# Nonlocal Boundary Value Problems forq-Difference Equations and Inclusions

Authors: Sotiris K. Ntouyas; Jessada Tariboon
Journal: International Journal of Differential Equations
 (2015)

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## Abstract

We study boundary value problems forq-difference equations and inclusions with nonlocal and integral boundary conditions which have different quantum numbers. Some new existence and uniqueness results are obtained by using fixed point theorems. Examples are given to illustrate the results.


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## Body

## 1. Introduction

In this paper we introduce a new class of boundary value problems forq-difference equations with nonlocal boundary conditions given by
(1)Dq2xt=ft,xt,t∈IqT,xξ=gx,αDrxη+β∫ηT‍xsdps=0,0<ξ<η<T,
where f∈C(IqT×R,R) is such that f(t,x) is continuous at t=0,T, IqT=Iq∩[0,T], Iq=qn:n∈Z∪0, q∈(0,1) is a fixed constant, ξ,η∈IqT∖{0,T}∶=(0,T)q, 0<p, r<1, and α, β are given constants such that ξβ(T-η)≠α+βT2-η2/(1+p).The study ofq-difference equations, initiated by Jackson [1, 2], Carmichael [3], Mason [4], and Adams [5] in the first quarter of 20th century, has been developed over the years; for instance, see [6–8]. In recent years, the topic has attracted the attention of several researchers and a variety of new results can be found in the papers [9–21].Nonlocal conditions were initiated by Bitsadze [22]. As remarked by Byszewski [23, 24], the nonlocal condition can be more useful than the standard initial condition to describe some physical phenomena. For example, g(x) may be given by g(x)=∑i=1p‍cix(ti) where ci,i=1,…,p, are given constants and 0<t1<⋯<tp≤T.In Section3 we give some sufficient conditions for the existence and uniqueness of solutions and for the existence of at least one solution of problem (1). The first result is based on Banach’s contraction principle and the second on a fixed point theorem due to O’Regan [25]. Concrete examples are also provided to illustrate the possible applications of the established analytical results.In Section4, we extend the results to cover the multivalued case, considering the following boundary value problem for q-difference inclusions with nonlocal and integral boundary conditions:
(2)Dq2xt∈Ft,xt,t∈IqT,xξ=gx,αDrxη+β∫ηT‍xsdps=0,
where F:IqT×R→P(R) is a multivalued map and P(R) is the family of all nonempty subsets of R.We give an existence result for the problem (2) in the case when the right hand side is convex valued by using the nonlinear alternative for contractive maps.The paper is organized as follows: in Section2 we recall some preliminary facts that we need in the sequel, in Section 3 we prove our main results for single-valued case, and in Section 4 we prove our main results for multivalued case.


## 2. Preliminaries

Let us recall some basic concepts ofq-calculus [7, 8, 26].Let0<q<1 and f a function defined on a q-geometric set A; that is, qt∈A for all t∈A. The q-difference operator is defined by
(3)Dqft=f(t)-fqt1-qt,t∈A∖0,lim⁡n→∞ftqn-f0tqn,t=0,
provided that the limit exists and does not depend on t. The higher order q-derivatives are given by
(4)Dq0ft=ft,Dqnft=DqDqn-1ft,n∈N.The Jacksonq-integration [1] is
(5)∫0a‍ftdqt=a1-q∑n=0∞‍qnfaqn,∫ab‍ftdqt=∫0b‍ftdqt-∫0a‍ftdqt,
where a,b∈A, provided that the series converge. Here we remark that the integral ∫ab‍f(t)dq(t) is understood as a right inverse of the q-derivative.For0∈A, f is called q-regular at zero if lim⁡n→∞f(tqn)=f(0) for every t∈A, t≠0. It is important to note that continuity at zero implies q-regularity at zero but the converse is not true (see an example on page 7 in [26]).Definition 1.
Letf be a function defined on a q-geometric set A. Then f is q-integrable on A if and only if ∫0t‍f(μ)dqμ exists for all t∈A.Theq-integration by parts rule is
(6)∫ab‍uqtDqvtdqt=ubvb-uava+∫ab‍Dqutvtdqt,
provided that u and v are q-regular at zero functions.Letf be a q-regular at zero function defined on a q-geometric set A containing zero. Then
(7)Fz=∫cz‍fsdqs,z∈A
is q-regular at zero, where c is a fixed point in A. Furthermore, DqF(z) exists for every z∈A and
(8)DqFz=fz,z∈A.
