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MULTIPLICITY AND CONCENTRATION BEHAVIOUR OF POSITIVE SOLUTIONS FOR SCHRÖDINGER-KIRCHHOFF TYPE EQUATIONS INVOLVING THE p-LAPLACIAN IN RN∗

更新时间:2016-07-05

1 Introduction

In this article,we consider the multiplicity and concentration behavior of positive solutions for the following Schrödinger-Kirchhofftype problem

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involving the p-Laplacian,where 1<p<N,M:R+→R+,V:RN→R+are continuous function,∈is a small positive parameter,and Δpu=div(|∇u|p−2∇u)is the p-Laplacian of u.We assume that the potential V satisfies

(V1)V∈C(RN,R)and

(V2)for each δ> 0,there is an open and bounded set Λ = Λ(δ) ⊂ RNdepending on δ such that

and

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Before stating our main result,we make the following hypotheses:

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Problem(Q)is a natural extension of two classes of problems of great importance in applications,namely,Kirchhofftype problems and Schrödinger type problems.

(a)When ∈=1,p=2,and V=0,problem(Q)becomes the following problem

which represents the stationary case of Kirchhoffmodel for small transverse vibrations of an elastic string by considering the effects of the changes in the length of the string during the vibrations.

(b)When M ≡1 and p=2,(Q)becomes

which arises in different models,for example,to get a standing wave,that is,a solution of the form Ψ(x,t)=exp(−iEt/∈)u(x)of the following nonlinear Schrödinger equation

where f(t)=|t|s−2t,N > 2,and 2< s< 2=,and it will led to the study of(1.2).Many studies about the existence and concentration of positive solutions for problem(1.2)appeared in the past decade;see[1,4,16]and the references therein.

Recently,the following Kirchhofftype equation

has been studied extensively by many researchers,where f∈C(R3×R,R),and a,b>0 are constants.

X.He and W.Zou in[7]studied(1.4)under the conditions that f(x,u):=f(u)∈C1(R+,R+)satisfies the Ambrosetti-Rabinowitz condition((AR)condition in short):

(f2)there exists θ> 2p such that for all t> 0;

Similarly,J.Wang et al[15],Y.He,G.Li,and S.Peng[9],and G.Li and H.Ye[11]used the same arguments as in[7]to prove the existence of a positive ground state solution for(1.4)when f(x,u):=λf(u)+|u|4u,which exhibits a critical growth,wheref(u)u≥0,f(u)/u3is strictly increasing for u>0,and |f(u)|≤C(1+|u|q)for some 3<q<5,that is,f(x,u)∼ λ|u|p−2u+|u|4u(4< p< 6).

(F6)there is a q∈(4,6)such that

Recently,Y.He and G.Li in[8]studied the following Kirchhofftype equation with critical nonlinearity:

where ε is a small positive parameter,a,b> 0,λ > 0,and 2< p≤ 4.The potential V satisfies(V3)V∈C(R3,R)and

(V4)there is a bounded domain Λ such that

concentrating around a local minima of V under the conditions(V3),(V4),and the following conditions on f and M:

M.del Pino and P.Felmer in[4]studied(1.2)with the conditions on V replaced by(V3)and(V4).They proved that(1.2)possesses a positive bound state solution for ε> 0 small which concentrates around the local minima of V in Λ as ε→ 0.

In[6],G.Figueiredo,N.Ikoma,and J.Junior obtained the existence of positive solutions of the following equation

They constructed a family of positive solutions uε∈ H1(R3)which concentrates around a local minimum of V as ε→ 0.

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(F1)f∈C(R,R),f(s)=0 if s≤0;

(F2)−∞<

(F3)when N ≥ 3,f(s)/s(N+2)/(N−2)→ 0 as s→ ∞ and when N=2,f(s)/eαs2→ 0 as s→∞for any α>0;

(F4)there exists an s0>0 such thatwhere F(s):=f(t)dt when N≥2,and when N=1,−and there exists a>0 such that M(t)≥>0 for any t≥2;then,there holds→0 as t→∞;

the function M(t)is nondecreasing in[0,∞);

the function M(t)/t2/(N−2)is nonincreasing in[0,∞).

