Abstract

The main purpose of this paper is to obtain the unique solution of the constant coefficient homogeneous linear fractional differential equations 𝐷𝑞𝑡0𝑋(𝑡)=𝑃𝑋(𝑡),𝑋(𝑎)=𝐵 and the constant coefficient nonhomogeneous linear fractional differential equations 𝐷𝑞𝑡0𝑋(𝑡)=𝑃𝑋(𝑡)+𝐷,𝑋(𝑎)=𝐵 if 𝑃 is a diagonal matrix and 𝑋(𝑡)𝐶1𝑞[𝑡0,𝑇]×𝐶1𝑞[𝑡0,𝑇]××𝐶1𝑞[𝑡0,𝑇] and prove the existence and uniqueness of these two kinds of equations for any 𝑃𝐿(𝑅𝑚) and 𝑋(𝑡)𝐶1𝑞[𝑡0,𝑇]×𝐶1𝑞[𝑡0,𝑇]××𝐶1𝑞[𝑡0,𝑇]. Then we give two examples to demonstrate the main results.

1. Introduction

System of fractional differential equations has gained a lot of interest because of the challenges it offers compared to the study of system of ordinary differential equations. Numerous applications of this system in different areas of physics, engineering, and biological sciences have been presented in [13]. The differential equations involving the Riemman-Liouville differential operators of fractional order 0<𝑞<1 appear to be more important in modeling several physical phenomena and therefore seem to deserve an independent study of their theory parallel to the well-known theory of ordinary differential equations. The existence and uniqueness of solution for fractional differential equations with any 𝑋(𝑡)𝐶[𝑡0,𝑇]×𝐶[𝑡0,𝑇]××𝐶[𝑡0,𝑇] have been studied in many papers, see [428]. In [4] Daftradar-Gejji and Babakhani have studied the existence and uniqueness of 𝐷𝑞0𝑋(𝑡)𝑋0=𝑃𝑋(𝑡),(1.1) where 𝐷𝑞0 denotes the standard Riemman-Liouville fractional derivative, 0<𝑞<1,𝑋(𝑡)=(𝑥1(𝑡),𝑥2(𝑡),,𝑥𝑚(𝑡))𝑇, 𝑋(0)=𝑋0=(𝑥10,𝑥20,,𝑥𝑚0)𝑇, 𝑃𝐿(𝑅𝑚) which is an m dimensional linear space. They have obtained that the system (1.1) has a unique solution defined on [0,𝑇] if 𝑃𝐿(𝑅𝑚) and 𝑋(𝑡)𝐶[𝑡0,𝑇]×𝐶[𝑡0,𝑇]××𝐶[𝑡0,𝑇]. In [17] Belmekki et al. have studied the existence of periodic solution for some linear fractional differential equation in 𝐶1𝑞[0,1]. In [21] Ahmad and Nieto have studied the Riemann-Liouwille fractional differential equations with fractional boundary conditions. In comparison with the earlier results of this type we get more general assumptions. We assume 𝑋(𝑡)𝐶1𝑞[𝑡0,𝑇]×𝐶1𝑞[𝑡0,𝑇]××𝐶1𝑞[𝑡0,𝑇] instead of 𝑋(𝑡)𝐶[𝑡0,𝑇]×𝐶[𝑡0,𝑇]××𝐶[𝑡0,𝑇] and consider the following system of fractional differential equations: 𝐷𝑞𝑡0𝐷𝑋(𝑡)=𝑃𝑋(𝑡),𝑋(𝑎)=𝐵,𝑞𝑡0𝑋(𝑡)=𝑃𝑋(𝑡)+𝐷,𝑋(𝑎)=𝐵,(1.2) where 𝐷𝑞0 denotes the standard Riemman-Liouville fractional derivative, 0<𝑞<1, 𝑃𝐿(𝑅𝑚), 𝑋(𝑡)𝐶1𝑞𝑡0,𝑇×𝐶1𝑞𝑡0,𝑇××𝐶1𝑞𝑡0,,𝑇(1.3)(𝑎,𝐵)(𝑡0,𝑇]×𝑅𝑚 and 𝐷 is a constant vector. We completely generalize the results in [4] and obtain the new results if 𝑋(𝑡)𝐶1𝑞[𝑡0,𝑇]×𝐶1𝑞[𝑡0,𝑇]××𝐶1𝑞[𝑡0,𝑇]. Furthermore, we also obtain some results of the unique solution of the homogeneous and nonhomogeneous initial value problems with the classical Mittag-Leffler special function [5] which is similar to the ordinary differential equations. Now we introduce the first Mittag-Leffler function 𝑒𝑞(𝑡𝑡0) defined by 𝑒𝑞𝑡𝑡0=+𝑘=1𝑡𝑡0𝑘𝑞1.Γ(𝑘𝑞)(1.4) The function 𝑒𝑞(𝑡𝑡0) belongs to 𝐶1𝑞[𝑡0,𝑇]. Indeed, taking the norm in 𝐶1𝑞[𝑡0,𝑇], we have 𝑒𝑞𝑡𝑡01𝑞+𝑘=1𝑇𝑡0(𝑘1)𝑞Γ(𝑘𝑞)<+.(1.5) The formula remains valid for 𝑞1. In this case, 𝑒1(𝑡𝑡0)=exp(𝑡𝑡0). Then we introduce the second Mittag-Leffler function 𝐸𝑞(𝑡𝑡0) defined by 𝐸𝑞𝑡𝑡0=+𝑘=1𝑡𝑡0𝑘𝑞.Γ(𝑘𝑞+1)(1.6) The formula remains also valid for 𝑞1. In this case, 𝐸1(𝑡𝑡0)=exp(𝑡𝑡0)1.

The paper is organized as follows. In Section 2 we recall the definitions of fractional integral and derivative and related basic properties and preliminary results used in the text. In Section 3 we obtain the unique solution of the constant coefficient homogeneous and nonhomogeneous linear fractional differential equations for 𝑃 being the diagonal matrix. In Section 4 we prove the existence and uniqueness of these two kinds of equations for any 𝑃𝐿(𝑅𝑚). In Section 5 we give some specific examples to illustrate the results.

