# Impulsive Fractional Differential Equations withp-Laplacian Operator in Banach Spaces

Authors: Jingjing Tan; Kemei Zhang; Meixia Li
Journal: Journal of Function Spaces
 (2018)

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## Abstract

In this paper, we study a class of boundary value problem (BVP) with multiple point boundary conditions of impulsive p-Laplacian operator fractional differential equations. We establish the sufficient conditions for the existence of solutions in Banach spaces. Our analysis relies on the Kuratowski noncompactness measure and the Sadovskii fixed point theorem. An example is given to demonstrate the main results.


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## Body

## 1. Introduction

With the development of the theory of fractional order calculus, fractional differential equations are getting more and more extensively used (see[1–15]). For instance, the impulsive fractional differential equations are widely used in various scientific fields, such as the problem of dynamics of populations subject to abrupt changes, harvesting, diseases, and so on. Lakshmikantham et al. [16], Bainov and Simeonov [17], and Benchohra et al. [18] have done in-depth studies on this issue. Moreover, the p-Laplacian operator differential equation was first proposed by Leibenson [19] in order to study the problem of turbulent flow in a porous medium. He converted this problem into the existence of solution of the following differential equation:(1)φpu′t′=ft,ut,t∈0,1.φp is the p-Laplacian operator, φq is the inverse function of φp with φq(s)=sq-2s, and p,q satisfy 1/p+1/q=1. In recent years, many results about the solutions of the p-Laplacian operator fractional differential equation BVP have been obtained (see [20–26]). The research of the solutions of the BVP with p-Laplacian operator and with impulsive has been attracting increasing interest.Zhao and Gong [27] study the solution of the following impulsive fractional differential equations with generalized periodic boundary conditions:(2)DqCut=ft,ut,Δutt=tk=Ikutk,k=1,2,⋯,m,Δu′tt=tk=Jkutk,k=1,2,⋯,m,au0-bu1=0,au′0-bu′1=0.By means of the Schauder and Guo- Krasnosel’skii fixed point theorem, they get the existence of single and multiple positive solutions of the above BVP. By the technique of the Guo-Krasnosel’skii fixed point theorem and the Leggett-Williams theorem, Wang et al. [28] obtained the results of the following BVP.(3)D0+ηϕpD0+αut=ft,ut,0<α≤2,0<η≤1,u0=0,u1=auξ,D0+αu0=0,0<ξ<1,0≤a≤1The results about the BVP with multiple point boundary conditions of impulsive p-Laplacian operator fractional differential equations are few, especially in Banach space.In this paper, we study the following BVP with multiple point boundary conditions of impulsive p-Laplacian operator fractional differential equations in Banach spaceE:(4)D0+βφpD0+αxt=ft,xt,x′t,1<α≤2,0<β≤1,Δxtt=tk=Ikxtk,Δx′tt=tk=Jkxtk,x0=x′0=∫01a1xsds,x1=x′1=∫01a2xsds,D0+αx0=θ, where θ is the zero element of E, D0+α and D0+β are the fractional derivatives of order α(α>0), I=[0,1], t∈I′=I∖{t1,t2,⋯,tm}, φp(s)=sp-2s with p>1. f:I×E×E→E, Ik,Jk:E→E, and a1,a2:E→E are continuous. The impulsive point set {tk}k=1m satisfies 0=t0<t1<⋯<tm<tm+1=1; we denote T0=[t0,t1], Tk=(tk,tk+1],1≤k≤m. We denote by x(tk+) and x(tk-) the right and left limits of x(t) at the point t=tk, i.e., x(tk+)=limh→0+x(tk++h) and x(tk-)=limh→0-x(tk++h). We establish the existence of solution to BVP (4). by the technique of the Kuratowski noncompactness measure and the Sadovskii fixed point theorem. The main innovations of this paper are as follows. Firstly, we study the impulsive fractional differential equations with the p-Laplacian Operator. Secondly, we study the BVP in Banach space. Thirdly, the nonlinear term contains the derivatives x′(t).The paper is organized as follows. In Section2, we recall some definitions and lemmas. In Section 3, the main results of this paper are discussed. Finally, one example is given in Section 4.


