# Impulsive Problems for Fractional Differential Equations with Functional Boundary Value Conditions at Resonance

Authors: Bingzhi Sun; Shuqin Zhang; Weihua Jiang
Journal: Journal of Function Spaces
 (2019)

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## Abstract

We establish novel results on the existence of impulsive problems for fractional differential equations with functional boundary value conditions at resonance withdim⁡KerL=1. Our results are based on the degree theory due to Mawhin, which requires appropriate Banach spaces and suitable projection schemes.


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## Body

## 1. Introduction

Impulsive differential equations serve as basic models to study the dynamics of processes that are subject to sudden changes in their states. Recent development in this field has been motivated by many applied problems arising in control theory, population dynamics, and medicines. For some recent works on the theory of impulsive differential equations, we refer the reader to [1–3].Fractional differential equations have proved valuable and effective tools in the modeling of many phenomena in various fields of biology, medicine, mechanics, engineering, viscoelasticity, and so forth; see [4, 5]. In order to improve the mathematical modeling of several concepts arising in these areas, many researchers have paid a considerable attention to the subject of impulsive fractional differential equations in the recent literature [6–13].A boundary value problem is said to be at resonance if the corresponding homogeneous boundary value problem has a nontrivial solution. In [14, 15], the authors studied the existence of solutions for functional boundary value problems with a linear differential operator by using Mawhin continuation theorem [16]. In [15], N.Kosmatov and Jiang extended the results of [14] to new resonance scenarios.To the best of the author knowledge, the solvability of the functional boundary value problems for impulsive fractional differential equation at resonance has not been well studied till now. We will fill this gap in the literature. Inspired by the above excellent results [14–24], in this paper, we investigate the following boundary value problem for impulsive fractional equations at resonance:(1)Dc0+αut=ft,ut,u′t,1<α≤2,t∈J′=0,1∖t1,t2,…,tm,Δutk=Ikutk,Δu′tk=Ik¯u′tk,tk∈0,1,k=1,2,…,m,B1u=0,B2u=0.where J=[0,1], Ik,Ik¯:R→R are continuous functions with Ik(0)=Ik¯(0)=0,k=1,2,…,m and B1,B2:C[0,1]→R are continuous linear functionals with the resonance condition: B1(t)B2(1)=B1(1)B2(t). Δu(tk)=u(tk+)-u(tk-) and Δu′(tk)=u′(tk+)-u′(tk-), with u(tk+)=limh→0+⁡u(tk+h), u(tk-)=limh→0-⁡u(tk+h), u′(tk+)=limh→0+⁡u′(tk+h), u′(tk-)=limh→0-⁡u′(tk+h), k=1,2,…,m,0=t0<t1<t2⋯tm<tm+1=1.


## 2. Preliminaries

For the convenience of the reader, we introduce the theoretical foundations of the method.Definition 1 (see [4, 5]).
The Riemann-Liouville fractional integrals of orderα>0 of a function y:[a,+∞)→R is given by (2)Ia+αyt=1Γα∫att-sα-1ysds,where the right side is pointwise defined on (a,+∞).Definition 2 (see [4, 5]).
The Caputo fractional derivatives of orderα>0 of a function y:[a,+∞)→R is given by(3)Dca+αyt=1Γn-α∫att-sn-α-1ynsds,where n=[α]+1,[α] denotes the integer part of number α, and this derivative is called the right side which is pointwise defined on (a,+∞).Lemma 3 (see [4, 5]).
Letα>0. If u(t)∈ACn[a,b] or u(t)∈Cn[a,b], then the fractional differential equation Dca+αu(t)=0 (or Dcb-αu(t)=0) has solution(4)ut=c0+c1t-a+c2t-a2+⋯+cn-1t-an-1where ci=u(i)(a)/i!∈R,i=0,1,…,n-1, and n=[α]+1.Lemma 4 (see [4, 5]).
