# A Study of Nonlinear Fractionalq-Difference Equations with Nonlocal Integral Boundary Conditions

Authors: Ahmed Alsaedi; Bashir Ahmad; Hana Al-Hutami
Journal: Abstract and Applied Analysis
 (2013)

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## Abstract

This paper is concerned with the existence and uniqueness of solutions for a boundary value problem of nonlinear fractionalq-difference equations with nonlocal integral boundary conditions. The existence results are obtained by applying some well-known fixed point theorems and illustrated with examples.


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## Body

## 1. Introduction

Several kinds of boundary value problems of fractional-order have recently been investigated by many researchers. Fractional derivatives appear naturally in the mathematical modelling of dynamical systems involving fractals and chaos. In fact, the concept of fractional calculus has played a key role in improving the work based on integer-order (classical) calculus in several diverse disciplines of science and engineering. This might have been due to the fact that fractional-differential operators help to understand the hereditary phenomena in many materials and processes in a better way than the corresponding integer-order differential operators. Examples include physics, chemistry, biology, biophysics, blood flow phenomena, control theory, signal and image processing, and economics [1–4]. For some recent results on the topic, see a series of papers [5–12] and the references therein.Fractionalq-difference equations, regarded as fractional analogue of q-difference equations, have been studied by several researchers. For some earlier work on the subject, we refer to [13, 14], whereas the recent development on the existence theory of fractional q-difference equations can be found in [15–25]. In a recent paper [26], the authors investigated a nonlocal boundary value problem of nonlinear fractional q-difference equations:
(1)Dqαcx(t)=f(t,x(t)),0≤t≤1,1<α≤2,α1x(0)-β1Dqx(0)=γ1x(η1),α2x(1)+β2Dqx(1)=γ2x(η2),
where f∈C([0,1]×ℝ,ℝ),cDqα is the fractional q-derivative of the Caputo type, Dq is q-derivative, and αi,βi,γi∈ℝ,ηi∈(0,1),i=1,2.The purpose of the present paper is to study the following nonlocal boundary value problem of nonlinear fractionalq-difference equations:
(2)cDqβx(t)+λx(t)=f(t,x(t)),t∈[0,1],1<β≤2,0<q<1,x(0)=0,x(1)=Iqx(η)=∫0ηx(s)dqs,0<η<1,
where f∈C([0,1]×ℝ,ℝ),cDqβ is the fractional q-derivative of the Caputo type, and λ is a real number.The paper is organized as follows. Section2 contains some necessary background material on the topic, while the main results are presented in Section 3. We make use of Banach’s contraction principle, Krasnoselskii’s fixed point theorem, and Leray-Schauder nonlinear alternative to establish the existence results for the problem at hand.


