Uniqueness and Existence of Positive Solutions for Singular Differential Systems with Coupled Integral Boundary Value Problems

Abstract

This paper provides sufficient conditions for the existence and uniqueness of positive solutions to a singular differential system with integral boundary value. The emphasis here is that the boundary conditions are coupled and this is where the main novelty of this work lies. By mixed monotone method, the existence and uniqueness results of the problem are established. An example is given to demonstrate the main results.

1. Introduction

In recent years, differential system has been studied extensively in the literature (see, for instance, [1–17] and their references). Most of the results told us that the equations had at least single and multiple positive solutions. In papers [6], the authors obtained some of the newest results for differential system with four-point coupled boundary conditions. But there is no result on the uniqueness of solution in them.

In this paper, we discuss the existence and uniqueness of the positive solutions for a new class of boundary value problems of singular differential system. Precisely, we consider the following problem: −𝑥^′′(𝑡)=𝑓𝑡,𝑥(𝑡),𝑦(𝑡),𝑡∈(0,1),−𝑦^′′(𝑡)(𝑡)(𝑡)(),𝑥(0)==𝑔𝑡,𝑥,𝑦,𝑡∈0,1₁₀𝑦(𝑡)𝑑𝛼(𝑡),𝑦(0)=¹₀𝑥(𝑡)(𝑡),𝑥(1)(1)𝑑𝛽=𝑦=0, (1) where 𝛼 and 𝛽 are right continuous on [0,1), left continuous at 𝑡=1, and nondecreasing on [0,1], with 𝛼(0)=𝛽(0)=0; ∫¹₀𝑢(𝑠)𝑑𝛼(𝑠) and ∫¹₀𝑢(𝑠)𝑑𝛽(𝑠) denote the Riemann-Stieltjes integrals of 𝑢 with respect to 𝛼 and 𝛽, respectively; 𝑓∈𝐶((0,1)×[0,+∞)×(0,+∞),[0,+∞)), 𝑔∈𝐶((0,1)×(0,+∞)×[0,+∞),[0,+∞)); that is, 𝑓(𝑡,𝑥,𝑦) may be singular at 𝑡=0, 𝑡=1, and 𝑦=0 and 𝑔(𝑡,𝑥,𝑦) may be singular at 𝑡=0, 𝑡=1, and 𝑥=0. By a positive solution of the system (1), we mean that (𝑥,𝑦)∈(𝐶[0,1]∩𝐶²(0,1))×(𝐶[0,1]∩𝐶²(0,1)), (𝑥,𝑦) satisfies (1), and 𝑥>0 and 𝑦>0 on [0,1).

2. Preliminaries

For each 𝑢∈𝐸∶=𝐶[0,1], we write ‖𝑢‖=max{|𝑢(𝑡)|∶𝑡∈[0,1]}. Clearly, (𝐸,‖⋅‖) is a Banach space. Similarly, for each (𝑥,𝑦)∈𝐸×𝐸, we write ‖(𝑥,𝑦)‖₁=max{‖𝑥‖,‖𝑦‖}. Clearly, (𝐸×𝐸,‖⋅‖₁) is a Banach space.

Throughout this paper, we shall use the following notation: 𝑘()=𝑡()𝑠()𝑡,𝑠1−𝑠,0≤𝑡≤𝑠≤1,1−𝑡,0≤𝑠≤𝑡≤1. (2) It is well known that 𝑘(𝑡,𝑠) is the Green function of the following second order boundary value problem: −𝑥^′′(𝑡)𝑥(0)(1)=0,0<𝑡<1,=𝑥=0, (3) and 𝑘(𝑡,𝑠) is nonnegative continuous function. It is easy to verify that for 𝑡,𝑠∈[0,1]×[0,1], 𝑘()𝑘()()𝑠()()((.𝑡,𝑡𝑠,𝑠=𝑡1−𝑡1−𝑠≤𝑘𝑡,𝑠≤𝑡1−𝑡)(or𝑠1−𝑠)) (4)

We first list the following assumptions for convenience.(𝐻₁)𝑓∈𝐶((0,1)×[0,+∞)×(0,+∞),[0,+∞)), 𝑓(𝑡,𝑥,𝑦) is nondecreasing in 𝑥 and nonincreasing in 𝑦, and there exist 𝜆₁,  𝜇₁∈[0,1) such that 𝑐^𝜆1𝑓(),𝑓𝑡,𝑥,𝑦≤𝑓𝑡,𝑐𝑥,𝑦,∀𝑥,𝑦>0,𝑐∈0,1𝑡,𝑥,𝑐𝑦≤𝑐^−𝜇1𝑓𝑡,𝑥,𝑦,∀𝑥,𝑦>0,𝑐∈(0,1); (5)(6) 𝑔∈𝐶((0,1)×(0,+∞)×[0,+∞),[0,+∞)), 𝑔(𝑡,𝑥,𝑦) is nonincreasing in 𝑥 and nondecreasing in 𝑦, and there exist 𝜆₂, 𝜇₂∈[0,1) such that 𝑐^𝜆2𝑔(),𝑔𝑡,𝑥,𝑦≤𝑔𝑡,𝑥,𝑐𝑦,∀𝑥,𝑦>0,𝑐∈0,1𝑡,𝑐𝑥,𝑦≤𝑐^−𝜇2𝑔𝑡,𝑥,𝑦,∀𝑥,𝑦>0,𝑐∈(0,1). (7)(8)(𝐻₂)0<∫¹₀𝑓(𝑡,1,1−𝑡)𝑑𝑡<+∞,  0<∫¹₀𝑔(𝑡,1−𝑡,1)𝑑𝑡<+∞.(𝐻₃)𝜅₁>0,  𝜅₂>0,  𝜅>0, where 𝜅₁=¹₀(1−𝑡)𝑑𝛼(𝑡),𝜅₂=¹₀(1−𝑡)𝑑𝛽(𝑡),𝜅=1−𝜅₁𝜅₂. (9)

