Hindawi Publishing Corporation e Scientific World Journal Volume 2014, Article ID 817542, 11 pages http://dx.doi.org/10.1155/2014/817542
Research Article Multiple Positive Solutions to Nonlinear Boundary Value Problems of a System for Fractional Differential Equations Chengbo Zhai and Mengru Hao School of Mathematical Sciences, Shanxi University, Taiyuan, Shanxi 030006, China Correspondence should be addressed to Chengbo Zhai; [email protected] Received 5 August 2013; Accepted 3 November 2013; Published 23 January 2014 Academic Editors: L. Acedo and N. Hussain Copyright © 2014 C. Zhai and M. Hao. 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. By using Krasnoselskii’s fixed point theorem, we study the existence of at least one or two positive solutions to a system of fractional ] ] boundary value problems given by −𝐷01+ 𝑦1 (𝑡) = 𝜆 1 𝑎1 (𝑡)𝑓(𝑦1 (𝑡), 𝑦2 (𝑡)), −𝐷02+ 𝑦2 (𝑡) = 𝜆 2 𝑎2 (𝑡)𝑔(𝑦1 (𝑡), 𝑦2 (𝑡)), where 𝐷0]+ is the standard Riemann-Liouville fractional derivative, ]1 , ]2 ∈ (𝑛 − 1, 𝑛] for 𝑛 > 3 and 𝑛 ∈ 𝑁, subject to the boundary conditions 𝑦1(𝑖) (0) = 0 = 𝑦2(𝑖) (0), for 0 ≤ 𝑖 ≤ 𝑛 − 2, and [𝐷0𝛼+ 𝑦1 (𝑡)]𝑡=1 = 0 = [𝐷0𝛼+ 𝑦2 (𝑡)]𝑡=1 , for 1 ≤ 𝛼 ≤ 𝑛 − 2, or 𝑦1(𝑖) (0) = 0 = 𝑦2(𝑖) (0), for 0 ≤ 𝑖 ≤ 𝑛 − 2, and [𝐷0𝛼+ 𝑦1 (𝑡)]𝑡=1 = 𝜙1 (𝑦1 ), [𝐷0𝛼+ 𝑦2 (𝑡)]𝑡=1 = 𝜙2 (𝑦2 ), for 1 ≤ 𝛼 ≤ 𝑛 − 2, 𝜙1 , 𝜙2 ∈ 𝐶([0, 1], 𝑅). Our results are new and complement previously known results. As an application, we also give an example to demonstrate our result.
[𝐷0𝛼+ 𝑦1 (𝑡)]𝑡=1 = 𝜙1 (𝑦1 ) ,
1. Introduction
[𝐷0𝛼+ 𝑦2 (𝑡)]𝑡=1 = 𝜙2 (𝑦2 ) ,
The purpose of this paper is to consider the existence of multiple positive solutions for the following system of nonlinear fractional differential equations:
1 ≤ 𝛼 ≤ 𝑛 − 2, (3)
−
] 𝐷01+ 𝑦1
(𝑡) = 𝜆 1 𝑎1 (𝑡) 𝑓 (𝑦1 (𝑡) , 𝑦2 (𝑡)) ,
−
] 𝐷02+ 𝑦2
(𝑡) = 𝜆 2 𝑎2 (𝑡) 𝑔 (𝑦1 (𝑡) , 𝑦2 (𝑡)) ,
(1)
where 𝑡 ∈ (0, 1), ]1 , ]2 ∈ (𝑛 − 1, 𝑛] for 𝑛 > 3 and 𝑛 ∈ 𝑁, and 𝜆 1 , 𝜆 2 > 0, subject to a couple of boundary conditions. In particular, we first consider (1) subject to 𝑦1(𝑖) (0) = 0 = 𝑦2(𝑖) (0) ,
0 ≤ 𝑖 ≤ 𝑛 − 2,
[𝐷0𝛼+ 𝑦1 (𝑡)]𝑡=1 = 0 = [𝐷0𝛼+ 𝑦2 (𝑡)]𝑡=1 ,
1 ≤ 𝛼 ≤ 𝑛 − 2,
(2)
where 𝑓, 𝑔 ∈ 𝐶([0, ∞) × [0, ∞), [0, ∞)), and 𝑎1 , 𝑎2 ∈ 𝐶([0, 1], [0, ∞)). We then consider the case in which the boundary conditions are changed to 𝑦1(𝑖) (0) = 0 = 𝑦2(𝑖) (0) ,
0 ≤ 𝑖 ≤ 𝑛 − 2,
where 𝜙1 , 𝜙2 ∈ 𝐶([0, 1], 𝑅). Fractional differential equations arise in many fields, such as physics, mechanics, chemistry, economics, and engineering and biological sciences; see [1–11] for example. In recent years, the study of positive solutions for fractional differential equation boundary value problems has attracted considerable attention, and fruits from research into it emerge continuously. For a small sample of such work, we refer the reader to [12–20] and the references therein. The situation of at least one positive solution has been studied in many excellent monograph; see [12–19, 21] and other references therein. In [22], by means of Schauder fixed point theorem, Su investigated the existence of one positive solution to the following boundary value problem for a coupled system of nonlinear fractional differential equations: 𝜇
𝐷0𝛼+ 𝑦1 (𝑡) = 𝑓 (𝑡, 𝑦2 (𝑡) , 𝐷0+ 𝑦2 (𝑡)) , 𝛽
−𝐷0+ 𝑦2 (𝑡) = 𝑔 (𝑡, 𝑦1 (𝑡) , 𝐷0]+ 𝑦1 (𝑡)) ,
0 < 𝑡 < 1, 0 < 𝑡 < 1,
2
The Scientific World Journal 𝑦1 (0) = 𝑦1 (1) = 𝑦2 (0) = 𝑦2 (1) = 0,
(ii) 𝐺(𝑡, 𝑠) ≥ 0 for each (𝑡, 𝑠) ∈ [0, 1] × [0, 1]; (4)
where 1 < 𝛼, 𝛽 < 2, 𝜇, ] > 0, 𝛼 − ] ≥ 1, 𝛽 − 𝜇 ≥ 1. In [21], Goodrich established the existence of one positive solution to problems (1)-(2) and (1), (3) by using Krasnoselskii’s fixed point theorem. Different from the above works mentioned, in this paper we will present the existence of at least two positive solutions to problems (1)-(2) and (1), (3) by using the similar method presented in [21]. Moreover, under different conditions, we also present the existence of at least one positive solution to problems (1)-(2) and (1), (3) with 𝜆 1 = 𝜆 2 = 1.
2. Preliminaries For the convenience of the reader, we present here some definitions, lemmas, and basic results that will be used in the proofs of our theorems. Definition 1 (see [23]). Let ] > 0 with ] ∈ 𝑅. Suppose that 𝑦 : [𝑎, +∞) → 𝑅. Then the ]th Riemann-Liouville fractional integral is defined to be 𝐷𝑎−]+ 𝑦 (𝑡) :=
𝑡
1 ∫ 𝑦 (𝑠) (𝑡 − 𝑠)]−1 𝑑𝑠, Γ (]) 𝑎
(5)
whenever the right-hand side is defined. Similarly, with ] > 0 and ] ∈ 𝑅, we define the ]th Riemann-Liouville fractional derivative to be 𝐷𝑎] + 𝑦 (𝑡) :=
𝑦 (𝑠) 𝑑𝑛 𝑡 1 𝑑𝑠, ∫ 𝑛 Γ (𝑛 − ]) 𝑑𝑡 𝑎 (𝑡 − 𝑠)]+1−𝑛
(6)
where 𝑛 ∈ 𝑁 is the unique positive integer satisfying 𝑛 − 1 ≤ ] < 𝑛 and 𝑡 > 𝑎. Lemma 2 (see [24]). Let 𝑔 ∈ 𝐶𝑛 ([0, 1]) be given. Then the unique solution to problem −𝐷0]+ 𝑦(𝑡) = 𝑔(𝑡) together with the boundary conditions 𝑦(𝑖) (0) = 0 = [𝐷0𝛼+ 𝑦(𝑡)]𝑡=1 , where 1 ≤ 𝛼 ≤ 𝑛 − 2 and 0 ≤ 𝑖 ≤ 𝑛 − 2, is 1
𝑦 (𝑡) = ∫ 𝐺 (𝑡, 𝑠) 𝑔 (𝑠) 𝑑𝑠, 0
(7)
(iii) max𝑡∈[0,1] 𝐺(𝑡, 𝑠) = 𝐺(1, 𝑠), for each 𝑠 ∈ [0, 1]. Lemma 4 (see [24]). Let 𝐺(𝑡, 𝑠) be as given in the statement of Lemma 2. Then there exists a constant 𝛾 ∈ (0, 1) such that min 𝐺 (𝑡, 𝑠) ≥ 𝛾 max 𝐺 (𝑡, 𝑠) = 𝛾𝐺 (1, 𝑠) .
𝑡∈[(1/2),1]
To prove our results, we need the following Krasnoselskii’s fixed point theorem which can be seen in Guo and Lakshmikantham [25]. Lemma 5 (see [25]). Let 𝐸 be a Banach space, and let 𝑃 be a cone. Assume that Ω1 , Ω2 are open bounded subsets of 𝐸 with 0 ∈ Ω1 , Ω1 ⊂ Ω2 , and let 𝑇 : 𝑃 ⋂(Ω2 \ Ω1 ) → 𝑃 be a completely continuous operator such that (i) ‖𝑇𝑢‖ ≤ ‖𝑢‖, ∀𝑢 ∈ 𝑃 ⋂ 𝜕Ω1 , and ‖𝑇𝑢‖ ≥ ‖𝑢‖, ∀𝑢 ∈ P ⋂ 𝜕Ω2 ; or (ii) ‖𝑇𝑢‖ ≥ ‖𝑢‖, ∀𝑢 ∈ 𝑃 ⋂ 𝜕Ω1 , and ‖𝑇𝑢‖ ≤ ‖𝑢‖, ∀𝑢 ∈ 𝑃 ⋂ 𝜕Ω2 . Then 𝑇 has a fixed point in 𝑃 ⋂(Ω2 \ Ω1 ).
3. Main Results In this section, we apply Lemma 5 to study problems (1)-(2) and (1), (3), and we obtain some new results on the existence of multiple positive solutions. 3.1. Problem (1)-(2) in the General Case. In our considerations, let 𝐸 represent the Banach space of 𝐶([0, 1]) when equipped with the usual supremum norm, ‖ ⋅ ‖. Then put 𝑋 := 𝐸 × 𝐸, where 𝑋 is equipped with the norm ‖(𝑦1 , 𝑦2 )‖ := ‖𝑦1 ‖ + ‖𝑦2 ‖ for (𝑦1 , 𝑦2 ) ∈ 𝑋. Observe that 𝑋 is also a Banach space (see [26]). In addition, we define two operators 𝑇1 , 𝑇2 : 𝑋 → 𝐸 by 1
𝑇1 (𝑦1 , 𝑦2 ) (𝑡) := 𝜆 1 ∫ 𝐺1 (𝑡, 𝑠) 𝑎1 (𝑠) 𝑓 (𝑦1 (𝑠) , 𝑦2 (𝑠)) 𝑑𝑠, 0
1
𝑇2 (𝑦1 , 𝑦2 ) (𝑡) := 𝜆 2 ∫ 𝐺2 (𝑡, 𝑠) 𝑎2 (𝑠) 𝑔 (𝑦1 (𝑠) , 𝑦2 (𝑠)) 𝑑𝑠, 0
where 𝑡]−1 (1 − 𝑠)]−𝛼−1 − (𝑡 − 𝑠)]−1 { , 0≤𝑠≤𝑡≤1 { { { Γ (]) 𝐺 (𝑡, 𝑠) = { ]−1 { { 𝑡 (1 − 𝑠)]−𝛼−1 { , 0≤𝑡≤𝑠≤1 Γ (]) { (8) is the Green function for this problem. Lemma 3 (see [24]). Let 𝐺(𝑡, 𝑠) be as given in the statement of Lemma 2. Then one finds that (i) 𝐺(𝑡, 𝑠) is a continuous function on the unit square [0, 1] × [0, 1];
(9)
𝑡∈[0,1]
(10)
where 𝐺1 (𝑡, 𝑠) is the Green function of Lemma 2 with ] replaced by ]1 and, likewise, 𝐺2 (𝑡, 𝑠) is the Green function of Lemma 2 with ] replaced by ]2 . Now, we define an operator 𝑆 : 𝑋 → 𝑋 by 𝑆 (𝑦1 , 𝑦2 ) (𝑡) := (𝑇1 (𝑦1 , 𝑦2 ) (𝑡) , 𝑇2 (𝑦1 , 𝑦2 ) (𝑡)) 1
= (𝜆 1 ∫ 𝐺1 (𝑡, 𝑠) 𝑎1 (𝑠) 𝑓 (𝑦1 (𝑠) , 𝑦2 (𝑠)) 𝑑𝑠, 0
1
𝜆 2 ∫ 𝐺2 (𝑡, 𝑠) 𝑎2 (𝑠) 𝑔 (𝑦1 (𝑠) , 𝑦2 (𝑠)) 𝑑𝑠) . 0
(11)
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We claim that whenever (𝑦1 , 𝑦2 ) ∈ 𝑋 is a fixed point of the operator defined in (11), it follows that 𝑦1 (𝑡) and 𝑦2 (𝑡) solve problems (1)-(2). That is, a pair of functions 𝑦1 , 𝑦2 ∈ 𝑋 is a solution of problems (1)-(2) if and only if 𝑦1 , 𝑦2 is a fixed point of the operator 𝑆 defined in (11) (see [26]). In the following, we will look for fixed points of the operator 𝑆, because these fixed points coincide with solutions of problems (1)-(2). For use in the sequel, let 𝛾1 and 𝛾2 be the constants given by Lemma 4 associated, respectively, with the Green functions 𝐺1 and 𝐺2 , and define 𝛾̃ by 𝛾̃ := min{𝛾1 , 𝛾2 }, and notice that 𝛾̃ ∈ (0, 1). For the sake of convenience, we set 𝑓0 = 𝑔0 =
1
≤ 𝜆 1 ∫ 𝐺1 (1, 𝑠) 𝑎1 (𝑠) 𝑓 (𝑦1 (𝑠) , 𝑦2 (𝑠)) 𝑑𝑠 0
0
1
𝑓 (𝑦1 , 𝑦2 ) , + + (𝑦1 ,𝑦2 ) → (0 ,0 ) 𝑦1 + 𝑦2 lim
(𝑦1 ,𝑦2 ) → (0+ ,0+ )
≤ 𝐿Φ2 ∫ 𝐺1 (1, 𝑠) 𝑑𝑠 < +∞. 0
𝑔 (𝑦1 , 𝑦2 ) , 𝑦1 + 𝑦2
lim
(𝑦1 ,𝑦2 ) → (∞,∞)
(15)
1
≤ 𝜆 1 ∫ 𝐺1 (1, 𝑠) 𝐿𝑑𝑠
lim
(12)
𝑓 (𝑦1 , 𝑦2 ) lim , 𝑓∞ = (𝑦1 ,𝑦2 ) → (∞,∞) 𝑦1 + 𝑦2 𝑔∞ =
Let Ω ⊆ 𝐾 be bounded; that is, there exists a positive constant 𝑀 > 0 such that ‖(𝑦1 , 𝑦2 )‖ ≤ 𝑀, for all (𝑦1 , 𝑦2 ) ∈ Ω. Let 𝐿 = max0≤𝑡≤1, 0≤‖(𝑦1 ,𝑦2 )‖≤𝑀|𝑎1 (𝑡)𝑓(𝑦1 (𝑡), 𝑦2 (𝑡))| + 1; then, for (𝑦1 , 𝑦2 ) ∈ Ω, we have 𝑇1 (𝑦1 , 𝑦2 ) (𝑡)
𝑔 (𝑦1 , 𝑦2 ) . 𝑦1 + 𝑦2
Now we list some assumptions: (F 1 ) 𝑓0 , 𝑔0 ∈ (0, +∞); (F 2 ) 𝑓∞ , 𝑔∞ ∈ (0, +∞); (F 3 ) there are numbers Φ1 , Φ2 , where
Hence, 𝑇1 (Ω) is bounded. On the other hand, given 𝜀 > 0, setting 𝛿 = min{(1/2)(Γ(]1 )𝜀/𝐿Φ2 )1/(]1 −1) , 𝜀Γ(]1 )/(]1 − 1)𝐿Φ2 }, then, for each (𝑦1 , 𝑦2 ) ∈ Ω, 𝑡1 , 𝑡2 ∈ [0, 1], 𝑡1 < 𝑡2 , and 𝑡2 − 𝑡1 < 𝛿, one has |𝑇1 (𝑦1 , 𝑦2 )(𝑡2 ) − 𝑇1 (𝑦1 , 𝑦2 )(𝑡1 )| < 𝜀. That is to say, 𝑇(Ω) is equicontinuity. In fact, 𝑇1 (𝑦1 , 𝑦2 ) (𝑡2 ) − 𝑇1 (𝑦1 , 𝑦2 ) (𝑡1 ) 1 = 𝜆 1 ∫ 𝐺1 (𝑡2 , 𝑠) 𝑎1 (𝑠) 𝑓 (𝑦1 (𝑠) , 𝑦2 (𝑠)) 𝑑𝑠 0 1 −𝜆 1 ∫ 𝐺1 (𝑡1 , 𝑠) 𝑎1 (𝑠) 𝑓 (𝑦1 (𝑠) , 𝑦2 (𝑠)) 𝑑𝑠 0 𝑡1 = 𝜆 1 ∫ [𝐺1 (𝑡2 , 𝑠) − 𝐺1 (𝑡1 , 𝑠)] 𝑎1 (𝑠) 𝑓 (𝑦1 (𝑠) , 𝑦2 (𝑠)) 𝑑𝑠 0
−1
1 1 Φ1 := max { [∫ 𝛾̃𝐺1 (1, 𝑠) 𝑎1 (𝑠) 𝑓∞ 𝑑𝑠] , 2 1/2
𝑡2
+ 𝜆 1 ∫ [𝐺1 (𝑡2 , 𝑠) − 𝐺1 (𝑡1 , 𝑠)] 𝑎1 (𝑠)
−1
1 1 [∫ 𝛾̃𝐺 (1, 𝑠) 𝑎2 (𝑠) 𝑔∞ 𝑑𝑠] } , 2 1/2 2 −1
1 1 Φ2 := min { [∫ 𝐺1 (1, 𝑠) 𝑎1 (𝑠) 𝑓0 𝑑𝑠] , 2 0
𝑡1
× 𝑓 (𝑦1 (s) , 𝑦2 (𝑠)) 𝑑𝑠
(13)
1
+ 𝜆 1 ∫ [𝐺1 (𝑡2 , 𝑠) − 𝐺1 (𝑡1 , 𝑠)] 𝑎1 (𝑠) 𝑡2
× 𝑓 (𝑦1 (𝑠) , 𝑦2 (𝑠)) 𝑑𝑠
−1
1 1 [∫ 𝐺 (1, 𝑠) 𝑎2 (𝑠) 𝑔0 𝑑𝑠] } , 2 0 2 ≤
such that Φ1 < 𝜆 1 , 𝜆 2 < Φ2 . Next, we define the cone 𝐾 by
𝑡1 Φ2 𝐿 ] −1 ] −1 [ ∫ (1 − 𝑠)]1 −𝛼−1 (𝑡21 − 𝑡11 ) 𝑑𝑠 Γ (]1 ) 0 1
𝐾 := {(𝑦1 , 𝑦2 ) ∈ 𝑋 : 𝑦1 , 𝑦2 ≥ 0, min [𝑦1 (𝑡) + 𝑦2 (𝑡)] ≥ 𝛾̃ (𝑦1 , 𝑦2 )} .
