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

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.

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, [311]. Since (1) is set on R 3 , it is well known that the Sobolev embedding H 1 ( R 3 ) L s ( R 3 ) (2s 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 ) VC( 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)dy, 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 ) dx 1 4 R 3 | ϕ k | 2 dx+ 1 2 R 3 ϕ k u k 2 dx R 3 F(x, u k )dx 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 1.2 Assume that V satisfies

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

and f satisfies 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 xΩ and |u|ζ, where meas{Ω}>0, F(x,u):= 0 u f(x,y)dy;

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

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

as k.

By Theorem 1.2, we obtain the following corollary.

Corollary 1.3 Assume that L satisfies ( V 1 ) and W satisfies

( W 6 ) F(x,u)=a(x) | u | r , where F(x,u)= 0 u f(x,y)dy, 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 xΩ, 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 ) dx 1 4 R 3 | ϕ k | 2 dx+ 1 2 R 3 ϕ k u k 2 dx R 3 F(x, u k )dx 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 1.5 Assume that L satisfies ( V 1 ) and W satisfies ( 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 [216]. 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 n3. 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 [317].

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 ) dx

and the corresponding norm

u 2 =(u,u).

Note that the following embeddings

E L s ( R 3 ) ,2s 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 uE. Here L p ( R 3 ) (2p 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 2.1 Suppose that assumption ( V 1 ) holds. Then the embedding of E in 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 KE, there exists a positive constant M 0 such that u M 0 for all uK. 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,vK, 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 uK,

| x | T ε a(x) | u u i | r dx<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 2.2 Assume that assumptions ( V 1 ), ( W 1 ), ( W 2 ) and ( W 3 ) hold and u n u in E. 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 xS. 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)dx,

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 uR. 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 nN 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 nN. 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 ) dx<, R 3 b M 2 (x) | u | 2 ( r 3 1 ) dx<.
(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 dx0
(19)

as n. Now we have proved the lemma. □

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

Lemma 2.3 Assume that G R 3 is an open set. Then, for any closed set HG, 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 xH and 0ϕ(x)1 for all xGH.

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)dx=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(xy) α d 0 4 (y)dy,

then φ(x)=1 for all xH 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 0φ(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 2.4 Suppose 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 uE such that

u k u

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

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]

Let E be 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 cR. 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 )0as  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

  1. (i)

    a k := inf u Z k , u = ρ k Φ(u)0,

  2. (ii)

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

  3. (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 dx+ 1 2 R 3 ϕ u 2 dx R 3 F(x,u)dx.
(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 uE, consider the linear functional T u : D 1 , 2 ( R 3 )R defined as

T u (v)= R 3 u 2 vdx.

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 uvdx

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 uE, 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 vdx= R 3 ϕ u vdx

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

R 3 ϕ u vdx= R 3 vΔ ϕ u dx

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 | dy

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

It follows from (23) that

R 3 ϕ u u 2 dx= R 3 ϕ u (Δ ϕ u )dx= R 3 | ϕ u | 2 dx,
(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 dx 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:ER 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 dx R 3 F(x,u)dx.
(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 uE, 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 ] dx.

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

Lemma 3.1 Under 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 )0as  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 uR and x R 3 , which together with R 3 ϕ u n j u n j 2 dx0 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 uE and a constant d>0 such that

sup n j N u n j d,ud
(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)dx0

as n j , which together with (34) implies

R 3 ( ϕ u n j u n j ϕ u u)( u n j u)dx0
(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 u0 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 nN. The rest of the proof is the same as that of Lemma 3.1.

Proof of Theorem 1.2 For any kN, 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.

  1. (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 dx0, 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 u1 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.
  1. (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 ) implies

F(x, λ i v i )η | λ i v i | r 4
(44)

for all x Ω i and i=1,,k. Combining (24), (41), (42), (43) and (44), 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 + C 4 u 4 i = 1 k G i F ( x , λ i v i ) d x 1 2 u 2 i = 1 k [ G i Ω i F ( x , λ i v i ) d x + Ω i F ( x , λ i v i ) d x ] 1 2 u 2 + C 4 u 4 + 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 η i = 1 k | λ i | r 4 Ω i | v i | r 4 d x = 1 2 u 2 + C 4 u 4 + 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 η ( u L r 4 r 4 i = 1 k | λ i | r 4 φ i r 4 ε ) 1 2 u 2 + C 4 u 4 η C ˜ r 4 u r 4 + 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 + η i = 1 k | λ i | r 4 φ i r 4 ε = 1 2 i = 1 k λ i 2 + C 4 ( i = 1 k λ i 2 ) 2 η C ˜ r 4 ( i = 1 k λ i 2 ) r 4 / 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 + η i = 1 k | λ i | r 4 φ i r 4 ε = 1 2 γ k 2 + C 4 γ k 4 η ( C ˜ γ k ) r 4 + 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 + η i = 1 k | λ i | r 4 φ i r 4 ε γ k 2 + C 4 γ k 4 η ( C ˜ γ k ) r 4

