# Three Solutions for Fourth-Order Impulsive Differential Inclusions via Nonsmooth Critical Point Theory

Authors: Dongdong Gao; Jianli Li
Journal: Journal of Function Spaces
 (2018)

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## Abstract

An existence of at least three solutions for a fourth-order impulsive differential inclusion will be obtained by applying a nonsmooth version of a three-critical-point theorem. Our results generalize and improve some known results.


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## Body

## 1. Introduction

In this paper, we will consider the following fourth-order differential inclusion with impulsive effects:(1)uivt+Pu′′t+Qut∈λ∂Fut+μ∂Gut,ut,t∈0,T,t≠tj,j=1,2,…,m,Δu′′tj=I1ju′tj,j=1,2,…,m,Δu′′′tj=I2jutj,j=1,2,…,m,u0=uT=u′′0+=u′′T-=0,where P and Q are real constants, λ and μ are real parameters, I1j,I2j∈C(R,R) for 1≤j≤m,0=t0<t1<t2<⋯tm<tm+1=T, the impulsive functions satisfy that Δ(u′′(tj))=u′′(tj+)-u′′(tj-),Δ(u′′′(tj))=u′′′(tj+)-u′′′(tj-), and the left and right limits of u′′(tj),u′′′(tj) are represented by u′′(tj-),u′′′(tj-) and u′′(tj+),u′′′(tj+), at t=tj,j=1,2,⋯m, respectively. Here, F is a locally Lipschitz function defined on R satisfying the following: (F1)
F(0)=0 and there exists a Lipschitz positive constant L1 such that (2)∫0tFusds-∫0tFvsds≤L1u-vfor anyu,v∈R.(F2)
There exist constantsa>0 and 1≤r<2, such that ζ≤a(1+ur-1) for all u∈R,ζ∈∂F(u).G is defined on [0,T]×R satisfying(G1)
G(·,u):[0,T]×R→R is measurable for each u∈R; G(t,·):[0,T]×R→R is locally Lipschitz for each t∈[0,T].(G2)
There exist constantsb>0 and 1≤r<2, such that ζ≤b(1+ur-1) for a.e.t∈[0,T],u∈R,ζ∈∂uG(t,u).In the last decades, the existence of solutions for boundary value problems has been dramatically investigated, by which a number of physics phenomena, population dynamics, optimal control, ecology, and biotechnology are vividly described and some classical tools have been applied to study this problems, for instance, coincidence degree theory [1], the method of upper and lower solutions [2], and some classical fixed point theorems [3].After that it is becoming a trend to study boundary value problems with impulses or without impulses by variational methods and some critical point theory in recent years; see [4–8] and references therein. As far as we know, the difficulties dealing with this problems are that their states are discontinuous. In the literature few papers considered the higher-order boundary value problems with impulsive effects. However, the fourth-order boundary value problems are essential in describing a class of elastic deflection; because of this crucial reason, the existence of solutions for fourth-order boundary value problems has attracted much attention of many authors.More precisely, in [9], the authors have investigated the periodic solutions for semilinear fourth-order differential inclusion without impulsive effects as follows: (3)uiv-pu′′+axu∈λ∂Fu+μ∂Gux,u,0<x<L,u0=u′′0=uL=0,and they proved that the above system has at least three periodic solutions under certain conditions by using a nonsmooth critical point theorem.In [10], the authors considered the fourth-order boundary value problem with impulsive effects as follows: (4)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 have been obtained.In [11], the following boundary value problem for a fourth-order impulsive differential equation has been studied: (5)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 that the boundary value problem possesses at least one solution and infinitely many solutions via some existing critical point theorems.From above several examples, we can clearly see that either the system is a differential equation with impulsive effects, or the system is a differential inclusion without impulsive effects. To the best of our knowledge, there are few papers which have concerned the existence of three solutions for fourth-order differential inclusion with impulsive effects so far. Motivated by above, the main aim in this paper is intended as an attempt to establish some criteria of existence of at least three solutions for the fourth-order differential inclusion by using a nonsmooth version of a three-critical-point theorem.The arrangement of the rest paper is as follows. In Section2, some preliminaries and results which are applied in the later paper are presented. In Section 3, we establish a variational approach for problem (1). In Section 4, the main proof of prospective results will be showed. In Section 5, a meaningful example is given to illustrate the obtained results in Section 4.


