Open Access

Biharmonic equations with improved subcritical polynomial growth and subcritical exponential growth

Boundary Value Problems20142014:162

DOI: 10.1186/s13661-014-0162-y

Received: 11 February 2014

Accepted: 17 June 2014

Published: 12 July 2014

Abstract

The main purpose of this paper is to establish the existence of two nontrivial solutions and the existence of infinitely many solutions for a class of fourth-order elliptic equations with subcritical polynomial growth and subcritical exponential growth by using a suitable version of the mountain pass theorem and the symmetric mountain pass theorem.

Keywords

mountain pass theorem Adams-type inequality subcritical polynomial growth subcritical exponential growth

Introduction

Consider the following Navier boundary value problem:
{ 2 u ( x ) + c u = f ( x , u ) , in  Ω ; u = u = 0 , in  Ω , https://static-content.springer.com/image/art%3A10.1186%2Fs13661-014-0162-y/MediaObjects/13661_2014_Article_162_Equ1_HTML.gif
(1)
where 2 is the biharmonic operator and Ω is a bounded smooth domain in R N ( N 4 ).
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 , https://static-content.springer.com/image/art%3A10.1186%2Fs13661-014-0162-y/MediaObjects/13661_2014_Article_162_Equ2_HTML.gif
(2)
where u + = max { u , 0 } and b R . 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 2 k 1 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 ) = b g ( 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  t R  and  x Ω , https://static-content.springer.com/image/art%3A10.1186%2Fs13661-014-0162-y/MediaObjects/13661_2014_Article_162_Equ3_HTML.gif
where p = 2 N / ( N 4 ) 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 ) ( 1 q < 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 ) d x Ω F ( x , u ) d x , where  F ( x , u ) = 0 u f ( x , t ) d t . https://static-content.springer.com/image/art%3A10.1186%2Fs13661-014-0162-y/MediaObjects/13661_2014_Article_162_Equ4_HTML.gif
(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 https://static-content.springer.com/image/art%3A10.1186%2Fs13661-014-0162-y/MediaObjects/13661_2014_Article_162_Equ5_HTML.gif
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 < θ F ( x , t ) f ( x , t ) t for all  | t | γ  and  x Ω . https://static-content.springer.com/image/art%3A10.1186%2Fs13661-014-0162-y/MediaObjects/13661_2014_Article_162_Equ6_HTML.gif
(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 ) t 0 for all x Ω , t R ;

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

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

(H4): f ( x , t ) | t | is nondecreasing in t R for any x Ω .

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 . https://static-content.springer.com/image/art%3A10.1186%2Fs13661-014-0162-y/MediaObjects/13661_2014_Article_162_Equ7_HTML.gif
We also define E = H 2 ( Ω ) H 0 1 ( Ω ) .

Theorem 1.1

Let N > 4 and 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

Let N > 4 and assume that f has the improved subcritical polynomial growth on Ω (condition (SCPI)), is odd in t and satisfies (H3) and (H4). If f ( 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 d x C | Ω | . https://static-content.springer.com/image/art%3A10.1186%2Fs13661-014-0162-y/MediaObjects/13661_2014_Article_162_Equ8_HTML.gif
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 x Ω 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

Let N = 4 and 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

Let N = 4 and assume that f has the subcritical exponential growth on Ω (condition (SCE)), is odd in t and satisfies (H3) and (H4). If f ( 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 , D I ( x n ) 0 in  E , https://static-content.springer.com/image/art%3A10.1186%2Fs13661-014-0162-y/MediaObjects/13661_2014_Article_162_Equ9_HTML.gif
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 , https://static-content.springer.com/image/art%3A10.1186%2Fs13661-014-0162-y/MediaObjects/13661_2014_Article_162_Equ10_HTML.gif
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 that I C 1 ( E , R ) satisfies the condition
max { I ( 0 ) , I ( u 1 ) } α < β inf u = ρ I ( u ) https://static-content.springer.com/image/art%3A10.1186%2Fs13661-014-0162-y/MediaObjects/13661_2014_Article_162_Equ11_HTML.gif
for some α < β , ρ > 0 and u 1 E with u 1 > ρ . Let c β be characterized by
c = inf γ Γ max 0 t 1 I ( γ ( t ) ) , https://static-content.springer.com/image/art%3A10.1186%2Fs13661-014-0162-y/MediaObjects/13661_2014_Article_162_Equ12_HTML.gif
where Γ = { γ C ( [ 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 } E such that
I ( u n ) c β and ( 1 + u n ) I ( u n ) E 0 as  n . https://static-content.springer.com/image/art%3A10.1186%2Fs13661-014-0162-y/MediaObjects/13661_2014_Article_162_Equ13_HTML.gif
Consider the following problem:
{ 2 u + c u = f + ( x , u ) , x Ω , u | Ω = u | Ω = 0 , https://static-content.springer.com/image/art%3A10.1186%2Fs13661-014-0162-y/MediaObjects/13661_2014_Article_162_Equ14_HTML.gif
where
f + ( x , t ) = { f ( x , t ) , t > 0 , 0 , t 0 . https://static-content.springer.com/image/art%3A10.1186%2Fs13661-014-0162-y/MediaObjects/13661_2014_Article_162_Equ15_HTML.gif
Define a functional I + : E R by
I + ( u ) = 1 2 Ω ( | u | 2 c | u | 2 ) d x Ω F + ( x , u ) d x , https://static-content.springer.com/image/art%3A10.1186%2Fs13661-014-0162-y/MediaObjects/13661_2014_Article_162_Equ16_HTML.gif
where F + ( x , t ) = 0 t f + ( x , s ) d s , then I + C 1 ( E , R ) .

