Positive Solutions for Coupled Nonlinear Fractional Differential Equations

Abstract

We consider the existence of positive solutions for a coupled system of nonlinear fractional differential equations with integral boundary values. Assume the nonlinear term is superlinear in one equation and sublinear in the other equation. By constructing two cones 𝐾₁, 𝐾₂ and computing the fixed point index in product cone 𝐾₁×𝐾₂, we obtain that the system has a pair of positive solutions. It is remarkable that it is established on the Cartesian product of two cones, in which the feature of two equations can be opposite.

1. Introduction

Fractional calculus is a very old concept dating back to 17th century; it involves fractional integration and fractional differentiation. At the first stage, fractional calculus theory is mainly focused on pure mathematical fields. In the last few decades, fractional differential equations and fractional integration equations have found many applications in various fields, such as science and engineering, physics, chemistry, biology, economics, and signal and image processing; for details, see [1–5]. In recent years, fractional differential equations have attracted increasing interests for their extensive applications, which leads to intensive development of the theory of fractional calculus. And more and more results about the existence and uniqueness of solutions appear. We can refer to [6] for the latest studies of fractional calculus.

The existence theory for initial value problems has been paid considerable attentions by many authors; see [3, 7, 8] and the references cited therein. Also the existence of positive solutions for boundary value problem of nonlinear fractional differential equation has attracted attentions from many researches; see [9–14]. By using the cone extension method Feng et al. [15] studied the existence of solutions for higher-order nonlinear fractional differential equation with integral boundary conditions: 𝐷^𝛼₀₊𝑢(𝑡)(𝑡)𝑓((𝑡𝑢(0)+𝑔𝑡,𝑢))=0,0<𝑡<1,=𝑢^′(0)=⋅⋅⋅=𝑢^(𝑛−2)(0)𝑢(1)==0,₁₀ℎ(𝑡)𝑢(𝑡)𝑑𝑡, (1) where 𝐷^𝛼₀₊ is the standard Riemann-Liouville fractional derivative of order 𝑛−1<𝛼<𝑛, 𝑛≥3, 𝑔∈𝒞((0,1),(0,+∞)) and 𝑔 may be singular at 𝑡=0 or 𝑡=1, ℎ∈𝐿¹(0,1) is nonnegative, and 𝑓∈𝒞((0,1)×(0,+∞),(0,+∞)).

In recent years, many authors have established the existence and uniqueness for solutions of some systems of nonlinear fractional differential equations; readers can see [16–21]. For example, by using the alternative of Leray-Schauder theorem method Wang et al. [22] obtained the existence and uniqueness of positive solution of nonzero boundary values problem for a coupled system of nonlinear fractional differential equations: 𝐷^𝛼₀₊𝑢(𝑡)((𝑡𝐷=𝑓𝑡,𝑣)),0<𝑡<1,_𝛽0+𝑣(𝑡)((𝑡𝑢(0)(1)(𝜉),𝑣(0)(1)(𝜉),=𝑔𝑡,𝑢)),0<𝑡<1,=0,𝑢=𝑎𝑢=0,𝑣=𝑏𝑣 (2) where 1<𝛼, 𝛽<2, 0≤𝑎≤𝑏≤1, 0<𝜉<1, 𝑓,𝑔∶(0,1)×ℝ×ℝ→ℝ are given continuous functions and 𝐷^𝛼₀₊ is the standard Riemann-Liouville fractional derivative. By using the cone extension and cone compression Zhao et al. [23] studied the existence and nonexistence of positive solutions for a class of third order boundary value problem with integral boundary conditions in Banach spaces: 𝑥^′′′((𝑡𝑥(0)+𝑓𝑡,𝑥))=0,𝑡∈𝐽,=0,𝑥^′′(0)=0,𝑥(1)=¹₀𝑔(𝑡)𝑥(𝑡)𝑑𝑡. (3) Or 𝑥^′′′((𝑡𝑥(0)=+𝑓𝑡,𝑥))=0,𝑡∈𝐽,₁₀𝑔(𝑡)𝑥(𝑡)𝑑𝑡,𝑥^′′(0)=0,𝑥(1)=0, (4) where 𝐽=(0,1), 𝑓∈𝒞((0,1)×𝑃,𝑃), 𝑔∈𝐿¹(0,1) is nonnegative and 𝐷^𝛼₀₊ is the standard Riemann-Liouville fractional derivative; 𝑃 is a cone.

