# Attainability to Solve Fractional Differential Inclusion on the Half Line at Resonance

Authors: Ahmed Salem; Faris Alzahrani; Aeshah Al-Dosari
Journal: Complexity
 (2020)

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## Abstract

The presented article is deduced about the positive solutions of the fractional differential inclusion at resonance on the half line. The fractional derivative used is in the sense of Riemann–Liouville and the problem is supplemented by unseparated conditions. The existence results are illustrated in view of Leggett–Williams theorem due to O’Regan and Zima on unbounded domain.


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## Body

## 1. Introduction

In this paper, we study the possibility of finding a positive solution on unbounded domain with unseparated conditions for the following fractional differential inclusion:(1)DRακτντ∈Aτ,ντ,τ∈0,∞,(2)νi0=0,i=0,…,n−2;n∈ℕ,(3)DRα−1κ0ν0=limτ⟶∞DRα−1κτντ,which is contingent with Aτ,ντ=qτBτ,ντ for which B:0,∞×ℝ⟶ℝ is multivalued map, n−1<α<n,n≥2,κτ,qτ∈BC0,∞,κτ≥1, and DRα is the symbol of the Riemann–Liouville fractional derivative.Riemann–Liouville fractional derivative and corresponding integral and many other types of fractional derivatives and integrals with singular and nonsingular kernels are used in fractional calculus as generalizations of the ordinary differentiations and integrations. Thus, many scientific teams are attracted to link their works with fractional calculus, which has been effectively applied in the other scientific fields. As a result, there are some books talking about fractional calculus and some of its applications with fractional equations and inclusions, see [1–3]. In the existence and uniqueness fields, there are many articles concerned with different types of fractional differential equations or inclusions. For fractional differential equations, there are a huge number of contributions (see [4–9] and the references given therein). For fractional differential inclusions, Salem and Al-Dosari [10] have studied the existence results of solution for fractional SturmLiouville inclusion associated with composition with multimaps. Chen and Tang [11] have investigated probability of the positive solution for the fractional equation below on unbounded domain at resonance:(4)DRαντ=gτ,ντ,τ∈0,∞,νi0=0,i=0,1,2,3<α<4,DRα−1ν0=limτ⟶∞DRα−1ντ.Jiang and Yang [12] have studied the positive solution for the second ordinary boundary value problems:(5)ν″τ+gτ,ντ=0,τ∈0,∞,ν0=0,ν′+∞=∑i=1m−1αiν′ζi,where αi>0,∑i=1m−1αi=1,0=ζ1<,…,<ζm−1<∞ and gt,0 is not always equal to zero.On a bounded domain, Chen et al. [13] have explored sufficient conditions to obtain a positive solution for the problem:(6)DRαντ∈gτ,ντ,τ∈0,1,νi0=0,i=0,1,2,3<α<4,DRα−1ν0=DRα−1ν1.Furthermore, Hu [14] has given some results of positive solution at resonance for the boundary value problem formulated as follows:(7)DCαντ=gτ,ντ,τ∈0,1,νi0=0,i=1,2,….,n−1,ν0=ν1,n−1<α<n.By reducing nonperturbed boundary value problems at resonance, Wang and Liu [15] have presented their results about positive solution of the problem:(8)DRαντ+gτ,ντ=0,τ∈0,1,ν0=0,1<α<2,ν1=βνζ,βζα−1=1.About an overall summary of resonance and its applications, Haller [16] has mentioned that “resonances are regions in the phase space of a dynamical system in which the frequencies of some angular variables become nearly commensurate. Such regions have a profound effect on the dynamics of the system, since they are rich sources of highly complex motions. In molecular dynamics, resonances are known to give rise to chaotic patterns, multiple time scales, and apparent irreversibility in the transfer of energy between different oscillatory states of molecules. In engineering structures, interactions among resonant modes are responsible for most complicated dynamical phenomena, which again include energy transfer, multitime-scale behavior, and chaotic motions. It is of great practical importance to understand the common mechanism behind these irregular features, both qualitatively and quantitatively. The circle of further applications ranges from nonlinear optics through celestial and fluid mechanics to electromagnetism.”Chaotic patterns, oscillatory states of molecules, complicated dynamical phenomena, and multitime-scale behavior are all given a strong cause to study differential problems at resonance associated with multivalued maps, which are not presented until now. So, our aim is to show the possibility of having at least one positive solution at resonance. For the sake of that, inSection 2, we give some basic notations and results of fractional differentiations and integrations, some facts of multimaps and their arguments, and some needed lemmas and fixed point theorems. Section 3 contains some required results to the existence of positive solution for the given problem. Before the conclusion, Section 4 shows a quintessence as an application of our outcomes.


## 2. Precursory Notations

This section is devoted to provide some notations and definitions for the fractional calculus, multivalued functions, and Fredholm operator.


### 2.1. Fractional Calculus

We introduce some definitions and fundamental rules on the fractional calculus that are needed for the new results [1, 17, 18].Definition 1.
Forζ>0, the Riemann–Liouville fractional integral of a continuous function hτ:0,∞⟶ℝ is defined by(9)Iζhτ=1Γζ∫0ττ−tζ−1htdt.Definition 2.
Forn∈ℕ and n−1≤ζ<n, the Riemann–Liouville fractional derivative of an nth absolutely continuous function hτ:0,∞⟶ℝ is given by(10)DRζhτ=1Γn−ζddτn∫0ττ−tn−ζ−1htdt.Lemma 1.
Lethτ∈L10,∞ and ζ1,ζ2,ζ>0. Then,(11)Iζ1Iζ2hτ=Iζ1+ζ2hτ,DRζIζhτ=hτ.Lemma 2.
LetDRζhτ∈L10,∞,ζ>0. Then,(12)IζDRζhτ=hτ−∑m=1nbmτζ−m,τ∈0,∞,for some bm∈ℝ,m=1,2,…,n,n∈ℕ.


### 2.2. Multivalued Function

In this section, we state some necessary facts on multivalued maps [2, 19–21].LetE,.andH,. be Banach spaces and A:E⟶PclE be a multivalued map. The mapping A is(i)
Convex (closed) ifAe is convex (closed), for every e∈E.(ii)
Completely continuous ifAB is relatively compact, for every B∈PbE.(iii)
Upper semicontinuous ifA−1W is closed subset of E, for each closed subset W⊂E. That is, the set e∈E:Ae⊆O is open for all open sets O⊂E.(iv)
Lower semicontinuous ifA−1Z is open subset of E, for each open subset Z⊂E. In the other words, the set e∈E:Ae∩O≠∅ is open for all open sets O⊂E.(v)
Measurable multivalued if for everye∈E, the function τ⟶de,Aτ=infde,a:a∈Aτ is ℒ− measurable function.Definition 3.
A multivalued mapA:0,∞×ℝ⟶Pℝ is known as a Caratheodory if(1)
For allr∈ℝ,τ⟶Aτ,r is measurable.(2)
For a.eτ∈0,∞,r⟶Aτ,r is upper semicontinuous.
In addition to assumptions (1) and (2), the mapA is L1-Caratheodory if for each k>0,∃ϕk∈C0,∞∩L10,∞ satisfying supτ≥0ϕkτ<+∞ and ϕk>0, that is,(13)Aτ,1+τα−1r=supa:a∈Aτ,1+τα−1r≤ϕkτ,for all r≤k,τ∈0,∞.


