# Some Existence Results for High Order Fractional Impulsive Differential Equation on Infinite Interval

Authors: Zhaocai Hao; Tian Wang
Journal: Mathematical Problems in Engineering
 (2020)

---

## Abstract

In this paper, we consider the high order impulsive differential equation on infinite intervalD0+αut+ft,ut,J0+βut,D0+α−1ut=0,t∈0,∞∖tkk=1m△utk=Ikutk,t=tk,k=1,…,mu0=u′0=⋯=un−20=0,D0+α−1u∞=u0 By applying Schauder fixed points and Altman fixed points, we obtain some new results on the existence of solutions. The nonlinear term of the equation contains fractional integral operator Jβut and lower order derivative operator D0+α−1ut. An example is presented to illustrate our results.


---

## Body

## 1. Introduction

In this paper, we are concerned with the following impulsive differential equation on infinite interval:(1)D0+αut+ft,ut,J0+βut,D0+α−1ut=0,t∈0,∞∖tkk=1m,△utk=Ikutk,t=tk,k=1,…,m,u0=u′0=⋯=un−20=0,D0+α−1u∞=u0.where u0∈R, α,β∈n−1,n, n>2, D0+α is the standard Riemann–Liouville fractional derivative, 0=t0<t1<t2<⋯<tm<∞, △utk=utk+−utk−, utk−=utk, utk+=limh⟶0+utk+h and utk−=limh⟶0−utk−h represent the right and left limits of ut at t=tk, and D0+α−1u∞=limt⟶∞D0+α−1ut. Also, f∈C0,+∞×R×R×R,R, Ik∈CR,R.During the past decades, fractional differential equations have drawn wide concerns. Compared with integer order differential equations, fractional differential equations have more extensive application range, such as control theory, physics, aerodynamics, polymer rheology, chemistry, biology, and so forth. There are many papers focused on the existence of positive solutions for fractional differential equations (see [1–3]).Since the last century, the dynamics of populations subject to abrupt changes was described by impulsive differential system. And other phenomena, for instance, harvesting, diseases, and so on, also have been described by using impulsive differential systems. Impulsive differential equations of fractional order play an important role in fractional differential equations theory and applications. Recently, impulsive fractional differential equations have been studied extensively. For example, Wang et al. studied the existence and multiplicity of solutions for impulsive fractional boundary value problem with p-Laplacian in [4], and Liu considered fractional impulsive differential equations using bifurcation techniques in [5]. For more articles related to impulsive fractional differential equations, refer to [6–12].Recently, in [13], Liu investigated the existence of solutions for higher order impulsive fractional differential equations given by(2)D0+qcxt=Ft,xt,t∈ti,ti+1,i∈N0,△x|t=ti=Iti,xti,i∈N,x0=x0,D0+qxt=Gt,xt,t∈ti,ti+1,i∈N0,limt⟶ti+t−ti1−αxt=Jti,xti,i∈N,limt⟶0+t1−qxt=x0,where q∈0,1, t∈0,T, 0=t0<t1<t2<⋯<tm<tm+1<T, I,J:tk:k∈N×R⟶R are discrete Carathéodory functions, and F,G:0,T×R⟶R are strong Carathéodory functions. By using Schauder’s fixed-point theorem, Liu established some existence results.In [10], Liu and Ahmad studied the following problems:(3)D0+αcxt=qtft,xt,cD0+pxt,t∈0,∞,△xtk=Iktk,xtk,k=1,2,…,x0=x0,D∗αcxt=qtft,xt,cD∗pxt,t∈0,∞,△xtk=Iktk,xtk,k=1,2,…,x0=x0,where x0∈R, α∈0,1, 0<p<α, 0=t0<t1<t2<⋯ with limk⟶∞tk=∞, q:0,∞⟶R satisfies that there exists l>−α such that qt≤tl for all t∈0,∞, and q may be singular at t=0. And f:0,∞×R2⟶R is a Carathéodory function, Ik:0,∞×R⟶Rk=1,2,…,, Ik is a Carathéodory function sequence, and △xtk=limt⟶tk+xt−limt⟶tk−xt, k=1,2,…,. By using Schauder’s fixed-point theorem, the authors studied the existence of solution. And the authors also considered the uniqueness of solution under some appropriate conditions.In [9], Zhao and Ge considered the following boundary value problem:(4)D0+αut+ft,ut=0,t∈0,∞,t≠tk,k=1,2,…,m,utk+−utk−=−Ikutk,k=1,2,…,m,u0=0,D0+αu∞=0,where α is a real number with 1<α≤2, D0+α is the standard Riemann–Liouville fractional derivative, t0=0, 1<t1<t2<⋯<tm<∞, utk+=limh⟶0+utk+h, utk−=limh⟶0+utk−h, D0+α−1u∞=limt⟶∞D0+α−1ut, ft,1+tαu:0,∞×0,∞⟶0,∞ is continuous, and Ik:0,∞⟶0,∞k=1,2,…,m are continuous. Wang and Ge proved that the problem they studied has at least three positive solutions.Motivated by the aforementioned work, we studied existence of solution of problem (1) by Schauder’s fixed-point theorem and Altman’s fixed-point theorem. The main features of this paper are as follows. Firstly, the nonlinear term not only involved fractional order derivative but also contained fractional integral. Compared with [9, 10, 13], our nonlinear terms are more general. Many articles contain derivatives for nonlinear terms, but few articles contain both derivatives and integrals. Secondly, we studied the problem on the infinite interval. To the best of our knowledge, there are few articles involving the impulsive fractional order differential equations on the infinite interval. If the nonlinear term contained fractional integral and t∈0,∞, it will bring new obstacles to solve the problem. For this purpose, we overcome obstacles by constructing a special cone. Thirdly, our problem is higher order impulsive fractional equation. Compared with [9], we allowed α∈n−1,n, where n>2. It is obvious that our problem is more general.This paper is organized as follows. In Section2, we introduce some definitions and lemmas. In Section 3, we give our main results by fixed-point theorem. In Section 4, one example is presented to illustrate the main results.


