# Applications of Morse Theory to Fourth-Order Impulsive Differential Equations∗

Authors: Dongdong Gao
Journal: Mathematical Problems in Engineering
 (2022)

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## Abstract

This paper is devoted to studying the nontrivial solutions for fourth-order impulsive differential equations. By applying Morse theory and combining it with local linking theory, some new criteria to guarantee that the fourth-order impulsive differential equation has at least one nontrivial solution will be given. Then, we give an example to illustrate the effectiveness and feasibility of the obtained results in our paper. Moreover, some previous corresponding results will be significantly improved in our paper.


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## Body

## 1. Introduction

In this paper, we will consider the following fourth-order differential equation with impulsive effects:(1)uivt+Au″t+But=ft,ut,a.e.t∈0,T,Δu″tj=I1ju′tj,j=1,2,…m,−Δu″′tj=I2jutj,j=1,2,…m,u0=uT=u″0+=u″T−=0,where A and B are the real constants, I1j,I2j∈CR,R for 1≤j≤m,0=t0<t1<t2<⋯tm<tm+1=T and Δu″tj=u″tj+−u″tj−,Δu″′tj=u″′tj+−u″′tj−, where u″tj−,u″′tj− and u″tj+,u″′tj+, denote the left and the right limits, respectively, of at u″tj,u′′tjt=tj,j=1,2,⋯m.In the last decade, the existence of solutions for boundary value problems with impulses has been dramatically investigated, and the methods to deal with these problems are coincidence degree theory, the method of upper and lower solutions, the monotone iterative technique, and some classical fixed point theorems (see [1–9] and references therein). It is becoming a trend to study boundary value problems with impulses by variational methods and some critical point theory in recent years (see [10–16] and references therein). Especially, the existence of solutions for fourth-order impulsive differential equations has attracted attention of many authors (see [17–22] and references therein).More precisely, in [17], Afrouzi et al. considered a fourth-order impulsive differential equation with two control parameters as follows:(2)uivt+Au″t+But=λft,ut+μgt,ut,a.e.t∈0,1,Δu″tj=I1ju′tj,j=1,2,…n,−Δu″′tj=I2jutj,j=1,2,…n,u0=u1=u″0=u″1=0,where A and B are two real constants and λ>0 and μ≥0 are two control parameters; they obtained that system (2) has at least three solutions under certain conditions by employing a critical point theorem.In [18], Sun et al. investigated the fourth-order differential equation with impulsive effects as follows:(3)uivt+Au″t+But=ft,ut,a.e.t∈0,T,Δu″tj=I1ju′tj,j=1,2,…l,Δu″′tj=I2jutj,j=1,2,…l,u0=uT=u″0+=u″T−=0..By applying the variational methods and critical point theory, one nontrivial solution and infinitely many distinct solutions for system (3) have been obtained.Also, Yue et al. [21] further considered the existence of infinitely many solutions for the above system (3) involving oscillatory behaviors of nonlinearity at infinity.In [19], Xie and Luo studied the following fourth-order impulsive differential equation:(4)uivt+Au″t+But=λft,ut,t≠tj,t∈0,1,Δu″tj=I1ju′tj,j=1,2,…m,−Δu″′tj=I2jutj,j=1,2,…m,u0=u1=u″0=u″1=0..They obtained at least one solution and infinitely many solutions for system (4) via some existing critical point theorems.In [22], Kong considered the following nonperiodic fourth-order differential equation:(5)u4x+ωu″x+axux−fx,ux=0,x∈R,Δu″xj=Pju′xj,j=1,2,…,Δu″′xj=Qjuxj,j=1,2,…..By using the mountain pass theorem, at least two nontrivial homoclinic solutions for system (5) have been obtained.Motivated by above facts, it is clear to see the methods to deal with these fourth-order boundary value problems are variational methods and some critical point theory. But as far as we know, Morse theory is an important method in nonlinear analysis which can also be a powerful tool to study the existence of solutions for differential equations (see [23–25] and references therein). However, Morse theory has not received so much attention as variational method and critical point theory, and also it should be pointed out that there is no published paper which has discussed the existence of solutions for fourth-order boundary value problems by using Morse theory. So, the aim in this paper is to establish some criteria of existence of at least one nontrivial solution for the fourth-order differential equations by using Morse theory.The highlights and major contributions of this paper are reflected in the subsequent key aspects. (1) We firstly consider the existence of solutions for fourth-order impulsive differential equation by employing Morse theory coupled with local linking arguments, which will effectively fill in the blank of Morse theory in fourth-order differential equations. (2) We construct a new suitable inequality in Lemma1, which will significantly improve some previous results [17–21]. In fact, the restrictive conditions in [17–21] have room for improvement which is why we give a more accurate condition to ensure the feasibility of Lemma 1. (3) Our model is more general than the previous paper [26, 27]. In fact, when we remove the impulsive functions I1j and I2j, system (1) is reduced to the form in [26]; when we remove the impulsive functions I1j and I2j and let A=B=0, system (1) is reduced to the form in [27].The rest of the paper is organized as follows. In Section2, some preliminaries and results which are applied later in the paper are presented. Section 3 is devoted to studying the existence of at least one nontrivial solution to (1). In Section 4, an example is presented to illustrate the effectiveness and feasibility of the obtained results in Section 3.


