Hindawi Publishing Corporation Abstract and Applied Analysis Volume 2014, Article ID 230850, 6 pages http://dx.doi.org/10.1155/2014/230850
Research Article Solving Fractional Difference Equations Using the Laplace Transform Method Li Xiao-yan and Jiang Wei School of Mathematical Science, Anhui University, Hefei, Anhui 230601, China Correspondence should be addressed to Li Xiao-yan;
[email protected] Received 22 September 2013; Accepted 17 January 2014; Published 26 February 2014 Academic Editor: Stefan Siegmund Copyright © 2014 L. Xiao-yan and J. Wei. 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. We discuss the Laplace transform of the Caputo fractional difference and the fractional discrete Mittag-Leffer functions. On these bases, linear and nonlinear fractional initial value problems are solved by the Laplace transform method.
1. Introduction The study of continuous fractional calculus and equations has seen tremendous growth over the past few decades involving many aspects [1–4], such as initial value problem (IVP), boundary value problems (BVP), and stability of fractional equations. Compared with the continuous fractional calculus and fractional order differential equations, we can see that the research about the discrete fractional calculus and fractional difference equations has seen slower progress, but in recent years, a number of papers have appeared, and the study of the discrete fractional calculus and fractional difference equations has been arising. For example, Podlubny et al. [5], Holm [6], and Abdeljawad [7] have explored the definitions of fractional sum and difference operators and obtained many of their properties. Also, Atici and Eloe considered discrete fractional IVPs in paper [8]; moreover, discrete fractional BVPs were discussed in papers by Goodrich [9–11]. We know that the Laplace transform method has played an important role in solving basic problems of differential equations. Holm [6] developed properties of the Laplace transform in a discrete and applied the Laplace transform to solve a fractional initial value problem, which can be described as Δ]𝑎+]−𝑁𝑦 (𝑡) 𝑖
Δ 𝑦 (𝑎 + ] − 𝑁) = 𝐴 𝑖 ,
= 𝑓 (𝑡) ,
𝑡 ∈ 𝑁𝑎 ,
𝑖 = {0, 1, 2 . . . , 𝑁 − 1} ; 𝐴 𝑖 ∈ 𝑅. (1)
In this paper, we will discuss the Laplace transform of the Caputo fractional difference and the fractional discrete Mittag-Leffler functions and use the Laplace transform method to solve another kind of discrete fractional IVPs.
2. Preliminaries Let us start with some definitions and preliminaries. Definition 1 (see [8]). The generalized falling function is defined by 𝑡(𝜇) =
Γ (𝑡 + 1) , Γ (𝑡 + 1 − 𝜇)
for 𝑡, 𝜇 ∈ 𝑅,
(2)
where Γ denotes the special gamma function and 𝑡(𝜇) = 0 whenever 𝑡 + 1 − 𝜇 ∈ (−𝑁0 ). Here are some of the properties of the above fractional function: (i) Δ𝑡(𝜇) = 𝜇𝑡(𝜇−1) ; (ii) (𝑡 − 𝜇)𝑡(𝜇) = 𝜇𝑡(𝜇+1) ; (iii) 𝑡(𝜇+𝛼) = (𝑡 − 𝛼)(𝜇) 𝑡(𝛼) ; (iv) 𝜇(𝜇) = Γ(𝜇 + 1).
