In this paper, we intend to consider a kind of nonlinear Klein-Gordon equation coupled with Born-Infeld theory. By using critical point theory and the method of Nehari manifold, we obtain two existing results of infinitely many high-energy radial solutions and a ground-state solution for this kind of system, which improve and generalize some related results in the literature.
In this paper, we intend to consider the following Klein-Gordon equation coupled with Born-Infeld theory:
( − Δ u + V ( x ) u − ( 2 ω + ϕ ) ϕ u = f ( u ) , in ℝ 3 , Δ ϕ + β Δ 4 ϕ = 4 π ( ω + ϕ ) u 2 , in ℝ 3 , (1.1)
where β > 0 , ω > 0 , u and ϕ are unknowns, V : ℝ 3 → ℝ is a potential function and f satisfies some superlinear conditions. The Born-Infeld electromagnetic theory [
In recent years, some researchers considered the Klein-Gordon equation coupled with Born-Infeld theory by using variational methods. We recall some of them as follows.
In [
( − Δ u + ( m 2 − w 2 ) u − ( 2 ω + ϕ ) ϕ u = | u | p − 2 u , in ℝ 3 , Δ ϕ + β Δ 4 ϕ = 4 π ( ω + ϕ ) u 2 , in ℝ 3 , (1.2)
when p ∈ ( 4,6 ) and 0 < ω < m , they obtained some existing results of infinitely many radially symmetric solutions for system (1.2). After this, Mugnai [
( − Δ u + ( m 2 − w 2 ) u − ( 2 ω + ϕ ) ϕ u = | u | p − 2 u + | u | 2 ∗ − 2 u , in ℝ 3 , Δ ϕ + β Δ 4 ϕ = 4 π ( ω + ϕ ) u 2 , in ℝ 3 , (1.3)
they admits a nontrivial solution for problem (1.3) when m > ω > 0 or ( 1 2 − 1 p ) m 2 > 1 2 ω 2 .
Chen and Li [
1) | m | > ω > 0 , 4 < p < 6 ; 2) p 2 − 1 | m | > ω > 0 , 2 < p ≤ 4
In [
( − Δ u + a ( x ) u − ( 2 ω + ϕ ) ϕ u = λ k ( x ) | u | q − 2 u + g ( x ) | u | p − 2 , in ℝ 3 , Δ ϕ + β Δ 4 ϕ = 4 π ( ω + ϕ ) u 2 , in ℝ 3 , (1.4)
where 1 < q < 2 < p < 6 , a, k, and g satisfy some appropriate assumptions.
In [
( − Δ u + ( m 2 − w 2 ) u − ( 2 ω + ϕ ) ϕ u = μ | u | p − 2 u + | u | 2 ∗ − 2 u , in ℝ 3 , Δ ϕ + β Δ 4 ϕ = 4 π ( ω + ϕ ) u 2 , in ℝ 3 , (1.5)
they showed that problem (1.5) has at least a nontrivial radial ground-state solution, under one of the following conditions:
1) p ∈ ( 4,6 ) and m > ω > 0 for μ > 0 ;
2) p ∈ ( 3,4 ] and m > ω > 0 for sufficient large μ > 0 ;
3) p ∈ ( 2,3 ] and ( p − 2 ) ( 4 − p ) | m | > ω > 0 for sufficient large μ > 0 .
Wen, Tang and Chen [
( − Δ u + V ( x ) u − ( 2 ω + ϕ ) ϕ u = f ( x , u ) , in ℝ 3 , Δ ϕ + β Δ 4 ϕ = 4 π ( ω + ϕ ) u 2 , in ℝ 3 , (1.6)
they obtained infinitely many solutions and a least energy solution for problem (1.6) under different assumptions on V and f. In [
Motivated by the above works, in this paper, to certify the boundedness of Palais-Smale sequence for case of 2 < u < 6 , we use Pohožaev identity of (1.1). By applying the ideas employed by Ref. [
In this paper, we make the following assumptions:
(V1) V ∈ C ( ℝ 3 , ℝ ) is a radial function, which satisfies V 0 = inf x ∈ ℝ 3 V ( x ) > 0 . And there is a constant r > 0 such that:
l i m | y | → + ∞ meas ( { x ∈ ℝ 3 : | x − y | ≤ r , V ( x ) ≤ M } ) = 0, ∀ M > 0,
where meas ( ⋅ ) denotes the Lebesgue measure.
(V2) ( ∇ V ( x ) , x ) ≥ 0 for all x ∈ ℝ 3 and there exists θ ∈ [ 0,1 ) such that ( ∇ V ( x ) , x ) ≤ θ 2 | x | 2 for all x ∈ ℝ 3 \ { 0 } .
