Analytic study on linear neutral fractional differential equations

Applied Mathematics and Computation xxx (2015) xxx–xxx

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Applied Mathematics and Computation journal homepage: www.elsevier.com/locate/amc

Analytic study on linear neutral fractional differential equations q Xian-Feng Zhou ⇑, Fuli Yang, Wei Jiang School of Mathematical Sciences, Anhui University, Hefei 230039, China

a r t i c l e

i n f o

a b s t r a c t This paper is devoted to investigating linear neutral fractional differential equations with constant coefficients. The existence, uniqueness and iterative formula of the solution are obtained. Meanwhile, dependence of the solution on initial value and the general solution represented by a fundamental solution matrix are discussed. Several examples are given to illustrate the applications of our results. Some conclusions in the literature are extended greatly. Ó 2014 Elsevier Inc. All rights reserved.

Keywords: Linear fractional differential equation Neutral Existence Uniqueness Iterative formula

1. Introduction Fractional differential equations have gained increasing attention because of their varied applications in various fields of applied sciences and engineering. For example, memory and hereditary of various material and process can be properly described as fractional differential systems. At present, a great deal of effort were spent in linear and nonlinear fractional differential equations. Most of the works focused on the existence, uniqueness of solutions of initial value problem for fractional differential equations, stability, controllability and observability for fractional differential systems, etc, see the monograph [1,2,4,5,3], the papers [16,8,9,11,6,7,13,14,12,15,17–23,10,24] and the references therein. Usually, the existence and uniqueness of the solution for nonlinear fractional differential equations are obtained by the fixed point theory [6,7], the classical method which is involved of comparison principle [8,9], variational iteration and numerical approach [10,11], Cauchy problem for evolution equations were discussed in [13,14,12], etc. Linear fractional differential equations (or systems) with Caputo’s derivative have been discussed in [11,15,17,16,18–21]. In [18], the authors studied the existence, uniqueness and dependence of the solution of the fractional differential equation

Da0 yðtÞ ¼ ayðt  sÞ þ byðtÞ þ f ðtÞ; yðtÞ ¼ /ðtÞ;

t > 0;

ð1:1Þ

t 2 ½s; 0;

ð1:2Þ a

where a; b are constants, f is continuous on ½0; T; T > 0, the initial function / is continuous on ½s; 0, and D0 is the Caputo’s derivative of order a. In [19], the author considered the general solution of the commensurate fractional order equation

q This paper is supported by Natural Science Foundation of China (11371027 and 11471015), Program of Natural Science Research in Anhui Universities (KJ2011A020 and KJ2012A032), Research Fund for Doctoral Program of Higher Education of China (20123401120001 and 20133401120013). ⇑ Corresponding author. E-mail address: [email protected] (X.-F. Zhou).

http://dx.doi.org/10.1016/j.amc.2014.12.056 0096-3003/Ó 2014 Elsevier Inc. All rights reserved.

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X.-F. Zhou et al. / Applied Mathematics and Computation xxx (2015) xxx–xxx

a

d a xðtÞ ¼ AxðtÞ; dt

0 < t 6 a;

ð1:3Þ

and the existence, uniqueness of the incommensurate fractional order linear Eq. (1.3), where x 2 Rn ; a > 0, A is n  n matrix. da is the Caputo fractional derivative of order a and 0 < a 6 1. dt a Motivated by the papers [18,19], this paper is concerned with the linear neutral fractional differential equation

Da0 xðtÞ ¼ AxðtÞ þ Bxðt  sÞ þ C  Da0 xðt  sÞ þ f ðtÞ;

t P 0;

ð1:4Þ

s 6 t 6 0;

xðtÞ ¼ /ðtÞ;

ð1:5Þ

where 0 < a < 1; xðtÞ 2 Rn is a n dimensional vector, A; B; C are n  n constant matrices, C – 0, the initial function /ðtÞ 2 Cð½s; 0; Rn Þ is continuous, s > 0 is delay, Da0 denotes the Caputo’s derivative of order a with the lower limit 0, f ðtÞ 2 Rn and is continuous on ½0; þ1Þ or f ðtÞ 2 Cð½0; 1; Rn Þ. Obviously, if n ¼ 1 and C ¼ 0, then Eqs. (1.4) and (1.5) can be reduced to Eqs. (1.1) and (1.2); if s ¼ 0; B ¼ 0 and C ¼ 0, then Eqs. (1.4) and (1.5) can be reduced to Eq. (1.3). Throughout this paper, Dat0 f ðtÞ or c Dat0 f ðtÞ denotes the Caputo’s derivative of order a with the lower limit t 0 for function f ;r Dat0 f ðtÞdenotes the Riemann–Liouville’s derivative of order a with the lower limit t 0 for function f ; Iat0 f ðtÞ denotes the integral of order a with lower limit t 0 for function f ; f ðsÞ ¼ L½f ðtÞ; s denotes the Laplace transform of the function f ðtÞ, and ‘‘j  j’’ denotes the norm of ‘‘’’ in Rn or in Rnn . 2. Preliminaries In this section, we mainly recall some definitions and lemmas which will be used later. Definition 2.1 [1]. The fractional integral of order a with the lower limit 0 for function f is defined as

Ia0 f ðtÞ ¼

1 CðaÞ

Z

t

ðt  sÞa1 f ðsÞds;

t > 0;

a > 0;

ð2:1Þ

0

provided the right-hand side is pointwise defined on ½0; þ1Þ, where C is the gamma function. Definition 2.2 [1]. The Caputo’s derivative of order a with the lower limit 0 for a function f : ½0; 1Þ ! R can be written as 1 Da0 f ðtÞ ¼ Cðn aÞ

¼

Rt 0

ðt  sÞna1 f

a ðnÞ ðtÞ; In 0 f

ðnÞ

ðsÞds

ð2:2Þ

0 < n  1 < a 6 n:

t > 0;

Particularly, when 0 < a < 1, it holds

Da0 f ðtÞ ¼

1 Cð1  aÞ

Z 0

t

a ðt  sÞa f ðsÞds ¼ I1 0 f ðtÞ; 0

0

t > 0:

ð2:3Þ

The Laplace transform of the Caputo’s fractional derivative Da0 f ðtÞ is

L½Da0 f ðtÞ; s ¼

Z

þ1

0

est ðDa0 f ðtÞÞdt ¼ sa f ðsÞ 

n1 X ðkÞ sak1 f ð0Þ;

n  1 < a 6 n;

ð2:4Þ

k¼0

where f ðsÞ is the Laplace transform of f ðtÞ. Especially, for 0 < a < 1, it holds

