Existence results for a coupled system of nonlinear singular fractional differential equations with impulse effects

Hindawi Publishing Corporation Mathematical Problems in Engineering Volume 2013, Article ID 498781, 21 pages http://dx.doi.org/10.1155/2013/498781 Research Article Existence Results for a Coupled System of Nonlinear Singular Fractional Differential Equations with Impulse Effects Yuji Liu,1Juan J. Nieto,2,3 and Óscar Otero-Zarraquiños2 1Department of Mathematics, Guangdong University of Business Studies, Guangzhou 510320, China 2Departamento de An´ alisis Matem´ atico, Facultad de Matem´ aticas, Universidad de Santiago de Compostela, 15782 Santiago de Compostela, Spain 3Department of Mathematics, Faculty of Science, King Abdulaziz University, P.O. Box 80203, Jeddah 21589, Saudi Arabia Correspondence should be addressed to Juan J. Nieto; juanjose.nieto[email protected] Received 2 October 2012; Accepted 15 February 2013 Academic Editor: Jocelyn Sabatier Copyright © 2013 Yuji Liu et al. 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. 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 twopoint 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].Somerecent results on impulsive initial value problems or boundary value problems for fractional differential equations on a finite intervalcanbefoundin[21–23] and references therein. The memory property of fractional calculus makes studies more complicated. This paper is motivated by [24]inwhichthefollowing boundary value problem for the fractional differential equation 𝐷𝛼 0+𝑥(𝑡)=𝑓(𝑡,𝑦(𝑡),𝐷𝑝 0+𝑦(𝑡)), 𝑡∈(0,1), 𝐷𝛽 0+𝑦(𝑡)=𝑔(𝑡,𝑥(𝑡),𝐷𝑞 0+𝑥(𝑡)), 𝑡∈(0,1), 𝑥(0)=0, 𝑦(0)=0, 𝑥(1)−𝛾𝑥(𝜂)=0, 𝑦(1)−𝛾𝑦(𝜂)=0 (1) was studied, where 1<𝛼,𝛽<2,0<𝑝≤𝛽−1and 0< 𝑞≤𝛼−1,𝛾>0,1>𝛾𝜂 𝛼−1,1>𝛾𝜂 𝛽−1 and 𝑓,𝑔 : [0,1]× 𝑅2→𝑅are continuous functions, and 𝐷0+is the RiemannLiouville fractional derivative. An existence result was proved for BVP (1)in[24]. The growth assumptions imposed on 𝑓 and 𝑔are sublinear cases (see [25, Theorem 3.1]); that is, there exist functions 𝑎,𝑏∈𝐿1(0,1), nonnegative constants 𝜖1,𝜖2> 0,𝛿1,𝛿2≥0and 𝜌1,𝜌2,𝜎1,𝜎2∈(0,1)such that 󵄨󵄨󵄨󵄨𝑓(𝑡,𝑥,𝑦)󵄨󵄨󵄨󵄨≤𝑎(𝑡)+𝜖1|𝑥|𝜌1+𝜖2󵄨󵄨󵄨󵄨𝑦󵄨󵄨󵄨󵄨𝜌2, 󵄨󵄨󵄨󵄨𝑔(𝑡,𝑥,𝑦)󵄨󵄨󵄨󵄨≤𝑏(𝑡)+𝛿1|𝑥|𝜎1+𝛿2󵄨󵄨󵄨󵄨𝑦󵄨󵄨󵄨󵄨𝜎2.(2) 2Mathematical Problems in Engineering In [25], the following boundary value problem for the fractional differential equation 𝐷𝛼 0+𝑥(𝑡)=𝑓(𝑡,𝑦(𝑡),𝐷𝑝 0+𝑦(𝑡)), 𝑡∈(0,1), 𝐷𝛽 0+𝑦(𝑡)=𝑔(𝑡,𝑥(𝑡),𝐷𝑞 0+𝑥(𝑡)), 𝑡∈(0,1), 𝑥(0)=0, 𝑦(0)=0, 𝑥(1)=0, 𝑦(1)=0 (3) was studied, where 1<𝛼,𝛽<2,0<𝑝≤𝛽−1and 0<𝑞≤ 𝛼−1,and𝑓,𝑔 : [0,1]×𝑅2→𝑅are continuous functions, and 𝐷0+is the Riemann-Liouville fractional derivative. The growth assumptions imposed on 𝑓and 𝑔are 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 󵄨󵄨󵄨󵄨𝑓(𝑡,𝑥,𝑦)󵄨󵄨󵄨󵄨≤𝑎(𝑡)+𝜖1|𝑥|𝜌1+𝜖2󵄨󵄨󵄨󵄨𝑦󵄨󵄨󵄨󵄨𝜌2, 󵄨󵄨󵄨󵄨𝑔(𝑡,𝑥,𝑦)󵄨󵄨󵄨󵄨≤𝑏(𝑡)+𝛿1|𝑥|𝜎1+𝛿2󵄨󵄨󵄨󵄨𝑦󵄨󵄨󵄨󵄨𝜎2,(4) or sublinear cases, that is, there exist nonnegative constants 𝜖1,𝜖2>0,𝛿1,𝛿2≥0and 𝜌1,𝜌2,𝜎1,𝜎2∈(1,∞)such that 󵄨󵄨󵄨󵄨𝑓(𝑡,𝑥,𝑦)󵄨󵄨󵄨󵄨≤𝜖1|𝑥|𝜌1+𝜖2󵄨󵄨󵄨󵄨𝑦󵄨󵄨󵄨󵄨𝜌2, 󵄨󵄨󵄨󵄨𝑔(𝑡,𝑥,𝑦)󵄨󵄨󵄨󵄨≤𝛿1|𝑥|𝜎1+𝛿2󵄨󵄨󵄨󵄨𝑦󵄨󵄨󵄨󵄨𝜎2.(5) We find that in the superlinear cases, BVP (3)hasapairof 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)isnotstudied when 𝑞>𝛼−1or 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 𝐷𝛼 0+𝑥(𝑡)=𝜙(𝑡)𝑓(𝑡,𝑦(𝑡),𝐷𝑝 0+𝑦(𝑡)), 𝑡∈(0,1),𝑡=𝑡1, 𝐷𝛽 0+𝑦(𝑡)=𝜓(𝑡)𝑔(𝑡,𝑥(𝑡),𝐷𝑞 0+𝑥(𝑡)), 𝑡∈(0,1),𝑡=𝑡1, lim 𝑡→0𝑡2−𝛼𝑥(𝑡)=∫1 0𝑢(𝑠)𝐺(𝑠,𝑦(𝑠),𝐷𝑝 0+𝑦(𝑠))𝑑𝑠, lim 𝑡→0𝑡2−𝛽𝑦(𝑡)=∫1 0 V(𝑠)𝐻(𝑠,𝑥(𝑠),𝐷𝑞 0+𝑥(𝑠))𝑑𝑠, 𝑥(1)=∫1 0𝑚(𝑠)𝑀(𝑠,𝑦(𝑠),𝐷𝑝 0+𝑦(𝑠))𝑑𝑠, (6) 𝑦(1)=∫1 0𝑛(𝑠)𝑁(𝑠,𝑥(𝑠),𝐷𝑞 0+𝑥(𝑠))𝑑𝑠, Δ𝑥(𝑡1)=lim 𝑡→𝑡+ 1𝑥(𝑡)−lim 𝑡→𝑡− 1𝑥(𝑡)=𝐼(𝑡1,𝑦(𝑡1),𝐷𝑝 0+𝑦(𝑡1)), Δ𝑦(𝑡1)=lim 𝑡→𝑡+ 1𝑦(𝑡)−lim 𝑡→𝑡− 1𝑦(𝑡)=𝐽(𝑡1,𝑥(𝑡1),𝐷𝑞 0+𝑥(𝑡1)), Δ𝐷𝑞 0+𝑥(𝑡1)=lim 