WEI YUAN-HONG,CHANG XIAO-JUNAND LU¨ YUE

(1.Institute of Mathematics,Academy of Mathematics and Systems Science, Chinese Academy of Sciences,Beijing,100080)

(2.College of Mathematics,Jilin University,Changchun,130012)

(3.Chern Institute of Mathematics,Nankai University,Tianjin,300071)

Superlinear Fourth-order Elliptic Problem without Ambrosetti and Rabinowitz Growth Condition*

WEI YUAN-HONG1,CHANG XIAO-JUN2,3AND LU¨ YUE2

(1.Institute of Mathematics,Academy of Mathematics and Systems Science, Chinese Academy of Sciences,Beijing,100080)

(2.College of Mathematics,Jilin University,Changchun,130012)

(3.Chern Institute of Mathematics,Nankai University,Tianjin,300071)

Communicated by Li Yong

This paper deals with superlinear fourth-order elliptic problem under Navier boundary condition.By using the mountain pass theorem and suitable truncation,a multiplicity result is established for allλ＞ 0and some previous result is extended.

fourth-order elliptic problem,variational method,mountain pass theorem,Navier boundary condition

1 Introduction and Main Results

Fourth-order elliptic problems are usually used to describe some phenomena appeared in dif f erent physical,engineering and other sciences.Lazer and McKenna[1]studied the problem of nonlinear oscillation in a suspension bridge and they presented a mathematical model for the bridge that took account of the fact that the coupling provided by the stays connecting the suspension cable to the deck of the road bed is basically nonlinear.Also,Liu and Feng[2]pointed out that this kind of problem furnishes a good model to the static def l ection of an elastic plate in a f l uid.Ahmed and Harbi[3]indicated that this problem also arises in such as communication satellites,space shuttles,and space stations,which are equipped with large antennas mounted on long f l exible masts(beams).Fourth-order elliptic problems have been studied extensively in recent years,and we refer the reader to[4–9]and the referencestherein.

Consider the following fourth-order elliptic problem:

where Δ2is the biharmonic operator,c is a constant,Ω?RNis a bounded smooth domain and f(x,s)is a continuous function on×R.

Denote∫ We assume that f(x,s)satis fi es the following hypotheses:

(H2) Th ere exist positive constants Cand Csuch that12(H3)uniformly for a.e.x∈Ω;

(H4)There exists a C*＞0 such that

for all 0＜t＜s or s＜t＜0,x∈Ω.

To obtain nontrivial solutions of the problem(1.1)by applying variational method,one often uses the Ambrosetti-Rabinowitz condition(see[10]),i.e.,

(AR)There are constants θ＞0 and s0＞0 such that

This condition ensures the compactness of the corresponding functional,however,it eliminates many nonlinearities.To avoid the condition(AR),many approaches were developed. Costa and Magalh?aes[11]studied the problem(1.1)via replacing the condition(AR)byuniformly for a.e.x∈Ω, whereμ≥μ0＞0.Willem and Zou[12]assumed that H(x,s)is increasing in s and

whereμ＞2 and C0＞0,in place of the condition(AR).Recently,by using the assumptions (H1)–(H4),Miyagaki and Souto[13]obtained a nontrivial weak solution in the case of secondorder elliptic problem.

For the fourth-order problem(1.1),Zhang and Li[14]obtained at least two nontrivial solutions by means of Morse theory and local linking when f is sublinear at in fi nity.By using the linking theorem,Qian and Li[15]obtained one nontrivial solution if f is superlinear and satis fi es the Ambrosetti-Rabinowitz condition,and two nontrivial solutions if f is asymptotically linear as s is large enough.An and Liu[2]also established the existence of at leastone nontrivial solution if f is asymptotically linear at in fi nity.In this paper,we consider the fourth-order problem(1.1)when f is superlinear but the Ambrosetti-Rabinowitz condition is not required.Applying the mountain pass theorem,we obtain at least two nontrivial solutions for all λ＞0.

Our main result is as follows.

Theorem 1.1 Assume that(H1)–(H4)hold and c＜λ1,where λ1denotes the fi rst eigenvalue of-Δ in(Ω).Then,for all λ＞0,the problem(1.1)has at least two nontrivial solutions,one of which is positive and the other is negative.

Remark 1.1 Note that the condition(H3)is weaker than(AR)(see[13]).Let

It is easy to see that F satis fi es assumptions(H1)–(H4)but not(AR)condition.

