Advances in Difference Equations
Volume 2010 (2010), Article ID 340349, 10 pages
doi:10.1155/2010/340349
Research Article

Solutions to Fractional Differential Equations with Nonlocal Initial Condition in Banach Spaces

Zhi-Wei Lv,1 Jin Liang,2 and Ti-Jun Xiao3

1Department of Mathematics, University of Science and Technology of China, Hefei, Anhui 230026, China
2Department of Mathematics, Shanghai Jiao Tong University, Shanghai 200240, China
3Shanghai Key Laboratory for Contemporary Applied Mathematics, School of Mathematical Sciences, Fudan University, Shanghai 200433, China

Received 4 January 2010; Accepted 8 February 2010

Academic Editor: Gaston Mandata N'Guerekata

Copyright © 2010 Zhi-Wei Lv 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.

Abstract

A new existence and uniqueness theorem is given for solutions to differential equations involving the Caputo fractional derivative with nonlocal initial condition in Banach spaces. An application is also given.

1. Introduction

Fractional differential equations have played a significant role in physics, mechanics, chemistry, engineering, and so forth. In recent years, there are many papers dealing with the existence of solutions to various fractional differential equations; see, for example, [1–6].

In this paper, we discuss the existence of solutions to the nonlocal Cauchy problem for the following fractional differential equations in a Banach space 𝐸 :

𝑐 𝐷 𝛼 𝑥 ( 𝑡 ) = 𝑓 ( 𝑡 , 𝑥 ( 𝑡 ) ) , 0 ≤ 𝑡 ≤ 1 , 𝑥 ( 0 ) = 1 0 𝑔 ( 𝑠 ) 𝑥 ( 𝑠 ) 𝑑 𝑠 , ( 1 . 1 ) where 𝑐 𝐷 𝛼 is the standard Caputo's derivative of order 0 < 𝛼 < 1 , 𝑔 ∈ 𝐿 1 ( [ 0 , 1 ] , 𝑅 + ) , 𝑔 ( 𝑡 ) ∈ [ 0 , 1 ) , and 𝑓 is a given 𝐸 -valued function.

2. Basic Lemmas

Let 𝐸 be a real Banach space, and 𝜃 the zero element of 𝐸 . Denote by 𝐶 ( [ 0 , 1 ] , 𝐸 ) the Banach space of all continuous functions 𝑥 ∶ [ 0 , 1 ] → 𝐸 with norm ‖ 𝑥 ‖ 𝑐 = s u p 𝑡 ∈ [ 0 , 1 ] ‖ 𝑥 ( 𝑡 ) ‖ . Let 𝐿 1 ( [ 0 , 1 ] , 𝐸 ) be the Banach space of measurable functions 𝑥 ∶ [ 0 , 1 ] → 𝐸 which are Lebesgue integrable, equipped with the norm ‖ 𝑥 ‖ 𝐿 1 = ∫ 1 0 ‖ 𝑥 ( 𝑠 ) ‖ 𝑑 𝑠 . Let 𝑅 + = [ 0 , + ∞ ) , 𝑅 + ∫ = ( 0 , + ∞ ) , a n d 𝜇 = 1 0 𝑔 ( 𝑠 ) 𝑑 𝑠 . 𝐴 function 𝑥 ∈ 𝐶 ( [ 0 , 1 ] , 𝐸 ) is called a solution of (1.1) if it satisfies (1.1).

Recall the following defenition

Definition 2.1. Let 𝐵 be a bounded subset of a Banach space 𝑋 . The Kuratowski measure of noncompactness of 𝐵 is defined as 𝛼 ( 𝐵 ) ∶ = i n f { 𝛾 > 0 ; 𝐵 a d m i t s a fi n i t e c o v e r b y s e t s o f d i a m e t e r ≤ 𝛾 } . ( 2 . 1 )

Clearly, 0 ≤ 𝛼 ( 𝐵 ) < ∞ . For details on properties of the measure, the reader is referred to [2].

Definition 2.2 (see [7, 8]). The fractional integral of order 𝑞 with the lower limit 𝑡 0 for a function 𝑓 is defined as 𝐼 𝑞 1 𝑓 ( 𝑡 ) = Γ ( 𝑞 ) 𝑡 𝑡 0 ( 𝑡 − 𝑠 ) 𝑞 − 1 𝑓 ( 𝑠 ) 𝑑 𝑠 , 𝑡 > 𝑡 0 , 𝑞 > 0 , ( 2 . 2 ) where Γ is the gamma function.

