Existence Results for a Coupled System of Nonlinear Singular Fractional Differential Equations with Impulse Effects

Abstract

A boundary value problem for the singular fractional differential system with impulse effects is presented. By applying Schauder's fixed point theorem in a suitably Banach space, we obtain the existence of at least one solution for this problem. Two examples are presented to illustrate the main theorem.

1. Introduction

Fractional differential equations have received increasing attention during recent years since the behavior of many physical, chemical, and engineering processes can be properly described by using fractional differential equations theory; see the books [1–3], papers [4, 5] and references therein. For details on the geometric and physical interpretation of the derivatives of noninteger order, see, for example, [6–11]. For some recent works with applications to engineering we refer the reader to [12–15].

For an introduction of the basic theory of impulsive differential equation, we refer the reader to [16]. Among previous research, little is concerned with differential equations with fractional order with impulses [17]. Ahmad and Sivasundaram [18, 19] gave some existence results for two-point boundary value problems involving nonlinear impulsive hybrid differential equations of fractional order 1<𝛼≤2. Ahmad and Nieto in [20] establish sufficient conditions for the existence of solutions of the antiperiodic boundary value problem for impulsive differential equations with the Caputo derivative of order 𝑞∈(1,2]. Some recent results on impulsive initial value problems or boundary value problems for fractional differential equations on a finite interval can be found in [21–23] and references therein. The memory property of fractional calculus makes studies more complicated.

This paper is motivated by [24] in which the following boundary value problem for the fractional differential equation 𝐷^𝛼₀⁺𝑥(𝑡)=𝑓𝑡,𝑦(𝑡),𝐷^𝑝₀₊𝑦(𝑡(),𝐷),𝑡∈0,1_𝛽0+𝑦(𝑡)=𝑔𝑡,𝑥(𝑡),𝐷^𝑞₀₊𝑥(𝑡(),𝑥(0)(0)(1)𝜂𝑦(1)𝜂),𝑡∈0,1=0,𝑦=0,𝑥−𝛾𝑥=0,−𝛾𝑦=0 (1) was studied, where 1<𝛼, 𝛽<2, 0<𝑝≤𝛽−1 and 0<𝑞≤𝛼−1, 𝛾>0, 1>𝛾𝜂^𝛼−1, 1>𝛾𝜂^𝛽−1 and 𝑓,𝑔∶[0,1]×𝑅²→𝑅 are continuous functions, and 𝐷₀⁺ is the Riemann-Liouville fractional derivative. An existence result was proved for BVP (1) in [24]. The growth assumptions imposed on 𝑓 and 𝑔 are sublinear cases (see [24, Theorem 2.1]); that is, there exist functions 𝑎,𝑏∈𝐿¹(0,1), nonnegative constants 𝜖₁,𝜖₂>0, 𝛿₁,𝛿₂≥0 and 𝜌₁,𝜌₂,𝜎₁,𝜎₂∈(0,1) such that 𝑓𝑡,𝑥,𝑦≤𝑎(𝑡)+𝜖₁|𝑥|^𝜌1+𝜖₂𝑦^𝜌2,𝑔𝑡,𝑥,𝑦≤𝑏(𝑡)+𝛿₁|𝑥|^𝜎1+𝛿₂𝑦^𝜎2. (2)

In [25], the following boundary value problem for the fractional differential equation 𝐷^𝛼₀⁺𝑥(𝑡)=𝑓𝑡,𝑦(𝑡),𝐷^𝑝₀₊𝑦(𝑡(),𝐷),𝑡∈0,1_𝛽0+𝑦(𝑡)=𝑔𝑡,𝑥(𝑡),𝐷^𝑞₀₊𝑥(𝑡(),𝑥(0)(0)(1)(1)),𝑡∈0,1=0,𝑦=0,𝑥=0,𝑦=0 (3) was studied, where 1<𝛼, 𝛽<2, 0<𝑝≤𝛽−1 and 0<𝑞≤𝛼−1, and 𝑓,𝑔∶[0,1]×𝑅²→𝑅 are continuous functions, and 𝐷₀⁺ is the Riemann-Liouville fractional derivative. The growth assumptions imposed on 𝑓 and 𝑔 are sublinear cases (see [25, Theorem 3.1]), that is, there exist functions 𝑎,𝑏∈𝐿¹(0,1), nonnegative constants 𝜖₁,𝜖₂>0, 𝛿₁,𝛿₂≥0, and 𝜌₁,𝜌₂,𝜎₁,𝜎₂∈(0,1] such that 𝑓𝑡,𝑥,𝑦≤𝑎(𝑡)+𝜖₁|𝑥|^𝜌1+𝜖₂𝑦^𝜌2,𝑔𝑡,𝑥,𝑦≤𝑏(𝑡)+𝛿₁|𝑥|^𝜎1+𝛿₂𝑦^𝜎2, (4)or sublinear cases, that is, there exist nonnegative constants 𝜖₁,𝜖₂>0, 𝛿₁,𝛿₂≥0 and 𝜌₁,𝜌₂,𝜎₁,𝜎₂∈(1,∞) such that 𝑓𝑡,𝑥,𝑦≤𝜖₁|𝑥|^𝜌1+𝜖₂𝑦^𝜌2,𝑔𝑡,𝑥,𝑦≤𝛿₁|𝑥|^𝜎1+𝛿₂𝑦^𝜎2. (5) We find that in the superlinear cases, BVP (3) has a pair of solutions (𝑥,𝑦)=(0,0) without needing any other assumptions. Hence, these cases are trivial ones discussed in [25].

It is interesting to consider the solvability of BVP (1) when the growth assumptions imposed on 𝑓,𝑔 are superlinear cases. Furthermore, the solvability of BVP (1) is not studied when 𝑞>𝛼−1 or 𝑝>𝛽−1.

In this paper we consider the following nonlinear boundary value problem for the singular multiterm fractional differential equation with impulse effects whose boundary conditions are of integral form 𝐷^𝛼₀⁺𝑥(𝑡)=𝜙(𝑡)𝑓𝑡,𝑦(𝑡),𝐷^𝑝₀₊𝑦(𝑡,())𝑡∈0,1,𝑡≠𝑡₁,𝐷^𝛽₀₊𝑦(𝑡)=𝜓(𝑡)𝑔𝑡,𝑥(𝑡),𝐷^𝑞₀₊𝑥(𝑡),𝑡∈(0,1),𝑡≠𝑡₁,lim_𝑡→0𝑡^2−𝛼𝑥(𝑡)=¹₀𝑢(𝑠)𝐺𝑠,𝑦(𝑠),𝐷^𝑝₀₊𝑦(𝑠)𝑑𝑠,lim_𝑡→0𝑡^2−𝛽𝑦(𝑡)=¹₀𝑣(𝑠)𝐻𝑠,𝑥(𝑠),𝐷^𝑞₀₊𝑥(𝑠)𝑥(1)=𝑑𝑠,₁₀𝑚(𝑠)𝑀𝑠,𝑦(𝑠),𝐷^𝑝₀₊𝑦(𝑠)𝑦(1)=𝑑𝑠,₁₀𝑛(𝑠)𝑁𝑠,𝑥(𝑠),𝐷^𝑞₀₊𝑥(𝑠)𝑡𝑑𝑠,Δ𝑥₁=lim_{𝑡→𝑡+1}𝑥(𝑡)−lim_{𝑡→𝑡−1}𝑥(𝑡)=𝐼𝑡₁,𝑦𝑡₁,𝐷^𝑝₀₊𝑦𝑡₁,Δ𝑦𝑡₁=lim_{𝑡→𝑡+1}𝑦(𝑡)−lim_{𝑡→𝑡−1}𝑦(𝑡)=𝐽𝑡₁,𝑥𝑡₁,𝐷^𝑞₀₊𝑥𝑡₁,Δ𝐷^𝑞₀₊𝑥𝑡₁=lim_{𝑡→𝑡+1}𝐷^𝑞₀₊𝑥(𝑡)−lim_{𝑡→𝑡−1}𝐷^𝑞₀₊𝑥(𝑡)=𝐼₁𝑡₁,𝑦𝑡₁,𝐷^𝑝₀₊𝑦𝑡₁,Δ𝐷^𝑝₀₊𝑦𝑡₁=lim_{𝑡→𝑡+1}𝐷^𝑝₀₊𝑦(𝑡)−lim_{𝑡→𝑡−1}𝐷^𝑝₀₊𝑦(𝑡)=𝐽₁𝑡₁,𝑥𝑡₁,𝐷^𝑞₀₊𝑥𝑡₁, (6)(7) where(a)1<𝛼, 𝛽≤2, 0<𝑝<𝛽 and 0<𝑞<𝛼, 𝐷₀⁺ is the Riemann-Liouville fractional derivative,(b)𝜙,𝜓∶(0,1)→𝑅,𝑓,𝑔 defined on (0,1)×𝑅²,(c)𝑚,𝑛,𝑢,𝑣∶(0,1)→𝑅 with 𝑚,𝑛,𝑢,𝑣∈𝐿¹(0,1), 𝐺,𝐻,𝑀,𝑁 defined on (0,1)×𝑅²,(d)0=𝑡₀<𝑡₁<𝑡₂=1,(e)𝐼,𝐼₁,𝐽,𝐽₁∶(0,1)×𝑅²→𝑅.

A pair of functions (𝑥,𝑦) defined on (0,1) is called a solution of BVP (1) and BVP (3), if 𝑥|_{(𝑡𝑘,𝑡𝑘+1]}, 𝐷^𝑞₀₊𝑥|_{(𝑡𝑘,𝑡𝑘+1]} and 𝑦|_{(𝑡𝑘,𝑡𝑘+1]}, 𝐷^𝑝₀₊𝑦|_{(𝑡𝑘,𝑡𝑘+1]}(𝑘=0,1) are continuous, there exists the limits lim_{𝑡→𝑡+𝑘}𝑡^2−𝛼𝑥(𝑡),lim_{𝑡→𝑡+𝑘}𝑡^2−𝛽𝑦(𝑡),lim_{𝑡→𝑡+𝑘}𝑡^{2+𝑞−𝛼}𝐷^𝑞₀₊𝑥(𝑡),lim_{𝑡→𝑡+𝑘}𝑡^{2+𝑝−𝛽}𝐷^𝑝₀₊𝑦(𝑡),𝑘=0,1, (8)𝐷^𝛼₀⁺𝑥, 𝐷^𝛽₀₊𝑦∈𝐿¹(0,1) and (𝑥,𝑦) satisfies all equations in (6) and (7).

The novelty of this paper is as follows: first, the fractional differential equations in (6) are multiterm ones and their nonlinearities 𝑓,𝑔 depend on the lower fractional derivatives; second, both 𝜙 and 𝜓 may be singular at 𝑡=0 and 𝑡=1, that is, 𝜙(𝑡)𝑓(𝑡,𝑥,𝑦) and 𝜓(𝑡)𝑔(𝑡,𝑥,𝑦) may be not continuous functions on [0,1]×𝑅², the boundary conditions are integral boundary conditions, and we obtain the results on the existence of at least one solution of BVP (6)-(7); third, 0<𝑝<𝛽 and 0<𝑞<𝛼 are supposed; the growth assumptions imposed on 𝑓,𝑔,𝐺,𝐻,𝑀,𝑁 and 𝐼,𝐼₁,𝐽,𝐽₁ are allowed to be sublinear cases. Finally, two examples are given to illustrate the efficiency of the main theorem.

The remainder of this paper is as follows: in Section 2, we present preliminary results. In Section 3, the main theorem and its proof are given. In Section 4, two examples are given to illustrate the main results.

2. Preliminaries

In this section, we present some background definitions and preliminary results.

Definition 1 (see [1]). The Riemann-Liouville fractional integral of order 𝛼>0 of a function 𝑔∶(0,∞)→𝑅 is given by 𝐼^𝛼₀⁺𝑔(𝑡)=1Γ(𝛼)^𝑡₀(𝑡−𝑠)^𝛼−1𝑔(𝑠)𝑑𝑠, (9) provided that the right-hand side exists.

Definition 2 (see [1]). The Riemann-Liouville fractional derivative of order 𝛼>0 of a continuous function 𝑔∶(0,∞)→𝑅 is given by 𝐷^𝛼₀⁺𝑔(𝑡)=1Γ()𝑑𝑛−𝛼𝑑𝑡^𝑛𝑛𝑡₀𝑔(𝑠)(𝑡−𝑠)^{𝛼−𝑛+1}𝑑𝑠, (10) where 𝑛−1≤𝛼<𝑛, provided that the right-hand side is pointwise defined on (0,∞).

Definition 3. 𝐾∶(0,1)×𝑅²→𝑅 is called a 𝛽-Caratheodory function if 𝐾 satisfies that(i)𝑡→𝐾(𝑡,𝑡^𝛽−2𝑈,𝑡^{𝛽−𝑝−2}𝑉) is continuous on (𝑡_𝑘,𝑡_𝑘+1](𝑘=0,1) for every (𝑈,𝑉)∈𝑅²;(ii)(𝑈,𝑉)→𝐾(𝑡,𝑡^𝛽−2𝑈,𝑡^{𝛽−𝑝−2}𝑉) is continuous on 𝑅² for every 𝑡∈(0,1);(iii)for each 𝑟>0 there exists a constant 𝐴_𝑟>0 such that |𝐾(𝑡,𝑡^𝛽−2𝑈,𝑡^{𝛽−𝑝−2}𝑉)|≤𝐴_𝑟, 𝑡∈(0,1), |𝑈|,|𝑉|≤𝑟.

Definition 4. 𝑄∶(0,1)×𝑅²→𝑅 is called a 𝛼-Caratheodory function if 𝑄 satisfies that(i)𝑡→𝑄(𝑡,𝑡^𝛼−2𝑈,𝑡^{𝛼−𝑞−2}𝑉) is continuous on (𝑡_𝑘,𝑡_𝑘+1](𝑘=0,1) for every (𝑈,𝑉)∈𝑅²;(ii)(𝑈,𝑉)→𝑄(𝑡,𝑡^𝛼−2𝑈,𝑡^{𝛼−𝑞−2}𝑉) is continuous on 𝑅² for every 𝑡∈(0,1);(iii)for each 𝑟>0 there exists a constant 𝐵_𝑟>0 such that |𝑄(𝑡,𝑡^𝛼−2𝑈,𝑡^{𝛼−𝑞−2}𝑉)|≤𝐵_𝑟, 𝑡∈(0,1), |𝑈|,|𝑉|≤𝑟.

Lemma 5 (the Leray-Schauder nonlinear alternative [23]). Let 𝑋 be a Banach space and 𝑇∶𝑋→𝑋 be a completely continuous operator. Suppose Ω is a nonempty open subset of 𝑋 centered at zero. Then either there exists 𝑥∈𝜕Ω and 𝜆∈(0,1) such that 𝑥=𝜆𝑇𝑥 or there exists 𝑥∈Ω such that 𝑥=𝑇𝑥.

Let the gamma and beta functions Γ(𝛼) and 𝐁(𝑝,𝑞) be defined by Γ(𝛼)=₀^+∞𝑥^𝛼−1𝑒^−𝑥𝐁=𝑑𝑥,𝑝,𝑞¹₀𝑥^𝑝−1(1−𝑥)^𝑞−1‖𝑚‖𝑑𝑥,₁=¹₀|𝑚(𝑠)|𝑑𝑠for𝑚∈𝐿¹(0,1). (11) Choose 𝑋=𝑥|_{(𝑡𝑘,𝑡𝑘+1]}∈𝐶⁰𝑡_𝑘,𝑡_𝑘+1),𝐷(𝑘=0,1_𝑞0+𝑥|_{(𝑡𝑘,𝑡𝑘+1]}∈𝐶⁰𝑡_𝑘,𝑡_𝑘+1),(](𝑘=0,1𝑥∶0,1⟶𝑅thereexistthelimitslim_{𝑡→𝑡+𝑘}𝑡^2−𝛼𝑥(𝑡),lim_{𝑡→𝑡+𝑘}𝑡^{2+𝑞−𝛼}𝐷^𝑞₀₊𝑥(𝑡),𝑌=𝑦|_{(𝑡𝑘,𝑡𝑘+1]}∈𝐶⁰𝑡_𝑘,𝑡_𝑘+1),𝐷(𝑘=0,1_𝑝0+𝑥|_{(𝑡𝑘,𝑡𝑘+1]}∈𝐶⁰𝑡_𝑘,𝑡_𝑘+1),(](𝑘=0,1𝑦∶0,1⟶𝑅thereexistthelimitslim_{𝑡→𝑡+𝑘}𝑡^2−𝛽𝑦(𝑡),lim_{𝑡→𝑡+𝑘}𝑡^{2+𝑝−𝛽}𝐷^𝑝₀₊𝑦(𝑡). (12) For 𝑥∈𝑋, define the norm by ‖𝑥‖=‖𝑥‖_𝑋=maxsup_𝑡∈(0,1)𝑡^2−𝛼|𝑥(𝑡)|,sup_()𝑡∈0,1𝑡^{2+𝑞−𝛼}𝐷^𝑞₀₊𝑥(𝑡). (13) It is easy to show that 𝑋 is a real Banach space. For 𝑦∈𝑌, define the norm by 𝑦=𝑦_𝑌=maxsup_()𝑡∈0,1𝑡^2−𝛽𝑦(𝑡),sup_()𝑡∈0,1𝑡^{2+𝑝−𝛽}𝐷^𝑝₀₊𝑦(𝑡). (14) It is easy to show that 𝑌 is a real Banach space. Thus, (𝑋×𝑌,||⋅||) is a Banach space with the norm defined by ||(𝑥,𝑦)||=max{||𝑥||_𝑋,||𝑦||_𝑌}for(𝑥,𝑦)∈𝑋×𝑌.

