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 πΌ, π½ 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: π
π·πΌ π₯ (π‘) + π (π‘, π₯ (π‘) , π¦ (π‘)) = 0,
π‘ β [0, π] ,
π
π·π½ π¦ (π‘) + π (π‘, π₯ (π‘) , π¦ (π‘)) = 0,
π‘ β [0, π] ,
π₯(π) (0) = βπ₯(π) (π) ,
π = 0, 1, 2, 3, 4,
π¦(π) (0) = βπ¦(π) (π) ,
π = 0, 1, 2, 3, 4,
π
πΌ
(1)
where 4 < πΌ, π½ β€ 5, and π· denotes the Caputo fractional derivative of order πΌ. Here our nonlinearity π, π : [0, π] 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 [1β4]. 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 [5β14]. The study of a coupled system of fractional order is also very significant because this kind of system can often occur in applications; see [15β17]. 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, 18β22]. In [5], Alsaedi et al. study an Antiperiodic boundary value problem of nonlinear fractional differential equations of order π β (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 πΌ > 0 of a function π¦ : (0, β) β R is given by
where πΊ(π‘, π ) is the Green function given by πΊ (π‘, π ) 2(π‘ β π )πβ1 β (π β π )πβ1 (π β 2π‘) (π β π )πβ2 + 2Ξ (π) 4Ξ (π β 1) 2 πβ3 (6ππ‘ β 4π‘3 β π3 ) (π β π )πβ4 π‘ (π β π‘) (π β π ) + + 4Ξ (π β 2) 48Ξ (π β 3) (2ππ‘3 β π‘π3 β π‘4 ) (π β π )πβ5 + , 0 < π < π‘ < π, 48Ξ (π β 4) (π β π )πβ1 (π β 2π‘) (π β π )πβ2 β + 2Ξ (π) 4Ξ (π β 1) πβ4 2 3 3 π‘ (π β π‘) (π β π )πβ3 (6ππ‘ β 4π‘ β π ) (π β π ) + + 4Ξ (π β 2) 48Ξ (π β 3) (2ππ‘3 β π‘π3 β π‘4 ) (π β π )πβ5 + , 0 < π‘ < π < π. 48Ξ (π β 4) (6)
{ { { { { { { { { { { { { { { { { { { { { { { ={ { { { { { { { { { { { { { { { { { { { { { { { Let
πΌπΌ π¦ (π‘) =
π‘ 1 β« (π‘ β π )πΌβ1 π¦ (π ) ππ , Ξ (πΌ) 0
(2)
provided the right hand side is pointwise defined on (0, β). Definition 2 (see [2, 3]). The Caputo fractional derivative of order πΌ > 0 of a continuous function π¦ : (0, β) β R is given by
π
π·πΌ π¦ (π‘) =
π‘ π¦(π) (π ) 1 ππ , β« Ξ (π β πΌ) 0 (π‘ β π )πΌβπ+1
(3)
where π = [πΌ] + 1 and [πΌ] denotes the integer part of number πΌ, provided that the right side is pointwise defined on (0, β). πΌ
Lemma 3 (see [19]). Consider πΌπΌ π π· π₯(π‘) = π₯(π‘) + πΆ0 + πΆ1 π‘ + πΆ2 π‘2 + β
β
β
+ πΆπβ1 π‘πβ1 , for some ππ β R, π = 0, 1, 2, . . . , π β 1, where π = [πΌ] + 1 and [πΌ] denotes the integer part of number πΌ. Lemma 4 (see [5]). For any π¦ β πΆ[0, π], the unique solution of the boundary value problem π
