Abstract

We obtain the analytical general solution of the linear fractional differential equations with constant coefficients by Adomian decomposition method under nonhomogeneous initial value condition, which is in the sense of the Caputo fractional derivative.

1. Introduction

Fractional differential equations are hot topics both in mathematics and physics. Recently, the fractional differential equations have been the subject of intensive research. There are several methods to obtain the solution, such as the Laplace transform method, power series method, and Green function method. Many remarkable results for the fractional differential equations can be found in the literature [111]. In particular, the Adomian decomposition method has attracted the attention of many mathematicians [1215].

For a better understanding of the fractional derivatives and for a physical understanding of the fractional equations, the readers can refer to the recent publications in [16, 17]. Ebaid [18] suggested a modification of the Adomian method, and a few iterations lead to exact solution. Das [19] compared the variational iteration method with the Adomian method for fractional equations and found that the variational iteration method is much more effective. For other methods of the fractional differential equations, especially the homotopy perturbation method, variational iteration method and differential transform method were presented in [20, 21].

Consider the following 𝑛-term fractional differential equation with constant coefficients:𝑎𝑛𝐶𝐷𝛽𝑛𝑦(𝑡)+𝑎𝑛1𝐶𝐷𝛽𝑛1𝑦(𝑡)++𝑎1𝐶𝐷𝛽1𝑦(𝑡)+𝑎0𝐶𝐷𝛽0𝑦(𝑡)=𝑓(𝑡),(1.1) where 𝑛+1>𝛽𝑛>𝛽𝑛1>>𝛽1>𝛽0 and 𝑎𝑖(𝑖=0,1,,𝑛) is a real constant. In [12], the authors obtain the particular solution of (1.1) of the homogeneous initial value problem of the form 𝑦𝑖(0)=0,𝑖=0,1,,𝑛.(1.2)

However, it seems also more meaningful and more complicated for solving general solution of (1.1) under nonhomogeneous initial value condition. Therefore, in this paper, we will remove the restriction of the homogeneous initial value, consider the nonhomogeneous initial value problems of the form 𝑦𝑗𝑖(0)=𝑐𝑖𝑗𝑖,𝑖=0,1,,𝑛,𝑗𝑖=1,2,,𝑙𝑖,𝑙𝑖1𝛽𝑖<𝑙𝑖,(1.3) and obtain the analytical general solution of (1.1), which generalizes the result in [12].

We organize the paper as follows. In Section 2, we give some basic definitions and properties. In Section 3, we obtain the analytical general solution of the linear fractional differential equations by Adomian decomposition method. Some explicit examples are given in Section 4.

2. Basic Definitions and Notations

Definition 2.1 (see [1]). The Riemann-Liouville integral of order 𝑝 is defined by 𝑎𝐷𝑡𝑝1𝑓(𝑡)=Γ(𝑝)𝑡𝑎(𝑡𝜏)𝑝1𝑓(𝜏)𝑑𝜏,𝑝>0.(2.1) From Definition 2.1, we clearly see that 𝑎𝐷𝑡𝛼(𝑡𝑎)𝜈=Γ(1+𝜈)Γ(1+𝜈+𝛼)(𝑡𝑎)𝜈+𝛼,(2.2)𝑎𝐷𝛼𝑡(𝑡𝑎)𝜈=Γ(1+𝜈)Γ(1+𝜈𝛼)(𝑡𝑎)𝜈𝛼,(2.3) where 𝛼>0 and 𝜈 is a real number.

Definition 2.2 (see [1]). For 𝑓(𝑡)𝐶𝑚(𝑚𝑁), the Caputo fractional derivative of 𝑓(𝑡) is defined by 𝐶𝑎𝐷𝑝𝑡𝐷𝑓(𝑡)=𝑚𝑝𝑓(𝑚)𝑑(𝑡),𝑚1<𝑝<𝑚,(𝑚)𝑑𝑡𝑚𝑓(𝑡),𝑝=𝑚,𝑎𝐷𝑝𝑡𝑓(𝑡),𝑝0.(2.4) Therefore, 𝑐𝑎𝐷𝑡𝛼(𝑡𝑎)𝜈=𝑎𝐷𝑡𝛼(𝑡𝑎)𝜈=Γ(1+𝜈)Γ(1+𝜈+𝛼)(𝑡𝑎)𝜈+𝛼.(2.5)

Lemma 2.3 (see [1]). If 𝑓(𝑡) is continuous, then 𝐶𝑎𝐷𝑡𝑛𝑓(𝑛)(𝑡)=𝑎𝐷𝑡𝑛𝑓(𝑛)(𝑡)=𝑓(𝑡)𝑛1𝑗=0𝑓(𝑗)(𝑎)(𝑡𝑎)𝑗Γ(𝑗+1),𝑛𝑍.(2.6)

Lemma 2.4 (see [1]). If 𝑓(𝑡) is continuous, then 𝐶𝑎𝐷𝑡𝑝𝐶𝑎𝐷𝑡𝑞𝑓(𝑡)=𝑎𝐷𝑡𝑝𝑎𝐷𝑡𝑞𝑓(𝑡)=𝑎𝐷𝑡𝑝𝑞𝑓(𝑡)=𝐶𝑎𝐷𝑡𝑝𝑞𝑓(𝑡)=𝐶𝑎𝐷𝑡𝑞𝐶𝑎𝐷𝑡𝑝𝑓(𝑡).(2.7)