### 2.3. Fredholm Operator

We recall some preliminaries about the operators specially Fredholm type where we can get them from [3, 22–24].LetL:E⟶H be any considered operator. Then, we define the following.(1)
The kernel ofL by(14)KerL=e∈E:LE=0H.(2)
The image (rang) ofL by(15)ImL=RL=Le:e∈E⊆H.(3)
The cokernel ofL by(16)CokerL=h:h∈H/ImL.(4)
The index ofF by(17)IndL=dimKerL−dimCokerL.Note that(18)dimCokerL=codimImL.The operatorL is called Fredholm operator if it is a bounded linear operator with finite dimensional kernel and cokernel and closed rang. It is said to be Fredholm of index zero if and only if IndL=0.LetL:domL⊂E⟶H be Fredholm of index zero and P:E⟶E,Q:H⟶H be continuous projections. In the case of(19)ImP=KerL∧ImL=KerQ,E=KerL⊕KerP∧H=ImL⊕ImQ,then LdomL∩KerP is invertible, and so we can define(20)KP=LdomL∩KerP−1.In fact, there is an isomorphic map J:ImQ⟶KerL.Consider the inclusion(21)Lν∈Nν,which is equivalent to(22)ν∈P+JQNν+KpI−QNν.Let C⊂E be a convex and a closed subset. Then, C is called a cone if(i)
ρν∈C,∀ν∈C,ρ≥0(ii)
ν,−ν∈C⟹ν=0Consequently, there is an induced partial order≼ in E subject to(23)o≼ν⟺ν−o∈C.Lemma 3.
LetE be a Banach space contains a cone C. Then, for every η∈C/0, there exists a positive number ση which satisfies(24)ν+η≥σην,ν∈C.Defineμ:E⟶C to be a retraction (continuous mapping constrained with μη=η,∀η∈C). Define the map Θ as(25)Θ=P+JQN+KpI−QN.Then,(i)
Lν∈Nν⟺ν∈P+JQNν+KpI−QNν⟺ν∈Θν(ii)
Θμ=Θ°μTheorem 1.
(O’Regan and Zima). LetC⊂E be a cone, and Ω1andΩ2 be open bounded subsets of E which satisfy Ω1¯⊂Ω2, and C∩Ω2¯/Ω1≠∅. Assume that(1)
L:E⟶H is a Fredholm operator of index zero.(2)
N:E⟶2H is an upper-semicontinuous mapping with nonempty compact convex values.(3)
QN:E⟶2H is bounded on bounded subsets of C, and KpI−QN:E⟶2E is compact on every bounded subset of C.(4)
Lν∉γNν,∀ν∈C∩∂Ω2∩domL and γ∈0,1.(5)
μ subsets of Ω2¯ into bounded subsets of C.(6)
degI−P+JQNμKerL,KerL∩Ω2,0≠0.(7)
There isν0∈C/0 such that(26)ν≤σν0η,∀ν∈Cν0∩∂Ω1,η∈Θν,where(27)Cν0=ν∈C∣ρν0≼ν,for someρ>0,and σν0 is such that ν+ν0≥σν0ν,ν∈C.(8)
P+JQNμ∂Ω2⊂C.(9)
ΘμΩ2¯/Ω1⊂C.(10)
ν∉P+JQNμν,ν∈KerL∩∂Ω2.
Then, the inclusionLν∈Nν can be solvable in the set C∩Ω2¯/Ω1.Remark 1.
Let(28)E=νν∈C0,∞,limτ⟶∞ντ1+τα−1exists,with the norm(29)νE=supτ≥0ντ1+τα−1.
And let(30)H=h∣hτ∈C0,∞∩L10,∞∩Aτ,ντ,supτ≥0|hτ|<+∞,with the norm(31)hH=supτ≥0hτ+∫0∞hτdτ.
Then,E,·EandH,·H are Banach spaces, see [11].
Expand the operatorL, for which(32)domL=ν∈E∣DRακτντ∈C0,∞∩L10,∞,νi0=0,i=0,..,n−2,DRα−1κτνττ=0=limτ⟶∞DRα−1κτντ,where(33)L:domL⊂E⟶H,ν↦DRακτντ,(34)N:E⟶H,ν⟼aτ,ντ.
Then, the multivalue problems (1)–(3) are equivalent to the inclusion:(35)Lν∈Nν,ν∈domL.
We can understand the proof of the following lemma if we see the compactness hypothesis in [11, 25, 26].Lemma 4.
LetΩ be subset of E; then, Ω is a relatively compact subset if and only if the next conditions hold.(i)
The subsetΩ is uniformly bounded, i.e.,(36)∀ν∈Ω,∃R>0∋νE≤R.(ii)
LetI be compact subinterval of 0,∞ and let for all ϵ>0 there exists δ=δϵ>0 such that τ1−τ2<δ for τ1,τ2∈I. Then,(37)ντ11+τ1α−1−ντ21+τ2α−1<ϵ,∀ν∈Ω,(iii)
Forϵ>0 be given, there exists λ=λϵ in since(38)νω11+ω1α−1−νω21+ω2α−1<ϵ,∀ν∈Ω,ω1,ω2>λ.


## 2.1. Fractional Calculus

We introduce some definitions and fundamental rules on the fractional calculus that are needed for the new results [1, 17, 18].Definition 1.
Forζ>0, the Riemann–Liouville fractional integral of a continuous function hτ:0,∞⟶ℝ is defined by(9)Iζhτ=1Γζ∫0ττ−tζ−1htdt.Definition 2.
Forn∈ℕ and n−1≤ζ<n, the Riemann–Liouville fractional derivative of an nth absolutely continuous function hτ:0,∞⟶ℝ is given by(10)DRζhτ=1Γn−ζddτn∫0ττ−tn−ζ−1htdt.Lemma 1.
Lethτ∈L10,∞ and ζ1,ζ2,ζ>0. Then,(11)Iζ1Iζ2hτ=Iζ1+ζ2hτ,DRζIζhτ=hτ.Lemma 2.
LetDRζhτ∈L10,∞,ζ>0. Then,(12)IζDRζhτ=hτ−∑m=1nbmτζ−m,τ∈0,∞,for some bm∈ℝ,m=1,2,…,n,n∈ℕ.


## 2.2. Multivalued Function

In this section, we state some necessary facts on multivalued maps [2, 19–21].LetE,.andH,. be Banach spaces and A:E⟶PclE be a multivalued map. The mapping A is(i)
Convex (closed) ifAe is convex (closed), for every e∈E.(ii)
Completely continuous ifAB is relatively compact, for every B∈PbE.(iii)
Upper semicontinuous ifA−1W is closed subset of E, for each closed subset W⊂E. That is, the set e∈E:Ae⊆O is open for all open sets O⊂E.(iv)
Lower semicontinuous ifA−1Z is open subset of E, for each open subset Z⊂E. In the other words, the set e∈E:Ae∩O≠∅ is open for all open sets O⊂E.(v)
Measurable multivalued if for everye∈E, the function τ⟶de,Aτ=infde,a:a∈Aτ is ℒ− measurable function.Definition 3.
A multivalued mapA:0,∞×ℝ⟶Pℝ is known as a Caratheodory if(1)
For allr∈ℝ,τ⟶Aτ,r is measurable.(2)
For a.eτ∈0,∞,r⟶Aτ,r is upper semicontinuous.
In addition to assumptions (1) and (2), the mapA is L1-Caratheodory if for each k>0,∃ϕk∈C0,∞∩L10,∞ satisfying supτ≥0ϕkτ<+∞ and ϕk>0, that is,(13)Aτ,1+τα−1r=supa:a∈Aτ,1+τα−1r≤ϕkτ,for all r≤k,τ∈0,∞.


