Patanarapeelert and Sitthiwirattham Boundary Value Problems (2018) 2018:41 https://doi.org/10.1186/s13661-018-0962-6

RESEARCH

Open Access

On four-point fractional q-integrodifference boundary value problems involving separate nonlinearity and arbitrary fractional order Nichaphat Patanarapeelert1 and Thanin Sitthiwirattham2* *

Correspondence: [email protected] 2 Mathematics Department, Faculty of Science and Technology, Suan Dusit University, Bangkok, Thailand Full list of author information is available at the end of the article

Abstract In this paper, we study a sequential Caputo fractional q-integrodifference equation with fractional q-integral and Riemann–Liouville fractional q-derivative boundary value conditions. Our problem contains 2(M + N + 1) different orders and six different numbers of q in derivatives and integrals. The problem contains separate nonlinear functions. To examine existence and uniqueness results of the problem, Banach’s contraction principle and the Leray–Schauder nonlinear alternative are employed. An illustrative example is also provided. MSC: 39A05; 39A13 Keywords: Existence; q-derivative; q-integral; q-integrodifference equation

1 Introduction In the 20th century, q-difference calculus and fractional q-difference calculus play an important role in the areas of mathematics and applications [1–3] such as the applications to orthogonal polynomials and mathematical control theories. Essentially, for q-difference calculus, basic definitions and properties have been presented in Ref. [4]. For the fractional q-difference calculus proposed by Al-Salam [5] and Agarwal [6], see [7]. Recently, many researchers have extensively studied in q-difference equations and fractional q-difference equations (see [8–34]). However, there is a lack of research in boundary value problem of nonlinear q-difference equations. In what follows, we fill up this gap. In 2014, Ahmad et al. [15] studied the existence of solutions for the Caputo fractional q-difference integral equation with nonlocal boundary conditions, ⎧ ξ C β C γ ⎪ ⎪ ⎨ Dq ( Dq + λ)x(t) = pf (t, x(t)) + kIq g(t, x(t)), ⎪ ⎪ ⎩

α1 x(0) – β1 (t

(1–γ )

Dq x(0))t=0 = σ1 x(η1 ),

t ∈ [0, 1), (1.1)

α2 x(1) – β2 Dq (1) = σ2 x(η2 ),

where β, γ , ξ ∈ (0, 1), f , g are given continuous functions, λ, p, k are real constants and αi , βi , σi ∈ R, ηi ∈ (0, 1), i = 1, 2. © The Author(s) 2018. This article is distributed under the terms of the Creative Commons Attribution 4.0 International License (http://creativecommons.org/licenses/by/4.0/), which permits unrestricted use, distribution, and reproduction in any medium, provided you give appropriate credit to the original author(s) and the source, provide a link to the Creative Commons license, and indicate if changes were made.

Patanarapeelert and Sitthiwirattham Boundary Value Problems (2018) 2018:41

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In 2016, Sitthiwirattham [31] examined the existence results of solutions to a fractional q-difference equation and a fractional q-integrodifference equation   Dαq x(t) = f t, x(t), Dνw x(t) ,   Dαq x(t) = f t, x(t), wγ x(t) ,

(1.2) t ∈ [0, T],

(1.3)

with nonlocal three-point fractional p-integral boundary conditions of the form x(η) = ρ(x),

Ipβ g(T)x(T) = 0,

where p, q, w ∈ (0, 1), α ∈ (1, 2], ν ∈ (0, 1], β, γ > 0 and η ∈ (0, T) are given constants, f ∈ C([0, T] × R × R, R), g ∈ C([0, T], R+ ) are given functions, and ρ ∈ C([0, T], R) → R is a t γ γ given functional. For ϕ ∈ C([0, T] × [0, T], [0, ∞)), define w x(t) := (Iw ϕx)(t) = w1(γ ) 0 (t – ws)(γ –1) ϕ(t, s)x(s) dw s. Recently, Patanarapeelert et al. [33] considered a sequential q-integrodifference boundary value problem involving two different orders and six different numbers of q in derivatives and integrals of the form ⎧ γ ⎪ Dq [ρ(t)Dp (κ + Do )]x(t) = f (t, x(t), Dw [eκt ⎪ o x(t)], ν x(t)), ⎪ ⎪ ⎪ ⎨x(0) = x(T),

(1.4)

κT ⎪ ⎪ (Do [eκt o x(t)])t=0 = Do [eo x(T)], ⎪ ⎪ ⎪θ ⎩ Ir σ (t)x(T) = 0,

where t ∈ IαT := {α k T : k ∈ N} ∪ {0, T}, γ , θ ∈ (0, 1], p = pp12 , q = qq12 , o = oo12 , r = rr12 , w = ww12 , ν = ν1 , and α = LCM(p2 ,q2 1,o2 ,r2 ,w2 ,ν2 ) are proper fractions with w ≤ o, LCM is the lest common ν2 multiple, κ ≤ T1 , ρ, σ ∈ C(IαT , R+ ) and f ∈ C(IαT × R × R × R, R) are given functions. In this paper, we aim to develop an understanding of nonlinear q-integrodifference equations. Particulary, our attention is to analyze existence and uniqueness for a four-point Riemann–Liouville fractional q-integrodifference boundary value problem for a sequential Caputo fractional q-integrodifference equation of the form   Dαq C Dβp u(t) = λ1 F t, u(t), Dγr u(t), Dγr –1 u(t), . . . , Dγr –M+1 u(t)   + λ2 H t, u(t), wθ u(t), wθ–1 u(t), . . . , wθ–N+1 u(t) , ⎧ β+j k ⎪ ⎪ ⎨Dp u(0) = Dp u(0) = 0, k ∈ N0,N–2 , j ∈ N0,M–2 , Dνm u(0) = μDνm u(T), ⎪ ⎪ ⎩ u(ξ ) = τ Inϑ g(η)u(η),

C

t ∈ IχT ,

(1.5)

(1.6)

where IχT := {χ k T : k ∈ N} ∪ {0, T}; M, N ∈ N2 := {2, 3, . . .}; α ∈ (M – 1, M); β ∈ (N – 1, N); γ ∈ (M – 2, M – 1); θ ∈ (N – 2, N – 1); ν, ϑ ∈ (0, 1); λ1 , λ2 , μ, τ > 0; ξ , η ∈ IχT – {0, T}, ξ > η; 1 , n = nn12 are simplest form of proper fractions and χ = p = pp12 , q = qq12 , r = rr12 , w = ww12 , m = m m2 1 , LCM is the least common multiple; g ∈ C(IχT , R+ ), F ∈ C(IχT ×RM+2 , R), LCM(p2 ,q2 ,r2 ,w2 ,m2 ,n2 ) T N+2 H ∈ C(Iχ × R , R) are given functions; and, for ϕ ∈ C(IχT × IχT , [0, ∞)), define wθ u(t) := t (Iwθ ϕu)(t) = w1(θ) 0 (t – ws)(θ–1) ϕ(t, s)u(s) dw s.

