Research

# Existence and multiplicity of solutions for a class of sublinear Schrödinger-Maxwell equations

Ying Lv

Author Affiliations

School of Mathematics and Statistics, Southwest University, Chongqing, 400715, People’s Republic of China

Boundary Value Problems 2013, 2013:177  doi:10.1186/1687-2770-2013-177

 Received: 24 September 2012 Accepted: 5 July 2013 Published: 30 July 2013

This is an Open Access article distributed under the terms of the Creative Commons Attribution License (http://creativecommons.org/licenses/by/2.0), which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

### Abstract

In this paper I consider a class of sublinear Schrödinger-Maxwell equations, and new results about the existence and multiplicity of solutions are obtained by using the minimizing theorem and the dual fountain theorem respectively.

##### Keywords:
Schrödinger-Maxwell equations; sublinear; minimizing theorem; dual fountain theorem

### 1 Introduction and main result

Consider the following semilinear Schrödinger-Maxwell equations:

{ Δ u + V ( x ) u + ϕ u = f ( x , u ) , in  R 3 , Δ ϕ = u 2 , lim | x | ϕ ( x ) = 0 , in  R 3 . (1)

Such a system, also known as the nonlinear Schrödinger-Poisson system, arises in an interesting physical context. Indeed, according to a classical model, the interaction of a charge particle with an electromagnetic field can be described by coupling the nonlinear Schrödinger and the Maxwell equations (we refer to [1,2] for more details on the physical aspects and on the qualitative properties of the solutions). In particular, if we are looking for electrostatic-type solutions, we just have to solve (1).

In recent years, system (1), with V ( x ) 1 or being radially symmetric, has been widely studied under various conditions on f; see, for example, [3-11]. Since (1) is set on R 3 , it is well known that the Sobolev embedding H 1 ( R 3 ) L s ( R 3 ) ( 2 s 2 = 6 ) is not compact, and then it is usually difficult to prove that a minimizing sequence or a sequence that satisfies the ( PS ) condition, briefly a Palais-Smale sequence, is strongly convergent if we seek solutions of (1) by variational methods. If V ( x ) is radial (for example, V ( x ) 1 ), we can avoid the lack of compactness of Sobolev embedding by looking for solutions of (1) in the subspace of radial functions of H 1 ( R 3 ) , which is usually denoted by H r 1 ( R 3 ) , since the embedding H r 1 ( R 3 ) L s ( R 3 ) ( 2 < s < 6 ) is compact. Specially, Ruiz [11] dealt with (1) under the assumption that V ( x ) 1 and f ( u ) = u p ( 1 < p < 5 ) and got some general existence, nonexistence and multiplicity results.

Moreover, in [12] the authors considered system (1) with periodic potential V ( x ) , and the existence of infinitely many geometrically distinct solutions was proved by the nonlinear superposition principle established in [13].

There are also some papers treating the case with nonradial potential V ( x ) . More precisely, Wang and Zhou [14] got the existence and nonexistence results of (1) when f ( u ) is asymptotically linear at infinity. Chen and Tang [15] proved that (1) has infinitely many high energy solutions under the condition that f ( x , u ) is superlinear at infinity in u by the fountain theorem. Soon after, Li, Su and Wei [16] improved their results.

Up to now, there have been few works concerning the case that V ( x ) is nonradial potential and f ( x , u ) is sublinear at infinity in u. Very recently, Sun [17] treated the above case based on the variant fountain theorem established in Zou [18].

Theorem 1.1[17]

Assume that the following conditions hold:

( V 1 ) V C ( R 3 , R ) satisfies inf x R 3 V ( x ) a > 0 , where a > 0 is a constant. For every M > 0 , meas { x R 3 : v ( x ) M } < .

( H 1 ) F ( x , u ) = a ( x ) | u | r , where F ( x , u ) = 0 u f ( x , y ) d y , a : R 3 R + is a positive function such that a L 2 2 r ( R 3 ) and 1 < r < 2 .

Then problem (1) has infinitely many nontrivial solutions { ( u k , ϕ k ) } satisfying

1 2 R 3 ( | u k | 2 + V ( x ) u k 2 ) d x 1 4 R 3 | ϕ k | 2 d x + 1 2 R 3 ϕ k u k 2 d x R 3 F ( x , u k ) d x 0

as k .

In the present paper, based on the dual fountain theorem, we can prove the same result under a more generic condition, which generalizes the result in [17]. Our first result can be stated as follows.

Theorem 1.2Assume thatVsatisfies

( V 1 ) V C ( R 3 , R ) and inf x R 3 V ( x ) > 0 ;

andfsatisfies the following conditions.