Definition 2.3 (see [7, 8]). Caputo's derivative of order 𝑞 with the lower limit 𝑡 0 for a function 𝑓 can be written as 𝑐 𝐷 𝑞 1 𝑓 ( 𝑡 ) = Γ ( 𝑛 − 𝑞 ) 𝑡 𝑡 0 ( 𝑡 − 𝑠 ) 𝑛 − 𝑞 − 1 𝑓 ( 𝑛 ) ( 𝑠 ) 𝑑 𝑠 , 𝑡 > 𝑡 0 [ 𝑞 ] , 𝑞 > 0 , 𝑛 = + 1 . ( 2 . 3 )

Remark. Caputo's derivative of a constant is equal to 𝜃 .

Lemma (see [7]). Let 𝛼 > 0 . Then we have 𝑐 𝐷 𝑞 ( 𝐼 𝑞 𝑓 ( 𝑡 ) ) = 𝑓 ( 𝑡 ) . ( 2 . 4 )

Lemma (see [7]). Let 𝛼 > 0 and 𝑛 = [ 𝛼 ] + 1 . Then 𝐼 𝛼 ( 𝑐 𝐷 𝛼 𝑓 ( 𝑡 ) ) = 𝑓 ( 𝑡 ) − 𝑛 − 1 𝑘 = 0 𝑓 ( 𝑘 ) ( 0 ) 𝑡 𝑘 ! 𝑘 . ( 2 . 5 )

Lemma(see [9]). If 𝐻 ⊂ 𝐶 ( [ 0 , 1 ] , 𝐸 ) is bounded and equicontinuous, then (a) 𝛼 𝐶 ( 𝐻 ) = 𝛼 ( 𝐻 ( [ 0 , 1 ] ) ) ;(b) 𝛼 ( 𝐻 ( [ 0 , 1 ] ) ) = m a x 𝑡 ∈ [ 0 , 1 ] 𝛼 ( 𝐻 ( 𝑡 ) ) , where 𝐻 ( [ 0 , 1 ] ) = { 𝑥 ( 𝑡 ) ∶ 𝑥 ∈ 𝐻 , 𝑡 ∈ [ 0 , 1 ] } .

Lemma. 𝑄 ( 𝜏 ) ∫ Γ ( 𝛼 ) < 𝑒 , 𝑡 0 ( 𝑡 − 𝑠 ) 𝛼 − 1 𝑑 𝑠 Γ ( 𝛼 ) < 𝑒 , ( 2 . 6 ) where ∫ 𝑄 ( 𝜏 ) = 1 𝜏 𝑔 ( 𝑠 ) ( 𝑠 − 𝜏 ) 𝛼 − 1 𝑑 𝑠 , 𝑡 , 𝜏 ∈ [ 0 , 1 ] .

Proof. A direct computation shows 𝑄 ( 𝜏 ) Γ = ∫ ( 𝛼 ) 1 𝜏 𝑔 ( 𝑠 ) ( 𝑠 − 𝜏 ) 𝛼 − 1 𝑑 𝑠 ∫ ∞ 0 𝑠 𝛼 − 1 𝑒 − 𝑠 < ∫ 𝑑 𝑠 1 𝜏 ( 𝑠 − 𝜏 ) 𝛼 − 1 𝑑 𝑠 ∫ ∞ 0 𝑠 𝛼 − 1 𝑒 − 𝑠 = ∫ 𝑑 𝑠 0 1 − 𝜏 𝑠 𝛼 − 1 𝑑 𝑠 ∫ ∞ 0 𝑠 𝛼 − 1 𝑒 − 𝑠 ≤ 𝑒 ∫ 𝑑 𝑠 0 1 − 𝜏 𝑠 𝛼 − 1 𝑒 − 𝑠 𝑑 𝑠 ∫ ∞ 0 𝑠 𝛼 − 1 𝑒 − 𝑠 𝑑 𝑠 < 𝑒 ( 2 . 7 ) and ∫ 𝑡 0 ( 𝑡 − 𝑠 ) 𝛼 − 1 𝑑 𝑠 = ∫ Γ ( 𝛼 ) 𝑡 0 𝑠 𝛼 − 1 𝑑 𝑠 ∫ ∞ 0 𝑠 𝛼 − 1 𝑒 − 𝑠 ≤ 𝑒 ∫ 𝑑 𝑠 𝑡 0 𝑠 𝛼 − 1 𝑒 − 𝑠 𝑑 𝑠 ∫ ∞ 0 𝑠 𝛼 − 1 𝑒 − 𝑠 𝑑 𝑠 < 𝑒 . ( 2 . 8 )