Z

þ1

0

est ðDa0 f ðtÞÞdt ¼ sa f ðsÞ  sa1 f ð0Þ:

ð2:5Þ

Definition 2.3 [1]. The two-parameter Mittag–Leffler function is defined as

Ea;b ðzÞ ¼

1 X

zk ; Cðak þ bÞ k¼0

a > 0; b > 0:

ð2:6Þ

Especially,

Ea;1 ðzÞ ¼ Ea ðzÞ ¼

1 X

zk ; Cðak þ 1Þ k¼0

a > 0; b > 0:

ð2:7Þ

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The Laplace transform of Mittag–Leffler function is

h i Z ðkÞ L t akþb1 Ea;b ðata Þ; s ¼

1

0

k!sab

ðkÞ

est takþb1 Ea;b ðat a Þdt ¼

ðsa  aÞkþ1

;

1

ReðsÞ > jaja :

ð2:8Þ

where ReðsÞ denotes the real part of s. In addition, the Laplace transform of ta1 is

L½t a1 ; s ¼ CðaÞsa ;

ReðsÞ > 0:

ð2:9Þ

Definition 2.4. A function xðtÞ is said to be a solution of initial value problem (IVP in short) Eqs. (1.4) and (1.5) on ½s; T 0 , if there exists a constant T 0 > 0 such that (i) xðtÞ 2 Cð½s; T 0 ; Rn Þ for t 2 ½s; T 0 ; (ii) xðtÞ satisfies

Da0 ðxðtÞ  Cxðt  sÞÞ ¼ AxðtÞ þ Bxðt  sÞ þ f ðtÞ;

t 2 ½0; T 0 

ð2:10Þ

and xðtÞ ¼ /ðtÞ for t 2 ½s; 0 almost everywhere.

Lemma 2.1 [4]. Let 0 < a < 1. Then it holds

Ia0 ðc Da0 xðtÞÞ ¼ xðtÞ  xð0Þ:

ð2:11Þ

Lemma 2.2 [4]. There exists a link between the Riemann–Liouville and the Caputo’s fractional derivative of order a . Namely, c

n1 X

ðkÞ

f ðt 0 Þ ðt  t 0 Þka ; C ðk  a þ 1Þ k¼0

Dat0 f ðtÞ¼r Dat0 f ðtÞ 

t > t0 ;

n  1 < ReðaÞ < n:

ð2:12Þ

t a ;

ð2:13Þ

Especially, for 0 < a < 1 and t0 ¼ 0, it holds c

Da0 f ðtÞ ¼

Z

1 Cð1  aÞ

t 0

ðt  sÞa f ðsÞds¼r Da0 f ðtÞ  0

f ð0Þ

Cð1  aÞ

t > 0:

Lemma 2.3 [25]. Suppose that b > 0; aðtÞ is a nonnegative nondecreasing function locally integrable on 0 6 t < T ðT 6 þ1Þ and gðtÞ is nonnegative nondecreasing continuous function defined on 0 6 t < T; gðtÞ 6 M (constant), and suppose uðtÞ is nonnegative and locally integrable on 0 6 t < T with

uðtÞ 6 aðtÞ þ gðtÞ

Z

t

ðt  uÞb1 uðsÞds;

ð2:14Þ

0

on this interval. Then

  uðtÞ 6 aðtÞEb gðtÞCðbÞtb :

ð2:15Þ

3. Existence, uniqueness and iterative formula of the solution In this section, we will study the existence, uniqueness of the initial value problem (1.4) and (1.5). First, we give a lemma. Lemma 3.1. Eqs. (1.4) and (1.5) is equivalent to the following integral equation

xðtÞ ¼ /ð0Þ  C/ðsÞ þ Cxðt  sÞ þ xðtÞ ¼ /ðtÞ;

s 6 t 6 0:

1 CðaÞ

Z

t

ðt  uÞa1 ½AxðuÞ þ Bxðu  sÞ þ f ðuÞdu;

t P 0;

ð3:1Þ

0

ð3:2Þ

Proof. On one hand, if xðtÞ is a solution of Eqs. (1.4) and (1.5), then applying Ia0 on both side of Eq. (1.4) and using Lemma 2.1 yields

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  Ia0 Da0 ðxðtÞ  Cxðt  sÞ ¼ Ia0 ðAxðtÞ þ Bxðt  sÞ þ f ðtÞÞ:

ð3:3Þ

That is

xðtÞ ¼ /ð0Þ  C/ðsÞ þ Cxðt  sÞ þ

1 CðaÞ

Z

t

ðt  uÞa1 ½AxðuÞ þ Bxðu  sÞ þ f ðuÞdu;

t P 0:

ð3:4Þ

0

Thus xðtÞ is a solution of Eq. (3.1) and (3.2). On the other hand, if xðtÞ is a solution of Eqs. (3.1) and (3.2), then applying c Da0 on both side of Eq. (3.1) yields

Da0 xðtÞ  Da0 ðCxðt  sÞÞ n o Rt ¼ Da0 Cð1aÞ 0 ðt  uÞa1 ½AxðuÞ þ Bxðu  sÞ þ f ðuÞdu   ¼ Da0 Ia0 ½AxðtÞ þ Bxðt  sÞ þ f ðtÞ :

ð3:5Þ

By the formula (2.13), we have

Da0 xðtÞ  Da0 ðCxðt  sÞÞ ¼

r

  t a Da0 ðIa0 ½AxðtÞ þ Bxðt  sÞ þ f ðtÞÞ  Ia0 ðAxðtÞ þ Bxðt  sÞ þ f ðtÞÞ t¼0  Cð1 aÞ

ð3:6Þ

t ¼ AxðtÞ þ Bxðt  sÞ þ f ðtÞ  ½Ia0 ðAxðtÞ þ Bxðt  sÞ þ f ðtÞÞt¼0  Cð1 aÞ : a

Since xðtÞ is continuous t 2 ½s; 1Þ and f ðtÞ is continuous on ½0; 1Þ, thus ½Ia0 ðAxðtÞ þ Bxðt  sÞ þ f ðtÞÞt¼0 ¼ 0. Hence, by (3.6), xðtÞ is also a solution of Eqs. (1.4) and (1.5). This completes the proof. h The explicit solution of Eqs. (1.4) and (1.5) is given by the following theorem. Theorem 3.1. If xðtÞ is a solution of Eqs. (1.4) and (1.5), then xðtÞ can be written as

xðtÞ ¼ Ea ðAta Þ½/ð0Þ  C/ðsÞ þ Cxðt  sÞ þ þ

Z

Z

t

Aðt  uÞa1 Ea;a ðAðt  uÞa ÞCxðu  sÞdu

0 t

ðt  uÞa1 Ea;a ðAðt  uÞa Þ½Bxðu  sÞ þ f ðuÞdu:

