Boundary value problems for a class of sequential integrodifferential equations of fractional order

Hindawi Publishing Corporation Journal of Function Spaces and Applications Volume 2013, Article ID 149659, 8pages http://dx.doi.org/10.1155/2013/149659 Research Article Boundary Value Problems for a Class of Sequential Integrodifferential Equations of Fractional Order Bashir Ahmad1and Juan J. Nieto1,2 1Department of Mathematics, Faculty of Science, King Abdulaziz University, P.O. Box 80203, Jeddah 21589, Saudi Arabia 2Departamento de An´ alisis Matem´ atico, Facultad de Matem´ aticas, Universidad de Santiago de Compostela, 15782 Santiago de Compostela, Spain Correspondence should be addressed to Juan J. Nieto; juanjose.nieto[email protected] Received 16 January 2013; Accepted 13 March 2013 Academic Editor: Jose Luis Sanchez Copyright © 2013 B. Ahmad and J. J. Nieto. 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. We investigate the existence of solutions for a sequential integrodifferential equation of fractional order with some boundary conditions. The existence results are established by means of some standard tools of fixed point theory. An illustrative example is also presented. 1. Introduction Nonlinear boundary value problems of fractional differential equationshavereceivedaconsiderableattentioninthe last few decades. One can easily find a variety of results ranging from theoretical analysis to asymptotic behavior and numerical methods for fractional equations in the literature on the topic. The interest in the subject has been mainly due to the extensive applications of fractional calculus in the mathematical modeling of several real-world phenomena occurring in physical and technical sciences; see, for example, [1–4]. An important feature of a fractional order differential operator, distinguishing it from an integer-order differential operator, is that it is nonlocal in nature. It means that the future stateofadynamicalsystemorprocessbasedonafractional operator depends on its current state as well as its past states. Thus, differential equations