# Ak-Dimensional System of Fractional Finite Difference Equations

Authors: Dumitru Baleanu; Shahram Rezapour; Saeid Salehi
Journal: Abstract and Applied Analysis
 (2014)

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## Abstract

We investigate the existence of solutions for ak-dimensional system of fractional finite difference equations by using the Kranoselskii’s fixed point theorem. We present an example in order to illustrate our results.


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## Body

## 1. Introduction

The fractional calculus revealed during the last decade its huge potential applications in many branches of science and engineering (see, e.g., [1–9]). A new and promising direction within fractional calculus is the discrete fractional calculus (see [6, 7, 10–14]). The advantages of this type of calculus are that it treats better phenomena with memory effect (see [10, 11, 14]). We recall that some researchers have been investigating discrete fractional calculus for special equations via very definite boundary conditions (see, e.g., [12, 13, 15–24] and the references therein). Many researchers could focus on this field by considering natural potential of fractional finite difference equations. In this paper, we investigate the existence of solutions for k-dimensional system of fractional finite difference equations:
(1)Δν1y1(t)+f1(y1(t+ν1-1),y2(t+ν2-1),…,yk(t+νk-1))=0,Δν2y2(t)+f2(y1(t+ν1-1),y2(t+ν2-1),…,yk(t+νk-1))=0,⋮Δνkyk(t)+fk(y1(t+ν1-1),y2(t+ν2-1),…,yk(t+νk-1))=0,y1(ν1-2)=Δy1(ν1+b)=0,y2(ν2-2)=Δy2(ν2+b)=0,⋮yk(νk-2)=Δyk(νk+b)=0,
where b∈ℕ0, 1<νi≤2, and fi:ℝk→ℝ are continuous functions for i=1,2,…,k. One-dimensional version of the problem has been studied by Goodrich [18]. Also, Pan et al. studied two-dimensional version of the problem [24]. We show that the problem (1) is equivalent to a summation equation and by using Krasnoselskii's fixed point theorem we investigate solutions of the problem. In this way, we present an example to illustrate our result.