## 2. Preliminaries and Lemmas

Letu:0,∞⟶ℝ, J=0,∞, J0=0,t1, Jm=tm,∞, Jk=tk,tk+1, k=1,…,m−1. For k=1,2,…,m, define the function ukt=ut. Let CJ,R be the Banach space of continuous functions from J to ℝ. Let us to introduce the Banach spaces(5)PCJ,ℝ=u:uk∈CJk,ℝ,k=0,1,…,m,utk+andutk−exist,utk=utk−,limt⟶∞ut1+tα−1exsits,with the norm(6)uPC=supt∈0,∞ut1+tα−1,PC1J,ℝ=u∈PCJ,R:Dα−1ut∈CJk,ℝ,k=0,1,…,m,Dα−1utk+andDα−1utk−exist,Dα−1utk−=Dα−1utk,limt⟶∞Dα−1utexists,with the norm(7)uPC1=maxsupt∈Jut1+tα−1,supt∈JDα−1ut.Definition 1.
The Riemann–Liouville fractional integral of orderα>0 of a function f:0,∞⟶ℝ is given by(8)J0+αft=1Γα∫0tt−sα−1fsds,where the right side is pointwise defined on 0,∞.Definition 2.
The Riemann–Liouville fractional derivative of orderα>0 of a function f:0,∞⟶ℝ is given by(9)D0+αft=1Γn−αddtn∫0tt−sn−α−1fsds,where n is the smallest integer greater than or equal to α and the right side is pointwise defined on 0,∞. In particular, for α=n, D0+αft=fnt.Lemma 1.
Letα>0, and n denotes the smallest integer greater than or equal to α. For all t∈a,b,(10)J0+αD0+αut=ut+c1tα−1+c2tα−2+⋯+cntα−n,where cj∈ℝ, j=1,2,…,n.Lemma 2 (see [2]).
LetΩ⊆PC1. Then, Ω is relatively compact in PC1 if the following conditions hold:(1)
Ω is bounded in PC1(2)
For anyut∈Ω, ut/1+tα−1 and Dα−1ut are equicontinuous on any interval Jk(3)
Givenε>0, there exists a constant N=Nε>0 such that(11)ut11+t1α−1−ut21+t2α−1<ε,Dα−1ut1−Dα−1ut2<ε,for any t1,t2≥N and ut∈Ω.Theorem 1 (Schauder fixed-point theorem).
IfU is a closed bounded convex subset of a Banach space X and T:U⟶U is completely continuous, then T has at least one fixed point in U.Theorem 2 (Altman theorem).
LetΩ be an open bounded subset of a Banach space E with 0∈Ω and T:Ω¯⟶E be a completely continuous operator. Then, T has a fixed point in Ω¯, provided that(12)Tx−x2≥Tx2−x2,∀x∈∂Ω.Lemma 3.
For a giveny∈CJ,ℝ, a function u∈PC1J,ℝ is a solution of the following boundary value problem:(13)D0+αut+yt=0,t∈0,∞∖tkk=1m,△utk=Ikutk,t=tk,u0=u′0=⋯=un−20=0,D0+α−1u∞=u0,if and only if u∈PC1J,ℝ is a solution of the impulsive fractional integral equation(14)ut=−1Γα∫0tt−sα−1ysds+tα−1Γα∫0∞ysds+tα−1Γαu0−tα−1∑t<tiIiti1−α.Proof.
