# Solving Fractional Difference Equations Using the Laplace Transform Method

Authors: Li Xiao-yan; Jiang Wei
Journal: Abstract and Applied Analysis
 (2014)

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## Abstract

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.


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## Body

## 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
(1)Δa+ν-Nνy(t)=f(t),t∈Na,Δiy(a+ν-N)=Ai,i={0,1,2…,N-1};Ai∈R.
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(2)t(μ)=Γ(t+1)Γ(t+1-μ),fort,μ∈R,
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).Definition 2 (see [4]).
Theαth fractional sum of a function f, for α>0, is defined by
(3)Δ-αf(t)=1Γ(α)∑s=at-α‍(t-s-1)(α-1)f(s),
for t∈a+α, a+α+1,…=:Na+α.Definition 3 (see [7]).
Theα-order Caputo left fractional difference is defined by
(4)ΔCαf(t)=Δ-(n-α)Δnf(s)=1Γ(n-α)∑s=at-(n-α)(t-s-1)(n-α-1)Δnf(s),
where n=[α]+1, t∈Na+n-α. If α=n∈N, then ΔCαf(t)=ΔCnf(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 functionf on the time scale
(5)Na∶=N0+a={a,a+1,a+2,…}(a∈Rfixed),
is represented by
(6)La{f}(s)=∑k=0∞‍f(k+a)(s+1)k+1.Definition 5 (see [6]).
One says that a functionf:Na→R is of exponential order r>0, if there exists a constant A>0 such that
(7)|f(t)|≤Art,forsufficientlylarget∈Na.
Via a geometric series, it is straightforward to show that if f:Na→R is of exponential order r>0, then
(8)L{f}(s)existforalls∈C∖B-1(r)¯.Letm∈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)La-m{f}(s)=1(s+1)mLa{f}(s)+∑k=0m-1f(k+a-m)(s+1)k+1,(10)La+m{f}(s)=(s+1)mLa{f}(s)-∑k=0m-1(s+1)m-1-k‍g(k+a).Definition 6 (see [6]).
Forf,g:Na→R, define the convolution of f and g by
(11)(f*g)(t)=∑r=0t-1‍f(r)g(t-1-r+a),fort∈Na.
By a standard convolution on sums, it is understood that (f*g)(a)=0.


## 3. The Laplace Transform of Caputo Fractional Difference

Lemma 7 (see [6]).
Supposef:Na→R is of exponential order r≥1 and let α>0 be given with N-1<α≤N. Then both La+α-N{Δ-αf}(s) and La-α+N{Δαf}(s) converge for all s∈C∖B-1¯(r), and
(12)La+α-N{Δ-αf}(s)=(s+1)α-NsαLa{f}(s),La+N-α{Δαf}(s)=(s+1)N-αsα-∑j=0N-1‍sjΔaα-1-jf(a+N-α).Theorem 8.
Supposef:Na→R is of exponential order r≥1 and let α>0 be given. Then for each fixed ϵ>0, ΔCαf is of exponential order r+ϵ and La+N-α{ΔCαf}(s) converge for all s∈C∖B-1¯(r).Proof.
Consider the relationship between Caputo fractional difference and Riemann-Liouville difference(13)ΔCαf(t)=Δαf(t)-∑k=0N-1‍(t-a)(k-α)Γ(k-α+1)Δkf(a),
and in Lemma 7 we have
(14)|ΔCαf(t)|=|Δαf(t)-∑k=0N-1(t-a)(k-α)Γ(k-α+1)Δkf(a)|≤|Δαf(t)|+|∑k=0N-1(t-a)(k-α)Γ(k-α+1)Δkf(a)|≤|Δαf(t)|+|∑k=0N-1Δkf(a)Γ(k-α+1)(t-a)(k-α)|≤∑k=0N(-1)k(Nk)(r+ϵ)N-k(r+ϵ)t+|∑k=0N-1Δkf(a)Γ(k-α+1)|(t-a)N-1-α,
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
(15)|ΔCαf(t)|<|∑k=0N(-1)k(Nk)(r+ϵ)N-k+∑k=0N-1Δkf(a)Γ(k-α+1)|(r+ϵ)t=A(r+ϵ)t.
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 ΔCαf(t) is of exponential order r+ϵ0, so it follows from (7) that La+N-α{ΔCαf}(s) is well defined.Theorem 9.
Supposef: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)¯(16)La+N-α{ΔCαf}(s)=(s+1)N-αsαLa{f}(s)-∑j=0N-1sj+α-N(s+1)α-NΔN-1-jf(a).Proof.
SinceΔCαf(t)=Δ-(N-α)ΔNf(t), then
(17)La+N-α{ΔCαf}(s)=La+N-α{Δ-(N-α)(ΔNf)}(s).
Because La+N-αΔ-(N-α){g}(s)=((s+1)N-α/sN-α)La{g}(s) and
(18)La{ΔNf}(s)=sNLa{f}(s)-∑j=0N-1‍sjΔN-1-jf(a),
we get
(19)La+N-α{ΔCαf}(s)=(s+1)N-αsN-αLa{ΔNf}(s)=(s+1)N-αsN-α(sNLa{ΔNf}(s)-∑j=0N-1‍sjΔN-1-jf(a))=(s+1)N-αsαLa{f}(s)-∑j=0N-1‍sj+α-N(s+1)α-NΔN-1-jf(a).


