# Three-Point Boundary Value Problems of Nonlinear Second-Orderq-Difference Equations Involving Different Numbers of q

Authors: Thanin Sitthiwirattham; Jessada Tariboon; Sotiris K. Ntouyas
Journal: Journal of Applied Mathematics
 (2013)

---

## Abstract

We study a new class of three-point boundary value problems of nonlinear second-orderq-difference equations. Our problems contain different numbers of q in derivatives and integrals. By using a variety of fixed point theorems (such as Banach’s contraction principle, Boyd and Wong fixed point theorem for nonlinear contractions, Krasnoselskii’s fixed point theorem, and Leray-Schauder nonlinear alternative) and Leray-Schauder degree theory, some new existence and uniqueness results are obtained. Illustrative examples are also presented.


---

## Body

## 1. Introduction

Theq-difference calculus or quantum calculus is an old subject that was initially developed by Jackson [1], Carmichael [2], Mason [3], and Adams [4], in the first quarter of 20th century, has been developed over the years, for instance, see [5–14] and the references therein. In fact, q-calculus has a rich history, and the details of its basic notions, results, and methods can be found in the text [15]. In recent years, the topic has attracted the attention of several researchers, and a variety of new results can be found in the papers [16–28] and the references cited therein.In [24], Ahmad et al. studied a boundary value problem of nonlinear q-difference equations with nonlocal boundary conditions given by
(1)Dq2x(t)=f(t,x(t)),t∈Iq1,α1x(0)-β1Dqx(0)=γ1x(η1),α2x(1)+β2Dqx(1)=γ2x(η2),
where f∈C(Iq1×ℝ,ℝ), Iq1={qn:n∈ℕ}∪{0,1}, and q∈(0,1) is a fixed constant. The existence of solutions for problem (1) is shown by means of a variety of fixed point theorems such as Banach's contraction principle, Krasnoselskii’s fixed point theorem, and Leray-Schauder nonlinear alternative.Yu and Wang [28] considered a boundary value problem with the nonlinear second-order q-difference equation,
(2)Dq2u(t)+f(t,u(t),Dqu(t))=0,t∈Iq1,Dqu(0)=0,Dqu(1)=αu(1),
where f∈C(Iq1×ℝ2,ℝ) and α≠0 is a fixed number. Existence and uniqueness of the solutions are obtained by means of Banach’s contraction principle, Leray-Schauder nonlinear alternative, and Leray-Schauder continuation theorem.Pongarm et al. [29] considered sequential derivative of nonlinear q-difference equation with three-point boundary conditions,
(3)Dq(Dp+λ)u(t)=f(t,u(t)),t∈IqT=[0,T]∩Iq1,u(0)=0,u(T)=α∫0η‍u(s)drs,
where 0<p,q,r<1, f∈C(IqT×ℝ,ℝ), 0<η<T, and λ, β are given constants. Existence results are proved based on Banach’s contraction mapping principle, Krasnoselskii’s fixed point theorem, and Leray-Schauder degree theory.We note that in the above-mentioned papers [24, 28] the q-numbers in the equation and the boundary conditions are the same. As far as we know the paper by Pongarm et al. [29] is the first paper which has different values of the q-numbers in q-derivative and q-integral.In this paper, we discuss the existence of solutions for the following nonlinearq-difference equation with three-point integral boundary condition
(4)Dq2x(t)=f(t,x(t)),t∈IqT,αx(η)+βDrx(η)=0,∫0T‍x(s)dps=0,0<η<T,
where f∈C(IqT×ℝ,ℝ), IqT=Iq1∩[0,T], Iq1={qn:n∈ℕ}∪{0,1}, q∈(0,1) is a fixed constant, and η∈IqT∖{0,T}∶=(0,T)q. Also, 0<p,q,r<1, and α, β are given constants such that β≠α((T/(1+p))-η).It is noteworthy that, in the above problem (4), we have three different values of the q-numbers, in q-derivatives and the q-integral. Moreover, we emphasize the fact that, instead the value x(0) is usually used in the literature, we use the values of the function and its derivative in an intermediate point η∈(0,T).The aim of this paper is to prove some existence and uniqueness results for the boundary value problem (4). Our results are based on Banach’s contraction mapping principle, nonlinear contraction, Krasnoselskii’s fixed point theorem, Leray-Schauder nonlinear alternative, and Leray-Schauder degree theory.The rest of the paper is organized as follows. In Section2, we provide some basic definitions, preliminaries facts, and a lemma, which are used later. The main results are given in Section 3. In the end, Section 4, some results illustrating the results established in this paper are also presented.


