# Sequential Derivatives of Nonlinearq-Difference Equations with Three-Point q-Integral Boundary Conditions

Authors: Nittaya Pongarm; Suphawat Asawasamrit; Jessada Tariboon
Journal: Journal of Applied Mathematics
 (2013)

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## Abstract

This paper studies sufficient conditions for the existence of solutions to the problem of sequential derivatives of nonlinearq-difference equations with three-point q-integral boundary conditions. Our results are concerned with several quantum numbers of derivatives and integrals. By using Banach's contraction mapping, Krasnoselskii's fixed-point theorem, and Leray-Schauder degree theory, some new existence results are obtained. Two examples illustrate our results.


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## Body

## 1. Introduction

The study ofq-calculus or quantum calculus was initiated by the pioneer works of Jackson [1], Carmichael [2], Mason [3], Adams [4], Trjitzinsky [5], and so forth. Since then, in the last few decades, this subject has evolved into a multidisciplinary research area with many applications; for example, see [6–14]. For some recent works, we refer the reader to [15–21] and references therein. However, the theory of boundary value problems for nonlinear q-difference equations is still in the beginning stages and it needs to be explored further.In [22], Ahmad investigated the existence of solutions for a nonlinear boundary value problem of third-order q-difference equation:
(1)Dq3u(t)=f(t,u(t)),0≤t≤1,u(0)=0,Dqu(0)=0,u(1)=0.
Using Leray-Schauder degree theory and standard fixed-point theorems, some existence results were obtained. Moreover, he showed that if q→1, then his results corresponded to the classical results. Ahmad et al. [23] studied a boundary value problem of a nonlinear second-order q-difference equation with nonseparated boundary conditions
(2)Dq2u(t)=f(t,u(t)),t∈[0,T],u(0)=ηu(T),Dqu(0)=ηDqu(T).
They proved the existence and uniqueness theorems of the problem (2) using the Leray-Schauder nonlinear alternative and some standard fixed-point theorems. For some very recent results on nonlocal boundary value problems of nonlinear q-difference equations and inclusions, see [24–26].In this paper, we discuss the existence of solutions for the following nonlinearq-difference equation with three-point integral boundary condition:
(3)Dq(Dp+λ)x(t)=f(t,x(t)),t∈[0,T],x(0)=0,β∫0η‍x(s)drs=x(T),
where 0<p, q, r<1, f∈C([0,T]×ℝ, ℝ), β≠T(1+r)/η2, η∈(0,T) is a fixed point, and λ is a given constant.The aim of this paper is to prove some existence and uniqueness results for the boundary value problem (3). Our results are based on Banach’s contraction mapping, Krasnoselskii's fixed-point theorem, and Leray-Schauder degree theory. Since the problem (3) has different values of the quantum numbers of the q-derivative and the q-integral, the existence results of such problem are also new.