] −1
− 𝑡11 ) 𝑑𝑠
] −1
− 𝑡11 ) 𝑑𝑠]
+ ∫ (1 − 𝑠)]1 −𝛼−1 (𝑡21 𝑡2
(14)
𝑡2
+ ∫ (1 − 𝑠)]1 −𝛼−1 (𝑡21 𝑡1
𝑡∈[(1/2),1]
Lemma 6 (see [21]). Let 𝑆 be the operator defined by (11). Then 𝑆 : 𝐾 → 𝐾. Lemma 7. 𝑆 is a completely continuous operator. Proof. The operator 𝑇1 : 𝐾 → 𝐸 is continuous in view of nonnegativeness and continuity of 𝐺1 (𝑡, 𝑠), 𝑓(𝑦1 , 𝑦2 ) and 𝑎1 (𝑡).
=
Φ2 𝐿 1 ] −1 ] −1 (𝑡 1 − 𝑡11 ) Γ (]1 ) ]1 − 𝛼 2
≤
Φ2 𝐿 ]1 −1 ]1 −1 − 𝑡1 ) . (𝑡 Γ (]1 ) 2
] −1
] −1
(16) In the following, we divide the proof into two cases.
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Case 1. If 𝛿 ≤ 𝑡1 < 𝑡2 < 1, then we have
(𝑃1 ) there are numbers Λ 1 , Λ 2 , where
𝑇1 (𝑦1 , 𝑦2 ) (𝑡2 ) − 𝑇1 (𝑦1 , 𝑦2 ) (𝑡1 ) ≤
Φ2 𝐿 ]1 −1 ]1 −1 − 𝑡1 ) (𝑡 Γ (]1 ) 2
=
Φ2 𝐿 ] −2 (] − 1) (𝑡2 − 𝑡1 ) 𝑡𝜉1 Γ (]1 ) 1
0, such that 𝑓 (𝑦1 , 𝑦2 ) , 𝑔 (𝑦1 , 𝑦2 ) < 𝐴−1 1 𝜌1
for 0 ≤ (𝑦1 , 𝑦2 ) ≤ 𝜌1 ; (19)
(𝐻2 ) there exist constants 𝜌2 , 𝐴 2 > 0, such that 𝑓 (𝑦1 , 𝑦2 ) , 𝑔 (𝑦1 , 𝑦2 ) ≥
𝐴−1 2 𝜌2
for 𝛾̃𝜌2 ≤ (𝑦1 , 𝑦2 ) ≤ 𝜌2 ; (20)
1 1 [∫ 𝐺 (1, 𝑠) 𝑎2 (𝑠) 𝑑𝑠] } , 2 0 2
such that Λ 3 < 𝜆 1 , 𝜆 2 < Λ 4 . Theorem 9. Suppose that 𝑓0 = 𝑓∞ = 𝑔0 = 𝑔∞ = ∞ and (𝐻1 ), (𝑃1 ) are satisfied. Then problem (1)-(2) has at least two positive solutions (𝑦10 , 𝑦20 ), (𝑦1 , 𝑦2 ), such that 0 < ‖(𝑦10 , 𝑦20 )‖ < 𝜌1 < ‖(𝑦1 , 𝑦2 )‖. Proof. From Lemma 7, 𝑆 is a completely continuous operator. At first, in view of 𝑓0 = 𝑔0 = ∞, we have 𝑓(𝑦1 , 𝑦2 ) ≥ 𝑀(𝑦1 + 𝑦2 ), for 0 < ‖(𝑦1 , 𝑦2 )‖ < 𝑟1∗ < 𝜌1 ; 𝑔(𝑦1 , 𝑦2 ) ≥ 𝑀(𝑦1 + 𝑦2 ), for 0 < ‖(𝑦1 , 𝑦2 )‖ < 𝑟2∗ < 𝜌1 , where 𝑀 satisfies 𝑀 ≥ 1. Set 𝜌0 := min{𝑟1∗ , 𝑟2∗ }. So we define Ω𝜌0 by Ω𝜌0 := {(𝑦1 , 𝑦2 ) ∈ 𝑋 : ‖(𝑦1 , 𝑦2 )‖ < 𝜌0 }. Then for each (𝑦1 , 𝑦2 ) ∈ 𝐾 ⋂ 𝜕Ω𝜌0 , we find that 1
𝑇1 (𝑦1 , 𝑦2 ) (1) = 𝜆 1 ∫ 𝐺1 (1, 𝑠) 𝑎1 (𝑠) 𝑓 (𝑦1 (𝑠) , 𝑦2 (𝑠)) 𝑑𝑠 0
≥ 𝜆1 ∫
1
1/2
𝐺1 (1, 𝑠) 𝑎1 (𝑠) 𝑓 (𝑦1 (𝑠) , 𝑦2 (𝑠)) 𝑑𝑠
The Scientific World Journal ≥ 𝜆1 ∫
1
1/2
≥ 𝜆1 ∫
1
1/2
≥
5
𝐺1 (1, 𝑠) 𝑎1 (𝑠) 𝑀 (𝑦1 (𝑠) + 𝑦2 (𝑠)) 𝑑𝑠 𝛾̃𝐺1 (1, 𝑠) 𝑎1 (𝑠) 𝑀 (𝑦1 , 𝑦2 ) 𝑑𝑠
1
𝑇1 (𝑦1 , 𝑦2 ) = 𝜆 1 ∫ 𝐺1 (1, 𝑠) 𝑎1 (𝑠) 𝑓 (𝑦1 (𝑠) , 𝑦2 (𝑠)) 𝑑𝑠 0
𝑀 1 (𝑦1 , 𝑦2 ) ≥ (𝑦1 , 𝑦2 ) . 2 2 (23)
So ‖𝑇1 (𝑦1 , 𝑦2 )‖ ≥ (1/2)‖(𝑦1 , 𝑦2 )‖ for (𝑦1 , 𝑦2 ) ∈ 𝐾 ⋂ 𝜕Ω𝜌0 . Similarly, we find that ‖𝑇2 (𝑦1 , 𝑦2 )‖ ≥ (1/2)‖(𝑦1 , 𝑦2 )‖ for (𝑦1 , 𝑦2 ) ∈ 𝐾 ⋂ 𝜕Ω𝜌0 . Consequently, 𝑆 (𝑦1 , 𝑦2 ) = (𝑇1 (𝑦1 , 𝑦2 ) , 𝑇2 (𝑦1 , 𝑦2 )) = 𝑇1 (𝑦1 , 𝑦2 ) + 𝑇2 (𝑦1 , 𝑦2 ) ≥ (𝑦1 , 𝑦2 ) ,
(24)
whenever (𝑦1 , 𝑦2 ) ∈ 𝐾 ⋂ 𝜕Ω𝜌0 . Thus, 𝑆 is cone expansion on 𝐾 ⋂ 𝜕Ω𝜌0 . Next, since 𝑓∞ = 𝑔∞ = ∞, we have 𝑓(𝑦1 , 𝑦2 ) ≥ 𝑀1 (𝑦1 + 𝑦2 ) for 𝑦1 + 𝑦2 ≥ 𝑟1∗∗ > 𝜌1 ; 𝑔(𝑦1 , 𝑦2 ) ≥ 𝑀1 (𝑦1 + 𝑦2 ) for 𝑦1 + 𝑦2 ≥ 𝑟2∗∗ > 𝜌1 , where 𝑀1 satisfies 𝑀1 ≥ 1. Set 𝛾)} and Ω𝜌0∗ := 𝜌10 := max{𝑟1∗∗ , 𝑟2∗∗ }. Let 𝜌0∗ = max{2𝜌1 , (𝜌10 /̃ {(𝑦1 , 𝑦2 ) ∈ 𝑋 : ‖(𝑦1 , 𝑦2 )‖ < 𝜌0∗ }. Then (𝑦1 , 𝑦2 ) ∈ 𝐾 ⋂ 𝜕Ω𝜌0∗ implies 𝑦1 (𝑡) + 𝑦2 (𝑡) ≥
≥ 𝛾̃ (𝑦1 , 𝑦2 ) = 𝛾̃𝜌0∗ ≥ 𝜌10 .
(25)
So we obtain 1
𝑇1 (𝑦1 , 𝑦2 ) (1) = 𝜆 1 ∫ 𝐺1 (1, 𝑠) 𝑎1 (𝑠) 𝑓 (𝑦1 (𝑠) , 𝑦2 (𝑠)) 𝑑𝑠 0
1
≥ 𝜆1 ∫
1/2 1
≥ 𝜆1 ∫
1/2 1
≥ 𝜆1 ∫
1/2
𝐺1 (1, 𝑠) 𝑎1 (𝑠) 𝑓 (𝑦1 (𝑠) , 𝑦2 (𝑠)) 𝑑𝑠 𝐺1 (1, 𝑠) 𝑎1 (𝑠) 𝑀1 (𝑦1 (𝑠) + 𝑦2 (𝑠)) 𝑑𝑠 𝛾̃𝐺1 (1, 𝑠) 𝑎1 (𝑠) 𝑀1 (𝑦1 , 𝑦2 ) 𝑑𝑠
𝑀 1 ≥ 1 (𝑦1 , 𝑦2 ) ≥ (𝑦1 , 𝑦2 ) . 2 2
1 𝐴 ≤ 1 [∫ 𝐺1 (1, 𝑠) 𝑎1 (𝑠) 𝑑𝑠] 2 0
−1
1
× ∫ 𝐺1 (1, 𝑠) 𝑎1 (𝑠) 𝑑𝑠𝐴−1 1 𝜌1 0
=
𝜌1 1 = (𝑦1 , 𝑦2 ) . 2 2
(27)
Similarly, we find that ‖𝑇2 (𝑦1 , 𝑦2 )‖ ≤ (1/2)‖(𝑦1 , 𝑦2 )‖ for (𝑦1 , 𝑦2 ) ∈ 𝐾 ⋂ 𝜕Ω𝜌1 . Consequently, ‖𝑆(𝑦1 , 𝑦2 )‖ ≤ ‖(𝑦1 , 𝑦2 )‖, whenever (𝑦1 , 𝑦2 ) ∈ 𝐾 ⋂ 𝜕Ω𝜌1 . Thus, 𝑆 is cone compression on 𝐾 ⋂ 𝜕Ω𝜌1 . So, from Lemma 5, 𝑆 has a fixed point (𝑦10 , 𝑦20 ) ∈ 𝐾 ⋂(Ω𝜌1 \ Ω𝜌0 ) and a fixed point (𝑦1 , 𝑦2 ) ∈ 𝐾 ⋂(Ω𝜌0∗ \ Ω𝜌1 ). Both are positive solutions of BVP (1)-(2) with 0 < (𝑦10 , 𝑦20 ) < 𝜌1 < (𝑦1 , 𝑦2 ) .
(28)
The proof is complete.
min [𝑦1 (𝑡) + 𝑦2 (𝑡)]
𝑡∈[(1/2),1]
Finally, let Ω𝜌1 := {(𝑦1 , 𝑦2 ) ∈ 𝑋 : ‖(𝑦1 , 𝑦2 )‖ < 𝜌1 }. For (𝑦1 , 𝑦2 ) ∈ 𝐾 ⋂ 𝜕Ω𝜌1 , from (𝐻1 ), (𝑃1 ), we have
Theorem 10. Suppose that 𝑓0 = 𝑓∞ = 𝑔0 = 𝑔∞ = 0 and (𝐻2 ), (𝑃2 ) are satisfied. Then problem (1)-(2) has at least two positive solutions (𝑦10 , 𝑦20 ), (𝑦1 , 𝑦2 ), such that 0 < ‖(𝑦10 , 𝑦20 )‖ < 𝜌2 < ‖(𝑦1 , 𝑦2 )‖. Proof. At first, in view of 𝑓0 = 𝑔0 = 0, we have 𝑓(𝑦1 , 𝑦2 ) < 𝜀(𝑦1 + 𝑦2 ), 𝑔(𝑦1 , 𝑦2 ) < 𝜀(𝑦1 + 𝑦2 ), for 0 < ‖(𝑦1 , 𝑦2 )‖ ≤ 𝜌 < 𝜌2 , where 𝜀 satisfies 𝜀 ≤ 1. Let Ω𝜌 := {(𝑦1 , 𝑦2 ) ∈ 𝑋 : ‖(𝑦1 , 𝑦2 )‖ < 𝜌}. Then for each (𝑦1 , 𝑦2 ) ∈ 𝐾 ⋂ 𝜕Ω𝜌 , we find that 1
𝑇1 (𝑦1 , 𝑦2 ) = 𝜆 1 ∫ 𝐺1 (1, 𝑠) 𝑎1 (𝑠) 𝑓 (𝑦1 (𝑠) , 𝑦2 (𝑠)) 𝑑𝑠 0 1
≤ 𝜆 1 ∫ 𝐺1 (1, 𝑠) 𝑎1 (𝑠) 𝜀 (𝑦1 (𝑠) + 𝑦2 (𝑠)) 𝑑𝑠 0
1
(26)
So ‖𝑇1 (𝑦1 , 𝑦2 )‖ ≥ (1/2)‖(𝑦1 , 𝑦2 )‖ for (𝑦1 , 𝑦2 ) ∈ 𝐾 ⋂ 𝜕Ω𝜌0∗ . Similarly, we find that ‖𝑇2 (𝑦1 , 𝑦2 )‖ ≥ (1/2)‖(𝑦1 , 𝑦2 )‖ for (𝑦1 , 𝑦2 ) ∈ 𝐾 ⋂ 𝜕Ω𝜌0∗ . Consequently, ‖𝑆2 (𝑦1 , 𝑦2 )‖ ≥ ‖(𝑦1 , 𝑦2 )‖, whenever (𝑦1 , 𝑦2 ) ∈ 𝐾 ⋂ 𝜕Ω𝜌0∗ . Thus, S is cone expansion on 𝐾 ⋂ 𝜕Ω𝜌0∗ .
≤ 𝜆 1 ∫ 𝐺1 (1, 𝑠) 𝑎1 (𝑠) 𝜀 (𝑦1 , 𝑦2 ) 𝑑𝑠 0
≤
𝜀 1 (𝑦 , 𝑦 ) ≤ (𝑦 , 𝑦 ) . 2 1 2 2 1 2
(29)
Like Theorem 9, we get ‖𝑆(𝑦1 , 𝑦2 )‖ ≤ ‖(𝑦1 , 𝑦2 )‖ for (𝑦1 , 𝑦2 ) ∈ 𝐾 ⋂ 𝜕Ω𝜌 .
6
The Scientific World Journal
Next, in view of 𝑓∞ = 𝑔∞ = 0, we have 𝑓(𝑦1 , 𝑦2 ) < 𝜀1 (𝑦1 + 𝑦2 ), 𝑔(𝑦1 , 𝑦2 ) < 𝜀1 (𝑦1 + 𝑦2 ), for 𝑦1 + 𝑦2 ≥ 𝜌 > 𝜌2 , where 𝜀1 satisfies 𝜀1 ≤ 1. We consider two cases. Case 1. Suppose that 𝑓 is unbounded; there exists 𝜌∗ > 𝜌 such that 𝑓 (𝑦1 , 𝑦2 ) ≤ 𝑓 (𝑦1∗ , 𝑦2∗ ) for 0 ≤ (𝑦1 , 𝑦2 ) ≤ 𝜌∗ , (30) ∗ ∗ ∗ (𝑦1 , 𝑦2 ) = 𝜌 . Since 𝜌∗ > 𝜌 , one has 𝑓(𝑦1 , 𝑦2 ) ≤ 𝑓(𝑦1∗ , 𝑦2∗ ) < 𝜀1 (𝑦1∗ +𝑦2∗ ) for 0 ≤ ‖(𝑦1 , 𝑦2 )‖ ≤ 𝜌∗ . Then, for (𝑦1 , 𝑦2 ) ∈ 𝐾 and ‖(𝑦1 , 𝑦2 )‖ = 𝜌∗ , we obtain 1
𝑇1 (𝑦1 , 𝑦2 ) = 𝜆 1 ∫ 𝐺1 (1, 𝑠) 𝑎1 (𝑠) 𝑓 (𝑦1 (𝑠) , 𝑦2 (𝑠)) 𝑑𝑠 0 1
0
≤ 𝜆 1 ∫ 𝐺1 (1, 𝑠) 𝑎1 (𝑠) 𝜀1 (𝑦1 , 𝑦2 ) 𝑑𝑠 0 1 (𝑦1 , 𝑦2 ) ≤ (𝑦1 , 𝑦2 ) . 2
(31)
1
1
≤ 𝜆 1 ∫ 𝐺1 (1, 𝑠) 𝑎1 (𝑠) 𝐿 1 𝑑𝑠 0
𝐵2 := min {2 ∫
1/2
𝐺1 (1, 𝑠) 𝑎1 (𝑠) 𝑑𝑠, 2 ∫
1
1/2
𝐺2 (1, 𝑠) 𝑎2 (𝑠) 𝑑𝑠} .
(𝐻4 ) 𝑓(𝑦1 , 𝑦2 ), 𝑔(𝑦1 , 𝑦2 ) ≥ 𝐵2−1 𝜌2 , for 𝛾̃𝜌2 ≤ ‖(𝑦1 , 𝑦2 )‖ ≤ 𝜌2 .