G.Figueiredo and J.Santos in[5]studied the multiplicity and concentration behavior of positive solutions of the following elliptic problem

where∈is a small positive parameter,the potential V satisfies(V1)and(V2),and the function M:[0,∞)→R+satisfies

(M1)M∈C()andM(t)≥ m0> 0,where m0> 0 is a constant;

(M2)the function t→M(t)is increasing on[0,+∞);

(M3)for all t1≥t2>0,

and f satisfies

(F5)

For the case f(x,u)=|u|p−2u(3< p≤ 4),G.Li and H.Ye[10]used the constrained minimization on a new manifold,which is obtained by combining the Nehari manifold and the corresponding Pohozaev’s identity,to get a positive ground state solution to(1.4).

(F7)there is a θ∈(4,6)such that

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(F8)the application

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is nondecreasing in(0,∞).

Motivated by the results in[5],we study the existence,multiplicity,and concentration behavior of positive solutions of the problem(Q)by using the variational method and the penalization method.

Problem(Q)is of nonlocal because of the presence of the termwhich implies that the equation in(Q)is no longer a pointwise identity.

(f1)f ∈ C(R,R)and there exist q,with p< q< p,a1> 0 such that|f(t)|≤ a1(1+|t|q−1)for all t>0;

f(u)/|u|3=0,f(u)/|u|q=0 for some 3< q< 5,and f(u)/u3is strictly increasing for u> 0,that is,f(u)behaves like|u|p−2u(4< p< 6).They showed that the Mountain Pass Theorem and the Nehari manifold can be used directly to obtain a positive ground state solution to(1.4).

(f3)f(t)=o(|t|p−1)as t→ 0;

(f4)the function is increasing for t> 0.

We define

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with the norm

It is easy to see that(E,‖ ·‖E)is a real Banach space.

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For I∈C1(E,R),we say that(un)⊂E is a Palais-Smale(PS)sequence at level c(henceforth denoted(PS)c)for I if(un)satisfies

Moreover,I satisfies the(PS)ccondition if any(PS)csequence possesses a convergent subsequence.

We recall that,if A is a closed set of a topological space of X,CatX(A)is the Ljusternik-Schnirelmann category of A in X,namely,the least number of closed and contractible sets in X which cover A.

Our main result is as follows:

再来观察遮挡物的位置对阴影形成的影响。在图9 中,面光源在遮挡物的影响下,在物体表面形成的阴影大小为△S。而在图10 中,在其他条件不变的情况下,将遮挡物向更靠近物体表面的方向移动一定距离,形成的阴影大小变为了△S’。对比两幅图,可以明显看出△S>△S’。由此可得:在其他条件保持不变的情况下,当遮挡物与被照射物体之间的距离越小时,面光源在遮挡物的影响下在物体表面形成的阴影面积就越小(即半影的面积越小),半影的面积越小说明物体被遮挡得越严密,相应地,遮挡系数就应该越大。因此,遮挡系数和遮挡物与被照射物体之间的距离成反相关。

Theorem 1.1 Suppose that the function M satisfies(M1)–(M3),the potential V satisfies(V1)–(V2),and the function f satisfies(f1)–(f4).Then,for any δ> 0,there exists a ∈δ > 0 such that,for any ∈∈ (0,∈δ),the problem(Q)has at least CatΠδ(Π)positive solutions.Moreover,if u denotes one of these positive solutions and η ∈ RNits global maximum,then

Remark 1.2 Our main result for the problem(Q)when p=2 includes the main result of[5]as a special case.A typical example of function f satisfying the conditions(f1)–(f4)is given bywith λi≥ 0 not all null and 2p < qi< p for all i∈ {1,2,···,k}.Any function of the formsatisfies the hypotheses(M1)–(M3)for all i∈ {1,2,···,k},where m0 > 0,b> 0,bi≥ 0,and γi∈ (0,1).

The proof of Theorem 1.1 is based on the frame work used in[5]which uses the Lusternik-Schnirelmann theory and abstract minimax theorems(see[14,17]).The main difficulties are the appearance of the non-local term and the lack of compactness because of the unboundedness of the domain RN.Moreover,as f is only continuous,we cannot use standard arguments on the Nehari manifold.As we will see later,the competing effect of the nonlocal term with the nonlinearity f(u)and the lack of compactness of the Sobolev’s embedding prevent us from using the variational methods in a standard way.

Now,we outline the sketch of the proof of our main result.The problem(Q)is equivalent to the following problem

by using the change of variable v(x)=u(∈x).The corresponding energy functional associated with problem()is defined by

whereandwhich are well defined on the space Egiven by

We call u ∈ E a weak solution to(Q)if for any ϕ ∈ E,it holds that

The norm of u ∈ Eis defined asand Eis a Banach space under the norm ‖ ·‖given above.

