# Impulsive Antiperiodic Boundary Value Problems for Nonlinearqk-Difference Equations

Authors: Lihong Zhang; Bashir Ahmad; Guotao Wang
Journal: Abstract and Applied Analysis
 (2014)

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## Abstract

We show the existence and uniqueness of solutions for an antiperiodic boundary value problem of nonlinear impulsiveqk-difference equations by applying some well-known fixed point theorems. An example is presented to illustrate the main results.


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## Body

## 1. Introduction

The subject ofq-calculus (also known as quantum calculus) is regarded as ordinary calculus without the idea of limit. The systematic development of q-calculus started with the work of Jackson [1] at the beginning of the twentieth century. The application of q-calculus covers a variety of topics such as special functions, particle physics and supersymmetry, combinatorics, initial and boundary value problems of q-difference equations, operator theory, and control theory. For details of the advancement of q-calculus, we refer the reader to the texts [2–4] and papers [5–13].One of the advantages for consideringq-difference equations is that these equations are always completely controllable and appear in the q-optimal control problem [14]. The variational q-calculus is regarded as a generalization of the continuous variational calculus due to the presence of an extra parameter q whose nature may be physical or economical. The variational calculus on the q-uniform lattice includes the study of the q-Euler equations and its applications to the isoperimetric and Lagrange problem and commutation equations. In other words, it suffices to solve the q-Euler-Lagrange equation for finding the extremum of the functional involved instead of solving the Euler-Lagrange equation [15]. Further details can be found in [16–22].Impulsive differential equations have extensively been studied in the past two decades. In particular, initial and boundary value problems of impulsive fractional differential equations have attracted the attention of many researchers; for instance, see [23–34] and references therein. In a recent work [32], the authors discussed the existence and uniqueness of solutions for impulsive qk-difference equations.Motivated by [32], we investigate the existence and uniqueness of solutions for an antiperiodic boundary value problem of nonlinear impulsive qk-difference equation in this paper. Precisely, we consider
(1)Dqku(t)=f(t,u(t)),0<qk<1,t∈J′,Δu(tk)=Ik(u(tk)),k=1,2,…,m,u(0)=-u(T),
where Dqk are qk-derivatives (k=0,1,2,…,m), f∈C(J×ℝ,ℝ), Ik∈C(ℝ,ℝ),J=[0,T](T>0), 0=t0<t1<⋯<tk<⋯<tm<tm+1=T, J′=J∖{t1,t2,…,tm}, and Δu(tk)=u(tk+)-u(tk-), where u(tk+) and u(tk-) denote the right and the left limits of u(t) at t=tk(k=1,2,…,m), respectively.Here, we remark that the classicalq-difference equations cannot be considered with impulses as the definition of q-derivative fails to work when an impulse point tk∈(qt,t) for some k∈ℕ. On the other hand, this situation does not arise for impulsive problems on q-time scale as the points t and qt are consecutive points. In quantum calculus on finite intervals, the points t and qkt+(1-qk)tk are considered only in an interval [tk,tk+1]. Therefore, the problems with impulses at fixed times can be considered in the framework of qk-calculus. For more details, see [32].