## 2. Preliminaries

Let us recall some basic concepts of quantum calculus [15].For0<q<1, we define the q-derivative of a real-valued function f as
(4)Dqf(t)=f(t)-f(qt)(1-q)t,Dqf(0)=limt→0Dqf(t).
The higher-order q-derivatives are given by
(5)Dq0f(t)=f(t),Dqnf(t)=DqDqn-1f(t),n∈ℕ.
The q-integral of a function f defined on the interval [0,T] is given by
(6)∫at‍f(s)dqs:=∑n=0∞‍(1-q)qn×[tf(tqn)-af(qna)],t∈[0,T],
and for a = 0, we denote
(7)Iqf(t)=∫0t‍f(s)dqs=∑n=0∞‍t(1-q)qnf(tqn),
provided the series converges. If a∈[0,T] and f is defined on the interval [0,T], then
(8)∫ab‍f(s)dqs=∫0b‍f(s)dqs-∫0a‍f(s)dqs.
Similarly, we have
(9)Iq0f(t)=f(t),Iqnf(t)=IqIqn-1f(t),n∈ℕ.
Observe that
(10)DqIqf(t)=f(t),
and if f is continuous at t=0, then
(11)IqDqf(t)=f(t)-f(0).
In q-calculus, the product rule and integration by parts formula are
(12)Dq(gh)(t)=(Dqg(t))h(t)+g(qt)Dqh(t),∫0t‍f(s)Dqg(s)dqs=[f(s)g(s)]0t-∫0t‍Dqf(s)g(qs)dqs.
In the limit q→1, the q-calculus corresponds to the classical calculus. The above results are also true for quantum numbers p, r such that 0<p<1 and 0<r<1.Lemma 1.
LetT(1+r)≠βη2, 0<p, q, r<1, and let λ be a constant. Then for any h∈C[0,T], the boundary value problem
(13)Dq(Dp+λ)x(t)=h(t),t∈[0,T],(14)x(0)=0,β∫0η‍x(s)drs=x(T),0<η<T,
is equivalent to the integral equation
(15)x(t)=∫0t‍∫0s‍h(u)dqudps-λ∫0t‍x(s)dps+β(1+r)tT(1+r)-βη2×∫0η‍∫0v‍(∫0s‍h(u)dqu-λx(s))dpsdrv+λ(1+r)tT(1+r)-βη2∫0T‍x(s)dps-(1+r)tT(1+r)-βη2∫0T‍∫0s‍h(u)dqudps.Proof.
Fort∈[0,T], q-integrating (13) from 0 to t, we obtain
(16)(Dp+λ)x(t)=∫0t‍h(s)dqs+c1.
Equation (16) can be written as
(17)Dpx(t)=∫0t‍h(s)dqs-λx(t)+c1.
For t∈[0,T], p-integrating (17) from 0 to t, we have
(18)x(t)=∫0t‍∫0s‍h(u)dqudps-λ∫0t‍x(s)dps+c1t+c2.
From the first condition of (14), it follows that c2=0. For t∈[0,T], r-integrating equation (18) from 0 to t, we get
(19)∫0t‍x(v)drv=∫0t‍∫0v‍∫0s‍h(u)dqudpsdrv-λ∫0t‍∫0v‍x(s)dpsdrv+c1t21+r.
The second boundary condition (14) implies that
(20)β∫0η‍x(v)drv=β∫0η‍∫0v‍∫0s‍h(u)dqudpsdrv-βλ∫0η‍∫0v‍x(s)dpsdrv+c1βη21+r=β∫0η‍∫0v‍(∫0s‍h(u)dqu-λx(s))dpsdrv+c1βη21+r=∫0T‍∫0s‍h(u)dqudps-λ∫0T‍x(s)dps+c1T.
Therefore,
(21)c1=β(1+r)T(1+r)-βη2∫0η‍∫0v‍(∫0s‍h(u)dqu-λx(s))dpsdrv+λ(1+r)T(1+r)-βη2∫0T‍x(s)dps-1+rT(1+r)-βη2∫0T‍∫0s‍h(u)dqudps.
Substituting the values of c1 and c2 in (18), we obtain (15). This completes the proof.For the forthcoming analysis, let𝒞=C([0,T],ℝ) denote the Banach space of all continuous functions from [0,T] to ℝ endowed with the norm defined by ∥x∥=sup{|x(t)|,t∈[0,T]}.In the following, for the sake of convenience, we set(22)Ω=11+p×(T2+|β|Tη3(1+r)|T(1+r)-βη2|(1+r+r2)+T3(1+r)|T(1+r)-βη2|),(23)Φ=|λ|T+|β||λ|Tη2|T(1+r)-βη2|+|λ|T2(1+r)|T(1+r)-βη2|.


