# Weak Solutions for ap-Laplacian Impulsive Differential Equation

Authors: Wei Dong; Jiafa Xu; Xiaoyan Zhang
Journal: Abstract and Applied Analysis
 (2013)

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## Abstract

By the virtue of variational method and critical point theory, we give some existence results of weak solutions for ap-Laplacian impulsive differential equation with Dirichlet boundary conditions.


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## Body

## 1. Introduction

In this paper, we shall consider the following problem:(1)-(|u′(t)|p-2u′(t))′=f(t,u(t)),a.e.t∈[0,T],ΔΦ(u′(tj))=|u′(tj+)|p-2u′(tj+)-|u′(tj-)|p-2u′(tj-)=Ij(u(tj)),j=1,2,…,m,u(0)=u(T)=0,
where p>1, 0=t0<t1<t2<⋯<tm<tm+1=T, f:[0,T]×ℝ→ℝ and Ij:ℝ→ℝ, (j=1,2,…,m) are continuous.Many evolution processes are characterized by the fact that, at certain moments of time, they experience a change of state abruptly. These processes are subject to short-term perturbations whose duration is negligible in comparison with the duration of the process. Consequently, it is natural to assume that these perturbations act instantaneously, that is, in the form of impulses. Thus impulsive differential equations appear as a natural description of observed evolution phenomena of several real world problems.Recently, there have been many papers to study impulsive problems by variational method and critical point theory, such as [1–11] and the references therein.In [7], Nieto and O’Regan studied the linear Dirichlet impulsive problem
(2)-u′′(t)+λu(t)=σ(t),t≠tj,t∈[0,T],Δu′(tj)=dj,j=1,2,…,p,u(0)=u(T)=0
and the nonlinear Dirichlet impulsive problem
(3)-u′′(t)+λu(t)=f(t,u(t)),t≠tj,t∈[0,T],Δu′(tj)=Ij(u(tj-)),j=1,2,…,p,u(0)=u(T)=0.
In the paper, they have shown that the impulsive problem minimizes some (energy) functional, and the critical points of that functional are indeed solutions of the impulsive problem.In [3, 4], Sun et al. utilized some variant fountain theorems by [12] to consider the existence of infinitely many solutions for the following two impulsive problems:
(4)-u′′(t)+g(t)u(t)=f(t,u(t)),a.e.t∈[0,T],Δu′(tj)=Ij(u(tj)),j=1,2,…,p,u(0)=u(T)=0,-u¨+A(t)u=∇W(t,u),a.e.t∈[0,T],Δu˙(tj)=Iij(ui(tj)),i=1,2,…,N,j=1,2,…,l,u(0)-u(T)=u˙(0)-u˙(T)=0.
Admittedly, they obtained many perfect results.We also note that some people begin to studyp-Laplacian differential equations with impulsive effects; for example, see [1, 2, 8–11].In [1], Chen and Tang considered the p-Laplacian impulsive problem
(5)-(|u′(t)|p-2u′(t))′+g(t)|u(t)|p-2u(t)=f(t,u(t)),a.e.t∈[0,T],Δu′(tj)=Ij(u(tj)),j=1,2,…,m,u(0)=u(T)=0.
They established some existence theorems for one or infinitely many solutions under more relaxed assumptions on their nonlinearity f, which satisfies a kind of new superquadratic and subquadratic condition.In [8], Bogun discussed the existence of weak solutions for the p-Laplacian problem with superlinear impulses by the virtue of mountain pass theorem and symmetric mountain pass theorem
(6)-(|u′|p-2u′)′=f(t,u),inΩ,u(0)=u(1)=0,u(tj+)-u(tj-)=0,j=1,2,…,n,Δ|u′(tj)|p-2u′(tj)=Ij(u(tj)),j=1,2,…,n.In [9], Xu et al. reconsidered the previous problem by topological degree theory and Fountain theorem under Cerami condition.In [11], by the virtue of three critical points theorem obtained by Bai and Dai is studied the existence of at least three solutions for the following p-Laplacian boundary value problem:
(7)(ρ(t)ϕp(u′(t)))′-s(t)ϕp(u(t))+λf(t,u(t))=0,a.e.t∈(a,b),α1u′(a+)-α2u(a)=0,βu′(b-)+β2u(b)=0,Δ(ρ(tj)ϕp(u′(tj)))=Ij(u(tj)),j=1,2,…,l.Motivated by the previous facts, in this paper, our aim is to study the existence and multiplicity of weak solutions for impulsive problem (1) by using variational method and critical point theory. It is well known that the Ambrosetti-Rabinowitz type condition is to ensure the boundedness of all (PS) sequences of the corresponding functional. However, without it, it will become more complicated. Therefore, we will use new variant fountain theorems due to Zou [12] to overcome this difficulty and obtain infinitely many weak solutions for (1). On the other hand, for the superlinear at +∞ and asymptotically linear at -∞, we obtain a weak solution for (1) by the mountain pass theorem. The results obtained here improve some existing results in the literature.