## 2. Preliminaries

We will assume for the remainder of the paper thatA≤0≤B.In the Sobolev spaceX:=H20,T∩H0T0,T, we can consider the inner product(6)u,v=∫0Tu″tv′′tdt,∀u,v∈X,which induces the norm(7)u=∫0Tu″t2dt1/2.Then, we can define the norm ofX as follows:(8)uX=∫0Tu″t2−Au′t2+But2dt1/2.It is immediate that(9)∫0Tu″2dt1/2≤uX.The usual norm ofC10,T,Lp0,T is defined as follows:(10)u∞=maxmaxt∈0,Tut,maxt∈0,Tu,t,uLp=∫0Tuptdt1/p.Lemma 1.
Letu∈X; then, u∞≤M1uX, where(11)M1=maxT3/22π,T.Proof.
Fromu0=uT=0, we have(12)ut=∫0tu′sds=−∫tTu′sds.
Therefore,(13)ut≤∫0tu′sds,ut≤∫tTu′sds.
The above two equalities imply(14)2ut≤∫0Tu′sds≤T∫0Tu′s2ds1/2≤T3/2π∫0Tu″s2ds1/2≤T3/2πuX,which shows that(15)ut≤T3/22πuX.
On the other hand, sinceut∈C10,T and u0=uT, by the mean value theorem, there exists at least one τ∈0,T, such that u′τ=0. So, we have(16)u′t=∫τtu″sds≤∫0Tu″sds≤T∫0Tu″s2ds1/2≤TuX.
LetM1=maxT3/2/2π,T, and we obtain u∞≤M1uX.Remark 1.
It is clear thatM1 is more simple in our results than that in [18, 21], which is defined as M1=maxT3/2/πδ,0T/δ1+1/π, where δ:=1−σ and σ=:maxAT2/π2,−BT4/π4,AT2/π2−BT4/π4,0, and if we assume that A≤0≤B, then σ=0 and δ=1, so M1=maxT3/2/π,T1+1/π.Remark 2.
It is also obvious thatM1 is better in our results than that in [17, 19], which is defined as M1=1+1/π. However, M1=1 in our results if T=1. Thus, our results generalize and improve the results in [17, 19].Remark 3.
It is also clear thatM1 is better in our results than that in [20], which is defined as M1=1/T+T. When T=1, M1=2 in [20]; however, M1=1 in our results. Thus, our results generalize and improve the result in [20].Definition 1.
Letu∈X be a weak solution of system (1) if the equality(17)∫0Tu″v′′−Au′v′+Buvdt=−∑j=1mI1ju′tjv′tj−∑j=1mI2jutjvtj+∫0Tft,utvtdtholds for v∈X.
Then, we consider the functionalφ:X⟶R defined by(18)φu=12uX2+∑j=1m∫0u′tjI1jtdt+∑j=1m∫0utjI2jtdt−∫0TFt,u0dt,where Ft,u=∫0uft,sds.
Owing to the continuity off and Iij,i=1,2, we immediately deduce that φ is continuous and differential for all u∈X and(19)φ′uv=∫0Tu″v′′−Au′v′+Buvdt+∑j=1mI1ju′tjv′tj+∑j=1mI2jutjvtj−∫0Tft,utvtdt,for any v∈X. Then, we can clearly know that the critical points of φ are weak solutions of system (1).Lemma 2 (see [18]).
Ifu∈X is a weak solution of system (1), then u is a classic solution of (1).Definition 2.
The functionalφ satisfies the (C)-condition if for c∈R, any sequence un⊂X such that φun⟶c,1+unXφ′unX⟶0 has a convergent subsequence as n⟶∞.
The following lemmas are crucial for proving the existence results in Section3.Lemma 3 (see [28]).
IfQ=0, then Cqφ,∞=Cqφ,0 for every q∈N. It follows that if Cqφ,∞≠Cqφ,0 for some q∈N, then φ must have a nontrivial critical point.Lemma 4 (see [29]).
Letu=0 be a critical point of φ with φ0=0. Suppose that φ has a local linking at 0 with respect to E=V⊕W,dimV=k<∞, that is, there exists ρ>0 such that(20)φu≤0,u∈V,u≤ρ;φu>0,u∈W,0<u≤ρ..Then,Ckφ,0≠0, and hence 0 is a homological nontrivial critical point of φ.For more details about theq-th critical group of φ and the singular relative homology group Hq, we refer readers to [25, 30].