3. Main Results

( H 1 ) 𝑓 ∈ ( [ 0 , 1 ] × 𝐸 , 𝐸 ) , and there exist 𝑀 > 0 , 𝑝 𝑓 ( 𝑡 ) ≤ 𝑀 𝑓 𝑜 𝑟 𝑡 ∈ [ 0 , 1 ] , 𝑝 𝑓 ∈ 𝐿 1 ( [ 0 , 1 ] , 𝑅 + ) such that ‖ 𝑓 ( 𝑡 , 𝑥 ) ‖ ≤ 𝑝 𝑓 ( 𝑡 ) ‖ 𝑥 ‖ for 𝑡 ∈ [ 0 , 1 ] and each 𝑥 ∈ 𝐸 . ( H 2 ) For any 𝑡 ∈ [ 0 , 1 ] and 𝑅 > 0 , 𝑓 ( 𝑡 , 𝐵 𝑅 ) = { 𝑓 ( 𝑡 , 𝑥 ) ∶ 𝑥 ∈ 𝐵 𝑅 } is relatively compact in 𝐸 , where 𝐵 𝑅 = { 𝑥 ∈ 𝐶 ( [ 0 , 1 ] , 𝐸 ) , ‖ 𝑥 ‖ 𝐶 ≤ 𝑅 } and Λ 1 = ( 2 − 𝜇 ) 𝑒 1 − 𝜇 𝑀 < 1 . ( 3 . 1 )

Lemma. If ( H 1 ) holds, then the problem (1.1) is equivalent to the following equation: 1 𝑥 ( 𝑡 ) = ( 1 − 𝜇 ) Γ ( 𝛼 ) 1 0 1 𝑄 ( 𝜏 ) 𝑓 ( 𝜏 , 𝑥 ( 𝜏 ) ) 𝑑 𝜏 + Γ ( 𝛼 ) 𝑡 0 ( 𝑡 − 𝑠 ) 𝛼 − 1 𝑓 ( 𝑠 , 𝑥 ( 𝑠 ) ) 𝑑 𝑠 . ( 3 . 2 )

