Introduction

Consider the following Navier boundary value problem:

{ 2 u ( x ) + c u = f ( x , u ) , in  Ω ; u = u = 0 , in  Ω ,
(1)

where 2 is the biharmonic operator and Ω is a bounded smooth domain in R N (N4).

In problem (1), let f(x,u)=b[ ( u + 1 ) + 1], then we get the following Dirichlet problem:

{ 2 u ( x ) + c u = b [ ( u + 1 ) + 1 ] , in  Ω ; u = u = 0 , in  Ω ,
(2)

where u + =max{u,0} and bR. We let λ k (k=1,2,) denote the eigenvalues of −△ in H 0 1 (Ω).

Thus, fourth-order problems with N>4 have been studied by many authors. In [[1]], Lazer and McKenna pointed out that this type of nonlinearity furnishes a model to study traveling waves in suspension bridges. Since then, more general nonlinear fourth-order elliptic boundary value problems have been studied. For problem (2), Lazer and McKenna [[2]] proved the existence of 2k1 solutions when N=1, and b> λ k ( λ k c) by the global bifurcation method. In [[3]], Tarantello found a negative solution when b λ 1 ( λ 1 c) by a degree argument. For problem (1) when f(x,u)=bg(x,u), Micheletti and Pistoia [[4]] proved that there exist two or three solutions for a more general nonlinearity g by the variational method. Xu and Zhang [[5]] discussed the problem when f satisfies the local superlinearity and sublinearity. Zhang [[6]] proved the existence of solutions for a more general nonlinearity f(x,u) under some weaker assumptions. Zhang and Li [[7]] proved the existence of multiple nontrivial solutions by means of Morse theory and local linking. An and Liu [[8]] and Liu and Wang [[9]] also obtained the existence result for nontrivial solutions when f is asymptotically linear at positive infinity.

We noticed that almost all of works (see [[4]–[9]]) mentioned above involve the nonlinear term f(x,u) of a subcritical (polynomial) growth, say,

(SCP): there exist positive constants c 1 and c 2 and q 0 (1, p 1) such that

|f(x,t)| c 1 + c 2 | t | q 0 for all tR and xΩ,

where p =2N/(N4) denotes the critical Sobolev exponent. One of the main reasons to assume this condition (SCP) is that they can use the Sobolev compact embedding H 2 (Ω) H 0 1 (Ω) L q (Ω) (1q< p ). At that time, it is easy to see that seeking a weak solution of problem (1) is equivalent to finding a nonzero critical points of the following functional on H 2 (Ω) H 0 1 (Ω):

I(u)= 1 2 Ω ( | u | 2 c | u | 2 ) dx Ω F(x,u)dx,where F(x,u)= 0 u f(x,t)dt.
(3)

In this paper, stimulated by Lam and Lu [[10]], our first main results will be to study problem (1) in the improved subcritical polynomial growth

(SCPI): lim t f ( x , t ) | t | p 1 =0

which is much weaker than (SCP). Note that in this case, we do not have the Sobolev compact embedding anymore. Our work is to study problem (1) when nonlinearity f does not satisfy the (AR) condition, i.e., for some θ>2 and γ>0,

0<θF(x,t)f(x,t)tfor all |t|γ and xΩ.
(AR)

In fact, this condition was studied by Liu and Wang in [[11]] in the case of Laplacian by the Nehari manifold approach. However, we will use a suitable version of the mountain pass theorem to get the nontrivial solution to problem (1) in the general case N>4. We will also use the symmetric mountain pass theorem to get infinitely many solutions for problem (1) in the general case N>4 when nonlinearity f is odd.

Let us now state our results. In this paper, we always assume that f(x,t)C( Ω ¯ ×R). The conditions imposed on f(x,t) are as follows:

(H1): f(x,t)t0 for all xΩ, tR;

(H2): lim | t | 0 f ( x , t ) t = f 0 uniformly for xΩ, where f 0 is a constant;

(H3): lim | t | f ( x , t ) t =+ uniformly for xΩ;

(H4): f ( x , t ) | t | is nondecreasing in tR for any xΩ.