## 2. Preliminaries

It is known that the finite fractional difference theory is important in many branches of science and engineering (see, e.g., [13, 16, 18, 19, 21, 25, 26] and the references therein). The Gamma function is defined by Γ(z)=∫0∞‍e-ttz-1dt for the complex numbers z in which the real part of z is positive (see [8]). Note that the domain of the Gamma function is ℝ∖{0,-1,-2,-3,…} (see [8]). Now, we recall tν_∶=Γ(t+1)/Γ(t+1-ν) for all t,ν∈ℝ whenever the right-hand side is defined (see [16]). If t+1-ν is a pole of the Gamma function and t+1 is not a pole, then tν_=0 (see [16]). We recall that Δβtμ_=Γ(μ+1)tμ-β_/Γ(μ-β+1) (see [16]). One can verify that νν_=νν-1_=Γ(ν+1) and tν+1_/tν_=t-ν.In this paper, we use the standard notationsℕa={a,a+1,a+2,…} for all a∈ℝ and ℕab={a,a+1,a+2,…,b} for all real numbers a and b whenever b-a is a natural number. Let ν>0 with m-1<ν<m for some natural number m. Then, the νth fractional sum of f based at a is defined by
(2)Δa-νf(t)=1Γ(ν)∑k=at-ν(t-σ(k))ν-1_f(k)
for all t∈ℕa+ν, where σ(k)=k+1 is the forward jump operator (see [16]). Similarly, we define Δaνf(t)=(1/Γ(-ν))∑k=at+ν(t-σ(k))-ν-1_f(k) for all t∈ℕa+N-ν. Note that the domain of Δarf(t) is ℕa+N-r for r>0 and ℕa-r for r<0 (see [16]). Also, for the natural number ν=m, we have to recall the formula
(3)Δaνf(t)=Δmf(t)=∑i=0m(-1)i(mi)f(t+m-i).
We define the trivial sum Δa0f(t)=f(t) for all t∈ℕa.Lemma 1 (see [13]).
Leth:ℕa→ℝ be a mapping and m a natural number. Then, the general solution of the equation Δa+ν-mνy(t)=h(t) is given by y(t)=∑i=1m‍-Ci(t-a)ν-i_+Δa-νh(t) for all t∈ℕa+ν-m, where C1,…,Cm are arbitrary constants.Leth:ℕν-m×ℝ→ℝ be a mapping and m a natural number. By using a similar proof, one can check that the general solution of the equation Δν-mνy(t)=h(t+ν-m+1,y(t+ν-m+1)) is given by
(4)y(t)=∑i=1m-Citν-i_+Δ-νh(t+ν-m+1,y(t+ν-m+1))
for all t∈ℕν-m. In particular, the general solution has the following representation:
(5)y(t)=∑i=1m-Citν-i_+1Γ(ν)×∑s=0t-ν(t-σ(s))ν-1_h((ys+ν-m+1)s+ν-m+1,y(s+ν-m+1))
for all t∈ℕν-m. By considering the details, note that ∑k=ab‍(t-σ(k))-ν-1_f(k)=0 whenever b<a. Also for ν,μ>0 with m-1<ν≤m and n-1<μ≤n, the domain of the operator Δ is given by 𝒟{Δa-νf}=ℕa+ν, 𝒟{Δaνf}=ℕa+m-ν, 𝒟{Δa+n-μ-νΔaμf}=ℕa+n+ν-μ, 𝒟{Δa+μνΔa-μf}=ℕa+μ+m-ν, 𝒟{Δa+n-μνΔaμf}=ℕa+n-μ+m-ν, and 𝒟{Δa+μ-νΔa-μf}=ℕa+ν+μ (for more details see [13, 21, 22]). One can find next result about composing a difference with a sum in [12].Lemma 2.
Letf:ℕa→ℝ be a map, k∈ℕ0, and μ>k with n-1<μ≤n. Then ΔkΔa-μf(t)=Δak-μf(t) for all t∈ℕa+μ and ΔkΔaμf(t)=Δak+μf(t) for all t∈ℕa+n-μ.By using Lemma1 and last lemma for k=1, we get
(6)Δy(t)=∑i=1m-Ci(ν-i)tν-i-1_+1Γ(ν-1)×∑s=0t-ν+1(t-σ(s))ν-2_h(y(t+ν-m+1)t+ν-m+1,y(t+ν-m+1)).
We are going to use this in our main results. A nonempty, closed subset P≠{0} of a topological vector space E is called a cone whenever P∩(-P)={0} and ax+by∈P for all x,y∈P and nonnegative real numbers a,b (for more details and examples see [27] and references therein).Lemma 3 (see [28]).
LetX be a Banach space and K a cone in X. Assume that B1 and B2 are open subsets of X such that 0∈B1 and B1¯⊆B2. Suppose that T:K∩(B2¯∖B1)→K is a completely continuous operator. If either ∥Ty∥≤∥y∥ for all y∈K∩∂B1 and ∥Ty∥≥∥y∥ for all y∈K∩∂B2 or ∥Ty∥≥∥y∥ for all y∈K∩∂B1 and ∥Ty∥≤∥y∥ for all y∈K∩∂B2, then T has at least one fixed point in K∩(B2¯∖B1).