(1) Letα>0; if u(t)∈ACn[a,b] or u(t)∈Cn[a,b], then one has(5)Ia+αDc0+αut=ut-c0-c1t-a-c2t-a2-⋯-cn-1t-an-1,where ci=u(i)(a)/i!∈R,i=0,1,…,n-1, and n=[α]+1;
(2) The equalityDc0+βI0+βy=y holds for every β>0 and y∈L1[0,1].Definition 5 (see [16]).
LetX, Z be real Banach spaces and let L:domL⊂X→Z be a linear operator. X is said to be the Fredholm operator of index zero provided that (i)
ImL is a closed subset of Y;(ii)
dim⁡KerL=codim⁡ImL<+∞.LetP:X→X, Q:Z→Z be continuous projectors such that ImP=KerL, KerQ=ImL, X=KerL⊕KerP, and Z=ImL⊕ImQ. It follows that LdomL∩KerP:domL∩KerP→ImL is reversible. We denote the inverse of the mapping by KP(generalized inverse operator of L). If Ω is an open bounded subset of X such that domL∩Ω≠∅, the mapping N:X→Z will be called L-compact on Ω¯, if QN(Ω¯) and KP(I-Q)N:Ω¯→X are continuous and compact.Theorem 6 (see [16] Mawhin continuation theorem).
LetL:domL⊂X→Z be a Fredholm operator of index zero and N:X→Z is L-compact on Ω¯. Assume that the following conditions are satisfied:(i)
Lu≠λNu for every (u,λ)∈[(domL∖KerL)∩∂Ω]×(0,1);(ii)
Nu∉ImL for every u∈KerL∩∂Ω;(iii)
deg⁡(QNKerL,Ω∩KerL,0)≠0, where Q:Z→Z is a continuous projection such that ImL=KerQ.
Then the equationLu=Nu has at least one solution in domL∩Ω¯.In the following, we use the classical Banach spaceP C ( J , R ) =  u : J → R : u ∈ C ( ( t k , t k + 1 ] , R ),  k=0,1,2,…,m,u(tk+),u(tk-) exist with u(tk-)=u(tk),k=1,2,…,m with uPC=maxt∈J⁡u(t).X = u , u ′ ∈ P C J , R with the norm uX=maxuPC,u′PC.Y = L 1 [ 0,1 ] with the norm y1=∫01ytdt.LetZ=Y×R2m endowed with the norm (y,c1,…,cm,d1,…,dm)=max⁡y1,c1,…,cm,d1,…,dm. Then (X,·) and (Z,·) are Banach spaces.Define the operatorsL:domL⊂X→Z,N:X→Z as follows: (6)Lu=Dc0+αuΔut1⋯ΔutmΔu′t1⋯Δu′tm,Nu=ft,ut,u′tI1ut1-⋯Imutm-I1¯u′t1-⋯Im¯u′tm-,where domL=u∈X:u′′∈Y,Biu=0,i=1,2. For convenience, let (a,b)T≔ab. So problem (1) becomes Lu=Nu.


## 3. Main Results

In order to obtain the main results, we will always suppose the following condition holds.( H 0 ) The functionals B1,B2:X→R are linear continuous with respective norms β1,β2, that is, Biu≤βiuX,i=1,2 and satisfy B2(t)=b,B2(1)=a,B1(t)=kb,B1(1)=ka, where a2+b2≠0,k,a,b∈R.( C 0 ) There must exist a function h(t)∈Y such that(7)B1-kB2∫0tt-sα-1Γαhsds=1,and maxt∈[0,1]⁡(1-s)α-2h(ts)1≔Mh<+∞, where k is introduced in (H0).( C 1 ) Let f:[0,1]×R2→R satisfy the following conditions:(1)
f(·,u) is measurable for each fixed u∈R2; f(t,·) is continuous for a.e.t∈[0,1].(2)
∀r>0, and there must exist gr(t)∈Y such that ft,x,y≤gr(t),x≤r,y≤r.In addition,(8)maxt∈0,1⁡1-sα-2grts1<+∞.Remark 7.
We show that there must be ah(t)∈Y satisfying (C0).