Let 0< μ 1 < μ 2 << μ k < be the eigenvalues of ( 2 c, H 2 (Ω) H 0 1 (Ω)) and φ 1 (x)>0 be the eigenfunction corresponding to μ 1 . Let E μ k denote the eigenspace associated to μ k . In fact, μ k = λ k ( λ k c). Throughout this paper, we denote by | | p the L p (Ω) norm, c< λ 1 in 2 c and the norm of u in H 2 (Ω) H 0 1 (Ω) will be defined by

u:= ( Ω ( | u | 2 c | u | 2 ) d x ) 1 2 .

We also define E= H 2 (Ω) H 0 1 (Ω).

Theorem 1.1

LetN>4and assume that f has the improved subcritical polynomial growth on Ω (condition (SCPI)) and satisfies (H1)-(H4). If f 0 < μ 1 , then problem (1) has at least two nontrivial solutions.

Theorem 1.2

LetN>4and assume that f has the improved subcritical polynomial growth on Ω (condition (SCPI)), is odd in t and satisfies (H3) and (H4). Iff(x,0)=0, then problem (1) has infinitely many nontrivial solutions.

In the case of N=4, we have p =+. So it is necessary to introduce the definition of the subcritical (exponential) growth in this case. By the improved Adams inequality (see [[12]]) for the fourth-order derivative, namely,

sup u E , u 1 Ω e 32 π 2 u 2 dxC|Ω|.

So, we now define the subcritical (exponential) growth in this case as follows:

(SCE): f has subcritical (exponential) growth on Ω, i.e., lim t | f ( x , t ) | exp ( α t 2 ) =0 uniformly on xΩ for all α>0.

When N=4 and f has the subcritical (exponential) growth (SCE), our work is still to study problem (1) without the (AR) condition. Our results are as follows.

Theorem 1.3

LetN=4and assume that f has the subcritical exponential growth on Ω (condition (SCE)) and satisfies (H1)-(H4). If f 0 < μ 1 , then problem (1) has at least two nontrivial solutions.

Theorem 1.4

LetN=4and assume that f has the subcritical exponential growth on Ω (condition (SCE)), is odd in t and satisfies (H3) and (H4). Iff(x,0)=0, then problem (1) has infinitely many nontrivial solutions.

Preliminaries and auxiliary lemmas

Definition 2.1

Let (E, E ) be a real Banach space with its dual space ( E , E ) and I C 1 (E,R). For c R, we say that I satisfies the ( PS ) c condition if for any sequence { x n }E with

I( x n ) c ,DI( x n )0in  E ,

there is a subsequence { x n k } such that { x n k } converges strongly in E. Also, we say that I satisfies the ( C ) c condition if for any sequence { x n }E with

I( x n ) c , D I ( x n ) E ( 1 + x n E ) 0,

there is a subsequence { x n k } such that { x n k } converges strongly in E.

We have the following version of the mountain pass theorem (see [[13]]).

Proposition 2.1

Let E be a real Banach space and suppose thatI C 1 (E,R)satisfies the condition

max { I ( 0 ) , I ( u 1 ) } α<β inf u = ρ I(u)

for someα<β, ρ>0and u 1 Ewith u 1 >ρ. Let c βbe characterized by

c = inf γ Γ max 0 t 1 I ( γ ( t ) ) ,

whereΓ={γC([0,1],E),γ(0)=0,γ(1)= u 1 }is the set of continuous paths joining 0 and  u 1 . Then there exists a sequence{ u n }Esuch that

I( u n ) c βand ( 1 + u n ) I ( u n ) E 0as n.