## 2. Preliminaries

In this section, we recall some fundamental facts of critical point theory which will be used in the proofs of our main results. LetW be the Sobolev space W01,p(0,T) with the usual norm
(8)∥u∥=(∫0T‍|u′(t)|pdt)1/p,∀u∈W01,p(0,T).
It is clear that W01,p(0,T) is a reflexive Banach space. Next, we make a finite dimensional decomposition for W. In order to do this, we first need to consider the eigenvalue problem
(9)(|u′(t)|p-2u′(t))′+λ|u(t)|p-2u(t)=0,t∈[0,T],u(0)=u(T)=0.
It is well known that the set of all eigenvalues of the problem (9) is given by the sequence of positive numbers (see [1, 13–15])
(10)λk:=(p-1)(kπpT)p,fork=1,2,…,whereπp:=2πpsin(π/p).
We denote by φk the corresponding eigenfunctions associated with λk for all k, and φk∈W. Moreover, the first eigenvalue λ1 is simple and isolated, and φ1 is positive in [0,T]. Furthermore, the Poincaré inequality
(11)∫0T‍|u′(t)|pdt≥λ1∫0T‍|u(t)|pdt,∀u∈W01,p(0,T)
holds. Note that we can normalize φk such that
(12)∫0T‍|φk(t)|pdt=1,∀k.
Fix any k≥1 define Yk=span{φ1,…,φk} and
(13)Zk=⋂j=1k‍ker(ℒj)={u∈W:ℒ1(u)=⋯=ℒk(u)=0},
where
(14)ℒju=∫0T‍|φj(t)|p-2φj(t)u(t)dt.
By [16, Section 5], the conclusions are
(15)W=Yk⊕Zk,dimYk=k,λk+1∫0T‍|u(t)|pdt≤∫0T‍|u′(t)|pdt,∀u∈Zk,k≥1.
Note the definitions of λk and φk; by (9), we have
(16)(|φk′(t)|p-2φk′(t))′+λk|φk(t)|p-2φk(t)=0fort∈[0,T].
For each v∈W, multiply by v on both sides of (16) to obtain
(17)∫0T‍|φk′(t)|p-2φk′(t)v′(t)dt=λk∫0T‍|φk(t)|p-2φk(t)v(t)dt.
In particular, choosing v=φk, we see ∫0T‍|φk′(t)|pdt=λk for all k.We denote the norms inLr(0,T)(1<r<∞) and C[0,T] as follows:
(18)∥u∥r:=(∫0T‍|u(t)|rdt)1/r,∀u∈Lr(0,T),∥u∥∞:=maxt∈[0,T]|u(t)|,∀u∈C[0,T].
By the Sobolev embedding theorem, the embeddings W↪Lr(0,T) and W↪C[0,T] are compact. Consequently, we also find that there are two constants Cemb1>0 and Cemb2>0 such that
(19)∥u∥r≤Cemb1∥u∥,∥u∥∞≤Cemb2∥u∥,∀u∈W.Foru∈W1,p(0,T), we have that u and u′ are both absolutely continuous. Hence, Δu′(t)=u′(t+)-u′(t-)=0 for any t∈[0,T]. If u∈W01,p(0,T), then u is absolutely continuous. In this case, the one-sided derivatives u′(t+), u′(t-) may not exist. It leads to the impulsive effects. As a result, we need to introduce a different concept of solution. Suppose that u∈C[0,T] satisfies the Dirichlet condition u(0)=u(T)=0. Assume that, for every j=1,2,…,m, uj=u|(tj,tj+1) and uj∈W1,p(0,T). Let 0=t0<t1<t2<⋯<tm<tm+1=T.Takev∈W and multiply the two sides of the equality
(20)-(|u′(t)|p-2u′(t))′=f(t,u(t))
by v and integrate from 0 to T:
(21)∫0T‍-(|u′(t)|p-2u′(t))′v(t)dt=∫0T‍f(t,u(t))v(t)dt.
