑥 (𝜏 + ] − 1) − 𝑐8 ] ≥ (𝜅1

𝑠=0

𝜏=0

𝑏+2

− 𝑧 (𝑠 + ] − 1)) − 𝑐5 ≥ 𝛾1 ∑ 𝐺 (𝑡 + ] − 1, 𝑠) 𝑥 (𝑠 + ]

𝑏+2

+ 𝛾1 𝜅12 ) ∑ 𝑥 (𝑡 + ] − 1) 𝜑 (𝑡 + ] − 1) − 𝑐9 = (𝜅1

𝑠=0

𝑡=0

− 1) − 𝑐6 .

𝑏+]+1

+ 𝛾1 𝜅12 ) ∑ 𝑥 (𝑡) 𝜑 (𝑡) − 𝑐9 .

Thus

𝑡=]−1

𝑏+2

𝑝 (𝑦 (𝑡 + ] − 1)) ≥ 𝛾1 ∑ 𝐺 (𝑡 + ] − 1, 𝑠) 𝑥 (𝑠 + ] − 1) 𝑠=0

From (23), (39), and 𝑥0 ∈ 𝑃0 we have 𝑥 ∈ 𝑃0 . This implies (43)

− 𝑐7 ,

𝑏+]+1

𝑏+]+1

𝑡=]−1

𝑡=]−1

𝜅1 ‖𝑥‖ = ‖𝑥‖ ∑ 𝑞∗ (𝑡) 𝜑 (𝑡) ≤ ∑ 𝑥 (𝑡) 𝜑 (𝑡)

and, therefore,

(46)

𝑐9 ≤ , 𝜅1 (1 + 𝛾1 𝜅1 ) − 1

𝑏+2

𝑥 (𝑡) ≥ ∑ 𝐺 (𝑡, 𝑠) 𝑥 (𝑠 + ] − 1)

and

𝑠=0 𝑏+2

+ ∑ 𝐺 (𝑡, 𝑠) 𝑝 (𝑦 (𝑠 + ] − 1)) − 𝑐4 𝑠=0

𝑏+2

𝑏+2

𝑠=0

𝑠=0

≥ ∑ 𝐺 (𝑡, 𝑠) 𝑥 (𝑠 + ] − 1) + ∑ 𝐺 (𝑡, 𝑠) 𝑏+2

⋅ [𝛾1 ∑ 𝐺 (𝑠 + ] − 1, 𝜏) 𝑥 (𝜏 + ] − 1) − 𝑐7 ] − 𝑐4 𝜏=0

‖𝑥‖ ≤

𝜅1−1 𝑐9 . 𝜅1 (1 + 𝛾1 𝜅1 ) − 1

(47)

Note that, from (23), (39), and 𝑦0 ∈ 𝑃0 , we find 𝑦 ∈ 𝑃0 . Moreover, we may assume 𝑦(𝑡) ≢ 0, for 𝑡 ∈ []−1, 𝑏+]+1]N]−1 . Then ‖𝑦‖ > 0 and 𝑝(‖𝑦‖) > 0. Thus, from the concavity of 𝑝, we have


7

𝑏+]+1

𝑏+]+1

𝑡=]−1

𝑡=]−1

Hence, ‖𝑥‖ ≤ ‖𝐵1 (𝑥, 𝑦)‖ and ‖𝑦‖ ≤ ‖𝐵2 (𝑥, 𝑦)‖. However, from (H2) we have

󵄩 󵄩 󵄩 󵄩 𝜅1 󵄩󵄩󵄩𝑦󵄩󵄩󵄩 = 󵄩󵄩󵄩𝑦󵄩󵄩󵄩 ∑ 𝑞∗ (𝑡) 𝜑 (𝑡) ≤ ∑ 𝑦 (𝑡) 𝜑 (𝑡) 𝑏+2

𝑏+2

𝐵1 (𝑥, 𝑦) (𝑡) = ∑ 𝐺 (𝑡, 𝑠) 𝑓̃ (𝑠 + ] − 1, 𝑥 (𝑠 + ] − 1)