ð3:7Þ

0

Proof. By Lemma 3.1, Eqs. (1.4) and (1.5) is equivalent to the following integral equation

xðtÞ ¼ /ð0Þ  C/ðsÞ þ Cxðt  sÞ þ xðtÞ ¼ /ðtÞ;

1 CðaÞ

Z

t

ðt  uÞa1 ½AxðuÞ þ Bxðu  sÞ þ f ðuÞdu;

t > 0:

ð3:8Þ

0

t 2 ½s; 0:

ð3:9Þ

Since

Z

t

ðt  uÞa1 ½AxðuÞ þ Bxðu  sÞ þ f ðuÞdu ¼ t a1  ½AxðtÞ þ Bxðt  sÞ þ f ðtÞ;

ð3:10Þ

0

we get

xðtÞ ¼ /ð0Þ  C/ðsÞ þ Cxðt  sÞ þ

1

CðaÞ

t a1  ½AxðtÞ þ Bxðt  sÞ þ f ðtÞ;

ð3:11Þ

where ⁄ denotes convolution. Applying the Laplace transform on both side of Eq. (3.11) and using the formula (2.9) yields

xðsÞ ¼ L½/ð0Þ  C/ðsÞ þ Cxðt  sÞ; s þ

1

CðaÞ

 CðaÞsa ðAxðsÞ þ L½Bxðt  sÞ þ f ðtÞ; sÞ;

ð3:12Þ

where  xðsÞ is Laplace transform of xðtÞ. Reducing Eq. (3.12) yields

xðsÞ ¼ sa ðsa I  AÞ1 L½/ð0Þ  C/ðsÞ þ Cxðt  sÞ; s þ ðsa I  AÞ1 L½Bxðt  sÞ þ f ðtÞ; s ¼ L½/ð0Þ  C/ðsÞ þ Cxðt  sÞ; s þ Aðsa I  AÞ1 L½/ð0Þ  C/ðsÞ þ Cxðt  sÞ; s þ ðsa I  AÞ1 L½Bxðt  sÞ þ f ðtÞ; sÞ: ð3:13Þ By the formula (2.8), we get

xðsÞ ¼ L½/ð0Þ  C/ðsÞ þ Cxðt  sÞ; s þ L½Ata1 Ea;a ðAt a Þ; s  L½/ð0Þ  C/ðsÞ þ Cxðt  sÞ; s þ L½ta1 Ea;a ðAt a Þ; s  L½Bxðt  sÞ þ f ðtÞ; s:

ð3:14Þ

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Applying the Laplace inverse transform and Convolution Theorem yields

xðtÞ ¼ /ð0Þ  C/ðsÞ þ Cxðt  sÞ þ ðAt a1 Ea;a ðAta ÞÞ  ð/ð0Þ  C/ðsÞ þ Cxðt  sÞÞ þ ðta1 Ea;a ðAt a ÞÞ  ðBxðt  sÞ þ f ðtÞÞ

¼ /ð0Þ  C/ðsÞ þ Cxðt  sÞ þ þ

Z

t

Aðt  uÞa1 Ea;a ðAðt  uÞa Þð/ð0Þ  C/ðsÞ þ Cxðu  sÞÞdu

0

Z

t

ðt  uÞa1 Ea;a ðAðt  uÞa ÞðBxðu  sÞ þ f ðuÞÞdu:

0

  Z t Z t ¼ /ð0Þ I þ Aðt  uÞa1 Ea;a ðAðt  uÞa Þdu þ Cxðt  sÞ  C/ðsÞ I þ Aðt  uÞa1 Ea;a ðAðt  uÞa Þdu þ

0

Z

t

Aðt  uÞ

a1

a

Ea;a ðAðt  uÞ ÞCxðu  sÞdu þ

Z

0

0 t

ðt  uÞ

a1

Ea;a ðAðt  uÞa Þ½ðBxðu  sÞ þ f ðuÞdu:

ð3:15Þ

0

Noting that

Z

t

Aðt  uÞa1 Ea;a ðAðt  uÞa Þdu ¼

0

Z

t

Aðt  uÞa1

0

¼

k 1 X ðAðt  uÞa Þ k¼0

Cðka þ aÞ

du ¼

1 X

Akþ1 Cðka þ aÞ k¼0

Z

t

ðt  uÞkaþa1 du

0

1 X

1 X Akþ1 t ðkþ1Þa Ak t k a ; ¼ Cððk þ 1Þa þ 1Þ k¼1 Cðka þ 1Þ k¼0

ð3:16Þ

we get

Iþ

Z

t

Aðt  uÞa1 Ea;a ðAðt  uÞa Þdu ¼ I þ

0

1 X

1 X Ak t k a Ak t k a ¼ ¼ Ea;1 ðAta Þ ¼ Ea ðAt a Þ: Cðka þ 1Þ k¼0 Cðka þ 1Þ k¼1

Inserting (3.17) into (3.15) yields the formula (3.7). This completes the proof.

ð3:17Þ

h

Denote the solution of Eqs. (1.4) and (1.5) by: /ðtÞ ¼ /0 ðtÞ for t 2 ½s; 0; xðtÞ ¼ /i ðtÞ for t 2 ½ði  1Þs; is; the formula (3.7), we have, for t 2 ½0; s, that

Z t /1 ðtÞ ¼ Ea ðAta Þ½/ð0Þ  C/ðsÞ þ C/ðt  sÞ; t 2 ½0; s þ Aðt  uÞa1 Ea;a ðAðt  uÞa ÞC/ðu  sÞdu 0 Z t þ ðt  uÞa1 Ea;a ðAðt  uÞa Þ½B/ðu  sÞ þ f ðuÞdu:

i ¼ 1; 2; . . . ; k. By

ð3:18Þ

0

Furthermore, we have the following corollary. Corollary 3.1. Suppose that the solution of Eqs. (1.4) and (1.5) exists on ½s; ks; k is a positive integer and k P 2. Then for t 2 ½ðk  1Þs; ks, the iterative formula for the solution /i ðtÞ of Eqs. (1.4) and (1.5) defined on ½ði  1Þs; is; i ¼ 2; . . . k, can be written as

/k ðtÞ ¼ Ea ðAta Þ½/ð0Þ  C/ðsÞ þ C/k1 ðt  sÞ k2 X R ðjþ1Þs þ Aðt  uÞa1 Ea;a ðAðt  uÞa ÞC/j ðu  sÞdu js j¼0 k2 X   R ðjþ1Þs þ ðt  uÞa1 Ea;a ðAðt  uÞa Þ B/j ðu  sÞ þ f ðuÞ du js

