# Multiple Solution Results for Perturbed Fractional Differential Equations with Impulses

Authors: Peiluan Li; Liang Xu; Peiyu Li; Hui Wang
Journal: Journal of Function Spaces
 (2020)

---

## Abstract

The multiplicity of classical solutions for impulsive fractional differential equations has been studied by many scholars. Using Morse theory, Brezis and Nirenberg’s Linking Theorem, and Clark theorem, we aim to solve this kind of problems. By this way, we obtain the existence of at least three classical solutions andk distinct pairs of classical solutions. Finally, an example is presented to illustrate the feasibility of the main results in this paper.


---

## Body

## 1. Introduction

Consider the multiple solutions of fractional order impulsive systems as follows:(1)−12ddtD0t−β+DtT−βu′t=atut+∇Ft,ut,t≠tj,a.e.t∈0,T,ΔDtαutj=Ijutj,tj∈0,T,j=1,2,⋯,l,u0=uT=0,where β∈0,1,α=1−β/2∈1/2,1;D0t−β,DtT−β are the left and right Riemann-Liouville fractional integrals of order β, D0ctα,DtcTα are used to denote the left and right Caputo fractional derivatives of order α, 0=t0<t1<⋯<tl<tl+1=T, a∈C0,T, F:0,T×RN⟶R is a given function, ∇Ft,x is the gradient of F at x, there are constants a0, a1 with 0<a0≤at≤a1,
(2)Dtαut=12D0tα−1D0ctαu−DtTα−1DtcTαut,ΔDtαutj=12D0tα−1D0ctαu−DtTα−1DtcTαutj+−12D0tα−1D0ctαu−DtTα−1DtcTαutj−,D0tα−1D0ctαu−DtTα−1DtcTαutj+=limt→tj+D0tα−1D0ctαu−DtTα−1DtcTαut,D0tα−1D0ctαu−DtTα−1DtcTαutj−=limt→tj−D0tα−1D0ctαu−DtTα−1DtcTαut, for j=1,2,⋯,l.The problem (1) arises from the phenomena of advection dispersion and was first scrutinized by Erwin and Roop in [1]. From then on, more and more scholars began to pay attention to the problem in [1] and the related problems.Fractional calculus is different from integral calculus in nature. It has nonlocal characteristics and is very suitable for describing materials and processes with memory effect and genetic properties. Therefore, fractional differential equations are widely used in many domains, for instance, biomedicine, economic mathematics, and technology science [2, 3]. In recent years, the variational methods and critical point theory have been widely used to study fractional differential equations [4–8].In [8], the authors discussed the following fractional order differential systems:
(3)−12ddtD0t−β+DtT−βu′t=∇Ft,ut,a.e.t∈0,T,u0=uT=0.They used the critical point theory and other tools to verify the existence of solutions. From then on, a number of scholars began to use such methods for research, as shown in [9–11].In [12], the authors discussed the following problems:
