Positive Solution to Nonzero Boundary Values Problem for a Coupled System of Nonlinear Fractional Differential Equations

Abstract

We consider the existence and uniqueness of positive solution to nonzero boundary values problem for a coupled system of fractional differential equations. The differential operator is taken in the standard Riemann-Liouville sense. By using Banach fixed point theorem and nonlinear differentiation of Leray-Schauder type, the existence and uniqueness of positive solution are obtained. Two examples are given to demonstrate the feasibility of the obtained results.

1. Introduction

Fractional differential equation can describe many phenomena in various fields of science and engineering such as control, porous media, electrochemistry, viscoelasticity, and electromagnetic. There are many papers dealing with the existence and uniqueness of solution for nonlinear fractional differential equation; see, for example, [1–5]. In [1], the authors investigated a singular coupled system with initial value problems of fractional order. In [2], Su discussed a boundary value problem of coupled system with zero boundary values. By means of Schauder fixed point theorem, the existence of the solution is obtained. The nonzero boundary values problem of nonlinear fractional differential equations is more difficult and complicated. No contributions exist, as far as we know, concerning the existence of positive solution for coupled system of nonlinear fractional differential equations with nonzero boundary values.

In this paper, we consider the existence and uniqueness of positive solution to nonzero boundary values problem for a coupled system of nonlinear fractional differential equations: 𝐷^𝛼𝐷𝑢(𝑡)+𝑓(𝑡,𝑣(𝑡))=0,0<𝑡<1,_𝛽𝑣(𝑡)+𝑔(𝑡,𝑢(𝑡))=0,0<𝑡<1,𝑢(0)=0,𝑢(1)=𝑎𝑢(𝜉),𝑣(0)=0,𝑣(1)=𝑏𝑣(𝜉), (1.1) where 1<𝛼<2,1<𝛽<2,0⩽𝑎,𝑏⩽1,0<𝜉<1, 𝑓,𝑔∶[0,1]×[0,+∞)→[0,+∞) are given functions, and 𝐷 is the standard Riemann-Liouville differentiation. By using Banach fixed point theorem and nonlinear differentiation of Leray-Schauder type, some sufficient conditions for the existence and uniqueness of positive solution to the above coupled boundary values problem are obtained.

The rest of the paper is organized as follows. In Section 2, we introduce some basic definitions and preliminaries used in later. In Section 3, the existence and uniqueness of positive solution for the coupled boundary values problem (1.1) will be discussed, and examples are given to demonstrate the feasibility of the obtained results.

2. Basic Definitions and Preliminaries

In this section, we introduce some basic definitions and lemmas which are used throughout this paper.

Definition 2.1 (see [6, 7]). The fractional integral of order 𝛼(𝛼>0) of a function 𝑦∶(0,∞)→𝑅 is given by 𝐼^𝛼1𝑦(𝑡)=Γ(𝛼)^𝑡₀(𝑡−𝑠)^𝛼−1𝑦(𝑠)𝑑𝑠, (2.1) provided that the right side is pointwise defined on (0,∞).

Definition 2.2 (see [6, 7]). The fractional derivative of order 𝛼>0 of a continuous function 𝑦∶(0,∞)→𝑅 is given by 𝐷^𝛼1𝑦(𝑡)=(𝑑Γ(𝑛−𝛼))𝑑𝑡^𝑛𝑡₀(𝑡−𝑠)^{𝑛−𝛼−1}𝑦(𝑠)𝑑𝑠, (2.2) where 𝑛=[𝛼]+1 provided that the right side is pointwise defined on (0,∞).

Remark 2.3 (see [3]). The following properties are useful for our discussion:(1)𝐼^𝛼𝐷^𝛼𝑢(𝑡)=𝑢(𝑡)−∑^𝑁_𝑘=1𝐶_𝑘𝑡^𝛼−𝑘,𝐷^𝛼𝑢(𝑡)∈𝐶(0,1)⋂𝐿(0,1), 𝐶_𝑘∈𝑅, 𝑁=[𝛼]+1,(2)𝐷^𝛼𝐼^𝛼𝑢(𝑡)=𝑢(𝑡),(3)𝐷^𝛼𝑡^𝛾=Γ(𝛾+1)/Γ(𝛾+1−𝛼)𝑡^𝛾−𝛼,𝛼>0,𝛾>−1,𝛾>𝛼−1,𝑡>0.