Proof. By Lemma 2.6 and (1.1), we have 1 𝑥 ( 𝑡 ) = 𝑥 ( 0 ) + Γ ( 𝛼 ) 𝑡 0 ( 𝑡 − 𝑠 ) 𝛼 − 1 𝑓 ( 𝑠 , 𝑥 ( 𝑠 ) ) 𝑑 𝑠 . ( 3 . 3 ) Therefore, 𝑥 ( 0 ) = 1 0 = 𝑔 ( 𝑠 ) 𝑥 ( 𝑠 ) 𝑑 𝑠 1 0 𝑔 𝑥 1 ( 𝑠 ) ( 0 ) + Γ ( 𝛼 ) 𝑠 0 ( 𝑠 − 𝜏 ) 𝛼 − 1 𝑓 = ( 𝜏 , 𝑥 ( 𝜏 ) ) 𝑑 𝜏 𝑑 𝑠 1 0 1 𝑔 ( 𝑠 ) 𝑑 𝑠 𝑥 ( 0 ) + Γ ( 𝛼 ) 1 0 𝑔 ( 𝑠 ) 𝑠 0 ( 𝑠 − 𝜏 ) 𝛼 − 1 𝑓 ( 𝜏 , 𝑥 ( 𝜏 ) ) 𝑑 𝜏 𝑑 𝑠 . ( 3 . 4 ) So, 1 𝑥 ( 0 ) = ∫ 1 − 1 0 𝑔 ( 𝑠 ) 𝑑 𝑠 Γ ( 𝛼 ) 1 0 𝑔 ( 𝑠 ) 𝑠 0 ( 𝑠 − 𝜏 ) 𝛼 − 1 = 1 𝑓 ( 𝜏 , 𝑥 ( 𝜏 ) ) 𝑑 𝜏 𝑑 𝑠 ( 1 − 𝜇 ) Γ ( 𝛼 ) 1 0 𝑓 ( 𝜏 , 𝑥 ( 𝜏 ) ) 1 𝜏 ( 𝑠 − 𝜏 ) 𝛼 − 1 = 1 𝑔 ( 𝑠 ) 𝑑 𝑠 𝑑 𝜏 ( 1 − 𝜇 ) Γ ( 𝛼 ) 1 0 𝑄 ( 𝜏 ) 𝑓 ( 𝜏 , 𝑥 ( 𝜏 ) ) 𝑑 𝜏 , ( 3 . 5 ) and then 1 𝑥 ( 𝑡 ) = ( 1 − 𝜇 ) Γ ( 𝛼 ) 1 0 1 𝑄 ( 𝜏 ) 𝑓 ( 𝜏 , 𝑥 ( 𝜏 ) ) 𝑑 𝜏 + Γ ( 𝛼 ) 𝑡 0 ( 𝑡 − 𝑠 ) 𝛼 − 1 𝑓 ( 𝑠 , 𝑥 ( 𝑠 ) ) 𝑑 𝑠 . ( 3 . 6 )
Conversely, if 𝑥 is a solution of (3.2), then for every 𝑡 ∈ [ 0 , 1 ] , according to Remark 2.4 and Lemma 2.5, we have
𝑐 𝐷 𝛼 𝑥 ( 𝑡 ) = 𝑐 𝐷 𝛼 1 ( 1 − 𝜇 ) Γ ( 𝛼 ) 1 0 1 𝑄 ( 𝜏 ) 𝑓 ( 𝜏 , 𝑥 ( 𝜏 ) ) 𝑑 𝜏 + Γ ( 𝛼 ) 𝑡 0 ( 𝑡 − 𝑠 ) 𝛼 − 1 = 𝑓 ( 𝑠 , 𝑥 ( 𝑠 ) ) 𝑑 𝑠 𝑐 𝐷 𝛼 1 ( 1 − 𝜇 ) Γ ( 𝛼 ) 1 0 + 𝑄 ( 𝜏 ) 𝑓 ( 𝜏 , 𝑥 ( 𝜏 ) ) 𝑑 𝜏 𝑐 𝐷 𝛼 1 Γ ( 𝛼 ) 𝑡 0 ( 𝑡 − 𝑠 ) 𝛼 − 1 𝑓 ( 𝑠 , 𝑥 ( 𝑠 ) ) 𝑑 𝑠 = 𝜃 + 𝑐 𝐷 𝛼 ( 𝐼 𝛼 𝑓 ( 𝑡 , 𝑥 ( 𝑡 ) ) ) = 𝑓 ( 𝑡 , 𝑥 ( 𝑡 ) ) . ( 3 . 7 ) It is obvious that ∫ 𝑥 ( 0 ) = 1 0 𝑔 ( 𝑠 ) 𝑥 ( 𝑠 ) 𝑑 𝑠 . This completes the proof.

Theorem. If (H1) and (H2) hold, then the initial value problem (1.1) has at least one solution.