In this paper, we suppose the following:(A) 𝜙 satisfies that there exist constants 𝐿₁>0, 𝑘>−1, 𝛿∈(𝑞−𝛼,0] such that 𝛼+2𝛿−𝑞>0, 𝛼+𝑘+𝛿−𝑞≥0, and |𝜙(𝑡)|≤𝐿₁𝑡^𝑘(1−𝑡)^𝛿 for all 𝑡∈(0,1); 𝜓 satisfies that there exist constants 𝐿₂>0, 𝑙>−1, 𝜃∈(𝑝−𝛽,0] such that 𝛽+2𝜃−𝑝>0, 𝛽+𝑙+𝜃−𝑝≥0, and |𝜓(𝑡)|≤𝐿₂𝑡^𝑙(1−𝑡)^𝜃 for all 𝑡∈(0,1).(B) 𝑓,𝐺,𝑀,𝐼,𝐼₁ are 𝛽-Caratheodory functions and 𝑔,𝐻,𝑁,𝐽,𝐽₁ are 𝛼-Caratheodory functions.

Remark 6. Suppose that 𝑓 is a 𝛽-Caratheodory function. For example, 𝛼=7/4, 𝑞=1/8, choose 𝑘=−1/2, 𝛿=−3/4 and 𝜙(𝑡)=𝑡^𝑘(1−𝑡)^𝛿, then 𝑘>−1, 𝛿∈(−𝛼,0] such that 𝛼+2𝛿−𝑞>0, 𝛼+𝑘+𝛿−𝑞≥0, and |𝜙(𝑡)|≤𝑡^𝑘(1−𝑡)^𝛿 for all 𝑡∈(0,1). It is easy to see that 𝜙 is singular at 𝑡=0 and 𝑡=1.

Lemma 7. Suppose that 𝑦∈𝑌, and (a)–(e), (A)-(B) hold. Then 𝑥∈𝑋 is a solution of 𝐷^𝛼₀⁺𝑥(𝑡)=𝜙(𝑡)𝑓𝑡,𝑦(𝑡),𝐷^𝑝₀₊𝑦(𝑡),𝑡∈(0,1),𝑡≠𝑡₁,lim_𝑡→0𝑡^2−𝛼𝑥(𝑡)=¹₀𝑢(𝑠)𝐺𝑠,𝑦(𝑠),𝐷^𝑝₀₊𝑦(𝑠𝑥(1)=)𝑑𝑠,₁₀𝑚(𝑠)𝑀𝑠,𝑦(𝑠),𝐷^𝑝₀₊𝑦(𝑠)𝑡𝑑𝑠,Δ𝑥₁=𝐼𝑡₁,𝑦𝑡₁,𝐷^𝑝₀₊𝑦𝑡₁,Δ𝐷^𝑞₀₊𝑥𝑡₁=𝐼₁𝑡₁,𝑦𝑡₁,𝐷^𝑝₀₊𝑦𝑡₁, (15) if and only if 𝑥∈𝑋 satisfies the integral equation 𝑥(𝑡)=∫^𝑡₀(𝑡−𝑠)^𝛼−1Γ(𝛼)𝜙(𝑢)𝑓𝑢,𝑦(𝑢),𝐷^𝑝₀₊𝑦(𝑢)−𝑡𝑑𝑢_𝛼−1Γ(𝛼)×∫¹₀(1−𝑠)^𝛼−1𝜙(𝑠)𝑓𝑠,𝑦(𝑠),𝐷^𝑝₀₊𝑦(𝑠)𝑑𝑠+𝑡^𝛼−2∫¹₀𝑢(𝑠)𝐺𝑠,𝑦(𝑠),𝐷^𝑝₀₊𝑦(𝑠)𝑑𝑠+𝑡^𝛼−1∫¹₀𝑚(𝑠)𝑀𝑠,𝑦(𝑠),𝐷^𝑝₀₊𝑦(𝑠)+𝑡𝑑𝑠_𝛼−1Π×Γ(𝛼)Γ𝑡𝛼−𝑞^{1𝛼−𝑞−1}−Γ(𝛼−1)Γ𝑡𝛼−𝑞−2^{1𝛼−𝑞−2}×𝐼𝑡₁,𝑦𝑡₁,𝐷^𝑝₀₊𝑦𝑡₁+𝑡^𝛼−1𝑡₁^𝛼−2−𝑡₁^𝛼−1Π×𝐼₁𝑡₁,𝑦𝑡₁,𝐷^𝑝₀₊𝑦𝑡₁,𝑡∈0,𝑡₁,∫^𝑡₀(𝑡−𝑠)^𝛼−1Γ(𝛼)𝜙(𝑠)𝑓𝑠,𝑦(𝑠),𝐷^𝑝₀₊𝑦(𝑠)−𝑡𝑑𝑠_𝛼−1Γ(𝛼)×∫¹₀(1−𝑠)^𝛼−1𝜙(𝑠)𝑓𝑠,𝑦(𝑠),𝐷^𝑝₀₊𝑦(𝑠)+𝑡𝑑𝑠_𝛼−2−𝑡^𝛼−1×∫¹₀𝑢(𝑠)𝐺𝑠,𝑦(𝑠),𝐷^𝑝₀₊𝑦(𝑠)𝑑𝑠+𝑡^𝛼−1∫¹₀𝑚(𝑠)𝑀𝑠,𝑦(𝑠),𝐷^𝑝₀₊𝑦(𝑠)+𝑡𝑑𝑠_𝛼−1−𝑡^𝛼−2ΠΓ(𝛼)Γ𝑡𝛼−𝑞^{1𝛼−𝑞−1}×𝐼𝑡₁,𝑦𝑡₁,𝐷^𝑝₀₊𝑦𝑡₁+𝑡^𝛼−2−𝑡^𝛼−1Π𝑡₁^𝛼−1𝐼₁𝑡₁,𝑦𝑡₁,𝐷^𝑝₀₊𝑦𝑡₁,𝑡∈𝑡₁,1, (16) where Γ()Π=𝛼−1Γ−Γ(𝛼)𝛼−𝑞−1Γ𝑡𝛼−𝑞^{12𝛼−𝑞−3}. (17)

Proof. If 𝑦∈𝑌 is a solution of BVP (15), then 𝑦=maxsup_()𝑡∈0,1𝑡^2−𝛽𝑦(𝑡),sup_()𝑡∈0,1𝑡^{2+𝑝−𝛽}𝐷^𝑝₀₊𝑦(𝑡)=𝑟<+∞, (18) and 𝑥 satisfies all equations in (31) From (B), 𝑓 is a 𝛽-Caratheodory function, then there exists 𝐴_𝑟>0 such that 𝑓𝑡,𝑦(𝑡),𝐷^𝑝₀₊𝑦(𝑡=𝑓)𝑡,𝑡^𝛽−2𝑡^2−𝛽𝑦(𝑡),𝑡^{𝛽−𝑝−2}𝑡^{2+𝑝−𝛽}𝐷^𝑝₀₊𝑦(𝑡)≤𝐴_𝑟. (19) Similarly we get that there exist constants 𝐴^′_𝑟,𝐴_𝑟^′′,𝐵^′_𝑟,𝐵_𝑟^′′>0 such that 𝐺𝑡,𝑦(𝑡),𝐷^𝑝₀₊𝑦(𝑡)≤𝐴^′_𝑟,𝑀𝑡,𝑦(𝑡),𝐷^𝑝₀₊𝑦(𝑡)≤𝐴_𝑟^′′,(),𝐼𝑡𝑡∈0,1₁,𝑦𝑡₁,𝐷^𝑝₀₊𝑦𝑡₁≤𝐵^′_𝑟,𝐼₁𝑡₁,𝑦𝑡₁,𝐷^𝑝₀₊𝑦𝑡₁≤𝐵_𝑟^′′. (20)
It follows from (15) that, for 𝑡∈(𝑡_𝑘,𝑡_𝑘+1](𝑘=0,1), there exist constants 𝑐_𝑘,𝑑_𝑘∈𝑅 such that 𝑥(𝑡)=1Γ(𝛼)^𝑡₀(𝑡−𝑠)^𝛼−1𝜙(𝑠)𝑓𝑠,𝑦(𝑠),𝐷^𝑝₀₊𝑦(𝑠)𝑑𝑠+𝑐_𝑘𝑡^𝛼−1+𝑑_𝑘𝑡^𝛼−2,𝑡∈𝑡_𝑘,𝑡_𝑘+1,𝑘=0,1. (21)
From lim_𝑡→0𝑡^2−𝛼𝑥(𝑡)=∫¹₀𝑢(𝑠)𝐺(𝑠,𝑦(𝑠),𝐷^𝑝₀₊𝑦(𝑠))𝑑𝑠, we get 𝑑₀=¹₀𝑢(𝑠)𝐺𝑠,𝑦(𝑠),𝐷^𝑝₀₊𝑦(𝑠)𝑑𝑠. (22)
From 𝑥(1)=∫¹₀𝑚(𝑠)𝑀(𝑠,𝑦(𝑠),𝐷^𝑝₀₊𝑦(𝑠))𝑑𝑠, we get 1Γ(𝛼)¹₀(1−𝑠)^𝛼−1𝜙(𝑠)𝑓𝑠,𝑦(𝑠),𝐷^𝑝₀₊𝑦(𝑠)𝑑𝑠+𝑐₁+𝑑₁=¹₀𝑚(𝑠)𝑀𝑠,𝑦(𝑠),𝐷^𝑝₀₊𝑦(𝑠)𝑑𝑠. (23)
From Δ𝑥(𝑡₁)=𝐼(𝑡₁,𝑦(𝑡₁),𝐷^𝑝₀₊𝑦(𝑡₁)), we get 𝑐₁−𝑐₀𝑡₁^𝛼−1+𝑑₁−𝑑₀𝑡₁^𝛼−2=𝐼𝑡₁,𝑦𝑡₁,𝐷^𝑝₀₊𝑦𝑡₁. (24)
From Δ𝐷^𝑞₀₊𝑥(𝑡₁)=𝐼₁(𝑡₁,𝑦(𝑡₁),𝐷^𝑝₀₊𝑦(𝑡₁)), we get 𝑐₁−𝑐₀Γ(𝛼)Γ𝑡𝛼−𝑞^{1𝛼−𝑞−1}+𝑑₁−𝑑₀Γ(𝛼−1)Γ𝑡𝛼−𝑞−2^{1𝛼−𝑞−2}=𝐼₁𝑡₁,𝑦𝑡₁,𝐷^𝑝₀₊𝑦𝑡₁. (25)
It follows that 𝑐₁−𝑐₀=Γ(𝛼−1)Γ𝑡𝛼−𝑞−2^{1𝛼−𝑞−2}𝐼𝑡₁,𝑦𝑡₁,𝐷^𝑝₀₊𝑦𝑡₁−𝑡₁^𝛼−2𝐼₁𝑡₁,𝑦𝑡₁,𝐷^𝑝₀₊𝑦𝑡₁×(Π)⁻¹,𝑑₁−𝑑₀=𝑡₁^𝛼−1𝐼₁𝑡₁,𝑦𝑡₁,𝐷^𝑝₀₊𝑦𝑡₁−Γ(𝛼)Γ𝑡𝛼−𝑞^{1𝛼−𝑞−1}𝐼𝑡₁,𝑦𝑡₁,𝐷^𝑝₀₊𝑦𝑡₁×(Π)⁻¹. (26) Then 𝑑₁=𝑡₁^𝛼−1𝐼₁𝑡₁,𝑦𝑡₁,𝐷^𝑝₀₊𝑦𝑡₁−Γ(𝛼)Γ𝑡𝛼−𝑞^{1𝛼−𝑞−1}𝐼𝑡₁,𝑦𝑡₁,𝐷^𝑝₀₊𝑦𝑡₁×(Π)⁻¹+¹₀𝑢(𝑠)𝐺𝑠,𝑦(𝑠),𝐷^𝑝₀₊𝑦(𝑠)𝑑𝑠. (27) So 𝑐₁=¹₀𝑚(𝑠)𝑀𝑠,𝑦(𝑠),𝐷^𝑝₀₊𝑦(𝑠)−1𝑑𝑠Γ(𝛼)¹₀(1−𝑠)^𝛼−1𝜙(𝑠)𝑓𝑠,𝑦(𝑠),𝐷^𝑝₀₊𝑦(𝑠−𝑡)𝑑𝑠_1𝛼−1𝐼₁𝑡₁,𝑦𝑡₁,𝐷^𝑝₀₊𝑦𝑡₁−Γ(𝛼)Γ𝑡𝛼−𝑞^{1𝛼−𝑞−1}𝐼𝑡₁,𝑦𝑡₁,𝐷^𝑝₀₊𝑦𝑡₁×(Π)⁻¹−¹₀𝑢(𝑠)𝐺𝑠,𝑦(𝑠),𝐷^𝑝₀₊𝑦(𝑠)𝑐𝑑𝑠,₀=¹₀𝑚(𝑠)𝑀𝑠,𝑦(𝑠),𝐷^𝑝₀₊𝑦(𝑠)−1𝑑𝑠Γ(𝛼)¹₀(1−𝑠)^𝛼−1𝜙(𝑠)𝑓𝑠,𝑦(𝑠),𝐷^𝑝₀₊𝑦(𝑠)−𝑡𝑑𝑠_1𝛼−1𝐼₁𝑡₁,𝑦𝑡₁,𝐷^𝑝₀₊𝑦𝑡₁−Γ(𝛼)Γ𝑡𝛼−𝑞^{1𝛼−𝑞−1}𝐼𝑡₁,𝑦𝑡₁,𝐷^𝑝₀₊𝑦𝑡₁×(Π)⁻¹−¹₀𝑢(𝑠)𝐺𝑠,𝑦(𝑠),𝐷^𝑝₀₊𝑦(𝑠)−Γ()𝑑𝑠𝛼−1Γ𝑡𝛼−𝑞−2^{1𝛼−𝑞−2}𝐼𝑡₁,𝑦𝑡₁,𝐷^𝑝₀₊𝑦𝑡₁−𝑡₁^𝛼−2𝐼₁𝑡₁,𝑦𝑡₁,𝐷^𝑝₀₊𝑦𝑡₁×(Π)⁻¹. (28)
Hence, for 𝑡∈(0,𝑡₁], we have 𝑥(𝑡)=^𝑡₀(𝑡−𝑠)^𝛼−1Γ(𝛼)𝜙(𝑢)𝑓𝑢,𝑦(𝑢),𝐷^𝑝₀₊𝑦(𝑢)−𝑡𝑑𝑢_𝛼−1Γ(𝛼)¹₀(1−𝑠)^𝛼−1𝜙(𝑠)𝑓𝑠,𝑦(𝑠),𝐷^𝑝₀₊𝑦(𝑠)𝑑𝑠+𝑡^𝛼−21₀𝑢(𝑠)𝐺𝑠,𝑦(𝑠),𝐷^𝑝₀₊𝑦(𝑠)𝑑𝑠+𝑡^𝛼−11₀𝑚(𝑠)𝑀𝑠,𝑦(𝑠),𝐷^𝑝₀₊𝑦(𝑠)+𝑡𝑑𝑠_𝛼−1ΠΓ(𝛼)Γ𝑡𝛼−𝑞^{1𝛼−𝑞−1}−Γ(𝛼−1)Γ𝑡𝛼−𝑞−2^{1𝛼−𝑞−2}×𝐼𝑡₁,𝑦𝑡₁,𝐷^𝑝₀₊𝑦𝑡₁+𝑡^𝛼−1𝑡₁^𝛼−2−𝑡₁^𝛼−1Π𝐼₁𝑡₁,𝑦𝑡₁,𝐷^𝑝₀₊𝑦𝑡₁. (29) And for 𝑡∈(𝑡₁,1], we have 𝑥(𝑡)=^𝑡₀(𝑡−𝑠)^𝛼−1Γ(𝛼)𝜙(𝑠)𝑓𝑠,𝑦(𝑠),𝐷^𝑝₀₊𝑦(𝑠)−𝑡𝑑𝑠_𝛼−1Γ(𝛼)¹₀(1−𝑠)^𝛼−1𝜙(𝑠)𝑓𝑠,𝑦(𝑠),𝐷^𝑝₀₊𝑦(𝑠)+𝑡𝑑𝑠_𝛼−2−𝑡^𝛼−11₀𝑢(𝑠)𝐺𝑠,𝑦(𝑠),𝐷^𝑝₀₊𝑦(𝑠)𝑑𝑠+𝑡^𝛼−11₀𝑚(𝑠)𝑀𝑠,𝑦(𝑠),𝐷^𝑝₀₊𝑦(𝑠)+𝑡𝑑𝑠_𝛼−1−𝑡^𝛼−2ΠΓ(𝛼)Γ𝑡𝛼−𝑞^{1𝛼−𝑞−1}×𝐼𝑡₁,𝑦𝑡₁,𝐷^𝑝₀₊𝑦𝑡₁+𝑡^𝛼−2−𝑡^𝛼−1Π𝑡₁^𝛼−1𝐼₁𝑡₁,𝑦𝑡₁,𝐷^𝑝₀₊𝑦𝑡₁. (30) Hence, 𝑥∈𝑋 satisfies (16).
On the other hand, if 𝑦∈𝑌 and 𝑥∈𝑋 is a solution of (16), then we can prove that 𝑥∈𝑋 is a solution of BVP (6)-(7). The proof is completed.