## 2. Preliminaries on Fractionalq-Calculus

Here we recall some definitions and fundamental results on fractionalq-calculus.Definition 1 (see [13]).
Letβ≥0, let 0<q<1, and let f be a function defined on [0,1]. The fractional q-integral of the Riemann-Liouville type is (Iq0f)(t)=f(t) and
(3)(Iqβf)(t)=∫0t(t-qs)(β-1)Γq(β)f(s)dq(s),β>0,t∈[0,1],
where
(4)Γq(β)=(1-q)(β-1)(1-q)β-1,0<q<1.Recall thatΓq(β+1)=⌈β⌉qΓq(β), with
(5)⌈β⌉q=qβ-1q-1,(1-q)(0)=1,(1-q)(n)=∏k=0n-1‍(1-qk+1),n∈ℕ.More generally,  ifβ∈ℝ, then
(6)(1-q)(β)=∏i=0∞‍(1-qi+1)(1-qβ+i+1).Lemma 2 (see [27]).
Forβ∈ℝ+,   λ∈(-1,∞), the following is valid:
(7)Iqβ((t-a)(λ))=Γq(λ+1)Γq(β+λ+1)(t-a)(β+λ),0<a<t<b.In particular, forλ=0,a=0, using q-integration by parts, we have
(8)(Iqβ1)(t)=1Γq(β)∫0t(t-qs)(β-1)dqs=1Γq(β)∫0tDq((t-s)(β))-[β]qdqs=-1Γq(β+1)∫0tDq((t-s)(β))dqs=1Γq(β+1)t(β).For0<q<1,  we define the q-derivative of a real valued function f as
(9)Dqf(t)=f(t)-f(qt)(1-q)t,t≠0,Dqf(0)=limn→∞f(sqn)-f(0)sqn,s≠0.
For more details, see [28].Definition 3 (see [29]).
The fractionalq-derivative of the Riemann-Liouville type of order β≥0 is defined by (Dq0f)(t)=f(t) and
(10)(Dqβf)(t)=(Dq[β]Iq[β]-βf)(t),β>0,
where [β] is the smallest integer greater than or equal to β.Definition 4 (see [29]).
The fractionalq-derivative of the Caputo type of order β≥0 is defined by
(11)(Dcqβf)(t)=(Iq[β]-βDq[β]f)(t),β>0,
where [β] is the smallest integer greater than or equal to β.Now we state some known results involvingq-derivatives and q-integrals.Lemma 5 (see [29]).
Letβ,γ≥0, and let f be a function defined on [0,1]. Then(i)
(
I
q
β
I
q
γ
f
)
(
t
)
=
(
I
q
β
+
γ
f
)
(
t
),(ii)
(
D
q
β
I
q
β
f
)
(
t
)
=
f
(
t
),(iii)
(
I
q
β
D
c
q
β
f
)
(
t
)
=
f
(
t
)
-
∑
k
=
0
[
β
]
-
1
(
t
k
/
Γ
q
(
k
+
1
)
)
(
c
D
q
k
f
)
(
0
), β>0.Lemma 6 (see [19]).
Letβ≥0 and n∈ℕ. Then, the following equality holds:
(12)(IqβDqnf)(t)=DqnIqβf(t)-∑k=0[β]-1tβ-n+kΓq(β-n+k)(Dqkf)(0).In the forthcoming analysis, the following lemma plays a pivotal role.Lemma 7.
Leth∈C([0,1],ℝ) be a given function. Then the unique solution of the boundary value problem,
(13)Dcqβx(t)+λx(t)=h(t),t∈[0,1],1<β≤2,0<q<1,x(0)=0,x(1)=∫0ηx(s)dqs,0<η<1,
is given by
(14)x(t)=∫0t(t-qs)(β-1)Γq(β)(h(s)-λx(s))dqs+δ1t∫0η(η-qs)(β)Γq(β+1)(h(s)-λx(s))dqs-δ1t∫01(1-qs)(β-1)Γq(β)(h(s)-λx(s))dqs,
where
(15)δ1=1+q1+q-η2.Proof.
Using Lemma5, we can write the solution of fractional q-difference equation in (13) as
(16)x(t)=∫0t(t-qs)(β-1)Γq(β)(h(s)-λx(s))dqs+c0t+c1,t∈[0,1].
Using the boundary conditions of (13) in (16), we get c1=0 and
(17)c0=δ1[∫0η(η-qs)(β)Γq(β+1)(h(s)-λx(s))dqs--∫01(1-qs)(β-1)Γq(β)(h(s)-λx(s))dqs],
where δ1 is given by (15). Substituting the values of c0,c1 in (16), we obtain (14).Let𝒞:=C([0,1],ℝ) denote the Banach space of all continuous functions from [0,1] to ℝ endowed with the norm defined by ∥x∥=sup{|x(t)|:t∈[0,1]}.By virtue of Lemma7, we define an operator 𝒢:𝒞→𝒞 as
(18)(𝒢x)(t)=∫0t(t-qs)(β-1)Γq(β)(f(s,x(s))-λx(s))dqs+δ1t∫0η(η-qs)(β)Γq(β+1)(f(s,x(s))-λx(s))dqs-δ1t∫01(1-qs)(β-1)Γq(β)(f(s,x(s))-λx(s))dqs
and note that the given problem (2) has solutions only if the operator equation 𝒢x=x has fixed points.