2
Abstract and Applied Analysis
Definition 2 (see [4]). The 𝛼th fractional sum of a function 𝑓, for 𝛼 > 0, is defined by Δ−𝛼 𝑓 (𝑡) =
1 𝑡−𝛼 ∑ (𝑡 − 𝑠 − 1)(𝛼−1) 𝑓 (𝑠) , Γ (𝛼) 𝑠=𝑎
(3)
for 𝑡 ∈ 𝑎 + 𝛼, 𝑎 + 𝛼 + 1, . . . =: 𝑁𝑎+𝛼 . Definition 3 (see [7]). The 𝛼-order Caputo left fractional difference is defined by Δ𝛼𝐶𝑓 (𝑡) = Δ−(𝑛−𝛼) Δ𝑛 𝑓 (𝑠) 𝑡−(𝑛−𝛼)
1 = ∑ (𝑡 − 𝑠 − 1)(𝑛−𝛼−1) Δ𝑛 𝑓 (𝑠) , Γ (𝑛 − 𝛼) 𝑠=𝑎
(4)
where 𝑛 = [𝛼] + 1, 𝑡 ∈ 𝑁𝑎+𝑛−𝛼 . If 𝛼 = 𝑛 ∈ 𝑁, then Δ𝛼𝐶𝑓(𝑡) = Δ𝑛𝐶𝑓(𝑡). In this paper, we will mainly discuss the problems involving the Caputo left fractional difference. Definition 4 (see [6]). The Laplace transform of the function 𝑓 on the time scale 𝑁𝑎 := 𝑁0 + 𝑎 = {𝑎, 𝑎 + 1, 𝑎 + 2, . . .}
(𝑎 ∈ 𝑅 fixed) ,
∞
𝑓 (𝑘 + 𝑎)
𝑘=0 (𝑠
+ 1)𝑘+1
.
(6)
exist for all 𝑠 ∈ 𝐶 \ 𝐵−1 (𝑟).
(8)
Let 𝑚 ∈ 𝑁0 be given and suppose 𝑓 : 𝑁𝑎−𝑚 → 𝑅 and 𝑔 : 𝑁𝑎 → 𝑅 are of exponential order 𝑟 > 0. Then for 𝑠 ∈ 𝐶 \ 𝐵−1 (𝑟),
(9) 𝐿 𝑎+𝑚 {𝑓} (𝑠) = (𝑠 + 1)𝑚 𝐿 𝑎 {𝑓} (𝑠) 𝑚−1
(10)
𝑘=0
Definition 6 (see [6]). For 𝑓, 𝑔 : 𝑁𝑎 → 𝑅, define the convolution of 𝑓 and 𝑔 by 𝑡−1
(𝑓 ∗ 𝑔) (𝑡) = ∑𝑓 (𝑟) 𝑔 (𝑡 − 1 − 𝑟 + 𝑎) , 𝑟=0
(𝑠 + 1)𝛼−𝑁 𝐿 𝑎 {𝑓} (𝑠) , 𝑠𝛼
𝐿 𝑎+𝑁−𝛼 {Δ𝛼 𝑓} (𝑠) = (𝑠 + 1)𝑁−𝛼 𝑠𝛼
(12)
𝑁−1
− ∑ 𝑠𝑗 Δ𝛼−1−𝑗 𝑓 (𝑎 + 𝑁 − 𝛼) . 𝑎 𝑗=0
Theorem 8. Suppose 𝑓 : 𝑁𝑎 → 𝑅 is of exponential order 𝑟 ≥ 1 and let 𝛼 > 0 be given. Then for each fixed 𝜖 > 0, Δ𝛼𝐶𝑓 is of exponential order 𝑟 + 𝜖 and 𝐿 𝑎+𝑁−𝛼 {Δ𝛼𝐶𝑓}(𝑠) converge for all 𝑠 ∈ 𝐶 \ 𝐵−1 (𝑟). Proof. Consider the relationship between Caputo fractional difference and Riemann-Liouville difference 𝑁−1
(𝑡 − 𝑎)(𝑘−𝛼) 𝑘 Δ 𝑓 (𝑎) , Γ (𝑘 − 𝛼 + 1) 𝑘=0
and in Lemma 7 we have 𝛼 Δ 𝐶𝑓 (𝑡) 𝑁−1 (𝑡 − 𝑎)(𝑘−𝛼) 𝑘 = Δ𝛼 𝑓 (𝑡) − ∑ Δ 𝑓 (𝑎) Γ (𝑘 − 𝛼 + 1) 𝑘=0 𝑁−1 (𝑡 − 𝑎)(𝑘−𝛼) 𝑘 𝛼 ≤ Δ 𝑓 (𝑡) + ∑ Δ 𝑓 (𝑎) 𝑘=0 Γ (𝑘 − 𝛼 + 1) 𝑁−1 Δ𝑘 𝑓 (𝑎) ≤ Δ𝛼 𝑓 (𝑡) + ∑ 𝑘=0 Γ (𝑘 − 𝛼 + 1) (𝑡 − 𝑎)(𝑘−𝛼)
(13)
(14)
𝑁
𝑁 ≤ ∑ (−1)𝑘 ( ) (𝑟 + 𝜖)𝑁−𝑘 (𝑟 + 𝜖)𝑡 𝑘 𝑘=0
𝑚−1 𝑓 (𝑘 + 𝑎 − 𝑚) 1 𝐿 𝑎−𝑚 {𝑓} (𝑠) = , 𝑚 𝐿 𝑎 {𝑓} (𝑠) + ∑ 𝑘+1 (𝑠 + 1) 𝑘=0 (𝑠 + 1)
− ∑ (𝑠 + 1)𝑚−1−𝑘 𝑔 (𝑘 + 𝑎) .
