# New Existence Results for Fractional Integrodifferential Equations with Nonlocal Integral Boundary Conditions

Authors: Ahmed Alsaedi; Sotiris K. Ntouyas; Bashir Ahmad
Journal: Abstract and Applied Analysis
 (2015)

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## Abstract

We consider a boundary value problem of fractional integrodifferential equations with new nonlocal integral boundary conditions of the form:x(0)=βx(θ),x(ξ)=α∫η1‍x(s)ds, and 0<θ<ξ<η<1. According to these conditions, the value of the unknown function at the left end point t=0 is proportional to its value at a nonlocal point θ while the value at an arbitrary (local) point ξ is proportional to the contribution due to a substrip of arbitrary length (1-η). These conditions appear in the mathematical modelling of physical problems when different parts (nonlocal points and substrips of arbitrary length) of the domain are involved in the input data for the process under consideration. We discuss the existence of solutions for the given problem by means of the Sadovski fixed point theorem for condensing maps and a fixed point theorem due to O’Regan. Some illustrative examples are also presented.


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## Body

## 1. Introduction

We consider a boundary value problem of fractional differential equations with nonlocal integral boundary conditions given by(1)cDqx(t)=Af(t,x(t))+BIrg(t,x(t)),t∈[0,1],x0=βxθ,xξ=α∫η1‍xsds,hhhhhhhhhhhhhhhh0<θ<ξ<η<1,where cDq denotes the Caputo fractional derivative of order q, f:[0,1]×R→R is a given continuous function, 1<q≤2, 0<r<1, and α, β, A, B are real constants.Here we remark that the boundary conditions introduced in the problem (1) are of nonlocal strip type and describe the situation when the receptors at the end points of the boundary are influenced by the nonlocal contributions due to interior points and strips of the domain for the problem. For practical examples, see [1, 2]. The problem (1) can also be termed as a five-point nonlocal fractional boundary value problem.In recent years, several aspects of fractional boundary value problems, ranging from theoretical analysis to numerical simulation, have been investigated. The nonlocal nature of fractional order differential operators has significantly contributed to the popularity and development of the subject. As a matter of fact, this characteristic of such operators help to understand the memory and hereditary properties of many useful materials and processes. For details and applications of fractional differential equations in physical and technical sciences such as biology, physics, biophysics, chemistry, statistics, economics, blood flow phenomena, control theory, and signal and image processing, see [3–6]. For some recent works on nonlocal fractional boundary value problems, we refer the reader to the papers [7–11], while the results based on monotone method for such problems can be found in [12, 13]. In [14], the limit properties of positive solutions of fractional boundary value problems have been discussed. Fractional differential inclusions supplemented with different kinds of boundary conditions have also been studied by several researchers, for instance, see [15–20].The paper is organized as follows. In Section2, we recall some basic definitions from fractional calculus and establish an auxiliary lemma which plays a pivotal role in the sequel. Section 3.1 contains an existence result for the problem (1) which is established by applying Sadovskii’s fixed point theorem for condensing maps. In Section 3.2, we show the existence of solutions for the problem (1) by means of a fixed point theorem due to O’Regan.


## 2. Preliminaries

In this section, some basic definitions on fractional calculus and an auxiliary lemma are presented [3, 4].Definition 1.
The Riemann-Liouville fractional integral of orderq for a continuous function g is defined as(2)Iqg(t)=1Γ(q)∫0t‍g(s)t-s1-qds,q>0,provided the integral exists.Definition 2.
For at leastn-times continuously differentiable function g:[0,∞)→R, the Caputo derivative of fractional order q is defined as(3)cDqg(t)=1Γ(n-q)∫0t‍t-sn-q-1g(n)(s)ds,hhhhhhhhhhhhhn-1<q<n,n=[q]+1,where [q] denotes the integer part of the real number q.Lemma 3.
For anyy∈C([0,1],R) the unique solution of the linear fractional boundary value problem(4)cDqx(t)=y(t),t∈[0,1],1<q≤2,x(0)=βx(θ),x(ξ)=α∫η1‍x(s)ds,hhhhhhhhhhhhhiihhh0<θ<ξ<η<1,is(5)xt=∫0t‍t-sq-1Γ(q)y(s)ds+βξ-α/21-η2Q∫0θ‍θ-sq-1Γ(q)y(s)ds+αβθQ∫01‍1-sqΓ(q+1)y(s)ds-∫0η‍η-sqΓq+1y(s)ds-βθQ∫0ξ‍ξ-sqΓ(q)y(s)ds+tQα(1-β)∫01‍1-sqΓ(q+1)y(s)dshhhhhhhhhhhhhhi-∫0η‍η-sqΓ(q+1)y(s)dshhhhhi-(1-β)∫0ξ‍ξ-sqΓ(q)y(s)dshhhhhi-β1-α1-η∫0θ‍θ-sq-1Γqysds,where(6)Q=(1-β)(ξ-α2(1-η2))+βθ(1-α(1-η))≠0.Proof.
It is well known that the general solution of the fractional differential equation in (4) can be written as(7)x(t)=c0+c1t+∫0t‍t-sq-1Γ(q)y(s)ds,where c0,c1∈R are arbitrary constants.
Applying the given boundary conditions, we obtain the following system(8)(1-β)c0-βθc1=β∫0θ‍θ-sq-1Γ(q)y(s)ds(1-α(1-η))c0+(ξ-α2(1-η2))c1=α∫01‍1-sqΓq+1ysds-∫0η‍η-sqΓq+1ysds-∫0ξ‍ξ-sqΓ(q)y(s)ds,from which we get(9)c0=β(ξ-(α/2)(1-η2))Q∫0θ‍θ-sq-1Γ(q)y(s)ds+αβθQ∫01‍1-sqΓq+1ysds-∫0η‍η-sqΓq+1ysds-βθQ∫0ξ‍ξ-sqΓqysds,c1=1Qα(1-β)∫01‍1-sqΓq+1y(s)dshhhhhhhhhhhhi-∫0η‍η-sqΓ(q+1)y(s)dshhhi-(1-β)∫0ξ‍ξ-sqΓ(q)y(s)dshhhi-β(1-α(1-η))∫0θ‍θ-sq-1Γ(q)y(s)ds.Substituting the values of c0,c1 in (7), we get (5). This completes the proof.


## 3. Existence Results

We denote byC=C([0,1],R) the Banach space of all continuous functions from [0,1]→R endowed with the norm defined by x=sup⁡{|x(t)|:t∈[0,1]}. Also by L1([0,1],R) we denote the Banach space of measurable functions x:[0,1]→R which are Lebesgue integrable and normed by xL1=∫01‍|x(t)|dt.In the following we will give two existence results for the problem (1), one with the help of Sadovskii’s fixed point theorem and the other based on a fixed point theorem due to O’Regan in [21].