π·π π₯ (π‘) = π¦ (π‘) ,
π‘ β [0, π] , 4 < π β€ 5,
π₯(π) (0) = βπ₯(π) (π) ,
(4) π = 0, 1, 2, 3, 4
is π
π₯ (π‘) = β« πΊ (π‘, π ) π¦ (π ) ππ , 0
(5)
πΊ1 (π‘, π ) { { { { { { { { { { { { { { { { { { { { { { ={ { { { { { { { { { { { { { { { { { { { { { {
(π‘ β π )πΌβ1 β (1/2) (π β π )πΌβ1 (π β 2π‘) (π β π )πΌβ2 + Ξ (πΌ) 4Ξ (πΌ β 1) πΌβ4 2 3 3 π‘ (π β π‘) (π β π )πΌβ3 (6ππ‘ β 4π‘ β π ) (π β π ) + + 4Ξ (πΌ β 2) 48Ξ (πΌ β 3) 3 3 4 (2ππ‘ β π‘π β π‘ ) (π β π )πΌβ5 + , 0 < π < π‘ < π, 48Ξ (πΌ β 4) 1 (π β 2π‘) (π β π )πΌβ2 β (π β π )πΌβ1 + 2Ξ (πΌ) 4Ξ (πΌ β 1) 2 πΌβ3 (6ππ‘ β 4π‘3 β π3 ) (π β π )πΌβ4 π‘ (π β π‘) (π β π ) + + 4Ξ (πΌ β 2) 48Ξ (πΌ β 3) (2ππ‘3 β π‘π3 β π‘4 ) (π β π )πΌβ5 , 0 < π‘ < π < π, + 48Ξ (πΌ β 4)
πΊ2 (π‘, π ) { { { { { { { { { { { { { { { { { { { { { { { ={ { { { { { { { { { { { { { { { { { { { { { { {
(π‘ β π )π½β1 β (1/2) (π β π )π½β1 (π β 2π‘) (π β π )π½β2 + Ξ (π½) 4Ξ (π½ β 1) 2 3 π½β3 (6ππ‘ β 4π‘ β π3 ) (π β π )π½β4 π‘ (π β π‘) (π β π ) + + 4Ξ (π½ β 2) 48Ξ (π½ β 3) (2ππ‘3 β π‘π3 β π‘4 ) (π β π )π½β5 + , 0 < π < π‘ < π, 48Ξ (π½ β 4) 1 (π β 2π‘) (π β π )π½β2 β (π β π )π½β1 + 2Ξ (π½) 4Ξ (π½ β 1) 2 π½β3 (6ππ‘ β 4π‘3 β π3 ) (π β π )π½β4 π‘ (π β π‘) (π β π ) + + 4Ξ (π½ β 2) 48Ξ (π½ β 3) (2ππ‘3 β π‘π3 β π‘4 ) (π β π )π½β5 + , 0 < π‘ < π < π. 48Ξ (π½ β 4) (7)
We call (πΊ1 , πΊ2 ) Greenβs function for problem (1). We define the space π = {π₯(π‘) | π₯(π‘) β πΆ[0, π]} endowed with βπ₯βπ = maxπ‘β[0,π] |π₯(π‘)|; for (π₯, π¦) β π Γ π,
Abstract and Applied Analysis
3
let β(π₯, π¦)βπΓπ = βπ₯βπ + βπ¦βπ . Obviously, (π, β β
βπ ) is a Banach space, and the product space (π Γ π, β(β
, β
)βπΓπ ) is also a Banach space. Consider the following coupled system of the integral equations:
σ΅¨ π σ΅¨σ΅¨ σ΅¨ ππΊ (π‘, π ) σ΅¨σ΅¨σ΅¨ β« σ΅¨σ΅¨σ΅¨ 1 σ΅¨ ππ ππ‘ σ΅¨σ΅¨σ΅¨ 0 σ΅¨σ΅¨ β€
ππΌβ1 3 πΌ3 β 3πΌ2 + 14πΌ β 12 ( + ) = π3 , Ξ (πΌ) 2 48 π‘ β [0, π] ,
π
π₯ (π‘) = β« πΊ1 (π‘, π ) π (π , π₯ (π ) , π¦ (π )) ππ , 0
(8)
π
π¦ (π‘) = β« πΊ2 (π‘, π ) π (π , π₯ (π ) , π¦ (π )) ππ .
σ΅¨ π σ΅¨σ΅¨ σ΅¨ ππΊ (π‘, π ) σ΅¨σ΅¨σ΅¨ β« σ΅¨σ΅¨σ΅¨ 2 σ΅¨ ππ ππ‘ σ΅¨σ΅¨σ΅¨ 0 σ΅¨σ΅¨
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 π, π : [0, π] Γ R Γ R β R are continuous functions. Then (π₯, π¦) β (π, π) is a solution of (1) if and only if (π₯, π¦) β (π, π) is a solution of system (8). Proof. The proof is immediate from the discussion above, and we omit the details here.