## 3. Main Results

Theorem 1.
Suppose that the following conditions hold:(A1)
I1j0=0,I1juu≥0 and there exist c1j,d1j>0 and δ1j∈0,1 such that for any u∈R and j=1,2,…m, we have(21)I1ju≤c1j+d1juδ1j.(A2)
I2j0=0,I2juu≥0 and there exist c2j,d2j>0 and δ2j∈0,1 such that for any u∈R and j=1,2,…m, we have(22)I2ju≤c2j+d2juδ2j.(A3)
I1j^u:=I1j˜u−1/2I1ju′u′≥0 and I2j^u:=I2j˜u−1/2I2juu≥0, where I1j˜u=∫0u′I1jtdt and I2j˜u=∫0uI2jtdt, j=1,2,…m. There exist θ1j∈0,1,θ2j∈0,1 such that I1j^u≥I1j^θ1ju and I2j^u≥I2j^θ2ju for any u∈R.(H1)
ft,0=0,0≤Ft,u=ou2u⟶0 and there exist c1>0,0<r<r0,1<γ<maxδ1j+1,δ2j+1 such that(23)Ft,u≥c1uγ,r≤u≤r0,t∈0,T,whereFt,u=∫0uft,sds.(H2)
ft,uu≥0 for any t,u∈0,T×R and limu⟶+∞ft,u/u=+∞ uniformly for t∈0,T.(H3)
lim|u|⟶∞Ft,u/u2=+∞ uniformly for t∈0,T.(H4)
F^t,u:=Ft,u−1/2ft,uu≤0 and there exists a θ∈0,1 such that F^t,u≥F^t,θu for any t,u∈0,T×R.Then, system (1) has at least one nontrivial solution.
In order to complete the proof, we need to prove the following lemmas and compute the critical groups at zero and at infinity.Lemma 5.
Suppose thatφ satisfies (A1), (A2), (A3), (H1), (H2), and (H4); then, φ satisfies the (C)-condition.Proof.
Firstly, we will show the boundedness of (C) sequences. Suppose thatφ has an unbounded (C) sequence uk. Up to a subsequence, we may assume that(24)φuk⟶c,ukX⟶∞,φ′uk,uk⟶0,as k⟶∞ for some c∈R.
Letwk:=uk/ukX, and thus wkX=1; up to a subsequence, we have wk⟶w0 in Lp0,T and wk⟶w0, a.e. t∈0,T.
Ifw0=0, as the same proof of [31], we can extract a sequence ηk⊂0,1 such that(25)φηkuk=maxη∈0,1φηuk.
Letr=2b, where b is a positive integer, such that rukX−1∈0,1 as k⟶∞. Since wk⟶0 in Lp0,T and from (H1), we immediately obtain F.,rwk⟶0 in L10,T. That is,(26)limk⟶∞∫0TFt,rwkdt=0.
By (A1), (A2), and (18), we have(27)φηkuk≥φr‖uk‖X−1uk=φrwk=12‖rwk‖X2+∑j=1m∫0rwk′tjI1jtdt+∑j=1m∫0rwktjI2jtdt−∫0TFt,rwktdt≥b−∫0TFt,rwktdt,which implies that φηkuk⟶+∞.
But from another dimension, we knowφ0=0 and φuk⟶c; then,(28)φ′ηkuk,ηkuk=ηkukX2+∑j=1mI1jηkuk′tjηkuk′tj+∑j=1mI2jηkuktjηkuktj−∫0Tft,ηkuktηkuktdt=ηk·ddηφηuk|η=ηk=0.