## 3. Main Results

In the sequel, we assume that(A1)
f:[0,1]×ℝ→ℝ is a continuous function and that there exists a q-integrable function ζ:[0,1]→ℝ such that |f(t,x)-f(t,y)|≤ζ(t)|x-y|,t∈[0,1],x,y∈ℝ.For computational convenience, we introduce the notations:(19)Λ=μ1+|λ|μ2,
where
(20)μ1:=(1+|δ1|)(Iqβζ)(1)+|δ1|(Iqβ+1ζ)(η),μ2:=(1Γq(β+1)(1+|δ1|)+|δ1|ηβ+1Γq(β+2)).Our first existence result is based on Leray-Schauder nonlinear alternative.Lemma 8 (nonlinear alternative for single valued maps [30]).
LetE be a Banach space, C a closed, convex subset of E,  and  V an open subset of C with 0∈V. Suppose that 𝒢:V¯→C is a continuous, compact (i.e., 𝒢(V¯) is a relatively compact subset of C) map. Then either(i)
𝒢
has a fixed point in
V
¯,
or(ii) there is
x
∈
∂
V
(the boundary of
V
in
C
) and
ε
∈
(
0,1
)
with
x
=
ε
𝒢
(
x
).Theorem 9.
Suppose thatf:[0,1]×ℝ→ℝ is a continuous function. In addition, it is assumed that (A2)
there exist functionsh1,h2∈L1([0,1],ℝ+)  and a nondecreasing function ϑ:ℝ+→ℝ+ such that |f(t,x)|≤h1(t)ϑ(|x|)+h2(t),  for (t,x)∈[0,1]×ℝ.
Then the problem (2) has at least one solution on [0,1] if there exists a positive number ω such that
(21)ω>ϑ(ω)ν1+ν21-|λ|μ2,
where
(22)|λ|≠1μ2,νi:=(1+|δ1|)(Iqβhi)(1)+|δ1|(Iqβ+1hi)(η),i=1,2.Proof.
As a first step, it will be shown that the operator𝒢:𝒞→𝒞 (defined by (18)) maps bounded sets into bounded sets in C([0,1],ℝ). Notice that 𝒢 is continuous. For a positive number σ, let B-σ={x∈C([0,1],ℝ):∥x∥≤σ} be a bounded set in C([0,1],ℝ). Then, for any x∈B-σ, we find that
(23)|(𝒢x)(t)|≤∫0t(t-qs)(β-1)Γq(β)(|f(s,x(s))|+|λ||x(s)|)dqs+|δ1||t|∫0η(η-qs)(β)Γq(β+1)(|f(s,x(s))|+|λ||x(s)|)dqs+|δ1||t|∫01(1-qs)(β-1)Γq(β)(|f(s,x(s))|+|λ||x(s)|)dqs≤∫0t(t-qs)(β-1)Γq(β)[h1(s)ϑ(∥x∥)+h2(s)]dqs+|δ1||t|∫0η(η-qs)(β)Γq(β+1)[h1(s)ϑ(∥x∥)+h2(s)]dqs+|δ1||t|∫01(1-qs)(β-1)Γq(β)[h1(s)ϑ(∥x∥)+h2(s)]dqs≤ϑ(σ){∫01(1-qs)(β-1)Γq(β)h1(s)dqs+|δ1|∫0η(η-qs)(β)Γq(β+1)h1(s)dqs+|δ1|∫01(1-qs)(β-1)Γq(β)h1(s)dqs}+{∫01(1-qs)(β-1)Γq(β)h2(s)dqs++|δ1|∫0η(η-qs)(β)Γq(β+1)h2(s)dqs+|δ1|∫01(1-qs)(β-1)Γq(β)h2(s)dqs}+σ|λ|{∫01(1-qs)(β-1)Γq(β)dqs+|δ1|∫0η(η-qs)(β)Γq(β+1)dqs+|δ1|+|δ1|∫01(1-qs)(β-1)Γq(β)dqs}≤ϑ(σ)ν1+ν2+σ|λ|μ2.
This establishes our assertion.
Next, we show that the operator𝒢 maps bounded sets into equicontinuous sets of C([0,1],ℝ). Taking t1,t2∈[0,1] with t1<t2 and x∈B-σ together with the inequality: (t2-qs)(β-1)-(t1-qs)(β-1)≤(t2-t1) for 0<t1<t2 (see [19] p. 4), we obtain