β€
π
σ΅¨ σ΅¨ β« σ΅¨σ΅¨σ΅¨πΊ1 (π‘, π )σ΅¨σ΅¨σ΅¨ ππ 0 σ΅¨σ΅¨ π‘ σ΅¨ σ΅¨ σ΅¨ σ΅¨ (π‘ β π )πΌβ1 σ΅¨σ΅¨σ΅¨ 1 σ΅¨σ΅¨σ΅¨ π (π β π )πΌβ1 σ΅¨σ΅¨σ΅¨ β€ σ΅¨σ΅¨σ΅¨σ΅¨β« ππ σ΅¨σ΅¨σ΅¨ + σ΅¨σ΅¨σ΅¨β« ππ σ΅¨σ΅¨σ΅¨ Ξ (πΌ) σ΅¨σ΅¨ 0 Ξ (πΌ) σ΅¨σ΅¨ 2 σ΅¨σ΅¨ 0 σ΅¨σ΅¨ +
|π β 2π‘| π (π β π )πΌβ2 ππ β« 4 0 Ξ (πΌ β 1)
|π‘ (π β π‘)| π (π β π )πΌβ3 ππ β« 4 0 Ξ (πΌ β 2) σ΅¨σ΅¨σ΅¨6ππ‘2 β 4π‘3 β π3 σ΅¨σ΅¨σ΅¨ π σ΅¨σ΅¨ σ΅¨ (π β π )πΌβ4 ππ +σ΅¨ β« 48 0 Ξ (πΌ β 3) σ΅¨σ΅¨σ΅¨2ππ‘3 β π‘π3 β π‘4 σ΅¨σ΅¨σ΅¨ π σ΅¨σ΅¨ σ΅¨ (π β π )πΌβ5 ππ +σ΅¨ β« 48 0 Ξ (πΌ β 4) +
πΉ1 (π₯, π¦) (π‘) = β« πΊ1 (π‘, π ) π (π , π₯ (π ) , π¦ (π )) ππ , (9)
πΉ2 (π₯, π¦) (π‘) = β« πΊ2 (π‘, π ) π (π , π₯ (π ) , π¦ (π )) ππ . 0
It is obvious that a fixed point of the operator πΉ is a solution of problem (1). β€
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 πΊ1 (π‘, π ), πΊ2 (π‘, π ) satisfy the following estimates: π
σ΅¨ σ΅¨ β« σ΅¨σ΅¨σ΅¨πΊ1 (π‘, π )σ΅¨σ΅¨σ΅¨ ππ 0 β€
(13)
Proof. For any π‘ β [0, π],
π
π
ππ½β1 3 π½3 β 3π½2 + 14π½ β 12 ( + ) = π4 , 48 Ξ (π½) 2 π‘ β [0, π] .
Let πΉ : π Γ π β π Γ π be an operator defined as πΉ(π₯, π¦)(π‘) = (πΉ1 (π₯, π¦)(π‘), πΉ2 (π₯, π¦)(π‘)), where
0
(12)
ππΌ 3 5πΌ4 β 14πΌ3 + 55πΌ2 + 146πΌ ( + ) = π1 , Ξ (πΌ + 1) 2 768 π‘ β [0, π] , (10)
π
σ΅¨ σ΅¨ β« σ΅¨σ΅¨σ΅¨πΊ2 (π‘, π )σ΅¨σ΅¨σ΅¨ ππ 0 ππ½ 3 5π½4 β 14π½3 + 55π½2 + 146π½ β€ ( + ) = π2 , 768 Ξ (π½ + 1) 2 π‘ β [0, π] , (11)
ππΌ ππΌ ππΌ + + Ξ (πΌ + 1) 2Ξ (πΌ + 1) 4Ξ (πΌ) +
=
ππΌ 5ππΌ ππΌ + + 16Ξ (πΌ β 1) 48Ξ (πΌ β 2) 768Ξ (πΌ β 3)
ππΌ 3 5πΌ4 β 14πΌ3 + 55πΌ2 + 146πΌ ( + ). Ξ (πΌ + 1) 2 768
On the other hand, σ΅¨ π σ΅¨σ΅¨ σ΅¨ ππΊ (π‘, π ) σ΅¨σ΅¨σ΅¨ β« σ΅¨σ΅¨σ΅¨ 1 σ΅¨ ππ ππ‘ σ΅¨σ΅¨σ΅¨ 0 σ΅¨σ΅¨ (π‘ β π )πΌβ2 ππ 0 Ξ (πΌ β 1) σ΅¨ π σ΅¨σ΅¨ σ΅¨ (π β π )πΌβ2 (π β 2π‘) (π β π )πΌβ3 + β« σ΅¨σ΅¨σ΅¨σ΅¨β + 4Ξ (πΌ β 2) 0 σ΅¨σ΅¨ 2Ξ (πΌ β 1) σ΅¨
=β«
π‘
+ +
(ππ‘ β π‘2 ) (π β π )πΌβ4 4Ξ (πΌ β 3) σ΅¨ (6ππ‘2 β π3 β 4π‘3 ) (π β π )πΌβ5 σ΅¨σ΅¨σ΅¨ σ΅¨σ΅¨ ππ σ΅¨σ΅¨ 48Ξ (πΌ β 4) σ΅¨σ΅¨ σ΅¨
(14)
4
Abstract and Applied Analysis β€
β€