Lemma 2.4 (the nonlinear alternative of Leray and Schauder type [8]). Let 𝐸 be a Banach space with 𝐶⊆𝐸 closed and convex. Let 𝑈 be a relatively open subset of 𝐶 with 0∈𝑈 and let 𝑇∶𝑈→𝐶 be a continuous and compact mapping. Then either(a)the mapping 𝑇 has a fixed point in 𝑈, or(b)there exist 𝑢∈𝜕𝑈 and 𝜆∈(0,1) with 𝑢=𝜆𝑇𝑢. Consider 𝐷^𝛼𝑢(𝑡)+𝑦(𝑡)=0,0<𝑡<1,𝑢(0)=0,𝑢(1)=𝑎𝑢(𝜉), (2.3) then one has the following lemma.

Lemma 2.5. Let 𝑦∈𝐶[0,1] and 1<𝛼<2, then 𝑢(𝑡) is a solution of BVP (2.3) if and only if 𝑢(𝑡) is a solution of the integral equation: 𝑢(𝑡)=¹₀𝐺₁(𝑡,𝑠)𝑦(𝑠)𝑑𝑠, (2.4) where 𝐺₁(𝑡,𝑠)=[𝑡(1−𝑠)]^𝛼−1−𝑎𝑡^𝛼−1(𝜉−𝑠)^𝛼−1−(𝑡−𝑠)^𝛼−1(1−𝑎𝜉^𝛼−1)(1−𝑎𝜉^𝛼−1)Γ(𝛼),0⩽𝑠⩽𝑡⩽1,𝑠⩽𝜉,[𝑡(1−𝑠)]^𝛼−1−(𝑡−𝑠)^𝛼−1(1−𝑎𝜉^𝛼−1)(1−𝑎𝜉^𝛼−1)Γ(𝛼),0<𝜉⩽𝑠⩽𝑡⩽1,[𝑡(1−𝑠)]^𝛼−1−𝑎𝑡^𝛼−1(𝜉−𝑠)^𝛼−1(1−𝑎𝜉^𝛼−1)Γ(𝛼),0⩽𝑡⩽𝑠⩽𝜉⩽1,[𝑡(1−𝑠)]^𝛼−1(1−𝑎𝜉^𝛼−1)Γ(𝛼),0⩽𝑡⩽𝑠⩽1,𝜉⩽𝑠. (2.5)

Proof. Assume that 𝑢(𝑡) is a solution of BVP (2.3), then by Remark 2.3, we have 𝑢(𝑡)=−𝐼^𝛼𝑦(𝑡)+𝐶₁𝑡^𝛼−1+𝐶₂𝑡^𝛼−2=−^𝑡₀(𝑡−𝑠)^𝛼−1Γ(𝛼)𝑦(𝑠)𝑑𝑠+𝐶₁𝑡^𝛼−1+𝐶₂𝑡^𝛼−2. (2.6)
By (2.3), we have 𝐶₂=0,𝐶₁=¹₀(1−𝑠)^𝛼−1Γ(𝛼)(1−𝑎𝜉^𝛼−1)𝑦(𝑠)𝑑𝑠−𝑎^𝜉₀(𝜉−𝑠)^𝛼−1Γ(𝛼)(1−𝑎𝜉^𝛼−1)𝑦(𝑠)𝑑𝑠. (2.7)
Therefore, we obtain 𝑢(𝑡)=−^𝑡₀(𝑡−𝑠)^𝛼−1Γ(𝛼)𝑦(𝑠)𝑑𝑠+¹₀𝑡^𝛼−1(1−𝑠)^𝛼−1Γ(𝛼)(1−𝑎𝜉^𝛼−1)𝑦(𝑠)𝑑𝑠−𝑎^𝜉₀𝑡^𝛼−1(𝜉−𝑠)^𝛼−1Γ(𝛼)(1−𝑎𝜉^𝛼−1)=𝑦(𝑠)𝑑𝑠₁₀𝐺₁(𝑡,𝑠)𝑦(𝑠)𝑑𝑠. (2.8)
Conversely, if 𝑢(𝑡) is a solution of integral equation (2.4), using the relation 𝐷^𝛼𝑡^𝛼−𝑚=0,𝑚=1,2,…,𝑁, where 𝑁 is the smallest integer greater than or equal to 𝛼 [3, Remark  2.1], we have 𝐷^𝛼𝑢(𝑡)=−𝐷^𝛼(^𝑡₀(𝑡−𝑠)^𝛼−1Γ(𝛼)𝑦(𝑠)𝑑𝑠)+𝐷^𝛼𝑡^𝛼−1[¹₀(1−𝑠)^𝛼−1Γ(𝛼)(1−𝑎𝜉^𝛼−1)𝑦(𝑠)𝑑𝑠−𝑎^𝜉₀(𝜉−𝑠)^𝛼−1Γ(𝛼)(1−𝑎𝜉^𝛼−1)𝑦(𝑠)𝑑𝑠]=−𝐷^𝛼𝐼^𝛼𝑦(𝑡)=−𝑦(𝑡). (2.9) A simple computation showed 𝑢(0)=0,𝑢(1)=𝑎𝑢(𝜉). The proof is complete.