## 2. Preliminaries

Let us recall some basic concepts ofq-calculus [15, 18].Definition 1.
For0<q<1, one defines the q-derivative of a real valued function f as
(5)Dqf(t)=f(t)-f(qt)(1-q)t,t∈Iq1∖{0},Dqf(0)=limt→0Dqf(t).The higher-orderq-derivatives are given by
(6)Dq0f(t)=f(t),Dqnf(t)=DqDqn-1f(t),n∈ℕ.Forx≥0 one sets Jx={xqn:n∈ℕ∪{0}}∪{0} and define, the definite q-integral of a function f:Jx→ℝ by
(7)Iqf(x)=∫0x‍f(s)dqs=∑n=0∞‍x(1-q)qnf(xqn)
provided that the series converges.Fora,b∈Jx, one sets
(8)∫ab‍f(s)dqs=Iqf(b)-Iqf(a)=(1-q)∑n=0∞‍qn[bf(bqn)-af(aqn)].Note that fora,b∈Jx, one has a=xqn1, b=xqn2 for some n1,n2∈ℕ; thus, the definite integral ∫ab‍f(s)dqs is just a finite sum, so no question about convergence is raised.One notes that(9)DqIqf(x)=f(x),
while if f is continuous at x=0, then
(10)IqDqf(x)=f(x)-f(0).
In q-calculus, the product rule and integration by parts formula are
(11)Dq(gh)(t)=(Dqg(t))h(t)+g(qt)Dqh(t),∫0x‍f(t)Dqg(t)dqt=[f(t)g(t)]0x-∫0x‍Dqf(t)g(qt)dqt.
Further, reversing the order of integration is given by
(12)∫0t‍∫0s‍f(r)dqrdqs=∫0t‍∫qrt‍f(r)dqsdqr.
In the limit q→1, the above results correspond to their counterparts in standard calculus.Lemma 2.
Let0<p,q,r<1 and η∈(0,T)q. Then, for any y∈C(IqT,ℝ), the boundary value problem,
(13)Dq2x(t)=y(t),t∈IqT,(14)αx(η)+βDrx(η)=0,∫0T‍x(s)dps=0,
is equivalent to the integral equation
(15)x(t)=∫0t‍(t-qs)y(s)dqs-(1+p)t-TΩ×[α∫0η‍(η-qs)y(s)dqs+β∫0rη‍y(s)dqs+β(1-r)η∫rηη‍(η-qs)y(s)dqs]+1+pTΩ(α(t-η)-β)∫0T‍∫0s‍(s-qv)y(v)dqvdps,
where
(16)Ω=(αη+β)(1+p)-αT≠0.Proof.
Taking doubleq-integral for (13), we have
(17)x(t)=∫0t‍∫0s‍y(v)dqvdqs+c1t+c2.
By changing the order of q-integration, we have
(18)x(t)=∫0t‍∫qvt‍y(v)dqsdqv+c1t+c2=∫0t‍(t-qs)y(s)dqs+c1t+c2.
In particular, for t=η, we get
(19)x(η)=∫0η‍(η-qs)y(s)dqs+ηc1+c2.
Taking r-derivative for (18), for t≠0, we obtain
(20)Drx(t)=Dr[∫0t‍(t-qs)y(s)dqs+c1t+c2]=1(1-r)t[∫0t‍(t-qs)y(s)dqs-∫0rt‍(rt-qs)y(s)dqs]+c1=∫0rt‍y(s)dqs+∫rtt‍t-qs(1-r)ty(s)dqs+c1.
For t=0, we have
(21)Drx(0)=limt→0Drx(0)=limt→0t(1-q)1-r∑n=0∞‍qn(1-qn+1)×[h(tqn)-r2h(rtqn)]+c1=c1.
Therefore,
(22)Drx(η)=∫0rη‍y(s)dqs+∫rηη‍η-qs(1-r)ηy(s)dqs+c1.
Now, using the first condition of (14) with (19), (22), we have
(23)(αη+β)c1+αc2=-α∫0η‍(η-qs)y(s)dqs-β∫0rη‍y(s)dqs-β(1-r)η∫rηη‍(η-qs)y(s)dqs.
Taking the p-integral for (18) from 0 to t, we obtain
(24)∫0tx(s)dps=∫0t‍∫0s(s-qv)y(v)dqvdps+t21+pc1+tc2.
Substituting t=T in (24) and using the second condition of (14), we get
(25)T21+pc1+Tc2=-∫0T‍∫0s‍(s-qv)y(v)dqvdps.
Solving the system of linear equations (23) and (25) for the unknown constants c1 and c2, we have
(26)c1=-1+pΩ[α∫0η‍(η-qs)y(s)dqs+β∫0rη‍y(s)dqs+β(1-r)η×∫rηη‍(η-qs)y(s)dqs]+α(1+p)TΩ∫0T‍∫0s‍(s-qv)y(v)dqvdps,c2=-(αη+β)(1+p)TΩ∫0T‍∫0s‍(s-qv)y(v)dqvdps+TΩ[α∫0η‍(η-qs)y(s)dqs+β∫0rη‍y(s)dqs+β(1-r)η∫rηη‍(η-qs)y(s)dqs],
where Ω is defined by (16). Substituting the values of c1 and c2 in (18), we obtain (15). This completes the proof.Let𝒞=C(IqT,ℝ) denotes the Banach space of all the continuous functions from IqT to ℝ endowed with the norm defined by ∥x∥=sup{|x(t)|,t∈IqT}. Define an operator A:𝒞→𝒞 by
(27)(Ax)(t)=∫0t‍(t-qs)f(s,x(s))dqs-(1+p)t-TΩ×[α∫0η‍(η-qs)f(s,x(s))dqs+β∫0rη‍f(s,x(s))dqs+β(1-r)η∫rηη‍(η-qs)f(s,x(s))dqs]+(1+p)(α(t-η)-β)TΩ×∫0T‍∫0s‍(s-qv)f(v,x(v))dqvdps.
Observe that the problem (4) has solutions if and only if the operator A has fixed points.For the sake of convenience, we set a constantΛ as
(28)Λ=T21+q+pT|Ω|[|α|η21+q+|β|rη+|β|η(1+q(2+r))1+q]+(1+p)(|α|(T-η)+|β|)T2(1+q)(1+p+p2)|Ω|.


