Three-Point Fractional h-sum Boundary Value Problems for ... - PMF Niš

Filomat 31:18 (2017), 5727–5742 https://doi.org/10.2298/FIL1718727S

Published by Faculty of Sciences and Mathematics, University of Niˇs, Serbia Available at: http://www.pmf.ni.ac.rs/filomat

Three-Point Fractional h-sum Boundary Value Problems for Sequential Caputo Fractional h-sum-difference Equations Jarunee Soontharanona , Jiraporn Reunsumrita , Thanin Sitthiwiratthamb a Department

of Mathematics, Faculty of Applied Science, King Mongkut’s University of Technology North Bangkok, Bangkok 10800, Thailand b Mathematics Department, Faculty of Science and Technology, Suan Dusit University, Bangkok 10700, Thailand

Abstract. In this article, we study an existence and uniqueness results for a sequential nonlinear Caputo fractional h-sum-difference equation with three-point fractional h-sum boundary conditions, by using the Banach contraction principle and the Schauder’s fixed point theorem. Our problem contains different orders in three fractional difference operators and three fractional sums. Finally, we provide an example to displays the importance of these results.

1. Introduction Fractional difference equations can be used for describing many problems in the real-world phenomena such as physics, mechanics, chemistry, control systems, electrical networks, and flow in porous media. In particular, fractional calculus appears in the studied in biology, ecology and other areas (see [1]-[2]). Mathematicians have used this fractional calculus to model and solve various related problems. Boundary value problems for fractional difference equations, which have helped to build up some of the basic theory of this field can be seen in the textbooks [3] and the papers [4]-[36] and references cited therein. There is a development of boundary value problems for sequential fractional difference equations which shows an operation of the examinative function. The study may also have another function which is related to our interested one. These creations are incorporating with nonlocal conditions which are both extensive and more complex. For example, Goodrich [15] considered a discrete fractional boundary value problem for a sequential fractional difference equations of the forms    −∆µ1 ∆µ2 ∆µ3 y(t) = f (t + µ1 + µ2 + µ3 − 1, y(t + µ1 + µ2 + µ3 − 1)),    y(0) = 0 = y(b + 2),

(1)

where t ∈ N2−µ1 −µ2 −µ3 ,b+2−µ1 −µ2 −µ3 , 0 < µ1 , µ2 , µ3 < 1, 1 < µ2 + µ3 < 2, 1 < µ1 + µ2 + µ3 < 2 and f : N0 × R → [0, +∞) is a continuous function. 2010 Mathematics Subject Classification. Primary 39A05; Secondary 39A12 Keywords. Fractional h-sum-difference equations, boundary value problem, existence, uniqueness Received: 20 April 2017; Accepted: 27 September 2017 Communicated by Dragan S. Djordjevi´c Research supported by King Mongkut’s University of Technology North Bangkok. Contract no.KMUTNB-GOV-59-34 Email addresses: [email protected] (Jarunee Soontharanon), [email protected] (Jiraporn Reunsumrit), thanin [email protected] (Thanin Sitthiwirattham)

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Weidong [22] examined the sequential Caputo fractional boundary value problem with a p-Laplacian  β  ∆ [φp (∆αC x)](t) = f (t + α + β − 1, x(t + α + β − 1)), t ∈ N0,b ,     Cβ β (2) ∆C x(β − 1) + ∆C x(β + b) = 0,     x(α + β − 2) + x(α + β + b) = 0, where 0 < α, β ≤ 1, 1 < α + β ≤ 2, f : Nα+β−1,α+β+T−1 × R → R is a continuous function and φp is the p-Laplacian operator. Recently, Sitthiwirattham [29] investigated three-point fractional sum boundary value problems for sequential Caputo fractional difference equations of the forms  β α   ∆α (∆α+β−1 + λEβ )x(t) = f (t + α + β − 1, x(t + α + β − 1)), (3)   x(α + β − 2) = 0, x(α + β + T) = ρ∆−γ x(η + γ), α+β−1 where t ∈ N0,T , 0 < α, β ≤ 1, 1 < α+β ≤ 2, 0 < γ ≤ 1, η ∈ Nα+β−1,α+β+T−1 , ρ is a constant, f : Nα+β−2,α+β+T ×R → R is a continuous function, Eβ x(t) = x(t + β − 1). Presently, we focus on the difference operator with step 1. Our knowledge, there is a gap in the literature about the details of this operation. To make it more general and flexible in the sense that it has the freedom to choose the different step h. However, not much research has involved the development of h-sum and h-difference operators (see [37]-[43]). We find that the boundary value problem for h-difference equations has not been studied. The results mentioned above are the motivation for this research. In this paper, we study a sequential Caputo fractional h-sum-difference equation with three point fractional h-sum boundary value conditions of the form "          ω ω β α ω λ ω−1 h, u t + α + β − 1 + h , C ∆h C ∆h ∆C + (e − 1)∆C E u(t) = F t + α + β − 1 + h h  #   ω γ (Ψh u) t + α + β + γ − 1 + h , t ∈ (hN)0,Th h    ω u α+β−2+ h = u(η) = 0, h       ω ω ∆−θ 1 T + α + β + θ + h u T + α + β + θ + h = 0, (4) h h h where α, β, ω, γ, θ ∈ (0, 1], 2 < α + β + ω ≤ 3, (hN)0,Th := {0, h, 2h, . . . , Th}, η ∈ (hN)[α+β−1+ ω ]h,[T+α+β−1+ ω ]h , h h     Eu(t) = u(t + 1), F ∈ C (hN)[α+β−2+ ω−1 ]h,[T+α+β+ ω ]h × R2 , R , 1 ∈ C (hN)[α+β−2+ ω ]h,[T+α+β+ ω ]h , R+ , and for h h h h   ϕ ∈ C (hN)[α+β−2+ ω ]h,[T+α+β+ ω ]h × (hN)[α+β−2+ ω ]h,[T+α+β+ ω ]h , [0, ∞) , h

γ

h

h

−γ

Ψh u)(t) := [∆h ϕ u](t + γ) =

h Γh (γ)

h

t h

X s=α+β−2+ ωh

γ−1

(t − σ(hs))h

  ϕ t + γh, s u(hs).