By using the cone extension and cone compression method Yang et al. [24] established sufficient conditions for the existence and nonexistence of positive solutions for a general class with integral boundary value problem for a coupled system of fractional equations: 𝐷^𝛼₀₊𝑢(𝑡)(𝑡)𝑓((𝑡𝐷+𝑎𝑡,𝑣))=0,0<𝑡<1,_𝛽0+𝑣(𝑡)(𝑡)𝑔((𝑡𝑢(0)(1)=+𝑏𝑡,𝑢))=0,0<𝑡<1,=0,𝑢¹₀𝜙(𝑡)𝑢(𝑡)𝑣(0)(1)=𝑑𝑡,=0,𝑣¹₀𝜑(𝑡)𝑣(𝑡)𝑑𝑡, (5) where 1<𝛼, 𝛽≤2, 𝑎,𝑏∈𝒞((0,1),(0,+∞)), 𝑓,𝑔∈𝒞((0,1)×(0,+∞),(0,+∞)), and 𝜙,𝜑∈𝐿¹(0,1) are nonnegative and 𝐷^𝛼₀₊ is the standard Riemann-Liouville fractional derivative.

However, all these works are obtained by getting a fixed point of a completely continuous operator in a cone of product Banach spaces 𝑋₁×𝑋₂. Thus, there is no essential difference between we consider the positive solution of coupled equations and positive solution of single equation. Furthermore, the authors cannot guarantee the obtained pair of positive solutions are all positive; maybe one is positive and another is zero. The features of nonlinear term are the same, which means the growth properties of nonlinear term are similar.

Motivated by the above mentioned works, in this paper, we consider the existence of positive solution to boundary value problem for a coupled system of nonlinear fractional differential equations as follows: 𝐷^𝛼₀₊𝑢(𝑡)((𝑡)(𝑡𝐷+𝑓𝑡,𝑢,𝑣))=0,0<𝑡<1,_𝛽0+𝑣(𝑡)((𝑡)(𝑡𝑢(0)(1)=+𝑔𝑡,𝑢,𝑣))=0,0<𝑡<1,=0,𝑢¹₀𝜙(𝑡)𝑢(𝑡)𝑣(0)(1)=𝑑𝑡,=0,𝑣¹₀𝜑(𝑡)𝑣(𝑡)𝑑𝑡, (6) where 1<𝛼, 𝛽≤2, 𝑓,𝑔∈𝒞((0,1)×(0,+∞)×(0,+∞),(0,+∞)), and 𝜙,𝜑∈𝐿¹(0,1) are nonnegative and 𝐷^𝛼₀₊ is the standard Riemann-Liouville fractional derivative. 𝑓 and 𝑔 have different features. It is very difficult to deal with our problem if we directly use the usual method because the features of nonlinear term are different. In order to overcome the difficulties, by using the ideals in [25–27], we need to consider our problem on the Cartesian product of two cones in the space 𝒞(0,1); thus we can better exploit the feature of two equations. We choose a cone 𝐾₁×𝐾₂ in the Cartesian product of two cones in 𝒞(0,1)×𝒞(0,1). By computing the fixed point index in 𝐾₁×𝐾₂, we obtain the existence of positive solution (𝑢,𝑣) of problem (6) such that 𝑢>0, 𝑣>0.

2. Preliminary

In this section, we review some basic facts which are used throughout this paper. We can see [10, 16, 25–29] for more details.

Definition 1. For a function 𝑓∶[𝑎,+∞)→ℝ, the Riemann-Liouville fractional integral of order 𝛼>0 is defined by 𝐼^𝛼_𝑎+𝑓(𝑡)=1Γ(𝛼)^𝑡_𝑎(𝑡−𝑠)^𝛼−1𝑓(𝑠)𝑑𝑠, (7)and the right-hand side is pointwise defined on [𝑎,+∞), where 𝑎∈ℝ and Γ is the gamma function.

Definition 2. For a function 𝑓∶[𝑎,+∞)→ℝ, the Riemann-Liouville fractional derivative of order 𝛼>0 is given by 𝐷^𝛼_𝑎+𝑓(𝑡)=1Γ(𝛼)𝑑^𝑛𝑑𝑡^𝑛−1𝑡_𝑎(𝑡−𝑠)^{𝑛−𝛼−1}𝑓(𝑠)𝑑𝑠, (8) and the right-hand side is pointwise defined on [𝑎,+∞), where 𝑎∈ℝ, 𝑛=[𝛼]+1 and Γ is the gamma function.

Let 𝐸 be a real Banach space with norm ‖‖ and let 𝐴∶𝐸→𝐸 be an operator. A nonempty, closed, convex set 𝑃⊂𝐸 is said to be a cone provided the following: (i) if 𝑥∈𝑃 and 𝜆≥0, then 𝜆𝑥∈𝑃; (ii) if 𝑥∈𝑃 and −𝑥∈𝑃, then 𝑥=0 are satisfied. 𝑢 is said to be a fixed point of 𝐴, if 𝑢=𝐴𝑢. Ω is a bounded open subset of 𝐸 with boundary 𝜕Ω. Denote 𝑖(𝐴,𝑃∩Ω,𝑃) the fixed point index of 𝐴.

