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MULTIPLICITY OF POSITIVE SOLUTIONS FOR A CLASS OF CONCAVE-CONVEX ELLIPTIC EQUATIONS WITH CRITICAL GROWTH∗

更新时间:2016-07-05

1 Introduction and Main Result

In this article,we study the following concave-convex elliptic equations involving critical Sobolev exponent

where Ω⊂RN(N≥3)is an open bounded domain with smooth boundary,1<q<2,λ>0.is the critical Sobolev exponent for the embedding ofinto Lp(Ω)for everywhereis a Sobolev space equipped with the normThe coefficient function f ∈ Lr(Ω)is nonzero and nonnegative,whereAnd g ∈ C(Ω)is a positive function.

More precisely,we say that a functionis called a weak solution of problem(1.1),if for allthere holds

where u+=max{u,0}.

It is well known that the pioneer work is Brézis and Nirenberg[4],that is,the existence of positive solutions of semilinear elliptic equations involving critical exponent is related to the dimension of space.After that,semilinear elliptic problems with critical exponent were extensively considered(see[1,2,5,7–30,33,35–37,39–42]).Particularly,Ambrosetti,Brézis,and Cerami[2]studied the following problem

where 0<q<1< p≤ 2−1.They obtained the classic results by the sub-supersolution method and variational method,that is,there exists λ0 > 0 such that problem(1.3)has at least two positive solutions for λ ∈ (0,λ0),a positive solution for λ = λ0,and no positive solution for λ > λ0.After that,many authors considered the concave-convex-type elliptic problems(see[1,9,14,20,21,24–28,35,40]).Particularly,Korman in[22]considered the exact number of positive solutions for problem(1.3)in the unit ball B⊂RN(N≥3).By the bifurcation theory,it is obtained that there exists a critical number>0 such that problem(1.3)has two positive solutions for λ ∈(0,),exactly one positive solution for λ=,and no positive solution for λ>.As any positive solution to problem(1.3)in the unit ball B is radial,Tang in[35]proved this result of[22]by an ordinary differential equation method.In[1],Ambrosetti,Azorer,and Peral studied problem(1.3)in RN(N≥3),that is,

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Recently,Lin considered problem(1.1)under the following conditions in[28].

(h1)

(h2)There exist k points a1,a2,···,akin Ω such that

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and for some σ > N such that g(x)−g(ai)=O(|x− ai|σ)as x → aiuniformly in i.(h3)Choosing r0>0 such thatwhereThere exists a positive number d0such that f(x)≥d0>0 for any

For the convenience of the reader,we repeat the main results of[28]as follows.

Theorem 1.1 Assume thatand(h1)–(h3)hold,then problem(1.1)has k+1 positive solutions for λ>0 small enough.Moreover,one of the solutions is a ground state solution.

Moreover,Cao and Chabrowski in[7]considered problem(1.1)with q=1 and obtained the multiplicity of positive solutions.Very recently,we studied problem(1.1)with 2≤ q< 2and obtained some existence and multiplicity results by the variational method;see[27].

In this article,an interesting question is whether there exist k+1 positive solutions for problem(1.1)without constraining conditions of<q<2 and N >4.We are interested in the existence of k+1 positive solutions for problem(1.1)with 1<q<2 and N≥3.Wefind the reason of restricting <q<2 and N >4 in[28]is that the author obtained k positive solutions from the prolongation of the origin.Now,we try to get rid of the constraining conditions of<q<2 and N>4 via obtaining k positive solutions from the prolongation of the first solution.In fact,the first solution is a positive ground state solution of problem(1.1).

Lemma 2.2 Iλis coercive and bounded from below on N.

As well known,the function

is an extremal function for the minimization problem(1.4),that is,it is a positive solution of the following problem

Moreover,

For all u ∈ (Ω),associated with problem(1.1),we define the energy functional Iλ by

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From the result in[31],one knows that the functional Iλ is of class C1on(Ω).As well known that there exists a one to one correspondence between the nonnegative solutions of problem(1.1)and the critical points of Iλ on (Ω).

In this article,assume that the coefficient functions f,g satisfy the following conditions:

(f)f ∈ Lr(Ω)with f ≥ 0 and f/≡ 0,where

(g)g is continuous on Ω and g> 0.

(h2)There exist k points a1,a2,···,akin Ω satisfying

and moreover,as x→aiuniformly in i∈N+and 1≤i≤k.