## 2. Preliminaries

First, we recall some definitions and preliminary.Definition 1 (see [29]).
Letα>0, f∈C([0,∞),R).
( 1 ) The fractional integral of order α(α>0) of f is given by (5)I0+αfx=1Γα∫0xftx-t1-αdt;(2) the fractional derivative of order α(α>0) of f is given by (6)D0+αft=1Γn-αddtn∫0tfst-sα-n+1ds,where n=[α]+1, [α] denotes the integer part of α and the right side integral is pointwise defined on [0,∞).Lemma 2 (see[30, 31]).
Letf(x) be integrable, α>0, β>α>0. Then
(1)(7)I0+αD0+αfx=fx+c1xα-1+c2xα-2+⋯+cnxα-n, where ci∈R, i=1,2,3,⋯,N, N=⌈α⌉.
(2)(8)D0+αI0+βfx=I0+β-αfx,D0+αI0+αfx=fx.Definition 3 (see [32]).
LetS be a bounded set in a real Banach space E; the Kuratowski noncompactness measure of S is given by (9)αS=infδ>0:S=⋃i=1mSi,diamSi<δ,i=1,2,⋯,m, where diam(Si) denote the diameters of Si.Remark 4.
From the definition, it is obvious that0≤α(S)<∞.Lemma 5 (see [33]).
IfH⊂C(I) is bounded and equicontinuous, then αE(H(t)) is continuous on I and αC(H)=maxt∈IαE(H(t)), αE(∫Ix(t)dt:x∈H)≤∫IαE(H(t))dt, where H(t)={x(t):x∈H} for each t∈I.Definition 6 (see [32]).
LetE1 and E2 be real Banach spaces, D⊂E1, and A:D→E2 be a continuous and bounded operator; A is called a k-set contraction operator if there exists a constant k≥0, for any bounded set S in D, such that α(A(S))≤kα(S).Remark 7 (see [34]).
A is called a strict set contraction operator if k<1. It is clear that a strict set contraction operator is a condensing operator.
In the following, we define the basic space of this paper. Denote(10)PCI=x∈CI∣x∈CI′∩C′I′,xtk-,x′tk-,xtk+andx′tk+exist  withxtk-=xtkandx′tk-=x′tk,1≤k≤mQCI=x∈PCI:supt∈Ixt1+t<+∞,DCI=x∈PCI:supt∈Ixt1+t<+∞andsupt∈Ix′t<+∞,and it is easy to see that DC(I) is Banach space with the norm (11)xD=maxsupt∈Ixt1+t,x′C. The basic space used in this paper is DC(I). The Kuratowski noncompactness measures in E, C(I), and DC(I) are denoted by αE(·), αC(·), and αD(·), respectively.The following Sadovskii fixed point theorem is needed for the proof of our main results.Lemma 8 ((Sadovskii)(see [32])).
LetD be a bounded, closed, and convex subset of the Banach space E. If the operator A:D→D is condensing, then A has a fixed point in D.