Conversely, if a and b are two points in A, then
(9)∫ab‍Dqfsdqs=fb-fa.We denote byC=C(IqT,R) the Banach space of all continuous functions from IqT→R which are q-regular at zero.To define the solution for the problem (1), we find the solution for its associated linear problem.Lemma 2.
Lety∈C(IqT,R), be a continuous function such that it is continuous at 0 and T. The solution of the q-difference equation
(10)Dq2xt=yt,t∈IqT
subject to the boundary conditions,
(11)xξ=gx,αDrxη+β∫ηT‍xsdps=0,
is given by
(12)xt=∫0t‍t-qsysdqs+t-ξD·β∫ηT‍∫0s‍s-qτyτdqτdpsz+α∫0rη‍ysdqs+α1-rη∫rηη‍η-qsysdqs+1Dα+βT2-η21+p-βT-ηt·∫0ξ‍ξ-qsysdqs-1Dα+βT2-η21+p-βT-ηtgx,t∈IqT,
where
(13)D=ξβT-η-α-βT2-η21+p≠0.Proof.
Integrating twice the given equation and changing the order of integration, we get(14)xt=∫0t‍t-qsysdqs+c1t+c2,
for some constants c1,c2∈R (for functions not necessarily continuous at zero, the constants c0, c1 are q-periodic functions [26]).
In particular, fort=ξ, we get
(15)xξ=∫0ξ‍ξ-qsysdqs+c1ξ+c2.
Taking r-derivative for (14), for t≠0, we obtain
(16)Drxt=Dr∫0t‍t-qsysdqs+c1t+c2=11-rt∫0t‍t-qsysdqs-∫0rt‍rt-qsysdqs+c1=∫0rt‍ysdqs+∫rtt‍t-qs1-rtysdqs+c1.
Therefore,(17)Drxη=∫0rη‍ysdqs+∫rηη‍η-qs1-rηysdqs+c1.
Taking thep-integral for (14) from 0 to t, we obtain
(18)∫0t‍xsdps=∫0t‍∫0s‍s-qvyvdqvdps+t21+pc1+tc2.
Then we have(19)∫ηT‍xsdps=∫0T‍xsdps-∫0η‍xsdps=∫0T‍∫0s‍s-qvyvdqvdps+T21+pc1+Tc2-∫0η‍∫0s‍s-qvyvdqvdps-η21+pc1-ηc2.
By applying the boundary conditions we get to the system(20)α+βT2-η21+pc1+βT-ηc2=-β∫ηT‍∫0s‍s-qτyτdqτdps-α∫0rη‍ysdqs-α∫rηη‍η-qs1-rηysdqsc1ξ+c2=gx-∫0ξ‍ξ-qsysdqs,
from which we have
(21)c1=1Dgx-∫0ξ‍ξ-qsysdqsβ∫ηT‍∫0s‍s-qτyτdqτdps+α∫0rη‍ysdqs+α∫rηη‍η-qs1-rηysdps+βT-ηgx-∫0ξ‍ξ-qsysdqs,c2=-1Dξβ∫ηT‍∫0s‍s-qτyτdqτdps+α∫0rη‍ysdqs+α∫rηη‍η-qs1-rηysdqs+α+βT2-η21+p·gx-∫0ξ‍ξ-qsysdqs.
Substituting into (14) the values of c1 and c2 we obtain (12).


## 3. Existence Results: The Single-Valued Case

In view of Lemma2, we define an operator Q:C→C by
(22)Qxt=∫0t‍t-qsfs,xsdqs+t-ξDβ∫ηT‍∫0s‍s-qτfτ,xτdqτdps+α∫0rη‍fs,xsdqs+α1-rη∫rηη‍η-qsfs,xsdqs+1Dα+βT2-η21+p-βT-ηt·∫0ξ‍ξ-qsfs,xsdqs-1Dα+βT2-η21+p-βT-ηtgx,t∈IqT.For convenience we set(23)Λ=T21+q+ξ+TD·βT3-η31+q1+p+p2+αrη+αη1-qr1+q+k0ξ21+q,(24)k0=1Dα+βT2-η21+p+βTT-η.Theorem 3.