Assumeut satisfies (13). We denote the solution of (13) by ut≜ukt in Jkk=0,1,…,m.
Fort∈0,t1, applying Lemma 1, we have(15)u0t=−1Γα∫0tt−sα−1ysds+C01tα−1+C02tα−2+⋯+C0ntα−n.
Fromu0=u′0=⋯=un−20=0, we know C0n=⋯=C03=C02=0. So, we get(16)u0t=−1Γα∫0tt−sα−1ysds+C01tα−1,t∈0,t1,ut1−=−1Γα∫0t1t1−sα−1ysds+C01t1α−1.
Fort∈t1,t2, by applying Lemma 1, we know(17)u1t=−1Γα∫0tt−sα−1hsds+C11tα−1+C12tα−2+⋯+C1ntα−n.
In view ofu0=u′0=⋯=un−20=0, we have C1n=⋯=C13=C12=0. So, we know(18)u1t=−1Γα∫0tt−sα−1ysds+C11tα−1,ut1+=−1Γα∫0t1t1−sα−1ysds+C11t1α−1.
And from impulsive condition of (13), △ut1=ut1+−ut1−=I1ut1. Then,(19)−1Γα∫0t1t1−sα−1ysds+C11t1α−1−−1Γα∫0t1t1−sα−1ysds+C01t1α−1=I1ut1.
Thus,(20)C11=C01+t11−αI1ut1.
Then,(21)u1t=−1Γα∫0tt−sα−1ysds+tα−1C01+tα−1t11−αI1ut1,t∈t1,t2.
Fort∈t2,t3, by applying Lemma 1, we obtain(22)u2t=−1Γα∫0tt−sα−1hsds+C21tα−1+C22tα−2+⋯+C2ntα−n.
In view ofu0=u′0=⋯=un−20=0, we have C2n=⋯=C23=C22=0. So, we know(23)u2t=−1Γα∫0tt−sα−1ysds+C21tα−1,ut2+=−1Γα∫0t2t2−sα−1ysds+C21t2α−1.
And from impulsive condition,△ut2=ut2+−ut2−=I2ut2. Then,(24)−1Γα∫0t2t2−sα−1ysds+C21t2α−1−−1Γα∫0t2t2−sα−1ysds+C11t2α−1=I2ut2.
We get(25)C21=C11+t21−αI2ut2=C01+t11−αI1ut1+t21−αI2ut2=C01+∑i=12ti1−αIi.
Consequently,(26)u2t=−1Γα∫0tt−sα−1ysds+tα−1C01+tα−1∑i=12ti1−αIi,t∈t2,t3.
By the recurrent method and Lemma1, for t∈tk,tk+1, k=0,1,2,…,m, we can say that(27)utukt=−1Γα∫0tt−sα−1ysds+tα−1C01+tα−1∑i=1kti1−αIi.
Thus, fort∈tm,∞, we have(28)ut=umt=−1Γα∫0tt−sα−1ysds+tα−1C01+tα−1∑i=1mti1−αIi.
FromD0+α−1u∞=u0, we get(29)−∫0∞ysds+Γα∑i=1mti1−αIi+ΓαC01=u0.
So,(30)C01=1Γαu0+1Γα∫0∞ysds−∑i=1mti1−αIi.
Therefore, fort∈0,∞, we have(31)ut=−1Γα∫0tt−sα−1ysds+tα−1Γα∫0∞ysds+tα−1Γαu0−tα−1∑i=1mti1−αIi+tα−1∑ti<tti1−αIi=−1Γα∫0tt−sα−1ysds+tα−1Γα∫0∞ysds+tα−1Γαu0−tα−1∑t<titi1−αIi.
Conversely, assume thatut satisfies impulsive fractional integral equation (14). Obviously, we get u0=u′0=⋯=un−20=0, and D0+α−1u∞=u0. Using the fact D0+αtα−1=0, we obtain D0+αut=−yt. Also, we can easily show that △utk=Ikutk, k=1,2,…,m. Then, u is also the solution of problem (13).


