Existence and Uniqueness Results for a Coupled System of Nonlinear ...

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Dec 13, 2013 - This paper studies the existence and uniqueness of solutions for a coupled ... Leray-Schauder alternative; then we obtain the uniqueness.
Hindawi Publishing Corporation Abstract and Applied Analysis Volume 2014, Article ID 463517, 7 pages http://dx.doi.org/10.1155/2014/463517

Research Article Existence and Uniqueness Results for a Coupled System of Nonlinear Fractional Differential Equations with Antiperiodic Boundary Conditions Huina Zhang1,2 and Wenjie Gao1 1 2

Institute of Mathematics, Jilin University, Changchun 130012, China College of Science, China University of Petroleum (East China), Qingdao 266555, China

Correspondence should be addressed to Wenjie Gao; [email protected] Received 28 October 2013; Accepted 13 December 2013; Published 12 January 2014 Academic Editor: Douglas Anderson Copyright Β© 2014 H. Zhang and W. Gao. 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. This paper studies the existence and uniqueness of solutions for a coupled system of nonlinear fractional differential equations of order 𝛼, 𝛽 ∈ (4, 5] with antiperiodic boundary conditions. Our results are based on the nonlinear alternative of Leray-Schauder type and the contraction mapping principle. Two illustrative examples are also presented.

1. Introduction In this paper, we consider the existence and uniqueness of solutions for the following coupled system of nonlinear fractional differential equations: 𝑐

𝐷𝛼 π‘₯ (𝑑) + 𝑓 (𝑑, π‘₯ (𝑑) , 𝑦 (𝑑)) = 0,

𝑑 ∈ [0, 𝑇] ,

𝑐

𝐷𝛽 𝑦 (𝑑) + 𝑔 (𝑑, π‘₯ (𝑑) , 𝑦 (𝑑)) = 0,

𝑑 ∈ [0, 𝑇] ,

π‘₯(𝑖) (0) = βˆ’π‘₯(𝑖) (𝑇) ,

𝑖 = 0, 1, 2, 3, 4,

𝑦(𝑖) (0) = βˆ’π‘¦(𝑖) (𝑇) ,

𝑖 = 0, 1, 2, 3, 4,

𝑐

𝛼

(1)

where 4 < 𝛼, 𝛽 ≀ 5, and 𝐷 denotes the Caputo fractional derivative of order 𝛼. Here our nonlinearity 𝑓, 𝑔 : [0, 𝑇] Γ— R Γ— R β†’ R are given continuous functions. Fractional differential equations have recently been addressed by many researchers in various fields of science and engineering, such as rheology, porous media, fluid flows, chemical physics, and many other branches of science; see [1–4]. As a matter of fact, fractional-order models become more realistic and practical than the classical integer-order models; as a consequence, there are a large number of papers

and books dealing with the existence and uniqueness of solutions to nonlinear fractional differential equations; see [5–14]. The study of a coupled system of fractional order is also very significant because this kind of system can often occur in applications; see [15–17]. Antiperiodic boundary value problems arise in the mathematical modeling of a variety of physical process; many authors have paid much attention to such problems; for examples and details of Antiperiodic boundary conditions, see [5, 18–22]. In [5], Alsaedi et al. study an Antiperiodic boundary value problem of nonlinear fractional differential equations of order π‘ž ∈ (4, 5]. It should be noted that, in [23], Ntouyas and Obaid have researched a coupled system of fractional differential equations with nonlocal integral boundary conditions, but this paper researches a coupled system of fractional differential equations with Antiperiodic boundary conditions. On the other hand, in [5, 19], the authors have discussed some existence results of solutions for Antiperiodic boundary value problems of fractional differential equation but not the coupled system. The rest of the papers above for the coupled systems have been devoted to the case of

2

Abstract and Applied Analysis

Riemann-Liouville fractional derivatives but not the Caputo fractional derivatives. This paper is organized as follows. In Section 2, we recall some basic definitions and preliminary results. In Section 3, we give the existence results of (1) by means of the Leray-Schauder alternative; then we obtain the uniqueness of solutions for system (1) by the contraction mapping principle. We give two examples in Section 4 to illustrate the applicability of our results.

2. Background Materials For the convenience of the readers, we present here some necessary definitions and lemmas which are used throughout this paper. Definition 1 (see [2, 3]). The Riemann-Liouville fractional integral of order 𝛼 > 0 of a function 𝑦 : (0, ∞) β†’ R is given by

where 𝐺(𝑑, 𝑠) is the Green function given by 𝐺 (𝑑, 𝑠) 2(𝑑 βˆ’ 𝑠)π‘žβˆ’1 βˆ’ (𝑇 βˆ’ 𝑠)π‘žβˆ’1 (𝑇 βˆ’ 2𝑑) (𝑇 βˆ’ 𝑠)π‘žβˆ’2 + 2Ξ“ (π‘ž) 4Ξ“ (π‘ž βˆ’ 1) 2 π‘žβˆ’3 (6𝑇𝑑 βˆ’ 4𝑑3 βˆ’ 𝑇3 ) (𝑇 βˆ’ 𝑠)π‘žβˆ’4 𝑑 (𝑇 βˆ’ 𝑑) (𝑇 βˆ’ 𝑠) + + 4Ξ“ (π‘ž βˆ’ 2) 48Ξ“ (π‘ž βˆ’ 3) (2𝑇𝑑3 βˆ’ 𝑑𝑇3 βˆ’ 𝑑4 ) (𝑇 βˆ’ 𝑠)π‘žβˆ’5 + , 0 < 𝑠 < 𝑑 < 𝑇, 48Ξ“ (π‘ž βˆ’ 4) (𝑇 βˆ’ 𝑠)π‘žβˆ’1 (𝑇 βˆ’ 2𝑑) (𝑇 βˆ’ 𝑠)π‘žβˆ’2 βˆ’ + 2Ξ“ (π‘ž) 4Ξ“ (π‘ž βˆ’ 1) π‘žβˆ’4 2 3 3 𝑑 (𝑇 βˆ’ 𝑑) (𝑇 βˆ’ 𝑠)π‘žβˆ’3 (6𝑇𝑑 βˆ’ 4𝑑 βˆ’ 𝑇 ) (𝑇 βˆ’ 𝑠) + + 4Ξ“ (π‘ž βˆ’ 2) 48Ξ“ (π‘ž βˆ’ 3) (2𝑇𝑑3 βˆ’ 𝑑𝑇3 βˆ’ 𝑑4 ) (𝑇 βˆ’ 𝑠)π‘žβˆ’5 + , 0 < 𝑑 < 𝑠 < 𝑇. 48Ξ“ (π‘ž βˆ’ 4) (6)