## 3. Existence of Solutions

Before proceeding further, let us give some denotations as follows:(12)M=10m+4ΓαΓβq-1,l=M∫011+tl1t+l2tdt,Kρ=x∈DCI:xD≤ρ,Kr=x∈E:x≤r,Fs=1Γβ∫0ss-τβ-1fτ,xτ,x′τdτ,L=max⁡L1,L2,L3,L4,M1=maxmaxt∈Iat,maxt∈Ibt,maxt∈Ict,M2=maxi=1,⋯,ma1x,a2x,Iix,Jix.For simplicity of presentation, we list some conditions.( H 1 )   a , b , c ∈ C ( I ) are nonnegative functions and satisfy (13)∫0tfs,x,yds≤φpatx+bty+ct,∀x∈E,∫011+tat+btdt<M-1,∫01ctdt<+∞.( H 2 ) For any r>0, [α,β]⊂I, f(t,x,y) is uniformly continuous on [α,β]×Kr×Kr.( H 3 ) For any x,y∈E,ak(x)-ak(y)≤Lkx-y(k=1,2), Ii(x)-Ii(y)≤L3x-y, and Ji(x)-Ji(y)≤L4x-y, i=1,2,⋯,m.( H 4 ) For all t∈I and all bounded subsets D1,D2⊂E, there exist l1,l2∈L[0,∞) such that (14)∫0tαEfs,D1,D2ds≤φpl1tαED1+l2tαED2 with l<1.Lemma 9.
Giveny∈C(I), the following BVP,(15)D0+βφpD0+αxt=yt,Δxtt=tk=Ikxtk,Δx′tt=tk=Jkxtk,x0=x′0=∫01a1xsds,x1=x′1=∫01a2xsds,D0+αx0=θ, has a unique solution satisfying the following.(16)xt=1Γα∫0tt-sα-1φqI0+βysds+∑i=1m+1α2-tΓα∫ti-1titi-sα-1φqI0+βysds+1Γα∫tktt-sα-1φqI0+βysds+2-t∫01a1xs-a2xsds+∑i=1m2-t1-tmΓα-1∫ti-1titi-sα-2φqI0+βysds+Iixti+2-tmJixti+∑i=1m-12-ttm-tiΓα-1∫ti-1titi-sα-2φqI0+βysds+tm-tiJixti+∑i=1kt-tiΓα-1∫ti-1titi-sα-2φqI0+βysds+1Γα∫ti-1titi-sα-1φqI0+βysds+∑i=1kIixti+t-tkJixti+∑i=1k-1tk-tiJixtiProof.
From (15) and Lemma 2, we know(17)φpD0+αxt=I0+βyt+c0.Because of D0+αx(0)=θ, we can obtain that c0=0. According to the definition of the p-Laplacian operator it follows that(18)D0+αxt=φqI0+βyt.Therefore(19)xt=1Γα∫0tt-sα-1φqI0+βysds-c1-c2t,(20)x′t=1Γα-1∫0tt-sα-2φqI0+βysds-c2.If t∈T1, then(21)xt=1Γα∫t1tt-sα-1φqI0+βysds-d1-d2t-t1,x′t=1Γα-1∫t1tt-sα-2φqI0+βysds-d2.For d1 and d2∈R, we have(22)xt1-=1Γα∫0t1t1-sα-1φqI0+βysds-c1-c2t1,xt1+=-d1,x′t1-=1Γα-1∫0tt-sα-2φqI0+βysds-c2,x′t1+=-d2.When t∈Tk, we can also obtain(23)xt=1Γα∫tktt-sα-1φqI0+βysds+∑i=1k-1∫ti-1tititk-titi-sα-2Γα-1+tk-tiJixti+∑i=1k∫ti-1titi-sα-1Γα+t-tkti-sα-2Γα-1+Iixti+t-tkJixti-c1-c2t,and taking the boundary condition of (15) into consideration, some tedious manipulation yields the following.(24)c1=-∑i=1m+1∫ti-1titi-sα-1Γα+ti-sα-2Γα-1φqI0+βysds-∑i=1m∫ti-1ti1-tmti-sα-2Γα-1φqI0+βysds+Iixti+2-tmJixti-∑i=1m-1∫ti-1titm-titi-sα-2Γα-1φqI0+βysds+tm-tiJixti-∫012a1xs-a2xsds(25)c2=∑i=1m+1∫ti-1titi-sα-1Γα+ti-sα-2Γα-1φqI0+βysds+∑i=1m∫ti-1ti1-tmti-sα-2Γα-1φqI0+βysds+Iixti+2-tmJixti+∑i=1m-1∫ti-1titm-titi-sα-2Γα-1φqI0+βysds+tm-tiJixti+∫01a1xsdsSubstituting (24) and (25) into (19), we can get (16), which implies that the solution of BVP (15) is given by (16).From Lemma9, we can establish the following conclusion.Lemma 10.