## 3. Main Results

In this section, we will prove the existence of solution of (1) by using Schauder fixed-point theorem and Altman theorem.According to Lemma3, we obtain the following lemma first.Lemma 4.
u∈PC1J,ℝ is a solution of problem (1) if and only if u∈PC1J,ℝ is a solution of the impulsive fractional integral equation(32)ut=−1Γα∫0tt−sα−1fs,us,Jβus,Dα−1usds+tα−1Γα∫0∞fs,us,Jβus,Dα−1usds+tα−1Γαu0−tα−1∑t<tiIiti1−α,t∈J.Define an operatorT:PC1J,ℝ⟶PC1J,ℝ as follows:(33)Tut=−1Γα∫0tt−sα−1fs,us,Jβus,Dα−1usds+tα−1Γα∫0∞fs,us,Jβus,Dα−1usds+tα−1Γαu0−tα−1∑t<tiIiti1−α,t∈J.Then, problem (1) has a solution if and only if the operator T has a fixed point.Theorem 3.
Assume that following conditions hold:
H1 For f∈C0,+∞×R×R×R,R, there exist nonnegative functions at, bt, ct, et∈L1J such that(34)ft,x,y,z≤atx+bty+etz+ct,∫0+∞1+tα−1at+btdt<∞,∫0+∞ctdt<∞,∫0+∞1+tα−1tβΓβ+1etdt<∞.
H2 For Ik∈CR,R, for all u∈ℝ, there exist some constants Lk>0 such that Iku<Lk, k=1,2,…,m.Then, problem (1) has at least one solution ut in PC1J,ℝ.Proof.
We will use five steps to prove our conclusion. Firstly, we will showT:PC1J,ℝ⟶PC1J,ℝ is continuous. From (33), we know(35)Dα−1Tut=−∫0tfs,us,Jβus,Dα−1usds+∫0∞fs,us,Jβus,Dα−1usds+u0−Γα∑t<tiIiti1−α.
FromH1, we have(36)∫0∞fs,us,Jβus,Dα−1usds≤∫0+∞asus+bsDα−1us+esJβus+csds≤∫0+∞1+sα−1asuPC+bsDα−1us+1+sα−1sβΓβ+1esuPC+csds≤uPC1∫0+∞1+sα−1as+bs+1+sα−1sβΓβ+1esds+∫0+∞csds<∞.
Letun,u∈PC1J,ℝ be such that un⟶un⟶∞. Then, unPC1<∞ and uPC1<∞. By (36) and the Lebesgue dominated convergence theorem, we get(37)limn⟶∞∫0∞fs,uns,Jβuns,Dα−1unsds=∫0∞fs,us,Jβus,Dα−1usds.
ByH1,H2, and (36), we have(38)Tut1+tα−1=−1Γα∫0tt−sα−11+tα−1fs,us,Jβus,Dα−1usds+tα−11+tα−11Γα∫0∞fs,us,Jβus,Dα−1usds+tα−11+tα−11Γαu0−tα−11+tα−1∑t<tiIiti1−α≤2Γα∫0∞fs,us,Jβus,Dα−1usds+u0Γα+∑t<tiLiti1−α<∞,(39)Dα−1Tut=−∫0tfs,us,Jβus,Dα−1usds+∫0∞fs,us,Jβus,Dα−1us+u0−Γα∑t<tiIiti1−α≤∫0∞fs,us,Jβus,Dα−1usds+∫0∞fs,us,Jβus,Dα−1usds+u0+Γα∑t<tiLiti1−α<∞.