For convenience, setB=B1-kB2, and hn(t)=tn-α,n≥2. Then, there must exist n≥2 such that B(I0+αhn(t))≠0. If not, B(I0+αhn(t))=0. That is B(tn)=0. By (H0), B(1)=B(t)=0. Thus, B(p)=0 for every polynomial p. Since p is dense in X, we can get B(u)≡0,i.e.B1≡kB2,u∈X. That is a contradiction (since B1 and B2 are linearly independent on X).
There must be somei,i=2,3,… such that B(ti)≠0. For these i, take h(t)=hi(t)/B(I0+αhi(t)), we have B(I0+αh(t))=1 and (1-s)α-2h(ts)1=ti-αB(α-1,i-α+1)/B(I0+αhi(t)), where B is beta function. Obviously, maxt∈[0,1]⁡(1-s)α-2h(ts)1≔Mh<+∞. Thus, there exists h∈Y satisfying (C0).Lemma 8.
Assume that(H0) holds; then L:domL⊂X→Z is a Fredholm operator of index zero. Moreover,(9)KerL=cat-b:c∈R,dim⁡KerL=1.and(10)ImL=y,c1,…,cm,d1,…,dmT:y∈Y,ci,di∈R,i=1,2,…,m,B1-kB2∫0tt-sα-1Γαysds+∑ti<tci+∑ti<tdit-ti=0.Proof.
It is easy to see that(11)KerL=cat-b:c∈R,dim⁡KerL=1.For u∈domL, if Lu=(y,c1,…,ck,d1,…,dk)T, then(12)Dc0+αut=yt,Δuti=ci,Δu′ti=di,i=1,2,…,m.From (12) and Lemma 4, we have(13)ut=∫0tt-sα-1Γαysds+c0+d0t+∑ti<tci+∑ti<tdit-ti,t∈J′,where c0, d0 are two arbitrary constants. Substituting the boundary condition Bi(u)=0,i=1,2 into (13), one has(14)B1ut=c0B11+d0B1t+B1∫0tt-sα-1Γαysds+∑ti<tci+∑ti<tdit-ti=0,B2ut=c0B21+d0B2t+B2∫0tt-sα-1Γαysds+∑ti<tci+∑ti<tdit-ti=0.Based on the condition (H0), we have(15)B1-kB2∫0tt-sα-1Γαysds+∑ti<tci+∑ti<tdit-ti=0.Conversely, if (15) holds, setting(16)ut=-bt+aa2+b2B2I0+αyt+∑ti<tci+∑ti<tdit-ti+I0+αyt+∑ti<tci+∑ti<tdit-ti,then it is easy to check that u(t) is a solution of (12) and satisfies B1(u)=B2(u)=0.
Combining the above arguments, we obtain (10).
DefineQ:Z→Z as follows:(17)Qy,c1,…,cm,d1,…,dmT=B1-kB2I0+αyt+∑ti<tci+∑ti<tdit-tiht,0,…,0T,where h(t) is introduced in (C0).
Hence,(18)Q2z=B1-kB2I0+αyt+∑ti<tci+∑ti<tdit-tihtB1-kB2I0+αht,0,…,0T=B1-kB2I0+αyt+∑ti<tci+∑ti<tdit-tiht,0,…,0T=Qz,∀z∈Z,by (H0) and (C0), which implies the operator Q is a projector.
Now, It is obvious thatImL=KerQ and Z=ImQ+ImL. Moreover, it follows from KerQ=ImL that ImL∩ImQ=0. Thus, Z=ImQ⊕ImL and dim⁡KerL=codim⁡ImL=1. Observe that ImL is a closed subspace of Z and L is a Fredholm operator of index zero.TakeP:X→X as follows:(19)Put=au′0-bu0a2+b2at-b.It is easy to check that P2u=Pu,u∈X, and it is also elementary to confirm the identity ImP=KerL. So, X=KerL⊕KerP.Lemma 9.
The mappingKP:Z→domL∩KerP defined by(20)KPy,c1,…,cm,d1,…,dmT=-bt+aa2+b2B2I0+αyt+∑ti<tci+∑ti<tdit-ti+I0+αyt+∑ti<tci+∑ti<tdit-tiis the inverse of L.Proof.
