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Multiple solutions for a fourth-order nonlinear elliptic problem which is superlinear at +∞ and linear at −∞

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Abstract

We consider a semilinear fourth-order elliptic equation with a right-hand side nonlinearity which exhibits an asymmetric growth at +∞ and at −∞. Namely, it is linear at −∞ and superlinear at +∞. Combining variational methods with Morse theory, we show that the problem has at least two nontrivial solutions, one of which is negative.

1 Introduction

Consider the following Navier boundary value problem:

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

where 2 is the biharmonic operator, and Ω is a bounded smooth domain in R N (N>4), and c< λ 1 the first eigenvalue of − in H 0 1 (Ω).

The conditions imposed on f(x,t) are as follows:

(H1) f C 1 ( Ω ¯ ×R,R), f(x,0)=0 for all xΩ and f(x,t)<0 for all t<0 and all xΩ;

(H2) there exist r(2, p ) and A,B>0 such that | f t (x,t)|A+B | t | r 2 for all xΩ, and tR, where p = 2 N N 4 , if N>4;

(H3) lim t f ( x , t ) t =l uniformly for xΩ, where l is a nonnegative constant;

(H4) there exist β,ξR such that for F(x,t)= 0 t f(x,s)ds, we have

lim t sup ( 2 F ( x , t ) f ( x , t ) t ) βuniformly for all xΩ,
(2)
lim t + F ( x , t ) t 2 =+uniformly for all xΩ
(3)

and

lim t + inf ( f ( x , t ) t 2 F ( x , t ) ) ξuniformly for all xΩ;
(4)

(H5) there exist ϑ 1 , ϑ 2 L ( Ω ) + and an integer k2 such that

λ k ϑ 1 (x) ϑ 2 (x) λ k + 1 for all xΩ,
(5)

the first and the last inequality are strict on sets (not necessary the same) of positive measure, and

ϑ 1 (x) f t (x,0)= lim t 0 f ( x , t ) t ϑ 2 (x)uniformly for all xΩ.
(6)

In view of the conditions (H3) and equation (3) in (H4), it is clear that for all xΩ, f(x,t) is linear at −∞ and superlinear at +∞. Clearly, u=0 is a trivial solution of problem (1). It follows from (H1) and (H2) that the functional

I(u)= 1 2 Ω ( | u | 2 c | u | 2 ) dx Ω F(x,u)dx
(7)

is C 2 on the space H 0 1 (Ω) H 2 (Ω) with the norm

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

where F(x,t)= 0 t f(x,s)ds. Under the condition (H2), the critical points of I are solutions of problem (1). Let 0< λ 1 < λ 2 << λ k < be the eigenvalues of ( 2 +c, H 2 (Ω) H 0 1 (Ω)) and ϕ 1 (x)>0 be the eigenfunction corresponding to λ 1 . In fact, λ 1 = λ 1 ( λ 1 c). Let E λ k denote the eigenspace associated with λ k . Throughout this article, we denoted by | | p the L p (Ω) norm and E= H 2 (Ω) H 0 1 (Ω). The aim of this paper is to prove a multiplicity theorem for problem (1) when the nonlinearity term f(x,t) exhibits an asymmetric behavior as tR approaches +∞ and −∞. In the past, some authors studied the following elliptic problem:

u=f(x,u),u H 0 1 (Ω)
(8)

with asymmetric nonlinearities by using the Fučík spectrum of the operator (, H 0 1 (Ω)). This approach requires that f(x,t) exhibits linear growth at both +∞ and −∞ and that the limits lim t ± f ( x , t ) t exist and belong to . See the works of Các [1], Dancer and Zhang [2], Magalhães [3], de Paiva [4], Schechter [5] and the references therein. Equations with nonlinearities which are superlinear in one direction and linear in the other were investigated by Arcoya and Villegas [6] and Perera [7]. They let the nonlinearity f(x,t) be line at −∞ and satisfy the Ambrosetti-Rabinowitz condition at +∞. Particularly, it is worth noticing paper [8]. The authors relax several of the above restrictions on the nonlinearity f(x,t). Their nonlinearity is only measurable in xΩ. The limit as t of f ( x , t ) t need not exist and the growth at −∞ can be linear or sublinear. Furthermore, their nonlinearity f(x,t) does not satisfy the famous AR-condition. They use the truncated skill of first order weak derivative to verify the (PS) condition and obtain multiple solutions for problem (1) by combining variational methods and Morse theory.