## 2. Preliminaries

In this section, we give some definitions and results that we shall use in the rest of the paper. For more details, please refer to [6, 12, 13].Let(X,·X) be a Banach space, (X∗,·X∗) be its topological dual, and φ:X→R be a functional. We denote that φ is locally Lipschitz if there exist a neighborhood U of u for each u∈X and a real number L>0 such that (6)φu-φv≤Lu-vXfor all u,v∈U. If φ is locally Lipschitz and u∈X, the generalized directional derivative of φ at u along the direction v∈X is (7)φ0u;v=limsupω→u,τ→0+φω+τv-φωτ. Meanwhile, the generalized gradient of φ at u is the set (8)∂φu=u∗∈X∗:u∗,v≤φ0u;v for all v∈X. So ∂φ:X→2X∗ is a multifunction. We also denote that φ has compact gradient if ∂φ maps bounded subsets of X into relatively compact subsets of X∗.Lemma 1 (see [13, proposition 1.1]).
Letφ∈C1(X) be a functional, then φ is locally Lipschitz and (9)φ0u;v=φ′u,vfor all u,v∈X; (10)∂φu=φ′ufor all u∈X.Lemma 2 (see [13, proposition 1.3]).
Letφ:X→R be a locally Lipschitz functional, then for u,v∈X, φ0(u;·) is subadditive and positively homogeneous, and there exists a Lipschitz constant L>0 such that (11)φ0u;v≤Lvfor φ around u.Lemma 3 (see [13, proposition 1.6]).
Letφ,ψ:X→R be locally Lipschitz functional, then (12)∂λφu=λ∂φu,∂φ+ψu⊆∂φu+∂ψufor all u∈X,λ∈R.Lemma 4 (see [12, Lemma 6]).
Letφ:X→R be a locally Lipschitz functional with a compact gradient, then φ is sequentially weakly continuous.We say thatu∈X is a critical point of locally Lipschitz functional φ if 0∈∂φ(u).In order to prove our main results, we also need some basic definitions and lemmas.Definition 5 (see [6, definition 2.1]).
An operatorA:X→X∗ is of type (S)+, if, for any sequence (un) in X, un⇀u and limsupn〈A(un),un-u〉≤0 imply un→u.Theorem 6 (a particular case of [12, Theorem 14]).
Let(X,·) be a reflexive Banach space, I⊆R be an interval, N∈C1(X) be a sequentially weakly lower semicontinuous functional and its derivative is of type (S)+, F:X→R be a locally Lipschitz functional with a compact gradient, and ρ∈R. Assume that, for all λ∈I, we have (13)limu→∞Nu-λFu=+∞;(14)supλ∈Iinfu∈XNu-λρ-Fu<infu∈Xsupλ∈INu-λρ-FuThen, there exist α,β∈I(α<β) and r>0 with the following characters: for any λ∈[α,β] and any multifunction G:X→R with compact gradient, then there exists δ>0 such that the functional φλ,μ=N-λF-μG possesses at least three critical points with norms in X less than r for all μ∈[0,δ].The minimax inequality (14) can be derived from the following result.Lemma 7 (see [5, Proposition 3.1]).
LetX be a nonempty set, N,F:X→R be functions, uˇ,u^∈X, and τ>0, such that (15)Nuˇ=Fuˇ=0;(16)Nu^>τ;(17)supNu<τFu<τFu^Nu^.Then, there exists ρ∈R such that (14) holds.Definition 8.
A functionu∈X is a weak solution of problem (1) if there exists u∗∈λ∂F(u)+μ∂Gu(t,u) such that (18)∫0Tu′′v′′-Pu′v′+Quv-u∗vdt+∑j=1mI1ju′tjv′tj-∑j=1mI2jutjvtj=0.for all v∈X.Now, we state our main result.Theorem 9.
Assume that(F1),(F2) hold. Moreover, the following conditions also hold: (F3)
There existsτ>0 satisfying (19)τ∫0TFsin⁡πt/Tdtπ4/2T3-Pπ2/2T+QT/2+∑j=1m∫0π/Tcos⁡πtj/TI1jtdt-∑j=1m∫0sin⁡πtj/TI2jtdt>Tsups<M12τ1/2Fs,where(20)π42T3-Pπ22T+QT2>τandM1 is defined in Lemma 10.(F4)
I1j(0)=0,I2j(0)=0,sI1j(s)>0,sI2j(s)<0,s∈R∖{0},j=1,2,…,m.
Then, there exist a nondegenerate interval[α,β]⊂[0,+∞] and r>0 accompanied by the following property: for any λ∈[α,β] and any multifunction G:X→R with compact gradient, then there exists δ>0 such that the functional φλ,μ=N-λF-μG possesses at least three critical points with norms in X less than r for all μ∈[0,δ].