2. Definitions and Preliminary Results

Let us denote by 𝐶[𝑡0,𝑇] the space of all continuous real functions defined on [𝑡0,𝑇], which turns out to be a Banach space with the norm 𝑥=max𝑡[𝑡0,𝑇]||||.𝑥(𝑡)(2.1)

We define similarly another Banach space 𝐶1𝑞[𝑡0,𝑇], in which function 𝑥(𝑡) is continuous on (𝑡0,𝑇] and (𝑡𝑡0)1𝑞𝑥(𝑡) is continuous on [𝑡0,𝑇] with the norm: 𝑥1𝑞=max𝑡[𝑡0,𝑇]𝑡𝑡01𝑞||||.𝑥(𝑡)(2.2)

𝐿[𝑡0,𝑇] is the space of real functions defined on [𝑡0,𝑇] which are Lebesgue integrable on [𝑡0,𝑇].

Obviously 𝐶1𝑞[𝑡0,𝑇]𝐿(𝑡0,𝑇).

The definitions and results of the fractional calculus reported below are not exhaustive but rather oriented to the subject of this paper. For the proofs, which are omitted, we refer the reader to [6] or other texts on basic fractional calculus.

Definition 2.1 (see [6]). The fractional primitive of order 𝑞>0 of function 𝑥(𝑡)𝐶1𝑞[𝑡0,𝑇] is given by 𝐼𝑞𝑡01𝑥(𝑡)=Γ(𝑞)𝑡𝑡0(𝑡𝑠)𝑞1𝑥(𝑠)𝑑𝑠.(2.3)
From [17] we know 𝐼𝑞𝑡0𝑥(𝑡) exists for all 𝑞>0, when 𝑥𝐶1𝑞[𝑡0,𝑇]; consider also that when 𝑥𝐶[𝑡0,𝑇] then 𝐼𝑞𝑡0𝑥(𝑡)𝐶[𝑡0,𝑇] and moreover 𝐼𝑞𝑡0𝑥𝑡0=0.(2.4)

Definition 2.2 (see [6]). The fractional derivative of order 0<𝑞<1 of a function 𝑥(𝑡)𝐶1𝑞[𝑡0,𝑇] is given by 𝐷𝑞𝑡01𝑥(𝑡)=𝑑Γ(1𝑞)𝑑𝑥𝑡𝑡0(𝑡𝑠)𝑞𝑥(𝑠)𝑑𝑠.(2.5)
We have 𝐷𝑞𝑡0𝐼𝑞𝑡0𝑥(𝑡)=𝑥(𝑡) for all 𝑥(𝑡)𝐶1𝑞[𝑡0,𝑇].

Lemma 2.3 (see [6]). Let 0<𝑞<1. If one assumes 𝑥(𝑡)𝐶1𝑞[𝑡0,𝑇], then the fractional differential equation 𝐷𝑞𝑡0𝑥(𝑡)=0(2.6) has 𝑥(𝑡)=𝑐(𝑡𝑡0)𝑞1,𝑐𝑅, as solutions.

From this lemma we can obtain the following law of composition.

Lemma 2.4 (see [6]). Assume that 𝑥(𝑡)𝐶1𝑞[𝑡0,𝑇] with a fractional derivative of order 0<𝑞<1 that belongs to 𝐶1𝑞[𝑡0,𝑇]. Then 𝐼𝑞𝑡0𝐷𝑞𝑡0𝑥(𝑡)=𝑥(𝑡)+𝑐𝑡𝑡0𝑞1,(2.7) for some 𝑐𝑅. When the function 𝑥 is in 𝐶[𝑡0,𝑇], then 𝑐=0.

Lemma 2.5 (see [6]). Let 𝑈 be a nonempty closed subset of a Banach space 𝐸, and let 𝛼𝑛0 for every 𝑛 and such 𝑛=0𝛼𝑛 converges. Moreover, let the mapping 𝐴𝑈𝑈 satisfy the inequality 𝐴𝑛𝑢𝐴𝑛𝑣𝛼𝑛𝑢𝑣,(2.8) for every 𝑛𝑁 and any 𝑢,𝑣𝑈. Then, 𝐴 has a uniquely defined fixed point 𝑢. Furthermore, for any 𝑢0𝑈, the sequence (𝐴𝑛𝑢0)𝑛=1 converges to this fixed point 𝑢.

Lemma 2.6 (see [12]). Let 𝑃𝐿(𝑅𝑚) and have real eigenvalues 𝜆1,𝜆2,,𝜆𝑟. Then there exists a basis of 𝑅𝑚 in which the matrix representation of 𝑃 assumes Jordan form, that is, the matrix of 𝑃 is made of diagonal blocks of the form diag(𝐽1,𝐽2,,𝐽𝑟), where each 𝐽𝑖 consists of diagonal blocks of the form 𝜆𝑖0001𝜆𝑖000100001𝜆𝑖.(2.9)

Lemma 2.7 (see [12]). Let 𝑃𝐿(𝑅𝑚) and have complex eigenvalues 𝜇𝑗=𝛼𝑗+𝑖𝛽𝑗,𝑗=1,2,,𝑟, with multiplicity. Then there exists a basis of 𝑅𝑚, where 𝑃 has matrix form 𝐽diag(1,𝐽2𝐽,,𝑟), where each 𝐽𝑖 consists of diagonal blocks of the type 𝐼𝐷0002𝐷000𝐼20000𝐼2𝐷𝛼,𝐷=𝑖𝛽𝑖𝛽𝑖𝛼𝑖,𝐼2=1001.(2.10)

Lemma 2.8 (see [12]). Let 𝑃𝐿(𝑅𝑚). Then 𝑅𝑚 has a basis giving 𝑃 a matrix representation composed of diagonal blocks of type 𝐽𝑖 and/or matrices 𝐽𝑖, where 𝐽𝑖 and 𝐽𝑖 are as defined in the preceding lemmas.

Now, we will introduce Lemma 2.9 to prove the following Theorem 4.4 in Section 4.