Theorem 11. Suppose that (𝐻3 ) and (𝐻4 ) are satisfied. Then problem (1)-(2), in the case where 𝜆 1 = 𝜆 2 = 1, has at least one positive solution (𝑦10 , 𝑦20 ) such that ‖(𝑦10 , 𝑦20 )‖ between 𝜌1 and 𝜌2 . Proof. With loss of generality, we may assume that 𝜌1 < 𝜌2 . Let Ω𝜌1 := {(𝑦1 , 𝑦2 ) ∈ 𝑋 : ‖(𝑦1 , 𝑦2 )‖ < 𝜌1 }. For (𝑦1 , 𝑦2 ) ∈ 𝐾 ⋂ 𝜕Ω𝜌1 , one has 1
(32)
≤
𝜌 𝐵1 −1 1 𝐵 𝜌 = 1 = (𝑦1 , 𝑦2 ) . 2 1 1 2 2 (37)
Like Theorem 9, we get ‖𝑆(𝑦1 , 𝑦2 )‖ ≤ ‖(𝑦1 , 𝑦2 )‖ for (𝑦1 , 𝑦2 ) ∈ 𝐾 ⋂ 𝜕Ω𝜌1 . Now, set Ω𝜌2 := {(𝑦1 , 𝑦2 ) ∈ 𝑋 : ‖(𝑦1 , 𝑦2 )‖ < 𝜌2 }. Then for (𝑦1 , 𝑦2 ) ∈ 𝐾 ⋂ 𝜕Ω𝜌2 , one has 𝑦1 (𝑡) + 𝑦2 (𝑡) ≥
min [𝑦1 (𝑡) + 𝑦2 (𝑡)]
≥ 𝛾̃ (𝑦1 , 𝑦2 ) = 𝛾̃𝜌2 .
0
𝑇1 (𝑦1 , 𝑦2 ) = 𝜆 1 ∫ 𝐺1 (1, 𝑠) 𝑎1 (𝑠) 𝑓 (𝑦1 (𝑠) , 𝑦2 (𝑠)) 𝑑𝑠 0
Hence, in either case, we always may set Ω𝜌∗ := {(𝑦1 , 𝑦2 ) ∈ 𝑋 : ‖(𝑦1 , 𝑦2 )‖ < 𝜌∗ } such that ‖𝑇1 (𝑦1 , 𝑦2 )‖ ≤ (1/2)‖(𝑦1 , 𝑦2 )‖ for (𝑦1 , 𝑦2 ) ∈ 𝐾 ⋂ 𝜕Ω𝜌∗ . Like Theorem 9, we get ‖𝑆(𝑦1 , 𝑦2 )‖ ≤ ‖(𝑦1 , 𝑦2 )‖, for (𝑦1 , 𝑦2 ) ∈ 𝐾 ⋂ 𝜕Ω𝜌∗ . Finally, set Ω𝜌2 := {(𝑦1 , 𝑦2 ) ∈ 𝑋 : ‖(𝑦1 , 𝑦2 )‖ < 𝜌2 }. Then (𝑦1 , 𝑦2 ) ∈ 𝐾 ⋂ 𝜕Ω𝜌2 implies 𝑡∈[(1/2),1]
0
1
(𝐻3 ) 𝑓(𝑦1 , 𝑦2 ), 𝑔(𝑦1 , 𝑦2 ) ≤ 𝐵1−1 𝜌1 , for 0 ≤ ‖(𝑦1 , 𝑦2 )‖ ≤ 𝜌1 ;
𝑇1 (𝑦1 , 𝑦2 ) = 𝜆 1 ∫ 𝐺1 (1, 𝑠) 𝑎1 (𝑠) 𝑓 (𝑦1 (𝑠) , 𝑦2 (𝑠)) 𝑑𝑠 0
𝑦1 (𝑡) + 𝑦2 (𝑡) ≥
1
Assume that there exist two positive constants 𝜌1 ≠ 𝜌2 such that
Case 2. Suppose that 𝑓 is bounded; there exists 𝐿 1 such that 𝑓(𝑦1 , 𝑦2 ) ≤ 𝐿 1 for all (𝑦1 , 𝑦2 ) ∈ 𝐾. Taking 𝜌∗ ≥ max{2𝜌2 , 𝐿 1 }, for (𝑦1 , 𝑦2 ) ∈ 𝐾 and ‖(𝑦1 , 𝑦2 )‖ = 𝜌∗ , we obtain
𝜌∗ 1 𝐿 ≤ 1 ≤ = (𝑦1 , 𝑦2 ) . 2 2 2
1
(36)
1
𝜀1 2
3.2. Problem (1)–(3) in Case 𝜆 1 = 𝜆 2 = 1. In the following, for the sake of convenience, set 𝐵1 := max {2 ∫ 𝐺1 (1, 𝑠) 𝑎1 (𝑠) 𝑑𝑠, 2 ∫ 𝐺2 (1, 𝑠) 𝑎2 (𝑠) 𝑑𝑠} ,
≤ 𝜆 1 ∫ 𝐺1 (1, 𝑠) 𝑎1 (𝑠) 𝜀1 (𝑦1 (𝑠) + 𝑦2 (𝑠)) 𝑑𝑠
≤
Consequently, ‖𝑇1 (𝑦1 , 𝑦2 )‖ ≥ (1/2)‖(𝑦1 , 𝑦2 )‖ for (𝑦1 , 𝑦2 ) ∈ 𝐾 ⋂ 𝜕Ω𝜌2 . Like Theorem 9, we get ‖𝑆(𝑦1 , 𝑦2 )‖ ≥ ‖(𝑦1 , 𝑦2 )‖ for (𝑦1 , 𝑦2 ) ∈ 𝐾 ⋂ 𝜕Ω𝜌2 . So, from Lemma 5, 𝑆 has a fixed point (𝑦10 , 𝑦20 ) ∈ 𝐾 ⋂(Ω𝜌2 \ Ω𝜌 ) and a fixed point (𝑦1 , 𝑦2 ) ∈ 𝐾 ⋂(Ω𝜌∗ \ Ω𝜌2 ). Both are positive solutions of BVP (1)-(2) with (35) 0 < (𝑦10 , 𝑦20 ) < 𝜌2 < (𝑦1 , 𝑦2 ) , which complete the proof.
(33)
min [𝑦1 (𝑡) + 𝑦2 (𝑡)]
𝑡∈[(1/2),1]
≥ 𝛾̃ (𝑦1 , 𝑦2 ) = 𝛾̃𝜌2 .
(38)
Thus, we get
Hence we have 1
𝑇1 (𝑦1 , 𝑦2 ) (1) = 𝜆 1 ∫ 𝐺1 (1, 𝑠) 𝑎1 (𝑠) 𝑓 (𝑦1 (𝑠) , 𝑦2 (𝑠)) 𝑑𝑠 0
≥ 𝜆1 ∫
1
1/2
≥
𝐺1 (1, 𝑠) 𝑎1 (𝑠) 𝐴−1 2 𝜌2 𝑑𝑠
𝜌 𝐴 2 −1 1 𝐴 2 𝜌2 = 2 = (𝑦1 , 𝑦2 ) . 2 2 2
1
𝑇1 (𝑦1 , 𝑦2 ) (1) = 𝜆 1 ∫ 𝐺1 (1, 𝑠) 𝑎1 (𝑠) 𝑓 (𝑦1 (𝑠) , 𝑦2 (𝑠)) 𝑑𝑠 0
≥ 𝜆1 ∫
1
1/2
≥ (34)
𝐺1 (1, 𝑠) 𝑎1 (𝑠) 𝐵2−1 𝜌2 𝑑𝑠
𝜌 𝐵2 −1 1 𝐵2 𝜌2 = 2 = (𝑦1 , 𝑦2 ) . 2 2 2
(39)
The Scientific World Journal
7
Like Theorem 9, we get ‖𝑆(𝑦1 , 𝑦2 )‖ ≥ ‖(𝑦1 , 𝑦2 )‖ for (𝑦1 , 𝑦2 ) ∈ 𝐾 ⋂ 𝜕Ω𝜌2 . Hence, from Lemma 5, we complete the proof. Remark 12. In [21], problem (1)-(2) with 𝜆 1 = 𝜆 2 = 1 is not considered.
(𝑃3 ) There are numbers Λ 5 , Λ 6 , where −1
1 1 Λ 5 := max { [∫ 𝛾0 𝐺1 (1, 𝑠) 𝑎1 (𝑠) 𝑑s] , 2 1/2 −1
1 1 [∫ 𝛾 𝐺 (1, 𝑠) 𝑎2 (𝑠) 𝑑𝑠] } , 2 1/2 0 2
3.3. Problem (1), (3) in the General Case. Consider the following. Lemma 13 (see [21]). A pair of functions (𝑦1 , 𝑦2 ) ∈ 𝑋 is a solution of (1), (3) if and only if (𝑦1 , 𝑦2 ) is a fixed point of the operator 𝑈 : 𝑋 → 𝑋 defined by
−1
1 𝐴 Λ 6 := min { 1 [∫ 𝐺1 (1, 𝑠) 𝑎1 (𝑠) 𝑑𝑠] , 4 0 −1
𝐴1 1 [∫ 𝐺 (1, 𝑠) 𝑎2 (𝑠) 𝑑𝑠] } , 4 0 2
𝑈 (𝑦1 , 𝑦2 ) (𝑡)
such that Λ 5 < 𝜆 1 , 𝜆 2 < Λ 6 .
:= (𝑈1 (𝑦1 , 𝑦2 ) (𝑡) , 𝑈2 (𝑦1 , 𝑦2 ) (𝑡))
(𝑃4 ) There are numbers Λ 7 , Λ 8 , where
1
= (𝛽1 (𝑡) 𝜙1 (𝑦1 ) + 𝜆 1 ∫ 𝐺1 (𝑡, 𝑠) 𝑎1 (𝑠) 𝑓
−1
0
× (𝑦1 (𝑠) , 𝑦2 (𝑠)) 𝑑𝑠,
(40)
Λ 7 := max {
−1
𝛽2 (𝑡) 𝜙2 (𝑦2 ) + 𝜆 2 ∫ 𝐺2 (𝑡, 𝑠) 𝑎2 (𝑠) 𝑔 0
−1
1 1 Λ 8 := min { [∫ 𝐺1 (1, 𝑠) 𝑎1 (𝑠) 𝑑𝑠] , 4 0
× (𝑦1 (𝑠) , 𝑦2 (𝑠)) 𝑑𝑠) ,
𝛽2 (𝑡) :=
Γ (]2 − 𝛼) ]2 −1 𝑡 . Γ (]2 )
1 1 [∫ 𝐺 (1, 𝑠) 𝑎2 (𝑠) 𝑑𝑠] } , 4 0 2 (41)
Lemma 14 (see [21]). Each of 𝛽1 (𝑡) and 𝛽2 (𝑡) is strictly increasing in t and satisfies 𝛽1 (0) = 𝛽2 (0) = 0 and 𝛽1 (1), 𝛽2 (1) ∈ (0, 1). Moreover, there exist constants 𝑀𝛽1 and 𝑀𝛽2 satisfying 𝑀𝛽1 , 𝑀𝛽2 ∈ (0, 1) such that min𝑡∈[(1/2),1] 𝛽1 (𝑡) ≥ 𝑀𝛽1 ‖𝛽1 ‖ and min𝑡∈[(1/2),1] 𝛽2 (𝑡) ≥ 𝑀𝛽2 ‖𝛽2 ‖. Let one define a new cone 𝐾1 by 𝐾1 := {(𝑦1 , 𝑦2 ) ∈ 𝑋 : 𝑦1 , 𝑦2 ≥ 0, min [𝑦1 (𝑡) + 𝑦2 (𝑡)] ≥ 𝛾0 (𝑦1 , 𝑦2 )} ,
(44)
−1
where 𝛽1 , 𝛽2 : [0, 1] → [0, 1] are defined by Γ (]1 − 𝛼) ]1 −1 𝑡 , Γ (]1 )
𝐴2 1 [∫ 𝐺 (1, 𝑠) 𝑎1 (𝑠) 𝑑𝑠] , 2 1/2 1 𝐴2 1 [∫ 𝐺 (1, 𝑠) 𝑎2 (𝑠) 𝑑𝑠] } , 2 1/2 2
1
𝛽1 (𝑡) :=
(43)
(42)
𝑡∈[(1/2),1]
such that Λ 7 < 𝜆 1 , 𝜆 2 < Λ 8 . Theorem 16. Suppose that 𝑓0 = 𝑓∞ = 𝑔0 = 𝑔∞ = ∞ and (𝐻1 ), (𝐷1 ), (𝑃3 ) are satisfied. Then problem (1), (3) has at least two positive solutions (𝑦10 , 𝑦20 ), (𝑦1 , 𝑦2 ), such that 0 < (𝑦10 , 𝑦20 ) < 𝜌1 < (𝑦1 , 𝑦2 ) .
Proof. At first, in view of 𝑓0 = 𝑔0 = ∞, we have 𝑓(𝑦1 , 𝑦2 ) ≥ 𝑀(𝑦1 + 𝑦2 ), 𝑔(𝑦1 , 𝑦2 ) ≥ 𝑀(𝑦1 + 𝑦2 ), for 0 < ‖(𝑦1 , 𝑦2 )‖ ≤ 𝜌0 < 𝜌1 , where 𝑀 satisfies 𝑀 ≥ 1. Let Ω𝜌0 := {(𝑦1 , 𝑦2 ) ∈ 𝑋 : ‖(𝑦1 , 𝑦2 )‖ < 𝜌0 }. Then for each (𝑦1 , 𝑦2 ) ∈ 𝐾1 ⋂ 𝜕Ω𝜌0 , we find that 𝑈1 (𝑦1 , 𝑦2 ) (1) ≥ 𝜆 1 ∫
1
1/2
≥ 𝜆1 ∫
1
1/2
where 𝛾0 := min{̃ 𝛾, 𝑀𝛽1 , 𝑀𝛽2 }. It is obvious that 𝛾0 ∈ (0, 1). Lemma 15 (see [21]). 𝑈 : 𝐾1 → 𝐾1 is a completely continuous operator. Now, one assumes (𝐷1 ) 𝜙1 (𝑦1 ) ≤ ‖𝑦1 ‖/4, 𝜙2 (𝑦2 ) ≤ ‖𝑦2 ‖/4 for each (𝑦1 , 𝑦2 ) ∈ 𝐾1 ;
(45)
≥ 𝜆1 ∫
1
1/2
≥
𝐺1 (1, 𝑠) 𝑎1 (𝑠) 𝑓 (𝑦1 (𝑠) , 𝑦2 (𝑠)) 𝑑𝑠 𝐺1 (1, 𝑠) 𝑎1 (𝑠) 𝑀 (𝑦1 (𝑠) + 𝑦2 (𝑠)) 𝑑𝑠 𝛾0 𝐺1 (1, 𝑠) 𝑎1 (𝑠) 𝑀 (𝑦1 , 𝑦2 ) 𝑑𝑠
𝑀 1 (𝑦 , 𝑦 ) ≥ (𝑦 , 𝑦 ) . 2 1 2 2 1 2
(46)
So ‖𝑈1 (𝑦1 , 𝑦2 )‖ ≥ (1/2)‖(𝑦1 , 𝑦2 )‖ for (𝑦1 , 𝑦2 ) ∈ 𝐾1 ⋂ 𝜕Ω𝜌0 .
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The Scientific World Journal
Similarly, we find that ‖𝑈2 (𝑦1 , 𝑦2 )‖ ≥ (1/2)‖(𝑦1 , 𝑦2 )‖ for (𝑦1 , 𝑦2 ) ∈ 𝐾1 ⋂ 𝜕Ω𝜌0 . Consequently, 𝑈 (𝑦1 , 𝑦2 ) = (𝑈1 (𝑦1 , 𝑦2 ) , 𝑈2 (𝑦1 , 𝑦2 )) = 𝑈1 (𝑦1 , 𝑦2 ) + 𝑈2 (𝑦1 , 𝑦2 ) ≥ (𝑦1 , 𝑦2 ) ,
(47)
whenever (𝑦1 , 𝑦2 ) ∈ 𝐾1 ⋂ 𝜕Ω𝜌0 . Thus, 𝑈 is cone expansion on 𝐾1 ⋂ 𝜕Ω𝜌0 . Next, since 𝑓∞ = 𝑔∞ = ∞, we get 𝑓(𝑦1 , 𝑦2 ) ≥ 𝑀1 (𝑦1 + 𝑦2 ), 𝑔(𝑦1 , 𝑦2 ) ≥ 𝑀1 (𝑦1 + 𝑦2 ), for 𝑦1 + 𝑦2 ≥ 𝜌10 > 𝜌1 , where 𝑀1 satisfies 𝑀1 ≥ 1. Let 𝜌0∗ = max{2𝜌1 , (𝜌10 /𝛾0 )} and Ω𝜌0∗ := {(𝑦1 , 𝑦2 ) ∈ 𝑋 : ‖(𝑦1 , 𝑦2 )‖ < 𝜌0∗ }; then, (𝑦1 , 𝑦2 ) ∈ 𝐾 ⋂ 𝜕Ω𝜌0∗ implies 𝑦1 (𝑡) + 𝑦2 (𝑡) ≥
min [𝑦1 (𝑡) + 𝑦2 (𝑡)]
𝑡∈[(1/2),1]
≥ 𝛾0 (𝑦1 , 𝑦2 ) = 𝛾0 𝜌0∗ ≥ 𝜌10 .
(48)
So for (𝑦1 , 𝑦2 ) ∈ 𝐾1 ⋂ 𝜕Ω𝜌0∗ , we obtain 1
𝑈1 (𝑦1 , 𝑦2 ) (1) ≥ 𝜆 1 ∫
1/2 1
≥ 𝜆1 ∫
1/2
𝐺1 (1, 𝑠) 𝑎1 (𝑠) 𝑓 (𝑦1 (𝑠) , 𝑦2 (𝑠)) 𝑑𝑠
0 < (𝑦10 , 𝑦20 ) < 𝜌1 < (𝑦1 , 𝑦2 ) .
(51)
The proof is complete. Theorem 17. Suppose that 𝑓0 = 𝑓∞ = 𝑔0 = 𝑔∞ = 0 and (𝐻2 ), (𝐷1 ), (𝑃4 ) are satisfied. Then problem (1), (3) has at least two positive solutions (𝑦10 , 𝑦20 ), (𝑦1 , 𝑦2 ), such that 0 < (𝑦10 , 𝑦20 ) < 𝜌2 < (𝑦1 , 𝑦2 ) .