To the authors’ knowledge, there seem to be few results on problem (1) when f(x,t) is asymmetric nonlinearity at positive infinity and at negative infinity. However, the method in [8] cannot be applied directly to the biharmonic problems. For example, for the Laplacian problem, u H 0 1 (Ω) implies |u|, u + , u H 0 1 (Ω), where u + =max(u,0), u =max(u,0). We can use u + or u as a test function, which is helpful in proving a solution nonnegative. While for the biharmonic problems, this trick fails completely since u H 0 2 (Ω) does not imply u + , u H 0 2 (Ω) (see [[9], Remark 2.1.10] and [10, 11]). As far as this point is concerned, we will make use of the new methods to overcome it.

This fourth-order semilinear elliptic problem can be considered as an analogue of a class of second-order problems which have been studied by many authors. In [12], there was a survey of results obtained in this direction. In [13], Micheletti and Pistoia showed that ( P 1 ) admits at least two solutions by a variation of linking if f(x,u) is sublinear. Chipot [14] proved that the problem ( P 1 ) has at least three solutions by a variational reduction method and a degree argument. In [15], Zhang and Li showed that ( P 1 ) admits at least two nontrivial solutions by Morse theory and local linking if f(x,u) is superlinear and subcritical on u.

In this article, under the guidance of [8], we consider multiple solutions of problem (1) with the asymmetric nonlinearity by using variational methods and Morse theory.

2 Main result and auxiliary lemmas

Let us now state the main result.

Theorem 2.1 Assume conditions (H1)-(H5) hold. If l< λ 1 , then problem (1) has at least two nontrivial solutions.

Lemma 2.2 Under the assumptions of Theorem  2.1, then I satisfies the (PS) condition.

Proof Let { u n }E be a sequence such that for every nN,

| 1 2 Ω ( | u n | 2 c | u n | 2 ) dx Ω F(x, u n )dx|c,
(9)
| Ω ( u n vc u n v)dx Ω f(x, u n )vdx| ε n v,vE,
(10)

where c>0 is a positive constant and { ε n } R + is a sequence which converges to zero. By a standard argument, in order to prove that { u n } has a convergence subsequence, we have to show that it is a bounded sequence. To do this, we argue by contradiction assuming that for a subsequence, denoted by { u n }, we have

u n +as n.

Without loss of generality we can assume u n >1 for all nN and define z n = u n u n . Obviously, z n =1 nN and then it is possible to extract a subsequence (denoted also by { z n }) such that

z n z 0 in E,
(11)
z n z 0 in  L 2 (Ω),
(12)
z n (x)z(x),a.e. xΩ,
(13)
| z n (x)|q(x),a.e. xΩ,
(14)

where z 0 E and q L 2 (Ω). Dividing both sides of inequality (10) by u n , we obtain

| Ω ( z n vc z n v)dx Ω f ( x , u n ) u n vdx| ε n u n vfor all vE.

Passing to the limit we deduce from equation (11) that

lim n Ω f ( x , u n ) u n vdx= Ω ( z 0 vc z 0 v)dx
(15)

for all vE.

Now we claim that z 0 (x)0 a.e. xΩ. To verify this, let us observe that by choosing v= z 0 in equation (15) we have

lim n Ω + f ( x , u n ) u n z 0 dx<+,
(16)

where Ω + ={xΩ| z 0 (x)>0}. But, on the other hand, from (H3) and equation (3) in (H4), we have

f ( x , u n ( x ) ) u n z 0 (x) l u n ( x ) K 1 u n z 0 (x) ( l q ( x ) K 1 ) z 0 (x),a.e. xΩ

for some positive constant K 1 >0. Moreover, using lim n u n (x)=+ a.e. x Ω + , equation (13) and the superlinearity of f, we also deduce

lim n f ( x , u n ( x ) ) u n z 0 (x)= lim n f ( x , u n ( x ) ) u n z n (x) z 0 (x)=+,a.e. x Ω + .

Therefore, if | Ω + |>0 we will obtain by Fatou’s lemma that

lim n Ω + f ( x , u n ( x ) ) u n z 0 (x)dx=+

which contradicts inequality (16). Thus | Ω + |=0 and the claim is proved.