(24)|(𝒢x)(t2)-(𝒢x)(t1)|≤|∫0t2‍(t2-qs)(β-1)Γq(β)(f(s,x(s))-λx(s))dqs-∫0t1‍(t1-qs)(β-1)Γq(β)(f(s,x(s))-λx(s))dqs|+|δ1|(t2-t1)∫0η(η-qs)(β)Γq(β+1)+|δ1|(t2-t1)××(|f(s,x(s))|+|λ||x(s)|)dqs.+|δ1|(t2-t1)∫01(1-qs)(β-1)Γq(β)+|δ1|(t2-t1)××(|f(s,x(s))|+|λ||x(s)|)dqs.≤|∫0t1(t2-t1)Γq(β)[h1(s)ϑ(σ)+h2(s)+|λ|σ]dqs+∫t1t2(t2-qs)(β-1)Γq(β)[h1(s)ϑ(σ)+h2(s)+|λ|σ]dqs|+|δ1|(t2-t1)∫0η(η-qs)(β)Γq(β+1)+|δ1|(t2-t1)××[h1(s)ϑ(σ)+h2(s)+|λ|σ]dqs.+|δ1|(t2-t1)∫01(1-qs)(β-1)Γq(β)+|δ1|(t2-t1)××[h1(s)ϑ(σ)+h2(s)+|λ|σ]dqs.
It is obvious that the right hand side of the above inequality tends to zero independently of x∈B-σ as t2→t1. Therefore the operator 𝒢 is completely continuous by the Arzelá-Ascoli theorem.
Thus the operator𝒢 satisfies the hypothesis of Lemma 8 and hence by its conclusion, either condition (i) or condition (ii) holds. We claim that the conclusion (ii) is not possible.
LetV={x∈C([0,1],ℝ):∥x∥<ω} with ω given by (21). Then we will show that ∥𝒢x∥<ω. Indeed, by means of (A4), we get
(25)∥𝒢x∥≤ϑ(∥x∥){∫01(1-qs)(β-1)Γq(β)h1(s)dqs≤ϑ(∥x∥)+|δ1|+|δ1|∫0η(η-qs)(β)Γq(β+1)h1(s)dqs≤ϑ(∥x∥)+|δ1|+|δ1|∫01(1-qs)(β-1)Γq(β)h1(s)dqs}+{∫01(1-qs)(β-1)Γq(β)h2(s)dqs++|δ1|∫0η(η-qs)(β)Γq(β+1)h2(s)dqs++|δ1|∫01(1-qs)(β-1)Γq(β)h2(s)dqs}+ω|λ|{∫01(1-qs)(β-1)Γq(β)dqs+|δ1|∫0η(η-qs)(β)Γq(β+1)dqs+M|λ|+++|δ1|∫01(1-qs)(β-1)Γq(β)dqs}≤ϑ(ω)ν1+ν2+ω|λ|μ2<ω.
Assume that there exist  x∈∂V and   ε∈(0,1) such that x=ε𝒢x. Then for such a choice of x and ε, we get a contradiction:
(26)ω=∥x∥=ε∥𝒢x∥<ϑ(∥x∥)ν1+ν2+ω|λ|μ2=ϑ(ω)ν1+ν2+ω|λ|μ2<ω.
Thus it follows by Lemma 8 that 𝒢 has a fixed point x∈V¯ which is a solution of the problem (2). This completes the proof.Remark 10.
In case we takeh1,h2 in (A2) to be continuous, then νi=μ2∥hi∥,i=1,2, where μ2 is given by (20).Our next result deals with existence and uniqueness of solutions for the problem (2) and is based on Banach’s fixed point theorem.Theorem 11.
Suppose that the assumption(A1) holds and that Λ<1, where Λ is given by (19). Then the boundary value problem (2) has a unique solution.Proof.
FixN=supt∈[0,1]|f(t,0)|<∞,  r≥(Nμ2/1-Λ) (μ2 is given by (20)), and define B-r={x∈𝒞:∥x∥≤r}. We will show that 𝒢B-r⊂B-r, where 𝒢 is defined by (18). For x∈B-r,t∈[0,1], it follows by the assumption (A1) that
(27)|f(t,x(t))|≤|f(t,x(t))-f(t,0)|+|f(t,0)|≤ζ(t)|x(t)|+|f(t,0)|≤ζ(t)r+N.
Then, for x∈B-r,t∈[0,1], we have