Proof. Define operator 𝐴 ∶ 𝐶 ( [ 0 , 1 ] , 𝐸 ) → 𝐶 ( [ 0 , 1 ] , 𝐸 ) , by 1 ( 𝐴 𝑥 ) ( 𝑡 ) = ( 1 − 𝜇 ) Γ ( 𝛼 ) 1 0 1 𝑄 ( 𝜏 ) 𝑓 ( 𝜏 , 𝑥 ( 𝜏 ) ) 𝑑 𝜏 + Γ ( 𝛼 ) 𝑡 0 ( 𝑡 − 𝑠 ) 𝛼 − 1 𝑓 ( 𝑠 , 𝑥 ( 𝑠 ) ) 𝑑 𝑠 . ( 3 . 8 ) Clearly, the fixed points of the operator 𝐴 are solutions of problem (1.1).
It is obvious that 𝐵 𝑅 is closed, bounded, and convex.
Step 1. We prove that 𝐴 is continuous.
Let
𝑥 𝑛 , [ ] ‖ ‖ 𝑥 𝑥 ∈ 𝐶 ( 0 , 1 , 𝐸 ) , 𝑛 − 𝑥 ‖ ‖ 𝑐 ⟶ 0 ( 𝑛 ⟶ ∞ ) . ( 3 . 9 ) Then 𝑟 = s u p 𝑛 ‖ 𝑥 𝑛 ‖ 𝐶 < ∞ and ‖ 𝑥 ‖ 𝐶 ≤ 𝑟 . For each 𝑡 ∈ [ 0 , 1 ] , ‖ ‖ 𝐴 𝑥 𝑛 𝐴 ( 𝑡 ) − 𝑥 ‖ ‖ ≤ 𝑒 ( 𝑡 ) 1 − 𝜇 1 0 ‖ ‖ 𝑓 𝜏 , 𝑥 𝑛 ( 𝜏 ) − 𝑓 𝜏 , ‖ ‖ + 1 𝑥 ( 𝜏 ) 𝑑 𝜏 Γ ( 𝛼 ) 𝑡 0 ( 𝑡 − 𝑠 ) ( 𝛼 − 1 ) ‖ ‖ 𝑓 𝑠 , 𝑥 𝑛 ( 𝑠 ) − 𝑓 𝑠 , ‖ ‖ 𝑥 ( 𝑠 ) 𝑑 𝑠 . ( 3 . 1 0 ) It is clear that 𝑓 𝑡 , 𝑥 𝑛 ( 𝑡 ) ⟶ 𝑓 𝑡 , 𝑥 [ ] , ‖ ‖ 𝑓 ( 𝑡 ) , a s 𝑛 ⟶ ∞ , 𝑡 ∈ 0 , 1 𝑡 , 𝑥 𝑛 ( 𝑡 ) − 𝑓 𝑡 , ‖ ‖ 𝑥 ( 𝑡 ) ≤ 2 𝑀 𝑟 . ( 3 . 1 1 ) It follows from (3.11) and the dominated convergence theorem that ‖ ‖ ( 𝐴 𝑥 𝑛 ) − ( 𝐴 ‖ ‖ 𝑥 ) 𝐶 ⟶ 0 , a s 𝑛 ⟶ ∞ . ( 3 . 1 2 ) Step 2. We prove that 𝐴 ( 𝐵 𝑅 ) ⊂ 𝐵 𝑅 .
Let 𝑥 ∈ 𝐵 𝑅 . Then for each 𝑡 ∈ [ 0 , 1 ] , we have
‖ 1 ( 𝐴 𝑥 ) ( 𝑡 ) ‖ ≤ 1 − 𝜇 1 0 𝑄 ( 𝜏 ) 1 Γ ( 𝛼 ) ‖ 𝑓 ( 𝜏 , 𝑥 ( 𝜏 ) ) ‖ 𝑑 𝜏 + Γ ( 𝛼 ) 𝑡 0 ( 𝑡 − 𝑠 ) 𝛼 − 1 ≤ 1 ‖ 𝑓 ( 𝑠 , 𝑥 ( 𝑠 ) ) ‖ 𝑑 𝑠 1 − 𝜇 1 0 𝑄 ( 𝜏 ) 𝑝 Γ ( 𝛼 ) 𝑓 1 ( 𝜏 ) ‖ 𝑥 ( 𝜏 ) ‖ 𝑑 𝜏 + Γ ( 𝛼 ) 𝑡 0 ( 𝑡 − 𝑠 ) 𝛼 − 1 𝑝 𝑓 ≤ 𝑒 ( 𝑠 ) ‖ 𝑥 ( 𝑠 ) ‖ 𝑑 𝑠 1 − 𝜇 𝑀 + 𝑒 𝑀 ‖ 𝑥 ‖ 𝐶 < 𝑅 . ( 3 . 1 3 ) Step 3. We prove that 𝐴 ( 𝐵 𝑅 ) is equicontinuous.
Let 𝑡 1 , 𝑡 2 ∈ [ 0 , 1 ] , 𝑡 1 < 𝑡 2 , and 𝑥 ∈ 𝐵 𝑅 . We deduce that