Combining (A3) and (H4), we have(29)φηkuk−12φ′ηkuk,ηkuk=12ηkukX2+∑j=1m∫0ηkuk′tjI1jtdt+∑j=1m∫0ηkuktjI2jtdt−∫0TFt,ηkuktdt−12ηkukX2−12∑j=1mI1jηkuk′tjηkuk′tj−12∑j=1mI2jηkuktjηkuktj+12∫0Tft,ηkuktηkuktdt=∑j=1mI1j^ηkuk+∑j=1mI2j^ηkuk+∫0TF^t,ηkukt≤∑j=1mI1j^uk+∑j=1mI2j^uk+∫0TF^t,ukt=φuk−12φ′uk,uk⟶c,as k⟶∞. This contradicts φηkuk⟶+∞.
Ifw0≠0, we define A1:=t∈0,T:w0t≠0 and A2:=0,T∖A1. So, measA1>0 and ukt⟶∞ for a.e. t∈A1 as k⟶∞. By (24), we have(30)φ′uk,uk=ukX2+∑j=1mI1juk′tjuk′tj+∑j=1mI2juktjuktj−∫0Tft,uktuktdt=o1.
Then,(31)1=limk⟶∞ukX2−o1ukX2=limk⟶∞∫0Tft,uktuktukX2dt−∑j=1mI1juk′tjuk′tjukX2−∑j=1mI2juktjuktjukX2=limk⟶∞∫A1ft,uktuktuk2wk2dt+∫A2ft,uktuktuk2wk2dt−limk⟶∞∑j=1mI1juk′tjuk′tjukX2−limk⟶∞∑j=1mI2juktjuktjukX2≥limk⟶∞∫A1ft,uktuktuk2wk2dt−B1,where B1 is a positive constant. By (H2), we have(32)limk⟶∞ft,uktuktuk2wk2=+∞,for any t∈0,T. According to Fatou’s lemma, we immediately have(33)limk⟶∞∫A1ft,uktuktuk2wk2dt=+∞,which is also a contradiction with 1≥limk⟶∞∫A1ft,uktukt/uk2wk2dt−B1. So, we deduce that uk is bounded in X.
SinceX is a reflexive space, we may extract a weakly convergent subsequence, and we denote uk and uk⇀u in X. Then, we have uk⟶u in Lp0,T and uk⟶u, a.e. t∈0,T. By (19), we have(34)0←φ′uk−φ′u,uk−u=uk−uX2−∫0Tft,ukt−ft,utukt−utdt+∑j=1mI1juk′tj−I1ju′tjuk′tj−u′tj+∑j=1mI2juktj−I2jutjuktj−utj,since(35)∫0Tft,ukt−ft,utuk−udt⟶0∑j=1mI1juk′tj−I1ju′tjuk′tj−u′tj⟶0∑j=1mI2juktj−I2jutjuktj−utj⟶0,as k⟶∞, so we deduce that ‖uk−u‖X⟶0 for k⟶∞, that is, uk⟶u in X. Thus, φ satisfies the (C)-condition.Lemma 6.
Suppose thatφ satisfies (A1), (A2), (A3), (H3), and (H4); then, Cqφ,∞=0 for all q∈N.Proof.
It is similar to the proof of Theorem 1.1 in [31]. Let Ω=u∈X:uX=1. By (A1), (A2), (10), and Lemma 1, we have(36)∫0τu′tjI1jtdt≤c1jτu′tj+d1jδ1j+1τu′tjδ1j+1≤c1jM1τuX+d1jM1δ1j+1δ1j+1τδ1j+1uXδ1j+1=c1jM1τ+d1jM1δ1j+1δ1j+1τδ1j+1,and(37)∫0τutjI2jtdt≤c2jτutj+d2jδ2j+1τutjδ2j+1≤c2jM1τuX+d2jM1δ2j+1δ2j+1τδ2j+1uXδ2j+1=c2jM1τ+d2jM1δ2j+1δ2j+1τδ2j+1.