Lemma 2.5. If 𝑓(𝑡) is continuous, then 𝐶𝑎𝐷𝑡𝑝𝐶𝑎𝐷𝑞𝑡𝑓(𝑡)=𝑎𝐷𝑡𝑞𝑝𝑓(𝑡)𝑚1𝑗=0𝑓(𝑗)(𝑎)(𝑡𝑎)𝑗𝑞+𝑝Γ,(𝑗𝑞+𝑝+1)(2.8) where 𝑚1<𝑞<𝑚 and 𝑞𝑝.

Proof. From Definition 2.2 and Lemmas 2.32.4, we get 𝐶𝑎𝐷𝑡𝑝𝐶𝑎𝐷𝑞𝑡𝑓(𝑡)=𝑎𝐷𝑡𝑝𝑎𝐷𝑡𝑞𝑚𝑓(𝑚)(𝑡)=𝑎𝐷𝑡𝑝+𝑞𝑎𝐷𝑡𝑞𝑎𝐷𝑡𝑞𝑚𝑓(𝑚)=(𝑡)𝑎𝐷𝑡𝑝+𝑞𝑎𝐷𝑡𝑚(𝑡)𝑓(𝑚)=(𝑡)𝑎𝐷𝑡𝑝+𝑞𝑓(𝑡)𝑚1𝑗=0𝑓(𝑗)(𝑎)(𝑡𝑎)𝑗=Γ(𝑗+1)𝑎𝐷𝑡𝑞𝑝𝑓(𝑡)𝑚1𝑗=0𝑓(𝑗)(𝑎)(𝑡𝑎)𝑗𝑞+𝑝.Γ(𝑗𝑞+𝑝+1)(2.9) It is easy to see that 𝐶𝑎𝐷𝑡𝑝𝐶𝑎𝐷𝑝𝑡𝑓(𝑡)=𝑓(𝑡)𝑚1𝑗=0𝑓(𝑗)(𝑎)(𝑡𝑎)𝑗Γ.(𝑗+1)(2.10)

Proposition 2.6 (see [12]). One has 𝑘1=0𝑘𝑛=0𝑎𝑘1,𝑘2𝑘𝑛1𝑘𝑛=𝑚=0𝑘1,,𝑘𝑛1,𝑘𝑛𝑘01++𝑘𝑛1+𝑘𝑛=𝑚𝑎𝑘1,𝑘2𝑘𝑛1𝑘𝑛.(2.11)

Proposition 2.7 (see [12]). More over, one has 𝑚=0𝑘1,,𝑘𝑛1,𝑘𝑛𝑘01++𝑘𝑛1+𝑘𝑛=𝑚𝑎𝑘1,𝑘2𝑘𝑛1𝑘𝑛=𝑠=0𝑘1,,𝑘𝑛1𝑘01++𝑘𝑛1=𝑠𝑘𝑛=0𝑎𝑘1,𝑘2𝑘𝑛1𝑘𝑛.(2.12)

3. The Analytical Solution of the Linear Constant Coefficient Fractional Differential Equation

For simplicity, if 𝑎=0, then we denote 𝐶𝑎𝐷𝑝𝑡 or 𝐶𝑎𝐷𝑡𝑝 by 𝐶𝐷𝑝 or 𝐶𝐷𝑝.

In this section, we use Adomian decomposition method to discuss the general form of the linear fractional differential equations with constant coefficients, and apply and some basic transformation and integration to obtain the solution of the equations.

Let us consider the following 𝑛-term linear fractional differential equations with constant coefficients: 𝑎𝑛𝐶𝐷𝛽𝑛𝑦(𝑡)+𝑎𝑛1𝐶𝐷𝛽𝑛1𝑦(𝑡)++𝑎1𝐶𝐷𝛽1𝑦(𝑡)+𝑎0𝐶𝐷𝛽0𝑦𝑦(𝑡)=𝑓(𝑡),(𝑗𝑖)(0)=𝑐𝑖𝑗𝑖,𝑖=0,1,,𝑛,𝑗𝑖=1,2,,𝑙𝑖,𝑙𝑖1𝛽𝑖<𝑙𝑖,(3.1) where 𝑛+1>𝛽𝑛𝑛>𝛽𝑛1>𝛽1>𝛽0,𝑎𝑖and𝑐𝑖𝑗𝑖 are real constants, 𝐶𝐷𝑝=𝐶0𝐷𝑝𝑡 denotes Caputo fractional derivative of order 𝛼.