𝑡→𝑡+ 1𝐷𝑞 0+𝑥(𝑡)−lim 𝑡→𝑡− 1𝐷𝑞 0+𝑥(𝑡) =𝐼1(𝑡1,𝑦(𝑡1),𝐷𝑝 0+𝑦(𝑡1)), Δ𝐷𝑝 0+𝑦(𝑡1)=lim 𝑡→𝑡+ 1𝐷𝑝 0+𝑦(𝑡)−lim 𝑡→𝑡− 1𝐷𝑝 0+𝑦(𝑡) =𝐽1(𝑡1,𝑥(𝑡1),𝐷𝑞 0+𝑥(𝑡1)), (7) where (a) 1<𝛼,𝛽≤2,0<𝑝<𝛽and 0<𝑞<𝛼,𝐷0+is the Riemann-Liouville fractional derivative, (b) 𝜙,𝜓:(0,1) → 𝑅,𝑓,𝑔defined on (0,1)×𝑅2, (c) 𝑚,𝑛,𝑢,V: (0,1) → 𝑅with 𝑚,𝑛,𝑢,V∈𝐿 1(0,1), 𝐺,𝐻,𝑀,𝑁defined on (0,1)×𝑅2, (d) 0=𝑡0<𝑡1<𝑡2=1, (e) 𝐼,𝐼1,𝐽,𝐽1:(0,1)×𝑅2→𝑅. Apairoffunctions(𝑥,𝑦)defined on (0,1)is called a solution of BVP (1)andBVP(3), if 𝑥|(𝑡𝑘,𝑡𝑘+1],𝐷𝑞 0+𝑥|(𝑡𝑘,𝑡𝑘+1]and 𝑦|(𝑡𝑘,𝑡𝑘+1],𝐷𝑝 0+𝑦|(𝑡𝑘,𝑡𝑘+1](𝑘 = 0,1)are continuous, there exists the limits lim 𝑡→𝑡+ 𝑘𝑡2−𝛼𝑥(𝑡),lim 𝑡→𝑡+ 𝑘𝑡2−𝛽𝑦(𝑡), lim 𝑡→𝑡+ 𝑘𝑡2+𝑞−𝛼𝐷𝑞 0+𝑥(𝑡),lim 𝑡→𝑡+ 𝑘𝑡2+𝑝−𝛽𝐷𝑝 0+𝑦(𝑡), 𝑘=0,1, (8) 𝐷𝛼 0+𝑥,𝐷𝛽 0+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,𝑔depend on the lower fractional derivatives; second, both 𝜙and 𝜓may be singular at 𝑡=0and 𝑡=1,thatis,𝜙(𝑡)𝑓(𝑡,𝑥,𝑦)and 𝜓(𝑡)𝑔(𝑡,𝑥,𝑦)may be not continuous functions on [0,1]×𝑅2,theboundaryconditions are integral boundary conditions, and we obtain the results on the existence of at least one solution of BVP (6)-(7); third, 0<𝑝<𝛽and 0<𝑞<𝛼aresupposed;thegrowth assumptions imposed on 𝑓,𝑔,𝐺,𝐻,𝑀,𝑁 and 𝐼,𝐼1,𝐽,𝐽1are 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,themaintheorem and its proof are given. In Section 4, two examples are given to illustrate the main results. Mathematical Problems in Engineering 3 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>0of a function 𝑔:(0,∞)→𝑅is given by 𝐼𝛼 0+𝑔(𝑡)=1 Γ(𝛼)∫𝑡 0(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>0of a continuous function 𝑔: (0,∞) → 𝑅is given by 𝐷𝛼 0+𝑔(𝑡)=1 Γ(𝑛−𝛼)𝑑𝑛 𝑑𝑡𝑛∫𝑡 0𝑔(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→𝑅is called a 𝛽-Caratheodory function if 𝐾satisfies that (i) 𝑡→𝐾(𝑡,𝑡 𝛽−2𝑈,𝑡𝛽−𝑝−2𝑉) is continuous on (𝑡𝑘,𝑡𝑘+1](𝑘=0,1)for every (𝑈,𝑉)∈𝑅2; (ii) (𝑈,𝑉) → 𝐾(𝑡,𝑡𝛽−2𝑈,𝑡𝛽−𝑝−2𝑉)is continuous on 𝑅2 for every 𝑡∈(0,1); (iii) for each 𝑟>0there exists a constant 𝐴𝑟>0such that |𝐾(𝑡,𝑡𝛽−2𝑈,𝑡𝛽−𝑝−2𝑉)|≤𝐴𝑟,𝑡∈(0,1),|𝑈|,|𝑉|≤𝑟. Definition 4. 𝑄:(0,1)×𝑅2→𝑅is called a 𝛼-Caratheodory function if 𝑄satisfies that (i) 𝑡→𝑄(𝑡,𝑡 𝛼−2𝑈,𝑡𝛼−𝑞−2𝑉) is continuous on (𝑡𝑘,𝑡𝑘+1](𝑘=0,1)for every (𝑈,𝑉)∈𝑅2; (ii) (𝑈,𝑉) → 𝑄(𝑡,𝑡𝛼−2𝑈,𝑡𝛼−𝑞−2𝑉)is continuous on 𝑅2 for every 𝑡∈(0,1); (iii) for each 𝑟>0there exists a constant 𝐵𝑟>0such that |𝑄(𝑡,𝑡𝛼−2𝑈,𝑡𝛼−𝑞−2𝑉)|≤𝐵𝑟,𝑡∈(0,1),|𝑈|,|𝑉|≤𝑟. Lemma 5 (the Leray-Schauder nonlinear alternative [23]). Let 𝑋be a Banach space and 𝑇:𝑋→𝑋be a completely continuous operator. Suppose Ωis a nonempty open subset of 𝑋 centered at zero. Then either there exists 𝑥∈𝜕Ωand 𝜆∈(0,1) such that 𝑥=𝜆𝑇𝑥or there exists 𝑥∈Ωsuch that 𝑥=𝑇𝑥. Let the gamma and beta functions Γ(𝛼)and B(𝑝,𝑞)be defined by Γ(𝛼)=∫+∞ 0𝑥𝛼−1𝑒−𝑥𝑑𝑥, B(𝑝,𝑞)=∫1 0𝑥𝑝−1(1−𝑥)𝑞−1𝑑𝑥, ‖𝑚‖1=∫1 0|𝑚(𝑠)|𝑑𝑠 for 𝑚∈𝐿1(0,1). (11) Choose 𝑋 = { { { { { { { { { { { { { { { { { { { { { 𝑥|(𝑡𝑘,𝑡𝑘+1]∈𝐶0(𝑡𝑘,𝑡𝑘+1](𝑘=0,1), 𝐷𝑞 0+𝑥|(𝑡𝑘,𝑡𝑘+1]∈𝐶0(𝑡𝑘,𝑡𝑘+1](𝑘=0,1), 𝑥:(0,1]󳨀→ 𝑅 there exist the limits lim 𝑡→𝑡+ 𝑘𝑡2−𝛼𝑥(𝑡), lim 𝑡→𝑡+ 𝑘𝑡2+𝑞−𝛼𝐷𝑞 0+𝑥(𝑡) } } } } } } } } } } } } } } } } } } } } } , 𝑌 = { { { { { { { { { { { { { { { { { { { { { 𝑦|(𝑡𝑘,𝑡𝑘+1]∈𝐶0(𝑡𝑘,𝑡𝑘+1](𝑘=0,1), 𝐷𝑝 0+𝑥|(𝑡𝑘,𝑡𝑘+1]∈𝐶0(𝑡𝑘,𝑡𝑘+1](𝑘=0,1), 𝑦:(0,1]󳨀→ 𝑅 there exist the limits lim 𝑡→𝑡+ 𝑘𝑡2−𝛽𝑦(𝑡), lim 𝑡→𝑡+ 𝑘𝑡2+𝑝−𝛽𝐷𝑝 0+𝑦(𝑡) } } } } } } } } } } } } } } } } } } } } } . (12) For 𝑥∈𝑋, define the norm by ‖𝑥‖=‖𝑥‖𝑋 =max {sup 𝑡∈(0,1)𝑡2−𝛼 |𝑥(𝑡)|,sup 𝑡∈(0,1)𝑡2+𝑞−𝛼 󵄨󵄨󵄨󵄨󵄨𝐷𝑞 0+𝑥(𝑡)󵄨󵄨󵄨󵄨󵄨}. (13) It is easy to show that 𝑋is a real Banach space. For 𝑦∈𝑌, define the norm by 󵄩󵄩󵄩󵄩𝑦󵄩󵄩󵄩󵄩=󵄩󵄩󵄩󵄩𝑦󵄩󵄩󵄩󵄩𝑌 =max {sup 𝑡∈(0,1)𝑡2−𝛽 󵄨󵄨󵄨󵄨𝑦(𝑡)󵄨󵄨󵄨󵄨,sup 𝑡∈(0,1)𝑡2+𝑝−𝛽 󵄨󵄨󵄨󵄨󵄨𝐷𝑝 0+𝑦(𝑡)󵄨󵄨󵄨󵄨󵄨}. (14) It is easy to show that 𝑌is 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) 𝜙satisfies 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)|≤𝐿1𝑡𝑘(1−𝑡)𝛿for all 𝑡∈(0,1);𝜓satisfies 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)|≤𝐿2𝑡𝑙(1−𝑡)𝜃for all 𝑡∈(0,1). (B) 𝑓,𝐺,𝑀,𝐼,𝐼1are 𝛽-Caratheodory functions and 𝑔,𝐻, 𝑁,𝐽,𝐽1are 𝛼-Caratheodory functions. Remark 6. Suppose that 𝑓is a 𝛽-Caratheodory function. For example, 𝛼=7/4,𝑞=1/8,choose𝑘=−1/2,𝛿=−3/4 and 𝜙(𝑡) = 𝑡𝑘(1−𝑡)𝛿,then𝑘>−1,𝛿 ∈ (−𝛼,0]such that 𝛼+2𝛿−𝑞>0,𝛼+𝑘+𝛿−𝑞≥0,and|𝜙(𝑡)|≤𝑡𝑘(1−𝑡)𝛿for all 𝑡∈(0,1).Itiseasytoseethat𝜙is singular at 𝑡=0and 𝑡=1. 