Consider the following problem:

{ 2 u + c u = f + ( x , u ) , x Ω , u | Ω = u | Ω = 0 ,

where

f + (x,t)= { f ( x , t ) , t > 0 , 0 , t 0 .

Define a functional I + :ER by

I + (u)= 1 2 Ω ( | u | 2 c | u | 2 ) dx Ω F + (x,u)dx,

where F + (x,t)= 0 t f + (x,s)ds, then I + C 1 (E,R).

Lemma 2.1

LetN>4and φ 1 >0be a μ 1 -eigenfunction with φ 1 =1and assume that (H2), (H3) and (SCPI) hold. If f 0 < μ 1 , then:

  1. (i)

    There exist ρ,α>0 such that I + (u)α for all uE with u=ρ.

  2. (ii)

    I + (t φ 1 ) as t+.

Proof

By (SCPI), (H2) and (H3), for any ε>0, there exist A 1 = A 1 (ε), B 1 = B 1 (ε) and l>2 μ 1 such that for all (x,s)Ω×R,

F + (x,s) 1 2 ( f 0 +ε) s 2 + A 1 s p ,
(4)
F + (x,s) 1 2 l s 2 B 1 .
(5)

Choose ε>0 such that ( f 0 +ε)< μ 1 . By (4), the Poincaré inequality and the Sobolev inequality | u | p p K u p , we get

I + (u) 1 2 u 2 f 0 + ε 2 | u | 2 2 A 1 | u | p p 1 2 ( 1 f 0 + ε μ 1 ) u 2 A 1 K u p .

So, part (i) is proved if we choose u=ρ>0 small enough.

On the other hand, from (5) we have

I + (t φ 1 ) 1 2 ( 1 l μ 1 ) t 2 + B 1 |Ω|as t.

Thus part (ii) is proved. □

Lemma 2.2

(see [[12]])

LetΩ R 4 be a bounded domain. Then there exists a constantC>0such that

sup u E , u 1 Ω e 32 π 2 u 2 dxC|Ω|,

and this inequality is sharp.

Lemma 2.3

LetN=4and φ 1 >0be a μ 1 -eigenfunction with φ 1 =1and assume that (H2), (H3) and (SCE) hold. If f 0 < μ 1 , then:

  1. (i)

    There exist ρ,α>0 such that I + (u)α for all uE with u=ρ.

  2. (ii)

    I + (t φ 1 ) as t+.

Proof

By (SCE), (H2) and (H3), for any ε>0, there exist A 1 = A 1 (ε), B 1 = B 1 (ε), κ>0, q>2 and l>2 μ 1 such that for all (x,s)Ω×R,

F + (x,s) 1 2 ( f 0 +ε) s 2 + A 1 exp ( κ | s | 2 ) s q ,
(6)
F + (x,s) 1 2 l s 2 B 1 .
(7)

Choose ε>0 such that ( f 0 +ε)< μ 1 . By (6), the Holder inequality and Lemma 2.2, we get

I + ( u ) 1 2 u 2 f 0 + ε 2 | u | 2 2 A 1 Ω exp ( κ | u | 2 ) | u | q d x 1 2 ( 1 f 0 + ε μ 1 ) u 2 A 1 ( Ω exp ( κ r u 2 ( | u | u ) 2 ) d x ) 1 r ( Ω | u | r q d x ) 1 r 1 2 ( 1 f 0 + ε μ 1 ) u 2 C u q ,

where r>1 is sufficiently close to 1, uσ and κr σ 2 <32 π 2 . So, part (i) is proved if we choose u=ρ>0 small enough.

On the other hand, from (7) we have

I + (t φ 1 ) 1 2 ( 1 l μ 1 ) | t | 2 + B 1 |Ω|as t.

Thus part (ii) is proved. □

Lemma 2.4

For the functional I defined by (3), if condition (H4) holds, and for any{ u n }Ewith

I ( u n ) , u n 0as n,

then there is a subsequence, still denoted by{ u n }, such that

I(t u n ) 1 + t 2 2 n +I( u n )for all tR and nN.

