# Positive Solutions for a Higher-Order Semipositone Nonlocal Fractional Differential Equation with Singularities on Both Time and Space Variable

Authors: Kemei Zhang
Journal: Journal of Function Spaces
 (2019)

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## Abstract

In this paper, we consider the following higher-order semipositone nonlocal Riemann-Liouville fractional differential equationD0+αx(t)+f(t,x(t),D0+βx(t))+e(t)=0,0<t<1,D0+βx(0)=D0+β+1x(0)=⋯=D0+n+β-2x(0)=0, and D0+βx(1)=∑i=1m-2ηiD0+βx(ξi), where D0+α and D0+β are the standard Riemann-Liouville fractional derivatives. The existence results of positive solution are given by Guo-krasnosel’skii fixed point theorem and Schauder’s fixed point theorem.


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## Body

## 1. Introduction

In this paper, we devote to the investigation of the following nonlinear fractional differential equation(1)D0+αxt+ft,xt,D0+βxt+et=0,0<t<1,D0+βx0=D0+β+1x0=⋯=D0+n+β-2x0=0,D0+βx1=∑i=1m-2ηiD0+βxξi,where D0+α and D0+β are the standard Riemann-Liouville derivatives, α≥2,1≤α-β≤n-1,n-1<α≤n,0<ξ1<ξ2<⋯<ξm-2<1, 0<∑i=1mηiξiα-β-1<1. The nonlinear term f(t,u,v) is continuous and may be singular on both t=0,1 and v=0; e(t)∈L1([0,1],R) permits sign-changing.Differential equation models can describe many nonlinear phenomena in applied mathematics, economics, finance, engineering, and physical and biological processes [1, 2]. In recent years, there has been a great deal of research on the existence and/or uniqueness of solution in studying FDEs nonlocal problems for their wide applications in modeling some important physical laws (see [3–16], for instance).In [3], the authors were concerned with the existence of monotone positive solutions to the following fractional-order multipoint boundary value problems (2)D0+αut+qtft,ut,u′t=0,t∈0,1,u0=u′0=0,u′1=∑i=0mηiu′ξi,where 0<ξ1<ξ2<⋯<ξm<1, 2<α<3, ηi≥0, and ∑i=1mηiξiα-2<1. The authors obtained the existence of monotone positive solutions and establish iterative schemes for approximating the solutions.In [4], the authors investigated the existence of positive solutions of the following fractional differential equation multipoint boundary value problems with changing sign nonlinearity(3)D0+αut+λft,ut=0,t∈0,1,u0=u′0=⋯=un-2=0,ui1=∑j=0m-2ηju′ξj,where λ is a positive parameter, α≥2,n-1<α≤n,i∈N,1≤i≤n-2,ηj≥0(j=1,2,⋯,m-2), 0<ξ1<ξ2<⋯<ξm-2<1, and f may change sign and may be singular at t=0,1. By employing the cone expansion and compression fixed point theorem, the existence of positive solutions was obtained.In [5], the authors established the uniqueness of a positive solution to the following higher-order fractional differential equation: (4)D0+αut+qtft,ut,D0+μ1ut,⋯,D0+μn-2ut=0,t∈0,1,u0=uμ10=⋯=uμn-20=0,D0+μu1=∑i=1m-2ηiD0+βxξi,where f:[0,1]×(0,+∞)n-1→[0,+∞) is continuous and f(t,u1,…,un-1) may be singular at u1=0,…,un-1=0, and q(t):(0,1)→[0,+∞) is continuous and may be singular at t=0 and/or t=1. By using the fixed point theorem for the mixed monotone operator, the existence of unique positive solutions for above singular nonlocal boundary value problems of fractional differential equations is established. The nonlinear term f in [11] is nonnegative.In [11], the authors studied the existence of positive solutions for the following nonlocal fractional-order differential equations with sign-changing singular perturbation. (5)-D0+α+2yt+D0+αyt=ft,yt,D0+αyt+et,t∈0,1,aD0+αy0-bD0+α+1y0=∑j=1m-2ajD0+αyξj,cD0+αy1-dD0+α+1y1=∑j=1m-2bjD0+αyξj,where 0<α≤1,a,c≥0,b,d>0, 0<ξj<1,aj,bj∈[0,+∞), f:(0,1)×[0,∞)×(0,∞)→(0,∞) is continuous and may be singular near the zero for the third argument, and e(t)∈L1([0,1],R) may be sign-changing. By means of Schauder’s fixed point theorem, the conditions for the existence of positive solutions are established, respectively, for the cases where the nonlinearity is positive, negative, and semipositone.Motivated by the work mentioned above, we consider the fractional-order singular nonlocal BVP (1) and establish the existence results of positive solutions for (1). The main tools used in this paper are Guo-krasnosel’skii fixed point theorem and Schauder’s fixed point theorem. For the concepts and properties about the cone theory and the fixed point theorem, one can refer to [17–21].The rest of this paper is organized as follows: in Section2, we present some useful preliminaries and lemmas. The main results are given in Section 3 and Section 4, in which the singular cases with respect to the time variables and space variables are discussed, respectively.