Combining (H3) with Fatou’s lemma, we obtain(38)limτ⟶∞∫0TFt,τuτ2dt≥∫0Tlimτ⟶∞Ft,τuτu2u2dt=+∞.
Then,(39)φτu=12τ2+∑j=1m∫0τu′tjI1jtdt+∑j=1m∫0τutjI2jtdt−∫0TFt,τutdt≤12τ2−∫0TFt,τutτ2τ2dt+∑j=1mc1jM1τ+d1jM1δ1j+1δ1j+1τδ1j+1+∑j=1mc2jM1τ+d12jM1δ2j+1δ2j+1τδ2j+1=τ212−∫0TFt,τutτ2dt+M1∑j=1mc1j+c2jτ+∑j=1md1jM1δ1j+1δ1j+1τ1−δ1j+∑j=1md2jM1δ2j+1δ2j+1τ1−δ2j⟶−∞,as τ⟶∞. Let a<mininfuX≤1φu,0; there must exist τ0>1 such that φτu≤a for τ>τ0 and u∈Ω. Thus, we immediately obtain(40)12τ2≤a+∫0TFt,τutdt−∑j=1m∫0τu′tjI1jtdt−∑j=1m∫0τutjI2jtdt.
By (A3), (H4), and (40), we have(41)ddτφτu=1ττ2+∑j=1mI1jτu′tjτu′tj+∑j=1mI2jτutjτutj−∫0Tft,τututdt≤1τ2a+2∫0TFt,τutdt−∫0Tft,τututdt+1τ∑j=1mI1jτu′tjτu′tj−2∑j=1m∫0τu′tjI1jtdt+1τ∑j=1mI2jτutjτutj−2∑j=1m∫0τutjI2jtdt≤2τa+F^t,u−I1j^u−I2j^u<0.
So, we clearly know that there exists a uniqueS∈CΩ,R such that φSuu=a by the implicit function theorem. From [29], we can apply the function S to make a strong deformation retract from X∖0 to φa. Thus, we obviously obtain(42)Cqφ,∞=HqX,φa=HqX,X∖0=0,q∈N.
The proof is complete.
SinceX is a reflexive and separable Banach space, we can extract an orthogonal basis ej of X such that X=spanej:j=1,…,¯ and denote that Pj:=spanej, Vk=⊕j=1kPj, Yk=⊕j=k∞Pj¯, so X=Vk⊕Yk.Lemma 7.
Suppose thatφ satisfies (A1), (A2), and (H1); then, there exists k0∈N such that Ck0φ,0≠0.Proof.
From (A1), (A2), and (H1), we know thatI1j0=0, I2j0=0, and ft,0=0, so we know that the zero function is a critical point of φ. Then, it is enough to verify that the functional φ has a local linking at zero with respect to X=Vk⊕Yk. Since all the norms of finite dimensional normed space are equivalent, by (10), for any u∈Vk, we have(43)uLγ≔∫0Tuγ1/γ,which means ·Lγ is a norm of Vk and there exists M2>0 such that(44)M2uX≤·Lγ.Step 1.
Letu∈Vk; since Vk is finite dimensional, for given r0, there exists 0<ρ<1 such that(45)uX≤ρ⇒u≤r0,t∈0,T.