## 3. Main Results

Now, we are in the position to establish the main results. Our first result is based on Banach’s fixed point theorem.Theorem 3.
Assume thatf:IqT×ℝ→ℝ is a continuous function satisfying the conditions(H1)
|f(t,x)-f(t,y)|≤L|x-y|, for all t∈IqT and x,y∈ℝ,(H2)
LΛ<1,
whereL is a Lipschitz constant, and Λ is defined by (28).
Then, the boundary value problem (4) has a unique solution.Proof.
We transform the boundary value problem (4) into a fixed point problem x=Ax, where A:𝒞→𝒞 is defined by (27). Assume that supt∈IqT|f(t,0)|=M, and choose a constant R satisfying
(29)R≥MΛ1-LΛ.
Now, we will show that ABR⊂BR, where BR={x∈𝒞:∥x∥≤R}. For any x∈BR, we have
(30)∥Ax∥=supt∈IqT|∫0t‍(t-qs)f(s,x(s))dqs-(1+p)t-TΩ×[α∫0η‍(η-qs)f(s,x(s))dqs+β∫0rη‍f(s,x(s))dqs+β(1-r)η×∫rηη‍(η-qs)f(s,x(s))dqs∫0η‍]+(1+p)(α(t-η)-β)TΩ×∫0T‍∫0s‍(s-qv)f(v,x(v))dqvdps∫0T‍|≤supt∈IqT{∫0t‍(t-qs)×(|f(s,x(s))-f(s,0)|+|f(s,0)|)dqs+|(1+p)t-T||Ω|×[|α|∫0η‍(η-qs)(|f(s,x(s))-f(s,0)|+|f(s,0)|(s,x(s)))dqs+|β|∫0rη‍(|f(s,x(s))-f(s,0)|+|f(s,0)|)dqs+|β|(1-r)η×∫rηη‍(η-qs)(|f(s,x(s))-f(s,0)|+|f(s,0)|)dqs∫0rη‍]+(1+p)|α(t-η)-β|T|Ω|×∫0T‍∫0s‍(s-qv)×(|f(v,x(s))-f(v,0)|+|f(v,0)|)dqvdps∫0T‍∫0s‍(s-qv)}≤supt∈IqT{∫0t‍(t-qs)(LR+M)dqs+|(1+p)t-T||Ω|×[|(1+p)t-T||Ω||α|∫0η‍(η-qs)(LR+M)dqs+|β|×∫0rη‍(LR+M)dqs+|β|(1-r)η∫rηη‍(η-qs)(LR+M)dqs]+(1+p)|α(t-η)-β|T|Ω|×∫0T‍∫0s‍(s-qv)(LR+M)dqvdps|(1+p)t-T||Ω|∫0t‍}≤(LR+M){T21+q+pT|Ω|×[|α|η21+q+|β|rη+|β|η(1+q(2+r))1+q]+(1+p)(|α|(T-η)+|β|)T2(1+q)(1+p+p2)|Ω|}=(LR+M)Λ≤R.
Therefore, ABR⊂BR.
Next, we will show thatA is a contraction. For any x,y∈𝒞 and for each t∈IqT, we have
(31)∥Ax-Ay∥=supt∈IqT|(Ax)(t)-(Ay)(t)|≤supt∈IqT|∫0t‍(t-qs)|f(s,x(s))-f(s,y(s))|dqs-(1+p)t-TΩ×[α∫0η‍(η-qs)|f(s,x(s))-f(s,y(s))|dqs+β∫0rη‍|f(s,x(s))-f(s,x2(s))|dqs+β(1-r)η×∫rηη‍(η-qs)|f(s,x(s)-f(s,y(s))|dqs∫0t‍]+(1+p)(α(t-η)-β)TΩ×∫0T‍∫0s‍(s-qv)×|f(v,x(v))-f(v,y(v))|dqvdps∫0t‍|≤supt∈IqT{L∥x-y∥∫0t‍(t-qs)dqs+L∥x-y∥×|(1+p)t-T||Ω|[|α|∫0η‍(η-qs)dqs+|β|∫0rη‍dqs+|β|(1-r)η×∫rηη‍(η-qs)dqs]+(1+p)(|α|(t-η)+|β|)T|Ω|L∥x-y∥×∫0T‍∫0s‍(s-qv)dqvdps}≤L∥x-y∥{T21+q+pT|Ω|×[|α|η21+q+|β|rη+|β|η(1+q(2+r))1+q]+(1+p)(|α|(T-η)+|β|)T2(1+q)(1+p+p2)|Ω|}=LΛ∥x-y∥.