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In order to overcome the obstacle caused by the non-compactness because of the unboundness of the domain RN,following[4],we will modify the nonlinearity in a special way and to work with a auxiliary problem.

Let δ> 0 and the related bounded domain Λ be given as in(V2)and letwhere m0was given in(M1),and a>0 verifyingwhere V0>0 was given in(V1).We consider the following auxiliary problem

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where

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with

Here,χΛ denotes the characteristic function of the set Λ.

It is easy to see that under the assumptions(f1)–(f4),g(x,t)is a Caratheodory function satisfying the following assumptions:

(g1)g(x,t)=o(|t|p−1)as t→ 0 uniformly in x ∈ RN;

(g2)g(x,t)≤f(t)for all x∈RN,t>0;

(g3)0 < θG(x,t)≤ g(x,t)t,∀(x,t)∈ Λ×(0,∞),where θ is given in(f2);

(g4)

(g5)for each x∈ Λ,the application t→is increasing in(0,∞),and for each x ∈ RNΛ,the application t→is increasing in(0,a).

The energy functional I:E→ R associated with(Q∈,g)is given by

where

Using variational method,we can prove that the auxiliary functional Isatisfies the Palais-Smale condition,and the auxiliary problem(Q∈,g)has a positive ground state solution for all∈>0.

Because we are interested in giving a multiplicity result for the auxiliary problem(Q∈,g),we need to study the limit problem associated to(),namely,the following problem

with the energy functional

This functional is well defined on the spacewith the norm

for u0∈E0.Next,using the technique due to V.Benci and G.Cerami[2],we establish a relationship between the category of the set Π and the number of solutions for the auxiliary problem,and we will show that(Q∈,g)has multiple positive solutions by using Lusternik-Schnirelmann theory.Finally,we use the Moser iteration technique in[13]to prove that the solutions for the auxiliary problem(Q∈,g)are indeed solutions for the original problem.

We note that the function f in this article is only a continuous function,then we cannot apply standard arguments by using the Nehari manifold.We overcome this difficulty by the methods given in[14].

Throughout this article,we use standard notations.For simplicity,we writeRh to mean the Lebesgue integral of h(x)over a Lebesgue measurable set Ω⊂RN.For a Lebesgue measurable set A,we denote the Lebesgue measure of A by|A|.Lp=Lp(RN)(1<p≤∞)is the usual Lebesgue space with the standard norm|·|p.We use“→ ”and“⇀ ”to denote the strong and weak convergence in the related function spaces,respectively.C and Ciwill denote positive constants unless specified.BR(x):={y ∈ RNy − x|< R,x ∈ RN}. 〈·,·〉denote the dual pair for any Banach space and its dual space.

2 The Auxiliary Problem

By using the change of variable v(x)=u(∈x),we see that problem(Q)is equivalent to the following problem

In order to find positive solutions to(),without loss of generality,we shall assume that f(s)=0 for all s≤ 0.The corresponding energy functional associated with problem()is defined by

where andwhich is well defined on the Banach space E,

given by

in which the norm of u∈Eis given by

We see that J∈ C1(E,R)and

From now on,we will denote by E0the space Ewith∈=0.We know that Eis continuously embedded in Lν(RN)for ν ∈ [p,p];moreover,E is compactly embedded in Lν(A)for ν ∈[p,p)for any bounded measurable set A ⊂ RN.

In order to overcome the lack of compactness of the problem(eQ),we shall adapt the penalization method introduced by del Pino and Felmer in[4].

Let us choosewhere m0is given in(M1),and a>0 verifying

where V0>0 was given as in(V1).Using these numbers,we set the functions

and

where Λ was given as in(V2)for related fixed δ> 0,and χΛ denotes the characteristic function of the set Λ.

Using the above functions,we will study the existence of positive solution for the following auxiliary proble

where

Here,we recall that if u∈ E is a solution to prpblem(Q∈,g)with

where Λ = Λ/∈,then,u(x)is a solution of problem().