= ∑ 𝑦 (𝑡 + ] − 1) 𝜑 (𝑡 + ] − 1) 𝑡=0

󵄩󵄩 󵄩󵄩 𝑏+2 𝑦 (𝑡 + ] − 1) 󵄩󵄩 󵄩󵄩 󵄩𝑦󵄩 = 󵄩󵄩󵄩 󵄩󵄩󵄩 ∑ 𝑝 (󵄩󵄩𝑦󵄩󵄩) 𝜑 (𝑡 + ] − 1) 󵄩󵄩 󵄩󵄩 𝑝 (󵄩󵄩𝑦󵄩󵄩) 𝑡=0 󵄩󵄩𝑦󵄩󵄩 ≤

(48)

(55)

𝑏+2 𝑠=0

This implies that 𝑏+2

(49)

𝑡=0

From (40) and Lemma ?? we obtain 𝑏+2

𝑥 (𝑡) + 𝑐4 ≥ ∑ 𝐺 (𝑡, 𝑠) 𝑝 (𝑦 (𝑠 + ] − 1)) 𝑠=0

𝑏+2

≥ ∑ 𝑞∗ (𝑡) 𝜑 (𝑠 + ] − 1) 𝑝 (𝑦 (𝑠 + ] − 1))

− 𝑧 (𝑠 + ] − 1) , 𝑦 (𝑠 + ] − 1) − 𝑧 (𝑠 + ] − 1)) < ∑ 𝜑 (𝑠 + ] − 1) 𝑞0−1 𝑀 = 𝑞0−1 𝑀𝜅2 ,

󵄩󵄩 󵄩󵄩 𝑏+2 󵄩󵄩𝑦󵄩󵄩 󵄩 󵄩 ∑ 𝑝 (𝑦 (𝑡 + ] − 1)) 𝜑 (𝑡 + ] − 1) . 𝑝 (󵄩󵄩󵄩𝑦󵄩󵄩󵄩) 𝑡=0

󵄩 󵄩 𝑝 (󵄩󵄩󵄩𝑦󵄩󵄩󵄩) ≤ 𝜅1−1 ∑ 𝑝 (𝑦 (𝑡 + ] − 1)) 𝜑 (𝑡 + ] − 1) .

𝑠=0

(50)

𝑠=0 𝑏+2

for all 𝑡 ∈ [] − 1, 𝑏 + ] + 1]N]−1 . This implies ‖𝐵1 (𝑥, 𝑦)‖ < 𝑞0−1 𝑀𝜅2 . Similarly, ‖𝐵2 (𝑥, 𝑦)‖ < 𝑞0−1 𝑀𝜅2 . Thus, note that (𝑥, 𝑦) ∈ 𝜕(𝐵𝑞0−1 𝑀𝜅2 × 𝐵𝑞0−1 𝑀𝜅2 ) ∩ (𝑃 × 𝑃), and we have 󵄩󵄩 󵄩 󵄩 󵄩 󵄩󵄩(𝑥, 𝑦)󵄩󵄩󵄩 = max {‖𝑥‖ , 󵄩󵄩󵄩𝑦󵄩󵄩󵄩} 󵄩 󵄩 󵄩 󵄩 ≤ max {󵄩󵄩󵄩𝐵1 (𝑥, 𝑦)󵄩󵄩󵄩 , 󵄩󵄩󵄩𝐵2 (𝑥, 𝑦)󵄩󵄩󵄩} < 𝑞0−1 𝑀𝜅2 󵄩 󵄩 = 󵄩󵄩󵄩(𝑥, 𝑦)󵄩󵄩󵄩 .

Clearly, this is a contradiction. Thus (53) is true. It follows from Lemma 6 that 𝑖 (𝐵, (𝐵𝑞0−1 𝑀𝜅2 × 𝐵𝑞0−1 𝑀𝜅2 ) ∩ (𝑃 × 𝑃) , 𝑃 × 𝑃) = 1.