## 2. Preliminaries and Some Lemmas

Definition 1 (see [1, 2]).
The Riemann-Liouville fractional integral of orderα>0 of a function x(t):(0,∞)→R is given by(6)I0+αxt=1Γα∫0tt-sα-1xsdsprovided the right-hand side is pointwisely defined on (0,∞).Definition 2 (see [1, 2]).
The Riemann-Liouville fractional derivative of orderα>0 of a function x(t):(0,∞)→R is given by(7)D0+αxt=1Γn-αddtn∫0txst-sα-n+1dsprovided the right-hand side is pointwisely defined on (0,∞), where n=[α]+1, [α] denotes the integer part of the number α.Lemma 3 (see [1, 2]).
The unique solution of the following linear Riemann-Liouville fractional differential equation of orderα>0(8)D0+αxt=0is(9)xt=c1tα-1+c2tα-2+⋯+cntα-n,where ci=1,2,…,n,n=[α]+1, [α] denotes the integer part of the number α.Lemma 4 (see [1, 2]).
Ifx∈L1(0,1) and D0+αx∈L1(0,1), then(10)I0+αD0+αxt=xt+c1tα-1+c2tα-2+⋯+cntα-n,where ci∈R,i=1,2,…,n,n=[α]+1.The similar proof of the following three lemmas can be traced to [5, 12]; in order to be convenient for readers to read, we now give the detailed process of proof for Lemma 5; the proofs for other two lemmas are omitted here.Lemma 5.
Leth∈L1(0,1) and 0<∑i=1mηiξiα-β-1<1, then the unique solution of the problem(11)D0+α-βxt+ht=0,0<t<1,x0=x′0=⋯=xn-20=0,x1=∑i=1m-2ηixξican be expressed uniquely by(12)xt=∫01Gt,shsds,where(13)Gt,s=1Γα-βg0gst1-sα-β-1,0≤t≤s≤1,gst1-sα-β-1-t-sα-β-1g0,0≤s≤t≤1,(14)gs=1-∑s≤ξiηiξi-s1-sα-β-1,Proof.
By Lemma4, the solution of (13) can be written as(15)xt=-1Γα-β∫0tt-sα-β-1hsds+C1tα-β-1+C2tα-β-2+⋯+Cntα-β-n.It follows from x(0)=x′(0)=⋯=x(n-2)(0)=0 that C2=⋯=Cn=0, i.e.,(16)xt=-1Γα-β∫0tt-sα-β-1hsds+C1tα-β-1;thus(17)x1=-1Γα-β∫011-sα-β-1hsds+C1,xξi=-1Γα-β∫0ξiξi-sα-β-1hsds+C1ξiα-β-1,which, together with the boundary value condition x(1)=∑i=1m-2ηix(ξi), implies that(18)-1Γα-β∫011-sα-β-1hsds+C1=∑i=1m-2ηi-1Γα-β∫0ξiξi-sα-β-1hsds+C1ξiα-β-1,C11-∑i=1m-2ηiξiα-β-1=1Γα-β∫011-sα-β-1hsds-∑i=1m-2ηi∫0ξiξi-sα-β-1hsdsΓα-β,C1=∫011-sα-β-11-∑i=1m-2ηiξiα-β-1Γα-βhsds-∑i=1m-2ηi∫0ξiξi-sα-β-1hsdsΓα-β1-∑i=1m-2ηiξiα-β-1.i.e.,(19)C1=1g0Γα-β∫011-sα-β-1hsds-∑i=1m-2ηi∫0ξiξi-sα-β-1hsds=1g0Γα-β[∫011-sα-β-1gshsds;thus(20)xt=-1Γα-β∫0tt-sα-β-1hsds+C1tα-β-1=1g0Γα-β∫0ttα-β-11-sα-β-1gs-t-sα-β-1g0hsds-1g0Γα-β∫t1tα-β-11-sα-β-1hsds=∫01Gt,shsds.Lemma 6.
If0<∑i=1mηiξiα-β-1<1, then the function g satisfies the following conditions:
(1)g is a nondecreasing function on [0,1];
(2) there existM1≥m1≥0, such that m1t+g(0)≤g(t)≤M1t+g(0), for any t∈[0,1], where M1=sup0<t≤1⁡((g(t)-g(0))/t), m1=inf0<t≤1⁡gt-g0/t.Remark 7.
It is easy to prove thatm1>0.Lemma 8.
The functionG(t,s) defined by (13) has the following properties:(1)
G(t,s)>0 for any (t,s)∈(0,1);(2)
G(t,s)≥(m1/g(0)Γ(α-β))s(1-s)α-β-1tα-β-1 for any (t,s)∈[0,1];(3)
G(t,s)≤((M1+g(0)[α-β])/g(0)Γ(α-β))s(1-s)α-β-1tα-β-1 for any (t,s)∈[0,1];(4)
G(t,s)≤Ms(1-s)α-β-1,
where[α-β] denotes the integer part of the number α-β,(21)M=M1+g0α-β-1g0Γα-β.Setv(t)=D0+βx(t), then (1) can be transformed into the following form:(22)D0+α-βvt+ft,I0+βvt,vt+et=0,0<t<1,v0=v′0=⋯=vn-20=0,v1=∑i=1m-2ηivξi.From Lemma 5, we know that the solution v(t) of (22) satisfies(23)vt=∫01Gt,sfs,I0+βvs,vs+esds.Lemma 9 (see [17, 18]).
Suppose thatE is a Banach space and D⊂E is a bounded convex closed set, the operator A:D→D is completely continuous, then A has one fixed point on D.Lemma 10 (see [20] (Guo-krasnosel’skii fixed point theorem)).
LetΩ1 and Ω1 be two bounded open sets in Banach space E such that θ∈Ω1 and Ω1¯⊂Ω2, A:P⋂(Ω2¯∖Ω1)→P a completely continuous operator, where θ denotes the zero element of E and P a cone of E. Suppose that one of the following conditions(i)
Ax≥x,∀x∈P∩∂Ω1 and Ax≤x,∀x∈P∩∂Ω2;(ii)
Ax≤x,∀x∈P∩∂Ω1 and Ax≥x,∀x∈P∩∂Ω2
holds. ThenA has at least one fixed point in P∩(Ω2¯∖Ω1).