Remark 1. By (𝐻₁) and (𝐻₂), we can get 0<¹₀𝑓(𝑡,1−𝑡,1)𝑑𝑡<+∞,0<¹₀𝑔(𝑡,1,1−𝑡)𝑑𝑡<+∞. (10)

Remark 2. (i) (5) and (6) imply that 𝑓𝑡,𝑐𝑥,𝑦≤𝑐^𝜆1𝑓𝑓𝑡,𝑥,𝑦,∀𝑥,𝑦>0,𝑐>1,𝑡,𝑥,𝑦≤𝑐^𝜇1𝑓𝑡,𝑥,𝑐𝑦,∀𝑥,𝑦>0,𝑐>1. (11)(12) Conversely, (11) implies (5) and (12) implies (6).
(ii) (7) and (8) implies that 𝑔𝑡,𝑐𝑥,𝑦≤𝑐^𝜆2𝑔𝑔𝑡,𝑥,𝑦,∀𝑥,𝑦>0,𝑐>1,𝑡,𝑥,𝑦≤𝑐^𝜇2𝑔𝑡,𝑥,𝑐𝑦,∀𝑥,𝑦>0,𝑐>1. (13)(14) Conversely, (13) implies (7) and (14) implies (8).

Lemma 3. Assume that (𝐻₃) holds. Let 𝑢,𝑣∈𝐿[0,1]∩𝐶(0,1); then the system of BVPs −𝑥^′′(𝑡)=𝑢(𝑡),−𝑦^′′(𝑡)(𝑡)(),𝑥(0)==𝑣,𝑡∈0,1₁₀𝑦(𝑡)𝑑𝛼(𝑡),𝑦(0)=¹₀𝑥(𝑡)(𝑡),𝑥(1)(1)𝑑𝛽=𝑦=0 (15)has integral representation 𝑥(𝑡)=¹₀𝐺₁(𝑡,𝑠)𝑢(𝑠)𝑑𝑠+¹₀𝐻₁()𝑣(𝑠)𝑦(𝑡)=𝑡,𝑠𝑑𝑠,₁₀𝐺₂(𝑡,𝑠)𝑣(𝑠)𝑑𝑠+¹₀𝐻₂(𝑡,𝑠)𝑢(𝑠)𝑑𝑠, (16) where 𝐺₁()=𝑘𝑡,𝑠¹(1−𝑡)𝜅¹₀𝑘()(𝜏)(),𝐻𝑠,𝜏𝑑𝛽+𝑘𝑡,𝑠₁(𝑡,𝑠)=1−𝑡𝜅¹₀𝑘()(𝜏),𝐺𝑠,𝜏𝑑𝛼₂()=𝑘𝑡,𝑠²(1−𝑡)𝜅¹₀𝑘()(𝜏)(),𝐻𝑠,𝜏𝑑𝛼+𝑘𝑡,𝑠₂(𝑡,𝑠)=1−𝑡𝜅¹₀𝑘(𝑠,𝜏)𝑑𝛽(𝜏). (17)

Proof. It is easy to see that (15) is equivalent to the system of integral equations 𝑥(𝑡)=𝑥(0)(1−𝑡)+¹₀𝑘()𝑢(𝑠)[],𝑦(𝑡)(0)+𝑡,𝑠𝑑𝑠,𝑡∈0,1=𝑦)(1−𝑡¹₀𝑘(𝑡,𝑠)𝑣(𝑠)𝑑𝑠,𝑡∈[0,1]. (18)(19) Integrating (18) and (19) with respect to 𝑑𝛽(𝑡) and 𝑑𝛼(𝑡), respectively, on [0,1] gives ¹₀𝑥(𝑡)𝑑𝛽(𝑡)=𝑥(0)¹₀()(𝑡)+1−𝑡𝑑𝛽₁₀𝑘(𝑡,𝑠)𝑢(𝑠)𝑑𝑠𝑑𝛽₁₀(𝑡),𝑦(𝑡)𝑑𝛼(𝑡)=𝑦(0)¹₀()(𝑡)+1−𝑡𝑑𝛼₁₀𝑘(𝑡,𝑠)𝑣(𝑠)𝑑𝑠𝑑𝛼(𝑡). (20) Therefore, −𝜅₂11−𝜅₁𝑥(0)𝑦(0)=∬¹₀𝑘()𝑢(𝑠)(𝑡)∬𝑡,𝑠𝑑𝑠𝑑𝛽₁₀𝑘(𝑡,𝑠)𝑣(𝑠)𝑑𝑠𝑑𝛼(𝑡), (21) and so 𝑥(0)𝑦(0)=1𝜅−𝜅₁11−𝜅₂∬¹₀𝑘()𝑢(𝑠)(𝑡)∬𝑡,𝑠𝑑𝑠𝑑𝛽₁₀𝑘(𝑡,𝑠)𝑣(𝑠)𝑑𝑠𝑑𝛼(𝑡). (22) Substituting (22) into (18) and (19), we have 𝑥(𝑡)=𝜅₁(1−𝑡)𝜅¹₀𝑘()𝑢(𝑠)(𝑡)+𝑡,𝑠𝑑𝑠𝑑𝛽1−𝑡𝜅¹₀𝑘()𝑣(𝑠)(𝑡)+𝑡,𝑠𝑑𝑠𝑑𝛼₁₀𝑘()𝑢(𝑠)𝑦(𝑡)=𝑡,𝑠𝑑𝑠,1−𝑡𝜅¹₀𝑘()𝑢(𝑠)(𝑡)+𝜅𝑡,𝑠𝑑𝑠𝑑𝛽₂(1−𝑡)𝜅¹₀𝑘()𝑣(𝑠)(𝑡)+𝑡,𝑠𝑑𝑠𝑑𝛼₁₀𝑘(𝑡,𝑠)𝑣(𝑠)𝑑𝑠, (23) which is equivalent to the system (16).