For the left term, in view of impulsive effects, we find
(22)∫0T‍-(|u′(t)|p-2u′(t))′v(t)dt=∑j=0m‍∫tjtj+1‍-(|u′(t)|p-2u′(t))′v(t)dt=∑j=1m‍Ij(u(tj))v(tj)-v(T)|u′(T)|p-2u′(T)+v(0)|u′(0)|p-2u′(0)+∫0T‍|u′(t)|p-2u′(t)v′(t)dt=∑j=1m‍Ij(u(tj))v(tj)+∫0T‍|u′(t)|p-2u′(t)v′(t)dt.
Consequently,
(23)∫0T‍|u′(t)|p-2u′(t)v′(t)dt+∑j=1m‍Ij(u(tj))v(tj)=∫0T‍f(t,u(t))v(t)dt.
Considering the previous, we introduce the following concept for the solution for (1).Definition 1.
One says that a functionu∈W is a weak solution for (1) if the identity
(24)∫0T‍|u′(t)|p-2u′(t)v′(t)dt+∑j=1m‍Ij(u(tj))v(tj)=∫0T‍f(t,u(t))v(t)dt,∀v∈W.
Consider the functionalJ:W→ℝ defined by
(25)J(u)=1p∥u∥p+∑j=1m‍∫0u(tj)‍Ij(t)dt-∫0T‍F(t,u(t))dt,
where F(t,u)=∫0u‍f(t,s)ds. Note that for the continuity of f and Ij(j=1,2,…,m), we see J∈C1(W,ℝ). Furthermore, the derivative of J is
(26)(J′(u),v)=∫0T‍|u′(t)|p-2u′(t)v′(t)dt+∑j=1m‍Ij(u(tj))v(tj)-∫0T‍f(t,u(t))v(t)dt,∀u,v∈W.
Thus, we easily know that weak solutions of (1) coincide with the critical points of the C1-functional J.For the reader’s convenience, we now present some critical point theorems; one can refer to [12, 17–22] for more details.Definition 2.
LetX be a real Banach space and J∈C1(X,ℝ). For any sequence {un}⊂X, if {J(un)} is bounded and J′(un)→0 as n→∞ possesses a convergent subsequence, then we say that J satisfies the Palais-Smale condition (PS condition for short).Definition 3.
One says thatJ satisfies (PS)c condition if the existence of a sequence {un}⊂X such that J(un)→c and J′(un)→0 as n→∞ implies that {un} has a convergent subsequence.Lemma 4 (see [17]).
LetJ∈C1(X,ℝ) satisfy (PS) condition. Suppose that(i)
J(0)=0,(ii)
there existρ>0 and α>0 such that J(u)≥α for all u∈X, ∥u∥=ρ,(iii)
there existsu1∈X with ∥u1∥>ρ such that J(u1)<α.
ThenJ has a critical value c≥α. Moreover, c can be characterized as infg∈Γmaxu∈g([0,1])J(u), where Γ:={g∈C([0,1],X):g(0)=0,g(1)=u1}.
LetX be a Banach space equipped with the norm ∥·∥ and X=⨁j∈ℕXj¯, where dimXj<∞ for any j∈ℕ. Set Yk=⨁j=1kXj and Zk=⨁j=k∞Xj¯. In the following, one will introduce variant fountain theorems by Zou [12]. Let X and the subspaces Yk and Zk be defined as previously. Consider the following C1-functional Jλ:X→ℝ defined by
(27)Jλ(u):=A(u)-λB(u),λ∈[1,2].Lemma 5.