Notice that u is a weak solution of problem(1.1),then u satisfies the following equation

So,if such a solution exists,then it must lie in Nehari manifold N,which is defined by

To obtain the multiplicity of positive solutions,we split N=N+∪N0∪N with N+,N0,Ndefined as follows:

This article is organized as follows.In Section 2,we give some preliminaries.We give the proof of existence of a positive ground state solution of problem(1.1)in Section 3.In Section 4,we give the proof of Theorem 1.2.

Here is our main result.

According to Theorem 2.3 in[6],we have the following lemma.

Remark 1.3 To our best knowledge,our results are up to date.Our results complement the case of 1<q≤ in[28],and we also obtain the same results but without restricting N >4.It is worth noticing that the growth condition inthat is,as x→aiuniformly in i∈N+and 1≤i≤k,which is more general than the corresponding condition of(h2)in[28].Moreover,we get rid of the condition(h3)which is important for estimating the level value of functional Iλin Lemma 4.2 of[28].

When λ ∈ (0,Λ),we will prove that N ± /= Ø and N0={0}in Section 2.

where 0< q< 1,h ∈ L1(RN)∩L(RN).When h ≥ 0,h/≡ 0,and λ > 0 small enough,they obtained two positive solutions.Moreover,[9,14]and[20]considered the multiplicity of positive solutions for concave-convex p-Laplacian problems with critical Sobolev exponent.

2 Some Preliminary Results

In this section,we give some lemmas about properties of the functional Iλon N,which are valuable preparation for the proof of our main result.

为方便计算,偏差和偏差变化率的模糊集合采用三角隶属函数,由于输出量的变化范围较小,为了取得更佳的精确度,选用正态函数作为输出的隶属度函数。输入输出变量的语言赋值表见表1、表2。

Lemma 2.1 For eachsuch thatwe have the followings:

(i)Assume thatthen there exists a unique t =t(u)> tmaxsuch that tu∈N and

(ii)Assume thatthen there exist uniquesuch that t+u ∈ N+and tu ∈ N for all λ ∈ (0,Λ),and

Furthermore,N0={0}and Nis a closed set for all 0<λ<Λ.

Proof According to(g),there existssuch thatFor eachsuch thatFor all t≥0,we define k:R+ →R by

Clearly,we obtain k(0)=0 and k(t)→−∞ as t→∞.Because

then k(tmax)=0,k(t)> 0 for 0 < t< tmax,and k(t)< 0 for t> tmax,where

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Thus,k(t)achieves its maximum at tmax.Moreover,by the Sobolev embedding theorem,one has

Case(i):There is a unique t > tmax,such that

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Then,we have

and

which implies that tu∈N .It follows that

Consequently,one has

Case(ii):By the Hölder and Sobolev inequalities,(1.4),and(2.3),for all 0<λ<Λ,we obtain

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there exist unique t+and tsuch that

for 0<t+<tmax<t.We have t+u∈N+,tu∈N and

for each t∈ [t+,t],and Iλ(t+u)≤ Iλ(tu)for each t∈ [0,t+].Thus,

Next,we prove that N0={0}for all 0<λ<Λ.By contradiction,suppose that there exists u0∈N0with u0/=0.Obviously,u0∈N,it follows that

and

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Consequently,

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Then,according to(2.3),(2.4),(2.6),and(2.7),for all 0<λ<Λ,one has

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which is a contradiction.Thus,N0={0}for all 0<λ<Λ.

Finally,we prove that Nis a closed set for all 0<λ<Λ.Assume that{un}⊂Nsuch that un→u in as n→+∞,then we need prove that u∈N .As un∈N ,from the definition of N,one has

Consequently,as un→u in as n→+∞,it follows from(2.8)that

Consequently,for all u∈N,it follows from(2.9)that

which contradicts u=0.Thus,u∈N for all λ ∈(0,Λ).Therefore,N is a closed set infor all λ ∈ (0,Λ).This completes the proof of Lemma 2.1.

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We denote by|·|pthe usual Lp-norm.Let S be the best Sobolev constant and Λ be a constant,respectively,

Proof By the Hölder and Sobolev inequalities,one has

thus u∈N ∪N0.If u∈N0,because N0={0}for all λ ∈(0,Λ),one has u=0.However,from(2.8),byand(1.4),one has

because 1< q< 2< 2,which implies that Iλ is coercive and bounded from below on N.Then,the proof of Lemma 2.2 is completed.

For 0<λ<Λ,by Lemma 2.1,one has N=N+∪N ∪{0}.Consequently,from Lemma 2.2,the following definitions are well defined

Lemma 2.3 (i) α ≤ α+ <0 for all λ ∈(0,Λ);

(ii)there exists a positive constant c0depending on λ,N,S,and|f|r,such that α ≥ c0 > 0 for all

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Proof (i)For all u∈N+,we have

consequently,because of 1<q<2 and u/=0,it follows that

which implies that α+ <0.Thus,α ≤ α+ <0.