## 3. Main Result

In this section we provide the main results. For next result, consider the problem (1).Lemma 4.
The fractional finite difference equation(*)Δνiyi(t)+fi(yk(t+νk-1)y1(t+ν1-1),y2(t+ν2-1),…,yk(t+νk-1))=0
via the boundary conditions yi(νi-2)=Δyi(νi+b)=0 has a solution yi0 if and only if yi0 is a solution of the summation equation yi(t)=∑s=0b+1‍Gi(t,s)fi(y1(t+ν1-1),y2(t+ν2-1),…,yk(t+νk-1)), where the Green function Gi(t,s) is given by
(7)Gi(t,s)=1Γ(νi)×{tνi-1_(νi+b)νi-2_(νi+b-σ(s))νi-2_-(t-σ(s))νi-1_s≤t-ν≤b+1,tνi-1_(νi+b)νi-2_(νi+b-σ(s))νi-2_t-ν+1≤s≤b+1
for all s∈ℕ0b+1. Here, i∈{1,2,…,k} and (*) is one of equations of the system.Proof.
Lethi(t+νi-1):=fi(y1(t+ν1-1),y2(t+ν2-1),…,yk(t+νk-1)) and letyi0be a solution of the fractional finite difference equation Δνiyi(t)+fi(y1(t+ν1-1), y2(t+ν2-1),…,yk(t+νk-1))=0. By using Lemma 1, we get yi0(t)=C1tνi-1_+C2tνi-2_-(1/Γ(νi)) ∑s=0t-νi‍(t-σ(s))νi-1_hi(s+νi-1). By using the boundary condition yi0(νi-2)=0, we obtain
(8)0=C1(νi-2)νi-1_+C2(νi-2)νi-2_-1Γ(νi)∑s=0-2(νi-2-σ(s))νi-1_hi(s+νi-1)
and so C2=0. Now by using the boundary condition Δyi0(νi+b)=0, we get
(9)0=C1(νi-1)(νi+b)νi-2_-1Γ(νi-1)∑s=0b+1(νi+b-σ(s))νi-2_hi(s+νi-1).
Hence, C1=(1/((νi+b)νi-2_Γ(νi)))∑s=0b+1‍(νi+b-σ(s))νi-2_hi(s+νi-1) and so
(10)yi0(t)=tνi-1_(νi+b)νi-2_Γ(νi)×∑s=0b+1‍(νi+b-σ(s))νi-2_hi(s+νi-1)-1Γ(νi)∑s=0t-νi(t-σ(s))νi-1_hi(s+νi-1)=∑s=0b+1Gi(t,s)fi(y1(t+ν1-1),y2(t+ν2-1),…,yk(t+νk-1)).
Now, let yi0 be a solution of the fractional sum equation
(11)yi(t)=∑s=0b+1Gi(t,s)fi(y1(t+ν1-1),y2(t+ν2-1),…,yk(t+νk-1)).
Then, yi0(t)=(tνi-1_/(νi+b)νi-2_Γ(νi)) ∑s=0b+1‍(νi+b-σ(s))νi-2_hi(s+νi-1)-(1/Γ(νi)) ∑s=0t-νi‍(t-σ(s))νi-1_hi (s+νi-1). Since (νi-2)νi-1_=0, we get yi0(νi-2)=0. Also,
(12)Δyi0(t)=(νi-1)tνi-2_(νi+b)νi-2_Γ(νi)×∑s=0b+1(νi+b-σ(s))νi-2_hi(s+νi-1)-1Γ(νi-1)∑s=0t-νi+1(t-σ(s))νi-2_hi(s+νi-1).
Hence, we get
(13)Δyi0(νi+b)=(νi-1)(νi+b)νi-2_(νi+b)νi-2_Γ(νi)×∑s=0b+1‍(νi+b-σ(s))νi-2_hi(s+νi-1)-1Γ(νi-1)∑s=0b+1‍((νi+b)-σ(s))νi-2_×hi(s+νi-1)=0.
Moreover, Δνiyi0(t)=(Δνitνi-1_/(νi+b)νi-2_Γ(νi)) ∑s=0b+1‍(νi+b-σ(s))νi-2_hi(s+νi-1) -ΔνiΔ-νihi(s+νi-1). Since Δνitνi-1_= Γ(νi)tνi-1-νi_/Γ(νi-νi)=0 and ΔνiΔ-νihi(s+νi-1)=hi(s+νi-1), we get
(14)Δνiyi0(t)+fi(y1(t+ν1-1),y2(t+ν2-1),…,yk(t+νk-1))=0.