Patanarapeelert and Sitthiwirattham Boundary Value Problems (2018) 2018:41

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As is clear from our fractional q-integrodifference equation, there are 2(M + N + 1) different orders in derivatives and integral, and there are six different values of the q numbers consisting of q, p, r, m-derivatives and w, n-integrals. The rest of paper is organized as follows. Section 2 describes some basis definitions, some properties of the q-difference and fractional q-difference operators and lemma that are used to evaluate the results. In Sect. 3, we employ Banach’s contraction mapping principle and the Leray–Schauder nonlinear alternative to prove an existence and uniqueness of solution of the problem (1.5)–(1.6). Finally, using our main results, we provide an example in Sect. 4.

2 Preliminaries In this section, we provide some notations, definitions, and lemmas which are used in the main results. Let q ∈ (0, 1) and define [n]q :=

1 – qn = qn–1 + · · · + q + 1 and 1–q

[n]q ! :=

n 1 – qk k=1

1–q

,

n ∈ R.

The q-analogue of the power function (a – b)(n) with n ∈ N0 := [0, 1, . . .] is defined by (a – b)(0) := 1,

(a – b)(n) :=

n–1 

 a – bqk ,

a, b ∈ R.

k=0

Generally, if α ∈ R, then (a – b)(α) := aα

∞ 1 – ( ab )qn n=0

1 – ( ab )qα+n

,

a = 0.

Specifically, if b = 0 then a(α) = aα . In addition, 0(α) = 0 for α > 0. The q-gamma function is defined by (1 – q)(x–1) , (1 – q)x–1

q (x) :=

x ∈ R \ {0, –1, –2, . . .},

and satisfies q (x + 1) = [x]q q (x). For any x, s > 0, the q-beta function is defined by

1

t (x–1) (1 – qt)(s–1) dq t

Bq (x, s) := 0

= (1 – q)

∞ 

 (α–1)  n (x–1) q (x) q (s) q . qn 1 – qn+1 =

q (x + s) n=0

Definition 2.1 ([6]) For q ∈ (0, 1), the q-derivative of a real function f is defined by Dq f (t) =

f (t) – f (qt) (1 – q)t

and Dq f (0) = lim Dq f (t). t→0

The higher order q-derivatives of f is defined by Dnq f (t) = Dq Dn–1 q f (t),

n ∈ N and

D0q f (t) = f (t).

Patanarapeelert and Sitthiwirattham Boundary Value Problems (2018) 2018:41

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For function f defined on the interval [0, T], q-integral is defined as

t

Iq f (t) =

f (s) dq s = 0

∞ 

  t(1 – q)qn f tqn ,

n=0

where the infinite series is convergent. Definition 2.2 ([6]) For α ≥ 0 and f defined on [0, T], the fractional q-integral is defined by 

 Iqα f (x) :=

1

q (α)



x

(x – qt)(α–1) f (t) dq t 0 ∞

=

(α–1)  n  x(1 – q)  n  q x – xqn+1 f xq

q (α) n=0

=

(α–1)  n  xα (1 – q)  n  q 1 – qn+1 f xq .

q (α) n=0

∞

We note that (Iq0 f )(x) = f (x). Definition 2.3 ([8]) For α ≥ 0, m is the smallest integer such that m ≥ α and f defined on [0, T], the fractional q-derivative of the Riemann–Liouville type of order α is defined by 

   m–α Dαq f (x) := Dm f (x) = q Iq

1

q (–α)



x

(x – qt)(–α–1) f (t) dq t,

α > 0,

0

and 

 D0q f (x) = f (x),

the fractional q-derivative of the Caputo type of order α is defined by C

Dαq f



(x) :=



Iqm–α Dm qf

 (x) =

1

q (m – α)

0

x

(x – qt)(m–α–1) Dm q f (t) dq t,

α>0

and C

 D0q f (x) = f (x).

Lemma 2.1 ([6]) Let α, β ≥ 0 and f be a function defined on [0, T]. Then the following properties hold: (i) (ii)



   Iqβ Iqα f (x) = Iqα+β f (x),  α α  Dq Iq f (x) = f (x).

Lemma 2.2 ([8]) Let N – 1 < α ≤ N and N ∈ N. Then the following equality holds: 

N–1   Iqα Dαq f (x) = f (x) – k=0

 α–N+k  xα–N+k D f (0).

q (α + k – N + 1) q

Patanarapeelert and Sitthiwirattham Boundary Value Problems (2018) 2018:41

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Lemma 2.3 ([19]) Let N – 1 < α ≤ N and N ∈ N. Then the following equality holds: 

N–1   Iqα C Dαq f (x) = f (x) –

C



Dαq Iqα f

k=0

 k  xk D f (0),

q (k + 1) q

(x) = f (x).

Lemma 2.4 ([17]) Let α, β ≥ 0 and 0 < p, q < 1. Then the following formulas hold:

η

(η – qt)(α–1) t (β) dq t = ηα+β Bq (α, β + 1),

(i) 0



η s

(η – ps)(α–1) (s – qt)(β–1) dq t dp s =

(ii) 0



0

η s t

ηα+β Bp (α, β + 1), [β]q

(η – ps)(α–1) (s – qt)(β–1) (t – rv)(γ –1) dr v dq t dp s

(iii) 0

0

=

0

ηα+β+γ Bq (β, γ + 1)Bp (α, β + γ + 1). [γ ]r

We next provided a lemma dealing with a linear variant of the boundary value problem. This lemma is used to define the solution of the boundary value problem (1.5)–(1.6). Lemma 2.5 Let M, N ∈ N2 , α ∈ (M – 1, M), β ∈ (N – 1, N), γ ∈ (M – 2, M – 1), θ ∈ (N – 1 2, N – 1), p = pp12 , q = qq12 , m = m , n = nn12 be simplest proper fractions and φ = LCM(p2 ,q1 2 ,m2 ,n2 ) ; m2 λ1 , λ2 , μ, τ > 0. For f , h ∈ C(IφT , R) and g ∈ C(IφT , R+ ), the solution for the boundary value problem Dαq Dβp u(t) = λ1 f (t) + λ2 h(t), t ∈ IφT ⎧ β+j k ⎪ ⎪ ⎨Dp u(0) = Dp u(0) = 0, k ∈ N0,N–2 , j ∈ N0,M–2 , Dνm u(0) = μDνm u(T), ⎪ ⎪ ⎩ u(ξ ) = τ Inϑ g(η)u(η),