Noting from our problem, there are three sequential of two Caputo fractional h-difference operators of order α, β and a ω-order Caputo fractional difference operator, moreover, there exist two fractional h-sum of order γ, θ and a (1 − ω)-order fractional sum. So, this problem is complicated more than the previous works. The rest of paper is organized as follows. In Section 2 we provide some definitions and basic lemmas. Next, we convert the problem to an equivalent summation equation in order to derive a representation for the solution of (4). In Section 3, we prove existence and uniqueness results of the problem (4) by employing the Banach contraction principle and the Schauder’s theorem. Furthermore, we also show the existence of a positive solution to (4). Finally an illustrative example is given in Section 4.

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2. Preliminaries In the following, there are basic notations, definitions, and lemmas which are used to get the main results. Definition 2.1. For any t, α ∈ R and h > 0, the h-falling function is defined by   t Γ + 1 h α , th := hα  t Γ h +1−α where

t h

α

+ 1 < Z− ∪ {0}, and we use the convention that division at a pole yields zero. If h = 1, then th = tα .

Definition 2.2. For α, h > 0 and f define on (hN)a := {a, a + h, a + 2h, . . .}, the α-order fractional h-sum of f is defined by t h −α h X α−1 −α (t − σ(hs))h f (hs), ∆h f (t) := Γ(α) a s= h

where t ∈ Na+αh := {a + αh, a + (α + 1)h, a + (α + 2)h, ...} and σ(hs) = (s + 1)h. If h = 1, then ∆−α f (t) = ∆−α f (t). h Definition 2.3. For α > 0 and f define on (hN)a , the α-order Caputo fractional h-difference of f is defined by α C ∆h f (t)

:=

−(N−α) N ∆h ∆h

h f (t) = Γ(N − α)

t h −(N−α)

X

N−α−1

(t − σ(hs))h

∆N h f (sh),

s= ha

where t ∈ Na+(N−α)h and N ∈ N is chosen so that 0 ≤ N − 1 < α < N. f (t), and if h = 1 then C ∆αh f (t) = ∆αC f (t). If α = N then C ∆αh f (t) = ∆N h To present the solution of the boundary value problem (4), we need the following lemma that deals with a linear variant of the boundary value problem (4).   Lemma 2.4. Let α, β, ω, θ ∈ (0, 1], 2 < α+β+ω ≤ 3, η ∈ (hN)[α+β−1+ ω ]h,[T+α+β−1+ ω ]h , 1 ∈ C (hN)[α+β−2+ ω ]h,[T+α+β+ ω ]h , R+ h h h h   and f ∈ C (hN)[α+β−1+ ω ]h,[T+α+β−1+ ω ]h , R be given. Then for t ∈ (hN)0,Th , the problem h

h

     ω β α ω λ ω−1 h , t+α+β−1+ C ∆h C ∆h ∆C + (e − 1)∆C E u(t) = f h  h i    β − 2 + ωh h = u(η) = 0, u α + h i  h i    −θ ∆ 1 T + α + β + θ + ωh h u T + α + β + θ + ωh h = 0, h 

(5) (6)

has the unique solution eλt u(t) =

P[h] ΛΓ(ω − 1)

t−1 X

z−ω+1 X

eλz (z − σ(s))ω−2

z=[α+β−2+ ωh ]h s=α+β−2

hQ[h] − ΛΓ(ω − 1)Γ(β)

t−1 X

z−ω+1 X

s

h −β X

β−1

eλz (z − σ(s))ω−2 (s − σ(hv))h

z=[α+β−2+ ωh ]h s=α+β−2 v=α−1 s

v

t−1 z−ω+1 h −β X h −α X X X β−1 h2 α−1 + eλz (z − σ(s))ω−2 (s − σ(hv))h (v − σ(hx))h × Γ(ω − 1)Γ(β)Γ(α) z=[α+β−2+ ωh ]h s=α+β−2 v=α−1 x=0    ω f x+α+β−1+ h h

(7)

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where the functionals P[ f ] and Q[ f ] are defined as η−1 X

" P[ f ] =

z−ω+1 X

α+β−2+ ωh

z=[

"

r−1 X

X

r=[

ΛΓ(ω − 1)Γ(β)

]h s=α+β−2 v=α−1

T+α+β+ ωh α+β−2+ ωh

s β−1 # h −β X heλ(z−η) (z − σ(s))ω−2 (s − σ(hv))h

h i θ−1 s v h −β X h −α h3 eλ(z−r) X T + α + β + θ + ωh h − σ(hr)

z−ω+1 X

α+β−2+ ωh

]h z=[

×

h

Γ(θ)Γ(ω − 1)Γ(β)Γ(α)

]h s=α+β−2 v=α−1 x=0

×

   #  ω ω  h f x+α+β−1+ h (z − σ(s)) (s − σ(hv))h (v − 1 T+α+β+θ+ h h h i θ−1 ω s " T+α+β+ r−1 z−ω+1 h −β X X X X h T + α + β + θ + ωh h − σ(hr) eλ(z−r) h2 (z − σ(s))ω−2 h − × Γ(θ)Γ(ω − 1) r=α+β−2+ ωh z=[α+β−2+ ω ]h s=α+β−2 v=α−1 h s v β−1 η−1 z−ω+1  # " h −β X h −α X X X  (s − σ(hv))h (z − σ(s))ω−2 ω  1 T+α+β+θ+ h · × Γ(β) h Γ(ω − 1) z=[α+β−2+ ωh ]h s=α+β−2 v=α−1 x=0 β−1 α−1  #  h2 eλ(z−η) (s − σ(hv))h (v − σ(hx))h ω h , (8) f x+α+β−1+ Γ(β)Γ(α) h s ω v " η−1 z−ω+1 r−1 z−ω+1 h −β X h −α X X eλ(z−η) (z − σ(s))ω−2 # " T+α+β+ X h X X X Q[ f ] = · h3 eλ(z−r) × Γ(ω − 1) ω r=α+β−2+ h z=[α+β−2+ ω ]h s=α+β−2 v=α−1 x=0 z=[α+β−2+ ωh ]h s=α+β−2 h h i θ−1 β−1 α−1 ω  T + α + β + h h − σ(hr) (z − σ(s))ω−2 (s − σ(hv))h (v − σ(hx))h  ω  h 1 T+α+β+θ+ h× Γ(θ)Γ(ω − 1)Γ(β)Γ(α) h i θ−1 h ω r−1 z−ω+1    # " T+α+β+ X h X X T + α + β + θ + ωh h − σ(hr) ω h f x+α+β−1+ h − × h Γ(θ)Γ(β)Γ(ω − 1) ω r=α+β−2+ h z=[α+β−2+ ω ]h s=α+β−2 h s v # " η−1 z−ω+1   h −β X h −α X X X  ω  2 λ(z−r) ω−2 he (z − σ(s)) 1 T+α+β+θ+ h · h2 eλ(z−η) × h z=[α+β−2+ ωh ]h s=α+β−2 v=α−1 x=0 β−1 α−1   # (z − σ(s))ω−2 (s − σ(hv))h (v − σ(hx))h ω f x+α+β−1+ h , (9) Γ(ω − 1)Γ(β)Γ(α) h β−1