Lemma 8. Suppose that 𝑥∈𝑋, and (a)–(e), (A)-(B) hold. Then 𝑦∈𝑌 is a solution of 𝐷^𝛽₀₊𝑦(𝑡)=𝜓(𝑡)𝑔𝑡,𝑥(𝑡),𝐷^𝑞₀₊𝑥(𝑡),𝑡∈(0,1),𝑡≠𝑡₁,lim_𝑡→0𝑡^2−𝛽𝑦(𝑡)=¹₀𝑣(𝑠)𝐻𝑠,𝑥(𝑠),𝐷^𝑞₀₊𝑥(𝑠)𝑦(1)=𝑑𝑠,₁₀𝑛(𝑠)𝑁𝑠,𝑥(𝑠),𝐷^𝑞₀₊𝑥(𝑠𝑡)𝑑𝑠,Δ𝑦₁=𝐽𝑡₁,𝑥𝑡₁,𝐷^𝑞₀₊𝑥𝑡₁,Δ𝐷^𝑝₀₊𝑦𝑡₁=𝐽₁𝑡₁,𝑥𝑡₁,𝐷^𝑞₀₊𝑥𝑡₁, (31) if and only if 𝑦∈𝑌 satisfies the integral equation 𝑦(𝑡)=∫^𝑡₀(𝑡−𝑠)^𝛽−1Γ𝛽𝜓(𝑢)𝑔𝑢,𝑥(𝑢),𝐷^𝑞₀₊𝑥(𝑢−𝑡)𝑑𝑢_𝛽−1Γ𝛽×∫¹₀(1−𝑠)^𝛽−1𝜓(𝑠)𝑔𝑠,𝑥(𝑠),𝐷^𝑞₀₊𝑥(𝑠)𝑑𝑠+𝑡^𝛽−2×∫¹₀𝑣(𝑠)𝐻𝑠,𝑦(𝑠),𝐷^𝑝₀₊𝑦(𝑠)𝑑𝑠+𝑡^𝛽−1×∫¹₀𝑛(𝑠)𝑁𝑠,𝑥(𝑠),𝐷^𝑞₀₊𝑥(𝑠)+𝑡𝑑𝑠_𝛽−1Ξ×Γ𝛽Γ𝑡𝛽−𝑝^{1𝛽−𝑝−1}−Γ𝛽−1Γ𝑡𝛽−𝑝−2^{1𝛽−𝑝−2}×𝐽𝑡₁,𝑥𝑡₁,𝐷^𝑞₀₊𝑥𝑡₁+𝑡^𝛽−1𝑡₁^𝛽−2−𝑡₁^𝛽−1Ξ×𝐽₁𝑡₁,𝑥𝑡₁,𝐷^𝑞₀₊𝑥𝑡₁,𝑡∈0,𝑡₁,∫^𝑡₀(𝑡−𝑠)^𝛽−1Γ𝛽𝜓(𝑠)𝑔𝑠,𝑥(𝑠),𝐷^𝑞₀₊𝑥(𝑠)−𝑡𝑑𝑠_𝛽−1Γ𝛽×∫¹₀(1−𝑠)^𝛽−1𝜓(𝑠)𝑔𝑠,𝑥(𝑠),𝐷^𝑞₀₊𝑥(𝑠)+𝑡𝑑𝑠_𝛽−2−𝑡^𝛽−1×∫¹₀𝑣(𝑠)𝐻𝑠,𝑦(𝑠),𝐷^𝑝₀₊𝑦(𝑠)𝑑𝑠+𝑡^𝛽−1×∫¹₀𝑛(𝑠)𝑁𝑠,𝑥(𝑠),𝐷^𝑞₀₊𝑥(𝑠)+𝑡𝑑𝑠_𝛽−1−𝑡^𝛽−2ΞΓ𝛽Γ𝑡𝛽−𝑝^{1𝛽−𝑝−1}×𝐽𝑡₁,𝑥𝑡₁,𝐷^𝑞₀₊𝑥𝑡₁+𝑡^𝛽−2−𝑡^𝛽−1Ξ𝑡₁^𝛽−1×𝐽₁𝑡₁,𝑥𝑡₁,𝐷^𝑞₀₊𝑥𝑡₁,𝑡∈𝑡₁,1, (32) where ΓΞ=𝛽−1Γ−Γ𝛽𝛽−𝑝−1Γ𝑡𝛽−𝑝^{12𝛽−𝑝−3}. (33)

Proof. The proof is similar to that of the proof of Lemma 7 and is omitted.
Now, we define the operator 𝑇 on 𝑋×𝑌 by 𝑇(𝑥,𝑦)(𝑡)=((𝑇₁𝑦)(𝑡),(𝑇₂𝑥)(𝑡)) with 𝑇₁𝑦(𝑡)=∫^𝑡₀(𝑡−𝑠)^𝛼−1Γ(𝛼)𝜙(𝑢)𝑓𝑢,𝑦(𝑢),𝐷^𝑝₀₊𝑦(𝑢)−𝑡𝑑𝑢_𝛼−1Γ(𝛼)∫¹₀(1−𝑠)^𝛼−1𝜙(𝑠)𝑓𝑠,𝑦(𝑠),𝐷^𝑝₀₊𝑦(𝑠)𝑑𝑠+𝑡^𝛼−2∫¹₀𝑢(𝑠)𝐺𝑠,𝑦(𝑠),𝐷^𝑝₀₊𝑦(𝑠)𝑑𝑠+𝑡^𝛼−1∫¹₀𝑚(𝑠)𝑀𝑠,𝑦(𝑠),𝐷^𝑝₀₊𝑦(𝑠)+𝑡𝑑𝑠_𝛼−1ΠΓ(𝛼)Γ𝑡𝛼−𝑞^{1𝛼−𝑞−1}−Γ(𝛼−1)Γ𝑡𝛼−𝑞−2^{1𝛼−𝑞−2}×𝐼𝑡₁,𝑦𝑡₁,𝐷^𝑝₀₊𝑦𝑡₁+𝑡^𝛼−1𝑡₁^𝛼−2−𝑡₁^𝛼−1Π×𝐼₁𝑡₁,𝑦𝑡₁,𝐷^𝑝₀₊𝑦𝑡₁,𝑡∈0,𝑡₁,∫^𝑡₀(𝑡−𝑠)^𝛼−1Γ(𝛼)𝜙(𝑠)𝑓𝑠,𝑦(𝑠),𝐷^𝑝₀₊𝑦(𝑠)−𝑡𝑑𝑠_𝛼−1Γ(𝛼)∫¹₀(1−𝑠)^𝛼−1𝜙(𝑠)𝑓𝑠,𝑦(𝑠),𝐷^𝑝₀₊𝑦(𝑠)+𝑡𝑑𝑠_𝛼−2−𝑡^𝛼−1∫¹₀𝑢(𝑠)𝐺𝑠,𝑦(𝑠),𝐷^𝑝₀₊𝑦(𝑠)𝑑𝑠+𝑡^𝛼−1∫¹₀𝑚(𝑠)𝑀𝑠,𝑦(𝑠),𝐷^𝑝₀₊𝑦(𝑠+𝑡)𝑑𝑠_𝛼−1−𝑡^𝛼−2ΠΓ(𝛼)Γ𝑡𝛼−𝑞^{1𝛼−𝑞−1}𝐼𝑡₁,𝑦𝑡₁,𝐷^𝑝₀₊𝑦𝑡₁+𝑡^𝛼−2−𝑡^𝛼−1Π𝑡₁^𝛼−1𝐼₁𝑡₁,𝑦𝑡₁,𝐷^𝑝₀₊𝑦𝑡₁,𝑡∈𝑡₁,𝑇,1₂𝑥(𝑡)=∫^𝑡₀(𝑡−𝑠)^𝛽−1Γ𝛽𝜓(𝑢)𝑔𝑢,𝑥(𝑢),𝐷^𝑞₀₊𝑥(𝑢)−𝑡𝑑𝑢_𝛽−1Γ𝛽∫¹₀(1−𝑠)^𝛽−1𝜓(𝑠)𝑔𝑠,𝑥(𝑠),𝐷^𝑞₀₊𝑥(𝑠)𝑑𝑠+𝑡^𝛽−2∫¹₀𝑣(𝑠)𝐻𝑠,𝑦(𝑠),𝐷^𝑝₀₊𝑦(𝑠)𝑑𝑠+𝑡^𝛽−1∫¹₀𝑛(𝑠)𝑁𝑠,𝑥(𝑠),𝐷^𝑞₀₊𝑥(𝑠)+𝑡𝑑𝑠_𝛽−1ΞΓ𝛽Γ𝑡𝛽−𝑝^{1𝛽−𝑝−1}−Γ𝛽−1Γ𝑡𝛽−𝑝−2^{1𝛽−𝑝−2}×𝐽𝑡₁,𝑥𝑡₁,𝐷^𝑞₀₊𝑥𝑡₁+𝑡^𝛽−1𝑡₁^𝛽−2−𝑡₁^𝛽−1Ξ𝐽₁𝑡₁,𝑥𝑡₁,𝐷^𝑞₀₊𝑥𝑡₁,𝑡∈0,𝑡₁,∫^𝑡₀(𝑡−𝑠)^𝛽−1Γ𝛽𝜓(𝑠)𝑔𝑠,𝑥(𝑠),𝐷^𝑞₀₊𝑥(𝑠)−𝑡𝑑𝑠_𝛽−1Γ𝛽∫¹₀(1−𝑠)^𝛽−1𝜓(𝑠)𝑔𝑠,𝑥(𝑠),𝐷^𝑞₀₊𝑥(𝑠)+𝑡𝑑𝑠_𝛽−2−𝑡^𝛽−1∫¹₀𝑣(𝑠)𝐻𝑠,𝑦(𝑠),𝐷^𝑝₀₊𝑦(𝑠)𝑑𝑠+𝑡^𝛽−1∫¹₀𝑛(𝑠)𝑁𝑠,𝑥(𝑠),𝐷^𝑞₀₊𝑥(𝑠)+𝑡𝑑𝑠_𝛽−1−𝑡^𝛽−2ΞΓ𝛽Γ𝑡𝛽−𝑝^{1𝛽−𝑝−1}×𝐽𝑡₁,𝑥𝑡₁,𝐷^𝑞₀₊𝑥𝑡₁+𝑡^𝛽−2−𝑡^𝛽−1Ξ𝑡₁^𝛽−1𝐽₁𝑡₁,𝑥𝑡₁,𝐷^𝑞₀₊𝑥𝑡₁,𝑡∈𝑡₁,1. (34)(35)

Remark 9. By Lemmas 7 and 8, (𝑥,𝑦)∈𝑋×𝑌 is a solution of BVP (6)-(7) if and only if (𝑥,𝑦)∈𝑋×𝑌 is a fixed point of the operator 𝑇.

Lemma 10. Suppose that (a)–(e) and (A)-(B) hold. Then 𝑇∶𝑋×𝑌→𝑋×𝑌 is well defined and is completely continuous.