If the functionalJλ satisfies the following:(T1)
Jλ maps bounded sets to bounded sets uniformly for λ∈[1,2]. Moreover, Jλ(-u)=Jλ(u) for all (λ,u)∈[1,2]×X,(T2)
B(u)≥0 for all u∈X; moreover, A(u)→∞ or B(u)→∞ as ∥u∥→∞,(T3)
there existsrk>ρk>0 such that
(28)ak(λ)∶=infu∈Zk,∥u∥=ρkJλ(u)>bk(λ)∶=maxu∈Yk,∥u∥=rkJλ(u),∀λ∈[1,2],then
(29)ak(λ)≤ζk(λ)=infγ∈Γkmaxu∈BkJλ(γ(u)),∀λ∈[1,2],
where Bk={u∈Yk:∥u∥≤rk} and Γk={γ∈C(Bk,X):γisodd,γ|∂Bk=id}. Moreover, for a.e. λ∈[1,2], there exists a sequence {unk(λ)}n=1∞ such that
(30)supn∥unk(λ)∥<∞,Jλ′(unk(λ))⟶0,Jλ(unk(λ))→ζk(λ)asn⟶∞.Now, we list our assumptions onf and Ij(j=1,2,…,m).(H1)
There existμ1,μ2>0 such that |f(t,u)|≤μ1+μ2|u|q-1(1<q<∞), ∀u∈ℝ,t∈[0,T],(H2)
limu→0(f(t,u)/|u|p-2u)=f0∈[0,+∞) uniformly for t∈[0,T],(H3)
limu→-∞(f(t,u)/|u|p-2u)=f∞∈[0,+∞) uniformly for t∈[0,T],(H4)
There existδ>0 and θ∈(0,1/p) such that
(31)0<F(t,u)≤θuf(t,u),t∈[0,T],|u|≥δ,(H5)
F(t,u)≥0 for all (t,u)∈[0,T]×ℝ and lim|u|→∞(F(t,u)/|u|p)=+∞, uniformly on t∈[0,T],(H6)
there is a positive constantb>0 such that lim|u|→∞((-pF(t,u)+f(t,u)u)/|u|q)≥b, uniformly on t∈[0,T],(H7)
∫0u‍Ij(s)ds≥0, ∀u∈ℝ, j=1,2,…,m,(H8)
p∫0u‍Ij(s)ds-Ij(u)u≥0, ∀u∈ℝ, j=1,2,…,m,(H9)
There existσj,τj>0 and γj∈[1,p) such that |Ij(u)|≤σj+τj|u|γj-1, ∀u∈ℝ and j=1,2,…,m,(H10)
f(t,u) and Ij(u)(j=1,2,…,m) are odd functions about u, for all t∈[0,T].Remark 6.
(1) As known to all, (H4) implies thatlimu→∞(F(t,u)/|u|p)=+∞; that is, f(t,u) is superlinear at ∞ with respect to |u|p-2u. In view of (H3) and (H4), we see that f(t,u) is superlinear at +∞ and asymptotically linear at -∞; this is a new case. However, the nonlinearity f in [9] is asymptotically linear at ±∞.
(2) Condition (H6) is weaker than the well-known Ambrosetti-Rabinowitz condition (H4); also see condition(p2) in [8]. Indeed, by (H6), there is a δ>0 such that
(32)-pF(t,u)+uf(t,u)≥b|u|q≥0,∀|u|≥δ.
Consequently, pF(t,u)≤uf(t,u) for all t∈[0,T] and |u|≥δ, which is weaker than condition (H4).


## 3. Main Results

Theorem 7.
Suppose that (H1)–(H4), (H7), and (H9) hold,q∈(p,+∞), and f0<λ1<f∞ with f∞≠λk for all k. Then (1) has a weak solution.Proof.
From (H1)–(H4), for allε>0, there exist μ3>0,μ4>0 such that
(33)F(t,u)≤1p(f0+ε)|u|p+μ3|u|q,∀t∈[0,T],u∈ℝ,(34)F(t,u)≥1p(f∞-ε)|u|p-μ4,∀t∈[0,T],u∈ℝ.
Choose ε>0 such that (f0+ε)<λ1, together with (33), (11), (19), and (H7); we obtain
(35)J(u)=1p∥u∥p+∑j=1m‍∫0u(tj)‍Ij(t)dt-∫0T‍F(t,u(t))dt≥1p∥u∥p-∫0T‍F(t,u(t))dt≥1p∥u∥p-∫0T‍[1p(f0+ε)|u|p+μ3|u|q]dt≥1p∥u∥p-1pf0+ελ1∥u∥p-μ3Cemb1q∥u∥q.
If ρ is small enough, (ii) of Lemma 4 can be proved.
On the other hand, we can takeε>0 such that f∞-ε>λ1; by (34), (19), and (H9), noting that γj∈[1,p), we find
(36)J(sφ1)=1p∥sφ1∥p+∑j=1m‍∫0sφ1(tj)‍Ij(t)dt-∫0T‍F(t,sφ1(t))dt≤1p∥sφ1∥p+∑j=1m‍∫0sφ1(tj)‍[σj+τj|t|γj-1]dt-∫0T‍[1p(f∞-ε)|sφ1|p-μ4]dt≤1p∥sφ1∥p-1pf∞-ελ1∥sφ1∥p+∑j=1m‍[σj∥sφ1∥∞+τjγj∥sφ1∥∞γj]+μ4T≤1p∥sφ1∥p-1pf∞-ελ1∥sφ1∥p+∑j=1m‍[σjCemb2∥sφ1∥+τjγjCemb2γj∥sφ1∥γj]+μ4T⟶-∞ass⟶-∞.