Letf:IqT×R→R and g:C(IqT,R)→R be continuous functions. Assume that(A1)
f(t,x)-f(t,y)≤Lx-y, ∀t∈IqT, L>0, x,y∈R;(A2)
g(u)-g(v)≤lu-v, l<k0-1 for all u,v∈C(IqT,R);(A3)
γ=LΛ+k0l<1.Then the boundary value problem (1) has a unique solution.Proof.
Forx,y∈C and for each t∈IqT, from the definition of Q and assumptions (A1) and (A2), we obtain
(25)Qxt-Qyt≤∫0t‍t-qsfs,xs-fs,ysdqs+ξ+TD·β∫ηT‍∫0s‍s-qτfτ,xτ-fτ,yτdqτdps+α∫0rη‍fs,xs-fs,ysdqs+α1-rη∫rηη‍η-qsfs,xs-fs,ysdqs+1Dα+βT2-η21+p+βTT-η·∫0ξ‍ξ-qsfs,xs-fs,ysdqs+1Dα+βT2-η21+p+βTT-η·gx-gy≤Lx-yα+βT2-η21+p+βTT-ηT21+q+ξ+TD·βT3-η31+q1+p+p2+αrη+αη1-qr1+q+1D·α+βT2-η21+p+βTT-η·ξ21+qα+βT2-η21+p+βTT-η+1Dα+βT2-η21+p+βTT-ηlx-y.
Hence
(26)Qx-Qy≤γx-y.
As γ<1, by (A3), F is a contraction map from the Banach space C into itself. Thus, the conclusion of the theorem follows by the contraction mapping principle (Banach fixed point theorem).Example 4.
Consider the following nonlocal boundary value problem of nonlinearq-difference equation
(27)D1/22xt=e-2t2t+52·x1+x+12,t∈0,21/2,x14=130x18+13,23D3/4x12-13∫1/22‍xsd1/4s=0.
Here, q=1/2, r=3/4, p=1/4, T=2, ξ=1/4, η=1/2, α=2/3, β=-1/3, g(x)=(1/30)x+(1/3), and f(t,x)=(e-2tx)/(2(t+5)2(1+x))+(1/2). We find that D=5/24, k0=192/15, and Λ=981/45.
Asf(t,x)-f(t,y)≤(1/50)x-y and g(x)-g(y)≤(1/30)x-y, therefore, (A1) and (A2) are satisfied with L=1/50 and l=1/30<15/192=k0-1, respectively. Hence γ=LΛ+k0l≈0.86267<1. By the conclusion of Theorem 3, the boundary value problem (27) has a unique solution on [0,2]1/2.Next, we introduce the fixed point theorem which was established by O’Regan in [25]. This theorem will be adopted to prove the next main result.Lemma 5.
LetU be an open set in a closed, convex set C of a Banach space E. Assume 0∈U. Also assume that F(U¯) is bounded and that F:U¯→C is given by F=F1+F2, in which F1:U¯→E is continuous and completely continuous and F2:U¯→E is a nonlinear contraction (i.e., there exists a continuous nondecreasing function ϑ:[0,∞)→[0,∞) satisfying ϑ(z)<z for z>0, such that F2(x)-F2(y)≤ϑ(x-y) for all x,y∈U¯). Then, either(C1)
F has a fixed point u∈U¯ or(C2)
there exist a pointu∈∂U and κ∈(0,1) with u=κF(u), where U¯ and ∂U, respectively, represent the closure and boundary of U on C.In the sequel, we will use Lemma5 by taking C to be E. For more details of such fixed point theorems, we refer a paper [27] by Petryshyn.Let(28)Ωr=x∈CIqT,R:x<r.Theorem 6.
Letf:IqT×R→R be a continuous function. Suppose that (A2) holds. In addition we assume that(A4)
g(0)=0;(A5)
there exists a nonnegative functionp∈C(IqT,R) and a nondecreasing function ψ:[0,∞)→(0,∞) such that
(29)ft,u≤ptψuforanyt,u∈IqT×R;(A6)
sup⁡r∈0,∞(r/Λ∥p∥ψr)>1/(1-k0l), where Λ and k0 are defined in (23) and (24), respectively.Then the boundary value problem (1) has at least one solution on IqT.Proof.