𝐿 𝑎+𝛼−𝑁 {Δ−𝛼 𝑓} (𝑠) =
Δ𝛼𝐶𝑓 (𝑡) = Δ𝛼 𝑓 (𝑡) − ∑
Definition 5 (see [6]). One says that a function 𝑓 : 𝑁𝑎 → 𝑅 is of exponential order 𝑟 > 0, if there exists a constant 𝐴 > 0 such that 𝑡 (7) 𝑓 (𝑡) ≤ 𝐴𝑟 , for sufficiently large 𝑡 ∈ 𝑁𝑎 . Via a geometric series, it is straightforward to show that if 𝑓 : 𝑁𝑎 → 𝑅 is of exponential order 𝑟 > 0, then 𝐿 {𝑓} (𝑠)
Lemma 7 (see [6]). Suppose 𝑓 : 𝑁𝑎 → 𝑅 is of exponential order 𝑟 ≥ 1 and let 𝛼 > 0 be given with 𝑁 − 1 < 𝛼 ≤ 𝑁. Then both 𝐿 𝑎+𝛼−𝑁{Δ−𝛼 𝑓}(𝑠) and 𝐿 𝑎−𝛼+𝑁{Δ𝛼 𝑓}(𝑠) converge for all 𝑠 ∈ 𝐶 \ 𝐵−1 (𝑟), and
(5)
is represented by 𝐿 𝑎 {𝑓} (𝑠) = ∑
3. The Laplace Transform of Caputo Fractional Difference
for 𝑡 ∈ 𝑁𝑎 . (11)
By a standard convolution on sums, it is understood that (𝑓 ∗ 𝑔)(𝑎) = 0.
𝑁−1 Δ𝑘 𝑓 (𝑎) 𝑁−1−𝛼 + ∑ , (𝑡 − 𝑎) 𝑘=0 Γ (𝑘 − 𝛼 + 1) since for sufficiently large 𝑇2 when 𝑡 > 𝑇2 , (𝑟 + 𝜖)𝑡 will eventually grow larger than the function (𝑡 − 𝑎)𝑁−1−𝛼 . So we get 𝛼 Δ 𝐶𝑓 (𝑡) 𝑁 𝑁−1 Δ𝑘 𝑓 (𝑎) 𝑁 𝑡 < ∑ (−1)𝑘 ( ) (𝑟 + 𝜖)𝑁−𝑘 + ∑ (𝑟 + 𝜖) 𝑘 𝑘=0 Γ − 𝛼 + 1) (𝑘 𝑘=0 = 𝐴(𝑟 + 𝜖)𝑡 . (15)
Abstract and Applied Analysis
3