(52)
Proof. At first, in view of 𝑓0 = 𝑔0 = 0, we have 𝑓(𝑦1 , 𝑦2 ) < 𝜀(𝑦1 +𝑦2 ), 𝑔(𝑦1 , 𝑦2 ) < 𝜀(𝑦1 +𝑦2 ) for ‖(𝑦1 , 𝑦2 )‖ ≤ 𝜌 < 𝜌2 , where 𝜀 satisfies 𝜀 ≤ 1. Let Ω𝜌 := {(𝑦1 , 𝑦2 ) ∈ 𝑋 : ‖(𝑦1 , 𝑦2 )‖ < 𝜌}. Then for each (𝑦1 , 𝑦2 ) ∈ 𝐾1 ⋂ 𝜕Ω𝜌 , we find that 𝑈1 (𝑦1 , 𝑦2 ) 1
𝐺1 (1, 𝑠) 𝑎1 (𝑠) 𝑀1 (𝑦1 (𝑠) + 𝑦2 (𝑠)) 𝑑𝑠
1
≥ 𝜆 1 ∫ 𝛾0 𝐺1 (1, 𝑠) 𝑎1 (𝑠) 𝑀1 (𝑦1 , 𝑦2 ) 𝑑𝑠 1/2 ≥
So, from Lemma 5, 𝑈 has a fixed point (𝑦10 , 𝑦20 ) ∈ 𝐾1 ⋂ (Ω𝜌1 \Ω𝜌0 ) and a fixed point (𝑦1 , 𝑦2 ) ∈ 𝐾1 ⋂ (Ω𝜌0∗ \Ω𝜌1 ). Both are positive solutions of BVP (1), (3) with
𝑀1 1 (𝑦 , 𝑦 ) ≥ (𝑦 , 𝑦 ) . 2 1 2 2 1 2
(49)
That is, ‖𝑈1 (𝑦1 , 𝑦2 )‖ ≥ (1/2)‖(𝑦1 , 𝑦2 )‖ for (𝑦1 , 𝑦2 ) ∈ 𝐾1 ⋂ 𝜕Ω𝜌0∗ . Similarly, we find that ‖𝑈2 (𝑦1 , 𝑦2 )‖ ≥ (1/2)‖(𝑦1 , 𝑦2 )‖ for (𝑦1 , 𝑦2 ) ∈ 𝐾1 ⋂ 𝜕Ω𝜌0∗ . Consequently, ‖𝑈(𝑦1 , 𝑦2 )‖ ≥ ‖(𝑦1 , 𝑦2 )‖, whenever (𝑦1 , 𝑦2 ) ∈ 𝐾1 ⋂ 𝜕Ω𝜌0∗ . Thus, 𝑈 is cone expansion on 𝐾1 ⋂ 𝜕Ω𝜌0∗ . Finally, let Ω𝜌1 := {(𝑦1 , 𝑦2 ) ∈ 𝑋 : ‖(𝑦1 , 𝑦2 )‖ < 𝜌1 }. For (𝑦1 , 𝑦2 ) ∈ 𝐾 ⋂ 𝜕Ω𝜌1 , from (𝐻1 ), (𝐷1 ), and (𝑃3 ), we have
≤ 𝜙1 (𝑦1 ) + 𝜆 1 ∫ 𝐺1 (𝑡, 𝑠) 𝑎1 (𝑠) 𝑓 (𝑦1 (𝑠) , 𝑦2 (𝑠)) 𝑑𝑠 (53) 0 𝑦 𝜀 1 ≤ 1 + (𝑦1 , 𝑦2 ) ≤ (𝑦1 , 𝑦2 ) . 4 4 2 Like Theorem 16, we get ‖𝑈(𝑦1 , 𝑦2 )‖ ≤ ‖(𝑦1 , 𝑦2 )‖ for (𝑦1 , 𝑦2 ) ∈ 𝐾1 ⋂ 𝜕Ω𝜌 . Next, in view of 𝑓∞ = 𝑔∞ = 0, we have 𝑓(𝑦1 , 𝑦2 ) < 𝜀1 (𝑦1 + 𝑦2 ), 𝑔(𝑦1 , 𝑦2 ) < 𝜀1 (𝑦1 + 𝑦2 ), for 𝑦1 + 𝑦2 ≥ 𝜌 > 𝜌2 , where 𝜀1 satisfies 𝜀1 ≤ 1. We consider two cases. Case 1. Suppose that 𝑓 is unbounded and there exists 𝜌∗ > 𝜌 such that 𝑓 (𝑦1 , 𝑦2 ) ≤ 𝑓 (𝑦1∗ , 𝑦2∗ )
for 0 ≤ (𝑦1 , 𝑦2 ) ≤ 𝜌∗ , ∗ ∗ ∗ (𝑦1 , 𝑦2 ) = 𝜌 .
(54)
1
𝑈1 (𝑦1 , 𝑦2 ) ≤ 𝜙1 (𝑦1 ) + 𝜆 1 ∫ 𝐺1 (𝑡, 𝑠) 𝑎1 (𝑠) 0 × 𝑓 (𝑦1 (𝑠) , 𝑦2 (𝑠)) 𝑑𝑠 𝑦 𝐴 𝜌 ≤ 1 + 1 𝐴−1 4 4 1 1 𝜌 1 ≤ 1 = (𝑦1 , 𝑦2 ) . 2 2
(50)
Similarly, we find that ‖𝑈2 (𝑦1 , 𝑦2 )‖ ≤ (1/2)‖(𝑦1 , 𝑦2 )‖ for (𝑦1 , 𝑦2 ) ∈ 𝐾1 ⋂ 𝜕Ω𝜌1 . Consequently, ‖𝑈(𝑦1 , 𝑦2 )‖ ≤ ‖(𝑦1 , 𝑦2 )‖, whenever (𝑦1 , 𝑦2 ) ∈ 𝐾1 ⋂ 𝜕Ω𝜌1 . Thus, 𝑈 is cone compression on 𝐾1 ⋂ 𝜕Ω𝜌1 .
Since 𝜌∗ > 𝜌 , one has 𝑓(𝑦1 , 𝑦2 ) ≤ 𝑓(𝑦1∗ , 𝑦2∗ ) < 𝜀1 (𝑦1∗ + 𝑦2∗ ) for 0 ≤ ‖(𝑦1 , 𝑦2 )‖ ≤ 𝜌∗ . Then, for (𝑦1 , 𝑦2 ) ∈ 𝐾1 and ‖(𝑦1 , 𝑦2 )‖ = 𝜌∗ , we obtain 𝑈1 (𝑦1 , 𝑦2 ) 1
≤ 𝜙1 (𝑦1 ) + 𝜆 1 ∫ 𝐺1 (𝑡, 𝑠) 𝑎1 (𝑠) 𝑓 (𝑦1 (𝑠) , 𝑦2 (𝑠)) 𝑑𝑠 0
𝑦 𝜀 1 ≤ 1 + 1 (𝑦1 , 𝑦2 ) ≤ (𝑦1 , 𝑦2 ) . 4 4 2
(55)
The Scientific World Journal
9
Case 2. Suppose that 𝑓 is bounded; there, exists 𝐿 1 such that 𝑓(𝑦1 , 𝑦2 ) ≤ 𝐿 1 for all (𝑦1 , 𝑦2 ) ∈ 𝐾1 . Taking 𝜌∗ ≥ max{2𝜌2 , 𝐿 1 }, for (𝑦1 , 𝑦2 ) ∈ 𝐾1 and ‖(𝑦1 , 𝑦2 )‖ = 𝜌∗ , we obtain 𝑈1 (𝑦1 , 𝑦2 ) 1
≤ 𝜙1 (𝑦1 ) + 𝜆 1 ∫ 𝐺1 (𝑡, 𝑠) 𝑎1 (𝑠) 𝑓 (𝑦1 (𝑠) , 𝑦2 (𝑠)) 𝑑𝑠 0
1
≤ 𝜙1 (𝑦1 ) + 𝜆 1 ∫ 𝐺1 (𝑡, 𝑠) 𝑎1 (𝑠) 𝐿 1 𝑑𝑠
(56)
𝜌∗ 1 𝑦 𝐿 ≤ 1 + 1 ≤ = (𝑦1 , 𝑦2 ) . 4 4 2 2
≤
(60)
𝜌1 𝐵3 −1 𝜌 1 + 𝐵3 𝜌1 = 1 = (𝑦1 , 𝑦2 ) . 4 4 2 2
Like Theorem 16, we get ‖𝑈(𝑦1 , 𝑦2 )‖ ≤ ‖(𝑦1 , 𝑦2 )‖ for (𝑦1 , 𝑦2 ) ∈ 𝐾1 ⋂ 𝜕Ω𝜌1 . Now, set Ω𝜌2 := {(𝑦1 , 𝑦2 ) ∈ 𝑋 : ‖(𝑦1 , 𝑦2 )‖ < 𝜌2 }. For (𝑦1 , 𝑦2 ) ∈ 𝐾1 ⋂ 𝜕Ω𝜌2 , one has 𝑦1 (𝑡) + 𝑦2 (𝑡) ≥
min [𝑦1 (𝑡) + 𝑦2 (𝑡)]
≥ 𝛾0 (𝑦1 , 𝑦2 ) = 𝛾0 𝜌2 .
1
× 𝑓 (𝑦1 (𝑠) , 𝑦2 (𝑠)) 𝑑𝑠
Hence, in either case, we always may set Ω𝜌∗ := {(𝑦1 , 𝑦2 ) ∈ 𝑋 : ‖(𝑦1 , 𝑦2 )‖ < 𝜌∗ } such that ‖𝑈1 (𝑦1 , 𝑦2 )‖ ≤ (1/2)‖(𝑦1 , 𝑦2 )‖ for (𝑦1 , 𝑦2 ) ∈ 𝐾1 ⋂ 𝜕Ω𝜌∗ . Like Theorem 16, we get ‖𝑈(𝑦1 , 𝑦2 )‖ ≤ ‖(𝑦1 , 𝑦2 )‖, for (𝑦1 , 𝑦2 ) ∈ 𝐾1 ⋂ 𝜕Ω𝜌∗ . Finally, set Ω𝜌2 := {(𝑦1 , 𝑦2 ) ∈ 𝑋 : ‖(𝑦1 , 𝑦2 )‖ < 𝜌2 }. Then (𝑦1 , 𝑦2 ) ∈ 𝐾1 ⋂ 𝜕Ω𝜌2 implies 𝑡∈[(1/2),1]
Proof. With loss of generality, we may assume that 𝜌1 < 𝜌2 . Let Ω𝜌1 := {(𝑦1 , 𝑦2 ) ∈ 𝑋 : ‖(𝑦1 , 𝑦2 )‖ < 𝜌1 }. For (𝑦1 , 𝑦2 ) ∈ 𝐾1 ⋂ 𝜕Ω𝜌1 , from (𝐻7 ), (𝐷1 ), one has 𝑈1 (𝑦1 , 𝑦2 ) ≤ 𝜙1 (𝑦1 ) + ∫ 𝐺1 (𝑡, 𝑠) 𝑎1 (𝑠) 0
0
𝑦1 (𝑡) + 𝑦2 (𝑡) ≥
Theorem 18. Suppose that (𝐻5 ), (𝐻6 ), and (𝐷1 ) are satisfied. Then problem (1), (3), in the case where 𝜆 1 = 𝜆 2 = 1, has at least one positive solution (𝑦10 , 𝑦20 ) such that ‖(𝑦1 , 𝑦2 )‖ between 𝜌1 and 𝜌2 .
(57)
min [𝑦1 (𝑡) + 𝑦2 (𝑡)]
𝑡∈[(1/2),1]
(61)
≥ 𝛾0 (𝑦1 , 𝑦2 ) = 𝛾0 𝜌2 . Thus, from (𝐻8 ), we get
Hence we have 𝑈1 (𝑦1 , 𝑦2 ) (1) ≥ 𝜆 1 ∫
1
1/2
𝐺1 (1, 𝑠) 𝑎1 (𝑠) 𝐴−1 2 𝜌2 𝑑𝑠
𝜌2 1 𝐴 ≥ 2 𝐴−1 = (𝑦1 , 𝑦2 ) . 2 𝜌2 = 2 2 2
𝑈1 (𝑦1 , 𝑦2 ) (1) ≥ 𝜆 1 ∫
0
0
𝐵3 := max {4 ∫ 𝐺1 (1, 𝑠) 𝑎1 (𝑠) 𝑑𝑠, 4 ∫ 𝐺2 (1, 𝑠) 𝑎2 (𝑠) 𝑑𝑠} . (59)
Assume that there exist two positive constants 𝜌1 ≠ 𝜌2 such that (𝐻5 ) 𝑓(𝑦1 , 𝑦2 ), 𝑔(𝑦1 , 𝑦2 ) ≤ 𝐵3−1 𝜌1 for 0 ≤ ‖(𝑦1 , 𝑦2 )‖ ≤ 𝜌1 ; (𝐻6 ) 𝑓(𝑦1 , 𝑦2 ), 𝑔(𝑦1 , 𝑦2 ) ≥ 𝜌2 .
𝐵2−1 𝜌2
𝐺1 (1, 𝑠) 𝑎1 (𝑠) 𝐵2−1 𝜌2 𝑑𝑠
𝜌 𝐵 1 ≥ 2 𝐵2−1 𝜌2 = 2 = (𝑦1 , 𝑦2 ) . 2 2 2
3.4. Problem (1), (3) in Case 𝜆 1 = 𝜆 2 = 1. In [21], the author obtained that problem (1), (3) with 𝜆 1 = 𝜆 2 = 1 having at least one positive solution. In the following, we also establish the existence of one positive solution to problem (1), (3) with 𝜆 1 = 𝜆 2 = 1 under different conditions. For the sake of convenience, set 1
1/2
(58)
So, ‖𝑈1 (𝑦1 , 𝑦2 )‖ ≥ (1/2)‖(𝑦1 , 𝑦2 )‖ for (𝑦1 , 𝑦2 ) ∈ 𝐾1 ⋂ 𝜕Ω𝜌2 . Like Theorem 16, we get ‖𝑈(𝑦1 , 𝑦2 )‖ ≥ ‖(𝑦1 , 𝑦2 )‖ for (𝑦1 , 𝑦2 ) ∈ 𝐾1 ⋂ 𝜕Ω𝜌2 . So, from Lemma 5, 𝑈 has a fixed point (𝑦10 , 𝑦20 ) ∈ 𝐾1 ⋂(Ω𝜌2 \ Ω𝜌 ) and a fixed point (𝑦1 , 𝑦2 ) ∈ 𝐾1 ⋂(Ω𝜌∗ \ Ω𝜌2 ). Both are positive solutions of BVP (1), (3) with 0 < ‖(𝑦10 , 𝑦20 )‖ < 𝜌2 < ‖(𝑦1 , 𝑦2 )‖, which complete the proof.
1
1
for 𝛾0 𝜌2 ≤ ‖(𝑦1 , 𝑦2 )‖ ≤
(62)
Like Theorem 16, we get ‖𝑈(𝑦1 , 𝑦2 )‖ ≥ ‖(𝑦1 , 𝑦2 )‖ for (𝑦1 , 𝑦2 ) ∈ 𝐾1 ⋂ 𝜕Ω𝜌2 . Hence, from Lemma 5, we complete the proof.
4. An Example To illustrate how our main results can be used in practice, we present one example. Example 19. Consider the following BVP, for 𝑡 ∈ (0, 1): 1/2
−2𝑡 − 𝐷05.2 [(𝑦1 (𝑡) + 𝑦2 (𝑡)) + 𝑦1 (𝑡) = 164500𝑒
2
+ (𝑦1 (𝑡) + 𝑦2 (𝑡)) ] , −3𝑡 [(𝑦1 (𝑡) + 𝑦2 (𝑡)) − 𝐷05.95 + 𝑦2 (𝑡) = 164000𝑒
(63)
1/3 3
+ (𝑦1 (𝑡) + 𝑦2 (𝑡)) ] , subject to the boundary conditions 𝑦1(𝑖) (0) = 𝑦2(𝑖) (0) = 0,
0 ≤ 𝑖 ≤ 4,
1.5 1.5 𝐷0+ [𝑦1 (𝑡)]𝑡=1 = 𝐷0+ [𝑦2 (𝑡)]𝑡=1 = 0.
(64) (65)
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The Scientific World Journal
Obviously, problem (63)–(65) fits the framework of problem (1)-(2) with ]1 := 5.2,
]2 := 5.95,
𝜆 1 = 164500,
𝛼 = 1.5,
𝜆 2 = 164000,
𝑛 = 6.
(66)
In addition, we have set 1/2
+ (𝑦1 + 𝑦2 ) ,
1/3
+ (𝑦1 + 𝑦2 ) ,
𝑓 (𝑦1 , 𝑦2 ) := (𝑦1 + 𝑦2 ) 𝑔 (𝑦1 , 𝑦2 ) := (𝑦1 + 𝑦2 ) 𝑎1 (𝑡) := 𝑒−2𝑡 ,
2
3
(67)
𝑎2 (𝑡) := 𝑒−3𝑡 .
We can see that 𝑓, 𝑔 : [0, +∞) × [0, +∞) → [0, +∞) and are continuous. The functions 𝑎1 (𝑡) and 𝑎2 (𝑡) are obviously nonnegative. Now, observe that 𝑓0 = 𝑓∞ = 𝑔0 = 𝑔∞ = ∞ holds. Again set 𝐴 1 = 1/850, because 𝑓(𝑦1 , 𝑦2 ), 𝑔(𝑦1 , 𝑦2 ) is monotone increasing function for (𝑦1 , 𝑦2 ) ≥ 0, taking 𝜌1 = 4; then, when ‖(𝑦1 , 𝑦2 )‖ ∈ [0, 𝜌1 ], we get 1/2 2 𝑓 (𝑦1 , 𝑦2 ) ≤ (𝑦1 , 𝑦2 ) + (𝑦1 , 𝑦2 ) ≤ 2 + 16 = 18 < 4 × 850 = 𝐴−1 1 𝜌1 1/3 3 𝑔 (𝑦1 , 𝑦2 ) ≤ (𝑦1 , 𝑦2 ) + (𝑦1 , 𝑦2 )
(68)
≤ 41/3 + 64 < 4 × 850 = 𝐴−1 1 𝜌1 , which implies that (𝐻1 ) holds. On the other hand, to calculate the admissible range of the eigenvalues 𝜆 1 , 𝜆 2 , as given by condition (𝑃1 ), observe by numerical approximation, we find that Λ 1 ≈ 163530,
Λ 2 ≈ 164547.25.
(69)
Thus, for any 𝜆 1 , 𝜆 2 satisfying 163530 < 𝜆 1 , 𝜆 2 < 164547.25, condition (𝑃1 ) will be satisfied. Consequently, by Theorem 9, problem (63)–(65) has at least two positive solutions.
Conflict of Interests The authors declare that they have no competing interests.
Authors’ Contribution The authors declare that the study was realized in collaboration with the same responsibility. All authors read and approved the final manuscript.
Acknowledgments The first author was supported financially by the Youth Science Foundations of China (11201272) and Shanxi Province (2010021002-1).