Let 𝐺₂(𝑡,𝑠)=[𝑡(1−𝑠)]^𝛽−1−𝑏𝑡^𝛽−1(𝜉−𝑠)^𝛽−1−(𝑡−𝑠)^𝛽−1(1−𝑏𝜉^𝛽−1)(1−𝑏𝜉^𝛽−1)Γ(𝛽),0⩽𝑠⩽𝑡⩽1,𝑠⩽𝜉,[𝑡(1−𝑠)]^𝛽−1−(𝑡−𝑠)^𝛽−1(1−𝑏𝜉^𝛽−1)(1−𝑏𝜉^𝛽−1)Γ(𝛽),0<𝜉⩽𝑠⩽𝑡⩽1,[𝑡(1−𝑠)]^𝛽−1−𝑏𝑡^𝛽−1(𝜉−𝑠)^𝛽−1(1−𝑏𝜉^𝛽−1)Γ(𝛽),0⩽𝑡⩽𝑠⩽𝜉⩽1,[𝑡(1−𝑠)]^𝛽−1(1−𝑏𝜉^𝛽−1)Γ(𝛽),0⩽𝑡⩽𝑠⩽1,𝜉⩽𝑠, (2.10) we call 𝐺(𝑡,𝑠)=(𝐺₁(𝑡,𝑠),𝐺₂(𝑡,𝑠)) Green's function of the boundary value problem (1.1).

Lemma 2.6. Let 0⩽𝑎,𝑏⩽1, then the function 𝐺(𝑡,𝑠) is continuous and satisfies(1)𝐺(𝑡,𝑠)>0,for𝑡,𝑠∈(0,1),(2)𝐺(𝑡,𝑠)⩽𝐺(𝑠,𝑠),for𝑡,𝑠∈(0,1).