## 3. Variational Approach for the Problem (1)

We will assume for the remainder of the paper thatP≤0≤Q.TakingX≔H20,T∩H0T(0,T), we can define the following norm of X: (21)uX=∫0Tu′′2-Pu′2+Qu2dt1/2. It is immediate that (22)∫0Tu′′2dt1/2≤uX. The usual norm of C1([0,T]),L2(0,T) is defined as follows: (23)u=max⁡maxt∈0,Tut,maxt∈0,Tu′t,uL2=∫0Tu2tdt1/2.Lemma 10.
Letu∈X, then u≤M1uX, where (24)M1=max⁡T3/22π,T.Proof.
Fromu(0)=u(T)=0, we have (25)ut=∫0tu′sds=-∫tTu′sds;therefore (26)ut≤∫0tu′sds,ut≤∫tTu′sds.The above two equalities imply (27)2ut≤∫0Tu′sds≤T∫0Tu′s2ds1/2≤T3/2π∫0Tu′′s2ds1/2≤T3/2πuX, which shows (28)ut≤T3/22πuX. On the other hand, since u(t)∈C1[0,T] and u(0)=u(T), by the mean value theorem, there exists at least one τ∈(0,T), such that u′(τ)=0. So we have (29)u′t=∫τtu′′sds≤∫0Tu′′sds≤T∫0Tu′′s2ds1/2≤TuX.Let M1=max⁡{T3/2/2π,T}; we obtain u≤M1uX.Remark 11.
It is clear thatM1 is more simple in our results than that in [10], which is defined as M1=max⁡{T3/2/πδ,T/δ(1+1/π)}, where δ≔1-σ and σ≕max⁡{AT2/π2,-BT4/π4,AT2/π2-BT4/π4,0}. It is easy to see that if we assume A≤0≤B, then σ=0 and δ=1, so M1=max⁡{T3/2/π,T(1+1/π)}.Remark 12.
It is also obvious thatM1 is better in our results than that in [11], which is defined as M1=1+1/π. However, M1=1 in our results if T=1. Thus our results generalize and improve some known results.Next we are intended to look for suitable weak solution for system (1). Firstly, we assume that there exists a u∗∈λ∂F(u)+μ∂Gu(t,u) such that the first inclusion of (1) becomes the following form: (30)uivt+Pu′′t+Qut=u∗. Then, let t0=0<t1<⋯,<tm<tm+1=T. Take v∈X and multiply (30) by v and integrate from 0 to T; we have (31)∫0Tuivvdt+P∫0Tu′′vdt+Q∫0Tuvdt=∫0Tu∗vdt. One has (32)∫0Tuivvdt=∑j=0m∫tjtj+1uivvdt and (33)∫tjtj+1uivvdt=(u′′′tj+1-vtj+1--u′′′tj+vtj++u′′tj+v′tj+-u′′tj+1-v′tj+1-+∫tjtj+1u′′v′′dt. From above, we have (34)∫0Tuivvdt=-∑j=1mΔu′′′tjvtj+∑j=1mΔu′′tjv′tj+∫0Tu′′v′′dt=∑j=1mI1ju′tjv′tj-∑j=1mI2jutjvtj+∫0Tu′′v′′dt and (35)P∫0Tu′′vdt=-P∫0Tu′v′dt. So the weak solutions of (1) is defined as follows: (36)∫0Tu′′v′′-Pu′v′+Quvdt=-∑j=1mI1ju′tjv′tj+∑j=1mI2jutjvtj+∫0Tu∗vdt, where u∗∈λ∂F(u)+μ∂Gu(t,u) and v∈X.Lemma 13.
Ifu∈X is a weak solution of (1), then u is a classic solution of (1).Proof.
Firstly, it is easy to see thatu(0)=u(T) for u∈X. Since u∈X is a weak solution of (1), we have (37)∫0Tu′′v′′-Pu′v′+Quvdt=-∑j=1mI1ju′tjv′tj+∑j=1mI2jutjvtj+∫0Tu∗vdt for any v∈X. Then, choose any j∈{0,1,…,m} and v∈X such that v(0)=0 if t∈[tj,tj+1] for i≠j. Hence v′(tj)=0,j=1,…,m. Equation (37) implies that (38)∫tjtj+1u′′v′′-Pu′v′+Quvdt=∫tjtj+1u∗vtdt, and this also implies that for, any η∈Xj≔H2tj,tj+1∩H01(tj,tj+1), (39)∫tjtj+1uj′′η′′-Puj′η′+Qujηdt=∫tjtj+1u∗ηtdt, so uj∈Xj is a weak solution of the first equation of (1), and obviously it is a classic solution of (1).
By previous equality (37) and using integration by parts, we obtain (40)∫0Tu′′v′′-Pu′v′+Quvdt+∑j=1mI1ju′tjv′tj-∑j=1mI2jutjvtj-∫0Tu∗vdt=∑j=1m+1u′′v′tj+tj+1--∑j=1mu′′′vtj+tj+1-+∑j=1mI1ju′tjv′tj-∑j=1mI2jutjvtj+∫0Tuiv+Pu′′+Quvdt-∫0Tu∗vdt=∫0Tuiv+Pu′′+Qu-u∗vdt+∑j=1mI1ju′tj-Δu′′tjv′tj-∑j=1mI2jutj-Δu′′′tjvtj+u′′T-v′T-u′′0+v′0=0; since (41)uiv+Pu′′+Qu-u∗=0, we obtain (42)∑j=1mI1ju′tj-Δu′′tjv′tj-∑j=1mI2jutj-Δu′′′tjvtj+u′′T-v′T-u′′0+v′0=0.Next we will verify the second condition of (1); if not, we assume that there exists i∈{1,2…,m} such that (43)I2iuti-Δu′′′ti≠0. Let (44)v2t=∏j=0,j≠im+1tj-tit-tj2, and taking a derivation of (44), we have (45)v2′t=∑k=0,k≠im+12tk-tit-tk×∏j=0,j≠i,km+1tj-tit-tj2; obviously, v2∈X, v2(tj)=0,j=1,2,…,i-1,i+1,…,m,v2(ti)≠0, and v2′(tj)=0,j=0,1,2,…,m,m+1, which together with (42) yields (46)∑j=1mI1ju′tj-Δu′′tjv′tj-∑j=1mI2jutj-Δu′′′tjvtj+u′′T-v′T-u′′0+v′0=I2juti-Δu′′′tivti=I2juti-Δu′′′ti∏j=0,j≠im+1tj-ti3=0; since tj-ti≠0, we get (47)I2juti-Δu′′′ti=0, which contradicts with (43), so u satisfies the second impulsive condition of (1). From (47), we know that (42) can be described as the following form:(48)∑j=1mI1ju′tj-Δu′′tjv′tj+u′′T-v′T-u′′0+v′0=0. Let (49)v1t=∏j=0,j≠im+1t-tj2, then (50)v1′t=∑k=0,k≠im+12t-tk×∏j=0,j≠i,km+1t-tj2; by the same skills of the above proof, we also obtain (51)I1ju′tj-Δu′′tj=0. So, the third impulsive condition of (1) holds and (42) becomes the following form: (52)u′′T-v′T-u′′0+v′0=0; since v′(0),v′(T) are arbitrary, then we have u′′(0+)=u′′(T-)=0. Therefore, u is a classical solution of (1).