Hence, according to (37)–(39) and Lebesgue dominated convergence theorem, we can easily get(40)Tun−TuPC1⟶0n⟶∞.
Therefore,T:PC1J,ℝ⟶PC1J,ℝ is continuous.
Secondly, chooser such that(41)r≥2∫0∞csds+u0+Γα∑t<tiLiti1−α1−2∫0∞1+sα−1as+bs+1+sα−1sβ/Γβ+1esds,and let Br=‖u∈PC1uPC1≤r⊂PC1J,ℝ.
For anyut∈Br, by (41) and condition (H1), we have(42)Tut1+tα−1≤2Γα∫0∞fs,us,Jβus,Dα−1usds+u0Γα+∑t<tiLiti1−α≤2uPC1Γα∫0+∞1+sα−1as+bsds+2Γα∫0+∞csds+2uPC1Γα∫0+∞1+sα−1sβΓβ+1esds+u0Γα+∑t<tiLiti1−α≤r,Dα−1Tut≤2∫0∞fs,us,Jβus,Dα−1usds+u0+Γα∑t<tiLiti1−α≤2uPC1∫0+∞1+sα−1as+bs+1+sα−1sβΓβ+1esds+2∫0+∞csds+u0+Γα∑t<tiLiti1−α≤r.
So,TuPC1≤r and T:Br⟶Br.
Thirdly, we show thatTBr is uniformly bounded. From (38) and (39), we know(43)supt∈JTut1+tα−1<∞,supt∈JDα−1Tut<∞.
So, foru∈Br, it is easy to know that TuPC1<∞. Hence, TBr is uniformly bounded.
Fourth, we prove that for anyut∈Br, Tut/1+tα−1 and Dα−1Tut are equicontinuous on any interval Jk.
For anyut∈Br, t1,t2∈Jkk=0,1,2,…,m, t1<t2, we have(44)Tut21+t2α−1−Tut11+t1α−1≤1Γα∫0t1t2−sα−11+t2α−1−t1−sα−11+t1α−1fs,us,Jβus,Dα−1usds+1Γα∫t1t2t2−sα−11+t2α−1fs,us,Jβus,Dα−1usds+1Γα∫0∞fs,us,Jβus,Dα−1usdst2α−11+t2α−1−t1α−11+t1α−1+u0Γαt2α−11+t2α−1−t1α−11+t1α−1+∑t<t2Liti1−αΓαt2α−11+t2α−1−t1α−11+t1α−1⟶0ift2⟶t1,Dα−1Tut2−Dα−1Tut1≤∫t1t2fs,us,Jβus,Dα−1usds+Γα∑t2<tiIiti1−α−Γα∑t1<tiIiti1−α⟶0ift2⟶t1.
Therefore, for anyut∈Br, Tut/1+tα−1 and Dα−1Tut are equicontinuous on any interval Jk.
Fifth, we need to verify that condition (3) in Lemma 2 is satisfied. It means that we need to verify Tut/1+tα−1 and Dα−1Tut are equiconvergent at t=Jkk=1,2,…,m,…, and t=∞ for any u∈Br. We have(45)limt⟶∞Tut1+tα−1≤limt⟶∞2Γα∫0∞fs,us,Jβus,Dα−1usdstα−11+tα−1+u0Γαtα−11+tα−1+∑t<tiLiti1−αtα−11+tα−1≤2uPC1Γα∫0∞1+sα−1as+bs+1+sα−1sβΓβ+1esds+2Γα∫0∞csds+u0Γα+∑t<tiLiti1−αlimt⟶∞tα−11+tα−1<∞,limt⟶∞Dα−1Tut<limt⟶∞∫0∞fs,us,Jβus,Dα−1usds+∫0∞fs,us,Jβus,Dα−1usds+u0+Γα∑t<tiLiti1−α<limt⟶∞2uPC1∫0∞1+sα−1as+bs+1+sα−1sβΓβ+1esds+2∫0∞csds+u0+Γα∑t<tiLiti1−α<∞.
