Skip to main content

The existence of solutions for some fractional finite difference equations via sum boundary conditions

Abstract

In this manuscript we investigate the existence of the fractional finite difference equation (FFDE) Δ μ 2 μ x(t)=g(t+μ1,x(t+μ1),Δx(t+μ1)) via the boundary condition x(μ2)=0 and the sum boundary condition x(μ+b+1)= k = μ 1 α x(k) for order 1<μ2, where g: N μ 1 μ + b + 1 ×R×RR, α N μ 1 μ + b , and t N 0 b + 2 . Along the same lines, we discuss the existence of the solutions for the following FFDE: Δ μ 3 μ x(t)=g(t+μ2,x(t+μ2)) via the boundary conditions x(μ3)=0 and x(μ+b+1)=0 and the sum boundary condition x(α)= k = γ β x(k) for order 2<μ3, where g: N μ 2 μ + b + 1 ×RR, b N 0 , t N 0 b + 3 , and α,β,γ N μ 2 μ + b with γ<β<α.

MSC:34A08.

1 Introduction

By the late 19th century, combined efforts made by several mathematicians led to a fairly solid understanding of fractional calculus in the continuous setting but significantly less is still known about discrete fractional calculus (see for example [1, 2] and [3] and the references therein). Recently, there has been a strong interest in this subject but still little progress was made in developing the theory of fractional finite difference equations (see [411] and [12] and the references therein).

Discrete fractional calculus is a powerful tool for the processes which appears in nature, e.g. biology, ecology and other areas (see for example [13] and [14] and the references therein), where the discrete models have to be considered in order to describe properly the complexity of the dynamical processes with memory effect. We notice that the existence of solutions for fractional finite difference equations is a hot topic of the fractional calculus with direct implications in modeling of some real world phenomena which have only discrete behaviors.

Motivated by the above mentioned results, in this paper we investigate the fractional finite difference equation

Δ μ 2 μ x(t)=g ( t + μ 1 , x ( t + μ 1 ) , Δ x ( t + μ 1 ) )

via the boundary conditions x(μ2)=0 and x(μ+b+1)= k = μ 1 α x(k), where b N 0 , t N 0 b + 2 , 1<μ2, g: N μ 1 b + μ + 1 ×R×RR, and α N μ 1 b + μ .

Moreover, we investigate the FFDE given by

Δ μ 3 μ x(t)=g ( t + μ 2 , x ( t + μ 2 ) )

via the boundary conditions x(μ3)=0, x(μ+b+1)=0, and x(α)= k = γ β x(k), where b N 0 , t N 0 b + 3 , 2<μ3, g: N μ 2 μ + b + 1 ×RR and α,β,γ N μ 2 μ + b with γ<β<α.

In the following we present the basic definitions and theorems used in this manuscript. In Section 3 we present the main result. The manuscript ends with our conclusions.

2 Preliminaries

As you know, the gamma function is defined by

Γ(z)= 0 e t t z 1 dt,

which converges in the right half of the complex plane Re(z)>0. It is well known that Γ(z+1)=zΓ(z) and Γ(n)=(n1)! for all nN. Now, we define

t μ ̲ := Γ ( t + 1 ) Γ ( t + 1 μ )

for all t,μR [15]. If t+1μ is a pole of the gamma function and t+1 is not a pole, then we define t μ ̲ =0 [16]. For example, we have ( μ 2 ) μ 1 ̲ =0. Also, one can verify that μ μ ̲ = μ μ 1 ̲ =Γ(μ+1) and t μ + 1 ̲ t μ ̲ =tμ.

In this paper, we use the notations N p ={p,p+1,p+2,} for all pR and N p q ={p,p+1,p+2,,q} for all real numbers p and q whenever qp is a natural number.

Let μ>0 with m1<μ<m for some natural number m. The μ th fractional sum of f based at a is defined as [3]

Δ a μ f(t)= 1 Γ ( μ ) r = a t μ ( t σ ( r ) ) μ 1 ̲ f(r)

for all t N a + μ , where σ(r)=r+1 is the forward jump operator. Similarly, we define

Δ a μ f(t)= 1 Γ ( μ ) r = a t + μ ( t σ ( r ) ) μ 1 ̲ f(r)

for all t N a + m μ [17]. Note that the domain of Δ a r f is N a + m r for r>0 and N a r for r<0. Also, for the natural number μ=m, we have the known formula [16]

Δ a μ f(t)= Δ m f(t)= i = 0 m ( 1 ) i ( m i ) f(t+mi).

We define Δ a 0 f(t)=f(t) for all t N a , too.

Lemma 2.1 [16]

Let g: N a R be a mapping and m a natural number. Then the general solution of the equation Δ a + μ m μ x(t)=g(t) is given by

x(t)= i = 1 m C i ( t a ) μ i ̲ + Δ a μ g(t)

for all t N a + μ m , where C 1 ,, C m are arbitrary constants.

Let g: N μ 1 b + 1 + μ ×R×RR be a mapping and m a natural number. By using a similar proof, one can check that the general solution of the equation Δ μ m μ x(t)=g(t+μm+1,x(t+μm+1),Δx(t+μm+1)) is given by

x(t)= i = 1 m C i t μ i ̲ + Δ μ g ( t + μ m + 1 , x ( t + μ m + 1 ) , Δ x ( t + μ m + 1 ) )

for all t N μ m . In particular, the general solution has the following representation:

x ( t ) = i = 1 m C i t μ i ̲ + 1 Γ ( μ ) r = 0 t μ ( t σ ( r ) ) μ 1 ̲ × g ( r + μ m + 1 , x ( r + μ m + 1 ) , Δ x ( r + μ m + 1 ) )
(2.1)

for all t N μ m . The next theorem plays an important role in our main results.

Theorem 2.2 [18]

Every continuous function from a compact, convex, nonempty subset of a Banach space to itself has a fixed point.

3 Main results

In the following, we are ready to provide the main results. First, we investigate the FFDE

Δ μ 2 μ x(t)=g ( t + μ 1 , x ( t + μ 1 ) , Δ x ( t + μ 1 ) )

via the boundary conditions x(μ2)=0 and x(μ+b+1)= k = μ 1 α x(k), where b N 0 , t N 0 b + 2 , 1<μ2, g: N μ 1 b + μ + 1 ×R×RR, and α N μ 1 b + μ .