(2.1)

(2.2)

is represented by u(t) =

P [f + g] N–1 Q[f + g] t (t – ps)(β–1) M–1 t s – dp s  

p (β) 0

t s

(s – qv)(α–1) λ1 f (t) + λ2 h(t) dq v dp s, (t – ps)(β–1) +

p (β) q (α) 0 0

where the functionals P [f + g] and Q[f + g] are defined by

P [f + g]

=μ

T



s

(T – ms)(–ν–1) (s – pv)(β–1) M–1 v dp v dm s

p (β) m (–ν) 0 0  η s v (η – ns)(ϑ–1) (s – pv)(β–1) (v – qx)(α–1) × τ

p (β) q (α) n (ϑ) 0 0 0

(2.3)

Patanarapeelert and Sitthiwirattham Boundary Value Problems (2018) 2018:41



× g(η) λ1 f (x) + λ2 h(x) dq x dp v dns 

ξ s

(ξ – ps)(β–1) (s – qv)(α–1) λ1 f (v) + λ2 h(v) dq v dp s –

p (β) q (α) 0 0  ξ (ξ – ps)(β–1) M–1 s dp s +

p (β) 0 

η s (η – ns)(ϑ–1) (s – pv)(β–1) –τ g(η)vM–1 dp v dn s

p (β) n (ϑ) 0 0

T s v (T – ms)(–ν–1) (s – pv)(β–1) (v – qx)(α–1) ×μ

p (β) q (α) m (–ν) 0 0 0

× λ1 f (x) + λ2 h(x) dq x dp v dm s,

Q[f + g]

=μ

Page 6 of 20

(2.4)

T

(T – ms)(–ν–1) M–1 s dm s

m (–ν) 0  η s v (η – ns)(ϑ–1) (s – pv)(β–1) (v – qx)(α–1) × τ

p (β) q (α) n (ϑ) 0 0 0

× g(η) λ1 f (x) + λ2 h(x) dq x dp v dn s 

ξ s

(ξ – ps)(β–1) (s – qv)(α–1) – λ1 f (v) + λ2 h(v) dq v dp s

p (β) q (α) 0 0  

η (η – ns)(ϑ–1) N–1 N–1 g(η)s –τ dn s + ξ

n (ϑ) 0

T s v (T – ms)(–ν–1) (s – pv)(β–1) (v – qx)(α–1) ×μ

p (β) q (α) m (–ν) 0 0 0

× λ1 f (x) + λ2 h(x) dq x dp v dm s,

(2.5)

and the constant

T



s

(T – ms)(–ν–1) (s – px)(β–1) M–1 v dp x dm s

p (β) m (–ν) 0 0  

η (η – ns)(ϑ–1) × ξ N–1 – τ g(η)sN–1 dn s

n (ϑ) 0  ξ (β–1) (ξ – ps) sM–1 dp s –

p (β) 0 

η s (η – ns)(ϑ–1) (s – pv)(β–1) –τ g(η)vM–1 dp v dn s

p (β) n (ϑ) 0 0

T (T – ms)(–ν–1) M–1 s dm s. ×μ

m (–ν) 0

=μ

(2.6)

Proof Using the q-integral of order α for (2.1), we obtain

Dβp u(t) =

M–1  i=0

i

Ci t + 0

t

(t – qs)(α–1) λ1 f (s) + λ2 h(s) dq s.

q (α)

(2.7)

Patanarapeelert and Sitthiwirattham Boundary Value Problems (2018) 2018:41

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Then we take the p-integral of order β for (2.7). We have

u(t) =

N–1 

CM+i t i +

M–1 

i=0

i=0

t

s

+ 0

=

N–1 

0



t

Ci 0

(t – ps)(β–1) i s dp s

p (β)

(t – ps)(β–1) (s – qv)(α–1) λ1 f (v) + λ2 h(v) dq v dp s

p (β) q (α) 1  Ci B(i + 1, β)t β+i

p (β) i=0 M–1

CM+i t i +

i=0

t

s

+ 0

0

(t – ps)(β–1) (s – qv)(α–1) λ1 f (v) + λ2 h(v) dq v dp s.

p (β) q (α)

(2.8)

Next, taking the p-derivative of order k ∈ N0,N–2 for (2.8) where t ∈ IφT , we get

Dkp u(t) =

N–1 



t

CM+i 0

i=0

(t – ps)(–k–1) i s dp s

p (–k)

1  + Ci B(i + 1, β)

p (β) i=0 M–1

t s



t 0

(t – ps)(–k–1) β+i s dp s

p (–k)

x

(t – ps)(–k–1) (s – px)(β–1) (x – qv)(α–1)

p (–k) p (β) q (α) 0 0 0

× λ1 f (v) + λ2 h(v) dq v dp x dp s +

=

N–1 

p (i + 1) i–k 

p (i + 1) t + t β+i–k Ci

p (i + 1 – k)

(β + i + 1 – k) p i=0 M–1

CM+i

i=0

t s

+

x

(t – ps)(–k–1) (s – px)(β–1) (x – qv)(α–1)

p (–k) p (β) q (α) 0 0 0

× λ1 f (v) + λ2 h(v) dq v dp x dp s.

(2.9)

Letting t = 0 in (2.9), and by the first conditions of (2.2) for k ∈ N0,N–2 , we get CM = CM+1 = CM+2 = · · · = CM+N–2 = 0 for k ∈ N0,N–2 , respectively. Substituting the constants Ci , i ∈ NM,M+N–2 into (2.8), we obtain 1  + Ci B(i + 1, β)t β+i

p (β) i=0 M–1

u(t) = CM+N–1 t

t

s

+ 0

N–1

0

(t – ps)(β–1) (s – qv)(α–1) λ1 f (v) + λ2 h(v) dq v dp s.

p (β) q (α)

Next, taking the p-derivative of order β + j, j ∈ N0,M–2 for (2.10), we get

Dβ+j p u(t) = CM+N–1

t 0

(t – ps)(–β–j–1) N–1 s dp s

p (–β – j)

(2.10)

Patanarapeelert and Sitthiwirattham Boundary Value Problems (2018) 2018:41

1  + Ci B(i + 1, β)

p (β) i=0 M–1

t s



t 0

Page 8 of 20

(t – ps)(–β–j–1) β+i s dp s

p (–β – j)

x

(t – ps)(–β–j–1) (s – px)(β–1) (x – qv)(α–1)

p (–β – j) p (β) q (α) 0 0 0

× λ1 f (v) + λ2 h(v) dq v dp x dp s

+



p (N)

p (i + 1) i–j t N–β–j–1 + t Ci

p (N – β – j)

p (i + 1 – j) i=0 M–1

= CM+N–1

t s

x

(t – ps)(–β–j–1) (s – px)(β–1) (x – qv)(α–1)

p (–β – j) p (β) q (α) 0 0 0

× λ1 f (v) + λ2 h(v) dq v dp x dp s.