4Mathematical Problems in Engineering Lemma 7. Suppose that 𝑦∈𝑌, and (a)–(e), (A)-(B) hold. Then 𝑥∈𝑋is a solution of 𝐷𝛼 0+𝑥(𝑡)=𝜙(𝑡)𝑓(𝑡,𝑦(𝑡),𝐷𝑝 0+𝑦(𝑡)), 𝑡∈(0,1),𝑡=𝑡1, lim 𝑡→0𝑡2−𝛼𝑥(𝑡)=∫1 0𝑢(𝑠)𝐺(𝑠,𝑦(𝑠),𝐷𝑝 0+𝑦(𝑠))𝑑𝑠, 𝑥(1)=∫1 0𝑚(𝑠)𝑀(𝑠,𝑦(𝑠),𝐷𝑝 0+𝑦(𝑠))𝑑𝑠, Δ𝑥(𝑡1)=𝐼(𝑡1,𝑦(𝑡1),𝐷𝑝 0+𝑦(𝑡1)), Δ𝐷𝑞 0+𝑥(𝑡1)=𝐼1(𝑡1,𝑦(𝑡1),𝐷𝑝 0+𝑦(𝑡1)), (15) if and only if 𝑥∈𝑋satisfies the integral equation 𝑥(𝑡)= { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { ∫𝑡 0(𝑡−𝑠)𝛼−1 Γ(𝛼)𝜙(𝑢)𝑓(𝑢,𝑦(𝑢),𝐷𝑝 0+𝑦(𝑢))𝑑𝑢 −𝑡𝛼−1 Γ(𝛼) ×∫1 0(1−𝑠)𝛼−1𝜙(𝑠)𝑓(𝑠,𝑦(𝑠),𝐷𝑝 0+𝑦(𝑠))𝑑𝑠 +𝑡𝛼−2 ∫1 0𝑢(𝑠)𝐺(𝑠,𝑦(𝑠),𝐷𝑝 0+𝑦(𝑠))𝑑𝑠 +𝑡𝛼−1 ∫1 0𝑚(𝑠)𝑀(𝑠,𝑦(𝑠),𝐷𝑝 0+𝑦(𝑠))𝑑𝑠 +𝑡𝛼−1 Π ×( Γ(𝛼) Γ(𝛼−𝑞)𝑡𝛼−𝑞−1 1−Γ(𝛼−1) Γ(𝛼−𝑞−2)𝑡𝛼−𝑞−2 1) ×𝐼(𝑡1,𝑦(𝑡1),𝐷𝑝 0+𝑦(𝑡1)) +𝑡𝛼−1 (𝑡𝛼−2 1−𝑡𝛼−1 1) Π ×𝐼1(𝑡1,𝑦(𝑡1),𝐷𝑝 0+𝑦(𝑡1)), 𝑡∈(0,𝑡1], ∫𝑡 0(𝑡−𝑠)𝛼−1 Γ(𝛼)𝜙(𝑠)𝑓(𝑠,𝑦(𝑠),𝐷𝑝 0+𝑦(𝑠))𝑑𝑠 −𝑡𝛼−1 Γ(𝛼) ×∫1 0(1−𝑠)𝛼−1𝜙(𝑠)𝑓(𝑠,𝑦(𝑠),𝐷𝑝 0+𝑦(𝑠))𝑑𝑠 +(𝑡𝛼−2 −𝑡𝛼−1) ×∫1 0𝑢(𝑠)𝐺(𝑠,𝑦(𝑠),𝐷𝑝 0+𝑦(𝑠))𝑑𝑠 +𝑡𝛼−1 ∫1 0𝑚(𝑠)𝑀(𝑠,𝑦(𝑠),𝐷𝑝 0+𝑦(𝑠))𝑑𝑠 +𝑡𝛼−1 −𝑡𝛼−2 ΠΓ(𝛼) Γ(𝛼−𝑞)𝑡𝛼−𝑞−1 1 ×𝐼(𝑡1,𝑦(𝑡1),𝐷𝑝 0+𝑦(𝑡1)) +𝑡𝛼−2 −𝑡𝛼−1 Π𝑡𝛼−1 1𝐼1(𝑡1,𝑦(𝑡1),𝐷𝑝 0+𝑦(𝑡1)), 𝑡∈(𝑡1,1], (16) where Π=( Γ(𝛼−1) Γ(𝛼−𝑞−1)−Γ(𝛼) Γ(𝛼−𝑞))𝑡2𝛼−𝑞−3 1.(17) Proof. If 𝑦∈𝑌is a solution of BVP (15), then 󵄩󵄩󵄩󵄩𝑦󵄩󵄩󵄩󵄩=max {sup 𝑡∈(0,1)𝑡2−𝛽 󵄨󵄨󵄨󵄨𝑦(𝑡)󵄨󵄨󵄨󵄨,sup 𝑡∈(0,1)𝑡2+𝑝−𝛽 󵄨󵄨󵄨󵄨󵄨𝐷𝑝 0+𝑦(𝑡)󵄨󵄨󵄨󵄨󵄨} =𝑟<+∞, (18) and 𝑥satisfies all equations in (31)From(B),𝑓is a 𝛽Caratheodory function, then there exists 𝐴𝑟>0such that 󵄨󵄨󵄨󵄨󵄨𝑓(𝑡,𝑦(𝑡),𝐷𝑝 0+𝑦(𝑡))󵄨󵄨󵄨󵄨󵄨 =󵄨󵄨󵄨󵄨󵄨𝑓(𝑡,𝑡𝛽−2𝑡2−𝛽𝑦(𝑡),𝑡𝛽−𝑝−2𝑡2+𝑝−𝛽𝐷𝑝 0+𝑦(𝑡))󵄨󵄨󵄨󵄨󵄨≤𝐴𝑟. (19) Similarly we get that there exist constants 𝐴󸀠 𝑟,𝐴󸀠󸀠 𝑟,𝐵󸀠 𝑟,𝐵󸀠󸀠 𝑟>0 such that 󵄨󵄨󵄨󵄨󵄨𝐺(𝑡,𝑦(𝑡),𝐷𝑝 0+𝑦(𝑡))󵄨󵄨󵄨󵄨󵄨≤𝐴󸀠 𝑟, 󵄨󵄨󵄨󵄨󵄨𝑀(𝑡,𝑦(𝑡),𝐷𝑝 0+𝑦(𝑡))󵄨󵄨󵄨󵄨󵄨≤𝐴󸀠󸀠 𝑟, 𝑡∈(0,1), 󵄨󵄨󵄨󵄨󵄨𝐼(𝑡1,𝑦(𝑡1),𝐷𝑝 0+𝑦(𝑡1))󵄨󵄨󵄨󵄨󵄨≤𝐵󸀠 𝑟, 󵄨󵄨󵄨󵄨󵄨𝐼1(𝑡1,𝑦(𝑡1),𝐷𝑝 0+𝑦(𝑡1))󵄨󵄨󵄨󵄨󵄨≤𝐵󸀠󸀠 𝑟. (20) It follows from (15)that,for𝑡∈(𝑡𝑘,𝑡𝑘+1](𝑘=0,1),there exist constants 𝑐𝑘,𝑑𝑘∈𝑅such that 𝑥(𝑡)=1 Γ(𝛼)∫𝑡 0(𝑡−𝑠)𝛼−1𝜙(𝑠)𝑓(𝑠,𝑦(𝑠),𝐷𝑝 0+𝑦(𝑠))𝑑𝑠 +𝑐𝑘𝑡𝛼−1 +𝑑𝑘𝑡𝛼−2,𝑡∈(𝑡 𝑘,𝑡𝑘+1],𝑘=0,1. (21) From lim𝑡→0𝑡2−𝛼𝑥(𝑡) = ∫1 0𝑢(𝑠)𝐺(𝑠,𝑦(𝑠),𝐷𝑝 0+𝑦(𝑠))𝑑𝑠,we get 𝑑0=∫1 0𝑢(𝑠)𝐺(𝑠,𝑦(𝑠),𝐷𝑝 0+𝑦(𝑠))𝑑𝑠. (22) From 𝑥(1)=∫1 0𝑚(𝑠)𝑀(𝑠,𝑦(𝑠),𝐷𝑝 0+𝑦(𝑠))𝑑𝑠,weget 1 Γ(𝛼)∫1 0(1−𝑠)𝛼−1𝜙(𝑠)𝑓(𝑠,𝑦(𝑠),𝐷𝑝 0+𝑦(𝑠))𝑑𝑠+𝑐1+𝑑1 =∫1 0𝑚(𝑠)𝑀(𝑠,𝑦(𝑠),𝐷𝑝 0+𝑦(𝑠))𝑑𝑠. (23) From Δ𝑥(𝑡1)=𝐼(𝑡1,𝑦(𝑡1),𝐷𝑝 0+𝑦(𝑡1)),weget (𝑐1−𝑐0)𝑡𝛼−1 1+(𝑑1−𝑑0)𝑡𝛼−2 1=𝐼(𝑡1,𝑦(𝑡1),𝐷𝑝 0+𝑦(𝑡1)). (24) Mathematical Problems in Engineering 5 From Δ𝐷𝑞 0+𝑥(𝑡1)=𝐼1(𝑡1,𝑦(𝑡1),𝐷𝑝 0+𝑦(𝑡1)),weget (𝑐1−𝑐0)Γ(𝛼) Γ(𝛼−𝑞)𝑡𝛼−𝑞−1 1+(𝑑1−𝑑0)Γ(𝛼−1) Γ(𝛼−𝑞−2)𝑡𝛼−𝑞−2 1 =𝐼1(𝑡1,𝑦(𝑡1),𝐷𝑝 0+𝑦(𝑡1)). (25) It follows that 𝑐1−𝑐0=( Γ(𝛼−1) Γ(𝛼−𝑞−2)𝑡𝛼−𝑞−2 1𝐼(𝑡1,𝑦(𝑡1),𝐷𝑝 0+𝑦(𝑡1)) −𝑡𝛼−2 1𝐼1(𝑡1,𝑦(𝑡1),𝐷𝑝 0+𝑦(𝑡1)))×(Π)−1, 𝑑1−𝑑0=(𝑡 𝛼−1 1𝐼1(𝑡1,𝑦(𝑡1),𝐷𝑝 0+𝑦(𝑡1)) −Γ(𝛼) Γ(𝛼−𝑞)𝑡𝛼−𝑞−1 1𝐼(𝑡1,𝑦(𝑡1),𝐷𝑝 0+𝑦(𝑡1))) ×(Π)−1.