(ii)For u ∈ N ,byand the Sobolev embedding theorem,one has

this implies that

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Consequently,from(2.10)and(2.11),we obtain

for allThis implies that there exists a constant c0=c0(λ,N,S,|f|r)> 0 such that α ≥ c0> 0 for allThus,the proof of Lemma 2.3 is completed.

Theorem 1.2 Assume that 1 < q < 2,N ≥ 3,and f,g satisfy(f),(g)andthen there exists Λ > 0 such that for λ ∈ (0,Λ),problem(1.1)has at least k+1 positive solutions.Moreover,there exists one solution uλ which is a ground state solution with Iλ(uλ) → 0 and‖uλ‖ → 0 as λ → 0+.

Lemma 2.4 Suppose that u0is a local minimizer of Iλon N and 0< λ < Λ,then

Proof Suppose that u0is a local minimizer of Iλon N,then u0is a solution of the optimization problem

where

Furthermore,by the theory of Lagrange multipliers,there exists θ∈R such thatAs u0∈N,we get

As 0< λ < Λ,from Lemma 2.3,one has u0/∈N0.Consequently,θ=0 and inThis completes the proof of Lemma 2.4.

3 Existence of a Positive Ground State Solution

According to[28]or[38],we show the existence of a(PS)α-sequence and a(PS)α−-sequence in

Lemma 3.1 (i)For all λ ∈ (0,Λ),there exists a(PS)α-sequence{un} ⊂ N in for Iλ.

(ii)For all then we have α > 0,and there exists a(PS)α−-sequence infor Iλ.

The proof of Lemma 3.1 is similar to the proof of Proposition 9 in[38];the reader can refer to[38].Now,we have the following proposition.

Proposition 3.2 Let λ ∈ (0,Λ),then there exists uλ ∈ N+such that

(i)Iλ(uλ)= α = α+;

(ii)uλ is a positive ground state solution of problem(1.1).Moreover,one has Iλ(uλ)→ 0 and‖uλ‖ → 0 as λ → 0+.

Proof By Lemma 3.1,there exists a minimizing sequence{un}⊂N such that

From Lemma 2.1,then Iλ is coercive on N.It follows that{un}is bounded onGoing if necessary to a subsequence,still denoted by{un},there exists such that

From(3.1),we haveConsequently,combining with(3.2),it follows that

which implies that uλ is a solution of problem(1.1).Particularly,choosing ϕ =uλ in(3.3),one has uλ∈N.Because{un}⊂N,one has

Letting n→∞in(3.4),with(3.1),(3.2),and the fact α<0,we obtain

Therefore,uλ∈N is a nontrivial solution of problem(1.1).

Next,we prove that un → uλ strongly in as n → ∞ and Iλ(uλ)= α.By Vitali’s theorem(see[32]pp:133),we claim that

Indeed,we only need to prove thatis equi-absolutely-continuous.Noting that{un}is bounded,by the Sobolev embedding theorem,so there exists a constant C1 > 0 such that|un|2∗ ≤ C1 < ∞.By the Hölder inequality,for every ε> 0,setting δ> 0,when E⊂Ω with mesE<δ,we have

where the last inequality is from the absolutely-continuity ofRThus,our claim is proved to be true.Because of uλ∈N,by the Fatou Lemma,it follows from(3.5)that

which implies that Iλ(uλ)= α andCombining with(3.2),un ⇀ uλ as n → ∞ in it shows that un → uλ strongly in Moreover,we can claim that uλ ∈ N+.Indeed,if uλ ∈ N ,by Lemma 2.1,there exist uniqueandsuch thatBecause of

there existssuch thatAccording Lemma 2.1,one obtains

which is contradiction.Thus,according to Lemma 2.3(i),Iλ(uλ)= α,and uλ ∈ N+,one has Iλ(uλ)= α = α+.

Finally,we prove that uλ is a positive solution of problem(1.1).In fact,we choosein(3.3),one hasthis implies that =0,so uλ ≥ 0.Consequently,as uλ0,by Lemma 2.4 and the strong maximum principle,then uλis a positive solution of problem(1.1).Therefore,uλis a positive ground state solution of problem(1.1).Moreover,from(2.10)and Lemma 2.4,we obtain

which implies that Iλ(uλ)= α → 0 as λ → 0+.As uλ ∈ N+,from(2.9),one has

which implies.Therefore, ‖uλ‖ → 0 as λ → 0+.Then,the proof of Proposition 3.2 is completed.