## 3. Main Result I:f Is Singular with Respect to the Time Variables

LetE=C[0,1], u=max0≤t≤1⁡ut, then (E,·) is a Banach space. Set(24)P=v∈E:vt≥m1M1+g0α-βtα-β-1v,t∈0,1,where [α-β] denotes the integer part of the number α-β. Then P⊂E is a positive cone of E. For convenience, we list some conditions which will be used in this section.(H1) For any (t,x,y)∈(0,1)×[0,+∞)×[0,+∞),(25)0≤ft,x,y≤ϕtρx+hy,where ϕ∈C(0,1), ϕ(t)>0 on (0,1), ρ(x)>0 is continuous and increasing on [0,+∞), h(x)>0 is continuous and decreasing on [0,+∞).(H2)(26)0<∫01s1-sα-β-1ϕs+e+s+e-sds<+∞,where e+=max⁡{e(t),0}, e-=-min⁡et,0 are the positive part and negative part of e(t), respectively.(H3) There exists r1>0, such that(27)ρr1+wΓβ∫01s1-sα-β-1ϕsds+∫01s1-sα-β-1h0ϕs+e+sds<r1M,where w(t) is the solution of the following linear equation(28)D0+α-βvt+e-t=0,0<t<1,v0=v′0=⋯=vn-20=0,v1=∑i=1m-2ηivξi,i.e., w(t)=∫01G(t,s)e-(s)ds.(H4) There exists [a,b]⊂(0,1) such that(29)limx+y→+∞⁡ft,x,yx+y>Luniformly holds for t∈[a,b], where(30)L=2Γα-βg0M1+g0α-βm12aα-β-1∫absα-β1-sα-β-1ds.For anyv∈P, let(31)vt-wt∗=max⁡vt-wt,0,and define operator(32)Fvt=∫01Gt,sfs,I0+βvs-ws∗,vs-ws∗+e+sds,From condition (H1) and (H2), it is easy to know that F is well defined.Lemma 11.
F : P → P is a completely continuous operator.Proof.
For anyv∈P, it follows from Lemma 8 that(33)Fvt=∫01Gt,sfs,I0+βvs-ws∗,vs-ws∗+e+sds≤M1+g0α-βg0Γα-βtα-β-1∫01s1-sα-β-1fs,I0+βvs-ws∗,vs-ws∗+e+sds≤M1+g0α-βg0Γα-β∫01s1-sα-β-1fs,I0+βvs-ws∗,vs-ws∗+e+sds,Fvt=∫01Gt,sfs,I0+βvs-ws∗,vs-ws∗+e+sds≥m1g0Γα-βtα-β-1∫01s1-sα-β-1fs,I0+βvs-ws∗,vs-ws∗+e+sds,which deduce that Fv(t)≥(m1/(M1+g(0)[α-β]))tα-β-1Fv, i.e., F:P→P.
LetB⊂P be a bounded set, i.e., there exists L1>0 such that v≤L1 for any v∈B, then(34)0≤vt-wt∗≤L1+∫01Gt,se-sds≤L1+∫01Ms1-sα-β-1e-sds≐L~.(35)Fvt=∫01Gt,sfs,I0+βvs-ws∗,vs-ws∗+e+sds≤∫01Gt,sϕsρI0+βL~+h0+e+sds≤MρL~βΓβ+h0∫01s1-sα-β-1ϕsds+M∫01s1-sα-β-1e+sds<+∞,therefore(36)Fv≤MρL~βΓβ+h0∫01s1-sα-β-1ϕsds+M∫01s1-sα-β-1e+sdsfor any v∈B, which implies that F is uniformly bounded.
From(H2), the absolutely continuity of integral and the uniformly continuity of G(t,s) on [0,1], we know that for any ε>0, ∃δ>0, such that(37)∫0δs1-sα-β-1ϕsds<ε6MρL~/βΓβ+h0,(38)∫1-δδs1-sα-β-1ϕsds<ε6MρL~/βΓβ+h0,and(39)Gt1,s-Gt2,s<ε3lρL~/βΓβ+h0