Consider the operatorQ:C→C as that defined in (30). We decompose Q into a sum of two operators
(30)Qxt=Q1xt+Q2xt,t∈IqT,
where
(31)Q1xt=∫0t‍t-qsfs,xsdqs+t-ξDβ∫ηT‍∫0s‍s-qτfτ,xτdqτdps+α∫0rη‍f(s,x(s))dqs+α(1-r)η∫rηη‍(η-qs)f(s,x(s))dqs+1Dα+βT2-η21+p-βT-ηt·∫0ξ‍ξ-qsfs,xsdqs,t∈IqT,Q2xt=-1Dα+βT2-η21+p-βT-ηtgx,t∈IqT.
From(A6) there exists a number r0>0 such that
(32)r0Λpψr0>11-k0l.
We will prove that the operators Q1 and Q2 satisfy all the conditions in Lemma 5.
Step 1. The operator Q1 is continuous and completely continuous. We first show that Q1(Ω¯r0) is bounded. For any x∈Ω¯r0 we have
(33)Q1x≤∫0t‍t-qsfs,xsdqs+t-ξDβ∫ηT‍∫0s‍s-qτfτ,xτdqτdps+α∫0rη‍fs,xsdqs+α1-rη∫rηη‍η-qsfs,xsdqs+1Dα+βT2-η21+p+βTT-η·∫0ξ‍ξ-qsfs,xsdqs≤pψr0T21+q+ξ+TD·βT3-η31+q1+p+p2+αrη+αη1-qr1+q+1D·α+βT2-η21+p+βTT-η·ξ21+q.
This proves that Q1(Ω¯r0) is uniformly bounded.
In addition for anyt1,t2∈IqT, t1<t2, we have
(34)Q1xt2-Q1xt1≤∫0t2‍t2-qsfs,xsdqs-∫0t1‍t1-qsfs,xsdqs+t2-t1Dβ∫ηT‍∫0s‍s-qτfτ,xτdqτdps+α∫0rη‍fs,xsdqs+α1-rη∫rηη‍η-qsfs,xsdqs+βT-ηt2-t1D∫0ξ‍ξ-sfs,xsdqs≤pψr0βT-ηt2-t1Dξ21+q∫0t1‍t2-t1dqs+∫t1t2‍t2-qsdqs+t2-t1DβT3-η31+q1+p+p2+αrη+αη1-qr1+qβT3-η31+q1+p+p2+βT-ηt2-t1Dξ21+q,
which is independent of x and tends to zero as t2-t1→0. Thus, Q1 is equicontinuous. Hence, by the Arzelá-Ascoli Theorem, Q1(Ω¯r0) is a relatively compact set. Now, let xn⊂Ω¯r0 with xn-x→0. Then the limit xn(t)-x(t)→0 uniformly valid on IqT. From the uniform continuity of f(t,x) on the compact set IqT×[-r0,r0] it follows that f(t,xn(t))-f(t,x(t))→0 is uniformly valid on IqT. Hence Q1xn-Q1x→0 as n→∞ which proves the continuity of Q1. Hence Step  1 is completely proved.
Step 2. The operator Q2:Ω¯r0→C(IqT,R) is contractive. This is a consequence of (A2). Indeed, we have
(35)Q2xt-Q2yt=1Dα+βT2-η21+p-βT-ηtgx-gy≤k0lx-y,
or
(36)Q2x-Q2y≤L0x-y,L0=k0l<1.
Step 3. The set F(Ω¯r0) is bounded. By (A2) and (A4) we imply that
(37)Q2x≤k0lr0,
for any x∈Ω¯r0. This, with the boundedness of the set Q1(Ω¯r0), implies that the set Q(Ω¯r0) is bounded.
Step 4. Finally, it is to show that the case (C2) in Lemma 5 does not occur. To this end, we suppose that (C2) holds. Then, we have that there exist λ∈(0,1) and x∈∂Ωr0 such that x=λQx. So, we have x=r0 and
(38)xt=λα+βT2-η21+p-βT-ηt∫0t‍t-qsfs,xsdqs+t-ξD·β∫ηT‍∫0s‍s-qτfτ,xτdqτdps+α∫0rη‍fs,xsdqs+α1-rη∫rηη‍η-qsfs,xsdqs+1Dα+βT2-η21+p-βT-ηt·∫0ξ‍ξ-qsfs,xsdqs-1Dα+βT2-η21+p-βT-ηtgx,t∈IqT.