≥ 𝑞0 ∑ 𝑝 (𝑦 (𝑡 + ] − 1)) 𝜑 (𝑡 + ] − 1) .

(56)

(57)

𝑡=0

From (52) and (57) we have

Combining the above two inequalities, we get −1 󵄩 󵄩 𝑝 (󵄩󵄩󵄩𝑦󵄩󵄩󵄩) ≤ (𝜅1 𝑞0 ) (𝑥 (𝑡) + 𝑐4 )

𝜅1−1 𝑐9 ≤ (𝜅1 𝑞0 ) [ + 𝑐 ]. 𝜅1 (1 + 𝛾1 𝜅1 ) − 1 4 −1

𝑖 (𝐵, ((𝐵𝑅 × 𝐵𝑅 ) \ (𝐵𝑞0−1 𝑀𝜅2 × 𝐵𝑞0−1 𝑀𝜅2 )) ∩ (𝑃 × 𝑃) , 𝑃 (51)

× 𝑃) = 0 − 1 = −1.

(58)

From (H1), lim𝑧󳨀→+∞ 𝑝(𝑧) = +∞, and thus there exists M1 > 0 such that ‖𝑦‖ ≤ M1 . Hence, we have ‖𝑥‖ ≤ 𝜅1−1 𝑐9 /(𝜅1 (1 + 𝛾1 𝜅1 ) − 1) and ‖𝑦‖ ≤ M1 . As a result, choosing 𝑅 > max{𝑞0−1 𝑀𝜅2 , 𝜅1−1 𝑐9 /(𝜅1 (1 + 𝛾1 𝜅1 )−1), M1 } we have a contradiction (recall in general 𝜕(𝐴× 𝐵) = (𝜕𝐴 × 𝐵) ∪ (𝐴 × 𝜕𝐵)). Thus (38) is true. Consequently Lemma 5 (with 𝑅 chosen above) implies

Therefore the operator 𝐵 has at least one fixed point (𝑥, 𝑦) in ((𝐵𝑅 × 𝐵𝑅 ) \ (𝐵𝑞0−1 𝑀𝜅2 × 𝐵𝑞0−1 𝑀𝜅2 )) ∩ (𝑃 × 𝑃) with ‖𝑥‖, ‖𝑦‖ ≥ 𝑞0−1 𝑀𝜅2 , and then (𝑥 − 𝑧, 𝑦 − 𝑧)(𝑡) is a positive solution for (1). This completes the proof.

𝑖 (𝐵, (𝐵𝑅 × 𝐵𝑅 ) ∩ (𝑃 × 𝑃) , 𝑃 × 𝑃) = 0.

Proof. We show there exists a positive constant 𝑅 > 𝑞0−1 𝑀𝜅2 such that

(52)

Theorem 11. Suppose that (H0), (H3), and (H4) hold. Then (1) has at least one positive solution.

Now we show that (𝑥, 𝑦) ≠ 𝜆𝐵 (𝑥, 𝑦) ,

(𝑥, 𝑦) ≠ 𝜆𝐵 (𝑥, 𝑦) , ∀ (𝑥, 𝑦) ∈ 𝜕 (𝐵𝑞0−1 𝑀𝜅2 × 𝐵𝑞0−1 𝑀𝜅2 ) ∩ (𝑃 × 𝑃) , 𝜆 ∈ [0, 1] .

(53)

Suppose not. Then there exist (𝑥, 𝑦) ∈ 𝜕(𝐵𝑞0−1 𝑀𝜅2 × 𝐵𝑞0−1 𝑀𝜅2 ) ∩ (𝑃 × 𝑃), 𝜆 0 ∈ [0, 1] such that (𝑥, 𝑦) = 𝜆 0 𝐵(𝑥, 𝑦). This implies that 𝑥 (𝑡) ≤ 𝐵1 (𝑥, 𝑦) (𝑡) ,

∀ (𝑥, 𝑦) ∈ 𝜕 (𝐵𝑅 × 𝐵𝑅 ) ∩ (𝑃 × 𝑃) , 𝜆 ∈ [0, 1] .