(28)|(𝒢x)(t)|≤∫0t‍(t-qs)(β-1)Γq(β)[ζ(s)r+N]dqs+r|λ|∫0t‍(t-qs)(β-1)Γq(β)dqs+|t||δ1|×∫0η‍(η-qs)(β)Γq(β+1)[ζ(s)r+N]dqs+r|λ|∫0η‍(η-qs)(β)Γq(β+1)dqs+|t||δ1|×∫01‍(1-qs)(β-1)Γq(β)[ζ(s)r+N]dqs+r|λ|∫01‍(1-qs)(β-1)Γq(β)dqs≤r{∫01‍(1-qs)(β-1)Γq(β)ζ(s)dqs++|δ1|∫0η‍(η-qs)(β)Γq(β+1)ζ(s)dqs++|δ1|∫01‍(1-qs)(β-1)Γq(β)ζ(s)dqs}+N{∫01‍(1-qs)(β-1)Γq(β)dqs+|δ1|∫0η‍(η-qs)(β)Γq(β+1)dqs|δ1|++|δ1|∫01‍(1-qs)(β-1)Γq(β)dqs}+r|λ|{∫01‍(1-qs)(β-1)Γq(β)dqs+|δ1|∫0η‍(η-qs)(β)Γq(β+1)dqs|δ1|++|δ1|∫01‍(1-qs)(β-1)Γq(β)dqs}≤r{(Iqβζ)(1)+|δ1|(Iqβ+1ζ)(η)+|δ1|(Iqβζ)(1)}+N{1Γq(β+1)+|δ1|ηβ+1Γq(β+2)+|δ1|Γq(β+1)}+r|λ|{1Γq(β+1)+|δ1|ηβ+1Γq(β+2)+|δ1|Γq(β+1)},
which, in view of (19), implies that
(29)∥𝒢x∥≤Nμ2+rΛ≤r.
This shows that 𝒢B-r⊂B-r.
Now, forx,y∈𝒞, we have
(30)∥(𝒢x)-(𝒢y)∥≤supt∈[0,1]{∫0t‍(t-qs)(β-1)Γq(β)××(|f(s,x(s))-f(s,y(s))|×+|λ||x(s)-y(s)|)dqs≤supt∈[0,1]++|δ1||t|∫0η(η-qs)(β)Γq(β+1)(|f(s,x(s))-f(s,y(s))|+|δ1||t|(s)-y(s)+++|λ||x(s)-y(s)|)dqs+|δ1||t|∫01‍(1-qs)(β-1)Γq(β)≤supt∈[0,1]×××(|f(s,x(s))-f(s,y(s))|≤supt∈[0,1]+|λ||x(s)-y(s)|)dqs∫0t‍(t-qs)(β-1)Γq(β)}≤∥x-y∥supt∈[0,1]{∫0t‍(t-qs)(β-1)Γq(β)ζ(s)dqs≤∥x-y∥supt∈[0,1]++|δ1||t|∫0η‍(η-qs)(β)Γq(β+1)ζ(s)dqs≤∥x-y∥supt∈[0,1]++|δ1||t|∫01‍(1-qs)(β-1)Γq(β)ζ(s)dqs≤∥x-y∥supt∈[0,1]++|λ|(∫0t‍(t-qs)(β-1)Γq(β)dqs≤∥x-y∥supt∈[0,1]++|λ|+|δ1||t|∫0η‍(η-qs)(β)Γq(β+1)dqs≤∥x-y∥supt∈[0,1]++|λ|+|δ1||t|∫01‍(1-qs)(β-1)Γq(β)dqs)},
which, by (19), takes the form
(31)∥(𝒢x)(t)-(𝒢y)(t)∥≤Λ∥x-y∥.
As Λ<1 (the given assumption), therefore 𝒢 is a contraction. Hence, by Banach’s contraction mapping principle, the problem (2) has a unique solution.If we takeζ(t)=L (L is a positive constant), the condition Λ<1 becomes L<((1/μ2)-|λ|) and Theorem 11 can be phrased as follows.Corollary 12.
Let there exists a constantL∈(0,(1/μ2)-|λ|) with μ2 given by (20), and let f:[0,1]×ℝ→ℝ be a continuous function satisfying the assumption: |f(t,x)-f(t,y)|≤L|x-y|, t∈[0,1], x,y∈ℝ. Then the boundary value problem (2) has a unique solution.Our last result relies on Krasnoselskii’s fixed point theorem [31].Lemma 13 (Krasnoselskii).
LetY be a closed, bounded, convex, and nonempty subset of a Banach space X. Let Q1,Q2 be two operators such that (i) Q1x+Q2y∈Y whenever x,y∈Y; (ii) Q1 is compact and continuous; (iii) Q2 is a contraction mapping. Then there exists z∈Y such that z=Q1z+Q2z.Theorem 14.
Letf:[0,1]×ℝ→ℝ be a continuous function satisfying (A1). Furthermore, it is assumed that  (A3)
there exist a functionξ∈C([0,1],ℝ+) and a nondecreasing function χ∈C([0,1],ℝ+) such that