Associated with(Q∈,g),we have the energy functional I:E → R given by

where

which is well defined on the space Eand I∈ C1(E,R).Also,

Using the definition of g,it follows that

(g1)g(x,t)=o(|t|p−1)as t→ 0 uniformly in x ∈ RN;

(g2)g(x,t)≤f(t)for all x∈RN,t>0;

(g3)0 < θG(x,t)≤ g(x,t)t,∀(x,t)∈ Λ×(0,∞),where θ is given in(f2);

(g4)0 ≤ 2pG(x,t)≤ g(x,t)t≤,∀(x,t)∈ (Λ)c× (0,∞);

(g5)for each x∈ Λ,the application t→is increasing in(0,∞)and for each x ∈ RNΛ,the application t→is increasing in(0,a).

Lemma 2.1 The functional Isatisfies the following conditions:

(i)There are α0 > 0,ρ0 > 0 such that

(ii)There is an e∈ Ewith ‖v‖> ρ0and I(e)< 0.

Proof (i)From(g1)and(g2),for every∈>0,we have

Taking∈=min{1,m0}/2pCppand setting

we see that there exists an ρ0 > 0 such thatas q > p > 1 by(f1),we obtain I(v)≥ α0= ρ0η(ρ0)> 0 for all v ∈ E,with ‖v‖= ρ0.

(ii)Fix u ∈ Ewith suppu ⊂ Λ,and ‖u‖=1.By assumption(M3),we see that there is a γ1 > 0 such that M(t)≤ γ1(1+t)for all t≥ 0.

Then,for t>0,we have

because θ> 2p> p> 1 by(f2).The Lemma 2.1(ii)is then proved by taking v=t0u,with t0>0 large enough.

The main feature of the auxiliary functional is that it satisfies the Palais-Smale condition as we can see from the next result.

Proposition 2.2 The functional Iverifies the(PS)ccondition in Efor any c>0.

Proof (1)Assume that{un}is a(PS)csequence for I,then,

By using the hypothesis(M3),we deduce that for all t≥0.By(g3)and(g4),we obtain

therefore,{un}is bounded in E.Passing to a subsequence,for some u ∈ E,we obtain

Next,we will prove that{un}has a convergent subsequence in E.Firstly,we give two claims,

Claim 1 For all R>0,

Indeed,we assume that‖un → t0as n → ∞,we have‖u‖ ≤ t0.Take ηρ ∈ C(RN)such that

with

For each R > 0 fixed,denoting a(ξ,η)=(|ξ|p−2ξ− |η|p−2η,ξ− η),it is well known that a(ξ,η)> 0 for any p > 1 and ξ,η ∈ RNwith ξ/= η.Choosing ρ > R,we obtain

Setting

we have

Observe that

As{unηρ}is bounded in E,we have〈I(un),(unηρ)〉=on(1).Moreover,from a straight forward computation,

thus,

On the other hand,from the weak convergence,

We see that

By(2.4)andwe have

We conclude that

From(2.3),(2.5)–(2.8),we obtain

Hence,we obtain

By Lemma 2.1 of[3],for p ≥ 2 and ξ,η ∈ RN,

for 1 < p < 2 and ξ,η ∈ RN,

here,d1,d2,d3,and d4are some constants.

Using the above inequalities,for 1<p<2,we know that

For p≥2,

and consequently,

Claim 2 For each ξ> 0,there is an R=R(ξ)> 0 such that

For R >0,let ηR ∈C(RN)be such that

with 0 ≤ ηR(x)≤ 1,|∇ηR|≤,and C is a constant independent on R.Because{ηRun}is bounded in E,it follows fromthat

Therefore, fixing R > 0 such that Λ ⊂ BR2(0)and by using(M1)and(g3),we have

Thus,

By using Cauchy-Schwarz and Hölder’s inequality,we have

As{un}and{unηR}are bounded in E,we obtain

where R=R(ξ)>

Therefore,for each ξ> 0,there is an R=R(ξ)> 0 such that

where R=R(ξ)>.Taking ξ→0,we get R→ ∞,which implies that

hence,‖un → ‖u‖,and un → u in E.

Theorem 2.3 Suppose that the conditions(M1)–(M3),(V1)–(V2),and(f1)–(f4)are satisfied.Then,the auxiliary problem(Q∈,g)has a positive ground state solution for all∈> 0.

Proof This result follows from Lemma 2.1,Proposition 2.2,and the maximum principle.

Next,we introduce some tools needed for the multiplicity result of the auxiliary problem(Q∈,g).

We denote by Nthe associated Nehari manifold of Igiven by

Set Γu+={x∈RN|u+(x)/=0},letbe the open subset of Egiven by

and where Sis the unit sphere of E.Note that is a incomplete C1-manifold of codimension one in E(for example[14]).Hence,E=⊕Ru for each u∈,where

The idea of the proof of the next two results comes from[14](See also[5]).