(F1) l i m | t | → ∞ f ( t ) | t | 5 = 0 .
(F2) l i m | t | → 0 f ( t ) | t | = 0 .
(F3) There exists μ > 2 such that f ( t ) t ≥ μ F ( t ) ≥ 0 for all t ∈ ℝ , where F ( t ) : = ∫ 0 t f ( s ) d s .
(F4) f ( − t ) = − f ( t ) , for all t ∈ ℝ .
Now, we present two main results:
Theorem 1.1. Assume that (V1), (V2), (F1) - (F4) hold. If the following condition holds:
1) μ ∈ [ 3,6 ) ; or 2) μ ∈ ( 2,3 ) and ω ∈ ( 0, ( μ − 2 ) ( 4 − μ ) V 0 / ( 3 − μ ) ) , then, problem (1.1) possesses infinitely many solutions { ( u n , ϕ n ) } ⊂ E × D r 1,2 ( ℝ 3 ) satisfying:
1 2 ∫ ℝ 3 ( | ∇ u | 2 + [ V ( x ) − ( 2 ω + ϕ ) ϕ ] u 2 ) d x − 1 8 π ∫ ℝ 3 | ∇ ϕ | 2 d x − β 16 π ∫ ℝ 3 | ∇ ϕ | 4 d x − ∫ ℝ 3 F ( u ) d x → ∞ .
Theorem 1.2. Assume that (V1), (V2), (F1) - (F3) hold. If the following condition holds:
1) μ ∈ [ 3,6 ) ; or 2) μ ∈ ( 2,3 ) and ω ∈ ( 0, ( μ − 2 ) ( 4 − μ ) V 0 / ( 3 − μ ) ) , then, the problem (1.1) has a ground-state solution.
Remark 1.3. We consider the variable potential V and generalized nonlinearity f, which brings some difficulties such as the proof of boundedness of Palais-Smale sequence ((PS)-sequence for short). To conquer the boundedness of (PS)-sequence, we use some analytical methods. Besides, when μ ∈ [ 3,6 ) , we do not need any restriction on ω , and when μ ∈ ( 2,3 ) , we get a more delicate range for ω . Hence, Theorem 1.1 and Theorem 1.2 can be seen as improvements of the relative results in the literature. To the best of our knowledge, similar results for this kind of Klein-Gordon equation coupled with Born-Infeld theory by using analytical methods in this paper can not be found.
The rest of this paper is organized as follows: in Section 2, some preliminaries are given; in Section 3, we give out the proofs of Theorem 1.1 and Theorem 1.2. We denote C i as different positive constants.
Henceforth, the following notations will be used.
〈 , 〉 denote dual inner products between workspaces.
⇀ denote weak convergence.
d i s t ( x , y ) denote Euclidean distance between x and y.
∂ S denote boundary of S.
a.e. almost everywhere.
ℝ N denote N-dimensional Euclidean space.
X : = Y denote define X as Y.
C , C 1 , C 2 , ⋯ denote various positive constants.
B r ( x ) : = { y ∈ ℝ : | y − x | < r } , ∀ u ∈ H 1 ( ℝ N ) , r > 0 .
u t ( x ) : = u ( x / t ) , ∀ u ∈ H 1 ( ℝ N ) \ { 0 } , t > 0 .
D ( ℝ 3 ) denote the complete space of C 0 ∞ ( ℝ 3 ) .
D r ( ℝ 3 ) : = { u ∈ D ( ℝ 3 ) : u ( x ) = u ( | x | ) } .
D 1,2 ( ℝ 3 ) : = { u ∈ L 6 ( ℝ 3 ) : | ∇ u | ∈ L 2 ( ℝ 3 ) } .
D r 1,2 ( ℝ 3 ) : = { u ∈ D 1,2 ( ℝ 3 ) : u ( x ) = u ( | x | ) } .
H 1 ( ℝ 3 ) : = { u ∈ L 2 ( ℝ 3 ) : | ∇ u | ∈ L 2 ( ℝ 3 ) } .
H r 1 ( ℝ 3 ) : = { u ∈ H 1 ( ℝ 3 ) : u ( x ) = u ( | x | ) } .
‖ u ‖ D = ( ∫ ℝ 3 | ∇ u | 2 d x ) 1 2 + ( ∫ ℝ 3 | ∇ u | 4 d x ) 1 4 .
‖ u ‖ D r 1 , 2 = ( ∫ ℝ 3 | ∇ u | 2 d x ) 1 2 .