Lemma 2.9. Let 0<𝑞<1. Assume that 𝑥(𝑡) and 𝑓(𝑡) belong to 𝐶1𝑞[𝑡0,𝑇]. Then For the initial value problem 𝐷𝑞𝑡0𝑥(𝑡)=𝜆𝑥(𝑡)+𝑓(𝑡),𝑥(𝑎)=𝑏(2.11) has a unique solution 𝑥(𝑡)𝐶1𝑞[𝑡0,𝑇] provided 𝑡0<𝑎<𝑎0, where 𝑎0 is a suitable constant depending on 𝑡0, 𝑞, and 𝜆.

Proof. The initial value problem (2.11) will be solved in two steps. (1) Local existence.Our problem is equivalent to the problem of determination of fixed points of the following operator: 𝐴𝑥(𝑡)=𝑐𝑡𝑡0𝑞1+1Γ(𝑞)𝑡𝑡0(𝑡𝑠)𝑞1(𝜆𝑥(𝑠)+𝑓(𝑠))𝑑𝑠,(2.12)with 1𝑐=𝑏Γ(𝑞)𝑎𝑡0(𝑎𝑠)𝑞1(𝜆𝑥(𝑠)+𝑓(𝑠))𝑑𝑠𝑎𝑡01𝑞.(2.13)It is immediate to verify that 𝐴𝐶1𝑞[𝑡0,𝑇]𝐶1𝑞[𝑡0,𝑇] is also well defined. Indeed, |||𝑡𝑡01𝑞|||||𝜆||||||1𝐴𝑥(𝑡)Γ(𝑞)𝑡𝑡0(𝑡𝑠)𝑞1𝑠𝑡0𝑞1𝑠𝑡01𝑞||||+||||1𝑥(𝑠)𝑑𝑠Γ(𝑞)𝑡𝑡0(𝑡𝑠)𝑞1𝑠𝑡0𝑞1𝑠𝑡01𝑞||||||𝜆||𝑓(𝑠)𝑑𝑠|𝑐|+𝑥1𝑞|||𝐼𝑞𝑡𝑡0𝑞1|||+𝑓1𝑞|||𝐼𝑞𝑡𝑡0𝑞1|||||𝜆|||𝑐|+𝑥1𝑞Γ(𝑞)Γ(2𝑞)𝑡𝑡0𝑞+𝑓1𝑞Γ(𝑞)Γ(2𝑞)𝑡𝑡0𝑞,(2.14)for 𝑥(𝑡) and 𝑓(𝑡) belong to 𝐶1𝑞[𝑡0,𝑇].Then we can also prove 𝐴 is a contraction operator. Indeed, 𝑡𝑡01𝑞||||||𝜆||𝐴𝑥(𝑡)𝐴𝑦(𝑡)Γ(𝑞)Γ(2𝑞)𝑎𝑡0𝑞𝑥𝑦1𝑞+||𝜆||Γ(𝑞)Γ(2𝑞)𝑡𝑡0𝑞𝑥𝑦1𝑞||𝜆||Γ(𝑞)Γ(2𝑞)𝑎𝑡0𝑞𝑥𝑦1𝑞+||𝜆||Γ(𝑞)Γ(2𝑞)𝑇𝑡0𝑞𝑥𝑦1𝑞,(2.15)for all 𝑥(𝑡),𝑦(𝑡)𝐶1𝑞[𝑡0,𝑇]. Let us assume ||𝜆||Γ(𝑞)Γ(2𝑞)𝑎𝑡0𝑞<12,(2.16)that is, 𝑎<𝑎0=Γ(2𝑞)2||𝜆||Γ(𝑞)1/𝑞+𝑡0.(2.17)Taking 𝑇𝑎>0 sufficiently small, we also have ||𝜆||Γ(𝑞)Γ(2𝑞)𝑇𝑡0𝑞<12,(2.18) and then 𝐴𝑥(𝑡)𝐴𝑦(𝑡)1𝑞𝐿𝑥𝑦1𝑞,(2.19)with 𝐿<1. Therefore 𝐴 is a contraction operator. This shows that initial problem (2.11) has a unique solution. (2) Continuation of solution.Since we know the value of 𝑥(𝑡) on (𝑡0,𝑎], then we can compute 𝑐=1𝑏Γ(𝑞)𝑎𝑡0(𝑎𝑠)𝑞1(𝜆𝑥(𝑠)+𝑓(𝑠))𝑑𝑠𝑎𝑡01𝑞.(2.20)We can solve the integral problem 𝑦(𝑡)=𝑐𝑡𝑡0𝑞1+1Γ(𝑞)𝑡𝑡0(𝑡𝑠)𝑞1(𝜆𝑦(𝑠)+𝑓(𝑠))𝑑𝑠,(2.21)obtaining a unique solution 𝑦(𝑡)𝐶1𝑞[𝑡0,𝑇] for all 𝑇>𝑡0. Now 𝑥(𝑡) and 𝑦(𝑡) agree on (𝑡0,𝑎]. Thus the solution admits 𝑦(𝑡) as its continuation. Hence the proof of Lemma 2.9 is complete.

3. Initial Value Problem: Continuous Solutions on (𝑡0,𝑇]

We open this section with some basic examples, concerning the case when the solutions in 𝐶1𝑞[𝑡0,𝑇]×𝐶1𝑞[𝑡0,𝑇]××𝐶1𝑞[𝑡0,𝑇] are submitted to an initial condition.

Theorem 3.1. Let 0<𝑞<1. For all (𝑎,𝐵)(𝑡0,𝑇]×𝑅𝑚 the initial value problem 𝐷𝑞𝑡0𝑋(𝑡)=0,𝑋(𝑎)=𝐵(3.1) admits 𝑋(𝑡)=𝐵𝑎𝑡01𝑞𝑡𝑡0𝑞1,(3.2) as unique solution in 𝐶1𝑞[𝑡0,𝑇]×𝐶1𝑞[𝑡0,𝑇]××𝐶1𝑞[𝑡0,𝑇].

Proof. According to Lemma 2.4, the initial value problem (3.1) is equivalent to the following equations: 𝑋(𝑡)=𝐶𝑡𝑡0𝑞1,𝐶=𝐵𝑎𝑡01𝑞.(3.3) Hence the proof of Theorem 3.1 is complete.