Proof. It is easy to prove that 𝐺(𝑡,𝑠) is continuous on [0,1]×[0,1], here we omit it. Now we prove 𝐺₁(𝑡,𝑠)>0. Let 𝑔₁(𝑡,𝑠)=[𝑡(1−𝑠)]^𝛼−1−𝑎𝑡^𝛼−1(𝜉−𝑠)^𝛼−1−(𝑡−𝑠)^𝛼−1(1−𝑎𝜉^𝛼−1)(1−𝑎𝜉^𝛼−1𝑔)Γ(𝛼),0<𝑠⩽𝑡⩽1,𝑠⩽𝜉,₂(𝑡,𝑠)=[𝑡(1−𝑠)]^𝛼−1−(𝑡−𝑠)^𝛼−1(1−𝑎𝜉^𝛼−1)(1−𝑎𝜉^𝛼−1𝑔)Γ(𝛼),0<𝜉⩽𝑠⩽𝑡⩽1,₃(𝑡,𝑠)=[𝑡(1−𝑠)]^𝛼−1−𝑎𝑡^𝛼−1(𝜉−𝑠)^𝛼−1(1−𝑎𝜉^𝛼−1𝑔)Γ(𝛼),0<𝑡⩽𝑠⩽𝜉⩽1,₄(𝑡,𝑠)=[𝑡(1−𝑠)]^𝛼−1(1−𝑎𝜉^𝛼−1)Γ(𝛼),0<𝑡⩽𝑠⩽1,𝜉⩽𝑠. (2.11) We only need to prove 𝑔₁(𝑡,𝑠)>0,0<𝑠⩽𝑡⩽1,𝑠⩽𝜉. Since [𝑡(1−𝑠)]^𝛼−1−𝑎𝑡^𝛼−1(𝜉−𝑠)^𝛼−1−(𝑡−𝑠)^𝛼−1(1−𝑎𝜉^𝛼−1)=𝑡^𝛼−1[(1−𝑠)]^𝛼−1−𝑎(𝜉−𝑠)^𝛼−1𝑠−(1−𝑡)^𝛼−1(1−𝑎𝜉^𝛼−1), (2.12) set 𝑔(𝑡)=(1−𝑠)^𝛼−1−𝑎(𝜉−𝑠)^𝛼−1−(1−𝑠/𝑡)^𝛼−1(1−𝑎𝜉^𝛼−1), we have 𝑔^′𝑠(𝑡)=−(𝛼−1)(1−𝑡)^𝛼−2𝑠𝑡²(1−𝑎𝜉^𝛼−1)⩽0,for0<𝑠<𝑡⩽1,𝑠⩽𝜉. (2.13)
Then 𝑔(𝑡) is decreasing on (0,1). Meanwhile, 𝑔(1)=(1−𝑠)^𝛼−1−𝑎(𝜉−𝑠)^𝛼−1−(1−𝑠)^𝛼−1(1−𝑎𝜉^𝛼−1)=𝑎𝜉^𝛼−1[(1−𝑠)^𝛼−1𝑠−(1−𝜉)^𝛼−1]>0,0<𝑠<𝑡⩽1,𝑠⩽𝜉. (2.14)
Therefore, 𝑔₁(𝑡,𝑠)>0, for 0<𝑠<𝑡⩽1,𝑠⩽𝜉. Clearly 𝑔₁(𝑡,𝑠)>0,𝑡=𝑠, so 𝑔₁(𝑡,𝑠)>0,𝑠,𝑡∈(0,1). It is easy to show that 𝑔₂(𝑡,𝑠)>0,𝑔₃(𝑡,𝑠)>0, 𝑔₄(𝑡,𝑠)>0. Hence, 𝐺₁(𝑡,𝑠)>0,𝑠,𝑡∈(0,1).
Similarly, 𝐺₂(𝑡,𝑠)>0,𝑠,𝑡∈(0,1). The proof of (1) is completed.
Let 𝑔₂(𝑡)=[𝑡(1−𝑠)]^𝛼−1−(𝑡−𝑠)^𝛼−1(1−𝑎𝜉^𝛼−1)(1−𝑎𝜉^𝛼−1)Γ(𝛼),0<𝜉⩽𝑠⩽𝑡⩽1, (2.15) then, 𝑔^′₂(𝑡)=(𝛼−1)𝑡^𝛼−2[(1−𝑠)]^𝛼−1−(1−𝑠/𝑡)^𝛼−2(1−𝑎𝜉^𝛼−1)(1−𝑎𝜉^𝛼−1)Γ(𝛼),0<𝜉⩽𝑠<𝑡⩽1,[(1−𝑠)]^𝛼−1−(1−𝑠/𝑡)^𝛼−2(1−𝑎𝜉^𝛼−1)⩽[(1−𝑠)]^𝛼−1−(1−𝑠)^𝛼−2(1−𝑎𝜉^𝛼−1)=[(1−𝑠)]^𝛼−2(𝑎𝜉^𝛼−1−𝑠)⩽0,0<𝜉⩽𝑠<𝑡⩽1, (2.16) therefore, 𝑔^′₂(𝑡)⩽0,0<𝜉⩽𝑠<𝑡⩽1. (2.17) So, 𝑔₂(𝑡,𝑠) is decreasing with respect to 𝑡. Similarly, 𝑔₁(𝑡,𝑠) is decreasing with respect to 𝑡. Also 𝑔₃(𝑡,𝑠) and 𝑔₄(𝑡,𝑠) are increasing with respect to 𝑡. We obtain that 𝐺₁(𝑡,𝑠) is decreasing with respect to 𝑡 for 𝑠⩽𝑡 and increasing with respect to 𝑡 for 𝑡⩽𝑠.
With the use of the monotonicity of 𝐺₁(𝑡,𝑠), we have max_0⩽𝑡⩽1𝐺₁(𝑡,𝑠)=𝐺₁(𝑠,𝑠)=[𝑠(1−𝑠)]^𝛼−1−𝑎[𝑠(𝜉−𝑠)]^𝛼−1Γ(𝛼)(1−𝑎𝜉^𝛼−1),𝑠∈(0,𝜉],[𝑠(1−𝑠)]^𝛼−1Γ(𝛼)(1−𝑎𝜉^𝛼−1),𝑠∈[𝜉,1). (2.18) Similarly, max_0⩽𝑡⩽1𝐺₂(𝑡,𝑠)=𝐺₂(𝑠,𝑠)=[𝑠(1−𝑠)]^𝛽−1−𝑏[𝑠(𝜉−𝑠)]^𝛽−1Γ(𝛽)(1−𝑏𝜉^𝛽−1),𝑠∈(0,𝜉],[𝑠(1−𝑠)]^𝛽−1Γ(𝛽)(1−𝑏𝜉^𝛽−1),𝑠∈[𝜉,1). (2.19) The proof of (2) is completed.