This completes the proof.Hereafter, for simplicity we use the notationsIi:=ℕνi-1νi+b+1 and Ji:=[((νi+b)/4),((3(νi+b))/4)] for all i=1,2,…,k.Lemma 5 (see [18]).
The Green function (7) satisfies Gi(t,s)≥0 for all t∈Ii and s∈ℕ0b+1 and max⁡t∈IiGi(t,s)=Gi(s+νi-1,s) for all s∈ℕ0b and there exist λi∈(0,1) such that
(15)min⁡t∈JiGi(t,s)≥λimax⁡t∈IiGi(t,s)=λiGi(s+νi-1,s)
for all s∈ℕ0b+1.Goodrich showed thatλi=min⁡{γ1i,γ2i} (see [18]), where γ1i=((b+νi)/4)νi-1_/(b+νi)νi-1_ and
(16)γ2i=1(3(b+νi)/4)νi-1_×[(b+1)((3(b+νi)/4)-1)νi-1_Γ(νi+b+1)Γ(b+3)(νi+b-1)νi-1_(3(b+νi)4)νi-1_-(b+1)(3(b+νi)/4-1)νi-1_Γ(νi+b+1)Γ(b+3)(νi+b-1)νi-1_].
Note that γ2i can be written in the simple form γ2i=(νi+2)/3(b+2), because(**)γ2i=1(3(b+νi)/4)νi-1_×[(3(b+νi)4)νi-1_-(b+1)(3(b+νi)/4-1)νi-1_Γ(νi+b+1)Γ(b+3)(νi+b-1)νi-1_]=1-(b+1)(3(b+νi)/4-1)νi-1_Γ(νi+b+1)(3(b+νi)/4)νi-1_Γ(b+3)(νi+b-1)νi-1_=1-(b+1)(3(b+νi)/4-νi+1)Γ(νi+b+1)(3(b+νi)/4)Γ(b+3)(Γ(νi+b)/Γ(b+1))=1-(b+1)(3(b+νi)/4-νi+1)(νi+b)Γ(νi+b)(3(b+νi)/4)(b+1)(b+2)Γ(b+1)(Γ(νi+b)/Γ(b+1))=νi+23(b+2).Note that (**) hold because (a-1)b_/ab_=(a-b)/a. Suppose that 𝒜i is the Banach space of the maps u:ℕνi-2νi+b→ℝ via the usual maximum norm ∥u∥=max⁡{|u(t)|:t∈ℕνi-2νi+b}. Consider the space 𝒳=𝒜1×𝒜2×⋯×𝒜k via the norm ∥(y1,y2,…,yk)∥𝒳=∥y1∥+∥y2∥+⋯+∥yk∥. It is clear that (𝒳,∥·∥𝒳) is a Banach space (see [29]). Now, define the map T:𝒳→𝒳 by
(17)T(y1,y2,…,yk)(t1,t2,…,tk)=(T1(y1,y2,…,yk)(t1)T2(y1,y2,…,yk)(t2)⋮Tk(y1,y2,…,yk)(tk)),
where Ti(y1,y2,…,yk)(t)=∑s=0b+1‍Gi(t,s)fi(y1(s+ν1-1),y2(s+ν2-1),…,yk(s+νk-1)) for i=1,2,…,k. Also, consider the cone 𝒦 defined by
(18)𝒦={∥(y1,y2,…,yk)∥𝒳(y1,y2,…,yk)∈𝒳:yi≥0,min⁡(t1,t2,…,tk)∈J1×J2×⋯×Jk[y1(t1)+y2(t2)+⋯+yk(tk)]≥λ∥(y1,y2,…,yk)∥𝒳},
where λ=min⁡1≤i≤kλi. First, for the operator T we show that T(𝒦)⊆𝒦 whenever the functions fi are nonnegative for i=1,2,…,k. Let (y1,y2,…,yk)∈𝒦. Then, we have
(19)min⁡(t1,t2,…,tk)∈J1×J2×⋯×Jk∑n=1kTn(y1,y2,…,yk)(tn)≥∑n=1kmin⁡tn∈JnTn(y1,y2,…,yk)(tn)=∑n=1kmin⁡tn∈Jn∑s=0b+1Gn(tn,s)fn(y1(s+ν1-1)y2(s+ν2-1)⋮yk(s+νk-1))≥∑n=1kλnmax⁡tn∈In∑s=0b+1Gn(tn,s)fn(y1(s+ν1-1)y2(s+ν2-1)⋮yk(s+νk-1))=∑n=1kλn∥Tn(y1,y2,…,yk)∥≥λ∑n=1k∥Tn(y1,y2,…,yk)∥=λ∥T(y1,y2,…,yk)∥𝒳,