Lemma 2.4 Suppose that the conditions(M1)–(M3),(V1)–(V2),and(f1)–(f4)are satisfied.Then,we have the following results:

(A1)For each u∈and by hu(t)=I(tu),there exists a unique tu>0 such that>0 in(0,tu)and< 0 in(tu,∞).

(A2)There is a τ>0 independent on u such that tu> τ for all u∈Moreover,for each compact set W⊂,there is a CW>0 such that tu≤CWfor all u∈W.

(A3)The mapgiven by is continuous andis a homeomorphism betweenand N.Moreover,

(A4)If there is a sequence(un)⊂such that dist(un,)→0,then,‖r(un)‖→ ∞and I(r(un))→ ∞ as n → ∞ for each ∈> 0.

Proof It is clear that,by Lemma 2.1,hu(0)=0,hu(t)>0 for t>0 small,and hu(t)<0 for t> 0 large.Thus,there is a global maximum point tu> 0 of husuch that=0,that is,tuu∈ N.We claim that there is a unique tu> 0 such that=0.Indeed,suppose that there exist t1>t2>0 withThen,for i=1,2,

Therefore,

By using(M3)and the definition of g,we obtain

Multiplying both sides byand using the hypothesis t1>t2,we get

So,

As u/=0,which is absurd in view of min{m0,1}≤<min{m0,1}.Thus,(A1)is proved.

Suppose that u∈,from(M1),(g1),(g2),and the Sobolev embeddings,we get

which implies that tu≥τ for some τ>0.If W⊂is compact,then argue by contradiction that there is a sequence{un}⊂W such that tun→∞.As W is compact,there is a u∈W with un → u in E.By the process of the proof of Lemma 2.1,we know that I(tunun)→ −∞.Note that if v∈ N,then,by(M3),(g3),and(g4),we obtain

Because{tunun}⊂ N,which contradicts to the fact that{tunun}⊂ Nand I(tunun)→−∞.This proves(A2).

We note that,r,andare well defined.Indeed,we have the fact by(A1)that,for each u∈,there is a uniqueOn the other hand,if u∈N,then u∈Otherwise,we have|supp(u+)∩Λ|=0,and by(g4),we obtain

So,by(M1),we have

which is impossible.Thus,is well defined and continuous.From

we conclude that ris a bijection between and N.

Now,we prove thatis continuous.Letsuch that un → u in E.By(A2),we can suppose that→ t0 > 0.We conclude fromthat

Letting n→∞,we have

which implies thatandThis shows thatin E.So,and r are continuous functions,and(A3)is proved.

Finally,letbe a sequence such that dist(un,)→0 as n→ ∞.Then,for each v∈ and n∈N,we havea.e.in x∈ Λ.Therefore,we conclude from(V1),(V2)and Sobolev embedding that there is a C(s)>0 such that

It follows from(f1),(g3),and(g4)that

for each t>0.Thus,

By using the definition of r,we get

As t>0 is arbitrary,we have

We conclude from(M3)that

for each n ∈ N,so,‖r(un)‖ → ∞ as n → ∞.The Lemma is proved.

Now,we define the applications

by

The next proposition can be found in[[14],Corollary 2.3],and we omit the proof of it.

Proposition 2.5 Suppose that the conditions(M1)–(M3),(V1)–(V2),and(f1)–(f4)are satisfied.Then,the following properties are obtained:(a)and

(b)and

(c)If{un}is a(PS)csequence of Ψ,then,{r(un)}is a(PS)csequence for I.If{un} ⊂ Nis a bounded(PS)csequence of I,then,{(un)}is a(PS)csequence for Ψ.

(d)u is a critical point of Ψ if and only if r(u)is a nontrivial critical point of I.Moreover,the corresponding critical values Ψ and I coincide,and

Remark 2.6 As in[14],we have the following minimax characterization of the in fimum of Iover N,

Corollary 2.7([5],Corollary 2.1) The functional Ψgiven in Proposition 2.5 satisfies the(PS)ccondition on.

3 Multiplicity of Solutions of Auxiliary Problem

3.1 The autonomous problem

In this section,we prove a multiplicity result for problem(Q∈,g).We first consider the associated limit problem of(),which is given as

with the corresponding energy functional

whereendowed with the norm

We denote by N0the associated Nehari manifold of J0given by

Letbe the open subset of E0given by

and =S0,where S0is the unit sphere of E0.