If(H1) is satisfied, then BVP (4) has a unique solution satisfying(26)xt=1Γα∫0tt-sα-1φqFsds+∑i=1m+1α2-tΓα∫ti-1titi-sα-1φqFsds+1Γα∫tktt-sα-1φqFsds+2-t∫01a1xs-a2xsds+∑i=1m2-t1-tmΓα-1∫ti-1titi-sα-2φqFsds+Iixti+2-tmJixti+∑i=1m-12-ttm-tiΓα-1∫ti-1titi-sα-2φqFsds+tm-tiJixti+∑i=1kt-tiΓα-1∫ti-1titi-sα-2φqFsds+1Γα∫ti-1titi-sα-1φqFsds+∑i=1kIixti+t-tkJixti+∑i=1k-1tk-tiJixti.Proof.
The proof is almost identical to Lemma9, with the major change being the substitution of y(t) for f(t,x(t),x′(t)).Remark 11.
From Lemma10, we can get the conclusion that the solutions to the BVP (4) are equivalent to the fixed point of the following operator:(27)Txt=1Γα∫0tt-sα-1φqFsds+∑i=1m+1α2-tΓα∫ti-1titi-sα-1φqFsds+1Γα∫tktt-sα-1φqFsds+2-t∫01a1xs-a2xsds+∑i=1m2-t1-tmΓα-1∫ti-1titi-sα-2φqFsds+Iixti+2-tmJixti+∑i=1m-12-ttm-tiΓα-1∫ti-1titi-sα-2φqFsds+tm-tiJixti+∑i=1kt-tiΓα-1∫ti-1titi-sα-2φqFsds+1Γα∫ti-1titi-sα-1φqFsds+∑i=1kIixti+t-tkJixti+∑i=1k-1tk-tiJixti,∀x∈DCI. Taking derivative to both sides of (27), we have(28)T′xt=1Γα-1∫0tt-sα-2φqFsds+1Γα-1∫tktt-sα-2φqFsds-∫01a1xs-a2xsds-∑i=1m+1αΓα∫ti-1titi-sα-1φqFsds-∑i=1m1-tmΓα-1∫ti-1titi-sα-2φqFsds+Iixti+2-tmJixti-∑i=1m-1tm-tiΓα-1∫ti-1titi-sα-2φqFsds+tm-tiJixti+∑i=1k1Γα-1∫ti-1titi-sα-2φqFsds+Jixti,∀x∈DCI.Lemma 12.
If(H1), (H2), and (H3) are satisfied, then operator T:DC(I)→DC(I) is continuous and bounded.Proof.
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Step 1. For any x∈DC(I), we prove that T is well defined and (Tx)(t)∈DC(I). From condition (H1), we have(29)∫01φqFsds=∫01φq1Γβ∫0ss-τβ-1fτ,xτ,x′τdτds≤∫01φq1Γβ∫0sfτ,xτ,x′τdτ≤φq∫011Γβφpasx+bsx′+csds≤Γβ1-qM13xD+1.Together with the definition of operator T, we have(30)Txt1+t≤11+t1Γα∫0tt-sα-1φqFsds+∑i=1m+1α2-tΓα∫ti-1titi-sα-1φqFsds+1Γα∫tktt-sα-1φqFsds+2-t∫01a1xs-a2xsds+∑i=1m2-t1-tmΓα-1∫ti-1titi-sα-2φqFsds+Iixti+2-tmJixti+∑i=1m-12-ttm-tiΓα-1∫ti-1titi-sα-2φqFsds+tm-tiJixti+∑i=1kt-tiΓα-1∫ti-1titi-sα-2φqFsds+1Γα∫ti-1titi-sα-1φqFsds+∑i=1kIixti+t-tkJixti+∑i=1k-1tk-tiJixti≤10m+4Γα∫01φqFsds+11m+1M2<+∞.As above, from (28), a tedious calculation gives(31)Tx′t≤5m+3Γα∫01φqFsds+5m+1M2<+∞.From (30) and (31), we have that (Tx)(t) is well defined and (Tx)(t)∈DC(I) for any x∈DC(I).
Step 2. We prove that T is bounded. For any x∈Kρ, from (29)−(31), we get(32)Txt1+t≤3ρ+1MM1+11m+1M2,Tx′t≤3ρ+1MM1+5m+1M2.So T is bounded operator.
