ISSN: 1311-1728 (printed version); ISSN: 1314-8060 (on-line version) doi:http://dx.doi.org/10.12732/ijam.v33i2.7
EXISTENCE AND UNIQUENESS OF SOLUTIONS FOR A DISCRETE FRACTIONAL BOUNDARY
VALUE PROBLEM
A. George Maria Selvam1§, R. Dhineshbabu2 1,2Department of Mathematics
Sacred Heart College (Autonomous) Tirupattur, 635601, Tamil Nadu, INDIA
Abstract: This present work discusses existence and uniqueness of solutions for the following discrete fractional antiperiodic boundary value problem of the form
C
0∆αkx(k) =f(k+α−1, x(k+α−1)),
for k ∈ [0, ℓ+ 2]N0 = {0,1, ..., ℓ+ 2}, with boundary conditions x(α−3) =
−x(α+ℓ), ∆x(α−3) =−∆x(α+ℓ), ∆2x(α−3) =−∆2x(α+ℓ), wheref : [α− 2, α+ℓ]Nα−2×R→Ris continuous and
C
0∆αk is the Caputo fractional difference
operator with order 2 < α ≤ 3. Finally, the main results are illustrated by suitable examples.
AMS Subject Classification: 34A12, 34B15, 39A12
Key Words: existence, uniqueness, discrete, fractional order, boundary value problem
1. Introduction
The theory of fractional calculus is one of the branches of study which involves integrals and derivatives of an arbitrary order. In the past few decades, frac-tional calculus is one of the most novel types of tools applied to problems in engineering, science and technology (see [16], [17]). It also played a key role for
the improvement of the specialized research in mathematical modeling of several phenomena such as enzyme kinetics, nonlinear oscillation of earthquake, blood flow problems, aerodynamics, cancer modeling being among the popular ones, control theory, electrical circuits, fluid-dynamic traffic model, regular variation in thermodynamics, polymer rheology, arterial and heart disease modeling, etc (see [13], [14]). In recent years, the theory of discrete fractional calculus has been developed by a very few researchers (see [8], [9], [11], [15]) and they ob-tained many important properties and interesting fruitful results.
Anti-periodic boundary value problems (classical and fractional), recently have received great attention as they occur in the mathematical modeling of a diversity of physical processes (see [2], [3], [4], [7]). As a matter of fact, many numerical problems converge faster when anti-periodic boundary conditions are used instead of periodic boundary conditions. However, the concept of non local anti-periodic boundary conditions has not been addressed yet. In the last two decades, many researchers have focused their attention on the study of stability of solutions, existence and uniqueness, positive and multiplicity of solutions of boundary value problems (BVP’s) for nonlinear fractional differential and difference equations by using fixed point techniques such as Krasnoselskii fixed point theorem, Brower fixed point theorem and contraction mapping principle (see [1], [3], [5], [6], [11], [12], [15]).
Although the discrete boundary value problems with fractional order 0< α≤1 and 1< α≤2 have been studied by researchers, very little is known in the literature about discrete fractional equation with order 2< α ≤3. Motivated by the aforesaid work, in this present work, we investigate the existence and uniqueness of solutions for the following discrete fractional order antiperiodic boundary value problem of the form
C
0∆αkx(k) =f(k+α−1, x(k+α−1)), (1)
fork∈[0, ℓ+ 2]N0 ={0,1, ..., ℓ+ 2}, subject to the conditions
(i) x(α−3) =−x(α+ℓ), (ii) ∆x(α−3) =−∆x(α+ℓ), and
(iii) ∆2x(α−3) =−∆2x(α+ℓ), (2)
where f : [α−2, α+ℓ]Nα−2 ×R → R is a continuous, C
0∆αk is the Caputo
fractional difference operator with order 2< α≤3.
contraction mapping principle and Brouwer fixed point theorem. In Section 5, suitable examples to illustrate the main results are presented and the paper ends with a conclusion.