Proof

This lemma is essentially due to [[14]]. We omit it here. □

Proofs of the main results

Proof of Theorem 1.1

By Lemma 2.1 and Proposition 2.1, there exists a sequence { u n }E such that

I + ( u n )= 1 2 u n 2 Ω F + (x, u n )dx= c +o(1),
(8)
( 1 + u n ) I + ( u n ) E 0as n.
(9)

Clearly, (9) implies that

I + ( u n ) , u n = u n 2 Ω f + ( x , u n ( x ) ) u n dx=o(1).
(10)

To complete our proof, we first need to verify that { u n } is bounded in E. Assume u n + as n. Let

s n = 2 c u n , w n = s n u n = 2 c u n u n .
(11)

Since { w n } is bounded in E, it is possible to extract a subsequence (denoted also by { w n }) such that

w n w 0 in  E , w n + w 0 + in  L 2 ( Ω ) , w n + ( x ) w 0 + ( x ) a.e.  x Ω , | w n + ( x ) | h ( x ) a.e.  x Ω ,

where w n + =max{ w n ,0}, w 0 E and h L 2 (Ω).

We claim that if u n + as n+, then w + (x)0. In fact, we set Ω 1 ={xΩ: w + =0}, Ω 2 ={xΩ: w + >0}. Obviously, by (11), u n + + a.e. in Ω 2 , noticing condition (H3), then for any given K>0, we have

lim n + f ( x , u n + ) u n + ( w n + ( x ) ) 2 K w + ( x ) 2 for a.e. x Ω 2 .
(12)

From (10), (11) and (12), we obtain

4 c = lim n + w n 2 = lim n + Ω f ( x , u n + ) u n + ( w n + ) 2 d x Ω 2 lim n + f ( x , u n + ) u n + ( w n + ) 2 d x K Ω 2 ( w + ) 2 d x .

Noticing that w + >0 in Ω 2 and K>0 can be chosen large enough, so | Ω 2 |=0 and w + 0 in Ω. However, if w + 0, then lim n + Ω F(x, w n + )dx=0 and consequently

I + ( w n )= 1 2 w n 2 +o(1)=2 c +o(1).
(13)

By u n + as n+ and in view of (11), we observe that s n 0, then it follows from Lemma 2.4 and (8) that

I + ( w n )= I + ( s n u n ) 1 + s n 2 2 n + I + ( u n ) c >0as n+.
(14)

Clearly, (13) and (14) are contradictory. So { u n } is bounded in E.

Next, we prove that { u n } has a convergence subsequence. In fact, we can suppose that

u n u in  E , u n u in  L q ( Ω ) , 1 q < p , u n ( x ) u ( x ) a.e.  x Ω .

Now, since f has the improved subcritical growth on Ω, for every ε>0, we can find a constant C(ε)>0 such that

f + (x,s)C(ε)+ε | s | p 1 ,(x,s)Ω×R,

then

| Ω f + ( x , u n ) ( u n u ) d x | C ( ε ) Ω | u n u | d x + ε Ω | u n u | | u n | p 1 d x C ( ε ) Ω | u n u | d x + ε ( Ω ( | u n | p 1 ) p p 1 d x ) p 1 p ( Ω | u n u | p ) 1 p C ( ε ) Ω | u n u | d x + ε C ( Ω ) .

Similarly, since u n u in E, Ω | u n u|dx0. Since ε>0 is arbitrary, we can conclude that

Ω ( f + ( x , u n ) f + ( x , u ) ) ( u n u)dx0as n.
(15)

By (10), we have

I + ( u n ) I + ( u ) , ( u n u ) 0as n.
(16)

From (15) and (16), we obtain

Ω [ | ( u n u ) | 2 c | ( u n u ) | 2 ] dx0as n.