### 3.1. Existence Results via Sadovskii’s Fixed Point Theorem

Definition 4.
LetM be a bounded set in metric space (X,d); then Kuratowskii measure of noncompactness, α(M), is defined as inf⁡{ϵ:M covered by finitely many sets such that the diameter of each set ≤ϵ}.Definition 5 (see [22]).
LetΦ:D(Φ)⊆X→X be a bounded and continuous operator on Banach space X. Then Φ is called a condensing map if α(Φ(B))<α(B) for all bounded sets B⊂D(Φ), where α denotes the Kuratowski measure of noncompactness.Lemma 6 (see [23, Example 11.7]).
The mapK+C is a k-set contraction with 0≤k<1 and is thus condensing, if(i)
K,C:D⊆X→X are operators on the Banach space X;(ii)
K is k-contractive; that is,(10)Kx-Ky≤kx-yfor all x,y∈D and fixed k∈[0,1);(iii)
C is compact.Theorem 7 (see [24]).
LetB be a convex, bounded, and closed subset of a Banach space X and let Φ:B→B be a condensing map. Then Φ has a fixed point.In view of Lemma3, we define an operator P:C→C by(11)(Px)(t)=(P1x)(t)+(P2x)(t),t∈[0,1],where(12)P1xt=A∫0t‍t-sq-1Γ(q)f(s,x(s))ds+Aβ(ξ-(α/2)(1-η2))Q∫0θ‍θ-sq-1Γ(q)f(s,x(s))ds+AαβθQ∫01‍1-sqΓ(q+1)f(s,x(s))dshhhhhhhhi-∫0η‍η-sqΓ(q+1)f(s,x(s))ds-AβθQ∫0ξ‍ξ-sqΓ(q)f(s,x(s))ds+AtQα(1-β)∫01‍1-sqΓ(q+1)f(s,x(s))dshhhhhhhhhhhhhhi-∫0η‍η-sqΓ(q+1)f(s,x(s))dshhhhhh-(1-β)∫0ξ‍ξ-sqΓ(q)f(s,x(s))dshhhhih-β1-α1-η∫0θ‍θ-sq-1Γqfs,xsds,(13)(P2x)(t)=B∫0t‍t-sq-1Γq∫0s‍s-ur-1Γrgu,xududs+Bβ(ξ-(α/2)(1-η2))Q×∫0θ‍θ-sq-1Γ(q)∫0s‍s-ur-1Γ(r)g(u,x(u))duds+BαβθQ∫01‍1-sqΓ(q+1)∫0s‍s-ur-1Γ(r)g(u,x(u))dudshhhhhhhhi-∫0η‍η-sqΓ(q+1)∫0s‍s-ur-1Γ(r)g(u,x(u))duds-BβθQ∫0ξ‍ξ-sqΓ(q)∫0s‍s-ur-1Γ(r)g(u,x(u))duds+BtQ∫01‍1-sqΓ(q+1)∫0s‍s-ur-1Γ(r)α1-βhhhhhh×∫01‍1-sqΓ(q+1)∫0s‍s-ur-1Γ(r)g(u,x(u))dudshhhhhhhhhhi-∫0η‍η-sqΓq+1∫0s‍s-ur-1Γ(r)hhhhhhhhhhhhhhhhhhhhhhhhh×g(u,x(u))duds∫0η‍η-sqΓq+1hhhhhh-(1-β)∫0ξ‍ξ-sqΓq∫0s‍s-ur-1Γrhhhhhhhhhhhhhhhhhhhihhhh×g(u,x(u))dudshhhhhh-β(1-α(1-η))hhhhhh×∫0θ‍θ-sq-1Γ(q)∫0s‍s-ur-1Γ(r)g(u,x(u))duds.Theorem 8.
Letf,g:[0,1]×R→R be continuous functions satisfying the following conditions.
(H
1
)
f satisfies the Lipschitz condition:(14)ft,x-ft,y≤Lx-y,L>0,∀t,x,t,y∈0,1×R,
(H
2
)
there exist a functionm∈C([0,1],R+) and a nondecreasing function ψ:R+→R+ such that(15)gt,x≤mtψx,∀(t,x)∈[0,1]×R.Then the boundary value problem (1) has at least one solution provided that(16)γ≔AL1Γ(q+1)+βξ-α/21-η2θq|Q|Γ(q+1)hhhhhii+αβθQ1Γq+2+ηq+1Γq+2hhhhhii+βθξqQΓq+1+1|Q|hhhhhii×α1-β1Γq+2+ηq+1Γq+21Γ(q+1)+βξ-α/21-η2θq|Q|Γ(q+1)hhhhhhhhii+1Γ(q+1)+βξ-α/21-η2θq|Q|Γ(q+1)1-βξqΓq+1+β1-α1-ηθqΓq+1<1.Proof.
LetBν={x∈C:x≤ν} be a closed bounded and convex subset of C:=C([0,1],R), where ν will be fixed later. We define a map P:Bν→C as(17)Pxt=P1xt+P2xt,t∈0,1,where P1 and P2 are defined by (12) and (13), respectively. Notice that the problem (1) is equivalent to a fixed point problem P(x)=x.
Step 1
(
(
P
x
)
(
B
ν
)
⊂
B
ν
). For that, we set M=sup⁡t∈[0,1]|f(t,0)| and select ν≥ω/(1-γ), where(18)ω=AMβ1-α1-ηθqΓ(q+1)1Γ(q+1)iiiiiiiiiii+βξ-α/21-η2θq|Q|Γ(q+1)iiiiiiiiiii+αβθQ1Γq+2+ηq+1Γq+2iiiiiiiiiii+|β|θξq|Q|Γ(q+1)+1|Q|iiiiiiiiiii×β1-α1-ηθqΓ(q+1)α1-β1Γq+2+ηq+1Γq+2hhhhhhhh+|1-β|ξqΓ(q+1)+β1-α1-ηθqΓ(q+1)+Bmψr×1Γ(q+r+1)+|β(ξ-(α/2)(1-η2))|θq+r|Q|Γ(q+r+1)hhh+αβθQ1Γq+r+2+ηq+r+1Γq+r+2hhh+|β|θξq+r|Q|Γ(q+r+1)+1|Q|hhh×α1-ββ1-α1-ηθq+rΓq+r+1hhhhhh×1Γq+r+2+ηq+r+1Γq+r+2hhhhhh+|1-β|ξq+rΓ(q+r+1)hhhhhih+β1-α1-ηθq+rΓq+r+11Γ(q+r+1)+|β(ξ-(α/2)(1-η2))|θq+r|Q|Γ(q+r+1).