2. Preliminaries
This section presents some required definitions and theorems which are needed throughout this paper.
Definition 1. (see [8], [10]) Let α > 0. The αth fractional sum of f : Na→R is defined by
∆−αf(k) = 1 Γ(α)
k−α
X
s=a
(k−s−1)α−1f(s), (3)
for all k∈ {a+α, a+α+ 1, . . .}=:Na+α and kα:= Γ(Γ(k+1−k+1)α).
Definition 2. (see [8]) Let α > 0 and set µ = n−α. Then, the αth fractional Caputo difference operator is defined as
C
0∆αkf(k) = ∆−µ(∆nf(k)) =
1 Γ(µ)
k−µ
X
s=a
(k−s−1)µ−1 ∆nf(s), (4)
for all k∈ {a+µ, a+µ+ 1, . . .}=: Na+µ and n−1< α < n, where n=⌈α⌉,
⌈.⌉is the ceiling function.
Theorem 3. (see [8]) Assume thatα >0and f is defined on Na. Then C
0∆αk∆−αf(k) = f(k)− n−1 X
j=0
(k−a)j j! ∆
jf(a)
= f(k) +c0+c1k+...+cn−1kn−1, (5)
for someci ∈R, wherei= 1,2, ..., n−1.
Theorem 4. (see [8]) Letf :Na+α×Na→R be given. Then
∆
k−α
X
s=a
f(k, s) !
=
k−α
X
s=a
∆kf(k, s) +f(k+ 1, k+ 1−α), (6)
3. Solution for antiperiodic BVP
In order to discuss the main results, now we state and prove important theorems which will aid to obtain a form of the solution of (1)-(2), provided the solution exists.
Theorem 5. Let2< α≤3andf : [α−2,· · · , α+ℓ]Nα−2 →Rbe given. A
functionx(k) is a solution of a discrete fractional antiperiodic boundary value problem
C
0∆αkx(k) =f(k+α−1),
x(α−3) =−x(α+ℓ), ∆x(α−3) =−∆x(α+ℓ), ∆2x(α−3) =−∆2x(α+ℓ),
(7)
wherek∈[0, ℓ+ 2]N0, if and only if x(k), fork∈[α−3, α+ℓ]Nα−3 has the form
x(k) = 1 Γ(α)
k−α
X
s=0
(k−s−1)α−1f(s+α−1)
− 1
2Γ(α)
ℓ
X
s=0
(α+ℓ−s−1)α−1f(s+α−1)
+ A(k)
4Γ(α−1)
ℓ+1 X
s=0
(α+ℓ−s−1)α−2f(s+α−1)
+ B(k)
8Γ(α−2)
ℓ+2 X
s=0
(α+ℓ−s−1)α−3f(s+α−1), (8)
whereA(k) = [(2α−3) + (ℓ−2k)]andB(k) = [2(ℓk−k2) + 4k(α−1)−2α(α+ ℓ−2) + (5ℓ+ 3)].
Proof. Suppose that x(k) defined on [α−3, α+ℓ]Nα−3 is a solution of (7). From Theorem 3, we obtain a general solution for (7) as
x(k) = ∆−αf(k+α−1)−c
0−c1k−c2k2
= 1
Γ(α)
k−α
X
s=0
for somec0,c1,c2 ∈R. Using Theorem 4, we obtain
∆x(k) = 1 Γ(α−1)
k−α+1 X
s=0
(k−s−1)α−2f(s+α−1)−c1−2c2k,
∆2x(k) = 1 Γ(α−2)
k−α+2 X
s=0
(k−s−1)α−3f(s+α−1)−2c 2.