( W 1 ) There exist constants δ > 0 , r 1 ( 1 , 2 ) and a function a 1 L 2 2 r 1 ( R 3 , [ 0 , + ) ) such that

| f ( x , u ) | a 1 ( x ) | u | r 1 1

for all x R 3 and | u | δ ;

( W 2 ) There exist constants M > 0 , r 2 ( 1 , 2 ) and a function a 2 L 2 2 r 2 ( R 3 , [ 0 , + ) ) such that

| f ( x , u ) | a 2 ( x ) | u | r 2 1

for all x R 3 and | u | M ;

( W 3 ) For every m > δ , there exist a constant r 3 ( 1 , 2 ) and a function b m L 2 2 r 3 ( R 3 , [ 0 , + ) ) such that

| f ( x , u ) | b m ( x )

for all x R 3 and | u | m ;

( W 4 ) There exist constants r 4 ( 1 , 2 ) , η > 0 and ζ > 0 such that

F ( x , u ) η | u | r 4

for all x Ω and | u | ζ , where meas { Ω } > 0 , F ( x , u ) : = 0 u f ( x , y ) d y ;

( W 5 ) F ( x , u ) = F ( x , u ) for all x R 3 and u R .

Then problem (1) has infinitely many nontrivial solutions { ( u k , ϕ k ) } satisfying

1 2 R 3 ( | u k | 2 + V ( x ) u k 2 ) d x 1 4 R 3 | ϕ k | 2 d x + 1 2 R 3 ϕ k u k 2 d x R 3 F ( x , u k ) d x 0

as k .

By Theorem 1.2, we obtain the following corollary.

Corollary 1.3Assume thatLsatisfies ( V 1 ) andWsatisfies

( W 6 ) F ( x , u ) = a ( x ) | u | r , where F ( x , u ) = 0 u f ( x , y ) d y , 1 < r < 2 is a constant and a : R 3 R is a function such that a L 2 2 r ( R 3 ) and a ( x ) > 0 for x Ω , where meas { Ω } > 0 .

Then problem (1) has infinitely many nontrivial solutions { ( u k , ϕ k ) } satisfying

1 2 R 3 ( | u k | 2 + V ( x ) u k 2 ) d x 1 4 R 3 | ϕ k | 2 d x + 1 2 R 3 ϕ k u k 2 d x R 3 F ( x , u k ) d x 0

as k .

Remark 1.4 In Theorem 1.2, infinitely many solutions for problem (1) are obtained under the symmetry condition ( W 5 ) by using the dual fountain theorem. As a special case of Theorem 1.2, Corollary 1.3 generalizes and improves Theorem 1.1. To show this, it suffices to compare ( V 1 ) and ( V 1 ), ( H 1 ) and ( W 6 ). Firstly, it is clear that ( V 1 ) is really weaker than ( V 1 ). Secondly, in ( H 1 ) a is assumed to be positive, while in ( W 6 ) we assume that a is indefinite.

Moreover, under all the conditions of Theorem 1.2 except ( W 5 ) we obtain an existence result.

Theorem 1.5Assume thatLsatisfies ( V 1 ) andWsatisfies ( W 1 ), ( W 2 ), ( W 3 ), ( W 4 ). Then problem (1) possesses a nontrivial solution.

Remark 1.6 In Theorem 1.5 we obtain the existence of solutions for problem (1) under the assumption that f ( x , u ) is indefinite and without any coercive assumptions respect to V such as ( V 1 ). There are functions V and f which satisfy Theorem 1.5, but do not satisfy the corresponding results in [2-16]. For example,

V ( x ) 1 , f ( x , u ) = a ˜ ( x ) | u | 3 2 (2)

and

a ˜ ( x ) = { ( 1 ) n n 3 ( | x | n ) for  n | x | n + 1 n 2 , 0 else , (3)

in which n 3 . It is clear that a ˜ C ( R 3 , R ) is indefinite. Denoting by π the area of the unit ball in R 3 , we obtain

R 3 a ˜ 4 ( x ) d x = n = 3 ( n n + 1 n 2 n 12 r 2 ( r n ) 4 d r + n + 1 n 2 n + 2 n 2 n 12 r 2 ( n + 2 n 2 r ) 4 d r ) π = π n = 3 2 n 12 0 1 n 2 r 6 d x = 2 π 7 n = 3 n 2 < , (4)

which means that a ˜ L 2 2 3 2 ( R 3 ) . So, (2) satisfies our results, but does not satisfy the results in [3-17].

### 2 Preliminary results

In order to establish our results via critical point theory, we firstly describe some properties of the space H 1 ( R 3 ) , on which the variational functional associated with problem (1) is defined. Define the function space

H 1 ( R 3 ) : = { u L 2 ( R 3 ) : u ( L 2 ( R 3 ) ) 3 }

equipped with the norm

u H 1 : = ( R 3 ( | u | 2 + u 2 ) d x ) 1 / 2

and the function space

D 1 , 2 ( R 3 ) : = { u L 2 : u ( L 2 ( R 3 ) ) 3 }

with the norm

u D 1 , 2 = ( R 3 | u | 2 d x ) 1 / 2 .