{ { { { { { { { { { { { { { { { { { { { { { { ={ { { { { { { { { { { { { { { { { { { { { { { { Let

𝐼𝛼 𝑦 (𝑑) =

𝑑 1 ∫ (𝑑 βˆ’ 𝑠)π›Όβˆ’1 𝑦 (𝑠) 𝑑𝑠, Ξ“ (𝛼) 0

(2)

provided the right hand side is pointwise defined on (0, ∞). Definition 2 (see [2, 3]). The Caputo fractional derivative of order 𝛼 > 0 of a continuous function 𝑦 : (0, ∞) β†’ R is given by

𝑐

𝐷𝛼 𝑦 (𝑑) =

𝑑 𝑦(𝑛) (𝑠) 1 𝑑𝑠, ∫ Ξ“ (𝑛 βˆ’ 𝛼) 0 (𝑑 βˆ’ 𝑠)π›Όβˆ’π‘›+1

(3)

where 𝑛 = [𝛼] + 1 and [𝛼] denotes the integer part of number 𝛼, provided that the right side is pointwise defined on (0, ∞). 𝛼

Lemma 3 (see [19]). Consider 𝐼𝛼 𝑐 𝐷 π‘₯(𝑑) = π‘₯(𝑑) + 𝐢0 + 𝐢1 𝑑 + 𝐢2 𝑑2 + β‹… β‹… β‹… + πΆπ‘›βˆ’1 π‘‘π‘›βˆ’1 , for some 𝑐𝑖 ∈ R, 𝑖 = 0, 1, 2, . . . , 𝑛 βˆ’ 1, where 𝑛 = [𝛼] + 1 and [𝛼] denotes the integer part of number 𝛼. Lemma 4 (see [5]). For any 𝑦 ∈ 𝐢[0, 𝑇], the unique solution of the boundary value problem 𝑐

π·π‘ž π‘₯ (𝑑) = 𝑦 (𝑑) ,

𝑑 ∈ [0, 𝑇] , 4 < π‘ž ≀ 5,

π‘₯(𝑖) (0) = βˆ’π‘₯(𝑖) (𝑇) ,

(4) 𝑖 = 0, 1, 2, 3, 4

is 𝑇

π‘₯ (𝑑) = ∫ 𝐺 (𝑑, 𝑠) 𝑦 (𝑠) 𝑑𝑠, 0

(5)

𝐺1 (𝑑, 𝑠) { { { { { { { { { { { { { { { { { { { { { { ={ { { { { { { { { { { { { { { { { { { { { { {

(𝑑 βˆ’ 𝑠)π›Όβˆ’1 βˆ’ (1/2) (𝑇 βˆ’ 𝑠)π›Όβˆ’1 (𝑇 βˆ’ 2𝑑) (𝑇 βˆ’ 𝑠)π›Όβˆ’2 + Ξ“ (𝛼) 4Ξ“ (𝛼 βˆ’ 1) π›Όβˆ’4 2 3 3 𝑑 (𝑇 βˆ’ 𝑑) (𝑇 βˆ’ 𝑠)π›Όβˆ’3 (6𝑇𝑑 βˆ’ 4𝑑 βˆ’ 𝑇 ) (𝑇 βˆ’ 𝑠) + + 4Ξ“ (𝛼 βˆ’ 2) 48Ξ“ (𝛼 βˆ’ 3) 3 3 4 (2𝑇𝑑 βˆ’ 𝑑𝑇 βˆ’ 𝑑 ) (𝑇 βˆ’ 𝑠)π›Όβˆ’5 + , 0 < 𝑠 < 𝑑 < 𝑇, 48Ξ“ (𝛼 βˆ’ 4) 1 (𝑇 βˆ’ 2𝑑) (𝑇 βˆ’ 𝑠)π›Όβˆ’2 βˆ’ (𝑇 βˆ’ 𝑠)π›Όβˆ’1 + 2Ξ“ (𝛼) 4Ξ“ (𝛼 βˆ’ 1) 2 π›Όβˆ’3 (6𝑇𝑑 βˆ’ 4𝑑3 βˆ’ 𝑇3 ) (𝑇 βˆ’ 𝑠)π›Όβˆ’4 𝑑 (𝑇 βˆ’ 𝑑) (𝑇 βˆ’ 𝑠) + + 4Ξ“ (𝛼 βˆ’ 2) 48Ξ“ (𝛼 βˆ’ 3) (2𝑇𝑑3 βˆ’ 𝑑𝑇3 βˆ’ 𝑑4 ) (𝑇 βˆ’ 𝑠)π›Όβˆ’5 , 0 < 𝑑 < 𝑠 < 𝑇, + 48Ξ“ (𝛼 βˆ’ 4)

𝐺2 (𝑑, 𝑠) { { { { { { { { { { { { { { { { { { { { { { { ={ { { { { { { { { { { { { { { { { { { { { { { {