ππΌβ1 ππΌβ1 |π β 2π‘| ππΌβ2 + + Ξ (πΌ) 2Ξ (πΌ) 4Ξ (πΌ β 1) σ΅¨ σ΅¨σ΅¨ σ΅¨ σ΅¨ σ΅¨σ΅¨ππ‘ β π‘2 σ΅¨σ΅¨σ΅¨ ππΌβ3 σ΅¨σ΅¨σ΅¨6ππ‘2 β π3 β 4π‘3 σ΅¨σ΅¨σ΅¨ ππΌβ4 σ΅¨ σ΅¨ σ΅¨ σ΅¨ + + 4Ξ (πΌ β 2) 48Ξ (πΌ β 3)
Then for any (π₯, π¦) β Ξ©, according to the inequalities (10) and (11), we have σ΅¨ π σ΅¨σ΅¨ σ΅¨ σ΅¨ σ΅¨σ΅¨ σ΅¨σ΅¨ σ΅¨σ΅¨πΉ1 (π₯, π¦) (π‘)σ΅¨σ΅¨σ΅¨ = σ΅¨σ΅¨σ΅¨σ΅¨β« πΊ1 (π‘, π ) π (π , π₯ (π ) , π¦ (π )) ππ σ΅¨σ΅¨σ΅¨σ΅¨ β€ πΎ1 π1 . σ΅¨σ΅¨ 0 σ΅¨σ΅¨ (20)
ππΌβ1 ππΌβ1 ππΌβ1 + + Ξ (πΌ) 2Ξ (πΌ) 4Ξ (πΌ β 1)
Similarly, we get σ΅¨ π σ΅¨σ΅¨σ΅¨ σ΅¨σ΅¨ σ΅¨ σ΅¨σ΅¨ σ΅¨σ΅¨πΉ2 (π₯, π¦) (π‘)σ΅¨σ΅¨σ΅¨ = σ΅¨σ΅¨σ΅¨σ΅¨β« πΊ2 (π‘, π ) π (π , π₯ (π ) , π¦ (π )) ππ σ΅¨σ΅¨σ΅¨σ΅¨ β€ πΎ2 π2 . σ΅¨σ΅¨ σ΅¨σ΅¨ 0 (21)
ππΌβ1 ππΌβ1 + + 16Ξ (πΌ β 2) 48Ξ (πΌ β 3) β€
ππΌβ1 3 πΌ3 β 3πΌ2 + 14πΌ β 12 ( + ). Ξ (πΌ) 2 48 (15)
Step 3. We show that πΉ 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 πΉ : πΈ β πΈ be a completely continuous operator (i.e., a map that is restricted to any bounded set in πΈ is compact). Let π (πΉ) = {π₯ β πΈ : π₯ = ππΉ (π₯) πππ π πππ 0 < π < 1} .
(16)
Then either the set π(πΉ) is unbounded or πΉ has at least one fixed point. Theorem 8. Let π and π satisfy the following growth condition: σ΅¨ σ΅¨ σ΅¨ σ΅¨σ΅¨ σ΅¨σ΅¨π (π‘, π₯, π¦)σ΅¨σ΅¨σ΅¨ β€ π0 + π1 |π₯| + π2 σ΅¨σ΅¨σ΅¨π¦σ΅¨σ΅¨σ΅¨ , ππ β₯ 0
(π = 1, 2) ,
π0 > 0,
σ΅¨ σ΅¨ σ΅¨ σ΅¨σ΅¨ σ΅¨σ΅¨π (π‘, π₯, π¦)σ΅¨σ΅¨σ΅¨ β€ π0 + π1 |π₯| + π2 σ΅¨σ΅¨σ΅¨π¦σ΅¨σ΅¨σ΅¨ , ππ β₯ 0
(π = 1, 2) ,
(18)
Then problem (1) has at least one solution. For sake of convenience, we set π0 = min{1 β (π1 π1 + π2 π1 ), 1 β (π1 π2 + π2 π2 )}. Proof. First, we show that operator πΉ : π Γ π β π Γ π is completely continuous. Step 1. In view of the continuity of π, π, π₯(π‘), π¦(π‘) and πΊ1 (π‘, π ), πΊ2 (π‘, π ), it is obvious that the operator πΉ is continuous. Step 2. Let Ξ© β π Γ π be bounded; then there exist positive constants πΎ1 and πΎ2 such that σ΅¨ σ΅¨σ΅¨ σ΅¨σ΅¨π (π‘, π₯ (π‘) , π¦ (π‘))σ΅¨σ΅¨σ΅¨ β€ πΎ1 ,
σ΅¨ σ΅¨σ΅¨ σ΅¨σ΅¨π (π‘, π₯ (π‘) , π¦ (π‘))σ΅¨σ΅¨σ΅¨ β€ πΎ2 , β (π₯, π¦) β Ξ©, π‘ β [0, π] .