## 2.3. Fredholm Operator

We recall some preliminaries about the operators specially Fredholm type where we can get them from [3, 22–24].LetL:E⟶H be any considered operator. Then, we define the following.(1)
The kernel ofL by(14)KerL=e∈E:LE=0H.(2)
The image (rang) ofL by(15)ImL=RL=Le:e∈E⊆H.(3)
The cokernel ofL by(16)CokerL=h:h∈H/ImL.(4)
The index ofF by(17)IndL=dimKerL−dimCokerL.Note that(18)dimCokerL=codimImL.The operatorL is called Fredholm operator if it is a bounded linear operator with finite dimensional kernel and cokernel and closed rang. It is said to be Fredholm of index zero if and only if IndL=0.LetL:domL⊂E⟶H be Fredholm of index zero and P:E⟶E,Q:H⟶H be continuous projections. In the case of(19)ImP=KerL∧ImL=KerQ,E=KerL⊕KerP∧H=ImL⊕ImQ,then LdomL∩KerP is invertible, and so we can define(20)KP=LdomL∩KerP−1.In fact, there is an isomorphic map J:ImQ⟶KerL.Consider the inclusion(21)Lν∈Nν,which is equivalent to(22)ν∈P+JQNν+KpI−QNν.Let C⊂E be a convex and a closed subset. Then, C is called a cone if(i)
ρν∈C,∀ν∈C,ρ≥0(ii)
ν,−ν∈C⟹ν=0Consequently, there is an induced partial order≼ in E subject to(23)o≼ν⟺ν−o∈C.Lemma 3.
LetE be a Banach space contains a cone C. Then, for every η∈C/0, there exists a positive number ση which satisfies(24)ν+η≥σην,ν∈C.Defineμ:E⟶C to be a retraction (continuous mapping constrained with μη=η,∀η∈C). Define the map Θ as(25)Θ=P+JQN+KpI−QN.Then,(i)
Lν∈Nν⟺ν∈P+JQNν+KpI−QNν⟺ν∈Θν(ii)
Θμ=Θ°μTheorem 1.
(O’Regan and Zima). LetC⊂E be a cone, and Ω1andΩ2 be open bounded subsets of E which satisfy Ω1¯⊂Ω2, and C∩Ω2¯/Ω1≠∅. Assume that(1)
L:E⟶H is a Fredholm operator of index zero.(2)
N:E⟶2H is an upper-semicontinuous mapping with nonempty compact convex values.(3)
QN:E⟶2H is bounded on bounded subsets of C, and KpI−QN:E⟶2E is compact on every bounded subset of C.(4)
Lν∉γNν,∀ν∈C∩∂Ω2∩domL and γ∈0,1.(5)
μ subsets of Ω2¯ into bounded subsets of C.(6)
degI−P+JQNμKerL,KerL∩Ω2,0≠0.(7)
There isν0∈C/0 such that(26)ν≤σν0η,∀ν∈Cν0∩∂Ω1,η∈Θν,where(27)Cν0=ν∈C∣ρν0≼ν,for someρ>0,and σν0 is such that ν+ν0≥σν0ν,ν∈C.(8)
P+JQNμ∂Ω2⊂C.(9)
ΘμΩ2¯/Ω1⊂C.(10)
ν∉P+JQNμν,ν∈KerL∩∂Ω2.
Then, the inclusionLν∈Nν can be solvable in the set C∩Ω2¯/Ω1.Remark 1.
Let(28)E=νν∈C0,∞,limτ⟶∞ντ1+τα−1exists,with the norm(29)νE=supτ≥0ντ1+τα−1.
And let(30)H=h∣hτ∈C0,∞∩L10,∞∩Aτ,ντ,supτ≥0|hτ|<+∞,with the norm(31)hH=supτ≥0hτ+∫0∞hτdτ.
Then,E,·EandH,·H are Banach spaces, see [11].
Expand the operatorL, for which(32)domL=ν∈E∣DRακτντ∈C0,∞∩L10,∞,νi0=0,i=0,..,n−2,DRα−1κτνττ=0=limτ⟶∞DRα−1κτντ,where(33)L:domL⊂E⟶H,ν↦DRακτντ,(34)N:E⟶H,ν⟼aτ,ντ.
Then, the multivalue problems (1)–(3) are equivalent to the inclusion:(35)Lν∈Nν,ν∈domL.
We can understand the proof of the following lemma if we see the compactness hypothesis in [11, 25, 26].Lemma 4.
LetΩ be subset of E; then, Ω is a relatively compact subset if and only if the next conditions hold.(i)
The subsetΩ is uniformly bounded, i.e.,(36)∀ν∈Ω,∃R>0∋νE≤R.(ii)
LetI be compact subinterval of 0,∞ and let for all ϵ>0 there exists δ=δϵ>0 such that τ1−τ2<δ for τ1,τ2∈I. Then,(37)ντ11+τ1α−1−ντ21+τ2α−1<ϵ,∀ν∈Ω,(iii)
Forϵ>0 be given, there exists λ=λϵ in since(38)νω11+ω1α−1−νω21+ω2α−1<ϵ,∀ν∈Ω,ω1,ω2>λ.


## 3. Acquired Upshots

In this section, we discuss some results for the sake of seeing the attainability of solution by apply (O’Regan and Zima, Theorem1) on the half line. Consequently, in view of the arguments of Fredholm operator, we first need to explore some lemmas and facts as follows.