With hypotheses (A4)–(A6), and follows the computations of Step  1, we have
(39)xt≤∫0t‍t-qsfs,xsdqs+t-ξDβ∫ηT‍∫0s‍s-qτfτ,xτdqτdps+α∫0rη‍fs,xsdqs+α1-rη∫rηη‍η-qsfs,xsdqs+1Dα+βT2-η21+p+βTT-η·∫0ξ‍ξ-qsfs,xsdqs+1Dα+βT2-η21+p+βTT-ηgx≤pψr0α+βT2-η21+p+βTT-ηT21+q+ξ+TDβT3-η31+q1+p+p2+αrη+αη1-qr1+qβT3-η31+q1+p+p2+1Dα+βT2-η21+p+βTT-η·ξ21+qα+βT2-η21+p+βTT-η+1Dα+βT2-η21+p+βTT-ηgx,
which implies
(40)r0≤Λpψr0+k0lr0.
Thus,
(41)r0Λpψr0≤11-k0l,
which contradicts to (32). Consequently, we have proved that the operators Q1 and Q2 satisfy all the conditions in Lemma 5. Hence, the operator Q has at least one fixed point x∈Ω¯r0, which is the solution of the boundary value problem (1). The proof is completed.Example 7.
Consider the following nonlocal boundary value problem of nonlinearq-difference equation
(42)D1/22xt=t25x+x+1x+2,t∈0,11/2,x18=133sinx14,-34D2/3x12+45∫1/21‍xsd1/5s=0.
Here, q=1/2, r=2/3, p=1/5, T=1, ξ=1/8, η=1/2, α=-3/4, β=4/5, g(x)=(1/33)sinx, and f(t,x)=(t/25)(x+((x+1)/(x+2))). We find that D=3/10, k0=11/2, and Λ=22009/5952.
As|g(x)-g(y)|≤(1/33)|x-y| with l=(1/33)<(2/11)=k0-1 and g(0)=0; therefore, (A2) and (A4) are satisfied, respectively. Since |f(t,x)|=|(t/25)(|x|+((|x|+1)/(|x|+2)))|≤(t/25)(x2+3|x|+1), we choose p(t)=t/25 and ψ(|x|)=x2+3|x|+1. We can show that
(43)sup⁡r∈(0,∞)rΛpψr≈1.35240>65=11-k0l.
Therefore, by Theorem 6, the boundary value problem (42) has at least one solution on [0,1]1/2.


## 4. Existence Results: The Multivalued Case

Let us recall some basic definitions on multivalued maps [28, 29].For a normed space(X,·), let Pcl⁡(X)={Y∈P(X):Yisclosed}, Pb(X)={Y∈P(X):Yisbounded}, Pcp(X)={Y∈P(X):Yiscompact}, and Pcp,c(X)={Y∈P(X):Yiscompactandconvex}. A multivalued map G:X→P(X) is convex (closed) valued if G(x) is convex (closed) for all x∈X. The map G is bounded on bounded sets if G(B)=∪x∈BG(x) is bounded in X for all B∈Pb(X) (i.e., sup⁡x∈B{sup⁡{|y|:y∈G(x)}}<∞). G is called upper semicontinuous (u.s.c.) on X if for each x0∈X, the set G(x0) is a nonempty closed subset of X, and if, for each open set N of X containing G(x0), there exists an open neighborhood N0 of x0 such that G(N0)⊆N. G is said to be completely continuous if G(B) is relatively compact for every B∈Pb(X). If the multivalued map G is completely continuous with nonempty compact values, then G is u.s.c. if and only if G has a closed graph; that is, xn→x*,yn→y*, yn∈G(xn) imply y*∈G(x*). G has a fixed point if there is x∈X such that x∈G(x). The fixed point set of the multivalued operator G will be denoted by FixG. A multivalued map G:[0;1]→Pcl⁡(R) is said to be measurable if, for every y∈R, the function,
(44)t⟼dy,Gt=inf⁡y-z:z∈Gt,
is measurable.LetL1(IqT,R) denote the space of all functions f defined on IqT such that xL1=∫0T‍x(t)dt<∞.Definition 8.