ω−2

α−1 σ(hx))h

and " Λ =

T+α+β+ ωh

r−1 X

X

z−ω+1 X

h i θ−1 s h −β h2 eλ(z−r) X (z − σ(s))ω−2 T + α + β + θ + ωh h − σ(hr) h

Γ(θ)Γ(ω − 1)Γ(β)

r=α+β−2+ ωh z=[α+β−2+ ω ]h s=α+β−2 v=α−1 h β−1

(s − σ(hv))h " −

η−1 X

 # "  ω  h · 1 T+α+β+θ+ h z−ω+1 X

η−1 X z=[α+β−2+ ωh ]h

z−ω+1 X

e−λz (z − σ(s))ω−2 Γ(ω − 1) s=α+β−2

s β−1 # h −β X he−λz (z − σ(s))ω−2 (s − σ(hv))h

z=[α+β−2+ ωh ]h s=α+β−2 v=α−1

ΛΓ(ω − 1)Γ(β)

×

#

×

J. Soontharanon et al. / Filomat 31:18 (2017), 5727–5742

"

T+α+β+ ωh

r−1 X

X

r=α+β−2+ ωh

z−ω+1 X

5731

i θ−1 h h2 eλ(z−r) T + α + β + θ + ωh h − σ(hr) (z − σ(s))ω−2 h

Γ(θ)Γ(β)Γ(ω − 1)

z=[α+β−2+ ωh ]h s=α+β−2

×

 #  ω  h . 1 T+α+β+θ+ h

(10)

Proof. We use the fractional h-sum of order α for (5) to obtain β C ∆h

h

t α−1    h −α X i (t − σ(hs))h ω ω λ ω−1 ∆C + (e − 1)∆C E u(t) = C0 + h h , f s+α+β−1+ Γ(α) h s=0

(11)

for t ∈ N(α−1)h,(T+α)h . In addition using the fractional h-sum of order β for (11), we get λ ω−1 ∆ω C u(t) + (e − 1)∆C Eu(t) β−1

t

= C1 + C0 h

h −β X (t − σ(hs))

h

Γ(β)

s=0

t

+h

2

s

h −β X h −α X (t − σ(hs))

β−1 h

α−1

(s − σ(hv))h

Γ(β)Γ(α)

s=0 v=0

 f

v+α+β−1+

  ω h , h

(12)

for t ∈ N(α+β−2)h,(T+α+β)h . Taking the fractional sum of order ω for (12), we get u(t) + (eλ − 1)∆−1 u(t + 1) β−1

s

t−ω t−ω h −β X X X (t − σ(s))ω−1 (s − σ(vh))h (t − σ(s))ω−1 = C2 + C1 + C0 h Γ(ω) Γ(ω)Γ(β) s=α+β−2 s=α+β−2 v=α−1

s h −β

X X h2 (t − σ(s))ω−1 (s − σ(vh))

t−ω X

+

v h −α

β−1 h

α−1

(v − σ(xh))h

Γ(ω)Γ(β)Γ(α)

s=α+β−2 v=α−1 x=0

 f

x+α+β−1+

  ω h , h

(13)

for t ∈ N[α+β−2+ ω ]h,[T+α+β+ ω ]h . Next, taking the forward difference ∆ for (13), we have h

h

∆u(t) + (eλ − 1)u(t + 1) β−1

s

t−ω+1 h −β X X h (t − σ(s))ω−2 (s − σ(vh))h (t − σ(s))ω−2 = C1 + C0 Γ(ω − 1) Γ(ω − 1)Γ(β) s=α+β−2 s=α+β−2 t−ω+1 X

v=α−1

+

t−ω+1 X

s h −β

β−1

v h −α

X X h2 (t − σ(s))ω−2 (s − σ(vh)) h

s=α+β−2 v=α−1 x=0

α−1

(v − σ(xh))h

Γ(ω − 1)Γ(β)Γ(α)

f

   ω x+α+β−1+ h , h

(14)

multiply the equation (14) by eλt , we obtain h i ∆ eλt u(t) = eλt ∆u(t) + (eλ − 1)eλt u(t + 1) s β−1 ( t−ω+1 t−ω+1 h −β X (t − σ(s))ω−2 X X h (t − σ(s))ω−2 (s − σ(vh))h λt = e C1 + C0 Γ(ω − 1) Γ(ω − 1)Γ(β) s=α+β−2 s=α+β−2 v=α−1

+

t−ω+1 X

s h −β

v h −α

X X h2 (t − σ(s))ω−2 (s − σ(vh))

s=α+β−2 v=α−1 x=0

β−1 h

α−1

(v − σ(xh))h

Γ(ω − 1)Γ(β)Γ(α)

  ) ω f x+α+β−1+ h , h

(15)