Remark 4. From (4) and (𝐻₃), for 𝑡∈[0,1], we have 𝐺_𝑖()()(),𝐻𝑡,𝑠≤𝜌𝑠1−𝑠or𝜌1−𝑡_𝑖()()(),𝐺𝑡,𝑠≤𝜌𝑠1−𝑠or𝜌1−𝑡𝑖=1,2,_𝑖()()𝑠(),𝐻𝑡,𝑠≥𝜈1−𝑡1−𝑠_𝑖(𝑡,𝑠)≥𝜈(1−𝑡)𝑠(1−𝑠),𝑖=1,2, (24) where 𝜅𝜌=max¹𝜅¹₀(𝜏)𝜅𝑑𝛽+1,²𝜅¹₀(𝜏)1𝑑𝛼+1,𝜅¹₀(𝜏),1𝑑𝛽𝜅¹₀(𝜏,𝜅𝑑𝛼)𝜈=min¹𝜅¹₀𝜏()(𝜏),𝜅1−𝜏𝑑𝛽₂𝜅¹₀𝜏()(𝜏),11−𝜏𝑑𝛼𝜅¹₀𝜏()(𝜏),11−𝜏𝑑𝛽𝜅¹₀𝜏(1−𝜏)𝑑𝛼(𝜏). (25)

Denote 𝑃=𝑥,𝑦∈𝐸×𝐸∶𝑥(𝑡)≥𝛾(1−𝑡)𝑥,𝑦₁,𝑦(𝑡)≥𝛾(1−𝑡)𝑥,𝑦₁,𝑡∈[0,1], (26) where 𝛾=𝜈/𝜌∈(0,1). It can be easily seen that 𝑃 is a cone in 𝐸×𝐸. For any real constant 𝑟>0, define 𝑃_𝑟={(𝑥,𝑦)∈𝑃∶‖(𝑥,𝑦)‖₁<𝑟}.

Define an operator 𝑇∶𝑃⧵{𝜃}→𝑃 by 𝑇𝑥,𝑦=𝑇₁𝑥,𝑦,𝑇₂𝑥,𝑦, (27) where operators 𝑇₁,𝑇₂∶𝑃⧵{𝜃}→𝑄={𝑢∈𝐸∣𝑢(𝑡)≥0,𝑡∈[0,1]} are defined by 𝑇₁𝑥,𝑦(𝑡)=¹₀𝐺₁()𝑓(𝑠)(𝑠)+𝑡,𝑠𝑠,𝑥,𝑦𝑑𝑠₁₀𝐻₁()𝑔(𝑠)(𝑠[],𝑇𝑡,𝑠𝑠,𝑥,𝑦)𝑑𝑠,𝑡∈0,1₂𝑥,𝑦(𝑡)=¹₀𝐺₂()𝑔(𝑠)(𝑠)+𝑡,𝑠𝑠,𝑥,𝑦𝑑𝑠₁₀𝐻₂(𝑡,𝑠)𝑓𝑠,𝑥(𝑠),𝑦(𝑠)𝑑𝑠,𝑡∈[0,1]. (28) Now we claim that 𝑇(𝑥,𝑦) is well defined for (𝑥,𝑦)∈𝑃⧵{𝜃}. In fact, since (𝑥,𝑦)∈𝑃⧵{𝜃}, we can see that 𝛾(1−𝑡)𝑥,𝑦₁(𝑡),𝑦(𝑡)≤≤𝑥𝑥,𝑦₁,𝑡∈[0,1]. (29)

Let 𝑐 be a positive number such that ‖(𝑥,𝑦)‖₁/𝑐<1 and 𝑐>1. From (𝐻₁) and Remark 2, we have 𝑓𝑡,𝑥(𝑡),𝑦(𝑡)≤𝑓𝑡,𝑐,𝛾𝑥,𝑦₁(1−𝑡)≤𝑐^𝜆1𝑓𝛾𝑡,1,𝑥,𝑦₁𝑐(1−𝑡)≤𝑐^𝜆1+𝜇1𝛾𝑥,𝑦₁^−𝜇1𝑓(),𝑔(𝑡)(𝑡)𝑡,1,1−𝑡𝑡,𝑥,𝑦≤𝑐^𝜆2+𝜇2𝛾𝑥,𝑦₁^−𝜇2𝑓(𝑡,1,1−𝑡). (30) Hence, for any 𝑡∈[0,1], by Remark 4 and equation (30), we get 𝑇_𝑖𝑥,𝑦(𝑡)≤𝜌¹₀𝑓𝑠,𝑥(𝑠),𝑦(𝑠)𝑑𝑠+𝜌¹₀𝑔𝑠,𝑥(𝑠),𝑦(𝑠)𝑑𝑠≤𝜌𝑐^𝜆1+𝜇1𝛾𝑥,𝑦₁^−𝜇11₀𝑓(𝑠,1,1−𝑠)𝑑𝑠+𝜌𝑐^𝜆2+𝜇2𝛾𝑥,𝑦₁^−𝜇21₀𝑔(𝑠,1−𝑠,1)𝑑𝑠<+∞,𝑖=1,2. (31) Thus, 𝑇 is well defined on 𝑃⧵{𝜃}.

Lemma 5. Assume that (𝐻₁), (𝐻₂), and (𝐻₃) hold. Then, for any 0<𝑎<𝑏<+∞, 𝑇∶(𝑃_𝑏⧵𝑃_𝑎)→𝑃 is a completely continuous operator.