3. Main Result

In this section, we will discuss the existence and uniqueness of positive solution for boundary value problem (1.1).

We define the space 𝑋={𝑢(𝑡)∣𝑢(𝑡)∈𝐶[0,1]} endowed with ‖𝑢‖_𝑋=max_0⩽𝑡⩽1|𝑢(𝑡)|, 𝑌={𝑣(𝑡)∣𝑣(𝑡)∈𝐶[0,1]} endowed with ‖𝑢‖_𝑌=max_0⩽𝑡⩽1|𝑣(𝑡)|.

For (𝑢,𝑣)∈𝑋×𝑌, let ‖(𝑢,𝑣)‖_𝑋×𝑌=max{‖𝑢‖_𝑋,‖𝑣‖_𝑌}.

Define 𝑃={(𝑢,𝑣)∈𝑋×𝑌∣𝑢(𝑡)⩾0,𝑣(𝑡)⩾0}, then the cone 𝑃⊂𝑋×𝑌.

From Lemma 2.5 in Section 2, we can obtain the following lemma.

Lemma 3.1. Suppose that 𝑓(𝑡,𝑣) and 𝑔(𝑡,𝑢) are continuous, then (𝑢,𝑣)∈𝑋×𝑌 is a solution of BVP (1.1) if and only if (𝑢,𝑣)∈𝑋×𝑌 is a solution of the integral equations 𝑢(𝑡)=¹₀𝐺₁(𝑡,𝑠)𝑓(𝑠,𝑣(𝑠))𝑑𝑠,𝑣(𝑡)=¹₀𝐺₂(𝑡,𝑠)𝑔(𝑠,𝑢(𝑠))𝑑𝑠. (3.1) Let 𝑇∶𝑋×𝑌→𝑋×𝑌 be the operator defined as 𝑇(𝑢,𝑣)(𝑡)=(¹₀𝐺₁(𝑡,𝑠)𝑓(𝑠,𝑣(𝑠))𝑑𝑠,¹₀𝐺₂(𝑡,𝑠)𝑔(𝑠,𝑢(𝑠))𝑑𝑠)=∶(𝑇₁𝑣(𝑡),𝑇₂𝑢(𝑡)), (3.2) then by Lemma 3.1, the fixed point of operator 𝑇 coincides with the solution of system (1.1).

Lemma 3.2. Let 𝑓(𝑡,𝑣) and 𝑔(𝑡,𝑢) be continuous on [0,1]×[0,∞)→[0,∞), then 𝑇∶𝑃→𝑃 defined by (3.2) is completely continuous.

Proof. Let (𝑢,𝑣)∈𝑃, in view of nonnegativeness and continuity of functions 𝐺(𝑡,𝑠), 𝑓, and 𝑔, we conclude that 𝑇∶𝑃→𝑃 is continuous.
Let Ω∈𝑃 be bounded, that is, there exists a positive constant ℎ>0 such that ‖(𝑢,𝑣)‖⩽ℎ for all (𝑢,𝑣)∈Ω.
Let 𝑀=max{|𝑓(𝑡,𝑣(𝑡))|+1∶0⩽𝑡⩽1,0⩽𝑣⩽ℎ},𝑁=max{|𝑔(𝑡,𝑢(𝑡))|+1∶0⩽𝑡⩽1,0⩽𝑢⩽ℎ}, (3.3) then we have |𝑇₁𝑣(𝑡)|=|¹₀𝐺₁(𝑡,𝑠)𝑓(𝑠,𝑣(𝑠))𝑑𝑠|⩽𝑀¹₀𝐺₁(𝑠,𝑠)𝑑𝑠,|𝑇₂𝑢(𝑡)|=|¹₀𝐺₂(𝑡,𝑠)𝑔(𝑠,𝑢(𝑠))𝑑𝑠|⩽𝑁¹₀𝐺₂(𝑠,𝑠)𝑑𝑠. (3.4) Hence, ‖𝑇(𝑢,𝑣)‖⩽max{𝑀∫¹₀𝐺₁(𝑠,𝑠)𝑑𝑠,𝑁∫¹₀𝐺₂(𝑠,𝑠)𝑑𝑠}. 𝑇(Ω) is uniformly bounded.
Since 𝐺₁(𝑡,𝑠) is continuous on [0,1]×[0,1], it is uniformly continuous on [0,1]×[0,1]. Thus, for fixed 𝑠∈[0,1] and for any 𝜀>0, there exists a constant 𝛿>0, such that any 𝑡₁,𝑡₂∈[0,1] and |𝑡₁−𝑡₂|<𝛿,|𝐺₁(𝑡₁,𝑠)−𝐺₁(𝑡₂,𝑠)|<𝜀/𝑀. (3.5)
Then |𝑇₁(𝑣)(𝑡₂)−𝑇₁(𝑣)(𝑡₁)|⩽𝑀¹₀|𝐺₁(𝑡₂,𝑠)−𝐺₁(𝑡₁,𝑠)|𝑑𝑠<𝜀. (3.6)
Similarly, |𝑇₂(𝑢)(𝑡₂)−𝑇₂(𝑢)(𝑡₁)|⩽𝑁¹₀|𝐺₂(𝑡₂,𝑠)−𝐺₂(𝑡₁,𝑠)|𝑑𝑠<𝜀. (3.7) For the Euclidean distance 𝑑 on 𝑅², we have that if 𝑡₁,𝑡₂∈[0,1] are such that |𝑡₂−𝑡₁|<𝛿, then 𝑑(𝑇(𝑢,𝑣)(𝑡₂),𝑇(𝑢,𝑣)(𝑡₁))=(𝑇₁𝑣(𝑡₂)−𝑇₁𝑣(𝑡₁))²+(𝑇₂𝑢(𝑡₂)−𝑇₂𝑢(𝑡₁))²<√2𝜀. (3.8) That is to say, 𝑇(𝑃) is equicontinuous. By the means of the Arzela-Ascoli theorem, we have 𝑇∶𝑃→𝑃 is completely continuous. The proof is completed.