Therefore, (iii) of Lemma 4 is also proved, as required.
Now, we only claim thatJ satisfies (PS) condition. Supposing that {un}⊂W is a (PS) sequence, for all n∈ℕ, we have
(37)|1p∥un∥p+∑j=1m‍∫0un(tj)‍Ij(t)dt-∫0T‍F(t,un(t))dt|≤ζ,(38)|∫0T‍|un′(t)|p-2un′(t)v′(t)dt+∑j=1m‍Ij(un(tj))v(tj)-∫0T‍f(t,un(t))v(t)dt|≤εn∥v∥,∀v∈W,
where ζ>0 is a constant and εn→0+ as n→∞. Next, we will show that {un} is a bounded sequence in W. If not, there is a subsequence of {un}, still denoted by {un}, such that
(39)∥un∥⟶∞,n⟶∞.
Define zn=un/∥un∥, then ∥zn∥=1,∀n∈ℕ, and thus it has a subsequence, still denoted {zn}, such that zn⇀z0 weakly in W, zn→z0 strongly in Lp(0,T), zn(t)→z0(t)a.e.t∈[0,T], and |zn(t)|≤η(t), a.e.t∈[0,T], where z0∈W, η∈Lp(0,T).
Divide (38) by ∥un∥p-1 to get
(40)|∫0T‍|zn′(t)|p-2zn′(t)v′(t)dt+∑j=1m‍Ij(un(tj))∥un∥p-1v(tj)-∫0T‍f(t,un(t))∥un∥p-1v(t)dt|≤εn∥un∥p-1∥v∥,∀v∈W.
Passing to the limit in (40), we see
(41)limn→∞∫0T‍f(t,un(t))∥un∥p-1v(t)dt=∫0T‍|z0′(t)|p-2z0′(t)v′(t)dt,∀v∈W,
with the fact that
(42)|∑j=1m‍Ij(un(tj))∥un∥p-1v(tj)|≤∑j=1m‍|Ij(un(tj))|∥un∥p-1|v(tj)|≤∑j=1m‍σj+τjCemb2γj-1∥un∥γj-1∥un∥p-1Cemb2∥v∥⟶0,∀v∈W.
Now, we claim that z0(t)≤0, a.e.t∈[0,T]. Indeed, in (41), taking v=z0+=max{z0,0}, we arrive at
(43)limn→∞∫Ω+‍f(t,un(t))∥un∥p-1z0(t)dt=∫Ω+‍|z0′(t)|pdt<+∞,
where Ω+:={t∈[0,T]:z0(t)>0}. However, by (H3) and (H4), there is μ5>0 such that
(44)f(t,un(t))∥un∥p-1z0(t)≥(-f∞|η(t)|p-2η(t)-μ5)z0(t),a.e.t∈[0,T].
In addition, note that limn→∞un(t)=+∞a.e.t∈Ω+ and the super-linearity of f we have
(45)limn→∞f(t,un(t))∥un∥p-1z0(t)=limn→∞f(t,un(t))unp-1znp-1(t)z0(t)=+∞,a.e.t∈Ω+.
Consequently, if |Ω+|>0, by the Fatou theorem, we get
(46)limn→∞∫Ω+‍f(t,un(t))∥un∥p-1z0(t)dt=+∞,
which contradicts the fact of (43).
Obviously,z0(t)≢0. From (H2) and (H3), there exists μ6>0 such that |f(t,un)|/|un|p-1≤μ6a.e.t∈[0,T]. By (41), Lebesgue’s dominated convergence theorem enables us to see
(47)∫0T‍|z0′(t)|p-2z0′(t)v′(t)dt=f∞∫0T‍|z0(t)|p-2z0(t)v(t)dt,∀v∈W.
This contradicts f∞≠λk for all k. Therefore, {un} is bounded, as required. Going, if necessary, to a subsequence, we can assume that un⇀u weakly in W; then
(48)(J′(un)-J′(u),un-u)=∫0T‍(|un′|p-2un′-|u′|p-2u′)(un′-u′)dt+∑j=1m‍(Ij(un(tj))-Ij(u(tj)))(un(tj)-u(tj))-∫0T‍(f(t,un)-f(t,u))(un-u)dt.W↪↪C[0,T] enables us to obtain that
(49)∑j=1m‍(Ij(un(tj))-Ij(u(tj)))(un(tj)-u(tj))⟶0,∫0T‍(f(t,un)-f(t,u))(un-u)dt⟶0asn⟶∞.