(𝑑 βˆ’ 𝑠)π›½βˆ’1 βˆ’ (1/2) (𝑇 βˆ’ 𝑠)π›½βˆ’1 (𝑇 βˆ’ 2𝑑) (𝑇 βˆ’ 𝑠)π›½βˆ’2 + Ξ“ (𝛽) 4Ξ“ (𝛽 βˆ’ 1) 2 3 π›½βˆ’3 (6𝑇𝑑 βˆ’ 4𝑑 βˆ’ 𝑇3 ) (𝑇 βˆ’ 𝑠)π›½βˆ’4 𝑑 (𝑇 βˆ’ 𝑑) (𝑇 βˆ’ 𝑠) + + 4Ξ“ (𝛽 βˆ’ 2) 48Ξ“ (𝛽 βˆ’ 3) (2𝑇𝑑3 βˆ’ 𝑑𝑇3 βˆ’ 𝑑4 ) (𝑇 βˆ’ 𝑠)π›½βˆ’5 + , 0 < 𝑠 < 𝑑 < 𝑇, 48Ξ“ (𝛽 βˆ’ 4) 1 (𝑇 βˆ’ 2𝑑) (𝑇 βˆ’ 𝑠)π›½βˆ’2 βˆ’ (𝑇 βˆ’ 𝑠)π›½βˆ’1 + 2Ξ“ (𝛽) 4Ξ“ (𝛽 βˆ’ 1) 2 π›½βˆ’3 (6𝑇𝑑 βˆ’ 4𝑑3 βˆ’ 𝑇3 ) (𝑇 βˆ’ 𝑠)π›½βˆ’4 𝑑 (𝑇 βˆ’ 𝑑) (𝑇 βˆ’ 𝑠) + + 4Ξ“ (𝛽 βˆ’ 2) 48Ξ“ (𝛽 βˆ’ 3) (2𝑇𝑑3 βˆ’ 𝑑𝑇3 βˆ’ 𝑑4 ) (𝑇 βˆ’ 𝑠)π›½βˆ’5 + , 0 < 𝑑 < 𝑠 < 𝑇. 48Ξ“ (𝛽 βˆ’ 4) (7)

We call (𝐺1 , 𝐺2 ) Green’s function for problem (1). We define the space 𝑋 = {π‘₯(𝑑) | π‘₯(𝑑) ∈ 𝐢[0, 𝑇]} endowed with β€–π‘₯‖𝑋 = maxπ‘‘βˆˆ[0,𝑇] |π‘₯(𝑑)|; for (π‘₯, 𝑦) ∈ 𝑋 Γ— 𝑋,

Abstract and Applied Analysis

3

let β€–(π‘₯, 𝑦)‖𝑋×𝑋 = β€–π‘₯‖𝑋 + ‖𝑦‖𝑋 . Obviously, (𝑋, β€– β‹… ‖𝑋 ) is a Banach space, and the product space (𝑋 Γ— 𝑋, β€–(β‹…, β‹…)‖𝑋×𝑋 ) is also a Banach space. Consider the following coupled system of the integral equations:

󡄨 𝑇 󡄨󡄨 󡄨 πœ•πΊ (𝑑, 𝑠) 󡄨󡄨󡄨 ∫ 󡄨󡄨󡄨 1 󡄨 𝑑𝑠 πœ•π‘‘ 󡄨󡄨󡄨 0 󡄨󡄨 ≀

π‘‡π›Όβˆ’1 3 𝛼3 βˆ’ 3𝛼2 + 14𝛼 βˆ’ 12 ( + ) = π‘ˆ3 , Ξ“ (𝛼) 2 48 𝑑 ∈ [0, 𝑇] ,

𝑇

π‘₯ (𝑑) = ∫ 𝐺1 (𝑑, 𝑠) 𝑓 (𝑠, π‘₯ (𝑠) , 𝑦 (𝑠)) 𝑑𝑠, 0

(8)

𝑇

𝑦 (𝑑) = ∫ 𝐺2 (𝑑, 𝑠) 𝑔 (𝑠, π‘₯ (𝑠) , 𝑦 (𝑠)) 𝑑𝑠.

󡄨 𝑇 󡄨󡄨 󡄨 πœ•πΊ (𝑑, 𝑠) 󡄨󡄨󡄨 ∫ 󡄨󡄨󡄨 2 󡄨 𝑑𝑠 πœ•π‘‘ 󡄨󡄨󡄨 0 󡄨󡄨

0

As a result, differential problem (1) turns into integral problem (8), and here is a conclusion about the relationship between their solutions. Lemma 5. Assume that 𝑓, 𝑔 : [0, 𝑇] Γ— R Γ— R β†’ R are continuous functions. Then (π‘₯, 𝑦) ∈ (𝑋, 𝑋) is a solution of (1) if and only if (π‘₯, 𝑦) ∈ (𝑋, 𝑋) is a solution of system (8). Proof. The proof is immediate from the discussion above, and we omit the details here.

≀

𝑇

󡄨 󡄨 ∫ 󡄨󡄨󡄨𝐺1 (𝑑, 𝑠)󡄨󡄨󡄨 𝑑𝑠 0 󡄨󡄨 𝑑 󡄨 󡄨 󡄨 󡄨 (𝑑 βˆ’ 𝑠)π›Όβˆ’1 󡄨󡄨󡄨 1 󡄨󡄨󡄨 𝑇 (𝑇 βˆ’ 𝑠)π›Όβˆ’1 󡄨󡄨󡄨 ≀ σ΅„¨σ΅„¨σ΅„¨σ΅„¨βˆ« 𝑑𝑠󡄨󡄨󡄨 + σ΅„¨σ΅„¨σ΅„¨βˆ« 𝑑𝑠󡄨󡄨󡄨 Ξ“ (𝛼) 󡄨󡄨 0 Ξ“ (𝛼) 󡄨󡄨 2 󡄨󡄨 0 󡄨󡄨 +

|𝑇 βˆ’ 2𝑑| 𝑇 (𝑇 βˆ’ 𝑠)π›Όβˆ’2 𝑑𝑠 ∫ 4 0 Ξ“ (𝛼 βˆ’ 1)

|𝑑 (𝑇 βˆ’ 𝑑)| 𝑇 (𝑇 βˆ’ 𝑠)π›Όβˆ’3 𝑑𝑠 ∫ 4 0 Ξ“ (𝛼 βˆ’ 2) 󡄨󡄨󡄨6𝑇𝑑2 βˆ’ 4𝑑3 βˆ’ 𝑇3 󡄨󡄨󡄨 𝑇 󡄨󡄨 󡄨 (𝑇 βˆ’ 𝑠)π›Όβˆ’4 𝑑𝑠 +󡄨 ∫ 48 0 Ξ“ (𝛼 βˆ’ 3) 󡄨󡄨󡄨2𝑇𝑑3 βˆ’ 𝑑𝑇3 βˆ’ 𝑑4 󡄨󡄨󡄨 𝑇 󡄨󡄨 󡄨 (𝑇 βˆ’ 𝑠)π›Όβˆ’5 𝑑𝑠 +󡄨 ∫ 48 0 Ξ“ (𝛼 βˆ’ 4) +