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The Scientific World Journal [18] M. El-Shahed, “Positive solutions for boundary value problems of nonlinear fractional differential equation,” Abstract and Applied Analysis, vol. 2007, Article ID 10368, 8 pages, 2007. [19] S. H. Liang and J. H. Zhang, “Existence and uniqueness of strictly nondecreasing and positive solution for a fractional three-point boundary value problem,” Computers and Mathematics with Applications, vol. 62, no. 3, pp. 1333–1340, 2011. [20] Y. G. Zhao, S. R. Sun, Z. L. Han, and Q. P. Li, “The existence of multiple positive solutions for boundary value problems of nonlinear fractional differential equations,” Communications in Nonlinear Science and Numerical Simulation, vol. 16, no. 4, pp. 2086–2097, 2011. [21] C. S. Goodrich, “Existence of a positive solution to systems of differential equations of fractional order,” Computers and Mathematics with Applications, vol. 62, no. 3, pp. 1251–1268, 2011. [22] X. W. Su, “Boundary value problem for a coupled system of nonlinear fractional differential equations,” Applied Mathematics Letters, vol. 22, no. 1, pp. 64–69, 2009. [23] I. Podlubny, Fractional Differential Equations, Academic Press, New York, NY, USA, 1999. [24] C. S. Goodrich, “Existence of a positive solution to a class of fractional differential equations,” Applied Mathematics Letters, vol. 23, no. 9, pp. 1050–1055, 2010. [25] D. Guo and V. Lakshmikantham, Nonlinear Problems in Abstract Cones, Academic Press, New York, NY, USA, 1988. [26] D. R. Dunninger and H. Wang, “Existence and multiplicity of positive solutions for elliptic systems,” Nonlinear Analysis, Theory, Methods and Applications, vol. 29, no. 9, pp. 1051–1060, 1997.
11
Research Article Multiple Positive Solutions to Nonlinear Boundary Value Problems of a System for Fractional Differential Equations Chengbo Zhai and Mengru Hao School of Mathematical Sciences, Shanxi University, Taiyuan, Shanxi 030006, China Correspondence should be addressed to Chengbo Zhai; [email protected] Received 5 August 2013; Accepted 3 November 2013; Published 23 January 2014 Academic Editors: L. Acedo and N. Hussain Copyright © 2014 C. Zhai and M. Hao. 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. By using Krasnoselskii’s fixed point theorem, we study the existence of at least one or two positive solutions to a system of fractional ] ] boundary value problems given by −𝐷01+ 𝑦1 (𝑡) = 𝜆 1 𝑎1 (𝑡)𝑓(𝑦1 (𝑡), 𝑦2 (𝑡)), −𝐷02+ 𝑦2 (𝑡) = 𝜆 2 𝑎2 (𝑡)𝑔(𝑦1 (𝑡), 𝑦2 (𝑡)), where 𝐷0]+ is the standard Riemann-Liouville fractional derivative, ]1 , ]2 ∈ (𝑛 − 1, 𝑛] for 𝑛 > 3 and 𝑛 ∈ 𝑁, subject to the boundary conditions 𝑦1(𝑖) (0) = 0 = 𝑦2(𝑖) (0), for 0 ≤ 𝑖 ≤ 𝑛 − 2, and [𝐷0𝛼+ 𝑦1 (𝑡)]𝑡=1 = 0 = [𝐷0𝛼+ 𝑦2 (𝑡)]𝑡=1 , for 1 ≤ 𝛼 ≤ 𝑛 − 2, or 𝑦1(𝑖) (0) = 0 = 𝑦2(𝑖) (0), for 0 ≤ 𝑖 ≤ 𝑛 − 2, and [𝐷0𝛼+ 𝑦1 (𝑡)]𝑡=1 = 𝜙1 (𝑦1 ), [𝐷0𝛼+ 𝑦2 (𝑡)]𝑡=1 = 𝜙2 (𝑦2 ), for 1 ≤ 𝛼 ≤ 𝑛 − 2, 𝜙1 , 𝜙2 ∈ 𝐶([0, 1], 𝑅). Our results are new and complement previously known results. As an application, we also give an example to demonstrate our result.
[𝐷0𝛼+ 𝑦1 (𝑡)]𝑡=1 = 𝜙1 (𝑦1 ) ,
1. Introduction
[𝐷0𝛼+ 𝑦2 (𝑡)]𝑡=1 = 𝜙2 (𝑦2 ) ,
The purpose of this paper is to consider the existence of multiple positive solutions for the following system of nonlinear fractional differential equations:
1 ≤ 𝛼 ≤ 𝑛 − 2, (3)
−
] 𝐷01+ 𝑦1
(𝑡) = 𝜆 1 𝑎1 (𝑡) 𝑓 (𝑦1 (𝑡) , 𝑦2 (𝑡)) ,
−
] 𝐷02+ 𝑦2
(𝑡) = 𝜆 2 𝑎2 (𝑡) 𝑔 (𝑦1 (𝑡) , 𝑦2 (𝑡)) ,
(1)
where 𝑡 ∈ (0, 1), ]1 , ]2 ∈ (𝑛 − 1, 𝑛] for 𝑛 > 3 and 𝑛 ∈ 𝑁, and 𝜆 1 , 𝜆 2 > 0, subject to a couple of boundary conditions. In particular, we first consider (1) subject to 𝑦1(𝑖) (0) = 0 = 𝑦2(𝑖) (0) ,
0 ≤ 𝑖 ≤ 𝑛 − 2,
[𝐷0𝛼+ 𝑦1 (𝑡)]𝑡=1 = 0 = [𝐷0𝛼+ 𝑦2 (𝑡)]𝑡=1 ,
1 ≤ 𝛼 ≤ 𝑛 − 2,
(2)
where 𝑓, 𝑔 ∈ 𝐶([0, ∞) × [0, ∞), [0, ∞)), and 𝑎1 , 𝑎2 ∈ 𝐶([0, 1], [0, ∞)). We then consider the case in which the boundary conditions are changed to 𝑦1(𝑖) (0) = 0 = 𝑦2(𝑖) (0) ,
0 ≤ 𝑖 ≤ 𝑛 − 2,
where 𝜙1 , 𝜙2 ∈ 𝐶([0, 1], 𝑅). Fractional differential equations arise in many fields, such as physics, mechanics, chemistry, economics, and engineering and biological sciences; see [1–11] for example. In recent years, the study of positive solutions for fractional differential equation boundary value problems has attracted considerable attention, and fruits from research into it emerge continuously. For a small sample of such work, we refer the reader to [12–20] and the references therein. The situation of at least one positive solution has been studied in many excellent monograph; see [12–19, 21] and other references therein. In [22], by means of Schauder fixed point theorem, Su investigated the existence of one positive solution to the following boundary value problem for a coupled system of nonlinear fractional differential equations: 𝜇
𝐷0𝛼+ 𝑦1 (𝑡) = 𝑓 (𝑡, 𝑦2 (𝑡) , 𝐷0+ 𝑦2 (𝑡)) , 𝛽
−𝐷0+ 𝑦2 (𝑡) = 𝑔 (𝑡, 𝑦1 (𝑡) , 𝐷0]+ 𝑦1 (𝑡)) ,
0 < 𝑡 < 1, 0 < 𝑡 < 1,
2
The Scientific World Journal 𝑦1 (0) = 𝑦1 (1) = 𝑦2 (0) = 𝑦2 (1) = 0,
(ii) 𝐺(𝑡, 𝑠) ≥ 0 for each (𝑡, 𝑠) ∈ [0, 1] × [0, 1]; (4)
where 1 < 𝛼, 𝛽 < 2, 𝜇, ] > 0, 𝛼 − ] ≥ 1, 𝛽 − 𝜇 ≥ 1. In [21], Goodrich established the existence of one positive solution to problems (1)-(2) and (1), (3) by using Krasnoselskii’s fixed point theorem. Different from the above works mentioned, in this paper we will present the existence of at least two positive solutions to problems (1)-(2) and (1), (3) by using the similar method presented in [21]. Moreover, under different conditions, we also present the existence of at least one positive solution to problems (1)-(2) and (1), (3) with 𝜆 1 = 𝜆 2 = 1.
2. Preliminaries For the convenience of the reader, we present here some definitions, lemmas, and basic results that will be used in the proofs of our theorems. Definition 1 (see [23]). Let ] > 0 with ] ∈ 𝑅. Suppose that 𝑦 : [𝑎, +∞) → 𝑅. Then the ]th Riemann-Liouville fractional integral is defined to be 𝐷𝑎−]+ 𝑦 (𝑡) :=
𝑡
1 ∫ 𝑦 (𝑠) (𝑡 − 𝑠)]−1 𝑑𝑠, Γ (]) 𝑎
(5)
whenever the right-hand side is defined. Similarly, with ] > 0 and ] ∈ 𝑅, we define the ]th Riemann-Liouville fractional derivative to be 𝐷𝑎] + 𝑦 (𝑡) :=
𝑦 (𝑠) 𝑑𝑛 𝑡 1 𝑑𝑠, ∫ 𝑛 Γ (𝑛 − ]) 𝑑𝑡 𝑎 (𝑡 − 𝑠)]+1−𝑛
(6)
where 𝑛 ∈ 𝑁 is the unique positive integer satisfying 𝑛 − 1 ≤ ] < 𝑛 and 𝑡 > 𝑎. Lemma 2 (see [24]). Let 𝑔 ∈ 𝐶𝑛 ([0, 1]) be given. Then the unique solution to problem −𝐷0]+ 𝑦(𝑡) = 𝑔(𝑡) together with the boundary conditions 𝑦(𝑖) (0) = 0 = [𝐷0𝛼+ 𝑦(𝑡)]𝑡=1 , where 1 ≤ 𝛼 ≤ 𝑛 − 2 and 0 ≤ 𝑖 ≤ 𝑛 − 2, is 1
𝑦 (𝑡) = ∫ 𝐺 (𝑡, 𝑠) 𝑔 (𝑠) 𝑑𝑠, 0
(7)
(iii) max𝑡∈[0,1] 𝐺(𝑡, 𝑠) = 𝐺(1, 𝑠), for each 𝑠 ∈ [0, 1]. Lemma 4 (see [24]). Let 𝐺(𝑡, 𝑠) be as given in the statement of Lemma 2. Then there exists a constant 𝛾 ∈ (0, 1) such that min 𝐺 (𝑡, 𝑠) ≥ 𝛾 max 𝐺 (𝑡, 𝑠) = 𝛾𝐺 (1, 𝑠) .
𝑡∈[(1/2),1]
To prove our results, we need the following Krasnoselskii’s fixed point theorem which can be seen in Guo and Lakshmikantham [25]. Lemma 5 (see [25]). Let 𝐸 be a Banach space, and let 𝑃 be a cone. Assume that Ω1 , Ω2 are open bounded subsets of 𝐸 with 0 ∈ Ω1 , Ω1 ⊂ Ω2 , and let 𝑇 : 𝑃 ⋂(Ω2 \ Ω1 ) → 𝑃 be a completely continuous operator such that (i) ‖𝑇𝑢‖ ≤ ‖𝑢‖, ∀𝑢 ∈ 𝑃 ⋂ 𝜕Ω1 , and ‖𝑇𝑢‖ ≥ ‖𝑢‖, ∀𝑢 ∈ P ⋂ 𝜕Ω2 ; or (ii) ‖𝑇𝑢‖ ≥ ‖𝑢‖, ∀𝑢 ∈ 𝑃 ⋂ 𝜕Ω1 , and ‖𝑇𝑢‖ ≤ ‖𝑢‖, ∀𝑢 ∈ 𝑃 ⋂ 𝜕Ω2 . Then 𝑇 has a fixed point in 𝑃 ⋂(Ω2 \ Ω1 ).
3. Main Results In this section, we apply Lemma 5 to study problems (1)-(2) and (1), (3), and we obtain some new results on the existence of multiple positive solutions. 3.1. Problem (1)-(2) in the General Case. In our considerations, let 𝐸 represent the Banach space of 𝐶([0, 1]) when equipped with the usual supremum norm, ‖ ⋅ ‖. Then put 𝑋 := 𝐸 × 𝐸, where 𝑋 is equipped with the norm ‖(𝑦1 , 𝑦2 )‖ := ‖𝑦1 ‖ + ‖𝑦2 ‖ for (𝑦1 , 𝑦2 ) ∈ 𝑋. Observe that 𝑋 is also a Banach space (see [26]). In addition, we define two operators 𝑇1 , 𝑇2 : 𝑋 → 𝐸 by 1
𝑇1 (𝑦1 , 𝑦2 ) (𝑡) := 𝜆 1 ∫ 𝐺1 (𝑡, 𝑠) 𝑎1 (𝑠) 𝑓 (𝑦1 (𝑠) , 𝑦2 (𝑠)) 𝑑𝑠, 0
1
𝑇2 (𝑦1 , 𝑦2 ) (𝑡) := 𝜆 2 ∫ 𝐺2 (𝑡, 𝑠) 𝑎2 (𝑠) 𝑔 (𝑦1 (𝑠) , 𝑦2 (𝑠)) 𝑑𝑠, 0
where 𝑡]−1 (1 − 𝑠)]−𝛼−1 − (𝑡 − 𝑠)]−1 { , 0≤𝑠≤𝑡≤1 { { { Γ (]) 𝐺 (𝑡, 𝑠) = { ]−1 { { 𝑡 (1 − 𝑠)]−𝛼−1 { , 0≤𝑡≤𝑠≤1 Γ (]) { (8) is the Green function for this problem. Lemma 3 (see [24]). Let 𝐺(𝑡, 𝑠) be as given in the statement of Lemma 2. Then one finds that (i) 𝐺(𝑡, 𝑠) is a continuous function on the unit square [0, 1] × [0, 1];
(9)
𝑡∈[0,1]
(10)
where 𝐺1 (𝑡, 𝑠) is the Green function of Lemma 2 with ] replaced by ]1 and, likewise, 𝐺2 (𝑡, 𝑠) is the Green function of Lemma 2 with ] replaced by ]2 . Now, we define an operator 𝑆 : 𝑋 → 𝑋 by 𝑆 (𝑦1 , 𝑦2 ) (𝑡) := (𝑇1 (𝑦1 , 𝑦2 ) (𝑡) , 𝑇2 (𝑦1 , 𝑦2 ) (𝑡)) 1
= (𝜆 1 ∫ 𝐺1 (𝑡, 𝑠) 𝑎1 (𝑠) 𝑓 (𝑦1 (𝑠) , 𝑦2 (𝑠)) 𝑑𝑠, 0
1
𝜆 2 ∫ 𝐺2 (𝑡, 𝑠) 𝑎2 (𝑠) 𝑔 (𝑦1 (𝑠) , 𝑦2 (𝑠)) 𝑑𝑠) . 0
(11)
The Scientific World Journal
3
We claim that whenever (𝑦1 , 𝑦2 ) ∈ 𝑋 is a fixed point of the operator defined in (11), it follows that 𝑦1 (𝑡) and 𝑦2 (𝑡) solve problems (1)-(2). That is, a pair of functions 𝑦1 , 𝑦2 ∈ 𝑋 is a solution of problems (1)-(2) if and only if 𝑦1 , 𝑦2 is a fixed point of the operator 𝑆 defined in (11) (see [26]). In the following, we will look for fixed points of the operator 𝑆, because these fixed points coincide with solutions of problems (1)-(2). For use in the sequel, let 𝛾1 and 𝛾2 be the constants given by Lemma 4 associated, respectively, with the Green functions 𝐺1 and 𝐺2 , and define 𝛾̃ by 𝛾̃ := min{𝛾1 , 𝛾2 }, and notice that 𝛾̃ ∈ (0, 1). For the sake of convenience, we set 𝑓0 = 𝑔0 =
1
≤ 𝜆 1 ∫ 𝐺1 (1, 𝑠) 𝑎1 (𝑠) 𝑓 (𝑦1 (𝑠) , 𝑦2 (𝑠)) 𝑑𝑠 0
0
1
𝑓 (𝑦1 , 𝑦2 ) , + + (𝑦1 ,𝑦2 ) → (0 ,0 ) 𝑦1 + 𝑦2 lim
(𝑦1 ,𝑦2 ) → (0+ ,0+ )
≤ 𝐿Φ2 ∫ 𝐺1 (1, 𝑠) 𝑑𝑠 < +∞. 0
𝑔 (𝑦1 , 𝑦2 ) , 𝑦1 + 𝑦2
lim
(𝑦1 ,𝑦2 ) → (∞,∞)
(15)
1
≤ 𝜆 1 ∫ 𝐺1 (1, 𝑠) 𝐿𝑑𝑠
lim
(12)
𝑓 (𝑦1 , 𝑦2 ) lim , 𝑓∞ = (𝑦1 ,𝑦2 ) → (∞,∞) 𝑦1 + 𝑦2 𝑔∞ =
Let Ω ⊆ 𝐾 be bounded; that is, there exists a positive constant 𝑀 > 0 such that ‖(𝑦1 , 𝑦2 )‖ ≤ 𝑀, for all (𝑦1 , 𝑦2 ) ∈ Ω. Let 𝐿 = max0≤𝑡≤1, 0≤‖(𝑦1 ,𝑦2 )‖≤𝑀|𝑎1 (𝑡)𝑓(𝑦1 (𝑡), 𝑦2 (𝑡))| + 1; then, for (𝑦1 , 𝑦2 ) ∈ Ω, we have 𝑇1 (𝑦1 , 𝑦2 ) (𝑡)
𝑔 (𝑦1 , 𝑦2 ) . 𝑦1 + 𝑦2
Now we list some assumptions: (F 1 ) 𝑓0 , 𝑔0 ∈ (0, +∞); (F 2 ) 𝑓∞ , 𝑔∞ ∈ (0, +∞); (F 3 ) there are numbers Φ1 , Φ2 , where
Hence, 𝑇1 (Ω) is bounded. On the other hand, given 𝜀 > 0, setting 𝛿 = min{(1/2)(Γ(]1 )𝜀/𝐿Φ2 )1/(]1 −1) , 𝜀Γ(]1 )/(]1 − 1)𝐿Φ2 }, then, for each (𝑦1 , 𝑦2 ) ∈ Ω, 𝑡1 , 𝑡2 ∈ [0, 1], 𝑡1 < 𝑡2 , and 𝑡2 − 𝑡1 < 𝛿, one has |𝑇1 (𝑦1 , 𝑦2 )(𝑡2 ) − 𝑇1 (𝑦1 , 𝑦2 )(𝑡1 )| < 𝜀. That is to say, 𝑇(Ω) is equicontinuity. In fact, 𝑇1 (𝑦1 , 𝑦2 ) (𝑡2 ) − 𝑇1 (𝑦1 , 𝑦2 ) (𝑡1 ) 1 = 𝜆 1 ∫ 𝐺1 (𝑡2 , 𝑠) 𝑎1 (𝑠) 𝑓 (𝑦1 (𝑠) , 𝑦2 (𝑠)) 𝑑𝑠 0 1 −𝜆 1 ∫ 𝐺1 (𝑡1 , 𝑠) 𝑎1 (𝑠) 𝑓 (𝑦1 (𝑠) , 𝑦2 (𝑠)) 𝑑𝑠 0 𝑡1 = 𝜆 1 ∫ [𝐺1 (𝑡2 , 𝑠) − 𝐺1 (𝑡1 , 𝑠)] 𝑎1 (𝑠) 𝑓 (𝑦1 (𝑠) , 𝑦2 (𝑠)) 𝑑𝑠 0
−1
1 1 Φ1 := max { [∫ 𝛾̃𝐺1 (1, 𝑠) 𝑎1 (𝑠) 𝑓∞ 𝑑𝑠] , 2 1/2
𝑡2
+ 𝜆 1 ∫ [𝐺1 (𝑡2 , 𝑠) − 𝐺1 (𝑡1 , 𝑠)] 𝑎1 (𝑠)
−1
1 1 [∫ 𝛾̃𝐺 (1, 𝑠) 𝑎2 (𝑠) 𝑔∞ 𝑑𝑠] } , 2 1/2 2 −1
1 1 Φ2 := min { [∫ 𝐺1 (1, 𝑠) 𝑎1 (𝑠) 𝑓0 𝑑𝑠] , 2 0
𝑡1
× 𝑓 (𝑦1 (s) , 𝑦2 (𝑠)) 𝑑𝑠
(13)
1
+ 𝜆 1 ∫ [𝐺1 (𝑡2 , 𝑠) − 𝐺1 (𝑡1 , 𝑠)] 𝑎1 (𝑠) 𝑡2
× 𝑓 (𝑦1 (𝑠) , 𝑦2 (𝑠)) 𝑑𝑠
−1
1 1 [∫ 𝐺 (1, 𝑠) 𝑎2 (𝑠) 𝑔0 𝑑𝑠] } , 2 0 2 ≤
such that Φ1 < 𝜆 1 , 𝜆 2 < Φ2 . Next, we define the cone 𝐾 by
𝑡1 Φ2 𝐿 ] −1 ] −1 [ ∫ (1 − 𝑠)]1 −𝛼−1 (𝑡21 − 𝑡11 ) 𝑑𝑠 Γ (]1 ) 0 1
𝐾 := {(𝑦1 , 𝑦2 ) ∈ 𝑋 : 𝑦1 , 𝑦2 ≥ 0, min [𝑦1 (𝑡) + 𝑦2 (𝑡)] ≥ 𝛾̃ (𝑦1 , 𝑦2 )} .