of arbitrary order are capable of describing memory and hereditary properties of some important and useful materials and processes. This feature has fascinated many researchers, and they have shifted their focus to fractional order models from the classical integerorder models. For some recent work on the topic, we refer, for instance, to [5–9]. Recently, in [10], the authors studied sequential fractional differential equations with three-point boundary conditions. In this paper, we consider a nonlinear Dirichlet boundary value problem of sequential fractional integrodifferential equations given by (𝑐𝐷𝛼+𝑘𝑐𝐷𝛼−1)𝑢(𝑡)=𝑝𝑓(𝑡,𝑢(𝑡))+𝑞𝐼𝛽𝑔(𝑡,𝑢(𝑡)), 0<𝑡<1, (1) 𝑢(0)=0, 𝑢(1)=0, (2) where 𝑐𝐷𝛼denotes the Caputo fractional derivative of order 𝛼,1<𝛼≤2,𝐼𝛽(⋅)denotes Riemann-Liouville integral with 0<𝛽<1,𝑓,𝑔 are given continuous functions, 𝑘 =0,and 𝑝,𝑞arerealconstants.Wealsostudythefractionalintegrodifferential equation (1)subjecttothefollowingboundary conditions: 𝑢󸀠(0)+𝑘𝑢(0)=𝑎, 𝑢(1)=𝑏, 𝑎,𝑏∈R,(3) 𝑢(0)=𝑎, 𝑢󸀠(0)=𝑢󸀠(1),𝑎∈R.(4) 2. Linear Fractional Differential Equations For 𝛼 ∈ (1,2], we consider the following linear fractional differential equation: (𝑐𝐷𝛼+𝑘𝑐𝐷𝛼−1)𝑢(𝑡)=ℎ(𝑡),(5) 2 Journal of Function Spaces and Applications where 𝑐𝐷𝛼denotes the Caputo fractional derivative of order 𝛼.Rewriting(1)as 𝑐𝐷𝛼(𝑢(𝑡) + 𝑘𝑐𝐷−1𝑢(𝑡)) = ℎ(𝑡),wecan write its solution as 𝑢(𝑡)+𝑘𝑐𝐷−1𝑢(𝑡)=1 Γ(𝛼)∫𝑡 0(𝑡−𝑠)𝛼−1ℎ(𝑠)𝑑𝑠+𝑐0+𝑐1𝑡, (6) where 𝑐0,𝑐1are arbitrary constants. Now, (6) can be expressed as 𝑢(𝑡)=−𝑘 ∫𝑡 0𝑢(𝑠)𝑑𝑠+ 1 Γ(𝛼)∫𝑡 0(𝑡−𝑠)𝛼−1ℎ(𝑠)𝑑𝑠 +𝑐0+𝑐1𝑡. (7) Differentiating (7), we obtain 𝑢󸀠(𝑡)=−𝑘𝑢(𝑡)+1 Γ(𝛼−1)∫𝑡 0(𝑡−𝑠)𝛼−2ℎ(𝑠)𝑑𝑠+𝑐1,(8) which can alternatively be written as (𝑢(𝑡)𝑒𝑘𝑡)󸀠=𝑒𝑘𝑡 (1 Γ(𝛼−1)∫𝑡 0(𝑡−𝑠)𝛼−2ℎ(𝑠)𝑑𝑠+𝑐1). (9) Integrating from 0to 𝑡,wehave 𝑢(𝑡)=𝐴𝑒−𝑘𝑡 +∫𝑡 0𝑒−𝑘(𝑡−𝑠)𝐼𝛼−1ℎ(𝑠)𝑑𝑠+𝐵, (10) where 𝐴and 𝐵are arbitrary constants, and 𝐼𝛼−1ℎ(𝑡)=∫𝑡 0(𝑡−𝑥)𝛼−2 Γ(𝛼−1)ℎ(𝑥)𝑑𝑥. (11) Lemma 1. The unique solution of the linear equation (5) subject to the Dirichlet boundary conditions (2)is given by 𝑢(𝑡)=(1−𝑒−𝑘𝑡) (𝑒−𝑘 −1)∫1 0𝑒−𝑘(1−𝑠) (∫𝑠 0(𝑠−𝑥)𝛼−2 Γ(𝛼−1)ℎ(𝑥)𝑑𝑥)𝑑𝑠 +∫𝑡 0𝑒−𝑘(𝑡−𝑠) (∫𝑠 0(𝑠−𝑥)𝛼−2 Γ(𝛼−1)ℎ(𝑥)𝑑𝑥)𝑑𝑠. (12) Proof. Observe