(59)

Suppose not. Then there exist (𝑥, 𝑦) ∈ 𝜕(𝐵𝑅 × 𝐵𝑅 ) ∩ (𝑃 × 𝑃), 𝜆 0 ∈ [0, 1] such that (𝑥, 𝑦) = 𝜆 0 𝐵(𝑥, 𝑦). This implies that 𝑥 (𝑡) ≤ 𝐵1 (𝑥, 𝑦) (𝑡) ,

𝑦 (𝑡) ≤ 𝐵2 (𝑥, 𝑦) (𝑡) ,

(54) for 𝑡 ∈ [] − 1, 𝑏 + ] + 1]N]−1 .

𝑦 (𝑡) ≤ 𝐵2 (𝑥, 𝑦) (𝑡) ,

(60) for 𝑡 ∈ [] − 1, 𝑏 + ] + 1]N]−1 .

8


From (H3) we have

Consequently, we obtain

𝑏+2

𝑏+]+1

𝑥 (𝑡) ≤ ∑ 𝐺 (𝑡, 𝑠) [𝑒1 (𝑥 (𝑠 + ] − 1) − 𝑧 (𝑠 + ] − 1))

1 − 𝑒1 𝜅2

𝑠=0

(

+ 𝑒2 (𝑦 (𝑠 + ] − 1) − 𝑧 (𝑠 + ] − 1)) + 𝑐2 ] 𝑏+2

𝑡=]−1

(61)

≤ ∑ 𝐺 (𝑡, 𝑠) [𝑒1 𝑥 (𝑠 + ] − 1) + 𝑒2 𝑦 (𝑠 + ] − 1)]

(65)

𝜅2 𝑐10 ). ≤( 𝜅2 𝑐10

𝑠=0

+ 𝑐10 ,

−𝑒3 𝜅2

∑ 𝑥 (𝑡) 𝜑 (𝑡) 𝑡=]−1 ) (𝑏+]+1 ) 1 − 𝑒4 𝜅2 ∑ 𝑦 (𝑡) 𝜑 (𝑡) −𝑒2 𝜅2

From (H3)(i) we have

∀ (𝑡, 𝑥, 𝑦) ∈ [] − 1, 𝑏 + ] + 1]N]−1 × R+ × R+ .

𝑏+]+1

∑ 𝑥 (𝑡) 𝜑 (𝑡)

Similarly, we have 𝑏+2

𝑡=]−1 (𝑏+]+1

𝑠=0

𝑡=]−1

𝑦 (𝑡) ≤ ∑ 𝐺 (𝑡, 𝑠) [𝑒3 𝑥 (𝑠 + ] − 1) + 𝑒4 𝑦 (𝑠 + ] − 1)]

∑ 𝑦 (𝑡) 𝜑 (𝑡)

(62)

+ 𝑐10 , +

≤

+

∀ (𝑡, 𝑥, 𝑦) ∈ [] − 1, 𝑏 + ] + 1]N]−1 × R × R .

𝑏+2

𝑡=]−1

𝑡=0

∑ 𝑥 (𝑡) 𝜑 (𝑡) = ∑ 𝑥 (𝑡 + ] − 1) 𝜑 (𝑡 + ] − 1)

𝜅2 𝑐10 1 1 − 𝑒4 𝜅2 𝑒2 𝜅2 )( ) ( 𝜅 𝑒3 𝜅2 1 − 𝑒1 𝜅2 𝜅2 𝑐10

(66)