### 3.1. Basic Requirements

Here, by using the general form of solution, we prove several facts in order to explore that(ℋ1) L is Fredholm operator of index zero.Also, we explain that this operator is invertible in the setdomL∩KerP to compute the formula of its inverse Kp. Finally, under the version of Lemma 3 (i) and by the known maps J,Kp,Q,andI, we calculate the operator JQ+KPI−Q via green functions and show some related inequalities.Lemma 5.
Considerhτ∈H. Then, ν∈E is a solution of the boundary value problem:(39)DRακτντ=hτ,τ∈0,∞,νi0=0,i=0,…,n−2,DRα−1κ0ν0=limτ⟶∞DRα−1κτντ,if and only if(40)ντ=bτα−1κτ+1Γακτ∫0ττ−tα−1htdt,where b∈ℝ and(41)∫0∞hτdτ=0.Proof.
The proof can be completely seen if we are backing to both Lemmas1 and 2 and applying accompanying conditions.
Now, using (33)–(35), we obtain(42)Lν=DRακτντ,Nν=aaτ,ντ=qτβτ,ντ,βτ∈Bτ,ντ,which implies that(43)KerL=νν=bτα−1κτ,τ≥0,b∈ℝ,(44)ImL=hτ∈H∣∫0∞hτdτ=0⊂H.
By using the equation,(45)κτντ=bτα−1,we can define a continuous projection P on E as in the next Lemma.Lemma 6.
LetP:E⟶E be a map defined as(46)Pντ=1Γα∫0∞e−tκtνtdtτα−1κτ.Then,P is a continuous projection on E.Proof.
Letν∈E and P be defined as in (46). Then,(47)P2ντ=PPντ,=P1Γα∫0∞e−tκtνtdtτα−1κτ,=1Γα∫0∞e−tκtνtdtPτα−1κτ,=1Γα∫0∞e−tκtνtdtτα−1κτ,=Pντ,which completes the proof.Remark 2.
We can recall that(48)KerP=ν∣∫0∞e−tκtνtdt=0,(49)ImP=KerL.
Immediately, we have(50)E=KerP⊕ImP=KerP⊕KerL.
According to both integrals∫0∞e−tdt=1 and ∫0∞htdt=0 in (41), we can explain and prove that there exists a continuous projection Q on the space H. See the next lemma.Lemma 7.
LetQ:H⟶H be a mapping formed by(51)Qhτ=e−τ∫0∞htdt,where τ≥0,hτ∈H. Then, Q is a continuous projection on H.Remark 3.
We can make sure that(52)ImQ=be−τ,b∈ℝ,τ≥0,(53)KerQ=ImL.
Claim thatH=KerQ⊕ImQ=ImL⊕ImQ. For that, take hτ∈H. Then, we have(54)hτ=hτ−Qhτ+Qhτ=I−Qhτ+Qhτ,where Qhτ∈ImQ. Moreover,(55)QI−Qhτ=Qhτ−Q2hτ,=Qhτ−Qhτ=0.
Thus,H=KerQ+ImQ. Assume that hτ∈KerQ∩ImQ. Then,(56)hτ=be−τ,Qhτ=0,⟹be−τ∫0∞e−τdτ=0,⟹b=0,⟹hτ=0.
Hence,H=ImL⊕ImQ. It remains to prove that L is of index zero. Due to the previous results of Q, we find that(57)1=dimℝ=dimImQ=dimCokerQ=codimImL<+∞.
SinceImL is closed in H, it implies that L is the Fredholm operator of index zero.
Furthermore, we can see clearly thatKerL≅ImQ if we define the isomorphism J:ImQ⟶KerL by(58)Jbe−τ=bτα−1κτ,τ≥0,b∈ℝ,as long as ImQ defined by (52).
Once we haveL is the Fredholm operator of index zero, then it is invertible in the set domL∩KerP. Create L∣domL∩KerP=LP:domL∩KerP⟶ImL and LP−1=KP. Let ντ∈domL∩KerP, which follows that ντ is defined by (40) and satisfying the condition in (48). By substituting (40) in the equation given in (48), the constant b will be subject to(59)b=−1Γα∫0∞e−t∫0tt−sα−1Γαhsdsdt.
Therefore,(60)KPhτ=−1Γα∫0∞e−t∫0tt−sα−1Γαhsdsdtτα−1κτ+1κτ∫0ττ−tα−1Γαhtdt,hτ∈ImL.Lemma 8.
LetL be the Fredholm operator of index zero. That is,(61)Lν=DRακτντ,L∣domL∩KerP=LP,where domL,P and KerP are defined by (32), (46), and (48), respectively. Let KP be elected such as in (60). Then,(62)KPLP=LPKP=I.Proof.
Optingντ∈domL∩KerP drives(63)KPLPντ=KPLPντ=−1Γα∫0∞e−t∫0tt−sα−1ΓαLνsdsdtτα−1κτ+1κτ∫0ττ−tα−1ΓαLνtdt=−1Γα∫0∞e−t∫0tt−sα−1ΓαDRακsνsdsdt×τα−1κτ+bΓα∫0∞e−ttα−1dtτα−1κτ+ντ−bτα−1κτ=ντ.
And use (43) to verify that(64)LPKPhτ=LPKPhτ=L−1Γα∫0∞e−t∫0tt−sα−1Γαhsdsdtτα−1κτ+1κτ∫0ττ−tα−1Γαhtdt=DRακτ1κτ∫0ττ−tα−1Γαhtdt=hτ.
Therefore,KPLP=LPKP=I.
According to Lemma3 (i), we need to compute the value:(65)JQ+KPI−Qhτ,hτ∈Nν,which can be introduced by(66)JQ+KPI−Qhτ=h∗∈H∣h∗τ=τα−1Γακτ∫0∞G1τ,thtdt−∫0∞∫0τG2τ,thtdtdτ,ht∈Nν,where(67)G1τ,t=0,τ=0,Γα−∫0ττ−zα−1τα−1e−zdz+∫0∞∫0tG2t,me−mdmdt+τ−tα−1τα−1,τ≠0,0≤t≤τ<+∞,Γα−∫0ττ−zα−1τα−1e−zdz+∫0∞∫0tG2t,me−mdmdt,0<τ≤t<+∞,G2τ,t=0,0<τ≤t<+∞,τ−tα−1τα−1e−τ,0≤t≤τ<+∞.Lemma 9.
We have(68)0≤G2τ,t≤τα−1Γαe−τ,(69)0<Γα−1+12α+1≤G1τ,t≤1+α+Γα,∀α>1.Proof.
Basically, we have the following inequalities:(70)0<∫0ττ−zα−1τα−1e−zdz≤∫0∞e−zdz=1,0≤G2τ,t≤τα−1Γαe−τ.
In order to prove the L.H.S of (69), we note that, for all positive reals m<t⟹e−m>e−t, to imply that(71)∫0∞∫0tG2t,me−mdmdt≥∫0∞∫0tG2t,me−tdmdt=1Γα+1∫0∞tαe−2tdt=12α+1,which leads to(72)Γα−1+12α+1≤G1τ,t.
By analytical methods, we can see that(73)0<Γα−1+12α+1,∀α>1.
Now, to make sure about the R.H.S of (69), use(74)−∫0ττ−zα−1τα−1e−zdz<0.
Then,(75)G1τ,t≤Γα+1+∫0∞∫0tG2t,me−mdmdt≤Γα+1+∫0∞tα−1Γαe−t∫0te−mdmdt≤Γα+1+∫0∞tαΓαe−tdt=Γα+1+α.