Theorem 3.3. Assume that 𝑓(𝑡,𝑣) and 𝑔(𝑡,𝑢) are continuous on [0,1]×[0,∞)→[0,∞), and there exist two positive functions 𝑚(𝑡),𝑛(𝑡) that satisfy(𝐻₁)|𝑓(𝑡,𝑣₂)−𝑓(𝑡,𝑣₁)|⩽𝑚(𝑡)|𝑣₂−𝑣₁|,for𝑡∈[0,1],𝑣₁,𝑣₂∈[0,∞),(𝐻₂)|𝑔(𝑡,𝑢₂)−𝑔(𝑡,𝑢₁)|⩽𝑛(𝑡)|𝑢₂−𝑢₁|,for𝑡∈[0,1],𝑢₁,𝑢₂∈[0,∞).
Then system (1.1) has a unique positive solution if 𝜌=¹₀𝐺₁(𝑠,𝑠)𝑚(𝑠)𝑑𝑠<1,𝜃=¹₀𝐺₂(𝑠,𝑠)𝑛(𝑠)𝑑𝑠<1. (3.9)

Proof. For all (𝑢,𝑣)∈𝑃, by the nonnegativeness of 𝐺(𝑡,𝑠) and 𝑓(𝑡,𝑣),𝑔(𝑡,𝑢), we have 𝑇(𝑢,𝑣)(𝑡)⩾0. Hence, 𝑇(𝑃)⊆𝑃.‖𝑇₁𝑣₂−𝑇₁𝑣₁‖=max_𝑡∈[0,1]|𝑇₁𝑣₂−𝑇₁𝑣₁|=max_𝑡∈[0,1]|¹₀𝐺₁(𝑡,𝑠)[𝑓(𝑠,𝑣₂(𝑠))−𝑓(𝑠,𝑣₁⩽(𝑠))]𝑑𝑠|₁₀𝐺₁(𝑠,𝑠)𝑚(𝑠)𝑑𝑠‖𝑣₂−𝑣₁‖⩽𝜌‖𝑣₂−𝑣₁‖. (3.10) Similarly, ‖𝑇₂𝑢₂−𝑇₂𝑢₁‖⩽𝜃‖𝑢₂−𝑢₁‖. (3.11) We have, ‖𝑇(𝑢₂,𝑣₂)−𝑇(𝑢₁,𝑣₁)‖⩽max(𝜌,𝜃)‖(𝑢₂,𝑣₂)−(𝑢₁,𝑣₁)‖. (3.12) From Lemma 3.2, 𝑇 is completely continuous, by Banach fixed point theorem, the operator 𝑇 has a unique fixed point in 𝑃, which is the unique positive solution of system (1.1). This completes the proof.