for anyt1,t2,s∈[0,1] with |t1-t2|<δ.
(37)-(39), together with Lemma 8, imply that(40)Fvt1-Fvt2≤∫01Gt1,s-Gt2,sfs,I0+βvs-ws∗,vs-ws∗ds=∫0δGt1,s-Gt2,sfs,I0+βvs-ws∗,vs-ws∗ds+∫1-δ1Gt1,s-Gt2,sfs,I0+βvs-ws∗,vs-ws∗ds+∫δ1-δGt1,s-Gt2,sfs,I0+βvs-ws∗,vs-ws∗ds≤2M∫0δs1-sα-β-1ϕsρI0+βL~+h0ds+2M∫1-δ1s1-sα-β-1ϕsρI0+βL~+h0ds+∫δ1-δGt1,s-Gt2,slρI0+βL~+h0ds<ε3+ε3+ε3=εfor any t1,t2∈[0,1] with t1-t2<δ and any v∈B, where l=maxδ≤t≤1-δ⁡ϕ(t), which deduces that F is equicontinuous on [0,1]. Thus, according to Ascoli-Arzela theorem, we know that FB is a relatively compact set, and that F is a completely continuous operator.Theorem 12.
Suppose that(H1)-(H4) hold, then the FVP (1) has at least one positive solution.Proof.
For anyv∈∂Pr1, where Pr1=v∈P∣v<r1, by (32), Lemma 8 and condition (H3), one can get that(41)Fvt=∫01Gt,sfs,I0+βvs-ws∗,vs-ws∗+e+sds≤∫01Gt,sϕs(ρI0+βvs-ws+h0+e+sds≤∫01Gt,sϕsρI0+βv+w+h0+e+sds=M∫01s1-sα-β-1ϕsρI0+βr1+w+h0+e+sds=M∫01s1-sα-β-1ϕsρr1+wΓβ+h0+e+sds<r1=v,i.e., Fv≤v,v∈∂Pr1.
By condition(H4), ∃X>0, such that(42)ft,x,y>Lx+yfor any x>X and any t∈[a,b]. Choose r2 such that(43)r2>max⁡r1,2c1,2M1+g0α-βXm1aα-β-1,where(44)c1=M1+g0α-β2·∫011-sα-β-1e-sdsm1Γα-βg0.For any v∈∂Pr2, where Pr2=v∈P∣v<r2. Because(45)wt=∫01Gt,se-sds≤M1+g0α-βΓα-βg0·tα-β-1∫011-sα-β-1e-sds≤M1+g0α-β2·∫011-sα-β-1e-sdsm1Γα-βg0r2·vt=c1r2vt,so we have(46)vt-wt≥1-c1r2vt≥12vt,and then for t∈[a,b],(47)vt-wt≥12vt≥12·m1aα-β-1M1+g0α-β·v=12·m1aα-β-1M1+g0α-β·r2>Xfollows from (43) and the definition of cone P.
From (42), (43), and (47), one can obtain that(48)Fvt=∫01Gt,sfs,I0+βvs-ws∗,vs-ws∗+e+sds≥∫abGt,sLI0+βvs-ws+vs-ws+e+s]ds≥L∫abGt,s12vsds≥L2·m1Γα-βg0·∫abs1-sα-β-1tα-β-1vsds≥L2·m1aα-β-1Γα-βg0·m1M1+g0α-β·∫abs1-sα-β-1sα-β-1ds·v=L2·m1aα-β-1Γα-βg0·m1M1+g0α-β·∫ab1-sα-β-1sα-βds·v≥L2·m1aα-β-1Γα-βg0·m1M1+g0α-β·∫ab1-sα-β-1sα-βds·r2=r2=v,i.e., Fv≥v,v∈∂Pr2.
It follows from Lemma10 that F has at least fixed point v1∈Pr2¯∖Pr1, i.e., v1 satisfies(49)D0+α-βv1t+ft,I0+βv1t-wt,v1t-wt+et=0,0<t<1,v10=v1′0=⋯=v1n-20=0,v11=∑i=1m-2ηiv1ξi.Set v1¯(t)=v1(t)-w(t), noticing that v1(t),w(t) are the solutions of BVP (32) and (49), respectively; therefore we can conclude that v1¯(t) is a positive solution of (22). Let x1(t)=I0+βv1¯(t), then x1(t) is a positive solution of the nonlinear fractional differential equations (1).