As in Section 2,is a incomplete C1,1-manifold of codimension one in E0.Hence,E0= for each u∈,where

Lemma 3.1 Suppose that the conditions(M1)–(M3),(V1)–(V2),and(f1)–(f4)are satisfied.Then,we obtain the following results:

(A1)For each u∈and by gu(t)=J0(tu),there is a unique tu>0 such that >0 in(0,tu)and <0 in(tu,∞).

(A2)There is a τ>0 independent on u such that tu≥ τ for all u∈ .Moreover,for each compact set W⊂there is a CW>0 such that tu≤CWfor all u∈W.

(A3)The map:→N0given by=tuu is continuous,and r=is a homeomor

phism betweenand N0.Moreover,r−1(u)=

(A4)If there is a sequence{un} ⊂ such that dist(un,)→ 0,then,‖r(un)‖0→ ∞ and J0(r(un))→∞as n→∞for each∈>0.

Proof The proof of Lemma 3.1 is similar to that of Lemma 2.4.

We set the applications

byand

Proposition 3.2 Suppose that the conditions(M1)–(M3),(V1)–(V2),and(f1)–(f4)are satisfied.So,we have the following properties:and

(b)and

(c)If{un}is a(PS)csequence of Ψ0,then,{r(un)}is a(PS)csequence of J0.If{un} ⊂ N0 is a bounded(PS)csequence of J0,then,{r−1(un)}is a(PS)csequence of Ψ0.

(d)u is a critical point of Ψ0if and only if r(u)is a nontrivial critical point of J0.Moreover,corresponding critical values coincide and

Proof The proof of Proposition 3.2 is similar to that of Proposition 2.5.

Remark 3.3 Similar to Remark 2.6,we have

Lemma 3.4 Assume that{un}⊂E0is a(PS)csequence for J0satisfying un⇀0 in E0.Then,one and only one of the following alternatives holds,

(a)un→0 in E0,or

(b)there exist constants R,β>0 and sequence{yn}⊂RNsuch that

Proof Suppose that case(b)is flase,we would have

for all R > 0.Since{un}is bounded in E0,again using Lion’s vanishing Lemma(See[12],Lemma 1.1),we have

By(M1),(f1),and(f3),we obtain

therefore,un → 0 in E0,which contradicts our assumption I∈n(un)→ c0.Thus,(a)holds. □

Theorem 3.5 Assume that{un}⊂E0is a(PS)csequence for J0,where c0is given in Remark 3.3.Then,problem(Q0)has a positive ground state solution.

Proof Arguing as in the proof of Proposition 2.2,we see that{un}is bounded in E0.Now,up to a subsequence,for some u∈E0,we have

By Lemma 3.4 and c0> 0,we know that u/=0 in E.There exists a t0> 0 such that

Step 1 We claim∇un→∇u a.e.x∈RN.

Forsince un⇀u in E0and →0 inobviously,→0 as n→∞.Hence,

By(3.1)–(3.3)and{un}being bounded in E0,we see easily that

as n→∞.

Because is bounded in R,from(3.8)and(3.9),we obtain

hence,by(M1),we have

For 1<p≤2,from(2.10),we know that

For p≥2,from(2.9),we obtain

The above limits(3.12)and(3.13)imply that for some subsequence of{un},we have

So,from(3.14),we conclude that

On the other hand,by Vitali’s theorem,it results that

Step 2 We claim thatwhere t0is given in(3.4).

Now,by(3.1)and(3.3),we have

and from(M2),it follows thatJust suppose thatwe obtain

This inequality implies thattherefore,there exists a∈(0,1)such thatCombining this information with the characterization c0,we derive

On the other hand,from(M3),we getby Fatou’s Lemma,we obtain

Thus,which is absurd.So,we obtainand the theorem now follows.

Lemma 3.6([5],Lemma 3.3) Assume that{un}is a sequence in W1,p(RN)such that,as n→∞,J0(un)→c0and{un}⊂N0.Then,{un}has a convergent subsequence in W1,p(RN).

Proof This result can be proven by using,with suitable modifications,the same arguments of[5].