Step3. It is time to prove that T is continuous. Let xn,x∈DC(I), and for any s∈I, ε>0, there exists N1>0, when n≥N1, satisfying that xn-xD<ε/(22m+2)L. So {xn} is a bounded subset of DC(I); let η>0 such that xnD≤η; taking limit, we obtain xD≤η. From (H2), we know that, for any ε>0 and s∈I, there exists N2>0; when n≥N2, we have(33)fs,xns,xn′s-fs,xs,x′s≤M1/1-qε21/q-1.Taking N=max{N1,N2}, for any ε>0, n≥N, and t∈I, according to (27), (28), we have(34)Txnt1+t-Txt1+t≤10m+4Γα1+t∫01φq1Γβ∫0sfτ,xnτ,xn′τ-fτ,xτ,x′τdτds+2-t1+t∫01a1xns-a1xs+a1xns-a1xsds+11+t3mL3+8m-3L4xns-xs≤ε21+t+11m+11+tLxns-xs≤ε21+t+ε21+t<ε,(35)Txn′t-Tx′t≤5mΓα∫01φq1Γβ∫0sfτ,xnτ,x′τ-fτ,xτ,x′τdτds+∫01a1xns-a1xs+a1xns-a1xsds+mL3+4m-1L4xns-xs≤5m+3ε10m+4+5m+1ε22m+2<ε.Thus Txnt/1+t-Txt/1+tD<ε. So T:DC(I)→DC(I) is continuous.Lemma 13.
If(H1)-(H3) are satisfied, U is a bounded subset of DC(I). Then (TU)(t)/(1+t) and TU′(t) are equicontinuous on I.Proof.
In fact, from the boundedness ofU, that is, for any x∈U, there exists η>0 such that xD≤η. Suppose that t′,t′′∈I with t′<t′′; by the mean value theorem, we have the following.(36)Txt′′1+t′′-Txt′1+t′≤1Γα∫0t′′t′′-sα-11+t′′φqFsds-∫0t′t′-sα-11+t′φqFsds+1Γα∫tkt′′t′′-sα-11+t′′φqFsds-∫tkt′t′-sα-11+t′φqFsds+3m+2Γα-1∫01φqFsds+3mM12-t′′1+t′′-2-t′1+t′+mΓα-1t′′1+t′′-t′1+t′∫01φqFsds+mM1t′′-t′≤1Γα∫0t′t′′-sα-11+t′′φqFsds+∫t′t′′t′′-sα-11+t′′φqFsds-∫0t′t′-sα-11+t′φqFsds+1Γα∫tkt′t′′-sα-11+t′′φqFsds+∫t′t′′t′′-sα-11+t′′φqFsds-∫tkt′t′-sα-11+t′φqFsds+33m+2Γα-1∫01φqFsds+3mM1t′′-t′+mΓα-1∫01φqFsdst′′-t′+mM1t′′-t′≤1Γα∫0t′t′′-sα-11+t′′-t′-sα-11+t′φqFsds+1Γα∫t′t′′t′′-sα-11+t′′φqFsds+1Γα∫tkt′t′′-sα-11+t′′-t′-sα-11+t′φqFsds+1Γα∫t′t′′t′′-sα-11+t′′φqFsds+33m+2Γα-1∫01φqFsds+3mM1t′′-t′+mΓα-1∫01φqFsdst′′-t′+mM1t′′-t′≤2Γαt′′α1+t′′-t′α1+t′-t′′-t′α1+t′′∫0t′φqFsds+2Γα∫01φqFsdst′′-t′+33m+2Γα-1∫01φqFsds+3mM1t′′-t′+mΓα-1∫01φqFsdst′′-t′+mM1t′′-t′≤10m+18Γα∫01φqFsds+10mM1t′′-t′Let (37)δ=10m+183η+1M1ΓαΓβq-1+10mM1-1·ε2.From (37),(38)Txt′′1+t′′-Txt′1+t′<ε.For the case of t′≥t′′, using the same methods we can also get (38). So (TU)(t)/(1+t) is equicontinuous on I. An argument similar to the one used in (TU)(t)/(1+t) shows that(39)Tx′t′′-Tx′t′<ε.So TU′(t) is equicontinuous on I.Lemma 14 (see [35]).
Let condition(H1) be satisfied and U be a bounded subset of DC(I). Then (40)αDTU=maxsupt∈IαETUt1+t,supt∈IαETU′t.Theorem 15.
Let conditions(H1)-(H4) be satisfied. Then BVP (4) has at least one solution belonging to DC(I).Proof.
By Remark11, we need only to show that the operator T has a fixed point in DC(I). Let (41)KR=x∈DCI:xD≤R,where (42)R>11m+1M-1M2+∫01csds·M-1-∫011+tat+btdt-1.
Step 1. we prove that TKR⊂KR. In fact, for any x∈KR,t∈I, by (27), we have(43)Txt1+t≤10m+4Γα∫01φqFsds+11m+1M2≤M∫011+tat+btdtxD+∫01ctdt+11m+1M2≤M∫011+tat+btdtR+RM-1-∫011+tat+btdt-11m+1M-1M2+11m+1M2<R,(44)Tx′t≤5m+3Γα∫01φqFsds+5m+1M2<M∫011+tat+btdtxD+∫01ctdt+5m+1M2≤M∫011+tat+btdtR+RM-1-∫011+tat+btdt-11m+1M-1M2+5m+1M2<R.From (43) and (44), we know TxD<R. Thus, from Lemma 12, TKR⊂KR follows.