4 Proof of Theorem 1.2

In this part,we want to obtain k positive solutions of problem(1.1).To get rid of the constraining conditions of<q<2 and N >4 in[28],we try to obtain k positive solutions from the prolongation of the first solution uλin N .Before proving Theorem 1.2,we give some important lemmas.

Lemma 4.1 For allIλ satisfies the(PS)c-condition in

ProofLet{un}⊂be a(PS)c-sequence satisfying Iλ(un)=c+o(1)and=o(1).We claim that{un}is bounded in In fact,for n large enough,one has

which implies that{un}is bounded in So,our claim is true.Therefore,‖un‖ is bounded.Set vn=un − u.By Brézis-Lieb’s Lemma,we have

consequently,from(3.5),one has

and

Becausewith(4.1)–(4.3),we obtain

Now,we can assume that

as n→∞.Applying the Sobolev inequality,one obtains

Then,l≥Sl22∗,which implies that either l=0 or l≥ SN2.If l≥ SN2,by(4.4)and(4.5),we have

which is contradicts the definition of c.Therefore,l=0 and un→ ustrongly inThis completes the proof of Lemma 4.1.

We consider the following critical exponent problem

Associated with(4.6),we consider the energy functional I inthat is

Now,following the methods of[5],let ηibe a radially symmetric function such that 0 ≤ ηi≤ 1,|∇ηi|≤ C and for 1 ≤ i≤ k,we define

and

where U(x)is defined as(1.6).

Lemma 4.2 Suppose that(f),(g),andhold,then

where 1≤i≤k.

Proof According to[4]or[16],we can easy to obtain the following classical results

Moreover,one has

where Cl>0(l=2,3,4,5,6,7)are positive constants.As uλis a positive solution of problem(1.1),one has

Moreover,by a standard method,we get uλ ∈ C1(Ω,R+)and there exists a positive constant C8(C8independent of x)such that uλ < C8.Now,we give the following two elementary inequalities:

For any m>2,there exists a positive constant C9=C9(m)such that

where M>0 is a positive constant.

Then,combining with(4.10),(4.12)and(4.11),we easily obtain

where we can assume that≥1 for all t≥and|x − ai|≤ .We denote

Notice that Φε(0)=0, uniformly for all ε.On the one hand,whenone hasThus,we have

uniformly in i.On the other hand,whenthen and it attains for some tε >0.So,there exist two constants such that T0<tε<T0.In fact,from uniformly for all ε,we choosethen there exists T0>0 such that

According to the monotonicity of Φε near t=0,we have tε > T0.Similarly,we can obtainLet

Then,we have

Let=0,one has

such that for all 0 < t< tmaxand for all t> tmax,so Φε,1(t)attains its maximum at tmax.By(g)and,letting ε→ 0+,one has

In fact,for all ε> 0,it follows that

Because ofthat is,for all ε > 0,there exists δ> 0 such that

When ε> 0 small enough,for δ> ε12,it follows from(4.14),(g),andthat

where cN=[N(N−2)]N2,and C,C> 0 are constants.Consequently,one has

which implies that

Then,from the arbitrariness of η,we obtain(4.13).Combining with(4.7)and(4.13),one has

Consequently,from(4.8),it follows that

thus

From the boundness ofwith g > 0,and uλ ∈ C1(Ω),there exist three positive constants C10,C11,and C12such that

and from(4.9),one obtains

Consequently,it follows thatTherefore,

where C13>0 is a constant and 1≤i≤k.

While for all 0<t<,it follows from(g),(4.7),(4.8),and(4.11)that

This completes the proof of Lemma 4.2.

According to[36],we have the following lemma.

Lemma 4.3For any 1≤i≤k,there exists such thatN for all λ ∈(0,Λ).Moreover,

Proof According to Lemma 2.1,there exists u∈ H10(Ω){0},consequently,there exists unique t(u)> 0 such that t(u)u ∈ N .Set

and

Then,Thus,and N+⊂H1.In particular,uλ∈N+⊂H1.By simple computations,then there exists a suitable constant M > 0 such that 0 < t(u)< M for u ∈ with ‖u‖ =1.Letwe claim that

for ε > 0 small.Indeed,from(4.10),(4.8),and(4.9),one has

for ε>0 small.Define h:there existssuch thatMoreover,according to Lemma 4.2,one has

This completes the proof of Lemma 4.3.

From(g)andwe can choose r0 > 0 such that

andwhereLetSuppose thatfor some ρ0 > 0.According to[3],let φ :be a barycenter map defined by

where χ:RN →RN

From Lemma 4.3,there exists such thatfor any 1≤i≤k.Then,we have the following lemma.