Remark 1.2 Note that(H4)is weaker than the following condition:

(i) There exists an s0＞0 such thatis increasing in s＞sand decreasing in

0s＜-s0for all x∈Ω.

In previous works,many authors(see[16–17])used the condition(i)to assure that the corresponding energy functional satis fi es the Cerami condition.In this paper,our arguments show that the condition(i)implies that the energy functional satis fi es Palais-Smale condition.

2 Preliminary Results

Let H=H2(Ω)∩(Ω)be a Hilbert space equipped with the inner product

and the deduced norm

Let λk(k∈N)be the eigenvalues and φk(k∈N)be the corresponding eigenfunctions of the eigenvalue problem

where each eigenvalue λkis repeated according to the multiplicity.Recall that 0＜λ1＜λ2≤λ3≤···≤λk→+∞and φ1＞0 for x∈Ω.It is easily seen that

are eigenvalues of the problem

and the corresponding eigenfunctions are still φk.

Assume that c＜λ1.We denote by‖·‖the norm in H which is given by

It is easy to show that the norm‖·‖is an equivalent norm on H and the following Poincarˊe inequality holds:

We say that u∈H is a weak solution to problem(1.1),if u satis fi es

where H*is the dual space of H.

It is well known that the weak solution of problem(1.1)is equivalent to the critical point of the Euler-Lagrange functional

Obviously,Iλ∈C1(H,R)and

Let

Consider the problem

where

De fi ne the corresponding functional:H→R as follows:

where

Similarly,we can de fi ne

and

where

Now we prove that the functionalsandhave the mountain pass geometry.

Lemma 2.1 Under the assumption(H3),andare unbounded from below. Proof. (H3)implies that for all M ＞0 there exists CM＞0 such that

Taking φ∈H with φ＞0,from(2.6)we obtain

where|Ω|denotes the Lebesgue measure of Ω.Let

Then

The proof is completed.

Lemma 2.2 Assume that(H1)and(H2)hold.Then there exist ρ,R＞0 such that

if

Proof. We just considerthe case of.The case ofcan be dealt with similarly.

?

Combining(2.10)and the Poincarˊe inequality as well as the Sobolve embedding,we have

where Csis a positive constant.In(2.11),taking?＞0 such that

and choosing

small enough,we can fi nd an R＞0 such that

if

This completes the proof.

Now,we prove that every Palais-Smale sequence ofis relatively compact.

We recall that a sequence{un}?H is said to be a Palais-Smale sequence of the functional Φ provided that Φ(un)is bounded and Φ′(un)→0 in H*.

Lemma 2.3 Suppose that(H2)–(H4)hold.Then for all λ＞0,every Palais-Smale sequence ofhas a convergent subsequence.

Proof. We just prove the case of.The arguments for the case ofare similar.

Since Ω is bounded and(H2)holds,if{un}is bounded in H,by using the Sobolve embedding and the standard procedures,we can get a subsequence converges strongly.So we need only to show that{un}is bounded in H.

Assume that{un}?H is a Palais-Smale sequence of,i.e.,

We suppose,by contradiction,that passing to a subsequence,if necessary,

Set

Then

Passing to a subsequence,if necessary,we may assume that there exists an ω∈H such that

We claim that

In fact,we denote

If

then for x∈Ω*,

By(H3)we have

The Fatou Lemma and(2.12)imply

Hence Ω*has zero measure.Consequently,

As in[18],we take tn∈[0,1]such that

which implies that

Since

together with(2.16)it follows that

Hence,by(H4)we obtain

On the other hand,for all R0＞0,

which contradicts(2.17)for R0and n large.This completes the proof.

3 Proof of the Main Result

Proof of Theorem 1.1 By(H1),it is easily seen that

From Lemma 2.1 we know that there exists an

such that

In addition,Lemma 2.2 implies that there exist ρ,R＞0 such that

De fi ne

and

By Lemma 2.3 we can see thatsatis fi es the Palais-Smale condition.By the mountain pass theorem,we know thatis a critical value ofand there is at least one nontrivial critical point uλ,+∈H such that

Clearly,

Then the strong maximum principle implies

Thus uλ,+is a positive solution of the problem(1.1).By an analogous argument we know that there exists at least one negative solution uλ,-∈H of the problem(1.1),which is a nontrivial critical point of.Hence,the problem(1.1)admits at least one positive solution and one negative solution.

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tion:35J30,35J35

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1674-5647(2013)01-0023-09

*Received date:March 15,2010.

The 985 Program of Jilin University and the Science Research Foundation for Excellent Young Teachers of College of Mathematics at Jilin University.