In view of x(α−3) =−x(α+ℓ), ∆x(α−3) =−∆x(α+ℓ) and ∆2x(α−3) = −∆2x(α+ℓ), the value ofc
0,c1 and c2 as follows:
c0 = 1 2Γ(α)
ℓ
X
s=0
(α+ℓ−s−1)α−1f(s+α−1)
−(2α+ℓ−3) 4Γ(α−1)
ℓ+1 X
s=0
(α+ℓ−s−1)α−2f(s+α−1)
+[2α(α+ℓ−2)−(5ℓ+ 3)] 8Γ(α−2)
ℓ+2 X
s=0
(α+ℓ−s−1)α−3f(s+α−1),
c1 = 1 2Γ(α−1)
ℓ+1 X
s=0
(α+ℓ−s−1)α−2f(s+α−1)
−(2α+ℓ−3) 4Γ(α−2)
ℓ+2 X
s=0
(α+ℓ−s−1)α−3f(s+α−1), and
c2 = 1 4Γ(α−2)
ℓ+2 X
s=0
(α+ℓ−s−1)α−3f(s+α−1).
Substituting the values ofc0,c1 andc2 into (9), we obtain (8). This shows that if (7) is a solution, then it can be represented by (8).
Conversely, every function of the form (8) is a solution of (7). The proof is completed.
Theorem 6. If α andk are any numbers then
k−α
X
s=0
(k−s−1)α−1 = Γ(k+ 1) αΓ(k−α+ 1).
Proof. Foru > m,u, m∈R,m >−1, u >−1, we have
Γ(u+ 1)
Γ(m+ 1)Γ(u−m+ 1) =
Γ(u+ 2)
Γ(m+ 2)Γ(u−m+ 1) −
That is Γ(u+ 1) Γ(u−m+ 1) =
1 m+ 1
Γ(u+ 2)
Γ(u−m+ 1)−
Γ(u+ 1) Γ(u−m)
.
Then,
k−α X
s=0
(k−s−1)α−1= k−α−1
X
s=0
(k−s−1)α−1+ Γ(α). (10)
Now we find
k −α−1
X
s=0
(k−s−1)α−1= 1 α
Γ(k+ 1)
Γ(k−α+ 1) −
Γ(α+ 1) Γ(1)
. (11)
Substituting equation (11) in (10), we get
k −α X
s=0
(k−s−1)α−1= 1 α
Γ(k+ 1) Γ(k−α+ 1).
Corollary 7. The following results are immediate consequences of Theo-rem 6:
1. Pℓ
s=0
(α+ℓ−s−1)α−1= Γ(α+ℓ+ 1) αΓ(ℓ+ 1) .
2.
ℓ+1 P
s=0
(α+ℓ−s−1)α−2= Γ(α+ℓ+ 1) (α−1)Γ(ℓ+ 2).
3. ℓP+2
s=0
(α+ℓ−s−1)α−3= Γ(α+ℓ+ 1) (α−2)Γ(ℓ+ 3).
4. Existence and uniqueness of solution
only if x(k) is a fixed point of the operator T :Rℓ+4→R, where
(T x)(k) = 1 Γ(α)
k−α
X
s=0
(k−s−1)α−1f(s+α−1, x(s+α−1))
− 1
2Γ(α)
ℓ
X
s=0
(α+ℓ−s−1)α−1f(s+α−1, x(s+α−1))
+ A(k) 4Γ(α−1)
ℓ+1 X
s=0
(α+ℓ−s−1)α−2f(s+α−1, x(s+α−1))
+ B(k) 8Γ(α−2)
ℓ+2 X
s=0
(α+ℓ−s−1)α−3f(s+α−1, x(s+α−1)),
(12)
for k ∈ [α−3, α+ℓ]Nα−3, where A(k) = [(2α −3) + (ℓ−2k)] and B(k) = [2(ℓk−k2) + 4k(α−1)−2α(α+ℓ−2) + (5ℓ+ 3)].