Proof. Firstly, we show that 𝑇(𝑃_𝑏⧵𝑃_𝑎)⊂𝑃. By Remark 4, for 𝜏,𝑡,𝑠∈[0,1], we obtain 𝐺_𝑖(𝑡,𝑠)≥𝛾(1−𝑡)𝐺_𝑖(),𝐻𝜏,𝑠_𝑖(𝑡,𝑠)≥𝛾(1−𝑡)𝐻_𝑖(),𝐻𝜏,𝑠𝑖=1,2,₁(𝑡,𝑠)≥𝛾(1−𝑡)𝐺₂(),𝐺𝜏,𝑠₁(𝑡,𝑠)≥𝛾(1−𝑡)𝐻₂(),𝐻𝜏,𝑠₂(𝑡,𝑠)≥𝛾(1−𝑡)𝐺₁(),𝐺𝜏,𝑠₂(𝑡,𝑠)≥𝛾(1−𝑡)𝐻₁(𝜏,𝑠). (32) Hence, for (𝑥,𝑦)∈𝑃_𝑏⧵𝑃_𝑎, 𝑡∈[0,1], we have 𝑇₁𝑥,𝑦(𝑡)=¹₀𝐺₁()𝑓(𝑠)(𝑠)+𝑡,𝑠𝑠,𝑥,𝑦𝑑𝑠₁₀𝐻₁()𝑔(𝑠)(𝑠()𝑡,𝑠𝑠,𝑥,𝑦)𝑑𝑠≥𝛾1−𝑡¹₀𝐺₁()𝑓(𝑠)(𝑠)()𝜏,𝑠𝑠,𝑥,𝑦𝑑𝑠+𝛾1−𝑡¹₀𝐻₁()𝑔(𝑠)(𝑠()𝑇𝜏,𝑠𝑠,𝑥,𝑦)𝑑𝑠=𝛾1−𝑡₁(𝜏)[],𝑇𝑥,𝑦,∀𝜏∈0,1₁𝑥,𝑦(𝑡)=¹₀𝐺₁()𝑓(𝑠)(𝑠)+𝑡,𝑠𝑠,𝑥,𝑦𝑑𝑠₁₀𝐻₁()𝑔(𝑠)(𝑠)()𝑡,𝑠𝑠,𝑥,𝑦𝑑𝑠≥𝛾1−𝑡¹₀𝐻₂()𝑓(𝑠)(𝑠)()𝜏,𝑠𝑠,𝑥,𝑦𝑑𝑠+𝛾1−𝑡¹₀𝐺₂()𝑔(𝑠)(𝑠)()𝑇𝜏,𝑠𝑠,𝑥,𝑦𝑑𝑠=𝛾1−𝑡₂𝑥,𝑦(𝜏),∀𝜏∈[0,1]. (33) Then 𝑇₁(𝑥,𝑦)(𝑡)≥𝛾(1−𝑡)‖𝑇₁(𝑥,𝑦)‖ and 𝑇₁(𝑥,𝑦)(𝑡)≥𝛾(1−𝑡)‖𝑇₂(𝑥,𝑦)‖; that is, 𝑇₁(𝑥,𝑦)(𝑡)≥𝛾(1−𝑡)‖(𝑇₁(𝑥,𝑦),𝑇₂(𝑥,𝑦))‖₁. In the same way, we can prove that 𝑇₂(𝑥,𝑦)(𝑡)≥𝛾(1−𝑡)‖(𝑇₁(𝑥,𝑦),𝑇₂(𝑥,𝑦))‖₁. Therefore, 𝑇(𝑃_𝑏⧵𝑃_𝑎)⊂𝑃.
Next, we prove that 𝑇 is a compact operator. That is, for any bounded subset 𝑉⊂𝑃_𝑏⧵𝑃_𝑎, we show that 𝑇(𝑉) is relatively compact in 𝐸×𝐸. Since 𝑉⊂𝑃_𝑏⧵𝑃_𝑎 is a bounded subset, there exists a constant 𝑐>1 such that ‖(𝑥,𝑦)‖₁=max{‖𝑥‖,‖𝑦‖}≤𝑐 for all (𝑥,𝑦)∈𝑉. Notice that, for any (𝑥,𝑦)∈𝑉, we have 𝑇𝑥,𝑦₁=max𝑇₁𝑥,𝑦,𝑇₂𝑥,𝑦 (34) and from (𝐻₁), (𝐻₂), Remarks 2 and 4, (16), and (18), we obtain 𝑇_𝑖𝑥,𝑦(𝑡)≤𝜌¹₀𝑓𝑠,𝑥(𝑠),𝑦(𝑠)𝑑𝑠+𝜌¹₀𝑔𝑠,𝑥(𝑠),𝑦(𝑠)𝑑𝑠≤𝜌𝑐^𝜆1+𝜇1𝛾𝑎^−𝜇11₀𝑓(𝑠,1,1−𝑠)𝑑𝑠+𝜌𝑐^𝜆2+𝜇2𝛾𝑎^−𝜇21₀𝑔(𝑠,1−𝑠,1)𝑑𝑠<+∞,𝑖=1,2. (35) Therefore, 𝑇(𝑉) is uniformly bounded.
In the following, we shall show that 𝑇(𝑉) is equicontinuous on [0,1].
For (𝑥,𝑦)∈𝑉, 𝑡∈[0,1], using Lemma 3, we have 𝑇₁(𝑡)=𝑥,𝑦₁₀𝐺₁()𝑓(𝑠)(𝑠)+𝑡,𝑠𝑠,𝑥,𝑦𝑑𝑠₁₀𝐻₁()𝑔(𝑠)(𝑠)=𝜅𝑡,𝑠𝑠,𝑥,𝑦𝑑𝑠₁(1−𝑡)𝜅¹₀¹₀𝑘()(𝜏)𝑓(𝑠)(𝑠)+𝑠,𝜏𝑑𝛽𝑠,𝑥,𝑦𝑑𝑠_𝑡0𝑠()𝑓(𝑠)(𝑠)+1−𝑡𝑠,𝑥,𝑦𝑑𝑠_1𝑡𝑡()𝑓(𝑠)(𝑠)+1−𝑠𝑠,𝑥,𝑦𝑑𝑠1−𝑡𝜅¹₀¹₀𝑘(𝑠,𝜏)𝑑𝛼(𝜏)𝑔𝑠,𝑥(𝑠),𝑦(𝑠)𝑑𝑠. (36) Differentiating with respect to 𝑡 and combining (𝐻₁) and (𝐻₂), we obtain 𝑇₁𝑥,𝑦^′(𝑡=−𝜅₁)𝜅¹₀¹₀𝑘()(𝜏)𝑓(𝑠)(𝑠)−𝑠,𝜏𝑑𝛽𝑠,𝑥,𝑦𝑑𝑠_𝑡0𝑠𝑓𝑠,𝑥(𝑠),𝑦(𝑠)𝑑𝑠+¹_𝑡()𝑓(𝑠)(𝑠)−11−𝑠𝑠,𝑥,𝑦𝑑𝑠𝜅¹₀¹₀𝑘()(𝜏𝑔(𝑠)(𝑠≤𝜅𝑠,𝜏𝑑𝛼)𝑠,𝑥,𝑦)𝑑𝑠₁𝜅¹₀¹₀𝑘()(𝜏)𝑓(𝑠)(𝑠)+𝑠,𝜏𝑑𝛽𝑠,𝑥,𝑦𝑑𝑠_𝑡0(𝑠)(𝑠+𝑠𝑓𝑠,𝑥,𝑦)𝑑𝑠_1𝑡()𝑓(𝑠)(𝑠)+11−𝑠𝑠,𝑥,𝑦𝑑𝑠𝜅¹₀¹₀𝑘(𝑠,𝜏)𝑑𝛼(𝜏)𝑔𝑠,𝑥(𝑠),𝑦(𝑠)𝑑𝑠≤𝑐^𝜆1+𝜇1𝛾𝑎^−𝜇1𝜌𝜅₁𝜅¹₀𝑓()+𝑠,1,1−𝑠𝑑𝑠_𝑡0()+𝑠𝑓𝑠,1,1−𝑠𝑑𝑠_1𝑡()𝑓()+𝜌1−𝑠𝑠,1,1−𝑠𝑑𝑠𝜅𝑐^𝜆2+𝜇2𝛾𝑎^−𝜇21₀𝑔(𝑠,1−𝑠,1)𝑑𝑠=∶𝐾(𝑡). (37) Exchanging the integral order, we have ¹₀𝐾(𝑡)𝑑𝑡=𝑐^𝜆1+𝜇1𝛾𝑎^−𝜇1𝜌𝜅₁𝜅¹₀𝑓(𝑠,1,1−𝑠)𝑑𝑠+2¹₀𝑠()𝑓()+𝜌1−𝑠𝑠,1,1−𝑠𝑑𝑠𝜅𝑐^𝜆2+𝜇2𝛾𝑎^−𝜇21₀𝑔(𝑠,1−𝑠,𝑠)𝑑𝑠<+∞. (38) From