+

(2.11)

Letting t = 0 in (2.11), and by the first conditions of (2.2) for j ∈ N0,M–2 , we get C0 = C1 = C2 = · · · = CM–2 = 0

for j ∈ N0,M–2 , respectively.

Substituting the constants Ci , i ∈ N0,M–2 into (2.10), we obtain

u(t) = CM+N–1 t N–1 + CM–1 0

t

(t – ps)(β–1) M–1 dp s s

p (β)

(t – ps) (s – qv)(α–1) λ1 f (v) + λ2 h(v) dq v dp s.

p (β) q (α)

s

(β–1)

+ 0

t

0

(2.12)

Next, taking the m-derivative of order ν for u(t) where t ∈ IφT , we get Dνm u(t)



t

(t – ms)(–ν–1) N–1 s dm s

m (–ν) 0

t s (t – ms)(–ν–1) (s – pv)(β–1) M–1 + CM–1 v dp v dm s

m (–ν) p (β) 0 0

t s v (t – ms)(–ν–1) (s – pv)(β–1) (v – qx)(α–1) +

m (–ν) p (β) q (α) 0 0 0

× λ1 f (x) + λ2 h(x) dq x dp v dm s.

= CM+N–1

(2.13)

Letting t = 0, T in (2.9), and by the second conditions of (2.2), we get

T

(T – ms)(–ν–1) N–1 s dm s

m (–ν) 0

T s (T – ms)(–ν–1) (s – pv)(β–1) M–1 + CM–1 μ v dp v dm s

m (–ν) p (β) 0 0

T s v (T – ms)(–ν–1) (s – pv)(β–1) (v – qx)(α–1) = –μ

m (–ν) p (β) q (α) 0 0 0

× λ1 f (x) + λ2 h(x) dq x dp v dm s.

CM+N–1 μ

(2.14)

Patanarapeelert and Sitthiwirattham Boundary Value Problems (2018) 2018:41

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Taking the n-integral of order ϑ for (2.12) where t ∈ IφT , we have

t

(t – ns)(ϑ–1) N–1 s dn s

n (ϑ) 0

t s (t – ns)(ϑ–1) (s – pv)(β–1) M–1 + CM–1 v dp v dn s

n (ϑ) p (β) 0 0

t s v (t – ns)(ϑ–1) (s – pv)(β–1) (v – qx)(α–1) +

n (ϑ) p (β) q (α) 0 0 0

× λ1 f (x) + λ2 h(x) dq x dp v dn s.

Inϑ u(t) = CM+N–1

(2.15)

Letting t = ξ in (2.11), and by the third conditions of (2.2), we get 

CM+N–1 ξ N–1 – τ 0



ξ

0

η

s

–τ 

= τ

0

0

η s

0

(η – ns)(ϑ–1) g(η)sN–1 dn s

n (ϑ)



(ξ – ps)(β–1) M–1 s dp s

p (β)

+ CM–1

η s

(η – ns)(ϑ–1) (s – pv)(β–1) g(η)vM–1 dp v dn s

n (ϑ) p (β)



v

(η – ns)(ϑ–1) (s – pv)(β–1) (v – qx)(α–1)

n (ϑ) p (β) q (α) 0 0 0

× λ1 f (x) + λ2 h(x) g(η) dq x dp v dn s 

ξ s

(ξ – ps)(β–1) (s – qv)(α–1) – λ1 f (v) + λ2 h(v) dq v dp s .

p (β) q (α) 0 0

(2.16)

Finally, solving the system of equations (2.14) and (2.12), we obtain CM–1 = –

Q[f + h] 

and CM+N–1 =

P [f + h] , 

where P [f + g], Q[f + g] and  are defined by (2.4)–(2.6), respectively. After substituting the constants CM–1 , CM+N–1 into (2.12), we obtain (2.3). The proof is complete. 

3 Main results In order to obtain the main results, we first transform the boundary value problem (1.5)– (1.6) into a fixed point problem. Let C = C(IχT , R) be a Banach space of all continuous funcγ –i tions from IχT to R such that Dr u(t) exists for i ∈ N0,M–1 , γ ∈ (M –2, M –1). Define a norm by u C = max

i∈N0,M–1

   u , Dγr –i u , γ –i

γ –i

where u = supt∈IχT |u(t)| and Dr u = supt∈IχT |Dr u(t)|. Denote 

 F t, u(t), Dγr u(t), Dγr –1 u(t), . . . , Dγr –M+1 u(t) := F t, u(t), Dγr –i u(t) ,

Patanarapeelert and Sitthiwirattham Boundary Value Problems (2018) 2018:41

Page 10 of 20

 

H t, u(t), wθ u(t), wθ–1 u(t), . . . , wθ–N+1 u(t) := H t, u(t), wθ–j u(t) , for i ∈ N0,M–1 and j ∈ N0,N–1 . Define the operator A : C → C by (Au)(t)

P [F(u) + G(u)] N–1 Q[F(u) + G(u)] t (t – ps)(β–1) M–1 – dp s = t s  

p (β) 0

t s

(s – qv)(α–1)  λ1 F t, u(t), Dγr –i u(t) (t – ps)(β–1) +

(β) (α) p q 0 0

 θ–j + λ2 H t, u(t), w u(t) dq v dp s,

(3.1)

where the functionals P [F(u) + G(u)] and Q[F(u) + G(u)] are defined by

P F(u) + H(u)