Hence, for 0 β€ π‘1 β€ π‘2 β€ π, by the inequalities (12) and (13), we have σ΅¨ σ΅¨σ΅¨ σ΅¨σ΅¨πΉ1 (π₯, π¦) (π‘2 ) β πΉ1 (π₯, π¦) (π‘1 )σ΅¨σ΅¨σ΅¨ π‘2 (23) σ΅¨ σ΅¨ σΈ σ΅¨ σ΅¨ β€ β« σ΅¨σ΅¨σ΅¨σ΅¨(πΉ1 (π₯, π¦))π (π )σ΅¨σ΅¨σ΅¨σ΅¨ ππ β€ πΎ1 π3 σ΅¨σ΅¨σ΅¨π‘2 β π‘1 σ΅¨σ΅¨σ΅¨ . π‘ 1
Analogously, we can obtain π‘2 σ΅¨ σ΅¨ σΈ σ΅¨σ΅¨ σ΅¨ σ΅¨σ΅¨πΉ2 (π₯, π¦) (π‘2 ) β πΉ2 (π₯, π¦) (π‘1 )σ΅¨σ΅¨σ΅¨ β€ β« σ΅¨σ΅¨σ΅¨σ΅¨(πΉ2 (π₯, π¦))π (π )σ΅¨σ΅¨σ΅¨σ΅¨ ππ π‘ 1
π0 > 0.
π1 π2 + π2 π2 < 1.
σ΅¨σ΅¨ π σ΅¨σ΅¨ σ΅¨σ΅¨ ππΊ1 (π‘, π ) σ΅¨σ΅¨ σ΅¨σ΅¨β« σ΅¨σ΅¨ π (π , π₯ , π¦ ππ (π ) (π )) σ΅¨σ΅¨ 0 σ΅¨σ΅¨ ππ‘ σ΅¨ σ΅¨ (22) σ΅¨ π σ΅¨σ΅¨ σ΅¨ ππΊ π ) (π‘, σ΅¨ σ΅¨σ΅¨ β€ πΎ1 β« σ΅¨σ΅¨σ΅¨ 1 σ΅¨ ππ β€ πΎ1 π3 . ππ‘ σ΅¨σ΅¨σ΅¨ 0 σ΅¨σ΅¨
σ΅¨σ΅¨ σ΅¨ σ΅¨σ΅¨(πΉ1 (π₯, π¦))σΈ π‘ (π‘)σ΅¨σ΅¨σ΅¨ = σ΅¨ σ΅¨
(17)
In addition, it is assumed that π1 π1 + π2 π1 < 1,
Thus, it follows from the above inequalities that the operator πΉ is uniformly bounded.
(19)
σ΅¨ σ΅¨ β€ πΎ2 π4 σ΅¨σ΅¨σ΅¨π‘2 β π‘1 σ΅¨σ΅¨σ΅¨ .
(24)
Therefore, the operator πΉ(π₯, π¦) is equicontinuous, and thus the operator πΉ(π₯, π¦) is completely continuous. Finally, it will be verified that the set π = {(π₯, π¦) β π Γ π : (π₯, π¦) = ππΉ(π₯, π¦), 0 β€ π β€ 1} is bounded. Let (π₯, π¦) β π, then (π₯, π¦) = ππΉ(π₯, π¦). For any π‘ β [0, π], we have π₯(π‘) = ππΉ1 (π₯, π¦)(π‘), π¦(π‘) = ππΉ2 (π₯, π¦)(π‘); then, σ΅¨ σ΅¨ σ΅¨ σ΅¨ |π₯ (π‘)| β€ σ΅¨σ΅¨σ΅¨πΉ1 (π₯, π¦) (π‘)σ΅¨σ΅¨σ΅¨ β€ (π0 + π1 |π₯ (π‘)| + π2 σ΅¨σ΅¨σ΅¨π¦ (π‘)σ΅¨σ΅¨σ΅¨) π1 , σ΅¨ σ΅¨ σ΅¨ σ΅¨σ΅¨ σ΅¨σ΅¨π¦ (π‘)σ΅¨σ΅¨σ΅¨ β€ (π0 + π1 |π₯ (π‘)| + π2 σ΅¨σ΅¨σ΅¨π¦ (π‘)σ΅¨σ΅¨σ΅¨) π2 .