Applying 𝐶𝐷𝛽𝑛 to both sides of (1.1) and utilizing Lemma 2.5, we get 𝑦𝑎(𝑡)+𝑛1𝑎𝑛𝐶𝐷𝛽𝑛1𝛽𝑛𝑦𝑎(𝑡)++0𝑎𝑛𝐶𝐷𝛽0𝛽𝑛𝑦=1(𝑡)𝑎𝑛𝐶𝐷𝛽𝑛𝑓1(𝑡)+𝑎𝑛𝑛𝑖=0𝑎𝑖𝑙𝑖1𝑗𝑖=0𝑦(𝑗𝑖)𝑡(0)𝛽𝑛𝛽𝑖+𝑗𝑖Γ1+𝛽𝑛𝛽𝑖+𝑗𝑖.(3.2)

By the Adomian decomposition method, we obtain the recursive relationship as follows: 𝑦01(𝑡)=𝑎𝑛𝐶𝐷𝛽𝑛1𝑓(𝑡)+𝑎𝑛𝑛𝑖=0𝑎𝑖𝑙𝑖1𝑗𝑖=0𝑦(𝑗𝑖)𝑡(0)𝛽𝑛𝛽𝑖+𝑗𝑖Γ1+𝛽𝑛𝛽𝑖+𝑗𝑖,𝑦1𝑎(𝑡)=𝑛1𝑎𝑛𝐶𝐷𝛽𝑛1𝛽𝑛𝑎++0𝑎𝑛𝐶𝐷𝛽0𝛽𝑛𝑦0𝑦(𝑡),2(𝑡)=(1)2𝑎𝑛1𝑎𝑛𝐶𝐷𝛽𝑛1𝛽𝑛𝑎++0𝑎𝑛𝐶𝐷𝛽0𝛽𝑛2𝑦0𝑦(𝑡),𝑠(𝑡)=(1)𝑠𝑎𝑛1𝑎𝑛𝐶𝐷𝛽𝑛1𝛽𝑛𝑎++0𝑎𝑛𝐶𝐷𝛽0𝛽𝑛𝑠𝑦0(𝑡),(3.3)

By Adomian decomposition method, adding all terms of the recursion, we obtain the solution of (1.1) as 𝑦(𝑡)=𝑠=0𝑦𝑠(=𝑡)𝑠=0(1)𝑠𝑎𝑛1𝑎𝑛𝐶𝐷𝛽𝑛1𝛽𝑛𝑎++0𝑎𝑛𝐶𝐷𝛽0𝛽𝑛𝑠𝑦0=1(𝑡)𝑎𝑛𝑠=0(1)𝑠𝑎𝑛1𝑎𝑛𝐷𝛽𝑛1𝛽𝑛𝑎++0𝑎𝑛𝐷𝛽0𝛽𝑛𝑠𝐷𝛽𝑛+1𝑓(𝑡)𝑎𝑛𝑠=0(1)𝑠𝑎𝑛1𝑎𝑛𝐷𝛽𝑛1𝛽𝑛𝑎++0𝑎𝑛𝐷𝛽0𝛽𝑛𝑠𝑛𝑖=0𝑎𝑖𝑙𝑖1𝑗𝑖=0𝑦(𝑗𝑖)𝑡(0)𝛽𝑛𝛽𝑖+𝑗𝑖Γ1+𝛽𝑛𝛽𝑖+𝑗𝑖.(3.4) Let 𝐼1=1𝑎𝑛𝑠=0(1)𝑠𝑎𝑛1𝑎𝑛𝐷𝛽𝑛1𝛽𝑛𝑎++0𝑎𝑛𝐷𝛽0𝛽𝑛𝑠𝐷𝛽𝑛𝐼𝑓(𝑡)2=1𝑎𝑛𝑠=0(1)𝑠𝑎𝑛1𝑎𝑛𝐷𝛽𝑛1𝛽𝑛𝑎++0𝑎𝑛𝐷𝛽0𝛽𝑛𝑠𝑛𝑖=0𝑎𝑖𝑙𝑖1𝑗𝑖=0𝑦(𝑗𝑖)(𝑡0)𝛽𝑛𝛽𝑖+𝑗𝑖Γ1+𝛽𝑛𝛽𝑖+𝑗𝑖.(3.5) Then,𝑦(𝑡)=𝐼1+𝐼2.(3.6)

Next, we estimate 𝐼1 and 𝐼2, respectively.

For 𝐼1, by [12] we obtain 1𝑎𝑛𝑠=0(1)𝑠𝑎𝑛1𝑎𝑛𝐷𝛽𝑛1𝛽𝑛𝑎++0𝑎𝑛𝐷𝛽0𝛽𝑛𝑠𝐷𝛽𝑛=𝑓(𝑡),𝑡01𝑎𝑛𝑚=0(1)𝑚𝑚!𝑘0,𝑘1,,𝑘𝑛2𝑘00+𝑘1++𝑘𝑛2=𝑚𝑚;𝑘0,𝑘1,,𝑘𝑛2×𝑛2𝑝=0𝑎𝑝𝑎𝑛𝑘𝑝(𝑡𝜏)(𝛽𝑛𝛽𝑛1)𝑚+𝛽𝑛+𝑛2𝑗=0(𝛽𝑛1𝛽𝑗)𝑘𝑗1×𝐸𝛽(𝑚)𝑛𝛽𝑛1,𝛽𝑛+𝑛2𝑗=0𝛽𝑛1𝛽𝑗𝑘𝑗𝑎𝑛1𝑎𝑛𝐷𝛽𝑛𝛽𝑛1𝑓(𝜏)𝑑𝜏.(3.7)