the absolute continuity of the integral, we know that 𝑇₁(𝑉) is equicontinuous on [0,1]. Thus, according to the Ascoli-Arzela theorem, 𝑇₁(𝑉) is a relatively compact set. In the same way, we can prove that 𝑇₂(𝑉) is relatively compact. Therefore, 𝑇(𝑉) is relatively compact.
Finally, it remains to show that 𝑇 is continuous. We need to prove only that 𝑇₁,𝑇₂∶𝑃_𝑏⧵𝑃_𝑎→𝑄 are continuous. Suppose that (𝑥_𝑚,𝑦_𝑚),(𝑥₀,𝑦₀)∈𝑃_𝑏⧵𝑃_𝑎, and ‖(𝑥_𝑚,𝑦_𝑚)−(𝑥₀,𝑦₀)‖₁→0(𝑚→∞). Let 𝐿=sup{‖(𝑥_𝑚,𝑦_𝑚)‖₁,𝑚=0,1,2,…}. Then we may still choose positive constants 𝑐 such that 𝐿/𝑐<1 and 𝑐>1. From (𝐻₁) and Remark 2, we get 𝑓𝑡,𝑥_𝑚(𝑡),𝑦_𝑚(𝑡)≤𝑐^𝜆1+𝜇1𝛾𝑎^−𝜇1𝑓(),𝑔𝑡,1,1−𝑡𝑚=0,1,2…,𝑡,𝑥_𝑚(𝑡),𝑦_𝑚(𝑡)≤𝑐^𝜆2+𝜇2𝛾𝑎^−𝜇2𝑔(),𝑇𝑡,1−𝑡,1𝑚=0,1,2…,₁𝑥_𝑚,𝑦_𝑚(𝑡)−𝑇₁𝑥₀,𝑦₀(𝑡)≤𝜌¹₀𝑓𝑠,𝑥_𝑚(𝑠),𝑦_𝑚(𝑠)−𝑓𝑠,𝑥₀(𝑠),𝑦₀(𝑠)𝑑𝑠+𝜌¹₀𝑔𝑠,𝑥_𝑚(𝑠),𝑦_𝑚(𝑠)−𝑔𝑠,𝑥₀(𝑠),𝑦₀(𝑠)𝑑𝑠. (39)
For any 𝜖>0, by (𝐻₂), there exists a positive number 𝛿∈(0,1/2) such that _{[0,𝛿]∪[1−𝛿,1]}𝜌𝑐^𝜆1+𝜇1𝛾𝑎^−𝜇1𝑓()𝜀𝑠,1,1−𝑠𝑑𝑠<4_{[0,𝛿]∪[1−𝛿,1]},𝜌𝑐^𝜆2+𝜇2𝛾𝑎^−𝜇2𝑔()𝜀𝑠,1−𝑠,1𝑑𝑠<4. (40) On the other hand, for (𝑥,𝑦)∈𝑃_𝑏⧵𝑃_𝑎 and 𝑡∈[𝛿,1−𝛿], we have 𝑎𝛾𝛿≤𝑥(𝑡),𝑦(𝑡)≤𝑏. (41) Since 𝑓(𝑡,𝑥,𝑦) and 𝑔(𝑡,𝑥,𝑦) are uniformly continuous in [𝛿,1−𝛿]×[𝑎𝛾𝛿,𝑏]×[𝑎𝛾𝛿,𝑏], we have lim_𝑚→+∞𝑓𝑡,𝑥_𝑚(𝑡),𝑦_𝑚(𝑡)−𝑓𝑡,𝑥₀(𝑡),𝑦₀(𝑡)=lim_𝑚→+∞𝑔𝑡,𝑥_𝑚(𝑡),𝑦_𝑚(𝑡)−𝑔𝑡,𝑥₀(𝑡),𝑦₀(𝑡)=0 (42) holds uniformly on 𝑡∈[𝛿,1−𝛿]. Then the Lebesgue dominated convergence theorem yields that _𝛿^1−𝛿𝑓𝑠,𝑥_𝑚(𝑠),𝑦_𝑚(𝑠)−𝑓𝑠,𝑥₀(𝑠),𝑦₀(𝑠)𝑑𝑠→0_𝛿1−𝛿𝑔𝑠,𝑥_𝑚(𝑠),𝑦_𝑚(𝑠)−𝑔𝑠,𝑥₀(𝑠),𝑦₀(𝑠)𝑑𝑠→0,𝑚→+∞. (43) Thus, for above 𝜀>0, there exists a natural number 𝑁, for 𝑚>𝑁; we have _𝛿^1−𝛿𝜌𝑓𝑠,𝑥_𝑚(𝑠),𝑦_𝑚(𝑠)−𝑓𝑠,𝑥₀(𝑠),𝑦₀(𝑠)𝜀𝑑𝑠<4_𝛿1−𝛿,𝜌𝑔𝑠,𝑥_𝑚(𝑠),𝑦_𝑚(𝑠)−𝑔𝑠,𝑥₀(𝑠),𝑦₀(𝑠)𝜀𝑑𝑠<4. (44) It follows from (39)–(44) that when 𝑚>𝑁𝑇₁𝑥_𝑚,𝑦_𝑚−𝑇₁𝑥₀,𝑦₀≤𝜌¹₀𝑓𝑠,𝑥_𝑚(𝑠),𝑦_𝑚(𝑠)−𝑓𝑠,𝑥₀(𝑠),𝑦₀(𝑠)𝑑𝑠+𝜌¹₀𝑔𝑠,𝑥_𝑚(𝑠),𝑦_𝑚(𝑠)−𝑔𝑠,𝑥₀(𝑠),𝑦₀(𝑠)𝑑𝑠≤_{[0,𝛿]∪[1−𝛿,1]}𝜌𝑐^𝜆1+𝜇1𝛾𝑎^−𝜇1𝑓()+𝑠,1,1−𝑠𝑑𝑠_{[0,𝛿]∪[1−𝛿,1]}𝜌𝑐^𝜆2+𝜇2𝛾𝑎^−𝜇2𝑔(𝑠,1−𝑠,1)𝑑𝑠+_𝛿^1−𝛿𝜌𝑓𝑠,𝑥_𝑚(𝑠),𝑦_𝑚(𝑠)−𝑔𝑠,𝑥₀(𝑠),𝑦₀(𝑠)+𝑑𝑠_𝛿1−𝛿𝜌𝑔𝑠,𝑥_𝑚(𝑠),𝑦_𝑚(𝑠)−𝑔𝑠,𝑥₀(𝑠),𝑦₀(𝑠)𝑑𝑠<𝜀. (45) This implies that 𝑇₁∶𝑃_𝑏⧵𝑃_𝑎→𝑄 is continuous. Similarly, we can prove that 𝑇₂∶𝑃_𝑏⧵𝑃_𝑎→𝑄 is continuous. So, 𝑇∶𝑃_𝑏⧵𝑃_𝑎→𝑃 is continuous. Summing up, 𝑇∶𝑃_𝑏⧵𝑃_𝑎→𝑃 is completely continuous.