## 4. The Laplace Transform of Discrete Mittag-Leffler Function

In paper [8], the discrete Mittag-Leffler function is introduced as the following form.Definition 10.
For any constantλ∈R and α,β,t∈C with Re(α)>0, the discrete Mittag-Leffler functions are defined by
(20)Eα,β(λ,t)=∑k=0∞λk(t+(k-1)(α-1))(kα)(t+k(α-1))(β-1)Γ(kα+β).
For β=1, it is written that
(21)Eα,1(λ,t)=∑k=0∞‍λk(t+(k-1)(α-1))(kα)Γ(kα+1).Theorem 11.
We assumeα>0 and N-1<α≤N in (20); then for any fixed r>1(22)|Eα,1(λ,t)|≤Artforsufficientlarget∈Na,A>0.
And so La{Eα,1(λ,t)}s exists for s∈C∖B-1¯(r).Proof.
ForN-1<α<N, we have
(23)(t+(k-1)(α-1))(kα)Γ(kα+β)<(t+(k-1)(N-1))(KN)Γ(kN+1).
Moreover, for sufficiently large t∈Na we get
(24)(t+(k-1)(N-1))(kN)Γ(kN+1)=(t-k-N+1)⋯((t+(k-1)(N-1)))Γ(kN+1)<(t-N+1)⋯(t-N+1+KN)Γ(kN+1)<(t-N+1)⋯(t-N+1+KN)Γ(kN+1);
then,
(25)|Eα,1(λ,t)|=|∑k=0∞‍λk(t+(k-1)(α-1))(kα)Γ(kα+1)|<|∑k=0∞‍λk(t-N+1)⋯(t-N+1+KN-1)Γ(kN+1)|.
We know that
(26)(1-λ-1)-(t-N+1)=1+∑k=1∞‍(t-N+1)(t-N+2)⋯(t-N+k)k!λk=1+∑k=1∞‍N(t-N+1)N(t-N+1)⋯N(t-N+k)(kN)!λk>1+∑k=1∞‍(t-N+1)(t-N+1+N)⋯N(t-N+1+kN)(kN)!×λk.
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 La{Eα,1}(λ,s) exists for s∈C∖B-1(r)¯.
We will discuss the Laplace transform of the discrete Mittag-Leffler functionEα,1(λ,t).Theorem 12.
Letα>0, a=α-1; then one gets
(27)La{Eα(λ,t)}(s)=sα-1sα-λ(s+1)α-1,|λ(s+1)α-1|<|sα|.Proof.
Whena=α-1, we have
(28)La{Eα(λ,t)}(s)=Lα-1{∑k=0∞‍λk(t+(k-1)(α-1))(kα)Γ(kα+1)}(s)=∑k=0∞‍Lα-1{λk(t+(k-1))(α-1)(kα)Γ(kα+1)}(s).
From Lemma 7 we get
(29)Lk+α-1{λk(t+(k-1)(α-1))(kα)Γ(kα+1)}(s)=(s+1)kαskα+1;
by (10), we conclude that
(30)Lk+α-1{(t+(k-1)(α-1))(kα)Γ(kα+1)}(s)=(s+1)kLα-1{(t+(k-1)(α-1))(kα)Γ(kα+1)}(s)-∑l=0k-1(s+1)k-1-l(l+k(α-1))(kα)Γ(kα+1).
For 0≤l≤k-1 Definition 1 implies that
(31)(l+k(α-1))(kα)Γ(kα+1)=Γ(l+k(α-1)+1)Γ(l+1-kα)Γ(kα+1)=0.
So, we get
(32)Lk+α-1{(t+(k-1)(α-1))(kα)Γ(kα+1)}(s)=(s+1)kLα-1{(t+(k-1)(α-1))(kα)Γ(kα+1)}(s).
Recalling (28), we have
(33)Lα-1{(t+(k-1)(α-1))(kα)Γ(kα+1)}(s)=(s+1)k(α-1)skα+1.
Using (33), (27) can be rewritten as
(34)La{Eα,1(λ,t)}(s)=∑k=0∞‍λkLα-1{(t+(k-1)(α-1))(kα)Γ(kα+1)}(s)=∑k=0∞‍λk(s+1)k(α-1)skα+1=sα-1sα-λ(s+1)α-1.
When α=1, the result is L0{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 functionEα,α(λ,t); we will use this result in the following section.Theorem 13.
Lettingα>0, a=α-1, then one has
(35)La{Eα,α(λ,t)}(s)=1(s+1)1-αsα-λ.Proof.
We recall that(36)Eα,α(λ,t)=∑k=0∞‍λk(t+(k-1)(α-1))(kα)(t+k(α-1))(α-1)Γ(kα+α).
By property (iii) of the generalized falling function, we get
(37)Eα,α(λ,t)=∑k=0∞‍λk(t+k(α-1))(kα+α-1)Γ((k+1)α).
Then
(38)La{Eα,α(λ,t)}(s)=∑k=0∞‍λkLα-1{(t+k(α-1))(kα+α-1)Γ((k+1)α)}(s)=∑k=0∞‍λk(s+1)k(α-1)+α-1skα+α=(s+1)α-1sα∑k=0∞‍λk(s+1)k(α-1)skα=1(s+1)1-αsα-λ.