### 3.2. Essential Theorem

Apply all results in the first part to illustrate the main theorem with some needed hypothesis as follows.Theorem 2.
Reckoning in addition of (ℋ1) that the following.(ℋ2) Let B:0.∞×ℝ⟶BCCℝ be L1−Caratheodory (Definition 3) with SB,ν which has nonempty compact convex values conditional on which there exist nonnegative functions γiτ,i=1,2,3,ηjτ,j=1,2 and θτ satisfying(76)βτ,ντ≤−γ1τ|βτ,ντ|+γ2τντ1+τα−1+γ3τ,(77)−θτντ1+τα−1≤βτ,ντ≤−η1τντ1+τα−1+η2τ,for all βτ∈Bτ,ντ, and 0<ντ/1+τα−1<R.(ℋ3) For aτ,ντ=qτβτ,ντ, we have(78)∫0∞qτθτdτ<Γα+12Γα+1/2Γα+α+1,(79)eτqτθτ<1+τα−1Γα+α+1,and there exists δ>0 in which(80)δ=ΓαΓα+1/2+Γα+α+1Γα+1∫0∞qτθτdτ<1.(ℋ4) There exist positive constants q0,γ0,Γ0,andR0, where as(81)Γ0=2γ0supτ≥0γ2τ+e−τκτγ01+τα−1/2qτη1τ,(82)R0=maxq0Γ0Γα∫0∞η2τdτ,2q0γ0Γα∫0∞γ3τdτ,q0η0∫0∞η2τdτ,where q0=supτ≥0qτ,γ0=infτ≥0γ1τ, and(83)η0=∫0∞qtη1ttα−11+tα−1κtdt.Then, it is distinctly possible to find positive solution for problems (1)–(3) in C∩Ω2¯/Ω1.Proof.
LetE and H be Banach spaces defined as in Remark. Define the linear operator L and its property by (32), (33), (43), and (44), and the nonlinear operator N by (34) as well as (35) holds. Then, to show that if it is possible for the given problem to has a solution, we will prove all points of (O’Regan and Zima, Theorem 1-).Step 1: the operatorL will be the Fredholm operator of index zero by using (ℋ1).Step 2: by (ℋ2), the operator (N) is upper semicontinuous mapping with nonempty compact convex values.Step 3: according to (ℋ2), QN is bounded in the bounded set and(84)KPI−QNis compact, where P,Q,andKP are all conformed by (46), (51), and (60), respectively.Define the needed setsΩ1,Ω2,andC as follows:(85)Ω1=ν∈EϵνE<ντ1+τα−1<R1,R1∈0,R2,ϵ∈δ,1,Ω2=ν∈EνE<R2,R2∈R0,R,C=ν∈Eντ≥0,τ≥0⊂E.Clearly,Ω1¯⊂Ω2 and C forms a cone in E.Step 4: claim thatLν∉γNν,∀ν∈C∩∂Ω2∩domL,γ∈0,1.
Assume by contradiction thatν≥0,νE=R2 such that Lν∈γNν. Since(86)ν=ν−Pν+Pν=KPLI−Pν+Pν=KPLν+Pν,Lν=γhτ,hτ∈Nν,we have, by using Hölder’s inequality with p=1 and q=∞, (66) and (68) that(87)ντ1+τα−1=τα−1Γακτ1+τα−1∫0∞e−tκtνtdt−∫0∞e−tIαDRακtνtdt+1κτ1+τα−1IαDRακτντ≤1Γα∫0∞e−tκtνtdt+τα−1Γακτ1+τα−11Γα∫0∞e−ttα−1∫0tDRακsνsdsdt+∫0∞DRακtνtdt≤1Γα∫0∞e−tκtνtdt+2Γα∫0∞DRακtνtdt.
In case(88)DRακτντ=Lν=γaτ,ντ=γqτβτ,ντ,we can see by (ℋ2 (76) and (77)) that(89)DRακτντ=γqτβτ,ντ≤−γγ1τqτβτ,ντ+γγ2τqτντ1+τα−1+γγ3τqτ≤−γ1τDRακτντ+γ2τqτντ1+τα−1+γ3τqτ,(90)DRακτντ=γqτβτ,ντ≤−γη1τqτντ1+τα−1+γqτη2τ.
Now, we have(91)∫0∞DRακtνtdt=∫0∞DDRα−1κtνtdt=limt⟶∞DRα−1κtνt−DRα−1κtνt∣t=0=0.
Relations (89), (90), and (91) give the ensuing results:(92)0=∫0∞DRακtνtdt≤−∫0∞γ1tDRακtνtdt+∫0∞γ2tqtνt1+tα−1dt+∫0∞γ3tqtdt,(93)0=∫0∞DRακtνtdt≤−∫0∞γη1tqtνt1+tα−1dt+∫0∞γη2tqtdt,that are equivalent, respectively, to(94)∫0∞DRακtνtdt≤1γ0∫0∞γ2tqtνt1+tα−1dt+1γ0∫0∞γ3tqtdt,(95)∫0∞η1tqtνt1+tα−1dt≤∫0∞η2tqtdt.
From (95), it can be seen that(96)∫0∞qtη1tνt1+tα−1dt≤q0∫0∞η2tdt.
The relations (81) and (94)–(96) together with (ℋ4) show that (87) will be equivalent to(97)ντ1+τα−1≤2q0γ0Γα∫0∞γ2tνt1+tα−1dt+2q0γ0Γα∫0∞γ3tdt+∫0∞e−tκtνtdt≤Γ0Γα∫0∞qtη1tνt1+tα−1dt+2q0γ0Γα∫0∞γ3tdt≤q0Γ0Γα∫0∞η2tdt+2q0γ0Γα∫0∞γ3tdt<R0<R2<R,which is contradiction to ν∈C∩∂Ω2∩domL.Step 5: to recall thatμ maps subsets of Ω2¯ into bounded sets of C, let ν∈Ω2¯. Since μντ=ντ=ντ,∀ν∈C, then(98)μνC=μνE=νE≤R2,∀ν∈Ω2¯.Step 6: this step will show that(99)degI−P+JQNμKerL,KerL∩Ω2,0≠0.
Define the mapH^ν,ρ by(100)H^ν,ρ=I−ρP+JQNμν,ρ∈0,1.
Letν∈KerL∩Ω2¯, and consider by contradiction that 0∈H^ν,ρ. Since ν∈KerL, then it is given by the form(101)ντ=bτα−1κτ,τ≥0,b∈ℝ.
Consequently, we have(102)H^ν,ρ=I−ρP+JQNμbτα−1κτ=b−ρb−ρ∫0∞qtft,btα−1κtdtτα−1κτ,where b∈−R2,R2,ρ∈0,1. Now, expound a map(103)J1:KerL∩Ω2¯⟶ℝ,such as(104)J1bτα−1κτ=b.
Then,J1,J (J as in (58)) are used to see that(105)dBH^bτα−1κτ,ρ,KerL∩Ω2,0=dBJ1H^J1−1b,ρ,J1KerL∩Ω2,0.so that(106)0∈J1H^J1−1b,ρ.
Thus, for allaτ,ντ=qτβτ,ντ∈Aτ,ντ, we have(107)b−ρb−ρ∫0∞qtβt,btα−1κtdt=0⟺b=ρb+∫0∞qtβt,btα−1κtdt.
Using (ℋ2 (77)) and (ℋ3 (78)) to explain that b≥0, take b0∈J1KerL∩Ω2, then b0=R2. Suppose 0∈J1H^J1−1b,ρ, which leads to b0=R2. Together with (ℋ2 (77)) and (ℋ4 (82)), we work out that ∃aτ,ντ=qτβτ,R2τα−1/κτ∈Aτ,R2τα−1/κτ, satisfying(108)R2=ρR2+∫0∞qtβt,R2tα−1κtdt≤ρR21−∫0∞qtη1ttα−11+tα−1κtdt+∫0∞qtη2tdt≤ρR2≤R2.
Ifρ=0, then R2=0 is impossible. This is the cause of 0∉J1H^J1−1b,ρ for all b∈J1KerL∩Ω2,ρ∈0,1. Therefore,(109)degI−P+JQNμKerL,KerL∩Ω2,0=degH^.,1,KerL∩Ω2,0=dBJ1H^J1−1b,1,J1KerL∩Ω2,0=dBI,J1KerL∩Ω2,0=1≠0.Step 7: in this step, our aim is proving that there isν0∈C/0, where(110)ν≤σν0η,∀ν∈Cν0∩∂Ω1,η∈Θν,where(111)Cν0=ν∈C∣ρν0≼ν,for someρ>0,and σν0 that is ν+ν0≥σν0ν,ν∈C.
In order to see that takeν0τ=1+τα−1,σν0=1, and(112)Cν0=ν∈C∣infτ≥0ντ1+τα−1>0.
Now, let(113)τ0>0∋τ0α−1κτ01+τ0α−1>Γα+1/2Γα+1.
Then, forν∈Cν0∩∂Ω1, we have(114)νE≤R1<R2,ντ1+τα−1≥ϵνE,ϵ∈δ,1.
And use (66), (68), (69), (ℋ2 (77)), and (ℋ4 (78), (80)) to show that(115)Θντ01+τ0α−1=11+τ0α−1Pντ0+JQN+KPI−QNντ0=τ0α−1Γακτ01+τ0α−1∫0∞e−tκtνtdt+∫0∞G1τ0,tqtβt,νtdt−∫0∞∫0τ0G2τ0,tqtβt,νtdtdτ0≥τ0α−1κτ01+τ0α−1ϵνE1+ΓαΓα−νE1+α+ΓαΓα∫0∞qtθtdt≥Γα+1/2Γα+1νEδ1+ΓαΓα−1+α+ΓαΓα∫0∞qtθtdt=νE.
Then, we conclude that(116)νE≤Θντ01+τ0α−1≤Θντ0≤ΘνH≤σν0ΘνH.Step 8: here, we need to explain thatP+JQNμ∂Ω2⊂C is held, i.e., ∀ν∈∂Ω2, and we have P+JQNντ∈C:(117)P+JQNντ=τα−1κτ1Γα∫0∞e−tκtνtdt+∫0∞qtβt,νtdt≥τα−1κτ∫0∞e−t1+tα−1−qtθtνt1+tα−1dt≥0.
That makesP+JQNμ∂Ω2⊂C.Step 9: by this step,ΘμΩ2¯/Ω1⊂C will be satisfied. For that take ν∈Ω2¯/Ω1. It follows that R1≤νt≤R2. Now, use (ℋ4 (79)), we obtain(118)Θνt≥τα−1Γακτ∫0∞e−t1+tα−1−Γα+α+1qtθtνt1+tα−1dt≥0,which means that ΘμΩ2¯/Ω1⊂C is hold.Step 10: this step proves thatν∉P+JQNμν,∀ν∈KerL∩∂Ω2 will be true. Consider ν∈KerL∩∂Ω2, then b=R2. Moreover,(119)P+JQNbτα−1κτ=τα−1κτ1Γα∫0∞e−tκtbtα−1κtdt+∫0∞qtβt,btα−1κtdt≤τα−1κτR2−R2∫0∞qtη1ttα−11+tα−1κtdt+∫0∞qtη2tdt≤R2.
So,ν∉P+JQNμν,ν∈KerL∩∂Ω2 is true.
Hence, problem (39), the steps (Step 1–Step 10), and (O’Regan and Zima, Theorem 1) together show that our problem (1)–(3) can attain a positive solution at resonance in the set C∩Ω2¯/Ω1.