(25)
Hence, we have σ΅© σ΅© βπ₯β β€ (π0 + π1 βπ₯β + π2 σ΅©σ΅©σ΅©π¦σ΅©σ΅©σ΅©) π1 , σ΅© σ΅© σ΅©σ΅© σ΅©σ΅© σ΅©σ΅©π¦σ΅©σ΅© β€ (π0 + π1 βπ₯β + π2 σ΅©σ΅©σ΅©π¦σ΅©σ΅©σ΅©) π2 ,
(26)
which imply that σ΅© σ΅© βπ₯β + σ΅©σ΅©σ΅©π¦σ΅©σ΅©σ΅© β€ (π1 π0 + π2 π0 ) + (π1 π1 + π2 π1 ) βπ₯β σ΅© σ΅© + (π1 π2 + π2 π2 ) σ΅©σ΅©σ΅©π¦σ΅©σ΅©σ΅© .
(27)
Abstract and Applied Analysis
5 σ΅© σ΅© β€ (πΏ 1 π1 + π 1 π2 ) σ΅©σ΅©σ΅©π₯2 β π₯1 σ΅©σ΅©σ΅© σ΅© σ΅© + (πΏ 2 π1 + π 2 π2 ) σ΅©σ΅©σ΅©π¦2 β π¦1 σ΅©σ΅©σ΅© σ΅© σ΅© σ΅© σ΅© β€ πΏ (σ΅©σ΅©σ΅©π₯2 β π₯1 σ΅©σ΅©σ΅© + σ΅©σ΅©σ΅©π¦2 β π¦1 σ΅©σ΅©σ΅©) σ΅© σ΅© = πΏ σ΅©σ΅©σ΅©(π₯2 , π¦2 ) β (π₯1 , π¦1 )σ΅©σ΅©σ΅© .
As a result, σ΅© π π + π2 π0 σ΅©σ΅© , σ΅©σ΅©(π₯, π¦)σ΅©σ΅©σ΅© β€ 1 0 π 0
(28)
for any π‘ β [0, π], which proves that π is bounded; thus by Lemma 7, the operator πΉ 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 π and π satisfy the following growth conditions: (H 1 ) there exist two constants πΏ π > 0 and π π > 0, π = 1, 2 such that σ΅¨ σ΅¨ σ΅¨ σ΅¨ σ΅¨ σ΅¨σ΅¨ σ΅¨σ΅¨π (π‘, π₯1 , π¦1 ) β π (π‘, π₯2 , π¦2 )σ΅¨σ΅¨σ΅¨ β€ πΏ 1 σ΅¨σ΅¨σ΅¨π₯1 β π₯2 σ΅¨σ΅¨σ΅¨ + πΏ 2 σ΅¨σ΅¨σ΅¨π¦1 β π¦2 σ΅¨σ΅¨σ΅¨ , σ΅¨σ΅¨ σ΅¨ σ΅¨ σ΅¨ σ΅¨ σ΅¨ σ΅¨σ΅¨π (π‘, π₯1 , π¦1 ) β π (π‘, π₯2 , π¦2 )σ΅¨σ΅¨σ΅¨ β€ π 1 σ΅¨σ΅¨σ΅¨π₯1 β π₯2 σ΅¨σ΅¨σ΅¨ + π 2 σ΅¨σ΅¨σ΅¨π¦1 β π¦2 σ΅¨σ΅¨σ΅¨ ,
(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 π
π‘ β [0, π] , π₯1 , π₯2 , π¦1 , π¦2 β R; (29)
π
1 (π‘ + π₯ (π‘) + π¦ (π‘)) = 0, 2
0 < π‘ < 1,
3 (βπ‘ + π₯ (π‘) + π¦ (π‘)) = 0, 2
0 < π‘ < 1,
π·17/4 π₯ (π‘) +
π·9/2 π¦ (π‘) + (π)
(H 2 ) Consider max {πΏ 1 π1 + π 1 π2 , πΏ 2 π1 + π 2 π2 } = πΏ < 1.