Lemma 3.1 Let b N 0 , t N 0 b + 2 , 1<μ2, g: N μ 1 b + μ + 1 ×R×RR, and α N μ 1 b + μ . Then x 0 is a solution of the problem

Δ μ 2 μ x(t)=g ( t + μ 1 , x ( t + μ 1 ) , Δ x ( t + μ 1 ) )

via the boundary conditions x(μ+b+1)= k = μ 1 α x(k) and x(μ2)=0 if and only if x 0 is a solution of the fractional sum equation

x(t)= r = 0 b + 1 G(t,r,α)g ( r + μ 1 , x ( r + μ 1 ) , Δ x ( r + μ 1 ) ) ,

where

G ( t , r , α ) = t μ 1 ̲ k = r + μ α ( k σ ( r ) ) μ 1 ̲ ( ( μ + b + 1 ) μ 1 ̲ 1 μ ( α + 1 ) μ ̲ ) Γ ( μ ) ( μ + b + 1 σ ( r ) ) μ 1 ̲ 1 μ ( α + 1 ) μ ̲ t μ 1 ̲ ( μ + b + 1 ) μ 1 ̲ Γ ( μ ) ( ( μ + b + 1 ) μ 1 ̲ 1 μ ( α + 1 ) μ ̲ ) ( μ + b + 1 σ ( r ) ) μ 1 ̲ t μ 1 ̲ Γ ( μ ) ( μ + b + 1 ) μ 1 ̲ + ( t σ ( r ) ) μ 1 ̲ Γ ( μ )

whenever r<tμαμ or rαμ<tμ,

G ( t , r , α ) = ( μ + b + 1 σ ( r ) ) μ 1 ̲ 1 μ ( α + 1 ) μ ̲ t μ 1 ̲ ( μ + b + 1 ) μ 1 ̲ Γ ( μ ) ( ( μ + b + 1 ) μ 1 ̲ 1 μ ( α + 1 ) μ ̲ ) ( μ + b + 1 σ ( r ) ) μ 1 ̲ t μ 1 ̲ Γ ( μ ) ( μ + b + 1 ) μ 1 ̲ + ( t σ ( r ) ) μ 1 ̲ Γ ( μ )

whenever αμ<rtμ,

G ( t , r , α ) = t μ 1 ̲ k = r + μ α ( k σ ( r ) ) μ 1 ̲ ( ( μ + b + 1 ) μ 1 ̲ 1 μ ( α + 1 ) μ ̲ ) Γ ( μ ) ( μ + b + 1 σ ( r ) ) μ 1 ̲ 1 μ ( α + 1 ) μ ̲ t μ 1 ̲ ( μ + b + 1 ) μ 1 ̲ Γ ( μ ) ( ( μ + b + 1 ) μ 1 ̲ 1 μ ( α + 1 ) μ ̲ ) ( μ + b + 1 σ ( r ) ) μ 1 ̲ t μ 1 ̲ Γ ( μ ) ( μ + b + 1 ) μ 1 ̲

whenever tμrαμ and

G ( t , r , α ) = ( μ + b + 1 σ ( r ) ) μ 1 ̲ 1 μ ( α + 1 ) μ ̲ t μ 1 ̲ ( μ + b + 1 ) μ 1 ̲ Γ ( μ ) ( ( μ + b + 1 ) μ 1 ̲ 1 μ ( α + 1 ) μ ̲ ) ( μ + b + 1 σ ( r ) ) μ 1 ̲ t μ 1 ̲ Γ ( μ ) ( μ + b + 1 ) μ 1 ̲

whenever tμαμ<r or αμtμ<r.

Proof Let x 0 be a solution of the problem

Δ μ 2 μ x(t)=g ( t + μ 1 , x ( t + μ 1 ) , Δ x ( t + μ 1 ) )

via the boundary conditions x(μ+b+1)= k = μ 1 α x(k) and x(μ2)=0. By using Lemma 2.1, we get

x 0 ( t ) = C 1 t μ 1 ̲ + C 2 t μ 2 ̲ + Δ μ g ( t + μ 1 , x 0 ( t + μ 1 ) , Δ x 0 ( t + μ 1 ) ) = C 1 t μ 1 ̲ + C 2 t μ 2 ̲ + 1 Γ ( μ ) r = 0 t μ ( t σ ( r ) ) μ 1 ̲ × g ( r + μ 1 , x 0 ( r + μ 1 ) , Δ x 0 ( r + μ 1 ) ) .

Since x 0 (μ2)=0, we have

0 = C 1 ( μ 2 ) μ 1 ̲ + C 2 ( μ 2 ) μ 2 ̲ + 1 Γ ( μ ) r = 0 ( μ 2 ) μ ( ( μ 2 ) σ ( r ) ) μ 1 ̲ × g ( r + μ 1 , x 0 ( r + μ 1 ) , Δ x 0 ( r + μ 1 ) ) .

Since ( μ 2 ) μ 1 ̲ =0 and

r = 0 2 ( ( μ 2 ) σ ( r ) ) μ 1 ̲ g ( r + μ 1 , x 0 ( r + μ 1 ) , Δ x 0 ( r + μ 1 ) ) =0,

C 2 =0. On the other hand, we have x 0 (μ+b+1)= k = μ 1 α x 0 (k). Thus,

k = μ 1 α x 0 ( k ) = C 1 ( μ + b + 1 ) μ 1 ̲ + 1 Γ ( μ ) r = 0 b + 1 ( μ + b + 1 σ ( r ) ) μ 1 ̲ × g ( r + μ 1 , x 0 ( r + μ 1 ) , Δ x 0 ( r + μ 1 ) ) .

Hence,

C 1 = 1 ( μ + b + 1 ) μ 1 ̲ [ k = μ 1 α x 0 ( k ) 1 Γ ( μ ) r = 0 b + 1 ( μ + b + 1 σ ( r ) ) μ 1 ̲ × g ( r + μ 1 , x 0 ( r + μ 1 ) , Δ x 0 ( r + μ 1 ) ) ]

and so

x 0 (t)= 1 ( μ + b + 1 ) μ 1 ̲ [ k = μ 1 α x 0 (k)
(3.1)
1 Γ ( μ ) r = 0 b + 1 ( μ + b + 1 σ ( r ) ) μ 1 ̲ g ( r + μ 1 , x 0 ( r + μ 1 ) , Δ x 0 ( r + μ 1 ) ) ] t μ 1 ̲ + 1 Γ ( μ ) r = 0 t μ ( t σ ( r ) ) μ 1 ̲ g ( r + μ 1 , x 0 ( r + μ 1 ) , Δ x 0 ( r + μ 1 ) ) .
(3.2)

To calculate k = μ 1 α x 0 (k), taking the summation k = μ 1 α on both sides of the above relation gives us

k = μ 1 α x 0 ( k ) = k = μ 1 α k μ 1 ̲ ( μ + b + 1 ) μ 1 ̲ k = μ 1 α x 0 ( k ) k = μ 1 α k μ 1 ̲ ( μ + b + 1 ) μ 1 ̲ Γ ( μ ) × r = 0 b + 1 ( μ + b + 1 σ ( r ) ) μ 1 ̲ g ( r + μ 1 , x 0 ( r + μ 1 ) , Δ x 0 ( r + μ 1 ) ) + k = μ 1 α 1 Γ ( μ ) r = 0 k μ ( k σ ( r ) ) μ 1 ̲ g ( r + μ 1 , x 0 ( r + μ 1 ) , Δ x 0 ( r + μ 1 ) ) .