Hence,TBr is equiconvergent at infinity.
Then, we prove thatTut/1+tα−1 and Dα−1Tut are equiconvergent at t⟶tk+k=0,1,2,…,. We have(46)limt⟶tk+Tut1+tkα−1+1Γα∫0tktk−sα−11+tkα−1fs,us,Jβus,Dα−1usds−tkα−1Γα1+tkα−1∫0∞fs,us,Jβus,Dα−1usds−tkα−1Γα1+tkα−1u0+tkα−11+tkα−1∑tk<tiIiti1−α=0,limt⟶tk+Dα−1Tut+∫0tkfs,us,Jβus,Dα−1usds−∫0∞fs,us,Jβus,Dα−1usds−u0+Γα∑tk<tiIiti1−α=0.
Therefore,Tut/1+tα−1 and Dα−1Tut are equiconvergent at t=Jkk=1,2,…,m,…, and t=∞ for any u∈Br. By using Lemma 2, we obtain that TBr is relatively compact, that is, T is a compact operator.
Therefore, Schauder’s fixed-point theorem implies that problem (1) has at least one solution in Br.
Our second result is based on Altman fixed-point theorem.Theorem 4.
AssumeH2 and the following condition hold:
H3 For f∈C0,+∞×R×R×R,R, there exist nonnegative functions at, bt, ct defined on 0,∞ and constants p,q,l≥0 such that(47)ft,x,y,z≤at+btxp+ctyq+etzl,∫0+∞atdt=a∗<∞,∫0+∞1+tα−1pbtdt=b∗<+∞,∫0+∞ctdt=c∗<+∞,∫0+∞1+tα−1tβΓβ+1letdt=e∗<∞.
If0≤p,q,l<1, then problem (1) has at least one solution ut in PC1J,ℝ.Proof.
Let us choose(48)R≥max12a∗,12b∗1/1−p,12c∗1/1−q,12e∗1/1−l,6u0,6Γα∑t<tiLiti1−α,and define U=u∈PC1|uPC1≤R. According to Theorem 3, we know T:U⟶U is a completely continuous operator. For any u∈∂U, by (H3), we have(49)Tut1+tα−1≤2Γα∫0∞fs,us,Jβus,Dα−1usds+u0Γα+∑t<tiLiti1−α≤2Γα∫0∞as+bsusp+csDα−1usq+esJβuslds+u0Γα+∑t<tiLiti1−α≤2Γαa∗+∫0∞bs1+sα−1pusp1+sα−1pds+∫0∞csuPC1qds+∫0+∞es1+sα−1sβΓβ+1luPC1lds+u0Γα+∑t<tiLiti1−α≤2Γαa∗+b∗uPC1p+c∗uPC1q+e∗uPC1l+u0Γα+∑t<tiLiti1−α≤2ΓαR12+R12+R12+R12+R6Γα+R6Γα<R,(50)Dα−1Tut≤2∫0∞fs,us,Jβus,Dα−1usds+u0+Γα∑t<tiLiti1−α≤2a∗+b∗uPC1p+c∗uPC1q+e∗uPC1l+u0+Γα∑t<tiLiti1−α≤2R12+R12+R12+R12+R6+R6=R.
Thus, from (49) and (50), we have TU⊂U and TuPC1≤uPC1, ∀u∈∂U. So, by Theorem 2, we know that problem (1) has at least one solution.Theorem 5.