Theorem 3.4. Assume that 𝑓(𝑡,𝑣) and 𝑔(𝑡,𝑢) are continuous on [0,1]×[0,∞)→[0,∞) and satisfy(𝐻₃)|𝑓(𝑡,𝑣(𝑡))|⩽𝑎₁(𝑡)+𝑎₂(𝑡)|𝑣(𝑡)|,(𝐻₄)|𝑔(𝑡,𝑢(𝑡))|⩽𝑏₁(𝑡)+𝑏₂(𝑡)|𝑢(𝑡)|,(𝐻₅)𝐴₁=∫¹₀𝐺₁(𝑠,𝑠)𝑎₂(𝑠)𝑑𝑠<1,0<𝐵₁=∫¹₀𝐺₁(𝑠,𝑠)𝑎₁(𝑠)𝑑𝑠<∞,(𝐻₆)𝐴₂=∫¹₀𝐺₂(𝑠,𝑠)𝑏₂(𝑠)𝑑𝑠<1,0<𝐵₂=∫¹₀𝐺₂(𝑠,𝑠)𝑏₁(𝑠)𝑑𝑠<∞.
Then the system (1.1) has at least one positive solution (𝑢,𝑣) in 𝐵𝐶={(𝑢,𝑣)∈𝑃∣‖(𝑢,𝑣)‖<min(¹1−𝐴₁,𝐵₂1−𝐴₂)}. (3.13)

Proof. Let 𝐶={(𝑢,𝑣)∈𝑋×𝑌∶‖(𝑢,𝑣)‖<𝑟} with 𝑟=min(𝐵₁/(1−𝐴₁),𝐵₂/(1−𝐴₂)), define the operator 𝑇∶𝐶→𝑃 as (3.2).
Let (𝑢,𝑣)∈𝐶, that is, ‖(𝑢,𝑣)‖<𝑟. Then ‖𝑇₁𝑣‖=max_𝑡∈[0,1]|¹₀𝐺₁⩽(𝑡,𝑠)𝑓(𝑠,𝑣(𝑠))𝑑𝑠|₁₀𝐺₁(𝑠,𝑠)(𝑎₁(𝑠)+𝑎₂≤(𝑠)|𝑣(𝑠)|)𝑑𝑠₁₀𝐺₁(𝑠,𝑠)𝑎₁(𝑠)𝑑𝑠+¹₀𝐺₁(𝑠,𝑠)𝑎₂(𝑠)𝑑𝑠‖𝑣‖=𝐵₁+𝐴₁‖𝑣‖⩽𝑟. (3.14)
Similarly, ‖𝑇₂𝑢‖⩽𝑟, so ‖𝑇(𝑢,𝑣)‖⩽𝑟, 𝑇(𝑢,𝑣)⊆𝐶. From Lemma 3.2  𝑇∶𝐶→𝐶 is completely continuous.
Consider the eigenvalue problem (𝑢,𝑣)=𝜆𝑇(𝑢,𝑣),𝜆∈(0,1). (3.15)
Under the assumption that (𝑢,𝑣) is a solution of (3.15) for a 𝜆∈(0,1), one obtains ‖𝑢‖=‖𝜆𝑇₁𝑣‖=𝜆max_𝑡∈[0,1]|¹₀𝐺₁<(𝑡,𝑠)𝑓(𝑠,𝑣(𝑠))𝑑𝑠|₁₀𝐺₁(𝑠,𝑠)(𝑎₁(𝑠)+𝑎₂=(𝑠)|𝑣(𝑠)|)𝑑𝑠₁₀𝐺₁(𝑠,𝑠)𝑎₁(𝑠)𝑑𝑠+¹₀𝐺₁(𝑠,𝑠)𝑎₂(𝑠)𝑑𝑠‖𝑣‖=𝐵₁+𝐴₁‖𝑣‖⩽𝑟. (3.16) Similarly, ‖𝑣‖=‖𝜆𝑇₂𝑢‖<𝑟, so ‖(𝑢,𝑣)‖<𝑟, which shows that (𝑢,𝑣)∉𝜕𝐶. By Lemma 2.4, 𝑇 has a fixed point in 𝐶. We complete the proof of Theorem 3.4.