] −1
− 𝑡11 ) 𝑑𝑠
] −1
− 𝑡11 ) 𝑑𝑠]
+ ∫ (1 − 𝑠)]1 −𝛼−1 (𝑡21 𝑡2
(14)
𝑡2
+ ∫ (1 − 𝑠)]1 −𝛼−1 (𝑡21 𝑡1
𝑡∈[(1/2),1]
Lemma 6 (see [21]). Let 𝑆 be the operator defined by (11). Then 𝑆 : 𝐾 → 𝐾. Lemma 7. 𝑆 is a completely continuous operator. Proof. The operator 𝑇1 : 𝐾 → 𝐸 is continuous in view of nonnegativeness and continuity of 𝐺1 (𝑡, 𝑠), 𝑓(𝑦1 , 𝑦2 ) and 𝑎1 (𝑡).
=
Φ2 𝐿 1 ] −1 ] −1 (𝑡 1 − 𝑡11 ) Γ (]1 ) ]1 − 𝛼 2
≤
Φ2 𝐿 ]1 −1 ]1 −1 − 𝑡1 ) . (𝑡 Γ (]1 ) 2
] −1
] −1
(16) In the following, we divide the proof into two cases.
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Case 1. If 𝛿 ≤ 𝑡1 < 𝑡2 < 1, then we have
(𝑃1 ) there are numbers Λ 1 , Λ 2 , where
𝑇1 (𝑦1 , 𝑦2 ) (𝑡2 ) − 𝑇1 (𝑦1 , 𝑦2 ) (𝑡1 ) ≤
Φ2 𝐿 ]1 −1 ]1 −1 − 𝑡1 ) (𝑡 Γ (]1 ) 2
=
Φ2 𝐿 ] −2 (] − 1) (𝑡2 − 𝑡1 ) 𝑡𝜉1 Γ (]1 ) 1
0, such that 𝑓 (𝑦1 , 𝑦2 ) , 𝑔 (𝑦1 , 𝑦2 ) < 𝐴−1 1 𝜌1
for 0 ≤ (𝑦1 , 𝑦2 ) ≤ 𝜌1 ; (19)
(𝐻2 ) there exist constants 𝜌2 , 𝐴 2 > 0, such that 𝑓 (𝑦1 , 𝑦2 ) , 𝑔 (𝑦1 , 𝑦2 ) ≥
𝐴−1 2 𝜌2
for 𝛾̃𝜌2 ≤ (𝑦1 , 𝑦2 ) ≤ 𝜌2 ; (20)
1 1 [∫ 𝐺 (1, 𝑠) 𝑎2 (𝑠) 𝑑𝑠] } , 2 0 2
such that Λ 3 < 𝜆 1 , 𝜆 2 < Λ 4 . Theorem 9. Suppose that 𝑓0 = 𝑓∞ = 𝑔0 = 𝑔∞ = ∞ and (𝐻1 ), (𝑃1 ) are satisfied. Then problem (1)-(2) has at least two positive solutions (𝑦10 , 𝑦20 ), (𝑦1 , 𝑦2 ), such that 0 < ‖(𝑦10 , 𝑦20 )‖ < 𝜌1 < ‖(𝑦1 , 𝑦2 )‖. Proof. From Lemma 7, 𝑆 is a completely continuous operator. At first, in view of 𝑓0 = 𝑔0 = ∞, we have 𝑓(𝑦1 , 𝑦2 ) ≥ 𝑀(𝑦1 + 𝑦2 ), for 0 < ‖(𝑦1 , 𝑦2 )‖ < 𝑟1∗ < 𝜌1 ; 𝑔(𝑦1 , 𝑦2 ) ≥ 𝑀(𝑦1 + 𝑦2 ), for 0 < ‖(𝑦1 , 𝑦2 )‖ < 𝑟2∗ < 𝜌1 , where 𝑀 satisfies 𝑀 ≥ 1. Set 𝜌0 := min{𝑟1∗ , 𝑟2∗ }. So we define Ω𝜌0 by Ω𝜌0 := {(𝑦1 , 𝑦2 ) ∈ 𝑋 : ‖(𝑦1 , 𝑦2 )‖ < 𝜌0 }. Then for each (𝑦1 , 𝑦2 ) ∈ 𝐾 ⋂ 𝜕Ω𝜌0 , we find that 1
𝑇1 (𝑦1 , 𝑦2 ) (1) = 𝜆 1 ∫ 𝐺1 (1, 𝑠) 𝑎1 (𝑠) 𝑓 (𝑦1 (𝑠) , 𝑦2 (𝑠)) 𝑑𝑠 0
≥ 𝜆1 ∫
1
1/2
𝐺1 (1, 𝑠) 𝑎1 (𝑠) 𝑓 (𝑦1 (𝑠) , 𝑦2 (𝑠)) 𝑑𝑠
The Scientific World Journal ≥ 𝜆1 ∫
1
1/2
≥ 𝜆1 ∫
1
1/2
≥
5
𝐺1 (1, 𝑠) 𝑎1 (𝑠) 𝑀 (𝑦1 (𝑠) + 𝑦2 (𝑠)) 𝑑𝑠 𝛾̃𝐺1 (1, 𝑠) 𝑎1 (𝑠) 𝑀 (𝑦1 , 𝑦2 ) 𝑑𝑠
1
𝑇1 (𝑦1 , 𝑦2 ) = 𝜆 1 ∫ 𝐺1 (1, 𝑠) 𝑎1 (𝑠) 𝑓 (𝑦1 (𝑠) , 𝑦2 (𝑠)) 𝑑𝑠 0
𝑀 1 (𝑦1 , 𝑦2 ) ≥ (𝑦1 , 𝑦2 ) . 2 2 (23)
So ‖𝑇1 (𝑦1 , 𝑦2 )‖ ≥ (1/2)‖(𝑦1 , 𝑦2 )‖ for (𝑦1 , 𝑦2 ) ∈ 𝐾 ⋂ 𝜕Ω𝜌0 . Similarly, we find that ‖𝑇2 (𝑦1 , 𝑦2 )‖ ≥ (1/2)‖(𝑦1 , 𝑦2 )‖ for (𝑦1 , 𝑦2 ) ∈ 𝐾 ⋂ 𝜕Ω𝜌0 . Consequently, 𝑆 (𝑦1 , 𝑦2 ) = (𝑇1 (𝑦1 , 𝑦2 ) , 𝑇2 (𝑦1 , 𝑦2 )) = 𝑇1 (𝑦1 , 𝑦2 ) + 𝑇2 (𝑦1 , 𝑦2 ) ≥ (𝑦1 , 𝑦2 ) ,
(24)
whenever (𝑦1 , 𝑦2 ) ∈ 𝐾 ⋂ 𝜕Ω𝜌0 . Thus, 𝑆 is cone expansion on 𝐾 ⋂ 𝜕Ω𝜌0 . Next, since 𝑓∞ = 𝑔∞ = ∞, we have 𝑓(𝑦1 , 𝑦2 ) ≥ 𝑀1 (𝑦1 + 𝑦2 ) for 𝑦1 + 𝑦2 ≥ 𝑟1∗∗ > 𝜌1 ; 𝑔(𝑦1 , 𝑦2 ) ≥ 𝑀1 (𝑦1 + 𝑦2 ) for 𝑦1 + 𝑦2 ≥ 𝑟2∗∗ > 𝜌1 , where 𝑀1 satisfies 𝑀1 ≥ 1. Set 𝛾)} and Ω𝜌0∗ := 𝜌10 := max{𝑟1∗∗ , 𝑟2∗∗ }. Let 𝜌0∗ = max{2𝜌1 , (𝜌10 /̃ {(𝑦1 , 𝑦2 ) ∈ 𝑋 : ‖(𝑦1 , 𝑦2 )‖ < 𝜌0∗ }. Then (𝑦1 , 𝑦2 ) ∈ 𝐾 ⋂ 𝜕Ω𝜌0∗ implies 𝑦1 (𝑡) + 𝑦2 (𝑡) ≥
≥ 𝛾̃ (𝑦1 , 𝑦2 ) = 𝛾̃𝜌0∗ ≥ 𝜌10 .
(25)
So we obtain 1
𝑇1 (𝑦1 , 𝑦2 ) (1) = 𝜆 1 ∫ 𝐺1 (1, 𝑠) 𝑎1 (𝑠) 𝑓 (𝑦1 (𝑠) , 𝑦2 (𝑠)) 𝑑𝑠 0
1
≥ 𝜆1 ∫
1/2 1
≥ 𝜆1 ∫
1/2 1
≥ 𝜆1 ∫
1/2
𝐺1 (1, 𝑠) 𝑎1 (𝑠) 𝑓 (𝑦1 (𝑠) , 𝑦2 (𝑠)) 𝑑𝑠 𝐺1 (1, 𝑠) 𝑎1 (𝑠) 𝑀1 (𝑦1 (𝑠) + 𝑦2 (𝑠)) 𝑑𝑠 𝛾̃𝐺1 (1, 𝑠) 𝑎1 (𝑠) 𝑀1 (𝑦1 , 𝑦2 ) 𝑑𝑠
𝑀 1 ≥ 1 (𝑦1 , 𝑦2 ) ≥ (𝑦1 , 𝑦2 ) . 2 2
1 𝐴 ≤ 1 [∫ 𝐺1 (1, 𝑠) 𝑎1 (𝑠) 𝑑𝑠] 2 0
−1
1
× ∫ 𝐺1 (1, 𝑠) 𝑎1 (𝑠) 𝑑𝑠𝐴−1 1 𝜌1 0
=
𝜌1 1 = (𝑦1 , 𝑦2 ) . 2 2
(27)
Similarly, we find that ‖𝑇2 (𝑦1 , 𝑦2 )‖ ≤ (1/2)‖(𝑦1 , 𝑦2 )‖ for (𝑦1 , 𝑦2 ) ∈ 𝐾 ⋂ 𝜕Ω𝜌1 . Consequently, ‖𝑆(𝑦1 , 𝑦2 )‖ ≤ ‖(𝑦1 , 𝑦2 )‖, whenever (𝑦1 , 𝑦2 ) ∈ 𝐾 ⋂ 𝜕Ω𝜌1 . Thus, 𝑆 is cone compression on 𝐾 ⋂ 𝜕Ω𝜌1 . So, from Lemma 5, 𝑆 has a fixed point (𝑦10 , 𝑦20 ) ∈ 𝐾 ⋂(Ω𝜌1 \ Ω𝜌0 ) and a fixed point (𝑦1 , 𝑦2 ) ∈ 𝐾 ⋂(Ω𝜌0∗ \ Ω𝜌1 ). Both are positive solutions of BVP (1)-(2) with 0 < (𝑦10 , 𝑦20 ) < 𝜌1 < (𝑦1 , 𝑦2 ) .
(28)
The proof is complete.
min [𝑦1 (𝑡) + 𝑦2 (𝑡)]
𝑡∈[(1/2),1]
Finally, let Ω𝜌1 := {(𝑦1 , 𝑦2 ) ∈ 𝑋 : ‖(𝑦1 , 𝑦2 )‖ < 𝜌1 }. For (𝑦1 , 𝑦2 ) ∈ 𝐾 ⋂ 𝜕Ω𝜌1 , from (𝐻1 ), (𝑃1 ), we have
Theorem 10. Suppose that 𝑓0 = 𝑓∞ = 𝑔0 = 𝑔∞ = 0 and (𝐻2 ), (𝑃2 ) are satisfied. Then problem (1)-(2) has at least two positive solutions (𝑦10 , 𝑦20 ), (𝑦1 , 𝑦2 ), such that 0 < ‖(𝑦10 , 𝑦20 )‖ < 𝜌2 < ‖(𝑦1 , 𝑦2 )‖. Proof. At first, in view of 𝑓0 = 𝑔0 = 0, we have 𝑓(𝑦1 , 𝑦2 ) < 𝜀(𝑦1 + 𝑦2 ), 𝑔(𝑦1 , 𝑦2 ) < 𝜀(𝑦1 + 𝑦2 ), for 0 < ‖(𝑦1 , 𝑦2 )‖ ≤ 𝜌 < 𝜌2 , where 𝜀 satisfies 𝜀 ≤ 1. Let Ω𝜌 := {(𝑦1 , 𝑦2 ) ∈ 𝑋 : ‖(𝑦1 , 𝑦2 )‖ < 𝜌}. Then for each (𝑦1 , 𝑦2 ) ∈ 𝐾 ⋂ 𝜕Ω𝜌 , we find that 1
𝑇1 (𝑦1 , 𝑦2 ) = 𝜆 1 ∫ 𝐺1 (1, 𝑠) 𝑎1 (𝑠) 𝑓 (𝑦1 (𝑠) , 𝑦2 (𝑠)) 𝑑𝑠 0 1
≤ 𝜆 1 ∫ 𝐺1 (1, 𝑠) 𝑎1 (𝑠) 𝜀 (𝑦1 (𝑠) + 𝑦2 (𝑠)) 𝑑𝑠 0
1
(26)
So ‖𝑇1 (𝑦1 , 𝑦2 )‖ ≥ (1/2)‖(𝑦1 , 𝑦2 )‖ for (𝑦1 , 𝑦2 ) ∈ 𝐾 ⋂ 𝜕Ω𝜌0∗ . Similarly, we find that ‖𝑇2 (𝑦1 , 𝑦2 )‖ ≥ (1/2)‖(𝑦1 , 𝑦2 )‖ for (𝑦1 , 𝑦2 ) ∈ 𝐾 ⋂ 𝜕Ω𝜌0∗ . Consequently, ‖𝑆2 (𝑦1 , 𝑦2 )‖ ≥ ‖(𝑦1 , 𝑦2 )‖, whenever (𝑦1 , 𝑦2 ) ∈ 𝐾 ⋂ 𝜕Ω𝜌0∗ . Thus, S is cone expansion on 𝐾 ⋂ 𝜕Ω𝜌0∗ .
≤ 𝜆 1 ∫ 𝐺1 (1, 𝑠) 𝑎1 (𝑠) 𝜀 (𝑦1 , 𝑦2 ) 𝑑𝑠 0
≤
𝜀 1 (𝑦 , 𝑦 ) ≤ (𝑦 , 𝑦 ) . 2 1 2 2 1 2
(29)
Like Theorem 9, we get ‖𝑆(𝑦1 , 𝑦2 )‖ ≤ ‖(𝑦1 , 𝑦2 )‖ for (𝑦1 , 𝑦2 ) ∈ 𝐾 ⋂ 𝜕Ω𝜌 .
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Next, in view of 𝑓∞ = 𝑔∞ = 0, we have 𝑓(𝑦1 , 𝑦2 ) < 𝜀1 (𝑦1 + 𝑦2 ), 𝑔(𝑦1 , 𝑦2 ) < 𝜀1 (𝑦1 + 𝑦2 ), for 𝑦1 + 𝑦2 ≥ 𝜌 > 𝜌2 , where 𝜀1 satisfies 𝜀1 ≤ 1. We consider two cases. Case 1. Suppose that 𝑓 is unbounded; there exists 𝜌∗ > 𝜌 such that 𝑓 (𝑦1 , 𝑦2 ) ≤ 𝑓 (𝑦1∗ , 𝑦2∗ ) for 0 ≤ (𝑦1 , 𝑦2 ) ≤ 𝜌∗ , (30) ∗ ∗ ∗ (𝑦1 , 𝑦2 ) = 𝜌 . Since 𝜌∗ > 𝜌 , one has 𝑓(𝑦1 , 𝑦2 ) ≤ 𝑓(𝑦1∗ , 𝑦2∗ ) < 𝜀1 (𝑦1∗ +𝑦2∗ ) for 0 ≤ ‖(𝑦1 , 𝑦2 )‖ ≤ 𝜌∗ . Then, for (𝑦1 , 𝑦2 ) ∈ 𝐾 and ‖(𝑦1 , 𝑦2 )‖ = 𝜌∗ , we obtain 1
𝑇1 (𝑦1 , 𝑦2 ) = 𝜆 1 ∫ 𝐺1 (1, 𝑠) 𝑎1 (𝑠) 𝑓 (𝑦1 (𝑠) , 𝑦2 (𝑠)) 𝑑𝑠 0 1
0
≤ 𝜆 1 ∫ 𝐺1 (1, 𝑠) 𝑎1 (𝑠) 𝜀1 (𝑦1 , 𝑦2 ) 𝑑𝑠 0 1 (𝑦1 , 𝑦2 ) ≤ (𝑦1 , 𝑦2 ) . 2
(31)
1
1
≤ 𝜆 1 ∫ 𝐺1 (1, 𝑠) 𝑎1 (𝑠) 𝐿 1 𝑑𝑠 0
𝐵2 := min {2 ∫
1/2
𝐺1 (1, 𝑠) 𝑎1 (𝑠) 𝑑𝑠, 2 ∫
1
1/2
𝐺2 (1, 𝑠) 𝑎2 (𝑠) 𝑑𝑠} .
(𝐻4 ) 𝑓(𝑦1 , 𝑦2 ), 𝑔(𝑦1 , 𝑦2 ) ≥ 𝐵2−1 𝜌2 , for 𝛾̃𝜌2 ≤ ‖(𝑦1 , 𝑦2 )‖ ≤ 𝜌2 .