## 3.1. Basic Requirements

Here, by using the general form of solution, we prove several facts in order to explore that(ℋ1) L is Fredholm operator of index zero.Also, we explain that this operator is invertible in the setdomL∩KerP to compute the formula of its inverse Kp. Finally, under the version of Lemma 3 (i) and by the known maps J,Kp,Q,andI, we calculate the operator JQ+KPI−Q via green functions and show some related inequalities.Lemma 5.
Considerhτ∈H. Then, ν∈E is a solution of the boundary value problem:(39)DRακτντ=hτ,τ∈0,∞,νi0=0,i=0,…,n−2,DRα−1κ0ν0=limτ⟶∞DRα−1κτντ,if and only if(40)ντ=bτα−1κτ+1Γακτ∫0ττ−tα−1htdt,where b∈ℝ and(41)∫0∞hτdτ=0.Proof.
The proof can be completely seen if we are backing to both Lemmas1 and 2 and applying accompanying conditions.
Now, using (33)–(35), we obtain(42)Lν=DRακτντ,Nν=aaτ,ντ=qτβτ,ντ,βτ∈Bτ,ντ,which implies that(43)KerL=νν=bτα−1κτ,τ≥0,b∈ℝ,(44)ImL=hτ∈H∣∫0∞hτdτ=0⊂H.
By using the equation,(45)κτντ=bτα−1,we can define a continuous projection P on E as in the next Lemma.Lemma 6.
LetP:E⟶E be a map defined as(46)Pντ=1Γα∫0∞e−tκtνtdtτα−1κτ.Then,P is a continuous projection on E.Proof.
Letν∈E and P be defined as in (46). Then,(47)P2ντ=PPντ,=P1Γα∫0∞e−tκtνtdtτα−1κτ,=1Γα∫0∞e−tκtνtdtPτα−1κτ,=1Γα∫0∞e−tκtνtdtτα−1κτ,=Pντ,which completes the proof.Remark 2.
We can recall that(48)KerP=ν∣∫0∞e−tκtνtdt=0,(49)ImP=KerL.
Immediately, we have(50)E=KerP⊕ImP=KerP⊕KerL.
According to both integrals∫0∞e−tdt=1 and ∫0∞htdt=0 in (41), we can explain and prove that there exists a continuous projection Q on the space H. See the next lemma.Lemma 7.
LetQ:H⟶H be a mapping formed by(51)Qhτ=e−τ∫0∞htdt,where τ≥0,hτ∈H. Then, Q is a continuous projection on H.Remark 3.
We can make sure that(52)ImQ=be−τ,b∈ℝ,τ≥0,(53)KerQ=ImL.
Claim thatH=KerQ⊕ImQ=ImL⊕ImQ. For that, take hτ∈H. Then, we have(54)hτ=hτ−Qhτ+Qhτ=I−Qhτ+Qhτ,where Qhτ∈ImQ. Moreover,(55)QI−Qhτ=Qhτ−Q2hτ,=Qhτ−Qhτ=0.
Thus,H=KerQ+ImQ. Assume that hτ∈KerQ∩ImQ. Then,(56)hτ=be−τ,Qhτ=0,⟹be−τ∫0∞e−τdτ=0,⟹b=0,⟹hτ=0.
Hence,H=ImL⊕ImQ. It remains to prove that L is of index zero. Due to the previous results of Q, we find that(57)1=dimℝ=dimImQ=dimCokerQ=codimImL<+∞.
SinceImL is closed in H, it implies that L is the Fredholm operator of index zero.
Furthermore, we can see clearly thatKerL≅ImQ if we define the isomorphism J:ImQ⟶KerL by(58)Jbe−τ=bτα−1κτ,τ≥0,b∈ℝ,as long as ImQ defined by (52).
Once we haveL is the Fredholm operator of index zero, then it is invertible in the set domL∩KerP. Create L∣domL∩KerP=LP:domL∩KerP⟶ImL and LP−1=KP. Let ντ∈domL∩KerP, which follows that ντ is defined by (40) and satisfying the condition in (48). By substituting (40) in the equation given in (48), the constant b will be subject to(59)b=−1Γα∫0∞e−t∫0tt−sα−1Γαhsdsdt.
Therefore,(60)KPhτ=−1Γα∫0∞e−t∫0tt−sα−1Γαhsdsdtτα−1κτ+1κτ∫0ττ−tα−1Γαhtdt,hτ∈ImL.Lemma 8.
LetL be the Fredholm operator of index zero. That is,(61)Lν=DRακτντ,L∣domL∩KerP=LP,where domL,P and KerP are defined by (32), (46), and (48), respectively. Let KP be elected such as in (60). Then,(62)KPLP=LPKP=I.Proof.
Optingντ∈domL∩KerP drives(63)KPLPντ=KPLPντ=−1Γα∫0∞e−t∫0tt−sα−1ΓαLνsdsdtτα−1κτ+1κτ∫0ττ−tα−1ΓαLνtdt=−1Γα∫0∞e−t∫0tt−sα−1ΓαDRακsνsdsdt×τα−1κτ+bΓα∫0∞e−ttα−1dtτα−1κτ+ντ−bτα−1κτ=ντ.
And use (43) to verify that(64)LPKPhτ=LPKPhτ=L−1Γα∫0∞e−t∫0tt−sα−1Γαhsdsdtτα−1κτ+1κτ∫0ττ−tα−1Γαhtdt=DRακτ1κτ∫0ττ−tα−1Γαhtdt=hτ.
Therefore,KPLP=LPKP=I.
According to Lemma3 (i), we need to compute the value:(65)JQ+KPI−Qhτ,hτ∈Nν,which can be introduced by(66)JQ+KPI−Qhτ=h∗∈H∣h∗τ=τα−1Γακτ∫0∞G1τ,thtdt−∫0∞∫0τG2τ,thtdtdτ,ht∈Nν,where(67)G1τ,t=0,τ=0,Γα−∫0ττ−zα−1τα−1e−zdz+∫0∞∫0tG2t,me−mdmdt+τ−tα−1τα−1,τ≠0,0≤t≤τ<+∞,Γα−∫0ττ−zα−1τα−1e−zdz+∫0∞∫0tG2t,me−mdmdt,0<τ≤t<+∞,G2τ,t=0,0<τ≤t<+∞,τ−tα−1τα−1e−τ,0≤t≤τ<+∞.Lemma 9.
We have(68)0≤G2τ,t≤τα−1Γαe−τ,(69)0<Γα−1+12α+1≤G1τ,t≤1+α+Γα,∀α>1.Proof.
Basically, we have the following inequalities:(70)0<∫0ττ−zα−1τα−1e−zdz≤∫0∞e−zdz=1,0≤G2τ,t≤τα−1Γαe−τ.
In order to prove the L.H.S of (69), we note that, for all positive reals m<t⟹e−m>e−t, to imply that(71)∫0∞∫0tG2t,me−mdmdt≥∫0∞∫0tG2t,me−tdmdt=1Γα+1∫0∞tαe−2tdt=12α+1,which leads to(72)Γα−1+12α+1≤G1τ,t.
By analytical methods, we can see that(73)0<Γα−1+12α+1,∀α>1.
Now, to make sure about the R.H.S of (69), use(74)−∫0ττ−zα−1τα−1e−zdz<0.
Then,(75)G1τ,t≤Γα+1+∫0∞∫0tG2t,me−mdmdt≤Γα+1+∫0∞tα−1Γαe−t∫0te−mdmdt≤Γα+1+∫0∞tαΓαe−tdt=Γα+1+α.