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for t ∈ N[α+β−2+ ω ]h,[T+α+β+ ω ]h . Using the sum ∆−1 for (15), h

h

λt

e u(t) t−1 X

= C2 + C1

z=[α+β−2+ ωh ]h

+C0

eλz (z − σ(s))ω−2 Γ(ω − 1) s=α+β−2 z−ω+1 X

z−ω+1 X

t−1 X

β−1

s

h −β X heλz (z − σ(s))ω−2 (s − σ(vh))h

z=[α+β−2+ ωh ]h s=α+β−2 v=α−1 t−1 X

+

z−ω+1 X

(16)

s

Γ(ω − 1)Γ(β) β−1

v

h −β X h −α X h2 eλz (z − σ(s))ω−2 (s − σ(vh))h

Γ(ω − 1)Γ(β)Γ(α)

z=[α+β−2+ ωh ]h s=α+β−2 v=α−1 x=0

α−1

(v − σ(xh))h

 f

x+α+β−1+

  ω h , h

h i for t ∈ N[α+β−2+ ω ]h,[T+α+β+ ω ]h . By substituting t = α + β − 2 + ωh h and t = η into (16), and employing the h h first and second conditions of (6), we find that C2 = 0 and C1

η−1 X

z−ω+1 X

z=[α+β−2+ ωh ]h

+C0

η−1 X

eλ(z−η) (z − σ(s))ω−2 Γ(ω − 1) s=α+β−2 z−ω+1 X

=−

z−ω+1 X

β−1

s

h −β X heλ(z−η) (z − σ(s))ω−2 (s − σ(hv))h

z=[α+β−2+ ωh ]h s=α+β−2 v=α−1 η−1 X

(17)

ΛΓ(ω − 1)Γ(β)

s v    h −β X h −α X β−1 h2 eλ(z−η) (z − σ(s))ω−2 ω α−1 (s − σ(hv))h (v − σ(hx))h f x + α + β − 1 + h . Γ(ω − 1)Γ(β)Γ(α) h x=0

z=[α+β−2+ ωh ]h s=α+β−2 v=α−1

h i Using the fractional h-sum of order θ for (16), and substituting t = T + α + β + θ + ωh h, and employing the third condition of (6), this implies h i θ−1 T+α+β+ ωh r−1 z−ω+1 X X h2 eλ(z−r) T + α + β + θ + ωh h − σ(hr) X (z − σ(s))ω−2 h C1 × Γ(θ)Γ(β)Γ(ω − 1) r=α+β−2+ ωh z=[α+β−2+ ω ]h s=α+β−2 h    ω h (18) 1 T+α+β+θ+ h h i θ−1 s T+α+β+ ωh r−1 z−ω+1 h −β h2 eλ(z−r) X X X X T + α + β + θ + ωh h − σ(hr) (z − σ(s))ω−2 h +C0 × Γ(θ)Γ(ω − 1)Γ(β) r=α+β−2+ ωh z=[α+β−2+ ω ]h s=α+β−2 v=α−1 h    β−1 ω (s − σ(hv))h 1 T + α + β + θ + h h i θ−1 h s v T+α+β+ ωh r−1 z−ω+1 h −β X h −α h3 eλ(z−r) X X X X (z − σ(s))ω−2 T + α + β + θ + ωh h − σ(hr) h =− × Γ(θ)Γ(ω − 1)Γ(β)Γ(α) r=α+β−2+ ωh z=[α+β−2+ ω ]h s=α+β−2 v=α−1 x=0 h       β−1 ω ω α−1 (s − σ(hv))h (v − σ(hx))h 1 T + α + β + θ + h f x+α+β−1+ h . h h Solving the system of equations (17) and (18), we obtain C1 = P[ f ] and C0 = −Q[ f ] where P[ f ], Q[ f ] are defined by (8), (9), respectively. Substituting the constants C0 , C1 and C2 into (15), we obtain the solution (7).

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3. Main Results In this section, we wish to prove the existence solution to the problem (4). To accomplish this, let C =   C (hN)[α+β−2+ ω ]h,[T+α+β+ ω ]h , R be a Banach space of all continuous functions u from (hN)[α+β−2+ ω ]h,[T+α+β+ ω ]h h h h h to R, with the norm defined by kukC = max |u(t)|. t∈(hN) α+β−2+ ω h, T+α+β+ ω h [ h] [ h] We consider the operator A : C → C by (Au)(t) =

e−λt P[Fu ] ΛΓ(ω − 1)

z−ω+1 X

t−1 X

eλz (z − σ(s))ω−2

z=[α+β−2+ ωh ]h s=α+β−2

he−λt Q[Fu ] − ΛΓ(ω − 1)Γ(β)

t−1 X

s

h −β X

z−ω+1 X

β−1

eλz (z − σ(s))ω−2 (s − σ(hv))h

z=[α+β−2+ ωh ]h s=α+β−2 v=α−1

h2 e−λt + Γ(ω − 1)Γ(β)Γ(α)

z−ω+1 X

t−1 X

s

v

h −β X h −α X

β−1

eλz (z − σ(s))ω−2 (s − σ(hv))h ×

z=[α+β−2+ ωh ]h s=α+β−2 v=α−1 x=0

      ω ω h, u x + α + β − 1 + h , F x+ α+β−1+ h h    ! ω γ (Ψh u) x + α + β + γ − 1 + h , h α−1

(v − σ(hx))h

(19)

where the functionals P[Fu ] and Q[Fu ] are defined by η−1 X

" P[Fu ] = z=[

"

α+β−2+ ωh

T+α+β+ ωh

X

z−ω+1 X

s β−1 # h −β X heλ(z−η) (z − σ(s))ω−2 (s − σ(hv))h

ΛΓ(ω − 1)Γ(β)