3.2 Some technical results

In this subsection,we will establish the relation between the topology of the set Π and the number of positive solutions of(Q∈,g).We choose δ>0 such that Πδ ⊂ Λ.Let η ∈([0,∞))be such that 0≤η(t)≤ 1 with

We denote w to be a positive ground state solution of the problem(Q0)attained by Theorem 3.5 such that J0(w)=c0.For each y∈Π={x∈Λ:V(x)=V0},we define

with t > 0 satisfying

Let φ:Π →Nbe such that

Lemma 3.7 The function φ satisfies

Proof Assume the contrary that the lemma does not hold.We suppose that there exist δ0>0 and a sequence{yn}⊂Π such that

For each n∈N and for allwe have ∈nz ∈ Bδ(0).Then,

By using the change of variableand the definition of φ∈n(yn),we have

Since φ∈n(yn) ∈ N∈nand g=f in Λ,we get

and

Set

By the Lebesgue’s theorem,as n → ∞,we obtain

and

Now,we claim that,up to a subsequence,t∈n→ 1 as n → ∞.Indeed,in view of(3.19),we have

From continuity of w,there is asuch thatSo,from(f4),we get

Assume the contrary that there is a subsequence{t∈n}with t∈n → ∞.Thus,taking n → ∞in(3.24),by(M3)and(f3),we have the fact that the downside converges to in finity and the upside is bounded,which leads to a contradiction.Thus,{t∈n}is bounded,up to a subsequence,we have t∈n→t0≥0.From(3.19),(3.21),(M1),and(f4),we have t0>0.Thus,as n→∞in(3.19),we have

We get t0=1 from w∈N0.So,taking n→∞in(3.20)and combining(3.21)with(3.22),we get

which contradicts to(3.17).

Proposition 3.8 Assume that{un}is a sequence in W1,p(RN)such that,as n→∞,

and

with∈n→0.Then,there existssuch that the sequencehas a convergent subsequence in W1,p(RN).Moreover,up to a subsequence,forand some y∈Π,we have yn→y.

Proof Direct calculations shows that{un}is bounded in W1,p(RN),then we can argue as the proof of Lemma 3.4 to obtain a sequence()⊂RNand positive constants R and α such that

Denotethen,{vn}is also bounded in W1,p(RN),and going if necessary to a subsequence,we can assume that

Let tn>0 be such that(See Lemma 3.1(A1)).We have

which implies that

Thus,is bounded in W1,p(RN)and in W1,p(RN).We may assume that tn→t0>0.From the uniqueness of the weak limit,we have=t0v/=0.From Lemma 3.6,we obtain in E0,and so,vn=v in E0.Thus,

Now,we are going to prove that→y∈M.First,we will prove that{yn}is bounded,where yn=.Indeed,argue by contradiction that there exists a subsequence{yn}with|yn|→∞.Choose R>0 such that Λ⊂BR(0).Then,for n large enough,we have|yn|≥2R,and for each z∈(0),we have

Hence,using vn → v in E0,the above expression,the definition of g,and Lebesgue’s theorem,we obtain

and therefore,

which is a contradiction.Hence,{yn}is bounded and yn → y in RN.If y/∈ Λ,we can proceed as above to deduce that‖vn0≤ on(1).

Next,we prove that y∈Π.Arguing by contradiction again,we assume that V0<V(y).Consequently,from in E0,Fatou’s Lemma,and the invariance of RNby translations,we have

which does not make sense and completes the proof.

Remark 3.9 Proposition 3.8 is very important to show that the solution of the auxiliary problem are actually solution of the original problem,and it makes us possible to study the concentration behavior of the solutions.

Define

where h1:R+ → R+is a function such that lim∈→0h1(∈)=0.We can conclude from Lemma 3.7 that h1(∈)=|I(y))−c0|is such that h1(∈)→ 0 as∈→ 0+.By the definition of h1,we know that,for any y∈Π and for∈>0 small.

Let ρ > 0 be such that Πδ ⊂ Bρ(0),and consider the map χ :RN → RNdefined by

We define the barycenter map β:N→ RNby

By the definition of χ and Lebesgue’s theorem,we have

Lemma 3.10 For any δ> 0 and Πδ ={x ∈ RN:dist(x,Π)≤ δ},we have

Proof The proof of this lemma follows from well-known argument and can be found in([5],Lemma 5.5).

3.3 Multiplicity of solutions for(Q∈,g)

In this section,we present a relation between the topology of Π and the number of solutions of the auxiliary problem(Q∈,g).Asis not a complete metric space,the method of[1]can not be applied directly,but we can use the result in[14]to fulfill our task.