Using|f(t,x(t))|≤|f(t,x(t))-f(t,0)|+|f(t,0)|≤Lν+M, for x∈Bν,t∈[0,1], we get(19)P1xt≤A(Lν+M)×1Γ(q+1)+βξ-α/21-η2θq|Q|Γ(q+1)hhh+|αβ|θ|Q|(1Γ(q+2)+ηq+1Γ(q+2))hhh+|β|θξq|Q|Γ(q+1)+1|Q|hhh×α1-β1Γq+2+ηq+1Γq+21Γ(q+1)+βξ-α/21-η2θq|Q|Γ(q+1)hhhhhhh+|1-β|ξqΓ(q+1)hhhhhhh+1Γ(q+1)+βξ-α/21-η2θq|Q|Γ(q+1)β1-α1-ηθqΓ(q+1),P2xt≤Bmψ(ν)×1Γ(q+r+1)hhii+βξ-α/21-η2θq+r|Q|Γ(q+r+1)hhii+|αβ|θ|Q|(1Γ(q+r+2)+ηq+r+1Γ(q+r+2))hhii+|β|θξq+r|Q|Γ(q+r+1)+1|Q|hhii×α1-β1Γq+r+2+ηq+r+1Γq+r+2hhhhhhi+|1-β|ξq+rΓ(q+r+1)hhhhhhi+β1-α1-ηθq+rΓ(q+r+1),where we have used the following relations:(20)∫0t‍t-sq-1Γq∫0s‍s-ur-1Γrduds=∫0t‍t-sq-1Γ(q)srΓ(r+1)ds=tq+rΓ(q)Γ(r+1)∫01‍1-vq-1vrdv=tq+rΓ(q+r+1),∫01‍1-uα-1uβdu=Γ(α)Γ(β+1)Γ(α+β+1).Consequently(21)Pxt≤P1xt+P2xt≤ALν+M×1Γq+1+βξ-α/21-η2θqQΓq+11Γq+1+βξ-α/21-η2θqQΓq+1hhh+αβθQ1Γq+2+ηq+1Γq+2hhh+βθξqQΓq+1+1Qhhh×α1-β1Γq+2+ηq+1Γq+21Γq+1+βξ-α/21-η2θqQΓq+1hhhhhh+1Γq+1+βξ-α/21-η2θqQΓq+11-βξqΓq+1+β1-α1-ηθqΓq+1+Bmψν×1Γq+r+1+βξ-α/21-η2θq+rQΓq+r+11Γq+1+βξ-α/21-η2θqQΓq+1hhh+αβθQ1Γq+r+2+ηq+r+1Γq+r+2hhh+βθξq+r+rQΓq+r+1+1Qhhh×α1-β1Γq+r+2+ηq+r+1Γq+r+21Γq+r+1+βξ-α/21-η2θq+rQΓq+r+1hhhhhh+1Γq+r+1+βξ-α/21-η2θq+rQΓq+r+11-βξq+rΓq+r+1+β1-α1-ηθq+rΓq+r+1≤γν+ω≤ν,which implies that (Px)(Bν)⊂Bν.
Step 2 (
P
1
is continuous and
γ
-contractive). To show the continuity of P1 for t∈[0,1], let us consider a sequence xn converging to x. Then, by the assumption (H1), we have(22)P1xnt-P1xt≤AL1Γq+1+βξ-α/21-η2θqQΓq+1hhhhh+αβθQ1Γq+2+ηq+1Γq+2hhhhh+βθξqQΓ(q+1)+1Qhhhhh×α1-β1Γq+2+ηq+1Γq+21Γq+1+βξ-α/21-η2θqQΓq+1hhhhhhhih+1Γq+1+βξ-α/21-η2θqQΓq+11-βξqΓ(q+1)+β1-α1-ηθqΓ(q+1)×xn-x.Next, we show that P1 is γ-contractive. For x,y∈Bν, we get(23)P1xt-P1yt≤AL1Γ(q+1)+βξ-α/21-η2θqQΓ(q+1)hhhhh+αβθQ1Γq+2+ηq+1Γq+2hhhhh+βθξqQΓ(q+1)+1Qhhhhh×α1-β1Γq+2+ηq+1Γq+21Γ(q+1)+βξ-α/21-η2θqQΓ(q+1)hhhhhhhh+1Γ(q+1)+βξ-α/21-η2θqQΓ(q+1)1-βξqΓ(q+1)+β1-α1-ηθqΓ(q+1)×x-y.By the given assumption(24)γ≔AL1Γ(q+1)hhhi+βξ-α/21-η2θqQΓ(q+1)hhhi+αβθQ1Γq+2+ηq+1Γq+2hhhi+βθξqQΓ(q+1)+1Qhhhi×α1-β1Γq+2+ηq+1Γq+21Γ(q+1)hhhhhhi+1Γ(q+1)1-βξqΓ(q+1)+β1-α1-ηθqΓ(q+1)<1,it follows that P1 is γ-contractive.
Step 3 (
P
2
is compact). In Step 1, it has been shown that P2 is uniformly bounded. Now we show that P2 maps bounded sets into equicontinuous sets of C([0,1],R). Let t1,t2∈[0,1] with t1<t2 and x∈Bν. Then we obtain(25)P2xt2-P2xt1≤Bψ(ν)Γ(q+r+1)×∫0t1‍t2-sq-1-t1-sq-1m(s)ds+BψνΓq+r+1∫t1t2‍t2-sq-1m(s)ds+Bψνmt2-t1Q×α1-β1Γq+r+2+ηq+r+1Γq+r+2hhh+1-βξq+rΓq+r+1+β1-α1-ηθq+rΓq+r+1.Obviously the right hand side of the above inequality tends to zero independently of x∈Bν as t2-t1→0. Therefore it follows by the Arzelá-Ascoli theorem that P2:C([0,1],R)→C([0,1],R) is completely continuous. Thus P2 is compact on [0,1].
Step 4 (
P
is condensing). Since P1 is continuous, γ-contractive and P2 is compact, by Lemma 6, P:Bν→Bν with P=P1+P2 is a condensing map on Bν.
Consequently, by Theorem7, the map P has a fixed point which, in turn, implies that the problem (1) has a solution.Example 9.
Consider a nonlocal integral boundary value problem of fractional integrodifferential equations given by(26)D3/2x(t)=f(t,x(t))+I3/4g(t,x(t)),t∈[0,1],x(0)=12x(14),x(13)=∫2/31‍x(s)ds,where q=3/2,A=B=1,r=3/4,θ=1/4,ξ=1/3,η=2/3,α=1,β=1/2,f(t,x)=(1/2+t3)tan-1x+t+1, and g(t,x)=(t2/(1+t3))(1+(x/(1+x))).ClearlyL=1/8 as |f(t,x)-f(t,y)|≤(1/8)|x-y|, and |g(t,x)|≤m(t)ψ(∥x∥) with m(t)=t2/(1+t3) and ψ(x)=2. Furthermore |Q|=1/9, and the condition (16) yields γ≃0.522371<1. Thus all the conditions of Theorem 8 are satisfied and consequently the problem (26) has a solution.