‖ ‖ 𝑡 ( 𝐴 𝑥 ) 2 − 𝑡 ( 𝐴 𝑥 ) 1 ‖ ‖ = 1 ‖ ‖ ‖ Γ ( 𝛼 ) 𝑡 2 0 𝑡 2 − 𝑠 𝛼 − 1 𝑓 ( 𝑠 , 𝑥 ( 𝑠 ) ) 𝑑 𝑠 − 𝑡 1 0 𝑡 1 − 𝑠 𝛼 − 1 𝑓 ‖ ‖ ‖ ≤ 1 ( 𝑠 , 𝑥 ( 𝑠 ) ) 𝑑 𝑠 Γ ( 𝛼 ) 𝑡 1 0 | | | 𝑡 2 − 𝑠 𝛼 − 1 − 𝑡 1 − 𝑠 𝛼 − 1 | | | + 1 ‖ 𝑓 ( 𝑠 , 𝑥 ( 𝑠 ) ) ‖ 𝑑 𝑠 Γ ( 𝛼 ) 𝑡 2 𝑡 1 𝑡 2 − 𝑠 𝛼 − 1 ‖ ≤ 𝑓 ( 𝑠 , 𝑥 ( 𝑠 ) ) ‖ 𝑑 𝑠 𝑡 1 0 | | | 𝑡 2 − 𝑠 𝛼 − 1 − 𝑡 1 − 𝑠 𝛼 − 1 | | | 𝑑 𝑠 + 𝑡 2 𝑡 1 𝑡 2 − 𝑠 𝛼 − 1 𝑑 𝑠 𝑀 𝑅 ≤ 2 𝑡 Γ ( 𝛼 ) 2 − 𝑡 1 𝛼 + 𝑡 𝛼 2 − 𝑡 𝛼 1 𝑀 𝑅 . Γ ( 𝛼 + 1 ) ( 3 . 1 4 ) As 𝑡 1 → 𝑡 2 , the right-hand side of the above inequality tends to zero.Step 4. We prove that 𝐴 ( 𝐵 𝑅 ) is relatively compact.
Let 5 ⊂ 𝐵 𝑅 be arbitrarily given. Using the formula
𝑏 𝑎 𝑦 ( 𝑡 ) 𝑑 𝑡 ∈ ( 𝑏 − 𝑎 ) [ ] } c o { 𝑦 ( 𝑡 ) ∶ 𝑡 ∈ 0 , 1 ( 3 . 1 5 ) for 𝑦 ∈ 𝐶 ( [ 𝑎 , 𝑏 ] , 𝐸 ) and ( H 2 ) , we obtain 𝛼 ( ( 𝐴 𝑉 ) ( 𝑡 ) ) ≤ 𝛼 c o 𝑄 ( 𝑠 ) [ ] ( 1 − 𝑢 ) Γ ( 𝛼 ) 𝑓 ( 𝑠 , 𝑥 ( 𝑠 ) ) ∶ 𝑠 ∈ 0 , 1 , 𝑥 ∈ 𝑉 + 𝛼 c o ( 𝑡 − 𝑠 ) 𝛼 − 1 [ ] [ ] ≤ Γ ( 𝛼 ) 𝑓 ( 𝑠 , 𝑥 ( 𝑠 ) ) ∶ 𝑠 ∈ 0 , 𝑡 , 𝑡 ∈ 0 , 1 , 𝑥 ∈ 𝑉 𝑄 ( 𝑠 ) [ ] + ( 1 − 𝑢 ) Γ ( 𝛼 ) 𝛼 ( 𝑓 ( 𝑠 , 𝑉 ( 𝑠 ) ) ) ∶ 𝑠 ∈ 0 , 1 ( 𝑡 − 𝑠 ) 𝛼 − 1 [ ] [ ] Γ ( 𝛼 ) 𝛼 ( 𝑓 ( 𝑠 , 𝑉 ( 𝑠 ) ) ) ∶ 𝑠 ∈ 0 , 𝑡 , 𝑡 ∈ 0 , 1 = 0 . ( 3 . 1 6 ) It follows from (3.16) that 𝛼 ( ( 𝐴 𝑉 ) ( 𝑡 ) ) = 0 𝑓 𝑜 𝑟 𝑡 ∈ [ 0 , 1 ] . This, together with Lemma 2.7, yields that 𝛼 𝐶 ( 𝐴 𝑉 ) = 0 . ( 3 . 1 7 ) From (3.17), we see that 𝐴 ( 𝐵 𝑅 ) is relatively compact. Hence, 𝐴 ∶ 𝐵 𝑅 → 𝐵 𝑅 is completely continuous. Finally, the Schauder fixed point theorem guarantees that 𝐴 has a fixed point in 𝐵 𝑅 .