## 4. Proof of Theorem9

In order to prove our main results, we will first show some related lemmas. Then, the functional is defined as follows:(53)Nu=12uX2+∑j=1m∫0u′tjI1jtdt-∑j=1m∫0utjI2jtdt, so (54)N′u,v=∫0Tu′′v′′-Pu′v′+Quvdt+∑j=1mI1ju′tjv′tj-∑j=1mI2jutjvtj, where u,v∈X. (55)Fu=∫0TFutdt,G=∫0TGt,utdt,(56)φλ,μ=N-λF-μG.Lemma 14.
N ′ : X → X ∗ is of type ( S ) + .Proof.
Assumeun⇀u and limsupn〈N′(un),un-u〉≤0. According to Definition 5, we need to verify that un→u. In fact, (57)0=limn→∞N′u,un-u=limn→∞∫0Tu′′un-u′′-Pu′un-u′+Quun-udt+∑j=1mI1ju′tjun-u′tj-∑j=1mI2jutjun-utj, and (58)N′un,un-u=∫0Tun′′un-u′′-Pun′un-u′+Qunun-udt+∑j=1mI1jun′tjun-u′tj-∑j=1mI2juntjun-utj≤0.Equations (57) and (58) imply (59)limn→∞∫0Tun-u′′2-Pun-u′)2+Qun-u2dt+∑j=1mI1jun′tj-I1ju′tjun-u′tj-∑j=1mI2juntj-I2jutjun-utj≤0. Since un⇀u in X, we have un→u in C1[0,T], so ∑j=1mI1j(un′(tj))-I1j(u′(tj)))(un-u)′(tj)→0 and ∑j=1mI2j(un(tj))-I2j(u(tj)))(un-u)(tj)→0 as n→∞, and thus we have (60)limn→∞∫0Tun-u′′2-Pun-u′)2+Qun-u2dt=0; then from the definition of uX and the equivalence of uX with u, we have un-u→0 when n→∞, which implies un→u.Lemma 15.
The functionalIλ,μ:X→R is locally Lipschitz; moreover, for each critical point u∈X of Iλ,μ, u is a weak solution of (1).Proof.
LetIλ,μ=N(u)-I1(u), where I1(u)=λF(u)+μG(u). Firstly, we will show Iλ,μ is locally Lipschitz on X. Since N∈C1(X), according to Lemma 1, we know N is locally Lipschitz on X. On the other hand, from (F2) and (G2), we get that I1(u) is locally Lipschitz on Lr[0,T], since X is compactly embedded into Lr[0,T], so I1 is also locally Lipschitz on X. Then, by Lemma 3, we have (61)∂I1u⊂λ∫0T∂Fudt+μ∫0T∂Gut,udt. That is to say, for every u∗∈∂I1(u), we immediately have u∗∈λ∫0T∂F(u)dt+μ∫0T∂Gu(t,u)dt for a.e.t∈[0,T] with the character that the functional u∗v∈L1[0,T] and (u∗,v)=∫0Tu∗vdt for every v∈X, so Iλ,μ is locally Lipschitz on X.
Next we will prove that each critical pointu of Iλ,μ is a weak solution of (1). Let u∈X be a critical point of Iλ,μ, so (62)0∈∂Iλ,μ=u∗∈X∗:u∗,v≤Iλ,μ0u;v. According to (62) and Lemma 1, we have (63)N′u-u∗=0, where u∗∈∂I1(u), u∗(t)∈λ∂F(u(t))+μ∂Gu(t,u(t))a.e. on [0,T]. Thus, by (54) and (63), we have (64)∫0Tu′′v′′-Pu′v′+Quv-u∗vdt+∑j=1mI1ju′tjv′tj-∑j=1mI2jutjvtj=0 for v∈X, so u is a weak solution of (1).Lemma 16.
∂ F : X → 2 X ∗ is compact.Proof.
Let us define a bounded sequenceun in X and un∗∈∂F(un) for all n∈N, then we choose L>0 and un∗(t)∈∂F(un(t)) such that un≤L and 〈un∗,v〉=∫0Tun∗vdt, respectively. From (F2), we have (65)un∗,v≤∫0Tun∗vdt≤a∫0Tvdt+∫0Tunr-1vdt≤avL1+aunLrr-1vLr,so it is clear to see the sequence {un∗} is bounded, up to a subsequence; if un∗⇀u∗, we can prove un∗→u∗.