ð3:19Þ

j¼0

þ

Rt

þ

Rt

ðk1Þs

Aðt  uÞa1 Ea;a ðAðt  uÞa ÞC/k1 ðu  sÞdu

ðk1Þs

ðt  uÞa1 Ea;a ðAðt  uÞa Þ½B/k1 ðu  sÞ þ f ðuÞdu;

where /0 ðtÞ ¼ /ðtÞ for t 2 ½s; 0. Proof. For t 2 ½s; 2s, we have by the formula (3.7) that

xðtÞ ¼ /2 ðtÞ ¼ Ea ðAta Þ½/ð0Þ  C/ðsÞ þ C/1 ðt  sÞ Rs þ 0 Aðt  uÞa1 Ea;a ðAðt  uÞa ÞC/ðu  sÞdu Rs þ 0 ðt  uÞa1 Ea;a ðAðt  uÞa Þ½B/ðu  sÞ þ f ðuÞdu Rt þ s Aðt  uÞa1 Ea;a ðAðt  uÞa ÞC/1 ðu  sÞdu Rt þ s ðt  uÞa1 Ea;a ðAðt  uÞa Þ½B/1 ðu  sÞ þ f ðuÞdu:

ð3:20Þ

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For t 2 ½ðk  1Þs; ks, k P 1, we have,

xðtÞ ¼ /k ðtÞ

¼ Ea ðAta Þ½/ð0Þ þ C/ðsÞ þ C/k1 ðt  sÞ Rt þ 0 Aðt  uÞa1 Ea;a ðAðt  uÞa ÞCxðu  sÞdu Rt þ 0 ðt  uÞa1 Ea;a ðAðt  uÞa Þ½Bxðu  sÞ þ f ðuÞdu

¼ Ea ðAta Þ½/ð0Þ þ C/ðsÞ þ C/k1 ðt  sÞ Rs þ 0 Aðt  uÞa1 Ea;a ðAðt  uÞa ÞCxðu  sÞdu Rs þ 0 ðt  uÞa1 Ea;a ðAðt  uÞa Þ½Bxðu  sÞ þ f ðuÞdu R 2s þ s Aðt  uÞa1 Ea;a ðAðt  uÞa ÞCxðu  sÞdu R 2s þ s ðt  uÞa1 Ea;a ðAðt  uÞa Þ½Bxðu  sÞ þ f ðuÞdu

ð3:21Þ

þ R ðk1Þs þ ðk2Þs Aðt  uÞa1 Ea;a ðAðt  uÞa ÞCxðu  sÞdu R ðk1Þs þ ðk2Þs ðt  uÞa1 Ea;a ðAðt  uÞa Þ½Bxðu  sÞ þ f ðuÞdu Rt þ ðk1Þs Aðt  uÞa1 Ea;a ðAðt  uÞa ÞCxðu  sÞdu Rt þ ðk1Þs ðt  uÞa1 Ea;a ðAðt  uÞa Þ½Bxðu  sÞ þ f ðuÞdu: It is easy to obtain that

Rs ¼

R0s 0

Rs ¼

R0s 0

Aðt  uÞa1 Ea;a ðAðt  uÞa ÞCxðu  sÞdu Aðt  uÞa1 Ea;a ðAðt  uÞa ÞC/0 ðu  sÞdu; ðt  uÞa1 Ea;a ðAðt  uÞa Þ½Bxðu  sÞ þ f ðuÞdu ðt  uÞa1 Ea;a ðAðt  uÞa Þ½B/0 ðu  sÞ þ f ðuÞdu;

R 2s ¼

s

Aðt  uÞa1 Ea;a ðAðt  uÞa Þ½Cxðu  sÞdu

s

Aðt  uÞa1 Ea;a ðAðt  uÞa ÞC/1 ðu  sÞdu;

R 2s R 2s s

¼

R 2s

ðt  uÞa1 Ea;a ðAðt  uÞa Þ½B/1 ðu  sÞ þ f ðuÞdu; ......... s

ðk2Þ

Aðt  uÞa1 Ea;a ðAðt  uÞa ÞCxðu  sÞdu

ðk2Þ

Aðt  uÞa1 Ea;a ðAðt  uÞa ÞC/k2 ðu  sÞdu;

R ðk1Þs R ðk1Þs

¼

ðk2Þ

ðt  uÞa1 Ea;a ðAðt  uÞa Þ½Bxðu  sÞ þ f ðuÞdu

ðk2Þ

ðt  uÞa1 Ea;a ðAðt  uÞa Þ½B/k2 ðu  sÞ þ f ðuÞdu;

R ðk1Þs Rt

¼

ðk1Þs

Aðt  uÞa1 Ea;a ðAðt  uÞa ÞCxðu  sÞdu

ðk1Þs

Aðt  uÞa1 Ea;a ðAðt  uÞa ÞC/k1 ðu  sÞdu:

Rt Rt

¼

ðk1Þs

ðt  uÞa1 Ea;a ðAðt  uÞa Þ½Bxðu  sÞ þ f ðuÞdu

ðk1Þs

ðt  uÞa1 Ea;a ðAðt  uÞa Þ½B/k1 ðu  sÞ þ f ðuÞdu:

Rt

ð3:23Þ

ð3:24Þ

ðt  uÞa1 Ea;a ðAðt  uÞa Þ½Bxðu  sÞ þ f ðuÞdu

R ðk1Þs ¼

ð3:22Þ

ð3:25Þ

ð3:26Þ

ð3:27Þ

ð3:28Þ

ð3:29Þ

Inserting (3.22)–(3.29) into (3.21) yields the iterative formula (3.19). This completes the proof. h Remark 3.1. (i) For the case C ¼ 0 in Eqs. (1.4) and (1.5), Theorem 3.1 and Corollary 3.1 are also valid. (ii) By the formulae (3.7) and (3.19), we know if the initial function /0 ðtÞ defined on ½s; 0 and f ðtÞ defined on ½0; 1 are continuous, then /k ðtÞ is continuous on ½ðk  1Þs; ks for any positive integer k. By Corollary 3.1, we can easily get the following existence theorem. Please cite this article in press as: X.-F. Zhou et al., Analytic study on linear neutral fractional differential equations, Appl. Math. Comput. (2015), http://dx.doi.org/10.1016/j.amc.2014.12.056

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X.-F. Zhou et al. / Applied Mathematics and Computation xxx (2015) xxx–xxx