Let

E : = { u H 1 ( R 3 ) : R 3 V ( x ) u 2 d x < + }

equipped with the inner product

( u , v ) = R 3 ( u v + V ( x ) u v ) d x

and the corresponding norm

u 2 = ( u , u ) .

Note that the following embeddings

E L s ( R 3 ) , 2 s 2 , D 1 , 2 ( R 3 ) L 2 ( R 3 )

are continuous, where 2 = 6 is the critical exponent for the Sobolev embeddings in dimension 3. Therefore, there exist constants C p and C such that

u L p C p u , u L 2 C u D 1 , 2 (5)

for all u E . Here L p ( R 3 ) ( 2 p 2 ) denotes the Banach spaces of a function on R 3 with values in R under the norm

u L p = ( R 3 | u ( x ) | p d x ) 1 / p .

Let

L a r ( R 3 ) : = { u : R 3 R : R 3 a ( x ) | u | r d x < + } ,

where a ( x ) > 0 for a.e. x R 3 . Then L a r ( R 3 ) is a Banach space with the norm

u L a r = ( R 3 a ( x ) | u | r d x ) 1 / r .

Lemma 2.1Suppose that assumption ( V 1 ) holds. Then the embedding ofEin L a r ( R 3 ) is compact, where r ( 1 , 2 ) , a L 2 2 r ( R 3 ) is positive for a.e. x R 3 .

Proof For any bounded set K E , there exists a positive constant M 0 such that u M 0 for all u K . We claim that K is precompact in L a r ( R 3 ) . In fact, since a L 2 2 r ( R 3 ) , for any ε > 0 , there exists T ε > 0 such that

( | x | T ε a ( x ) 2 2 r d x ) ( 2 r ) / 2 < ε .

For any u , v K , applying the Hölder inequality for r such that r 2 + 2 r 2 = 1 and the first inequality in (5), we have

| x | T ε a ( x ) | u v | r d x ( | x | T ε a ( x ) 2 2 r d x ) ( 2 r ) / 2 ( | x | T ε | u v | 2 d x ) r / 2 u v L 2 r ( | x | T ε a ( x ) 2 2 r d x ) ( 2 r ) / 2 C 2 r u v r ε 2 C 2 r M 0 r ε . (6)

Besides, since E ( B T ε ( 0 ) ) H 1 ( B T ε ( 0 ) ) is compactly embedded in L a r ( B T ε ( 0 ) ) , where B T ε ( 0 ) = { x R 3 : | x | T ε } , there are u 1 , u 2 , , u m K such that for any u K ,

| x | T ε a ( x ) | u u i | r d x < ε . (7)

Now it follows from (6) and (7) that K is precompact in L a r ( R 3 ) . Obviously, we have E is compact embedded in L a r ( R 3 ) , where r ( 1 , 2 ) , a L 2 2 r ( R 3 ) is positive for a.e. x R 3 . □

Lemma 2.2Assume that assumptions ( V 1 ), ( W 1 ), ( W 2 ) and ( W 3 ) hold and u n u inE. Then

f ( x , u n ) f ( x , u )

in L 2 ( R 3 ) .

Proof Assume that u n u in E. Then, by Lemma 2.1,

u n u

in L a r ( R 3 ) , where r ( 1 , 2 ) , a L 2 2 r ( R 3 ) is positive for a.e. x R 3 . Passing to a subsequence if necessary, it can be assumed that

n = 1 u n u L a r < + .

It is clear that

h k ( x ) : = n = 1 k | u n ( x ) u ( x ) | L a r ( R 3 ) (8)

and

h g h l L a r n = l g u n u L a r (9)

for all g > l N + . Since { u n } is a Cauchy sequence in L a r ( R 3 ) , so by (9) we know that { h k } is also a Cauchy sequence in L a r ( R 3 ) . Therefore, by the completeness of L a r ( R 3 ) , there exists h L a r ( R 3 ) such that h k h in L a r ( R 3 ) . Now we show that

h k ( x ) h ( x ) (10)

for all k N + and almost every x R 3 . If not, there exist k 0 N + and S R 3 , with meas { S } > 0 , such that

h k 0 ( x ) > h ( x )

for all x S . Then there exist a constant c > 0 and S 0 S , with meas { S 0 } > 0 , such that

h k 0 ( x ) h ( x ) + c

for all x S 0 . By the definition of h k , we have

h k ( x ) h k 0 ( x ) h ( x ) + c

for all k k 0 and x S 0 . Therefore, one has

R 3 a ( x ) | h k h | r d x S 0 a ( x ) | h k h | r d x c r S 0 a ( x ) d x .