Lemma 2.1

Let N > 4 and φ 1 > 0 be a μ 1 -eigenfunction with φ 1 = 1 and 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 u E 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 , https://static-content.springer.com/image/art%3A10.1186%2Fs13661-014-0162-y/MediaObjects/13661_2014_Article_162_Equ17_HTML.gif
(4)
F + ( x , s ) 1 2 l s 2 B 1 . https://static-content.springer.com/image/art%3A10.1186%2Fs13661-014-0162-y/MediaObjects/13661_2014_Article_162_Equ18_HTML.gif
(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 . https://static-content.springer.com/image/art%3A10.1186%2Fs13661-014-0162-y/MediaObjects/13661_2014_Article_162_Equ19_HTML.gif
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 . https://static-content.springer.com/image/art%3A10.1186%2Fs13661-014-0162-y/MediaObjects/13661_2014_Article_162_Equ20_HTML.gif
Thus part (ii) is proved. □

Lemma 2.2

(see [[12]])

Let Ω R 4 be a bounded domain. Then there exists a constant C > 0 such that
sup u E , u 1 Ω e 32 π 2 u 2 d x C | Ω | , https://static-content.springer.com/image/art%3A10.1186%2Fs13661-014-0162-y/MediaObjects/13661_2014_Article_162_Equ21_HTML.gif
and this inequality is sharp.

Lemma 2.3

Let N = 4 and φ 1 > 0 be a μ 1 -eigenfunction with φ 1 = 1 and 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 u E 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 , https://static-content.springer.com/image/art%3A10.1186%2Fs13661-014-0162-y/MediaObjects/13661_2014_Article_162_Equ22_HTML.gif
(6)
F + ( x , s ) 1 2 l s 2 B 1 . https://static-content.springer.com/image/art%3A10.1186%2Fs13661-014-0162-y/MediaObjects/13661_2014_Article_162_Equ23_HTML.gif
(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 , https://static-content.springer.com/image/art%3A10.1186%2Fs13661-014-0162-y/MediaObjects/13661_2014_Article_162_Equ24_HTML.gif
where r > 1 is sufficiently close to 1, u σ and κ r 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 . https://static-content.springer.com/image/art%3A10.1186%2Fs13661-014-0162-y/MediaObjects/13661_2014_Article_162_Equ25_HTML.gif
Thus part (ii) is proved. □

Lemma 2.4

For the functional I defined by (3), if condition (H4) holds, and for any { u n } E with
I ( u n ) , u n 0 as  n , https://static-content.springer.com/image/art%3A10.1186%2Fs13661-014-0162-y/MediaObjects/13661_2014_Article_162_Equ26_HTML.gif
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  t R  and  n N . https://static-content.springer.com/image/art%3A10.1186%2Fs13661-014-0162-y/MediaObjects/13661_2014_Article_162_Equ27_HTML.gif