𝜅2 𝑐10 (1 + 𝑒2 𝜅2 − 𝑒4 𝜅2 ) 𝜅 ). =( 𝜅2 𝑐10 (1 + 𝑒3 𝜅2 − 𝑒1 𝜅2 ) 𝜅

Consequently, for all 𝑡 ∈ [0, 𝑏+2]N0 , multiply both sides of the above two inequalities by 𝜑(𝑡) and sum from ] − 1 to 𝑏 + ] + 1 and together with (11) we obtain 𝑏+]+1

)

Note that 𝑥, 𝑦 ∈ 𝑃0 from the fact that 𝐵𝑖 (𝑃×𝑃) ⊂ 𝑃0 (𝑖 = 1, 2). This implies

𝑏+2

≤ ∑ 𝜑 (𝑡 + ] − 1)

𝑏+]+1

𝑏+]+1

𝑡=]−1

𝑡=]−1

‖𝑥‖ ∑ 𝑞∗ (𝑡) 𝜑 (𝑡) ≤ ∑ 𝑥 (𝑡) 𝜑 (𝑡)

𝑡=0 𝑏+2

⋅ [ ∑ 𝐺 (𝑡, 𝑠) [𝑒1 𝑥 (𝑠 + ] − 1) + 𝑒2 𝑦 (𝑠 + ] − 1)]

(63)

≤

𝑠=0

𝑏+]+1 𝑡=]−1

𝑏+]+1

𝑡=]−1

𝑡=]−1

𝑏+]+1

𝑏+]+1

𝑡=]−1

𝑡=]−1

(67)

󵄩󵄩 󵄩󵄩 ∗ 󵄩󵄩𝑦󵄩󵄩 ∑ 𝑞 (𝑡) 𝜑 (𝑡) ≤ ∑ 𝑦 (𝑡) 𝜑 (𝑡)

+ 𝑐10 ] ≤ 𝑒1 𝜅2 ∑ 𝑥 (𝑡) 𝜑 (𝑡) 𝑏+]+1

𝜅2 𝑐10 (1 + 𝑒2 𝜅2 − 𝑒4 𝜅2 ) , 𝜅

≤

+ 𝑒2 𝜅2 ∑ 𝑦 (𝑡) 𝜑 (𝑡) + 𝑐10 ∑ 𝜑 (𝑡) .

𝜅2 𝑐10 (1 + 𝑒3 𝜅2 − 𝑒1 𝜅2 ) . 𝜅

Hence

Similarly, we have 𝑏+]+1

𝑏+2

𝑡=]−1

𝑡=0

‖𝑥‖ ≤

∑ 𝑦 (𝑡) 𝜑 (𝑡) = ∑ 𝑦 (𝑡 + ] − 1) 𝜑 (𝑡 + ] − 1)

𝜅2 𝑐10 (1 + 𝑒2 𝜅2 − 𝑒4 𝜅2 ) , 𝜅1 𝜅

󵄩󵄩 󵄩󵄩 𝜅2 𝑐10 (1 + 𝑒3 𝜅2 − 𝑒1 𝜅2 ) . 󵄩󵄩𝑦󵄩󵄩 ≤ 𝜅𝜅

𝑏+2

≤ ∑ 𝜑 (𝑡 + ] − 1)

(68)

1

𝑡=0 𝑏+2

⋅ [ ∑ 𝐺 (𝑡, 𝑠) [𝑒3 𝑥 (𝑠 + ] − 1) + 𝑒4 𝑦 (𝑠 + ] − 1)] 𝑠=0

(64)

Thus if we choose 𝑅 > max{𝑞0−1 𝑀𝜅2 , 𝜅2 𝑐10 (1+𝑒2 𝜅2 −𝑒4 𝜅2 )/𝜅1 𝜅, and 𝜅2 𝑐10 (1 + 𝑒3 𝜅2 − 𝑒1 𝜅2 )/𝜅1 𝜅} we have a contradiction. Thus (59) is true. Lemma 6 (with 𝑅 chosen above) implies 𝑖 (𝐵, (𝐵𝑅 × 𝐵𝑅 ) ∩ (𝑃 × 𝑃) , 𝑃 × 𝑃) = 1.