## 4. Main Result II:f Is Singular with Respect to Both the Time Variables and the Space Variable

In this section, we always suppose that the following condition holds.(H5) f(t,u,v):(0,1)×[0,∞)×(0,∞)→[0,∞) is continuous, there exist ε∈(0,1) and μ1,μ2∈C+[0,1], μ1(t)≢0 for t∈[0,1] such that(50)μ1tx+yε≤ft,x,y≤μ2tx+yεfor any (x,y)∈[0,∞)×(0,∞),t∈(0,1), where(51)C+0,1=xt∈C0,1∣xt≥0,t∈0,1Set(52)φt=∫01Gt,sesds,t∈0,1,it follows from Lemma 5 that φ(t) is the solution of the following linear equation (53)D0+α-βvt+et=0,0<t<1,v0=v′0=⋯=vn-20=0,v1=∑i=1m-2ηivξi.Denote(54)a1t=∫01Gt,sμ1sds,a2t=∫01Gt,sμ2sds,φ∗=inf0≤t≤1⁡φt,φ∗=sup0≤t≤1⁡φt,a1∗=min0≤t≤1⁡a1t,a1∗=max0≤t≤1⁡a1t,a2∗=min0≤t≤1⁡a2t,a2∗=max0≤t≤1⁡a2t.Clearly, aj∗≥aj∗>0,j=1,2.Theorem 13.
Suppose that the condition (H5) holds and φ∗≥0. Then the FVP (1) has at least one positive solution.Proof.
Becauseφ∗≥0, so we can choose R>0 large enough such that(55)rRε+φ∗≥1R,Rεa2∗+φ∗≤R,where r=a1∗/[1+1/(1+βΓ(β))]ε. In fact, since(56)limx→+∞xεx=0<min⁡r,1a2∗+φ∗,there exists X1>0 such that xε/x<min⁡r,1/(a2∗+φ∗) for any x>X1, i.e.,(57)rxε≥1x,xεa2∗+φ∗≤xfor any x>X1. If φ∗>0, then from limx→+∞[1/x-r/xε]=0<φ∗, one can get that there exists X2>0 such that 1/x-r/xε<φ∗ for any x>X2, i.e.,(58)rxε+φ∗≥1xfor any x>X2. By (57) (58), we can choose R>max⁡X1,X2 such that R satisfies (55). Set(59)D=v∈C+0,1:1R≤vt≤R,t∈0,1.For any v∈D1, from (23) we have(60)I0+βvt=1Γβ∫0tt-sβ-1vsds≤RΓβ∫0tt-sβ-1ds=RtββΓβ≤RβΓβ,(61)I0+βvt=1Γβ∫0tt-sβ-1vsds≥1RΓβ∫0tt-sβ-1ds=tβRβΓβ.It follows from (60), (61), and (H5) that(62)μ1tRε1+1/βΓβε≤μ1tvt+I0+βvtε≤ft,I0+βvt,vt,ft,I0+βvt,vt≤μ2tvt+I0+βvtε≤μ2t1/R+tβ/RβΓβε≤Rεμ2t,i.e.,(63)μ1tRε1+1/βΓβε≤ft,I0+βvt,vt≤Rεμ2t.And then(64)Tvt=∫01Gt,sfs,I0+βvs,vs+esds=∫01Gt,sfs,I0+βvs,vsds+φt≤∫01Gt,sfs,I0+βvs,vsds+φ∗≤Rε∫01Gt,sμ2sds+φ∗<+∞,which deduces that the operator T is well defined.