Our main tool of this paper is the following cone compression and expansion fixed point theorem.

Lemma 6 (see [18]). Let 𝐸 be a Banach space and 𝑃 a cone in 𝐸. Suppose that Ω₁ and Ω₂ are two bounded open subsets of 𝐸 with 𝜃∈Ω₁, Ω₁⊂Ω₂. If  𝑇∶𝑃∩(Ω₂⧵Ω₁)→𝑃 is a completely continuous operator satisfying ‖𝑇𝑥‖≥‖𝑥‖,𝑓𝑜𝑟𝑥∈𝑃∩𝜕Ω₁,‖𝑇𝑥‖≤‖𝑥‖,𝑓𝑜𝑟𝑥∈𝑃∩𝜕Ω₂, (46) then 𝑇 has a fixed point in 𝑃∩(Ω₂⧵Ω₁).

3. Main Results

In this section, we present our main results.

Theorem 7. Suppose that conditions (𝐻₁), (𝐻₂), and (𝐻₃) hold. Then, if 𝜆₁+𝜇₁<1 and 𝜆₂+𝜇₂<1, the differential system (1) has a unique positive solution (𝑥^∗,𝑦^∗).

Proof. We divide the rather long proof into three steps. (i) The differential system (1) has at least one positive solution (𝑥^∗,𝑦^∗).
Choose r, 𝑅 such that 𝜈0<𝑟≤min4𝛾^𝜆11₀𝑠()()1−𝑠×𝑓𝑠,1−𝑠,1𝑑𝑠^{1/(1−𝜆1)},12,𝑅≥max𝜌¹₀𝑓(𝑠,1,1−𝑠)𝑑𝑠+𝜌¹₀𝑔(𝑠,1−𝑠,1)𝑑𝑠^{1/(1−max{𝜆1,𝜆2})},1𝛾,2. (47)
Clearly 0<𝑟<1<𝑅. By Lemma 5, 𝑇∶𝑃_𝑅⧵𝑃_𝑟→𝑃 is completely continuous.
Extend 𝑇 (denote 𝑇 yet) to 𝑇∶𝑃_𝑅→𝑃 which is completely continuous.
Then, for (𝑥,𝑦)∈𝜕𝑃_𝑟, we have 𝑟𝛾(1−𝑡)≤𝑥(𝑡),𝑦(𝑡)≤𝑟,𝑡∈[0,1]. (48) By Remarks 1 and 2, (𝐻₁), and (𝐻₂), we get 𝑇_𝑖(𝑡)≥𝜈𝑥,𝑦4¹₀𝑠()𝑓()≥𝜈1−𝑠𝑠,𝛾𝑟1−𝑠,𝑟𝑑𝑠4¹₀𝑠()𝑓()≥𝜈1−𝑠𝑠,𝛾𝑟1−𝑠,1𝑑𝑠4𝛾^𝜆1𝑟^𝜆11₀𝑠(1−𝑠)𝑓(𝑠,1−𝑠,1)𝑑𝑠≥𝑟=𝑥,𝑦₁3,𝑖=1,2,𝑡∈0,4. (49) This guarantees that 𝑇𝑥,𝑦₁≥𝑥,𝑦₁,∀𝑥,𝑦∈𝜕𝑃_𝑟. (50) On the other hand, for (𝑥,𝑦)∈𝜕𝑃_𝑅, we have 𝑅𝛾(1−𝑡)≤𝑥(𝑡),𝑦(𝑡)≤𝑅,𝑡∈[0,1]. (51) Therefore, 𝑇_𝑖𝑥,𝑦(𝑡)≤𝜌¹₀𝑓𝑠,𝑅,𝛾𝑅(1−𝑠)𝑑𝑠+𝜌¹₀𝑔𝑠,𝛾𝑅(1−𝑠),𝑅𝑑𝑠≤𝜌¹₀𝑓(𝑠,𝑅,1−𝑠)𝑑𝑠+𝜌¹₀𝑔(𝑠,1−𝑠,𝑅)𝑑𝑠≤𝜌𝑅^𝜆11₀𝑓(𝑠,1,1−𝑠)𝑑𝑠+𝜌𝑅^𝜆21₀𝑔(𝑠,1−𝑠,1)𝑑𝑠≤𝜌𝑅^{max{𝜆1,𝜆2}1}₀𝑓()+𝑠,1,1−𝑠𝑑𝑠₁₀𝑔(𝑠,1−𝑠,1)𝑑𝑠≤𝑅=𝑥,𝑦₁,𝑖=1,2,𝑡∈[0,1]. (52)This guarantees that 𝑇𝑥,𝑦₁≤𝑥,𝑦₁,∀𝑥,𝑦∈𝜕𝑃_𝑅. (53) By the complete continuity of 𝑇, (50) and (53), and Lemma 6, we obtain that 𝑇 has a fixed point (𝑥^∗,𝑦^∗) in 𝑃_𝑅⧵𝑃_𝑟. Consequently, (1) has a positive solution (𝑥^∗,𝑦^∗) in 𝑃_𝑅⧵𝑃_𝑟. (ii) Suppose that (𝑥,𝑦) is a positive solution of the differential system (1).