Hence,

k = μ 1 α x 0 ( k ) [ 1 k = μ 1 α k μ 1 ̲ ( μ + b + 1 ) μ 1 ̲ ] = k = μ 1 α k μ 1 ̲ ( μ + b + 1 ) μ 1 ̲ Γ ( μ ) r = 0 b + 1 ( μ + b + 1 σ ( r ) ) μ 1 ̲ × g ( r + μ 1 , x 0 ( r + μ 1 ) , Δ x 0 ( r + μ 1 ) ) + k = μ 1 α 1 Γ ( μ ) r = 0 k μ ( k σ ( r ) ) μ 1 ̲ g ( r + μ 1 , x 0 ( r + μ 1 ) , Δ x 0 ( r + μ 1 ) ) ,

and so by interchanging the order of summations, we have

k = μ 1 α x 0 ( k ) = r = 0 α μ k = r + μ α ( k σ ( r ) ) μ 1 ̲ Γ ( μ ) 1 k = μ 1 α k μ 1 ̲ ( μ + b + 1 ) μ 1 ̲ g ( r + μ 1 , x 0 ( r + μ 1 ) , Δ x 0 ( r + μ 1 ) ) r = 0 b + 1 ( μ + b + 1 σ ( r ) ) μ 1 ̲ k = μ 1 α k μ 1 ̲ ( μ + b + 1 ) μ 1 ̲ Γ ( μ ) 1 k = μ 1 α k μ 1 ̲ ( μ + b + 1 ) μ 1 ̲ × g ( r + μ 1 , x 0 ( r + μ 1 ) , Δ x 0 ( r + μ 1 ) ) .
(3.3)

Since

k = μ 1 α k μ 1 ̲ = k = μ 1 α Δ k ( k μ ̲ μ ) = 1 μ ( ( α + 1 ) μ ̲ ( μ 1 ) μ ̲ ) = 1 μ ( α + 1 ) μ ̲ ,

by replacing (3.3) in (3.1), we get

x 0 ( t ) = r = 0 α μ t μ 1 ̲ k = r + μ α ( k σ ( r ) ) μ 1 ̲ ( ( μ + b + 1 ) μ 1 ̲ 1 μ ( α + 1 ) μ ̲ ) Γ ( μ ) g ( r + μ 1 , x 0 ( r + μ 1 ) , Δ x 0 ( r + μ 1 ) ) r = 0 b + 1 ( μ + b + 1 σ ( r ) ) μ 1 ̲ 1 μ ( α + 1 ) μ ̲ t μ 1 ̲ ( μ + b + 1 ) μ 1 ̲ Γ ( μ ) ( ( μ + b + 1 ) μ 1 ̲ 1 μ ( α + 1 ) μ ̲ ) × g ( r + μ 1 , x 0 ( r + μ 1 ) , Δ x 0 ( r + μ 1 ) ) r = 0 b + 1 ( μ + b + 1 σ ( r ) ) μ 1 ̲ t μ 1 ̲ Γ ( μ ) ( μ + b + 1 ) μ 1 ̲ g ( r + μ 1 , x 0 ( r + μ 1 ) , Δ x 0 ( r + μ 1 ) ) + r = 0 t μ ( t σ ( r ) ) μ 1 ̲ Γ ( μ ) g ( r + μ 1 , x 0 ( r + μ 1 ) , Δ x 0 ( r + μ 1 ) ) = r = 0 b + 1 G ( t , r , α ) g ( r + μ 1 , x 0 ( r + μ 1 ) , Δ x 0 ( r + μ 1 ) ) .

Now, let x 0 be a solution of the fractional sum equation

x(t)= r = 0 b + 1 G(t,r,α)g ( r + μ 1 , x ( r + μ 1 ) , Δ x ( r + μ 1 ) ) .

Then x 0 is a solution of the equation

x ( t ) = 1 ( μ + b + 1 ) μ 1 ̲ [ k = μ 1 α x ( k ) 1 Γ ( μ ) r = 0 b + 1 ( μ + b + 1 σ ( r ) ) μ 1 ̲ g ( r + μ 1 , x ( r + μ 1 ) , Δ x ( r + μ 1 ) ) ] t μ 1 ̲ + 1 Γ ( μ ) r = 0 t μ ( t σ ( r ) ) μ 1 ̲ g ( r + μ 1 , x ( r + μ 1 ) , Δ x ( r + μ 1 ) ) = 1 ( μ + b + 1 ) μ 1 ̲ [ k = μ 1 α x ( k ) 1 Γ ( μ ) r = 0 b + 1 ( μ + b + 1 σ ( r ) ) μ 1 ̲ × g ( r + μ 1 , x ( r + μ 1 ) , Δ x ( r + μ 1 ) ) ] t μ 1 ̲ + Δ μ g ( t + μ 1 , x ( t + μ 1 ) , Δ x ( t + μ 1 ) ) .

It is easy to check that x 0 (μ2)=0. Also, we have

x 0 ( μ + b + 1 ) = 1 ( μ + b + 1 ) μ 1 ̲ [ k = μ 1 α x 0 ( k ) 1 Γ ( μ ) r = 0 b + 1 ( μ + b + 1 σ ( r ) ) μ 1 ̲ × g ( r + μ 1 , x 0 ( r + μ 1 ) , Δ x 0 ( r + μ 1 ) ) ] ( μ + b + 1 ) μ 1 ̲ + 1 Γ ( μ ) r = 0 b + 1 ( μ + b + 1 σ ( r ) ) μ 1 ̲ g ( r + μ 1 , x 0 ( r + μ 1 ) , Δ x 0 ( r + μ 1 ) ) = k = μ 1 α x 0 ( k ) .

Moreover, we have Δ μ 2 μ x 0 (t)=g(t+μ1, x 0 (t+μ1),Δ x 0 (t+μ1)). This completes the proof. □

Some authors tried to find the maximum or exact value of r = 0 b + 1 |G(t,r,α)| in some papers (see for example [16, 19] and [17]). Now, we show that r = 0 b + 1 G(t,r,α) is bounded, where G(t,r,α) is the Green function of the last result.

Lemma 3.2 For each t N μ 2 b + 1 + μ and α N μ 1 b + μ , we have

| r = 0 b + 1 G(t,r,α)| r = 0 b + 1 |G(t,r,α)| M G

for some positive number M G <.