(32)|f(t,x)|≤ξ(t)χ(|x|),(t,x)∈[0,1]×ℝ.
If(33)|δ1|[∫0η‍(η-qs)(β)Γq(β+1)ζ(s)dqs+∫01‍(1-qs)(β-1)Γq(β)ζ(s)dqs]+|λ||δ1|[ηβ+1Γq(β+2)+1Γq(β+1)]<1,
with |λ|μ2<1, then the boundary value problem (2) has at least one solution on [0,1].Proof.
Let us consider the setB-r¯={x∈𝒞:∥x∥≤r¯}, where r¯ is given by
(34)r¯≥χ(r¯)∥ξ∥1-|λ|μ2,∥ξ∥=supt∈[0,t]|ξ(t)|,
and introduce the operators 𝒢1 and 𝒢2 on B-r¯ as
(35)(𝒢1x)(t)=∫0t‍(t-qs)(β-1)Γq(β)(f(s,x(s))-λx(s))dqs,λx(s)dqs=∫0t‍(t-qs)(β-1)Γq(β)t∈[0,1],(𝒢2x)(t)=δ1t∫0η‍(η-qs)(β)Γq(β+1)(f(s,x(s))-λx(s))dqs-δ1t∫01‍(1-qs)(β-1)Γq(β)(f(s,x(s))-λx(s))dqs,λx(s)dqs-δ1t-λx∫01‍(1-qs)(β-1)Γq(β)11t∈[0,1].
In order to show the hypothesis of Krasnoselskii’s fixed point theorem, we proceed as follows. (i)
Forx,y∈B-r¯, we find that
(36)|(𝒢1x+𝒢2y)(t)|≤∫0t‍(t-qs)(β-1)Γq(β)(ξ(s)χ(|x(s)|)+|λ||x(s)|)dqs+|δ1|∫0η‍(η-qs)(β)Γq(β+1)(ξ(s)χ(|y(s)|)+|λ||y(s)|)dqs+|δ1|∫01‍(1-qs)(β-1)Γq(β)(ξ(s)χ(|y(s)|)+|λ||y(s)|)dqs≤χ(r¯)∥ξ∥[1Γq(β+1)(1+|δ1|)+|δ1|ηβ+1Γq(β+2)]+r¯|λ|[1Γq(β+1)(1+|δ1|)+|δ1|ηβ+1Γq(β+2)]≤χ(r¯)∥ξ∥μ2+r¯|λ|μ2≤r¯.
This implies that 𝒢1x+𝒢2y∈B-r¯.(ii)
From the continuity off, it follows that the operator 𝒢1 is continuous. Also, 𝒢1 is uniformly bounded on B-r¯ as
(37)∥𝒢1x∥≤χ(r¯)∥ξ∥Γq(β+1)+|λ|r¯Γq(β+1).Next, for anyx∈B-r¯, and τ1,τ2∈[0,1] with τ1<τ2, we have
(38)|(𝒢1x)(τ2)-(𝒢1x)(τ1)|=|∫0τ2‍(τ2-qs)(β-1)Γq(β)(f(s,x(s))-λx(s))dqs-∫0τ1‍(τ1-qs)(β-1)Γq(β)(f(s,x(s))-λx(s))dqs|≤|∫0τ1‍(τ2-qs)(β-1)-(τ1-qs)(β-1)Γq(β)××(f(s,x(s))-λx(s))dqs∫0τ1‍(τ2-qs)(β-1)-(τ1-qs)(β-1)Γq(β)|+|∫τ1τ2‍(τ2-qs)(β-1)Γq(β)(f(s,x(s))-λx(s))dqs|≤(χ(r¯)∥ξ∥+|λ|r¯)×[∫0τ1‍(τ2-qs)(β-1)-(τ1-qs)(β-1)Γq(β)dqs+∫τ1τ2‍(τ2-qs)(β-1)Γq(β)dqs]which is independent ofx and tends to zero as τ2→τ1. Thus, 𝒢1 is equicontinuous. So 𝒢1 is relatively compact on B-r¯. Hence, by the Arzelá-Ascoli theorem, 𝒢1 is compact on B-r¯.(iii)
From(A1) and (33) it follows that 𝒢2 is a contraction mapping.
Thus all the conditions of Lemma13 are satisfied. Hence, by the conclusion of Lemma 13, the problem (2) has at least one solution on [0,1].As a special case, forχ(x)≡1, there always exists a positive r¯ such that (34) holds true. In consequence, we have the following corollary.Corollary 15.
Letf:[0,1]×ℝ→ℝ be a continuous function satisfying (A1). Furthermore,  |f(t,x)|≤ξ(t), ∀(t,x)∈[0,1]×ℝ, and ξ∈C([0,1],ℝ+). Then the boundary value problem (2) has at least one solution on [0,1] provided (33) holds.