Then there exist real numbers 0<𝑚<1 such that 𝑚()(𝑡)≤11−𝑡≤𝑥𝑚(),𝑚()(𝑡)≤11−𝑡1−𝑡≤𝑦𝑚(1−𝑡),∀𝑡∈[0,1]. (54) From Lemma 5, we know that (𝑥,𝑦)∈𝑃⧵{𝜃}. So, we have 𝛾𝑥,𝑦₁(1−𝑡)≤𝑥(𝑡),𝑦(𝑡)≤𝑥,𝑦₁. (55) Let 𝑐 be a constant such that ‖(𝑥,𝑦)‖₁/𝑐<1 and 𝑐>1/𝛾. By Lemma 3, we get 𝑥(𝑡)≤𝜌(1−𝑡)¹₀𝑓𝛾𝑠,𝑐,𝑥,𝑦₁𝑐()()1−𝑠𝑑𝑠+𝜌1−𝑡¹₀𝑔𝛾𝑠,𝑥,𝑦₁𝑐(1−𝑠),𝑐𝑑𝑠≤𝑐^𝜆1+𝜇1𝛾𝑥,𝑦₁^−𝜇1𝜌(1−𝑡)¹₀𝑓(𝑠,1,1−𝑠)𝑑𝑠+𝑐^𝜆2+𝜇2𝛾𝑥,𝑦₁^−𝜇2𝜌(1−𝑡)¹₀𝑔()()[].𝑠,1−𝑠,1𝑑𝑠=∶𝐶1−𝑡,𝑡∈0,1 (56) In the same way, we can prove that 𝑦(𝑡)≤𝐶(1−𝑡), 𝑡∈[0,1]. Then we may pick out 𝑚 such that 𝑚=min{𝛾‖(𝑥,𝑦)‖₁,1/𝐶,1/2}, which implies that (54) holds. (iii) The differential system (1) has a unique positive solution (𝑥^∗,𝑦^∗).
Assuming the contrary, we find that the differential system (1) has a positive solution (𝑥_∗,𝑦_∗) different from (𝑥^∗,𝑦^∗). By (54), there exist 𝛿₁,𝛿₂>0, such that 𝛿₁(1−𝑡)≤𝑥^∗(𝑡),𝑦^∗(𝑡)≤1𝛿₁()[],𝛿1−𝑡,∀𝑡∈0,1₂(1−𝑡)≤𝑥_∗(𝑡),𝑦_∗(𝑡)≤1𝛿₂(1−𝑡),∀𝑡∈[0,1]. (57) Hence, we have 𝛿₁𝛿₂𝑥_∗(𝑡)≤𝑥^∗(𝑡)≤1𝛿₁𝛿₂𝑥_∗(𝑡),𝛿₁𝛿₂𝑦_∗(𝑡)≤𝑦^∗(𝑡)≤1𝛿₁𝛿₂𝑦_∗(𝑡),∀𝑡∈[0,1]. (58) Clearly, 𝛿₁𝛿₂≠1. Put 𝛿^∗=sup𝛿∣𝛿𝑥_∗(𝑡)≤𝑥^∗(𝑡)≤1𝛿𝑥_∗(𝑡),𝛿𝑦_∗(𝑡)≤𝑦^∗(𝑡)≤1𝛿𝑦_∗(𝑡),∀𝑡∈[0,1]. (59) It is easy to see that 1>𝛿^∗≥𝛿₁𝛿₂>0, and 𝛿^∗𝑥_∗(𝑡)≤𝑥^∗(𝑡)≤1𝛿^∗𝑥_∗(𝑡),𝛿^∗𝑦_∗(𝑡)≤𝑦^∗(𝑡)≤1𝛿^∗𝑦_∗(𝑡),∀𝑡∈[0,1]. (60) So, by (𝐻₁), we have 𝑓𝑡,𝑥^∗(𝑡),𝑦^∗(𝑡)≥𝑓𝑡,𝛿^∗𝑥_∗(𝑡),1𝛿^∗𝑦_∗(𝑡)≥𝛿^{∗𝜆1+𝜇1}𝑓𝑡,𝑥_∗(𝑡),𝑦_∗(𝑡)≥𝛿^∗𝜎𝑓𝑡,𝑥_∗(𝑡),𝑦_∗(𝑡),𝑔𝑡,𝑥^∗(𝑡),𝑦^∗(𝑡)≥𝛿^{∗𝜆2+𝜇2}𝑔𝑡,𝑥_∗(𝑡),𝑦_∗(𝑡)≥𝛿^∗𝜎𝑔𝑡,𝑥_∗(𝑡),𝑦_∗(𝑡), (61) where 𝜎=max{𝜆₁+𝜇₁,𝜆₂+𝜇₂} such that 𝜎<1. Therefore, we have 𝑥^∗(𝑡)=𝑇₁𝑥^∗,𝑦^∗𝑡)=₁₀(𝐺₁(𝑡,𝑠)𝑓𝑠,𝑥^∗(𝑠),𝑦^∗(𝑠)+𝑑𝑠₁₀𝐻₁(𝑡,𝑠)𝑔𝑠,𝑥^∗(𝑠),𝑦^∗(𝑠)𝑑𝑠≥𝛿^∗𝜎1₀𝐺₁(𝑡,𝑠)𝑓𝑠,𝑥_∗(𝑠),𝑦_∗(𝑠)𝑑𝑠+¹₀𝐻₁(𝑡,𝑠)𝑔𝑠,𝑥_∗(𝑠),𝑦_∗(𝑠)𝑑𝑠=𝛿^∗𝜎𝑇₁𝑥_∗,𝑦_∗(𝑡)=𝛿^∗𝜎𝑥_∗(𝑡). (62) Similarly, we can get 𝑦^∗(𝑡)≥𝛿^∗𝜎𝑦_∗(𝑡),𝑥_∗(𝑡)≥𝛿^∗𝜎𝑥^∗(𝑡),𝑦_∗(𝑡)≥𝛿^∗𝜎𝑦^∗(𝑡). (63) Noticing that 0<𝛿^∗, 𝜎<1, we get to a contradiction with the maximality of 𝛿^∗. Thus, the differential system (1) has a unique positive solution (𝑥^∗,𝑦^∗). This completes the proof of Theorem 7.