Theorem 11. Suppose that (𝐻3 ) and (𝐻4 ) are satisfied. Then problem (1)-(2), in the case where 𝜆 1 = 𝜆 2 = 1, has at least one positive solution (𝑦10 , 𝑦20 ) such that ‖(𝑦10 , 𝑦20 )‖ between 𝜌1 and 𝜌2 . Proof. With loss of generality, we may assume that 𝜌1 < 𝜌2 . Let Ω𝜌1 := {(𝑦1 , 𝑦2 ) ∈ 𝑋 : ‖(𝑦1 , 𝑦2 )‖ < 𝜌1 }. For (𝑦1 , 𝑦2 ) ∈ 𝐾 ⋂ 𝜕Ω𝜌1 , one has 1
(32)
≤
𝜌 𝐵1 −1 1 𝐵 𝜌 = 1 = (𝑦1 , 𝑦2 ) . 2 1 1 2 2 (37)
Like Theorem 9, we get ‖𝑆(𝑦1 , 𝑦2 )‖ ≤ ‖(𝑦1 , 𝑦2 )‖ for (𝑦1 , 𝑦2 ) ∈ 𝐾 ⋂ 𝜕Ω𝜌1 . Now, set Ω𝜌2 := {(𝑦1 , 𝑦2 ) ∈ 𝑋 : ‖(𝑦1 , 𝑦2 )‖ < 𝜌2 }. Then for (𝑦1 , 𝑦2 ) ∈ 𝐾 ⋂ 𝜕Ω𝜌2 , one has 𝑦1 (𝑡) + 𝑦2 (𝑡) ≥
min [𝑦1 (𝑡) + 𝑦2 (𝑡)]
≥ 𝛾̃ (𝑦1 , 𝑦2 ) = 𝛾̃𝜌2 .
0
𝑇1 (𝑦1 , 𝑦2 ) = 𝜆 1 ∫ 𝐺1 (1, 𝑠) 𝑎1 (𝑠) 𝑓 (𝑦1 (𝑠) , 𝑦2 (𝑠)) 𝑑𝑠 0
Hence, in either case, we always may set Ω𝜌∗ := {(𝑦1 , 𝑦2 ) ∈ 𝑋 : ‖(𝑦1 , 𝑦2 )‖ < 𝜌∗ } such that ‖𝑇1 (𝑦1 , 𝑦2 )‖ ≤ (1/2)‖(𝑦1 , 𝑦2 )‖ for (𝑦1 , 𝑦2 ) ∈ 𝐾 ⋂ 𝜕Ω𝜌∗ . Like Theorem 9, we get ‖𝑆(𝑦1 , 𝑦2 )‖ ≤ ‖(𝑦1 , 𝑦2 )‖, for (𝑦1 , 𝑦2 ) ∈ 𝐾 ⋂ 𝜕Ω𝜌∗ . Finally, set Ω𝜌2 := {(𝑦1 , 𝑦2 ) ∈ 𝑋 : ‖(𝑦1 , 𝑦2 )‖ < 𝜌2 }. Then (𝑦1 , 𝑦2 ) ∈ 𝐾 ⋂ 𝜕Ω𝜌2 implies 𝑡∈[(1/2),1]
0
1
(𝐻3 ) 𝑓(𝑦1 , 𝑦2 ), 𝑔(𝑦1 , 𝑦2 ) ≤ 𝐵1−1 𝜌1 , for 0 ≤ ‖(𝑦1 , 𝑦2 )‖ ≤ 𝜌1 ;
𝑇1 (𝑦1 , 𝑦2 ) = 𝜆 1 ∫ 𝐺1 (1, 𝑠) 𝑎1 (𝑠) 𝑓 (𝑦1 (𝑠) , 𝑦2 (𝑠)) 𝑑𝑠 0
𝑦1 (𝑡) + 𝑦2 (𝑡) ≥
1
Assume that there exist two positive constants 𝜌1 ≠ 𝜌2 such that
Case 2. Suppose that 𝑓 is bounded; there exists 𝐿 1 such that 𝑓(𝑦1 , 𝑦2 ) ≤ 𝐿 1 for all (𝑦1 , 𝑦2 ) ∈ 𝐾. Taking 𝜌∗ ≥ max{2𝜌2 , 𝐿 1 }, for (𝑦1 , 𝑦2 ) ∈ 𝐾 and ‖(𝑦1 , 𝑦2 )‖ = 𝜌∗ , we obtain
𝜌∗ 1 𝐿 ≤ 1 ≤ = (𝑦1 , 𝑦2 ) . 2 2 2
1
(36)
1
𝜀1 2
3.2. Problem (1)–(3) in Case 𝜆 1 = 𝜆 2 = 1. In the following, for the sake of convenience, set 𝐵1 := max {2 ∫ 𝐺1 (1, 𝑠) 𝑎1 (𝑠) 𝑑𝑠, 2 ∫ 𝐺2 (1, 𝑠) 𝑎2 (𝑠) 𝑑𝑠} ,
≤ 𝜆 1 ∫ 𝐺1 (1, 𝑠) 𝑎1 (𝑠) 𝜀1 (𝑦1 (𝑠) + 𝑦2 (𝑠)) 𝑑𝑠
≤
Consequently, ‖𝑇1 (𝑦1 , 𝑦2 )‖ ≥ (1/2)‖(𝑦1 , 𝑦2 )‖ for (𝑦1 , 𝑦2 ) ∈ 𝐾 ⋂ 𝜕Ω𝜌2 . Like Theorem 9, we get ‖𝑆(𝑦1 , 𝑦2 )‖ ≥ ‖(𝑦1 , 𝑦2 )‖ for (𝑦1 , 𝑦2 ) ∈ 𝐾 ⋂ 𝜕Ω𝜌2 . So, from Lemma 5, 𝑆 has a fixed point (𝑦10 , 𝑦20 ) ∈ 𝐾 ⋂(Ω𝜌2 \ Ω𝜌 ) and a fixed point (𝑦1 , 𝑦2 ) ∈ 𝐾 ⋂(Ω𝜌∗ \ Ω𝜌2 ). Both are positive solutions of BVP (1)-(2) with (35) 0 < (𝑦10 , 𝑦20 ) < 𝜌2 < (𝑦1 , 𝑦2 ) , which complete the proof.
(33)
min [𝑦1 (𝑡) + 𝑦2 (𝑡)]
𝑡∈[(1/2),1]
≥ 𝛾̃ (𝑦1 , 𝑦2 ) = 𝛾̃𝜌2 .
(38)
Thus, we get
Hence we have 1
𝑇1 (𝑦1 , 𝑦2 ) (1) = 𝜆 1 ∫ 𝐺1 (1, 𝑠) 𝑎1 (𝑠) 𝑓 (𝑦1 (𝑠) , 𝑦2 (𝑠)) 𝑑𝑠 0
≥ 𝜆1 ∫
1
1/2
≥
𝐺1 (1, 𝑠) 𝑎1 (𝑠) 𝐴−1 2 𝜌2 𝑑𝑠
𝜌 𝐴 2 −1 1 𝐴 2 𝜌2 = 2 = (𝑦1 , 𝑦2 ) . 2 2 2
1
𝑇1 (𝑦1 , 𝑦2 ) (1) = 𝜆 1 ∫ 𝐺1 (1, 𝑠) 𝑎1 (𝑠) 𝑓 (𝑦1 (𝑠) , 𝑦2 (𝑠)) 𝑑𝑠 0
≥ 𝜆1 ∫
1
1/2
≥ (34)
𝐺1 (1, 𝑠) 𝑎1 (𝑠) 𝐵2−1 𝜌2 𝑑𝑠
𝜌 𝐵2 −1 1 𝐵2 𝜌2 = 2 = (𝑦1 , 𝑦2 ) . 2 2 2
(39)
The Scientific World Journal
7
Like Theorem 9, we get ‖𝑆(𝑦1 , 𝑦2 )‖ ≥ ‖(𝑦1 , 𝑦2 )‖ for (𝑦1 , 𝑦2 ) ∈ 𝐾 ⋂ 𝜕Ω𝜌2 . Hence, from Lemma 5, we complete the proof. Remark 12. In [21], problem (1)-(2) with 𝜆 1 = 𝜆 2 = 1 is not considered.
(𝑃3 ) There are numbers Λ 5 , Λ 6 , where −1
1 1 Λ 5 := max { [∫ 𝛾0 𝐺1 (1, 𝑠) 𝑎1 (𝑠) 𝑑s] , 2 1/2 −1
1 1 [∫ 𝛾 𝐺 (1, 𝑠) 𝑎2 (𝑠) 𝑑𝑠] } , 2 1/2 0 2
3.3. Problem (1), (3) in the General Case. Consider the following. Lemma 13 (see [21]). A pair of functions (𝑦1 , 𝑦2 ) ∈ 𝑋 is a solution of (1), (3) if and only if (𝑦1 , 𝑦2 ) is a fixed point of the operator 𝑈 : 𝑋 → 𝑋 defined by
−1
1 𝐴 Λ 6 := min { 1 [∫ 𝐺1 (1, 𝑠) 𝑎1 (𝑠) 𝑑𝑠] , 4 0 −1
𝐴1 1 [∫ 𝐺 (1, 𝑠) 𝑎2 (𝑠) 𝑑𝑠] } , 4 0 2
𝑈 (𝑦1 , 𝑦2 ) (𝑡)
such that Λ 5 < 𝜆 1 , 𝜆 2 < Λ 6 .
:= (𝑈1 (𝑦1 , 𝑦2 ) (𝑡) , 𝑈2 (𝑦1 , 𝑦2 ) (𝑡))
(𝑃4 ) There are numbers Λ 7 , Λ 8 , where
1
= (𝛽1 (𝑡) 𝜙1 (𝑦1 ) + 𝜆 1 ∫ 𝐺1 (𝑡, 𝑠) 𝑎1 (𝑠) 𝑓
−1
0
× (𝑦1 (𝑠) , 𝑦2 (𝑠)) 𝑑𝑠,
(40)
Λ 7 := max {
−1
𝛽2 (𝑡) 𝜙2 (𝑦2 ) + 𝜆 2 ∫ 𝐺2 (𝑡, 𝑠) 𝑎2 (𝑠) 𝑔 0
−1
1 1 Λ 8 := min { [∫ 𝐺1 (1, 𝑠) 𝑎1 (𝑠) 𝑑𝑠] , 4 0
× (𝑦1 (𝑠) , 𝑦2 (𝑠)) 𝑑𝑠) ,
𝛽2 (𝑡) :=
Γ (]2 − 𝛼) ]2 −1 𝑡 . Γ (]2 )
1 1 [∫ 𝐺 (1, 𝑠) 𝑎2 (𝑠) 𝑑𝑠] } , 4 0 2 (41)
Lemma 14 (see [21]). Each of 𝛽1 (𝑡) and 𝛽2 (𝑡) is strictly increasing in t and satisfies 𝛽1 (0) = 𝛽2 (0) = 0 and 𝛽1 (1), 𝛽2 (1) ∈ (0, 1). Moreover, there exist constants 𝑀𝛽1 and 𝑀𝛽2 satisfying 𝑀𝛽1 , 𝑀𝛽2 ∈ (0, 1) such that min𝑡∈[(1/2),1] 𝛽1 (𝑡) ≥ 𝑀𝛽1 ‖𝛽1 ‖ and min𝑡∈[(1/2),1] 𝛽2 (𝑡) ≥ 𝑀𝛽2 ‖𝛽2 ‖. Let one define a new cone 𝐾1 by 𝐾1 := {(𝑦1 , 𝑦2 ) ∈ 𝑋 : 𝑦1 , 𝑦2 ≥ 0, min [𝑦1 (𝑡) + 𝑦2 (𝑡)] ≥ 𝛾0 (𝑦1 , 𝑦2 )} ,
(44)
−1
where 𝛽1 , 𝛽2 : [0, 1] → [0, 1] are defined by Γ (]1 − 𝛼) ]1 −1 𝑡 , Γ (]1 )
𝐴2 1 [∫ 𝐺 (1, 𝑠) 𝑎1 (𝑠) 𝑑𝑠] , 2 1/2 1 𝐴2 1 [∫ 𝐺 (1, 𝑠) 𝑎2 (𝑠) 𝑑𝑠] } , 2 1/2 2
1
𝛽1 (𝑡) :=
(43)
(42)
𝑡∈[(1/2),1]
such that Λ 7 < 𝜆 1 , 𝜆 2 < Λ 8 . Theorem 16. Suppose that 𝑓0 = 𝑓∞ = 𝑔0 = 𝑔∞ = ∞ and (𝐻1 ), (𝐷1 ), (𝑃3 ) are satisfied. Then problem (1), (3) has at least two positive solutions (𝑦10 , 𝑦20 ), (𝑦1 , 𝑦2 ), such that 0 < (𝑦10 , 𝑦20 ) < 𝜌1 < (𝑦1 , 𝑦2 ) .
Proof. At first, in view of 𝑓0 = 𝑔0 = ∞, we have 𝑓(𝑦1 , 𝑦2 ) ≥ 𝑀(𝑦1 + 𝑦2 ), 𝑔(𝑦1 , 𝑦2 ) ≥ 𝑀(𝑦1 + 𝑦2 ), for 0 < ‖(𝑦1 , 𝑦2 )‖ ≤ 𝜌0 < 𝜌1 , where 𝑀 satisfies 𝑀 ≥ 1. Let Ω𝜌0 := {(𝑦1 , 𝑦2 ) ∈ 𝑋 : ‖(𝑦1 , 𝑦2 )‖ < 𝜌0 }. Then for each (𝑦1 , 𝑦2 ) ∈ 𝐾1 ⋂ 𝜕Ω𝜌0 , we find that 𝑈1 (𝑦1 , 𝑦2 ) (1) ≥ 𝜆 1 ∫
1
1/2
≥ 𝜆1 ∫
1
1/2
where 𝛾0 := min{̃ 𝛾, 𝑀𝛽1 , 𝑀𝛽2 }. It is obvious that 𝛾0 ∈ (0, 1). Lemma 15 (see [21]). 𝑈 : 𝐾1 → 𝐾1 is a completely continuous operator. Now, one assumes (𝐷1 ) 𝜙1 (𝑦1 ) ≤ ‖𝑦1 ‖/4, 𝜙2 (𝑦2 ) ≤ ‖𝑦2 ‖/4 for each (𝑦1 , 𝑦2 ) ∈ 𝐾1 ;
(45)
≥ 𝜆1 ∫
1
1/2
≥
𝐺1 (1, 𝑠) 𝑎1 (𝑠) 𝑓 (𝑦1 (𝑠) , 𝑦2 (𝑠)) 𝑑𝑠 𝐺1 (1, 𝑠) 𝑎1 (𝑠) 𝑀 (𝑦1 (𝑠) + 𝑦2 (𝑠)) 𝑑𝑠 𝛾0 𝐺1 (1, 𝑠) 𝑎1 (𝑠) 𝑀 (𝑦1 , 𝑦2 ) 𝑑𝑠
𝑀 1 (𝑦 , 𝑦 ) ≥ (𝑦 , 𝑦 ) . 2 1 2 2 1 2
(46)
So ‖𝑈1 (𝑦1 , 𝑦2 )‖ ≥ (1/2)‖(𝑦1 , 𝑦2 )‖ for (𝑦1 , 𝑦2 ) ∈ 𝐾1 ⋂ 𝜕Ω𝜌0 .
8
The Scientific World Journal
Similarly, we find that ‖𝑈2 (𝑦1 , 𝑦2 )‖ ≥ (1/2)‖(𝑦1 , 𝑦2 )‖ for (𝑦1 , 𝑦2 ) ∈ 𝐾1 ⋂ 𝜕Ω𝜌0 . Consequently, 𝑈 (𝑦1 , 𝑦2 ) = (𝑈1 (𝑦1 , 𝑦2 ) , 𝑈2 (𝑦1 , 𝑦2 )) = 𝑈1 (𝑦1 , 𝑦2 ) + 𝑈2 (𝑦1 , 𝑦2 ) ≥ (𝑦1 , 𝑦2 ) ,
(47)
whenever (𝑦1 , 𝑦2 ) ∈ 𝐾1 ⋂ 𝜕Ω𝜌0 . Thus, 𝑈 is cone expansion on 𝐾1 ⋂ 𝜕Ω𝜌0 . Next, since 𝑓∞ = 𝑔∞ = ∞, we get 𝑓(𝑦1 , 𝑦2 ) ≥ 𝑀1 (𝑦1 + 𝑦2 ), 𝑔(𝑦1 , 𝑦2 ) ≥ 𝑀1 (𝑦1 + 𝑦2 ), for 𝑦1 + 𝑦2 ≥ 𝜌10 > 𝜌1 , where 𝑀1 satisfies 𝑀1 ≥ 1. Let 𝜌0∗ = max{2𝜌1 , (𝜌10 /𝛾0 )} and Ω𝜌0∗ := {(𝑦1 , 𝑦2 ) ∈ 𝑋 : ‖(𝑦1 , 𝑦2 )‖ < 𝜌0∗ }; then, (𝑦1 , 𝑦2 ) ∈ 𝐾 ⋂ 𝜕Ω𝜌0∗ implies 𝑦1 (𝑡) + 𝑦2 (𝑡) ≥
min [𝑦1 (𝑡) + 𝑦2 (𝑡)]
𝑡∈[(1/2),1]
≥ 𝛾0 (𝑦1 , 𝑦2 ) = 𝛾0 𝜌0∗ ≥ 𝜌10 .
(48)
So for (𝑦1 , 𝑦2 ) ∈ 𝐾1 ⋂ 𝜕Ω𝜌0∗ , we obtain 1
𝑈1 (𝑦1 , 𝑦2 ) (1) ≥ 𝜆 1 ∫
1/2 1
≥ 𝜆1 ∫
1/2
𝐺1 (1, 𝑠) 𝑎1 (𝑠) 𝑓 (𝑦1 (𝑠) , 𝑦2 (𝑠)) 𝑑𝑠
0 < (𝑦10 , 𝑦20 ) < 𝜌1 < (𝑦1 , 𝑦2 ) .
(51)
The proof is complete. Theorem 17. Suppose that 𝑓0 = 𝑓∞ = 𝑔0 = 𝑔∞ = 0 and (𝐻2 ), (𝐷1 ), (𝑃4 ) are satisfied. Then problem (1), (3) has at least two positive solutions (𝑦10 , 𝑦20 ), (𝑦1 , 𝑦2 ), such that 0 < (𝑦10 , 𝑦20 ) < 𝜌2 < (𝑦1 , 𝑦2 ) .