A functionx∈C is a solution of the problem (2) if x(ξ)=g(x), αDrx(η)+β∫ηT‍x(s)dqs=0, and there exists a function f∈L1(IqT,R) such that it is continuous at t=0, T and f(t)∈F(t,x(t)) on IqT and
(45)xt=∫0t‍t-qsfsdqs+t-ξDβ∫ηT‍∫0s‍s-qτfτdqτdps+α∫0rη‍fsdqs+α1-rη∫rηη‍η-qsfsdqs+1Dα+βT2-η21+p-βT-ηt·∫0ξ‍ξ-qsfsdqs-1Dα+βT2-η21+p-βT-ηtgx,t∈IqT.Definition 9.
A multivalued mapF:IqT×R→P(R) is said to be Carathéodory (in the sense of q-calculus) if x↦F(t,x) is upper semicontinuous on IqT. Further a Carathéodory function F is called L1—Carathéodory if there exists φα∈L1(IqT,R+) such that F(t,x)=sup⁡{v:v∈F(t,x)}≤φα(t) for all x≤α on IqT for each α>0.For eachy∈C, define the set of selections of F by
(46)SF,y∶=v∈C:vt∈Ft,yton⁡IqT.The following lemma will be used in the sequel.Lemma 10 (see [30]).
LetX be a Banach space. Let F:J×R→Pcp,c(X) be an L1—Carathéodory multivalued map and let Θ be a linear continuous mapping from L1(J,X) to C(J,X). Then the operator
(47)Θ∘SF:CJ,X⟶Pcp,cCJ,X,x⟼Θ∘SFx=ΘSF,x,
is a closed graph operator in C(J,X)×C(J,X).To prove our main result in this section we will use the following form of the nonlinear alternative for contractive maps [31, Corollary 3.8].Theorem 11.
LetX be a Banach space and D a bounded neighborhood of 0∈X. Let Z1:X→Pcp,c(X) and Z2:D¯→Pcp,c(X) two multivalued operators satisfying the following:(a)
Z1 is contraction,(b)
Z2 is u.s.c and compact. Then, if G=Z1+Z2, either(i)
G has a fixed point in D¯ or(ii)
there is a pointu∈∂D and λ∈(0,1) with u∈λG(u).Theorem 12.
Assume that(A2) holds. In addition we suppose that(H1)
F:IqT×R→Pcp,c(R) is such that x→F(t,x) is u.s.c. on IqT;(H2)
there exists a continuous nondecreasing functionψ:[0,∞)→(0,∞) and a function p∈L1(IqT,R+) such that
(48)Ft,xP∶=sup⁡y:y∈Ft,x≤ptψxforeacht,x∈IqT×R;(H3)
there exists a numberM>0 such that
(49)1-k0lMΛpψM>1,where Λ and k0 are defined in (23) and (24), respectively.
Then the boundary value problem (2) has at least one solution on Iq.Proof.
Transform the problem (2) into a fixed point problem. Consider the operator N:C(IqT,R)→P(C(IqT,R)) defined by
(50)Nx=1Dα+βT2-η21+p-βT-ηth∈CIqT,R:ht=∫0t‍t-qsfsdqs+t-ξD·β∫ηT‍∫0s‍s-qτfτdqτdps+α∫0rη‍fsdqs+α1-rη∫rηη‍η-qsfsdqs+1Dα+βT2-η21+p-βT-ηt·∫0ξ‍ξ-qsfsdqs-1Dα+βT2-η21+p-βT-ηtgx,
for f∈SF,x.
Now, we define two operators as follows.A:C(IqT,R)→C(IqT,R) by
(51)Axt=-1Dα+βT2-η21+p-βT-ηtgx,
and the multivalued operator B:C(IqT,R)→P(C(IqT,R)) by
(52)Bx=∫0ξ‍ξ-qsfsdqs.h∈CIqT,R:ht=∫0t‍t-qsfsdqs+t-ξD·β∫ηT‍∫0s‍s-qτfτdqτdps+α∫0rη‍fsdqs+α1-rη∫rηη‍η-qsfsdqs+1Dα+βT2-η21+p-βT-ηt·∫0ξ‍ξ-qsfsdqs.