4. An Example

In this section, we give an example to illustrate the usefulness of our main results. Let us consider the singular differential system with couple boundary value problem −𝑥^′′=√𝑥₃𝑦𝑡(1−𝑡),−𝑦^′′=³√𝑦√𝑥,𝑥(1)(1)(0)1=𝑦=0,𝑥=𝑦31+𝑦2,𝑦(0)=¹₀𝑥(𝑡)𝑑𝑡². (64)

Let 𝑓=√𝑡,𝑥,𝑦₃𝑥()=𝑦𝑡1−𝑡,𝑔𝑡,𝑥,𝑦³√𝑦√𝑥,𝛼(𝑡)=10,𝑡∈0,3,11,𝑡∈3,12,12,𝑡∈2,𝛽(𝑡),1=𝑡²,𝜆₁=𝜇₂=12,𝜆₂=𝜇₁=13; (65) then 𝜅₁=76,𝜅₂=13,𝜅=1−𝜅₁𝜅₂=11,18₁₀𝑓()2𝑠,1,1−𝑠𝑑𝑠=𝐵3,16₁₀,𝑔()1𝑠,1−𝑠,1𝑑𝑠=𝐵1,2. (66) So all conditions of Theorem 7 are satisfied for (64), and our conclusion follows from Theorem 7.

Acknowledgments

The authors sincerely thank the reviewers for their valuable suggestions and useful comments that have led to the present improved version of the original paper. This project is supported by the National Natural Science Foundation of China (11371221, 11071141), the Specialized Research Foundation for the Doctoral Program of Higher Education of China (20123705110001), the Program for Scientific Research Innovation Team in Colleges and Universities of Shandong Province, the Postdoctoral Science Foundation of Shandong Province, and Foundation of SDUST.