(30)
Proof. Let (π₯1 , π¦1 ), (π₯2 , π¦2 ) β π Γ π; then σ΅¨ σ΅¨σ΅¨ σ΅¨σ΅¨πΉ1 (π₯2 , π¦2 ) (π‘) β πΉ1 (π₯1 , π¦1 ) (π‘)σ΅¨σ΅¨σ΅¨
σ΅¨σ΅¨ π σ΅¨ = σ΅¨σ΅¨σ΅¨σ΅¨β« πΊ1 (π‘, π ) π (π , π₯2 (π ) , π¦2 (π )) ππ σ΅¨σ΅¨ 0
(31)
π
σ΅¨σ΅¨ σ΅¨ β€ β« σ΅¨σ΅¨σ΅¨πΊ1 (π‘, π )σ΅¨σ΅¨σ΅¨ σ΅¨σ΅¨σ΅¨π (π , π₯2 (π ) , π¦2 (π )) 0 σ΅¨ βπ (π , π₯1 (π ) , π¦1 (π ))σ΅¨σ΅¨σ΅¨ ππ
σ΅©σ΅© σ΅© σ΅©σ΅©πΉ (π₯2 , π¦2 ) β πΉ (π₯1 , π¦1 )σ΅©σ΅©σ΅©
σ΅© σ΅© = σ΅©σ΅©σ΅©πΉ1 (π₯2 , π¦2 ) β πΉ1 (π₯1 , π¦1 )σ΅©σ΅©σ΅©
σ΅© σ΅© + σ΅©σ΅©σ΅©πΉ2 (π₯2 , π¦2 ) β πΉ2 (π₯1 , π¦1 )σ΅©σ΅©σ΅©
π¦(π) (0) = βπ¦(π) (1) ,
π = 0, 1, 2, 3, 4,
ππΌ 3 5πΌ4 β 14πΌ3 + 55πΌ2 + 146πΌ ( + ) β 0.1229, Ξ (πΌ + 1) 2 768
π2 =
ππ½ 3 5π½4 β 14π½3 + 55π½2 + 146π½ ( + ) β 0.0920. 768 Ξ (π½ + 1) 2 (35)
Then π1 π1 + π2 π1 < 1 and π1 π2 + π2 π2 < 1; by Theorem 8, the existence of the solution for the system (34) is obvious. Example 2. Consider the system π
Analogously,
Thus,
π = 0, 1, 2, 3, 4,
π1 =
σ΅© σ΅© σ΅© σ΅© β€ π1 (πΏ 1 σ΅©σ΅©σ΅©π₯2 β π₯1 σ΅©σ΅©σ΅© + πΏ 2 σ΅©σ΅©σ΅©π¦2 β π¦1 σ΅©σ΅©σ΅©) .
σ΅¨σ΅¨ σ΅¨ σ΅¨σ΅¨πΉ2 (π₯2 , π¦2 ) (π‘) β πΉ2 (π₯1 , π¦1 ) (π‘)σ΅¨σ΅¨σ΅¨ σ΅© σ΅© σ΅© σ΅© β€ π2 (π 1 σ΅©σ΅©σ΅©π₯2 β π₯1 σ΅©σ΅©σ΅© + π 2 σ΅©σ΅©σ΅©π¦2 β π¦1 σ΅©σ΅©σ΅©) .
π₯ (0) = βπ₯ (1) ,
(34)
where π = 1, |π(π‘, π₯, π¦)| β€ 1 + |π₯| + |π¦|, |π(π‘, π₯, π¦)| β€ 2 + 2|π₯| + 2|π¦|, πΌ = 17/4, π½ = 9/2, π0 = π1 = π2 = 1, π0 = π1 = π2 = 2. Consider
Then problem (1) has a unique solution.