## 2. Preliminaries

Let us setJ0=[0,t1],J1=(t1,t2],…,Jm-1=(tm-1,tm],Jm=(tm,T] and introduce the space as follows:
(2)PC(J,ℝ)={u:J→ℝ∣u∈C(Jk),k=0,1,…,m,andu(tk+)exist,k=1,2,…,m}
with the norm ∥u∥=sup⁡t∈J|u(t)|. Then, PC(J,ℝ) is a Banach space.Let us recall some basic concepts ofqk-calculus [32].For0<qk<1 and t∈Jk, we define the qk-derivatives of a real valued continuous function f as
(3)Dqkf(t)=f(t)-f(qkt+(1-qk)tk)(1-qk)(t-tk),Dqkf(tk)=lim⁡t→tkDqkf(t).
Higher order qk-derivatives are given by
(4)Dqk0f(t)=f(t),Dqknf(t)=DqkDqkn-1f(t),n∈ℕ,t∈Jk.
The qk-integral of a function f is defined by
(5)tkIqkf(t)∶=∫tkt‍f(s)dqks=(1-qk)(t-tk)×∑n=0∞‍qknf(qknt+(1-qkn)tk),0000000000000000000000000000000i000t∈Jk,
provided that the series converges. If a∈(tk,t) and f is defined on the interval (tk,t), then
(6)∫at‍f(s)dqks=∫tkt‍f(s)dqks-∫tka‍f(s)dqks.
Observe that
(7)Dqk(tkIqkf(t))=Dqk∫tkt‍f(s)dqks=f(t),tkIqk(Dqkf(t))=∫tktDqkf(s)dqks=f(t),aIqk(Dqkf(t))=∫atDqkf(s)dqks=f(t)-f(a),000000000000000000000000000000a∈(tk,t).Note that iftk=0 and qk=q in (3) and (5), then Dqkf=Dqf and tkIqkf=0Iqf, where Dq and 0Iq are the well-known q-derivative and q-integral of the function f(t) defined by
(8)Dqf(t)=f(t)-f(qt)(1-q)t,0Iqf(t)=∫0t‍f(s)dqs=∑n=0∞‍t(1-q)qnf(tqn).Lemma 1.
A functionu∈C(J,ℝ) is a solution of the impulsive antiperiodic boundary value problem (1) if and only if it is a solution of the following impulsive qk-integral equation:
(9)u(t)={∫0t‍f(s,u(s))dq0s-12[∑i=0m‍∫titi+1‍f(s,u(s))dqis+∑i=1m‍Ii(u(ti))],t∈J0;∫tkt‍f(s,u(s))dqks+∑0<tk<t‍[∫tk-1tk‍f(s,u(s))dqk-1s+Ik(u(tk))]-12[∑i=0m‍∫titi+1‍f(s,u(s))dqis+∑i=1m‍Ii(u(ti))],t∈Jk.Proof.
Letu be a solution of (1). For t∈J0, q0-integrating both sides of (1), we get
(10)u(t)=u(0)+t0Iq0f(t,u(t))=u(0)+∫0t‍f(s,u(s))dq0s.
Thus, we have
(11)u(t1-)=u(0)+∫0t1‍f(s,u(s))dq0s.
For t∈J1, q1-integrating both sides of (1), we obtain
(12)u(t)=u(t1+)+∫t1t‍f(s,u(s))dq1s.
In view of Δu(t1)=u(t1+)-u(t1-)=I1(u(t1)), it follows that
(13)u(t)=u(0)+∫t1t‍f(s,u(s))dq1s+∫0t1‍f(s,u(s))dq0s+I1(u(t1)),∀t∈J1.
Similarly, we get
(14)u(t)=u(0)+∫tkt‍f(s,u(s))dqks+∑0<tk<t‍[∫tk-1tk‍f(s,u(s))dqk-1s+Ik(u(tk))],000000000000000000000000000000t∈Jk.
Using the antiperiodic boundary value condition u(0)=-u(T), we obtain (9). Conversely, assume that u is a solution of the impulsive qk-integral equation (9); then by a direct computation, it follows that the solution given by (9) satisfies qk-difference equation (1). This completes the proof.