that the general solution of (5)isgivenby (10). Using the given boundary conditions in (10), we find that 𝐴=−𝐵= 1 (1−𝑒−𝑘)∫1 0𝑒−𝑘(1−𝑠) (∫𝑠 0(𝑠−𝑥)𝛼−2 Γ(𝛼−1)ℎ(𝑥)𝑑𝑥)𝑑𝑠. (13) Substituting the values of 𝐴and 𝐵in (10)yieldsthesolution (12). This completes the proof. In the next two lemmas, we present the unique solutions of (5) with different kinds of boundary conditions. We do not provide the proofs for these lemmas as they are similar to that of Lemma 1. Lemma 2. The unique solution of the problem (5)–(3)is given by 𝑢(𝑡)=𝑒𝑘(1−𝑡) ×[(𝑏𝑘−𝑎) 𝑘−∫1 0𝑒−𝑘(1−𝑠) (∫𝑠 0(𝑠−𝑥)𝛼−2 Γ(𝛼−1)ℎ(𝑥)𝑑𝑥)𝑑𝑠] +∫𝑡 0𝑒−𝑘(𝑡−𝑠) (∫𝑠 0(𝑠−𝑥)𝛼−2 Γ(𝛼−1)ℎ(𝑥)𝑑𝑥)𝑑𝑠+𝑎 𝑘. (14) Lemma 3. The unique solution of (5)with the boundary conditions (4)is 𝑢(𝑡)=−(1−𝑒−𝑘𝑡) 𝑘(1−𝑒−𝑘) ×[𝑘∫1 0𝑒−𝑘(1−𝑠) (∫𝑠 0(𝑠−𝑥)𝛼−2 Γ(𝛼−1)ℎ(𝑥)𝑑𝑥)𝑑𝑠 −∫1 0(1−𝑠)𝛼−2 Γ(𝛼−1)ℎ(𝑠)𝑑𝑠] +∫𝑡 0𝑒−𝑘(𝑡−𝑠) (∫𝑠 0(𝑠−𝑥)𝛼−2 Γ(𝛼−1)ℎ(𝑥)𝑑𝑥)𝑑𝑠+𝑎. (15) 3. Existence Results for the Nonlinear Problems Let P= 𝐶([0,1],R)denote the Banach space of all continuous functions from [0,1]into Rendowed with the usual norm defined by ‖𝑥‖=sup{|𝑥(𝑡)|,𝑡∈[0,1]}. In view of Lemma 1,wetransformproblem(1)-(2)toan equivalent fixed point problem as 𝑢=V𝑢, (16) where V:P→Pis defined by (V𝑢)(𝑡) =(1−𝑒−𝑘𝑡) (𝑒−𝑘 −1)∫1 0𝑒−𝑘(1−𝑠) (𝑝∫𝑠 0(𝑠−𝑥)𝛼−2 Γ(𝛼−1)𝑓(𝑥,𝑢(𝑥))𝑑𝑥 +𝑞∫𝑠 0(𝑠−𝑥)𝛼+𝛽−2 Γ(𝛼+𝛽−1) ×𝑔(𝑥,𝑢(𝑥))𝑑𝑥)𝑑𝑠 +∫𝑡 0𝑒−𝑘(𝑡−𝑠) (𝑝∫𝑠 0(𝑠−𝑥)𝛼−2 Γ(𝛼−1)𝑓(𝑥,𝑢(𝑥))𝑑𝑥 +𝑞∫𝑠 0(𝑠−𝑥)𝛼+𝛽−2 Γ(𝛼+𝛽−1) ×𝑔(𝑥,𝑢(𝑥))𝑑𝑥)𝑑𝑠. (17) Journal of Function Spaces and Applications 3 In a similar manner, we can define a fixed point operator V 1:P→Pfor the nonlinear problem (1)–(3) as follows: (V 1𝑢)(𝑡) =𝑒𝑘(1−𝑡) [(𝑏𝑘−𝑎) 𝑘−∫1 0𝑒−𝑘(1−𝑠) ×(𝑝∫𝑠 0(𝑠−𝑥)𝛼−2 Γ(𝛼−1)𝑓(𝑥,𝑢(𝑥))𝑑𝑥 +𝑞∫𝑠 0(𝑠−𝑥)𝛼+𝛽−2 Γ(𝛼+𝛽−1) ×𝑔(𝑥,𝑢(𝑥))𝑑𝑥)𝑑𝑠] +∫𝑡 0𝑒−𝑘(𝑡−𝑠) (𝑝∫𝑠 0(𝑠−𝑥)𝛼−2 Γ(𝛼−1)𝑓(𝑥,𝑢(𝑥))𝑑𝑥 +𝑞∫𝑠 0(𝑠−𝑥)𝛼+𝛽−2 Γ(𝛼+𝛽−1) ×𝑔(𝑥,𝑢(𝑥))𝑑𝑥)𝑑𝑠+𝑎 𝑘. (18) A fixed point operator V 2:P→Pfor the nonlinear problem (1)–(4) is defined by (V 2𝑢)(𝑡) =−(1−𝑒−𝑘𝑡) 𝑘(1−𝑒−𝑘) ×[𝑘∫1 0𝑒−𝑘(1−𝑠) (𝑝∫𝑠 0(𝑠−𝑥)𝛼−2 Γ(𝛼−1) ×𝑓(𝑥,𝑢(𝑥))𝑑𝑥 +𝑞∫𝑠 0(𝑠−𝑥)𝛼+𝛽−2 Γ(𝛼+𝛽−1) ×𝑔(𝑥,𝑢(𝑥))𝑑𝑥)𝑑𝑠 −𝑝∫1 0(1−𝑠)𝛼−2 Γ(𝛼−1)𝑓(𝑠,𝑢(𝑠))𝑑𝑠 −𝑞∫1 0(1−𝑠)𝛼+𝛽−2 Γ(𝛼+𝛽−1)𝑔(𝑠,𝑢(𝑠))𝑑𝑠] +∫𝑡 0𝑒−𝑘(𝑡−𝑠) (𝑝∫𝑠 0(𝑠−𝑥)𝛼−2 Γ(𝛼−1)𝑓(𝑥,𝑢(𝑥))𝑑𝑥 +𝑞∫𝑠 