Now, we shall prove thatT:D→D. For v∈D, it is easy to see that Tv(t)∈C+[0,1], and by (55) (63) we can obtain that(65)Tvt=∫01Gt,sfs,I0+βvs,vs+esds=∫01Gt,sfs,I0+βvs,vsds+φt≥1Rε1+1/βΓβε∫01Gt,sμ1sds+φt≥1Rε1+1/βΓβεa1∗+φ∗≥1R,(66)Tvt=∫01Gt,sfs,I0+βvs,vs+esds=∫01Gt,sfs,I0+βvs,vsds+φt≤Rε∫01Gt,sμ2sds+φ∗≤Rεa2∗+φ∗≤Rεa2∗+φ∗≤R.i.e., T:D→D.
Next, let us prove thatT:D→D is completely continuous.
For any{vn}⊂D,v0∈D, and vn→v0. The continuity of f deduces that(67)ft,I0+βvnt,vnt→ft,I0+βv0t,v0t,n→∞,and it follows from (63) that(68)ft,I0+βvnt,vnt-ft,I0+βv0t,v0t≤2Rϵmax⁡max0≤t≤1⁡μ1t,max0≤t≤1⁡μ2t.By using the Lebesgue dominated convergence theorem, we obtain that(69)limn→∞⁡Tvn-Tv0=limn→∞⁡max0≤t≤1⁡∫01Gt,sfs,I0+βvns,vns-fs,I0+βv0s,v0sds≤limn→∞⁡max0≤t≤1⁡∫01Gt,sfs,I0+βvns,vns-fs,I0+βv0s,v0sds=max0≤t,s≤1⁡Gt,s·∫01limn→∞⁡fs,I0+βvns,vns-fs,I0+βv0s,v0sds=0,and this implies that T is a continuous operator.
Now, we shall prove thatT:D→D is compact. For any v∈D, Tv∈D, which deduces that 1/R≤Tv(t)≤R, for t∈[0,1], i.e., T is uniformly bounded.
SinceG(t,s)∈C([0,1]×[0,1]), it is also uniformly continuous on [0,1]×[0,1], and then for any ε>0, ∃δ>0, s.t., for any (t1,s),(t2,s)∈[0,1]×[0,1] with |t1-t2|<δ, we always have(70)Gt1,s-Gt2,s<εRε∫01μ2sds+∫01esds.Thus, one can obtain by virtue of (63) (70) that(71)Tvt1-Tvt2=∫01Gt1,s-Gt2,sfs,Iβvs,vs+esds≤∫01Gt1,s-Gt2,sfs,I0+βvs,vs+esds<Rε∫01μ2sds+∫01esds·εRε∫01μ2sds+∫01esds=εfor any t1,t2∈[0,1] with t1-t2<δ and any v∈D1, which shows that T:D→D is equicontinuous. Thus, Arzela-Ascoli theorem guarantees that T:D→D is completely continuous. Existence of at least one fixed point v1∈D follows from Lemma 6, i.e., u1(t)=I0+βv1(t) is a positive solution of differential equation (1), which satisfies(72)1RΓαtα≤u1t=1Γα∫0tt-sα-1v1sds≤RΓαtα.



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*Source: 7161894-2019-02-03.xml*