Choose 𝑠0 ∈ 𝐶 \ 𝐵−1 (𝑟); there exists an 𝜖0 > 0 small enough so that 𝑠0 ∈ 𝐶 \ 𝐵−1 (𝑟 + 𝜖0 ), and Lemma 7 tells us that Δ𝛼𝐶𝑓(𝑡) is of exponential order 𝑟 + 𝜖0 , so it follows from (7) that 𝐿 𝑎+𝑁−𝛼 {Δ𝛼𝐶𝑓}(𝑠) is well defined. Theorem 9. Suppose 𝑓 : 𝑁𝑎 → 𝑅 is of exponential order 𝑟 ≥ 1 and let 𝛼 > 0 be given with 𝑁 − 1 < 𝛼 ≤ 𝑁. Then for 𝑠 ∈ 𝐶 \ 𝐵−1 (𝑟) 𝐿 𝑎+𝑁−𝛼 {Δ𝛼𝐶𝑓} (𝑠)
= (𝑠 + 1)
𝑁−𝛼 𝛼
𝑠 𝐿 𝑎 {𝑓} (𝑠)
𝑁−1
𝑠𝑗+𝛼−𝑁 Δ𝑁−1−𝑗 𝑓 (𝑎) . 𝛼−𝑁 𝑗=0 (𝑠 + 1)
−∑
(16)
Because 𝐿 𝑎+𝑁−𝛼 Δ and
−(𝑁−𝛼)
{𝑔}(𝑠) = ((𝑠 + 1)
𝑁−𝛼
𝑁−𝛼
/𝑠
And so 𝐿 𝑎 {𝐸𝛼,1 (𝜆, 𝑡)}𝑠 exists for 𝑠 ∈ 𝐶 \ 𝐵−1 (𝑟). Proof. For 𝑁 − 1 < 𝛼 < 𝑁, we have (𝑡 + (𝑘 − 1) (𝛼 − 1))(𝑘𝛼) (𝑡 + (𝑘 − 1) (𝑁 − 1))(𝐾𝑁) . (23) < Γ (𝑘𝑁 + 1) Γ (𝑘𝛼 + 𝛽) Moreover, for sufficiently large 𝑡 ∈ 𝑁𝑎 we get
Proof. Since Δ𝛼𝐶𝑓(𝑡) = Δ−(𝑁−𝛼) Δ𝑁𝑓(𝑡), then 𝐿 𝑎+𝑁−𝛼 {Δ𝛼𝐶𝑓} (𝑠) = 𝐿 𝑎+𝑁−𝛼 {Δ−(𝑁−𝛼) (Δ𝑁𝑓)} (𝑠) .
Theorem 11. We assume 𝛼 > 0 and 𝑁 − 1 < 𝛼 ≤ 𝑁 in (20); then for any fixed 𝑟 > 1 𝑡 for sufficient large 𝑡 ∈ 𝑁𝑎 , 𝐴 > 0. 𝐸𝛼,1 (𝜆, 𝑡) ≤ 𝐴𝑟 (22)
(17)
(𝑡 + (𝑘 − 1) (𝑁 − 1))(𝑘𝑁) Γ (𝑘𝑁 + 1) =
)𝐿 𝑎 {𝑔}(𝑠)
(𝑡 − 𝑁 + 1) ⋅ ⋅ ⋅ (𝑡 − 𝑁 + 1 + 𝐾𝑁) < Γ (𝑘𝑁 + 1)
𝑁−1
𝐿 𝑎 {Δ𝑁𝑓} (𝑠) = 𝑠𝑁𝐿 𝑎 {𝑓} (𝑠) − ∑ 𝑠𝑗 Δ𝑁−1−𝑗 𝑓 (𝑎) ,
(18)
𝑗=0
0, the discrete Mittag-Leffler functions are defined by ∞
𝑘=0
>1 ∞
(𝑡 − 𝑁 + 1) (𝑡 − 𝑁 + 1 + 𝑁) ⋅ ⋅ ⋅ 𝑁 (𝑡 − 𝑁 + 1 + 𝑘𝑁) (𝑘𝑁)! 𝑘=1
+∑
× 𝜆𝑘 .