### 3.2. Existence Results via O’Regan’s Fixed Point Theorem

Our next existence result relies on a fixed point theorem due to O’Regan in [21].Lemma 10.
Denote byU 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 nonnegative nondecreasing function ϕ:[0,∞)→[0,∞) satisfying ϕ(z)<z for z>0, such that F2(x)-F2(y)≤ϕ(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 λ∈(0,1) with u=λF(u), where U¯ and ∂U, respectively, represent the closure and boundary of  U.For convenience we set(27)p0=A1Γ(q+1)+βξ-α/21-η2θq|Q|Γ(q+1)hhhii+αβθQ1Γq+2+ηq+1Γq+2hhhii+βθξqQΓ(q+1)+1Qhhhii×α1-β1Γq+2+ηq+1Γq+21Γ(q+1)+βξ-α/21-η2θq|Q|Γ(q+1)hhhhhhi+1Γ(q+1)+βξ-α/21-η2θq|Q|Γ(q+1)1-βξqΓ(q+1)+β1-α1-ηθqΓ(q+1),k0=B1Γ(q+r+1)+βξ-α/21-η2θq+rQΓ(q+r+1)hhhi+αβθQ1Γq+r+2+ηq+r+1Γq+r+2hhhi+βθξq+rQΓ(q+r+1)+1Qhhhi×α1-β1Γq+r+2+ηq+r+1Γq+r+21Γ(q+r+1)+βξ-α/21-η2θq+rQΓ(q+r+1)hhhhhhi+1Γ(q+r+1)+βξ-α/21-η2θq+rQΓ(q+r+1)1-βξq+rΓ(q+r+1)+β1-α1-ηθq+rΓ(q+r+1).Let(28)Ωσ=x∈C0,1,R:x<σand denote the maximum number by(29)Mσ=max⁡ft,x:t,x∈0,1×-σ,σ.Theorem 11.
Letf,g:[0,1]×R→R be continuous functions. Assume that(A1)
there exist a nonnegative functionp∈C([0,1],R) and a nondecreasing function ζ:[0,∞)→(0,∞) such that(30)ft,u≤ptζuforany(t,u)∈[0,1]×R;(A2)
there exist a positive constantl<k0-1 and a continuous function ϕ:[0,∞)→(0,∞) such that ϕ(z)≤lz and |g(t,u)-g(t,v)|≤ϕ(∥u-v∥) for all t∈[0,1] and u,v∈R;(A3)
sup⁡r∈0,∞(r/(k0K+p0ζ(r)p))>(1/1-k0l), where K=sup⁡t∈[0,1]|g(t,0)|.Then the boundary value problem (1) has at least one solution on [0,1].Proof.
By the assumption (A3), there exists a number r0>0 such that(31)r0k0K+p0ζ(r0)p>11-k0l.We will show that the operators P1 and P2 defined by (12) and (13), respectively, satisfy all the conditions of Lemma 10. The proof consists of a series of steps.
Step 1 (the operator
P
1
is continuous and completely continuous). We first show that P1(Ω¯r0) is bounded. For any x∈Ω¯r0, we have(32)P1x≤AMσ×1Γ(q+1)+βξ-(α/2)1-η2θqQΓ(q+1)hhh+αβθQ1Γq+2+ηq+1Γq+2hhh+βθξqQΓ(q+1)+1Qhhh×α1-β1Γq+2+ηq+1Γq+21Γ(q+1)+βξ-(α/2)1-η2θqQΓ(q+1)hhhhhh+1Γ(q+1)+βξ-(α/2)1-η2θqQΓ(q+1)1-βξqΓ(q+1)+β1-α1-ηθqΓ(q+1).Thus the operator P1(Ω¯r0) is uniformly bounded. For any t1,t2∈[0,1],t1<t2, we have(33)P1xt2-P1xt1≤AMr∫0t1‍t2-sq-1-t1-sq-1dshhhhhhh+∫t1t2‍t2-sq-1ds+AMrt2-t1Q×α1-β1Γq+2+ηq+1Γq+2hhhh+1-βξqΓ(q+1)+β1-α1-ηθqΓ(q+1),which is independent of x and tends to zero as t2-t1→0. Thus, P1 is equicontinuous. Hence, by the Arzelá-Ascoli theorem, P1(Ω¯r0) is a relatively compact set. Now, let xn∈Ω¯r0 with xn-x→0. Then the limit xn(t)-x(t)→0 is uniformly valid on [0,1]. From the uniform continuity of f(t,x) on the compact set [0,1]×[-r0,r0], it follows that f(t,xn(t))-f(t,x(t))→0 is uniformly valid on [0,1]. Hence P1xn-P1x→0 as n→∞. This shows the continuity of P1.
Step 2 (the operator
P
2
:
Ω
¯
r
0
→
C
(
[
0,1
]
,
R
)
is contractive). Consider(34)P2xt-P2yt≤B1Γ(q+r+1)hhhh+βξ-(α/2)1-η2θq+r|Q|Γ(q+r+1)hhhh+αβθQ1Γq+r+2+ηq+r+1Γq+r+2hhhh+βθξq+rQΓ(q+r+1)+1Qhhhh×α1-β1Γq+r+2+ηq+r+1Γq+r+21Γ(q+r+1)hhhhhhh+1Γ(q+r+1)1-βξq+rΓq+r+1+β1-α1-ηθq+rΓq+r+1×ϕ(x-y).This, together with (A2), implies that(35)(P2x)-(P2y)≤ϕ(x-y),so P2:Ω¯r0→C([0,1],R) is a nonlinear contraction.
Step 3 (the set
P
(
Ω
¯
r
0
)
is bounded). Using the inequality(36)gt,x≤gt,x-gt,0+gt,0≤ϕx+K≤lr0+K,we have(37)P2x≤B(lr0+K)×1Γ(q+r+1)hhh+βξ-α/21-η2θq+rQΓ(q+r+1)hhh+αβθQ1Γq+r+2+ηq+r+1Γq+r+2hhh+βθξq+rQΓ(q+r+1)+1Qhhh×α1-β1Γq+r+2+ηq+r+1Γq+r+21Γ(q+r+1)hhhhhh+1Γ(q+r+1)1-βξq+rΓ(q+r+1)+β1-α1-ηθq+rΓ(q+r+1),for any x∈Ω¯r0. This, with the boundedness of the set P1(Ω¯r0), implies that the set P(Ω¯r0) is bounded.
Step 4 (finally, it will be shown that the case (C2) in Lemma10 does not hold). On the contrary, we suppose that (C2) holds. Then, we have that there exist λ∈(0,1) and x∈∂Ωr0 such that x=λPx. So, we have x=r0 and(38)xt=λA∫0t‍t-sq-1Γ(q)f(s,x(s))ds+λAβ(ξ-(α/2)(1-η2))Q×∫0θ‍θ-sq-1Γ(q)f(s,x(s))ds+λAαβθQ∫01‍1-sqΓ(q+1)f(s,x(s))dshhhhhhhhhh-∫0η‍η-sqΓ(q+1)f(s,x(s))ds-λAβθQ∫0ξ‍ξ-sqΓ(q)f(s,x(s))ds+λAtQ×α1-β∫01‍1-sqΓq+1fs,xsdshhhhhhhhhhhhh-∫0η‍η-sqΓ(q+1)f(s,x(s))dshhh-(1-β)∫0ξ‍ξ-sqΓ(q)f(s,x(s))dshhh-β1-α1-η∫0θ‍θ-sq-1Γqhhhhhhhhhhhhhhhhihh×f(s,x(s))ds∫01‍1-sqΓq+1fs,xsds+λB∫0t‍t-sq-1Γ(q)∫0s‍s-ur-1Γ(r)g(u,x(u))duds+λBβ(ξ-(α/2)(1-η2))Q×∫0θ‍θ-sq-1Γq∫0s‍s-ur-1Γrgu,xududs+λBαβθQ∫01‍1-sqΓq+1∫0s‍s-ur-1Γrhhhhhhhhhhhhhhhhhhhh×g(u,x(u))dudshhhhhhhhh-∫0η‍η-sqΓq+1∫0s‍s-ur-1Γrhhhhhhhhhhhhhhhhhhhhhh×g(u,x(u))duds∫01‍1-sqΓq+1-λBβθQ∫0ξ‍ξ-sqΓq∫0s‍s-ur-1Γrgu,xududs+λBtQ∫0s‍s-ur-1Γ(r)g(u,x(u))dudsα(1-β)hhhhhhh×∫01‍1-sqΓq+1∫0s‍s-ur-1Γrhhhhhhhhhhhhhhhhhhhh×gu,xududshhhhhhhhhhh-∫0η‍η-sqΓq+1∫0s‍s-ur-1Γrhhhhhhhhhhhhhhhihihhhihh×g(u,x(u))duds∫01‍1-sqΓq+1hhhhhh-(1-β)∫0ξ‍ξ-sqΓq∫0s‍s-ur-1Γrhhhhhhhhhhhhhhhhhhhhhhh×g(u,x(u))dudshhhhhh-β1-α1-ηhhhhhh×∫0θ‍θ-sq-1Γq∫0s‍s-ur-1Γr.hhhhhhhhhhhhhhhhhhh×gu,xududs∫0ξ‍ξ-sqΓq.Using the assumptions (A1) and (A2), we get(39)r0≤k0K+p0ζ(r0)p+k0lr0,which leads to a contradiction:(40)r0k0K+p0ζ(r0)p≤11-k0l.Thus the operators P1 and P2 satisfy all the conditions of Lemma 10. Hence, the operator P has at least one fixed point x∈Ω¯r0, which is the solution of the problem (1). This completes the proof.Example 12.
Consider a nonlocal integral boundary value problem of fractional integrodifferential equations given by(41)D3/2x(t)=f(t,x(t))+I3/4g(t,x(t)),t∈[0,1],x(0)=12x(14),x(13)=∫2/31‍x(s)ds,where q=3/2,A=B=1,r=3/4,θ=1/4,ξ=1/3,η=2/3,α=1,β=1/2,f(t,x)=(1/27)(21+t-1)sinx, and g(t,x)=(1/2+t3)(3+(x/(1+x))).Observe that|f(t,x)|≤(1/27)(21+t-1)x,|g(t,x)-g(t,y)|≤(1/8)|x-y|, and sup⁡t∈[0,1]g(t,0)=3/8. Further, we set p(t)=(1/3)(2(1+t)-1),ζ(x)=x/9,l=1/8, and K=3/8. With the given data, it is found that k0≃1.495606,p0≃4.178968,|Q|=1/9,p=(22-1)/3, and(42)sup⁡r∈(0,∞)rk0K+p0ζrp≃3.533597,11-k0l≃1.229953.Clearly, all the conditions of Theorem 11 are satisfied and hence there exists a solution for the problem (41).