Since LΛ<1, A is a contraction. Thus, the conclusion of the theorem follows by Banach’s contraction mapping principle. This completes the proof.Next, we can still deduce the existence and uniqueness of a solution to the boundary value problem (4). We will use nonlinear contraction to accomplish this.Definition 4.
LetE be a Banach space and let F:E→E be a mapping. F is said to be a nonlinear contraction if there exists a continuous nondecreasing function Ψ:R+→R+ such that Ψ(0)=0 and Ψ(ρ)<ρ for all ρ>0 with the following property:
(32)∥Fx-Fy∥≤Ψ(∥x-y∥),∀x,y∈E.Lemma 5 (Boyd and Wong [30]).
LetE be a Banach space and let F:E→E be a nonlinear contraction. Then, F has a unique fixed point in E.Theorem 6.
Suppose that(H3)
there exists a continuous functionh:IqT→ℝ+ such that
(33)|f(t,x)-f(t,y)|≤h(t)|x-y|G+|x-y|
for all t∈IqT and x,y≥0, where
(34)G=∫0T‍(T-qs)h(s)dqs+pT|Ω|×[∫0η‍|α|∫0η‍(η-qs)h(s)dqs+|β|∫0rη‍h(s)dqs+|β|(1-r)η∫rηη‍(η-qs)h(s)dqs∫0η‍]+(1+p)(|α|(T-η)+|β|)T|Ω|×∫0T‍∫0s‍(s-qv)h(v)dqvdpsandΩ is defined in (16).
Then, the boundary value problem (4) has a unique solution.Proof.
Let the operatorA:𝒞→𝒞 be defined as (27). We define a continuous nondecreasing function Ψ:ℝ+→ℝ+ by
(35)Ψ(ρ)=GρG+ρ,∀ρ≥0,
such that Ψ(0)=0 and Ψ(ρ)<ρ, for all ρ>0.
Letx,y∈𝒞. Then, we get
(36)|f(s,x(s))-f(s,y(s))|≤h(s)GΨ(∥x-y∥).
Thus,
(37)|Ax(t)-Ay(t)|≤∫0t‍(t-qs)h(s)|x(s)-y(s)|G+|x(s)-y(s)|dqs+|(1+p)t-T||Ω|[|α|∫0η‍(η-qs)×h(s)|x(s)-y(s)|G+|x(s)-y(s)|dqs+|β|∫0rη‍h(s)|x(s)-y(s)|G+|x(s)-y(s)|dqs+|β|(1-r)η∫rηη‍(η-qs)×h(s)|x(s)-y(s)|G+|x(s)-y(s)|dqs]+(1+p)|α(t-η)-β|T|Ω|×∫0T‍∫0s‍(s-qv)×h(v)|x(v)-y(v)|G*+|x(v)-y(v)|dqvdps≤{∫0T‍(T-qs)h(s)dqs+pT|Ω|×[|β|(1-r)η∫0η‍|α|∫0η‍(η-qs)h(s)dqs+|β|∫0rη‍h(s)dqs+|β|(1-r)η∫rηη‍(η-qs)h(s)dqs∫0η‍]+(1+p)(|α|(T-η)+|β|)T|Ω|×∫0T‍∫0s‍(s-qv)h(v)dqvdps∫0T‍}×∥x-y∥G+∥x-y∥=G∥x-y∥G+∥x-y∥,∀t∈IqT.
This implies that ∥Ax-Ay∥≤Ψ(∥x-y∥). Hence, A is a nonlinear contraction. Therefore, by Lemma 5, the operator A has a unique fixed point in 𝒞, which is a unique solution of problem (4).The third result is based on the following Krasnoselskii fixed point theorem [31].Theorem 7.