Obviously,LKPz=z for all z∈Z. For v=(y,c1,…,ck,d1,…,dk)T∈ImL, it is clear that (LKPv)(t)=v(t).
For eachu∈domL∩KerP, noting that I0+αDc0+αu(t)=u(t)-u(0)-u′(0)t-∑ti<tci-∑ti<tdi(t-ti), where ci=Δu(ti),di=Δu′(ti), we get (21)KPLut=KPDc0+αu,c1,…,cm,d1,…,dmTt=-bt+aa2+b2B2I0+αDc0+αut+∑ti<tci+∑ti<tdit-ti+I0+αDc0+αut+∑ti<tci+∑ti<tdit-ti=-bt+aa2+b2B2ut-u0-u′0t+ut-u0-u′0t=-bt+aa2+b2-u0a-u′0b+ut-u0-u′0t=ut-au′0-bu0a2+b2at-b=ut.From u∈KerP, we obtain au′(0)-bu(0)/a2+b2=0, i.e., (KPLu)(t)=u(t).
Thus,(22)KP=LdomL∩KerP-1.The next lemma provides norm-estimates needed for the main results.Lemma 10.
Forz=(y(t),c1,…,cm,d1,…,dm)T∈Z,(23)KPzZ=KPyt,c1,…,cm,d1,…,dmTZ≤Amax⁡y1Γα+1,maxt∈0,1⁡1-sα-2yts1Γα-1+∑i=1mci+di,where A=1+a+b/a2+b2β2 and maxt∈[0,1]⁡(1-s)α-2y(ts)1<+∞.Proof.
It is easy to see that(24)I0+αyt+∑ti<tci+∑ti<tdit-tiPC=max⁡I0+αyt+∑ti<tci+∑ti<tdit-ti≤y1Γα+∑i=1mci+∑i=1mdi,I0+αyt+∑ti<tci+∑ti<tdit-ti′tPC=max⁡I0+α-1yt+∑ti<tdi≤maxt∈0,1⁡1-sα-2yts1Γα-1+∑i=1mdi,so that(25)I0+αyt+∑ti<tci+∑ti<tdit-tiX≤max⁡y1Γα,maxt∈0,1⁡1-sα-2yts1Γα-1+∑i=1mci+di.By representation of KP and observing that due to B2u≤β2uX, (26)KPzt=KPyt,c1,…,cm,d1,…,dmT=-bt+aa2+b2B2I0+αyt+∑ti<tci+∑ti<tdit-ti+I0+αyt+∑ti<tci+∑ti<tdit-ti≤a+ba2+b2β2+1I0+αyt+∑ti<tci+∑ti<tdit-tiX≤a+ba2+b2β2+1max⁡y1Γα,maxt∈0,1⁡1-sα-2yts1Γα-1+∑i=1mci+di,similarly, (27)KPz′t≤ba2+b2β2I0+αyt+∑ti<tci+∑ti<tdit-tiX+I0+αyt+∑ti<tci+∑ti<tdit-tiX≤ba2+b2β2+1max⁡y1Γα,maxt∈0,1⁡1-sα-2yts1Γα-1+∑i=1mci+di.Finally, (23) follows from above arguments.Lemma 11.
Assume(C0-C1) hold, Ω⊂X is an open bounded set, and domL∩Ω=∅. For any r>0, there exist a function gr(t)∈Y and two positive constants M1r,M2r such that ft,ut,u′t≤gr(t), maxt∈[0,1]⁡(1-s)α-2gr(ts)1<+∞,t∈[0,1],ut<r,u′t<r,Iiuti≤M1r,Ii¯u′ti≤M2r,uti≤r,u′ti≤r,i=1,2,…,m. Then N is L-compact on Ω¯.Proof.
SinceΩ⊂X is bounded, there exists a constant r>0 such that uX≤r,u∈Ω¯.