Suppose the assertion of conclusion is false, assume that there existsϵ>0 such that un∗-u∗X∗>ϵ for all n∈N, and then choose vn∈X with vn<1 such that (66)un∗-u∗,vn>ϵ. Passing if necessary to a subsequence, we suppose that vn⇀v in X, then vn→v in L1[0,T] and Lr[0,T]. By (F2), we have (67)un∗-u∗,vn=un∗,vn-v+un∗-u∗,v+u∗,v-vn≤avn-vL1+vn-vLrunLrr-1+un∗-u∗,v+u∗,v-vn→0 as n→0, which contradicts with (66).Proof of Theorem9.
We mainly use Theorem6 to prove Theorem 9. By the equivalence between usual norm and ·X which is defined in (21), we can easily obtain that the functional N defined in (53) is sequentially weak lower semicontinuous; according to above Lemma 14, we have that N′:X→X∗ is of type (S)+ and F:X→R is a locally Lipschitz functional with compact gradient by Lemmas 15 and 16.
Firstly, we will verify condition (13) in Theorem 6.
By(F4), for all u∈X we have (68)∫0utjI2jtdt<0,∫0u′tjI1jtdt>0, then Lemma 10, (F1), and (68) imply (69)Nu-λFu≥12uX2+∑j=1m∫0u′tjI1jtdt-∑j=1m∫0utjI2jtdt-λ∫0TL1udt≥12uX2-λa2uX, where a2=L1M1>0 and limuX→∞N(u)-λF(u)=+∞.
Next, we will verify condition (14) in Theorem 6. According to Lemma 7, choose uˇ(t)=0,u^(t)=sin⁡(πt/T),t∈[0,T]. Obviously, uˇ,u^∈X and N(uˇ)=F(uˇ)=0, so (15) holds. Then (70)Nu^=12∫0Tu^′′2-Pu^′2+Qu^2dt+∑j=1m∫0π/Tcos⁡πtj/TI1jtdt-∑j=1m∫0sin⁡πtj/TI2jtdt=π42T3-Pπ22T+QT2+∑j=1m∫0π/Tcos⁡πtj/TI1jtdt-∑j=1m∫0sin⁡πtj/TI2jtdt; from (F3) and (68), we have (71)Nu^>π42T3-Pπ22T+QT2>τ, which shows (16) holds.
Finally, we will verify condition (17) in Lemma 7. For any u∈X with N(u)<τ, from (F4) and Lemma 10, we get (72)u≤M1uX<M12τ1/2, then (73)supNu<τFu=supNu<τ∫0TFudt≤Tsups<M12τ1/2Fs and (74)τFu^Nu^=τ∫0TFsin⁡πt/T⁡dtπ4/2T3-Pπ2/2T+QT/2+∑j=1m∫0π/Tcos⁡πtj/TI1jtdt-∑j=1m∫0sin⁡πtj/TI2jtdt. By (F3), (73), and (74), we obtain (75)τFu^Nu^>supNu<τFu; thus we know that (17) in Lemma 7 holds; that is to say (14) in Theorem 6 holds for some ρ∈R.
With the same methods, the functionG satisfies (G1)-(G3), then according to Lemmas 15 and 16, the functional G:X→R is locally Lipschitz and ∂G is compact. From Theorem 6, there exist α,β∈I(α<β) and r>0 accompanied by the following property: for any λ∈[α,β] and any multifunction G:X→R with compact gradient, then there exists δ>0 such that the functional φλ,μ=N-λF-μG possesses at least three critical points with norms in X less than r for all μ∈[0,δ]. By Lemmas 13 and 15, we define the solutions as u0,u1,u2, which are ui<r,(1=0,1,2).