It follows from un⇀u weakly in W and (J′(un)-J′(u),un-u)→0 that
(50)∫0T‍(|un′|p-2un′-|u′|p-2u′)(un′-u′)dt⟶0asn⟶∞.
Note that
(51)∫0T‍(|un′|p-2un′-|u′|p-2u′)(un′-u′)dt≥(∥un∥p-1-∥u∥p-1)(∥un∥-∥u∥),
and thus ∥un-u∥→0 as n→∞. So, J satisfies (PS) condition. This completes the proof.In what follows, we will utilize Lemma5 to study (1). Now, we define a class of functionals on W by
(52)Jλ(u)=1p∥u∥p+∑j=1m‍∫0u(tj)‍Ij(t)dt-λ∫0T‍F(t,u(t))dt=A(u)-λB(u),λ∈[1,2].
It is easy to know that Jλ∈C1(W,ℝ) for all λ∈[1,2] and the critical points of J1 correspond to the weak solutions of problem (1). Note that J1=J, where J is the functional defined in (25).Theorem 8.
Assume that (H1) and (H5)–(H10) hold. Then (1) possesses infinitely many weak solutions.Proof.
We first prove that there is a positive integerk1 and two sequences rk>ρk→∞ as k→∞ such that
(53)ak(λ)=infu∈Zk,∥u∥=ρkJλ(u)>0,∀k≥k1,(54)bk(λ)=maxu∈Yk,∥u∥=rkJλ(u)<0,∀k∈ℕ,
where Yk=span{φ1,…,φk} and Zk=⋂j=1k‍ker(ℒj).
Step
1. We will show that (53) holds true.
From (H1), we see that there existμ7>0, μ8>0 such that
(55)F(t,u)≤μ7|u|+μ8|u|q,∀(t,u)∈[0,T]×ℝ.
Consequently, by (H5) and (H7), we obtain
(56)Jλ(u)=1p∥u∥p+∑j=1m‍∫0u(tj)‍Ij(t)dt-λ∫0T‍F(t,u(t))dt≥1p∥u∥p-2∫0T‍F(t,u(t))dt≥1p∥u∥p-2∫0T‍(μ7|u|+μ8|u|q)dt≥1p∥u∥p-2μ7T1-1/q∥u∥q-2μ8∥u∥qq.
Let νq(k)=supu∈Zk,∥u∥=1∥u∥q, ∀k∈ℕ. Then by [19, Lemma 3.8],  νq(k)→0 as k→∞. Since W↪↪Lq(0,T), we find
(57)Jλ(u)≥1p∥u∥p-2μ7T1-1/qνq(k)∥u∥-2μ8νqq(k)∥u∥q.
Let ρk=1/νq(k)→∞ as k→∞. Then there exists k1 such that (1/p)ρkp-2μ7T1-1/q-2μ8>0, ∀k≥k1. Therefore,
(58)ak(λ)=infu∈Zk,∥u∥=ρkJλ(u)≥1pρkp-2μ7T1-1/q-2μ8>0,∀k≥k1.Step2. We will show that (54) holds true.
We first prove that there existsε>0 such that
(59)meas(t∈[0,T]:|u(t)|≥ε∥u∥)≥ε,∀u∈𝒳∖{0},∀𝒳⊂W,dim𝒳<∞.
There exists otherwise a sequence {un}n∈ℕ⊂𝒳∖{0} such that
(60)meas(t∈[0,T]:|un(t)|≥∥un∥n)<1n,∀n∈ℕ.
For each n∈ℕ, let vn:=un/∥un∥∈𝒳⇒∥vn∥=1,∀n∈ℕ and
(61)meas(t∈[0,T]:|vn(t)|≥1n)<1n,∀n∈ℕ.
Passing to a subsequence if necessary, we may assume vn→v0 in W for some v0∈𝒳 since 𝒳 is of finite dimension. We easily find ∥v0∥=1. Consequently, there exists a constant σ0>0 such that
(62)meas(t∈[0,T]:|v0(t)|≥σ0)≥σ0.
Indeed, if not, then we have
(63)meas(t∈[0,T]:|v0(t)|≥1n)=0,i.e.,meas(t∈[0,T]:|v0(t)|<1n)=T,∀n∈ℕ,
which implies
(64)0<∫0T‍|v0(t)|pdt≤Tnp⟶0,asn⟶∞.