Clearly, z 0 (x)0, by (H3), there exists C>0 such that | f ( x , u n ) | | u n | C for a.e. xΩ. By using Lebesgue dominated convergence theorem in equation (15), we have

Ω ( z 0 vc z 0 v)dx Ω l z 0 vdx=0
(17)

for all vE. This contradicts l< λ 1 . □

Lemma 2.3 Let E=VW, where V= E λ 1 E λ 2 E λ k . If k0 is an integer, ϑ L ( Ω ) + , ϑ(x) λ k + 1 a.e. on Ω and the inequality is strict on a set of positive measure, then there exists γ>0 such that

u 2 Ω ϑ u 2 dxγ u 2

for all uW.

Proof We claim that there exists a constant ϑ 0 <1 such that

Ω ϑ(x) u 2 dx ϑ 0 u 2
(18)

for all uW. In fact, if not, there exists a sequence { u n } such that

Ω ϑ(x) | u n | 2 dx> ( 1 1 n ) u n 2

for all nN, which implies u n 0 for all n. By the homogeneity of the above inequality, we may assume that u n =1 and

Ω ϑ(x) | u n | 2 dx>1 1 n
(19)

for all n. It follows from the weak compactness of the unit ball of W that there exists a subsequence, say { u n }, such that u n weakly converges to u in W. Now Sobolev’s embedding theorem suggests that { u n } converges to u in L 2 (Ω). From inequality (19) we obtain

Ω ϑ(x) | u | 2 dx1.

Moreover one has

1 u 2 λ k + 1 | u | 2 2 Ω ϑ(x) | u | 2 dx1.

Hence we have

u 2 = λ k + 1 | u | 2 2

and

Ω ( λ k + 1 ϑ ( x ) ) u 2 dx=0

which implies that u E λ k + 1 {0} and u=0 on a positive measure subset. It contradicts the unique continuation property of the eigenfunction. □

3 Computation of the critical groups

It is well known that critical groups and Morse theory are the main tools in solving elliptic partial differential equation. Let us recall some results which will be used later. We refer the readers to the book [16] for more information on Morse theory.

Let H be a Hilbert space and I C 1 (H,R) be a functional satisfying the (PS) condition or (C) condition, and H q (X,Y) be the q th singular relative homology group with integer coefficients. Let u 0 be an isolated critical point of I with I( u 0 )=c, cR, and U be a neighborhood of u 0 . The group

C q (I, u 0 ):= H q ( I c U , I c U { u 0 } ) ,qZ

is said to be the q th critical group of I at u 0 , where I c ={uH:I(u)c}.

Let K:={uH: I (u)=0} be the set of critical points of I and a<infI(K), the critical groups of I at infinity are formally defined by [17]

C q (I,):= H q ( H , I a ) ,qZ.

From the deformation theorem, we see that the above definition is independent of the particular choice of c<infI(K). If c<infI(K) then

C q (I,):= H q ( H , I ˙ a ) ,qZ.
(20)

For the convenience of our proof, we first recall two interesting results and prove two important propositions.

Proposition 3.1 [18]

Under (H2), if uE:= H 2 (Ω) H 0 1 (Ω) is an isolated critical point of I, then C (I,u) C (I | C 0 3 ( Ω ) ,u).

Proposition 3.2 [19]

If D 1 D E 0 E 1 X and for some integer k0 we have H k (E,D)0 and H k ( E 1 , D 1 )=0, then either H k + 1 ( E 1 ,E)0 or H k 1 (D, D 1 )0.

Proposition 3.3 If the assumptions of Theorem  2.1 hold, then

C k (I,)=0for all integers k0.

Proof Under the guidance of [8] and [18], we begin to prove this result. Let I 1 =I | C 0 3 ( Ω ¯ ) . Indeed, it follows from above Proposition 3.1 that I and I 1 have same critical set. Since C 0 3 ( Ω ¯ ) is dense in E, invoking Proposition 16 of Palais [20], we have

H k ( E , I ˙ a ) = H k ( C 0 3 ( Ω ¯ ) , I ˙ 1 a ) for all aR and all integers k0.
(21)

From equations (20) and (21), we see that in order to prove the proposition, it suffices to show that

H k ( C 0 3 ( Ω ¯ ) , I 1 a ) =0for all a<0 with |a| large and all integers k0.
(22)

In order to prove equation (22), we proceed as follows. We define the sets

B 1 c = { u C 0 3 ( Ω ¯ ) : u C 0 3 ( Ω ¯ ) = 1 }

and

B 1 , + c = { u B 1 c : u ( x ) > 0  for some  x Ω } .