Denotef(t)≔ft,ut,u′t,Ii(·)≔Ii(u(ti)),Ii¯·≔Ii¯u′ti,(28)K≔B1-kB2I0+αf·+∑ti<tIi·+∑ti<tIi¯·t-ti,ba2+b2B2I0+αf·-Kht+∑ti<tIi·+∑ti<tIi¯·t-ti+∑ti<x2Ii¯·≔C<+∞,QNuZ=Qft,ut,u′t,I1ut1,…,Imutm,I1¯u′t1,…,Im¯u′tmTZ=B1-kB2I0+αft+∑ti<tIi·+∑ti<tIi¯·t-tiht,0,…,0TZ=B1-kB2I0+αft+∑ti<tIi·+∑ti<tIi¯·t-tih1≤β1+kβ2max⁡gr1Γα,maxt∈0,1⁡1-sα-2grts1Γα-1+mM1r+M2rh1<∞.Since (29)I0+αft+∑ti<tIi·+∑ti<tIi¯·t-tiPC≤gr1Γα+mM1r+M2r,I0+α-1ft+∑ti<tIi¯·PC≤maxt∈0,1⁡1-sα-2grts1Γα-1+mMr2,hence QN(Ω¯) is bounded.
Foru∈Ω¯, we have (30)KPI-QNut=bt+aa2+b2B2I0+αft,ut,u′t-Kht+∑ti<tIi·+∑ti<tIi¯·t-ti+I0+αft,ut,u′t-Kht+∑ti<tIi·+∑ti<tIi¯·t-ti.This, together with (C0) and (C1) means that(31)KPI-QNuX≤1+a+ba2+b2β2I0+αft,ut,u′t-Kht+∑ti<tIi·+∑ti<tIi¯·t-tiX<∞.Since (32)I0+αft,ut,u′t-Kht+∑ti<tIi·+∑ti<tIi¯·t-tiX≤max⁡gr1+Kh1Γα+1,Mg+KMhΓα-1+mM1r+M2r<+∞,where(33)maxt∈0,1⁡1-tα-2grts1≔Mg<+∞,maxt∈0,1⁡1-sα-2hts1≔Mh,f,h∈Y,then KP(I-Q)N(Ω¯) is uniformly bounded.
For eachu∈Ω¯,0≤tk<x1<x2≤tk+1≤1,k=0,1,…,m. Then (34)KPI-QNux2-KPI-QNux1=bx2-x1a2+b2B2I0+αf·-Kht+∑ti<tIi·+∑ti<tIi¯·t-ti+∑ti<x2Ii¯·x2-x1+1Γα∫0x2x2-sα-1(fs,us,u′s-Khsds-1Γα∫0x1x1-sα-1(fs,us,u′s-Khsds≤Cx2-x1+1Γα∫0x1x2-sα-1-x1-sα-1f(s,us,u′s-Khsds+1Γα∫x1x2x2-sα-1f(s,us,u′s-Khsds,and(35)KPI-QNu′x2-KPI-QNu′x1=α-1Γα∫0x2x2-sα-2f(s,us,u′s-Khsds-α-1Γα∫0x1x1-sα-2f(s,us,u′s-Khsds≤1Γα-1∫x1x2x2-sα-2f(s,us,u′s-Khsds+1Γα-1∫0x1x2-sα-2-x1-sα-2f(s,us,u′s-KhsdsFor tk<x1<x2≤tk+1, by the mean value theorem of differentiation, we have (36)x2-sα-1-x1-sα-1=α-1ξ1-sα-2x2-x1,x2-sα-2-x1-sα-2=2-αξ2-sα-3x2-x1,ξi∈x1,x2,s∈0,x1and absolute continuity of integral. Thus, we obtain that KP(I-Q)N(Ω¯) is equicontinuous. By the Arzela-Ascoli theorem, KP(I-Q)N is completely continuous. The proof is completed.Theorem 12.