Proof. The proof is very long, so we list the steps. First, we prove that 𝑇 is well defined; second, we prove that 𝑇 is continuous, and, finally, we prove that 𝑇 is compact. So 𝑇 is completely continuous. Thus, the proof is divided into three steps.
Step  1. Prove that 𝑇∶𝑋×𝑌→𝑋×𝑌 is well defined.
For (𝑥,𝑦)∈𝑋×𝑌, we have ||(𝑥,𝑦)||=𝑟>0. Then maxsup_𝑡∈(0,1)𝑡^2−𝛼|𝑥(𝑡)|,sup_()𝑡∈0,1𝑡^{2+𝑞−𝛼}𝐷^𝑞₀₊𝑥(𝑡)≤𝑟<+∞,maxsup_𝑡∈(0,1)𝑡^2−𝛽𝑦(𝑡),sup_()𝑡∈0,1𝑡^{2+𝑝−𝛽}𝐷^𝑝₀₊𝑦(𝑡)≤𝑟<+∞. (36) From (B), 𝑓,𝐺,𝑀,𝐼,𝐼₁ are 𝛽-Caratheodory functions, then there exist constants 𝐴_𝑟>0 such that 𝑓𝑡,𝑦(𝑡),𝐷^𝑝₀₊𝑦(𝑡)≤𝐴_𝑟(),𝐺(𝑡),𝑡∈0,1𝑡,𝑦,𝐷^𝑝₀₊𝑦(𝑡)≤𝐴_𝑟(),𝑀(𝑡),𝑡∈0,1𝑡,𝑦,𝐷^𝑝₀₊𝑦(𝑡)≤𝐴_𝑟(),𝐼𝑡,𝑡∈0,1₁,𝑦𝑡₁,𝐷^𝑝₀₊𝑦𝑡₁≤𝐴_𝑟(),𝐼,𝑡∈0,1₁𝑡₁,𝑦𝑡₁,𝐷^𝑝₀₊𝑦𝑡₁≤𝐴_𝑟,𝑡∈(0,1). (37) Hence, ^𝑡₀(𝑡−𝑠)^𝛼−1Γ(𝛼)𝜙(𝑢)𝑓𝑢,𝑦(𝑢),𝐷^𝑝₀₊𝑦(𝑢)≤𝑑𝑢_𝑡0(𝑡−𝑠)^𝛼−1Γ(𝛼)𝜙(𝑢)𝑓𝑢,𝑦(𝑢),𝐷^𝑝₀₊𝑦(𝑢)𝑑𝑢≤𝐴_𝑟𝐿₁𝐁(𝛼+𝛿,𝑘+1)Γ(𝛼)<∞. (38) From (34), (37), and (38), we see that (𝑇₁𝑦)(𝑡) is defined on (0,1], continuous on (0,𝑡₁] and (𝑡₁,1], respectively. One sees that lim_𝑡→0𝑡^2−𝛼𝑇₁𝑦(𝑡)=lim_𝑡→0𝑡^2−𝛼𝑡₀(𝑡−𝑠)^𝛼−1Γ(𝛼)𝜙(𝑢)𝑓𝑢,𝑦(𝑢),𝐷^𝑝₀₊𝑦(𝑢)−𝑡𝑑𝑢Γ(𝛼)¹₀(1−𝑠)^𝛼−1𝜙(𝑠)𝑓𝑠,𝑦(𝑠),𝐷^𝑝₀₊𝑦(𝑠+)𝑑𝑠₁₀𝑢(𝑠)𝐺𝑠,𝑦(𝑠),𝐷^𝑝₀₊𝑦(𝑠)𝑑𝑠+𝑡¹₀𝑚(𝑠)𝑀𝑠,𝑦(𝑠),𝐷^𝑝₀₊𝑦(𝑠)+𝑡𝑑𝑠ΠΓ(𝛼)Γ𝑡𝛼−𝑞^{1𝛼−𝑞−1}−Γ(𝛼−1)Γ𝑡𝛼−𝑞−2^{1𝛼−𝑞−2}×𝐼𝑡₁,𝑦𝑡₁,𝐷^𝑝₀₊𝑦𝑡₁+𝑡𝑡₁^𝛼−2−𝑡₁^𝛼−1Π𝐼₁𝑡₁,𝑦𝑡₁,𝐷^𝑝₀₊𝑦𝑡₁=¹₀𝑢(𝑠)𝐺𝑠,𝑦(𝑠),𝐷^𝑝₀₊𝑦(𝑠)𝑑𝑠, (39) and there exits the limit lim_𝑡→𝑡⁺₁(𝑇₁𝑦)(𝑡).
On the other hand, we have 𝐷^𝑞₀₊𝑇₁𝑦(𝑡)=∫^𝑡₀(𝑡−𝑠)^{𝛼−𝑞−1}Γ𝜙(𝑢)𝑓(𝑢)𝛼−𝑞𝑢,𝑦,𝐷^𝑝₀₊𝑦(𝑢)−𝑡𝑑𝑢_{𝛼−𝑞−1}Γ𝛼−𝑞×∫¹₀(1−𝑠)^𝛼−1𝜙(𝑠)𝑓𝑠,𝑦(𝑠),𝐷^𝑝₀₊𝑦(𝑠)𝑑𝑠+𝑡^{𝛼−𝑞−2}Γ(𝛼−1)Γ𝛼−𝑞−1×∫¹₀𝑢(𝑠)𝐺𝑠,𝑦(𝑠),𝐷^𝑝₀₊𝑦(𝑠)𝑑𝑠+𝑡^{𝛼−𝑞−1}Γ(𝛼)Γ𝛼−𝑞×∫¹₀𝑚(𝑠)𝑀𝑠,𝑦(𝑠),𝐷^𝑝₀₊𝑦(𝑠+𝑡)𝑑𝑠_{𝛼−𝑞−1}ΠΓ(𝛼)Γ×Γ(𝛼)𝛼−𝑞Γ𝑡𝛼−𝑞^{1𝛼−𝑞−1}−Γ(𝛼−1)Γ𝑡𝛼−𝑞−2^{1𝛼−𝑞−2}×𝐼𝑡₁,𝑦𝑡₁,𝐷^𝑝₀₊𝑦𝑡₁+Γ(𝛼)Γ𝑡𝛼−𝑞^{𝛼−𝑞−1}𝑡₁^𝛼−2−𝑡₁^𝛼−1Π×𝐼₁𝑡₁,𝑦𝑡₁,𝐷^𝑝₀₊𝑦𝑡₁,𝑡∈0,𝑡₁,∫^𝑡₀(𝑡−𝑠)^{𝛼−𝑞−1}Γ𝜙(𝑠)𝑓(𝑠)𝛼−𝑞𝑠,𝑦,𝐷^𝑝₀₊𝑦(𝑠)−𝑡𝑑𝑠_{𝛼−𝑞−1}Γ𝛼−𝑞×∫¹₀(1−𝑠)^𝛼−1𝜙(𝑠)𝑓𝑠,𝑦(𝑠),𝐷^𝑝₀₊𝑦(𝑠)+𝑡𝑑𝑠_{𝛼−𝑞−2}Γ(𝛼−1)Γ𝛼−𝑞−1−𝑡^{𝛼−𝑞−1}Γ(𝛼)Γ𝛼−𝑞×∫¹₀𝑢(𝑠)𝐺𝑠,𝑦(𝑠),𝐷^𝑝₀₊𝑦(𝑠)𝑑𝑠+𝑡^{𝛼−𝑞−1}Γ(𝛼)Γ∫𝛼−𝑞¹⁰𝑚(𝑠)𝑀𝑠,𝑦(𝑠),𝐷^𝑝₀₊𝑦(𝑠)+1𝑑𝑠Π𝑡^{𝛼−𝑞−1}Γ(𝛼)Γ𝛼−𝑞−𝑡^{𝛼−𝑞−2}Γ(𝛼−1)Γ×Γ(𝛼)𝛼−𝑞−1Γ𝑡𝛼−𝑞^{1𝛼−𝑞−1}𝐼𝑡₁,𝑦𝑡₁,𝐷^𝑝₀₊𝑦𝑡₁+1Π𝑡^{𝛼−𝑞−2}Γ(𝛼−1)Γ𝛼−𝑞−1−𝑡^{𝛼−𝑞−1}Γ(𝛼)Γ𝛼−𝑞×𝑡₁^𝛼−1𝐼₁𝑡₁,𝑦𝑡₁,𝐷^𝑝₀₊𝑦𝑡₁,𝑡∈𝑡₁,𝐷,1_𝑝0+𝑇₂𝑥(𝑡)=∫^𝑡₀(𝑡−𝑠)^{𝛽−𝑝−1}Γ𝜓(𝑢)𝑔(𝑢)𝛽−𝑝𝑢,𝑥,𝐷^𝑞₀₊𝑥(𝑢−𝑡)𝑑𝑢_{𝛽−𝑝−1}Γ𝛽−𝑝×∫¹₀(1−𝑠)^𝛽−1𝜓(𝑠)𝑔𝑠,𝑥(𝑠),𝐷^𝑞₀₊𝑥(𝑠)𝑑𝑠+𝑡^{𝛽−𝑝−2}Γ𝛽−1Γ𝛽−𝑝−1×∫¹₀𝑣(𝑠)𝐻𝑠,𝑦(𝑠),𝐷^𝑝₀₊𝑦(𝑠)𝑑𝑠+𝑡^{𝛽−𝑝−1}Γ𝛽Γ𝛽−𝑝×∫¹₀𝑛(𝑠)𝑁𝑠,𝑥(𝑠),𝐷^𝑞₀₊𝑥(𝑠)+𝑡𝑑𝑠_{𝛽−𝑝−1}ΞΓ𝛽Γ×Γ𝛽𝛽−𝑝Γ𝑡𝛽−𝑝^{1𝛽−𝑝−1}−Γ𝛽−1Γ𝑡𝛽−𝑝−2^{1𝛽−𝑝−2}×𝐽𝑡₁,𝑥𝑡₁,𝐷^𝑞₀₊𝑥𝑡₁+𝑡^{𝛽−𝑝−1}𝑡₁^𝛽−2−𝑡₁^𝛽−1ΞΓ𝛽Γ𝛽−𝑝×𝐽₁𝑡₁,𝑥𝑡₁,𝐷^𝑞₀₊𝑥𝑡₁,𝑡∈0,𝑡₁,∫^𝑡₀(𝑡−𝑠)^{𝛽−𝑝−1}Γ𝜓(𝑠)𝑔(𝑠)𝛽−𝑝𝑠,𝑥,𝐷^𝑞₀₊𝑥(𝑠)−𝑡𝑑𝑠_{𝛽−𝑝−1}Γ𝛽Γ𝛽Γ×∫𝛽−𝑝₁₀(1−𝑠)^𝛽−1𝜓(𝑠)𝑔𝑠,𝑥(𝑠),𝐷^𝑞₀₊𝑥(𝑠)+𝑡𝑑𝑠_{𝛽−𝑝−2}Γ𝛽−1Γ𝛽−𝑝−1−𝑡^{𝛽−𝑝−1}Γ𝛽Γ𝛽−𝑝×∫¹₀𝑣(𝑠)𝐻𝑠,𝑦(𝑠),𝐷^𝑝₀₊𝑦(𝑠)𝑑𝑠+𝑡^{𝛽−𝑝−1}Γ𝛽Γ∫𝛽−𝑝¹⁰𝑛(𝑠)𝑁𝑠,𝑥(𝑠),𝐷^𝑞₀₊𝑥(𝑠)+1𝑑𝑠Ξ𝑡^{𝛽−𝑝−1}Γ𝛽Γ𝛽−𝑝−𝑡^{𝛽−𝑝−2}Γ𝛽−1Γ×Γ𝛽𝛽−𝑝−1Γ𝑡𝛽−𝑝^{1𝛽−𝑝−1}𝐽𝑡₁,𝑥𝑡₁,𝐷^𝑞₀₊𝑥𝑡₁+1Ξ𝑡^{𝛽−𝑝−2}Γ𝛽−1Γ𝛽−𝑝−1−𝑡^{𝛽−𝑝−1}Γ𝛽Γ𝛽−𝑝×𝑡₁^𝛽−1𝐽₁𝑡₁,𝑥𝑡₁,𝐷^𝑞₀₊𝑥𝑡₁,𝑡∈𝑡₁,1. (40)
It is easy to see that ^𝑡₀(𝑡−𝑠)^{𝛼−𝑞−1}Γ𝜙(𝑢)𝑓(𝑢)𝛼−𝑞𝑢,𝑦,𝐷^𝑝₀₊𝑦(𝑢)𝑑𝑢≤𝐴_𝑟𝐿₁𝐁𝛼+𝛿−𝑞,𝑘+1Γ𝛼−𝑞<∞. (41) From (37) and (41), we see that 𝐷^𝑞₀₊(𝑇₁𝑦)(𝑡) is defined on (0,1], continuous on (0,𝑡₁] and (𝑡₁,1], respectively. One sees that lim_𝑡→0𝑡^{2+𝑞−𝛼}𝐷^𝑞₀₊𝑇₁𝑦(𝑡)=lim_𝑡→0𝑡^{2+𝑞−𝛼}×^𝑡₀(𝑡−𝑠)^{𝛼−𝑞−1}Γ𝜙(𝑢)𝑓(𝑢)𝛼−𝑞𝑢,𝑦,𝐷^𝑝₀₊𝑦(𝑢)−𝑡𝑑𝑢Γ×𝛼−𝑞₁₀(1−𝑠)^𝛼−1𝜙(𝑠)𝑓𝑠,𝑦(𝑠),𝐷^𝑝₀₊𝑦(𝑠)+Γ()𝑑𝑠𝛼−1Γ𝛼−𝑞−1¹₀𝑢(𝑠)𝐺𝑠,𝑦(𝑠),𝐷^𝑝₀₊𝑦(𝑠)Γ(𝛼)𝑑𝑠+𝑡Γ𝛼−𝑞¹₀𝑚(𝑠)𝑀𝑠,𝑦(𝑠),𝐷^𝑝₀₊𝑦(𝑠)+𝑡𝑑𝑠ΠΓ(𝛼)Γ×Γ(𝛼)𝛼−𝑞Γ𝑡𝛼−𝑞^{1𝛼−𝑞−1}−Γ(𝛼−1)Γ𝑡𝛼−𝑞−2^{1𝛼−𝑞−2}×𝐼𝑡₁,𝑦𝑡₁,𝐷^𝑝₀₊𝑦𝑡₁+Γ(𝛼)Γ𝑡𝑡𝛼−𝑞^1𝛼−2−𝑡₁^𝛼−1Π×𝐼₁𝑡₁,𝑦𝑡₁,𝐷^𝑝₀₊𝑦𝑡₁=Γ(𝛼−1)Γ𝛼−𝑞−1¹₀𝑢(𝑠)𝐺𝑠,𝑦(𝑠),𝐷^𝑝₀₊𝑦(𝑠)𝑑𝑠, (42) and there exits the limit lim_𝑡→𝑡⁺₁𝐷^𝑞₀₊(𝑇₁𝑦)(𝑡).
From the above discussion, we have (𝑇₁𝑦)∈𝑋. Similarly, we can show that (𝑇₂𝑥)∈𝑌. Hence, ((𝑇₁𝑦),(𝑇₂𝑥))∈𝑋×𝑌. Then 𝑇∶𝑋×𝑌→𝑋×𝑌 is well defined.
Step  2. We prove that 𝑇 is continuous. Let (𝑥_𝑛,𝑦_𝑛)∈𝑋×𝑌 with (𝑥_𝑛,𝑦_𝑛)→(𝑥₀,𝑦₀) as 𝑛→∞. We will show that 𝑇(𝑥_𝑛,𝑦_𝑛)→𝑇(𝑥₀,𝑦₀) as 𝑛→∞, that is, prove that 𝑇₁𝑦_𝑛→𝑇₁𝑦₀ and 𝑇₂𝑥_𝑛→𝑇₂𝑥₀ as 𝑛→∞.
In fact, we have 𝑟>0 such that ||(𝑥_𝑛,𝑦_𝑛)||=𝑟>0. Then maxsup_𝑡∈(0,1)𝑡^2−𝛼𝑥_𝑛(𝑡),sup_()𝑡∈0,1𝑡^{2+𝑞−𝛼}𝐷^𝑞₀₊𝑥_𝑛(𝑡)≤𝑟<+∞,𝑛=0,1,2,…,maxsup_𝑡∈(0,1)𝑡^2−𝛽𝑦_𝑛(𝑡),sup_()𝑡∈0,1𝑡^{2+𝑝−𝛽}𝐷^𝑝₀₊𝑦_𝑛(𝑡)≤𝑟<+∞,𝑛=0,1,2,…. (43)
From (B), 𝑓,𝐺,𝑀,𝐼,𝐼₁ are 𝛽-Caratheodory functions, then there exist constants 𝐴_𝑟>0 such that 𝑓𝑡,𝑦_𝑛(𝑡),𝐷^𝑝₀₊𝑦_𝑛(𝑡)≤𝐴_𝑟,()𝐺𝑡∈0,1,𝑛=0,1,2,…,𝑡,𝑦_𝑛(𝑡),𝐷^𝑝₀₊𝑦_𝑛(𝑡)≤𝐴_𝑟,()𝑀𝑡∈0,1,𝑛=0,1,2,…,𝑡,𝑦_𝑛(𝑡),𝐷^𝑝₀₊𝑦_𝑛(𝑡)≤𝐴_𝑟,()𝐼𝑡𝑡∈0,1,𝑛=0,1,2,…,₁,𝑦_𝑛𝑡₁,𝐷^𝑝₀₊𝑦_𝑛𝑡₁≤𝐴_𝑟,()𝐼𝑡∈0,1,𝑛=0,1,2,…,₁𝑡₁,𝑦_𝑛𝑡₁,𝐷^𝑝₀₊𝑦_𝑛𝑡₁≤𝐴_𝑟,𝑡∈(0,1),𝑛=0,1,2,…,sup_𝑡∈(0,1)𝑡^2−𝛼𝑥_𝑛(𝑡)−𝑥₀(𝑡)⟶0,sup_()𝑡∈0,1𝑡^2−𝛽𝑦_𝑛(𝑡)−𝑦₀(𝑡),sup_𝑡∈(0,1)𝑡^{2+𝑞−𝛼}𝐷^𝑞₀₊𝑥_𝑛(𝑡)−𝐷^𝑞₀₊𝑥₀(𝑡)⟶0,sup_𝑡∈(0,1)𝑡^{2+𝑝−𝛽}𝐷^𝑝₀₊𝑦_𝑛(𝑡)−𝐷^𝑝₀₊𝑦₀(𝑡)⟶0, (44) as 𝑛→∞. We have 𝐷^𝑞₀₊𝑇₁𝑦_𝑛(𝑡)=∫^𝑡₀(𝑡−𝑠)^{𝛼−𝑞−1}Γ𝜙(𝑢)𝑓𝛼−𝑞𝑢,𝑦_𝑛(𝑢),𝐷^𝑝₀₊𝑦_𝑛(𝑢)−𝑡𝑑𝑢_{𝛼−𝑞−1}Γ𝛼−𝑞×∫¹₀(1−𝑠)^𝛼−1𝜙(𝑠)𝑓𝑠,𝑦_𝑛(𝑠),𝐷^𝑝₀₊𝑦_𝑛(𝑠)𝑑𝑠+𝑡^{𝛼−𝑞−2}Γ(𝛼−1)Γ𝛼−𝑞−1×∫¹₀𝑢(𝑠)𝐺𝑠,𝑦_𝑛(𝑠),𝐷^𝑝₀₊𝑦_𝑛(𝑠)𝑑𝑠+𝑡^{𝛼−𝑞−1}Γ(𝛼)Γ𝛼−𝑞×∫¹₀𝑚(𝑠)𝑀𝑠,𝑦_𝑛(𝑠),𝐷^𝑝₀₊𝑦_𝑛(𝑠+𝑡)𝑑𝑠_{𝛼−𝑞−1}ΠΓ(𝛼)Γ×Γ(𝛼)𝛼−𝑞Γ𝑡𝛼−𝑞^{1𝛼−𝑞−1}−Γ(𝛼−1)Γ𝑡𝛼−𝑞−2^{1𝛼−𝑞−2}×𝐼𝑡₁,𝑦_𝑛𝑡₁,𝐷^𝑝₀₊𝑦_𝑛𝑡₁+Γ(𝛼)Γ𝑡𝛼−𝑞^{𝛼−𝑞−1}𝑡₁^𝛼−2−𝑡₁^𝛼−1Π×𝐼₁𝑡₁,𝑦_𝑛𝑡₁,𝐷^𝑝₀₊𝑦_𝑛𝑡₁,𝑡∈0,𝑡₁,∫^𝑡₀(𝑡−𝑠)^{𝛼−𝑞−1}Γ𝜙(𝑠)𝑓𝛼−𝑞𝑠,𝑦_𝑛(𝑠),𝐷^𝑝₀₊𝑦_𝑛(𝑠)−𝑡𝑑𝑠_{𝛼−𝑞−1}Γ𝛼−𝑞×∫¹₀(1−𝑠)^𝛼−1𝜙(𝑠)𝑓𝑠,𝑦(𝑠),𝐷^𝑝₀₊𝑦(𝑠)+𝑡𝑑𝑠_{𝛼−𝑞−2}Γ(𝛼−1)Γ𝛼−𝑞−1−𝑡^{𝛼−𝑞−1}Γ(𝛼)Γ𝛼−𝑞×∫¹₀𝑢(𝑠)𝐺𝑠,𝑦_𝑛(𝑠),𝐷^𝑝₀₊𝑦_𝑛(𝑠)𝑑𝑠+𝑡^{𝛼−𝑞−1}Γ(𝛼)Γ×∫𝛼−𝑞₁₀𝑚(𝑠)𝑀𝑠,𝑦_𝑛(𝑠),𝐷^𝑝₀₊𝑦_𝑛(𝑠)+1𝑑𝑠Π𝑡^{𝛼−𝑞−1}Γ(𝛼)Γ𝛼−𝑞−𝑡^{𝛼−𝑞−2}Γ(𝛼−1)Γ×Γ(𝛼)𝛼−𝑞−1Γ𝑡𝛼−𝑞^{1𝛼−𝑞−1}𝐼𝑡₁,𝑦_𝑛𝑡₁,𝐷^𝑝₀₊𝑦_𝑛𝑡₁+1Π𝑡^{𝛼−𝑞−2}Γ(𝛼−1)Γ𝛼−𝑞−1−𝑡^{𝛼−𝑞−1}Γ(𝛼)Γ𝛼−𝑞×𝑡₁^𝛼−1𝐼₁𝑡₁,𝑦_𝑛𝑡₁,𝐷^𝑝₀₊𝑦_𝑛𝑡₁,𝑡∈𝑡₁,1. (45) From the Lebesgue dominated convergence theorem, we get sup_𝑡∈(0,1)𝑡^2−𝛽𝑇₁𝑦_𝑛(𝑡)−𝑇₁𝑦₀(𝑡),sup_𝑡∈(0,1)𝑡^{2+𝑝−𝛽}𝐷^𝑝₀₊𝑇₁𝑦_𝑛(𝑡)−𝐷^𝑝₀₊𝑇₁𝑦₀(𝑡)⟶0, (46) as 𝑛→∞. Similarly, we can show that sup_𝑡∈(0,1)𝑡^2−𝛼𝑇₂𝑥_𝑛(𝑡)−𝑇₂𝑥₀(𝑡)⟶0,sup_𝑡∈(0,1)𝑡^{2+𝑞−𝛼}𝐷^𝑞₀₊𝑇₂𝑥_𝑛(𝑡)−𝐷^𝑞₀₊𝑇₂𝑥₀(𝑡)⟶0, (47) as 𝑛→∞. It follows from (46) and (47) that 𝑇 is continuous.