(4)DtTaatD0tαux=λut∇Ft,ut,a.e.t∈0,T,u0=uT=0.They proved that there are at leastk pairs of weak solutions and two weak solutions by using the Clark Theorem and other methods.An impulsive phenomenon is a common phenomenon in nature and engineering applications. The models reflected in mathematics are impulsive differential equations. The most prominent feature of impulsive differential equation is that it can fully consider the impact of instantaneous mutation on the state. Therefore, in recent decades, impulsive differential equation theory has been widely used in biological mathematics, theoretical mechanics, biomedicine, and economic mathematics (see [13–18]).For the past few years, very few scholars used the variational method and critical point theory to discuss impulsive fractional differential equations and their boundary value problems. Moreover, few papers discuss the fractional order system by using Morse theory (see [19–24]).In [23], the authors discussed the following problems:
(5)DtTαD0ctαut+ktut=ft,ut,0<t<T,t≠tj,ΔDtTα−1D0ctαiutj=Ijutj,j=1,2,⋯,m,u0=uT=0.The multiple solutions of this problem are verified with Morse theory and the Clark theorem by the authors.In [25], the sufficient conditions for the existence of infinite solutions to the system (1) are obtained by using the variational method.Based on the above literatures, in the present paper, we will discuss the existence of multiple classical solutions for (1) by using Morse theory, Clark theorem, and Brezis and Nirenberg’s Linking Theorem.First of all, we give some assumptions.(H1)Ij∈C0,T,R,Ij0=0, there exist some constants ej,γj∈0,1, aj,bj>0, such that limu→0Iju/uej=bj, Iju≤ajuγj,j=1,⋯,l, and ∫0uIjsds≥0, for ∀u∈R(H2)F∈C0,T,R and Ft,0=0, limu→∞supFt,u/u2<1/2Γ2α+1cosπα/T2α−a1 uniformly on t∈0,T(H3)limu→0supFt,u/u2<Γ2α+1/2T2α uniformly for t∈0,T. There exist four constants C>0,r,r0,γ∈1,maxj∈1,2,⋯,lγj+1 such that Ft,u≥Cuγ, r≤u≤r0, a.e.t∈0,T(H4)Ft,−u=Ft,u and Ij−u=−Ijuj=1,2,⋯,l, for u∈RThe key outcomes are as follows.Theorem 1.
Let (H1)–(H3) hold. Then, the problem (1) has at least three classical solutions.Theorem 2.
Let (H1)–(H4) hold. Then, the problem (1) has at least k distinct pairs of classical solutions.Note that the methods in this article are distinct from [25] and our results are richer. The problems in this paper we studied are different from the problems in [23]. Compared with [23], classical solutions are investigated in this paper.The structure of this article is as below. In Section2, we provide some preliminary knowledge, which are helpful to the proof the key outcomes. We prove the key outcomes in Section 3. Finally, an example is given to illustrate the main results.