ThenN=A+B. We will show that the operators A and B satisfy all the conditions of Theorem 11 on IqT. For better readability, we break the proof into a sequence of steps and claims.
Step 1. We show that Ais a contraction onC(IqT,R). The proof is similar to the one for the operator Q2 in Step  2 of  Theorem 6.
Step 2. We will show that the operator Bis compact and convex valued and it is completely continuous. This will be given in several claims.
Claim I. Bmaps bounded sets into bounded sets inC(IqT,R). To see this, let Br={x∈C(IqT,R):x≤r} be a bounded set in C(IqT,R). Then, for each h∈B(x), x∈Bρ, there exists f∈SF,x such that
(53)ht=∫0t‍t-qsfsdqs+t-ξD·β∫ηT‍∫0s‍s-qτfτdqτdps+α∫0rη‍fsdqs+α1-rη∫rηη‍η-qsfsdqs+1Dα+βT2-η21+p-βT-ηt·∫0ξ‍ξ-qsfsdqs.
Then fort∈IqT we have
(54)ht≤∫0t‍t-qsfs,xsdqs+t-ξDβ∫ηT‍∫0s‍s-qτfτdqτdps+α∫0rη‍fsdqs+α1-rη∫rηη‍η-qsfsdqs+1Dα+βT2-η21+p+βTT-η·∫0ξ‍ξ-qsfsdqs≤pψx·T21+q+ξ+TDβT3-η31+q1+p+p2+αrη+αη1-qr1+q+1D·α+βT2-η21+p+βTT-ηξ21+q.
Thus,
(55)h≤ψ(ρ)pT21+q+ξ+TD·βT3-η31+q1+p+p2+αrη+αη1-qr1+q+1D·α+βT2-η21+p+βTT-η·ξ21+q.
Claim II. Next we show that Bmaps bounded sets into equicontinuous sets. Let t1,t2∈Iq with t1<t2 and x∈Bρ. For each h∈B(x), we obtain
(56)ht2-ht1≤∫0t2‍t2-qsfsdqs-∫0t1‍t1-qsfsdqs+t2-t1Dβ∫0T‍∫0s‍s-qτfτdqτdps+α∫0rη‍fsdqs+α1-rη∫rηη‍η-qsfsdqs+βT-ηt2-t1D∫0ξ‍ξ-sfsdqs≤pψρβT3-η31+q1+p+p2∫0t1‍t2-t1dqs+∫t1t2‍t2-qsdqs+t2-t1DβT3-η31+q1+p+p2+αrη+αη1-qr1+qβT3-η31+q1+p+p2+βT-ηt2-t1Dξ21+qαrη+αη1-qr1+qβT3-η31+q1+p+p2.
Obviously the right hand side of the above inequality tends to zero independently ofx∈Bρ as t2-t1→0. As B satisfies the above three assumptions; therefore, it follows by the Ascoli-Arzelá theorem that B:C(IqT,R)→P(C(IqT,R)) is completely continuous.
Claim III. Next we prove that Bhas a closed graph. Let xn→x*,hn∈B(xn), and hn→h*. Then we need to show that h*∈B(x*). Associated with hn∈B(xn), there exists fn∈SF,xn such that, for each t∈IqT,
(57)hnt=∫0t‍t-qsfnsdqs+t-ξDβ∫ηT‍∫0s‍s-qτfnτdqτdps+α∫0rη‍fnsdqs+α1-rη∫rηη‍η-qsfnsdqs+1Dα+βT2-η21+p-βT-ηt·∫0ξ‍ξ-qsfnsdqs.
Thus it suffices to show that there existsf*∈SF,x* such that for each t∈IqT,
(58)h*t=∫0t‍t-qsf*sdqs+t-ξD·β∫ηT‍∫0s‍s-qτf*τdqτdps+α∫0rη‍f*sdqs+α1-rη∫rηη‍η-qsf*sdqs+1Dα+βT2-η21+p-βT-ηt·∫0ξ‍ξ-qsf*sdqs.