𝐸𝛼,𝛽 (𝜆, 𝑡) = ∑ 𝜆𝑘
𝑁 (𝑡 − 𝑁 + 1) 𝑁 (𝑡 − 𝑁 + 1) ⋅ ⋅ ⋅ 𝑁 (𝑡 − 𝑁 + 𝑘) 𝑘 𝜆 (𝑘𝑁)! 𝑘=1
=1+∑
(𝑡 + (𝑘 − 1) (𝛼 − 1))(𝑘𝛼) (𝑡 + 𝑘 (𝛼 − 1))(𝛽−1) . Γ (𝑘𝛼 + 𝛽)
(20)
For 𝛽 = 1, it is written that ∞
𝐸𝛼,1 (𝜆, 𝑡) = ∑ 𝜆𝑘 𝑘=0
(𝑡 + (𝑘 − 1) (𝛼 − 1))(𝑘𝛼) . Γ (𝑘𝛼 + 1)
(21)
(26) Now, it is easy to see that |𝐸𝛼,1 (𝜆, 𝑡)| ≤ 𝐴𝑟𝑡 for sufficiently large 𝑡 ∈ 𝑁𝑎 , where 𝐴 > 1 is a constant and 𝑟 > |𝜆/(𝜆 − 1)|, so the function 𝐸𝛼 (𝜆, 𝑡) converges and 𝐿 𝑎 {𝐸𝛼,1 }(𝜆, 𝑠) exists for 𝑠 ∈ 𝐶 \ 𝐵−1 (𝑟). We will discuss the Laplace transform of the discrete Mittag-Leffler function 𝐸𝛼,1 (𝜆, 𝑡).
4
Abstract and Applied Analysis
Theorem 12. Let 𝛼 > 0, 𝑎 = 𝛼 − 1; then one gets 𝐿 𝑎 {𝐸𝛼 (𝜆, 𝑡)} (𝑠) =
𝛼−1
𝑠𝛼
𝑠 , − 𝜆(𝑠 + 1)𝛼−1
𝜆(𝑠 + 1)𝛼−1 < 𝑠𝛼 . (27)
Proof. When 𝑎 = 𝛼 − 1, we have
Theorem 13. Letting 𝛼 > 0, 𝑎 = 𝛼 − 1, then one has
𝐿 𝑎 {𝐸𝛼 (𝜆, 𝑡)} (𝑠) ∞
= 𝐿 𝛼−1 { ∑ 𝜆𝑘 𝑘=0
∞
= ∑ 𝐿 𝛼−1 {𝜆𝑘 𝑘=0
(𝑡 + (𝑘 − 1) (𝛼 − 1)) Γ (𝑘𝛼 + 1)
(𝑘𝛼)
𝐿 𝑎 {𝐸𝛼,𝛼 (𝜆, 𝑡)} (𝑠) = } (𝑠)
(28)
(𝑡 + (𝑘 − 1)) (𝛼 − 1)(𝑘𝛼) } (𝑠) . Γ (𝑘𝛼 + 1)
𝐿 𝑘+𝛼−1 {𝜆
∞
= ∑ 𝜆𝑘 𝑘=0
+ (𝑘 − 1) (𝛼 − 1))(𝑘𝛼) (𝑠 + 1)𝑘𝛼 } (𝑠) = 𝑘𝛼+1 ; Γ (𝑘𝛼 + 1) 𝑠
𝑙=0
∞
𝐸𝛼,𝛼 (𝜆, 𝑡) = ∑ 𝜆𝑘
𝐿 𝑎 {𝐸𝛼,𝛼 (𝜆, 𝑡)} (𝑠) = ∑ 𝜆𝑘 𝐿 𝛼−1 {
(𝑙 + 𝑘 (𝛼 − 1)) Γ (𝑘𝛼 + 1)
𝑘=0
. = (31)
So, we get
= (𝑠 + 1)𝑘 𝐿 𝛼−1 {
(𝑘𝛼)
(32) } (𝑠) .
Recalling (28), we have 𝐿 𝛼−1 {
(𝑡 + (𝑘 − 1) (𝛼 − 1)) Γ (𝑘𝛼 + 1)
} (𝑠) =
𝑘(𝛼−1)
(𝑠 + 1) 𝑠𝑘𝛼+1
.