where λ=min⁡1≤n≤kλn. Hence, T(y1,y2,…,yk)∈𝒦 and so T(𝒦)⊆𝒦. For providing our main result, we use similar conditions which have been given by Goodrich in [18] and Henderson et al. in [30].Theorem 6.
Suppose thatfi∈C([0,∞)k) for all i=1,2,…,k:
(20)lim⁡(y1,y2,…,yk)→(0+,0+,…,0+)fi(y1,y2,,…,yk)y1+y2+⋯+yk=fi*,lim⁡(y1,y2,…,yk)→(+∞,+∞,…,+∞)fi(y1,y2,,…,yk)y1+y2+⋯+yk=fi**
such that ∑s=0b+1‍Gi(s+νi-1,s)(fi*+ϵ)≤1/k and ∑s=0b+1‍λGi(s+νi-1,s)(fi**-ϵ)≥1/k for some
(21)0<ϵ<min⁡{fi**:1≤i≤k},
where Gi is the Green function (7) and λ=min⁡1≤i≤kλi. Then the k-dimensional system of fractional finite difference equations (1) has at least one solution.Proof.
Consider the operatorT:𝒦→𝒦 defined by (17) and the cone 𝒦. It is clear that T is completely continuous because it is a summation operator on a finite set. Choose δ1>0 such that
(22)fi(y1,y2,…,yk)≤(fi*+ϵ)(y1+y2+⋯+yk)
for all ∥(y1,y2,…,yk)∥𝒳<δ1. Put ℬ1={(y1,y2,…,yk)∈𝒳:∥(y1,y2,…,yk)∥𝒳<δ1}. Then, 0∈ℬ1 and ∥(y1,y2,…,yk)∥𝒳=δ1 for all (y1,y2,…,yk)∈𝒦∩∂ℬ1. Also, we have
(23)∥Ti(y1,y2,…,yk)∥=max⁡ti∈Ii∑s=0b+1Gi(ti,s)fi(y1(s+ν1-1),y2(s+ν2-1),…,yk(s+νk-1))≤∑s=0b+1Gi(s+ν-1,s)(fi*+ϵ)(y1+y2+⋯+yk)≤∥(y1,y2,…,yk)∥𝒳∑s=0b+1Gi(s+ν-1,s)(fi*+ϵ)≤1k∥(y1,y2,…,yk)∥𝒳
for all (y1,y2,…,yk)∈𝒦∩∂ℬ1. Hence,
(24)∥T(y1,y2,…,yk)∥𝒳=∑i=1k∥Ti(y1,y2,…,yk)∥≤k×1k∥(y1,y2,…,yk)∥𝒳=∥(y1,y2,…,yk)∥𝒳
for all (y1,y2,…,yk)∈𝒦∩∂ℬ1. Now, choose β∈ℝ such that β>δ1 and
(25)fi(y1,y2,…,yk)≥(fi**-ϵ)(y1+y2+⋯+yk)
for all ∥(y1,y2,…,yk)∥𝒳≥β. Also, choose δ2 such that (1/k)β≤δ2≤λβmin⁡1≤i≤k ∑s=0b+1‍Gi(s+ν-1,s)(fi**-ϵ). Now, put ℬ2={(y1,y2,…,yk)∈𝒳:∥(y1,y2,…,yk)∥𝒳<kδ2}. Then, ℬ1¯⊆ℬ2 and
(26)y1(t1)+y2(t2)+⋯+yk(tk)≥min⁡(t1,t2,⋯,tk)∈J1×J2×⋯×Jk[y1(t1)+y2(t2)+⋯+yk(tk)]≥λ∥(y1,y2,…,yk)∥𝒳
for all (y1,y2,…,yk)∈𝒦∩∂ℬ2. Thus, by using (25) we get
(27)∥Ti(y1,y2,…,yk)∥=max⁡ti∈Ii∑s=0b+1‍Gi(ti,s)fi(yk(s+νk-1)y1(s+ν1-1),y2(s+ν2-1),…,yk(s+νk-1))≥∑s=0b+1Gi(s+ν-1,s)(fi**-ϵ)(y1+y2+⋯+yk)≥λ∥(y1,y2,…,yk)∥𝒳∑s=0b+1Gi(s+ν-1,s)(fi**-ϵ)≥1k∥(y1,y2,…,yk)∥𝒳
for all (y1,y2,…,yk)∈𝒦∩∂ℬ2. Hence,
(28)∥T(y1,y2,…,yk)∥𝒳=∑i=1k∥Ti(y1,y2,…,yk)∥≥k×1k∥(y1,y2,…,yk)∥𝒳=∥(y1,y2,…,yk)∥𝒳
for all (y1,y2,…,yk)∈𝒦∩∂ℬ2. By using Lemma 3, T has at least one fixed point (y10,y20,…,yk0) in 𝒦∩(ℬ2¯∖ℬ1) and so by using Lemma 4, the k-dimensional system of fractional finite difference equations (1) has at least one solution.