Letting k , we get

0 c r S 0 a ( x ) d x ,

which contradicts the fact that a ( x ) > 0 for a.e. x R 3 . Now we have proved (10). It follows from ( W 2 ) that there exists M > 0 such that

| f ( x , u ) | a 2 ( x ) | u | r 2 1 (11)

for all x R 3 and | u | M . By ( W 1 ), there exists δ > 0 such that

| f ( x , u ) | a 1 ( x ) | u | r 1 1 (12)

for all x R 3 and | u | δ , which together with ( W 3 ) shows there exists b M L 2 2 r 3 ( R 3 ) such that

| f ( x , u ) | a 1 ( x ) | u | r 1 1 + b M ( x ) δ r 3 1 | u | r 3 1 (13)

for all x R 3 and | u | M . Combining (11) and (13), we have

| f ( x , u ) | a 1 ( x ) | u | r 1 1 + a 2 ( x ) | u | r 2 1 + b M δ r 3 1 | u | r 3 1 (14)

for all x R 3 and u R . Hence, by (10) one has

| f ( x , u n ) f ( x , u ) | a 1 ( x ) ( | u n | r 1 1 + | u | r 1 1 ) + a 2 ( x ) ( | u n | r 2 1 + | u | r 2 1 ) + b M ( x ) δ r 3 1 ( | u n | r 3 1 + | u | r 3 1 ) a 1 ( x ) ( | u n u | r 1 1 + 2 | u ( x ) | r 1 1 ) + a 2 ( x ) ( | u n u | r 2 1 + 2 | u | r 2 1 ) + b M ( x ) δ r 3 1 ( | u n u | r 3 1 + 2 | u | r 3 1 ) a 1 ( x ) ( | h | r 1 1 + 2 | u | r 1 1 ) + a 2 ( x ) ( | h | r 2 1 + 2 | u | r 2 1 ) + b M ( x ) δ r 3 1 ( | h | r 3 1 + 2 | u | r 3 1 )

for all n N and x R 3 . It follows that

| f ( x , u n ) f ( x , u ) | 2 d x 6 a 1 2 ( x ) ( | h | 2 ( r 1 1 ) + 4 | u | 2 ( r 1 1 ) ) d x + 6 a 2 2 ( x ) ( | h | 2 ( r 2 1 ) + 4 | u | 2 ( r 2 1 ) ) d x + 6 b M 2 ( x ) δ 2 ( r 3 1 ) ( | h | 2 ( r 3 1 ) + 4 | u | 2 ( r 3 1 ) ) d x = : ϱ ( x ) (15)

for all n N . By the Hölder inequality, we have

R 3 a 1 2 ( x ) | h | 2 ( r 1 1 ) d x ( R 3 a 1 ( x ) 2 2 r 1 d x ) 2 r 1 r 1 ( R 3 a 1 ( x ) | h | r 1 d x ) 2 ( r 1 1 ) r 1 = a 1 L 2 2 r 1 2 r 1 h L a 1 r 1 2 ( r 1 1 ) < . (16)

Similarly, we can prove

R 3 a 1 2 ( x ) | u | 2 ( r 1 1 ) d x < , R 3 a 2 2 ( x ) | h | 2 ( r 2 1 ) d x < , R 3 a 2 2 ( x ) | u | 2 ( r 2 1 ) d x < , (17)

also

R 3 b M 2 ( x ) | h | 2 ( r 3 1 ) d x < , R 3 b M 2 ( x ) | u | 2 ( r 3 1 ) d x < . (18)

It follows from (15), (16), (17) and (18) that

ϱ L 1 ( R 3 ) ,

which together with Lebesgue’s convergence theorem shows

R 3 | f ( x , u n ) f ( x , u ) | 2 d x 0 (19)

as n . Now we have proved the lemma. □

In the proof of Theorem 1.2, the following lemma is needed.

Lemma 2.3Assume that G R 3 is an open set. Then, for any closed set H G , there exists a function φ C 0 ( R 3 ) such that φ ( x ) = 0 for all x R 3 G , φ ( x ) = 1 for all x H and 0 ϕ ( x ) 1 for all x G H .

Proof Letting

α ˜ ( x ) = { e 1 | x | 2 1 , | x | < 1 , 0 , | x | 1 ,

then α ˜ C 0 ( R 3 ) and supp α ˜ = B 1 ( 0 ) . For any given ε > 0 , defining α and α ε as follows,

α ( x ) = α ˜ ( x ) R 3 α ˜ ( x ) d x , α ε ( x ) = 1 ε 3 α ( x ε ) ,

one has α ε C 0 ( R 3 ) , supp α ε = { x : | x | ε } and R 3 α ε ( x ) d x = 1 . Denoting

d 0 = inf x H , y G d ( x , y )

and

G θ : = { x G , d ( x , G ) θ } ,

it is clear that d 0 > 0 and H G d 0 . Lastly, we define

ψ ( x ) = { 1 , x G d 0 2 , 0 , x R G d 0 2

and

φ ( x ) = R 3 ψ ( x y ) α d 0 4 ( y ) d y ,

then φ ( x ) = 1 for all x H and φ ( x ) = 0 for all x G d 0 4 . Moreover, by the definition of α ε , we have φ C 0 ( R 3 ) and 0 φ ( x ) 1 . □

Since E is a Hilbert space, then there exists a basis { v n } X such that X = j 1 X j ¯ , where X j = span { v j } . Letting Y k = j = 1 k X j , Z k = j k X j ¯ , now we show the following lemma, which will be used in the proof of Theorem 1.2.