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 existsz∈K such that z=Az+Bz.Theorem 8.
Assume that(H1) and (H2) hold. In addition one supposes that: (H4)
|f(t,x)|≤μ(t), for all (t,x)∈IqT×ℝ, with μ∈L1(IqT,ℝ+).
If(38)Λ<1,
where Λ is given by (28), then the boundary value problem (4) has at least one solution on IqT.Proof.
Settingmaxt∈IqT|μ(t)|=∥μ∥ and choosing a constant
(39)R≥∥μ∥Λ,
we consider BR={x∈𝒞:∥x∥≤R}.
In view of Lemma2, we define the operators ℱ1 and ℱ2 on the ball BR as
(40)(ℱ1x)(t)=∫0t‍(t-qs)f(s,x(s))dqs,(ℱ2x)(t)=-(1+p)t-TΩ[α∫0η‍(η-qs)f(s,x(s))dqs+β∫0rη‍f(s,x(s))dqs+β(1-r)η×∫rηη‍(η-qs)f(s,x(s))dqs∫0η‍]+(1+p)(α(t-η)-β)TΩ×∫0T‍∫0s‍(s-qv)f(v,x(v))dqvdps.
For x,y∈BR, by computing directly, we have
(41)∥ℱ1x+ℱ2y∥≤∥μ∥∫0t‍(t-qs)dqs+∥μ∥|(1+p)t-T||Ω|×[|α|∫0η‍(η-qs)dqs+|β|∫0rη‍dqs+|β|(1-r)η∫rηη‍(η-qs)dqs]+∥μ∥(1+p)|α(t-η)-β)|T|Ω|×∫0T‍∫0s‍(s-qv)dqvdps≤∥μ∥Λ≤R.
Therefore, ℱ1x+ℱ2y∈BR. Condition (38) 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)∈IqT×BR|f(t,x)|=fmax<∞. For t1,t2∈IqT with t1≤t2 and x∈BR, we have
(42)|ℱ1x(t2)-ℱ1x(t1)|=|∫0t2‍(t2-qs)f(s,x(s))dqs-∫0t1‍(t1-qs)f(s,x(s))dqs|=|∫0t1‍(t2-t1)f(s,x(s))dqs+∫t1t2‍(t2-qs)f(s,x(s))dqs|≤|t22-t12|(1+2q1+q)fmax.
Actually, as t2-t1→0, the right-hand side of the above inequality tends to be 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 7 are satisfied, and the conclusion of Theorem 7 implies that the boundary value problem (4) has at least one solution on IqT. This completes the proof.As the fourth result, we prove the existence of solutions of (4) by using Leray-Schauder nonlinear alternative.Theorem 9 (Nonlinear Alternative for Single Valued Maps [32]).
LetE 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 au∈∂U (the boundary of U in C) and λ∈(0,1) with u=λF(u).Theorem 10.
Assume that:(H5)
there exists a continuous nondecreasing functionψ:[0,∞)→(0,∞) and a function z∈L1(IqT,ℝ+) such that
(43)|f(t,u)|≤z(t)ψ(∥u∥),foreach(t,u)∈IqT×ℝ;(H6)
there exists a constantM>0 such that