(33)
𝑘=0
(𝑡 + (𝑘 − 1) (𝛼 − 1))(𝑘𝛼) } (𝑠) Γ (𝑘𝛼 + 1)
∞
𝑘(𝛼−1)
= ∑ 𝜆𝑘 𝑘=0
=
(𝑠 + 1) 𝑠𝑘𝛼+1 𝛼−1
𝑠 . 𝑠𝛼 − 𝜆(𝑠 + 1)𝛼−1
(𝑠 + 1)𝑘(𝛼−1)+𝛼−1 𝑠𝑘𝛼+𝛼
(𝑠 + 1)𝛼−1 ∞ 𝑘 (𝑠 + 1)𝑘(𝛼−1) ∑𝜆 𝑠𝛼 𝑠𝑘𝛼 𝑘=0 1 (𝑠 +
1)1−𝛼 𝑠𝛼
−𝜆
. (38)
5. Laplace Transform Method for Solving Fractional Difference Equation with Caputo Fractional Difference
Δ𝛼𝐶 𝑦 (𝑡) = 𝜆𝑦 (𝑡 + 𝛼 − 1) ,
𝑦 (𝑎) = 𝑎0 , 𝑡 ∈ 𝑁0 ,
(39)
where 𝜆 ∈ 𝑅, 0 < 𝛼 ≤ 1, and 𝑎 = 𝛼 − 1. The solution of (39) is given by Atici and Eloe in [8] using the method of successive approximation; we will give the solution of (39) by the method of Laplace transform.
𝐿 𝑎 {𝐸𝛼,1 (𝜆, 𝑡)} (𝑠) ∞
=
(𝑡 + 𝑘 (𝛼 − 1))(𝑘𝛼+𝛼−1) } (𝑠) Γ ((𝑘 + 1) 𝛼)
In this section, we first consider the following Caputo fractional difference equations:
Using (33), (27) can be rewritten as
= ∑ 𝜆𝑘 𝐿 𝛼−1 {
(37)
} (𝑠)
(𝑡 + (𝑘 − 1) (𝛼 − 1)) Γ (𝑘𝛼 + 1) (𝑘𝛼)
∞
= ∑ 𝜆𝑘
Γ (𝑙 + 𝑘 (𝛼 − 1) + 1) (𝑙 + 𝑘 (𝛼 − 1))(𝑘𝛼) = = 0. Γ (𝑘𝛼 + 1) Γ (𝑙 + 1 − 𝑘𝛼) Γ (𝑘𝛼 + 1) (𝑡 + (𝑘 − 1) (𝛼 − 1)) Γ (𝑘𝛼 + 1)
𝑘=0
(30)
For 0 ≤ 𝑙 ≤ 𝑘 − 1 Definition 1 implies that
𝐿 𝑘+𝛼−1 {
(𝑡 + 𝑘 (𝛼 − 1))(𝑘𝛼+𝛼−1) . Γ ((𝑘 + 1) 𝛼)
∞
(𝑘𝛼)
(𝑘𝛼)
(35)
Then
(𝑡 + (𝑘 − 1) (𝛼 − 1))(𝑘𝛼) } (𝑠) Γ (𝑘𝛼 + 1)
− ∑ (𝑠 + 1)𝑘−1−𝑙
.
(𝑡 + (𝑘 − 1) (𝛼 − 1))(𝑘𝛼) (𝑡 + 𝑘 (𝛼 − 1))(𝛼−1) (36) . Γ (𝑘𝛼 + 𝛼)
𝑘=0
(𝑡 + (𝑘 − 1) (𝛼 − 1))(𝑘𝛼) } (𝑠) Γ (𝑘𝛼 + 1)
𝑘−1
−𝜆
By property (iii) of the generalized falling function, we get
by (10), we conclude that
= (𝑠 + 1)𝑘 𝐿 𝛼−1 {
(𝑠 +
Proof. We recall that
(29)
𝐿 𝑘+𝛼−1 {
1 1)1−𝛼 𝑠𝛼
𝐸𝛼,𝛼 (𝜆, 𝑡)
From Lemma 7 we get 𝑘 (𝑡
When 𝛼 = 1, the result is 𝐿 0 {𝐸1,1 (𝜆, 𝑡)}(𝑠) = 1/(𝑠 − 𝜆), which coincided with integer order. With this in mind, let us discuss the Laplace transform of the Mittag-Leffler function 𝐸𝛼,𝛼 (𝜆, 𝑡); we will use this result in the following section.