For 𝐼2, by the initial conditions (2.10) we get 1𝑎𝑛𝑠=0(1)𝑠𝑎𝑛1𝑎𝑛𝐷𝛽𝑛1𝛽𝑛𝑎++0𝑎𝑛𝐷𝛽0𝛽𝑛𝑠𝑛𝑖=0𝑎𝑖𝑙𝑖1𝑗𝑖=0𝑦(𝑗𝑖)𝑡(0)𝛽𝑛𝛽𝑖+𝑗𝑖Γ1+𝛽𝑛𝛽𝑖+𝑗𝑖=1𝑎𝑛𝑠=0(1)𝑠𝑘0,𝑘1,,𝑘𝑛1𝑘00+𝑘1++𝑘𝑛1=𝑠𝑠!𝑘0!𝑘1!𝑘𝑛1!𝑎𝑛1𝑎𝑛𝑘𝑛1𝑎𝑛2𝑎𝑛𝑘𝑛2𝑎0𝑎𝑛𝑘0×𝐷𝑘𝑛1(𝛽𝑛1𝛽𝑛)+𝑘𝑛2(𝛽𝑛2𝛽𝑛)++𝑘0(𝛽0𝛽𝑛)𝑛𝑖=0𝑎𝑖𝑙𝑖1𝑗𝑖=0𝑐𝑖𝑗𝑖𝑡𝛽𝑛𝛽𝑖+𝑗𝑖Γ1+𝛽𝑛𝛽𝑖+𝑗𝑖.(3.8)

Using formulas (2.2) and (2.3), the above expression can be written as1𝑎𝑛𝑠=0(1)𝑠𝑘0,𝑘1,,𝑘𝑛1𝑘00+𝑘1++𝑘𝑛1=𝑠𝑠!𝑘0!𝑘1!𝑘𝑛1!𝑎𝑛1𝑎𝑛𝑘𝑛1𝑎𝑛2𝑎𝑛𝑘𝑛2𝑎0𝑎𝑛𝑘0×𝑛𝑖=0𝑎𝑖𝑙𝑖1𝑗𝑖=0𝑐𝑖𝑗𝑖Γ(𝒜)𝑡𝛽𝑛𝛽𝑖+𝑗𝑖+𝑘𝑛1(𝛽𝑛𝛽𝑛1)+𝑘𝑛2(𝛽𝑛𝛽𝑛2)++𝑘0(𝛽𝑛𝛽0)Γ(𝒜)Γ𝒜+𝑘𝑛1𝛽𝑛𝛽𝑛1+𝑘𝑛2𝛽𝑛𝛽𝑛2++𝑘0𝛽𝑛𝛽0=1𝑎𝑛𝑠=0(1)𝑠𝑘0,𝑘1,,𝑘𝑛2𝑘00+𝑘1++𝑘𝑛1=𝑠𝑠!𝑘0!𝑘1!𝑘𝑛1!𝑎𝑛1𝑎𝑛𝑘𝑛1𝑎𝑛2𝑎𝑛𝑘𝑛2𝑎0𝑎𝑛𝑘0×𝑛𝑖=0𝑎𝑖𝑙𝑖1𝑗𝑖=0𝑐𝑖𝑗𝑖𝑡𝛽𝑛𝛽𝑖+𝑗𝑖+𝑘𝑛1(𝛽𝑛𝛽𝑛1)+𝑘𝑛2(𝛽𝑛𝛽𝑛2)++𝑘0(𝛽𝑛𝛽0)Γ𝒜+𝑘𝑛1𝛽𝑛𝛽𝑛1+𝑘𝑛2𝛽𝑛𝛽𝑛2++𝑘0𝛽𝑛𝛽0=𝑛𝑖=0𝑎𝑖𝑙𝑖1𝑗𝑖=0𝑐𝑖𝑗𝑖𝐷(𝛽𝑖𝑗𝑖1)1𝑎𝑛𝑠=0(1)𝑠𝑘0,𝑘1,,𝑘𝑛1𝑘00+𝑘1++𝑘𝑛1=𝑠𝑠!𝑘0!𝑘1!𝑘𝑛1!×𝑎𝑛1𝑎𝑛𝑘𝑛1𝑎𝑛2𝑎𝑛𝑘𝑛2𝑎0𝑎𝑛𝑘0×𝑡𝛽𝑛1+𝑘𝑛1(𝛽𝑛𝛽𝑛1)+𝑘𝑛2(𝛽𝑛𝛽𝑛2)++𝑘0(𝛽𝑛𝛽0)Γ𝛽𝑛+𝑘𝑛1𝛽𝑛𝛽𝑛1+𝑘𝑛2𝛽𝑛𝛽𝑛2++𝑘0𝛽𝑛𝛽0,(3.9) where 𝒜 denotes 1+𝛽𝑛𝛽𝑖+𝑗𝑖.