Consider the map h + :[0,1]× B 1 , + c B 1 , + c defined by

h + (t,u)= ( 1 t ) u + t ϕ 1 ( 1 t ) u + t ϕ 1 C 0 3 ( Ω ¯ ) for all (t,u)[0,1]× B 1 , + c .

Clearly, h + is a continuous homotopy and h(1,u)= ϕ 1 for all x B 1 , + c . Therefore, B 1 , + c is contractible in itself.

By equation (3) in (H4), given any γ>0, we can find C=C(γ)>0 such that

F(x,t) γ 2 t 2 for all xΩ and all tC.
(23)

Similarly, from condition (H3), and by choosing C>0 even bigger if necessary, we observe that there is a number γ 0 >0 such that

F(x,t) γ 0 2 t 2 for all xΩ and all tC(γ).
(24)

Moreover, by condition (H2), we have

|F(x,t)| C 3 for all xΩ and all |t|C(γ)
(25)

for some C 3 >0.

Let u B 1 , + c . By inequalities (23), (24), and (25), for all t>0 we have

I ( t u ) = t 2 2 u 2 Ω F ( x , t u ) d x = t 2 2 u 2 t u C F ( x , t u ) d x t u C F ( x , t u ) d x | t u | C F ( x , t u ) d x t 2 2 [ ( 1 + γ 0 λ 1 ) u 2 γ t u C u 2 d x ] + C 3 | Ω | .
(26)

Recalling that γ>0 is arbitrary, from (26), we have

I(tu)as t.
(27)

Using formula (4) in condition (H4), we see that there exist constants ξ 0 and M>0 such that

f(x,t)t2F(x,t) ξ 0 for all xΩ and all tM.
(28)

By (H2) and formula (2) in condition (H4), we have

2F(x,t)f(x,t)tCfor all xΩ and all t<M
(29)

for some C>0. By inequalities (28) and (29), for any uE we have

Ω ( 2 F ( x , t ) f ( x , t ) t ) dxC,
(30)

where C is a positive constant. Let i: C 0 3 ( Ω ¯ )E be the continuous embedding map. Let , 0 denote the duality brackets for the pair ( C 0 3 ( Ω ¯ ) , C 0 3 ( Ω ¯ )). We let I 2 =Ii, and so

I 2 ( u ) = i I ( i ( u ) ) for all  u C 0 3 ( Ω ¯ ) , d d t I 2 ( t u ) = I 2 ( t u ) , u 0 = t u 2 Ω f ( x , t u ) u d x 1 t ( 2 I ( t u ) + C ) .
(31)

Then, from equation (27), we obtain

d d t I 2 (tu)<0for all t>0 large such that I(tu)< C 2 .
(32)

From conditions (H2) and (H3), we see that given ϵ>0, we can find M>0 such that

F(x,t) 1 2 (l+ϵ) t 2 +Mfor all xΩ and all t0.
(33)

Using inequality (33), we have

I ( u ) 1 2 ( u 2 l | u | 2 2 ϵ | u | 2 2 ) M C u 2 M

for u C + , where C + is defined as

{ u C 0 3 ( Ω ) : u ( x ) 0  for all  x Ω }

and C>0 is a positive constant. So I | C + is coercive, thus we find C >0 such that I | C + C . We pick

a<min { C 2 , C , inf B 1 c I 2 } .

Then inequality (32) implies that we can find k(u)>1 such that

{ I 2 ( t u ) > a if  t [ 0 , k ( u ) ) , I 2 ( t u ) = a if  t = k ( u ) , I 2 ( t u ) < a if  t > k ( u ) .

Moreover, the implicit function theorem implies that kC( B 1 , + c ,[1,+)).

By the choice of a, we have

I 2 a = { t u : u B 1 , + c , t k ( u ) } .
(34)

We define the set E + ={tu:u B 1 , + c ,t1}. The map h ˆ + :[0,1]× E + E + defined by

h ˆ + (s,tu)={ ( 1 s ) t u + s k ( u ) u if  1 t < k ( u ) , t u if  t k ( u ) , s[0,1],
(35)

is a continuous deformation of E + , h ˆ + (1, E + ) I 2 a and h ˆ + (s,) | I 2 a =id | I 2 a for all s[0,1] (see equations (34) and (35)). Therefore, I 2 a is a strong deformation retract of E + . Hence we have

H k ( C 0 3 ( Ω ¯ ) , I 2 a ) = H k ( C 0 3 ( Ω ¯ ) , E + ) = H k ( C 0 3 ( Ω ¯ ) , B 1 , + c ) for all k0.
(36)

Recalling that in the first part of the proof, we established that B 1 , + c is contractible. This yields

H k ( C 0 3 ( Ω ¯ ) , B 1 , + c ) =0for all integers k0.