Assume that conditions (H2) and (H3) are satisfied. If p=q=l=1,1+Γαb∗+c∗<Γα, then problem (1) has at least one solution.Proof.
Let us take(51)R>u0+Γα∑t<tiLiti1−α+2a∗1−2b∗+c∗+e∗,and define U=u∈PC1|uPC1<R.
For anyu∈∂U, we have(52)Tut1+tα−1≤2Γα∫0∞fs,us,Jβus,Dα−1usds+u0Γα+∑t<tiLiti1−α≤2Γα∫0∞as+bsus+csDα−1us+esJβusds+u0Γα+∑t<tiLiti1−α≤2Γαa∗+∫0∞bs1+sα−1us1+sα−1ds+∫0∞csuPC1ds+∫0∞esus1+sα−11+sα−1sβΓβ+1+u0Γα+∑t<tiLiti1−α≤2Γαa∗+b∗uPC1+c∗uPC1+e∗uPC1+u0Γα+∑t<tiLiti1−α≤2Γαa∗+b∗R+c∗R+e∗R+u0Γα+∑t<tiLiti1−α<R,(53)Dα−1Tut≤2∫0∞fs,us,Jβus,Dα−1usds+u0+Γα∑t<tiLiti1−α≤2a∗+b∗uPC1+c∗uPC1+e∗uPC1+u0+Γα∑t<tiLiti1−α≤2a∗+b∗R+c∗R+e∗R+u0+Γα∑t<tiLiti1−α<R.
Thus, from (52) and (53), we have TU⊂U and TuPC1≤uPC1, ∀u∈∂U. So, by Theorem 2, we know that problem (1) has at least one solution.Remark 1.
If we use other conditions instead of the condition“p=q=1”, for example, 0≤p<1,q=1 or p>1,q=1 or 0≤q<1,p=1 or q>1,p=1 or p,q>1 or 0≤p<1,q>1 or 0≤q<1,p>1, and choose proper R, respectively, then we can obtain the same result. The proof is similar to Theorem 4 or Theorem 5, so we omit it.


## 4. Example

In this section, we give an example to illustrate of our main result.Example 1.
Consider the following impulsive boundary value problem of fractional order:(54)D0+3/2ut+ln1+D0+1/2ut201+t2+utD0+1/2ut20et+J3/2ut20et=0,t∈0,∞∖12,△u12=Iu12,t=12,u0=u′0=0,D0+1/2u∞=u0,where α=3/2, ft,x,y,z=ln1+y/201+t2+xy/20et+z/20et, k=1,t1=1/2.
LetIu=1/u+1/u. Then, we have(55)ft,x,y≤140etx+1201+t2+140ety+120etz,Iu=1u+1/u≤1.
By computing, we know that(56)∫0+∞1+tα−1at+btdt=∫0+∞1+t1/2140et+1201+t2+140etdt=15+π40≈0.2785<∞,∫0+∞1+tα−1tβΓβ+1etdt=∫0+∞1+t1/2t3/2Γ5/2120et≈64.5850<∞.
Thus, the conditions of Theorem3 are satisfied, and hence problem (54) has at least one solution.Remark 2.
By theorems in [9, 10, 13], this problem could not be solved.



---

*Source: 6018273-2020-10-22.xml*