0(𝑠−𝑥)𝛼+𝛽−2 Γ(𝛼+𝛽−1) ×𝑔(𝑥,𝑢(𝑥))𝑑𝑥)𝑑𝑠+𝑎. (19) We only present the existence results for the problem (1)- (2). Observe that problem (1)-(2) has solutions if the operator equation (16) has fixed points. For computational convenience, we introduce the following constant: 𝑄=2󵄨󵄨󵄨󵄨󵄨1−𝑒−𝑘󵄨󵄨󵄨󵄨󵄨[󵄨󵄨󵄨󵄨𝑝󵄨󵄨󵄨󵄨Γ(𝛼+𝛽)+󵄨󵄨󵄨󵄨𝑞󵄨󵄨󵄨󵄨Γ(𝛼)] |𝑘|Γ(𝛼+𝛽)Γ(𝛼).(20) Theorem 4. Assume that 𝑓,𝑔 : [0,1] × R→Rare continuous functions satisfying the following condition: (𝐴1)󵄨󵄨󵄨󵄨𝑓(𝑡,𝑢)−𝑓(𝑡,V)󵄨󵄨󵄨󵄨≤𝐿1|𝑢−V|, 󵄨󵄨󵄨󵄨𝑔(𝑡,𝑢)−𝑔(𝑡,V)󵄨󵄨󵄨󵄨≤𝐿2|𝑢−V|,∀𝑡∈ [0,1], 𝐿1,𝐿2>0,𝑢,V∈R. (21) Then, the boundary value problem (1)-(2)has a unique solution if 𝐿<1/𝑄,where𝐿=max{𝐿1,𝐿2}and 𝑄is given by (20). Proof. Let us define 𝑀=max{𝑀1,𝑀2},where𝑀1,𝑀2 are finite numbers given by sup𝑡∈[0,1]|𝑓(𝑡,0)| = 𝑀1, sup𝑡∈[0,1]|𝑔(𝑡,0)| = 𝑀2.Selecting𝑟 ≥ (𝑄𝑀)/(1−𝐿𝑄),we show that V𝐵𝑟⊂𝐵 𝑟,where𝐵𝑟={𝑢∈P:‖𝑢‖≤𝑟}.For 𝑢∈𝐵𝑟,wehave ‖(V𝑢)‖≤sup 𝑡∈[0,1]{󵄨󵄨󵄨󵄨󵄨󵄨󵄨󵄨󵄨1−𝑒−𝑘𝑡 1−𝑒−𝑘 󵄨󵄨󵄨󵄨󵄨󵄨󵄨󵄨󵄨 ×∫1 0𝑒−𝑘(1−𝑠) (󵄨󵄨󵄨󵄨𝑝󵄨󵄨󵄨󵄨∫𝑠 0(𝑠−𝑥)𝛼−2 Γ(𝛼−1) ×󵄨󵄨󵄨󵄨𝑓(𝑥,𝑢(𝑥))󵄨󵄨󵄨󵄨𝑑𝑥 +󵄨󵄨󵄨󵄨𝑞󵄨󵄨󵄨󵄨∫𝑠 0(𝑠−𝑥)𝛼+𝛽−2 Γ(𝛼+𝛽−1) ×󵄨󵄨󵄨󵄨𝑔(𝑥,𝑢(𝑥))󵄨󵄨󵄨󵄨𝑑𝑥)𝑑𝑠 +∫𝑡 0𝑒−𝑘(𝑡−𝑠) ×(󵄨󵄨󵄨󵄨𝑝󵄨󵄨󵄨󵄨∫𝑠 0(𝑠−𝑥)𝛼−2 Γ(𝛼−1) ×󵄨󵄨󵄨󵄨𝑓(𝑥,𝑢(𝑥))󵄨󵄨󵄨󵄨𝑑𝑥 +󵄨󵄨󵄨󵄨𝑞󵄨󵄨󵄨󵄨∫𝑠 0(𝑠−𝑥)𝛼+𝛽−2 Γ(𝛼+𝛽−1) ×󵄨󵄨󵄨󵄨𝑔(𝑥,𝑢(𝑥))󵄨󵄨󵄨󵄨𝑑𝑥)𝑑𝑠} 4 Journal of Function Spaces and Applications ≤sup 𝑡∈[0,1]{󵄨󵄨󵄨󵄨󵄨󵄨󵄨󵄨󵄨1−𝑒−𝑘𝑡 1−𝑒−𝑘 󵄨󵄨󵄨󵄨󵄨󵄨󵄨󵄨󵄨∫1 0𝑒−𝑘(1−𝑠) ×(󵄨󵄨󵄨󵄨𝑝󵄨󵄨󵄨󵄨∫𝑠 0(𝑠−𝑥)𝛼−2 Γ(𝛼−1) ×(󵄨󵄨󵄨󵄨𝑓(𝑥,𝑢(𝑥))−𝑓(𝑥,0)󵄨󵄨󵄨󵄨 +󵄨󵄨󵄨󵄨𝑓(𝑥,0)󵄨󵄨󵄨󵄨)𝑑𝑥 +󵄨󵄨󵄨󵄨𝑞󵄨󵄨󵄨󵄨∫𝑠 0(𝑠−𝑥)𝛼+𝛽−2 Γ(𝛼+𝛽−1) ×(󵄨󵄨󵄨󵄨𝑔(𝑥,𝑢(𝑥))−𝑔(𝑥,0)󵄨󵄨󵄨󵄨 +󵄨󵄨󵄨󵄨𝑔(𝑥,0)󵄨󵄨󵄨󵄨)𝑑𝑥)𝑑𝑠 +∫𝑡 0𝑒−𝑘(𝑡−𝑠) ×(󵄨󵄨󵄨󵄨𝑝󵄨󵄨󵄨󵄨∫𝑠 0(𝑠−𝑥)𝛼−2 Γ(𝛼−1) ×(󵄨󵄨󵄨󵄨𝑓(𝑥,𝑢(𝑥))−𝑓(𝑥,0)󵄨󵄨󵄨󵄨 +󵄨󵄨󵄨󵄨𝑓(𝑥,0)󵄨󵄨󵄨󵄨)𝑑𝑥 +󵄨󵄨󵄨󵄨𝑞󵄨󵄨󵄨󵄨∫𝑠 0(𝑠−𝑥)𝛼+𝛽−2 Γ(𝛼+𝛽−1) ×(󵄨󵄨󵄨󵄨𝑔(𝑥,𝑢(𝑥))−𝑔(𝑥,0)󵄨󵄨󵄨󵄨 +󵄨󵄨󵄨󵄨𝑔(𝑥,0)󵄨󵄨󵄨󵄨)𝑑𝑥)𝑑𝑠} ≤󵄨󵄨󵄨󵄨𝑝󵄨󵄨󵄨󵄨(𝐿1𝑟+𝑀1) ×sup 𝑡∈[0,1]{󵄨󵄨󵄨󵄨󵄨󵄨󵄨󵄨󵄨1−𝑒−𝑘𝑡 1−𝑒−𝑘 󵄨󵄨󵄨󵄨󵄨󵄨󵄨󵄨󵄨 ×∫1 0𝑒−𝑘(1−𝑠) (∫𝑠 0(𝑠−𝑥)𝛼−2 Γ(𝛼−1)𝑑𝑥)𝑑𝑠 +∫𝑡 0𝑒−𝑘(𝑡−𝑠) (∫𝑠 0(𝑠−𝑥)𝛼−2 Γ(𝛼−1)𝑑𝑥)𝑑𝑠} +󵄨󵄨󵄨󵄨𝑞󵄨󵄨󵄨󵄨(𝐿2𝑟+𝑀2) ×sup 𝑡∈[0,1]{󵄨󵄨󵄨󵄨󵄨󵄨󵄨󵄨󵄨1−𝑒−𝑘𝑡 1−𝑒−𝑘 󵄨󵄨󵄨󵄨󵄨󵄨󵄨󵄨󵄨 ×∫1 0𝑒−𝑘(1−𝑠) (∫𝑠 0(𝑠−𝑥)𝛼+𝛽−2 Γ(𝛼+𝛽−1)𝑑𝑥)𝑑𝑠 +∫𝑡 0𝑒−𝑘(𝑡−𝑠) ×(∫𝑠 0(𝑠−𝑥)𝛼+𝛽−2 Γ(𝛼+𝛽−1)𝑑𝑥)𝑑𝑠} ≤(𝐿𝑟+𝑀)2󵄨󵄨󵄨󵄨󵄨1−𝑒−𝑘󵄨󵄨󵄨󵄨󵄨[󵄨󵄨󵄨󵄨𝑝󵄨󵄨󵄨󵄨Γ(𝛼+𝛽)+󵄨󵄨󵄨󵄨𝑞󵄨󵄨󵄨󵄨Γ(𝛼)] |𝑘|Γ(𝛼+𝛽)Γ(𝛼) =(𝐿𝑟+𝑀)𝑄≤𝑟, (22) which means that V𝐵𝑟⊂𝐵𝑟. Now, for 𝑢,V∈P,weobtain ‖V𝑢−VV‖ ≤sup 𝑡∈[0,1]{󵄨󵄨󵄨󵄨󵄨󵄨󵄨󵄨󵄨1−𝑒−𝑘𝑡 1−𝑒−𝑘 󵄨󵄨󵄨󵄨󵄨󵄨󵄨󵄨󵄨 ×∫1 0𝑒−𝑘(1−𝑠) (󵄨󵄨󵄨󵄨𝑝󵄨󵄨󵄨󵄨∫𝑠 