Combining with equation (36) leads to equation (22), which completes the proof. □

Proposition 3.4 If the assumptions of Theorem  2.1 hold, then

C d (I,0)0,

where d=dimV (V being defined in Lemma  2.3).

Proof By condition (H5), given ϵ>0, we can find δ >0 such that

1 2 ( ϑ 1 ( x ) ϵ ) t 2 F(x,t)for all xΩ and all |t| δ .
(37)

Since V is finite dimensional, all norms are equivalent. Thus we can find ρ>0 small such that

uρ u δ
(38)

for all uV. Taking inequalities (37) and (38) into account, for all uV with uρ we have

I(u) 1 2 Ω ( λ k ϑ 1 ( x ) ) u 2 dx+ ϵ 2 | u | 2 2 .
(39)

Similar to the proof of Lemma 2.3, there exists C>0 such that

I(u)(C+ϵ) u 2 0
(40)

for all uV and uρ.

On the other hand, for given ϵ>0, it follows from (H2) and (H5) that

F(x,t) ϑ 2 ( x ) + ϵ 2 t 2 + C ϵ | t | r
(41)

for all xΩ and tR. By (41) and Lemma 2.3, we have

I(u) C 4 u 2 C 5 u r
(42)

for all uW. From inequality (42), we infer that for ρ small enough we have

I(u)>0for all uW with 0<uρ.
(43)

From inequalities (40) and (43), we know that I has a local linking at 0. Then invoking Proposition 2.3 of Bartsch and Li [17], we obtain C d (I,0)0. □

4 Proof of the main result

Proof of Theorem 2.1 We 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 :E= H 2 (Ω) H 0 1 (Ω)R 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 2 (E,R). Obviously, by conditions (H1) and (H3), we know that I is coercive and boundedness from below. Thus we can find v 0 E such that

I ( v 0 )=inf I =: m .
(44)

Next, we claim that v 0 0. By condition (H5), given ϵ(0, λ k λ 1 ), there exists δ>0 such that

1 2 ( ϑ 1 ( x ) ϵ ) t 2 F (x,t)for all xΩ,t[δ,0].
(45)

For s small enough, it follows from inequality (45) that

I ( s ϕ 1 ) = s 2 2 λ 1 Ω F ( x , s ϕ 1 ) d x s 2 2 ( λ 1 λ k + ϵ ) < 0 ,

and thus, by equation (44), I ( v 0 ) I (s ϕ 1 )<0, so v 0 0. From condition (H1) and strong maximum principle, we have v 0 <0 and

C k ( I , v 0 )= δ k , 0 Zfor all integers k0.

Since v 0 is an interior point of C + , from Proposition 3.1, we know

C k (I, v 0 )= C k ( I , v 0 )= δ k , 0 Zfor all integers k0.
(46)

Let θR, ϵ>0 be such that θ< m =I( v 0 )<ϵ. We consider the sublevel sets

I θ I ϵ I ϵ E.

Suppose that 0 and v 0 are the only critical points of I. Otherwise, we have a second nontrivial smooth solution and so we are done. By Proposition 3.3, we have

H k ( E , I θ ) = C k (I,)=0for all integers k0.
(47)

We know that I satisfies the (PS) condition (see Lemma 2.2). Hence choosing ϵ>0 small enough, we have

H d ( I ϵ , I ϵ ) = C d (I,0)0
(48)

(see Proposition 3.4). Because of equations (47) and (48), using Proposition 3.2, we obtain

H d + 1 ( E , I ϵ ) 0or H d 1 ( I ϵ , I θ ) 0.

If H d + 1 (E, I ϵ )0, then there is a critical point v E of I such that

I ( v ) >ϵ>0and v 0, v 0 .

If H d 1 ( I ϵ , I θ )0, then there is a critical point v E of I such that

C d 1 ( I , v ) 0.
(49)

Since d2, from equations (46) and (49), we see that v v 0 . It is obvious that v 0. Therefore v 0 and v are two solutions of problem (1). □

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Acknowledgements

The authors would like to thank the referees for valuable comments and suggestions in improving this article. 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).

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Correspondence to Ruichang Pei.

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Keywords

  • fourth-order elliptic boundary value problems
  • multiple solutions
  • critical groups
  • Morse theory