## 4. Example

Here, we provide an example to illustrate our last result.Example 1.
Consider the 5-dimensional fractional finite difference equation system:(29)Δ1.2y1(t)+f1(y1(t+0.2),y2(t+0.4),y3(t+0.5),y4(t+0.6),y5(t+0.8))=0,Δ1.4y2(t)+f2(y1(t+0.2),y2(t+0.4),y3(t+0.5),y4(t+0.6),y5(t+0.8))=0,Δ1.5y2(t)+f3(y1(t+0.2),y2(t+0.4),y3(t+0.5),y4(t+0.6),y5(t+0.8))=0,Δ1.6y2(t)+f4(y1(t+0.2),y2(t+0.4),y3(t+0.5),y4(t+0.6),y5(t+0.8))=0,Δ1.8y2(t)+f5(y1(t+0.2),y2(t+0.4),y3(t+0.5),y4(t+0.6),y5(t+0.8))=0,y1(-0.8)=Δy1(9.2)=0,y2(-0.6)=Δy2(9.4)=0,y3(-0.5)=Δy3(9.5)=0,y4(-0.4)=Δy4(9.6)=0,y5(-0.2)=Δy5(9.8)=0.
We show that the problem has at least one solution, where
(30)f1(y1,y2,y3,y4,y5)=(y1+y2+cos⁡y3)(y1+y2+y3+y4+y5)y1+y2+1000,f2(y1,y2,y3,y4,y5)=3e-10/(y1+y2+y3+1)(y1+y2+y3+y4+y5),f3(y1,y2,y3,y4,y5)=(y1+y2+y3+y4+y5){5y1+11000y1<1,2.001+3y1y1≥1,f4(y1,y2,y3,y4,y5)=(3y3-siny52y3+1+11000)(y1+y2+y3+y4+y5),f5(y1,y2,y3,y4,y5)=(y1+y2+y3+y4+y5){e-8sin(y2)/y2y2>0,e-8y2=0.
Let ν1=1.2, ν2=1.4, ν3=1.5, ν4=1.6, ν5=1.8, b=8, and k=5. Thus, the system (29) is a special case of the system (1). It is easy to check that fi∈C([0,∞)5) for i=1,2,3,4,5. Put γ2i=(νi+2)/3(8+2),
(31)γ1i=((8+νi)/4)νi-1_(8+νi)νi-1_=Γ(3+νi/4)×Γ(10)Γ(4-3νi/4)×Γ(9+νi),
and λi=min⁡{γ1i,γ2i} for i=1,2,3,4,5. Then, by a calculation we get λ1=0.1066, λ2=0.1133, λ3=0.1166, λ4=0.1200, and λ5=0.1266. Thus, λ=min⁡{λi:i=1,2,3,4,5}=0.1066. On the other hand by calculation of some limits, one can get that f1*=10-3, f1**=1, f2*=3e-10, f2**=3, f3*=10-3, f3**=2.001, f4*=10-3, f4**=1.501, f5*=e-8, and f5**=1. Moreover, we have