Proof Since Γ(α)>0 for all α>0, we have

( μ + b + 1 ) μ 1 ̲ = Γ ( μ + b + 2 ) ( b + 2 ) ! >0

for all μ>0 and b>0. Thus, G(t,r,α) is a (finite) real number, for all t N μ 2 b + 1 + μ , α N μ 1 b + μ and r N 0 b + 1 . Consequently, both sums in the statement are finite, because N 0 b + 1 is finite. □

Theorem 3.3 Let g: N μ 1 b + μ + 1 ×R×RR be bounded and continuous in its second and third variables. Then the fractional finite difference equation Δ μ 2 μ x(t)=g(t+μ1,x(t+μ1),Δx(t+μ1)) via the boundary conditions x(μ+b+1)= k = μ 1 α x(k) and x(μ2)=0 has a solution x 0 with x 0 (t)[ M G , M G ], for all admissible t.

Proof Since g is bounded, there exists a constant C such that |g(u,v,w)|C for all u N μ 1 b + μ + 1 and v,wR. Let X be the Banach space of real valued functions defined on N μ 1 μ + b + 1 via the norm

x=max { | x ( t ) | : t N μ 1 μ + b + 1 }

and K={xX:xC M G }. One can check easily that K is a compact, convex, and nonempty subset of X. Now, define the map T on K by

Tx(t)= r = 0 b + 1 G(t,r,α)g ( r + μ 1 , x ( r + μ 1 ) , Δ x ( r + μ 1 ) )

for all t N μ 1 μ + b + 1 . First, we show that T(K)K. Let xK and t N μ 1 μ + b + 1 . Then

| T x ( t ) | = | r = 0 b + 1 G ( t , r , α ) g ( r + μ 1 , x ( r + μ 1 ) , Δ x ( r + μ 1 ) ) | r = 0 b + 1 | G ( t , r , α ) | | g ( r + μ 1 , x ( r + μ 1 ) , Δ x ( r + μ 1 ) ) | C M G .

Since t N μ 2 μ + b + 1 was arbitrary, TxC M G and so T(K)K. Now, we show that T is continuous. Let ϵ>0 be given. Since g is continuous in its second and third variables, it is uniformly continuous in its second and third variables on [C M G ,C M G ] and so there exists δ>0 such that |g(t, u 1 , u 2 )g(t, v 1 , v 2 )|< ϵ M G for all t N μ 1 μ + b + 1 and u 1 , u 2 , v 1 , v 2 [C M G ,C M G ] with | u 1 v 1 |<δ and | u 2 v 2 |<δ. Thus, we get

| T y ( t ) T x ( t ) | = | r = 0 b + 1 G ( t , r , α ) g ( r + μ 1 , y ( r + μ 1 ) , Δ y ( r + μ 1 ) ) r = 0 b + 1 G ( t , r , α ) g ( r + μ 1 , x ( r + μ 1 ) , Δ x ( r + μ 1 ) ) | r = 0 b + 1 | G ( t , r , α ) | | g ( r + μ 1 , y ( r + μ 1 ) , Δ y ( r + μ 1 ) ) g ( r + μ 1 , x ( r + μ 1 ) , Δ x ( r + μ 1 ) ) | r = 0 b + 1 | G ( t , r , α ) | ϵ M G M G ϵ M G = ϵ

for all t N μ 1 μ + b + 1 . Hence, TxTy<ϵ and so T is continuous on K. By using Theorem 2.2, T has a fixed point x 0 and so, by using Lemma 3.1, the fractional finite difference equation

Δ μ 2 μ x(t)=g ( t + μ 1 , x ( t + μ 1 ) , Δ x ( t + μ 1 ) )

via the boundary conditions x(μ+b+1)= k = μ 1 α x(k) and x(μ2)=0 has a solution in [ M G , M G ]. □

Now, we consider the fractional finite difference equation Δ μ 3 μ x(t)=g(t+μ2,x(t+μ2)) via the boundary conditions x(μ3)=0, x(μ+b+1)=0 and x(α)= k = γ β x(k), where 2<μ3 and α,β,γ N μ 2 μ + b with γ<β<α.

Lemma 3.4 Let b N 0 , t N 0 b + 3 , 2<μ3, g: N μ 2 b + μ + 1 ×RR, and α,β,γ N μ 2 μ + b with γ<β<α. Then x 0 is a solution of the problem Δ μ 3 μ x(t)=g(t+μ2,x(t+μ2)) via the boundary conditions x(μ+b+1)=0, x(α)= k = γ β x(k), and x(μ3)=0 if and only if x 0 is a solution of the fractional sum equation

x(t)= r = 0 b + 1 G(t,r,β,α)g ( r + μ 2 , x ( r + μ 2 ) ) ,

where

G ( t , r , β , α ) = ( t μ 2 ̲ t μ 1 ̲ ) ( μ + b r ) μ 1 ̲ ( α μ 1 ̲ ( b α + μ + 1 ) 1 μ 1 ( ( β + 1 ) μ 1 ̲ γ μ 1 ̲ ) 1 μ ( ( β + 1 ) μ ̲ γ μ ̲ ) ) × ( 1 μ 1 ( ( β + 1 ) μ 1 ̲ γ μ 1 ̲ ) ( α μ + 2 ) 1 μ ( ( β + 1 ) μ ̲ γ μ ̲ ) ) ( b α + μ + 1 ) ( μ + b + 1 ) μ 2 ̲ Γ ( μ ) + ( t μ 2 ̲ ( α μ + 2 ) t μ 1 ̲ ) ( μ + b r ) μ 1 ̲ ( b α + μ + 1 ) ( μ + b + 1 ) μ 2 ̲ Γ ( μ ) + ( t μ 2 ̲ t μ 1 ̲ ) ( α r 1 ) μ 1 ̲ ( α μ 1 ̲ ( b α + μ + 1 ) ( 1 μ 1 ( ( β + 1 ) μ 1 ̲ γ μ 1 ̲ ) 1 μ ( ( β + 1 ) μ ̲ γ μ ̲ ) ) ) × ( 1 μ ( ( β + 1 ) μ ̲ γ μ ̲ ) ( b + 3 ) 1 μ 1 ( ( β + 1 ) μ 1 ̲ γ μ 1 ̲ ) ) α μ 1 ̲ ( b α + μ + 1 ) Γ ( μ ) + ( t μ 1 ̲ ( b + 3 ) t μ 2 ̲ ) ( α r 1 ) μ 1 ̲ α μ 1 ̲ ( b α + μ + 1 ) Γ ( μ ) + ( t μ 2 ̲ t μ 1 ̲ ) k = r + μ β ( k σ ( r ) ) μ 1 ̲ ( α μ 1 ̲ ( b α + μ + 1 ) ( 1 μ 1 ( ( β + 1 ) μ 1 ̲ γ μ 1 ̲ ) 1 μ ( ( β + 1 ) μ ̲ γ μ ̲ ) ) ) Γ ( μ ) + ( t σ ( r ) ) μ 1 ̲ Γ ( μ )