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 ) d x = c + o ( 1 ) , https://static-content.springer.com/image/art%3A10.1186%2Fs13661-014-0162-y/MediaObjects/13661_2014_Article_162_Equ28_HTML.gif
(8)
( 1 + u n ) I + ( u n ) E 0 as  n . https://static-content.springer.com/image/art%3A10.1186%2Fs13661-014-0162-y/MediaObjects/13661_2014_Article_162_Equ29_HTML.gif
(9)
Clearly, (9) implies that
I + ( u n ) , u n = u n 2 Ω f + ( x , u n ( x ) ) u n d x = o ( 1 ) . https://static-content.springer.com/image/art%3A10.1186%2Fs13661-014-0162-y/MediaObjects/13661_2014_Article_162_Equ30_HTML.gif
(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 . https://static-content.springer.com/image/art%3A10.1186%2Fs13661-014-0162-y/MediaObjects/13661_2014_Article_162_Equ31_HTML.gif
(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 , https://static-content.springer.com/image/art%3A10.1186%2Fs13661-014-0162-y/MediaObjects/13661_2014_Article_162_Equ32_HTML.gif
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 = { x Ω : w + = 0 } , Ω 2 = { x Ω : 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 . https://static-content.springer.com/image/art%3A10.1186%2Fs13661-014-0162-y/MediaObjects/13661_2014_Article_162_Equ33_HTML.gif
(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 . https://static-content.springer.com/image/art%3A10.1186%2Fs13661-014-0162-y/MediaObjects/13661_2014_Article_162_Equ34_HTML.gif
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 + ) d x = 0 and consequently
I + ( w n ) = 1 2 w n 2 + o ( 1 ) = 2 c + o ( 1 ) . https://static-content.springer.com/image/art%3A10.1186%2Fs13661-014-0162-y/MediaObjects/13661_2014_Article_162_Equ35_HTML.gif
(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 > 0 as  n + . https://static-content.springer.com/image/art%3A10.1186%2Fs13661-014-0162-y/MediaObjects/13661_2014_Article_162_Equ36_HTML.gif
(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 . https://static-content.springer.com/image/art%3A10.1186%2Fs13661-014-0162-y/MediaObjects/13661_2014_Article_162_Equ37_HTML.gif
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 , https://static-content.springer.com/image/art%3A10.1186%2Fs13661-014-0162-y/MediaObjects/13661_2014_Article_162_Equ38_HTML.gif
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 ) . https://static-content.springer.com/image/art%3A10.1186%2Fs13661-014-0162-y/MediaObjects/13661_2014_Article_162_Equ39_HTML.gif
Similarly, since u n u in E, Ω | u n u | d x 0 . Since ε > 0 is arbitrary, we can conclude that
Ω ( f + ( x , u n ) f + ( x , u ) ) ( u n u ) d x 0 as  n . https://static-content.springer.com/image/art%3A10.1186%2Fs13661-014-0162-y/MediaObjects/13661_2014_Article_162_Equ40_HTML.gif
(15)
By (10), we have
I + ( u n ) I + ( u ) , ( u n u ) 0 as  n . https://static-content.springer.com/image/art%3A10.1186%2Fs13661-014-0162-y/MediaObjects/13661_2014_Article_162_Equ41_HTML.gif
(16)
From (15) and (16), we obtain
Ω [ | ( u n u ) | 2 c | ( u n u ) | 2 ] d x 0 as  n . https://static-content.springer.com/image/art%3A10.1186%2Fs13661-014-0162-y/MediaObjects/13661_2014_Article_162_Equ42_HTML.gif
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 , https://static-content.springer.com/image/art%3A10.1186%2Fs13661-014-0162-y/MediaObjects/13661_2014_Article_162_Equ43_HTML.gif
where a ( 0 , 1 ) is defined by
1 p = a ( 1 2 1 N ) + ( 1 a ) 1 2 . https://static-content.springer.com/image/art%3A10.1186%2Fs13661-014-0162-y/MediaObjects/13661_2014_Article_162_Equ44_HTML.gif
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 https://static-content.springer.com/image/art%3A10.1186%2Fs13661-014-0162-y/MediaObjects/13661_2014_Article_162_Equ45_HTML.gif
(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 . https://static-content.springer.com/image/art%3A10.1186%2Fs13661-014-0162-y/MediaObjects/13661_2014_Article_162_Equ46_HTML.gif
(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 , x Ω , t R . https://static-content.springer.com/image/art%3A10.1186%2Fs13661-014-0162-y/MediaObjects/13661_2014_Article_162_Equ47_HTML.gif
So, for any u E { 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 + . https://static-content.springer.com/image/art%3A10.1186%2Fs13661-014-0162-y/MediaObjects/13661_2014_Article_162_Equ48_HTML.gif
Hence, for every finite dimension subspace E ˜ E , there exists R = R ( E ˜ ) such that
I ( u ) 0 , u E ˜ B R ( E ˜ ) https://static-content.springer.com/image/art%3A10.1186%2Fs13661-014-0162-y/MediaObjects/13661_2014_Article_162_Equ49_HTML.gif
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 . https://static-content.springer.com/image/art%3A10.1186%2Fs13661-014-0162-y/MediaObjects/13661_2014_Article_162_Equ50_HTML.gif
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 . https://static-content.springer.com/image/art%3A10.1186%2Fs13661-014-0162-y/MediaObjects/13661_2014_Article_162_Equ51_HTML.gif
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 . https://static-content.springer.com/image/art%3A10.1186%2Fs13661-014-0162-y/MediaObjects/13661_2014_Article_162_Equ52_HTML.gif
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. □

Declarations

Acknowledgements

This study was supported by the National NSF (Grant No. 11101319) of China and Planned Projects for Postdoctoral Research Funds of Jiangsu Province (Grant No. 1301038C).

Authors’ Affiliations

(1)
School of Mathematics and Statistics, Tianshui Normal University
(2)
School of Mathematics and Computer Sciences, Nanjing Normal University

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© Pei and Zhang; licensee Springer. 2014

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