0(𝑠−𝑥)𝛼−2 Γ(𝛼−1) ×󵄨󵄨󵄨󵄨𝑓(𝑠,𝑢(𝑠)) −𝑓(𝑠,V(𝑠))󵄨󵄨󵄨󵄨𝑑𝑥 +󵄨󵄨󵄨󵄨𝑞󵄨󵄨󵄨󵄨∫𝑠 0(𝑠−𝑥)𝛼+𝛽−2 Γ(𝛼+𝛽−1) ×󵄨󵄨󵄨󵄨𝑔(𝑠,𝑢(𝑠)) −𝑔(𝑠,V(𝑠))󵄨󵄨󵄨󵄨𝑑𝑥)𝑑𝑠 +∫𝑡 0𝑒−𝑘(𝑡−𝑠) ×(󵄨󵄨󵄨󵄨𝑝󵄨󵄨󵄨󵄨∫𝑠 0(𝑠−𝑥)𝛼−2 Γ(𝛼−1) ×󵄨󵄨󵄨󵄨𝑓(𝑠,𝑢(𝑠))−𝑓(𝑠,V(𝑠))󵄨󵄨󵄨󵄨𝑑𝑥 +󵄨󵄨󵄨󵄨𝑞󵄨󵄨󵄨󵄨∫𝑠 0(𝑠−𝑥)𝛼+𝛽−2 Γ(𝛼+𝛽−1) ×󵄨󵄨󵄨󵄨𝑔(𝑠,𝑢(𝑠))−𝑔(𝑠,V(𝑠))󵄨󵄨󵄨󵄨𝑑𝑥)𝑑𝑠} ≤𝐿2󵄨󵄨󵄨󵄨󵄨1−𝑒−𝑘󵄨󵄨󵄨󵄨󵄨[󵄨󵄨󵄨󵄨𝑝󵄨󵄨󵄨󵄨Γ(𝛼+𝛽)+󵄨󵄨󵄨󵄨𝑞󵄨󵄨󵄨󵄨Γ(𝛼)] |𝑘|Γ(𝛼+𝛽)Γ(𝛼)‖𝑢−V‖ =𝐿𝑄‖𝑢−V‖.(23) By the given assumption, 𝐿<1/𝑄,Vis a contraction. Thus, the conclusion of the theorem follows by the contraction mapping principle (Banach fixed point theorem). Our next existence result relies on Krasnoselskii’s fixed point theorem. Lemma 5 (Krasnoselskii, see [11]). Let 𝑀be a closed, convex, bounded,andnonemptysubsetofaBanachspace𝑋.Let𝐴,𝐵 be the operators such that (i) 𝐴𝑥+𝐵𝑦∈𝑀whenever 𝑥,𝑦∈𝑀, (ii) 𝐴is compact, and continuous, and (iii) 𝐵is a contraction mapping. Then, there exists 𝑧∈𝑀such that 𝑧=𝐴𝑧+𝐵𝑧. Journal of Function Spaces and Applications 5 Theorem 6. Let 𝑓,𝑔 : [0,1] × R→Rbe continuous functions satisfying assumption (𝐴1),and (𝐴2)󵄨󵄨󵄨󵄨𝑓(𝑡,𝑢)󵄨󵄨󵄨󵄨≤𝜇1(𝑡),󵄨󵄨󵄨󵄨𝑔(𝑡,𝑢)󵄨󵄨󵄨󵄨≤𝜇2(𝑡), ∀(𝑡,𝑢)∈[0,1]×R,𝜇 1,𝜇2∈𝐶([0,1],R+). (24) Then, the problem (1)-(2)has at least one solution on [0,1] provided that 󵄨󵄨󵄨󵄨󵄨1−𝑒−𝑘󵄨󵄨󵄨󵄨󵄨[󵄨󵄨󵄨󵄨𝑝󵄨󵄨󵄨󵄨󵄩 󵄩 󵄩 󵄩𝜇1󵄩 󵄩 󵄩 󵄩Γ(𝛼+𝛽)+󵄨󵄨󵄨󵄨𝑞󵄨󵄨󵄨󵄨󵄩 󵄩 󵄩 󵄩𝜇2󵄩 󵄩 󵄩 󵄩Γ(𝛼)] |𝑘|Γ(𝛼+𝛽)Γ(𝛼)<1, (25) where sup𝑡∈[0,1]|𝜇𝑖(𝑡)|=‖𝜇𝑖‖,𝑖=1,2. Proof. Let us fix 𝑟≥2󵄨󵄨󵄨󵄨󵄨1−𝑒−𝑘󵄨󵄨󵄨󵄨󵄨[󵄨󵄨󵄨󵄨𝑝󵄨󵄨󵄨󵄨󵄩 󵄩 󵄩 󵄩𝜇1󵄩 󵄩 󵄩 󵄩Γ(𝛼+𝛽)+󵄨󵄨󵄨󵄨𝑞󵄨󵄨󵄨󵄨󵄩 󵄩 󵄩 󵄩𝜇2󵄩 󵄩 󵄩 󵄩Γ(𝛼)] |𝑘|Γ(𝛼+𝛽)Γ(𝛼)(26) and consider 𝐵𝑟={𝑢∈P:‖𝑢‖≤𝑟}. We define the operators 𝜓1and 𝜓2on 𝐵𝑟as (𝜓1𝑢)(𝑡)=∫𝑡 0𝑒−𝑘(𝑡−𝑠) (𝑝∫𝑠 0(𝑠−𝑥)𝛼−2 Γ(𝛼−1)𝑓(𝑥,𝑢(𝑥))𝑑𝑥 +𝑞∫𝑠 0(𝑠−𝑥)𝛼+𝛽−2 Γ(𝛼+𝛽−1) ×𝑔(𝑥,𝑢(𝑥))𝑑𝑥)𝑑𝑠, 𝑡∈[0,1], (𝜓2𝑢)(𝑡)=(1−𝑒−𝑘𝑡) (𝑒−𝑘 −1) ×∫1 0𝑒−𝑘(1−𝑠) (𝑝∫𝑠 0(𝑠−𝑥)𝛼−2 Γ(𝛼−1)𝑓(𝑥,𝑢(𝑥))𝑑𝑥 +𝑞∫𝑠 0(𝑠−𝑥)𝛼+𝛽−2 Γ(𝛼+𝛽−1) ×𝑔(𝑥,𝑢(𝑥))𝑑𝑥)𝑑𝑠, 𝑡∈[0,1]. (27) For 𝑢,V∈𝐵𝑟,wefindthat 󵄩 󵄩 󵄩 󵄩𝜓1𝑢+𝜓2V󵄩 󵄩 󵄩 󵄩 ≤2󵄨󵄨󵄨󵄨󵄨1−𝑒−𝑘󵄨󵄨󵄨󵄨󵄨[󵄨󵄨󵄨󵄨𝑝󵄨󵄨󵄨󵄨󵄩 󵄩 󵄩 󵄩𝜇1󵄩 󵄩 󵄩 󵄩Γ(𝛼+𝛽)+󵄨󵄨󵄨󵄨𝑞󵄨󵄨󵄨󵄨󵄩 󵄩 󵄩 󵄩𝜇2󵄩 󵄩 󵄩 󵄩Γ(𝛼)] |𝑘|Γ(𝛼+𝛽)Γ(𝛼) ≤𝑟. (28) Thus, 𝜓1𝑢+𝜓 2V∈𝐵 𝑟. It follows from assumption (𝐴1)together with (25)that𝜓2is a contraction mapping. Continuities of 𝑓and 𝑔imply that the operator 𝜓1is continuous. Also, 𝜓1is uniformly bounded on 𝐵𝑟as 󵄩 󵄩 󵄩 󵄩𝜓1𝑢󵄩 󵄩 󵄩 󵄩 ≤󵄨󵄨󵄨󵄨󵄨1−𝑒−𝑘󵄨󵄨󵄨󵄨󵄨[󵄨󵄨󵄨󵄨𝑝󵄨󵄨󵄨󵄨󵄩 󵄩 󵄩 󵄩𝜇1󵄩 󵄩 󵄩 󵄩Γ(𝛼+𝛽)+󵄨󵄨󵄨󵄨𝑞󵄨󵄨󵄨󵄨󵄩 󵄩 󵄩 󵄩𝜇2󵄩 󵄩 󵄩 󵄩Γ(𝛼)] |𝑘|Γ(𝛼+𝛽)Γ(𝛼).