Step  3. We prove that 𝑇 is compact, that is, for each nonempty open bounded subset Ω of 𝑋×𝑌, prove that 𝑇(Ω) is relatively compact. We must prove that 𝑇(Ω) is uniformly bounded, equicontinuous on each subinterval [𝑎,𝑏]⊆(𝑡_𝑘,𝑡_𝑘+1](𝑘=0,1), 𝑇(Ω) is equiconvergent as 𝑡→0, and equiconvergent as 𝑡→𝑡₁.
Let Ω be a bounded open subset of 𝑌. We have 𝑟>0 such that maxsup_𝑡∈(0,1)𝑡^2−𝛽𝑦(𝑡),sup_()𝑡∈0,1𝑡^{2+𝑝−𝛽}𝐷^𝑝₀₊𝑦(𝑡)≤𝑟<+∞,𝑦∈Ω. (48)
From (B), 𝑓,𝐺,𝑀,𝐼,𝐼₁ are 𝛽-Caratheodory functions, then there exist constants 𝐴_𝑟>0 such that 𝑓𝑡,𝑦(𝑡),𝐷^𝑝₀₊𝑦(𝑡)≤𝐴_𝑟(),𝐺(𝑡),𝑡∈0,1𝑡,𝑦,𝐷^𝑝₀₊𝑦(𝑡)≤𝐴_𝑟(),𝑀(𝑡),𝑡∈0,1𝑡,𝑦,𝐷^𝑝₀₊𝑦(𝑡)≤𝐴_𝑟(),𝐼𝑡,𝑡∈0,1₁,𝑦𝑡₁,𝐷^𝑝₀₊𝑦𝑡₁≤𝐴_𝑟(),𝐼,𝑡∈0,1₁𝑡₁,𝑦𝑡₁,𝐷^𝑝₀₊𝑦𝑡₁≤𝐴_𝑟,𝑡∈(0,1). (49)
Substep  3.1. Prove that 𝑇(Ω) is uniformly bounded.
In fact, for 𝑡∈(0,𝑡₁], use (49), we have 𝑡^2−𝛼𝑇₁𝑦(𝑡)≤𝑡^2−𝛼𝑡₀(𝑡−𝑠)^𝛼−1Γ(𝛼)𝜙(𝑢)𝑓𝑢,𝑦(𝑢),𝐷^𝑝₀₊𝑦(𝑢)+𝑡𝑑𝑢Γ(𝛼)¹₀(1−𝑠)^𝛼−1𝜙(𝑠)𝑓𝑠,𝑦(𝑠),𝐷^𝑝₀₊𝑦(𝑠)+𝑑𝑠₁₀𝑢(𝑠)𝐺𝑠,𝑦(𝑠),𝐷^𝑝₀₊𝑦(𝑠)𝑑𝑠+𝑡¹₀𝑚(𝑠)𝑀𝑠,𝑦(𝑠),𝐷^𝑝₀₊𝑦(𝑠)+𝑡𝑑𝑠ΠΓ(𝛼)Γ𝑡𝛼−𝑞^{1𝛼−𝑞−1}−Γ(𝛼−1)Γ𝑡𝛼−𝑞−2^{1𝛼−𝑞−2}×𝐼𝑡₁,𝑦𝑡₁,𝐷^𝑝₀₊𝑦𝑡₁+𝑡𝑡₁^𝛼−2−𝑡₁^𝛼−1Π𝐼₁𝑡₁,𝑦𝑡₁,𝐷^𝑝₀₊𝑦𝑡₁≤𝐴_𝑟𝐿₁𝐁(𝛼+𝛿,𝑘+1)Γ(𝛼)+𝐴_𝑟𝐿₁Γ(𝛼)𝐁(𝛼+𝛿,𝑘+1)+𝐴_𝑟‖𝑢‖₁+𝐴_𝑟‖𝑚‖₁+𝐴_𝑟ΠΓ(𝛼)Γ𝑡𝛼−𝑞^{1𝛼−𝑞−1}−Γ(𝛼−1)Γ𝑡𝛼−𝑞−2^{1𝛼−𝑞−2}+𝑡₁^𝛼−2−𝑡₁^𝛼−1Π𝐴_𝑟<∞. (50) Similarly, we can get for 𝑡∈(𝑡₁,1] that 𝑡^2−𝛼𝑇₁𝑦(𝑡)≤𝐴_𝑟𝐿₁𝐁(𝛼+𝛿,𝑘+1)Γ(𝛼)+𝐴_𝑟𝐿₁Γ(𝛼)𝐁(𝛼+𝛿,𝑘+1)+𝐴_𝑟‖𝑢‖₁+𝐴_𝑟‖𝑚‖₁+𝐴_𝑟ΠΓ(𝛼)Γ𝑡𝛼−𝑞^{1𝛼−𝑞−1}+𝐴_𝑟Π𝑡₁^𝛼−1<∞. (51) Furthermore, we have for 𝑡∈(0,𝑡₁] that 𝑡^{2+𝑞−𝛼}𝐷^𝑞₀₊𝑇₁𝑦(𝑡)≤𝐴_𝑟𝐿₁𝐁𝛼+𝛿−𝑞,𝑘+1Γ+𝐴𝛼−𝑞_𝑟𝐿₁Γ𝐁()+𝐴𝛼−𝑞𝛼+𝛿,𝑘+1_𝑟Γ(𝛼−1)Γ‖𝑢‖𝛼−𝑞−1¹+𝐴_𝑟Γ(𝛼)Γ‖𝑚‖𝛼−𝑞¹+𝐴_𝑟ΠΓ(𝛼)Γ×Γ(𝛼)𝛼−𝑞Γ𝑡𝛼−𝑞^{1𝛼−𝑞−1}−Γ(𝛼−1)Γ𝑡𝛼−𝑞−2^{1𝛼−𝑞−2}+Γ(𝛼)Γ𝐴𝛼−𝑞^𝑟𝑡₁^𝛼−2−𝑡₁^𝛼−1Π<∞, (52) and for 𝑡∈(𝑡₁,1] that 𝑡^{2+𝑞−𝛼}𝐷^𝑞₀₊𝑇₁𝑦(𝑡)≤𝐴_𝑟𝐿₁𝐁𝛼+𝛿−𝑞,𝑘+1Γ+𝐴𝛼−𝑞^𝑟𝐿₁Γ𝐁()+Γ()𝛼−𝑞𝛼+𝛿,𝑘+1𝛼−1Γ+Γ(𝛼)𝛼−𝑞−1Γ𝐴𝛼−𝑞^𝑟‖𝑢‖₁+Γ(𝛼)Γ𝐴𝛼−𝑞^𝑟‖𝑚‖₁+1ΠΓ(𝛼)Γ+Γ()𝛼−𝑞𝛼−1ΓΓ(𝛼)𝛼−𝑞−1Γ𝑡𝛼−𝑞^{1𝛼−𝑞−1}𝐴_𝑟+1ΠΓ(𝛼−1)Γ+Γ(𝛼)𝛼−𝑞−1Γ𝑡𝛼−𝑞^1𝛼−1𝐴_𝑟<∞. (53) Hence, maxsup_𝑡∈(0,1)𝑡^2−𝛼𝑇𝑦(𝑡),sup_𝑡∈(0,1)𝑡^{2+𝑞−𝛼}𝐷^𝑞₀₊𝑇𝑦(𝑡)<+∞,𝑦∈Ω. (54) Similarly, we can show that maxsup_𝑡∈(0,1)𝑡^2−𝛽|(𝑇𝑥)(𝑡)|,sup_()𝑡∈0,1𝑡^{2+𝑝−𝛽}𝐷^𝑝₀₊(𝑇𝑥)(𝑡)<+∞,𝑦∈Ω. (55) It is easy to see that 𝑇(Ω) is uniformly bounded.
Substep  3.2. Prove that 𝑇(Ω) is equicontinuous on each subinterval [𝑎,𝑏]⊆(𝑡_𝑘,𝑡_𝑘+1](𝑘=0,1).
For each [𝑎,𝑏]⊆(𝑡₀,𝑡₁], and 𝑠₁,𝑠₂∈[𝑎,𝑏] with 𝑠₂<𝑠₁, we have 𝑠₁^2−𝛼𝑇𝑦𝑠₁−𝑠₂^2−𝛼𝑇𝑦𝑠₂≤𝑠₁^{2−𝛼𝑠1}₀𝑠₁−𝑠^𝛼−1Γ(𝛼)𝜙(𝑢)𝑓𝑢,𝑦(𝑢),𝐷^𝑝₀₊𝑦(𝑢)𝑑𝑢−𝑠₂^{2−𝛼𝑠2}₀𝑠₂−𝑠^𝛼−1Γ(𝛼)𝜙(𝑢)𝑓𝑢,𝑦(𝑢),𝐷^𝑝₀₊𝑦(𝑢)+𝑠𝑑𝑢₁−𝑠₂𝐴_𝑟×𝐿₁𝐁(𝛼+𝛿,𝑘+1)Γ(𝛼)+‖𝑚‖₁+1ΠΓ(𝛼)Γ𝑡𝛼−𝑞^{1𝛼−𝑞−1}−Γ(𝛼−1)Γ𝑡𝛼−𝑞−2^{1𝛼−𝑞−2}+𝑡₁^𝛼−2−𝑡₁^𝛼−1Π. (56) Note that |𝜏^𝜚₁−𝜏^𝜚₂|≤|𝜏₁−𝜏₂|^𝜚 for all 𝜏₁,𝜏₂≥0 and 𝜚∈(0,1).
Since 𝑠₁^{2−𝛼𝑠1}₀𝑠₁−𝑠^𝛼−1Γ(𝛼)𝜙(𝑢)𝑓𝑢,𝑦(𝑢),𝐷^𝑝₀₊𝑦(𝑢)𝑑𝑢−𝑠₂^{2−𝛼𝑠2}₀𝑠₂−𝑠^𝛼−1Γ(𝛼)𝜙(𝑢)𝑓𝑢,𝑦(𝑢),𝐷^𝑝₀₊𝑦(𝑢)≤𝑠𝑑𝑢_12−𝛼−𝑠₂^2−𝛼𝐴_𝑟𝐿₁𝐁(𝛼+𝛿,𝑘+1)Γ(𝛼)+𝑏^2−𝛼𝐴_𝑟𝐿₁𝑠₁^{𝛼+𝑘+𝛿1}_𝑠2/𝑠1(1−𝑤)^{𝛼+𝛿−1}𝑤^𝑘𝑑𝑤Γ(𝛼)+𝑠₁−𝑠₂^𝛼−1𝐴_𝑟𝐿₁𝑏^2−𝛼Γ(𝛼)^𝑏₀𝑠^𝑘(1−𝑠)^𝛿𝑑𝑠⟶0uniformlyas𝑠₁⟶𝑠₂. (57) It follows that 𝑠₁^2−𝛼𝑇𝑦𝑠₁−𝑠₂^2−𝛼𝑇𝑦𝑠₂⟶0uniformlyas𝑠₁⟶𝑠₂. (58) For [𝑎,𝑏]⊆(𝑡₁,1], and 𝑠₁,𝑠₂∈[𝑎,𝑏] with 𝑠₂<𝑠₁, we can prove similarly that 𝑠₁^2−𝛼𝑇𝑦𝑠₁−𝑠₂^2−𝛼𝑇𝑦𝑥𝑠₂⟶0uniformlyas𝑠₁⟶𝑠₂. (59) On the other hand, for [𝑎,𝑏]⊆(𝑡₀,𝑡₁], and 𝑠₁,𝑠₂∈[𝑎,𝑏] with 𝑠₂<𝑠₁, we have 𝑠₁^{2+𝑞−𝛼}𝐷^𝑞₀₊𝑇₁𝑦𝑠₁−𝑠₂^{2+𝑞−𝛼}𝐷^𝑞₀₊𝑇₁𝑦𝑠₂≤𝑠₁^{2+𝑞−𝛼𝑠1}₀𝑠₁−𝑠^{𝛼−𝑞−1}Γ𝜙(𝑢)𝑓(𝑢)𝛼−𝑞𝑢,𝑦,𝐷^𝑝₀₊𝑦(𝑢)𝑑𝑢−𝑠₂^{2+𝑞−𝛼𝑠2}₀𝑠₂−𝑠^{𝛼−𝑞−1}Γ𝜙(𝑢)𝑓(𝑢)𝛼−𝑞𝑢,𝑦,𝐷^𝑝₀₊𝑦(𝑢)+𝑠𝑑𝑢₂−𝑠₁𝐴_𝑟𝐿₁𝐵(𝛼+𝛿,𝑘+1)Γ+Γ(𝛼𝑚‖𝛼−𝑞)‖₁Γ+1𝛼−𝑞ΠΓ(𝛼)Γ×Γ(𝛼)𝛼−𝑞Γ𝑡𝛼−𝑞^{1𝛼−𝑞−1}−Γ(𝛼−1)Γ𝑡𝛼−𝑞−2^{1𝛼−𝑞−2}+Γ(𝛼)Γ𝑡𝛼−𝑞^1𝛼−2−𝑡₁^𝛼−1Π. (60) It is easy to see that 𝑠₁^{2+𝑞−𝛼}−𝑠₂^{2+𝑞−𝛼}𝐴_𝑟𝐿₁¹₀(1−𝑤)^{𝛼+𝛿−𝑞−1}𝑤^𝑘𝑑𝑠Γ𝛼−𝑞⟶0uniformlyas𝑠₁⟶𝑠₂,𝑏^{2+𝑞−𝛼}𝐴_𝑟𝐿₁¹_𝑠2/𝑠1(1−𝑤)^{𝛼−𝑞−1}𝑤^𝑘𝑑𝑤Γ𝛼−𝑞⟶0uniformlyas𝑠₁⟶𝑠₂. (61)(62) For the third term, if 𝛼−𝑞−1≥0, use |𝜏^𝜚₁−𝜏^𝜚₂|≤|𝜏₁−𝜏₂|^𝜚, then ^𝑠2₀𝑠₁−𝑠^{𝛼−𝑞−1}−𝑠₂−𝑠^{𝛼−𝑞−1}Γ𝑠𝛼−𝑞^𝑘(1−𝑠)^𝛿≤𝑠𝑑𝑠₁−𝑠₂^{𝛼−𝑞−1𝑏}₀1Γ𝑠𝛼−𝑞^𝑘(1−𝑠)^𝛿𝑑𝑠⟶0uniformlyas𝑠₁⟶𝑠₂. (63) If 𝛼−𝑞−1<0, use |𝜏^𝜚₁−𝜏^𝜚₂|≤|𝜏₁−𝜏₂|^𝜚, then ^𝑠2₀𝑠₁−𝑠^{𝛼−𝑞−1}−𝑠₂−𝑠^{𝛼−𝑞−1}Γ𝑠𝛼−𝑞^𝑘(1−𝑠)^𝛿=𝑑𝑠_𝑠20𝑠₂−𝑠^{𝛼−𝑞−1}−𝑠₁−𝑠^{𝛼−𝑞−1}Γ𝑠𝛼−𝑞^𝑘𝑠₁−𝑠^𝛿≤𝑠𝑑𝑠_{2𝛼+𝛿+𝑘−𝑞}−𝑠₁^{𝛼+𝛿+𝑘−𝑞}𝐁𝛼+𝛿+𝑘−𝑞,𝑘+1𝛾𝛼−𝑞+𝑏^{𝛼+𝑘+𝛿−𝑞}𝐁𝛼+2𝛿−𝑞,𝑘+1Γ𝑠𝛼−𝑞1−¹𝑠₂^−𝛿⟶0uniformlyas𝑠₁⟶𝑠₂. (64) For [𝑎,𝑏]⊆(𝑡₁,1], and 𝑠₁,𝑠₂∈[𝑎,𝑏] with 𝑠₂<𝑠₁, we can prove similarly that 𝑠₁^{2+𝑞−𝛼}𝐷^𝑞₀₊𝑇₁𝑦𝑠₁−𝑠₂^{2+𝑞−𝛼}𝐷^𝑞₀₊𝑇₁𝑦𝑠₂⟶0uniformlyas𝑠₁⟶𝑠₂. (65) Similarly, we can show that for each [𝑎,𝑏]⊆(𝑡₀,𝑡₁], and 𝑠₁,𝑠₂∈[𝑎,𝑏] with 𝑠₂<𝑠₁, we have 𝑠₁^2−𝛽(𝑇𝑥)𝑠₁−𝑠₂^2−𝛽(𝑇𝑥)𝑠₂⟶0uniformlyas𝑠₁⟶𝑠₂. (66) For [𝑎,𝑏]⊆(𝑡₁,1], and 𝑠₁,𝑠₂∈[𝑎,𝑏] with 𝑠₂<𝑠₁, we can prove similarly that 𝑠₁^2−𝛽(𝑇𝑥)𝑠₁−𝑠₂^2−𝛽(𝑇𝑥)𝑠₂⟶0uniformlyas𝑠₁⟶𝑠₂. (67) For each [𝑎,𝑏]⊆(𝑡₀,𝑡₁], and 𝑠₁,𝑠₂∈[𝑎,𝑏] with 𝑠₂<𝑠₁, we have 𝑠₁^{2+𝑝−𝛽}𝐷^𝑝₀₊(𝑇𝑥)𝑠₁−𝑠₂^{2+𝑝−𝛽}𝐷^𝑝₀₊(𝑇𝑥)𝑠₂⟶0uniformlyas𝑠₁⟶𝑠₂. (68) For [𝑎,𝑏]⊆(𝑡₁,1], and 𝑠₁,𝑠₂∈[𝑎,𝑏] with 𝑠₂<𝑠₁, we can prove similarly that 𝑠₁^{2+𝑝−𝛽}𝐷^𝑝₀₊(𝑇𝑥)𝑠₁−𝑠₂^{2+𝑝−𝛽}𝐷^𝑝₀₊(𝑇𝑥)𝑠₂⟶0uniformlyas𝑠₁⟶𝑠₂. (69) So 𝑇(Ω) is equicontinuous on each subinterval [𝑎,𝑏]⊆(𝑡_𝑘,𝑡_𝑘+1](𝑘=0,1).
Substep  3.3. Prove that 𝑇(Ω) is equiconvergent as 𝑡→0, and equiconvergent as 𝑡→𝑡₁.
We have 𝑡^2−𝛼𝑇𝑦(𝑡)−¹₀𝑢(𝑠)𝐺𝑠,𝑦(𝑠),𝐷^𝑝₀₊𝑦(𝑠)𝑑𝑠≤𝐴_𝑟𝐿₁𝑡^2+𝑘+𝛿𝐁(𝛼+𝛿,𝑘+1)Γ(𝛼)+𝑡Γ(𝛼)𝐴_𝑟𝐿₁𝐁(𝛼+𝛿,𝑘+1)+𝑡𝐴_𝑟‖𝑚‖₁+𝑡ΠΓ(𝛼)Γ𝑡𝛼−𝑞^{1𝛼−𝑞−1}−Γ(𝛼−1)Γ𝑡𝛼−𝑞−2^{1𝛼−𝑞−2}𝐴_𝑟+𝑡𝑡₁^𝛼−2−𝑡₁^𝛼−1Π𝐴_𝑟. (70) It follows that 𝑡^2−𝛼𝑇𝑦(𝑡)−¹₀𝑢(𝑠)𝐺𝑠,𝑦(𝑠),𝐷^𝑝₀₊𝑦(𝑠)𝑑𝑠⟶0uniformlyas𝑡⟶0. (71) Similarly, we can show that 𝑡^2−𝛼(𝑇𝑦)(𝑡) is equiconvergent at 𝑡=𝑡₁. On the other hand, we have 𝑡^{2+𝑞−𝛼}𝐷^𝑞₀₊(𝑡)−Γ()𝑇𝑦𝛼−1Γ×𝛼−𝑞−1₁₀𝑢(𝑠)𝐺𝑠,𝑦(𝑠),𝐷^𝑝₀₊𝑦(𝑠)𝑑𝑠≤𝐴_𝑟𝐿₁𝑡^2+𝑘+𝛿𝐁𝛼+𝛿−𝑞,𝑘+1Γ+𝑡𝛼−𝑞Γ𝐴𝛼−𝑞^𝑟𝐿₁𝐁()Γ(𝛼)𝛼+𝛿,𝑘+1+𝑡Γ𝐴𝛼−𝑞^𝑟‖𝑚‖₁+𝑡ΠΓ(𝛼)Γ×Γ(𝛼)𝛼−𝑞Γ𝑡𝛼−𝑞^{1𝛼−𝑞−1}−Γ(𝛼−1)Γ𝑡𝛼−𝑞−2^{1𝛼−𝑞−2}𝐴_𝑟+Γ(𝛼)Γ𝑡𝑡𝛼−𝑞^1𝛼−2−𝑡₁^𝛼−1Π𝐴_𝑟. (72) It follows that 𝑡^{2+𝑞−𝛼}𝐷^𝑞₀₊(𝑡)−Γ()𝑇𝑦𝛼−1Γ×𝛼−𝑞−1₁₀𝑢(𝑠)𝐺𝑠,𝑦(𝑠),𝐷^𝑝₀₊𝑦(𝑠)𝑑𝑠⟶0uniformlyas𝑡⟶0. (73) Similarly, we can show that 𝑡^{2+𝑞−𝛼}𝐷^𝑞₀₊(𝑇𝑦)(𝑡) is equiconvergent at 𝑡=𝑡₁.
Similarly we can prove that 𝑡^2−𝛽(𝑇𝑥)(𝑡)−¹₀𝑣(𝑠)𝐻𝑠,𝑥(𝑠),𝐷^𝑞₀₊𝑥(𝑠)𝑑𝑠⟶0uniformlyas𝑡⟶0, (74) and 𝑡^{2+𝑝−𝛽}𝐷^𝑝₀₊(𝑇𝑥)(𝑡) is equiconvergent at 𝑡=0, both 𝑡^2−𝛽(𝑇𝑥)(𝑡) and 𝑡^{2+𝑝−𝛽}𝐷^𝑝₀₊(𝑇𝑥)(𝑡) are equiconvergent at 𝑡=𝑡₁.
Hence, 𝑇(Ω) is equiconvergent as 𝑡→0 and 𝑇(Ω) is equiconvergent as 𝑡→𝑡₁.
So 𝑇(Ω) is relatively compact. Then 𝑇 is completely continuous. The proofs are completed.