𝐹1 (π‘₯, 𝑦) (𝑑) = ∫ 𝐺1 (𝑑, 𝑠) 𝑓 (𝑠, π‘₯ (𝑠) , 𝑦 (𝑠)) 𝑑𝑠, (9)

𝐹2 (π‘₯, 𝑦) (𝑑) = ∫ 𝐺2 (𝑑, 𝑠) 𝑔 (𝑠, π‘₯ (𝑠) , 𝑦 (𝑠)) 𝑑𝑠. 0

It is obvious that a fixed point of the operator 𝐹 is a solution of problem (1). ≀

3. Main Results In this section, we will discuss the existence and uniqueness of solutions for problem (1). Lemma 6. One can conclude that the Green functions 𝐺1 (𝑑, 𝑠), 𝐺2 (𝑑, 𝑠) satisfy the following estimates: 𝑇

󡄨 󡄨 ∫ 󡄨󡄨󡄨𝐺1 (𝑑, 𝑠)󡄨󡄨󡄨 𝑑𝑠 0 ≀

(13)

Proof. For any 𝑑 ∈ [0, 𝑇],

𝑇

𝑇

π‘‡π›½βˆ’1 3 𝛽3 βˆ’ 3𝛽2 + 14𝛽 βˆ’ 12 ( + ) = π‘ˆ4 , 48 Ξ“ (𝛽) 2 𝑑 ∈ [0, 𝑇] .

Let 𝐹 : 𝑋 Γ— 𝑋 β†’ 𝑋 Γ— 𝑋 be an operator defined as 𝐹(π‘₯, 𝑦)(𝑑) = (𝐹1 (π‘₯, 𝑦)(𝑑), 𝐹2 (π‘₯, 𝑦)(𝑑)), where

0

(12)

𝑇𝛼 3 5𝛼4 βˆ’ 14𝛼3 + 55𝛼2 + 146𝛼 ( + ) = π‘ˆ1 , Ξ“ (𝛼 + 1) 2 768 𝑑 ∈ [0, 𝑇] , (10)

𝑇

󡄨 󡄨 ∫ 󡄨󡄨󡄨𝐺2 (𝑑, 𝑠)󡄨󡄨󡄨 𝑑𝑠 0 𝑇𝛽 3 5𝛽4 βˆ’ 14𝛽3 + 55𝛽2 + 146𝛽 ≀ ( + ) = π‘ˆ2 , 768 Ξ“ (𝛽 + 1) 2 𝑑 ∈ [0, 𝑇] , (11)

𝑇𝛼 𝑇𝛼 𝑇𝛼 + + Ξ“ (𝛼 + 1) 2Ξ“ (𝛼 + 1) 4Ξ“ (𝛼) +

=

𝑇𝛼 5𝑇𝛼 𝑇𝛼 + + 16Ξ“ (𝛼 βˆ’ 1) 48Ξ“ (𝛼 βˆ’ 2) 768Ξ“ (𝛼 βˆ’ 3)

𝑇𝛼 3 5𝛼4 βˆ’ 14𝛼3 + 55𝛼2 + 146𝛼 ( + ). Ξ“ (𝛼 + 1) 2 768

On the other hand, 󡄨 𝑇 󡄨󡄨 󡄨 πœ•πΊ (𝑑, 𝑠) 󡄨󡄨󡄨 ∫ 󡄨󡄨󡄨 1 󡄨 𝑑𝑠 πœ•π‘‘ 󡄨󡄨󡄨 0 󡄨󡄨 (𝑑 βˆ’ 𝑠)π›Όβˆ’2 𝑑𝑠 0 Ξ“ (𝛼 βˆ’ 1) 󡄨 𝑇 󡄨󡄨 󡄨 (𝑇 βˆ’ 𝑠)π›Όβˆ’2 (𝑇 βˆ’ 2𝑑) (𝑇 βˆ’ 𝑠)π›Όβˆ’3 + ∫ σ΅„¨σ΅„¨σ΅„¨σ΅„¨βˆ’ + 4Ξ“ (𝛼 βˆ’ 2) 0 󡄨󡄨 2Ξ“ (𝛼 βˆ’ 1) 󡄨

=∫

𝑑

+ +

(𝑇𝑑 βˆ’ 𝑑2 ) (𝑇 βˆ’ 𝑠)π›Όβˆ’4 4Ξ“ (𝛼 βˆ’ 3) 󡄨 (6𝑇𝑑2 βˆ’ 𝑇3 βˆ’ 4𝑑3 ) (𝑇 βˆ’ 𝑠)π›Όβˆ’5 󡄨󡄨󡄨 󡄨󡄨 𝑑𝑠 󡄨󡄨 48Ξ“ (𝛼 βˆ’ 4) 󡄨󡄨 󡄨

(14)

4

Abstract and Applied Analysis ≀

≀

π‘‡π›Όβˆ’1 π‘‡π›Όβˆ’1 |𝑇 βˆ’ 2𝑑| π‘‡π›Όβˆ’2 + + Ξ“ (𝛼) 2Ξ“ (𝛼) 4Ξ“ (𝛼 βˆ’ 1) 󡄨 󡄨󡄨 󡄨 󡄨 󡄨󡄨𝑇𝑑 βˆ’ 𝑑2 󡄨󡄨󡄨 π‘‡π›Όβˆ’3 󡄨󡄨󡄨6𝑇𝑑2 βˆ’ 𝑇3 βˆ’ 4𝑑3 󡄨󡄨󡄨 π‘‡π›Όβˆ’4 󡄨 󡄨 󡄨 󡄨 + + 4Ξ“ (𝛼 βˆ’ 2) 48Ξ“ (𝛼 βˆ’ 3)

Then for any (π‘₯, 𝑦) ∈ Ξ©, according to the inequalities (10) and (11), we have 󡄨 𝑇 󡄨󡄨 󡄨 󡄨 󡄨󡄨 󡄨󡄨 󡄨󡄨𝐹1 (π‘₯, 𝑦) (𝑑)󡄨󡄨󡄨 = σ΅„¨σ΅„¨σ΅„¨σ΅„¨βˆ« 𝐺1 (𝑑, 𝑠) 𝑓 (𝑠, π‘₯ (𝑠) , 𝑦 (𝑠)) 𝑑𝑠󡄨󡄨󡄨󡄨 ≀ 𝐾1 π‘ˆ1 . 󡄨󡄨 0 󡄨󡄨 (20)