(29) Now, we prove the compactness of the operator 𝜓1.Inviewof (𝐴1), we define sup (𝑡,𝑢)∈[0,1]×𝐵𝑟 󵄨󵄨󵄨󵄨𝑓(𝑡,𝑢)󵄨󵄨󵄨󵄨=𝑓, sup (𝑡,𝑢)∈[0,1]×𝐵𝑟 󵄨󵄨󵄨󵄨𝑔(𝑡,𝑢)󵄨󵄨󵄨󵄨=𝑔. (30) Consequently, we have 󵄩 󵄩 󵄩 󵄩(𝜓1𝑢)(𝑡2)−(𝜓1𝑢)(𝑡1)󵄩 󵄩 󵄩 󵄩 ≤󵄨󵄨󵄨󵄨󵄨𝑒−𝑘𝑡2−𝑒−𝑘𝑡1󵄨󵄨󵄨󵄨󵄨∫𝑡1 0𝑒𝑘𝑠 (󵄨󵄨󵄨󵄨𝑝󵄨󵄨󵄨󵄨𝑓∫𝑠 0(𝑠−𝑥)𝛼−2 Γ(𝛼−1)𝑑𝑥 +󵄨󵄨󵄨󵄨𝑞󵄨󵄨󵄨󵄨𝑔∫𝑠 0(𝑠−𝑥)𝛼+𝛽−2 Γ(𝛼+𝛽−1)𝑑𝑥)𝑑𝑠 ≤∫𝑡2 𝑡1𝑒−𝑘(𝑡2−𝑠) ×(󵄨󵄨󵄨󵄨𝑝󵄨󵄨󵄨󵄨𝑓∫𝑠 0(𝑠−𝑥)𝛼−2 Γ(𝛼−1)𝑑𝑥+󵄨󵄨󵄨󵄨𝑞󵄨󵄨󵄨󵄨𝑔∫𝑠 0(𝑠−𝑥)𝛼+𝛽−2 Γ(𝛼+𝛽−1)𝑑𝑥)𝑑𝑠 (31) which is independent of 𝑢and tends to zero as 𝑡2→𝑡 1.Thus, 𝜓1is relatively compact on 𝐵𝑟.Hence,bytheArzel ´ a-Ascoli theorem, 𝜓1is compact on 𝐵𝑟. Thus, all the assumptions of Lemma 5 are satisfied. So, by the conclusion of Lemma 5, problem (1)-(2) has at least one solution on [0,1]. Now, we show the existence of solutions for the problem (1)-(2) via Leray-Schauder alternative. Lemma 7 (nonlinear alternative for single valued maps, see [12]). Let 𝐸be a Banach space, 𝐶aclosed,convexsubsetof 𝐸, 𝑈an open subset of 𝐶,and0∈𝑈.Supposethat𝐹:𝑈→𝐶 is a continuous, compact (that is, 𝐹(𝑈)is a relatively compact subset of 𝐶)map.Then,either (i) 𝐹has a fixed point in 𝑈,or (ii) there is a 𝑢∈𝜕𝑈(the boundary of 𝑈in 𝐶)and𝜆∈ (0,1)with 𝑢=𝜆𝐹(𝑢). Theorem 8. Let 𝑓,𝑔 : [0,1] × R→Rbe continuous functions and the following assumptions hold. (𝐴3)There exist functions 𝜎1,𝜎2∈ 𝐶([0,1],R+),and nondecreasing functions 𝜓1,𝜓2:R+→R+ such that |𝑓(𝑡,𝑢)| ≤ 𝜎1(𝑡)𝜓1(‖𝑢‖),|𝑔(𝑡,𝑢)| ≤ 𝜎2(𝑡)𝜓2(‖𝑢‖),forall(𝑡,𝑢)∈[0,1]×R. 6 Journal of Function Spaces and Applications (𝐴4)There exists a constant 𝑀>0such that 𝑀×((2󵄨󵄨󵄨󵄨󵄨1−𝑒−𝑘󵄨󵄨󵄨󵄨󵄨[󵄨󵄨󵄨󵄨𝑝󵄨󵄨󵄨󵄨𝜓1(‖𝑢‖)Γ(𝛼+𝛽)󵄩 󵄩 󵄩 󵄩𝜎1󵄩 󵄩 󵄩 󵄩 +󵄨󵄨󵄨󵄨𝑞󵄨󵄨󵄨󵄨𝜓2(‖𝑢‖)󵄩 󵄩 󵄩 󵄩𝜎2󵄩 󵄩 󵄩 󵄩Γ(𝛼)]) ×(|𝑘|Γ(𝛼+𝛽)Γ(𝛼))−1)−1 >1. (32) Then, the boundary value problem (1)-(2)has at least one solution on [0,1]. Proof. Consider the operator V:P→Pwith 𝑢=V𝑢, where (V𝑢)(𝑡)=(1−𝑒−𝑘𝑡) (𝑒−𝑘 −1) ×∫1 0𝑒−𝑘(1−𝑠) (𝑝∫𝑠 0(𝑠−𝑥)𝛼−2 Γ(𝛼−1) ×𝑓(𝑥,𝑢(𝑥))𝑑𝑥 +𝑞∫𝑠 0(𝑠−𝑥)𝛼+𝛽−2 Γ(𝛼+𝛽−1) ×𝑔(𝑥,𝑢(𝑥))𝑑𝑥)𝑑𝑠 +∫𝑡 0𝑒−𝑘(𝑡−𝑠) (𝑝∫𝑠 0(𝑠−𝑥)𝛼−2 Γ(𝛼−1)𝑓(𝑥,𝑢(𝑥))𝑑𝑥 +𝑞∫𝑠 0(𝑠−𝑥)𝛼+𝛽−2 Γ(𝛼+𝛽−1) ×𝑔(𝑥,𝑢(𝑥))𝑑𝑥)𝑑𝑠. (33) We show that Vmaps bounded sets into bounded sets in 