]h s=α+β−2 v=α−1 r−1 X

z−ω+1 X

s

v

h −β X h −α X

×

h i θ−1 T + α + β + θ + ωh h − σ(hr)

h × Γ(θ)Γ(ω − 1)Γ(β)Γ(α) r=α+β−2+ ωh z=[α+β−2+ ω ]h s=α+β−2 v=α−1 x=0 h    β−1 ω α−1 h3 eλ(z−r) (z − σ(s))ω−2 (s − σ(hv))h (v − σ(hx))h 1 T + α + β + θ + h × h           !# ω ω ω γ F x+ α+β−1+ h, u x + α + β − 1 + h , (Ψh u) x + α + β + γ − 1 + h h h h h i θ−1 ω s " T+α+β+ r−1 z−ω+1 h −β X h X X X T + α + β + θ + ωh h − σ(hr) eλ(z−r) h − × Γ(θ) r=α+β−2+ ωh z=[α+β−2+ ω ]h s=α+β−2 v=α−1 h β−1   # h2 (z − σ(s))ω−2 (s − σ(hv))h ω 1 T+α+β+θ+ h × Γ(ω − 1)Γh (β) h s v β−1 " α−1 η−1 z−ω+1 h −β X h −α X X X h2 eλ(z−η) (z − σ(s))ω−2 (s − σ(hv))h (v − σ(hx))h × Γ(ω − 1)Γ(β)Γ(α) z=[α+β−2+ ωh ]h s=α+β−2 v=α−1 x=0           !# ω ω ω γ F x+ α+β−1+ h, u x + α + β − 1 + h , (Ψh u) x + α + β + γ − 1 + h , h h h

(20)

J. Soontharanon et al. / Filomat 31:18 (2017), 5727–5742

" Q[Fu ] =

η−1 X z=[α+β−2+ ωh ]h

"

T+α+β+ ωh

X

5734

z−ω+1 X

# eλ(z−η) (z − σ(s))ω−2 × Γ(ω − 1) s=α+β−2 r−1 X

z−ω+1 X

s

v

h −β X h −α X

i θ−1 h T + α + β + θ + ωh h − σ(hr)

h × Γ(θ)Γ(ω − 1)Γ(β)Γ(α) z=[α+β−2+ ωh ]h s=α+β−2 v=α−1 x=0    β−1 ω α−1 3 λ(z−r) ω−2 h e (z − σ(s)) (s − σ(hv))h (v − σ(hx))h 1 T + α + β + θ + h × h          !#  ω ω ω γ h, u x + α + β − 1 + h , (Ψh u) x + α + β + γ − 1 + h F x+ α+β−1+ h h h i θ−1 h ω " T+α+β+ r−1 z−ω+1 X X h2 T + α + β + θ + ωh h − σ(hr) X h eλ(z−r) h − × Γ(θ)Γ(β) r=α+β−2+ ωh z=[α+β−2+ ω ]h s=α+β−2 h  # ω−2  (z − σ(s)) ω 1 T+α+β+θ+ h × Γ(ω − 1) h s v β−1 " α−1 η−1 z−ω+1 h −β X h −α X X X h2 eλ(z−η) (z − σ(s))ω−2 (s − σ(hv))h (v − σ(hx))h × Γ(ω − 1)Γ(β)Γ(α) z=[α+β−2+ ωh ]h s=α+β−2 v=α−1 x=0           !# ω ω ω γ F x+ α+β−1+ h, u x + α + β − 1 + h , (Ψh u) x + α + β + γ − 1 + h , h h h r=α+β−2+ ωh

(21)

and Λ is defined by (10). We find that the problem (4) has solutions if and only if the operator A has fixed points. Next, we present the existence and uniqueness of a solution to the problem (4), by using the Banach contraction principle. Theorem 3.1. Assume that F : (hN)[α+β−2+ ω ]h,[T+α+β+ ω ]h ×R×R → R are continuous, ϕ : (hN)[α+β−2+ ω ]h,[T+α+β+ ω ]h × h h h h n (hN)[α+β−2+ ω ]h,[T+α+β+ ω ]h → [0, ∞) is continuous with ϕ0 = max ϕ(t+γh, s) : (t, s) ∈ (hN)[α+β−2+ ω ]h,[T+α+β+ ω ]h × h h h h o (hN)[α+β−2+ ω ]h,[T+α+β+ ω ]h . In addition, suppose that: h

h

(H1 ) There exist constants L1 , L2 > 0 such that for each t ∈ (hN)[α+β−2+ ω ]h,[T+α+β+ ω ]h and u, v ∈ R h

h

F(t, u, Ψγh u) − F(t, v, Ψγh v) ≤ L1 |u − v| + L2 |(Ψγ u) − (Ψγ v)|. (H2 ) 0 < 1(t) < G for each t ∈ (hN)[α+β−2+ ω ]h,[T+α+β+ ω ]h . h

h

 γ    ϕ0 (T+2)h    (H3 ) L1 + L2 Γ(γ+1) h  χ < 1.

Then the problem (4) has a unique solution on (hN)[α+β−2+ ω ]h,[T+α+β+ ω ]h , h

h

J. Soontharanon et al. / Filomat 31:18 (2017), 5727–5742

5735

where  θ  ω−1   β ω −λ T + β + h + (1 − h) ωh e−λ (η − α − β)ω−1 (η − ω + (1 − α)h)h e G h(T + θ + 2) h T + h Ω1 = · Γ(ω)Γ(β + 1) Γ(θ + 1)Γ(ω)Γ(β + 1)  h  iα  η−ω α " 1 T + α + (1 − h) ω + β # h h h −β h h + |Λ|Γ(α + 1) Γ(α + 1) ω−1  θ  " −λ # ω ω−1 e−λ G h(T + θ + 2) T + e (η − α − β) h h Ω2 = · × Γ(ω) Γ(θ + 1)Γ(ω)      " T + β + h + (1 − h) ω + α β (η − ω + (1 − α)h)β η−ω − β α # h h h h h + Γ(β + 1) Γ(β + 1)Γ(α + 1) β  ω−1  α (T + β + 1)h (T + α)h e−λ Th + ω + (h − 1)(α + β) h Ω3 = , Γ(ω)Γ(β + 1)Γ(α + 1) "

+α

 β

(22)

(23)

(24)

and χ :=

 ω−1 " e−λ Th + ω + (h − 1)(α + β) |Λ|Γ(ω)

 Ω1 + Ω2

β (T + β + 1)h # h

Γ(β + 1)

+ Ω3 .