## 5. An Example

In this part, we will give a meaningful examples to illustrate the main results in our paper, but in [9], the authors do not give the corresponding examples.We consider the following differential inclusion:(76)uivt+Pu′′t+Qut∈λ∂Fut+μ∂Gut,ut,t∈0,T,t≠t1,Δu′′t1=I11u′t1,Δu′′′t1=I21ut1,u0=uT=u′′0+=u′′T-=0;let T=1,P=-1,Q=1,t1=2/3; let I11(s)=(1/4)s,I21(s)=-(1/5)s,F(s)=0 as s<1/2 and F(s)=s2 as s≥1/2, by Lemma 10; we obtain M1=1; obviously, the assumptions (F1),(F2), and (F4) hold; then we only need to verify (F3). Let τ=1/8; by a simple calculation, we have (77)π42T3-Pπ22T+QT2=π4+π2+12>τ,Tsups<M12τ1/2Fs=sups<1/2Fs=0and(78)∫01Fsin⁡πtdt=∫01/6Fsin⁡πtdt+∫1/65/6Fsin⁡πt)dt+∫5/61Fsin⁡πt)dt=∫1/65/6Fsin⁡πt)dt=∫1/65/6sin2⁡πtdt=13+34π;thus (79)τ∫0TFsin⁡πtdtπ4/2T3-Pπ2/2T+QT/2+∑j=1m∫0π/Tcos⁡πtj/TI1jtdt-∑j=1m∫0sin⁡πtj/TI2jtdt>0. So (F3) holds. Let Gt,s=-tcos⁡s+(2/3)s3/2, then ∂Gst,s=tsin⁡s+s1/2; clearly, the assumptions (G1),(G2) all hold. By Theorem 9, we can obviously deduce that (76) possesses at least three solutions.



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*Source: 1871453-2018-09-06.xml*