Using Propositions 2.6 and 2.7, the above solution is equivalent to the following form: 𝑛𝑖=0𝑎𝑖𝑙𝑖1𝑗𝑖=0𝑐𝑖𝑗𝑖𝐷(𝛽𝑖𝑗𝑖1)1𝑎𝑛𝑘𝑛1=0𝑚=0𝑘0,𝑘1,,𝑘𝑛2𝑘00+𝑘1++𝑘𝑛2=𝑚(1)𝑘𝑛1+𝑚𝑘𝑛1!+𝑚𝑘0!𝑘1!𝑘𝑛1!×𝑎𝑛1𝑎𝑛𝑘𝑛1𝑎𝑛2𝑎𝑛𝑘𝑛2𝑎0𝑎𝑛𝑘0×𝑡𝛽𝑛1+𝑘𝑛1(𝛽𝑛𝛽𝑛1)+𝑘𝑛2(𝛽𝑛𝛽𝑛2)++𝑘0(𝛽𝑛𝛽0)Γ𝛽𝑛+𝑘𝑛1𝛽𝑛𝛽𝑛1+𝑘𝑛2𝛽𝑛𝛽𝑛2++𝑘0𝛽𝑛𝛽0=𝑛𝑖=0𝑎𝑖𝑙𝑖1𝑗𝑖=0𝑐𝑖𝑗𝑖𝐷(𝛽𝑖𝑗𝑖1)1𝑎𝑛𝑚=0(1)𝑚𝑚!𝑘0,𝑘1,,𝑘𝑛2𝑘00+𝑘1++𝑘𝑛2=𝑚𝑚!𝑘0!𝑘1!𝑘𝑛2!×𝑛2𝑟=0𝑎𝑟𝑎𝑛𝑘𝑟𝑡(𝛽𝑛𝛽𝑛1)𝑚+𝑛2𝜉=0(𝛽𝑛1𝛽𝜉)𝑘𝜉+𝛽𝑛1×𝑘𝑛1=0(1)𝑘𝑛1𝑎𝑛1𝑎𝑛𝑘𝑛1𝑘𝑛1!+𝑚𝑘𝑛1!×𝑡𝑘𝑛𝑙(𝛽𝑛𝛽𝑛1)Γ𝑘𝑛1𝛽𝑛𝛽𝑛1+𝛽𝑛𝛽𝑛1𝑚+𝑛2𝛿=0𝛽𝑛1𝛽𝛿𝑘𝛿+𝛽𝑛=𝑛𝑖=0𝑎𝑖𝑙𝑖1𝑗𝑖=0𝑐𝑖𝑗𝑖𝐷(𝛽𝑖𝑗𝑖1)1𝑎𝑛𝑚=0(1)𝑚𝑚!𝑘0,𝑘1,,𝑘𝑛2𝑘00+𝑘1++𝑘𝑛2=𝑚𝑚;𝑘0,𝑘1,,𝑘𝑛2×𝑛2𝑟=0𝑎𝑟𝑎𝑛𝑘𝑟𝑡(𝛽𝑛𝛽𝑛1)𝑚+𝑛2𝜉=0(𝛽𝑛1𝛽𝜉)𝑘𝜉+𝛽𝑛1×𝐸𝛽(𝑚)𝑛𝛽𝑛1,𝑛2𝛿=0𝛽𝑛1𝛽𝑙𝑘𝛿+𝛽𝑛𝑎𝑛1𝑎𝑛𝑡𝛽𝑛𝛽𝑛1.(3.10)