whenever rtμ, rβμ,

G ( t , r , β , α ) = ( t μ 2 ̲ t μ 1 ̲ ) ( μ + b r ) μ 1 ̲ ( α μ 1 ̲ ( b α + μ + 1 ) 1 μ 1 ( ( β + 1 ) μ 1 ̲ γ μ 1 ̲ ) 1 μ ( ( β + 1 ) μ ̲ γ μ ̲ ) ) × ( 1 μ 1 ( ( β + 1 ) μ 1 ̲ γ μ 1 ̲ ) ( α μ + 2 ) 1 μ ( ( β + 1 ) μ ̲ γ μ ̲ ) ) ( b α + μ + 1 ) ( μ + b + 1 ) μ 2 ̲ Γ ( μ ) + ( t μ 2 ̲ ( α μ + 2 ) t μ 1 ̲ ) ( μ + b r ) μ 1 ̲ ( b α + μ + 1 ) ( μ + b + 1 ) μ 2 ̲ Γ ( μ ) + ( t μ 2 ̲ t μ 1 ̲ ) ( α r 1 ) μ 1 ̲ ( α μ 1 ̲ ( b α + μ + 1 ) ( 1 μ 1 ( ( β + 1 ) μ 1 ̲ γ μ 1 ̲ ) 1 μ ( ( β + 1 ) μ ̲ γ μ ̲ ) ) ) × ( 1 μ ( ( β + 1 ) μ ̲ γ μ ̲ ) ( b + 3 ) 1 μ 1 ( ( β + 1 ) μ 1 ̲ γ μ 1 ̲ ) ) α μ 1 ̲ ( b α + μ + 1 ) Γ ( μ ) + ( t μ 1 ̲ ( b + 3 ) t μ 2 ̲ ) ( α r 1 ) μ 1 ̲ α μ 1 ̲ ( b α + μ + 1 ) Γ ( μ ) + ( t μ 2 ̲ t μ 1 ̲ ) k = r + μ β ( k σ ( r ) ) μ 1 ̲ ( α μ 1 ̲ ( b α + μ + 1 ) ( 1 μ 1 ( ( β + 1 ) μ 1 ̲ γ μ 1 ̲ ) 1 μ ( ( β + 1 ) μ ̲ γ μ ̲ ) ) ) Γ ( μ )

whenever tμ<r, rβμ,

G ( t , r , β , α ) = ( t μ 2 ̲ t μ 1 ̲ ) ( μ + b r ) μ 1 ̲ ( α μ 1 ̲ ( b α + μ + 1 ) 1 μ 1 ( ( β + 1 ) μ 1 ̲ γ μ 1 ̲ ) 1 μ ( ( β + 1 ) μ ̲ γ μ ̲ ) ) × ( 1 μ 1 ( ( β + 1 ) μ 1 ̲ γ μ 1 ̲ ) ( α μ + 2 ) 1 μ ( ( β + 1 ) μ ̲ γ μ ̲ ) ) ( b α + μ + 1 ) ( μ + b + 1 ) μ 2 ̲ Γ ( μ ) + ( t μ 2 ̲ ( α μ + 2 ) t μ 1 ̲ ) ( μ + b r ) μ 1 ̲ ( b α + μ + 1 ) ( μ + b + 1 ) μ 2 ̲ Γ ( μ ) + ( t μ 2 ̲ t μ 1 ̲ ) ( α r 1 ) μ 1 ̲ ( α μ 1 ̲ ( b α + μ + 1 ) ( 1 μ 1 ( ( β + 1 ) μ 1 ̲ γ μ 1 ̲ ) 1 μ ( ( β + 1 ) μ ̲ γ μ ̲ ) ) ) × ( 1 μ ( ( β + 1 ) μ ̲ γ μ ̲ ) ( b + 3 ) 1 μ 1 ( ( β + 1 ) μ 1 ̲ γ μ 1 ̲ ) ) α μ 1 ̲ ( b α + μ + 1 ) Γ ( μ ) + ( t μ 1 ̲ ( b + 3 ) t μ 2 ̲ ) ( α r 1 ) μ 1 ̲ α μ 1 ̲ ( b α + μ + 1 ) Γ ( μ )

whenever tμ<r, βμ<r, rαμ,

G ( t , r , β , α ) = ( t μ 2 ̲ t μ 1 ̲ ) ( μ + b r ) μ 1 ̲ ( α μ 1 ̲ ( b α + μ + 1 ) 1 μ 1 ( ( β + 1 ) μ 1 ̲ γ μ 1 ̲ ) 1 μ ( ( β + 1 ) μ ̲ γ μ ̲ ) ) × ( 1 μ 1 ( ( β + 1 ) μ 1 ̲ γ μ 1 ̲ ) ( α μ + 2 ) 1 μ ( ( β + 1 ) μ ̲ γ μ ̲ ) ) ( b α + μ + 1 ) ( μ + b + 1 ) μ 2 ̲ Γ ( μ ) + ( t μ 2 ̲ ( α μ + 2 ) t μ 1 ̲ ) ( μ + b r ) μ 1 ̲ ( b α + μ + 1 ) ( μ + b + 1 ) μ 2 ̲ Γ ( μ ) + ( t σ ( r ) ) μ 1 ̲ Γ ( μ )

whenever rtμ, αμ<r and

G ( t , r , β , α ) = ( t μ 2 ̲ t μ 1 ̲ ) ( μ + b r ) μ 1 ̲ ( α μ 1 ̲ ( b α + μ + 1 ) 1 μ 1 ( ( β + 1 ) μ 1 ̲ γ μ 1 ̲ ) 1 μ ( ( β + 1 ) μ ̲ γ μ ̲ ) ) × ( 1 μ 1 ( ( β + 1 ) μ 1 ̲ γ μ 1 ̲ ) ( α μ + 2 ) 1 μ ( ( β + 1 ) μ ̲ γ μ ̲ ) ) ( b α + μ + 1 ) ( μ + b + 1 ) μ 2 ̲ Γ ( μ ) + ( t μ 2 ̲ ( α μ + 2 ) t μ 1 ̲ ) ( μ + b r ) μ 1 ̲ ( b α + μ + 1 ) ( μ + b + 1 ) μ 2 ̲ Γ ( μ )

whenever tμ<r, αμ<r.