T s (T – ms)(–ν–1) (s – px)(β–1) M–1 =μ x dp x dm s

p (β) m (–ν) 0 0  η s y (η – ns)(ϑ–1) (s – py)(β–1) (y – qx)(α–1) × τ g(η)

p (β) q (α) n (ϑ) 0 0 0 



 × λ1 F t, u(t), Dγr –i u(t) + λ2 H t, u(t), wθ–j u(t) dq x dp y dns

ξ s

(ξ – ps)(β–1) (s – qx)(α–1)  – λ1 F t, u(t), Dγr –i u(t)

p (β) q (α) 0 0 

 θ–j + λ2 H t, u(t), w u(t) dq x dp s 

ξ

(ξ – ps)(β–1) M–1 s dp s

p (β) 0 

η s (η – ns)(ϑ–1) (s – px)(β–1) g(η)xM–1 dp x dn s –τ

p (β) n (ϑ) 0 0

T s y (T – ms)(–ν–1) (s – py)(β–1) (y – qx)(α–1) ×μ

p (β) q (α) m (–ν) 0 0 0 



 × λ1 F t, u(t), Dγr –i u(t) + λ2 H t, u(t), wθ–j u(t) dq x dp y dm s +

(3.2)

and

Q F(u) + H(u)  η s y

T (T – ms)(–ν–1) M–1 (η – ns)(ϑ–1) (s – py)(β–1) (y – qx)(α–1) =μ dm s τ s

m (–ν)

p (β) q (α) n (ϑ) 0 0 0 0 



 × g(η) λ1 F t, u(t), Dγr –i u(t) + λ2 H t, u(t), wθ–j u(t) dq x dp y dn s

ξ s

(ξ – ps)(β–1) (s – qx)(α–1)  – λ1 F t, u(t), Dγr –i u(t)

(β) (α) p q 0 0 

 + λ2 H t, u(t), wθ–j u(t) dq x dp s

Patanarapeelert and Sitthiwirattham Boundary Value Problems (2018) 2018:41



+ ξ N–1 – τ

T

s

η 0

(η – ns)(ϑ–1) g(η)sN–1 dn s

n (ϑ)

Page 11 of 20



y

(T – ms)(–ν–1) (s – py)(β–1) (y – qx)(α–1)

p (β) q (α) m (–ν) 0 0 0 



 × λ1 F t, u(t), Dγr –i u(t) + λ2 H t, u(t), wθ–j u(t) dq x dp y dm s, ×μ

(3.3)

where  is defined by (2.6). Clearly, the problem (1.5)–(1.6) has solutions if and only if the operator F has fixed points. Theorem 3.1 Assume F : IχT × RM+2 → R, H : IχT × RN+2 → R, g : IχT → R+ and ϕ : IχT × IχT → [0, ∞) are continuous, let ϕ0 := sup(t,s)∈IχT ×IχT {ϕ(t, s)}. In addition, F, H and g satisfy the following conditions: (H1 ) there exist positive constants Li , i ∈ N0,M–1 such that, for all t ∈ IχT and u, v ∈ R M–1    



 F t, u, Dγ –i u – F t, v, Dγ –i v  ≤ LM |u – v| + Li Dγr –i u – Dγr –i v, r r i=0

(H2 ) there exist positive constants j , j ∈ N0,N–1 such that, for all t ∈ IχT and u, v ∈ R N–1    



 H t, u, θ–j u – H t, v, θ–j v  ≤ N |u – v| + j  wθ–j u – wθ–j v, w w j=0

(H3 ) 0 < g(t) < G for all t ∈ IχT , θ

T α+β–1 (α)

T α+β (α+1)

T p p 1 β–1 2 )] × { || T + || + p (α+β+1) }< (H4 )  := [λ1 (LM + L) + λ2 (N +  ϕw0(θ+1)

p (α+β) q (β+1) 1. Then the given boundary value problem (1.5)–(1.6) has a unique solution, where

L=

M–1  i=0

Li ,

=

N–1 

j ,

j=0

μT M+β–ν p (M) m (M + β + 1) p (α + 1)

p (M + β) m (M + β – ν) p (α + β + 1) q (β + 1)    τ Gηα+β+ϑ n (α + β + 1)  × ξ α+β –

n (α + β + ϑ + 1)     M+β–1 τ GηM+β+ϑ–1 n (M + β)    , – + ξ 

n (M + β + ϑ)    τ Gηα+β+ϑ n (α + β + 1)  m (α) M–ν–1  α+β – 2 = μT ξ

n (α + β + ϑ + 1)  m (α – ν)    N–1 τ GηN+ϑ–1 n (N)    + ξ –

n (N + ϑ)   T β+1 p (α) m (α + β + 1) p (α + 1) × .

p (α + β) m (α + β – ν + 1) p (α + β + 1) q (β + 1) 1 =

(3.4)

Patanarapeelert and Sitthiwirattham Boundary Value Problems (2018) 2018:41

Page 12 of 20

Proof We transform the boundary value problem (1.5)–(1.6) into a fixed point problem u = Au, where A : C → C is defined by (3.1). For t ∈ IχT , letting 

 



 F t, u, v, Dγ –i  := F t, u(t), Dγ –i u(t) – F t, v(t), Dγ –i v(t) , r r r

i ∈ N0,M–1

and 

 



 H t, u, v, θ–j  := H t, u(t), θ–j u(t) – H t, v(t), θ–j v(t) , w w w

j ∈ N0,N–1 ,

we find that 



 P F(u) + H(u) – P F(v) + H(v) 

T s (T – ms)(–ν–1) (s – px)(β–1) M–1 ≤μ dp x dm s x

p (β) m (–ν) 0 0  η s v  (η – ns)(ϑ–1) (s – py)(β–1) (y – qx)(α–1) × τ g(η)

p (β) q (α) n (ϑ) 0 0 0 



  × λ1 F x, u, v, Dγr –i  + λ2 H x, u, v, wθ–j  dq x dp y dn s

ξ s

 (ξ – ps)(β–1) (s – qx)(α–1)  – λ1 F x, u, v, Dγr –i 

(β) (α) p q 0 0   



+ λ2 H x, u, v, wθ–j  dq x dp s  ξ  (ξ – ps)(β–1) M–1 s dp s + 

p (β) 0 

η s  (η – ns)(ϑ–1) (s – px)(β–1) α–1 –τ x g(η) dp x dn Ms

p (β) n (ϑ) 0 0

T s y (T – ms)(–ν–1) (s – py)(β–1) (y – qx)(α–1) ×μ

p (β) q (α) m (–ν) 0 0 0 



  × λ1 F x, u, v, Dγr  + λ2 H x, u, v, wθ  dq x dp y dm s    M–1    γ –i γ –i ≤ λ1 LM |u – v| + Li D u – D v r