Lemma 4.4For any 1≤i≤k,thenMoreover,there exists ε0>0 such that for all 0< ε< ε0,thenfor each 1≤i≤k.

Proof According to the definition of φ,we have

as ε→0+.This implies that there exists ε0>0 such thatfor any 0< ε< ε0and each 1≤ i≤k.Then,the proof of Lemma 4.4 is completed.

For each 1≤i≤k,we define

Next,we shall show thatare(PS)-values.So,we give the following lemmas.

Lemma 4.5 For all cIλ satisfies the(PS)c-condition in

Proof Assuming the contrary,we can suppose that there exists a sequence{un}⊂N0 such thatand

Then,I has a(PS)-sequence{un}in(see[24]).It follows from Lemma 2.2 that there exist a subsequence{un}and u0such that un ⇀ u0weakly in H10(Ω).BecauseΩ is a bounded domain,is not achieved.Applying the Palais-Smale Decomposition Lemma(see Theorem 3.1 in Struwe[34]),there exist two sequences{zn}⊂ Ω,{Rn}⊂R+,z0∈Ω,and a positive solution v0∈H1(RN)of the critical problem(1.6)with I(v0)=1NSN2such that zn→z0,Rn→∞as n→∞and

Let and as n→∞,one hasΩn→RNas n→∞.Then,

Now,we will show that Because of

which impliesandwhich is a contradiction.This completes the proof of Lemma 4.5.

Lemma 4.6 There exists such that if 0<λ<Λand u∈Nwithis given in Lemma 4.5),then we have φ(u) ∈

Proof Similar to Lemma 2.1,then there is a unique positive number

such that tuu∈N0.Now,we want to prove that tu<C14for some constant C14>0(independent of u and λ).From(2.11),if u ∈ N ,then

For λ ∈ (0,),because Iλ is coercive on N andby(2.10),we obtain‖u‖2< C16for some constant C16> 0.Next,we claim that|u|2∗> C17> 0,where C17=If u ∈ N ,that is

consequently,fromone obtains

Thus,tu<C14for some constant C14>0.Thus,we have

From the above inequality,we deduce that

Hence,there exists such that tuu∈N0and

for 0< λ < Λ.Consequently,by Lemma 4.5,it follows that

for any 0< λ < Λ.The proof of Lemma 4.6 is completed.

From the above lemma,one has

By Lemma 4.3,we have

Lemma 4.7For u∈,then there exist τ>0 and a differentiable functionalsuch that ζ(0)=1,ζ(v)(u−v)∈for any v∈B(0;τ),and

for any ϕ ∈ (Ω).

ProofThe proof is almost the same as in[10]or[38].For each u∈,define a function Fu:R×→R,given by

Then,Fu(1,0)=〉=0 and

According to the implicit function theorem,there exist τ> 0 and a differentiable functionsuch that ζ(0)=1,

and

which is equivalent to

That is,ζ(v)(u−v)∈This completes the proof of Lemma 4.7.

Lemma 4.8 For each 1 ≤ i≤ k,λ ∈ (0,Λ),there is ainfor Iλ.

Proof For each 1≤i≤k,by(4.15)and(4.16),we obtain

Then,Letbe a minimizing sequence forApplying Ekeland’s variational principle,there exists a subsequencesuch thatand

From(4.17),we may assume thatfor sufficiently large n.By Lemma 4.7,there exist a> 0 and a differentiable functionalsuch that =1,for anyLet vσ=σv with‖v‖=1 andThen,andFrom(4.18)and by the mean value theorem,as σ → 0,we obtain

where t0 ∈ (0,1)is a constant and

as σ →0.Hence,

where o(1)→ 0 as σ → 0.From Lemma 4.7,then there exists a positive constant M0such thatfor all n and i.Then,strongly inas n→∞.This completes the proof of Lemma 4.8.

Proof of Theorem 1.2 By Lemma 4.8,there exists a(PS)βλi -sequence{un}⊂infor Iλ and each 1 ≤ i≤ k.As Iλ satisfies the(PS)β-condition forfrom(4.16),then Iλhas at least k critical points in N for 0< λ < Λ.It follows that problem(1.1)has k nonnegative solutions inApplying the strong maximum principle,problem(1.1)has at least k positive solutions.Combining with Proposition 3.2,we complete the proof of Theorem 1.2.

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Jiafeng,LIAO(廖家锋)Yang,PU(蒲洋),ChunleiTANG(唐春雷)
《Acta Mathematica Scientia(English Series)》2018年第2期文献

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