3. Main Result

In this section, we will establish the existence of at least one solution of BVP (6)-(7).

Definition 11 (see [26]). An odd homeomorphism Φ of the real line ℝ onto itself is called a pseudo-sub-multiplicative function if there exists a homeomorphism 𝜔 of [0,∞) onto itself which supports Φ in the sense that for all 𝑣₁,𝑣₂≥0 we have Φ(𝑣₁𝑣₂)≥𝜔(𝑣₁)Φ(𝑣₂).  𝜔 is called the supporting function of Φ.

Remark 12. Note that any submultiplicative function is a pseudo-submultiplicative function. Also any function of the form Φ(𝑢)∶=∑^𝑘_𝑗=0𝑐_𝑗|𝑢|^𝑗𝑢,𝑢∈ℝ is pseudo-sup-multiplicative, provided that 𝑐_𝑗≥0. Here, a supporting function is defined by 𝜔(𝑢)∶=min{𝑢^𝑘+1,𝑢}, 𝑢≥0.

Remark 13. It is clear that a pseudo-submultiplicative function Φ and any corresponding supporting function 𝜔 are increasing functions vanishing at zero; moreover, their inverses Φ⁻¹ and 𝜈, respectively, are increasing and for all 𝑣₁,𝑣₂≥0, we have Φ⁻¹(𝑣₁𝑣₂)≤𝜈(𝑣₁)Φ⁻¹(𝑣₂).

Theorem 14. Suppose that (a)–(e) and (A)-(B) hold, Φ∶𝑅→𝑅 is a submultiplicative-like function with the supporting function 𝜔, its inverse function is denoted by Φ⁻¹∶𝑅→𝑅 with the supporting function 𝜈. Furthermore, suppose that(i)there exist nonnegative numbers 𝐶_𝑓,𝐵_𝑓,𝐴_𝑓,𝐶_𝐺,𝐵_𝐺,𝐴_𝐺,𝐶_𝑀,𝐵_𝑀, and 𝐴_𝑀 such that 𝑓𝑡,𝑡^𝛼−2𝑈,𝑡^{𝛼−𝑞−2}𝑉≤𝐶_𝑓+𝐵_𝑓Φ⁻¹(|𝑈|)+𝐴_𝑓Φ⁻¹𝑉,𝐺(||)𝑡,𝑡^𝛼−2𝑈,𝑡^{𝛼−𝑞−2}𝑉≤𝐶_𝐺+𝐵_𝐺Φ⁻¹(|𝑈|)+𝐴_𝐺Φ⁻¹𝑉,𝑀(||)𝑡,𝑡^𝛼−2𝑈,𝑡^{𝛼−𝑞−2}𝑉≤𝐶_𝑀+𝐵_𝑀Φ⁻¹(|𝑈|)+𝐴_𝑀Φ⁻¹Φ⁻¹(|𝑈|), (75) holds for all (𝑈,𝑉)∈𝑅², 𝑡∈(0,1].(ii)there exist nonnegative numbers 𝐶_𝑔,𝐵_𝑔,𝐴_𝑔,𝐶_𝐻,𝐵_𝐻,𝐴_𝐻,𝐶_𝑁,𝐵_𝑁, and 𝐴_𝑁 such that 𝑔𝑡,𝑡^𝛽−2𝑈,𝑡^{𝛽−𝑝−2}𝑉≤𝐶_𝑔+𝐵_𝑔Φ(𝑈)+𝐴_𝑔Φ(𝑉),𝐻𝑡,𝑡^𝛽−2𝑈,𝑡^{𝛽−𝑝−2}𝑉≤𝐶_𝐻+𝐵_𝐻Φ(𝑈)+𝐴_𝐻Φ(𝑉),𝑁𝑡,𝑡^𝛽−2𝑈,𝑡^{𝛽−𝑝−2}𝑉≤𝐶_𝑁+𝐵_𝑁Φ(𝑈)+𝐴_𝑁Φ(𝑉), (76) hold for all (𝑈,𝑉)∈𝑅², 𝑡∈(0,1].(iii)there exist the nonnegative numbers 𝐶_𝐼,𝐵_𝐼,𝐴_𝐼,𝐶_1,𝐼,𝐵_1,𝐼, and 𝐴_1,𝐼 such that 𝐼𝑡₁,𝑡₁^𝛼−2𝑈,𝑡₁^{𝛼−𝑞−2}𝑉≤𝐶_𝐼+𝐵_𝐼Φ⁻¹(|𝑈|)+𝐴_𝐼Φ⁻¹𝑉,𝐼(||)₁𝑡₁,𝑡₁^𝛼−2𝑈,𝑡₁^{𝛼−𝑞−2}𝑉≤𝐶_1,𝐼+𝐵_1,𝐼Φ⁻¹(|𝑈|)+𝐴_1,𝐼Φ⁻¹(|𝑉|), (77) hold for all (𝑈,𝑉)∈𝑅².(iv)there exist the nonnegative numbers 𝐶_𝐽,𝐵_𝐽,𝐴_𝐽,𝐶_1,𝐽,𝐵_1,𝐽, and 𝐴_1,𝐽 such that 𝐽𝑡₁,𝑡₁^𝛽−2𝑈,𝑡₁^{𝛽−𝑝−2}𝑉≤𝐶_𝐽+𝐵_𝐽Φ(𝑈)+𝐴_𝐽Φ(𝑉),𝐽₁𝑡₁,𝑡₁^𝛽−2𝑈,𝑡₁^{𝛽−𝑝−2}𝑉≤𝐶_1,𝐽+𝐵_1,𝐽Φ(𝑈)+𝐴_1,𝐽Φ(𝑉), (78) hold for all (𝑈,𝑉)∈𝑅².
Then BVP (6)-(7) has at least one solution if maxΘ₂,Σ₂,Θ₄,Σ₄𝜈2maxΥ₂,Λ₂,Υ₄,Λ₄Υ<1𝑜𝑟max₂,Λ₂,Υ₄,Λ₄𝑤2max{Θ₂,Σ₂,Θ₄,Σ₄}⁻¹<1, (79) where Θ₁=𝐿₁𝐁(𝛼+𝛿,𝑘+1)Γ(𝛼)𝐶_𝑓+𝐿₁𝐁(𝛼+𝛿,𝑘+1)Γ(𝛼)𝐶_𝑓+‖𝑢‖₁𝐶_𝐺+‖𝑚‖₁𝐶_𝑀+1ΠΓ(𝛼)Γ𝑡𝛼−𝑞^{1𝛼−𝑞−1}−Γ(𝛼−1)Γ𝑡𝛼−𝑞−2^{1𝛼−𝑞−2}𝐶_𝐼+𝑡₁^𝛼−2−𝑡₁^𝛼−1Π𝐶_1,𝐼,Θ₂=𝐿₁𝐁(𝛼+𝛿,𝑘+1)Γ(𝛼)𝐵_𝑓+𝐴_𝑓+𝐿₁𝐁(𝛼+𝛿,𝑘+1)Γ(𝛼)𝐵_𝑓+𝐴_𝑓+‖𝑢‖₁𝐵_𝐺+𝐴_𝐺+‖𝑚‖₁𝐵_𝑀+𝐴_𝑀+1ΠΓ(𝛼)Γ𝑡𝛼−𝑞^{1𝛼−𝑞−1}−Γ(𝛼−1)Γ𝑡𝛼−𝑞−2^{1𝛼−𝑞−2}×𝐵_𝐼+𝐴_𝐼+𝑡₁^𝛼−2−𝑡₁^𝛼−1Π𝐵_1,𝐼+𝐴_1,𝐼,Θ₃=𝐿₁𝐁𝛼+𝛿−𝑞,𝑘+1Γ𝐶𝛼−𝑞^𝑓+𝐿₁Γ𝐁()𝐶𝛼−𝑞𝛼+𝛿,𝑘+1_𝑓+Γ(𝛼−1)Γ‖𝑢‖𝛼−𝑞−1¹𝐶_𝐺+Γ(𝛼)Γ‖𝑚‖𝛼−𝑞¹𝐶_𝑀+1ΠΓ(𝛼)Γ×Γ(𝛼)𝛼−𝑞Γ𝑡𝛼−𝑞^{1𝛼−𝑞−1}−Γ(𝛼−1)Γ𝑡𝛼−𝑞−2^{1𝛼−𝑞−2}𝐶_𝐼+Γ(𝛼)Γ𝑡𝛼−𝑞^1𝛼−2−𝑡₁^𝛼−1Π𝐶_1,𝐼,Θ₄=𝐿₁𝐁𝛼+𝛿−𝑞,𝑘+1Γ𝐵𝛼−𝑞^𝑓+𝐴_𝑓+𝐿₁Γ𝐁(𝐵𝛼−𝑞𝛼+𝛿,𝑘+1)_𝑓+𝐴_𝑓+Γ(𝛼−1)Γ‖𝑢‖𝛼−𝑞−1¹𝐵_𝐺+𝐴_𝐺+Γ(𝛼)Γ‖𝑚‖𝛼−𝑞¹𝐵_𝑀+𝐴_𝑀+1ΠΓ(𝛼)Γ×Γ(𝛼)𝛼−𝑞Γ𝑡𝛼−𝑞^{1𝛼−𝑞−1}−Γ(𝛼−1)Γ𝑡𝛼−𝑞−2^{1𝛼−𝑞−2}×𝐵_𝐼+𝐴_𝐼+Γ(𝛼)Γ𝑡𝛼−𝑞^1𝛼−2−𝑡₁^𝛼−1Π𝐵_1,𝐼+𝐴_1,𝐼,Σ₁=𝐿₁𝐁(𝛼+𝛿,𝑘+1)Γ(𝛼)𝐶_𝑓+𝐿₁𝐁(𝛼+𝛿,𝑘+1)Γ(𝛼)𝐶_𝑓+‖𝑢‖₁𝐶_𝐺+1ΠΓ(𝛼)Γ𝑡𝛼−𝑞^{1𝛼−𝑞−1}𝐶_𝐼+1Π𝑡₁^𝛼−1𝐶_1,𝐼,Σ₂=𝐿₁𝐁(𝛼+𝛿,𝑘+1)Γ(𝛼)𝐵_𝑓+𝐴_𝑓+𝐿₁𝐁(𝛼+𝛿,𝑘+1)Γ(𝛼)𝐵_𝑓+𝐴_𝑓+‖𝑢‖₁𝐵_𝐺+𝐴_𝐺+1ΠΓ(𝛼)Γ𝑡𝛼−𝑞^{1𝛼−𝑞−1}𝐵_𝐼+𝐴_𝐼+1Π𝑡₁^𝛼−1𝐵_1,𝐼+𝐴_1,𝐼,Σ₃=𝐿₁𝐁𝛼+𝛿−𝑞,𝑘+1Γ𝐶𝛼−𝑞^𝑓+𝐿₁Γ𝐁()𝐶𝛼−𝑞𝛼+𝛿,𝑘+1_𝑓+Γ(𝛼−1)Γ+Γ(𝛼)𝛼−𝑞−1Γ‖𝑢‖𝛼−𝑞¹𝐶_𝐺+Γ(𝛼)Γ‖𝑚‖𝛼−𝑞¹𝐶_𝑀+1ΠΓ(𝛼)Γ+Γ()𝛼−𝑞𝛼−1ΓΓ(𝛼)𝛼−𝑞−1Γ𝑡𝛼−𝑞^{1𝛼−𝑞−1}𝐶_𝐼+1ΠΓ(𝛼−1)Γ+Γ(𝛼)𝛼−𝑞−1Γ𝑡𝛼−𝑞^1𝛼−1𝐶_1,𝐼,Σ₄=𝐿₁𝐁𝛼+𝛿−𝑞,𝑘+1Γ𝐵𝛼−𝑞^𝑓+𝐴_𝑓+𝐿₁Γ𝐁(𝐵𝛼−𝑞𝛼+𝛿,𝑘+1)_𝑓+𝐴_𝑓+Γ(𝛼−1)Γ+Γ(𝛼)𝛼−𝑞−1Γ‖𝑢‖𝛼−𝑞¹𝐵_𝐺+𝐴_𝐺+Γ(𝛼)Γ‖𝑚‖𝛼−𝑞¹𝐵_𝑀+𝐴_𝑀+1ΠΓ(𝛼)Γ+Γ()𝛼−𝑞𝛼−1Γ×Γ(𝛼)𝛼−𝑞−1Γ𝑡𝛼−𝑞^{1𝛼−𝑞−1}𝐵_𝐼+𝐴_𝐼+1ΠΓ(𝛼−1)Γ+Γ(𝛼)𝛼−𝑞−1Γ𝛼−𝑞×𝑡₁^𝛼−1𝐵_1,𝐼+𝐴_1,𝐼,Υ₁=𝐿₂𝐁𝛽+𝜃,𝑙+1Γ𝛽𝐶_𝑔+𝐿₂𝐁𝛽+𝜃,𝑙+1Γ𝛽𝐶_𝑔+‖𝑣‖₁𝐶_𝐻+‖𝑛‖₁𝐶_𝑁+1ΞΓ𝛽Γ𝑡𝛽−𝑝^{1𝛽−𝑝−1}−Γ𝛽−1Γ𝑡𝛽−𝑝−2^{1𝛽−𝑝−2}𝐶_𝐽+𝑡₁^𝛽−2−𝑡₁^𝛽−1Ξ𝐶_1,𝐽,Υ₂=𝐿₂𝐁𝛽+𝜃,𝑙+1Γ𝛽𝐵_𝑔+𝐴_𝑔+𝐿₂𝐁𝛽+𝜃,𝑙+1Γ𝛽𝐵_𝑔+𝐴_𝑔+‖𝑏‖₁𝐵_𝐻+𝐴_𝐻+‖𝑛‖₁𝐵_𝑁+𝐴_𝑁+1ΞΓ𝛽Γ𝑡𝛽−𝑝^{1𝛽−𝑝−1}−Γ𝛽−1Γ𝑡𝛽−𝑝−2^{1𝛽−𝑝−2}×𝐵_𝐽+𝐴_𝐽+𝑡₁^𝛽−2−𝑡₁^𝛽−1Ξ𝐵_1,𝐽+𝐴_1,𝐽,Υ₃=𝐿₂𝐁𝛽+𝜃−𝑝,𝑙+1Γ𝐶𝛽−𝑝^𝑔+𝐿₂Γ𝐁𝐶𝛽−𝑝𝛽+𝜃,𝑙+1_𝑔+Γ𝛽−1Γ‖𝑣‖𝛽−𝑝−1¹𝐶_𝐻+Γ𝛽Γ‖𝑛‖𝛽−𝑝¹𝐶_𝑁+1ΞΓ𝛽Γ×Γ𝛽𝛽−𝑝Γ𝑡𝛽−𝑝^{1𝛽−𝑝−1}−Γ𝛽−1Γ𝑡𝛽−𝑝−2^{1𝛽−𝑝−2}𝐶_𝐽+Γ𝛽Γ𝑡𝛽−𝑝^1𝛽−2−𝑡₁^𝛽−1Ξ𝐶_1,𝐽,Υ₄=𝐿₂𝐁𝛽+𝜃−𝑝,𝑙+1Γ𝐵𝛽−𝑝^𝑔+𝐴_𝑔+𝐿₂Γ𝐁𝐵𝛽−𝑝𝛽+𝜃,𝑙+1_𝑔+𝐴_𝑔+Γ𝛽−1Γ‖𝑣‖𝛽−𝑝−1¹𝐵_𝐻+𝐴_𝐻+Γ𝛽Γ‖𝑛‖𝛽−𝑝¹𝐵_𝑁+𝐴_𝑁+1ΞΓ𝛽Γ×Γ𝛽𝛽−𝑝Γ𝑡𝛽−𝑝^{1𝛽−𝑝−1}−Γ𝛽−1Γ𝑡𝛽−𝑝−2^{1𝛽−𝑝−2}𝐵_𝐽+𝐴_𝐽+Γ𝛽Γ𝑡𝛽−𝑝^1𝛽−2−𝑡₁^𝛽−1Ξ𝐵_1,𝐽+𝐴_1,𝐽,Λ₁=𝐿₂𝐁𝛽+𝜃,𝑙+1Γ𝛽𝐶_𝑔+𝐿₂𝐁𝛽+𝜃,𝑙+1Γ𝛽𝐶_𝑔+‖𝑣‖₁𝐶_𝐻+1ΞΓ𝛽Γ𝑡𝛽−𝑝^{1𝛽−𝑝−1}𝐶_𝐽+1Ξ𝑡₁^𝛽−1𝐶_1,𝐽,Λ₂=𝐿₂𝐁𝛽+𝜃,𝑙+1Γ𝛽𝐵_𝑔+𝐴_𝑔+𝐿₂𝐁𝛽+𝜃,𝑙+1Γ𝛽𝐵_𝑔+𝐴_𝑔+‖𝑣‖₁𝐵_𝐻+𝐴_𝐻+1ΞΓ𝛽Γ𝑡𝛽−𝑝^{1𝛽−𝑝−1}𝐵_𝐽+𝐴_𝐽+1Ξ𝑡₁^𝛽−1𝐵_1,𝐽+𝐴_1,𝐽,Λ₃=𝐿₂𝐁𝛽+𝜃−𝑝,𝑙+1Γ𝐶𝛽−𝑝^𝑔+𝐿₂Γ𝐁𝐶𝛽−𝑝𝛽+𝜃,𝑙+1_𝑔+Γ𝛽−1Γ+Γ𝛽𝛽−𝑝−1Γ‖𝑣‖𝛽−𝑝¹𝐶_𝐻+Γ𝛽Γ‖𝑛‖𝛽−𝑝¹𝐶_𝑁+1ΞΓ𝛽Γ+Γ𝛽−𝑝𝛽−1ΓΓ𝛽𝛽−𝑝−1Γ𝑡𝛽−𝑝^{1𝛽−𝑝−1}𝐶_𝐽+1ΞΓ𝛽−1Γ+Γ𝛽𝛽−𝑝−1Γ𝑡𝛽−𝑝^1𝛽−1𝐶_1,𝐽,Λ₄=𝐿₂𝐁𝛽+𝜃−𝑝,𝑙+1Γ𝐵𝛽−𝑝^𝑔+𝐴_𝑔+𝐿₂Γ𝐁𝐵𝛽−𝑝𝛽+𝜃,𝑙+1_𝑔+𝐴_𝑔+Γ𝛽−1Γ+Γ𝛽𝛽−𝑝−1Γ‖𝑣‖𝛽−𝑝¹𝐵_𝐻+𝐴_𝐻+Γ𝛽Γ‖𝑛‖𝛽−𝑝¹𝐵_𝑁+𝐴_𝑁+1ΞΓ𝛽Γ+Γ𝛽−𝑝𝛽−1Γ×Γ𝛽𝛽−𝑝−1Γ𝑡𝛽−𝑝^{1𝛽−𝑝−1}𝐵_𝐽+𝐴_𝐽+1ΞΓ𝛽−1Γ+Γ𝛽𝛽−𝑝−1Γ𝑡𝛽−𝑝^1𝛽−1𝐵_1,𝐽+𝐴_1,𝐽. (80)(81)