‖ u ‖ s = ( ∫ ℝ N | u | s d x ) 1 / s , 1 < s < ∞ .
We define:
E : = { u ∈ H r 1 ( ℝ 3 ) : ∫ ℝ 3 ( | ∇ u | 2 + V ( x ) u 2 ) d x < ∞ } .
Then, E is a Hilbert space with the inner product:
( u , v ) E = ∫ ℝ 3 [ ∇ u ∇ v + V ( x ) u v ] d x
and the norm ‖ u ‖ : = ‖ u ‖ E = ( u , u ) E 1 / 2 . By (V1), (V2) and Poincaré inequality, we see that E ↪ H r 1 ( ℝ 3 ) is continuous. Then, for p ∈ [ 2,6 ] , there exists r p > 0 such that:
‖ u ‖ L p : = ( ∫ ℝ 3 | u | p d x ) 1 p ≤ r p ‖ u ‖ E , u ∈ E . (2.1)
Apparently, we know that a solution ( u , ϕ ) ∈ E × D r 1,2 ( ℝ 3 ) for the system (1.1) is a critical point of the energy functional J : ( u , ϕ ) ∈ E × D r 1,2 ( ℝ 3 ) → ℝ defined as:
J ( u , ϕ ) = 1 2 ∫ ℝ 3 [ | ∇ u | 2 + V ( x ) u 2 − ( 2 ω + ϕ ) ϕ u 2 ] d x − ∫ ℝ 3 F ( u ) d x − 1 8 π ∫ ℝ 3 | ∇ ϕ | 2 d x − β 16 π ∫ ℝ 3 | ∇ ϕ | 4 d x . (2.2)
We need the following lemma to reduce the functional J in the only variable u.
Lemma 2.1. [
1) There exists a unique ϕ = ϕ u ∈ D ( ℝ 3 ) , which solves equation:
Δ ϕ + β Δ 4 ϕ = 4 π ( ω + ϕ ) u 2 ;
2) − ω ≤ ϕ u ≤ 0 on the set { x : u ( x ) ≠ 0 } ;
3) If u is radially symmetric, then ϕ u is also radially symmetric;
4) u n ⇀ u in H 1 ( ℝ 3 ) ,then ϕ u n ⇀ ϕ u in D r ( ℝ 3 ) .
From the second equation in system (1.1), we get:
1 4 π ∫ ℝ 3 | ∇ ϕ | 2 d x + β 4 π ∫ ℝ 3 | ∇ ϕ | 4 d x = − ∫ ℝ 3 ( ω ϕ u + ϕ u 2 ) u 2 d x . (2.3)
From Lemma 2.1, we rewrite J ( u , ϕ ) as the functional I ( u ) : E → ℝ as follows:
I ( u ) = 1 2 ∫ ℝ 3 [ | ∇ u | 2 + V ( x ) u 2 − ( 2 ω + ϕ u ) ϕ u u 2 ] d x − ∫ ℝ 3 F ( u ) d x − 1 8 π ∫ ℝ 3 | ∇ ϕ u | 2 d x − β 16 π ∫ ℝ 3 | ∇ ϕ u | 4 d x . (2.4)
By (2.3) and (2.4), we obtain:
I ( u ) = 1 2 ∫ ℝ 3 [ | ∇ u | 2 + V ( x ) u 2 + ϕ u 2 u 2 ] d x − ∫ ℝ 3 ( ω ϕ u + ϕ u 2 ) u 2 d x − 1 8 π ∫ ℝ 3 | ∇ ϕ u | 2 d x − β 16 π ∫ ℝ 3 | ∇ ϕ u | 4 d x − ∫ ℝ 3 F ( u ) d x = 1 2 ∫ ℝ 3 [ | ∇ u | 2 + V ( x ) u 2 + ϕ u 2 u 2 ] d x + 1 8 π ∫ ℝ 3 | ∇ ϕ u | 2 d x + 3 β 16 π ∫ ℝ 3 | ∇ ϕ u | 4 d x − ∫ ℝ 3 F ( u ) d x = 1 2 ∫ ℝ 3 [ | ∇ u | 2 + V ( x ) u 2 − ω ϕ u u 2 ] d x + β 16 π ∫ ℝ 3 | ∇ ϕ u | 4 d x − ∫ ℝ 3 F ( u ) d x . (2.5)