Theorem 8. Define kxk = max
k∈[α−3,α+ℓ]Nα−3
|x(k)|. Assume that f(k, x)
is Lipschitz in x. That is, there exists L > 0 such that |f(k, x)−f(k, y)| ≤ L|x−y| whenever x, y ∈ R. Then it follows that discrete fractional antiperi-odic boundary value problem (1)-(2) has a unique solution provided that the condition
Λ =βL Γ(α+ℓ+ 1)
2Γ(α−1)Γ(ℓ+ 1) <1, (13)
whereβ =
3 α(α−1) +
(ℓ+ 3) 2(α−1)(ℓ+ 1)+
(ℓ+ 3) 4(ℓ+ 1)(ℓ+ 2)
holds.
Proof. We show thatT is a contraction mapping. To achieve this, we notice that for given x and y,
kT x−T yk ≤Lkx−yk max
k∈[α−3,α+ℓ]Nα
−3 1 Γ(α)
k−α
X
s=0
(k−s−1)α−1
+Lkx−yk max
k∈[α−3,α+ℓ]Nα−3
1 2Γ(α)
ℓ
X
s=0
+Lkx−yk max
k∈[α−3,α+ℓ]Nα−3
|A(k)| 4Γ(α−1)
ℓ+1 X
s=0
(α+ℓ−s−1)α−2
+Lkx−yk max
k∈[α−3,α+ℓ]Nα−3
|B(k)| 8Γ(α−2)
ℓ+2 X
s=0
(α+ℓ−s−1)α−3.
Since |A(k)|=|(2α−3) + (ℓ−2k)|= (ℓ+ 3) and|B(k)|=
2(ℓk−k2) + 4k(α−1)−2α(α+ℓ−2) + (5ℓ+ 3)
= (ℓ+ 3). By an applica-tion of Theorem 6 and Corollary 7, we obtain
kT x−T yk ≤ 3Lkx−yk 2
Γ(α+ℓ+ 1) Γ(α+ 1)Γ(ℓ+ 1)
+Lkx−yk(ℓ+ 3) 4
Γ(α+ℓ+ 1) Γ(α)Γ(ℓ+ 2)
+Lkx−yk(ℓ+ 3) 8
Γ(α+ℓ+ 1) Γ(α−1)Γ(ℓ+ 3)
.
Above inequality leads to the conclusion
kT x−T yk ≤
βL Γ(α+ℓ+ 1) 2Γ(α−1)Γ(ℓ+ 1)
kx−yk.
When the condition (13) holds, boundary value problem (1) - (2) has a unique solution, which completes the proof of the theorem.
Theorem 9. Assume that there exists a constantM >0such thatf(k, x)
satisfies the inequality
max
(k,x)∈[α−3,α+ℓ]Nα−3×[−M,M]
|f(k, x)| ≤ M β
Γ(α+ℓ+ 1) 2Γ(α−1)Γ(ℓ+ 1)
,
(14)
where β =
3 α(α−1) +
(ℓ+ 3) 2(α−1)(ℓ+ 1) +
(ℓ+ 3) 4(ℓ+ 1)(ℓ+ 2)
. Then it follows
that discrete fractional antiperiodic boundary value problem (1) - (2) has at least one solution x0 satisfying |x0(k)| ≤M for all k∈[α−3, α+ℓ]Nα−3.
Proof. Consider the Banach space B :=
theorem to obtain the conclusion. Suppose that inequality (14) holds for given f. For convenience, we assume
Φ := M
β
Γ(α+ℓ+ 1) 2Γ(α−1)Γ(ℓ+ 1)
, (15)
which is a strictly positive constant. Then we obtain
kT xk ≤ Φ Γ(α)
k−α
X
s=0
(k−s−1)α−1+ Φ 2Γ(α)
ℓ
X
s=0
(α+ℓ−s−1)α−1
+ Φ|A(k)| 4Γ(α−1)
ℓ+1 X
s=0
(α+ℓ−s−1)α−2
+ Φ|B(k)| 8Γ(α−2)
ℓ+2 X
s=0
(α+ℓ−s−1)α−3.