Theorem 14. Equation (39) has its solution given by (34)
𝑦 (𝑡) = 𝑎0 𝐸𝛼,1 (𝜆, 𝑡) ,
𝑡 ∈ 𝑁0 .
(40)
Proof. Both sides of (39) carried out Laplace transform; we get 𝐿 𝑎+𝑁−𝛼 {Δ𝛼𝐶𝑦} (𝑠) = 𝜆𝐿 𝑎+𝑁−𝛼 {𝑦 (𝑡 + 𝛼 − 1)} (𝑠) .
(41)
Abstract and Applied Analysis
5
Equations (6) and (9) and Theorem 9 imply that 𝑠𝛼−1 𝑎0 = 𝜆𝐿 𝑎 {𝑦} (𝑠) . (𝑠 + 1)𝛼−1
(𝑠 + 1)1−𝛼 𝑠𝛼 𝐿 𝑎 {𝑦} (𝑠) −
Carrying out Laplace inverse transform of both sides of (45), according to (10), (28), (33), and (35), we have (42)
𝑡−1
Then 𝐿 𝑎 {𝑦} (𝑠) =
𝑠𝛼−1 𝑎0 . 𝑠𝛼 − 𝜆(𝑠 + 1)𝛼−1
+ ∑ 𝐸𝛼,𝛼 (𝜆, 𝑡 − 1 − 𝑟 + 𝑎) 𝑓 (𝑟 + 1 − 𝛼) .
𝑡 ∈ 𝑁0 .
Letting 𝑠 = 𝑟 − 𝑎 = 𝑟 − 𝛼 + 1, formula (45) yields 𝑡−𝛼
(44)
This result coincides with the result of paper [7], which obtains the solution of (40) using the method of successive approximation. Next, we consider the solution of Caputo nonhomogeneous difference equation Δ𝛼𝐶𝑦 (𝑡) = 𝜆𝑦 (𝑡 + 𝛼 − 1) + 𝑓 (𝑡) ,
(51)
𝑟=𝑎
(43)
Since Theorem 11, we have the solution of (39): 𝑦 (𝑡) = 𝑎0 𝐸𝛼,1 (𝜆, 𝑡) ,
𝑦 (𝑡) = 𝑎0 𝐸𝛼,1 (𝜆, 𝑡)
𝑦 (𝑎) = 𝑎0 , 𝑡 ∈ 𝑁0 , (45)
𝑦 (𝑡) = 𝑎0 𝐸𝛼,1 (𝜆, 𝑡) + ∑ 𝐸𝛼,𝛼 (𝜆, 𝑡 − 1 − 𝑠) 𝑓 (𝑠) ,
(52)
𝑠=0
which is the expression of the Caputo nonhomogeneous difference equation (45). In our future research work, we will consider the solution of fractional difference equations (39) and (45) in general situation: 𝑁 − 1 < 𝛼 ≤ 𝑁, 𝑁 > 1.
Conflict of Interests
where 𝑓 : 𝑁0 → 𝑅, 0 < 𝛼 ≤ 1, 𝑎 = 𝛼 − 1. The following standard rule for composing the Laplace transform with the convolution is necessary for solving the fractional initial value problems (45).
The authors declare that they have no conflict of interests regarding the publication of this paper.
Lemma 15 (see [6]). Let 𝑓, 𝑔 : 𝑁𝑎 → 𝑅 be of exponential order 𝑟 > 0. Then
This research had been supported by the National Nature Science Foundation of China (no. 11371027), Starting Research Fund for Doctors of Anhui University (no. 023033190249), National Natural Science Foundation of China, Tian Yuan Special Foundation (no. 11326115), and the Special Research Fund for the Doctoral Program of the Ministry of Education of China (no. 20123401120001)
𝐿 𝑎 {(𝑓 ∗ 𝑔)} (𝑠) = 𝐿 𝑎 {𝑓} (𝑠) 𝐿 𝑎 {𝑔} (𝑠) , (46) 𝑓𝑜𝑟 𝑠 ∈ 𝐶 \ (𝐵−1 (𝑟)) . Theorem 16. Let 𝑓(𝑡) : 𝑁𝑎 → 𝑅 of be exponential order 𝑟 > 0; then (45) has its solution given by 𝑡−𝛼
𝑦 (𝑡) = 𝑎0 𝐸𝛼,1 (𝜆, 𝑡) + ∑ 𝐸𝛼,𝛼 (𝜆, 𝑡 − 1 − 𝑠) 𝑓 (𝑠) .