Lemma 3 (see [25–27]). Let 𝐸 be a Banach space and let 𝐾_𝑖⊂𝐸 (𝑖=1,2) be a closed convex cone in 𝐸. For 𝑟_𝑖>0 (𝑖=1,2), denote 𝐾_𝑟𝑖={𝑢∈𝐾_𝑖∣‖𝑢‖<𝑟_𝑖}, 𝜕𝐾_𝑖={𝑢∈𝐾_𝑖∣‖𝑢‖=𝑟_𝑖}. Suppose 𝐴_𝑖∶𝐾_𝑖→𝐾_𝑖 is completely continuous. If 𝑢_𝑖≠𝐴_𝑖𝑢_𝑖, ∀𝑢_𝑖∈𝜕𝐾_𝑖, then 𝑖𝐴,𝐾_𝑟1×𝐾_𝑟2,𝐾₁×𝐾₂=𝑖𝐴₁,𝐾_𝑟1,𝐾₁⋅𝑖𝐴₂,𝐾_𝑟2,𝐾₂, (9) where 𝐴(𝑢,𝑣)=(𝐴₁𝑢,𝐴₂𝑣) and for all (𝑢,𝑣)∈(𝐾₁×𝐾₂).

In order to prove the existence of solution for problem (6), we can translate it into obtaining a fixed point of operator. Then we use the topological method to deal with it.

Theorem 4. Let 𝐸 be a Banach space and let 𝐾₁,𝐾₂⊂𝐸 be a closed convex cone in 𝐸. Denote 𝐾_𝑟𝑖={𝑢∈𝐾_𝑖∣‖𝑢‖<𝑟_𝑖}, 𝜕𝐾_𝑖={𝑢∈𝐾_𝑖‖𝑢‖=𝑟_𝑖}, where 𝑟>0, 𝑖=1,2. Let 𝐷=(𝐾_𝑅1⧵𝐾_𝑟1)×(𝐾_𝑅2⧵𝐾_𝑟2). Suppose 𝑇∶𝐾₁×𝐾₂→𝐾₁×𝐾₂ is completely continuous and 𝑇(𝑢,𝑣)=(𝑇₁(𝑢,𝑣),𝑇₂(𝑢,𝑣)). For 𝜆∈[0,1], let 𝐻_𝜆,1(𝑢,𝑣)=𝜆𝑇₁(𝑢,𝑣)+(1−𝜆)𝑇₁(),𝐻𝑢,0_𝜆,2(𝑢,𝑣)=𝜆𝑇₂(𝑢,𝑣)+(1−𝜆)𝑇₂(0,𝑣). (10) If 𝐻_𝜆,1, 𝐻_𝜆,2 satisfy the following conditions: (i)𝐻_𝜆,1 is uniformly cone extension about 𝑣∈𝐾_𝑅2 at 𝐾_𝑅1⧵𝐾_𝑟1, that is to say, 𝐻_𝜆,1<‖𝑢‖,∀𝑢∈𝜕𝐾_𝑟1,𝐻_𝜆,1>‖𝑢‖,∀𝑢∈𝜕𝐾_𝑅1; (11)(ii)𝐻_𝜆,2 is uniformly cone compression about 𝑢∈𝐾_𝑅1 at 𝐾_𝑅2⧵𝐾_𝑟2, that is to say, 𝐻_𝜆,2>‖𝑣‖,∀𝑣∈𝜕𝐾_𝑟2,𝐻_𝜆,2<‖𝑣‖,∀𝑣∈𝜕𝐾_𝑅2, (12)then 𝑖(𝑇,𝐷,𝐾₁×𝐾₂)=−1. Thus the mapping 𝑇 in the region 𝐷 has a fixed point.