This leads to v0=0, contradicting to ∥v0∥=1. In view of W↪↪Lp(0,T) and the equivalence of any two norms on 𝒳, we have
(65)∫0T‍|vn-v0|pdt⟶0,asn⟶∞.
For every n∈ℕ, denote
(66)𝒩:={t∈[0,T]:|vn(t)|<1n},𝒩c:={t∈[0,T]:|vn(t)|≥1n},
and 𝒩0:={t∈[0,T]:|v0(t)|≥σ0}, where σ0 is defined by (62). Then for n large enough, by (62), we see
(67)meas(𝒩∩𝒩0)≥meas(𝒩0)-meas(𝒩c)≥σ0-1n≥σ02.
Consequently, for n large enough, we arrive immediately at
(68)∫0T‍|vn-v0|pdt≥∫𝒩∩𝒩0‍|vn-v0|pdt≥12p∫𝒩∩𝒩0‍|v0|pdt-∫𝒩∩𝒩0‍|vn|pdt≥(σ0p2p-1np)meas(𝒩∩𝒩0)≥σ0p+12p+2>0.
This contradicts (65). Therefore, (59) holds. For the ε given in (59), we let
(69)𝒩u:={t∈[0,T]:|u(t)|≥ε∥u∥},∀u∈𝒳∖{0}.
Then by (59), we find
(70)meas(𝒩u)≥ε,∀u∈𝒳∖{0}.
By (H5), for any k∈ℕ, there is a constant Sk>0 such that
(71)F(t,u)≥|u|pεp+1,∀|u|≥Sk,
where ε is determined in (59). Therefore,
(72)Jλ(u)=1p∥u∥p+∑j=1m‍∫0u(tj)‍Ij(t)dt-λ∫0T‍F(t,u(t))dt≤1p∥u∥p+∑j=1m‍∫0u(tj)‍[σj+τj|t|γj-1]dt-∫𝒩u‍|u|pεp+1dt≤(1p-1)∥u∥p+∑j=1m‍[σjCemb2∥u∥+τjγjCemb2γj∥u∥γj].
Now for any k∈ℕ, if we take rk>max{ρk,Sk/ε}, noting that p>1 and p>γj and ∥u∥=rk large enough, we have
(73)bk(λ)=maxu∈Yk,∥u∥=rkJλ(u)<0,∀k∈ℕ.Step3. The continuity f and Ij(j=1,2,…,m) and Jλ∈C1(W,ℝ) imply that Jλ maps bounded sets to bounded sets uniformly for λ∈[1,2]. In view of (H10), Jλ(-u)=Jλ(u) for all (λ,u)∈[1,2]×W. Thus condition (T1) of Lemma 5 holds. Besides, by (H7), we get
(74)A(u)=1p∥u∥p+∑j=1m‍∫0u(tj)‍Ij(t)dt≥1p∥u∥p⟶∞as∥u∥⟶∞
and B(u)≥0 since F(t,u)≥0. Thus the condition (T2) of Lemma 5 holds. For step 1 and step 2, the condition (T3) of Lemma 5 also holds for all k≥k1. Consequently, Lemma 5 implies that, for any k≥k1 and a.e. λ∈[1,2], there exists a sequence {unk(λ)}n=1∞ such that
(75)supn∥unk(λ)∥<∞,Jλ′(unk(λ))⟶0,Jλ(unk(λ))⟶ζk(λ)asn⟶∞,
where Bk={u∈Yk:∥u∥≤rk}, Γk={γ∈C(Bk,W):γisodd,γ|∂Bk=id} and ζk(λ)=infγ∈Γkmaxu∈BkJλ(γ(u)), ∀λ∈[1,2].
Furthermore, we easily have(76)ζk(λ)∈[a¯k,ζ¯k],∀k≥k1,
where ζ¯k:=maxu∈BkJλ(γ(u)) and a¯k:=(1/p)ρkp-2μ7T1-1/q-2μ8→∞ as k→∞.
Claim
1.
{
u
n
k
(
λ
)
}
n
=
1
∞
⊂
W possesses a strong convergent subsequence in W, for all λ∈[1,2] and k≥k1. In fact, by the boundedness of {unk(λ)}n=1∞, passing to a subsequence, as n→∞, we may assume unk(λ)⇀uk(λ) in W. By the method of Theorem 7, we easily prove that unk(λ)→uk(λ) strongly in W.