Theorem 3.2. Suppose that f ðtÞ is continuous on ½0; þ1Þ and /ðtÞ is continuous on ½s; 0. Then there exists a unique solution of Eqs. (1.4) and (1.5) defined on ½0; þ1Þ. Remark 3.2. Theorem 3.2 is an extension of Theorem 1 in [18], which weakens the conditions and extends the conclusion of Theorem 1 in [18]. Example 3.1. Consider the analytic solution of the neutral linear fractional differential equation 1

1

D20 xðtÞ ¼ xðt  1Þ  D20 xðt  1Þ þ t; 1

xðtÞ ¼ t3 ;

t P 0;

ð3:30Þ

t 2 ½1; 0:

ð3:31Þ 1 3

Comparing with Eqs. (1.4) and (1.5), it hold that a ¼ 12 ; s ¼ 1; A ¼ 0; B ¼ 1; C ¼ 1; f ðtÞ ¼ t and /ðtÞ ¼ t . 1 Note that /0 ðtÞ ¼ /ðtÞ ¼ t3 . By the formulae (3.7) and (3.19), we have

Rt 1 1 ¼ ½/ð0Þ  /ð1Þ  /0 ðt  1Þ þ 0 ðt  uÞ2 E1;1 ð0ðt  uÞ2 Þð/0 ðu  1Þ þ uÞdu 22 Rt Rt 1 1 1 1 ¼ 1  ðt  1Þ3 þ C1ð1Þ 0 ðt  uÞ2 ðu  1Þ3 du þ C1ð1Þ 0 ðt  uÞ2 udu

/1 ðtÞ

2

1 3

1 2

ð3:32Þ

2

1 3

3 2

¼ 1  ðt  1Þ þ I0 ðt  1Þ þ C1ð5Þ t ;

t 2 ½0; 1

2

and

/2 ðtÞ ¼ ½/ð0Þ  /ð1Þ  /1 ðt  1Þ þ

1 Cð12Þ

Z

1

1

ðt  uÞ2 ð/1 ðu  1Þ þ uÞdu þ

0

1 Cð12Þ

Z

t

1

ðt  uÞ2 ð/1 ðu  1Þ þ uÞdu; t 2 ½1; 2:

1

ð3:33Þ Remark 3.3. It is necessary to mention that the conditions of Theorem 3.2 is sufficient, not necessary. Example 3.2. Consider the neutral linear fractional differential equation 1

1

1

D20 xðtÞ ¼ xðt  1Þ þ D20 xðt  1Þ þ t6 ; xðtÞ ¼ t; here f ðtÞ ¼ t

16

t P 0;

ð3:34Þ

t 2 ½1; 0;

ð3:35Þ

is not continuous in ½0; 1. Using the iterative formula (3.19), we get the solution of Eq. (3.34), (3.35)

xðtÞ ¼ /1 ðtÞ ¼ t þ

1 3 1 1 Cð5Þ 1 t2  3 t 2 þ 64 t 3 ; 5 Cð 2 Þ Cð2Þ Cð3Þ

t 2 ½0; 1;

ð3:36Þ

which is continuous on ½0; 1. 4. Dependence of the solution on the initial value In this section, we consider the dependence on initial value for the solution of Eqs. (1.4) and (1.5). The main result is given by the following theorem. Theorem 4.1. Suppose that xðtÞ and yðtÞ are the solutions of Eq. (1.4) subject to initial value xðtÞ ¼ /ðtÞ and yðtÞ ¼ wðtÞ on ½s; 0, respectively, which exist on the interval ½0; ks for some k P 1 . Then it holds

jxðtÞ  yðtÞj 6 Ak j/  wj;

t 2 ½ðk  1Þs; ks;

h i Pk1 a a where A0 ¼ 1; Ak ¼ 1 þ jcj þ jcjAk1 þ CðajBjþ1Þ j¼0 ððk  jÞsÞ Aj Ea ðjAjðksÞ Þ, and j/  wj ¼ sup j/ðhÞ  wðhÞjRn .

ð4:1Þ

s6h60

Proof. By the assumptions, we have

Da0 yðtÞ ¼ AyðtÞ þ Byðt  sÞ þ CDa0 yðt  sÞ þ f ðtÞ; yðtÞ ¼ wðtÞ;

t P 0;

t 2 ½s; 0:

ð4:2Þ ð4:3Þ

Combining Eqs. (4.2) and (4.3) with Eqs. (1.4) and (1.5) yields

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X.-F. Zhou et al. / Applied Mathematics and Computation xxx (2015) xxx–xxx

Da0 ðxðtÞ  yðtÞÞ ¼ AðxðtÞ  yðtÞÞ þ Bðxðt  sÞ  yðt  sÞÞ þ C  Da0 ðxðtÞ  yðtÞÞ; xðtÞ  yðtÞ ¼ /ðtÞ  wðtÞ;

t P 0;

t 2 ½s; 0:

ð4:4Þ ð4:5Þ

Evidently, Eqs. (4.4) and (4.5) is equivalent to the following integral equation

xðtÞ  yðtÞ ¼ ð/ð0Þ  wð0ÞÞ  Cð/ðsÞ  wðsÞÞ þ Cðxðt  sÞ  yðt  sÞÞ þ

1 CðaÞ

Z

t

ðt  uÞa1 ½AðxðuÞ  yðuÞÞ

0

þ Bðxðu  sÞ  yðu  sÞÞdu:

ð4:6Þ

Now we prove the validity of inequality (4.1) by induction method on k. ðiÞ First, we prove that inequality (4.1) holds for k = 1. If t 2 ½0; s, then we have by Eq. (4.6) that

jxðtÞ  yðtÞj 6 j/  wj þ jCjj/  wj þ jCjj/  wj Rt þ Cð1aÞ 0 ðt  uÞa1 jAjjxðuÞ  yðuÞjdu Rt þ Cð1aÞ 0 ðt  uÞa1 jBjj/  wjdu: By the fact that

Rt 0

ð4:7Þ

ðt  uÞa1 du ¼ a1 ta 6 a1 sa for t 2 ½0; s, we have

jxðtÞ  yðtÞj 6 1 þ jcj þ jcjA0 þ

 Z t jBj jAj sa j/  wj þ ðt  uÞa1 jxðuÞ  yðuÞjdu: Cða þ 1Þ CðaÞ 0

ð4:8Þ

 jBj sa Ea ðjAjsa Þj/  wj; Cða þ 1Þ

ð4:9Þ

By Lemma 2.3, we have

jxðtÞ  yðtÞj 6 1 þ jcj þ jcjA0 þ

t 2 ½0; s:

Thus, the inequality (4.1) holds for k = 1. ðiiÞ Assume that the inequality (4.1) holds for 2;    ; k  1. That is

jxðtÞ  yðtÞj 6 Aj j/  wj;

j ¼ 2; 3; . . . ; k  1:

ð4:10Þ

Now we prove the inequality (4.1) holds for k. Now we prove that the inequality (4.1) holds for k. If t 2 ½ðk  1Þs; ks, we have by Eq. (4.6) that

xðtÞ  yðtÞ

¼ ð/ð0Þ  wð0ÞÞ  Cð/ðsÞ  wðsÞÞ þ Cðxðt  sÞ  yðt  sÞÞ Rt þ Cð1aÞ 0 ðt  uÞa1 AðxðuÞ  yðuÞÞdu Rt þ Cð1aÞ 0 ðt  uÞa1 Bðxðu  sÞ  yðu  sÞÞdu ¼ ð/ð0Þ  wð0ÞÞ  Cð/ðsÞ  wðsÞÞ þ Cðxðt  sÞ  yðt  sÞÞ Rt þ Cð1aÞ 0 ðt  uÞa1 AðxðuÞ  yðuÞÞdu Rs þ Cð1aÞ 0 ðt  uÞa1 Bðxðu  sÞ  yðu  sÞÞdu R 2s þ Cð1aÞ s ðt  uÞa1 Bðxðu  sÞ  yðu  sÞÞdu þ þ Cð1aÞ þ Cð1aÞ

R ðk1Þs ðk2Þs

ðt  uÞa1 Bðxðu  sÞ  yðu  sÞÞdu

Rt

ðk1Þs

ðt  uÞa1 Bðxðu  sÞ  yðu  sÞÞdu:

ð4:11Þ

Thus, it follows that

jxðtÞ  yðtÞj 6 ð1 þ jcjÞj/  wj þ jcjjxðt  sÞ  yðt  sÞj Rt þ Cð1aÞ 0 ðt  uÞa1 jAjjxðuÞ  yðuÞjdu Rs þ Cð1aÞ 0 ðt  uÞa1 jBjjxðu  sÞ  yðu  sÞjdu R 2s þ Cð1aÞ s ðt  uÞa1 jBjjxðu  sÞ  yðu  sÞjdu

ð4:12Þ

þ R ðk1Þs þ Cð1aÞ ðk2Þs ðt  uÞa1 jBjjxðu  sÞ  yðu  sÞjdu Rt þ Cð1aÞ ðk1Þs ðt  uÞa1 jBjjxðu  sÞ  yðu  sÞjdu: Since t 2 ½ðk  1Þs; ks, it holds that t  s 2 ½ðk  2Þs; ðk  1Þs. By the inequalities (4.9) and (4.10), we have

jxðt  sÞ  yðt  sÞj 6 Ak1 j/  wj for

t 2 ½ðk  1Þs; ks;

ð4:13Þ

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X.-F. Zhou et al. / Applied Mathematics and Computation xxx (2015) xxx–xxx

jxðu  sÞ  yðu  sÞj 6 j/  wj for

u 2 ½0; s;

jxðu  sÞ  yðu  sÞj 6 A1 j/  wj for

9

ð4:14Þ

u 2 ½s; 2s;

ð4:15Þ

 jxðu  sÞ  yðu  sÞj 6 Ak2 j/  wj for

u 2 ½ðk  2Þs; ðk  1Þs;

ð4:16Þ

jxðu  sÞ  yðu  sÞj 6 Ak1 j/  wj for

u 2 ½ðk  1Þs; ks:

ð4:17Þ

By computation, it holds, for t 2 ½ðk  1Þs; ks, that

Z s

ðt  uÞa1 du ¼

0

R 2s s

1

a

1

½ta  ðt  sÞa  6

a

ðksÞ ;

a

ð4:18Þ

ðt  uÞa1 du ¼ a1 ½ðt  sÞa  ðt  2sÞa  a

6 a1 ðt  sÞa 6 a1 ððk  1ÞsÞ ;

ð4:19Þ

 R ðk1Þs ðk2Þs

 a a ðt  uÞa1 du ¼ a1 ðt  ðk  2ÞsÞ  ðt  ðk  1ÞsÞ

ð4:20Þ

a

6 a1 ðt  ðk  2ÞsÞ 6 1a ð2sÞa ; Z

t

ðt  uÞa1 du ¼

ðk1Þs

1

a

a

ðt  ðk  1ÞsÞ 6

1

a

sa :

ð4:21Þ

Inserting the inequalities (4.13)–(4.21) into (4.12) yields

6

jxðtÞ  yðtÞj h  a a 1 þ jcj þ jcjAk1 þ CðajBjþ1Þ ðksÞ þ ððk  1ÞsÞ A1  þ    þ ð2sÞa Ak2 þ sa Ak1 j/  wj R t ðt  uÞa1 jxðuÞ  yðuÞjdu þ CjAj ða Þ 0

"

# Z t k1 X jBj jAj a ¼ 1 þ jcj þ jcjAk1 þ ððk  jÞsÞ Aj j/  wj þ ðt  uÞa1 jxðuÞ  yðuÞjdu: Cða þ 1Þ j¼0 CðaÞ 0

ð4:22Þ

By Lemma 2.3, we have

"

# k1 X jBj a a jxðtÞ  yðtÞj 6 1 þ jcj þ jcjAk1 þ ððk  jÞsÞ Aj Ea ðjAjðksÞ Þj/  wj: Cða þ 1Þ j¼0 Therefore, the inequality (4.1) is valid for k. This completes the proof.

ð4:23Þ

h

5. The general solution by a fundamental solution matrix In this section, the representation of the general solution of Eqs. (1.4) and (1.5) is given by a fundamental solution matrix of Eq. (1.4). For the case a ¼ 1, the general solution of Eqs. (1.4) and (1.5) is coincide with the general solution of first-order linear neutral differential equations [26,27]. Let’s begin with some definitions and Lemmas. Lemma 5.1. Let 0 < n  1 < a < n; s > 0 be a constant, xðtÞ be continuous on ½s; 1Þ and xðkÞ ðtÞ exist for k ¼ 1; 2;    ; n  1. Then the Laplace transform of c Da0 xðt  sÞ is given by

  L c Da0 xðt  sÞ; s ¼ sa ess xðsÞ þ sa ess

Z

0

est xðtÞdt 

s

n1 X   sak1 xðkÞ ðt  sÞ t¼0 ;

ð5:1Þ

k¼0

xðsÞ is the Laplace transform of xðtÞ. Namely, L½xðtÞ; s ¼  xðsÞ. where  Especially, when 0 < a < 1, it holds

L½c Da0 xðt  sÞ; s ¼ sa ess xðsÞ þ sa ess

Z

0

s

est xðtÞdt  sa1 xðt  sÞjt¼0 :