Lemma 3.11([14],Theorem 27)If there exist c≥cand a compact set K⊂such for some k∈N,wherethencontains at least k critical points of Ψ.

Theorem 3.12 Suppose that the conditions(M1)–(M3),(V1)–(V2),and(f1)–(f4)are satisfied.Then,for any δ>0,there exists a>0 such that the auxiliary problem(Q∈,g)has at least CatΠδ(Π)positive solutions for any∈∈(0,).

ProofFor each∈>0,we define the function ζ:Π →by

From Lemma 3.7,we obtain

Consider the set

where the function h1was already introduced in the definition of the setthen,for allthe setis nonempty.

From the above considerations,we can use Lemma 3.7,Lemma 2.4(A3),equality(3.28),and Lemma 3.10 to obtain>0,such that for any∈∈(0,),the diagram

is well defined.In view of(3.28),for allwe can denote β(y))=y+ϱ(∈,y)for all y ∈ Π,where|ϱ(∈,y)|<uniformly in y ∈ Π.Define H(t,y)=y+(1 − t)ϱ(∈,y).Thus,H:[0,1]× Π → Πδ is continuous.Obviously,H(0,y)= β(y)),H(1,y)=y for all y ∈ Π.That is,H(t,y)is a homotopy between β◦φ=(β◦r)◦ζand the inclusion map i:Π → Πδ.Therefore,

We can use Corollary 2.7 and Lemma 3.11,with c ≤ c0+h1(∈)=c and K= ζ(Π),to conclude that Ψhas at least Catζ∈(Π)ζ(Π)critical points onCombining Proposition 2.5(d)with(3.29),we deduce that I has at least CatΠδ(Π)critical points in

4 Proof of Theorem 1.1

In this section,our main purpose is to show that the solutions obtained in Theorem 3.12 are indeed solutions of the original problem().The idea is to prove u(x)< a for all x ∈ for∈small enough by using the Moser iterative method[13].The following lemma plays a fundamental role in the study of behavior of the maximum points of the solutions and can be found in[5].

Lemma 4.1Letbe a solution of problem(Q∈,g)with∈n→0+.Then,I∈n(un)→ c0and un ∈ L(RN).Moreover,for any given γ > 0,there exist R > 0 and n0∈N such that

whereis given by Proposition 3.8.

Proof This result can be proven by using,with suitable modifications,the same arguments of[5].

Proof of Theorem 1.1 We fix a small δ> 0 such that Πδ ⊂ Λ.

ClaimThere is a>0 such that for any solution u∈of the auxilary problem(Q∈,g),there holds

where 0 < ∈<

We argue by contradiction that there is a subsequence∈n→0+,unsuch that=0 and

It is clear that I∈n(un)→ c0as in Lemma 4.1.By Proposition 3.8,we can get a sequence⊂RNsuch that→y0∈Π.

For n large and letting r> 0 such that B2r(y0)⊂Λ,we get

and

for anyso,we haveTaking n0∈N such that for any n≥n0,there holds r/∈n > R.Following from Lemma 4.1 with γ =a,there holds

which contradicts(4.2)and completes the claim.

Set∈δ=min,and let∈∈(0,∈δ)be fixed.We conclude from Theorem 3.12 that the problem(Q∈,g)has CatΠδ(Π)nontrival solutions.From(4.1)and the definition of g,we obtain g(·,u)=f(u).Thus,any solution of the problem(Q∈,g)obtained in Theorem 3.12 is also a solution of the problem().It is clear thatis a solution of the original problem(Q).Then,(Q)has at least CatΠδ(Π)nontrivial solutions.

Let{un}⊂ E∈nbe a sequence of solutions of the problem()with ∈n → 0+.We now study the behavior of the maximum points of un.By(g1),there is a γ>0 such that

By Lemma 4.1,we obtain R>0 and()⊂RNsuch that

Up to a subsequence,we may also assume that

Indeed,if this is not the case,then,|un|L∞(RN)< γ,and therefore,it follows from =0 and(4.3)that

The above expression implies that ‖un∈n=0,which does not make sense,thus,(4.5)holds.

Let pn∈RNbe a maximum point of un,we conclude from(4.4)and(4.5)that pnbelongs toHence,for some qn∈BR(0).Sinceis a maximum point of.According to Proposition 3.8,we obtain

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HuifangJIA(贾,)GongbaoLI(李工宝)
《Acta Mathematica Scientia(English Series)》2018年第2期文献

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