## 2. Preliminaries

Similar to [25], we first convert system (1) into a new format as follows:
(6)ddt12D0tα−1D0ctαut−12DtTα−1DtcTαut+atut+∇Ft,ut=0,t≠tj,a.e.t∈0,T,ΔDtαutj=Ijutj,tj∈0,T,j=1,2,⋯,l,u0=uT=0.Remark 3.
Because of the equivalence of system (1) and system (6), we know that the solutions of system (6) are the solutions of system (1).We first build the function spaces as below, the goal of which is to establish the variational framework of system (6).Let us recall that for any fixed1≤p≤∞ and t∈0,T, u∞=maxt∈0,Tut,uLp=∫0Tuspds1/p.Definition 4 (see [22]).
Let0<α≤1, we define the fractional derivative space E0α by the closure of C0∞0,T,R with u0=uT under the norm
(7)uα=∫0TD0ctαut2dt+∫0Tut2dt1/2,∀u∈E0α.Lemma 5 (see [25]).
Let0<α≤1, and E0α is a Banach space with reflexive and separable.Definition 6 (see [25]).
We define that the functionu∈E0α is a weak solution of the system (6) if the following holds:
(8)∫0T−12D0ctαutDtcTαvt+DtcTαutD0ctαvt−atutvtdt+∑j=1lIjutjvtj−∫0T∇Ft,utvtdt=0for ∀v∈E0α.
We defineΦ:E0α⟶R as
(9)Φu=∫0T−12D0ctαutDtcTαutdt−12∫0Tatu2tdt−∫0TFt,utdt+∑j=1l∫0utjIjsds.
From (H1), (H2), we know the functional Φ is continuously differentiable. So ∀u,v∈E0α, we have
(10)Φ′u,v=∫0T−12D0ctαutDtcTαvt+DtcTαutD0ctαvtdt−∫0Tatutvtdt+∑j=1lIjutjvtj−∫0T∇Ft,utvtdt.Remark 7.
Obviously, from (10), we know that the critical points of functional Φ are the weak solutions of system (6).Definition 8 (see [25]).
We define(11)u∈u∈AC0,T:∫tjtj+1∫0TD0ctαut2+∫0Tut21/2dt<∞,j=0,⋯,l,as a classic solution of system (6) if it satisfies the following conditions:
(i)
1/2D0tα−1D0ctαu−DtTα−1DtcTαutj+=limt→tj+1/2D0tα−1D0ctαu−DtTα−1DtcTαut,(12)12D0tα−1D0ctαu−DtTα−1DtcTαutj−=limt→tj−12D0tα−1D0ctαu−DtTα−1DtcTαut,ΔDtαutj=Ijutj,u0=uT=0(ii)
u content system (1) a.e. on t∈0,T\t1,t2,⋯,tlLemma 9 (see [25]).
The functionu∈E0α is a classical solution of system (6) when u is a weak solution of system (6).Remark 10.
Combine Remarks3 and 7 and Lemma 9, we know that the critical point of functional Φ is the classical solution of the system (1). Therefore, we will directly discuss the critical point of Φ as below.Lemma 11 (see [8]).
Let1/2<α≤1 and 1<p<∞, for all u∈E0α, one has
(13)uLp≤TαΓα+1D0ctαuLp.
Moreover, ifα>1/p, 1/p+1/q=1, then
(14)u∞≤Tα−1/pΓαα−1q+11/qD0ctαuLp.
In particular, ifp=2, then
(15)uL22≤T2αΓ2α+1uα2,u∞2≤T2α−1Γ2α2α−1uα2.
It is easy to prove that the normuα=∫0TD0ctαut2dt+∫0Tut2dt1/2 is equivalent to uα=∫0TD0ctαut2dt1/2, ∀u∈E0α. Next, we will use uα=∫0TD0ctαut2dt1/2 as the norm in E0α.Lemma 12 (see [8]).
Let1/2<α≤1, ∀u∈E0α, have
(16)cosπαuα2≤−∫0TD0ctαutDtcTαutdt≤1cosπαuα2.Lemma 13 (see [8]).
Let1/2<α≤1. Assume the sequence un converges weakly to u in E0α. Then, un⟶u strongly in C0,T,R, i.e., un−u∞⟶0, as n⟶∞.Definition 14 (see [23]).
We say thatΦ satisfies the PS condition in E0α, if any unn∈N⊂E0α, for which Φunn∈N is bounded and Φ′un⟶0 as n⟶∞ owns a strongly convergent subsequence in E0α.Lemma 15 (see [26]).
LetE have a direct sum decomposition E=V⊕W, and k=dimV<∞. Let 0 be a critical point of Φ with Φ0=0, Φ is bounded below and satisfying PS condition. Suppose that, for some ρ>0,
(17)Φu≤0,∀u∈V,u≤ρ,Φu>0,∀u∈W,u≤ρ.Also, assume thatinfEΦ<0. Then, Φ has at least two nonzero critical points and CkΦ,0≅0.Lemma 16 (see [27]).
LetE be a real Banach space, Φ∈C1E,R; assume that Φ is even, bounded from below, and satisfying PS condition. Assume Φ0=0, there exists a set E′⊂E such that E′ is homeomorphic to Sk−1 by an odd map, and supE′Φ<0. Then, Φ has at least k distinct pairs of critical points.