Step 2. It is time to prove that T is a strict set contraction operator. Let D=co¯D(TKR). It is easy to see that D is nonempty, bounded, convex, and closed subset of KR. According to Lemma 13, (TKR)(t)/(1+t) and TKR′(t) are equicontinuous on I, so (D)(t)/(1+t) and D′(t) are equicontinuous on I. From the definition of D, we get D⊂KR and TKR⊂D, and by Lemma 12, T:D→D is bounded and continuous. From (H2), it follows that {f(s,x(s),x′(s)):x∈D} are equicontinuous on I. For any t∈I and U⊂D, by (H3) and Lemma 2, we obtain(45)αETUt1+t≤11+t1Γα∫0tt-sα-1φq1Γβ∫0ss-τβ-1αEfτ,xτ,x′τ:x∈Udτds+∑i=1m+1α2-tΓα∫ti-1titi-sα-1φq1Γβ∫0ss-τβ-1αEfτ,xτ,x′τ:x∈Udτds+1Γα∫tktt-sα-1φq1Γβ∫0ss-τβ-1αEfτ,xτ,x′τ:x∈Udτds+∑i=1m2-t1-tmΓα∫ti-1titi-sα-2φq1Γβ∫0ss-τβ-1αEfτ,xτ,x′τ:x∈Udτds+∑i=1kt-tiΓα-1∫ti-1titi-sα-2φq1Γβ∫0ss-τβ-1αEfτ,xτ,x′τ:x∈Udτds+∑i=1k1Γα∫ti-1titi-sα-1φq1Γβ∫0ss-τβ-1αEfτ,xτ,x′τ:x∈Udτds≤10m+41+t∫01φq1ΓβαEfτ,xτ,x′τ:x∈Udτds≤M1+t∫01l1sαEUs+l2sαEU′sds≤M1+t∫011+sl1s+l2sdsαDU.Since t is arbitrary, we obtain(46)supt∈IαETUt1+t≤lαDU.Using similar methods, we can also show that(47)supt∈IαETU′t≤lαDU.Form (46) and (47), using Lemma 14 we get (48)αDTU≤lαDU. Obviously, 0≤l<1, From Remark 7, T is a condensing operator too. It follows from the Sadovskii fixed point theorem that T has at least one fixed point in D; therefore, BVP(4) has at least one solution in DC(I).Remark 16.
IfE=[0,∞), we can get the following result.Corollary 17.
Letf∈C[I×[0,∞)×[0,∞),[0,∞)] and (H1)-(H4) be satisfied. Then BVP (4) has at least one solution in DC[I×[0,∞)×[0,∞),[0,∞)].Proof.
LettingE=[0,∞) in Theorem 15, we can prove the desired result.Corollary 18.
Suppose the following assumptions holds:
( H 1 )   a , b , c ∈ C ( I ) are nonnegative functions and satisfy (49)∫0tfs,xds≤φpatx+bt,∀x∈E,∫011+tatdt<M-1,∫01btdt<+∞;
( H 2 ) for any r>0, [α,β]⊂I, f(t,x) is uniformly continuous on [α,β]×Kr;
( H 3 ) for any x,y∈E,ak(x)-ak(y)≤Lkx-y(k=1,2), Ii(x)-Ii(y)≤L3x-y, and Ji(x)-Ji(y)≤L4x-y,i=1,2,⋯,m;
( H 4 ) for all t∈I and all bounded subsets D⊂E, there exists l1∈L[0,∞) such that (50)∫0tαEfs,Dds≤φpl1tαED with l=M∫01l1(t)dt<1. Then the following BVP,(51)D0+βφpD0+αxt=ft,xt,Δxtt=tk=Ikxtk,Δx′tt=tk=Jkxtk,x0=x′0=∫01a1xsds,x1=x′1=∫01a2xsds,D0+αx0=θ,has at least one solution belonging to QC(I).Proof.
The proofs are similar to Theorem15, except the need for substitution of f(t,x(t)) for nonlinear term f(t,x(t),x′(t)).