Proof. Following a similar procedure to the method given in [25–27], we present a proof. We suppose that 𝑇₁(𝑢,0)=𝐴₁𝑢 and 𝑇₂(0,𝑣)=𝐴₂𝑣. Let 𝐴(𝑢,𝑣)=(𝐴₁𝑢,𝐴₂𝑣). For 𝜆∈[0,1], for any (𝑢,𝑣)∈𝐾₁×𝐾₂, set 𝐻((𝑢,𝑣),𝜆)=𝜆𝑇(𝑢,𝑣)+(1−𝜆)𝐴(𝑢,𝑣). (13) Since 𝑇 is completely continuous, hence the mapping 𝐻 is also completely continuous. By Lemma 3 and homotopy invariance of fixed point index (see [7, 28]), we have 𝑖𝑇,𝐷,𝐾₁×𝐾₂=𝑖𝐴₁,𝐾_𝑅1⧵𝐾_𝑟1,𝐾₁⋅𝑖𝐴₂,𝐾_𝑅2⧵𝐾_𝑟2,𝐾₂. (14)
Next, we compute 𝑖(𝑇,𝐷,𝐾₁×𝐾₂). By assumptions (i) and (ii), also using the properties of fixed point index (see [7, 28]), we have 𝑖𝐴₁,𝐾_𝑟1,𝐾₁=𝑖𝐴₂,𝐾_𝑅2,𝐾₂𝑖𝐴=1,₁,𝐾_𝑅1,𝐾₁=𝑖𝐴₂,𝐾_𝑟2,𝐾₂=0. (15)
Thanks to (15), we obtain 𝑖𝑇,𝐷,𝐾₁×𝐾₂=𝑖𝐴₁,𝐾_𝑅1⧵𝐾_𝑟1,𝐾₁⋅𝑖𝐴₂,𝐾_𝑅2⧵𝐾_𝑟2,𝐾₂=𝑖𝐴₁,𝐾_𝑅1,𝐾₁−𝑖𝐴₁,𝐾_𝑟1,𝐾₁⋅𝑖𝐴₂,𝐾_𝑅2,𝐾₂−𝑖𝐴₂,𝐾_𝑟2,𝐾₂=(0−1)⋅(1−0)=−1. (16) The proof is complete.

Remark 5. If 𝐻_𝜆,1, 𝐻_𝜆,2 were all cone extension or cone compression, we can obtain the same conclusions.

The following lemmas are necessary to prove the existence of positive solution for problem (6).

Lemma 6 (see [15]). Assume that ∫¹₀𝜙(𝑡)𝑢(𝑡)𝑑𝑡≠1. Then for any 𝜓∈𝒞(0,1), the solution of boundary value problem 𝐷^𝛼₀₊𝑢(𝑡)(𝑡)𝑢(0)(1)=+𝜓=0,0<𝑡<1,=0,𝑢¹₀𝜙(𝑡)𝑢(𝑡)𝑑𝑡, (17) is given by 𝑢(𝑡)=∫¹₀𝐺_1𝛼(𝑡,𝑠)𝜓(𝑠)𝑑𝑠, where 𝐺_1𝛼(𝑡,𝑠)=𝐺_2𝛼(𝑡,𝑠)+𝐺_3𝛼(𝑡,𝑠), 𝐺_2𝛼()=𝑡𝑡,𝑠^𝛼−1(1−𝑠)^𝛼−1−(𝑡−𝑠)^𝛼−1Γ(𝛼)𝑡,0≤𝑠≤𝑡≤1._𝛼−1(1−𝑠)^𝛼−1Γ(𝛼)𝐺,0≤𝑡≤𝑠≤1._3𝛼()=𝑡𝑡,𝑠^𝛼−11−∫¹₀𝜙(𝑡)𝑡^𝛼−1𝑑𝑡¹⁰𝜙(𝑡)𝐺_2𝛼(𝑡,𝑠)𝑑𝑡. (18)

Lemma 7 (see [15]). If ∫¹₀𝜙(𝑡)𝑡^𝛼−1𝑑𝑡∈[0,1), then Green’s function 𝐺_1𝛼(𝑡,𝑠) has the following properties: (i)𝐺_1𝛼(𝑡,𝑠)≥0 is continuous for any 𝑡,𝑠∈[0,1], 𝐺_1𝛼(𝑡,𝑠)>0 for any 𝑡,𝑠∈(0,1);(ii)𝐺_1𝛼(𝑡,𝑠)≤𝐺_1𝛼(𝑠) for any 𝑡,𝑠∈[0,1] and min_{𝑡∈[𝜃,1−𝜃]} 𝐺_1𝛼(𝑡,𝑠)≥𝛾_𝜃𝐺_1𝛼(𝑠), where 𝐺_1𝛼(𝑠)=𝐺_2𝛼(𝑠,𝑠)+𝐺_3𝛼(1,𝑠), 𝛾_𝜃=𝜃^𝛼−1, 𝜃∈(0,1/2).

3. Main Results

In this section, we will discuss the existence of positive solutions for problem (6). For a coupled system, many researchers transformed the problem into computing the fixed point index of the composition operator on the single cone in the Cartesian product space and the nonlinear terms in two equations have similar features. However it is very difficult to deal with our problem directly. Denote 𝒞⁺[0,1] the continuous function in [0,1] with nonnegative value. In order to make full use of the different nature of the two equations, we construct single cone 𝐾_𝑖 and open set 𝐾_𝑅𝑖⧵𝐾_𝑟𝑖, 𝑖=1,2 in 𝒞⁺[0,1]. Thus we transform the problem into computing the fixed point index of operator that is defined in Cartesian product of two cones in open set 𝐷=(𝐾_𝑅1⧵𝐾_𝑟1)×(𝐾_𝑅2⧵𝐾_𝑟2) of product cone 𝐾₁×𝐾₂.