For any u , v ˜ ∈ E , we have:
〈 I ′ ( u ) , v ˜ 〉 = ∫ ℝ 3 [ ∇ u ∇ v ˜ + V ( x ) u v ˜ − ( 2 ω + ϕ u ) ϕ u u v ˜ ] d x − ∫ ℝ 3 f ( u ) v ˜ d x . (2.6)
For λ ∈ [ 1 / 2 ,1 ] , we define the family of functionals I λ : E → ℝ by:
I λ ( u ) = 1 2 ∫ ℝ 3 [ | ∇ u | 2 + V ( x ) u 2 − ( 2 ω + ϕ u ) ϕ u u 2 ] d x − λ ∫ ℝ 3 F ( u ) d x − 1 8 π ∫ ℝ 3 | ∇ ϕ u | 2 d x − β 16 π ∫ ℝ 3 | ∇ ϕ u | 4 d x . (2.7)
For any u , v ˜ ∈ E , we also have:
〈 I ′ λ ( u ) , v ˜ 〉 = ∫ ℝ 3 [ ∇ u ∇ v ˜ + V ( x ) u v ˜ − ( 2 ω + ϕ u ) ϕ u u v ˜ ] d x − λ ∫ ℝ 3 f ( u ) v ˜ d x . (2.8)
Let S : = { u ∈ H r 1 ( ℝ 3 ) \ { 0 } : I ′ ( u ) = 0 } be the critical points set. It is easy to see that any critical point u of I satisfies the following Pohožaev equality:
P ( u ) : = ∫ ℝ 3 | ∇ u | 2 d x + ∫ ℝ 3 [ 3 V ( x ) + ( ∇ V ( x ) , x ) − 5 ω ϕ u − 2 ϕ u 2 ] u 2 d x + 3 β 8 π ∫ ℝ 3 | ∇ ϕ u | 4 d x − 6 ∫ ℝ 3 F ( u ) d x = 0. (2.9)
For convenience, we also defined:
P λ ( u ) : = ∫ ℝ 3 | ∇ u | 2 d x + ∫ ℝ 3 [ 3 V ( x ) + ( ∇ V ( x ) , x ) − 5 ω ϕ u − 2 ϕ u 2 ] u 2 d x + 3 β 8 π ∫ ℝ 3 | ∇ ϕ u | 4 d x − 6 λ ∫ ℝ 3 F ( u ) d x = 0. (2.10)
Let:
G ( u ) : = 〈 I ′ ( u ) , u 〉 − 1 2 P ( u ) = 1 2 ∫ ℝ 3 | ∇ u | 2 d x − 1 2 ∫ ℝ 3 [ V ( x ) + ( ∇ V ( x ) , x ) − ω ϕ u ] u 2 d x − 3 β 16 π ∫ ℝ 3 | ∇ ϕ u | 4 d x + ∫ ℝ 3 [ 3 F ( u ) − f ( u ) u ] d x . (2.11)
Then, G ( u ) = 0 for any u ∈ S . We also define:
G λ ( u ) : = 〈 I ′ λ ( u ) , u 〉 − 1 2 P λ ( u ) = 1 2 ∫ ℝ 3 | ∇ u | 2 d x − 1 2 ∫ ℝ 3 [ V ( x ) + ( ∇ V ( x ) , x ) − ω ϕ u ] u 2 d x − 3 β 16 π ∫ ℝ 3 | ∇ ϕ u | 4 d x + λ ∫ ℝ 3 [ 3 F ( u ) − f ( u ) u ] d x . (2.12)
Lemma 2.2. Assume that (F1) - (F3) hold. Then, there exist some constants ζ λ , α λ > 0 , t λ > 0 and v λ = t λ u , λ ∈ [ 1 2 ,1 ] (see [
1) inf ‖ u ‖ ≤ ζ λ I λ ( u ) ≥ 0 and inf ‖ u ‖ = ζ λ I λ ( u ) ≥ α λ ;
2) ‖ v λ ‖ E > ζ λ and I λ ( v λ ) < 0 .