## 3.1. Existence Results via Sadovskii’s Fixed Point Theorem

Definition 4.
LetM be a bounded set in metric space (X,d); then Kuratowskii measure of noncompactness, α(M), is defined as inf⁡{ϵ:M covered by finitely many sets such that the diameter of each set ≤ϵ}.Definition 5 (see [22]).
LetΦ:D(Φ)⊆X→X be a bounded and continuous operator on Banach space X. Then Φ is called a condensing map if α(Φ(B))<α(B) for all bounded sets B⊂D(Φ), where α denotes the Kuratowski measure of noncompactness.Lemma 6 (see [23, Example 11.7]).
The mapK+C is a k-set contraction with 0≤k<1 and is thus condensing, if(i)
K,C:D⊆X→X are operators on the Banach space X;(ii)
K is k-contractive; that is,(10)Kx-Ky≤kx-yfor all x,y∈D and fixed k∈[0,1);(iii)
C is compact.Theorem 7 (see [24]).
LetB be a convex, bounded, and closed subset of a Banach space X and let Φ:B→B be a condensing map. Then Φ has a fixed point.In view of Lemma3, we define an operator P:C→C by(11)(Px)(t)=(P1x)(t)+(P2x)(t),t∈[0,1],where(12)P1xt=A∫0t‍t-sq-1Γ(q)f(s,x(s))ds+Aβ(ξ-(α/2)(1-η2))Q∫0θ‍θ-sq-1Γ(q)f(s,x(s))ds+AαβθQ∫01‍1-sqΓ(q+1)f(s,x(s))dshhhhhhhhi-∫0η‍η-sqΓ(q+1)f(s,x(s))ds-AβθQ∫0ξ‍ξ-sqΓ(q)f(s,x(s))ds+AtQα(1-β)∫01‍1-sqΓ(q+1)f(s,x(s))dshhhhhhhhhhhhhhi-∫0η‍η-sqΓ(q+1)f(s,x(s))dshhhhhh-(1-β)∫0ξ‍ξ-sqΓ(q)f(s,x(s))dshhhhih-β1-α1-η∫0θ‍θ-sq-1Γqfs,xsds,(13)(P2x)(t)=B∫0t‍t-sq-1Γq∫0s‍s-ur-1Γrgu,xududs+Bβ(ξ-(α/2)(1-η2))Q×∫0θ‍θ-sq-1Γ(q)∫0s‍s-ur-1Γ(r)g(u,x(u))duds+BαβθQ∫01‍1-sqΓ(q+1)∫0s‍s-ur-1Γ(r)g(u,x(u))dudshhhhhhhhi-∫0η‍η-sqΓ(q+1)∫0s‍s-ur-1Γ(r)g(u,x(u))duds-BβθQ∫0ξ‍ξ-sqΓ(q)∫0s‍s-ur-1Γ(r)g(u,x(u))duds+BtQ∫01‍1-sqΓ(q+1)∫0s‍s-ur-1Γ(r)α1-βhhhhhh×∫01‍1-sqΓ(q+1)∫0s‍s-ur-1Γ(r)g(u,x(u))dudshhhhhhhhhhi-∫0η‍η-sqΓq+1∫0s‍s-ur-1Γ(r)hhhhhhhhhhhhhhhhhhhhhhhhh×g(u,x(u))duds∫0η‍η-sqΓq+1hhhhhh-(1-β)∫0ξ‍ξ-sqΓq∫0s‍s-ur-1Γrhhhhhhhhhhhhhhhhhhhihhhh×g(u,x(u))dudshhhhhh-β(1-α(1-η))hhhhhh×∫0θ‍θ-sq-1Γ(q)∫0s‍s-ur-1Γ(r)g(u,x(u))duds.Theorem 8.
Letf,g:[0,1]×R→R be continuous functions satisfying the following conditions.
(H
1
)
f satisfies the Lipschitz condition:(14)ft,x-ft,y≤Lx-y,L>0,∀t,x,t,y∈0,1×R,
(H
2
)
there exist a functionm∈C([0,1],R+) and a nondecreasing function ψ:R+→R+ such that(15)gt,x≤mtψx,∀(t,x)∈[0,1]×R.Then the boundary value problem (1) has at least one solution provided that(16)γ≔AL1Γ(q+1)+βξ-α/21-η2θq|Q|Γ(q+1)hhhhhii+αβθQ1Γq+2+ηq+1Γq+2hhhhhii+βθξqQΓq+1+1|Q|hhhhhii×α1-β1Γq+2+ηq+1Γq+21Γ(q+1)+βξ-α/21-η2θq|Q|Γ(q+1)hhhhhhhhii+1Γ(q+1)+βξ-α/21-η2θq|Q|Γ(q+1)1-βξqΓq+1+β1-α1-ηθqΓq+1<1.Proof.
LetBν={x∈C:x≤ν} be a closed bounded and convex subset of C:=C([0,1],R), where ν will be fixed later. We define a map P:Bν→C as(17)Pxt=P1xt+P2xt,t∈0,1,where P1 and P2 are defined by (12) and (13), respectively. Notice that the problem (1) is equivalent to a fixed point problem P(x)=x.
Step 1
(
(
P
x
)
(
B
ν
)
⊂
B
ν
). For that, we set M=sup⁡t∈[0,1]|f(t,0)| and select ν≥ω/(1-γ), where(18)ω=AMβ1-α1-ηθqΓ(q+1)1Γ(q+1)iiiiiiiiiii+βξ-α/21-η2θq|Q|Γ(q+1)iiiiiiiiiii+αβθQ1Γq+2+ηq+1Γq+2iiiiiiiiiii+|β|θξq|Q|Γ(q+1)+1|Q|iiiiiiiiiii×β1-α1-ηθqΓ(q+1)α1-β1Γq+2+ηq+1Γq+2hhhhhhhh+|1-β|ξqΓ(q+1)+β1-α1-ηθqΓ(q+1)+Bmψr×1Γ(q+r+1)+|β(ξ-(α/2)(1-η2))|θq+r|Q|Γ(q+r+1)hhh+αβθQ1Γq+r+2+ηq+r+1Γq+r+2hhh+|β|θξq+r|Q|Γ(q+r+1)+1|Q|hhh×α1-ββ1-α1-ηθq+rΓq+r+1hhhhhh×1Γq+r+2+ηq+r+1Γq+r+2hhhhhh+|1-β|ξq+rΓ(q+r+1)hhhhhih+β1-α1-ηθq+rΓq+r+11Γ(q+r+1)+|β(ξ-(α/2)(1-η2))|θq+r|Q|Γ(q+r+1).