## 3. Proofs of Main Results

Lemma 17.
Suppose (H1), (H2) hold, if un is a PS sequence, then un is bounded.Proof.
Ifun is a PS sequence, that is,
(18)Φunisbounded,Φ′un⟶0,asn⟶∞.
From (H2), for some ξ>0 small enough, there is a constant Cξ>0, for any u∈R, t∈0,T such that
(19)Ft,u≤12Γ2α+1T2αcosπα−a1−ξu2+Cξ.
According to (19), for u∈E0α, u≠0, one has
(20)Φu=∫0T−12D0ctαutDtcTαutdt−12∫0Tatu2tdt−∫0TFt,utdt+∑j=1l∫0utjIjsds≥12cosπαuα2−12Γ2α+1T2αcosπα−a1−ξ∫0Tut2dt−a12∫0Tu2tdt−CξT=12cosπαuα2−12Γ2α+1T2αcosπα−ξuL22−CξT≥12cosπαuα2−12cosπα−T2αΓ2α+1ξuα2−CξT≥ξT2α2Γ2α+1uα2−CξT.
BecauseΦu is bounded, by (20), we can get un is bounded in E0α and Φ is bounded from below. The proof is completed.Lemma 18.
Assume (H1), (H2) hold, then Φ satisfies the PS condition.Proof.
Ifun is a PS sequence, from Lemma 17, we get un is a bounded sequence in E0α. By Lemma 5, we get un has a weakly convergent subsequence. Without loss of generality, we also assume that un converges weakly to u0 in E0α, then from (9) and (18), we know
(21)Φ′un−Φ′u0,un−u0≤Φ′un⋅un−u0−Φ′u0,un−u0⟶0,n⟶∞.
By Lemma13, we can obtain that un⟶u0 in C0,T,R, as n⟶∞, i.e.,
(22)un−u0∞⟶0,asn⟶∞.
From (10), we have
(23)Φ′un−Φ′u0,un−u0=∫0T−12D0ctαun−u0DtcTαun−u0+DtcTαun−u0D0ctαun−u0−∫0Tatun−u0un−u0dt−∫0T∇Ft,un−∇Ft,u0un−u0dt+∑j=1lIjuntj−Iju0tjuntj−u0tj≥12cosπαun−u0α2−∫0T∇Ft,un−∇Ft,u0un−u0∞+∑j=1lIjuntj−Iju0tjun−u0∞.
By (21), (22), and (23), we can infer that un−u0α2⟶0, as n⟶∞, i.e.,un strongly converges to u0. Therefore, Φ satisfies the PS condition.
By Lemma5, we can obtain that there is an orthogonal basis ei of E0α such that E0α=spanei:i=1,2,⋯¯. We define Xi≔spanei,Vk=⊕i=1kXi,Yk=⊕i=k∞Xi¯k=1,2,⋯. Then, E0α=Vk⊕Yk.Proof of Theorem 1.
From (H1), (H2), one knows Ft,0=0 and Ij0=0, j=1,⋯,l. We find out Φ has a critical point at 0. Therefore, we can get the linking E0α=Vk⊕Yk of Φ at 0.According to the equivalence of norm of normed space in finite dimension, there exist positive constantsM1,M2,M1′,M2′, such that
(24)M1uα≤u∞≤M2uα,M1′uα≤uLγ≤M2′uα,u∈Vk.First, letu∈Vk. Because Vk is finite dimensional, there exists 0<ρ1<1 small for r0>0, such that
(25)ut≤u∞≤M2uα<M2ρ1<r0,u∈Vk,uα<ρ1.For anyr∈0,r0, we set Ω1=t∈0,T:u≤r,Ω2=t∈0,T:r≤u≤r0, Ω3=t∈0,T:r0≤u, where 0,T=∪i=13Ωi and Ωii=1,2,3 are pairwise disjoint.SetF∗t,u=Ft,u−Cuγ, for uα<ρ1,u∈Vk, combine (H1), (H3) and Lemma 12, we have
(26)Φu=∫0T−12D0ctαutDtcTαutdt−12∫0Tatu2tdt−∫0TFt,utdt+∑j=1l∫0utjIjsds≤12cosπαuα2+∑j=1lajγj+1utjγj+1−∫0TCuγdt−∫Ω1F∗t,utdt−∫Ω2F∗t,utdt−∫Ω3F∗t,utdt≤12cosπαuα2+∑j=1lajAoγj+1γj+1uαγj+1−CuLγγ−∫Ω1F∗t,utdt−∫Ω2F∗t,utdt−∫Ω3F∗t,utdt,whereA0=Tα−1/2/Γα2α−1.According to (25) and the definition of Ω3 is empty set, we have ∫Ω3F∗t,utdt=0, for any u∈Vk. By (H3), one has ∫Ω2F∗t,utdt≥Cuγ−Cuγ=0. On Ω1, u<r. From (H3), we can get ∫Ω1F∗t,utdt⟶0, as r⟶0.Then,∀u∈Vk,r∈0,r0,uα≤ρ≤1, 1<γ<maxγj+1<2, according to (26), we can get