(44)Mψ(M)∥z∥L1Λ>1.Then, the boundary value problem (4) has at least one solution on IqT.Proof.
We will show thatAmaps bounded sets (balls) into bounded sets in𝒞. For a positive number ρ, let Bρ={x∈C(IqT,ℝ):∥x∥≤ρ} be a bounded ball in C(IqT,ℝ). Then, for t∈IqT, we have
(45)|(Ax)(t)|≤∫0t‍(t-qs)|f(s,x(s))|dqs+|(1+p)t-T||Ω|×[∫0η‍|α|∫0η‍(η-qs)|f(s,x(s))|dqs+|β|∫0rη‍|f(s,x(s))|dqs+|β|(1-r)η×∫rηη‍(η-qs)|f(s,x(s))|dqs]+(1+p)|α(t-η)-β|T|Ω|×∫0T‍∫0s‍(s-qv)|f(v,x(v))|dqvdps≤ψ(∥x∥)∫0t‍(t-qs)z(s)dqs+ψ(∥x∥)|(1+p)t-T||Ω|×[|β|(1-r)η∫0η‍|α|∫0η‍(η-qs)z(s)dqs+|β|∫0rη‍z(s)dqs+|β|(1-r)η∫rηη‍(η-qs)z(s)dqs∫0η‍]+ψ(∥x∥)(1+p)|α(t-η)-β|T|Ω|×∫0T‍∫0s‍(s-qv)z(s)dqvdps≤ψ(∥x∥)∥z∥L1∫0t‍(t-qs)dqs+ψ(∥x∥)∥z∥L1|(1+p)t-T||Ω|×[|α|∫0η‍(η-qs)dqs+|β|∫0rη‍dqs+|β|(1-r)η∫rηη‍(η-qs)dqs]+ψ(∥x∥)∥z∥L1(1+p)|α(t-η)-β|T|Ω|×∫0T‍∫0s‍(s-qv)dqvdps≤ψ(∥x∥)∥z∥L1T21+q+ψ(∥x∥)∥z∥L1pT|Ω|[|α|η21+q+|β|rη+|β|η(1+q(2+r))1+q]+ψ(∥x∥)∥z∥L1(1+p)(|α|(T-η)+|β|)T2(1+q)(1+p+p2)|Ω|=ψ(∥x∥)∥z∥L1Λ.
Consequently,(46)∥Ax∥≤ψ(∥x∥)∥z∥L1Λ.
Next, we will show thatAmaps bounded sets into equicontinuous sets ofC(IqT,ℝ). Let t1,t2∈IqT with t1≤t2 and x∈Bρ. Then, we have
(47)|(Ax)(t2)-(Ax)(t1)|≤|∫0t2‍(t2-qs)|f(s,x(s))|dqs-∫0t1‍(t1-qs)|f(s,x(s))|dqs|+(1+p)|t2-t1||Ω|[|α|∫0η‍(η-qs)|f(s,x(s))|dqs+|β|∫0rη‍|f(s,x(s))|dqs+|β|(1-r)η∫rηη‍(η-qs)×|f(s,x(s))|dqs∫0rη‍]+(1+p)|α||t2-t1|T|Ω|×∫0T‍∫0s‍(s-qv)|f(v,x(v))|dqvdps≤∫0t1‍|t2-t1|z(s)ψ(ρ)dqs+∫t1t2‍(t2-qs)z(s)ψ(ρ)dqs+(1+p)|t2-t1||Ω|[∫rηη‍|α|∫0η‍(η-qs)z(s)ψ(ρ)dqs+|β|∫0rη‍z(s)ψ(ρ)dqs+|β|(1-r)η×∫rηη‍(η-qs)z(s)ψ(ρ)dqs]+(1+p)|α||t2-t1|T|Ω|×∫0T‍∫0s‍(s-qv)z(s)ψ(ρ)dqvdps.
As t2-t1→0, the right-hand side of the above inequality tends to zero independently of x∈Bρ. As A satisfies the above assumptions; therefore, it follows by the Arzelá-Ascoli theorem that A:C(IqT,ℝ)→C(IqT,ℝ) is completely continuous.
Letx be a solution. Then, for t∈IqT and following the similar computations as in the first step, we have
(48)|x(t)|≤ψ(∥x∥)∥z∥L1Λ.
Consequently, we have(49)∥x∥ψ(∥x∥)∥z∥L1Λ≤1.
In view of (H5), there exists M such that ∥x∥≠M. Let us set
(50)U={x∈C(IqT,ℝ):∥x∥<M}.