Lemma 2.4Suppose r ( 1 , 2 ) and a L 2 2 r ( R 3 ) , then we have

β k ( a , r ) : = sup u Z k , u = 1 u L a r 0

as k .

Proof It is clear that 0 < β k + 1 ( a , r ) β k ( a , r ) , so there exists β ( a , r ) 0 such that

β k ( a , r ) β ( a , r ) (20)

as k . By the definition of β k ( a , r ) , there exists u k Z k with u k = 1 such that

u k L a r > β k ( a , r ) 2 . (21)

Since { u k } k N is bounded, then there exists u E such that

u k u

as k . Now, since { v j } is a basis of E, it follows that for all j N ,

0 = ( u k , v j ) k > j ( u , v j )

as k , which shows that u = 0 . By Lemma 2.1 we have

u k 0

in L a r ( R 3 ) for all r ( 1 , 2 ) and a L 2 2 r ( R 3 ) , which together with (20) and (21) implies that β ( a , r ) = 0 for all r ( 1 , 2 ) and a L 2 2 r ( R 3 ) . □

We obtain the existence of a solution for problem (1) by using the following standard minimizing argument.

Lemma 2.5[19]

LetEbe a real Banach space and Φ C 1 ( E , R ) satisfying the ( PS ) condition. If Φ is bounded from below,

c : = inf E Φ

is a critical value of Φ.

In order to prove the multiplicity of solutions, we will use the dual fountain theorem. Firstly, we introduce the definition of the ( PS ) c condition.

Definition 2.6 Let Φ C 1 ( E , R ) and c R . The function Φ satisfies the ( PS ) c condition if any sequence { u n j } E , such that

Φ ( u n j ) c , Φ | Y n j ( u n j ) 0 as  n j ,

contains a subsequence converging to a critical point of Φ.

Now we show the following dual fountain theorem.

Lemma 2.7[20]

If Φ ( u ) = Φ ( u ) and for every k k 0 , there exists ρ k > γ k > 0 such that

(i) a k : = inf u Z k , u = ρ k Φ ( u ) 0 ,

(ii) b k : = max u Y k , u = γ k Φ ( u ) < 0 ,

(iii) d k : = inf u Z k , u = ρ k Φ ( u ) 0 as k .

Moreover, if Φ C 1 ( X , R ) satisfies the ( PS ) c condition for all c [ d k 0 , 0 ) , then Φ has a sequence of critical points { u k } such that Φ ( u k ) 0 as k .

### 3 Proof of theorems

Define the functional I : E × D 1 , 2 ( R 3 ) R by

I ( u , ϕ ) = 1 2 u 2 1 4 R 3 | ϕ | 2 d x + 1 2 R 3 ϕ u 2 d x R 3 F ( x , u ) d x . (22)

It is easy to know that I exhibits a strong indefiniteness, namely it is unbounded both from below and from above on an infinitely dimensional subspace. This indefiniteness can be removed using the reduction method described in [1], by which we are led to study a variable functional that does not present such a strong indefinite nature.

Now we recall this method. For any u E , consider the linear functional T u : D 1 , 2 ( R 3 ) R defined as

T u ( v ) = R 3 u 2 v d x .

By the Hölder inequality and using the second inequality in (5), we have

R 3 u 2 v d x u 2 L 6 / 5 v L 6 u L 12 / 5 v L 6 C 12 / 5 C u 2 v D 1 , 2 .

So, T u is continuous on D 1 , 2 ( R 3 ) . Set

μ ( u , v ) = R 3 u v d x

for all u , v D 1 , 2 ( R 3 ) . Obviously, μ ( u , v ) is bilinear, bounded and coercive. Hence, the Lax-Milgram theorem implies that for every u E , there exists a unique ϕ u D 1 , 2 ( R 3 ) such that

T u ( v ) = μ ( ϕ u , v )

for any v D 1 , 2 ( R 3 ) , that is,

R 3 u 2 v d x = R 3 ϕ u v d x

for any v D 1 , 2 ( R 3 ) . Using integration by parts, we get

R 3 ϕ u v d x = R 3 v Δ ϕ u d x

for any v D 1 , 2 ( R 3 ) , therefore

Δ ϕ u = u 2 (23)

in a weak sense. We can write an integral expression for ϕ u in the form

ϕ u = 1 4 π R 3 u 2 ( y ) | x y | d y

for any u C 0 ( R 3 ) (see [21], Theorem 1); by density it can be extended for any u E (see Lemma 2.1 of [22]). Clearly, ϕ u 0 and ϕ u = ϕ u for all u E .