Let us consider the linear operatorΘ:L1(IqT,R)→C(IqT,R) given by
(59)f⟼Θft=∫0t‍t-qsfsdqs+t-ξD·β∫ηT‍∫0s‍s-qτfτdqτdps+α∫0rη‍fsdqs+α1-rη∫rηη‍η-qsfsdqs+1Dα+βT2-η21+p-βT-ηt·∫0ξ‍ξ-qsfsdqs.
Observe that(60)hnt-h*t=∫0t‍t-qsfns-f*sdqs+t-ξDβ∫ηT‍∫0s‍s-qτfnτ-f*τdqτdps+α∫0rη‍fns-f*sdqs+α1-rη·∫rηη‍η-qsfns-f*sdqs+1Dα+βT2-η21+p-βT-ηt·∫0ξ‍ξ-qsfns-f*sdqs.
Thus, it follows by Lemma10 that Θ∘SF is a closed graph operator. Further, we have hn(t)∈Θ(SF,xn). Since xn→x*, therefore, we have
(61)h*t=∫0t‍t-qsf*sdqs+t-ξD·β∫ηT‍∫0s‍s-qτfτdqτdps+α∫0rη‍f*sdqs+α1-rη∫rηη‍η-qsf*sdqs+1Dα+βT2-η21+p-βT-ηt·∫0ξ‍ξ-qsf*sdqs,
for some f*∈SF,x*. Hence B has a closed graph (and, therefore, has closed values). As a result B is compact valued.
Therefore, the operatorsA and B satisfy all the conditions of Theorem 11 and hence an application of it yields that either condition (i) or condition (ii) holds. We show that the conclusion (ii) is not possible. If x∈λA(x)+λB(x) for λ∈(0,1), then there exists f∈SF,x such that
(62)xt=λ∫0t‍t-qsfsdqs+λt-ξD·β∫ηT‍∫0s‍s-qτfτdqτdps+α∫0rη‍fsdqs+α1-rη∫rηη‍η-qsfsdqs+λ1Dα+βT2-η21+p-βT-ηt·∫0ξ‍ξ-qsfsdqs-λ1Dα+βT2-η21+p-βT-ηtgx,t∈IqT.
Consequently, we have
(63)x≤ψ(x)p·α+βT2-η21+p+βTT-ηT21+q+ξ+TDβT3-η31+q1+p+p2+αrη+αη1-qr1+qβT3-η31+q1+p+p2+1Dα+βT2-η21+p+βTT-η·ξ21+qα+βT2-η21+p+βTT-η+k0lx
or
(64)x≤Λpψx+k0lx.
If condition (ii) of Theorem 11 holds, then there exists λ∈(0,1) and x∈∂BM with x=λN(x). Then, x is a solution of (30) with x=M. Now, the previous inequality implies
(65)1-k0lMΛpψM≤1,
which contradicts to (49). Hence, N has a fixed point in IqT by Theorem 11, and consequently the boundary value problem (2) has a solution. This completes the proof.Example 13.
Consider the following nonlocal boundary value problem of nonlinearq-difference inclusion
(66)D1/22xt∈Ft,x,t∈0,121/2,x18=x1/1641+x1/16,12D1/3x14+35∫1/41/2‍xsd1/4s=0,
where F:[0,1/2]×R→P(R) is a multivalued map given by
(67)x⟶Ft,x=tx1+cos⁡2x1+x,t+1x+1e-x21+sin2x.
Here, q=1/2, r=1/3, p=1/4, T=1/2, ξ=1/8, η=1/4, α=1/2, β=3/5, and g(x)=(1/4)(x/(1+x)). We find that D=-457/800, k0=532/457, and Λ=11083/32904.Asg(x)-g(y)≤(1/4)x-y, therefore, (A2) is satisfied with l=(1/4)<(457/532)=k0-1. For f∈F, we have
(68)f≤max⁡tx1+cos⁡2x1+x,t+1x+1e-x21+sin2x≤t+1x+1,x∈R.
Thus
(69)Ft,xP∶=sup⁡y:y∈Ft,x≤ptψx,x∈R,
with p(t)=t+1 and ψ(x)=x+1. From the given data, it is found that M>2.47997. Clearly, all the conditions of Theorem 12 are satisfied. Hence, the nonlocal boundary value problem (66) has at least one solution on [0,1/2]1/2.



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*Source: 203715-2015-01-27.xml*