Assume(C0-C1),(H0) and the following conditions hold: H1
There exists a constante0>0 such that if u′t>e0 for t∈[0,1], then(37)B1-kB2∫0tt-sα-1Γαfs,us,u′sds+∑ti<tIiuti+∑ti<tIi¯u′tit-ti≠0.H2
There exists a constanta0>0 such that either(38)cB1-kB2∫0tt-sα-1Γαfs,ucs,acds+∑ti<tIiucti+∑ti<tIi¯act-ti>0,or(39)cB1-kB2∫0tt-sα-1Γαfs,ucs,acds+∑ti<tIiucti+∑ti<tIi¯act-ti<0,if|c|>a0, where uc(t)=c(at-b).H3
For anyr>0, there exist nonnegative functions pr,qr,gr∈Y, nonnegative constants ai,bi,ei,gi,i=1,2,…,m and positive constants Mp1,Mq1,Mw1 such that (40)ft,u,v≤prtu+qrtv+wrt,∀t,u,v∈0,1×R2,Iix≤aix+ei,Ii¯y≤biy+gi,i=1,2,…,m,x,y∈R,where(41)A+Bt-baXmax⁡pr1+qr1Γα,Mp1+Mq1Γα-1+A∑i=1mai+bi+∑i=1mBai+B+2bi+2Mp1+Mq1Γα-1t-baX<1,maxt∈0,1⁡1-sα-2prts1≔Mp1<+∞,maxt∈0,1⁡1-sα-2qrts1≔Mq1<+∞,maxt∈0,1⁡1-sα-2wrts1≔Mw1<+∞.
Then the functional boundary value problem (1) has at least one solution in X.In order to prove Theorem12, we show three lemmas.Lemma 13.
The setΩ1=u∈domL\KerL:Lu=λNu,λ∈0,1 is bounded if (H2) and (H3) are satisfied.Proof.
Foru∈Ω1, Nu∈ImL, i.e.,(42)B1-kB2∫0tt-sα-1Γαfs,us,u′sds+∑ti<tIiuti+∑ti<tIi¯u′tit-ti=0.By (H1), there exists a constant t0∈[0,1] such that u′t0≤e0.
It follows fromLu=λNu that(43)ut=u0+u′0t+∫0tλt-sα-1Γαysds+∑ti<tIiuti+∑ti<tIi¯u′tit-ti,and, subsequently,(44)u′t=u′0+∫0tλt-sα-2Γα-1fs,us,u′sds+∑ti<tIi¯u′ti,which implies(45)u′t≤u′x0+1Γα-1∫0tt-sα-2fs,us,u′sds+∑i=1mIi¯u′ti≤1Γα-1∫0tt-sα-2prsus+t-sα-2qrsu′sds+∫0tt-sα-2wrsusds+∑i=1mIi¯u′ti+u′0≤e0+2Mp1+Mq1Γα-1+∑i=1mbiuX+2Mw1Γα-1+2∑i=1mgi,since(46)u′0≤u′t0+1Γα-1∫0t0t0-sα-2fs,us,u′sds+∑i=1mIi¯u′ti≤e0+Mp1+Mq1Γα-1+∑i=1mbiuX+Mw1Γα-1+∑i=1mgi.
Letu=u1+u2, where u1=(I-P)u∈domL∩KerP and u2=Pu∈ImP. Then since u1=(I-P)u∈domL∩KerP, u1=KPLu1=KPL(I-P)u=KPLu=λKPNu. As in the proof of Lemma 10,(47)u1t<Amax⁡f1Γα,maxt∈0,1⁡1-sα-2fts,uts,u′ts1Γα-1+∑i=1mIiuti+Ii¯u′ti,(48)u1′t<Bmax⁡f1Γα,maxt∈0,1⁡1-sα-2fts,uts,u′ts1Γα-1+∑i=1mIiuti+Ii¯u′ti,and as in (23),(49)u1X<Amax⁡f1Γα,maxt∈0,1⁡1-sα-2fts,uts,u′ts1Γα-1+∑i=1mIiuti+Ii¯u′ti,where A=a+b/a2+b2β2+1, B=b/a2+b2β2+1. Now, u2=u-u1, so u2′=u′-u1′ and (50)u2′t≤u′t+u1′t<Bmax⁡f1Γα,maxt∈0,1⁡1-sα-2fts,uts,u′ts1Γα-1+∑i=1mIiuti+Ii¯u′ti+e0+2Mp1+Mq1Γα-1+∑i=1mbiuX+2Mw1Γα-1+∑i=1mgi≤Bmax⁡f1Γα,maxt∈0,1⁡1-sα-2fts,uts,u′ts1Γα-1+∑i=1mBai+B+2bi+2Mp1+Mq1Γα-1uX+∑i=1mBei+B+2gi+2Mw1Γα-1+e0,