(26) Then 𝑑1=(𝑡 𝛼−1 1𝐼1(𝑡1,𝑦(𝑡1),𝐷𝑝 0+𝑦(𝑡1)) −Γ(𝛼) Γ(𝛼−𝑞)𝑡𝛼−𝑞−1 1𝐼(𝑡1,𝑦(𝑡1),𝐷𝑝 0+𝑦(𝑡1)))×(Π)−1 +∫1 0𝑢(𝑠)𝐺(𝑠,𝑦(𝑠),𝐷𝑝 0+𝑦(𝑠))𝑑𝑠. (27) So 𝑐1=∫1 0𝑚(𝑠)𝑀(𝑠,𝑦(𝑠),𝐷𝑝 0+𝑦(𝑠))𝑑𝑠 −1 Γ(𝛼)∫1 0(1−𝑠)𝛼−1𝜙(𝑠)𝑓(𝑠,𝑦(𝑠),𝐷𝑝 0+𝑦(𝑠))𝑑𝑠 −(𝑡𝛼−1 1𝐼1(𝑡1,𝑦(𝑡1),𝐷𝑝 0+𝑦(𝑡1)) −Γ(𝛼) Γ(𝛼−𝑞)𝑡𝛼−𝑞−1 1𝐼(𝑡1,𝑦(𝑡1),𝐷𝑝 0+𝑦(𝑡1)))×(Π)−1 −∫1 0𝑢(𝑠)𝐺(𝑠,𝑦(𝑠),𝐷𝑝 0+𝑦(𝑠))𝑑𝑠, 𝑐0=∫1 0𝑚(𝑠)𝑀(𝑠,𝑦(𝑠),𝐷𝑝 0+𝑦(𝑠))𝑑𝑠 −1 Γ(𝛼)∫1 0(1−𝑠)𝛼−1𝜙(𝑠)𝑓(𝑠,𝑦(𝑠),𝐷𝑝 0+𝑦(𝑠))𝑑𝑠 −(𝑡𝛼−1 1𝐼1(𝑡1,𝑦(𝑡1),𝐷𝑝 0+𝑦(𝑡1)) −Γ(𝛼) Γ(𝛼−𝑞)𝑡𝛼−𝑞−1 1𝐼(𝑡1,𝑦(𝑡1),𝐷𝑝 0+𝑦(𝑡1)))×(Π)−1 −∫1 0𝑢(𝑠)𝐺(𝑠,𝑦(𝑠),𝐷𝑝 0+𝑦(𝑠))𝑑𝑠 −( Γ(𝛼−1) Γ(𝛼−𝑞−2)𝑡𝛼−𝑞−2 1𝐼(𝑡1,𝑦(𝑡1),𝐷𝑝 0+𝑦(𝑡1)) −𝑡𝛼−2 1𝐼1(𝑡1,𝑦(𝑡1),𝐷𝑝 0+𝑦(𝑡1)))×(Π)−1. (28) Hence, for 𝑡∈(0,𝑡1],wehave 𝑥(𝑡)=∫𝑡 0(𝑡−𝑠)𝛼−1 Γ(𝛼)𝜙(𝑢)𝑓(𝑢,𝑦(𝑢),𝐷𝑝 0+𝑦(𝑢))𝑑𝑢 −𝑡𝛼−1 Γ(𝛼)∫1 0(1−𝑠)𝛼−1𝜙(𝑠)𝑓(𝑠,𝑦(𝑠),𝐷𝑝 0+𝑦(𝑠))𝑑𝑠 +𝑡𝛼−2 ∫1 0𝑢(𝑠)𝐺(𝑠,𝑦(𝑠),𝐷𝑝 0+𝑦(𝑠))𝑑𝑠 +𝑡𝛼−1 ∫1 0𝑚(𝑠)𝑀(𝑠,𝑦(𝑠),𝐷𝑝 0+𝑦(𝑠))𝑑𝑠 +𝑡𝛼−1 Π(Γ(𝛼) Γ(𝛼−𝑞)𝑡𝛼−𝑞−1 1−Γ(𝛼−1) Γ(𝛼−𝑞−2)𝑡𝛼−𝑞−2 1) ×𝐼(𝑡1,𝑦(𝑡1),𝐷𝑝 0+𝑦(𝑡1)) +𝑡𝛼−1 (𝑡𝛼−2 1−𝑡𝛼−1 1) Π𝐼1(𝑡1,𝑦(𝑡1),𝐷𝑝 0+𝑦(𝑡1)). (29) And for 𝑡∈(𝑡1,1],wehave 𝑥(𝑡)=∫𝑡 0(𝑡−𝑠)𝛼−1 Γ(𝛼)𝜙(𝑠)𝑓(𝑠,𝑦(𝑠),𝐷𝑝 0+𝑦(𝑠))𝑑𝑠 −𝑡𝛼−1 Γ(𝛼)∫1 0(1−𝑠)𝛼−1𝜙(𝑠)𝑓(𝑠,𝑦(𝑠),𝐷𝑝 0+𝑦(𝑠))𝑑𝑠 +(𝑡𝛼−2 −𝑡𝛼−1)∫1 0𝑢(𝑠)𝐺(𝑠,𝑦(𝑠),𝐷𝑝 0+𝑦(𝑠))𝑑𝑠 +𝑡𝛼−1 ∫1 0𝑚(𝑠)𝑀(𝑠,𝑦(𝑠),𝐷𝑝 0+𝑦(𝑠))𝑑𝑠 +𝑡𝛼−1 −𝑡𝛼−2 ΠΓ(𝛼) Γ(𝛼−𝑞)𝑡𝛼−𝑞−1 1 ×𝐼(𝑡1,𝑦(𝑡1),𝐷𝑝 0+𝑦(𝑡1)) +𝑡𝛼−2 −𝑡𝛼−1 Π𝑡𝛼−1 1𝐼1(𝑡1,𝑦(𝑡1),𝐷𝑝 0+𝑦(𝑡1)). (30) Hence, 𝑥∈𝑋satisfies (16). On the other hand, if 𝑦∈𝑌and 𝑥∈𝑋is a solution of (16), then we can prove that 𝑥∈𝑋is a solution of BVP (6)-(7). The proof is completed. 6Mathematical Problems in Engineering Lemma 8. Suppose that 𝑥∈𝑋, and (a)–(e), (A)-(B) hold. Then 𝑦∈𝑌is a solution of 𝐷𝛽 0+𝑦(𝑡)=𝜓(𝑡)𝑔(𝑡,𝑥(𝑡),𝐷𝑞 0+𝑥(𝑡)), 𝑡∈(0,1),𝑡=𝑡1, lim 𝑡→0𝑡2−𝛽𝑦(𝑡)=∫1 0 V(𝑠)𝐻(𝑠,𝑥(𝑠),𝐷𝑞 0+𝑥(𝑠))𝑑𝑠, 𝑦(1)=∫1 0𝑛(𝑠)𝑁(𝑠,𝑥(𝑠),𝐷𝑞 0+𝑥(𝑠))𝑑𝑠, Δ𝑦(𝑡1)=𝐽(𝑡1,𝑥(𝑡1),𝐷𝑞 0+𝑥(𝑡1)), Δ𝐷𝑝 0+𝑦(𝑡1)=𝐽1(𝑡1,𝑥(𝑡1),𝐷𝑞 0+𝑥(𝑡1)), (31) if and only if 𝑦∈𝑌satisfies the integral equation 𝑦(𝑡)= { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { ∫𝑡 0(𝑡−𝑠)𝛽−1 Γ(𝛽) 𝜓(𝑢)𝑔(𝑢,𝑥(𝑢),𝐷𝑞 0+𝑥(𝑢))𝑑𝑢 −𝑡𝛽−1 Γ(𝛽) ×∫1 0(1−𝑠)𝛽−1𝜓(𝑠)𝑔(𝑠,𝑥(𝑠),𝐷𝑞 0+𝑥(𝑠))𝑑𝑠 +𝑡𝛽−2 ×∫1 0 V(𝑠)𝐻(𝑠,𝑦(𝑠),𝐷𝑝 0+𝑦(𝑠))𝑑𝑠 +𝑡𝛽−1 ×∫1 0𝑛(𝑠)𝑁(𝑠,𝑥(𝑠),𝐷𝑞 0+𝑥(𝑠))𝑑𝑠 +𝑡𝛽−1 Ξ ×( Γ(𝛽) Γ(𝛽−𝑝)𝑡𝛽−𝑝−1 1−Γ(𝛽−1) Γ(𝛽−𝑝−2)𝑡𝛽−𝑝−2 1) ×𝐽(𝑡1,𝑥(𝑡1),𝐷𝑞 0+𝑥(𝑡1)) +𝑡𝛽−1 (𝑡𝛽−2 1−𝑡𝛽−1 1) Ξ ×𝐽1(𝑡1,𝑥(𝑡1),𝐷𝑞 0+𝑥(𝑡1)), 𝑡∈(0,𝑡1], ∫𝑡 0(𝑡−𝑠)𝛽−1 Γ(𝛽) 𝜓(𝑠)𝑔(𝑠,𝑥(𝑠),𝐷𝑞 0+𝑥(𝑠))𝑑𝑠 −𝑡𝛽−1 Γ(𝛽) ×∫1 0(1−𝑠)𝛽−1𝜓(𝑠)𝑔(𝑠,𝑥(𝑠),𝐷𝑞 0+𝑥(𝑠))𝑑𝑠 +(𝑡𝛽−2 −𝑡𝛽−1) ×∫1 0 V(𝑠)𝐻(𝑠,𝑦(𝑠),𝐷𝑝 0+𝑦(𝑠))𝑑𝑠 +𝑡𝛽−1 ×∫1 0𝑛(𝑠)𝑁(𝑠,𝑥(𝑠),𝐷𝑞 0+𝑥(𝑠))𝑑𝑠 +𝑡𝛽−1 −𝑡𝛽−2 ΞΓ(𝛽) Γ(𝛽−𝑝)𝑡𝛽−𝑝−1 1 ×𝐽(𝑡1,𝑥(𝑡1),𝐷𝑞 0+𝑥(𝑡1)) +𝑡𝛽−2 −𝑡𝛽−1 Ξ𝑡𝛽−1 1 ×𝐽1(𝑡1,𝑥(𝑡1),𝐷𝑞 0+𝑥(𝑡1)), 𝑡∈(𝑡1,1], (32) where Ξ=( Γ(𝛽−1) Γ(𝛽−𝑝−1)−Γ(𝛽) Γ(𝛽−𝑝))𝑡2𝛽−𝑝−3 1.(33) Proof. The proof is similar to that of the proof of Lemma 7 and is omitted. Now, we define the operator 𝑇on 𝑋×𝑌by 𝑇(𝑥,𝑦)(𝑡)= ((𝑇1𝑦)(𝑡),(𝑇2𝑥)(𝑡))with (𝑇1𝑦)(𝑡) = { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { ∫𝑡 0(𝑡−𝑠)𝛼−1 Γ(𝛼)𝜙(𝑢)𝑓(𝑢,𝑦(𝑢),𝐷𝑝 0+𝑦(𝑢))𝑑𝑢 −𝑡𝛼−1 Γ(𝛼)∫1 0(1−𝑠)𝛼−1𝜙(𝑠)𝑓(𝑠,𝑦(𝑠),𝐷𝑝 0+𝑦(𝑠))𝑑𝑠 +𝑡𝛼−2 ∫1 0𝑢(𝑠)𝐺(𝑠,𝑦(𝑠),𝐷𝑝 0+𝑦(𝑠))𝑑𝑠 +𝑡𝛼−1 ∫1 0𝑚(𝑠)𝑀(𝑠,𝑦(𝑠),𝐷𝑝 0+𝑦(𝑠))𝑑𝑠 +𝑡𝛼−1 Π(Γ(𝛼) Γ(𝛼−𝑞)𝑡𝛼−𝑞−1 1−Γ(𝛼−1) Γ(𝛼−𝑞−2)𝑡𝛼−𝑞−2 1) ×𝐼(𝑡1,𝑦(𝑡1),𝐷𝑝 0+𝑦(𝑡1)) +𝑡𝛼−1 (𝑡𝛼−2 1−𝑡𝛼−1 1) Π ×𝐼1(𝑡1,𝑦(𝑡1),𝐷𝑝 0+𝑦(𝑡1)), 𝑡∈(0,𝑡1], ∫𝑡 0(𝑡−𝑠)𝛼−1 Γ(𝛼)𝜙(𝑠)𝑓(𝑠,𝑦(𝑠),𝐷𝑝 0+𝑦(𝑠))𝑑𝑠 −𝑡𝛼−1 Γ(𝛼)∫1 0(1−𝑠)𝛼−1𝜙(𝑠)𝑓(𝑠,𝑦(𝑠),𝐷𝑝 0+𝑦(𝑠))𝑑𝑠 +(𝑡𝛼−2 −𝑡𝛼−1)∫1 0𝑢(𝑠)𝐺(𝑠,𝑦(𝑠),𝐷𝑝 0+𝑦(𝑠))𝑑𝑠 +𝑡𝛼−1 ∫1 0𝑚(𝑠)𝑀(𝑠,𝑦(𝑠),𝐷𝑝 0+𝑦(𝑠))𝑑𝑠 +𝑡𝛼−1 −𝑡𝛼−2 ΠΓ(𝛼) Γ(𝛼−𝑞)𝑡𝛼−𝑞−1 1𝐼(𝑡1,𝑦(𝑡1),𝐷𝑝 0+𝑦(𝑡1)) +𝑡𝛼−2 −𝑡𝛼−1 Π𝑡𝛼−1 1𝐼1(𝑡1,𝑦(𝑡1),𝐷𝑝 0+𝑦(𝑡1)), 𝑡∈(𝑡1,1], (34) Mathematical Problems in Engineering 7 (𝑇2𝑥)(𝑡) = { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { ∫𝑡 0(𝑡−𝑠)𝛽−1 Γ(𝛽) 𝜓(𝑢)𝑔(𝑢,𝑥(𝑢),𝐷𝑞 0+𝑥(𝑢))𝑑𝑢 −𝑡𝛽−1 Γ(𝛽)∫1 0(1−𝑠)𝛽−1𝜓(𝑠)𝑔(𝑠,𝑥(𝑠),𝐷𝑞 0+𝑥(𝑠))𝑑𝑠 +𝑡𝛽−2 ∫1 0 V(𝑠)𝐻(𝑠,𝑦(𝑠),𝐷𝑝 0+𝑦(𝑠))𝑑𝑠 +𝑡𝛽−1 ∫1 0𝑛(𝑠)𝑁(𝑠,𝑥(𝑠),𝐷𝑞 0+𝑥(𝑠))𝑑𝑠 +𝑡𝛽−1 Ξ(Γ(𝛽) Γ(𝛽−𝑝)𝑡𝛽−𝑝−1 1−Γ(𝛽−1) Γ(𝛽−𝑝−2)𝑡𝛽−𝑝−2 1) ×𝐽(𝑡1,𝑥(𝑡1),𝐷𝑞 0+𝑥(𝑡1)) +𝑡𝛽−1 (𝑡𝛽−2 1−𝑡𝛽−1 1) Ξ𝐽1(𝑡1,𝑥(𝑡1),𝐷𝑞 0+𝑥(𝑡1)), 𝑡∈(0,𝑡1], ∫𝑡 0(𝑡−𝑠)𝛽−1 Γ(𝛽) 𝜓(𝑠)𝑔(𝑠,𝑥(𝑠),𝐷𝑞 0+𝑥(𝑠))𝑑𝑠 −𝑡𝛽−1 Γ(𝛽)∫1 0(1−𝑠)𝛽−1𝜓(𝑠)𝑔(𝑠,𝑥(𝑠),𝐷𝑞 0+𝑥(𝑠))𝑑𝑠 +(𝑡𝛽−2 −𝑡𝛽−1)∫1 0 V(𝑠)𝐻(𝑠,𝑦(𝑠),𝐷𝑝 0+𝑦(𝑠))𝑑𝑠 +𝑡𝛽−1 ∫1 0𝑛(𝑠)𝑁(𝑠,𝑥(𝑠),𝐷𝑞 0+𝑥(𝑠))𝑑𝑠 +𝑡𝛽−1 −𝑡𝛽−2 ΞΓ(𝛽) Γ(𝛽−𝑝)𝑡𝛽−𝑝−1 1 ×𝐽(𝑡1,𝑥(𝑡1),𝐷𝑞 0+𝑥(𝑡1)) +𝑡𝛽−2 −𝑡𝛽−1 Ξ𝑡𝛽−1 1𝐽1(𝑡1,𝑥(𝑡1),𝐷𝑞 0+𝑥(𝑡1)), 𝑡∈(𝑡1,1]. (35) Remark 9. By Lemmas 7and 8,(𝑥,𝑦)∈𝑋×𝑌is a solution of BVP (6)-(7)ifandonlyif(𝑥,𝑦)∈𝑋×𝑌is a fixed point of the operator 𝑇. Lemma 10. Suppose that (a)–(e) and (A)-(B) hold. Then 𝑇: 𝑋×𝑌→𝑋×𝑌is well defined and is completely continuous. Proof. The proof is very long, so we list the steps. First, we prove that 𝑇is well defined; second, we prove that 𝑇is continuous, and, finally, we prove that 𝑇is compact. So 𝑇is completely continuous. Thus, the proof is divided into three steps. Step 1. Prove that 𝑇:𝑋×𝑌→𝑋×𝑌is well defined. For (𝑥,𝑦)∈𝑋×𝑌,wehave||(𝑥,𝑦)||=𝑟>0.Then max {sup 𝑡∈(0,1)𝑡2−𝛼 |𝑥(𝑡)|,sup 𝑡∈(0,1)𝑡2+𝑞−𝛼 󵄨󵄨󵄨󵄨󵄨𝐷𝑞 0+𝑥(𝑡)󵄨󵄨󵄨󵄨󵄨}≤𝑟<+∞, max {sup 𝑡∈(0,1)𝑡2−𝛽 󵄨󵄨󵄨󵄨𝑦(𝑡)󵄨󵄨󵄨󵄨,sup 𝑡∈(0,1)𝑡2+𝑝−𝛽 󵄨󵄨󵄨󵄨󵄨𝐷𝑝 0+𝑦(𝑡)󵄨󵄨󵄨󵄨󵄨}≤𝑟<+∞. (36) From (B), 𝑓,𝐺,𝑀,𝐼,𝐼1are 𝛽-Caratheodory functions, then there exist constants 𝐴𝑟>0such that 󵄨󵄨󵄨󵄨󵄨𝑓(𝑡,𝑦(𝑡),𝐷𝑝 0+𝑦(𝑡))󵄨󵄨󵄨󵄨󵄨≤𝐴𝑟,𝑡∈ (0,1), 󵄨󵄨󵄨󵄨󵄨𝐺(𝑡,𝑦(𝑡),𝐷𝑝 0+𝑦(𝑡))󵄨󵄨󵄨󵄨󵄨≤𝐴𝑟,𝑡∈ (0,1), 󵄨󵄨󵄨󵄨󵄨𝑀(𝑡,𝑦(𝑡),𝐷𝑝 0+𝑦(𝑡))󵄨󵄨󵄨󵄨󵄨≤𝐴𝑟,𝑡∈ (0,1), 󵄨󵄨󵄨󵄨󵄨𝐼(𝑡1,𝑦(𝑡1),𝐷𝑝 0+𝑦(𝑡1))󵄨󵄨󵄨󵄨󵄨≤𝐴𝑟,𝑡∈ (0,1), 󵄨󵄨󵄨󵄨󵄨𝐼1(𝑡1,𝑦(𝑡1),𝐷𝑝 0+𝑦(𝑡1))󵄨󵄨󵄨󵄨󵄨≤𝐴𝑟,𝑡∈ (0,1). (37) Hence, 󵄨󵄨󵄨󵄨󵄨󵄨󵄨󵄨󵄨∫𝑡 0(𝑡−𝑠)𝛼−1 Γ(𝛼)𝜙(𝑢)𝑓(𝑢,𝑦(𝑢),𝐷𝑝 0+𝑦(𝑢))𝑑𝑢󵄨󵄨󵄨󵄨󵄨󵄨󵄨󵄨󵄨 ≤∫𝑡 0(𝑡−𝑠)𝛼−1 Γ(𝛼)󵄨󵄨󵄨󵄨󵄨𝜙(𝑢)𝑓(𝑢,𝑦(𝑢),𝐷𝑝 0+𝑦(𝑢))󵄨󵄨󵄨󵄨󵄨𝑑𝑢 ≤𝐴𝑟𝐿1 B(𝛼+𝛿,𝑘+1) Γ(𝛼)<∞. (38) From (34), (37), and (38), we see that (𝑇1𝑦)(𝑡)is defined on (0,1],continuouson(0,𝑡1]and (𝑡1,1],respectively.Onesees that lim 𝑡→0𝑡2−𝛼 (𝑇1𝑦)(𝑡) =lim 𝑡→0[𝑡2−𝛼 ∫𝑡 0(𝑡−𝑠)𝛼−1 Γ(𝛼)𝜙(𝑢)𝑓(𝑢,𝑦(𝑢),𝐷𝑝 0+𝑦(𝑢))𝑑𝑢 −𝑡 Γ(𝛼)∫1 0(1−𝑠)𝛼−1𝜙(𝑠)𝑓(𝑠,𝑦(𝑠),𝐷𝑝 0+𝑦(𝑠))𝑑𝑠 +∫1 0𝑢(𝑠)𝐺(𝑠,𝑦(𝑠),𝐷𝑝 0+𝑦(𝑠))𝑑𝑠 +𝑡∫1 0𝑚(𝑠)𝑀(𝑠,𝑦(𝑠),𝐷𝑝 0+𝑦(𝑠))𝑑𝑠 +𝑡 Π(Γ(𝛼) Γ(𝛼−𝑞)𝑡𝛼−𝑞−1 1−Γ(𝛼−1) Γ(𝛼−𝑞−2)𝑡𝛼−𝑞−2 1) ×𝐼(𝑡1,𝑦(𝑡1),𝐷𝑝 0+𝑦(𝑡1)) +𝑡(𝑡𝛼−2 1−𝑡𝛼−1 1) Π𝐼1(𝑡1,𝑦(𝑡1),𝐷𝑝 0+𝑦(𝑡1))] =∫1 0𝑢(𝑠)𝐺(𝑠,𝑦(𝑠),𝐷𝑝 0+𝑦(𝑠))𝑑𝑠, (39) andthereexitsthelimitlim 𝑡→𝑡+ 1(𝑇1𝑦)(𝑡). 8Mathematical Problems in Engineering On the other hand, we have 𝐷𝑞 0+(𝑇1𝑦)(𝑡) = { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { ∫𝑡 0(𝑡−𝑠)𝛼−𝑞−1 Γ(𝛼−𝑞) 𝜙(𝑢)𝑓(𝑢,𝑦(𝑢),𝐷𝑝 0+𝑦(𝑢))𝑑𝑢 −𝑡𝛼−𝑞−1 Γ(𝛼−𝑞) ×∫1 0(1−𝑠)𝛼−1𝜙(𝑠)𝑓(𝑠,𝑦(𝑠),𝐷𝑝 0+𝑦(𝑠))𝑑𝑠 +𝑡𝛼−𝑞−2 Γ(𝛼−1) Γ(𝛼−𝑞−1) ×∫1 0𝑢(𝑠)𝐺(𝑠,𝑦(𝑠),𝐷𝑝 0+𝑦(𝑠))𝑑𝑠 +𝑡𝛼−𝑞−1 Γ(𝛼) Γ(𝛼−𝑞) ×∫1 0𝑚(𝑠)𝑀(𝑠,𝑦(𝑠),𝐷𝑝 0+𝑦(𝑠))𝑑𝑠 +𝑡𝛼−𝑞−1 ΠΓ(𝛼) Γ(𝛼−𝑞) ×( Γ(𝛼) Γ(𝛼−𝑞)𝑡𝛼−𝑞−1 1−Γ(𝛼−1) Γ(𝛼−𝑞−2)𝑡𝛼−𝑞−2 1) ×𝐼(𝑡1,𝑦(𝑡1),𝐷𝑝 0+𝑦(𝑡1)) +Γ(𝛼) Γ(𝛼−𝑞)𝑡𝛼−𝑞−1 (𝑡𝛼−2 1−𝑡𝛼−1 1) Π ×𝐼1(𝑡1,𝑦(𝑡1),𝐷𝑝 0+𝑦(𝑡1)), 𝑡∈(0,𝑡1], ∫𝑡 0(𝑡−𝑠)𝛼−𝑞−1 Γ(𝛼−𝑞) 𝜙(𝑠)𝑓(𝑠,𝑦(𝑠),𝐷𝑝 0+𝑦(𝑠))𝑑𝑠 −𝑡𝛼−𝑞−1 Γ(𝛼−𝑞) ×∫1 0(1−𝑠)𝛼−1𝜙(𝑠)𝑓(𝑠,𝑦(𝑠),𝐷𝑝 0+𝑦(𝑠))𝑑𝑠 +(𝑡𝛼−𝑞−2 Γ(𝛼−1) Γ(𝛼−𝑞−1)−𝑡𝛼−𝑞−1 Γ(𝛼) Γ(𝛼−𝑞)) ×∫1 0𝑢(𝑠)𝐺(𝑠,𝑦(𝑠),𝐷𝑝 0+𝑦(𝑠))𝑑𝑠 +𝑡𝛼−𝑞−1 Γ(𝛼) Γ(𝛼−𝑞)∫1 0𝑚(𝑠)𝑀(𝑠,𝑦(𝑠),𝐷𝑝 0+𝑦(𝑠))𝑑𝑠 +1 Π(𝑡𝛼−𝑞−1 Γ(𝛼) Γ(𝛼−𝑞)−𝑡𝛼−𝑞−2 Γ(𝛼−1) Γ(𝛼−𝑞−1)) ×Γ(𝛼) Γ(𝛼−𝑞)𝑡𝛼−𝑞−1 1𝐼(𝑡1,𝑦(𝑡1),𝐷𝑝 0+𝑦(𝑡1)) +1 Π(𝑡𝛼−𝑞−2 Γ(𝛼−1) Γ(𝛼−𝑞−1)−𝑡𝛼−𝑞−1 Γ(𝛼) Γ(𝛼−𝑞)) ×𝑡𝛼−1 1𝐼1(𝑡1,𝑦(𝑡1),𝐷𝑝 0+𝑦(𝑡1)), 𝑡∈(𝑡1,1], 𝐷𝑝 0+(𝑇2𝑥)(𝑡) = { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { { ∫𝑡 0(𝑡−𝑠)𝛽−𝑝−1 Γ(𝛽−𝑝)𝜓(𝑢)𝑔(𝑢,𝑥(𝑢),𝐷𝑞 0+𝑥(𝑢))𝑑𝑢 −𝑡𝛽−𝑝−1 Γ(𝛽−𝑝) ×∫1 0(1−𝑠)𝛽−1𝜓(𝑠)𝑔(𝑠,𝑥(𝑠),𝐷𝑞 0+𝑥(𝑠))𝑑𝑠 +𝑡𝛽−𝑝−2 Γ(𝛽−1) Γ(𝛽−𝑝−1) ×∫1 0 V(𝑠)𝐻(𝑠,𝑦(𝑠),𝐷𝑝 0+𝑦(𝑠))𝑑𝑠 +𝑡𝛽−𝑝−1 Γ(𝛽) Γ(𝛽−𝑝) ×∫1 0𝑛(𝑠)𝑁(𝑠,𝑥(𝑠),𝐷𝑞 0+𝑥(𝑠))𝑑𝑠 +𝑡𝛽−𝑝−1 ΞΓ(𝛽) Γ(𝛽−𝑝) ×( Γ(𝛽) Γ(𝛽−𝑝)𝑡𝛽−𝑝−1 1−Γ(𝛽−1) Γ(𝛽−𝑝−2)𝑡𝛽−𝑝−2 1) ×𝐽(𝑡1,𝑥(𝑡1),𝐷𝑞 0+𝑥(𝑡1)) +𝑡𝛽−𝑝−1 (𝑡𝛽−2 1−𝑡𝛽−1 1) ΞΓ(𝛽) Γ(𝛽−𝑝) ×𝐽1(𝑡1,𝑥(𝑡1),𝐷𝑞 0+𝑥(𝑡1)), 𝑡∈(0,𝑡1], ∫𝑡 0(𝑡−𝑠)𝛽−𝑝−1 Γ(𝛽−𝑝)𝜓(𝑠)𝑔(𝑠,𝑥(𝑠),𝐷𝑞 0+𝑥(𝑠))𝑑𝑠 −𝑡𝛽−𝑝−1 Γ(𝛽) Γ(𝛽) Γ(𝛽−𝑝) ×∫1 0(1−𝑠)𝛽−1𝜓(𝑠)𝑔(𝑠,𝑥(𝑠),𝐷𝑞 0+𝑥(𝑠))𝑑𝑠 +(𝑡𝛽−𝑝−2 Γ(𝛽−1) Γ(𝛽−𝑝−1)−𝑡𝛽−𝑝−1 Γ(𝛽) Γ(𝛽−𝑝)) ×∫1 0 V(𝑠)𝐻(𝑠,𝑦(𝑠),𝐷𝑝 0+𝑦(𝑠))𝑑𝑠 +𝑡𝛽−𝑝−1 Γ(𝛽) Γ(𝛽−𝑝)∫1 0𝑛(𝑠)𝑁(𝑠,𝑥(𝑠),𝐷𝑞 0+𝑥(𝑠))𝑑𝑠 +1 Ξ(𝑡𝛽−𝑝−1 Γ(𝛽) Γ(𝛽−𝑝)−𝑡𝛽−𝑝−2 Γ(𝛽−1) Γ(𝛽−𝑝−1)) ×Γ(𝛽) Γ(𝛽−𝑝)𝑡𝛽−𝑝−1 1𝐽(𝑡1,𝑥(𝑡1),𝐷𝑞 0+𝑥(𝑡1)) +1 Ξ(𝑡𝛽−𝑝−2 Γ(𝛽−1) Γ(𝛽−𝑝−1)−𝑡𝛽−𝑝−1 Γ(𝛽) Γ(𝛽−𝑝)) ×𝑡𝛽−1 1𝐽1(𝑡1,𝑥(𝑡1),𝐷𝑞 0+𝑥(𝑡1)), 𝑡∈(𝑡1,1]. (40) Mathematical Problems in Engineering 9 It is easy to see that 󵄨󵄨󵄨󵄨󵄨󵄨󵄨󵄨󵄨∫𝑡 0(𝑡−𝑠)𝛼−𝑞−1 Γ(𝛼−𝑞)𝜙(𝑢)𝑓(𝑢,𝑦(𝑢),𝐷𝑝 0+𝑦(𝑢))𝑑𝑢󵄨󵄨󵄨󵄨󵄨󵄨󵄨󵄨󵄨 ≤𝐴𝑟𝐿1 