## 3.2. Essential Theorem

Apply all results in the first part to illustrate the main theorem with some needed hypothesis as follows.Theorem 2.
Reckoning in addition of (ℋ1) that the following.(ℋ2) Let B:0.∞×ℝ⟶BCCℝ be L1−Caratheodory (Definition 3) with SB,ν which has nonempty compact convex values conditional on which there exist nonnegative functions γiτ,i=1,2,3,ηjτ,j=1,2 and θτ satisfying(76)βτ,ντ≤−γ1τ|βτ,ντ|+γ2τντ1+τα−1+γ3τ,(77)−θτντ1+τα−1≤βτ,ντ≤−η1τντ1+τα−1+η2τ,for all βτ∈Bτ,ντ, and 0<ντ/1+τα−1<R.(ℋ3) For aτ,ντ=qτβτ,ντ, we have(78)∫0∞qτθτdτ<Γα+12Γα+1/2Γα+α+1,(79)eτqτθτ<1+τα−1Γα+α+1,and there exists δ>0 in which(80)δ=ΓαΓα+1/2+Γα+α+1Γα+1∫0∞qτθτdτ<1.(ℋ4) There exist positive constants q0,γ0,Γ0,andR0, where as(81)Γ0=2γ0supτ≥0γ2τ+e−τκτγ01+τα−1/2qτη1τ,(82)R0=maxq0Γ0Γα∫0∞η2τdτ,2q0γ0Γα∫0∞γ3τdτ,q0η0∫0∞η2τdτ,where q0=supτ≥0qτ,γ0=infτ≥0γ1τ, and(83)η0=∫0∞qtη1ttα−11+tα−1κtdt.Then, it is distinctly possible to find positive solution for problems (1)–(3) in C∩Ω2¯/Ω1.Proof.
LetE and H be Banach spaces defined as in Remark. Define the linear operator L and its property by (32), (33), (43), and (44), and the nonlinear operator N by (34) as well as (35) holds. Then, to show that if it is possible for the given problem to has a solution, we will prove all points of (O’Regan and Zima, Theorem 1-).Step 1: the operatorL will be the Fredholm operator of index zero by using (ℋ1).Step 2: by (ℋ2), the operator (N) is upper semicontinuous mapping with nonempty compact convex values.Step 3: according to (ℋ2), QN is bounded in the bounded set and(84)KPI−QNis compact, where P,Q,andKP are all conformed by (46), (51), and (60), respectively.Define the needed setsΩ1,Ω2,andC as follows:(85)Ω1=ν∈EϵνE<ντ1+τα−1<R1,R1∈0,R2,ϵ∈δ,1,Ω2=ν∈EνE<R2,R2∈R0,R,C=ν∈Eντ≥0,τ≥0⊂E.Clearly,Ω1¯⊂Ω2 and C forms a cone in E.Step 4: claim thatLν∉γNν,∀ν∈C∩∂Ω2∩domL,γ∈0,1.
Assume by contradiction thatν≥0,νE=R2 such that Lν∈γNν. Since(86)ν=ν−Pν+Pν=KPLI−Pν+Pν=KPLν+Pν,Lν=γhτ,hτ∈Nν,we have, by using Hölder’s inequality with p=1 and q=∞, (66) and (68) that(87)ντ1+τα−1=τα−1Γακτ1+τα−1∫0∞e−tκtνtdt−∫0∞e−tIαDRακtνtdt+1κτ1+τα−1IαDRακτντ≤1Γα∫0∞e−tκtνtdt+τα−1Γακτ1+τα−11Γα∫0∞e−ttα−1∫0tDRακsνsdsdt+∫0∞DRακtνtdt≤1Γα∫0∞e−tκtνtdt+2Γα∫0∞DRακtνtdt.
In case(88)DRακτντ=Lν=γaτ,ντ=γqτβτ,ντ,we can see by (ℋ2 (76) and (77)) that(89)DRακτντ=γqτβτ,ντ≤−γγ1τqτβτ,ντ+γγ2τqτντ1+τα−1+γγ3τqτ≤−γ1τDRακτντ+γ2τqτντ1+τα−1+γ3τqτ,(90)DRακτντ=γqτβτ,ντ≤−γη1τqτντ1+τα−1+γqτη2τ.
Now, we have(91)∫0∞DRακtνtdt=∫0∞DDRα−1κtνtdt=limt⟶∞DRα−1κtνt−DRα−1κtνt∣t=0=0.
Relations (89), (90), and (91) give the ensuing results:(92)0=∫0∞DRακtνtdt≤−∫0∞γ1tDRακtνtdt+∫0∞γ2tqtνt1+tα−1dt+∫0∞γ3tqtdt,(93)0=∫0∞DRακtνtdt≤−∫0∞γη1tqtνt1+tα−1dt+∫0∞γη2tqtdt,that are equivalent, respectively, to(94)∫0∞DRακtνtdt≤1γ0∫0∞γ2tqtνt1+tα−1dt+1γ0∫0∞γ3tqtdt,(95)∫0∞η1tqtνt1+tα−1dt≤∫0∞η2tqtdt.
From (95), it can be seen that(96)∫0∞qtη1tνt1+tα−1dt≤q0∫0∞η2tdt.
The relations (81) and (94)–(96) together with (ℋ4) show that (87) will be equivalent to(97)ντ1+τα−1≤2q0γ0Γα∫0∞γ2tνt1+tα−1dt+2q0γ0Γα∫0∞γ3tdt+∫0∞e−tκtνtdt≤Γ0Γα∫0∞qtη1tνt1+tα−1dt+2q0γ0Γα∫0∞γ3tdt≤q0Γ0Γα∫0∞η2tdt+2q0γ0Γα∫0∞γ3tdt<R0<R2<R,which is contradiction to ν∈C∩∂Ω2∩domL.Step 5: to recall thatμ maps subsets of Ω2¯ into bounded sets of C, let ν∈Ω2¯. Since μντ=ντ=ντ,∀ν∈C, then(98)μνC=μνE=νE≤R2,∀ν∈Ω2¯.Step 6: this step will show that(99)degI−P+JQNμKerL,KerL∩Ω2,0≠0.
Define the mapH^ν,ρ by(100)H^ν,ρ=I−ρP+JQNμν,ρ∈0,1.
Letν∈KerL∩Ω2¯, and consider by contradiction that 0∈H^ν,ρ. Since ν∈KerL, then it is given by the form(101)ντ=bτα−1κτ,τ≥0,b∈ℝ.
Consequently, we have(102)H^ν,ρ=I−ρP+JQNμbτα−1κτ=b−ρb−ρ∫0∞qtft,btα−1κtdtτα−1κτ,where b∈−R2,R2,ρ∈0,1. Now, expound a map(103)J1:KerL∩Ω2¯⟶ℝ,such as(104)J1bτα−1κτ=b.
Then,J1,J (J as in (58)) are used to see that(105)dBH^bτα−1κτ,ρ,KerL∩Ω2,0=dBJ1H^J1−1b,ρ,J1KerL∩Ω2,0.so that(106)0∈J1H^J1−1b,ρ.
Thus, for allaτ,ντ=qτβτ,ντ∈Aτ,ντ, we have(107)b−ρb−ρ∫0∞qtβt,btα−1κtdt=0⟺b=ρb+∫0∞qtβt,btα−1κtdt.
Using (ℋ2 (77)) and (ℋ3 (78)) to explain that b≥0, take b0∈J1KerL∩Ω2, then b0=R2. Suppose 0∈J1H^J1−1b,ρ, which leads to b0=R2. Together with (ℋ2 (77)) and (ℋ4 (82)), we work out that ∃aτ,ντ=qτβτ,R2τα−1/κτ∈Aτ,R2τα−1/κτ, satisfying(108)R2=ρR2+∫0∞qtβt,R2tα−1κtdt≤ρR21−∫0∞qtη1ttα−11+tα−1κtdt+∫0∞qtη2tdt≤ρR2≤R2.
Ifρ=0, then R2=0 is impossible. This is the cause of 0∉J1H^J1−1b,ρ for all b∈J1KerL∩Ω2,ρ∈0,1. Therefore,(109)degI−P+JQNμKerL,KerL∩Ω2,0=degH^.,1,KerL∩Ω2,0=dBJ1H^J1−1b,1,J1KerL∩Ω2,0=dBI,J1KerL∩Ω2,0=1≠0.Step 7: in this step, our aim is proving that there isν0∈C/0, where(110)ν≤σν0η,∀ν∈Cν0∩∂Ω1,η∈Θν,where(111)Cν0=ν∈C∣ρν0≼ν,for someρ>0,and σν0 that is ν+ν0≥σν0ν,ν∈C.
In order to see that takeν0τ=1+τα−1,σν0=1, and(112)Cν0=ν∈C∣infτ≥0ντ1+τα−1>0.
Now, let(113)τ0>0∋τ0α−1κτ01+τ0α−1>Γα+1/2Γα+1.
Then, forν∈Cν0∩∂Ω1, we have(114)νE≤R1<R2,ντ1+τα−1≥ϵνE,ϵ∈δ,1.
And use (66), (68), (69), (ℋ2 (77)), and (ℋ4 (78), (80)) to show that(115)Θντ01+τ0α−1=11+τ0α−1Pντ0+JQN+KPI−QNντ0=τ0α−1Γακτ01+τ0α−1∫0∞e−tκtνtdt+∫0∞G1τ0,tqtβt,νtdt−∫0∞∫0τ0G2τ0,tqtβt,νtdtdτ0≥τ0α−1κτ01+τ0α−1ϵνE1+ΓαΓα−νE1+α+ΓαΓα∫0∞qtθtdt≥Γα+1/2Γα+1νEδ1+ΓαΓα−1+α+ΓαΓα∫0∞qtθtdt=νE.
Then, we conclude that(116)νE≤Θντ01+τ0α−1≤Θντ0≤ΘνH≤σν0ΘνH.Step 8: here, we need to explain thatP+JQNμ∂Ω2⊂C is held, i.e., ∀ν∈∂Ω2, and we have P+JQNντ∈C:(117)P+JQNντ=τα−1κτ1Γα∫0∞e−tκtνtdt+∫0∞qtβt,νtdt≥τα−1κτ∫0∞e−t1+tα−1−qtθtνt1+tα−1dt≥0.
That makesP+JQNμ∂Ω2⊂C.Step 9: by this step,ΘμΩ2¯/Ω1⊂C will be satisfied. For that take ν∈Ω2¯/Ω1. It follows that R1≤νt≤R2. Now, use (ℋ4 (79)), we obtain(118)Θνt≥τα−1Γακτ∫0∞e−t1+tα−1−Γα+α+1qtθtνt1+tα−1dt≥0,which means that ΘμΩ2¯/Ω1⊂C is hold.Step 10: this step proves thatν∉P+JQNμν,∀ν∈KerL∩∂Ω2 will be true. Consider ν∈KerL∩∂Ω2, then b=R2. Moreover,(119)P+JQNbτα−1κτ=τα−1κτ1Γα∫0∞e−tκtbtα−1κtdt+∫0∞qtβt,btα−1κtdt≤τα−1κτR2−R2∫0∞qtη1ttα−11+tα−1κtdt+∫0∞qtη2tdt≤R2.
So,ν∉P+JQNμν,ν∈KerL∩∂Ω2 is true.
Hence, problem (39), the steps (Step 1–Step 10), and (O’Regan and Zima, Theorem 1) together show that our problem (1)–(3) can attain a positive solution at resonance in the set C∩Ω2¯/Ω1.