 + λ2

r

i=0

 N–1   θ–j  θ–j   N |u – v| + j w u – w v j=0

    ξ α+β p (α + 1) τ Gηα+β+ϑ p (α + 1) n (α + β + 1)   × –

p (α + β + 1) q (β + 1) p (α + β + 1) n (α + β + ϑ + 1) q (β + 1)  μT M+β–ν–1 p (M) m (M + β)

p (M + β) m (M + β – ν)    M–1    + λ1 LM |u – v| + Li Dγ –i u – Dγ –i v ×

r

 + λ2

i=0

r

 N–1   θ–j  θ–j   N |u – v| + j w u – w v j=0

Patanarapeelert and Sitthiwirattham Boundary Value Problems (2018) 2018:41

Page 13 of 20

  M+β–1 ξ

p (M) τ GηM+β+ϑ–1 p (M) n (M + β)  μT α+β–ν–1 p (α) m (α + β)  – ×

p (M + β)

p (M + β) n (M + β + ϑ)  p (α + β) m (α + β – ν)      ϕ0 T θ–j u – v ≤ λ1 (LM + L) u – v C + λ2 N +  max j∈N0,N–1 w (θ – j + 1)    α+β τ Gηα+β+ϑ n (α + β + 1)    – × ξ

n (α + β + ϑ + 1)  μT M+β–ν–1 p (M) m (M + β) p (α + 1)

p (M + β) m (M + β – ν) p (α + β + 1) q (β + 1)      ϕ0 T θ–j + λ1 (LM + L) u – v C + λ2 N +  max u – v j∈N0,N–1 w (θ – j + 1)    τ GηM+β+ϑ–1 n (M + β)  × ξ M+β–1 – 

n (M + β + ϑ)

×

μT α+β–ν p (M) m (α + β + 1) p (α + 1)

p (M + β) m (α + β – ν + 1) q (β + 1) p (α + β + 1)    ϕ0 T θ < u – v C λ1 (LM + L) + λ2 N + 

w (θ + 1) ×

μT M+β–ν p (M) m (M + β + 1) p (α + 1)

p (M + β) m (M + β – ν) p (α + β + 1) q (β + 1)      τ Gηα+β+ϑ n (α + β + 1)   M+β–1 τ GηM+β+ϑ–1 n (M + β)  × ξ α+β – – + ξ 

n (α + β + ϑ + 1)  

n (M + β + ϑ)    ϕ0 T θ = u – v C λ1 (LM + L) + λ2 N +  1 . (3.5)

w (θ + 1) ×

Similarly to above, we have 



 Q F(u) + H(u) – Q F(v) + H(v)   T  η s y  (T – ms)(–ν–1) M–1 (η – ns)(ϑ–1) (s – py)(β–1) (y – qx)(α–1)  ≤ μ dm s τ s

m (–ν)

p (β) q (α) n (ϑ) 0 0 0 0 



  γ –i θ–j × g(η) λ1 F x, u, v, Dr  + λ2 H x, u, v, w  dq x dp y dn s

ξ s

 (ξ – ps)(β–1) (s – qx)(α–1)  – λ1 F x, u, v, Dγr –i 

p (β) q (α) 0 0  

 θ–j   + λ2 H x, u, v, w dq x dp s 

N–1 –τ + ξ

T

×μ 0

s

0

(η – ns)(ϑ–1) N–1 s g(η) dn s

n (ϑ)

y

(T – ms)(–ν–1) (s – py)(β–1) (y – qx)(α–1)

p (β) q (α) m (–ν)

0

0



η

  



  × λ1 F x, u, v, Dγr –i  + λ2 H x, u, v, wθ–j  dq x dp y dm s    ≤ u – v C λ1 (LM + L) + λ2 N +  max j∈N0,N–1

ϕ0 T θ–j

w (θ – j + 1)

 μT M–ν–1

Patanarapeelert and Sitthiwirattham Boundary Value Problems (2018) 2018:41

Page 14 of 20

   α+β τ Gηα+β+ϑ n (α + β + 1)  m (M)   × ξ –

n (α + β + ϑ + 1)  m (M – ν)     N–1 τ GηN+ϑ–1 n (N)  T β+1 p (α) m (α + β + 1) p (α + 1)   + ξ –

n (N + ϑ)  p (α + β) m (α + β – ν + 1) p (α + β + 1) q (β + 1)    ϕ0 T θ < u – v C λ1 (LM + L) + λ2 N +  (3.6) 2 .

w (θ + 1) Therefore, we find that   (Au)(t) – (Av)(t)  N–1 t



P F(u) + H(u) – P F(v) + H(v) ≤  



1 Q F(u) + H(u) – Q F(v) + H(v) – 

t

t s (t – ps)(β–1) M–1 (s – qx)(α–1) × dp s + (t – ps)(β–1) s

p (β)

p (β) q (α) 0 0 0   



 γ –i θ–j × λ1 F x, u(x), Dr u(x) + λ2 H x, u(x), w u(x) dq x dp s    ϕ0 T θ < u – v C λ1 (LM + L) + λ2 N + 

w (θ + 1)   T α+β p (α + 1) 1 β–1 2 T α+β–1 p (α) T + + × || || p (α + β)

p (α + β + 1) q (β + 1) = u – v C .

(3.7)

Next, taking r-derivative of order γ for (3.1), we have 

 P [F(u) + H(u)] Dγr –i Au (t) = 



t

(t – rs)(i–γ –1) N–1 s dr s

r (i – γ ) 0



Q[F(u) + H(u)] t s (t – rs)(i–γ –1) (s – px)(β–1) M–1 – x dp x dr s 

r (i – γ )

p (β) 0 0

t s y (t – rs)(i–γ –1) (s – py)(β–1) (y – qx)(α–1) +

r (i – γ )

p (β) q (α) 0 0 0 



 × λ1 F x, u(x), Dγr –i u(x) + λ2 H x, u(x), wθ–j u(x) dq x dp y dr s. (3.8)

Further, for any u, v ∈ C and t ∈ IχT , we obtain  γ –i      D Au (t) – Dγ –i Av (t) r

r



1



t (t – rs)(i–γ –1) N–1 ≤  P F(u) + H(u) – P F(v) + H(v) s dr s 

r (i – γ ) 0



1 – Q F(u) + G(u) – Q F(v) + G(v) 

t s (t – rs)(i–γ –1) (s – px)(β–1) M–1 × x dp x dr s

r (i – γ )

p (β) 0 0

Patanarapeelert and Sitthiwirattham Boundary Value Problems (2018) 2018:41

t s

+ 0

0

0

y

Page 15 of 20

(t – rs)(i–γ –1) (s – py)(β–1) (y – qx)(α–1)

r (i – γ )

p (β) q (α)

 



 γ –i θ–j × λ1 F x, u(x), Dr u(x) + λ2 H x, u(x), w u(x) dq x dp y dr s 

   ϕ0 T θ 1 T N–γ –1 r (N) < u – v C λ1 (LM + L) + λ2 N + 

w (θ + 1) || r (N – γ ) T α+β–γ p (α + 1) r (α + β + 1) 2 T M+β–γ –1 p (M) r (M + β) + + || p (M + β) r (M + β – γ )

p (α + β + 1) q (β + 1) r (α + β – γ + 1) < u – v C .