## 5. Laplace Transform Method for Solving Fractional Difference Equation with Caputo Fractional Difference

In this section, we first consider the following Caputo fractional difference equations:(39)ΔCαy(t)=λy(t+α-1),y(a)=a0,t∈N0,
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.Theorem 14.
Equation (39) has its solution given by
(40)y(t)=a0Eα,1(λ,t),t∈N0.Proof.
Both sides of (39) carried out Laplace transform; we get
(41)La+N-α{ΔCαy}(s)=λLa+N-α{y(t+α-1)}(s).
Equations (6) and (9) and Theorem 9 imply that
(42)(s+1)1-αsαLa{y}(s)-sα-1(s+1)α-1a0=λLa{y}(s).
Then
(43)La{y}(s)=sα-1sα-λ(s+1)α-1a0.
Since Theorem 11, we have the solution of (39):
(44)y(t)=a0Eα,1(λ,t),t∈N0.
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(45)ΔCαy(t)=λy(t+α-1)+f(t),y(a)=a0,t∈N0,
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).Lemma 15 (see [6]).
Letf,g:Na→R be of exponential order r>0. Then
(46)La{(f*g)}(s)=La{f}(s)La{g}(s),La{(f*gg)}=fors∈C∖(B-1(r)).Theorem 16.
Letf(t):Na→R of be exponential order r>0; then (45) has its solution given by
(47)y(t)=a0Eα,1(λ,t)+∑s=0t-α‍Eα,α(λ,t-1-s)f(s).Proof.
Using Laplace transform on both sides of (45), we obtain
(48)La+N-α{ΔCαy}(s)=λLa+N-α{y(t+α-1)}(s)+La+N-α{f}(s),
because 0<α≤1; that is, N=1; similar to the above discussion, it is easy to obtain the following:
(49)(s+1)1-αsαLa{y}(s)-sα-1(s+1)α-1a0=λLa{y}(s)+La{f(t+1-α)}(s).
Then we obtain
(50)La{y}(s)=sα-1sα-λ(s+1)α-1a0+1(s+1)1-αsα-λLa{f(t+1-α)}(s).
Carrying out Laplace inverse transform of both sides of (45), according to (10), (28), (33), and (35), we have
(51)y(t)=a0Eα,1(λ,t)+∑r=at-1Eα,α(λ,t-1-r+a)f(r+1-α).
Letting s=r-a=r-α+1, formula (45) yields
(52)y(t)=a0Eα,1(λ,t)+∑s=0t-αEα,α(λ,t-1-s)f(s),
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.



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