3.1. The Transformation of the Problem

Firstly, we construct the relevant cone 𝐾 in Banach space 𝒞⁺[0,1]. Then we transform the problem into computing the fixed point index of Cartesian product of cone with the nature of analytical property of Green’s function.

Following, we introduce notations: 𝒞⁺[0,1]={𝑢∈𝒞[0,1]∣𝑢(𝑡)≥0,∀𝑡∈[0,1]}, (19) with the norm ‖𝑢‖=max_𝑡∈[0,1]{|𝑢(𝑡)|}. Clearly, it is a Banach space. Define 𝐾₁=𝑢∈𝒞⁺[](𝑡)≥10,1∣𝑢2^𝛼−1‖𝑢(𝑡,[]1)‖∀𝑡∈𝜃,1−𝜃,𝜃∈0,2,𝐾₂=𝑢∈𝒞⁺[](𝑡)≥10,1∣𝑢2^𝛽−1‖𝑢(𝑡,[]1)‖∀𝑡∈𝜃,1−𝜃,𝜃∈0,2. (20)

For 𝜆∈[0,1],𝑢,𝑣∈𝒞⁺[0,1], we define the operators 𝑇_𝜆,1(𝑢,𝑣)∶𝒞⁺[0,1]×𝒞⁺[0,1]⟶𝒞⁺[],𝑇0,1_𝜆,2(𝑢,𝑣)∶𝒞⁺[0,1]×𝒞⁺[0,1]⟶𝒞⁺[],𝑇0,1_𝜆(𝑢,𝑣)∶𝒞⁺[0,1]×𝒞⁺[0,1]⟶𝒞⁺[0,1]×𝒞⁺[0,1], (21) as follows: 𝑇_𝜆,1(𝑢,𝑣)=¹₀𝐺_1𝛼()((𝑠)(𝑠+()𝑓((𝑠))𝑇𝑡,𝑠𝜆𝑓𝑠,𝑢,𝑣))1−𝜆𝑠,𝑢,0𝑑𝑠,_𝜆,2(𝑢,𝑣)=¹₀𝐺_1𝛽()((𝑠)(𝑠+()𝑔((𝑠𝑇𝑡,𝑠𝜆𝑔𝑠,𝑢,𝑣))1−𝜆𝑠,0,𝑣))𝑑𝑠,_𝜆(𝑢,𝑣)(𝑡)=𝑇_𝜆,1(𝑢,𝑣)(𝑡),𝑇_𝜆,2(𝑢,𝑣)(𝑡). (22)

Then we translate that problem (6) possesses a pair of solutions (𝑢,𝑣) if and only if 𝑇_𝜆(𝑢,𝑣)(𝑡) have a fixed point in 𝐾₁×𝐾₂.

Lemma 8. The mapping 𝑇_𝜆∶𝐾₁×𝐾₂→𝐾₁×𝐾₂ as mentioned above is completely continuous.

Proof. For any (𝑢,𝑣)∈𝐾₁×𝐾₂, we want to prove 𝑇_𝜆(𝑢,𝑣)∈𝐾₁×𝐾₂, that is, to prove 𝑇_𝜆,1(𝑢,𝑣)∈𝐾₁ and 𝑇_𝜆,2(𝑢,𝑣)∈𝐾₂. By the preceding definition and the nature of Green's function, we have for 𝑡∈[𝜃,1−𝜃], 𝜃∈(0,1/2)𝑇_𝜆,1()=𝑢,𝑣₁₀𝐺_1𝛼()((𝑠)(𝑠+()𝑓((𝑠))≥𝑡,𝑠𝜆𝑓𝑠,𝑢,𝑣))1−𝜆𝑠,𝑢,0𝑑𝑠₁₀𝛾_𝜃𝐺_1𝛼(𝑠((𝑠)(𝑠+()𝑓((𝑠)))𝜆𝑓𝑠,𝑢,𝑣))1−𝜆𝑠,𝑢,0𝑑𝑠=𝛾_𝜃¹₀𝐺_1𝛼(𝑠)((𝑠)(𝑠+()𝑓((𝑠))≥1𝜆𝑓𝑠,𝑢,𝑣))1−𝜆𝑠,𝑢,0𝑑𝑠2^𝛼−11₀𝐺_1𝛼()((𝑠)(𝑠+()𝑓((𝑠))≥1𝑡,𝑠𝜆𝑓𝑠,𝑢,𝑣))1−𝜆𝑠,𝑢,0𝑑𝑠2^𝛼−1𝑇_𝜆,1(𝑢,𝑣). (23)
Similarly, 𝑇_𝜆,2(𝑢,𝑣)≥(1/2)^𝛽−1‖𝑇_𝜆,2(𝑢,𝑣)‖, 𝑡∈[𝜃,1−𝜃], 𝜃∈(0,1/2). Thus, 𝑇_𝜆,1(𝑢,𝑣)∈𝐾₁, 𝑇_𝜆,2(𝑢,𝑣)∈𝐾₂, and 𝑇_𝜆(𝐾₁×𝐾₂)⊂𝐾₁×𝐾₂.
It is easy to prove that 𝑇_𝜆(𝑢,𝑣) is continuous in 𝐾₁×𝐾₂.
So as Arzela-Ascoli theorem, we prove that 𝑇_𝜆∶𝐾₁×𝐾₂→𝐾₁×𝐾₂ is completely continuous.