(32)∑s=0b+1‍G1(s+ν1-1,s)=∑s=09G1(s+0.2,s)=∑s=09(s+0.2)0.2_9.2-0.8_(9.2-σ(s))-0.8_=∑s=09Γ(s+1.2)Γ(11)Γ(9.2-s)Γ(s+1)Γ(10.2)Γ(10-s)≥Γ(11)Γ(10.2)∑s=09Γ(9.2-s)Γ(10-s)≥6×6=36,∑s=0b+1‍G1(s+ν1-1,s)=∑s=09‍Γ(s+1.2)Γ(11)Γ(9.2-s)Γ(s+1)Γ(10.2)Γ(10-s)≤Γ(11)Γ(10.2)∑s=09‍Γ(s+1.2)Γ(s+1)≤6×13=78.
Similarly, we obtain
(33)∑s=0b+1G2(s+ν2-1,s)≥6×6=36,∑s=0b+1G2(s+ν2-1,s)≤6×19=114,∑s=0b+1G3(s+ν3-1,s)≥6×6=36,∑s=0b+1G3(s+ν3-1,s)≤6×22=132,∑s=0b+1‍G4(s+ν4-1,s)≥6×6=36,∑s=0b+1G4(s+ν4-1,s)≤6×27=162,∑s=0b+1G5(s+ν5-1,s)≥6×7=42,∑s=0b+1G5(s+ν5-1,s)≤6×38=228.
Now, let ϵ=0.0001. Then, 0<ϵ<min⁡{fi**:i=1,2,3,4,5} and we have
(34)∑s=0b+1‍λG1(s+ν1-1,s)(f1**-ϵ)≥0.1066×36×(1-0.0001)=3.8372≥15,∑s=0b+1G1(s+ν1-1,s)(f1*+ϵ)≤78×(10-3+0.0001)=0.0858≤15,∑s=0b+1‍λG2(s+ν2-1,s)(f2**-ϵ)≥0.1066×36×(3-0.0001)=11.5124≥15,∑s=0b+1‍G2(s+ν2-1,s)(f2*+ϵ)≤114×(3e-10+0.0001)=0.02692≤15,∑s=0b+1‍λG3(s+ν3-1,s)(f3**-ϵ)≥0.1066×36×(2.001-0.0001)=7.6786≥15,∑s=0b+1‍G3(s+ν3-1,s)(f3*+ϵ)≤132×(10-3+0.0001)=0.1452≤15,∑s=0b+1‍λG4(s+ν4-1,s)(f4**-ϵ)≥0.1066×36×(1.501-0.0001)=5.7598≥15,∑s=0b+1‍G4(s+ν4-1,s)(f4*+ϵ)≤162×(10-3+0.0001)=0.1782≤15,∑s=0b+1‍λG5(s+ν5-1,s)(f5**-ϵ)≥0.1066×37×(1-0.0001)=3.9438≥15,∑s=0b+1‍G5(s+ν5-1,s)(f5*+ϵ)≤228×(e-8+0.0001)=0.0992≤15.
Thus by using Theorem 6, the 5-dimensional system of fractional finite difference equations (29) has at least one solution.


## 5. Conclusions

In this paper, based on main idea of Goodrich we review the existence of solutions for ak-dimensional system of fractional finite difference equations. In fact we are going to extend the work of Goodrich in a sense. We give an example to illustrate our last result.



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*Source: 312578-2014-04-09.xml*