## 3. Main Results

Now, we are in the position to establish the main results. We transform the boundary value problem (3) into a fixed-point problem. In view of Lemma 1, for t∈[0,T], x∈𝒞, we define the operator A:𝒞→𝒞 as
(24)(Ax)(t)=∫0t‍∫0s‍f(u,x(u))dqudps-λ∫0t‍x(s)dps+β(1+r)tT(1+r)-βη2×∫0η‍∫0v‍(∫0s‍f(u,x(u))dqu-λx(s))dpsdrv+λ(1+r)tT(1+r)-βη2∫0T‍x(s)dps-(1+r)tT(1+r)-βη2∫0T‍∫0s‍f(u,x(u))dqudps.
Note that the problem (3) has solutions if and only if the operator equation Ax=x has fixed points.Our first result is based on Banach’s fixed-point theorem.Theorem 2.
Assume thatf:[0,T]×ℝ→ℝ is a jointly continuous function satisfying the conditions (H1)
|f(t,x)-f(t,y)|≤L|x-y|,  for all t∈[0,T], x,y∈ℝ;(H2)
Λ:=(Φ+LΩ)<1,where L is a Lipschitz constant, Ω and Φ are defined by (22) and (23), respectively.
Then, the boundary value problem (3) has a unique solution.Proof.
Assume thatsupt∈[0,T]|f(t,0)|=M0; we choose a constant
(25)R≥M0Ω1-Λ.
Now, we will show that ABR⊂BR, where BR={x∈𝒞:∥x∥≤R}. For any x∈BR, we have
(26)∥(Ax)∥=supt∈[0,T]|∫0t‍∫0s‍f(u,x(u))dqudps-λ∫0t‍x(s)dps+β(1+r)tT(1+r)-βη2×∫0η‍∫0v‍(∫0s‍f(u,x(u))dqu-λx(s))dpsdrv+λ(1+r)tT(1+r)-βη2∫0T‍x(s)dps-(1+r)tT(1+r)-βη2∫0T‍∫0s‍f(u,x(u))dqudps|≤supt∈[0,T]{∫0t‍∫0s‍(|f(u,x(u))-f(u,0)|+|f(u,0)|)dqudps+|λ|∫0t‍|x(s)|dps+|β|(1+r)t|T(1+r)-βη2|×(∫0η‍∫0v‍∫0s‍(|f(u,x(u))-f(u,0)|+|f(u,0)|)×dqudpsdrv+|λ|∫0η‍∫0v‍|x(s)|dpsdrv)+|λ|(1+r)t|T(1+r)-βη2|∫0T‍|x(s)|dps+(1+r)t|T(1+r)-βη2|×∫0T‍∫0s‍(|f(u,x(u))-f(u,0)|+|f(u,0)|)dqudps∫0t‍}≤supt∈[0,T]{∫0t‍∫0s‍(L|x(u)|+|f(u,0)|)dqudps+|λ|∫0t‍|x(s)|dps+|β|(1+r)t|T(1+r)-βη2|×(∫0η‍∫0v‍∫0s‍(L|x(u)|+|f(u,0)|)dqudpsdrv+|λ|∫0η‍∫0v‍|x(s)|dpsdrv)+|λ|(1+r)t|T(1+r)-βη2|∫0T‍|x(s)|dps+(1+r)t|T(1+r)-βη2|×∫0T‍∫0s‍(L|x(u)|+|f(u,0)|)dqudps}≤supt∈[0,T]{(L∥x∥+M0)∫0t‍∫0s‍dqudps+|λ|∥x∥∫0t‍dps+|β|(1+r)t|T(1+r)-βη2|×((L∥x∥+M0)∫0η‍∫0v‍∫0s‍dqudpsdrv+|λ|∥x∥∫0η‍∫0v‍dpsdrv)+|λ|∥x∥(1+r)t|T(1+r)-βη2|∫0T‍dps+(1+r)t|T(1+r)-βη2|(L∥x∥+M0)∫0T‍∫0s‍dqudps}=supt∈[0,T]{(L∥x∥+M0)t21+p+|λ|∥x∥t+|β|(1+r)t|T(1+r)-βη2|×((L∥x∥+M0)(1-r)η3(1+p)(1-r3)+|λ|∥x∥η21+r)+|λ|∥x∥(1+r)t|T(1+r)-βη2|T+(1+r)t|T(1+r)-βη2|(L∥x∥+M0)T21+p}=supt∈[0,T]{∥x∥(|λ|t+|β||λ|η2t|T(1+r)-βη2|+|λ|T(1+r)t|T(1+r)-βη2|)+(L∥x∥+M0)1+p×(t2+|β|(1+r)η3t|T(1+r)-βη2|(1+r+r2)+T2(1+r)t|T(1+r)-βη2|)}≤R(|λ|T+|β||λ|Tη2|T(1+r)-βη2|+|λ|T2(1+r)|T(1+r)-βη2|)+(LR+M0)1+p×(T2+|β|Tη3(1+r)|T(1+r)-βη2|(1+r+r2)+T3(1+r)|T(1+r)-βη2|)=RΦ+(LR+M)Ω≤R.
Next, we will show thatA is a contraction. For any x,y∈𝒞 and for each t∈[0,T], we have