(27)Φu≤12cosπαuα2+∑j=1lajAoγj+1γj+1uαγj+1−M′1γCuαγ≤uαγ12cosπαuα2−γ+∑j=1lajAoγj+1γj+1uαγj+1−γ−M′1γC≤0.Hence,(28)Φu≤0,∀u∈Vk,uα≤ρ1.Next, setu∈Yk. Because E0α⟶C0∞0,T,R is continuous compact embedding. Hence, for u∈Ykε>0, there exists 0<ρ2<1 small such that u≤u∞≤T2α−1uα/Γ2α2α−1<T2α−1ρ2/Γ2α2α−1<ε, for uα<ρ2.From (H3), ∀u<ε,u∈Yk,uα≤ρ2,t∈0,T, there is ξ∈0,cosπα, one has
(29)Ft,u≤cosπα−ξΓ2α+12T2αu2.From (H1), ∀u<ε,u∈Yk,uα≤ρ2, one has
(30)Ijuuej>12bj.Letb=minj=1lbj, e=maxj=1lej, ∀u<ε,u∈Yk,uα≤ρ2<1, by Lemmas 11 and 12 and (29) and (30), we obtain
(31)Φu=∫0T−12D0ctαutDtcTαutdt−12∫0Tatu2tdt−∫0TFt,utdt+∑j=1l∫0utjIjsds≥12cosπαuα2−cosπα−ξΓ2α+12T2αuL22−12a1Tu2+12ble+1ue+1≥12cosπαuα2−12cosπα−ξuα2−12a1Tu2+12ble+1ue+1≥12ξuα2+ue+1ble+1−a1Tu1−e>0.Hence,(32)Φu>0,∀u∈Yk,uα≤ρ2.Letρ=minρ1,ρ2, from (28) and (32), we obtain
(33)Φu≤0,∀u∈Vk,uα≤ρ,Φu>0,∀u∈Yk,uα≤ρ.It follows from Lemmas17 and 18 that Φ is bounded from below and satisfies the PS condition. Then, from Lemma 15, we can get Φ has at least two nonzero critical points, and CkΦ,0≅0, so u=0 is a homological nontrivial point of Φ. Hence, the system (1) has at least three classical solutions.Proof of Theorem 2.
According to (H4), we can deduce that Φ is even. By Lemmas 17 and 18, we know Φ is bounded from below and satisfies the PS condition. For given ρ>0, set E′=Sp=u∈Vk:u=ρ. By (27), if ρ is small enough, one has supE′Φu<0. Clearly, dimVk=k. Then, we can conclude that Φ has at least k distinct pairs of critical points from Lemma 16. Hence, the system (1) has at least k distinct pairs of classical solutions. We complete the proof.Example 19.
(34)−12ddtD0t−0.6+DtT−0.6u′t=25ut+∇Ft,ut,t≠tj,a.e.t∈0,1,ΔDt0.7utj=Ijutj,tj∈0,1,j=1,2,⋯,l,u0=uT=0.
According to (34), we can see that β=0.6, α=0.7, at=2/5, T=1.
LetIju=4/3u1/3, then the condition (H1) holds with aj=3/2, γj=1/3, bj=4/3, ej=1/3.
LetFt,ut=1/151+sin2tu21/lnu2/3+1.5, a1=9/20. By simple calculations, we can get Γ2α+1/T2α≈0.84474, cos0.7π≈0.58778,
(35)limu→∞supFt,uu2⟶0<12Γ2α+1T2αcosπα−a1≈0.02326;then, the condition (H2) holds.
By (H3), we know
(36)limu→0supFt,uu2≤0.32884<Γ2α+12T2α≈0.42237.
Letr=e−1.7, r0=e−1.5, C=2/25, γ=6/5; then, for e−1.7≤u≤e−1.5, we have
(37)Ft,u>0.14471u2≥225u6/5;then, the condition (H3) is satisfied.
It easy to see that the condition (H4) holds.According to Theorem1, the system (1) exists at least three classical solutions. According to Theorem 2, the system (1) possesses at least k distinct pairs of classical solutions.


## 4. Conclusions

In this work, we study perturbed fractional differential equation with impulses. We give sufficient conditions of the existence of at least three classical solutions and at leastk distinct pairs of classical solutions for problems (1), where k is the dimension of Vk.



---

*Source: 8512183-2020-11-12.xml*