So we have u n u in E which means that I + satisfies ( C ) c . Thus, from the strong maximum principle, we obtain that the functional I + has a positive critical point u 1 , i.e., u 1 is a positive solution of problem (1). Similarly, we also obtain a negative solution u 2 for problem (1). □

Proof of Theorem 1.2

It follows from the assumptions that I is even. Obviously, I C 1 (E,R) and I(0)=0. By the proof of Theorem 1.1, we easily prove that I(u) satisfies condition ( C ) c ( c >0). Now, we can prove the theorem by using the symmetric mountain pass theorem in [[15]–[17]].

Step 1. We claim that condition (i) holds in Theorem 9.12 (see [[16]]). Let V 1 = E μ 1 E μ 2 E μ k , V 2 =E V 1 . For all u V 2 , by (SCPI), we have

I ( u ) = 1 2 Ω ( | u | 2 c | u | 2 ) d x Ω F ( x , u ) d x 1 2 Ω ( | u | 2 c | u | 2 ) d x c 3 Ω | u | p d x c 4 u 2 ( 1 2 c 5 λ k + 1 ( 1 a ) p / 2 u p 2 ) c 6 ,

where a(0,1) is defined by

1 p =a ( 1 2 1 N ) +(1a) 1 2 .

Choose ρ=ρ(k)=u so that the coefficient of ρ 2 in the above formula is 1 4 . Therefore

I(u) 1 4 ρ 2 c 6
(17)

for u B ρ V 2 . Since λ k as k, ρ(k) as k. Choose k so that 1 4 ρ 2 >2 c 6 . Consequently

I(u) 1 8 ρ 2 α.
(18)

Hence, our claim holds.

Step 2. We claim that condition (ii) holds in Theorem 9.12 (see [[16]]). By (H3), there exists large enough M such that

F(x,t)M t 2 c 7 ,xΩ,tR.

So, for any uE{0}, we have

I ( t u ) = 1 2 t 2 Ω ( | u | 2 c | u | 2 ) d x Ω F ( x , t u ) d x 1 2 t 2 u 2 M t 2 Ω u 2 d x + c 7 | Ω | as  t + .

Hence, for every finite dimension subspace E ˜ E, there exists R=R( E ˜ ) such that

I(u)0,u E ˜ B R ( E ˜ )

and our claim holds. □

Proof of Theorem 1.3

By Lemma 2.3, the geometry conditions of the mountain pass theorem (see Proposition 2.1) for the functional I + hold. So, we only need to verify condition ( C ) c . Similar to the previous part of the proof of Theorem 1.1, we easily know that ( C ) c sequence { u n } is bounded in E. Next, we prove that { u n } has a convergence subsequence. Without loss of generality, suppose that

u n β , u n u in  E , u n u in  L q ( Ω ) , q 1 , u n ( x ) u ( x ) a.e.  x Ω .

Now, since f + has the subcritical exponential growth (SCE) on Ω, we can find a constant C β >0 such that

| f + (x,t)| C β exp ( 32 π 2 2 β 2 | t | 2 ) ,(x,t)Ω×R.

Thus, by the Adams-type inequality (see Lemma 2.2),

| Ω f + ( x , u n ) ( u n u ) d x | C ( Ω exp ( 32 π 2 β 2 | u n | 2 ) d x ) 1 2 | u n u | 2 C ( Ω exp ( 32 π 2 β 2 u n 2 | u n u n | 2 ) d x ) 1 2 | u n u | 2 C | u n u | 2 0 .

Similar to the last proof of Theorem 1.1, we have u n u in E, which means that I + satisfies ( C ) c . Thus, from the strong maximum principle, we obtain that the functional I + has a positive critical point u 1 , i.e., u 1 is a positive solution of problem (1). Similarly, we also obtain a negative solution u 2 for problem (1). □

Proof of Theorem 1.4

Combining the proof of Theorem 1.2 and Theorem 1.3, we easily prove it. □