(25)

Proof. We show that A is a contraction. For any u, v ∈ C, we have P[Fu ] − P[Fv ] s β−1 # " η−1 z−ω+1 h −β X X X heλ(z−η) (z − σ(s))ω−2 (s − σ(hv))h × = ΛΓ(ω − 1)Γ(β) ω s=α+β−2 v=α−1 z=[α+β−2+ h ]h h i θ−1 s v ω " T+α+β+ r−1 z−ω+1 h −β X h −α h3 eλ(z−r) X h X X X T + α + β + ωh h − σ(hr) (z − σ(s))ω−2 h

Γ(θ)Γ(ω − 1)Γ(β)Γ(α)

r=α+β−2+ ωh z=[α+β+θ−2+ ω ]h s=α+β−2 v=α−1 x=0 h β−1

(s − σ(hv))h (v − " −

T+α+β+ ωh

r−1 X

X

#        ω γ γ 1 T+α+β+θ+ h F x, u, (Ψh u) − F x, v, (Ψh v) h i θ−1 h s z−ω+1 h −β h2 eλ(z−r) X X T + α + β + θ + ωh h − σ(hr) (z − σ(s))ω−2

α−1 σ(hx))h

h

Γ(θ)Γ(ω − 1)Γ(β)

r=α+β−2+ ωh z=[α+β−2+ ω ]h s=α+β−2 v=α−1 h β−1

(s − σ(hv))h

β−1

×

   #" ω 1 T+α+β+θ+ h h α−1

(s − σ(hv))h (v − σ(hx))h

η−1 X

z−ω+1 X

s

×

v

h −β X h −α X h2 eλ(z−η) (z − σ(s))ω−2 × Γ(ω − 1) x=0

z=[α+β−2+ ωh ]h s=α+β−2 v=α−1

#     F x, u, (Ψγ u) − F x, v, (Ψγ v) h h

Γ(β)Γ(α)  γ   β# " −λ   ϕ t − ω − (α + β − 2)h e (η − α − β)ω−1 (η − ω + (1 − α)h)h 0   h ≤ L1 + L2 ×  ku − vkC   Γ(γ + 1) |Λ|Γ(ω)Γ(β + 1)  ω−1  β  h  iα  # " 1 ω T + β + h + (1 − h) ωh + α T + α + (1 − h) + β  θ T + ωh h h h h e−λ G h(T + θ + 2) h Γ(θ + 1)Γ(ω)Γ(β + 1)Γ(α + 1)

# h

×

J. Soontharanon et al. / Filomat 31:18 (2017), 5727–5742

5736

  γ   ϕ0 t − ω − (α + β − 2 h)h    ku − vkC × + L1 + L2   Γ(γ + 1)        " e−λ G h(T + θ + 2) θ T + ω ω−1 T + β + h + (1 − h) ω + α β # h h h h × Γ(θ + 1)Γ(ω)Γ(β + 1)   " e−λ (η − α − β)ω−1 (η − ω + (1 − α)h)β η−ω − β α # h h h

Γ(ω)Γ(β + 1)Γ(α + 1)   γ   ϕ t − ω − (α + β − 2 h)h  0   , = ku − vkC Ω1 L1 + L2   Γ(γ + 1) and Q[Fu ] − Q[Fv ] " η−1 z−ω+1 X eλ(z−η) (z − σ(s))ω−2 # X × = Γ(ω − 1) ω s=α+β−2 z=[α+β−2+ h ]h h i θ−1 s v ω " T+α+β+ z−ω+1 r−1 h −β X h −α h3 eλ(z−r) X h X X X T + α + β + θ + ωh h − σ(hr) (z − σ(s))ω−2 h

Γ(θ)Γ(ω − 1)Γ(β)Γ(α)

r=α+β−2+ ωh z=[α+β−2+ ω ]h s=α+β−2 v=α−1 x=0 h β−1

(s − σ(hv))h (v − "

T+α+β+ ωh

X

−

#       ω γ γ 1 T+α+β+θ+ h F x, u, (Ψh u) − F x, v, (Ψh v) h h i θ−1 z−ω+1 X h2 eλ(z−r) T + α + β + θ + ωh h − σ(hr) (z − σ(s))ω−2 h × Γ(θ)Γ(β)Γ(ω − 1) s=α+β−2

α−1 σ(hx))h

r−1 X

r=α+β−2+ ωh z=[α+β−2+ ω ]h h



 # " ω 1 T+α+β+θ+ h · h 

η−1 X

z−ω+1 X

s

β−1

v

h −β X h −α X h2 eλ(z−η) (z − σ(s))ω−2 (s − σ(hv))h

z=[α+β−2+ ωh ]h s=α+β−2 v=α−1 x=0

Γ(ω − 1)Γ(β)Γ(α)

#     γ γ (v − F x, u, (Ψh u) − F x, v, (Ψh v)  γ     ϕ t − ω − (α + β − 2)h 0   h ≤ L1 + L2  ku − vkC ×   Γ(γ + 1) α−1 σ(hx))h

 θ  ω−1   β #" −λ ω T + β + h + (1 − h) ωh + α # e−λ (η − α − β)ω−1 e G h(T + θ + 2) h T + h h Γ(ω) Γ(θ + 1)Γ(ω)Γ(β + 1)       γ  " e−λ G h(T + θ + 2) θ T + ω ω−1 #  ϕ0 t − ω − (α + β − 2 h)h  h  h  ku − vkC + L1 + L2 ×   Γ(γ + 1) Γ(θ + 1)Γ(ω)   " e−λ (η − α − β)ω−1 (η − ω + (1 − α)h)β η−ω − β α # h h "

h

Γ(ω)Γ(β + 1)Γ(α + 1)   γ   ϕ0 t − ω − (α + β − 2 h)h    . = ku − vkC Ω2 L1 + L2   Γ(γ + 1)