## 4. Examples

Example 1.
Consider the problem(39)cDq3/2x(t)+14x(t)=15x+cosx,0≤t≤1,x(0)=0,x(1)=Iqx(13),
where q=1/2,λ=1/4,η=1/3. It is easy to find that δ1=27/25, and
(40)|f(t,x)|=|15x+cosx|≤15|x|+1.
Obviously h1=1/5,h2=1,ϑ(ω)=ω. In consequence, ν1=0.3002935758,ν2=1.501467879, and condition (21) yields ω>4.629306104. Thus, the hypothesis of Theorem 9 is satisfied. Hence it follows by the conclusion of Theorem 9 that there exists at least one solution for the problem (39).Example 2.
Consider the nonlocal boundary value problem given by(41)cDq3/2x(t)+14x(t)=L(cost+tan-1x),0≤t≤1,x(0)=0,x(1)=Iqx(13).
Here q=1/2,λ=1/4,η=1/3 and L is a constant to be fixed later on. With the given data, it is found that δ1=27/25,   |f(t,x)-f(t,y)|≤L|x-y| and
(42)Λ=1Γ1/2(5/2)(1+273(42-1)+62253(42-1))(L+14).
Letting
(43)L<[1Γ1/2(5/2)(1+273(42-1)+62253(42-1))]-1-14,
all the conditions of Corollary 12 are satisfied. Therefore, the conclusion of Corollary 12 applies to the problem (41).



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*Source: 410505-2013-10-24.xml*