Recall thatu2(t)=Pu(t)=c(u)(at-b), where (51)cu=au′0-bu0a2+b2is introduced for the sake of brevity. Hence(52)u2′t=acu<Bmax⁡f1Γα,maxt∈0,1⁡1-sα-2fts,uts,u′ts1Γα-1+∑i=1mBai+B+2bi+2Mp1+Mq1Γα-1uX+∑i=1mBei+B+2gi+2Mw1Γα-1+e0≔Δ,that is, cu<1/a(Δ). Thus,(53)u2X=cuat-bX<t-baXΔ,for u∈domL∖KerL. By (49), (53), and (H3),(54)uX<u1X+u2X≤A+Bt-baXmax⁡pr1+qr1Γα,Mp1+Mq1Γα-1uX+∑i=1mAai+bi+∑i=1mBai+B+2bi+2Mp1+Mq1Γα-1t-baXuX+A+Bt-baXMw1Γα-1+wr1Γα+∑i=1mA+Bt-baXei+A+B+2t-baXgi+2Mw1Γα-1+e0t-baX,j=1,2,And, then, uX<∞, by (H3), i.e., Ω1 is bounded.Lemma 14.
The setΩ2=u∈KerL,Nu∈ImL is bounded if (H0) and (H2) hold.Proof.
Letu∈Ω2, and then uc(t)≡c(at-b),c∈R and QNuc(t)=(0,0,…,0)T. By (H2), we have c≤a0, so, uX≤a0at-bX; that is, Ω2 is bounded.
DefineJ:Z→X by Jz(t)=J(y(t),c1,c2,…,cm,d1,d2,…,dm)T=B1-kB2I0+αyt+∑ti<tci+∑ti<tdit-ti(at-b). Obviously, J:Z→KerL. If z(t)∈ImQ, then z(t)=(ch(t),0,0,…,0)T, where h is introduced in (C0), and (55)Jzt=Jcht=B1-kB2I0+αchtat-b=cat-b.Thus, J:ImQ→KerL is an isomorphism.Lemma 15.
The setΩ3=u∈KerL:ρλu+1-λJQNu=0,λ∈0,1 is bounded if conditions (H0) and (H2) are satisfied, where J:ImQ→KerL is defined above and(56)ρ=1,if39holds;-1.if42holds.Proof.
Suppose thatu∈Ω3, we have uc(t)=c(at-b),c∈R, and ρλuc+(1-λ)JQNuc=0. If λ=0, we have JQNuc=JQN[c(at-b)]=0. By Lemma 14, ucX≤a0at-bX. If λ=1, then uc=0. If λ∈(0,1), we suppose c>a0, and then (57)λc=-1-λρB1-kB2∫0tt-sα-1Γαfs,ucs,acds+∑ti<tIiucti+∑ti<tIi¯act-ti<0.Hence, (58)λc2=-1-λρcB1-kB2∫0tt-sα-1Γαfs,ucs,acds+∑ti<tIiucti+∑ti<tIi¯act-ti<0,which contradicts c2>0. So Ω3 is bounded.Proof of Theorem12.
SetΩ as a bounded open subset of X such that {0}∪⋃j=13Ω¯j⊂Ω. By Lemma 11, we know that N is L-compact on Ω¯. Then according to the above arguments, we have(i)
Lu≠λNu, for every (u,λ)∈[(domL∖KerL)∩∂Ω]×(0,1);(ii)
Nu∉ImL, for every u∈KerL∩∂Ω;
At last we will prove that (iii) of Theorem6 is satisfied. If I:KerL→KerL is the identity map, we define the homotopy H(u,λ)=±ρλIu+(1-λ)JQNu. If for every u∈∂Ω∩KerL, by the homotopic property of degree, we have (59)deg⁡JQNKerL,Ω∩KerL,0=deg⁡H·,0,Ω∩KerL,0=deg⁡H·,1,Ω∩KerL,0=deg⁡±I,Ω∩KerL,0≠0.The assumption (iii) of Theorem 6 is verified and the proof is completed.