Example 3.5. Consider the problem 𝐷^7/4𝐷𝑢(𝑡)+𝑓(𝑡,𝑣(𝑡))=0,0<𝑡<1,_3/21𝑣(𝑡)+𝑔(𝑡,𝑢(𝑡))=0,0<𝑡<1,𝑢(0)=0,𝑢(1)=21𝑢(23),𝑣(0)=0,𝑣(1)=41𝑣(2), (3.17) where 𝑓(𝑡,𝑣(𝑡))=𝑡𝑣(𝑡)𝑡(1+𝑡)(1+𝑣(𝑡)),𝑔(𝑡,𝑢(𝑡))=arctan1+𝑡|sin𝑢(𝑡)|. (3.18) Set 𝑣₁(𝑡),𝑣₂(𝑡),𝑢₁(𝑡),𝑢₂(𝑡)∈[0,∞) and 𝑡∈[0,1], then we have |𝑓(𝑡,𝑣₂(𝑡))−𝑓(𝑡,𝑣₁𝑡(𝑡))|⩽1+𝑡|𝑣₂(𝑡)−𝑣₁(𝑡)|,|𝑔(𝑡,𝑢₂(𝑡))−𝑔(𝑡,𝑢₁𝑡(𝑡))|⩽arctan1+𝑡|𝑢₂(𝑡)−𝑢₁(𝑡)|. (3.19) Therefore, 𝜌=¹₀𝐺₁(𝑠,𝑠)𝑚(𝑠)𝑑𝑠⩽¹₀𝐺₁=1(𝑠,𝑠)𝑑𝑠Γ(7/4)(1−(1/2)^7/4){₀^1/2[𝑠(1−𝑠)]^3/4𝑑𝑠−₀^1/2121[𝑠(2−𝑠)]^3/4𝑑𝑠+¹_1/2[𝑠(1−𝑠)]^3/4=𝑑𝑠}2(1+(1/2)^7/4)5⋅Γ(3/4)<4Γ(1/2)5<1,𝜃=¹₀𝐺₂𝜋(𝑠,𝑠)𝑛(𝑠)𝑑𝑠⩽4¹₀𝐺₂=𝜋(𝑠,𝑠)𝑑𝑠41Γ(3/2)(1−(3/4)(1/2)^1/2){₀^1/2[𝑠(1−𝑠)]^1/2𝑑𝑠−₀^1/2341[𝑠(2−𝑠)]^1/2𝑑𝑠+¹_1/2[𝑠(1−𝑠)]^1/2=𝜋𝑑𝑠}4[1−(3/4)(1/2)²][1−(3/4)(1/2)^1/2]⋅Γ(3/2)Γ(3)≈0.6018<1. (3.20) With the use of Theorem 3.3, BVP (3.17) has a unique positive solution.

Example 3.6. Consider the problem 𝐷^7/4𝐷𝑢(𝑡)+𝑓(𝑡,𝑣(𝑡))=0,0<𝑡<1,_3/21𝑣(𝑡)+𝑔(𝑡,𝑢(𝑡))=0,0<𝑡<1,𝑢(0)=0,𝑢(1)=21𝑢(23),𝑣(0)=0,𝑣(1)=41𝑣(2), (3.21) where 𝑓(𝑡,𝑣(𝑡))=𝑡²+𝑡𝑡1+𝑡ln(1+𝑣(𝑡)),𝑔(𝑡,𝑢(𝑡))=10+²20+𝑢(𝑡). (3.22) We have |𝑓(𝑡,𝑣(𝑡))|𝑡²+𝑡𝑡1+𝑡⋅|𝑣(𝑡)|,|𝑔(𝑡,𝑢(𝑡))|⩽(10+²20)+|𝑢(𝑡)|. (3.23) Hence, 𝐴₁=¹₀𝐺₁(𝑠,𝑠)𝑎₂(𝑠)𝑑𝑠⩽¹₀𝐺₁(𝑠,𝑠)𝑑𝑠=2(1+(1/2)^7/4)5⋅Γ(3/4)𝐵Γ(1/2)<1,₁=¹₀𝐺₁(𝑠,𝑠)𝑎₁(𝑠)𝑑𝑠=¹₀𝐺₁(𝑠,𝑠)⋅𝑠²𝐴𝑑𝑠<∞,₂=¹₀𝐺₂(𝑠,𝑠)𝑏₂(𝑠)𝑑𝑠=¹₀𝐺₂𝐵(𝑠,𝑠)𝑑𝑠≈0.7666<1,₂=¹₀𝐺₂(𝑠,𝑠)𝑏₁(𝑠)𝑑𝑠=¹₀𝐺₂𝑠(𝑠,𝑠)(10+²20)𝑑𝑠<∞. (3.24) By Theorem 3.4, BVP (3.21) has at least one positive solution in 𝐵𝐶={(𝑢,𝑣)∈𝑃∣‖𝑢,𝑣‖<min(¹1−𝐴₁,𝐵₂1−𝐴₂)}. (3.25)

Acknowledgments

This work was jointly supported by the Natural Science Foundation of Hunan Provincial Education Department under Grants 07A066 and 07C700, the Construct Program of the Key Discipline in Hunan Province, Aid Program for Science and Technology Innovative Research Team in Higher Educational Institutions of Hunan Province, and the Foundation of Xiangnan University.