Therefore, 𝑦(𝑡)=𝐼1+𝐼2=𝑡01𝑎𝑛𝑚=0(1)𝑚𝑚!𝑘0,𝑘1,,𝑘𝑛2𝑘00+𝑘1++𝑘𝑛2=𝑚𝑚;𝑘0,𝑘1,,𝑘𝑛2×𝑛2𝑝=0𝑎𝑝𝑎𝑛𝑘𝑝(𝑡𝜏)(𝛽𝑛𝛽𝑛1)𝑚+𝛽𝑛+𝑛2𝑗=0(𝛽𝑛1𝛽𝑗)𝑘𝑗1×𝐸𝛽(𝑚)𝑛𝛽𝑛1,𝛽𝑛+𝑛2𝑗=0(𝛽𝑛1𝛽𝑗)𝑘𝑗𝑎𝑛1𝑎𝑛𝐷𝛽𝑛𝛽𝑛1+𝑓(𝜏)𝑑𝜏𝑛𝑖=0𝑎𝑖𝑙𝑖1𝑗𝑖=0𝑐𝑖𝑗𝑖𝐷(𝛽𝑖𝑗𝑖1)1𝑎𝑛𝑚=0(1)𝑚𝑚!𝑘0,𝑘1,,𝑘𝑛2𝑘00+𝑘1++𝑘𝑛2=𝑚𝑚;𝑘0,𝑘1,,𝑘𝑛2×𝑛2𝑟=0𝑎𝑟𝑎𝑛𝑘𝑟𝑡(𝛽𝑛𝛽𝑛1)𝑚+𝑛2𝜉=0(𝛽𝑛1𝛽𝜉)𝑘𝜉+𝛽𝑛1×𝐸𝛽(𝑚)𝑛𝛽𝑛1,𝑛2𝛿=0𝛽𝑛1𝛽𝑙𝑘𝛿+𝛽𝑛𝑎𝑛1𝑎𝑛𝑡𝛽𝑛𝛽𝑛1,(3.11) where 𝑚;𝑘0,𝑘1,,𝑘𝑛2=𝑚!𝑘0!𝑘1!𝑘𝑛2!,(3.12) and 𝐸(𝑖)𝜆,𝜇(𝑦) is the Mittag-Leffler function 𝐸(𝑖)𝜆,𝜇(𝑑𝑦)=𝑖𝑑𝑦𝑖𝐸𝜆,𝜇(𝑦)=𝑗=0(𝑖+𝑗)!𝑦𝑗.𝑗!Γ(𝜆𝑗+𝜆𝑖+𝜇)(3.13) Substituting the Green function 𝐺𝑛(1𝑡)=𝑎𝑛𝑚=0(1)𝑚𝑚!𝑘0,𝑘1,,𝑘𝑛2𝑘00+𝑘1++𝑘𝑛2=𝑚𝑚;𝑘0,𝑘1,,𝑘𝑛2×𝑛2𝑝=0𝑎𝑝𝑎𝑛𝑘𝑝𝑡(𝛽𝑛𝛽𝑛1)𝑚+𝛽𝑛+𝑛2𝑗=0(𝛽𝑛1𝛽𝑗)𝑘𝑗1×𝐸𝛽(𝑚)𝑛𝛽𝑛1,𝛽𝑛+𝑛2𝑗=0(𝛽𝑛1𝛽𝑗)𝑘𝑗𝑎𝑛1𝑎𝑛𝐷𝛽𝑛𝛽𝑛1(3.14) into the above expression, we know that𝑦(𝑡)=𝑡0𝐺𝑛(𝑡𝜏)𝑓(𝜏)𝑑𝜏+𝑛𝑖=0𝑎𝑖𝑙𝑖1𝑗𝑖=0𝑐𝑖𝑗𝑖𝐺(𝛽𝑖𝑗𝑖𝑛1)(𝑡)(3.15) is the analytical general solution of (1.1).

4. Illustrative Examples

In order to verify our conclusions, we give some examples.

(1). Consider an initial value problem for the relaxation-oscillation equation (see [1]) 𝐶𝐷𝛼𝑦𝑦(𝑡)+𝐴𝑦(𝑡)=𝑓(𝑡),𝑡>0,(𝑗)(0)=𝑏𝑗,𝑗=1,2,,𝑚1,𝑚1<𝛼<𝑚,(4.1) where 𝑏𝑗 are real constants.

Utilizing Lemma 2.5 and applying 𝐶𝐷𝛼 to both sides of (4.1), we obtain 𝑦(𝑡)+𝐴𝐷𝛼𝑦(𝑡)=𝐷𝛼𝑓(𝑡)+𝑚1𝑗=0𝑦(𝑗)𝑡(0)𝑗Γ.(𝑗+1)(4.2)

According to the above procedure of solving the linear fractional differential equations with constant coefficients and using the Adomian decomposition method, let 𝑦0(𝑡)=𝐷𝛼𝑓(𝑡)+𝑚1𝑗=0𝑦(𝑗)𝑡(0)𝑗Γ,𝑦(𝑗+1)1(𝑡)=𝐴𝐷𝛼𝑦0𝑦(𝑡),2(𝑡)=𝐴𝐷𝛼𝑦1(𝑡)=(𝐴)2𝐷2𝛼𝑦0𝑦(𝑡),𝑠(𝑡)=𝐴𝐷𝛼𝑦𝑠1(𝑡)=(𝐴)𝑠𝐷𝑠𝛼𝑦0(𝑡),(4.3)

Adding all of the above terms, we obtain the solution of the equation by Adomian decomposition method as follows: 𝑦(𝑡)=𝑠=0𝑦𝑠(=𝑡)𝑠=0(𝐴)𝑠𝐷𝑠𝛼𝑦0=(𝑡)𝑠=0(𝐴)𝑠𝐷(𝑠1)𝛼𝑓(𝑡)+𝑠=0(𝐴)𝑠𝐷𝑠𝛼𝑚1𝑗=0𝑦(𝑗)𝑡(0)𝑗=Γ(𝑗+1)𝑠=0(𝐴)𝑠1Γ((𝑠+1)𝛼)𝑡0(𝑡𝜏)(𝑠+1)𝛼1𝑓(𝜏)𝑑𝜏+𝑚1𝑗=0𝑏𝑗𝑠=0(𝐴)𝑠𝑡𝑗+𝑠𝛼=Γ(𝑗+𝑠𝛼+1)𝑡0𝑠=0(𝑡𝜏)𝛼1[𝐴(𝑡𝜏)𝛼]𝑠Γ(𝑠𝛼+𝛼)𝑓(𝜏)𝑑𝜏+𝑚1𝑗=0𝑏𝑗𝐷(𝛼𝑗1)𝑠=0𝑡𝛼1(𝐴𝑡𝛼)𝑠=Γ(𝑠𝛼+𝛼)𝑡0(𝑡𝜏)𝛼1𝐸𝛼,𝛼(𝐴(𝑡𝜏)𝛼)𝑓(𝜏)𝑑𝜏+𝑚1𝑗=0𝑏𝑗𝐷(𝛼𝑗1)𝑡𝛼1𝐸𝛼,𝛼(𝐴𝑡𝛼)=𝑡0𝐺2(𝑡𝜏)𝑓(𝜏)𝑑𝜏+𝑚1𝑗=0𝑏𝑗𝐷(𝛼𝑗1)𝐺2(𝑡),(4.4) where 𝐺2(𝑡)=𝑡𝛼1𝐸𝛼,𝛼(𝐴𝑡𝛼).