Then by Theorem6, the functional boundary value problem (1) has at least one solution in X. The proof of Theorem 12 is also completed.In the next result the inequalityu′t>e0 is replaced with ut>M, which will lead to slight modifications of the proof of Lemma 13, we recall that a2+b2≠0.Theorem 16.
Assume thata=0, (C0-C1),(H0),(H2) and the following conditions hold.(H4)
There exists a constantM>0 such that if |u(t)|>M, then(60)B1-kB2∫0tt-sα-1Γαfs,us,u′sds+∑ti<tIiuti+∑ti<tIi¯u′tit-ti≠0.(H5)
For anyr>0, there exist nonnegative functions pr,qr,gr∈Y, nonnegative constants ai,bi,ei,gi,i=1,2,…,m, and positive constants Mp1,Mq1,Mw1 such that (61)ft,u,v≤prtu+qrtv+wrt,∀t,u,v∈0,1×R2,Iix≤aix+ei,Ii¯y≤biy+gi,i=1,2,…,m,x,y∈R,where(62)max⁡pr1+qr1Γα,Mp1+Mq1Γα-1+∑i=1mai+bi<12A,A=a+ba2+b2β2+1,maxt∈0,1⁡1-sα-2prts1≔Mp1<+∞,maxt∈0,1⁡1-sα-2qrts1≔Mq1<+∞,maxt∈0,1⁡1-sα-2wrts1≔Mw1<+∞.Then the boundary value problem (1) has at least one solution.Proof.
As in the proof of Lemma13, by (H4), there exists a constant M>0,t0′∈[0,1] such that ut0′≤M. Since u(t)=u1(t)+u2(t), where u1=(I-P)u∈domL∩KerP and u2=Pu∈ImP, dimilarly, u1=KPLu1=λKPNu. As in the proof of Lemma 13,(63)u1t<KPNut≤Amax⁡f1Γα,maxt∈0,1⁡1-sα-2fts,uts,u′ts1Γα-1+∑i=1mIiuti+Ii¯u′ti,where A=(a+b/a2+b2β2+1). Again, (49) holds. Now, u2=u-u1, so, by (63),(64)u2t0′≤ut0′+u1t0′<M+Amax⁡f1Γα,maxt∈0,1⁡1-sα-2fts,uts,u′ts1Γα-1+∑i=1mIiuti+Ii¯u′ti.As in the proof of Lemma 13, u2(t)=Pu(t)=c(u)(at-b).
Hence(65)u2x1=cuat0′-b<M+Amax⁡f1Γα,maxt∈0,1⁡1-sα-2fts,uts,u′ts1Γα-1+∑i=1mIiuti+Ii¯u′ti.Since a=0, we have(66)cu≤1bM+Amax⁡f1Γα,maxt∈0,1⁡1-sα-2fts,uts,u′ts1Γα-1+∑i=1mIiuti+Ii¯u′ti.Thus,(67)u2X=cuat-bX<M+Amax⁡f1Γα,maxt∈0,1⁡1-sα-2fts,uts,u′ts1Γα-1+∑i=1mIiuti+Ii¯u′ti,for u∈domL\KerL.
By (63) and (67),(68)uX<u1X+u2X≤2Amax⁡pr1+qr1Γα,Mp1+Mq1Γα-1uX+2A∑i=1mai+biuX+2AMw1Γα-1+wr1Γα+2A∑i=1mei+gi+M,where A=a+b/a2+b2β2+1, by (H5). Then, (69)uX<2AMw1/Γα-1+wr1/Γα+2A∑i=1mei+gi+M1-2Amax⁡pr1+qr1/Γα,Mp1+Mq1/Γα-1+2A∑i=1mai+bi.i.e., Ω1 is bounded. The rest of the proof repeats that of Theorem 12, which completes the proof of Theorem 16.



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*Source: 4208636-2019-06-02.xml*