(27)∥(Ax)-(Ay)∥=supt∈[0,T]|(Ax)(t)-(Ay)(t)|=supt∈[0,T]|∫0t‍∫0s‍(f(u,x(u))-f(u,y(u)))dqudps-λ∫0t‍(x(s)-y(s))dps+β(1+r)tT(1+r)-βη2×∫0η‍∫0v‍(∫0s‍(f(u,x(u))-f(u,y(u)))dqu-λ(x(s)-y(s))∫0s‍)dpsdrv+λ(1+r)tT(1+r)-βη2∫0T‍(x(s)-y(s))dqs-(1+r)t|T(1+r)-βη2|×∫0T‍∫0s‍(f(u,x(u))-f(u,y(u)))dqudps|≤supt∈[0,T]{L∥x-y∥∫0t‍∫0s‍dqudps+|λ|∥x-y∥∫0t‍dps+|β|(1+r)t|T(1+r)-βη2|×(L∥x-y∥∫0η‍∫0v‍∫0s‍dqudpsdrv+|λ|∥x-y∥∫0η‍∫0v‍dpsdrv)+|λ|(1+r)tT(1+r)-βη2∥x-y∥×∫0T‍dqs+(1+r)t|T(1+r)-βη2|×L∥x-y∥∫0T‍∫0s‍dqudps}=supt∈[0,T]{L∥x-y∥t21+p+|λ|∥x-y∥t+|β|(1+r)t|T(1+r)-βη2|×(L∥x-y∥(1-r)η3(1+p)(1-r3)+|λ|∥x-y∥η21+r)+|λ|∥x-y∥(1+r)t|T(1+r)-βη2|T+(1+r)t|T(1+r)-βη2|L∥x-y∥T21+p}≤∥x-y∥(|λ|T+|β||λ|Tη2|T(1+r)-βη2|+|λ|T2(1+r)|T(1+r)-βη2|)+L∥x-y∥1+p(T2+|β|η3T(1+r)|T(1+r)-βη2|(1+r+r2)+T3(1+r)|T(1+r)-βη2|)=(Φ+LΩ)∥x-y∥≤Λ∥x-y∥.
Since Λ<1, A is a contraction. Thus, the conclusion of the theorem follows by Banach's contraction mapping principle. This completes the proof.Our second result is based on the following Krasnoselskii’s fixed-point theorem [27].Theorem 3.
LetK be a bounded closed convex and nonempty subset of a Banach space X. Let A,B be operators such that (i)
Ax+By∈K whenever x,y∈K;(ii)
A is compact and continuous;(iii)
B is a contraction mapping.Then, there exists z∈K such that z=Az+Bz.Theorem 4.
Assume that(H1) and (H2) hold with
(
H
3
)
|
f
(
t
,
x
)
|
≤
μ
(
t
), for all (t,x)∈[0,T]×ℝ, with μ∈L1([0,T],ℝ+).
If(28)|β||λ|Tη2|T(1+r)-βη2|+|λ|T2(1+r)|T(1+r)-βη2|+11+p(|β|η3T(1+r)|T(1+r)-βη2|(1+r+r2)+T3(1+r)|T(1+r)-βη2|)<1,
then the boundary value problem (3) has at least one solution on [0,T].Proof.
Settingmaxt∈[0,T]|μ(t)|=||μ|| and choosing a constant
(29)R≥∥μ∥Ω1-Φ,
where Ω and Φ are given by (22) and (23), respectively, we consider that BR={x∈𝒞:||x||≤R}.
In view of Lemma1, we define the operators F1 and F2 on the set BR as
(30)(F1x)(t)=∫0t‍∫0s‍f(u,x(u))dqudps-λ∫0t‍x(s)dps,(F2x)(t)=β(1+r)tT(1+r)-βη2×∫0η‍∫0v‍(∫0s‍f(u,x(u))dqu-λx(s))dpsdrv+λ(1+r)tT(1+r)-βη2∫0T‍x(s)dps-(1+r)tT(1+r)-βη2∫0T‍∫0s‍f(u,x(u))dqudps,
for x,y∈BR. By computing directly, we have
(31)∥(F1x)+(F2y)∥≤∥μ∥∫0t‍∫0s‍dqudps+|λ|∥x∥∫0t‍dps+|β|(1+r)t|T(1+r)-βη2|×(∥μ∥∫0η‍∫0v‍∫0s‍dqudpsdrv+|λ|∥y∥∫0η‍∫0v‍dpsdrv)+|λ|(1+r)t|T(1+r)-βη2|∥y∥∫0T‍dps+(1+r)t|T(1+r)-βη2|∥μ∥∫0T‍∫0s‍dqudps,≤RΦ+∥μ∥Ω≤R.
Therefore, (F1x)+(F2y)∈BR. The condition (28) implies that F2 is a contraction mapping. Next, we will show that F1 is compact and continuous. Continuity of f coupled with the assumption (H3) implies that the operator F1 is continuous and uniformly bounded on BR. We define sup(t,x)∈[0,T]×BR|f(t,x)|=fmax<∞. For t1,t2∈[0,T] with t2<t1 and x∈BR, we have