## 3. Main Results

Define an operator𝒯:PC(J,ℝ)→PC(J,ℝ) as
(15)𝒯u(t)=∫tkt‍f(s,u(s))dqks+∑0<tk<t‍[∫tk-1tk‍f(s,u(s))dqk-1s+Ik(u(tk))]-12[∑i=0m‍∫titi+1‍f(s,u(s))dqis+∑i=1m‍Ii(u(ti))].
Obviously, problem (1) is equivalent to a fixed point problem u=𝒯u. In consequence, problem (1) has a solution if and only if the operator 𝒯 has a fixed point.Theorem 2.
Assume that there exist continuous functionsa(t),b(t) and a nonnegative constant L such that
(16)|f(t,u(t))|≤a(t)+b(t)|u(t)|,|Ik(u)|≤L,k=1,2,…,m.
Then, problem (1) has at least one solution.Proof.
Let us denotesup⁡t∈J|a(t)|=A and sup⁡t∈J|b(t)|=B. Take R≥3(AT+mL)/(2-3BT)>0 and define BR={u∈PC(J,ℝ)∣∥u∥≤R}. It is easy to verify that BR is a bounded, closed, and convex subset of PC(J,ℝ).
In order to show that there exists a solution for problem (1), we have to establish that the operator 𝒯 has a fixed point in BR. The proof consists of two steps:
(i)𝒯:BR→BR.
For anyu∈BR, we have
(17)|𝒯u(t)|≤∫tkt‍|f(s,u(s))|dqks+∑0<tk<t‍[∫tk-1tk|f(s,u(s))|dqk-1s+|Ik(u(tk))|]+12[∑i=0m‍∫titi+1|f(s,u(s))|dqis+∑i=1m‍|Ii(u(ti))|]≤∫tkt‍[a(s)+b(s)|u(s)|]dqks+∑0<tk<t‍[∫tk-1tk[a(s)+b(s)|u(s)|]dqk-1s+L]+12[∑i=0m‍∫titi+1[a(s)+b(s)|u(s)|]dqis+∑i=1m‍L]≤[A+B∥u∥](t-tk)+[A+B∥u∥]tk+mL+12[(A+B∥u∥)T+mL]≤3(AT+mL)2+3BT∥u∥2≤R,
which means ∥𝒯u∥≤R. So, 𝒯 is BR→BR. Consider the following:
(ii) the operator𝒯 is relatively compact.
Letsup⁡(t,u)∈J×BR|f(t,u)|=f¯. For any t′,t′′∈Jk(k=0,1,2,…,m) with t′<t′′, we have
(18)|𝒯u(t′′)-𝒯u(t′)|≤|∫tkt′′‍f(s,u(s))dqks-∫tkt′‍f(s,u(s))dqks|≤∫t′t′′‍|f(s,u(s))|dqks≤f¯|t′′-t′|
which is independent of u and tends to zero as t′′-t′→0. Thus, 𝒯 is equicontinuous. Hence, 𝒯BR is relatively compact as 𝒯BR⊂BR is uniformly bounded. Further, it is obvious that the operator 𝒯 is continuous in view of continuity of f and Ik. Therefore, the operator 𝒯:PC(J,ℝ)→PC(J,ℝ) is completely continuous on BR. By the application of Schauder fixed point theorem, we conclude that the operator 𝒯 has at least one fixed point in BR. This, in turn, implies that problem (1) has at least one solution.Theorem 3.
Assume that there exist a functionM(t)∈C(J,ℝ+) and a positive constant N such that 3(MT+mN)<2 and
(19)|f(t,u)-f(t,v)|≤M(t)|u-v|,|Ik(u)-Ik(v)|≤N|u-v|,
for t∈J, u,v∈ℝ and k=1,2,…,m, and M=sup⁡t∈J|M(t)|. Then, problem (1) has a unique solution.Proof.
For∀u,v∈PC(J,ℝ), we have
(20)|(𝒯u)(t)-(𝒯v)(t)|≤∫tkt‍|f(s,u(s))-f(s,v(s))|dqks+∑0<tk<t‍[∫tk-1tk‍|f(s,u(s))-f(s,v(s))|dqk-1s00000000+|Ik(u(tk))-Ik(v(tk))|∫tk-1tk‍]+12[∑i=0m‍∫titi+1‍|f(s,u(s))-f(s,v(s))|dqis0000+∑i=1m‍|Ii(u(ti))-Ii(v(ti))|]≤∫tkt‍M(s)|(u-v)(s)|dqks+∑0<tk<t‍[∫tk-1tk‍M(s)|(u-v)(s)|dqk-1s00000000+N|(u-v)(tk)|∫tk-1tk‍]+12[∑i=0m‍∫titi+1‍M(s)|(u-v)(s)|dqis0000+∑i=1m‍N|(u-v)(ti)|]≤{∫tkt‍M(s)dqks+∑0<tk<t‍[∫tk-1tk‍M(s)dqk-1s+N]+12[∑i=0m‍∫titi+1‍M(s)dqis+mN]}∥u-v∥≤{[MT+mN]+12[MT+mN]}∥u-v∥≤3(MT+mN)2∥u-v∥.
By the given condition,3(MT+mN)<2, it follows that ∥𝒯u-𝒯v∥<∥u-v∥. Therefore, 𝒯 is a contraction. By the contraction mapping principle, problem (1) has a unique solution.


## 4. Example

Example 1.
Consider impulsive antiperiodic boundary value problem of nonlinearqk-difference equation:
(21)D1/(5+k)u(t)=t210sinu(t)+2+et,t∈[0,1],t≠k3+k,Δu(k3+k)=e-u2(k/(3+k)),k=1,2,…,5,u(0)=-u(1),
where qk=1/(5+k)(k=0,1,2,…,5), tk=k/(3+k)(k=1,2,…,5), f(t,u)=(t2/10)sinu+2+et, and Ik(u)=e-u2. With a(t)=2+et, b(t)=t2/10, and L=1, it is easy to verify that all conditions of Theorem 2 hold. Thus, by Theorem 2, problem (21) has at least one solution.Example 2.
Consider impulsive antiperiodic boundary value problem of nonlinearqk-difference equation:
(22)D1/(4+k)u(t)=t315arctanu(t)+ln⁡(2+t),0000i0000t∈[0,1],t≠2+k3+k,Δu(2+k3+k)=et45sinu(2+k3+k),k=1,2,3,u(0)=-u(1),
where qk=1/(4+k)(k=0,1,2,3), tk=(2+k)/(3+k)(k=1,2,3), f(t,u)=(t3/15)arctanu+ln⁡(2+t), and Ik(u)=(et/45)sinu. With M(t)=t3/15, M=1/15, and N=e/45, combining with T=1 and m=3, it is easy to verify that all conditions of Theorem 3 hold. Thus, by Theorem 3, problem (22) has a unique solution.



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