𝐶([0,1],R).Forapositivenumber𝑟,let𝐵𝑟={𝑢∈ 𝐶([0,1],R):‖𝑢‖≤𝑟}be a bounded set in 𝐶([0,1],R).Then, |(V𝑢)(𝑡)| ≤󵄨󵄨󵄨󵄨󵄨󵄨󵄨󵄨󵄨1−𝑒−𝑘𝑡 1−𝑒−𝑘 󵄨󵄨󵄨󵄨󵄨󵄨󵄨󵄨󵄨∫1 0𝑒−𝑘(1−𝑠) ×(󵄨󵄨󵄨󵄨𝑝󵄨󵄨󵄨󵄨∫𝑠 0(𝑠−𝑥)𝛼−2 Γ(𝛼−1)󵄨󵄨󵄨󵄨𝑓(𝑥,𝑢(𝑥))󵄨󵄨󵄨󵄨𝑑𝑥 +󵄨󵄨󵄨󵄨𝑞󵄨󵄨󵄨󵄨∫𝑠 0(𝑠−𝑥)𝛼+𝛽−2 Γ(𝛼+𝛽−1)󵄨󵄨󵄨󵄨𝑔(𝑥,𝑢(𝑥))󵄨󵄨󵄨󵄨𝑑𝑥)𝑑𝑠 +∫𝑡 0𝑒−𝑘(𝑡−𝑠) (󵄨󵄨󵄨󵄨𝑝󵄨󵄨󵄨󵄨∫𝑠 0(𝑠−𝑥)𝛼−2 Γ(𝛼−1)󵄨󵄨󵄨󵄨𝑓(𝑥,𝑢(𝑥))󵄨󵄨󵄨󵄨𝑑𝑥 +󵄨󵄨󵄨󵄨𝑞󵄨󵄨󵄨󵄨∫𝑠 0(𝑠−𝑥)𝛼+𝛽−2 Γ(𝛼+𝛽−1) ×󵄨󵄨󵄨󵄨𝑔(𝑥,𝑢(𝑥))󵄨󵄨󵄨󵄨𝑑𝑥)𝑑𝑠 ≤󵄨󵄨󵄨󵄨󵄨󵄨󵄨󵄨󵄨1−𝑒−𝑘𝑡 1−𝑒−𝑘 󵄨󵄨󵄨󵄨󵄨󵄨󵄨󵄨󵄨 ×∫1 0𝑒−𝑘(1−𝑠) (󵄨󵄨󵄨󵄨𝑝󵄨󵄨󵄨󵄨∫𝑠 0(𝑠−𝑥)𝛼−2 Γ(𝛼−1)𝜎1(𝑥)𝜓1(‖𝑢‖)𝑑𝑥 +󵄨󵄨󵄨󵄨𝑞󵄨󵄨󵄨󵄨∫𝑠 0(𝑠−𝑥)𝛼+𝛽−2 Γ(𝛼+𝛽−1)𝜎2(𝑥)𝜓2(‖𝑢‖)𝑑𝑥)𝑑𝑠 +∫𝑡 0𝑒−𝑘(𝑡−𝑠) (󵄨󵄨󵄨󵄨𝑝󵄨󵄨󵄨󵄨∫𝑠 0(𝑠−𝑥)𝛼−2 Γ(𝛼−1)𝜎1(𝑥)𝜓1(‖𝑢‖)𝑑𝑥 +󵄨󵄨󵄨󵄨𝑞󵄨󵄨󵄨󵄨∫𝑠 0(𝑠−𝑥)𝛼+𝛽−2 Γ(𝛼+𝛽−1) ×𝜎 2(𝑥)𝜓2(‖𝑢‖)𝑑𝑥)𝑑𝑠 ≤󵄨󵄨󵄨󵄨𝑝󵄨󵄨󵄨󵄨𝜓1(𝑟)󵄩 󵄩 󵄩 󵄩𝜎1󵄩 󵄩 󵄩 󵄩 ×[󵄨󵄨󵄨󵄨󵄨󵄨󵄨󵄨󵄨1−𝑒−𝑘𝑡 1−𝑒−𝑘 󵄨󵄨󵄨󵄨󵄨󵄨󵄨󵄨󵄨∫1 0𝑒−𝑘(1−𝑠) (∫𝑠 0(𝑠−𝑥)𝛼−2 Γ(𝛼−1)𝑑𝑥)𝑑𝑠 +∫𝑡 0𝑒−𝑘(𝑡−𝑠) (∫𝑠 0(𝑠−𝑥)𝛼−2 Γ(𝛼−1)𝑑𝑥)𝑑𝑠] +󵄨󵄨󵄨󵄨𝑞󵄨󵄨󵄨󵄨󵄩 󵄩 󵄩 󵄩𝜎2󵄩 󵄩 󵄩 󵄩𝜓2(𝑟)[󵄨󵄨󵄨󵄨󵄨󵄨󵄨󵄨󵄨1−𝑒−𝑘𝑡 1−𝑒−𝑘 󵄨󵄨󵄨󵄨󵄨󵄨󵄨󵄨󵄨 ×∫1 0𝑒−𝑘(1−𝑠) (∫𝑠 0(𝑠−𝑥)𝛼+𝛽−2 Γ(𝛼+𝛽−1)𝑑𝑥)𝑑𝑠 +∫𝑡 0𝑒−𝑘(𝑡−𝑠) (∫𝑠 0(𝑠−𝑥)𝛼+𝛽−2 Γ(𝛼+𝛽−1)𝑑𝑥)𝑑𝑠] ≤(2󵄨󵄨󵄨󵄨󵄨1−𝑒−𝑘󵄨󵄨󵄨󵄨󵄨[󵄨󵄨󵄨󵄨𝑝󵄨󵄨󵄨󵄨𝜓1(‖𝑢‖)Γ(𝛼+𝛽)󵄩 󵄩 󵄩 󵄩𝜎1󵄩 󵄩 󵄩 󵄩 +󵄨󵄨󵄨󵄨𝑞󵄨󵄨󵄨󵄨𝜓2(‖𝑢‖)󵄩 󵄩 󵄩 󵄩𝜎2󵄩 󵄩 󵄩 󵄩Γ(𝛼)]) ×(|𝑘|Γ(𝛼+𝛽)Γ(𝛼))−1.(34) Consequently, ‖V𝑥‖≤(2󵄨󵄨󵄨󵄨󵄨1−𝑒𝑘󵄨󵄨󵄨󵄨󵄨 ×[󵄨󵄨󵄨󵄨𝑝󵄨󵄨󵄨󵄨𝜓1(‖𝑢‖)Γ(𝛼+𝛽)󵄩 󵄩 󵄩 󵄩𝜎1󵄩 󵄩 󵄩 󵄩 +󵄨󵄨󵄨󵄨𝑞󵄨󵄨󵄨󵄨𝜓2(‖𝑢‖)󵄩 󵄩 󵄩 󵄩𝜎2󵄩 󵄩 󵄩 󵄩Γ(𝛼)]) ×(|𝑘|Γ(𝛼+𝛽)Γ(𝛼))−1. (35) Next, we show that Vmaps bounded sets into equicontinuous sets of 𝐶([0,1],R).Let𝑡1,𝑡2∈[0,1]with 𝑡1<𝑡2and Journal of Function Spaces and Applications 7 𝑢∈𝐵 𝑟,where𝐵𝑟is a bounded set of 𝐶([0,1],R).Then,we obtain 󵄩 󵄩 󵄩 󵄩(V𝑢)(𝑡2)−(V𝑢)(𝑡1)󵄩 󵄩 󵄩 󵄩 ≤󵄨󵄨󵄨󵄨󵄨󵄨󵄨󵄨󵄨−𝑒−𝑘𝑡2+𝑒−𝑘𝑡1 1−𝑒−𝑘 󵄨󵄨󵄨󵄨󵄨󵄨󵄨󵄨󵄨 ×∫1 0𝑒−𝑘(1−𝑠) (󵄨󵄨󵄨󵄨𝑝󵄨󵄨󵄨󵄨𝜓1(𝑟)󵄩 󵄩 󵄩 󵄩𝜎1󵄩 󵄩 󵄩 󵄩 ×∫𝑠 0(𝑠−𝑥)𝛼−2 Γ(𝛼−1)𝑑𝑥 +󵄨󵄨󵄨󵄨𝑞󵄨󵄨󵄨󵄨𝜓2(𝑟)󵄩 󵄩 󵄩 󵄩𝜎2󵄩 󵄩 󵄩 󵄩 ×∫𝑠 0(𝑠−𝑥)𝛼+𝛽−2 Γ(𝛼+𝛽−1)𝑑𝑥)𝑑𝑠 +󵄨󵄨󵄨󵄨󵄨𝑒−𝑘𝑡2−𝑒−𝑘𝑡1󵄨󵄨󵄨󵄨󵄨 ×∫𝑡1 0𝑒𝑘𝑠 (󵄨󵄨󵄨󵄨𝑝󵄨󵄨󵄨󵄨𝜓1(𝑟)󵄩 󵄩 󵄩 󵄩𝜎1󵄩 󵄩 󵄩 󵄩 ×∫𝑠 0(𝑠−𝑥)𝛼−2 Γ(𝛼−1)𝑑𝑥 +󵄨󵄨󵄨󵄨𝑞󵄨󵄨󵄨󵄨𝜓2(𝑟)󵄩 󵄩 󵄩 󵄩𝜎2󵄩 󵄩 󵄩 󵄩 ×∫𝑠 0(𝑠−𝑥)𝛼+𝛽−2 Γ(𝛼+𝛽−1)𝑑𝑥)𝑑𝑠 +∫𝑡2 𝑡1𝑒−𝑘(𝑡2−𝑠) (󵄨󵄨󵄨󵄨𝑝󵄨󵄨󵄨󵄨𝜓1(𝑟)󵄩 󵄩 󵄩 󵄩𝜎1󵄩 󵄩 󵄩 󵄩∫𝑠 0(𝑠−𝑥)𝛼−2 Γ(𝛼−1)𝑑𝑥 +󵄨󵄨󵄨󵄨𝑞󵄨󵄨󵄨󵄨𝜓2(𝑟)󵄩 󵄩 󵄩 󵄩𝜎2󵄩 󵄩 󵄩 󵄩 ×∫𝑠 0(𝑠−𝑥)𝛼+𝛽−2 Γ(𝛼+𝛽−1)𝑑𝑥)𝑑𝑠. (36) Obviously, the right hand side of the previous inequality tends to zero independently of 𝑢∈𝐵 𝑟as 𝑡2−𝑡1→0.AsV satisfies the previous assumptions, therefore it follows by the Arzel´ a-Ascoli theorem