Proof. To apply Lemma 5, we should define an open bounded subset Ω of 𝑋×𝑌 centered at zero such that assumptions in Lemma 5 hold.
Let Ω₁={(𝑥,𝑦)∈𝑋×𝑌∶(𝑥,𝑦)=𝜆𝑇(𝑥,𝑦) for some 𝜆∈(0,1)}. We prove that Ω₁ is bounded. For (𝑥,𝑦)∈Ω₁, we get (𝑥,𝑦)=𝜆𝑇(𝑥,𝑦). It follows that 𝑥=𝜆𝑇₁𝑦 and 𝑦=𝜆𝑇₂𝑥.
For 𝑡∈(0,𝑡₁], we obtain 𝑡^2−𝛼|𝑥(𝑡)|≤𝑡^2−𝛼|(𝑇₁𝑦)(𝑡)|≤Θ₁+Θ₂Φ⁻¹(||𝑦||).
For 𝑡∈(𝑡₁,1], 𝑡^2−𝛼|𝑥(𝑡)|≤𝑡^2−𝛼𝑡₀(𝑡−𝑠)^𝛼−1Γ(𝛼)𝜙(𝑠)𝑓𝑠,𝑦(𝑠),𝐷^𝑝₀₊𝑦(𝑠)+𝑡𝑑𝑠Γ(𝛼)¹₀(1−𝑠)^𝛼−1𝜙(𝑠)𝑓𝑠,𝑦(𝑠),𝐷^𝑝₀₊𝑦(𝑠)+()𝑑𝑠1−𝑡¹₀𝑢(𝑠)𝐺𝑠,𝑦(𝑠),𝐷^𝑝₀₊𝑦(𝑠)𝑑𝑠+𝑡¹₀𝑚(𝑠)𝑀𝑠,𝑦(𝑠),𝐷^𝑝₀₊𝑦(𝑠)+𝑑𝑠1−𝑡ΠΓ(𝛼)Γ𝑡𝛼−𝑞^{1𝛼−𝑞−1}𝐼𝑡₁,𝑦𝑡₁,𝐷^𝑝₀₊𝑦𝑡₁+1−𝑡Π𝑡₁^𝛼−1𝐼₁𝑡₁,𝑦𝑡₁,𝐷^𝑝₀₊𝑦𝑡₁≤Σ₁+Σ₂Φ⁻¹𝑦. (82) It follows that sup_()𝑡∈0,1𝑡^2−𝛼|𝑥(𝑡)|≤maxΘ₁,Σ₁+maxΘ₂,Σ₂Φ⁻¹𝑦. (83) Similarly, we have for 𝑡∈(0,𝑡₁] that 𝑡^{𝑞+2−𝛼}𝐷^𝑞₀₊𝑥(𝑡)≤Θ₃+Θ₄Φ⁻¹𝑦 (84) and for 𝑡∈(0,𝑡₁]𝑡^{𝑞+2−𝛼}𝐷^𝑞₀₊𝑥(𝑡)≤Σ₃+Σ₄Φ⁻¹𝑦. (85)
It follows that sup_()𝑡∈0,1𝑡^{2+𝑞−𝛼}𝐷^𝑞₀₊𝑥(𝑡Θ)≤max₃,Σ₃+maxΘ₄,Σ₄Φ⁻¹𝑦. (86)
Hence, ‖𝑥‖≤maxΘ₁,Σ₁,Θ₃,Σ₃+maxΘ₂,Σ₂,Θ₄,Σ₄Φ⁻¹𝑦. (87)
Similar to the above discussion we can prove that 𝑦≤maxΥ₁,Λ₁,Υ₃,Λ₃+maxΥ₂,Λ₂,Υ₄,Λ₄Φ(‖𝑥‖). (88)
Case  1. Consider (max{Θ₂, Σ₂,Θ₄,Σ₄}𝜈  (2max{Υ₂,Λ₂,Υ₄, Λ₄})<1).
With out loss of generality, suppose that ‖𝑥‖≥Φ⁻¹maxΥ₁,Λ₁,Υ₃,Λ₃maxΥ₂,Λ₂,Υ₄,Λ₄. (89) Then use Remark 13, and the previous inequalities to get ‖𝑥‖≤maxΘ₁,Σ₁,Θ₃,Σ₃+maxΘ₂,Σ₂,Θ₄,Σ₄𝜈2maxΥ₂,Λ₂,Υ₄,Λ₄‖𝑥‖. (90) It follows that there exists a constant 𝑊>0 such that ||𝑥||≤𝑊. Thus ‖𝑥‖≤max𝑊,Φ⁻¹maxΥ₁,Λ₁,Υ₃,Λ₃maxΥ₂,Λ₂,Υ₄,Λ₄. (91) Then 𝑦≤maxΥ₁,Λ₁,Υ₃,Λ₃+maxΥ₂,Λ₂,Υ₄,Λ₄Φ×max𝑊,Φ⁻¹maxΥ₁,Λ₁,Υ₃,Λ₃maxΥ₂,Λ₂,Υ₄,Λ₄. (92) It follows that Ω₁ is bounded.
Case  2. Consider ((max{Υ₂,Λ₂,Υ₄,Λ₄}/𝑤((2max{Θ₂,Σ₂,Θ₄,Σ₄})⁻¹))   <1).
Without loss of generality, suppose that 𝑦Θ≥Φmax₁,Σ₁,Θ₃,Σ₃maxΘ₂,Σ₂,Θ₄,Σ₄. (93) Then using Remark 12 and the previous inequalities, we get 𝑦≤maxΥ₁,Λ₁,Υ₃,Λ₃+maxΥ₂,Λ₂,Υ₄,Λ₄𝑤2maxΘ₂,Σ₂,Θ₄,Σ₄⁻¹𝑦. (94) It follows that there exists a constant 𝑊>0 such that ||𝑦||≤𝑊. We get 𝑦Θ≤max𝑊,Φmax₁,Σ₁,Θ₃,Σ₃maxΘ₂,Σ₂,Θ₄,Σ₄. (95) Then ‖𝑥‖≤maxΘ₁,Σ₁,Θ₃,Σ₃+maxΘ₂,Σ₂,Θ₄,Σ₄Φ⁻¹×Θmax𝑊,Φmax₁,Σ₁,Θ₃,Σ₃maxΘ₂,Σ₂,Θ₄,Σ₄. (96) It follows that Ω₁ is bounded.
To apply Lemma 5, let Ω be a nonempty open bounded subset of 𝑋 such that Ω⊃Ω₁ centered at zero.
It is easy to see from Lemma 8 that 𝑇 is a completely continuous operator. One can see that 𝑥,𝑦≠𝜆𝑇𝑥,𝑦∀𝑥,𝑦∈𝜕Ω,𝜆∈(0,1). (97) Thus, from Lemma 5, (𝑥,𝑦)=𝑇(𝑥,𝑦) has at least one solution (𝑥,𝑦)∈Ω. So (𝑥,𝑦) is a pair of solutions of BVP (3) and BVP (6). The proof of Theorem 14 is complete.

4. Two Examples

To illustrate the usefulness of our main result, we present two examples that Theorem 14 can readily apply.