## 4. Illustrative Example

LetE=l∞={x=(x1,x2,⋯,xn,⋯),supnxn<+∞,t∈I} and equipped with the norm x=supnxn. The following BVP(52)D0+1/2φ3/2D0+3/2xnt=xnt81+t1+t2+sinx2nt2+sinxn′t236n3et1/2,Δxntt=1/2=xn1/2-2+xn1/2-,Δxn′tt=1/2=xn′1/2-2+xn′1/2-,xn0=xn′0=∫01xns4+xnsds,xn1=xn′1=∫01xns6+xnsds,D0+3/2xn0=0can be regarded as a problem with the form of BVP (4), where(53)ft,x,y=f1t,x,y,f2t,x,y,⋯,fnt,x,y,⋯,fnt,x,y=xnt81+t1+t2+sinx2nt2+sinynt236n3et1/2.Clearly L1=1/4, L2=1/6, L3=L4=1/2, and (54)∫0tfns,x,yds≤φ3/2181+t1+t2+136etx+136ety. Let a(t)=1/8(1+t)(1+t2)+1/36et, b(t)=1/36et, and c(t)=0. According to Γ(1/2)≈1.772 and Γ(3/2)≈0.8862, we have (55)M=10m+4ΓαΓβq-1=6.6979,∫011+tat+btdt=0.0242<M-1,∫01ctdt=0<+∞,l=M∫011+tl1t+l2tdt=0.162<1where l1(t)=1/8(1+t)(1+t2)+1/36et and l2(t)=1/36et. So, all the conditions of Theorem 15 are satisfied. Therefore, BVP (52) has at least one solution.


## 5. Conclusion

In this paper, we present some sufficient conditions which ensure the existence of solution to BVP (4) in Banach spaces. Through construction space DC(I), using the technique of the Kuratowski noncompactness measure and the Sadovskii fixed point theorem, we obtain some new existence criteria for BVP (4). As far as we know, only a few papers have dealt with the boundary value problem for impulsive p-Laplacian fractional differential equations, especially in Banach space.



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*Source: 2503915-2018-11-15.xml*