σ΅¨σ΅¨ π σ΅¨ β β« πΊ1 (π‘, π ) π (π , π₯1 (π ) , π¦1 (π )) ππ σ΅¨σ΅¨σ΅¨σ΅¨ σ΅¨σ΅¨ 0
(π)
π
(32)
σ΅¨ σ΅¨σ΅¨ σ΅¨π¦ (π‘)σ΅¨σ΅¨ π·17/4 π₯ (π‘) + π‘ + sin π₯ (π‘) + 2 σ΅¨ σ΅¨σ΅¨ σ΅¨ σ΅¨σ΅¨ = 0, 1 + σ΅¨σ΅¨π¦ (π‘)σ΅¨σ΅¨
π·9/2 π¦ (π‘) + π‘2 + 2
|π₯ (π‘)| + arctan π¦ (π‘) = 0, 1 + |π₯ (π‘)|
π₯(π) (0) = βπ₯(π) (1) ,
π = 0, 1, 2, 3, 4,
π¦(π) (0) = βπ¦(π) (1) ,
π = 0, 1, 2, 3, 4,
0 < π‘ < 1, 0 < π‘ < 1,
(36) where π = 1, π(π‘, π₯, π¦) = π‘ + sin π₯(π‘) + 2(|π¦(π‘)|/(1 + |π¦(π‘)|)), π(π‘, π₯, π¦) = π‘2 + 2(|π₯(π‘)|/(1 + |π₯(π‘)|)) + arctan π¦(π‘). πΌ = 17/4, π½ = 9/2.
6
Abstract and Applied Analysis Noting that σ΅¨σ΅¨ σ΅¨ σ΅¨σ΅¨(sin π₯)σΈ σ΅¨σ΅¨σ΅¨ = |cos π₯| β€ 1, σ΅¨ σ΅¨
1 σ΅¨σ΅¨ σ΅¨ σ΅¨σ΅¨(arctan π¦)σΈ σ΅¨σ΅¨σ΅¨ = σ΅¨ σ΅¨ 1 + π¦2 β€ 1, (37)
we have σ΅¨σ΅¨ σ΅¨ σ΅¨σ΅¨π (π‘, π₯2 , π¦2 ) β π (π‘, π₯1 , π¦1 )σ΅¨σ΅¨σ΅¨
σ΅¨σ΅¨ σ΅¨σ΅¨π¦ (π‘)σ΅¨σ΅¨ σ΅¨σ΅¨ σ΅¨σ΅¨ σ΅¨σ΅¨π¦ (π‘)σ΅¨σ΅¨ σ΅¨σ΅¨ σ΅¨ σ΅¨ σ΅¨ σ΅¨ σ΅¨ σ΅¨ σ΅¨ σ΅¨ σ΅¨ σ΅¨ = σ΅¨σ΅¨σ΅¨sin π₯2 (π‘) β sin π₯1 (π‘)σ΅¨σ΅¨σ΅¨ + 2 σ΅¨σ΅¨σ΅¨σ΅¨ σ΅¨ σ΅¨σ΅¨2 σ΅¨ σ΅¨σ΅¨ σ΅¨σ΅¨σ΅¨σ΅¨ β σ΅¨σ΅¨σ΅¨σ΅¨ σ΅¨ σ΅¨σ΅¨1 σ΅¨ σ΅¨σ΅¨ σ΅¨σ΅¨σ΅¨σ΅¨ σ΅¨σ΅¨ 1 + σ΅¨σ΅¨π¦2 (π‘)σ΅¨σ΅¨ σ΅¨σ΅¨ σ΅¨σ΅¨ 1 + σ΅¨σ΅¨π¦1 (π‘)σ΅¨σ΅¨ σ΅¨σ΅¨ σ΅© σ΅© σ΅© σ΅© β€ σ΅©σ΅©σ΅©π₯2 β π₯1 σ΅©σ΅©σ΅© + 2 σ΅©σ΅©σ΅©π¦2 β π¦1 σ΅©σ΅©σ΅© , σ΅© σ΅¨σ΅¨ σ΅© σ΅© σ΅© σ΅¨ σ΅¨σ΅¨π (π‘, π₯2 , π¦2 ) β π (π‘, π₯1 , π¦1 )σ΅¨σ΅¨σ΅¨ β€ 2 σ΅©σ΅©σ΅©π₯2 β π₯1 σ΅©σ΅©σ΅© + σ΅©σ΅©σ΅©π¦2 β π¦1 σ΅©σ΅©σ΅© ,
π1 =
ππΌ 3 5πΌ4 β 14πΌ3 + 55πΌ2 + 146πΌ ( + ) β 0.1229, Ξ (πΌ + 1) 2 768
π2 =
ππ½ 3 5π½4 β 14π½3 + 55π½2 + 146π½ ( + ) β 0.0920. 768 Ξ (π½ + 1) 2 (38)
Obviously, max {πΏ 1 π1 + π 1 π2 , πΏ 2 π1 + π 2 π2 } < 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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