Proof Let x 0 be a solution of the problem Δ μ 3 μ x(t)=g(t+μ2,x(t+μ2)) via the boundary conditions x(μ+b+1)=0, x(α)= k = γ β x(k), and x(μ3)=0. By using Lemma 2.1, we get

x 0 (t)= C 1 t μ 1 ̲ + C 2 t μ 2 ̲ + C 3 t μ 3 ̲ + 1 Γ ( μ ) r = 0 t μ ( t σ ( r ) ) μ 1 ̲ g ( r + μ 2 , x 0 ( r + μ 2 ) ) .

Similar to the proof of Lemma 3.1, by using the boundary value conditions we obtain C 3 =0,

C 1 = 1 α μ 1 ̲ ( b α + μ + 1 ) k = γ β x 0 ( k ) 1 ( b α + μ + 1 ) ( μ + b + 1 ) μ 1 ̲ Γ ( μ ) r = 0 b + 1 ( μ + b r ) μ 1 ̲ g ( r + μ 2 , x 0 ( r + μ 2 ) ) + 1 α μ 1 ̲ ( b α + μ + 1 ) Γ ( μ ) r = 0 α μ ( α r 1 ) μ 1 ̲ g ( r + μ 2 , x 0 ( r + μ 2 ) )

and

C 2 = b + 3 α μ 1 ̲ ( b α + μ + 1 ) k = γ β x 0 ( k ) + α μ + 2 ( b α + μ + 1 ) ( μ + b + 1 ) μ 2 ̲ Γ ( μ ) r = 0 b + 1 ( μ + b r ) μ 1 ̲ g ( r + μ 2 , x 0 ( r + μ 2 ) ) b + 3 α μ 1 ̲ ( b α + μ + 1 ) Γ ( μ ) r = 0 α μ ( α r 1 ) μ 1 ̲ g ( r + μ 2 , x 0 ( r + μ 2 ) ) .

Thus,

x 0 ( t ) = t μ 2 ̲ t μ 1 ̲ α μ 1 ̲ ( b α + μ + 1 ) k = γ β x 0 ( k ) + t μ 2 ̲ ( α μ + 2 ) t μ 1 ̲ ( b α + μ + 1 ) ( μ + b + 1 ) μ 2 ̲ Γ ( μ ) r = 0 b + 1 ( μ + b r ) μ 1 ̲ × g ( r + μ 2 , x 0 ( r + μ 2 ) ) + t μ 1 ̲ t μ 2 ̲ ( b + 3 ) α μ 1 ̲ ( b α + μ + 1 ) Γ ( μ ) r = 0 α μ ( α r 1 ) μ 1 ̲ × g ( r + μ 2 , x 0 ( r + μ 2 ) ) + 1 Γ ( μ ) r = 0 t μ ( t σ ( r ) ) μ 1 ̲ g ( r + μ 2 , x 0 ( r + μ 2 ) ) .
(3.4)

To calculate k = γ β x 0 (k), by taking the summation k = γ β on both sides of the above relation gives us

k = γ β x 0 ( k ) = k = γ β k μ 2 ̲ k = γ β k μ 1 ̲ α μ 1 ̲ ( b α + μ + 1 ) k = γ β x 0 ( k ) + k = γ β k μ 2 ̲ ( α μ + 2 ) k = γ β k μ 1 ̲ ( b α + μ + 1 ) ( μ + b + 1 ) μ 2 ̲ Γ ( μ ) r = 0 b + 1 ( μ + b r ) μ 1 ̲ × g ( r + μ 2 , x 0 ( r + μ 2 ) ) + k = γ β k μ 1 ̲ ( b + 3 ) k = γ β k μ 2 ̲ α μ 1 ̲ ( b α + μ + 1 ) Γ ( μ ) r = 0 α μ ( α r 1 ) μ 1 ̲ × g ( r + μ 2 , x 0 ( r + μ 2 ) ) + 1 Γ ( μ ) r = 0 β μ k = μ + s β ( k σ ( r ) ) μ 1 ̲ g ( r + μ 2 , x 0 ( r + μ 2 ) )

and so

k = γ β x 0 ( k ) = r = 0 b + 1 ( μ + b r ) μ 1 ̲ ( k = γ β k μ 2 ̲ ( α μ + 2 ) k = γ β k μ 1 ̲ ) ( 1 k = γ β k μ 2 ̲ k = γ β k μ 1 ̲ α μ 1 ̲ ( b α + μ + 1 ) ) ( b α + μ + 1 ) ( μ + b + 1 ) μ 2 ̲ Γ ( μ ) × g ( r + μ 2 , x 0 ( r + μ 2 ) ) + r = 0 α μ ( α r 1 ) μ 1 ̲ ( k = γ β k μ 1 ̲ ( b + 3 ) k = γ β k μ 2 ̲ ) ( 1 k = γ β k μ 2 ̲ k = γ β k μ 1 ̲ α μ 1 ̲ ( b α + μ + 1 ) ) α μ 1 ̲ ( b α + μ + 1 ) Γ ( μ ) × g ( r + μ 2 , x 0 ( r + μ 2 ) ) + r = 0 β μ k = μ + s β ( k σ ( r ) ) μ 1 ̲ ( 1 k = γ β k μ 2 ̲ k = γ β k μ 1 ̲ α μ 1 ̲ ( b α + μ + 1 ) ) Γ ( μ ) g ( r + μ 2 , x 0 ( r + μ 2 ) ) .
(3.5)

Since k = γ β k μ 1 ̲ = k = γ β Δ k ( k μ ̲ μ )= 1 μ ( ( β + 1 ) μ ̲ γ μ ̲ ) and

k = γ β k μ 2 ̲ = k = γ β Δ k ( k μ 1 ̲ μ 1 ) = 1 μ 1 ( ( β + 1 ) μ 1 ̲ γ μ 1 ̲ ) ,