(3.9)

Hence, we obtain Au – Av C ≤ u – v C .

(3.10)

We can conclude from (H4 ) that A is a contraction. The proof is completed by using Banach’s contraction mapping principle.  The following theorems show the existence of at least one solution to the boundary value problem (1.6) by employing the Leray–Schauder nonlinear alternative. Theorem 3.2 (Nonlinear alternative for single valued maps [35]) Let E be a Banach space, C a closed, convex subset of E, U an open subset of C and 0 ∈ U. Suppose that F : U → C is a continuous, compact [that is, F(U) is a relatively compact subset of C] map. Then either (i) F has a fixed point in U, or (ii) there is a u ∈ ∂U (the boundary of U in C) and σ ∈ (0, 1) with u = σ F(u). Theorem 3.3 Assume that (H3 ) holds and the functions F, H satisfy the following conditions: (H5 ) for i = 1, 2 there exist continuous nondecreasing functions ψ, φi : [0, ∞) → (0, ∞) and functions zi ∈ L1 (IχT , R+ ), such that for all (t, u) ∈ IχT × R   

 F t, u(t), Dγ –i u(t)  ≤ z1 (t)ψ1 u , i ∈ N0,M–1 and r   

 H t, u(t), θ–j u(t)  ≤ z2 (t)ψ2 u , j ∈ N0,N–1 , w (H6 ) there exists a constant K > 0 such that K > 1. (λ1 ψ1 (K) z1 L1 + λ2 ψ2 (K) z2 L1 ) Then the boundary value problem (1.5)–(1.6) has at least one solution on IχT . Proof To show that A maps bounded sets (balls) into bounded sets in C , the constructive proof is as follows. For a positive number ρ, let Bρ = {u ∈ C(IχT , R) : u C ≤ ρ} be a bounded ball in C(IχT , R). Then for t ∈ IχT we have 

 P F(u) + H(u) 

T s (T – ms)(–ν–1) (s – py)(β–1) α–1 ≤μ y dp y dm s

p (β) m (–ν) 0 0

Patanarapeelert and Sitthiwirattham Boundary Value Problems (2018) 2018:41



 × τ

Page 16 of 20

η s y

(η – ns)(ϑ–1) (s – py)(β–1) (y – qx)(α–1) g(η)

p (β) q (α) n (ϑ) 0 0 0        × λ1 ψ1 u z1 (x) + λ2 ψ2 u z2 (x) dq x dp y dn s 

ξ s        (ξ – ps)(β–1) (s – qy)(α–1)  – λ1 ψ1 u z1 (y) + λ2 ψ2 u z2 (y) dq y dp s

p (β) q (α) 0 0  ξ 

η s   (ξ – ps)(β–1) α–1 (η – ns)(ϑ–1) (s – py)(β–1) +  s dp s – τ g(η)yα–1 dp y dn s

p (β)

p (β) n (ϑ) 0 0 0

T s y (T – ms)(–ν–1) (s – py)(β–1) (y – qx)(α–1) ×μ

p (β) q (α) m (–ν) 0 0 0        × λ1 ψ1 u z1 (x) + λ2 ψ2 u z2 (x) dq x dp y dm s       ≤ λ1 ψ1 u z1 L1 + λ2 ψ2 u z2 L1 μT M+β–ν p (M) m (M + β) p (α + 1)

p (M + β) m (M + β – ν) p (α + β + 1) q (β + 1)    α+β τ Gηα+β+ϑ n (α + β + 1)    × ξ –

n (α + β + ϑ + 1)     τ GηM+β+ϑ–1 n (M + β)  + ξ M+β–1 – 

n (M + β + ϑ)       = λ1 ψ1 u z1 L1 + λ2 ψ2 u z2 L1 1 . ×

(3.11)

Using the same argument as above, we have 



 Q F(u) + H(u) – Q F(v) + H(v)        ≤ λ1 ψ1 u z1 L1 + λ2 ψ2 u z2 L1 μT M–ν–1      N–1 τ GηN+ϑ–1 n (N)   α+β τ Gηα+β+ϑ n (α + β + 1)  m (M)     + ξ × ξ – –

n (α + β + ϑ + 1)  m (M – ν) 

n (N + ϑ)   T β+1 p (α) m (α + β + 1) p (α + 1) ×

p (α + β) m (α + β – ν + 1) p (α + β + 1) q (β + 1)       = λ1 ψ1 u z1 L1 + λ2 ψ2 u z2 L1 2 . (3.12) Hence, we have         (Au)(t) ≤ λ1 ψ1 u z1 L1 + λ2 ψ2 u z2 L1   T α+β p (α + 1) 1 β–1 2 T α+β–1 p (α) T + + × || || p (α + β)

p (α + β + 1) q (β + 1)       := λ1 ψ1 u z1 L1 + λ2 ψ2 u z2 L1 , and, for i ∈ N0,M–1 ,  γ –i    D Au (t) r

      1 T N–γ –1 r (N)  ≤ λ1 ψ1 u z1 L1 + λ2 ψ2 u z2 L1 || r (N – γ )

(3.13)

Patanarapeelert and Sitthiwirattham Boundary Value Problems (2018) 2018:41

Page 17 of 20

 T α+β–γ p (α + 1) r (α + β + 1) 2 T M+β–γ –1 p (M) r (M + β) + || p (M + β) r (M + β – γ )

p (α + β + 1) q (β + 1) r (α + β – γ + 1)       ≤ λ1 ψ1 u z1 L1 + λ2 ψ2 u z2 L1 . (3.14) +

Consequently, Au C ≤ (λ1 ψ1 ( u ) z1 L1 + λ2 ψ2 ( u ) z2 L1 ). Further, we will show that A maps bounded sets into equicontinuous sets of C(IχT , R). Letting t1 , t2 ∈ IχT with t1 ≤ t2 and u ∈ Bρ , we have   (Au)(t2 ) – (Au)(t1 ) 