It is easy to see that 𝑦(𝑡)=𝑡0𝐺2(𝑡𝜏)𝑓(𝜏)𝑑𝜏+𝑏0𝐷(𝛼1)𝐺2(𝑡),(0<𝛼<1).(4.5)

(2). Consider an initial value problem for the nonhomogeneous Bagley-Torvik equation (see [5]) 𝐴𝑦(𝑡)+𝐵𝐶𝐷3/2𝑦𝑦(𝑡)+𝐶𝑦(𝑡)=𝑓(𝑡),𝑡>0,(𝑖)(0)=𝑎𝑖,𝑖=0,1,(4.6) where 𝑎𝑖are real constants.

Utilizing Lemma 2.5 and applying 𝐶𝐷2 to both sides of (4.6), we obtain 1𝑦(𝑡)+𝐴𝐵𝐷1/2𝑦(𝑡)+𝐶𝐷2=1𝑦(𝑡)𝐴𝐷2𝑓(𝑡)+1𝑖=0𝑦(𝑖)𝑡(0)𝑖+𝐵Γ(𝑖+1)𝐴1𝑖=0𝑦(𝑖)𝑡(0)1/2+𝑖=1Γ(3/2+𝑖)𝐴𝐷2𝑓(𝑡)+1𝑖=0𝑦(𝑖)𝑡(0)𝑖+𝐵Γ(𝑖+1)𝐴𝑡1/2+𝑖Γ(3/2+𝑖)(4.7)

According to the above procedure of solving the linear fractional differential equation with constant coefficients and using the Adomian decomposition method, let 𝑦01(𝑡)=𝐴𝐷2𝑓(𝑡)+1𝑖=0𝑦(𝑖)𝑡(0)𝑖Γ+𝐵(𝑖+1)𝐴𝑡1/2+𝑖Γ,𝑦(3/2+𝑖)11(𝑡)=𝐴𝐵𝐷1/2𝑦0(𝑡)+𝐶𝐷2𝑦0𝐶(𝑡)=𝐴𝐵𝐶𝐼+𝐷3/2𝐷1/2𝑦0𝑦(𝑡),2𝐶(𝑡)=𝐴𝐵𝐶𝐼+𝐷3/2𝐷1/2𝑦1𝐶(𝑡)=𝐴2𝐵𝐶𝐼+𝐷3/22𝐷2/2𝑦0𝑦(𝑡),𝑛𝐶(𝑡)=𝐴𝐵𝐶𝐼+𝐷3/2𝐷1/2𝑦𝑛1𝐶(𝑡)=𝐴𝑛𝐵𝐶𝐼+𝐷3/2𝑛𝐷𝑛/2𝑦0(𝑡),(4.8)