π‘‡π›Όβˆ’1 π‘‡π›Όβˆ’1 π‘‡π›Όβˆ’1 + + Ξ“ (𝛼) 2Ξ“ (𝛼) 4Ξ“ (𝛼 βˆ’ 1)

Similarly, we get 󡄨 𝑇 󡄨󡄨󡄨 󡄨󡄨 󡄨 󡄨󡄨 󡄨󡄨𝐹2 (π‘₯, 𝑦) (𝑑)󡄨󡄨󡄨 = σ΅„¨σ΅„¨σ΅„¨σ΅„¨βˆ« 𝐺2 (𝑑, 𝑠) 𝑔 (𝑠, π‘₯ (𝑠) , 𝑦 (𝑠)) 𝑑𝑠󡄨󡄨󡄨󡄨 ≀ 𝐾2 π‘ˆ2 . 󡄨󡄨 󡄨󡄨 0 (21)

π‘‡π›Όβˆ’1 π‘‡π›Όβˆ’1 + + 16Ξ“ (𝛼 βˆ’ 2) 48Ξ“ (𝛼 βˆ’ 3) ≀

π‘‡π›Όβˆ’1 3 𝛼3 βˆ’ 3𝛼2 + 14𝛼 βˆ’ 12 ( + ). Ξ“ (𝛼) 2 48 (15)

Step 3. We show that 𝐹 is equicontinuous:

Inequalities (11) and (13) can be proved in the same way. The first result is based on Leray-Schauder alternative. Lemma 7 (see [24]). Let 𝐹 : 𝐸 β†’ 𝐸 be a completely continuous operator (i.e., a map that is restricted to any bounded set in 𝐸 is compact). Let πœ€ (𝐹) = {π‘₯ ∈ 𝐸 : π‘₯ = πœ†πΉ (π‘₯) π‘“π‘œπ‘Ÿ π‘ π‘œπ‘šπ‘’ 0 < πœ† < 1} .

(16)

Then either the set πœ€(𝐹) is unbounded or 𝐹 has at least one fixed point. Theorem 8. Let 𝑓 and 𝑔 satisfy the following growth condition: 󡄨 󡄨 󡄨 󡄨󡄨 󡄨󡄨𝑓 (𝑑, π‘₯, 𝑦)󡄨󡄨󡄨 ≀ π‘š0 + π‘š1 |π‘₯| + π‘š2 󡄨󡄨󡄨𝑦󡄨󡄨󡄨 , π‘šπ‘– β‰₯ 0

(𝑖 = 1, 2) ,

π‘š0 > 0,

󡄨 󡄨 󡄨 󡄨󡄨 󡄨󡄨𝑔 (𝑑, π‘₯, 𝑦)󡄨󡄨󡄨 ≀ 𝑛0 + 𝑛1 |π‘₯| + 𝑛2 󡄨󡄨󡄨𝑦󡄨󡄨󡄨 , 𝑛𝑖 β‰₯ 0

(𝑖 = 1, 2) ,

(18)

Then problem (1) has at least one solution. For sake of convenience, we set π‘ˆ0 = min{1 βˆ’ (π‘ˆ1 π‘š1 + π‘ˆ2 𝑛1 ), 1 βˆ’ (π‘ˆ1 π‘š2 + π‘ˆ2 𝑛2 )}. Proof. First, we show that operator 𝐹 : 𝑋 Γ— 𝑋 β†’ 𝑋 Γ— 𝑋 is completely continuous. Step 1. In view of the continuity of 𝑓, 𝑔, π‘₯(𝑑), 𝑦(𝑑) and 𝐺1 (𝑑, 𝑠), 𝐺2 (𝑑, 𝑠), it is obvious that the operator 𝐹 is continuous. Step 2. Let Ξ© βŠ‚ 𝑋 Γ— 𝑋 be bounded; then there exist positive constants 𝐾1 and 𝐾2 such that 󡄨 󡄨󡄨 󡄨󡄨𝑓 (𝑑, π‘₯ (𝑑) , 𝑦 (𝑑))󡄨󡄨󡄨 ≀ 𝐾1 ,

󡄨 󡄨󡄨 󡄨󡄨𝑔 (𝑑, π‘₯ (𝑑) , 𝑦 (𝑑))󡄨󡄨󡄨 ≀ 𝐾2 , βˆ€ (π‘₯, 𝑦) ∈ Ξ©, 𝑑 ∈ [0, 𝑇] .

Hence, for 0 ≀ 𝑑1 ≀ 𝑑2 ≀ 𝑇, by the inequalities (12) and (13), we have 󡄨 󡄨󡄨 󡄨󡄨𝐹1 (π‘₯, 𝑦) (𝑑2 ) βˆ’ 𝐹1 (π‘₯, 𝑦) (𝑑1 )󡄨󡄨󡄨 𝑑2 (23) 󡄨 󡄨 σΈ€  󡄨 󡄨 ≀ ∫ 󡄨󡄨󡄨󡄨(𝐹1 (π‘₯, 𝑦))𝑠 (𝑠)󡄨󡄨󡄨󡄨 𝑑𝑠 ≀ 𝐾1 π‘ˆ3 󡄨󡄨󡄨𝑑2 βˆ’ 𝑑1 󡄨󡄨󡄨 . 𝑑 1

Analogously, we can obtain 𝑑2 󡄨 󡄨 σΈ€  󡄨󡄨 󡄨 󡄨󡄨𝐹2 (π‘₯, 𝑦) (𝑑2 ) βˆ’ 𝐹2 (π‘₯, 𝑦) (𝑑1 )󡄨󡄨󡄨 ≀ ∫ 󡄨󡄨󡄨󡄨(𝐹2 (π‘₯, 𝑦))𝑠 (𝑠)󡄨󡄨󡄨󡄨 𝑑𝑠 𝑑 1

𝑛0 > 0.

π‘ˆ1 π‘š2 + π‘ˆ2 𝑛2 < 1.