Since |A(k)|=|(2α−3) + (ℓ−2k)|= (ℓ+ 3) and|B(k)|=
2(ℓk−k2) + 4k(α−1)−2α(α+ℓ−2) + (5ℓ+ 3)
= (ℓ+ 3). By an applica-tion of Theorem 6 and Corollary 7, we get
kT xk ≤ 3Φ 2
Γ(α+ℓ+ 1) Γ(α+ 1)Γ(ℓ+ 1)
+Φ(ℓ+ 3) 4
Γ(α+ℓ+ 1) Γ(α)Γ(ℓ+ 2)
+Φ(ℓ+ 3) 8
Γ(α+ℓ+ 1) Γ(α−1)Γ(ℓ+ 3)
.
From the above inequality, we have
kT xk ≤β
Γ(α+ℓ+ 1) 2Γ(α−1)Γ(ℓ+ 1)
Φ. (16)
Finally, by the definition of Φ given in (15), we get that (16) implies that
kT xk ≤M. (17)
5. Examples
This section provides two examples to illustrate Theorems 8 and 9. We start with an example illustrating Theorem 8, subsequently by an example illustrat-ing Theorem 9.
Example 10. Suppose that α = 52 and ℓ= 5. Let f(k, x) = 1000+cosx(kk)2 and
L= 10001 . Then discrete fractional antiperiodic boundary value problem (1)-(2) becomes
C
0∆
5 2
kx(k) =
cos x k+32
1000 + k+322, k ∈[0,7], (18) subject to the conditions
x
−1 2
=−x
15
2
, ∆x
−1 2
=−∆x
15 2
,
∆2x
−1 2
=−∆2x
15 2
.
(19)
In this case, inequality (13) takes the form
Λ =βL Γ(α+ℓ+ 1)
2Γ(α−1)Γ(ℓ+ 1) ≤0.0853<1.
Therefore, from Theorem 8 we conclude that boundary value problem (18) -(19) has a unique solution. For the values of α ∈ (2,3] (with an increment of 0.1), the corresponding values of Λ are computed and tabulated in Table 1. Taking the values of α on the horizontal axis, the computed values of Λ are plotted in Figure 1. The resulting curve is nonlinear and shows monotonically increasing trend inα∈(2,3] (with an increment of 0.025).
α 2.1 2.2 2.3 2.4 2.5
Λ =βL2Γ(Γ(α−1)Γ(α+ℓ+1)ℓ+1) 0.0528 0.0597 0.0674 0.0759 0.0853
α 2.6 2.7 2.8 2.9 3
Λ =βL2Γ(Γ(α−1)Γ(α+ℓ+1)ℓ+1) 0.0956 0.1069 0.1194 0.1331 0.1480
Table 1: Illustration of 2< α≤3 and Λ.
2 2.2 2.4 2.6 2.8 3 0.04 0.06 0.08 0.1 0.12 0.14 0.16
α∈ (2,3]
Λ
Figure 1: α versus Λ.
Example 11. Suppose that α= 145, ℓ= 7 and M = 1000 withf(k, x) = 1
21ke−
1 10x
2
(k). Then discrete fractional antiperiodic boundary value problem (1)-(2) takes the form
C
0∆
14 5
k x(k) =
1 21
k+9
5
e−101x 2
(k+9
5), k∈[0,9], (20)
subject to the conditions
x
−1 5
=−x
49 5 , ∆x −1 5
=−∆x
49 5
,
∆2x
−1 5
=−∆2x 49 5 . (21)
The Banach space isB :=
x∈R11:kxk ≤1000 . We note that M
β
Γ(α+ℓ+ 1) 2Γ(α−1)Γ(ℓ+ 1)
=
1000
236.4345 ≈4.2295.
6. Conclusions
Arguments in this paper are based on a contraction mapping theorem and Brouwer fixed point theorem for Banach spaces. Conditions for the existence and uniqueness of solutions for the fractional order difference equations in the presence of antiperiodic boundary conditions are obtained. Results obtained in the main results are illustrated with suitable examples supported with numerical calculations.
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