Theorem 3.2. Let 0<𝑞<1. Assume 𝑓𝐹(𝑡)=1(𝑡),𝑓2(𝑡),,𝑓𝑚(𝑡)𝑇𝐶1𝑞𝑡0,𝑇×𝐶1𝑞𝑡0,𝑇××𝐶1𝑞𝑡0.,𝑇(3.4) Then for all (𝑎,𝐵)(𝑡0,𝑇]×𝑅𝑚 the initial value problem 𝐷𝑞𝑡0𝑋(𝑡)=𝐹(𝑡),𝑋(𝑎)=𝐵(3.5) has a unique solution in 𝐶1𝑞𝑡0,𝑇×𝐶1𝑞𝑡0,𝑇××𝐶1𝑞𝑡0,,𝑇(3.6) given by 𝑥𝑋(𝑡)=1(𝑡),𝑥2(𝑡),,𝑥𝑚(𝑡)𝑇,(3.7) with 𝑥𝑖(𝑏𝑡)=𝑖1Γ(𝑞)𝑎𝑡0(𝑎𝑠)𝑞1𝑓𝑖(𝑠)𝑑𝑠𝑎𝑡01𝑞𝑡𝑡0𝑞1+1Γ(𝑞)𝑡𝑡0(𝑡𝑠)𝑞1𝑓𝑖(𝑠)𝑑𝑠,(𝑖=1,2,,𝑚).(3.8)

Proof. According to Lemma 2.4, the initial value problem (3.5) is equivalent to the following equations: 𝑋(𝑡)=𝐶𝑡𝑡0𝑞1+1Γ(𝑞)𝑡𝑡0(𝑡𝑠)𝑞11𝐹(𝑠)𝑑𝑠,𝐶=𝐵Γ(𝑞)𝑎𝑡0(𝑎𝑠)𝑞1𝐹(𝑠)𝑑𝑠𝑎𝑡01𝑞.(3.9) Hence the proof of Theorem 3.2 is complete.

The result remains true even if 𝑞1. In this case, (3.5) is reduced to the ordinary differential equations 𝑋(𝑡)=𝐹(𝑡),𝑋(𝑎)=𝐵,(3.10) which have a unique solution in 𝐶𝑡0𝑡,𝑇×𝐶0𝑡,𝑇××𝐶0,𝑇,(3.11) given by 𝑥𝑋(𝑡)=1(𝑡),𝑥2(𝑡),,𝑥𝑚(𝑡)𝑇,(3.12) with 𝑥𝑖(𝑡)=𝑏𝑖𝑎𝑡0𝑓𝑖(𝑠)𝑑𝑠+𝑡𝑡0𝑓𝑖(𝑠)𝑑𝑠,(𝑖=1,2,,𝑚).(3.13)

Theorem 3.3. Let 0<𝑞<1. For all (𝑎,𝐵)(𝑡0,𝑇]×𝑅𝑚 the initial value problem 𝐷𝑞𝑡0𝑋(𝑡)=𝑃𝑋(𝑡),𝑋(𝑎)=𝐵,(3.14) where 𝜆𝑃=10000𝜆2000000000𝜆𝑚(3.15) has a unique solution in 𝐶1𝑞𝑡0,𝑇×𝐶1𝑞𝑡0,𝑇××𝐶1𝑞𝑡0,,𝑇(3.16) given by 𝑥𝑋(𝑡)=1(𝑡),𝑥2(𝑡),,𝑥𝑚(𝑡)𝑇,(3.17) with 𝑥𝑖(𝑡)=𝑏𝑖𝑒𝑞1𝜆𝑖1/𝑞𝑎𝑡0𝑒𝑞𝜆𝑖1/𝑞𝑡𝑡0,(𝑖=1,2,,𝑚).(3.18)

Proof. We can write (3.27) in the following form: 𝐷𝑞𝑡0𝑥1(𝑡)=𝜆1𝑥1𝐷(𝑡),𝑞𝑡0𝑥2(𝑡)=𝜆2𝑥2𝐷(𝑡),𝑞𝑡0𝑥𝑚(𝑡)=𝜆𝑚𝑥𝑚𝑥(𝑡),1(𝑎)=𝑏1,,𝑥𝑚(𝑎)=𝑏𝑚.(3.19) According to Lemma 2.4,𝐷𝑞𝑡0𝑥𝑖(𝑡)=𝜆𝑖𝑥𝑖(𝑡)(3.20) is equivalent to the following equations: 𝑥𝑖(𝑡)=𝑐𝑖𝑡𝑡0𝑞1+𝐼𝑞𝑡0𝜆𝑖𝑥𝑖,(𝑡)(3.21) for some 𝑐𝑖𝑅. From (3.21) we obtain, by iteration, 𝑥𝑖(𝑡)=𝑐𝑖Γ(𝑞)𝑡𝑡0𝑞1+𝜆Γ(𝑞)𝑖𝑡𝑡02𝑞1𝜆Γ(2𝑞)++𝑖𝑛1𝑡𝑡0𝑛𝑞1Γ(𝑛𝑞)+𝜆𝑛𝑖𝐼𝑡𝑛𝑞0𝑥𝑖(𝑡).(3.22) Letting 𝑛+, 𝜆𝑛𝑖𝐼𝑡𝑛𝑞0𝑥𝑖(𝑡)1𝑞0 if 𝑥𝑖(𝑡)𝐶1𝑞[𝑡0,𝑇]. Indeed, 𝜆𝑛𝑖𝐼𝑡𝑛𝑞0𝑥𝑖(𝑡)1𝑞||𝜆𝑛𝑖||𝑥𝑖(𝑡)1𝑞Γ(𝑞)Γ((𝑛+1)𝑞)𝑡𝑡0𝑛𝑞.(3.23)
On the other hand, 𝑒𝑞𝑡𝑡0=+𝑘=1𝑡𝑡0𝑘𝑞1,Γ(𝑘𝑞)(3.24) then we can obtain 𝑥𝑖(𝑡)=𝑐𝑖Γ(𝑞)𝜆𝑖(1/𝑞1)𝑒𝑞𝜆𝑖1/𝑞𝑡𝑡0.(3.25) Since 𝑥𝑖(𝑎)=𝑏𝑖, 𝑥𝑖(𝑡)=𝑏𝑖𝑒𝑞1𝜆𝑖1/𝑞𝑎𝑡0𝑒𝑞𝜆𝑖1/𝑞𝑡𝑡0,(𝑖=1,2,,𝑚).(3.26) Hence the proof of Theorem 3.3 is complete.
The result remains valid even if 𝑞1. In this case, 𝑋(𝑡)=𝑃𝑋(𝑡),𝑋(𝑎)=𝐵(3.27) has a unique solution in 𝐶𝑡0𝑡,𝑇×𝐶0𝑡,𝑇××𝐶0,𝑇,(3.28) given by 𝑥𝑋(𝑡)=1(𝑡),𝑥2(𝑡),,𝑥𝑚(𝑡)𝑇,(3.29) with 𝑥𝑖(𝑡)=𝑏𝑖exp𝜆𝑖𝑎𝑡0𝜆exp𝑖𝑡𝑡0,(𝑖=1,2,,𝑚).(3.30)