Note that the operator A:U¯→C(IqT,ℝ) is continuous and completely continuous. From the choice of U, there is no x∈∂U such that x=λAx for some λ∈(0,1). Consequently, by the nonlinear alternative of Leray-Schauder type (Theorem 9), we deduce that A has a fixed point x∈U¯ which is a solution of the problem (4). This completes the proof.Finally, we prove that problem (4) has at least one solution on IqT by using Leray-Schauder degree theory.Theorem 11.
Letf:IqT×ℝ→ℝ be a continuous function. Assume that: (H7)
there exist constants0≤κ<Λ-1, where Λ is given by (28) and N>0 such that |f(t,x)|≤κ|x|+N for all t∈IqT, x∈𝒞. Then, the boundary value problem (4) has at least one solution.Proof.
Let us define an operatorA:𝒞→𝒞 as (27). We wish to prove that there exists at least one solution x∈𝒞 of the fixed point equation
(51)x=Ax.
We define a ball BR⊂𝒞, with a constant radius R>0 given by
(52)BR={x∈𝒞:maxt∈IqT|x(t)|<R}.
Then, it is sufficient to show that A:B¯R→𝒞(IqT) satisfies
(53)x≠λAx,∀x∈∂BR,∀λ∈[0,1].
Now, we set
(54)H(λ,x)=λAx,x∈𝒞,λ∈[0,1].
Then, by the Arzelá-Ascoli theorem, we conclude that a continuous map hλ defined by hλ(x)=x-H(λ,x)=x-λAx is completely continuous. If (53) holds, then the following Leray-Schauder degrees are well defined. From the homotopy invariance of topological degree, it follows that
(55)deg(hλ,BR,0)=deg(I-λA,BR,0)=deg(h1,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=λAx for some λ∈[0,1]. Then, for all t∈IqT, we obtain
(56)|x(t)|=|λ(Ax)(t)|≤∫0t‍(t-qs)|f(s,x(s))|dqs+|(1+p)t-T||Ω|×[|α|∫0η‍(η-qs)|f(s,x(s))|dqs+|β|∫0rη‍|f(s,x(s))|dqs+|β|(1-r)η×∫rηη‍(η-qs)|f(s,x(s))|dqs]+(1+p)|α(t-η)-β|T|Ω|∫0T‍∫0s‍(s-qv)×|y(v,x(v))|dqvdps≤(κ|x|+N)∫0t‍(t-qs)dqs+(κ|x|+N)×|(1+p)t-T||Ω|×[|α|∫0η‍(η-qs)dqs+|β|∫0rη‍dqs+|β|(1-r)η∫rηη‍(η-qs)dqs]+(κ|x|+N)(1+p)|α(t-η)-β|T|Ω|×∫0T‍∫0s‍(s-qv)dqvdps≤(κ|x|+N){T21+q+pT|Ω|[|α|η21+q+|β|rη+|β|η(1+q(2+r))1+q]+(1+p)(|α|(T-η)+|β|)T2(1+q)(1+p+p2)|Ω|T21+q+pT|Ω|[|α|η21+q}=(κ|x|+N)Λ.
Taking norm supt∈IqT|x(t)|=∥x∥ and solving it for ∥x∥, this yields
(57)∥x∥≤NΛ1-κΛ.
Let R=(NΛ/(1-κΛ))+1, then (53) holds. This completes the proof