𝑏+]+1

+ 𝑐10 ] ≤ 𝑒3 𝜅2 ∑ 𝑥 (𝑡) 𝜑 (𝑡) 𝑡=]−1

𝑏+]+1

𝑏+]+1

𝑡=]−1

𝑡=]−1

+ 𝑒4 𝜅2 ∑ 𝑦 (𝑡) 𝜑 (𝑡) + 𝑐10 ∑ 𝜑 (𝑡) .

(69)

We next prove that (𝑥, 𝑦) ≠ 𝐵 (𝑥, 𝑦) + 𝜆 (𝑥0 , 𝑦0 ) , ∀ (𝑥, 𝑦) ∈ 𝜕 (𝐵𝑞0−1 𝑀𝜅2 × 𝐵𝑞0−1 𝑀𝜅2 ) ∩ (𝑃 × 𝑃) , 𝜆 ≥ 0,

(70)


9

where 𝑥0 , 𝑦0 ∈ 𝑃 are two fixed functions. Indeed, if not, there exist (𝑥, 𝑦) ∈ 𝜕(𝐵𝑞0−1 𝑀𝜅2 × 𝐵𝑞0−1 𝑀𝜅2 ) ∩ (𝑃 × 𝑃), 𝜆 0 ≥ 0 such that (𝑥, 𝑦) = 𝐵(𝑥, 𝑦) + 𝜆 0 (𝑥0 , 𝑦0 ). This implies that 𝑥 (𝑡) ≥ 𝐵1 (𝑥, 𝑦) (𝑡) , 𝑦 (𝑡) ≥ 𝐵2 (𝑥, 𝑦) (𝑡) ,

(71) for 𝑡 ∈ [] − 1, 𝑏 + ] + 1]N]−1 .

References

Hence, ‖𝑥‖ ≥ ‖𝐵1 (𝑥, 𝑦)‖ and ‖𝑦‖ ≥ ‖𝐵2 (𝑥, 𝑦)‖. However, from (H4) we have 𝑏+2

𝐵1 (𝑥, 𝑦) (𝑡) = ∑ 𝐺 (𝑡, 𝑠) 𝑓̃ (𝑠 + ] − 1, 𝑥 (𝑠 + ] − 1) 𝑠=0

− 𝑧 (𝑠 + ] − 1) , 𝑦 (𝑠 + ] − 1) − 𝑧 (𝑠 + ] − 1))

(72)

𝑏+2

> ∑ 𝑞∗ (𝑡) 𝜑 (𝑠 + ] − 1) 𝑞0−2 𝑀 ≥ 𝑞0−1 𝑀𝜅2 , 𝑠=0

for all 𝑡 ∈ [] − 1, 𝑏 + ] + 1]N]−1 . This implies ‖𝐵1 (𝑥, 𝑦)‖ > 𝑞0−1 𝑀𝜅2 . Similarly, ‖𝐵2 (𝑥, 𝑦)‖ > 𝑞0−1 𝑀𝜅2 . Thus, note that (𝑥, 𝑦) ∈ 𝜕(𝐵𝑞0−1 𝑀𝜅2 × 𝐵𝑞0−1 𝑀𝜅2 ) ∩ (𝑃 × 𝑃), and we have 󵄩󵄩 󵄩 󵄩 󵄩 󵄩󵄩(𝑥, 𝑦)󵄩󵄩󵄩 = max {‖𝑥‖ , 󵄩󵄩󵄩𝑦󵄩󵄩󵄩} 󵄩 󵄩 󵄩 󵄩 ≥ max {󵄩󵄩󵄩𝐵1 (𝑥, 𝑦)󵄩󵄩󵄩 , 󵄩󵄩󵄩𝐵2 (𝑥, 𝑦)󵄩󵄩󵄩} > 𝑞0−1 𝑀𝜅2 󵄩 󵄩 = 󵄩󵄩󵄩(𝑥, 𝑦)󵄩󵄩󵄩 .

and Technology Research Program of Chongqing Municipal Education Commission (Grant nos. KJ1703050, KJ1703043), Natural Science Foundation of Chongqing Normal University (Grant nos. 16XYY24, 15XLB011), and Graduate Student Scientific Research Innovation Projects of Chongqing (Grant no. CYS18296).