B(𝛼+𝛿−𝑞,𝑘+1) Γ(𝛼−𝑞) <∞. (41) From (37)and(41), we see that 𝐷𝑞 0+(𝑇1𝑦)(𝑡)is defined on (0,1],continuouson(0,𝑡1]and (𝑡1,1],respectively.Onesees that lim 𝑡→0𝑡2+𝑞−𝛼𝐷𝑞 0+(𝑇1𝑦)(𝑡) =lim 𝑡→0[𝑡2+𝑞−𝛼 ×∫𝑡 0(𝑡−𝑠)𝛼−𝑞−1 Γ(𝛼−𝑞)𝜙(𝑢)𝑓(𝑢,𝑦(𝑢),𝐷𝑝 0+𝑦(𝑢))𝑑𝑢 −𝑡 Γ(𝛼−𝑞) ×∫1 0(1−𝑠)𝛼−1𝜙(𝑠)𝑓(𝑠,𝑦(𝑠),𝐷𝑝 0+𝑦(𝑠))𝑑𝑠 +Γ(𝛼−1) Γ(𝛼−𝑞−1)∫1 0𝑢(𝑠)𝐺(𝑠,𝑦(𝑠),𝐷𝑝 0+𝑦(𝑠))𝑑𝑠 +𝑡 Γ(𝛼) Γ(𝛼−𝑞)∫1 0𝑚(𝑠)𝑀(𝑠,𝑦(𝑠),𝐷𝑝 0+𝑦(𝑠))𝑑𝑠 +𝑡 ΠΓ(𝛼) Γ(𝛼−𝑞) ×( Γ(𝛼) Γ(𝛼−𝑞)𝑡𝛼−𝑞−1 1−Γ(𝛼−1) Γ(𝛼−𝑞−2)𝑡𝛼−𝑞−2 1) ×𝐼(𝑡1,𝑦(𝑡1),𝐷𝑝 0+𝑦(𝑡1)) +Γ(𝛼) Γ(𝛼−𝑞)𝑡(𝑡𝛼−2 1−𝑡𝛼−1 1) Π ×𝐼1(𝑡1,𝑦(𝑡1),𝐷𝑝 0+𝑦(𝑡1))] =Γ(𝛼−1) Γ(𝛼−𝑞−1)∫1 0𝑢(𝑠)𝐺(𝑠,𝑦(𝑠),𝐷𝑝 0+𝑦(𝑠))𝑑𝑠, (42) and there exits the limit lim𝑡→𝑡+ 1𝐷𝑞 0+(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×𝑌is well defined. Step 2. We prove that 𝑇is continuous. Let (𝑥𝑛,𝑦𝑛)∈𝑋× 𝑌with (𝑥𝑛,𝑦𝑛)→(𝑥 0,𝑦0)as 𝑛→∞.Wewillshowthat 𝑇(𝑥𝑛,𝑦𝑛)→𝑇(𝑥 0,𝑦0)as 𝑛→∞,thatis,provethat𝑇1𝑦𝑛→ 𝑇1𝑦0and 𝑇2𝑥𝑛→𝑇 2𝑥0as 𝑛→∞. In fact, we have 𝑟>0such that ||(𝑥𝑛,𝑦𝑛)||=𝑟>0.Then max {sup 𝑡∈(0,1)𝑡2−𝛼 󵄨󵄨󵄨󵄨𝑥𝑛(𝑡)󵄨󵄨󵄨󵄨,sup 𝑡∈(0,1)𝑡2+𝑞−𝛼 󵄨󵄨󵄨󵄨󵄨𝐷𝑞 0+𝑥𝑛(𝑡)󵄨󵄨󵄨󵄨󵄨} ≤𝑟<+∞, 𝑛=0,1,2,..., max {sup 𝑡∈(0,1)𝑡2−𝛽 󵄨󵄨󵄨󵄨𝑦𝑛(𝑡)󵄨󵄨󵄨󵄨,sup 𝑡∈(0,1)𝑡2+𝑝−𝛽 󵄨󵄨󵄨󵄨󵄨𝐷𝑝 0+𝑦𝑛(𝑡)󵄨󵄨󵄨󵄨󵄨} ≤𝑟<+∞, 𝑛=0,1,2,.... (43) From (B), 𝑓,𝐺,𝑀,𝐼,𝐼1are 𝛽-Caratheodory functions, then there exist constants 𝐴𝑟>0such that 󵄨󵄨󵄨󵄨󵄨𝑓(𝑡,𝑦𝑛(𝑡),𝐷𝑝 0+𝑦𝑛(𝑡))󵄨󵄨󵄨󵄨󵄨≤𝐴𝑟, 𝑡∈(0,1), 𝑛=0,1,2,..., 󵄨󵄨󵄨󵄨󵄨𝐺(𝑡,𝑦𝑛(𝑡),𝐷𝑝 0+𝑦𝑛(𝑡))󵄨󵄨󵄨󵄨󵄨≤𝐴𝑟, 𝑡∈(0,1), 𝑛=0,1,2,..., 󵄨󵄨󵄨󵄨󵄨𝑀(𝑡,𝑦𝑛(𝑡),𝐷𝑝 0+𝑦𝑛(𝑡))󵄨󵄨󵄨󵄨󵄨≤𝐴𝑟, 𝑡∈(0,1), 𝑛=0,1,2,..., 󵄨󵄨󵄨󵄨󵄨𝐼(𝑡1,𝑦𝑛(𝑡1),𝐷𝑝 0+𝑦𝑛(𝑡1))󵄨󵄨󵄨󵄨󵄨≤𝐴𝑟, 𝑡∈(0,1), 𝑛=0,1,2,..., 󵄨󵄨󵄨󵄨󵄨𝐼1(𝑡1,𝑦𝑛(𝑡1),𝐷𝑝 0+𝑦𝑛(𝑡1))󵄨󵄨󵄨󵄨󵄨≤𝐴𝑟, 𝑡∈(0,1), 𝑛=0,1,2,..., sup 𝑡∈(0,1)𝑡2−𝛼 󵄨󵄨󵄨󵄨𝑥𝑛(𝑡)−𝑥0(𝑡)󵄨󵄨󵄨󵄨󳨀→ 0, sup 𝑡∈(0,1)𝑡2−𝛽 󵄨󵄨󵄨󵄨𝑦𝑛(𝑡)−𝑦0(𝑡)󵄨󵄨󵄨󵄨, sup 𝑡∈(0,1)𝑡2+𝑞−𝛼 󵄨󵄨󵄨󵄨󵄨𝐷𝑞 0+𝑥𝑛(𝑡)−𝐷𝑞 0+𝑥0(𝑡)󵄨󵄨󵄨󵄨󵄨󳨀→ 0, sup 𝑡∈(0,1)𝑡2+𝑝−𝛽 󵄨󵄨󵄨󵄨󵄨𝐷𝑝 0+𝑦𝑛(𝑡)−𝐷𝑝 0+𝑦0(𝑡)󵄨󵄨󵄨󵄨󵄨󳨀→ 0, (44) 16 Mathematical Problems in Engineering +Γ(𝛽) Γ(𝛽−𝑝)‖𝑛‖1[𝐵𝑁+𝐴𝑁] +1 ΞΓ(𝛽) Γ(𝛽−𝑝) ×󵄨󵄨󵄨󵄨󵄨󵄨󵄨󵄨󵄨Γ(𝛽) Γ(𝛽−𝑝)𝑡𝛽−𝑝−1 1−Γ(𝛽−1) Γ(𝛽−𝑝−2)𝑡𝛽−𝑝−2 1󵄨󵄨󵄨󵄨󵄨󵄨󵄨󵄨󵄨[𝐵𝐽+𝐴𝐽] +Γ(𝛽) Γ(𝛽−𝑝)󵄨󵄨󵄨󵄨󵄨󵄨𝑡𝛽−2 1−𝑡𝛽−1 1󵄨󵄨󵄨󵄨󵄨󵄨 Ξ[𝐵1,𝐽 +𝐴1,𝐽], Λ1=𝐿2 B(𝛽+𝜃,𝑙+1) Γ(𝛽) 𝐶𝑔+𝐿2B(𝛽+𝜃,𝑙+1) Γ(𝛽) 𝐶𝑔 +‖V‖1𝐶𝐻+1 ΞΓ(𝛽) Γ(𝛽−𝑝)𝑡𝛽−𝑝−1 1𝐶𝐽+1 Ξ𝑡𝛽−1 1𝐶1,𝐽, Λ2=𝐿2 B(𝛽+𝜃,𝑙+1) Γ(𝛽) [𝐵𝑔+𝐴𝑔] +𝐿2B(𝛽+𝜃,𝑙+1) Γ(𝛽) [𝐵𝑔+𝐴𝑔]+‖V‖1[𝐵𝐻+𝐴𝐻] +1 ΞΓ(𝛽) Γ(𝛽−𝑝)𝑡𝛽−𝑝−1 1[𝐵𝐽+𝐴𝐽]+1 Ξ𝑡𝛽−1 1[𝐵1,𝐽 +𝐴1,𝐽], Λ3=𝐿2 B(𝛽+𝜃−𝑝,𝑙+1) Γ(𝛽−𝑝) 𝐶𝑔 +𝐿2 Γ(𝛽−𝑝)B(𝛽+𝜃,𝑙+1)𝐶𝑔 +( Γ(𝛽−1) Γ(𝛽−𝑝−1)+Γ(𝛽) Γ(𝛽−𝑝))‖V‖1𝐶𝐻 +Γ(𝛽) Γ(𝛽−𝑝)‖𝑛‖1𝐶𝑁 +1 Ξ(Γ(𝛽) Γ(𝛽−𝑝)+Γ(𝛽−1) Γ(𝛽−𝑝−1))Γ(𝛽) Γ(𝛽−𝑝)𝑡𝛽−𝑝−1 1𝐶𝐽 +1 Ξ(Γ(𝛽−1) Γ(𝛽−𝑝−1)+Γ(𝛽) Γ(𝛽−𝑝))𝑡𝛽−1 1𝐶1,𝐽, Λ4=𝐿2 B(𝛽+𝜃−𝑝,𝑙+1) Γ(𝛽−𝑝) [𝐵𝑔+𝐴𝑔] +𝐿2 Γ(𝛽−𝑝)B(𝛽+𝜃,𝑙+1)[𝐵𝑔+𝐴𝑔] +( Γ(𝛽−1) Γ(𝛽−𝑝−1)+Γ(𝛽) Γ(𝛽−𝑝))‖V‖1[𝐵𝐻+𝐴𝐻] +Γ(𝛽) Γ(𝛽−𝑝)‖𝑛‖1[𝐵𝑁+𝐴𝑁] +1 Ξ(Γ(𝛽) Γ(𝛽−𝑝)+Γ(𝛽−1) Γ(𝛽−𝑝−1)) ×Γ(𝛽) Γ(𝛽−𝑝)𝑡𝛽−𝑝−1 1[𝐵𝐽+𝐴𝐽] +1 Ξ(Γ(𝛽−1) Γ(𝛽−𝑝−1)+Γ(𝛽) Γ(𝛽−𝑝))𝑡𝛽−1 1[𝐵1,𝐽 +𝐴1,𝐽]. (81) Proof. To apply Lemma 5, we should define an open bounded subset Ωof 𝑋×𝑌centered at zero such that assumptions in Lemma 5hold. Let Ω1={(𝑥,𝑦)∈𝑋×𝑌: (𝑥,𝑦)=𝜆𝑇(𝑥,𝑦)for some 𝜆∈(0,1)}.WeprovethatΩ1is bounded. For (𝑥,𝑦)∈Ω1,we get (𝑥,𝑦)=𝜆𝑇(𝑥,𝑦).Itfollowsthat𝑥=𝜆𝑇1𝑦and 𝑦=𝜆𝑇2𝑥. For 𝑡∈(0,𝑡 1],weobtain𝑡2−𝛼|𝑥(𝑡)| ≤ 𝑡2−𝛼|(𝑇1𝑦)(𝑡)| ≤ Θ1+Θ2Φ−1(||𝑦||). For 𝑡∈(𝑡1,1], 𝑡2−𝛼 |𝑥(𝑡)| ≤𝑡2−𝛼 ∫𝑡 0(𝑡−𝑠)𝛼−1 Γ(𝛼)󵄨󵄨󵄨󵄨󵄨𝜙(𝑠)𝑓(𝑠,𝑦(𝑠),𝐷𝑝 0+𝑦(𝑠))󵄨󵄨󵄨󵄨󵄨𝑑𝑠 +𝑡 Γ(𝛼)∫1 0(1−𝑠)𝛼−1 󵄨󵄨󵄨󵄨󵄨𝜙(𝑠)𝑓(𝑠,𝑦(𝑠),𝐷𝑝 0+𝑦(𝑠))󵄨󵄨󵄨󵄨󵄨𝑑𝑠 +(1−𝑡)∫1 0󵄨󵄨󵄨󵄨󵄨𝑢(𝑠)𝐺(𝑠,𝑦(𝑠),𝐷𝑝 0+𝑦(𝑠))󵄨󵄨󵄨󵄨󵄨𝑑𝑠 +𝑡∫1 0󵄨󵄨󵄨󵄨󵄨𝑚(𝑠)𝑀(𝑠,𝑦(𝑠),𝐷𝑝 0+𝑦(𝑠))󵄨󵄨󵄨󵄨󵄨𝑑𝑠 +1−𝑡 ΠΓ(𝛼) Γ(𝛼−𝑞)𝑡𝛼−𝑞−1 1󵄨󵄨󵄨󵄨󵄨𝐼(𝑡1,𝑦(𝑡1),𝐷𝑝 0+𝑦(𝑡1))󵄨󵄨󵄨󵄨󵄨 +1−𝑡 Π𝑡𝛼−1 1󵄨󵄨󵄨󵄨󵄨𝐼1(𝑡1,𝑦(𝑡1),𝐷𝑝 0+𝑦(𝑡1))󵄨󵄨󵄨󵄨󵄨 ≤Σ1+Σ2Φ−1 (󵄩󵄩󵄩󵄩𝑦󵄩󵄩󵄩󵄩). (82) It follows that sup 𝑡∈(0,1)𝑡2−𝛼 |𝑥(𝑡)|≤max {Θ1,Σ1}+max {Θ2,Σ2}Φ−1 (󵄩󵄩󵄩󵄩𝑦󵄩󵄩󵄩󵄩). (83) Similarly, we have for 𝑡∈(0,𝑡1]that 𝑡𝑞+2−𝛼 󵄨󵄨󵄨󵄨󵄨𝐷𝑞 0+𝑥(𝑡)󵄨󵄨󵄨󵄨󵄨≤Θ3+Θ4Φ−1 (󵄩󵄩󵄩󵄩𝑦󵄩󵄩󵄩󵄩)(84) and for 𝑡∈(0,𝑡1] 𝑡𝑞+2−𝛼 󵄨󵄨󵄨󵄨󵄨𝐷𝑞 0+𝑥(𝑡)󵄨󵄨󵄨󵄨󵄨≤Σ3+Σ4Φ−1 (󵄩󵄩󵄩󵄩𝑦󵄩󵄩󵄩󵄩). (85) It follows that sup 𝑡∈(0,1)𝑡2+𝑞−𝛼 󵄨󵄨󵄨󵄨󵄨𝐷𝑞 0+𝑥(𝑡)󵄨󵄨󵄨󵄨󵄨 ≤max {Θ3,Σ3}+max {Θ4,Σ4}Φ−1 (󵄩󵄩󵄩󵄩𝑦󵄩󵄩󵄩󵄩). (86) Mathematical Problems in Engineering 17 Hence, ‖𝑥‖≤max {Θ1,Σ1,Θ3,Σ3} +max {Θ2,Σ2,Θ4,Σ4}Φ−1 (󵄩󵄩󵄩󵄩𝑦󵄩󵄩󵄩󵄩). (87) Similartotheabovediscussionwecanprovethat 󵄩󵄩󵄩󵄩𝑦󵄩󵄩󵄩󵄩≤max {Υ1,Λ1,Υ3,Λ3}+max {Υ2,Λ2,Υ4,Λ4}Φ(‖𝑥‖). (88) Case 1. Consider (max{Θ2,Σ2,Θ4,Σ4}](2max{Υ2,Λ2,Υ4, Λ4})<1). Withoutlossofgenerality,supposethat ‖𝑥‖≥Φ−1 (max {Υ1,Λ1,Υ3,Λ3} max {Υ2,Λ2,Υ4,Λ4}). (89) Then use Remark 13, and the previous inequalities to get ‖𝑥‖≤max {Θ1,Σ1,Θ3,Σ3} +max {Θ2,Σ2,Θ4,Σ4}](2max {Υ2,Λ2,Υ4,Λ4})‖𝑥‖. (90) It follows that there exists a constant 𝑊>0such that ||𝑥||≤ 𝑊.Thus ‖𝑥‖≤max {𝑊,Φ−1 (max {Υ1,Λ1,Υ3,Λ3} max {Υ2,Λ2,Υ4,Λ4})}. (91) Then 󵄩󵄩󵄩󵄩𝑦󵄩󵄩󵄩󵄩≤max {Υ1,Λ1,Υ3,Λ3} +max {Υ2,Λ2,Υ4,Λ4}Φ ×(max {𝑊,Φ−1 (max {Υ1,Λ1,Υ3,Λ3} max {Υ2,Λ2,Υ4,Λ4})}). (92) It follows that Ω1is bounded. Case 2. Consider ((max{Υ2,Λ2,Υ4,Λ4}/𝑤((2max{Θ2,Σ2, Θ4,Σ4})−1)) <1). Without loss of generality, suppose that 󵄩󵄩󵄩󵄩𝑦󵄩󵄩󵄩󵄩≥Φ(max {Θ1,Σ1,Θ3,Σ3} max {Θ2,Σ2,Θ4,Σ4}). (93) Then using Remark 12 and the previous inequalities, we get 󵄩󵄩󵄩󵄩𝑦󵄩󵄩󵄩󵄩≤max {Υ1,Λ1,Υ3,Λ3} +max {Υ2,Λ2,Υ4,Λ4} 𝑤((2max {Θ2,Σ2,Θ4,Σ4})−1)󵄩󵄩󵄩󵄩𝑦󵄩󵄩󵄩󵄩.(94) It follows that there exists a constant 𝑊>0such that ||𝑦||≤ 𝑊.Weget 󵄩󵄩󵄩󵄩𝑦󵄩󵄩󵄩󵄩≤max {𝑊,Φ(max {Θ1,Σ1,Θ3,Σ3} max {Θ2,Σ2,Θ4,Σ4})}. (95) Then ‖𝑥‖≤max {Θ1,Σ1,Θ3,Σ3} +max {Θ2,Σ2,Θ4,Σ4}Φ−1 ×(max {𝑊,Φ(max {Θ1,Σ1,Θ3,Σ3} max {Θ2,Σ2,Θ4,Σ4})}). (96) It follows that Ω1is bounded. To apply Lemma 5,letΩbe a nonempty open bounded subset of 𝑋such that Ω⊃Ω1centered at zero. It is easy to see from Lemma 8that 𝑇is a completely continuous operator. One can see that (𝑥,𝑦) =𝜆𝑇(𝑥,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)andBVP (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: 𝐷8/5 0+𝑥(𝑡)=𝑡−1/5(1−𝑡)−1 ×(𝑐+𝑏𝑡6/5[𝑦(𝑡)]3+𝑎𝑡9/5[𝐷1/5 0+𝑦(𝑡)]3), 𝑡∈(0,1),𝑡=1 2, 𝐷9/5 0+𝑦(𝑡) =𝑡−1/5(1−𝑡)−1 ×(𝑐0+𝑏0𝑡1/15[𝑥(𝑡)]1/3 +𝑎0𝑡2/15[𝐷1/5 0+𝑥(𝑡)]1/3), 𝑡∈(0,1),𝑡=𝑡1, lim 𝑡→0𝑡2/5𝑥(𝑡)=𝐺, lim 𝑡→0𝑡1/5𝑦(𝑡)=𝐻, 𝑥(1)=𝑀, 𝑦(1)=𝑁, Δ𝑥(1 2)=𝑐𝐼,Δ𝑦( 1 2)=𝑐𝐽, Δ𝐷1 0+𝑥(1 2)=𝑐1,𝐼,Δ𝐷 1 0+𝑦(1 2)=𝑐1,𝐽,(98) where 𝑐,𝑏,𝑎,𝑐0,𝑏0,𝑎0,𝐺0,𝐻0,𝑀0,𝑁0,𝐶𝐼,𝐶𝐽,𝐶1,𝐼,𝐶1,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+𝑏𝑡6/5𝑈3+ 𝑎𝑡9/5𝑉3and 𝑔(𝑡,𝑈,𝑉)=𝑐0+𝑏0𝑡1/15𝑈1/3 +𝑎0𝑡2/15𝑉1/3 defined on (0,1)×𝑅2, 18 Mathematical Problems in Engineering (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)=𝑐𝐼,𝐼1(𝑡,𝑈,𝑉)=𝑐1,𝐼,𝐽(𝑡,𝑈,𝑉)=𝑐𝐽,𝐽1(𝑡, 𝑈,𝑉)=𝑐1,𝐽. It is easy to show that (A) 𝜙satisfies 𝛼+2𝛿−𝑞>0,𝛼+𝑘+𝛿−𝑞≥0,and |𝜙(𝑡)|≤𝐿1𝑡𝑘(1−𝑡)𝛿for all 𝑡∈(0,1)with 𝐿1=1and 𝑘=−(1/5)=𝛿; 𝜓satisfies 𝜂+2𝜃−𝑝>0,𝛽+𝑙+𝜃−𝑝≥0,and |𝜓(𝑡)|≤𝐿2𝑡𝑙(1−𝑡)𝜃for all 𝑡∈(0,1)with 𝐿2=1and 𝑙=−(1/5)=𝜃; (B) 𝑓,𝐺,𝑀,𝐼,𝐼1are 𝛽-Caratheodory functions and 𝑔,𝐻, 𝑁,𝐽,𝐽1are 𝛼-Caratheodory functions. Furthermore, we have Φ−1(𝑥)=𝑥3and Φ(𝑥)=𝑥1/3 with 𝑤(𝑥)=𝑥1/3 and ](𝑥)=𝑥3.Itiseasytoseethat (i) the inequalities 󵄨󵄨󵄨󵄨󵄨𝑓(𝑡,𝑡𝛼−2𝑈,𝑡𝛼−𝑞−2𝑉)󵄨󵄨󵄨󵄨󵄨≤𝐶𝑓+𝐵𝑓Φ−1 (|𝑈|)+𝐴𝑓Φ−1 (|𝑉|), 󵄨󵄨󵄨󵄨󵄨𝐺(𝑡,𝑡𝛼−2𝑈,𝑡𝛼−𝑞−2𝑉)󵄨󵄨󵄨󵄨󵄨≤𝐶𝐺+𝐵𝐺Φ−1 (|𝑈|)+𝐴𝐺Φ−1 (|𝑉|), 󵄨󵄨󵄨󵄨󵄨𝑀(𝑡,𝑡𝛼−2𝑈,𝑡𝛼−𝑞−2𝑉)󵄨󵄨󵄨󵄨󵄨 ≤𝐶𝑀+𝐵𝑀Φ−1 (|𝑈|)+𝐴𝑀Φ−1 (󵄨󵄨󵄨󵄨󵄨Φ−1 (|𝑈|)󵄨󵄨󵄨󵄨󵄨)(99) hold for all (𝑈,𝑉)∈𝑅2,𝑡∈(0,1]with 𝐶𝑓=|𝑐|,𝐵𝑓= |𝑏|,𝐴𝑓=|𝑎|,𝐶𝐺=|𝐺0|,𝐵𝐺=0,𝐴𝐺=0and 𝐶𝑀= |𝑀0|,𝐵𝑀=0,𝐴𝑀=0; (ii) the inequalities 󵄨󵄨󵄨󵄨󵄨𝑔(𝑡,𝑡𝛽−2𝑈,𝑡𝛽−𝑝−2𝑉)󵄨󵄨󵄨󵄨󵄨≤𝐶𝑔+𝐵𝑔Φ(𝑈)+𝐴𝑔Φ(𝑉), 󵄨󵄨󵄨󵄨󵄨𝐻(𝑡,𝑡𝛽−2𝑈,𝑡𝛽−𝑝−2𝑉)󵄨󵄨󵄨󵄨󵄨≤𝐶𝐻+𝐵𝐻Φ(𝑈)+𝐴𝐻Φ(𝑉), 󵄨󵄨󵄨󵄨󵄨𝑁(𝑡,𝑡𝛽−2𝑈,𝑡𝛽−𝑝−2𝑉)󵄨󵄨󵄨󵄨󵄨≤𝐶𝑁+𝐵𝑁Φ(𝑈)+𝐴𝑁Φ(𝑉) (100) hold for all (𝑈,𝑉) ∈ 𝑅2,𝑡 ∈ (0,1]with 𝐶𝑔=|𝑐 0|, 𝐵𝑔=|𝑏 0|,𝐴𝑔=|𝑎 0|,𝐶𝐻=|𝐻 0|,𝐵𝐻=𝐴 𝐻=0, 𝐶𝑁=|𝑁0|,𝐵𝑁=𝐴𝑁=0; (iii) the inequalities 󵄨󵄨󵄨󵄨󵄨𝐼(𝑡1,𝑡𝛼−2 1𝑈,𝑡𝛼−𝑞−2 1𝑉)󵄨󵄨󵄨󵄨󵄨≤𝐶𝐼+𝐵𝐼Φ−1 (|𝑈|)+𝐴𝐼Φ−1 (|𝑉|), 󵄨󵄨󵄨󵄨󵄨𝐼1(𝑡1,𝑡𝛼−2 1𝑈,𝑡𝛼−𝑞−2 1𝑉)󵄨󵄨󵄨󵄨󵄨 ≤𝐶1,𝐼 +𝐵1,𝐼Φ−1 (|𝑈|)+𝐴1,𝐼Φ−1 (|𝑉|)(101) hold for all (𝑈,𝑉)∈𝑅2with 𝐶𝐼=|𝑐 𝐼|,𝐵𝐼=𝐴𝐼=0, 𝐶1,𝐼 =|𝑐1,𝐼|,𝐵1,𝐼 =𝐴1,𝐼 =0; (iv) the inequalities 󵄨󵄨󵄨󵄨󵄨󵄨𝐽(𝑡1,𝑡𝛽−2 1𝑈,𝑡𝛽−𝑝−2 1𝑉)󵄨󵄨󵄨󵄨󵄨󵄨≤𝐶𝐽+𝐵𝐽Φ(𝑈)+𝐴𝐽Φ(𝑉), 󵄨󵄨󵄨󵄨󵄨󵄨𝐽1(𝑡1,𝑡𝛽−2 