It follows from (23) that

R 3 ϕ u u 2 d x = R 3 ϕ u ( Δ ϕ u ) d x = R 3 | ϕ u | 2 d x , (24)

and by the Hölder inequality, we have

ϕ u D 1 , 2 2 = R 3 ϕ u u 2 d x ( R 3 ϕ u 6 d x ) 1 / 6 ( R 3 | u | 12 5 ) 5 / 6 = C ϕ u D 1 , 2 u L 12 / 5 2 ,

and it follows that

ϕ u D 1 , 2 C u L 12 / 5 2 . (25)

Hence,

R 3 ϕ u u 2 d x C 2 u L 12 / 5 4 C 2 C 12 / 5 4 u 4 : = C u 4 . (26)

So, we can consider the functional Φ : E R defined by Φ ( u ) = I ( u , ϕ u ) . By (24), the reduced functional takes the form

Φ ( u ) = 1 2 u 2 + 1 4 R 3 ϕ u u 2 d x R 3 F ( x , u ) d x . (27)

By (12), we have

| F ( x , u ) | a 1 ( x ) r 1 | u | r 1 (28)

for all x R 3 and | u | δ , where r 1 ( 1 , 2 ) and a 1 L 2 2 r 1 ( R 3 ) . Let u E , then u C 0 ( R 3 ) , the space of continuous function u on R 3 , such that u ( x ) 0 as | x | . Therefore there exists T 1 > 0 such that

| u ( x ) | δ (29)

for all | x | > T 1 . Hence, one has

| x | > T 1 | F ( x , u ) | d x | x | > T 1 a 1 ( x ) r 1 | u ( x ) | r 1 d x 1 r 1 ( | x | T 1 a 1 ( x ) 2 2 r 1 d x ) ( 2 r 1 ) / 2 ( | x | T 1 | u ( x ) | 2 d x ) r 1 / 2 1 r 1 ( | x | T 1 a 1 ( x ) 2 2 r 1 d x ) ( 2 r 1 ) / 2 u L 2 r 1 1 r 1 C 2 r 1 u r 1 a 1 L 2 2 r 1 < ,

which together with (26) shows that Φ is well defined. Furthermore, it is well known that Φ is a C 1 functional with derivative given by

Φ ( u ) , v = R 3 [ ( u v ) + V ( x ) u v + ϕ u u v f ( x , u ) v ] d x .

It can be proved that ( u , ϕ ) E × D 1 , 2 ( R 3 ) is a solution of problem (1) if and only if u E is a critical point of the functional Φ and ϕ = ϕ u ; see, for instance, [1].

Lemma 3.1Under conditions ( V 1 ), ( W 1 ), ( W 2 ), ( W 3 ), Φ satisfies the ( PS ) c condition.

Proof Assume that { u n j } E is a sequence such that

Φ ( u n j ) c , Φ | Y n j ( u n j ) 0 as  n j .

Then there exists σ > 0 such that

| Φ ( u n j ) | σ , Φ | Y n j ( u n j ) E σ

for all n j N .

Firstly, we show that { u n j } is bounded. By (14), we have

| F ( x , u ) | a 1 ( x ) r 1 | u | r 1 + a 2 ( x ) r 2 | u | r 2 + b M ( x ) r 3 δ r 3 1 | u | r 3 (30)

for all u R and x R 3 , which together with R 3 ϕ u n j u n j 2 d x 0 implies

u n j 2 = 2 Φ ( u n j ) 1 2 R 3 ϕ u n j u n j 2 d x + 2 R 3 F ( x , u n j ) d x 2 σ + 2 r 1 R 3 a 1 ( x ) | u n j | r 1 d x + 2 r 2 R 3 a 2 ( x ) | u n j | r 2 d x + 2 r 3 δ r 3 1 R 3 b M ( x ) | u n j | r 3 d x 2 σ + 2 r 1 ( R 3 a 1 ( x ) 2 2 r 1 d x ) ( 2 r 1 ) / 2 ( R 3 | u n j | 2 d x ) r 1 / 2 + 2 r 2 ( R 3 a 2 ( x ) 2 2 r 2 d x ) ( 2 r 2 ) / 2 ( R 3 | u n j | 2 d x ) r 2 / 2 + 2 r 3 δ r 3 1 ( R 3 b M ( x ) 2 2 r 3 d x ) ( 2 r 3 ) / 2 ( R 3 | u n j | 2 d x ) r 3 / 2 2 σ + 2 r 1 C 2 r 1 a 1 L 2 2 r 1 u n j r 1 + 2 r 2 C 2 r 2 a 2 L 2 2 r 2 u n j r 2 + 2 r 3 δ r 3 1 C 2 r 3 b M L 2 2 r 3 u n j r 3 . (31)

Noting that r i < 2 for all i = 1 , 2 , 3 , so u n j is bounded.