Using|f(t,x(t))|≤|f(t,x(t))-f(t,0)|+|f(t,0)|≤Lν+M, for x∈Bν,t∈[0,1], we get(19)P1xt≤A(Lν+M)×1Γ(q+1)+βξ-α/21-η2θq|Q|Γ(q+1)hhh+|αβ|θ|Q|(1Γ(q+2)+ηq+1Γ(q+2))hhh+|β|θξq|Q|Γ(q+1)+1|Q|hhh×α1-β1Γq+2+ηq+1Γq+21Γ(q+1)+βξ-α/21-η2θq|Q|Γ(q+1)hhhhhhh+|1-β|ξqΓ(q+1)hhhhhhh+1Γ(q+1)+βξ-α/21-η2θq|Q|Γ(q+1)β1-α1-ηθqΓ(q+1),P2xt≤Bmψ(ν)×1Γ(q+r+1)hhii+βξ-α/21-η2θq+r|Q|Γ(q+r+1)hhii+|αβ|θ|Q|(1Γ(q+r+2)+ηq+r+1Γ(q+r+2))hhii+|β|θξq+r|Q|Γ(q+r+1)+1|Q|hhii×α1-β1Γq+r+2+ηq+r+1Γq+r+2hhhhhhi+|1-β|ξq+rΓ(q+r+1)hhhhhhi+β1-α1-ηθq+rΓ(q+r+1),where we have used the following relations:(20)∫0t‍t-sq-1Γq∫0s‍s-ur-1Γrduds=∫0t‍t-sq-1Γ(q)srΓ(r+1)ds=tq+rΓ(q)Γ(r+1)∫01‍1-vq-1vrdv=tq+rΓ(q+r+1),∫01‍1-uα-1uβdu=Γ(α)Γ(β+1)Γ(α+β+1).Consequently(21)Pxt≤P1xt+P2xt≤ALν+M×1Γq+1+βξ-α/21-η2θqQΓq+11Γq+1+βξ-α/21-η2θqQΓq+1hhh+αβθQ1Γq+2+ηq+1Γq+2hhh+βθξqQΓq+1+1Qhhh×α1-β1Γq+2+ηq+1Γq+21Γq+1+βξ-α/21-η2θqQΓq+1hhhhhh+1Γq+1+βξ-α/21-η2θqQΓq+11-βξqΓq+1+β1-α1-ηθqΓq+1+Bmψν×1Γq+r+1+βξ-α/21-η2θq+rQΓq+r+11Γq+1+βξ-α/21-η2θqQΓq+1hhh+αβθQ1Γq+r+2+ηq+r+1Γq+r+2hhh+βθξq+r+rQΓq+r+1+1Qhhh×α1-β1Γq+r+2+ηq+r+1Γq+r+21Γq+r+1+βξ-α/21-η2θq+rQΓq+r+1hhhhhh+1Γq+r+1+βξ-α/21-η2θq+rQΓq+r+11-βξq+rΓq+r+1+β1-α1-ηθq+rΓq+r+1≤γν+ω≤ν,which implies that (Px)(Bν)⊂Bν.
Step 2 (
P
1
is continuous and
γ
-contractive). To show the continuity of P1 for t∈[0,1], let us consider a sequence xn converging to x. Then, by the assumption (H1), we have(22)P1xnt-P1xt≤AL1Γq+1+βξ-α/21-η2θqQΓq+1hhhhh+αβθQ1Γq+2+ηq+1Γq+2hhhhh+βθξqQΓ(q+1)+1Qhhhhh×α1-β1Γq+2+ηq+1Γq+21Γq+1+βξ-α/21-η2θqQΓq+1hhhhhhhih+1Γq+1+βξ-α/21-η2θqQΓq+11-βξqΓ(q+1)+β1-α1-ηθqΓ(q+1)×xn-x.Next, we show that P1 is γ-contractive. For x,y∈Bν, we get(23)P1xt-P1yt≤AL1Γ(q+1)+βξ-α/21-η2θqQΓ(q+1)hhhhh+αβθQ1Γq+2+ηq+1Γq+2hhhhh+βθξqQΓ(q+1)+1Qhhhhh×α1-β1Γq+2+ηq+1Γq+21Γ(q+1)+βξ-α/21-η2θqQΓ(q+1)hhhhhhhh+1Γ(q+1)+βξ-α/21-η2θqQΓ(q+1)1-βξqΓ(q+1)+β1-α1-ηθqΓ(q+1)×x-y.By the given assumption(24)γ≔AL1Γ(q+1)hhhi+βξ-α/21-η2θqQΓ(q+1)hhhi+αβθQ1Γq+2+ηq+1Γq+2hhhi+βθξqQΓ(q+1)+1Qhhhi×α1-β1Γq+2+ηq+1Γq+21Γ(q+1)hhhhhhi+1Γ(q+1)1-βξqΓ(q+1)+β1-α1-ηθqΓ(q+1)<1,it follows that P1 is γ-contractive.
Step 3 (
P
2
is compact). In Step 1, it has been shown that P2 is uniformly bounded. Now we show that P2 maps bounded sets into equicontinuous sets of C([0,1],R). Let t1,t2∈[0,1] with t1<t2 and x∈Bν. Then we obtain(25)P2xt2-P2xt1≤Bψ(ν)Γ(q+r+1)×∫0t1‍t2-sq-1-t1-sq-1m(s)ds+BψνΓq+r+1∫t1t2‍t2-sq-1m(s)ds+Bψνmt2-t1Q×α1-β1Γq+r+2+ηq+r+1Γq+r+2hhh+1-βξq+rΓq+r+1+β1-α1-ηθq+rΓq+r+1.Obviously the right hand side of the above inequality tends to zero independently of x∈Bν as t2-t1→0. Therefore it follows by the Arzelá-Ascoli theorem that P2:C([0,1],R)→C([0,1],R) is completely continuous. Thus P2 is compact on [0,1].
Step 4 (
P
is condensing). Since P1 is continuous, γ-contractive and P2 is compact, by Lemma 6, P:Bν→Bν with P=P1+P2 is a condensing map on Bν.
Consequently, by Theorem7, the map P has a fixed point which, in turn, implies that the problem (1) has a solution.Example 9.
Consider a nonlocal integral boundary value problem of fractional integrodifferential equations given by(26)D3/2x(t)=f(t,x(t))+I3/4g(t,x(t)),t∈[0,1],x(0)=12x(14),x(13)=∫2/31‍x(s)ds,where q=3/2,A=B=1,r=3/4,θ=1/4,ξ=1/3,η=2/3,α=1,β=1/2,f(t,x)=(1/2+t3)tan-1x+t+1, and g(t,x)=(t2/(1+t3))(1+(x/(1+x))).ClearlyL=1/8 as |f(t,x)-f(t,y)|≤(1/8)|x-y|, and |g(t,x)|≤m(t)ψ(∥x∥) with m(t)=t2/(1+t3) and ψ(x)=2. Furthermore |Q|=1/9, and the condition (16) yields γ≃0.522371<1. Thus all the conditions of Theorem 8 are satisfied and consequently the problem (26) has a solution.