󡄨󡄨 𝑇 󡄨󡄨 󡄨󡄨 πœ•πΊ1 (𝑑, 𝑠) 󡄨󡄨 σ΅„¨σ΅„¨βˆ« 󡄨󡄨 𝑓 (𝑠, π‘₯ , 𝑦 𝑑𝑠 (𝑠) (𝑠)) 󡄨󡄨 0 󡄨󡄨 πœ•π‘‘ 󡄨 󡄨 (22) 󡄨 𝑇 󡄨󡄨 󡄨 πœ•πΊ 𝑠) (𝑑, 󡄨 󡄨󡄨 ≀ 𝐾1 ∫ 󡄨󡄨󡄨 1 󡄨 𝑑𝑠 ≀ 𝐾1 π‘ˆ3 . πœ•π‘‘ 󡄨󡄨󡄨 0 󡄨󡄨

󡄨󡄨 󡄨 󡄨󡄨(𝐹1 (π‘₯, 𝑦))󸀠𝑑 (𝑑)󡄨󡄨󡄨 = 󡄨 󡄨

(17)

In addition, it is assumed that π‘ˆ1 π‘š1 + π‘ˆ2 𝑛1 < 1,

Thus, it follows from the above inequalities that the operator 𝐹 is uniformly bounded.

(19)

󡄨 󡄨 ≀ 𝐾2 π‘ˆ4 󡄨󡄨󡄨𝑑2 βˆ’ 𝑑1 󡄨󡄨󡄨 .

(24)

Therefore, the operator 𝐹(π‘₯, 𝑦) is equicontinuous, and thus the operator 𝐹(π‘₯, 𝑦) is completely continuous. Finally, it will be verified that the set πœ€ = {(π‘₯, 𝑦) ∈ 𝑋 Γ— 𝑋 : (π‘₯, 𝑦) = πœ†πΉ(π‘₯, 𝑦), 0 ≀ πœ† ≀ 1} is bounded. Let (π‘₯, 𝑦) ∈ πœ€, then (π‘₯, 𝑦) = πœ†πΉ(π‘₯, 𝑦). For any 𝑑 ∈ [0, 𝑇], we have π‘₯(𝑑) = πœ†πΉ1 (π‘₯, 𝑦)(𝑑), 𝑦(𝑑) = πœ†πΉ2 (π‘₯, 𝑦)(𝑑); then, 󡄨 󡄨 󡄨 󡄨 |π‘₯ (𝑑)| ≀ 󡄨󡄨󡄨𝐹1 (π‘₯, 𝑦) (𝑑)󡄨󡄨󡄨 ≀ (π‘š0 + π‘š1 |π‘₯ (𝑑)| + π‘š2 󡄨󡄨󡄨𝑦 (𝑑)󡄨󡄨󡄨) π‘ˆ1 , 󡄨 󡄨 󡄨 󡄨󡄨 󡄨󡄨𝑦 (𝑑)󡄨󡄨󡄨 ≀ (𝑛0 + 𝑛1 |π‘₯ (𝑑)| + 𝑛2 󡄨󡄨󡄨𝑦 (𝑑)󡄨󡄨󡄨) π‘ˆ2 .

(25)

Hence, we have σ΅„© σ΅„© β€–π‘₯β€– ≀ (π‘š0 + π‘š1 β€–π‘₯β€– + π‘š2 󡄩󡄩󡄩𝑦󡄩󡄩󡄩) π‘ˆ1 , σ΅„© σ΅„© σ΅„©σ΅„© σ΅„©σ΅„© 󡄩󡄩𝑦󡄩󡄩 ≀ (𝑛0 + 𝑛1 β€–π‘₯β€– + 𝑛2 󡄩󡄩󡄩𝑦󡄩󡄩󡄩) π‘ˆ2 ,

(26)

which imply that σ΅„© σ΅„© β€–π‘₯β€– + 󡄩󡄩󡄩𝑦󡄩󡄩󡄩 ≀ (π‘ˆ1 π‘š0 + π‘ˆ2 𝑛0 ) + (π‘ˆ1 π‘š1 + π‘ˆ2 𝑛1 ) β€–π‘₯β€– σ΅„© σ΅„© + (π‘ˆ1 π‘š2 + π‘ˆ2 𝑛2 ) 󡄩󡄩󡄩𝑦󡄩󡄩󡄩 .

(27)

Abstract and Applied Analysis

5 σ΅„© σ΅„© ≀ (𝐿 1 π‘ˆ1 + πœ† 1 π‘ˆ2 ) σ΅„©σ΅„©σ΅„©π‘₯2 βˆ’ π‘₯1 σ΅„©σ΅„©σ΅„© σ΅„© σ΅„© + (𝐿 2 π‘ˆ1 + πœ† 2 π‘ˆ2 ) 󡄩󡄩󡄩𝑦2 βˆ’ 𝑦1 σ΅„©σ΅„©σ΅„© σ΅„© σ΅„© σ΅„© σ΅„© ≀ 𝐿 (σ΅„©σ΅„©σ΅„©π‘₯2 βˆ’ π‘₯1 σ΅„©σ΅„©σ΅„© + 󡄩󡄩󡄩𝑦2 βˆ’ 𝑦1 σ΅„©σ΅„©σ΅„©) σ΅„© σ΅„© = 𝐿 σ΅„©σ΅„©σ΅„©(π‘₯2 , 𝑦2 ) βˆ’ (π‘₯1 , 𝑦1 )σ΅„©σ΅„©σ΅„© .