Thus, for eachk≥k1, we can choose λl→1 such that the sequence {unk(λl)}n=1∞ obtained a convergent subsequence, and passing again to a subsequence, we may assume
(77)limn→∞unk(λl)=ulkinW,∀l∈ℕ,k≥k1.
Thus we obtain
(78)Jλl′(ulk)=0,Jλl(ulk)∈[a¯k,ζ¯k],∀l∈ℕ,k≥k1.Claim2.{ulk} is bounded in W and has a convergent subsequence with the limit uk∈W for all k≥k1. For convenience, we set ulk=ul for all l∈ℕ. If not, {ul} is unbounded in W; that is, ∥ul∥→∞. By (H6), there is μ9>0 such that
(79)-pF(t,u)+f(t,u)u≥b|u|q-μ9,∀(t,u)∈[0,T]×ℝ.
Combining this and (H8), we have
(80)pJλl(ul)-(Jλl′(ul),ul)≥-λlp∫0T‍F(t,ul(t))dt+λl∫0T‍f(t,ul(t))ul(t)dt≥λl∫0T‍(b|ul|q-μ9)dt=λlb∫0T‍|ul|qdt-λlμ9T.
This implies that
(81)∫0T‍|ul|qdt∥ul∥p⟶0asl⟶∞.
On the other hand, by (H1) and (H9), we see
(82)(Jλl′(ul),ul)=∥ul∥p+∑j=1m‍Ij(ul(tj))ul(tj)-λl∫0T‍f(t,ul(t))ul(t)dt≥∥ul∥p+∑j=1m‍(-σj-τj|ul(tj)|γj-1)|ul(tj)|-λl∫0T‍(μ1+μ2|ul(t)|q-1)|ul(t)|dt≥∥ul∥p-∑j=1m‍[σjCemb2∥ul∥+τjCemb2γj∥ul∥γj]-λlμ1T1-1/q(∫0T‍|ul(t)|qdt)1/q-λlμ2∫0T‍|ul(t)|qdt.
Consequently, noting that p>1 and p>γj, we have
(83)1=∥ul∥p∥ul∥p≤(Jλl′(ul),ul)∥ul∥p+∑j=1m[σjCemb2∥ul∥+τjCemb2γj∥ul∥γj]∥ul∥p+λlμ1T1-1/q(∫0T‍|ul(t)|qdt)1/q∥ul∥p+λlμ2∫0T‍|ul(t)|qdt∥ul∥p⟶0asl⟶∞.
This is a contradiction. Therefore, {ul}l=1∞ is bounded in W. By claim 1, we see that {ul}l=1∞ has a convergent subsequence, which converges to an element uk∈W for all k≥k1.
Hence, passing to the limit in (78), we see
(84)J1′(uk)=0,J1(uk)∈[a¯k,ζ¯k],∀l∈ℕ,k≥k1.
Since a¯k→∞ as k→∞, we get infinitely many nontrivial critical points of J1=J. Therefore (1) possesses infinitely many nontrivial solutions by Lemma 5. This completes the proof.Remark 9.
(1) Let(85)f(t,u)={g(u)|u|p-2u,u≤0,g(u)|u|p-2u+uq-1,u>0,
where g∈C(ℝ), g(0)=0, g(-∞)∈(λ1,+∞) with avoiding λk for all k. Then (H1)–(H4) hold true with q∈(p,+∞) and f0<λ1<f∞ with f∞≠λk for all k.
(2) Letp=5, q=6, f(t,u)=u5, and Ij(u)=u3 for all t∈[0,T] and u∈ℝ. Then F(t,u)=∫0u‍f(t,s)ds=(1/6)|u|6, ∫0u‍Ij(s)ds=(3/4)u43. Clearly, (H1), (H7), (H9), and (H10) are satisfied.(1)
lim|u|→∞(F(t,u)/|u|p)=lim|u|→∞(|u|6/6|u|5)=+∞, uniformly on t∈[0,T], and F(t,u)≥0 for all (t,u)∈[0,T]×ℝ. So, (H5) holds.(2)
lim|u|→∞((-pF(t,u)+f(t,u)u)/|u|q)=lim|u|→∞((-5×(1/6)|u|6+u5u)/|u|6)=1/6, uniformly on t∈[0,T], and hence (H6) holds.(3)
p∫0u‍Ij(s)ds-Ij(u)u=5×(3/4)u43-u3×u=(11/4)u43≥0, ∀u∈ℝ, j=1,2,…,m. Therefore, (H8) holds.



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*Source: 106023-2013-07-24.xml*