(52)
Proof. At first, in view of 𝑓0 = 𝑔0 = 0, we have 𝑓(𝑦1 , 𝑦2 ) < 𝜀(𝑦1 +𝑦2 ), 𝑔(𝑦1 , 𝑦2 ) < 𝜀(𝑦1 +𝑦2 ) for ‖(𝑦1 , 𝑦2 )‖ ≤ 𝜌 < 𝜌2 , where 𝜀 satisfies 𝜀 ≤ 1. Let Ω𝜌 := {(𝑦1 , 𝑦2 ) ∈ 𝑋 : ‖(𝑦1 , 𝑦2 )‖ < 𝜌}. Then for each (𝑦1 , 𝑦2 ) ∈ 𝐾1 ⋂ 𝜕Ω𝜌 , we find that 𝑈1 (𝑦1 , 𝑦2 ) 1
𝐺1 (1, 𝑠) 𝑎1 (𝑠) 𝑀1 (𝑦1 (𝑠) + 𝑦2 (𝑠)) 𝑑𝑠
1
≥ 𝜆 1 ∫ 𝛾0 𝐺1 (1, 𝑠) 𝑎1 (𝑠) 𝑀1 (𝑦1 , 𝑦2 ) 𝑑𝑠 1/2 ≥
So, from Lemma 5, 𝑈 has a fixed point (𝑦10 , 𝑦20 ) ∈ 𝐾1 ⋂ (Ω𝜌1 \Ω𝜌0 ) and a fixed point (𝑦1 , 𝑦2 ) ∈ 𝐾1 ⋂ (Ω𝜌0∗ \Ω𝜌1 ). Both are positive solutions of BVP (1), (3) with
𝑀1 1 (𝑦 , 𝑦 ) ≥ (𝑦 , 𝑦 ) . 2 1 2 2 1 2
(49)
That is, ‖𝑈1 (𝑦1 , 𝑦2 )‖ ≥ (1/2)‖(𝑦1 , 𝑦2 )‖ for (𝑦1 , 𝑦2 ) ∈ 𝐾1 ⋂ 𝜕Ω𝜌0∗ . Similarly, we find that ‖𝑈2 (𝑦1 , 𝑦2 )‖ ≥ (1/2)‖(𝑦1 , 𝑦2 )‖ for (𝑦1 , 𝑦2 ) ∈ 𝐾1 ⋂ 𝜕Ω𝜌0∗ . Consequently, ‖𝑈(𝑦1 , 𝑦2 )‖ ≥ ‖(𝑦1 , 𝑦2 )‖, whenever (𝑦1 , 𝑦2 ) ∈ 𝐾1 ⋂ 𝜕Ω𝜌0∗ . Thus, 𝑈 is cone expansion on 𝐾1 ⋂ 𝜕Ω𝜌0∗ . Finally, let Ω𝜌1 := {(𝑦1 , 𝑦2 ) ∈ 𝑋 : ‖(𝑦1 , 𝑦2 )‖ < 𝜌1 }. For (𝑦1 , 𝑦2 ) ∈ 𝐾 ⋂ 𝜕Ω𝜌1 , from (𝐻1 ), (𝐷1 ), and (𝑃3 ), we have
≤ 𝜙1 (𝑦1 ) + 𝜆 1 ∫ 𝐺1 (𝑡, 𝑠) 𝑎1 (𝑠) 𝑓 (𝑦1 (𝑠) , 𝑦2 (𝑠)) 𝑑𝑠 (53) 0 𝑦 𝜀 1 ≤ 1 + (𝑦1 , 𝑦2 ) ≤ (𝑦1 , 𝑦2 ) . 4 4 2 Like Theorem 16, we get ‖𝑈(𝑦1 , 𝑦2 )‖ ≤ ‖(𝑦1 , 𝑦2 )‖ for (𝑦1 , 𝑦2 ) ∈ 𝐾1 ⋂ 𝜕Ω𝜌 . Next, in view of 𝑓∞ = 𝑔∞ = 0, we have 𝑓(𝑦1 , 𝑦2 ) < 𝜀1 (𝑦1 + 𝑦2 ), 𝑔(𝑦1 , 𝑦2 ) < 𝜀1 (𝑦1 + 𝑦2 ), for 𝑦1 + 𝑦2 ≥ 𝜌 > 𝜌2 , where 𝜀1 satisfies 𝜀1 ≤ 1. We consider two cases. Case 1. Suppose that 𝑓 is unbounded and there exists 𝜌∗ > 𝜌 such that 𝑓 (𝑦1 , 𝑦2 ) ≤ 𝑓 (𝑦1∗ , 𝑦2∗ )
for 0 ≤ (𝑦1 , 𝑦2 ) ≤ 𝜌∗ , ∗ ∗ ∗ (𝑦1 , 𝑦2 ) = 𝜌 .
(54)
1
𝑈1 (𝑦1 , 𝑦2 ) ≤ 𝜙1 (𝑦1 ) + 𝜆 1 ∫ 𝐺1 (𝑡, 𝑠) 𝑎1 (𝑠) 0 × 𝑓 (𝑦1 (𝑠) , 𝑦2 (𝑠)) 𝑑𝑠 𝑦 𝐴 𝜌 ≤ 1 + 1 𝐴−1 4 4 1 1 𝜌 1 ≤ 1 = (𝑦1 , 𝑦2 ) . 2 2
(50)
Similarly, we find that ‖𝑈2 (𝑦1 , 𝑦2 )‖ ≤ (1/2)‖(𝑦1 , 𝑦2 )‖ for (𝑦1 , 𝑦2 ) ∈ 𝐾1 ⋂ 𝜕Ω𝜌1 . Consequently, ‖𝑈(𝑦1 , 𝑦2 )‖ ≤ ‖(𝑦1 , 𝑦2 )‖, whenever (𝑦1 , 𝑦2 ) ∈ 𝐾1 ⋂ 𝜕Ω𝜌1 . Thus, 𝑈 is cone compression on 𝐾1 ⋂ 𝜕Ω𝜌1 .
Since 𝜌∗ > 𝜌 , one has 𝑓(𝑦1 , 𝑦2 ) ≤ 𝑓(𝑦1∗ , 𝑦2∗ ) < 𝜀1 (𝑦1∗ + 𝑦2∗ ) for 0 ≤ ‖(𝑦1 , 𝑦2 )‖ ≤ 𝜌∗ . Then, for (𝑦1 , 𝑦2 ) ∈ 𝐾1 and ‖(𝑦1 , 𝑦2 )‖ = 𝜌∗ , we obtain 𝑈1 (𝑦1 , 𝑦2 ) 1
≤ 𝜙1 (𝑦1 ) + 𝜆 1 ∫ 𝐺1 (𝑡, 𝑠) 𝑎1 (𝑠) 𝑓 (𝑦1 (𝑠) , 𝑦2 (𝑠)) 𝑑𝑠 0
𝑦 𝜀 1 ≤ 1 + 1 (𝑦1 , 𝑦2 ) ≤ (𝑦1 , 𝑦2 ) . 4 4 2
(55)
The Scientific World Journal
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Case 2. Suppose that 𝑓 is bounded; there, exists 𝐿 1 such that 𝑓(𝑦1 , 𝑦2 ) ≤ 𝐿 1 for all (𝑦1 , 𝑦2 ) ∈ 𝐾1 . Taking 𝜌∗ ≥ max{2𝜌2 , 𝐿 1 }, for (𝑦1 , 𝑦2 ) ∈ 𝐾1 and ‖(𝑦1 , 𝑦2 )‖ = 𝜌∗ , we obtain 𝑈1 (𝑦1 , 𝑦2 ) 1
≤ 𝜙1 (𝑦1 ) + 𝜆 1 ∫ 𝐺1 (𝑡, 𝑠) 𝑎1 (𝑠) 𝑓 (𝑦1 (𝑠) , 𝑦2 (𝑠)) 𝑑𝑠 0
1
≤ 𝜙1 (𝑦1 ) + 𝜆 1 ∫ 𝐺1 (𝑡, 𝑠) 𝑎1 (𝑠) 𝐿 1 𝑑𝑠
(56)
𝜌∗ 1 𝑦 𝐿 ≤ 1 + 1 ≤ = (𝑦1 , 𝑦2 ) . 4 4 2 2
≤
(60)
𝜌1 𝐵3 −1 𝜌 1 + 𝐵3 𝜌1 = 1 = (𝑦1 , 𝑦2 ) . 4 4 2 2
Like Theorem 16, we get ‖𝑈(𝑦1 , 𝑦2 )‖ ≤ ‖(𝑦1 , 𝑦2 )‖ for (𝑦1 , 𝑦2 ) ∈ 𝐾1 ⋂ 𝜕Ω𝜌1 . Now, set Ω𝜌2 := {(𝑦1 , 𝑦2 ) ∈ 𝑋 : ‖(𝑦1 , 𝑦2 )‖ < 𝜌2 }. For (𝑦1 , 𝑦2 ) ∈ 𝐾1 ⋂ 𝜕Ω𝜌2 , one has 𝑦1 (𝑡) + 𝑦2 (𝑡) ≥
min [𝑦1 (𝑡) + 𝑦2 (𝑡)]
≥ 𝛾0 (𝑦1 , 𝑦2 ) = 𝛾0 𝜌2 .
1
× 𝑓 (𝑦1 (𝑠) , 𝑦2 (𝑠)) 𝑑𝑠
Hence, in either case, we always may set Ω𝜌∗ := {(𝑦1 , 𝑦2 ) ∈ 𝑋 : ‖(𝑦1 , 𝑦2 )‖ < 𝜌∗ } such that ‖𝑈1 (𝑦1 , 𝑦2 )‖ ≤ (1/2)‖(𝑦1 , 𝑦2 )‖ for (𝑦1 , 𝑦2 ) ∈ 𝐾1 ⋂ 𝜕Ω𝜌∗ . Like Theorem 16, we get ‖𝑈(𝑦1 , 𝑦2 )‖ ≤ ‖(𝑦1 , 𝑦2 )‖, for (𝑦1 , 𝑦2 ) ∈ 𝐾1 ⋂ 𝜕Ω𝜌∗ . Finally, set Ω𝜌2 := {(𝑦1 , 𝑦2 ) ∈ 𝑋 : ‖(𝑦1 , 𝑦2 )‖ < 𝜌2 }. Then (𝑦1 , 𝑦2 ) ∈ 𝐾1 ⋂ 𝜕Ω𝜌2 implies 𝑡∈[(1/2),1]
Proof. With loss of generality, we may assume that 𝜌1 < 𝜌2 . Let Ω𝜌1 := {(𝑦1 , 𝑦2 ) ∈ 𝑋 : ‖(𝑦1 , 𝑦2 )‖ < 𝜌1 }. For (𝑦1 , 𝑦2 ) ∈ 𝐾1 ⋂ 𝜕Ω𝜌1 , from (𝐻7 ), (𝐷1 ), one has 𝑈1 (𝑦1 , 𝑦2 ) ≤ 𝜙1 (𝑦1 ) + ∫ 𝐺1 (𝑡, 𝑠) 𝑎1 (𝑠) 0
0
𝑦1 (𝑡) + 𝑦2 (𝑡) ≥
Theorem 18. Suppose that (𝐻5 ), (𝐻6 ), and (𝐷1 ) are satisfied. Then problem (1), (3), in the case where 𝜆 1 = 𝜆 2 = 1, has at least one positive solution (𝑦10 , 𝑦20 ) such that ‖(𝑦1 , 𝑦2 )‖ between 𝜌1 and 𝜌2 .
(57)
min [𝑦1 (𝑡) + 𝑦2 (𝑡)]
𝑡∈[(1/2),1]
(61)
≥ 𝛾0 (𝑦1 , 𝑦2 ) = 𝛾0 𝜌2 . Thus, from (𝐻8 ), we get
Hence we have 𝑈1 (𝑦1 , 𝑦2 ) (1) ≥ 𝜆 1 ∫
1
1/2
𝐺1 (1, 𝑠) 𝑎1 (𝑠) 𝐴−1 2 𝜌2 𝑑𝑠
𝜌2 1 𝐴 ≥ 2 𝐴−1 = (𝑦1 , 𝑦2 ) . 2 𝜌2 = 2 2 2
𝑈1 (𝑦1 , 𝑦2 ) (1) ≥ 𝜆 1 ∫
0
0
𝐵3 := max {4 ∫ 𝐺1 (1, 𝑠) 𝑎1 (𝑠) 𝑑𝑠, 4 ∫ 𝐺2 (1, 𝑠) 𝑎2 (𝑠) 𝑑𝑠} . (59)
Assume that there exist two positive constants 𝜌1 ≠ 𝜌2 such that (𝐻5 ) 𝑓(𝑦1 , 𝑦2 ), 𝑔(𝑦1 , 𝑦2 ) ≤ 𝐵3−1 𝜌1 for 0 ≤ ‖(𝑦1 , 𝑦2 )‖ ≤ 𝜌1 ; (𝐻6 ) 𝑓(𝑦1 , 𝑦2 ), 𝑔(𝑦1 , 𝑦2 ) ≥ 𝜌2 .
𝐵2−1 𝜌2
𝐺1 (1, 𝑠) 𝑎1 (𝑠) 𝐵2−1 𝜌2 𝑑𝑠
𝜌 𝐵 1 ≥ 2 𝐵2−1 𝜌2 = 2 = (𝑦1 , 𝑦2 ) . 2 2 2
3.4. Problem (1), (3) in Case 𝜆 1 = 𝜆 2 = 1. In [21], the author obtained that problem (1), (3) with 𝜆 1 = 𝜆 2 = 1 having at least one positive solution. In the following, we also establish the existence of one positive solution to problem (1), (3) with 𝜆 1 = 𝜆 2 = 1 under different conditions. For the sake of convenience, set 1
1/2
(58)
So, ‖𝑈1 (𝑦1 , 𝑦2 )‖ ≥ (1/2)‖(𝑦1 , 𝑦2 )‖ for (𝑦1 , 𝑦2 ) ∈ 𝐾1 ⋂ 𝜕Ω𝜌2 . Like Theorem 16, we get ‖𝑈(𝑦1 , 𝑦2 )‖ ≥ ‖(𝑦1 , 𝑦2 )‖ for (𝑦1 , 𝑦2 ) ∈ 𝐾1 ⋂ 𝜕Ω𝜌2 . So, from Lemma 5, 𝑈 has a fixed point (𝑦10 , 𝑦20 ) ∈ 𝐾1 ⋂(Ω𝜌2 \ Ω𝜌 ) and a fixed point (𝑦1 , 𝑦2 ) ∈ 𝐾1 ⋂(Ω𝜌∗ \ Ω𝜌2 ). Both are positive solutions of BVP (1), (3) with 0 < ‖(𝑦10 , 𝑦20 )‖ < 𝜌2 < ‖(𝑦1 , 𝑦2 )‖, which complete the proof.
1
1
for 𝛾0 𝜌2 ≤ ‖(𝑦1 , 𝑦2 )‖ ≤
(62)
Like Theorem 16, we get ‖𝑈(𝑦1 , 𝑦2 )‖ ≥ ‖(𝑦1 , 𝑦2 )‖ for (𝑦1 , 𝑦2 ) ∈ 𝐾1 ⋂ 𝜕Ω𝜌2 . Hence, from Lemma 5, we complete the proof.
4. An Example To illustrate how our main results can be used in practice, we present one example. Example 19. Consider the following BVP, for 𝑡 ∈ (0, 1): 1/2
−2𝑡 − 𝐷05.2 [(𝑦1 (𝑡) + 𝑦2 (𝑡)) + 𝑦1 (𝑡) = 164500𝑒
2
+ (𝑦1 (𝑡) + 𝑦2 (𝑡)) ] , −3𝑡 [(𝑦1 (𝑡) + 𝑦2 (𝑡)) − 𝐷05.95 + 𝑦2 (𝑡) = 164000𝑒
(63)
1/3 3
+ (𝑦1 (𝑡) + 𝑦2 (𝑡)) ] , subject to the boundary conditions 𝑦1(𝑖) (0) = 𝑦2(𝑖) (0) = 0,
0 ≤ 𝑖 ≤ 4,
1.5 1.5 𝐷0+ [𝑦1 (𝑡)]𝑡=1 = 𝐷0+ [𝑦2 (𝑡)]𝑡=1 = 0.
(64) (65)
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The Scientific World Journal
Obviously, problem (63)–(65) fits the framework of problem (1)-(2) with ]1 := 5.2,
]2 := 5.95,
𝜆 1 = 164500,
𝛼 = 1.5,
𝜆 2 = 164000,
𝑛 = 6.
(66)
In addition, we have set 1/2
+ (𝑦1 + 𝑦2 ) ,
1/3
+ (𝑦1 + 𝑦2 ) ,
𝑓 (𝑦1 , 𝑦2 ) := (𝑦1 + 𝑦2 ) 𝑔 (𝑦1 , 𝑦2 ) := (𝑦1 + 𝑦2 ) 𝑎1 (𝑡) := 𝑒−2𝑡 ,
2
3
(67)
𝑎2 (𝑡) := 𝑒−3𝑡 .
We can see that 𝑓, 𝑔 : [0, +∞) × [0, +∞) → [0, +∞) and are continuous. The functions 𝑎1 (𝑡) and 𝑎2 (𝑡) are obviously nonnegative. Now, observe that 𝑓0 = 𝑓∞ = 𝑔0 = 𝑔∞ = ∞ holds. Again set 𝐴 1 = 1/850, because 𝑓(𝑦1 , 𝑦2 ), 𝑔(𝑦1 , 𝑦2 ) is monotone increasing function for (𝑦1 , 𝑦2 ) ≥ 0, taking 𝜌1 = 4; then, when ‖(𝑦1 , 𝑦2 )‖ ∈ [0, 𝜌1 ], we get 1/2 2 𝑓 (𝑦1 , 𝑦2 ) ≤ (𝑦1 , 𝑦2 ) + (𝑦1 , 𝑦2 ) ≤ 2 + 16 = 18 < 4 × 850 = 𝐴−1 1 𝜌1 1/3 3 𝑔 (𝑦1 , 𝑦2 ) ≤ (𝑦1 , 𝑦2 ) + (𝑦1 , 𝑦2 )
(68)
≤ 41/3 + 64 < 4 × 850 = 𝐴−1 1 𝜌1 , which implies that (𝐻1 ) holds. On the other hand, to calculate the admissible range of the eigenvalues 𝜆 1 , 𝜆 2 , as given by condition (𝑃1 ), observe by numerical approximation, we find that Λ 1 ≈ 163530,
Λ 2 ≈ 164547.25.
(69)
Thus, for any 𝜆 1 , 𝜆 2 satisfying 163530 < 𝜆 1 , 𝜆 2 < 164547.25, condition (𝑃1 ) will be satisfied. Consequently, by Theorem 9, problem (63)–(65) has at least two positive solutions.
Conflict of Interests The authors declare that they have no competing interests.
Authors’ Contribution The authors declare that the study was realized in collaboration with the same responsibility. All authors read and approved the final manuscript.
Acknowledgments The first author was supported financially by the Youth Science Foundations of China (11201272) and Shanxi Province (2010021002-1).
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