## 3.2. Existence Results via O’Regan’s Fixed Point Theorem

Our next existence result relies on a fixed point theorem due to O’Regan in [21].Lemma 10.
Denote byU 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 nonnegative nondecreasing function ϕ:[0,∞)→[0,∞) satisfying ϕ(z)<z for z>0, such that F2(x)-F2(y)≤ϕ(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 λ∈(0,1) with u=λF(u), where U¯ and ∂U, respectively, represent the closure and boundary of  U.For convenience we set(27)p0=A1Γ(q+1)+βξ-α/21-η2θq|Q|Γ(q+1)hhhii+αβθQ1Γq+2+ηq+1Γq+2hhhii+βθξqQΓ(q+1)+1Qhhhii×α1-β1Γq+2+ηq+1Γq+21Γ(q+1)+βξ-α/21-η2θq|Q|Γ(q+1)hhhhhhi+1Γ(q+1)+βξ-α/21-η2θq|Q|Γ(q+1)1-βξqΓ(q+1)+β1-α1-ηθqΓ(q+1),k0=B1Γ(q+r+1)+βξ-α/21-η2θq+rQΓ(q+r+1)hhhi+αβθQ1Γq+r+2+ηq+r+1Γq+r+2hhhi+βθξq+rQΓ(q+r+1)+1Qhhhi×α1-β1Γq+r+2+ηq+r+1Γq+r+21Γ(q+r+1)+βξ-α/21-η2θq+rQΓ(q+r+1)hhhhhhi+1Γ(q+r+1)+βξ-α/21-η2θq+rQΓ(q+r+1)1-βξq+rΓ(q+r+1)+β1-α1-ηθq+rΓ(q+r+1).Let(28)Ωσ=x∈C0,1,R:x<σand denote the maximum number by(29)Mσ=max⁡ft,x:t,x∈0,1×-σ,σ.Theorem 11.
Letf,g:[0,1]×R→R be continuous functions. Assume that(A1)
there exist a nonnegative functionp∈C([0,1],R) and a nondecreasing function ζ:[0,∞)→(0,∞) such that(30)ft,u≤ptζuforany(t,u)∈[0,1]×R;(A2)
there exist a positive constantl<k0-1 and a continuous function ϕ:[0,∞)→(0,∞) such that ϕ(z)≤lz and |g(t,u)-g(t,v)|≤ϕ(∥u-v∥) for all t∈[0,1] and u,v∈R;(A3)
sup⁡r∈0,∞(r/(k0K+p0ζ(r)p))>(1/1-k0l), where K=sup⁡t∈[0,1]|g(t,0)|.Then the boundary value problem (1) has at least one solution on [0,1].Proof.
By the assumption (A3), there exists a number r0>0 such that(31)r0k0K+p0ζ(r0)p>11-k0l.We will show that the operators P1 and P2 defined by (12) and (13), respectively, satisfy all the conditions of Lemma 10. The proof consists of a series of steps.
Step 1 (the operator
P
1
is continuous and completely continuous). We first show that P1(Ω¯r0) is bounded. For any x∈Ω¯r0, we have(32)P1x≤AMσ×1Γ(q+1)+βξ-(α/2)1-η2θqQΓ(q+1)hhh+αβθQ1Γq+2+ηq+1Γq+2hhh+βθξqQΓ(q+1)+1Qhhh×α1-β1Γq+2+ηq+1Γq+21Γ(q+1)+βξ-(α/2)1-η2θqQΓ(q+1)hhhhhh+1Γ(q+1)+βξ-(α/2)1-η2θqQΓ(q+1)1-βξqΓ(q+1)+β1-α1-ηθqΓ(q+1).Thus the operator P1(Ω¯r0) is uniformly bounded. For any t1,t2∈[0,1],t1<t2, we have(33)P1xt2-P1xt1≤AMr∫0t1‍t2-sq-1-t1-sq-1dshhhhhhh+∫t1t2‍t2-sq-1ds+AMrt2-t1Q×α1-β1Γq+2+ηq+1Γq+2hhhh+1-βξqΓ(q+1)+β1-α1-ηθqΓ(q+1),which is independent of x and tends to zero as t2-t1→0. Thus, P1 is equicontinuous. Hence, by the Arzelá-Ascoli theorem, P1(Ω¯r0) is a relatively compact set. Now, let xn∈Ω¯r0 with xn-x→0. Then the limit xn(t)-x(t)→0 is uniformly valid on [0,1]. From the uniform continuity of f(t,x) on the compact set [0,1]×[-r0,r0], it follows that f(t,xn(t))-f(t,x(t))→0 is uniformly valid on [0,1]. Hence P1xn-P1x→0 as n→∞. This shows the continuity of P1.
Step 2 (the operator
P
2
:
Ω
¯
r
0
→
C
(
[
0,1
]
,
R
)