(47)
𝑠=0
Proof. Using Laplace transform on both sides of (45), we obtain 𝐿 𝑎+𝑁−𝛼 {Δ𝛼𝐶𝑦} (𝑠) (48) = 𝜆𝐿 𝑎+𝑁−𝛼 {𝑦 (𝑡 + 𝛼 − 1)} (𝑠) + 𝐿 𝑎+𝑁−𝛼 {𝑓} (𝑠) , because 0 < 𝛼 ≤ 1; that is, 𝑁 = 1; similar to the above discussion, it is easy to obtain the following: 𝑠𝛼−1 𝑎0 (𝑠 + 1)𝛼−1 = 𝜆𝐿 𝑎 {𝑦} (𝑠) + 𝐿 𝑎 {𝑓 (𝑡 + 1 − 𝛼)} (𝑠) .
(𝑠 + 1)1−𝛼 𝑠𝛼 𝐿 𝑎 {𝑦} (𝑠) −
(49)
Then we obtain 𝐿 𝑎 {𝑦} (𝑠) =
𝑠𝛼 +
𝑠𝛼−1 𝑎0 − 𝜆(𝑠 + 1)𝛼−1 1 (𝑠 + 1)1−𝛼 𝑠𝛼 − 𝜆
(50) 𝐿 𝑎 {𝑓 (𝑡 + 1 − 𝛼)} (𝑠) .
Acknowledgments
References [1] I. Podlubny, Fractional Differential Equations, Academic Press, San Diego, Calif, USA, 1999. [2] A. A. Kilbas, H. M. Srivastava, and J. J. Trujillo, Theory and Applications of Fractional Differential Equations, Elsevier Science B.V., Amsterdam, The Netherlands, 2006. [3] V. Lakshmikantham, S. Leela, and J. V. Devi, Theory of Fractional Dynamic Systems, Cambridge Academic Publishers, Cambridge, UK, 2009. [4] K. S. Miller and B. Ross, An Introduction to the Fractional Calculus and Fractional Differential Equations, Wiley, New York, NY, USA, 1993. [5] I. Podlubny, A. Chechkin, T. Skovranek, Y. Chen, and B. M. Vinagre Jara, “Matrix approach to discrete fractional calculus. II. Partial fractional differential equations,” Journal of Computational Physics, vol. 228, no. 8, pp. 3137–3153, 2009. [6] M. T. Holm, “The Laplace transform in discrete fractional calculus,” Computers & Mathematics with Applications, vol. 62, no. 3, pp. 1591–1601, 2011. [7] T. Abdeljawad, “On Riemann and Caputo fractional differences,” Computers & Mathematics with Applications, vol. 62, no. 3, pp. 1602–1611, 2011. [8] F. M. Atici and P. W. Eloe, “Initial value problems in discrete fractional calculus,” Proceedings of the American Mathematical Society, vol. 137, no. 3, pp. 981–989, 2009.
6 [9] C. S. Goodrich, “Existence and uniqueness of solutions to a fractional difference equation with nonlocal conditions,” Computers Mathematics With Applications, vol. 62, pp. 1251– 1268, 2011. [10] C. S. Goodrich, “On discrete sequential fractional boundary value problems,” Journal of Mathematical Analysis and Applications, vol. 385, no. 1, pp. 111–124, 2012. [11] C. S. Goodrich, “Existence of a positive solution to a system of discrete fractional boundary value problems,” Applied Mathematics and Computation, vol. 217, no. 9, pp. 4740–4753, 2011.
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