by replacing (3.5) in (3.4), we get

x 0 ( t ) = r = 0 b + 1 [ ( t μ 2 ̲ t μ 1 ̲ ) ( μ + b r ) μ 1 ̲ ( α μ 1 ̲ ( b α + μ + 1 ) 1 μ 1 ( ( β + 1 ) μ 1 ̲ γ μ 1 ̲ ) 1 μ ( ( β + 1 ) μ ̲ γ μ ̲ ) ) × ( 1 μ 1 ( ( β + 1 ) μ 1 ̲ γ μ 1 ̲ ) ( α μ + 2 ) 1 μ ( ( β + 1 ) μ ̲ γ μ ̲ ) ) ( b α + μ + 1 ) ( μ + b + 1 ) μ 2 ̲ Γ ( μ ) + ( t μ 2 ̲ ( α μ + 2 ) t μ 1 ̲ ) ( μ + b r ) μ 1 ̲ ( b α + μ + 1 ) ( μ + b + 1 ) μ 2 ̲ Γ ( μ ) ] g ( r + μ 2 , x 0 ( r + μ 2 ) ) + r = 0 α μ [ ( t μ 2 ̲ t μ 1 ̲ ) ( α r 1 ) μ 1 ̲ ( α μ 1 ̲ ( b α + μ + 1 ) ( 1 μ 1 ( ( β + 1 ) μ 1 ̲ γ μ 1 ̲ ) 1 μ ( ( β + 1 ) μ ̲ γ μ ̲ ) ) ) × ( 1 μ ( ( β + 1 ) μ ̲ γ μ ̲ ) ( b + 3 ) 1 μ 1 ( ( β + 1 ) μ 1 ̲ γ μ 1 ̲ ) ) α μ 1 ̲ ( b α + μ + 1 ) Γ ( μ ) + ( t μ 1 ̲ ( b + 3 ) t μ 2 ̲ ) ( α r 1 ) μ 1 ̲ α μ 1 ̲ ( b α + μ + 1 ) Γ ( μ ) ] g ( r + μ 2 , x 0 ( r + μ 2 ) ) + r = 0 β μ ( t μ 2 ̲ t μ 1 ̲ ) k = r + μ β ( k σ ( r ) ) μ 1 ̲ ( α μ 1 ̲ ( b α + μ + 1 ) ( 1 μ 1 ( ( β + 1 ) μ 1 ̲ γ μ 1 ̲ ) 1 μ ( ( β + 1 ) μ ̲ γ μ ̲ ) ) ) Γ ( μ ) × g ( r + μ 2 , x 0 ( r + μ 2 ) ) + r = 0 t μ ( t σ ( r ) ) μ 1 ̲ Γ ( μ ) g ( r + μ 2 , x 0 ( r + μ 2 ) ) = r = 0 b + 1 G ( t , r , β , α ) g ( r + μ 2 , x 0 ( r + μ 2 ) ) .

Now, let x 0 be a solution of the fractional sum equation

x(t)= r = 0 b + 1 G(t,r,β,α)g ( r + μ 2 , x ( r + μ 2 ) ) .

Similar to proof of the Lemma 3.1, we conclude that x 0 is a solution to the problem

Δ μ 3 μ x(t)=g ( t + μ 2 , x ( t + μ 2 ) )

via the boundary conditions x(μ+b+1)=0, x(α)= k = γ β x(k), and x(μ3)=0. This completes the proof. □

By using similar proofs of Lemma 3.2 and Theorem 3.3, we obtain the next results.

Lemma 3.5 For each t N μ 2 b + 1 + μ and α,β N μ 2 b + μ , we have

| r = 0 b + 1 G(t,r,β,α)| r = 0 b + 1 |G(t,r,β,α)| M G

for some positive number M G <.

Theorem 3.6 Assume that g: N μ 2 b + μ + 1 ×RR is continuous and bounded in its second variable. Then the fractional finite difference equation Δ μ 3 μ x(t)=g(t+μ2,x(t+μ2)) via the boundary conditions x(μ3)=0, x(μ+b+1)=0, and x(α)= k = γ β x(k) has a solution x 0 with x 0 (t)[ M G , M G ], for all admissible t.

4 An example

Now, we provide an example for the first investigated problem.

Example 4.1 Consider the equation

Δ 2 3 4 3 x(t)=1+ e t + 1 3 +sin ( t + 1 3 + x ( t + 1 3 ) + Δ x ( t + 1 3 ) )
(4.1)

via the boundary value conditions x( 2 3 )=0 and x( 13 3 )= k = 1 3 4 3 x(k). We show that this equation has a solution x 0 with x 0 (t)[35.7073,35.7073] for all admissible t. Let μ= 4 3 , α= 4 3 , b=2, and

g(u,v,w)=1+ e u +sin(u+v+w)

in the first problem. Thus, we should investigate the fractional finite difference equation

Δ 2 3 4 3 x(t)=1+ e t + 1 3 +sin ( t + 1 3 + x ( t + 1 3 ) + Δ x ( t + 1 3 ) )

via the boundary value conditions x( 2 3 )=0 and x( 13 3 )= k = 1 3 4 3 x(k). Note that the map

g: N 1 3 13 3 ×R×RR

is continuous and bounded in its second and third variables. Now, we show that M G =35.7073. Also, the Green function is given by

G ( t , r , 4 3 ) = t 1 3 ̲ k = r + 4 3 4 3 ( k σ ( r ) ) 1 3 ̲ ( ( 13 3 ) 1 3 ̲ 3 4 ( 7 3 ) 4 3 ̲ ) Γ ( 4 3 ) ( 10 3 r ) 1 3 ̲ 3 4 ( 7 3 ) 4 3 ̲ t 1 3 ̲ ( 13 3 ) 1 3 ̲ Γ ( 4 3 ) ( ( 13 3 ) 1 3 ̲ 3 4 ( 7 3 ) 4 3 ̲ ) ( 10 3 r ) 1 3 ̲ t 1 3 ̲ Γ ( 4 3 ) ( 13 3 ) 1 3 ̲ + ( t σ ( r ) ) 1 3 ̲ Γ ( 4 3 )

whenever r=0, t> 4 3 ,

G ( t , r , 4 3 ) = ( 10 3 r ) 1 3 ̲ 3 4 ( 7 3 ) 4 3 ̲ t 1 3 ̲ ( 13 3 ) 1 3 ̲ Γ ( 4 3 ) ( ( 13 3 ) 1 3 ̲ 3 4 ( 7 3 ) 4 3 ̲ ) ( 10 3 r ) 1 3 ̲ t 1 3 ̲ Γ ( 4 3 ) ( 13 3 ) 1 3 ̲ + ( t σ ( r ) ) 1 3 ̲ Γ ( 4 3 )

whenever 0<rt 4 3 ,

G ( t , r , 4 3 ) = t 1 3 ̲ k = r + 4 3 4 3 ( k σ ( r ) ) 1 3 ̲ ( ( 13 3 ) 1 3 ̲ 3 4 ( 7 3 ) 4 3 ̲ ) Γ ( 4 3 ) ( 10 3 r ) 1 3 ̲ 3 4 ( 7 3 ) 4 3 ̲ t 1 3 ̲ ( 13 3 ) 1 3 ̲ Γ ( 4 3 ) ( ( 13 3 ) 1 3 ̲ 3 4 ( 7 3 ) 4 3 ̲ ) ( 10 3 r ) 1 3 ̲ t 1 3 ̲ Γ ( 4 3 ) ( 13 3 ) 1 3 ̲

whenever t= 4 3 , r=0, and

G ( t , r , 4 3 ) = ( 10 3 r ) 1 3 ̲ 3 4 ( 7 3 ) 4 3 ̲ t 1 3 ̲ ( 13 3 ) 1 3 ̲ Γ ( 4 3 ) ( ( 13 3 ) 1 3 ̲ 3 4 ( 7 3 ) 4 3 ̲ ) ( 10 3 r ) 1 3 ̲ t 1 3 ̲ Γ ( 4 3 ) ( 13 3 ) 1 3 ̲

for t 4 3 <r. Thus, for each r N 0 3 , one of the values of the Green function G satisfies