3.2. The Main Result and Its Proof

Now, we present our main result.

Theorem 9. Assume that the nonlinear terms of the problem (6) 𝑓 and 𝑔 meet the following conditions: 𝐻₁𝑎₁lim_𝑢→0+supmax_[]𝑡∈0,1𝑓(𝑡,𝑢,𝑣)𝑢=0𝑖𝑠𝑒𝑠𝑡𝑎𝑏𝑙𝑖𝑠ℎ𝑒𝑑𝑐𝑜𝑛𝑠𝑖𝑠𝑡𝑒𝑛𝑡𝑎𝑏𝑜𝑢𝑡𝑣∈ℝ⁺,𝑏₁lim_𝑢→+∞infmin_[]𝑡∈0,1𝑓(𝑡,𝑢,𝑣)𝑢=+∞𝑖𝑠𝑒𝑠𝑡𝑎𝑏𝑙𝑖𝑠ℎ𝑒𝑑𝑐𝑜𝑛𝑠𝑖𝑠𝑡𝑒𝑛𝑡𝑎𝑏𝑜𝑢𝑡𝑣∈ℝ⁺,𝐻₂𝑎₂lim_𝑣→0+infmin_[]𝑡∈0,1𝑔(𝑡,𝑢,𝑣)𝑣=+∞𝑖𝑠𝑒𝑠𝑡𝑎𝑏𝑙𝑖𝑠ℎ𝑒𝑑𝑐𝑜𝑛𝑠𝑖𝑠𝑡𝑒𝑛𝑡𝑎𝑏𝑜𝑢𝑡𝑢∈ℝ⁺,𝑏₂lim_𝑣→+∞supmax_[]𝑡∈0,1𝑔(𝑡,𝑢,𝑣)𝑣=0𝑖𝑠𝑒𝑠𝑡𝑎𝑏𝑙𝑖𝑠ℎ𝑒𝑑𝑐𝑜𝑛𝑠𝑖𝑠𝑡𝑒𝑛𝑡𝑎𝑏𝑜𝑢𝑡𝑢∈ℝ⁺,𝐻₃lim_𝑢→+∞supmax_[]𝑡∈0,1𝑔()(𝑣)[],𝑡,𝑢,𝑣=ℎ𝑖𝑠𝑒𝑠𝑡𝑎𝑏𝑙𝑖𝑠ℎ𝑒𝑑𝑐𝑜𝑛𝑠𝑖𝑠𝑡𝑒𝑛𝑡𝑎𝑏𝑜𝑢𝑡𝑣∈0,𝑀 (24) where ℎ(𝑣) is continuous in ℝ⁺ and 𝑀 is a constant.
Then the problem (6) has at least a pair of positive solutions.