ð5:2Þ

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10

X.-F. Zhou et al. / Applied Mathematics and Computation xxx (2015) xxx–xxx

Proof. Denote gðtÞ ¼ xðt  sÞ, then c

a ðnÞ Da0 xðt  sÞ ¼ In 0 g ðtÞ ¼

1 Cðn  aÞ

Z

t

ðt  uÞna1 g ðnÞ ðuÞdu ¼

0

1

Cðn  aÞ

tna1  g ðnÞ ðtÞ:

ð5:3Þ

By the Formula (2.9) and applying the Laplace transform on both side of (5.3), we get

  L½c Da0 xðt  sÞ; s ¼ sðnaÞ L g ðnÞ ðtÞ; s :

ð5:4Þ

Obviously, it holds n1 X   L g ðnÞ ðtÞ; s ¼ sn gðsÞ  snk1 g ðkÞ ð0Þ:

ð5:5Þ

k¼0

Then we have

gðsÞ ¼ L½gðtÞ; s ¼ L½xðt  sÞ; s ¼

Z

þ1

est xðt  sÞdt ¼

Z

þ1

esðuþsÞ xðuÞdu ¼ ess

s

0

Z

0

est xðtÞdt þ ess xðsÞ:

ð5:6Þ

s

Since g ðkÞ ð0Þ ¼ xðkÞ ðt  sÞjt¼0 , so we get

" # n1 X     L c Da0 xðt  sÞ; s ¼ sðnaÞ L g ðnÞ ðtÞ; s ¼ sðnaÞ sn gðsÞ  snk1 g ðkÞ ð0Þ (  Z ðnaÞ ¼s sn ess

k¼0

0 st

e

xðtÞdt þ e

s

¼ sa ess xðsÞ þ sa ess

) X n1 nk1 ðkÞ s x ðt  sÞjt¼0 xðsÞ 

ss 

k¼0

Z

0

est xðtÞdt 

s

n1 X

sak1 xðkÞ ðt  sÞjt¼0 :

ð5:7Þ

k¼0

If 0 < a < 1, then

  L c Da0 xðt  sÞ; s ¼ sa ess xðsÞ þ sa ess This completes the proof.

Z

0

s

est xðtÞdt  sa1 xðt  sÞjt¼0 :

ð5:8Þ

h

Applying the Laplace transform on both side of Eq. (1.4) yields

a  s I  A  Bess  Csa ess xðsÞ ¼ sa1 ðxð0Þ  CxðsÞÞ þ Bess Z 1 est f ðtÞdt: þ

Z

0

est xðtÞdt þ Csa ess

s

Z

0

est xðtÞdt

s

ð5:9Þ

0

Definition 5.1. The matrix ðsa I  A  Bess  Csa ess Þ in (5.9) is said to be the characteristic matrix of Eqs. (1.4) and (1.5), and the equation

detðsa I  A  Bess  Csa ess Þ ¼ 0;

ð5:10Þ

is said to be the characteristic equation of Eqs. (1.4) and (1.5). Definition 5.2. If n order square matrix XðtÞ satisfies

Da0 XðtÞ ¼ AXðtÞ þ BXðt  sÞ þ C  Da0 Xðt  sÞ;

t>0

ð5:11Þ

XðtÞ ¼ I;

t ¼ 0;

ð5:12Þ

XðtÞ ¼ 0;

t < 0;

ð5:13Þ

then we say XðtÞ is a fundamental solution matrix of Eq. (1.4), where I denotes the unit matrix with order n. Lemma 5.2. Let XðtÞ be a fundamental solution matrix of Eq. (1.4). Then it holds that Please cite this article in press as: X.-F. Zhou et al., Analytic study on linear neutral fractional differential equations, Appl. Math. Comput. (2015), http://dx.doi.org/10.1016/j.amc.2014.12.056

X.-F. Zhou et al. / Applied Mathematics and Computation xxx (2015) xxx–xxx

ðsa I  A  Bess  Csa ess ÞXðsÞ ¼ sa1 I;

11

ð5:14Þ

where XðsÞ is the Laplace transform of XðtÞ, and I denotes the unit matrix with order n. Proof. By Lemma 5.1 and applying the Laplace transform on both side of Eq. (5.11) yields the equality (5.14) immediately. h Now we will introduce a representation of the general solution of Eqs. (1.4) and (1.5) through a fundamental solution matrix of Eq. (1.4). Theorem 5.1. If XðtÞ is a fundamental solution matrix of Eq. (1.4), then the general solution of Eqs. (1.4) and (1.5) can be written as

xðtÞ ¼ XðtÞ½/ð0Þ  C/ðsÞ ih i R ts h a1 a I ^ xðhÞ þ f ðh þ sÞ dh þ s c D1 B/ðhÞ 0 Xðt  s  hÞ þ CðaÞ ðt  s  hÞ R0 _  h  sÞC/ðhÞdh þ C /ðt ^  sÞxðt  sÞ; t 2 ½0; þ1Þ; þ s Xðt where xðhÞ ¼



0; 1;

h P 0; ^ , /ðhÞ ¼ h < 0;



/ðhÞ; /ð0Þ;

ð5:15Þ

s 6 h 6 0; . h > 0:

Proof. Applying the Laplace transform on Eq. (1.4) yields



 sa I  A  Bess  Csa ess xðsÞ ¼ sa1 ðxð0Þ  CxðsÞÞ R0 R0 þBess s est xðtÞdt þ Csa ess s est xðtÞdt þ L½f ðtÞ; s:

ð5:16Þ

_ s þ Xð0Þ. By Lemma 5.2 and Denoting YðtÞ ¼ Cð1aÞ ta1  XðtÞ ¼ Ia0 XðtÞ, we Note that L½ta1 ; s ¼ CðaÞsa and sL½XðsÞ; s ¼ L½XðtÞ; have

 R R xðsÞ ¼ s1a XðsÞ sa1 ð/ð0Þ  C/ðsÞÞ þ Bess 0s est /ðtÞdt þ Csa ess 0s est /ðtÞdt þ L½f ðtÞ; s R0 R0 ¼ XðsÞð/ð0Þ  C/ðsÞÞ þ s1a XðsÞBess s est /ðtÞdt þ sXðsÞCess s est /ðtÞdt þ s1a XðsÞL½f ðtÞ; s h i R R0 0 ¼ XðsÞð/ð0Þ  C/ðsÞÞ þ sL Cð1aÞ t a1  XðtÞ;s Bess s est /ðtÞdt þ sXðsÞCess s est /ðtÞdt þ s1a XðsÞL½f ðtÞ; s

 n h i o

h i  R0 R0 _ _ _ ¼ XðsÞð/ð0Þ  C/ðsÞÞ þ L½YðtÞ; s þ Yð0Þ Bess s est /ðtÞdt þ L XðtÞ; s þ Xð0Þ Cess s est /ðtÞdt þ L YðtÞ; s þ Yð0Þ L½f ðtÞ;s:

ð5:17Þ

By the definition of YðtÞ, we have Yð0Þ ¼ 0. It follows

Z 0 Z 0 h i n h i o _ _ s Bess s þ Xð0Þ Cess xðtÞ ¼ XðsÞð/ð0Þ  C/ðsÞÞ þ L YðtÞ; est /ðtÞdt þ L XðtÞ; est /ðtÞdt s s h i _ s L½f ðtÞ; s: þ L YðtÞ;

ð5:18Þ

^ By the definitions of xðhÞ and /ðhÞ, we get

Bess

Z

0

est /ðtÞdt ¼ Bess

s

¼B

Z

Z

1

^ xðtÞdt ¼ Bess est /ðtÞ

s þ1

Z

þ1

^  sÞxðh  sÞdh esðhsÞ /ðh

0

h i ^  sÞxðh  sÞdh ¼ L B/ðh ^  sÞxðh  sÞ; s esh /ðh

ð5:19Þ

0

and

Cess

Z

0

h i ^  sÞxðh  sÞ; s : est /ðtÞdt ¼ L C /ðh

ð5:20Þ

s

Inserting (5.19) and (5.20) into (5.18) yields

h i h i _ ^  sÞxðh  sÞ þ f ðtÞ; s xðsÞ ¼ XðsÞð/ð0Þ  C/ðsÞÞ þ L YðtÞ; s  L B/ðh

 h i _ ^  sÞxðh  sÞ; s : þ L½XðtÞ; s þ Xð0Þ L C /ðh

ð5:21Þ

By the inverse Laplace transform and Convolution Theorem, we get

 h i _  B/ðt _  ½C /ðt ^  sÞxðt  sÞ þ f ðtÞ þ XðtÞ ^  sÞxðt  sÞ þ Xð0Þ C /ðt ^  sÞxðt  sÞ xðtÞ ¼ XðtÞð/ð0Þ  C/ðsÞÞ þ YðtÞ

 h i h i Rt _ R _ ^  u  sÞxðt  u  sÞ þ f ðt  uÞ du þ t Xðt ^  sÞxðu  sÞ du þ C /ðt ^  sÞxðt  sÞ : ¼ XðtÞð/ð0Þ  C/ðsÞÞ þ 0 YðuÞ B/ðt  uÞ C /ðu 0 ð5:22Þ

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X.-F. Zhou et al. / Applied Mathematics and Computation xxx (2015) xxx–xxx

In view of YðtÞ ¼ Cð1aÞ ta1  XðtÞ ¼ Ia0 XðtÞ and Yð0Þ ¼ 0, we have

d Xð0Þ a1 1 a c 1a _ YðtÞ ¼ ðIa0 XðtÞÞ ¼ r D1 t ¼ 0 XðtÞ ¼ D0 XðtÞ þ dt CðaÞ CðaÞ

Z

t

_ þ ðt  uÞa1 XðuÞdu

0

Xð0Þ

CðaÞ

ta1 :

ð5:23Þ

^ and xðhÞ yields Set u  s ¼ h. Using the definitions of /ðhÞ

h i _  uÞ C /ðu ^  sÞxðu  sÞ du Xðt h i R ts _  h  sÞ C /ðhÞ ^ xðhÞ dh ¼ s Xðt R0 _  h  sÞC/ðhÞdh: ¼ s Xðt Rt 0

ð5:24Þ

Inserting (5.23) and (5.24) into (5.22) yields

i Rt h a I a1  xðtÞ ¼ XðtÞ½/ð0Þ  C/ðsÞ þ 0 c D1 0 XðuÞ þ CðaÞ u h i R _ ^  u  sÞxðt  u  sÞ þ f ðt  uÞ du þ 0 Xðt ^  sÞxðt  sÞ:  h  sÞC/ðhÞdh þ C /ðt B/ðt s

ð5:25Þ

Setting t  s  u ¼ h, we have

R t hc

ih i a I a1 B/ðt ^  u  sÞxðt  u  sÞ þ f ðt  uÞ du D1 0 XðuÞ þ CðaÞ u ih i R ts h a1 a I ^ xðhÞ þ f ðh þ sÞ dh: ¼ s c D1 B/ðhÞ 0 Xðt  s  hÞ þ CðaÞ ðt  s  hÞ 0

ð5:26Þ

Inserting (5.26) into (5.25) yields

xðtÞ ¼ XðtÞ½/ð0Þ  C/ðsÞ ih i R ts h a1 a I ^ xðhÞ þ f ðh þ sÞ dh þ s c D1 B/ðhÞ 0 Xðt  s  hÞ þ CðaÞ ðt  s  hÞ R0 _  h  sÞC/ðhÞdh þ C /ðt ^  sÞxðt  sÞ: þ s Xðt This completes the proof.

ð5:27Þ

h

Remark 5.1. For the case a ¼ 1, the general solution of Eqs. (1.4) and (1.5) can be written as

xðtÞ ¼ XðtÞ½/ð0Þ  C/ðsÞ R ts þ s Xðt  s  hÞðB/ðhÞ þ f ðh þ sÞÞdh R0 _  h  sÞC/ðhÞdh þ C/ðt  sÞ; Xðt þ s

ð5:28Þ t 2 ½0; s

and

xðtÞ ¼ XðtÞ½/ð0Þ  C/ðsÞ R0 þ s Xðt  s  hÞ½B/ðhÞ þ f ðh þ sÞdh R ts þ 0 Xðt  s  hÞf ðh þ sÞdh R0 _  h  sÞC/ðhÞdh; t 2 ½s; þ1Þ: Xðt þ

ð5:29Þ

s

For t 2 ½s; þ1Þ and the case a ¼ 1, the solution (5.29) of Eqs. (1.4) and (1.5) is coincide with the general solution of first-order linear nonhomogeneous neutral differential equations [26,27]. However, for the case a ¼ 1 and c – 0, Theorem 11 in [28] is not coincide with the general solution. Acknowledgement The authors would like to thank the reviewers for their valuable comments on this paper, which helps to improve the quality of this paper. References [1] [2] [3] [4] [5] [6] [7] [8]

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Please cite this article in press as: X.-F. Zhou et al., Analytic study on linear neutral fractional differential equations, Appl. Math. Comput. (2015), http://dx.doi.org/10.1016/j.amc.2014.12.056