(32)|F1x(t1)-F2x(t2)|≤sup(t,x)∈[0,T]×BR|∫0t1‍∫0s‍f(u,x(u))dqudps-λ∫0t1‍x(s)dps-∫0t2‍∫0s‍f(u,x(u))dqudps+λ∫0t2‍x(s)dps|=sup(t,x)∈[0,T]×BR|∫t2t1‍∫0s‍f(u,x(u))dqudps-λ∫t2t1‍x(s)dps|≤fmax|t12-t22|1+p+|λ|(t1-t2)R.
Actually, as t1-t2→0, the right-hand side of the above inequality tends to zero. So F1 is relatively compact on BR. Hence, by the Arzelá-Ascoli theorem, F1 is compact on BR. Therefore, all the assumptions of Theorem 5 are satisfied and the conclusion of Theorem 5 implies that the three-point integral boundary value problem (3) has at least one solution on [0,T]. This completes the proof.As the third result, we prove the existence of solutions of (3) by using Leray-Schauder degree theory.Theorem 5.
Letf:[0,T]×ℝ→ℝ. Assume that there exist constants 0≤κ<(1-Φ)Ω-1, where Ω and Φ are given by (22) and (23), respectively, and M>0 such that |f(t,x)|≤κ|x|+M for all t∈[0,T], x∈𝒞. Then, the boundary value problem (3) has at least one solution.Proof.
Let us define an operatorA:𝒞→𝒞 as in (24). We will prove that there exists at least one solution x∈𝒞 of the fixed-point equation
(33)x=Ax.
We define a ball BR⊂𝒞, with a constant radius R>0, given by
(34)BR={x∈𝒞:maxt∈[0,T]|x(t)|<R}.
Then, it is sufficient to show that A:B¯R→𝒞 satisfies
(35)x≠θAx,∀x∈∂BR,∀θ∈[0,1].
Now, we set
(36)H(θ,x)=θAx,x∈𝒞,θ∈[0,1].
Then, by the Arzelá-Ascoli theorem, we get that hθ(x)=x-H(θ,x)=x-θAx is completely continuous. If (35) holds, then the following Leray-Schauder degrees are well defined. From the homotopy invariance of topological degree, it follows that
(37)deg(hθ,BR,0)=deg(I-θA,BR,0)=deg(h,BR,0)=deg(h0,BR,0)=deg(I,BR,0)=1≠0,0∈BR,
where I denotes the unit operator. By the nonzero property of Leray-Schauder degree, h1(x)=x-Ax=0 for at least one x∈BR. Let us assume that x=θAx for some θ∈[0,1]. Then, for all t∈[0,T], we obtain
(38)|x(t)|=|θ(Ax)(t)|≤∫0t‍∫0s‍|f(u,x(u))|dqudps+|λ|∫0t‍|x(s)|dps+|β|(1+r)t|T(1+r)-βη2|×∫0η‍∫0v‍(∫0s‍|f(u,x(u))|dqu+|λx(s)|)dpsdrv+|λ|(1+r)t|T(1+r)-βη2|×∫0T‍|x(s)|dps+(1+r)t|T(1+r)-βη2|×∫0T‍∫0s‍|f(u,x(u))|dqudps≤(κ|x|+M)∫0t‍∫0s‍dqudps+|λ||x|∫0t‍dps+|β|(1+r)t|T(1+r)-βη2|×((κ|x|+M)∫0η‍∫0v‍∫0s‍dqudpsdrv+|λ||x|∫0η‍∫0v‍dpsdrv)+|λ||x|(1+r)t|T(1+r)-βη2|×∫0T‍dps+(1+r)t|T(1+r)-βη2|(κ|x|+M)×∫0T‍∫0s‍dqudps≤(κ|x|+M)T21+p+|λ||x|T+|β|(1+r)T|T(1+r)-βη2|×((κ|x|+M)(1-r)η3(1+p)(1-r3)+|λ||x|η21+r)+|λ||x|(1+r)T2|T(1+r)-βη2|+(κ|x|+M)(1+r)T3|T(1+r)-βη2|(1+p)=|x|(|λ|T+|β||λ|Tη2|T(1+r)-βη2|+|λ|T2(1+r)|T(1+r)-βη2|)+(κ|x|+M)1+p(T2+|β|Tη3(1+r)|T(1+r)-βη2|(1+r+r2)+T3(1+r)|T(1+r)-βη2|)=|x|Φ+(κ|x|+M)Ω.
Taking norm supt∈[0,T]|x(t)|=∥x∥ and solving for ∥x∥, this yields
(39)∥x∥≤MΩ1-(Φ+κΩ).
Let R=MΩ/(1-(Φ+κΩ))+1, then (35) holds. This completes the proof.