×

×

J. Soontharanon et al. / Filomat 31:18 (2017), 5727–5742

5737

For each t ∈ (hN)[α+β−2+ ω−1 ]h,[T+α+β+ ω ]h , we have h

h

(Au)(t) − (Av)(t) −λt z−ω+1 t−1 e P[F ] − P[F ] X X u v eλz (z − σ(s))ω−2 = ΛΓ(ω − 1) ω z=[α+β−2+ h ]h s=α+β−2 s t−1 z−ω+1 h −β he−λt Q[Fu ] − Q[Fv ] X X X β−1 − eλz (z − σ(s))ω−2 (s − σ(hv))h ΛΓ(ω − 1)Γ(β) z=[α+β−2+ ωh ]h s=α+β−2 v=α−1 h2 e−λt + Γ(ω − 1)Γ(β)Γ(α)

t−1 X

z−ω+1 X

s

v

h −β X h −α X

β−1

α−1

eλz (z − σ(s))ω−2 (s − σ(hv))h (v − σ(hx))h ×

z=[α+β−2+ ωh ]h s=α+β−2 v=α−1 x=0

         ! ω ω ω γ h, u x + α + β − 1 + h , (Ψh u) x + α + β + γ − 1 + h F x+ α+β−1+ h h h   γ  ϕ0 t − ω − (α + β − 2 h)h  e−λ (t − α − β)ω−1   ≤ ku − vkC Ω1 L1 + L2   |Λ|Γ(ω) Γ(γ + 1)   γ β  ϕ0 t − ω − (α + β − 2 h)h  e−λ (t − α − β)ω−1 (t − ω + (1 − α)h)h  +ku − vkC Ω2 L1 + L2    |Λ|Γ(ω)Γ(β + 1) Γ(γ + 1)      γ " e−λ (t − α − β)ω−1 (t − ω + (1 − α)h)β t−ω − β α #   ϕ t − ω − (α + β − 2 h) 0  h h h  h + L1 + L2  ku − vkC   Γ(γ + 1) Γ(ω)Γ(β + 1)Γ(α + 1)   γ     β  ω−1    −λ        ϕ (T + 2)h (T + β + 1)h e Th + ω + (h − 1)(α + β) 0       h h Ω + Ω  ≤ ku − vkC L1 + L2 + Ω   1 2 3         Γ(γ + 1)   |Λ|Γ(ω) Γ(β + 1)     

= ku − vkC χ. From (H4 ), we have



(Au)(t) − (Av)(t)

< ku − vkC . C

Consequently, A is a contraction. By the Banach contraction principle, we hence get that A has a fixed point which is a unique solution of the problem (4) on t ∈ (hN)[α+β−2+ ω−1 ]h,[T+α+β+ ω ]h . h

h

The second result, we deduce the existence of at least of solution to (4) by using the following the Schauder’s fixed point theorem. Lemma 3.2. [44] (Arzel´a-Ascoli theorem) A set of functions in C[a, b] with the sup norm, is relatively compact if and only it is uniformly bounded and equicontinuous on [a, b]. Lemma 3.3. [44] If a set is closed and relatively compact then it is compact. Lemma 3.4. [45] (Schauder fixed point theorem) Let (D, d) be a complete metric space, U be a closed convex subset of D, and T : D → D be the map such that the set Tu : u ∈ U is relatively compact in D. Then the operator T has at least one fixed point u∗ ∈ U: Tu∗ = u∗ . Theorem 3.5. Suppose that (H1 ) − (H3 ) hold. Hence, problem (4) has at least one solution on Nα−n,T+α .

J. Soontharanon et al. / Filomat 31:18 (2017), 5727–5742

5738

Proof. We divide the proof into three steps. Step I. Verify A map bounded sets into bounded sets in BR = {u ∈ C : kukC ≤ R}. We consider n o   BR = u ∈ C (hN)[α+β−2+ ω−1 ]h,[T+α+β+ ω ]h : kukC ≤ R . h

Set

h

max |F(t, 0, 0)| = M and choose a constant h, T+α+β+ ω ]h [α+β−2+ ω−1 h ] [ h

t∈(hN)

χM  γ  ,   ϕ0 (T+2)h   1 − χ L1 + L2 Γ(γ+1) h 

R≥

(26)

where χ satisfies (H3). Denote that         !  ω ω ω γ S(t, u, 0) = F t + α + β − 1 + h, u t + α + β − 1 + h , (Ψh u) t + α + β − 1 + h h h h ! !     ω ω −F t+ α+β−1+ h, 0, 0 + F t + α + β − 1 + h, 0, 0 , h h for each u ∈ BR . We obtain

P[Fu ] =

" "

η−1 X

z−ω+1 X

s β−1 # h −β X heλ(z−η) (z − σ(s))ω−2 (s − σ(hv))h × ΛΓ(ω − 1)Γ(β)

z=[α+β−2+ ωh ]h s=α+β−2 v=α−1 T+α+β+ ωh

X

r=α+β−2+ ωh

r−1 X

z−ω+1 X

h i θ−1 s v h −β X h −α h3 eλ(z−r) X T + α + β + ωh h − σ(hr) h

z=[α+β+θ−2+ ωh ]h s=α+β−2 v=α−1 x=0 β−1

Γ(θ)Γ(ω − 1)Γ(β)Γ(α)

×

#    ω 1 T+α+β+θ+ h S(x, u, 0) h h i θ−1 s h −β X T + α + β + θ + ωh h − σ(hr) eλ(z−r) h × Γ(θ) α−1

(z − σ(s))ω−2 (s − σ(hv))h (v − σ(hx))h "

T+α+β+ ωh

X

−

r−1 X

z−ω+1 X

r=α+β−2+ ωh z=[α+β−2+ ω ]h s=α+β−2 v=α−1 h β−1

h2 (z − σ(s))ω−2 (s − σ(hv))h Γ(ω − 1)Γ(β) η−1 X

"

z−ω+1 X

s

  # ω 1 T+α+β+θ+ h × h

v

β−1

h −β X h −α X h2 eλ(z−η) (z − σ(s))ω−2 (s − σ(hv))

z=[α+β−2+ ωh ]h s=α+β−2 v=α−1 x=0

h

Γ(ω − 1)Γ(β)Γ(α)

 γ  "  # ϕ0 (T + 2)h   h  ≤ L1 + L2  kukC + M Ω1 .  Γ(γ + 1)  Similarly, we have

α−1

(v − σ(hx))h

 γ  "  # ϕ0 (T + 2)h   h Q[Fu ] ≤ L1 + L2  kukC + M Ω2 ,  Γ(γ + 1) 

S(x, u, 0)

#

J. Soontharanon et al. / Filomat 31:18 (2017), 5727–5742

and



Au

C

5739

 γ  "  # ϕ0 (T + 2)h   h ≤ L1 + L2  kukC + M χ.  Γ(γ + 1) 