Example 15. Consider the following impulsive boundary value problem: 𝐷₀^8/5₊𝑥(𝑡)=𝑡^−1/5(1−𝑡)⁻¹×𝑐+𝑏𝑡^6/5𝑦(𝑡)³+𝑎𝑡^9/5𝐷₀^1/5₊𝑦(𝑡)³,()1𝑡∈0,1,𝑡≠2,𝐷₀^9/5₊𝑦(𝑡)=𝑡^−1/5(1−𝑡)⁻¹×𝑐₀+𝑏₀𝑡^1/15[𝑥(𝑡)]^1/3+𝑎₀𝑡^2/15𝐷₀^1/5₊𝑥(𝑡)^1/3,𝑡∈(0,1),𝑡≠𝑡₁,lim_𝑡→0𝑡^2/5𝑥(𝑡)=𝐺,lim_𝑡→0𝑡^1/5𝑦(𝑡)𝑥(1)(1)1=𝐻,=𝑀,𝑦=𝑁,Δ𝑥2=𝑐_𝐼1,Δ𝑦2=𝑐_𝐽,Δ𝐷¹₀⁺𝑥12=𝑐_1,𝐼,Δ𝐷¹₀⁺𝑦12=𝑐_1,𝐽, (98) where 𝑐,𝑏,𝑎,𝑐₀,𝑏₀,𝑎₀,𝐺₀,𝐻₀,𝑀₀,𝑁₀,𝐶_𝐼,𝐶_𝐽,𝐶_1,𝐼,𝐶_1,𝐽 are constants.
Corresponding to BVP (1), we have(a)𝛼=8/5, 𝛽=9/5, 𝑝=𝑞=1/5,(b)𝜙(𝑡)=𝜓(𝑡)=𝑡^−1/5(1−𝑡)^−1/5, 𝑓(𝑡,𝑈,𝑉)=𝑐+𝑏𝑡^6/5𝑈³+𝑎𝑡^9/5𝑉³ and 𝑔(𝑡,𝑈,𝑉)=𝑐₀+𝑏₀𝑡^1/15𝑈^1/3 + 𝑎₀𝑡^2/15𝑉^1/3 defined on (0,1)×𝑅²,(c)𝑢(𝑡)=𝑣(𝑡)=𝑚(𝑡)=𝑛(𝑡)≡1, 𝐺(𝑡,𝑈,𝑉)=𝐺₀, 𝐻(𝑡,𝑈,𝑉)=𝐻₀, 𝑀(𝑡,𝑈,𝑉)=𝑀₀, 𝑁(𝑡,𝑈,𝑉) = 𝑁₀,(d)0=𝑡₀<𝑡₁=(1/2)<𝑡₂=1,(e)𝐼(𝑡,𝑈,𝑉)=𝑐_𝐼, 𝐼₁(𝑡,𝑈,𝑉)=𝑐_1,𝐼, 𝐽(𝑡,𝑈,𝑉)=𝑐_𝐽, 𝐽₁(𝑡,𝑈,𝑉)=𝑐_1,𝐽.
It is easy to show that(A) 𝜙 satisfies 𝛼+2𝛿−𝑞>0, 𝛼+𝑘+𝛿−𝑞≥0, and |𝜙(𝑡)|≤𝐿₁𝑡^𝑘(1−𝑡)^𝛿 for all 𝑡∈(0,1) with 𝐿₁=1 and 𝑘=−(1/5)=𝛿; 𝜓 satisfies 𝜂+2𝜃−𝑝>0, 𝛽+𝑙+𝜃−𝑝≥0, and |𝜓(𝑡)|≤𝐿₂𝑡^𝑙(1−𝑡)^𝜃 for all 𝑡∈(0,1) with 𝐿₂=1 and 𝑙=−(1/5)=𝜃;(B) 𝑓,𝐺,𝑀,𝐼,𝐼₁ are 𝛽-Caratheodory functions and 𝑔,𝐻,𝑁,𝐽,𝐽₁ are 𝛼-Caratheodory functions.
Furthermore, we have Φ⁻¹(𝑥)=𝑥³ and Φ(𝑥)=𝑥^1/3 with 𝑤(𝑥)=𝑥^1/3 and 𝜈(𝑥)=𝑥³. It is easy to see that(i)the inequalities 𝑓𝑡,𝑡^𝛼−2𝑈,𝑡^{𝛼−𝑞−2}𝑉≤𝐶_𝑓+𝐵_𝑓Φ⁻¹(|𝑈|)+𝐴_𝑓Φ⁻¹𝑉,𝐺(||)𝑡,𝑡^𝛼−2𝑈,𝑡^{𝛼−𝑞−2}𝑉≤𝐶_𝐺+𝐵_𝐺Φ⁻¹(|𝑈|)+𝐴_𝐺Φ⁻¹𝑉,𝑀(||)𝑡,𝑡^𝛼−2𝑈,𝑡^{𝛼−𝑞−2}𝑉≤𝐶_𝑀+𝐵_𝑀Φ⁻¹(|𝑈|)+𝐴_𝑀Φ⁻¹Φ⁻¹(|𝑈|) (99) hold for all (𝑈,𝑉)∈𝑅², 𝑡∈(0,1] with 𝐶_𝑓=|𝑐|,𝐵_𝑓=|𝑏|,𝐴_𝑓=|𝑎|, 𝐶_𝐺=|𝐺₀|,𝐵_𝐺=0,𝐴_𝐺=0 and 𝐶_𝑀=|𝑀₀|,𝐵_𝑀=0,𝐴_𝑀=0;(ii)the inequalities 𝑔𝑡,𝑡^𝛽−2𝑈,𝑡^{𝛽−𝑝−2}𝑉≤𝐶_𝑔+𝐵_𝑔Φ(𝑈)+𝐴_𝑔Φ(𝑉),𝐻𝑡,𝑡^𝛽−2𝑈,𝑡^{𝛽−𝑝−2}𝑉≤𝐶_𝐻+𝐵_𝐻Φ(𝑈)+𝐴_𝐻Φ(𝑉),𝑁𝑡,𝑡^𝛽−2𝑈,𝑡^{𝛽−𝑝−2}𝑉≤𝐶_𝑁+𝐵_𝑁Φ(𝑈)+𝐴_𝑁Φ(𝑉) (100) hold for all (𝑈,𝑉)∈𝑅², 𝑡∈(0,1] with 𝐶_𝑔=|𝑐₀|, 𝐵_𝑔=|𝑏₀|,𝐴_𝑔=|𝑎₀|, 𝐶_𝐻=|𝐻₀|,𝐵_𝐻=𝐴_𝐻=0, 𝐶_𝑁=|𝑁₀|,𝐵_𝑁=𝐴_𝑁=0;(iii)the inequalities 𝐼𝑡₁,𝑡₁^𝛼−2𝑈,𝑡₁^{𝛼−𝑞−2}𝑉≤𝐶_𝐼+𝐵_𝐼Φ⁻¹(|𝑈|)+𝐴_𝐼Φ⁻¹𝑉,𝐼(||)₁𝑡₁,𝑡₁^𝛼−2𝑈,𝑡₁^{𝛼−𝑞−2}𝑉≤𝐶_1,𝐼+𝐵_1,𝐼Φ⁻¹(|𝑈|)+𝐴_1,𝐼Φ⁻¹(|𝑉|) (101) hold for all (𝑈,𝑉)∈𝑅² with 𝐶_𝐼=|𝑐_𝐼|, 𝐵_𝐼=𝐴_𝐼=0, 𝐶_1,𝐼=|𝑐_1,𝐼|, 𝐵_1,𝐼=𝐴_1,𝐼=0;(iv)the inequalities 𝐽𝑡₁,𝑡₁^𝛽−2𝑈,𝑡₁^{𝛽−𝑝−2}𝑉≤𝐶_𝐽+𝐵_𝐽Φ(𝑈)+𝐴_𝐽Φ(𝑉),𝐽₁𝑡₁,𝑡₁^𝛽−2𝑈,𝑡₁^{𝛽−𝑝−2}𝑉≤𝐶_1,𝐽+𝐵_1,𝐽Φ(𝑈)+𝐴_1,𝐽Φ(𝑉) (102) hold for all (𝑈,𝑉)∈𝑅² with 𝐶_𝐽=|𝑐_𝐽|, 𝐵_𝐽=𝐴_𝐽=0, 𝐶_1,𝐽=|𝑐_1,𝐽|, 𝐵_1,𝐽=𝐴_1,𝐽=0.By direct computation, we know that Θ₂𝐁()=27/5,4/5Γ()𝑏|+|𝑎,Σ8/5[||]₂𝐁()=27/5,4/5Γ()𝑏|+|𝑎,Θ8/5[||]₄=𝐁(6/5,4/5)Γ()+𝐁()7/57/5,4/5Γ()𝑏|+|𝑎,Σ7/5[||]₄=𝐁(6/5,4/5)Γ()+𝐁()7/57/5,4/5Γ()𝑏|+|𝑎,Υ7/5[||]₂𝐁()=28/5,4/5Γ()𝑏9/5⁰+𝑎₀,Υ₄=𝐁(7/5,4/5)Γ()+𝐁()8/58/5,4/5Γ()𝑏8/5⁰+𝑎₀,Λ₂𝐁()=28/5,4/5Γ()𝑏9/5⁰+𝑎₀,Λ₄=𝐁(7/5,4/5)Γ()+𝐁()8/58/5,4/5Γ()𝑏8/5⁰+𝑎₀. (103) Then Theorem 14 implies that the existence of at least one solution if 2𝐁()max8/5,4/5Γ(),𝐁()9/57/5,4/5Γ()+𝐁()8/58/5,4/5Γ()×28/5max𝐁(7/5,4/5),Γ(8/5)𝐁(6/5,4/5)+Γ(7/5)𝐁(7/5,4/5)Γ(7/5)^1/3×𝑏₀+𝑎₀[|𝑏|+|𝑎|]^1/3<₃1√2. (104)

Example 16. Consider the following boundary value problem without impulse effects: 𝐷₀^7/4₊𝑥(𝑡)=𝑡^−1/4(1−𝑡)^−1/4×𝐶+𝐵𝑡^3/4𝑦(𝑡)³+𝐴𝑡^15/4𝐷¹₀⁺𝑦(𝑡)³,(),𝐷𝑡∈0,1_05/4+𝑦(𝑡)=𝑡^−1/8(1−𝑡)^−1/8×𝐶₀+𝐵₀𝑡^1/4[𝑥(𝑡)]^1/3+𝐴₀𝑡^7/12𝐷₀^1/4₊𝑥(𝑡)^1/3,𝑡∈(0,1),lim_𝑡→0𝑡^1/4𝑥(𝑡)=0,lim_𝑡→0𝑡^3/4𝑦(𝑡)𝑥(1)(1)=0,=0,𝑦=0, (105) where 𝐶, 𝐵, 𝐴, 𝐶₀, 𝐵₀, and 𝐴₀ are constants.
Corresponding to BVP (1), we have(a)𝛼=7/4, 𝛽=5/4, 𝑝=1 and 𝑞=1/4,(b)𝜙(𝑡)=𝑡^−1/4(1−𝑡)^−1/4, 𝜓(𝑡)=𝑡^−1/8(1−𝑡)^−1/8, 𝑓,𝑔 defined on (0,1)×𝑅², 𝑓(𝑡,𝑈,𝑉)=𝐶+𝐵𝑡^1/12𝑈³ + 𝐴𝑡^5/12𝑉³ and 𝑔(𝑡,𝑈,𝑉)=𝐶₀+𝐵₀𝑡^1/4𝑈^1/3+𝐴₀𝑡^7/12𝑉^1/3,(c)𝑚(𝑡)=𝑛(𝑡)=𝑢(𝑡)=𝑣(𝑡)≡0, 𝐺(𝑡,𝑈,𝑉)=𝐻(𝑡,𝑈,𝑉)=𝑀(𝑡,𝑈,𝑉)=𝑁(𝑡,𝑈,𝑉)≡0,(d)there exists no impulse point,(e)𝐼(𝑡,𝑈,𝑉)=𝐼₁(𝑡,𝑈,𝑉)=𝐽(𝑡,𝑈,𝑉)=𝐽₁(𝑡,𝑈,𝑉)≡0.
It is easy to show that(A) 𝜙 satisfies 𝛼+2𝛿−𝑞>0, 𝛼+𝑘+𝛿−𝑞>0, |𝜙(𝑡)|≤𝐿₁𝑡^𝑘(1−𝑡)^𝛿 for all 𝑡∈(0,1) with 𝐿₁=1, 𝑘=−(1/4)=𝛿; 𝜓 satisfies 𝛽+2𝜃−𝑝>0, 𝛽+𝑙+𝜃−𝑝≥0, and |𝜓(𝑡)|≤𝐿₂𝑡^𝑙(1−𝑡)^𝜃 for all 𝑡∈(0,1) with 𝐿₂=1, 𝑙=−(1/8)=𝜃;(B) 𝑓,𝐺,𝑀,𝐼,𝐼₁ are 𝛽-Caratheodory functions and 𝑔,𝐻,𝑁,𝐽,𝐽₁ are 𝛼-Caratheodory functions.Furthermore, Φ(𝑥)=𝑥^1/3 and Φ⁻¹(𝑥)=𝑥³, we have 𝑤(𝑥)=𝑥^1/3 and 𝜈(𝑥)=𝑥³, and(i)the inequalities 𝑓𝑡,𝑡^𝛼−2𝑈,𝑡^{𝛼−𝑞−2}𝑉≤𝐶_𝑓+𝐵_𝑓Φ⁻¹(|𝑈|)+𝐴_𝑓Φ⁻¹𝑉,𝐺(||)𝑡,𝑡^𝛼−2𝑈,𝑡^{𝛼−𝑞−2}𝑉≤𝐶_𝐺+𝐵_𝐺Φ⁻¹(|𝑈|)+𝐴_𝐺Φ⁻¹𝑉,𝑀(||)𝑡,𝑡^𝛼−2𝑈,𝑡^{𝛼−𝑞−2}𝑉≤𝐶_𝑀+𝐵_𝑀Φ⁻¹(|𝑈|)+𝐴_𝑀Φ⁻¹Φ⁻¹(|𝑈|) (106) hold for all (𝑈,𝑉)∈𝑅², 𝑡∈(0,1) with 𝐶_𝐺=𝐵_𝐺=𝐴_𝐺=𝐶_𝑀=𝐵_𝑀=𝐴_𝑀=0, 𝐶_𝑓=|𝐶|, 𝐵_𝑓=|𝐵| and 𝐴_𝑓=|𝐴|;(ii)the inequalities 𝑔𝑡,𝑡^𝛽−2𝑈,𝑡^{𝛽−𝑝−2}𝑉≤𝐶_𝑔+𝐵_𝑔Φ(𝑈)+𝐴_𝑔Φ(𝑉),𝐻𝑡,𝑡^𝛽−2𝑈,𝑡^{𝛽−𝑝−2}𝑉≤𝐶_𝐻+𝐵_𝐻Φ(𝑈)+𝐴_𝐻Φ(𝑉),𝑁𝑡,𝑡^𝛽−2𝑈,𝑡^{𝛽−𝑝−2}𝑉≤𝐶_𝑁+𝐵_𝑁Φ(𝑈)+𝐴_𝑁Φ(𝑉) (107) hold for all (𝑈,𝑉)∈𝑅², 𝑡∈(0,1) with 𝐶_𝐻=𝐵_𝐻=𝐴_𝐻=𝐶_𝑁=𝐵_𝑁=𝐴_𝑁=0, 𝐶_𝑔=|𝐶₀|, 𝐵_𝑔=|𝐵₀| and 𝐴_𝑔=|𝐴₀|;(iii)the inequalities 𝐼𝑡₁,𝑡₁^𝛼−2𝑈,𝑡₁^{𝛼−𝑞−2}𝑉≤𝐶_𝐼+𝐵_𝐼Φ⁻¹(|𝑈|)+𝐴_𝐼Φ⁻¹𝑉,𝐼(||)₁𝑡₁,𝑡₁^𝛼−2𝑈,𝑡₁^{𝛼−𝑞−2}𝑉≤𝐶_1,𝐼+𝐵_1,𝐼Φ⁻¹(|𝑈|)+𝐴_1,𝐼Φ⁻¹(|𝑉|) (108) hold for all (𝑈,𝑉)∈𝑅² with 𝐶_𝐼=𝐵_𝐼=𝐴_𝐼=𝐶_1,𝐼=𝐵_1,𝐼=𝐴_1,𝐼=0;(iv)there exist the nonnegative numbers 𝐴_𝑖,𝑘,𝐵_𝑖,𝑘, 𝐶_𝑖,𝑘(𝑖=1,2) such that 𝐽𝑡₁,𝑡₁^𝛽−2𝑈,𝑡₁^{𝛽−𝑝−2}𝑉≤𝐶_𝐽+𝐵_𝐽Φ(𝑈)+𝐴_𝐽Φ(𝑉),𝐽₁𝑡₁,𝑡₁^𝛽−2𝑈,𝑡₁^{𝛽−𝑝−2}𝑉≤𝐶_1,𝐽+𝐵_1,𝐽Φ(𝑈)+𝐴_1,𝐽Φ(𝑉) (109) hold for all (𝑈,𝑉)∈𝑅² with 𝐶_𝐽=𝐵_𝐽=𝐴_𝐽=𝐶_1,𝐽=𝐵_1,𝐽=𝐴_1,𝐽=0.
By direct computation, we know that Θ₂𝐁()=23/2,3/4Γ()𝐵|+|𝐴,Σ7/4[||]₂𝐁()=23/2,3/4Γ()𝐵|+|𝐴,Θ7/4[||]₄=𝐁(5/4,3/4)Γ()+𝐁()3/23/2,3/4Γ()𝐵|+|𝐴,Σ3/2[||]₄=𝐁(5/4,3/4)Γ()+𝐁()3/23/2,3/4Γ()𝐵|+|𝐴,Υ3/2[||]₂𝐁()=29/8,7/8Γ()𝐵5/4⁰+𝐴₀,Υ₄=𝐁(1/8,7/8)Γ+𝐁()𝛽−𝑝9/8,7/8Γ()𝐵1/4⁰+𝐴₀,Λ₂𝐁()=29/8,7/8Γ()𝐵5/4⁰+𝐴₀,Λ₄=𝐁(1/8,7/8)Γ()+𝐁()1/49/8,7/8Γ()𝐵1/4⁰+𝐴₀. (110)
Then Theorem 14 implies the existence of at least one solution if 2𝐁()max3/2,3/4Γ(),𝐁()7/45/4,3/4Γ()+𝐁()3/23/2,3/4Γ()×23/2max𝐁(9/8,7/8),Γ(5/4)𝐁(1/8,7/8)+Γ(𝛽−𝑝)𝐁(9/8,7/8)Γ(1/4)³×𝐵₀+𝐴₀³𝐵|+|𝐴<1[||]8. (111)

Remark 17. It is easy to see that the previous boundary value problems have at least one solution for sufficiently small |𝐵₁|,|𝐵₂| and |𝐴₀|,|𝐵₀|, |𝑎|,|𝑏|,|𝑎₀| and |𝑏₀|. They cannot be solved by the theorems in [24, 25].

Acknowledgments

This research is partially supported by the Natural Science Foundation of Guangdong province (no. S2011010001900) and the Guangdong Higher Education Foundation for High-Level Talents. This research is partially supported by Ministerio de Economía y Competitividad and EC fund FEDER, Project no. MTM2010-15314, Spain.