## 4. Examples

In this section, we give two examples to illustrate our results.Example 6.
Consider the following nonlinearq-difference equation with boundary value problem:
(40)D1/2(D1/3-27)x(t)=1(t+2)2·|x||x|+1,t∈[0,1],x(0)=0,x(1)+23∫03/4‍x(s)d1/4s=0.
Setq=1/2, p=1/3, r=1/4, T=1, λ=-2/7, η=3/4, β=-2/3, and f(t,x)=(1/(t+2)2)(∥x∥/(1+∥x∥)). Since |f(t,x)-f(t,y)|≤(1/4)∥x-y∥, then, (H1) and (H2) are satisfied with T(1+r)-βη2=13/8≠0,
(41)Ω=11+p(T2+|β|Tη3(1+r)|T(1+r)-βη2|(1+r+r2)+T3(1+r)|T(1+r)-βη2|)=13291,Φ=|λ|T+|β||λ|Tη2|T(1+r)-βη2|+|λ|T2(1+r)|T(1+r)-βη2|=47,L=1/4. Hence, Λ=:Φ+LΩ=85/91<1. Therefore, by Theorem 2, the boundary value problem (40) has a unique solution on [0,1].Example 7.
Consider the following nonlinearq-difference equation with boundary value problem:
(42)D2/3(D4/5+19)x(t)=sin(5πx)25π+|x||x|+1,t∈[0,32],x(0)=0,x(32)=14∫01‍x(s)d1/2s.
Setq=2/3, p=4/5, r=1/2, T=3/2, λ=1/9, η=1, and β=1/4. Here, |f(t,x)|=|sin(5πx)/25π+|x|/(1+|x|)|≤(|x|/5)+1. So, M=1, T(1+r)-βη2=2≠0, and
(43)Ω=11+p×(T2+|β|Tη3(1+r)|T(1+r)-βη2|(1+r+r2)+T3(1+r)|T(1+r)-βη2|)=615224,Φ=|λ|T+|β||λ|Tη2|T(1+r)-βη2|+|λ|T2(1+r)|T(1+r)-βη2|=38,κ=15<(1-Φ)Ω-1=28123.
Hence, by Theorem 5, the boundary value problem (42) has at least one solution on [0,3/2].



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*Source: 605169-2013-03-31.xml*