Using (26), we get that kAukC ≤ R. This implies that A is uniformly bounded. Step II. Show that A is continuous on BR . From the continuity of F and 1, imply that the operator A is continuous on BR . Step III. Examine A is equicontinuous with BR . For any  > 0, there exist a positive constant δ∗ = max{δ1 , δ2 , δ3 } such that for t1 , t2 ∈ (hN)[α+β−2+ ω−1 ]h,[T+α+β+ ω ]h , h

h

|Λ|Γ(ω) ω−1 ω−1 − (t1 − α − β) < , whenever |t2 − t1 | < δ1 , (t2 − α − β) Ω1 e−λ kFk |Λ|Γ(ω)Γ(β + 1) β β ω−1 ω−1 , (t2 − α − β) (t2 − ω + (1 − α)h)h − (t1 − α − β) (t1 − ω + (1 − α)h)h < Ω2 e−λ kFk whenever |t2 − t1 | < δ2 , and α α Γ(ω)Γ(β + 1)Γ(α + 1) β  t −ω β  t −ω ω−1 ω−1 2 1 , (t2 −α−β) (t2 −ω+(1−α)h)h h − β h −(t1 −α−β) (t1 −ω+(1−α)h)h h − β h < e−λ kFk whenever |t2 − t1 | < δ3 . Then we obtain (Au)(t2 ) − (Au)(t1 ) t1 −1 t2 −1 z−ω+1 z−ω+1 X X X X P[Fu ] λ(z−t2 ) ω−2 λ(z−t1 ) ω−2 − e (z − σ(s)) e (z − σ(s)) ≤ |Λ|Γ(ω − 1) z=[α+β−2+ ωh ]h s=α+β−2 z=[α+β−2+ ωh ]h s=α+β−2 s t2 −1 z−ω+1 h −β X X X β−1 hQ[Fu ] eλ(z−t2 ) (z − σ(s))ω−2 (s − σ(hv))h + |Λ|Γ(ω − 1)Γ(β) z=[α+β−2+ ωh ]h s=α+β−2 v=α−1 s t1 −1 z−ω+1 h −β X X X β−1 λ(z−t1 ) ω−2 − e (z − σ(s)) (s − σ(hv))h ω z=[α+β−2+ h ]h s=α+β−2 v=α−1 s v t2 −1 z−ω+1 h −β X h −α X X X β−1 h2 kFk α−1 eλ(z−t2 ) (z − σ(s))ω−2 (s − σ(hv))h (v − σ(hx))h + Γ(ω − 1)Γ(β)Γ(α) z=[α+β−2+ ωh ]h s=α+β−2 v=α−1 x=0 s v t1 −1 z−ω+1 h −β X h −α X X X β−1 α−1 λ(z−t1 ) ω−2 − e (z − σ(s)) (s − σ(hv))h (v − σ(hx))h ω x=0 s=α+β−2 v=α−1 z=[α+β−2+ h ]h Ω2 e−λ kFk Ω1 e−λ kFk ω−1 ω−1 − (t1 − α − β) + ≤ × (t2 − α − β) |Λ|Γ(ω)Γ(β + 1) |Λ|Γ(ω) β β e−λ kFk ω−1 ω−1 × (t2 − α − β) (t2 − ω + (1 − α)h)h − (t1 − α − β) (t1 − ω + (1 − α)h)h + Γ(ω)Γ(β + 1)Γ(α + 1)  α  α β t2 − ω β t1 − ω ω−1 ω−1 − β − (t1 − α − β) (t1 − ω + (1 − α)h)h − β (t2 − α − β) (t2 − ω + (1 − α)h)h h h h h    < + + = . 3 3 3

J. Soontharanon et al. / Filomat 31:18 (2017), 5727–5742

5740

Therefore, A(BR ) is an equicontinuous set. As a consequence of Steps I to III together with the Arzel´a-Ascoli theorem, we get that A : C → C is completely continuous. Using the Schauder’s fixed point theorem, we can conclude that problem (4) has at least one solution. Our proof is ended. 4. An example In this section, an example to illustrate our result is provided as follow. Consider the following fractional h-difference boundary value problem   1   − cos2 (2πt+ 16 !# " 15 ) u t + 16 + Ψ 2 u t + 16 e 1 15 15 2 4 5 − 16 2 2 3 5 6   , C D 1 C D 1 DC + (e − 1)DC E u(t) =  4   2 2 16 t + 61 1 + u t + 15 15       3 293 17 77 293 − u =u = 0, D 1 4 esin(2π 120 ) u = 0, (27) 30 30 120 2 

 where t ∈ 12 N

0,6

  − 1 2t− 12  1  X t − σ 2s 2 e− 2s + 14 )  s    and Ψ 21 u (t) = u . 2 (t + 2)2 2 2Γ 21 s= 17 15

Letting h = and F =

e− cos

2 (2πt)

1 2 5 1 5 3 77 −s , α = , β = , γ = , ω = , θ = , T = 6, λ = 2, η = , 1(t) = esin(2πt) , ϕ(t+γh, s) = e 9 2 2 3 6 2 6 4 30 (t+ 2 ) 1

|u(t)|+|Ψ 21 u(t)|

,

2

(t+3)4 [1+|u(t)|]

we can show that |Λ| = 0.0601, Ω1 = 15.0543, Ω2 = 1.9134, Ω3 = 0.8616   (H1 ) − (H2 ) hold, for each t ∈ 21 N 17 293 , because we obtain

and χ = 130.312.

30 , 120

1 1 F[t, u, Ψ 2 u] − F[t, v, Ψ 2 v] ≤ 1 1 2

Finally, we can show that

1 e

2

e 570 |u

1

− v| +

1 2 570 |Ψ 1 u

1

− Ψ 21 v|,

2

< 1(t) < e and ϕ0 ≤ 6e .

2

 γ     ϕ (T + 2)h 0   h L1 + L2  χ = 0.859 < 1.  Γ(γ + 1) 

Hence, by Theorem 3.1, the problem 27 has a unique solution on





1 2N

17 293 30 , 120

.



Acknowledgements: The authors express their gratitude to the referees for constructive and useful remarks. This research was funded by King Mongkut’s University of Technology North Bangkok. Contract no.KMUTNB-GOV-59-34 References [1] [2] [3] [4]

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