G ( 4 3 , 0 , 4 3 ) = ( 1 3 ) 1 3 ̲ ( 4 3 ) 1 3 ̲ ( ( 13 3 ) 1 3 ̲ 3 4 ( 7 3 ) 4 3 ̲ ) Γ ( 4 3 ) ( 10 3 ) 1 3 ̲ 3 4 ( 7 3 ) 4 3 ̲ ( 4 3 ) 1 3 ̲ ( 13 3 ) 1 3 ̲ Γ ( 4 3 ) ( ( 13 3 ) 1 3 ̲ 3 4 ( 7 3 ) 4 3 ̲ ) ( 10 3 ) 1 3 ̲ ( 4 3 ) 1 3 ̲ Γ ( 4 3 ) ( 13 3 ) 1 3 ̲ = Γ ( 4 3 ) Γ ( 7 3 ) ( Γ ( 16 3 ) 24 3 4 Γ ( 10 3 ) ) Γ ( 4 3 ) 3 4 Γ ( 13 3 ) 6 Γ ( 10 3 ) Γ ( 7 3 ) Γ ( 16 3 ) 24 ( Γ ( 16 3 ) 24 3 4 Γ ( 10 3 ) ) Γ ( 4 3 ) Γ ( 13 3 ) 6 Γ ( 7 3 ) Γ ( 4 3 ) Γ ( 16 3 ) 24 = 1.1089 .

Similar calculations give us the values of G summarized in Table 1.

Table 1 The values of the Green function

Thus, M G r = 0 3 |G(t,r,α)|=35.7073. Hence by using Theorem 3.6, (4.1) has a solution x 0 with x 0 (t)[35.7073,35.7073] for all admissible t.

5 Conclusions

In this manuscript based on a fixed point theorem we provided the existence results for two fractional finite difference equations in the presence of the sum boundary conditions. One example illustrates our results.

References

  1. 1.

    Baleanu D, Diethelm K, Scalas E, Trujillo JJ: Fractional Calculus Models and Numerical Methods. World Scientific, Singapore; 2012.

    Google Scholar 

  2. 2.

    Kilbas AA, Srivastava HM, Trujillo JJ: Theory and Applications of Fractional Differential Equations. Elsevier, New York; 2006.

    Google Scholar 

  3. 3.

    Podlubny I: Fractional Differential Equations. Academic Press, San Diego; 1999.

    Google Scholar 

  4. 4.

    Atici FM, Eloe PW: A transform method in discrete fractional calculus. Int. J. Differ. Equ. 2007, 2: 165-176.

    MathSciNet  Google Scholar 

  5. 5.

    Atici FM, Eloe PW: Discrete fractional calculus with the nabla operator. Electron. J. Qual. Theory Differ. Equ. 2009, 3: 1-12. special edition I

    MathSciNet  Article  Google Scholar 

  6. 6.

    Baleanu D, Rezapour S, Salehi S: A k -dimensional system of fractional finite difference equations. Abstr. Appl. Anal. 2014., 2014: Article ID 312578

    Google Scholar 

  7. 7.

    Elaydi SN: An Introduction to Difference Equations. Springer, Berlin; 1996.

    Google Scholar 

  8. 8.

    Goodrich CS: Some new existence results for fractional difference equations. Int. J. Dyn. Syst. Differ. Equ. 2011, 3: 145-162.

    MathSciNet  MATH  Google Scholar 

  9. 9.

    Holm M: Sum and differences compositions in discrete fractional calculus. CUBO 2011, 13: 153-184. 10.4067/S0719-06462011000300009

    MathSciNet  Article  MATH  Google Scholar 

  10. 10.

    Miller KS, Ross B: Fractional difference calculus. In Proc. International Symposium on Univalent Functions, Fractional Calculus and Their Applications. Nihon University, Koriyama; 1988.

    Google Scholar 

  11. 11.

    Mohan JJ, Deekshitulu GVSR: Fractional order difference equations. Int. J. Differ. Equ. 2012., 2012: Article ID 780619

    Google Scholar 

  12. 12.

    Pan Y, Han Z, Sun S, Zhao Y: The existence of solutions to a system of discrete fractional boundary value problems. Abstr. Appl. Anal. 2012., 2012: Article ID 707631

    Google Scholar 

  13. 13.

    Wu GC, Baleanu D: Discrete fractional logistic map and its chaos. Nonlinear Dyn. 2014, 75: 283-287. 10.1007/s11071-013-1065-7

    MathSciNet  Article  MATH  Google Scholar 

  14. 14.

    Wu GC, Baleanu D: Discrete chaos in fractional sine and standard maps. Phys. Lett. A 2013, 378: 484-487.

    MathSciNet  Article  Google Scholar 

  15. 15.

    Alyousef, K: Boundary value problems for discrete fractional equations. PhD thesis, University of Nebraska-Lincoln (2012)

    Google Scholar 

  16. 16.

    Awasthi, P: Boundary value problems for discrete fractional equations. PhD thesis, University of Nebraska-Lincoln, Ann Arbor, MI (2013)

    Google Scholar 

  17. 17.

    Holm, M: The theory of discrete fractional calculus: development and applications. PhD thesis, University of Nebraska-Lincoln, Ann Arbor, MI (2011)

    Google Scholar 

  18. 18.

    Garling DJH, Gorenstein D, Diesk TT, Walters P: Topics in Metric Fixed Point Theory. Cambridge University Press, Cambridge; 1990.

    Google Scholar 

  19. 19.

    Atici FM, Eloe PW: Initial value problems in discrete fractional calculus. Proc. Am. Math. Soc. 2009, 137: 981-989.

    MathSciNet  Article  MATH  Google Scholar 

Download references

Acknowledgements

Research of the third and fourth authors was supported by Azarbaijan Shahid Madani University.

Author information

Affiliations

Authors

Corresponding author

Correspondence to Dumitru Baleanu.

Additional information

Competing interests

The authors declare that they have no competing interests.

Authors’ contributions

All authors contributed equally to the writing of this paper. All authors read and approved the final version of the manuscript.

Rights and permissions

Reprints and Permissions

About this article

Verify currency and authenticity via CrossMark

Cite this article

Agarwal, R.P., Baleanu, D., Rezapour, S. et al. The existence of solutions for some fractional finite difference equations via sum boundary conditions. Adv Differ Equ 2014, 282 (2014). https://doi.org/10.1186/1687-1847-2014-282

Download citation

Keywords

  • fractional finite difference equation
  • fixed point