By the fact that { u n j } is bounded in E, there exists u E and a constant d > 0 such that

sup n j N u n j d , u d (32)

and

u n j u

in E as n j . It is obvious that

Φ ( u n j ) Φ ( u ) , u 0 (33)

and

ϕ u u ( u n j u ) 0 (34)

as n j . On the other hand, by ( V 1 ), (32) and Lemma 2.2, one has

| R 3 ( f ( x , u n j ) f ( x , u ) ) u n j d x | f ( x , u n j ) f ( x , u ) L 2 u n j L 2 C 2 f ( x , u n j ) f ( x , u ) L 2 u n j C 2 d f ( x , u n j ) f ( x , u ) L 2 0 (35)

as n j , which implies

Φ ( u n j ) Φ ( u ) , u n j 0 (36)

as n j . Summing up (33) and (36), we have

Φ ( u n j ) Φ ( u ) , u n j u 0 (37)

as n j . By the Hölder inequality and (25), one gets

R 3 ϕ u n j u n j ( u n j u ) d x ϕ u n j u n j L 2 u n j u L 2 ϕ u n j L 6 u n j L 3 u n j u L 2 C ϕ u n j D 1 , 2 u n j L 3 u n j u L 2 C 2 u n j L 12 / 5 2 u n j L 3 u n j u L 2 C 2 C 12 / 5 2 C 3 C 2 u n j 3 u n j u 2 C 2 C 12 / 5 2 C 3 C 2 d 4 < .

Then by Lebesgue’s convergence theorem, we have

R 3 ϕ u n j u n j ( u n j u ) d x 0

as n j , which together with (34) implies

R 3 ( ϕ u n j u n j ϕ u u ) ( u n j u ) d x 0 (38)

as n j . By Lemma 2.2 and (32), we get

| R 3 ( f ( x , u n j ) f ( x , u ) ) ( u n j u ) d x | f ( x , u n j ) f ( x , u ( x ) ) L 2 u n j u L 2 C 2 f ( x , u n j ) f ( x , u ) L 2 u n j u 2 C 2 d f ( x , u n j ) f ( x , u ) L 2 0

as n j . Moreover, an easy computation shows that

Φ ( u n j ) Φ ( u ) , u n j u = u n j u 2 + R 3 ( ϕ u n j u n j ϕ u u ) ( u n j u ) d x R 3 ( f ( x , u n j ) f ( x , u ) ) ( u n j u ) d x .

Consequently, u n j u 0 as n j . Φ satisfies the ( PS ) c condition. □

Remark 3.2 Under conditions ( V 1 ), ( W 1 ), ( W 2 ), ( W 3 ), Φ satisfies the ( PS ) condition. Assume that { u n } E is a sequence such that I ( u n ) is bounded and

I ( u n ) 0

as n . Then there exists σ > 0 such that

| I ( u n ) | σ , I ( u n ) E σ

for all n N . The rest of the proof is the same as that of Lemma 3.1.

Proof of Theorem 1.2 For any k N , we take k disjoint open sets { Ω i | i = 1 , , k } such that

i = 1 k Ω i Ω .

For any ε > 0 and Ω i , there exist a closed set H i and an open set G i such that H i Ω i G i and

meas { G i Ω i } < ε , meas { Ω i H i } < ε .

For every G i ( i = 1 , , k ), by Lemma 2.3 there exists φ i C 0 ( G i , R ) such that φ i | H i = 1 and 0 φ i 1 . Letting v i = φ i φ i , can be extended to be a basis { v n } X . Therefore X = j 1 X j ¯ , where X j = span { v j } . Now we define Y k : = j = 1 k X j , Z k : = j k X j ¯ .

By Lemma 3.1, Φ C 1 ( E , R ) satisfies the ( PS ) c condition and Φ ( u ) = Φ ( u ) . Hence, to prove Theorem 1.2, we should just show that Φ has the geometric property (i), (ii) and (iii) in Lemma 2.7.

(i) By Lemma 2.4

β k ( a , r ) = sup u Z k , u = 1 u L a r 0

as k for r ( 1 , 2 ) and a L 2 2 r ( R 3 ) . In view of (30) and the fact that R 3 ϕ u u 2 d x 0 , we have

Φ ( u ) = 1 2 u 2 + 1 4 R 3 ϕ u u 2 d x R 3 F ( x , u ) d x 1 2 | u | 2 R 3 F ( x , u ) d x 1 2 u 2 2 r 1 R 3 a 1 ( x ) | u | r 1 d x 2 r 2 R 3 a 2 ( x ) | u | r 2 d x 2 r 3 δ r 3 1 R 3 b M ( x ) | u | r 3 d x 1 2 u 2 2 u L a 1 r 1 r 1 r 1 2 u L a 2 r 2 r 2 r 2 2 u L a 3 r 3 r 3 r 3 δ r 3 1 1 2 u 2 2 β k ( a 1 , r 1 ) r 1 r 1 u r 1 2 β k ( a 2 , r 2 ) r 2 r 2 u r 2 2 β k ( b M , r 3 ) r 3 r 3 δ r 3 1 u r 3 . (39)