Theorem. Besides the hypotheses of Theorem 3.2, we suppose that there exists a constant 𝐿 such that 0 < 𝐿 < Λ 2 , ( 3 . 1 8 ) ‖ 𝑓 ( 𝑡 , 𝑢 ) − 𝑓 ( 𝑡 , 𝑤 ) ‖ ≤ 𝐿 ‖ 𝑢 − 𝑤 ‖ , f o r e v e r y 𝑢 , 𝑤 ∈ 𝐵 𝑅 , ( 3 . 1 9 ) where Λ 2 = 1 − 𝜇 . ( 2 − 𝜇 ) 𝑒 ( 3 . 2 0 ) Then, the solution 𝑥 ( 𝑡 ) of (1.1) is unique in 𝐵 𝑅 .

Proof. From Theorem 3.2, we know that there exists at least one solution 𝑥 ( 𝑡 ) in 𝐵 𝑅 . We suppose to the contrary that there exist two different solutions 𝑢 ( 𝑡 ) and 𝑤 ( 𝑡 ) in 𝐵 𝑅 . It follows from (3.8) that 𝑒 ‖ 𝑢 ( 𝑡 ) − 𝑤 ( 𝑡 ) ‖ ≤ 1 − 𝜇 1 0 + 1 ‖ 𝑓 ( 𝜏 , 𝑢 ( 𝜏 ) ) − 𝑓 ( 𝜏 , 𝑤 ( 𝜏 ) ) ‖ 𝑑 𝜏 Γ ( 𝛼 ) 𝑡 0 ( 𝑡 − 𝑠 ) 𝛼 − 1 ( ≤ 𝑒 ‖ 𝑓 𝑠 , 𝑢 ( 𝑠 ) ) − 𝑓 ( 𝑠 , 𝑤 ( 𝑠 ) ) ‖ 𝑑 𝑠 1 − 𝜇 1 0 + 1 𝐿 ‖ 𝑢 ( 𝜏 ) − 𝑤 ( 𝜏 ) ‖ 𝑑 𝜏 Γ ( 𝛼 ) 𝑡 0 ( 𝑡 − 𝑠 ) 𝛼 − 1 𝐿 ‖ 𝑢 ( 𝑠 ) − 𝑤 ( 𝑠 ) ‖ 𝑑 𝑠 . ( 3 . 2 1 ) Therefore, we get ‖ 𝑢 − 𝑤 ‖ 𝐶 ≤ 2 − 𝜇 1 − 𝜇 𝑒 𝐿 ‖ 𝑢 − 𝑤 ‖ 𝐶 . ( 3 . 2 2 ) By (3.18), we obtain ‖ 𝑢 − 𝑤 ‖ 𝐶 = 0 . So, the two solutions are identical in 𝐵 𝑅 .

4. Example

Let

𝐸 = 𝑐 0 = 𝑥 𝑥 = 1 , … , 𝑥 𝑛 , … ∶ 𝑥 𝑛 ⟶ 0 ( 4 . 1 ) with the norm ‖ 𝑥 ‖ = s u p 𝑛 | 𝑥 𝑛 | . Consider the following nonlocal Cauchy problem for the following fractional differential equation in 𝐸 :

𝑐 𝐷 𝛼 𝑥 𝑛 ( 𝑡 ) = 1 + 𝑡 1 0 0 𝑛 2 𝑥 𝑛 [ ] 𝑥 ( 𝑡 ) , 𝑡 ∈ 0 , 1 , 0 < 𝛼 < 1 , 𝑛 ( 0 ) = 1 0 1 2 𝑥 𝑛 ( 𝑠 ) 𝑑 𝑠 . ( 4 . 2 )

Conclusion 4. Problem (4.2) has only one solution on [ 0 , 1 ] .