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(73)

[5] C. S. Goodrich, “On a first-order semipositone discrete fractional boundary value problem,” Archiv der Mathematik, vol. 99, no. 6, pp. 509–518, 2012.

This is a contradiction. So (70) is true. It follows from Lemma 5 that

[6] C. S. Goodrich, “On semipositone discrete fractional boundary value problems with non-local boundary conditions,” Journal of Difference Equations and Applications, vol. 19, no. 11, pp. 1758– 1780, 2013.

𝑖 (𝐵, (𝐵𝑞0−1 𝑀𝜅2 × 𝐵𝑞0−1 𝑀𝜅2 ) ∩ (𝑃 × 𝑃) , 𝑃 × 𝑃) = 0.

(74)

[7] J. Xu, C. S. Goodrich, and Y. Cui, “Positive solutions for a system of first-order discrete fractional boundary value problems with semipositone nonlinearities,” Revista de la Real Academia de Ciencias Exactas, F´ısicas y Naturales. Serie A. Matemáticas.

(75)

[8] J. Xu and D. O’Regan, “Existence and uniqueness of solutions for a first-order discrete fractional boundary value problem,” Revista de la Real Academia de Ciencias Exactas, F´ısicas y Naturales. Serie A. Matemáticas.

From (69) and (74) we have 𝑖 (𝐵, ((𝐵𝑅 × 𝐵𝑅 ) \ (𝐵𝑞0−1 𝑀𝜅2 × 𝐵𝑞0−1 𝑀𝜅2 )) ∩ (𝑃 × 𝑃) , 𝑃 × 𝑃) = 1 − 0 = 1.

Therefore the operator 𝐵 has at least one fixed point (𝑥, 𝑦) in ((𝐵𝑅 × 𝐵𝑅 ) \ (𝐵𝑞0−1 𝑀𝜅2 × 𝐵𝑞0−1 𝑀𝜅2 )) ∩ (𝑃 × 𝑃) with ‖𝑥‖, ‖𝑦‖ ≥ 𝑞0−1 𝑀𝜅2 , and then (𝑥 − 𝑧, 𝑦 − 𝑧)(𝑡) is a positive solution for (1). This completes the proof.

Data Availability

[9] J. Xu, D. O’Regan, C. Hou, and Y. Cui, “Positive solutions for a class of fractional difference boundary value problems,” Differential Equations & Applications, vol. 9, no. 4, pp. 479–493, 2017. [10] C. Goodrich and A. C. Peterson, Discrete Fractional Calculus, Springer International, New York, NY, USA, 2015.

The data used to support the findings of this study are available from the corresponding author upon request.

[11] W. Yang, “Positive solutions for nonlinear semipositone fractional q-difference system with coupled integral boundary conditions,” Applied Mathematics and Computation, vol. 244, pp. 702–725, 2014.

Conflicts of Interest

Acknowledgments

[12] C. Yuan, “Two positive solutions for (n-1, 1)-type semipositone integral boundary value problems for coupled systems of nonlinear fractional differential equations,” Communications in Nonlinear Science and Numerical Simulation, vol. 17, no. 2, pp. 930–942, 2012.

Research supported by the National Natural Science Foundation of China (Grant no. 11601048), Natural Science Foundation of Chongqing (Grant no. cstc2016jcyjA0181), the Science

[13] J. Henderson and R. Luca, “Existence of positive solutions for a system of semipositone fractional boundary value problems,” Electronic Journal of Qualitative Theory of Differential Equations, Paper No. 22, 28 pages, 2016.

The authors declare that they have no conflicts of interest.

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