Then, we can divideΩ into three parts for 0<r<r0: Ω1=t∈0,T:ut≤r, Ω2=t∈0,T:r≤ut≤r0, and Ω3=t∈0,T:r0≤ut, that is, 0,T=∪i=13Ωi. For convenience, we let Gt,u=Ft,u−c1uγ. By (A1), (A2), (H1), (44), and Lemma 1, we have(46)φu=12uX2+∑j=1m∫0u′tjI1jtdt+∑j=1m∫0utjI2jtdt−∫0TFt,utdt=12uX2+∑j=1m∫0u′tjI1jtdt+∑j=1m∫0utjI2jtdt−c1∫0Tuγdt−∫Ω1Gt,utdt−∫Ω2Gt,utdt−∫Ω3Gt,utdt≤12uX2+M1∑j=1mc1j+c2juX+d1jM1δ1j+1δ1j+1uXδ1j+1+d2jM1δ2j+1δ2j+1uXδ2j+1−c1M2γuXγ−∫Ω1Gt,utdt.
Also, from Lemma1, we know that u≤M1uX≤M1ρ. If ρ<1/M1, then u<1. Since the norms on Vk are equivalent to each other and ∫Ω1Gt,utdt⟶0 as r⟶0, then for all u∈Vk and u≤ρ, we have φu≤0.Step 2.
Letu∈Yk; owing to the compactness of the continuous embedding X⟶C0∞0,T,R, for given ε>0, there exists 0<ρ<1 such that(47)uX≤ρ⇒u≤ε,t∈0,T.
Combining (A1), (A2), and (H1), we have(48)φu≥12uX2−∫0Tεu2dt≥12uX2−εuX2>0,in which ε is sufficiently small. So, we complete the proof owing to Lemma 4.
At last, we show the main proofs of Theorem1.Proof of Theorem 1.
LetV=Vk,W=Yk in Lemma 4. By Lemma 5, we obtain that φ satisfies the (C)-condition. Combining Lemma 6 with Lemma 7, we immediately have Ck0φ,∞≠Ck0φ,0 for some k0∈N. Thus, all conditions of Lemma 3 hold, that is, φ must have a nontrivial critical point which is also a classical solution of system (1).


## 4. An Example

In this part, we give corresponding example to illustrate the main results in our paper.Example 1.
LetT=B=1,A=−1; then, consider the following fourth-order impulsive differential equations:(49)uivt−u″t+ut=ft,ut,a.e.t∈0,T,Δu″tj=I1ju′tj,j=1,2,…m,−Δu″′tj=I2jutj,j=1,2,…m,u0=u1=u″0+=u″1−=0,where we denote(50)ft,u=et+12uln1+u+u|u|1+|u|,I1ju=43u1/3,I2ju=107u3/7.
By simple calculations, we have(51)Ft,u=et+1u2ln1+u,∫0u′I1jsds=u′4/3tj,∫0uI2jsds=u10/7tj.
Letc1j=c2j=1,d1j=4/3,δ1j=1/3,d2j=10/7,δ2j=3/7, c1=5/2,γ=29/21,r=e2−1,r0=e2, and we can clearly see that conditions (A1)–(A3) and (H1)–(H4) all hold. Then, system (49) possesses at least one nontrivial solution by Theorem 1 [32].



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