1𝑈,𝑡𝛽−𝑝−2 1𝑉)󵄨󵄨󵄨󵄨󵄨󵄨≤𝐶1,𝐽 +𝐵1,𝐽Φ(𝑈)+𝐴1,𝐽Φ(𝑉) (102) hold for all (𝑈,𝑉)∈𝑅2with 𝐶𝐽=|𝑐 𝐽|,𝐵𝐽=𝐴𝐽=0, 𝐶1,𝐽 =|𝑐1,𝐽|,𝐵1,𝐽 =𝐴1,𝐽 =0. By direct computation, we know that Θ2=2B(7/5,4/5) Γ(8/5)[|𝑏|+|𝑎|], Σ2=2B(7/5,4/5) Γ(8/5)[|𝑏|+|𝑎|], Θ4=(B(6/5,4/5) Γ(7/5)+B(7/5,4/5) Γ(7/5))[|𝑏|+|𝑎|], Σ4=(B(6/5,4/5) Γ(7/5)+B(7/5,4/5) Γ(7/5))[|𝑏|+|𝑎|], Υ2=2B(8/5,4/5) Γ(9/5)[󵄨󵄨󵄨󵄨𝑏0󵄨󵄨󵄨󵄨+󵄨󵄨󵄨󵄨𝑎0󵄨󵄨󵄨󵄨], Υ4=(B(7/5,4/5) Γ(8/5)+B(8/5,4/5) Γ(8/5))[󵄨󵄨󵄨󵄨𝑏0󵄨󵄨󵄨󵄨+󵄨󵄨󵄨󵄨𝑎0󵄨󵄨󵄨󵄨], Λ2=2B(8/5,4/5) Γ(9/5)[󵄨󵄨󵄨󵄨𝑏0󵄨󵄨󵄨󵄨+󵄨󵄨󵄨󵄨𝑎0󵄨󵄨󵄨󵄨], Λ4=(B(7/5,4/5) Γ(8/5)+B(8/5,4/5) Γ(8/5))[󵄨󵄨󵄨󵄨𝑏0󵄨󵄨󵄨󵄨+󵄨󵄨󵄨󵄨𝑎0󵄨󵄨󵄨󵄨]. (103) Then Theorem 14 implies that the existence of at least one solution if max {2B(8/5,4/5) Γ(9/5),B(7/5,4/5) Γ(8/5)+B(8/5,4/5) Γ(8/5)} ×(max {2B(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 <1 3 √2.(104) Example 16. Consider the following boundary value problem without impulse effects: 𝐷7/4 0+𝑥(𝑡)=𝑡−1/4(1−𝑡)−1/4 ×(𝐶+𝐵𝑡3/4[𝑦(𝑡)]3+𝐴𝑡15/4[𝐷1 0+𝑦(𝑡)]3), 𝑡∈(0,1), Mathematical Problems in Engineering 19 𝐷5/4 0+𝑦(𝑡) =𝑡−1/8(1−𝑡)−1/8 ×(𝐶0+𝐵0𝑡1/4[𝑥(𝑡)]1/3 +𝐴0𝑡7/12[𝐷1/4 0+𝑥(𝑡)]1/3), 𝑡∈(0,1), lim 𝑡→0𝑡1/4𝑥(𝑡)=0, lim 𝑡→0𝑡3/4𝑦(𝑡)=0, 𝑥(1)=0, 𝑦(1)=0, (105) where 𝐶,𝐵,𝐴,𝐶0,𝐵0,and𝐴0are constants. Corresponding to BVP (1), we have (a) 𝛼=7/4,𝛽=5/4,𝑝=1and 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 + 𝐵𝑡1/12𝑈3+𝐴𝑡5/12𝑉3and 𝑔(𝑡,𝑈,𝑉)=𝐶0+𝐵0𝑡1/4𝑈1/3+ 𝐴0𝑡7/12𝑉1/3, (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) 𝜙satisfies 𝛼+2𝛿−𝑞>0,𝛼+𝑘+𝛿−𝑞>0,|𝜙(𝑡)|≤ 𝐿1𝑡𝑘(1−𝑡)𝛿for all 𝑡∈(0,1)with 𝐿1=1,𝑘=−(1/4)= 𝛿; 𝜓satisfies 𝛽+2𝜃−𝑝>0,𝛽+𝑙+𝜃−𝑝≥0,and |𝜓(𝑡)| ≤ 𝐿2𝑡𝑙(1−𝑡)𝜃for all 𝑡 ∈ (0,1)with 𝐿2=1, 𝑙=−(1/8)=𝜃; (B) 𝑓,𝐺,𝑀,𝐼,𝐼1are 𝛽-Caratheodory functions and 𝑔,𝐻, 𝑁,𝐽,𝐽1are 𝛼-Caratheodory functions. Furthermore, Φ(𝑥)=𝑥1/3 and Φ−1(𝑥)=𝑥3,wehave𝑤(𝑥)= 𝑥1/3 and ](𝑥)=𝑥3,and (i) the inequalities 󵄨󵄨󵄨󵄨󵄨𝑓(𝑡,𝑡𝛼−2𝑈,𝑡𝛼−𝑞−2𝑉)󵄨󵄨󵄨󵄨󵄨≤𝐶𝑓+𝐵𝑓Φ−1 (|𝑈|)+𝐴𝑓Φ−1 (|𝑉|), 󵄨󵄨󵄨󵄨󵄨𝐺(𝑡,𝑡𝛼−2𝑈,𝑡𝛼−𝑞−2𝑉)󵄨󵄨󵄨󵄨󵄨≤𝐶𝐺+𝐵𝐺Φ−1 (|𝑈|)+𝐴𝐺Φ−1 (|𝑉|), 󵄨󵄨󵄨󵄨󵄨𝑀(𝑡,𝑡𝛼−2𝑈,𝑡𝛼−𝑞−2𝑉)󵄨󵄨󵄨󵄨󵄨 ≤𝐶𝑀+𝐵𝑀Φ−1 (|𝑈|)+𝐴𝑀Φ−1 (󵄨󵄨󵄨󵄨󵄨Φ−1 (|𝑈|)󵄨󵄨󵄨󵄨󵄨)(106) hold for all (𝑈,𝑉) ∈ 𝑅2,𝑡 ∈ (0,1)with 𝐶𝐺=𝐵 𝐺= 𝐴𝐺=𝐶𝑀=𝐵𝑀=𝐴𝑀=0,𝐶𝑓=|𝐶|,𝐵𝑓=|𝐵|and 𝐴𝑓=|𝐴|; (ii) the inequalities 󵄨󵄨󵄨󵄨󵄨𝑔(𝑡,𝑡𝛽−2𝑈,𝑡𝛽−𝑝−2𝑉)󵄨󵄨󵄨󵄨󵄨≤𝐶𝑔+𝐵𝑔Φ(𝑈)+𝐴𝑔Φ(𝑉), 󵄨󵄨󵄨󵄨󵄨𝐻(𝑡,𝑡𝛽−2𝑈,𝑡𝛽−𝑝−2𝑉)󵄨󵄨󵄨󵄨󵄨≤𝐶𝐻+𝐵𝐻Φ(𝑈)+𝐴𝐻Φ(𝑉), 󵄨󵄨󵄨󵄨󵄨𝑁(𝑡,𝑡𝛽−2𝑈,𝑡𝛽−𝑝−2𝑉)󵄨󵄨󵄨󵄨󵄨≤𝐶𝑁+𝐵𝑁Φ(𝑈)+𝐴𝑁Φ(𝑉)(107) hold for all (𝑈,𝑉)∈𝑅2,𝑡 ∈(0,1)with 𝐶𝐻=𝐵𝐻= 𝐴𝐻=𝐶𝑁=𝐵𝑁=𝐴𝑁=0,𝐶𝑔=|𝐶0|,𝐵𝑔=|𝐵0|and 𝐴𝑔=|𝐴0|; (iii) the inequalities 󵄨󵄨󵄨󵄨󵄨𝐼(𝑡1,𝑡𝛼−2 1𝑈,𝑡𝛼−𝑞−2 1𝑉)󵄨󵄨󵄨󵄨󵄨 ≤𝐶𝐼+𝐵𝐼Φ−1 (|𝑈|)+𝐴𝐼Φ−1 (|𝑉|), 󵄨󵄨󵄨󵄨󵄨𝐼1(𝑡1,𝑡𝛼−2 1𝑈,𝑡𝛼−𝑞−2 1𝑉)󵄨󵄨󵄨󵄨󵄨 ≤𝐶1,𝐼 +𝐵1,𝐼Φ−1 (|𝑈|)+𝐴1,𝐼Φ−1 (|𝑉|) (108) hold for all (𝑈,𝑉)∈𝑅2with 𝐶𝐼=𝐵𝐼=𝐴𝐼=𝐶1,𝐼 = 𝐵1,𝐼 =𝐴1,𝐼 =0; (iv) there exist the nonnegative numbers 𝐴𝑖,𝑘,𝐵𝑖,𝑘, 𝐶𝑖,𝑘 (𝑖=1,2)such that 󵄨󵄨󵄨󵄨󵄨󵄨𝐽(𝑡1,𝑡𝛽−2 1𝑈,𝑡𝛽−𝑝−2 1𝑉)󵄨󵄨󵄨󵄨󵄨󵄨≤𝐶𝐽+𝐵𝐽Φ(𝑈)+𝐴𝐽Φ(𝑉), 󵄨󵄨󵄨󵄨󵄨󵄨𝐽1(𝑡1,𝑡𝛽−2 1𝑈,𝑡𝛽−𝑝−2 1𝑉)󵄨󵄨󵄨󵄨󵄨󵄨≤𝐶1,𝐽 +𝐵1,𝐽Φ(𝑈)+𝐴1,𝐽Φ(𝑉) (109) hold for all (𝑈,𝑉)∈𝑅2with 𝐶𝐽=𝐵𝐽=𝐴𝐽=𝐶1,𝐽 = 𝐵1,𝐽 =𝐴1,𝐽 =0. By direct computation, we know that Θ2=2B(3/2,3/4) Γ(7/4)[|𝐵|+|𝐴|], Σ2=2B(3/2,3/4) Γ(7/4)[|𝐵|+|𝐴|], Θ4=(B(5/4,3/4) Γ(3/2)+B(3/2,3/4) Γ(3/2))[|𝐵|+|𝐴|], Σ4=(B(5/4,3/4) Γ(3/2)+B(3/2,3/4) Γ(3/2))[|𝐵|+|𝐴|], Υ2=2B(9/8,7/8) Γ(5/4)[󵄨󵄨󵄨󵄨𝐵0󵄨󵄨󵄨󵄨+󵄨󵄨󵄨󵄨𝐴0󵄨󵄨󵄨󵄨], Υ4=(B(1/8,7/8) Γ(𝛽−𝑝) +B(9/8,7/8) Γ(1/4))[󵄨󵄨󵄨󵄨𝐵0󵄨󵄨󵄨󵄨+󵄨󵄨󵄨󵄨𝐴0󵄨󵄨󵄨󵄨], Λ2=2B(9/8,7/8) Γ(5/4)[󵄨󵄨󵄨󵄨𝐵0󵄨󵄨󵄨󵄨+󵄨󵄨󵄨󵄨𝐴0󵄨󵄨󵄨󵄨], Λ4=(B(1/8,7/8) Γ(1/4)+B(9/8,7/8) Γ(1/4))[󵄨󵄨󵄨󵄨𝐵0󵄨󵄨󵄨󵄨+󵄨󵄨󵄨󵄨𝐴0󵄨󵄨󵄨󵄨]. (110) 20 Mathematical Problems in Engineering Then Theorem 14 implies the existence of at least one solution if max {2B(3/2,3/4) Γ(7/4),B(5/4,3/4) Γ(3/2)+B(3/2,3/4) Γ(3/2)} ×(max {2B(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󵄨󵄨󵄨󵄨+󵄨󵄨󵄨󵄨𝐴0󵄨󵄨󵄨󵄨]3[|𝐵|+|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|.Theycannotbe solved by the theorems in [24,25]. 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