Proof. We select an appropriate open set 𝐷=(𝐾_𝑅1⧵𝐾_𝑟1)×(𝐾_𝑅2⧵𝐾_𝑟2) in the product of cone 𝐾₁×𝐾₂; then we will verify that the solution operator meets the conditions of Theorem 4.
In the following, we separate our proof into four steps.
Step 1. Select 𝑟₁>0, in such a fashion that ‖𝑇_𝜆,1(𝑢,𝑣)‖<‖𝑢‖, where 𝜆∈[0,1], (𝑢,𝑣)∈𝜕𝐾_𝑟1×𝐾₁. Based on assuming (𝐻₁)(𝑎₁), for 𝜀₁=(∫¹₀𝐺_1𝛼(𝑠)𝑑𝑠)⁻¹>0, there exists 𝑟₁>0, such that 𝑓(𝑡,𝑢,𝑣)<𝜀₁𝑢, for all 𝑡∈[0,1], 0≤𝑢≤𝑟₁, 𝑣≥0. Then when 𝜆∈[0,1],(𝑢,𝑣)∈𝜕𝐾_𝑟1×𝐾₁, we have for 𝑡∈[0,1]𝑇_𝜆,1(𝑢,𝑣)=max_[]𝑡∈0,1¹₀𝐺_1𝛼(((𝑠)(𝑠+()𝑓((𝑠))𝑡,𝑠)𝜆𝑓𝑠,𝑢,𝑣))1−𝜆𝑠,𝑢,0𝑑𝑠<max_[]𝑡∈0,1¹₀𝐺_1𝛼(𝑠)𝜆𝜀₁𝑢(𝑠)+(1−𝜆)𝜀₁𝑢(𝑠)𝑑𝑠≤max_[]𝑡∈0,1¹₀𝐺_1𝛼(𝑠)𝜀₁|𝑢(𝑠)|𝑑𝑠≤𝜀₁¹₀𝐺_1𝛼(𝑠)max_[]𝑡∈0,1|𝑢(𝑠)|𝑑𝑠≤𝜀₁¹₀𝐺_1𝛼(𝑠)𝑑𝑠‖𝑢‖=‖𝑢‖. (25) That is to say, ‖𝑇_𝜆,1(𝑢,𝑣)‖<‖𝑢‖, where 𝜆∈[0,1], (𝑢,𝑣)∈𝜕𝐾_𝑟1×𝐾₁.
Step 2. Select 𝑅₁>𝑟₁; we claim that ‖𝑇_𝜆,1(𝑢,𝑣)‖>‖𝑢‖, where 𝜆∈[0,1], (𝑢,𝑣)∈𝜕𝐾_𝑅1×𝐾₁. In view of the assumption (𝐻₁)(𝑏₁), for 𝑁₁=(𝛾_𝜃∫¹₀𝐺_1𝛼(𝑠)𝑑𝑠)⁻¹>0, there exists 𝑅₀>0, such that 𝑓(𝑡,𝑢,𝑣)<𝑁₁𝑢, for all 𝑡∈[0,1], 𝑢≥𝑅₀, 𝑣≥0. Let 𝑅₁>max{𝑟₁,𝛾_𝜃𝑅₀}. Then when 𝜆∈[0,1],(𝑢,𝑣)∈𝜕𝐾_𝑅1×𝐾₁, we have 𝑡∈[0,1]𝑇_𝜆,1(𝑢,𝑣)=max_[]𝑡∈0,1¹₀𝐺_1𝛼(((𝑠)(𝑠+()𝑓((𝑠))𝑡,𝑠)𝜆𝑓𝑠,𝑢,𝑣))1−𝜆𝑠,𝑢,0𝑑𝑠>max_[]𝑡∈0,1¹₀𝛾_𝜃𝐺_1𝛼(𝑠)𝜆𝑁₁𝑢(𝑠)+(1−𝜆)𝑁₁𝑢(𝑠)𝑑𝑠≥max_[]𝑡∈0,1𝛾_𝜃¹₀𝐺_1𝛼(𝑠)𝑁₁|𝑢(𝑠)|𝑑𝑠≥𝑁₁𝛾_𝜃¹₀𝐺_1𝛼(𝑠)max_()𝑡∈0,1|𝑢(𝑠)|𝑑𝑠≥𝑁₁𝛾_𝜃¹₀𝐺_1𝛼(𝑠)𝑑𝑠‖𝑢‖=‖𝑢‖. (26) That is to say, ‖𝑇_𝜆,1(𝑢,𝑣)‖>‖𝑢‖, where 𝜆∈[0,1], (𝑢,𝑣)∈𝜕𝐾_𝑅1×𝐾₁.
Step 3. Select 𝑟₂>0, such that ‖𝑇_𝜆,2(𝑢,𝑣)‖>‖𝑣‖, where 𝜆∈[0,1], (𝑢,𝑣)∈𝜕𝐾_𝑟2×𝐾₂. In view of the assumption (𝐻₂)(𝑎₂), for 𝑁₂=(𝛾_𝜃∫¹₀𝐺_1𝛼(𝑠)𝑑𝑠)⁻¹>0, there exists 𝑟₂>0, such that 𝑔(𝑡,𝑢,𝑣)>𝑁₂𝑣, for all 𝑡∈[0,1], 0≤𝑣≤𝑟₂, 𝑢≥0. Hence when 𝜆∈[0,1], (𝑢,𝑣)∈𝜕𝐾_𝑟2×𝐾₂, we have for 𝑡∈[0,1]𝑇_𝜆,2(𝑢,𝑣)=max_[]𝑡∈0,1¹₀𝐺_1𝛽(((𝑠)(𝑠+()𝑔((𝑠𝑡,𝑠)𝜆𝑔𝑠,𝑢,𝑣))1−𝜆𝑠,0,𝑣))𝑑𝑠>max_[]𝑡∈0,1¹₀𝛾_𝜃𝐺_1𝛽(𝑠)𝜆𝑁₂𝑣(𝑠)+(1−𝜆)𝑁₂𝑣(𝑠)𝑑𝑠≥max_[]𝑡∈0,1𝛾_𝜃¹₀𝐺_1𝛽(𝑠)𝑁₂|𝑣(𝑠)|𝑑𝑠≥𝑁₂𝛾_𝜃¹₀𝐺_1𝛽(𝑠)max_()𝑡∈0,1|𝑣(𝑠)|𝑑𝑠≥𝑁₂𝛾_𝜃¹₀𝐺_1𝛼(𝑠)𝑑𝑠‖𝑣‖=‖𝑣‖. (27) That is to say, ‖𝑇_𝜆,2(𝑢,𝑣)‖>‖𝑣‖, where 𝜆∈[0,1], (𝑢,𝑣)∈𝜕𝐾_𝑟2×𝐾₂.