 1 P F(u) + G(u) t2N–1 – t1N–1  || 



t1 

 t2 (t2 – ps)(β–1) M–1 1 (t1 – ps)(β–1) M–1  + s s Q F(u) + G(u)  dp s – dp s ||

p (β)

p (β) 0 0       + λ1 ψ1 u z1 L1 + λ2 ψ2 u z2 L1  t2 s  (s – qy)(α–1) ×  dq y dp s (t2 – ps)(β–1)

p (β) q (α) 0 0 

t1 s  (s – qx)(α–1) – dq x dp s (t1 – ps)(β–1)

p (β) q (α) 0 0       1  N–1 N–1   t ≤ λ1 ψ1 u z1 L1 + λ2 ψ2 u z2 L1 – t1  || 2

≤

+ +

2 p (M)  M+β–1 M+β–1  t – t2 || p (M + β) 1

  α+β α+β 

p (α + 1)  , t – t 1

p (α + β + 1) q (β + 1) 2

(3.15)

and, for i ∈ N0,M–1 ,  γ –i      D Au (t2 ) – Dγ –i Au (t1 ) r



r

    ≤ λ1 ψ1 u z1 L1 + λ2 ψ2 u z2 L1  1 r (N)  N–γ –1 N–γ –1  × t – t1 || r (N – γ ) 2 + +



 M+β–γ –1 M+β–γ –1 

p (M) r (M + β) 2  t – t2 || p (M + β) r (M + β – γ ) 1

  α+β–γ

p (α + 1) r (α + β + 1) α+β–γ  t – t . 1

p (α + β + 1) q (β + 1) r (α + β – γ + 1) 2

(3.16)

As t2 – t1 → 0, the right hand side of (3.16) tends to zero independently of u ∈ Bρ . As A satisfies the above assumptions, it follows by the Arzelá–Ascoli theorem that A : C(IχT , R) → C(IχT , R) is completely continuous. Let u be a solution. Proceeding by similar computations to the first step for t ∈ IχT , we have         u(t) ≤ λ1 ψ1 u z1 L1 + λ2 ψ2 u z2 L1 .

Patanarapeelert and Sitthiwirattham Boundary Value Problems (2018) 2018:41

Page 18 of 20

Hence, u C ≤ 1. (λ1 ψ1 ( u ) z1 L1 + λ2 ψ2 ( u ) z2 L1 ) Under (H6 ), there exists K such that u C = K . We set     U = u ∈ C IχT , R : u C < K . Note that the operator A : U → C(IχT , R) is continuous and completely continuous. We find that there is no u ∈ ∂U such that u = σ Au for some σ ∈ (0, 1). Consequently, by the nonlinear alternative of Leray–Schauder type (Theorem 3.2), we can conclude that A has a fixed point u ∈ U which is a solution of the problem (1.5)–(1.6). This completes the proof. 

4 Example The following boundary value problem is an example illustrating our main result. Consider the second-order q-difference equation with q-integral boundary conditions

C

D

5 2C 1 2

– sin2 (2π t+5)

4 3 1 3

D u(t) =

e · 200 + ecos2 (2π t) 2 (2π t)

e– cos

4

4

1

|u(t)| + | 24 u(t)| ,

(t + 10)2 [1 + |u(t)|] 4

1

[1 + |u(t)|] 5

+

4

|u(t)| + |D 33 u(t)| + |D 33 u(t)|

7

10

3

3

(4.1)

u(0) = D 1 u(0) = D 31 u(0) = D 31 u(0) = D 13 u(0) = 0, 3

1 5 2 3

3

1 5 2 3

D u(0) = 10D u(6),

    3 π 1 π 4 cos( 36,000 ) = 2I 1 e , u u 10 36,000 4 1

1 n where t ∈ I 61 = {6( 60 ) : n ∈ N} ∪ {0, 3} and 24 u(t) = 60

5

1 , 2

1 , 3

3 , 4

1

2 ( 14 ) 3

t

2 , 5

e–s 0 (t+20)2 2 , 3

· u(s) d 2 s. 3

1 , 4

We apply Theorem 3.1 when q = p = r = w = m = n = M = 3, N = 2, 1 1 α = 52 , β = γ = 43 , θ = 14 , ν = 15 , ϑ = 34 , T = 6, λ1 = e–5 , λ2 = 1, μ = 10, τ = 2, χ = 60 , ξ = 10 ,η = 1 1 and ϕ0 = sup{ϕ(t, s)} = 400 . 36,000 Since      F t, u, Dγ u – F t, v, Dγ v  ≤ r r

    1 |u – v| + Dγr u – Dγr v + Dγr –1 u – Dγr –1 v , 201e        H t, u, θ u – H t, v, θ v  ≤ 1 |u – v| + 1  θ u – θ v, w w w w 100e 100 1 2 1 1 so (H1 )–(H2 ) are satisfied with L1 = L2 = L3 = 201e , L = 201e and 2 = 100e , 1 = 100 = . 1 In addition, since e ≤ g(t) ≤ e, then (H3 ) is satisfied with G = e. Moreover, we can show that

|| = 22.0923,

1 = 10.3252

and

2 = 28.9938.

Patanarapeelert and Sitthiwirattham Boundary Value Problems (2018) 2018:41

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We obtain  ≈ 0.0329 < 1. It implies that (H4 ) holds. From Theorem 3.1, we can conclude that the assigned problem (4.1) has a unique solution on I 61 . 60

5 Conclusion We have proved the existence results of the four-point fractional q-integral and Riemann– Liouville fractional q-derivative boundary value problem for a sequential Caputo fractional q-integrodifference equation involving separate nonlinearity (1.5)–(1.6), by using the Banach contraction mapping principle as regards the existence and uniqueness of a solution, and the Leray–Schauder nonlinear alternative for the existence of at least a solution. Our problem contains 2(M + N + 1) different orders and six different numbers of q in derivatives and integral, which is a new idea. Funding This research was funded by King Mongkut’s University of Technology North Bangkok. Contract no. KMUTNB-GOV-59-38. Competing interests The authors declare that they have no competing interests. Authors’ contributions All authors contributed equally and significantly in writing this article. All authors read and approved the final manuscript. Author details 1 Department of Mathematics, Faculty of Applied Science, King Mongkut’s University of Technology North Bangkok, Bangkok, Thailand. 2 Mathematics Department, Faculty of Science and Technology, Suan Dusit University, Bangkok, Thailand.

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