Theorem 3.4. Let 0<𝑞<1. For all (𝑎,𝐵)(𝑡0,𝑇]×𝑅𝑚 the initial value problem 𝐷𝑞𝑡0𝑋(𝑡)=𝑃𝑋(𝑡)+𝐷,𝑋(𝑎)=𝐵,(3.31) where 𝜆𝑃=10000𝜆2000000000𝜆𝑚,(3.32) and 𝐷=(𝑑1,𝑑2,,𝑑𝑚) has a unique solution in 𝐶1𝑞𝑡0,𝑇×𝐶1𝑞𝑡0,𝑇××𝐶1𝑞𝑡0,,𝑇(3.33) given by 𝑥𝑋(𝑡)=1(𝑡),𝑥2(𝑡),,𝑥𝑚(𝑡)𝑇,(3.34) with 𝑥𝑖𝑏(𝑡)=𝑖𝑑𝑖𝜆𝑖1𝐸𝑞𝜆𝑖1/𝑞𝑎𝑡0𝑒𝑞1𝜆𝑖1/𝑞𝑎𝑡0𝑒𝑞𝜆𝑖1/𝑞𝑡𝑡0+𝑑𝑖𝜆𝑖1𝐸𝑞𝜆𝑖1/𝑞𝑡𝑡0(𝑖=1,2,,𝑚).(3.35)

Proof. We can write (3.44) in the following form:𝐷𝑞𝑡0𝑥1(𝑡)=𝜆1𝑥1(𝑡)+𝑑1,𝐷𝑞𝑡0𝑥2(𝑡)=𝜆2𝑥2(𝑡)+𝑑2,𝐷𝑞𝑡0𝑥𝑚(𝑡)=𝜆𝑚𝑥𝑚(𝑡)+𝑑𝑚𝑥1(𝑎)=𝑏1,,𝑥𝑚(𝑎)=𝑏𝑚.(3.36)
According to Lemma 2.4, the equation 𝐷𝑞𝑡0𝑥𝑖(𝑡)=𝜆𝑖𝑥𝑖(𝑡)+𝑑𝑖(3.37) is equivalent to the following equations: 𝑥𝑖(𝑡)=𝑐𝑖𝑡𝑡0𝑞1+𝐼𝑞𝑡0𝜆𝑖𝑥𝑖(𝑡)+𝐼𝑞𝑡0𝑑𝑖,(3.38) for some 𝑐𝑖𝑅. From (3.38) we obtain, by iteration, 𝐼𝑞𝑡0𝜆𝑖𝑥𝑖(𝑡)=𝐼𝑞𝑡0𝑐𝑖𝜆𝑖𝑡𝑡0𝑞1+𝐼𝑡2𝑞0𝜆2𝑖𝑥𝑖(𝑡)+𝐼𝑡2𝑞0𝜆𝑖𝑑𝑖,𝑥𝑖(𝑡)=𝑐𝑖𝑡𝑡0𝑞1+𝐼𝑞𝑡0𝜆𝑖𝑐𝑖𝑡𝑡0𝑞1+𝐼𝑡2𝑞0𝜆2𝑖𝑥𝑖(𝑡)+𝐼𝑞𝑡0𝑑𝑖+𝐼𝑡2𝑞0𝜆𝑖𝑑𝑖,𝐼𝑞𝑡0𝑥𝑖(𝑡)=𝐼𝑞𝑡0𝑐𝑖𝑡𝑡0𝑞1+𝐼𝑡2𝑞0𝜆𝑖𝑥𝑖(𝑡)+𝐼𝑡2𝑞0𝑑𝑖,𝑥𝑖(𝑡)=𝑐𝑖Γ(𝑞)𝑡𝑡0𝑞1+𝜆Γ(𝑞)𝑖𝑡𝑡02𝑞1𝜆Γ(2𝑞)++𝑖𝑛1𝑡𝑡0𝑛𝑞1Γ(𝑛𝑞)+𝑑𝑖𝑡𝑡0𝑞+𝜆Γ(𝑞+1)𝑖𝑡𝑡02𝑞𝜆Γ(2𝑞+1)++𝑖𝑛1𝑡𝑡0𝑛𝑞Γ(𝑛𝑞+1)+𝜆𝑛𝑖𝐼𝑡𝑛𝑞0𝑥𝑖(𝑡).(3.39)
Letting 𝑛+, 𝜆𝑛𝑖𝐼𝑡𝑛𝑞0𝑥𝑖(𝑡)1𝑞0 if 𝑥𝑖(𝑡)𝐶1𝑞[𝑡0,𝑇]. Indeed, 𝜆𝑛𝑖𝐼𝑡𝑛𝑞0𝑥𝑖(𝑡)1𝑞=max𝑡𝑡0,𝑇𝑡𝑡01𝑞||||𝜆𝑛𝑖1Γ(𝑞)𝑡𝑡0(𝑡𝑠)𝑛𝑞1𝑥𝑖||||||𝜆(𝑠)𝑑𝑠𝑛𝑖||𝑥𝑖(𝑡)1𝑞Γ(𝑞)Γ((𝑛+1)𝑞)𝑡𝑡0𝑛𝑞.(3.40)
On the other hand, 𝑒𝑞𝑡𝑡0=+𝑘=1𝑡𝑡0𝑘𝑞1,𝐸Γ(𝑘𝑞)𝑞𝑡𝑡0=+𝑘=1𝑡𝑡0𝑘𝑞Γ.(𝑘𝑞+1)(3.41)
Then we can obtain 𝑥𝑖(𝑡)=𝑐𝑖Γ(𝑞)𝜆𝑖(1/𝑞1)𝑒𝑞𝜆𝑖1/𝑞𝑡𝑡0+𝑑𝑖𝜆𝑖1𝐸𝑞𝜆𝑖1/𝑞𝑡𝑡0.(3.42)
We know that 𝑑𝑖𝜆𝑖1𝐸𝑞(𝜆𝑖1/𝑞(𝑡𝑡0)) is satisfied for the fractional nonhomogeneous linear differential equation 𝐷𝑞𝑡0𝑥𝑖(𝑡)=𝜆1𝑥𝑖(𝑡)+𝑑𝑖. So we can also deduce that the general solution of the fractional nonhomogeneous linear differential equation is equal to the general solution of the corresponding homogeneous linear differential equation plus the special solution of the nonhomogeneous linear differential equation. If 𝑋(𝑎)=𝐵, 𝑥𝑖(𝑎)=𝑏𝑖, then 𝑥𝑖𝑏(𝑡)=𝑖𝑑𝑖𝜆𝑖1𝐸𝑞𝜆𝑖1/𝑞𝑎𝑡0𝑒𝑞1𝜆𝑖1/𝑞𝑎𝑡0𝑒𝑞𝜆𝑖1/𝑞𝑡𝑡0+𝑑𝑖𝜆𝑖1𝐸𝑞𝜆𝑖1/𝑞𝑡𝑡0(𝑖=1,2,,𝑚).(3.43) Hence the proof of Theorem 3.4 is complete.