that V:𝐶([0,1],R) → 𝐶([0,1],R) is completely continuous. The proof will be complete by the application of the LeraySchauder nonlinear alternative (Lemma 7)onceweestablish the boundedness of the set of all solutions to equations 𝑢= 𝜆V𝑢for 𝜆∈(0,1). Let 𝑢be a solution. Then, for 𝑡∈[0,1],andusingthe computations in proving that Vis bounded, we have |𝑢(𝑡)|=|𝜆(V𝑢)(𝑡)| ≤(2󵄨󵄨󵄨󵄨󵄨1−𝑒−𝑘󵄨󵄨󵄨󵄨󵄨 ×[󵄨󵄨󵄨󵄨𝑝󵄨󵄨󵄨󵄨𝜓1(‖𝑢‖)Γ(𝛼+𝛽)󵄩 󵄩 󵄩 󵄩𝜎1󵄩 󵄩 󵄩 󵄩 +󵄨󵄨󵄨󵄨𝑞󵄨󵄨󵄨󵄨𝜓2(‖𝑢‖)󵄩 󵄩 󵄩 󵄩𝜎2󵄩 󵄩 󵄩 󵄩Γ(𝛼)]) ×(|𝑘|Γ(𝛼+𝛽)Γ(𝛼))−1. (37) Consequently, we have ‖𝑢‖×((2󵄨󵄨󵄨󵄨󵄨1−𝑒−𝑘󵄨󵄨󵄨󵄨󵄨[󵄨󵄨󵄨󵄨𝑝󵄨󵄨󵄨󵄨𝜓1(‖𝑢‖)Γ(𝛼+𝛽)󵄩 󵄩 󵄩 󵄩𝜎1󵄩 󵄩 󵄩 󵄩 +󵄨󵄨󵄨󵄨𝑞󵄨󵄨󵄨󵄨𝜓2(‖𝑢‖)󵄩 󵄩 󵄩 󵄩𝜎2󵄩 󵄩 󵄩 󵄩Γ(𝛼)]) ×(|𝑘|Γ(𝛼+𝛽)Γ(𝛼))−1)−1 ≤1. (38) In view of (𝐴4), there exists 𝑀such that ‖𝑢‖ =𝑀.Letusset 𝑈={𝑢∈𝐶([0,1],R):‖𝑢‖<𝑀}.(39) Note that the operator V:𝑈 → 𝐶([0,1],R)is continuous and completely continuous. From the choice of 𝑈,thereis no 𝑢∈𝜕𝑈such that 𝑢=𝜆V(𝑢) for some 𝜆 ∈ (0,1). Consequently, by the nonlinear alternative of Leray-Schauder type (Lemma 7), we deduce that Vhas a fixed point 𝑢∈𝑈 which is a solution of the problem (1)-(2). This completes the proof. Example 9. Consider a boundary value problem of integrodifferential equations of fractional order given by (𝑐𝐷3/2 +2𝑐𝐷1/2)𝑢(𝑡) =1 2𝑓(𝑡,𝑢(𝑡))+𝐼1/2𝑔(𝑡,𝑢(𝑡)), 0<𝑡<1, 𝑢(0)=0, 𝑢(1)=0, (40) where 𝛼=3/2,𝑘=2,𝑝=1/2,𝑞=1,𝛽=1/2,𝑓(𝑡,𝑢) = (|𝑢|(2+|𝑢|))/(3(1+|𝑢|))+4𝑡,𝑔(𝑡,𝑢)=(1/4)tan−1𝑢+cos2𝑡+𝑡3+ 5. With the given data, it is found that 𝐿1=2/3,𝐿2=1/4as |𝑓(𝑡,𝑢)−𝑓(𝑡,V)|≤(2/3)|𝑢−V|,|𝑔(𝑡,𝑢)−𝑔(𝑡,V)|≤(1/4)|𝑢−V|, and 𝑄=2󵄨󵄨󵄨󵄨󵄨1−𝑒−𝑘󵄨󵄨󵄨󵄨󵄨[󵄨󵄨󵄨󵄨𝑝󵄨󵄨󵄨󵄨Γ(𝛼+𝛽)+󵄨󵄨󵄨󵄨𝑞󵄨󵄨󵄨󵄨Γ(𝛼)] |𝑘|Γ(𝛼+𝛽)Γ(𝛼)≃1.3525. (41) Clearly, 𝐿=max{𝐿1,𝐿2}=2/3and 𝐿<1/𝑄.Thus,all the assumptions of Theorem 4 are satisfied. Hence, by the conclusion of Theorem 4,theproblem(40)hasaunique solution. Acknowledgment The authors thank the anonymous referees for their valuable comments. The research of J. J. Nieto has been partially 8 Journal of Function Spaces and Applications supported by Ministerio de Economia y Competitividad (Spain), project MTM2010-15314, and co-financed by the European Community fund FEDER. References [1] S. G. 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