is contractive). Consider(34)P2xt-P2yt≤B1Γ(q+r+1)hhhh+βξ-(α/2)1-η2θq+r|Q|Γ(q+r+1)hhhh+αβθQ1Γq+r+2+ηq+r+1Γq+r+2hhhh+βθξq+rQΓ(q+r+1)+1Qhhhh×α1-β1Γq+r+2+ηq+r+1Γq+r+21Γ(q+r+1)hhhhhhh+1Γ(q+r+1)1-βξq+rΓq+r+1+β1-α1-ηθq+rΓq+r+1×ϕ(x-y).This, together with (A2), implies that(35)(P2x)-(P2y)≤ϕ(x-y),so P2:Ω¯r0→C([0,1],R) is a nonlinear contraction.
Step 3 (the set
P
(
Ω
¯
r
0
)
is bounded). Using the inequality(36)gt,x≤gt,x-gt,0+gt,0≤ϕx+K≤lr0+K,we have(37)P2x≤B(lr0+K)×1Γ(q+r+1)hhh+βξ-α/21-η2θq+rQΓ(q+r+1)hhh+αβθQ1Γq+r+2+ηq+r+1Γq+r+2hhh+βθξq+rQΓ(q+r+1)+1Qhhh×α1-β1Γq+r+2+ηq+r+1Γq+r+21Γ(q+r+1)hhhhhh+1Γ(q+r+1)1-βξq+rΓ(q+r+1)+β1-α1-ηθq+rΓ(q+r+1),for any x∈Ω¯r0. This, with the boundedness of the set P1(Ω¯r0), implies that the set P(Ω¯r0) is bounded.
Step 4 (finally, it will be shown that the case (C2) in Lemma10 does not hold). On the contrary, we suppose that (C2) holds. Then, we have that there exist λ∈(0,1) and x∈∂Ωr0 such that x=λPx. So, we have x=r0 and(38)xt=λA∫0t‍t-sq-1Γ(q)f(s,x(s))ds+λAβ(ξ-(α/2)(1-η2))Q×∫0θ‍θ-sq-1Γ(q)f(s,x(s))ds+λAαβθQ∫01‍1-sqΓ(q+1)f(s,x(s))dshhhhhhhhhh-∫0η‍η-sqΓ(q+1)f(s,x(s))ds-λAβθQ∫0ξ‍ξ-sqΓ(q)f(s,x(s))ds+λAtQ×α1-β∫01‍1-sqΓq+1fs,xsdshhhhhhhhhhhhh-∫0η‍η-sqΓ(q+1)f(s,x(s))dshhh-(1-β)∫0ξ‍ξ-sqΓ(q)f(s,x(s))dshhh-β1-α1-η∫0θ‍θ-sq-1Γqhhhhhhhhhhhhhhhhihh×f(s,x(s))ds∫01‍1-sqΓq+1fs,xsds+λB∫0t‍t-sq-1Γ(q)∫0s‍s-ur-1Γ(r)g(u,x(u))duds+λBβ(ξ-(α/2)(1-η2))Q×∫0θ‍θ-sq-1Γq∫0s‍s-ur-1Γrgu,xududs+λBαβθQ∫01‍1-sqΓq+1∫0s‍s-ur-1Γrhhhhhhhhhhhhhhhhhhhh×g(u,x(u))dudshhhhhhhhh-∫0η‍η-sqΓq+1∫0s‍s-ur-1Γrhhhhhhhhhhhhhhhhhhhhhh×g(u,x(u))duds∫01‍1-sqΓq+1-λBβθQ∫0ξ‍ξ-sqΓq∫0s‍s-ur-1Γrgu,xududs+λBtQ∫0s‍s-ur-1Γ(r)g(u,x(u))dudsα(1-β)hhhhhhh×∫01‍1-sqΓq+1∫0s‍s-ur-1Γrhhhhhhhhhhhhhhhhhhhh×gu,xududshhhhhhhhhhh-∫0η‍η-sqΓq+1∫0s‍s-ur-1Γrhhhhhhhhhhhhhhhihihhhihh×g(u,x(u))duds∫01‍1-sqΓq+1hhhhhh-(1-β)∫0ξ‍ξ-sqΓq∫0s‍s-ur-1Γrhhhhhhhhhhhhhhhhhhhhhhh×g(u,x(u))dudshhhhhh-β1-α1-ηhhhhhh×∫0θ‍θ-sq-1Γq∫0s‍s-ur-1Γr.hhhhhhhhhhhhhhhhhhh×gu,xududs∫0ξ‍ξ-sqΓq.Using the assumptions (A1) and (A2), we get(39)r0≤k0K+p0ζ(r0)p+k0lr0,which leads to a contradiction:(40)r0k0K+p0ζ(r0)p≤11-k0l.Thus the operators P1 and P2 satisfy all the conditions of Lemma 10. Hence, the operator P has at least one fixed point x∈Ω¯r0, which is the solution of the problem (1). This completes the proof.Example 12.
Consider a nonlocal integral boundary value problem of fractional integrodifferential equations given by(41)D3/2x(t)=f(t,x(t))+I3/4g(t,x(t)),t∈[0,1],x(0)=12x(14),x(13)=∫2/31‍x(s)ds,where q=3/2,A=B=1,r=3/4,θ=1/4,ξ=1/3,η=2/3,α=1,β=1/2,f(t,x)=(1/27)(21+t-1)sinx, and g(t,x)=(1/2+t3)(3+(x/(1+x))).Observe that|f(t,x)|≤(1/27)(21+t-1)x,|g(t,x)-g(t,y)|≤(1/8)|x-y|, and sup⁡t∈[0,1]g(t,0)=3/8. Further, we set p(t)=(1/3)(2(1+t)-1),ζ(x)=x/9,l=1/8, and K=3/8. With the given data, it is found that k0≃1.495606,p0≃4.178968,|Q|=1/9,p=(22-1)/3, and(42)sup⁡r∈(0,∞)rk0K+p0ζrp≃3.533597,11-k0l≃1.229953.Clearly, all the conditions of Theorem 11 are satisfied and hence there exists a solution for the problem (41).



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*Source: 205452-2015-03-22.xml*