The result remains valid even if 𝑞1. In this case, 𝑋(𝑡)=𝑃𝑋(𝑡)+𝐷,𝑋(𝑎)=𝐵,(3.44) where 𝜆𝑃=10000𝜆2000000000𝜆𝑚,(3.45) and 𝐷=(𝑑1,𝑑2,,𝑑𝑚) has a unique solution in 𝐶𝑡0𝑡,𝑇×𝐶0𝑡,𝑇××𝐶0,𝑇,(3.46) given by 𝑥𝑋(𝑡)=1(𝑡),𝑥2(𝑡),,𝑥𝑚(𝑡)𝑇,(3.47) with 𝑥𝑖𝑏(𝑡)=𝑖𝑑𝑖𝜆𝑖1𝐸1𝜆𝑖𝑎𝑡0𝑒11𝜆𝑖𝑎𝑡0𝑒1𝜆𝑖𝑡𝑡0+𝑑𝑖𝜆𝑖1𝐸1𝜆𝑖𝑡𝑡0=𝑏𝑖+𝑑𝑖𝜆𝑖1exp𝜆𝑎𝑡0𝜆exp𝑖𝑡𝑡0𝑑𝜆1,(𝑖=1,2,,𝑚).(3.48)

4. Existence and Uniqueness of the Solution

In Section 3 we have obtained the unique solution of the constant coefficient homogeneous and nonhomogeneous linear fractional differential equations for 𝑃 being the diagonal matrix. In the present section we will prove the existence and uniqueness of these two kinds of equations for any 𝑃𝐿(𝑅𝑚).

Theorem 4.1. Let 0<𝑞<1 and 𝑃𝐿(𝑅𝑚). If the matrix 𝑃 has distinct real eigenvalues, then for all (𝑎,𝐵)(𝑡0,𝑇]×𝑅𝑚 the initial value problem 𝐷𝑞𝑡0𝑋(𝑡)=𝑃𝑋(𝑡),𝑋(𝑎)=𝐵(4.1) has the unique solution 𝑋(𝑡)𝐶1𝑞[𝑡0,𝑇]×𝐶1𝑞[𝑡0,𝑇]××𝐶1𝑞[𝑡0,𝑇].

Proof. Since the matrix 𝑃 has distinct real eigenvalues, there exists an invertible matrix 𝑄 such that 𝑄1𝜆𝑃𝑄=10000𝜆2000000000𝜆𝑚,(4.2) where 𝜆1,𝜆2,,𝜆𝑚 are the eigenvalues of the matrix 𝑃. If we define 𝑌(𝑡)=𝑄1𝑋(𝑡), 𝐷𝑞𝑡0𝑌(𝑡)=𝐷𝑞𝑡0𝑄1𝑋(𝑡)=𝑄1𝐷𝑞𝑡0𝑋(𝑡)=𝑄1𝑃𝑋(𝑡)=𝑅𝑌(𝑡),(4.3) with 𝑅=diag(𝜆1,𝜆2,,𝜆𝑚). From the above Theorem 3.3 we know the initial value problem 𝐷𝑞𝑡0𝑌(𝑡)=𝑅𝑌(𝑡),𝑌(𝑎)=𝑄1𝐵(4.4) has a unique solution 𝑌(𝑡)𝐶1𝑞[𝑡0,𝑇]×𝐶1𝑞[𝑡0,𝑇]××𝐶1𝑞[𝑡0,𝑇] defined on [𝑡0,𝑇]. Then 𝑋(𝑡)=𝑄𝑌(𝑡) uniquely solves the equations (4.1), where 𝑡[𝑡0,𝑇]. Hence the proof of Theorem 4.1 is complete.

Theorem 4.2. Let 0<𝑞<1. For all (𝑎,𝐵)(𝑡0,𝑇]×𝑅2 the initial value problem 𝐷𝑞𝑡0𝑋(𝑡)=𝑃𝑋(𝑡),𝑋(𝑎)=𝐵,(4.5) where 𝑃=𝛼𝛽𝛽𝛼(4.6) has the unique solution 𝑋(𝑡)𝐶1𝑞[𝑡0,𝑇]×𝐶1𝑞[𝑡0,𝑇] defined on [𝑡0,𝑇].

Proof. Let us define 𝑍(𝑡)=𝑥1(𝑡)+𝑖𝑥2(𝑡),𝜇=𝛼+𝑖𝛽.(4.7) We can find that (4.5) is equivalent to the following equation 𝐷𝑞𝑡0𝑍(𝑡)=𝜇𝑍(𝑡),𝑍(𝑎)=𝑥1(𝑎)+𝑥2(𝑎)=𝑏1+𝑖𝑏2.(4.8) Obviously, 𝑍(𝑡)𝐶1𝑞[𝑡0,𝑇] if 𝑥1(𝑡) and 𝑥2(𝑡) belong to 𝐶1𝑞[𝑡0,𝑇]. From the above Theorem 3.3 in Section 3, we know the complex Equation (4.7) has a unique solution defined on [𝑡0,𝑇]. Hence the proof of Theorem 4.2 is complete.