## 4. Examples

In this section, we illustrate our main results with some examples. Let us consider the following boundary value problem of nonlinear second-orderq-difference equations with three-point boundary conditions
(58)D1/22x(t)=f(t,x(t)),t∈I1/21/2=I1/21∩[0,12],23x(18)-13D3/4x(18)=0,∫01/2‍x(s)d1/4s=0.
Here, we have q=1/2, p=1/4, r=3/4, T=1/2, α=2/3, β=-1/3, and η=1/8. We find that
(59)Λ=T21+q+pT|Ω|[|α|η21+q+|β|rη+|β|η(1+q(2+r))1+q]+(1+p)(|α|(T-η)+|β|)T2(1+q)(1+p+p2)|Ω|=16+631[1144+132+19288]+2801953≈0.33019713.(a)
Letf:I1/21/2×ℝ→ℝ be a continuous function given by
(60)f(t,x)=e-sin2t1+ecos2t·|x(t)|1+|x(t)|. Since, |f(t,x)-f(t,y)|≤(1/2)|x-y|, then (H1) is satisfied with L=1/2. We can find that
(61)LΛ≈0.16509857<1.
Hence, by Theorem 3, problem (58) with f(t,x) given by (60) has a unique solution on I1/21/2.(b)
Iff:I1/21/2×ℝ→ℝ is a continuous function given by
(62)f(t,x)=(t+1)|x|1+|x|.Choosingh(t)=t+1, we find that
(63)G=∫0T‍(T-qs)h(s)dqs+pT|Ω|×[|α|∫0η‍(η-qs)h(s)dqs+|β|∫0rη‍(s+1)dqs+|β|(1-r)η×∫rηη‍(η-qs)(s+1)dqs]+(1+p)(|α|(T-η)+|β|)T|Ω|×∫0T‍∫0s‍(s-qv)(v+1)dqvdps≈0.40987235.Here,(64)|f(t,x)-f(t,y)|≤(1+t)|x-y|0.40987235+|x-y|.Therefore, by Theorem6, the problem (58) with f(t,x) given by (62) has a unique solution on I1/21/2.(c)
Consider a continuous functionf:I1/21/2×ℝ→ℝ given by
(65)f(t,x)=sin2x+3e-x2+t+2.We can show that(66)|f(t,x)|=|sin2x+3e-x2+t+2|≤2∥x∥+32,with(67)κ=2<1Λ≈3.02849389,andN=3/2. By Theorem 11, the problem (58) with the f(t,x) given by (65) has at least one solution on I1/21/2.



---

*Source: 763786-2013-10-09.xml*