Let r : = min { r 1 , r 2 , r 3 } , β k : = max { β k ( a 1 , r 1 ) , β k ( a 2 , r 2 ) , β k ( b M , r 3 ) } , C : = max { 2 r 1 , 2 r 2 , 2 r 3 δ r 3 1 } , then β k 0 as k . Hence, we have

Φ ( u ) 1 2 u 2 3 C β k r u r (40)

when u 1 and β k 1 . Now we can choose ρ k = ( 12 β k r C ) 1 / ( 2 r ) , then ρ k 0 as k . When k is large enough, we have ρ k 1 , β k 1 , which together with (40) shows

a k : = inf u Z k , u = ρ k Φ ( u ) 1 4 ρ k 2 > 0 .

(ii) For any u Y k , there exists λ i = 1 , 2 , , k such that

u = i = 1 k λ i v i .

Then we have

u L r 4 r 4 = R 3 | u ( x ) | r 4 d x = i = 1 k | λ i | r 4 Ω i | v i ( x ) | r 4 d x + i = 1 k | λ i | r 4 G i Ω i | v i ( x ) | r 4 d x = i = 1 k | λ i | r 4 Ω i | v i ( x ) | r 4 d x + i = 1 k | λ i | r 4 G i Ω i | φ i ( x ) | r 4 φ i r 4 d x i = 1 k | λ i | r 4 Ω i | v i ( x ) | r 4 d x + i = 1 k | λ i | r 4 φ i r 4 meas { G i Ω i } i = 1 k | λ i | r 4 Ω i | v i ( x ) | r 4 + i = 1 k | λ i | r 4 φ i r 4 ε (41)

and also

u 2 = R 3 [ | u | 2 + V ( x ) u 2 ] d x = i = 1 k λ i 2 G i [ | v i | 2 + V ( x ) v i 2 ] d x = i = 1 k λ i 2 v i 2 = i = 1 k λ i 2 . (42)

Since all the norms of a finite dimensional space are equivalent, there is a constant C ˜ such that

C ˜ u u L r 4

for all u Y k . By (30), one has

F ( x , λ i v i ) a 1 ( x ) r 1 | λ i v i | r 1 a 2 ( x ) r 2 | λ i v i | r 2 b M ( x ) r 3 δ r 3 1 | λ i v i | r 3 .

Therefore, we have

i = 1 k G i Ω i F ( x , λ i v i ) d x i = 1 k G i Ω i | λ i | r 1 r 1 a 1 ( x ) | v i | r 1 d x i = 1 k G i Ω i | λ i | r 2 r 2 a 2 ( x ) | v i | r 2 d x i = 1 k G i Ω i | λ i | r 3 r 3 δ r 3 1 b M ( x ) | v i | r 3 d x i = 1 k | λ i | r 1 r 1 a 1 L 2 2 r 1 ( G i Ω i | v i | 2 d x ) r 1 / 2 i = 1 k | λ i | r 2 r 2 a 2 L 2 2 r 2 ( G i Ω i | v i | 2 d x ) r 2 / 2 i = 1 k | λ i | r 3 r 3 δ r 3 1 b M L 2 2 r 3 ( G i Ω i | v i | 2 d x ) r 3 / 2 i = 1 k | λ i | r 1 r 1 a 1 L 2 2 r 1 ( G i Ω i | φ i | 2 φ i 2 d x ) r 1 / 2 i = 1 k | λ i | r 2 r 2 a 2 L 2 2 r 2 ( G i Ω i | φ i | 2 φ i 2 d x ) r 2 / 2 i = 1 k | λ i | r 3 r 3 δ r 3 1 b M L 2 2 r 3 ( G i Ω i | φ i | 2 φ i 2 d x ) r 3 / 2 = 1 r 1 a 1 L 2 2 r 1 i = 1 k | λ i | r 1 φ i r 1 ( meas { G i Ω i } ) r 1 / 2 1 r 2 a 2 L 2 2 r 2 i = 1 k | λ i | r 2 φ i r 2 ( meas { G i Ω i } ) r 2 / 2 1 r 3 δ r 3 1 b M L 2 2 r 3 i = 1 k | λ i | r 3 φ i r 3 ( meas { G i Ω i } ) r 3 / 2 1 r 1 a 1 L 2 2 r 1 i = 1 k | λ i | r 1 φ i r 1 ε r 1 / 2 1 r 2 a 2 L 2 2 r 2 i = 1 k | λ i | r 2 φ i r 2 ε r 2 / 2 1 r 3 δ r 3 1 b M L 2 2 r 3 i = 1 k | λ i | r 3 φ i r 3 ε r 3 / 2 . (43)

For any u Y k with u = i = 1 k λ i 2 = γ k , we can choose γ k small enough such that | λ i v i ( x ) | < ζ for all x R 3 and i = 1 , , k , which together with ( W 4