Theorem 4.3. Let 0<𝑞<1 and 𝑃𝑅2. If 𝑃 has eigenvalues 𝛼±𝑖𝛽, for all (𝑎,𝐵)(𝑡0,𝑇]×𝑅2 the initial value problem 𝐷𝑞𝑡0𝑋(𝑡)=𝑃𝑋(𝑡),𝑋(𝑎)=𝐵(4.9) has a unique solution 𝑋(𝑡)𝐶1𝑞[𝑡0,𝑇]×𝐶1𝑞[𝑡0,𝑇].

Proof. Since 𝑃 has eigenvalues 𝛼±𝑖𝛽, there exists an invertible matrix 𝑄 such that 𝑃=𝑄𝑆𝑄1 where 𝑆=𝛼𝛽𝛽𝛼.(4.10) Define 𝑌(𝑡)=𝑄1𝑋(𝑡),(4.11) then 𝐷𝑞𝑡0𝑌(𝑡)=𝐷𝑞𝑡0𝑄1𝑋(𝑡)=𝑄1𝐷𝑞𝑡0𝑋(𝑡)=𝑄1𝑃𝑋(𝑡)=𝑆𝑌(𝑡).(4.12) From the above Theorem 4.2, we know the initial value problem 𝐷𝑞𝑡0𝑌(𝑡)=𝑆𝑌(𝑡),𝑌(𝑎)=𝑄1𝐵(4.13) has a unique solution defined on [𝑡0,𝑇]. Hence the proof of result is complete.

Theorem 4.4. Let 0<𝑞<1 and 𝑃𝑅𝑚 be an elementary Jordan matrix: 𝜆0001𝜆000100001𝜆.(4.14) The initial value problem 𝐷𝑞𝑡0𝑋(𝑡)=𝑃𝑋(𝑡),𝑋(𝑎)=𝐵(4.15) has a unique solution 𝑋(𝑡)𝐶1𝑞[𝑡0,𝑇]×𝐶1𝑞[𝑡0,𝑇]××𝐶1𝑞[𝑡0,𝑇] provided (𝑎,𝐵)(𝑡0,𝑎0]×𝑅𝑚, where 𝑎0 is a suitable constant depending on 𝑡0, 𝑞, and 𝜆.

Proof. From the (4.15), we can write the equations in the following form: 𝐷𝑞𝑡0𝑥1(𝑡)=𝜆𝑥1𝐷(𝑡),𝑞𝑡0𝑥2(𝑡)=𝑥1(𝑡)+𝜆𝑥2𝐷(𝑡),𝑞𝑡0𝑥𝑚(𝑡)=𝑥𝑚1(𝑡)+𝜆𝑥𝑚𝑥(𝑡),1(𝑎)=𝑏1,,𝑥𝑚(𝑎)=𝑏𝑚.(4.16)
Consider the first equation 𝐷𝑞𝑡0𝑥1(𝑡)=𝜆𝑥1(𝑡),𝑥1(𝑎)=𝑏1.(4.17)
We can obtain the solution of this equation 𝑥1(𝑡)=𝑏1𝑒𝑞1𝜆1/𝑞𝑎𝑡0𝑒𝑞𝜆1/𝑞𝑡𝑡0.(4.18)
Consider the second equation 𝐷𝑞𝑡0𝑥2(𝑡)=𝑥1(𝑡)+𝜆𝑥2(𝑡),𝑥2(𝑎)=𝑏2,(4.19) where now 𝑥1(𝑡)𝐶1𝑞[𝑡0,𝑇] is a known function. Since 𝑥1(𝑡),𝑥2(𝑡)𝐶1𝑞[𝑡0,𝑇], according to Lemma 2.9, (4.19) has a unique solution in 𝐶1𝑞[𝑡0,𝑇]. Now 𝑥1(𝑡) and 𝑥2(𝑡) are known functions which will be substituted in 𝐷𝑞𝑡0𝑥3(𝑡)=𝑥2(𝑡)+𝜆𝑥3(𝑡),𝑥3(𝑎)=𝑏3(4.20) and so on. Thus the system of equations given in (4.15) has unique solution in 𝐶1𝑞[𝑡0,𝑇]×𝐶1𝑞[𝑡0,𝑇]××𝐶1𝑞[𝑡0,𝑇].

Theorem 4.5. Let 0<𝑞<1 and 𝑃𝐿(𝑅𝑚). The initial value problem 𝐷𝑞𝑡0𝑋(𝑡)=𝑃𝑋(𝑡),𝑋(𝑎)=𝐵(4.21) has the unique solution 𝑋(𝑡)𝐶1𝑞[𝑡0,𝑇]×𝐶1𝑞[𝑡0,𝑇]××𝐶1𝑞[𝑡0,𝑇] provided (𝑎,𝐵)(𝑡0,𝑎0]×𝑅𝑚, where 𝑎0 is a suitable constant depending on 𝑡0, 𝑞, and 𝜆.

Proof. In view of Lemma 2.8, there exists an invertible matrix 𝑄 such that 𝑄1𝑃𝑄 is composed of diagonal blocks of the type 𝐽𝑖 and 𝐽𝑖, as defined in the preceding Lemmas 2.7 and 2.8. Let 𝐵=𝑄1𝑃𝑄 and 𝑌(𝑡)=𝑄1𝑋(𝑡). Consider the initial value problem: 𝐷𝑞𝑡0𝑌(𝑡)=𝐷𝑞𝑡0𝑄1𝑋(𝑡)=𝑄1𝐷𝑞𝑡0𝑋(𝑡)=𝑄1𝑃𝑋(𝑡)=𝐵𝑌(𝑡),𝑌(𝑎)=𝑄1𝑋(𝑎)=𝑄1𝐵.(4.22) Then in view of Theorems 4.14.5, (4.22) has a unique solution: 𝑌(𝑡)𝐶1𝑞[𝑡0,𝑇]×𝐶1𝑞[𝑡0,𝑇]××𝐶1𝑞[𝑡0,𝑇]. Therefore (4.21) has a unique solution 𝑄1𝑌(𝑡)𝐶1𝑞[𝑡0,𝑇]×𝐶1𝑞[𝑡0,𝑇]××𝐶1𝑞[𝑡0,𝑇].