for all u Y k with u= γ k , when ε and γ k are both small enough. Since r 4 <2, we can choose γ k < ρ k small enough such that

b k := max u Y k , u = γ k Φ(u)<0.
  1. (iii)

    By (40), for any u Z k with u= ρ k , we have

    Φ(u)3 C β k r u r .

Therefore

0 inf u Z k , u ρ k Φ(u)3 C β k r ρ k r .

Since β k , ρ k 0 as k, we have

d k := inf u Z k , u ρ k Φ(u)0

as k.

Hence, by Lemma 2.7, we obtain that problem (1) has infinitely many solutions {( u k , ϕ k )} satisfying

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

as k. □

Proof of Theorem 1.5 Similar to (31), there exist constants k i >0, i=1,2,3, such that

Φ(u) 1 2 u 2 i = 1 3 k i u r i
(45)

for all uE. Since 1< r i <2, it follows from (45) that the functional Φ is bounded from below. By Lemma 2.5 and Remark 3.2, Φ possesses a critical point u satisfying

Φ(u)= inf E Φ, Φ (u)=0.

It remains to show that u is nontrivial. For every ε>0, there exist an open set G and a closed set H such that HΩG and

meas{GΩ}<ε,meas{ΩH}<ε.

By Lemma 2.3, there exists a function φ C 0 ( R 3 ) such that 0φ(x)1 and φ | H (x)=1, φ | R G (x)=0, then φE. Choosing 0<λ<min{δ,ζ}, then |λφ(x)|<δ for all x R 3 , which together with (28) shows

F ( x , λ φ ( x ) ) a 1 ( x ) r 1 | λ φ ( x ) | r 1

for all x R 3 . Therefore, one has

G H F ( x , λ φ ) d x G H λ r 1 r 1 a 1 ( x ) φ r 1 d x λ r 1 r 1 ( G H a 1 ( x ) 2 2 r 1 d x ) ( 2 r 1 ) / 2 ( G H φ 2 d x ) r 1 / 2 λ r 1 r 1 a 1 L 2 2 r 1 ( G H 1 d x ) r 1 / 2 λ r 1 r 1 a 1 L 2 2 r 1 ( meas { G H } ) r 1 / 2 λ r 1 r 1 a 1 L 2 2 r 1 ( 2 ε ) r 1 / 2 .
(46)

In view of λ<ζ, we have |λφ(x)|<ζ for all x R 3 , which together with ( W 4 ) implies

F(x,λφ)η | λ φ | r 4
(47)

for all xΩ. It follows from (24), (46), (47) that

Φ ( λ φ ) = λ 2 2 φ 2 + 1 4 R 3 ϕ λ φ ( λ φ ) 2 d x R 3 F ( x , λ φ ) d x λ 2 2 φ 2 + C λ 4 φ 4 R 3 F ( x , λ φ ) d x λ 2 2 φ 2 + C λ 4 φ 4 G F ( x , λ φ ) d x = λ 2 2 φ 2 + C λ 4 φ 4 [ H F ( x , λ φ ) d x + G H F ( x , λ φ ) d x ] λ 2 2 φ 2 + C λ 4 φ 4 λ r 4 H η | φ | r 4 d x + λ r 1 r 1 a 1 L 2 2 r 1 ( 2 ε ) r 1 / 2 λ 2 φ 2 + C λ 4 φ 4 λ r 4 η meas { H } < 0

when ε and λ are both small enough. Since Φ(0)=0, then u0. Hence, (u, ϕ u ) is a nontrivial solution of problem (1). □

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Acknowledgements

The author is highly grateful for the referees’ careful reading and comments on this paper. This work is partially supported by the National Natural Science Foundation of China (No. 11071198) and Southwest University Doctoral Fund Project (SWU112107).

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Lv, Y. Existence and multiplicity of solutions for a class of sublinear Schrödinger-Maxwell equations. Bound Value Probl 2013, 177 (2013). https://doi.org/10.1186/1687-2770-2013-177

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