Step 4. Select 𝑅₂>𝑟₂, such that ‖𝑇_𝜆,2(𝑢,𝑣)‖<‖𝑣‖, where 𝜆∈[0,1], (𝑢,𝑣)∈𝜕𝐾_𝑅2×𝐾₂. Based on the condition (𝐻₂)(𝑏₂), for 𝜀₂=(∫¹₀𝐺_1𝛼(𝑠)𝑑𝑠)⁻¹>0, there exists 𝑅^′>0, such that 𝑔(𝑡,𝑢,𝑣)<𝜀₂𝑣, for all 𝑡∈[0,1], 𝑣≥𝑅^′, 𝑢≥0. Furthermore, paying attention to the assumptions (𝐻₃), we can select 𝑅₂>0,𝑅₂ which is sufficiently large and max_[]𝑡∈0,1𝑔(𝑡,𝑢,𝑣)∣0≤𝑣≤𝑅^′,𝑢≥0<𝜀₂𝑅₂. (28) Thus 𝑔(𝑡,𝑢,𝑣)<𝜀₂𝑅₂, where 𝑡∈[0,1], 0≤𝑣≤𝑅₂, 𝑢≥0. Hence when 𝜆∈[0,1], (𝑢,𝑣)∈𝜕𝐾_𝑅2×𝐾₂, we have for 𝑡∈[0,1]𝑇_𝜆,2(𝑢,𝑣)=max_[]𝑡∈0,1¹₀𝐺_1𝛽(((𝑠)(𝑠+()𝑔((𝑠𝑡,𝑠)𝜆𝑔𝑠,𝑢,𝑣))1−𝜆𝑠,0,𝑣))𝑑𝑠<max_[]𝑡∈0,1¹₀𝐺_1𝛽(𝑠)𝜆𝜀₂𝑅₂+(1−𝜆)𝜀₂𝑅₂𝑑𝑠≤max_[]𝑡∈0,1¹₀𝐺_1𝛽(𝑠)𝜀₂𝑅₂𝑑𝑠≤𝜀₂𝑅₂¹₀𝐺_1𝛽(𝑠)max_[]𝑡∈0,1𝑣(𝑠)𝑑𝑠≤𝜀₂𝑅₂¹₀𝐺_1𝛼(𝑠)𝑑𝑠=‖𝑣‖. (29) That is to say, ‖𝑇_𝜆,2(𝑢,𝑣)‖<‖𝑣‖, where 𝜆∈[0,1], (𝑢,𝑣)∈𝜕𝐾_𝑅2×𝐾₂.
Combining with (25)–(29), we know the conditions of Theorem 4 are all established. Thus the completely continuous operator 𝑇 has a fixed point in 𝐷; that is to say, at least a pair of positive solutions exist for the problem (6). The proof is complete.

Remark 10. Binding (27), (29) and the assumption (𝐻₁)(𝑎₁), we can get a nonnegative solution (0,𝑣) of the original equations (6). In fact, by (27) and (29), we know the fixed point index 𝑖(𝑇_0,2(𝑢,𝑣),𝐾_𝑅2⧵𝐾_𝑟2,𝐾₂)=−1; that is to say, the second equation of the original equations has a solution (0,𝑣). Further, by hypothesis (𝐻₁)(𝑎₁), we have 𝑓(𝑡,0,𝑣(𝑡))=0, so the solution of the second equation is also the solution of the first equation.

Conflict of Interests

The authors declare that there is no conflict of interests regarding the publication of this paper.

Acknowledgments

This work was partly supported by the Fundamental Research Funds for the Central Universities Grant (2012LWN56) and the Nature Science Foundation of Jiangsu Province Grant (BK20130170).