Proof. Write 𝑓 𝑛 ( 𝑡 , 𝑥 ) = 1 + 𝑡 1 0 0 𝑛 2 𝑥 𝑛 𝑓 , 𝑓 = 1 , … , 𝑓 𝑛 , 1 , … 𝑔 ( 𝑠 ) = 2 , 𝑝 𝑓 ( 𝑡 ) = 1 + 𝑡 . 1 0 0 𝑛 ( 4 . 3 ) Then it is clear that [ ] 𝑓 ∈ 𝐶 ( 0 , 1 × 𝐸 , 𝐸 ) , 𝑝 𝑓 1 ( 𝑡 ) ≤ 𝑝 5 0 = 𝑀 , 𝑓 [ ] ∈ 𝐿 0 , 1 , 𝑅 + , ‖ 𝑓 ( 𝑡 , 𝑥 ) ‖ ≤ 𝑝 𝑓 ‖ 𝑥 ‖ . ( 4 . 4 ) So, ( H 1 ) is satisfied.
In the same way as in Example 3 . 2 . 1 in [9], we can prove that 𝑓 ( 𝑡 , 𝐵 𝑅 ) is relatively compact in 𝑐 0 .
By a direct computation, we get
Λ 1 = ( 2 − 𝜇 ) 𝑒 1 − 𝜇 𝑀 ≤ ( 2 − 𝜇 ) 𝑒 1 1 − 𝜇 = 5 0 3 𝑒 5 0 < 1 . ( 4 . 5 ) Hence, condition ( H 2 ) is also satisfied.
Moreover, we have
| | 𝑓 𝑛 ( 𝑡 , 𝑢 ) − 𝑓 𝑛 | | = | | | ( 𝑡 , 𝑤 ) 1 + 𝑡 1 0 0 𝑛 2 𝑢 𝑛 − 1 + 𝑡 1 0 0 𝑛 2 𝑤 𝑛 | | | ≤ 1 | | 𝑢 5 0 𝑛 − 𝑤 𝑛 | | , ( 4 . 6 ) so 1 ‖ 𝑓 ( 𝑡 , 𝑢 ) − 𝑓 ( 𝑡 , 𝑤 ) ‖ ≤ 5 0 ‖ 𝑢 − 𝑤 ‖ . ( 4 . 7 ) Clearly, Λ 2 = 1 − 𝜇 = ( 2 − 𝜇 ) 𝑒 1 − 1 / 2 = 1 3 𝑒 / 2 . 3 𝑒 ( 4 . 8 ) Therefore, 𝐿 = 1 / 5 0 < 1 / 3 𝑒 . Thus, our conclusion follows from Theorem 3.3.

Acknowledgments

This work was supported partially by the NSF of China (10771202), the Research Fund for Shanghai Key Laboratory for Contemporary Applied Mathematics (08DZ2271900) and the Specialized Research Fund for the Doctoral Program of Higher Education of China (2007035805).

References

  1. S. Abbas and M. Benchohra, “Darboux problem for perturbed partial differential equations of fractional order with finite delay,” Nonlinear Analysis: Hybrid Systems, vol. 3, no. 4, pp. 597–604, 2009.
  2. J. Henderson and A. Ouahab, “Fractional functional differential inclusions with finite delay,” Nonlinear Analysis: Theory, Methods & Applications, vol. 70, no. 5, pp. 2091–2105, 2009.
  3. V. Lakshmikantham, “Theory of fractional functional differential equations,” Nonlinear Analysis: Theory, Methods & Applications, vol. 69, no. 10, pp. 3337–3343, 2008.
  4. V. Lakshmikantham and S. Leela, “Nagumo-type uniqueness result for fractional differential equations,” Nonlinear Analysis: Theory, Methods & Applications, vol. 71, no. 7-8, pp. 2886–2889, 2009.
  5. G. M. Mophou and G. M. N'Guérékata, “Existence of the mild solution for some fractional differential equations with nonlocal conditions,” Semigroup Forum, vol. 79, no. 2, pp. 315–322, 2009.
  6. X.-X. Zhu, “A Cauchy problem for abstract fractional differential equations with infinite delay,” Communications in Mathematical Analysis, vol. 6, no. 1, pp. 94–100, 2009.
  7. A. A. Kilbas, H. M. Srivastava, and J. J. Trujjllo, Theory and Applications of Fractional Differential Equations, North-Holland Mathematics Studies, vol. 204, Elsevier, Amsterdam, The Netherlands, 2006.
  8. I. Podlubny, Fractional Differential Equations, Academic Press, San Diego, Calif, USA, 1993.
  9. D. J. Guo, V. Lakshmikantham, and X. Z. Liu, Nonlinear Integral Equations in Abstract Spaces, Kluwer Academic Publishers, Dordrecht, The Netherlands, 1996.