As a result, σ΅„© π‘ˆ π‘š + π‘ˆ2 𝑛0 σ΅„©σ΅„© , σ΅„©σ΅„©(π‘₯, 𝑦)σ΅„©σ΅„©σ΅„© ≀ 1 0 π‘ˆ 0

(28)

for any 𝑑 ∈ [0, 𝑇], which proves that πœ€ is bounded; thus by Lemma 7, the operator 𝐹 has at least one fixed point. Hence the boundary value problem (1) has at least one solution; this completes the proof. In the second result, we prove uniqueness of solutions of the boundary value problem (1) via contraction mapping principle. Theorem 9. Let 𝑓 and 𝑔 satisfy the following growth conditions: (H 1 ) there exist two constants 𝐿 𝑖 > 0 and πœ† 𝑖 > 0, 𝑖 = 1, 2 such that 󡄨 󡄨 󡄨 󡄨 󡄨 󡄨󡄨 󡄨󡄨𝑓 (𝑑, π‘₯1 , 𝑦1 ) βˆ’ 𝑓 (𝑑, π‘₯2 , 𝑦2 )󡄨󡄨󡄨 ≀ 𝐿 1 󡄨󡄨󡄨π‘₯1 βˆ’ π‘₯2 󡄨󡄨󡄨 + 𝐿 2 󡄨󡄨󡄨𝑦1 βˆ’ 𝑦2 󡄨󡄨󡄨 , 󡄨󡄨 󡄨 󡄨 󡄨 󡄨 󡄨 󡄨󡄨𝑔 (𝑑, π‘₯1 , 𝑦1 ) βˆ’ 𝑔 (𝑑, π‘₯2 , 𝑦2 )󡄨󡄨󡄨 ≀ πœ† 1 󡄨󡄨󡄨π‘₯1 βˆ’ π‘₯2 󡄨󡄨󡄨 + πœ† 2 󡄨󡄨󡄨𝑦1 βˆ’ 𝑦2 󡄨󡄨󡄨 ,

(33) Hence, we conclude that problem (1) has a unique solution by (H 2 ) and the contraction mapping principle; this ends the proof.

4. Examples In this section, two examples are given in order to verify the validity of Theorems 8 and 9. Example 1. Consider the system 𝑐

𝑑 ∈ [0, 𝑇] , π‘₯1 , π‘₯2 , 𝑦1 , 𝑦2 ∈ R; (29)

𝑐

1 (𝑑 + π‘₯ (𝑑) + 𝑦 (𝑑)) = 0, 2

0 < 𝑑 < 1,

3 (βˆšπ‘‘ + π‘₯ (𝑑) + 𝑦 (𝑑)) = 0, 2

0 < 𝑑 < 1,

𝐷17/4 π‘₯ (𝑑) +

𝐷9/2 𝑦 (𝑑) + (𝑖)

(H 2 ) Consider max {𝐿 1 π‘ˆ1 + πœ† 1 π‘ˆ2 , 𝐿 2 π‘ˆ1 + πœ† 2 π‘ˆ2 } = 𝐿 < 1.

(30)

Proof. Let (π‘₯1 , 𝑦1 ), (π‘₯2 , 𝑦2 ) ∈ 𝑋 Γ— 𝑋; then 󡄨 󡄨󡄨 󡄨󡄨𝐹1 (π‘₯2 , 𝑦2 ) (𝑑) βˆ’ 𝐹1 (π‘₯1 , 𝑦1 ) (𝑑)󡄨󡄨󡄨

󡄨󡄨 𝑇 󡄨 = σ΅„¨σ΅„¨σ΅„¨σ΅„¨βˆ« 𝐺1 (𝑑, 𝑠) 𝑓 (𝑠, π‘₯2 (𝑠) , 𝑦2 (𝑠)) 𝑑𝑠 󡄨󡄨 0

(31)

𝑇

󡄨󡄨 󡄨 ≀ ∫ 󡄨󡄨󡄨𝐺1 (𝑑, 𝑠)󡄨󡄨󡄨 󡄨󡄨󡄨𝑓 (𝑠, π‘₯2 (𝑠) , 𝑦2 (𝑠)) 0 󡄨 βˆ’π‘“ (𝑠, π‘₯1 (𝑠) , 𝑦1 (𝑠))󡄨󡄨󡄨 𝑑𝑠

σ΅„©σ΅„© σ΅„© 󡄩󡄩𝐹 (π‘₯2 , 𝑦2 ) βˆ’ 𝐹 (π‘₯1 , 𝑦1 )σ΅„©σ΅„©σ΅„©

σ΅„© σ΅„© = 󡄩󡄩󡄩𝐹1 (π‘₯2 , 𝑦2 ) βˆ’ 𝐹1 (π‘₯1 , 𝑦1 )σ΅„©σ΅„©σ΅„©

σ΅„© σ΅„© + 󡄩󡄩󡄩𝐹2 (π‘₯2 , 𝑦2 ) βˆ’ 𝐹2 (π‘₯1 , 𝑦1 )σ΅„©σ΅„©σ΅„©

𝑦(𝑖) (0) = βˆ’π‘¦(𝑖) (1) ,

𝑖 = 0, 1, 2, 3, 4,

𝑇𝛼 3 5𝛼4 βˆ’ 14𝛼3 + 55𝛼2 + 146𝛼 ( + ) β‰ˆ 0.1229, Ξ“ (𝛼 + 1) 2 768

π‘ˆ2 =

𝑇𝛽 3 5𝛽4 βˆ’ 14𝛽3 + 55𝛽2 + 146𝛽 ( + ) β‰ˆ 0.0920. 768 Ξ“ (𝛽 + 1) 2 (35)

Then π‘ˆ1 π‘š1 + π‘ˆ2 𝑛1 < 1 and π‘ˆ1 π‘š2 + π‘ˆ2 𝑛2 < 1; by Theorem 8, the existence of the solution for the system (34) is obvious. Example 2. Consider the system 𝑐

Analogously,

Thus,

𝑖 = 0, 1, 2, 3, 4,

π‘ˆ1 =

σ΅„© σ΅„© σ΅„© σ΅„© ≀ π‘ˆ1 (𝐿 1 σ΅„©σ΅„©σ΅„©π‘₯2 βˆ’ π‘₯1 σ΅„©σ΅„©σ΅„© + 𝐿 2 󡄩󡄩󡄩𝑦2 βˆ’ 𝑦1 σ΅„©σ΅„©σ΅„©) .

󡄨󡄨 󡄨 󡄨󡄨𝐹2 (π‘₯2 , 𝑦2 ) (𝑑) βˆ’ 𝐹2 (π‘₯1 , 𝑦1 ) (𝑑)󡄨󡄨󡄨 σ΅„© σ΅„© σ΅„© σ΅„© ≀ π‘ˆ2 (πœ† 1 σ΅„©σ΅„©σ΅„©π‘₯2 βˆ’ π‘₯1 σ΅„©σ΅„©σ΅„© + πœ† 2 󡄩󡄩󡄩𝑦2 βˆ’ 𝑦1 σ΅„©σ΅„©σ΅„©) .