## 4. An Example

Example 1.
Taking problems (1)–(3) with respect to(120)α=52,κτ=2,Aτ,ντ=−130e−τντ+1λ21+τ3/21+τ2λ=1m,λ∈ℕ.
First, ifaτ,ντ∈Aτ,ντ, we obtain(121)aτ,ντ=−130e−τντ+1λ21+τ3/21+τ2=11+τ3/2−130e−τ1+τ3/22ντ1+τ3/2+1λ21+τ2.
From that we will take(122)qτ=11+τ3/2,q0=1,Bτ,ντ=−130e−τ1+τ3/22ντ1+τ3/2+1λ21+τ2λ=1m,λ∈ℕ,βτ,ντ=−130e−τ1+τ3/22ντ1+τ3/2+1λ21+τ2.
And then,(123)−130e−τ1+τ3/22ντ1+τ3/2≤βτ,ντ≤−130e−τ1+τ3/22ντ1+τ3/2+11+τ2,⟹θτ=η1τ=130e−τ1+τ3/22,η2τ=11+τ2.
Also,(124)βτ,ντ≤βτ,ντ≤η1τντ1+τ3/2+η2τ⟹−βτ,ντ|≥−η1τντ1+τ3/2−η2τ⟹βτ,ντ≤−2βτ,ντ+η1τντ1+τ3/2+3η2τ⟹γ1τ=γ0=2,γ2τ=η1τ,γ3τ=3η2τ.
So, (ℋ2) is held.
To explore (ℋ3), we need to compute the values hereinafter:(125)Γ52=34π,Γ52+72≃4.829<30,2Γ52+12≃3.659<30,2Γ52+12Γ52+72≃17.66901<30.
Now, we can illustrate that (ℋ3) is useful as follows:(126)qτθτ=130e−τ1+τ3/2⟹eτqτθτ=1301+τ3/2<1+τ3/2Γ5/2+7/2,∫0∞qτθτdτ=130Γ5/2+1<Γ5/2+12Γ5/2+1/2Γ5/2+7/2,δ≃0.887655<1.
At the last, by calculating the upcoming values, we see that (ℋ4) is verified:(127)η0=3π460=3π240,q0η0=2403π,q0=1,γ0=2,Γ0=61,q0Γ0Γ5/2=2443π,6q0γ0Γ5/2=33π,∫0∞η2tdt=π2,R0=π22443π=122π3.
By adopted upshots and Theorem2, the presented problem seems to be solvable.


## 5. Conclusion

Academically, we analyzed the potentialities of having a positive solution in unbounded domain (on the half line) for the presumed inclusion problems (1)–(3). As we see, getting the positive solution at resonance was contingent with using O’Regan and Zima, Theorem 1. In the way to see the importance of the new upshots, we spelled out by example. We picked out these ramifications, and fortunately we can verify them for the sake of finding some extent applications in the other scientific fields.



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*Source: 9609108-2020-07-23.xml*