Remark 4.6. All the above results are valid for 𝑞1. Moreover, we can also discuss the case if 𝑎=𝑡0, in this case, we cannot consider the usual initial condition 𝑥(𝑡0)=𝑏, but lim𝑡𝑡0(𝑡𝑡0)1𝑞𝑥(𝑡)=𝑏. We can also obtain some similar results by the same method, So we did not give the detailed process and conclusion in this paper.

5. Illustrative Examples

In this section, we give some specific examples to illustrate the above results.

Example 5.1. Consider the following system, where 0<𝑞<1, 𝑡[𝑡0,𝑇], (𝑎,𝐵)(𝑡0,𝑇]×𝑅3,𝐵=(𝑏1,𝑏2,𝑏3),𝐷𝑞𝑡0𝑥1(𝑡)=3𝑥1(𝑡)𝑥2(𝑡)+𝑥3𝐷(𝑡),𝑞𝑡0𝑥2(𝑡)=𝑥1(𝑡)+5𝑥2(𝑡)𝑥3𝐷(𝑡),𝑞𝑡0𝑥3(𝑡)=𝑥1(𝑡)𝑥2(𝑡)+3𝑥3𝑥(𝑡),1(𝑎)=𝑏1,𝑥2(𝑎)=𝑏2,𝑥3(𝑎)=𝑏3.(5.1) Here 𝑃=311151113,(5.2) having the eigenvalues 2, 3, and 6. Choose the eigenvectors  𝑔1=(1,0,1)𝑇,𝑔2=(1,1,1)𝑇, and 𝑔3=(1,2,1)𝑇. Then 200030006=𝑄1311151113𝑄,(5.3) where 𝑄1=12102131313161316.(5.4)
Define the 𝑌=𝑄1𝑋. Then the system of equation in 𝑌 is decoupled, namely, 𝐷𝑞𝑡0𝑦1(𝑡)=2𝑦1𝐷(𝑡),𝑞𝑡0𝑦2(𝑡)=3𝑦2𝐷(𝑡),𝑞𝑡0𝑦3(𝑡)=6𝑦3𝑦(𝑡),11(𝑎)=2𝑏112𝑏3,𝑦21(𝑎)=3𝑏1+13𝑏2+13𝑏3,𝑦31(𝑎)=6𝑏113𝑏2+16𝑏3.(5.5)
In view of (3.30), we can obtain 𝑦11(𝑡)=2𝑏112𝑏3𝑒𝑞121/𝑞𝑎𝑡0𝑒𝑞21/𝑞𝑡𝑡0,𝑦21(𝑡)=3𝑏1+13𝑏2+13𝑏3𝑒𝑞131/𝑞𝑎𝑡0𝑒𝑞31/𝑞𝑡𝑡0,𝑦31(𝑡)=6𝑏113𝑏2+16𝑏3𝑒𝑞161/𝑞𝑎𝑡0𝑒𝑞61/𝑞𝑡𝑡0.(5.6)
Hence 𝑥1(𝑡)=𝑦1(𝑡)+𝑦2(𝑡)+𝑦3𝑥(𝑡),2(𝑡)=𝑦2(𝑡)2𝑦3𝑥(𝑡),3(𝑡)=𝑦1(𝑡)+𝑦2(𝑡)+𝑦3(𝑡).(5.7)

Example 5.2. Consider the following system, where 0<𝑞<1, 𝑡[𝑡0,𝑇], (𝑎,𝐵)(𝑡0,𝑇]×𝑅2,𝐵=(𝑏1,𝑏2), 𝐷𝑞𝑡0𝑥1(𝑡)=2𝑥1(𝑡)𝑥2𝐷(𝑡),𝑞𝑡0𝑥2(𝑡)=13𝑥1(𝑡)+4𝑥2𝑥(𝑡),1(𝑎)=𝑏1,𝑥2(𝑎)=𝑏2.(5.8) Here 𝑃=21134(5.9) having the eigenvalues 1±2𝑖. Choose the eigenvectors   𝑔1=(1,32𝑖)𝑇, and 𝑔2=(1,3+2𝑖)𝑇, Then 1+2𝑖0012𝑖=𝑄121134𝑄,(5.10) where ,𝑄𝑄=1132𝑖3+2𝑖1=2+3𝑖4𝑖423𝑖4𝑖4.(5.11) Define the 𝑌=𝑄1𝑋. Then the system of equation in 𝑌 is decoupled, namely, 𝐷𝑞𝑡0𝑦1(𝑡)=(1+2𝑖)𝑦1(𝑡),𝑦11(𝑎)=2𝑏1+34𝑏1+14𝑏2𝑖𝐷𝑞𝑡0𝑦2(𝑡)=(12𝑖)𝑦2(𝑡),𝑦21(𝑎)=2𝑏134𝑏1+14𝑏2𝑖.(5.12) In view of (3.30), we can obtain 𝑦11(𝑡)=2𝑏1+34𝑏1+14𝑏2𝑖𝑒𝑞1(1+2𝑖)1/𝑞𝑎𝑡0𝑒𝑞(1+2𝑖)1/𝑞𝑡𝑡0𝑦21(𝑡)=2𝑏134𝑏1+14𝑏2𝑖𝑒𝑞1(12𝑖)1/𝑞𝑎𝑡0𝑒𝑞(12𝑖)1/𝑞𝑡𝑡0.(5.13) Hence 𝑥1(𝑡)=𝑦1(𝑡)+𝑦2𝑥(𝑡)2𝑦(𝑡)=31(𝑡)+𝑦2𝑦(𝑡)2𝑖1(𝑡)𝑦2.(𝑡)(5.14)

Acknowledgments

The authors are highly grateful for the referee's careful reading and comments on this paper. The present paper was supported by the NNSF of China Grants no. 11271087 and no. 61263006.