π‘₯ (0) = βˆ’π‘₯ (1) ,

(34)

where 𝑇 = 1, |𝑓(𝑑, π‘₯, 𝑦)| ≀ 1 + |π‘₯| + |𝑦|, |𝑔(𝑑, π‘₯, 𝑦)| ≀ 2 + 2|π‘₯| + 2|𝑦|, 𝛼 = 17/4, 𝛽 = 9/2, π‘š0 = π‘š1 = π‘š2 = 1, 𝑛0 = 𝑛1 = 𝑛2 = 2. Consider

Then problem (1) has a unique solution.

󡄨󡄨 𝑇 󡄨 βˆ’ ∫ 𝐺1 (𝑑, 𝑠) 𝑓 (𝑠, π‘₯1 (𝑠) , 𝑦1 (𝑠)) 𝑑𝑠󡄨󡄨󡄨󡄨 󡄨󡄨 0

(𝑖)

𝑐

(32)

󡄨 󡄨󡄨 󡄨𝑦 (𝑑)󡄨󡄨 𝐷17/4 π‘₯ (𝑑) + 𝑑 + sin π‘₯ (𝑑) + 2 󡄨 󡄨󡄨 󡄨 󡄨󡄨 = 0, 1 + 󡄨󡄨𝑦 (𝑑)󡄨󡄨

𝐷9/2 𝑦 (𝑑) + 𝑑2 + 2

|π‘₯ (𝑑)| + arctan 𝑦 (𝑑) = 0, 1 + |π‘₯ (𝑑)|

π‘₯(𝑖) (0) = βˆ’π‘₯(𝑖) (1) ,

𝑖 = 0, 1, 2, 3, 4,

𝑦(𝑖) (0) = βˆ’π‘¦(𝑖) (1) ,

𝑖 = 0, 1, 2, 3, 4,

0 < 𝑑 < 1, 0 < 𝑑 < 1,

(36) where 𝑇 = 1, 𝑓(𝑑, π‘₯, 𝑦) = 𝑑 + sin π‘₯(𝑑) + 2(|𝑦(𝑑)|/(1 + |𝑦(𝑑)|)), 𝑔(𝑑, π‘₯, 𝑦) = 𝑑2 + 2(|π‘₯(𝑑)|/(1 + |π‘₯(𝑑)|)) + arctan 𝑦(𝑑). 𝛼 = 17/4, 𝛽 = 9/2.

6

Abstract and Applied Analysis Noting that 󡄨󡄨 󡄨 󡄨󡄨(sin π‘₯)σΈ€  󡄨󡄨󡄨 = |cos π‘₯| ≀ 1, 󡄨 󡄨

1 󡄨󡄨 󡄨 󡄨󡄨(arctan 𝑦)σΈ€  󡄨󡄨󡄨 = 󡄨 󡄨 1 + 𝑦2 ≀ 1, (37)

we have 󡄨󡄨 󡄨 󡄨󡄨𝑓 (𝑑, π‘₯2 , 𝑦2 ) βˆ’ 𝑓 (𝑑, π‘₯1 , 𝑦1 )󡄨󡄨󡄨

󡄨󡄨 󡄨󡄨𝑦 (𝑑)󡄨󡄨 󡄨󡄨 󡄨󡄨 󡄨󡄨𝑦 (𝑑)󡄨󡄨 󡄨󡄨 󡄨 󡄨 󡄨 󡄨 󡄨 󡄨 󡄨 󡄨 󡄨 󡄨 = 󡄨󡄨󡄨sin π‘₯2 (𝑑) βˆ’ sin π‘₯1 (𝑑)󡄨󡄨󡄨 + 2 󡄨󡄨󡄨󡄨 󡄨 󡄨󡄨2 󡄨 󡄨󡄨 󡄨󡄨󡄨󡄨 βˆ’ 󡄨󡄨󡄨󡄨 󡄨 󡄨󡄨1 󡄨 󡄨󡄨 󡄨󡄨󡄨󡄨 󡄨󡄨 1 + 󡄨󡄨𝑦2 (𝑑)󡄨󡄨 󡄨󡄨 󡄨󡄨 1 + 󡄨󡄨𝑦1 (𝑑)󡄨󡄨 󡄨󡄨 σ΅„© σ΅„© σ΅„© σ΅„© ≀ σ΅„©σ΅„©σ΅„©π‘₯2 βˆ’ π‘₯1 σ΅„©σ΅„©σ΅„© + 2 󡄩󡄩󡄩𝑦2 βˆ’ 𝑦1 σ΅„©σ΅„©σ΅„© , σ΅„© 󡄨󡄨 σ΅„© σ΅„© σ΅„© 󡄨 󡄨󡄨𝑔 (𝑑, π‘₯2 , 𝑦2 ) βˆ’ 𝑔 (𝑑, π‘₯1 , 𝑦1 )󡄨󡄨󡄨 ≀ 2 σ΅„©σ΅„©σ΅„©π‘₯2 βˆ’ π‘₯1 σ΅„©σ΅„©σ΅„© + 󡄩󡄩󡄩𝑦2 βˆ’ 𝑦1 σ΅„©σ΅„©σ΅„© ,

π‘ˆ1 =

𝑇𝛼 3 5𝛼4 βˆ’ 14𝛼3 + 55𝛼2 + 146𝛼 ( + ) β‰ˆ 0.1229, Ξ“ (𝛼 + 1) 2 768

π‘ˆ2 =

𝑇𝛽 3 5𝛽4 βˆ’ 14𝛽3 + 55𝛽2 + 146𝛽 ( + ) β‰ˆ 0.0920. 768 Ξ“ (𝛽 + 1) 2 (38)

Obviously, max {𝐿 1 π‘ˆ1 + πœ† 1 π‘ˆ2 , 𝐿 2 π‘ˆ1 + πœ† 2 π‘ˆ2 } < 1;

(39)

then we can conclude from Theorem 9 that system (36) has a unique solution.

Conflict of Interests The authors declare that there is no conflict of interests regarding the publication of this paper.

Acknowledgments The authors express their thanks to the referee for his/her valuable suggestions. The second author was supported by NSF of China (11271154) and the first author was supported by the Fundamental Research Funds for the Central Universities (13cx02015A).

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