Adding all of the above terms, we obtain the solution of the equation by Adomian decomposition method as follows: 𝑦(𝑡)=𝑛=0𝑦𝑛(=𝑡)𝑛=0𝐶𝐴𝑛𝐵𝐶𝐼+𝐷3/2𝑛𝐷𝑛/2𝑦0=(𝑡)𝑛=0𝐶𝐴𝑛𝐵𝐶𝐼+𝐷3/2𝑛𝐷𝑛/21𝐴𝐷2+𝑓(𝑡)𝑛=0𝐶𝐴𝑛𝐵𝐶𝐼+𝐷3/2𝑛𝐷1𝑛/2𝑖=0𝑦(𝑖)(𝑡0)𝑖+𝐵Γ(𝑖+1)𝐴𝑡1/2+𝑖=1Γ(3/2+𝑖)𝐴𝑛=0𝐶𝐴𝑛𝑛𝑘=0𝐶𝑛𝑛𝑘𝐵𝐶𝐼𝑛𝑘𝐷3𝑘/2𝐷𝑛/2𝐷2+𝑓(𝑡)1𝑖=0𝑎𝑖𝑛=0𝐶𝐴𝑛𝑛𝑘=0𝐶𝑛𝑛𝑘𝐵𝐶𝐼𝑛𝑘𝐷3𝑘/2𝐷𝑛/2𝑡𝑖+𝐵Γ(𝑖+1)𝐴𝑡1/2+𝑖=1Γ(3/2+𝑖)𝐴𝑛=0𝐶𝐴𝑛𝑛𝑘=0𝑛!𝐵(𝑛𝑘)!𝑘!𝐶𝐼𝑛𝑘𝐷(3𝑘/2+𝑛/2+2)+𝑓(𝑡)1𝑖=0𝑎𝑖𝑛=0𝐶𝐴𝑛𝑛𝑘=0𝑛!𝐵(𝑛𝑘)!𝑘!𝐶𝐼𝑛𝑘𝐷(3𝑘/2+𝑛/2)𝑡𝑖+𝐵Γ(𝑖+1)𝐴𝑡1/2+𝑖=1Γ(3/2+𝑖)𝐴𝑡0𝑛=0𝐶𝐴𝑛𝑛𝑘=0𝑛!𝐵(𝑛𝑘)!𝑘!𝐶𝐼𝑛𝑘1Γ(3𝑘/2+𝑛/2+2)(𝑡𝜏)3𝑘/2+𝑛/2+21+𝑓(𝜏)𝑑𝜏1𝑖=0𝑎𝑖𝑛=0𝐶𝐴𝑛𝑛𝑘=0𝑛!𝐵(𝑛𝑘)!𝑘!𝐶𝐼𝑛𝑘𝑡𝑖+3𝑘/2+𝑛/2Γ+𝐵(𝑖+3𝑘/2+𝑛/2+1)𝐴𝑡1/2+𝑖+(3𝑘/2)+(𝑛/2)Γ=1(𝑖+3𝑘/2+𝑛/2+3/2)𝐴𝑡0𝑘=0𝑚=0𝐶𝐴𝑘+𝑚(𝑘+𝑚)!𝐵𝑚!𝑘!𝐶𝐼𝑚1Γ(3/2𝑘+(𝑘+𝑚/2)+2)×(𝑡𝜏)3𝑘/2+(𝑘+𝑚)/2+1+𝑓(𝜏)𝑑𝜏1𝑖=0𝑎𝑖𝐷(1𝑖)𝑘=0𝑚=0𝐶𝐴𝑘+𝑚(𝑘+𝑚)!𝐵𝑚!𝑘!𝐶𝐼𝑚𝑡3𝑘/2+(𝑘+𝑚)/2+1+Γ(3/2𝑘+(𝑘+𝑚)/2+2)1𝑖=0𝑎𝑖𝐷(1/2𝑖)𝑘=0𝑚=0𝐶𝐴𝑘+𝑚(𝑘+𝑚)!𝐵𝑚!𝑘!𝐶𝐼𝑚𝑡3𝑘/2+(𝑘+𝑚)/2+1=1Γ(3/2𝑘+(𝑘+𝑚)/2+2)𝐴𝑡0𝑘=0(𝐶/𝐴)𝑘𝑘!(𝑡𝜏)2𝑘+1𝑚=0(𝑘+𝑚)!𝑚!((𝐵/𝐴)(𝑡𝜏)1/2)𝑚Γ(𝑘/2+𝑚/2+3𝑘/2+2)𝑓(𝜏)𝑑𝜏+𝐴1𝑖=0𝑎𝑖𝐷(1𝑖)1𝐴𝑘=0(𝐶/𝐴)𝑘𝑘!(𝑡𝜏)2𝑘+1𝑚=0(𝑘+𝑚)!𝑚!(𝐵/𝐴)(𝑡𝜏)1/2𝑚Γ(𝑘/2+𝑚/2+3𝑘/2+2)+𝐴1𝑖=0𝑎𝑖𝐷(1/2𝑖)1𝐴𝑘=0(𝐶/𝐴)𝑘𝑘!(𝑡𝜏)2𝑘+1×𝑚=0(𝑘+𝑚)!𝑚!(𝐵/𝐴)(𝑡𝜏)1/2𝑚=1Γ((𝑘/2)+(𝑚/2)+(3𝑘/2)+2)𝐴𝑡0𝑘=0(𝐶/𝐴)𝑘𝑘!(𝑡𝜏)2𝑘+1𝐸(𝑘)1/2,3𝑘/2+2𝐵𝐴(𝑡𝜏)1/2𝑓(𝜏)𝑑𝜏+𝐴1𝑖=0𝑎𝑖𝐷(1𝑖)1𝐴𝑘=0(𝐶/𝐴)𝑘𝑡𝑘!2𝑘+1𝐸(𝑘)1/2,3𝑘/2+2𝐵𝐴𝑡1/2+𝐴1𝑖=0𝑎𝑖𝐷(1/2𝑖)1𝐴𝑘=0(𝐶/𝐴)𝑘𝑡𝑘!2𝑘+1𝐸(𝑘)1/2,3𝑘/2+2𝐵𝐴𝑡1/2=𝑡0𝐺3(𝑡𝜏)𝑓(𝜏)𝑑𝜏+𝐴1𝑖=0𝑎𝑖𝐷(1𝑖)𝐺3(𝑡)+𝐷(1/2𝑖)𝐺3,(𝑡)(4.9) where 𝐺3(𝑡)=(1/𝐴)𝑘=0((𝐶/𝐴)𝑘/𝑘!)𝑡2𝑘+1𝐸(𝑘)1/2,3𝑘/2+2((𝐵/𝐴)𝑡1/2).

Acknowledgment

The authors wish to thank the anonymous referees for their very careful reading of the manuscript and fruitful comments and suggestions. This work was supported by the National Natural Science Foundation of China (No. 60850005), the Natural Science Foundation of Zhejiang Province (Nos. D7080080 and Y607128) and the Innovation Team Foundation of the Department of Education of Zhejiang Province (No. T200924).