Proof. 1) From (F1) and (F2), for C 1 = 1 6 r 2 2 , there exists C 2 > 0 such that:
F ( t ) ≤ C 1 | t | 2 + C 2 | t | 6 , for all t ∈ ℝ . (2.13)
By (2.1), (2.3), (2.7), (2.13), − ω ≤ ϕ u ≤ 0 and Hölder inequality, we have:
I λ ( u ) = 1 2 ∫ ℝ 3 [ | ∇ u | 2 + V ( x ) u 2 − ( 2 ω + ϕ u ) ϕ u u 2 ] d x − 1 8 π ∫ ℝ 3 | ∇ ϕ u | 2 d x − β 16 π ∫ ℝ 3 | ∇ ϕ u | 4 d x − λ ∫ ℝ 3 F ( u ) d x = 1 2 ∫ ℝ 3 | ∇ u | 2 d x + 1 2 ∫ ℝ 3 V ( x ) u 2 d x − 1 2 ∫ ℝ 3 ω ϕ u u 2 d x + β 16 π ∫ ℝ 3 | ∇ ϕ u | 4 d x − λ ∫ ℝ 3 F ( u ) d x
≥ 1 2 ∫ ℝ 3 | ∇ u | 2 d x + 1 2 ∫ ℝ 3 V ( x ) u 2 d x − λ ∫ ℝ 3 F ( u ) d x ≥ 1 2 ∫ ℝ 3 | ∇ u | 2 d x + 1 2 ∫ ℝ 3 V ( x ) u 2 d x − λ ∫ ℝ 3 ( C 1 | u | 2 + C 2 | u | 6 ) d x ≥ 1 2 ‖ u ‖ E 2 − C 1 λ ‖ u ‖ L 2 2 − C 2 λ ‖ u ‖ L 6 6 ≥ 1 2 ‖ u ‖ E 2 − 1 6 ‖ u ‖ E 2 − λ C 2 r 6 6 ‖ u ‖ E 6 ≥ 1 3 ‖ u ‖ E 2 − λ C 2 r 6 6 ‖ u ‖ E 6 . (2.14)
From (2.14), there exist ζ λ , α λ > 0 such that inf ‖ u ‖ ≤ ζ λ I λ ( u ) ≥ 0 and inf ‖ u ‖ = ζ λ I λ ( u ) ≥ α λ .
2) From (F2) and (F3), there exists C 3 , C 4 > 0 such that:
F ( t ) ≥ C 3 | t | μ − C 4 | t | 2 , for all t ∈ ℝ . (2.15)
From (2.15), for u ∈ E \ { 0 } , we get:
I λ ( t λ u ) = t λ 2 2 ∫ ℝ 3 [ | ∇ u | 2 + V ( x ) u 2 − ( 2 ω + ϕ u ) ϕ u u 2 ] d x − 1 8 π ∫ ℝ 3 | ∇ ϕ t λ u | 2 d x − β 16 π ∫ ℝ 3 | ∇ ϕ t λ u | 4 d x − λ ∫ ℝ 3 F ( t λ u ) d x ≤ t λ 2 2 ∫ ℝ 3 [ | ∇ u | 2 + V ( x ) u 2 ] d x + ω 2 t λ 2 ∫ ℝ 3 u 2 d x − λ ∫ ℝ 3 F ( t λ u ) d x ≤ t λ 2 2 ∫ ℝ 3 [ | ∇ u | 2 + V ( x ) u 2 ] d x + ω 2 t λ 2 ∫ ℝ 3 u 2 d x − C 3 t λ μ λ ∫ ℝ 3 | u | μ d x + C 4 t λ 2 λ ∫ ℝ 3 u 2 d x → − ∞ , as t λ → ∞ . (2.16)
Hence, from (2.16), we can let v λ = t λ u with t λ > 0 large enough such that ‖ v λ ‖ E > ζ λ and I λ ( v λ ) < 0 .£
Lemma 2.3. Assume that (V1) and (F1) - (F3) hold. Let { u n } ⊂ E be a bounded (PS)c-sequence for I with c ∈ ( 0, ∞ ) , then { u n } has a strongly convergent subsequence in E.
Proof. Consider a sequence { u n } in E, which satisfies:
I ( u n ) → c , I ′ ( u n ) → 0. (2.17)
We may assume that, for any n ∈ ℕ , there exists a u ∈ E such that:
· u n ⇀ u in E;
· u n → u in L p ( ℝ 3 ) , for 2 < p < 6 ;
· u n → u a.e. in ℝ 3 .
By (2.6), we easily get:
‖ u n − u ‖ E = 〈 I ′ ( u n ) − I ′ ( u ) , u n − u 〉 + 2 ω ∫ ℝ 3 ( ϕ u n u n − ϕ u u ) ( u n − u ) d x + ∫ ℝ 3 ( ϕ u n 2 u n − ϕ u 2 u ) ( u n − u ) d x + ∫ ℝ 3 ( f ( u n ) − f ( u ) ) ( u n − u ) d x . (2.18)
It is clear that:
( I ′ ( u n ) − I ′ ( u ) , u n − u ) → 0, as n → ∞ . (2.19)
From (F1) and (F2), there exist C 6 , C 7 > 0 such that:
| f ( t ) | ≤ C 5 | t | + C 6 | t | 5 , ∀ t ∈ ℝ . (2.20)
By (2.20), one has:
∫ ℝ 3 [ f ( u n ) − f ( u ¯ ) ] ( u n − u ¯ ) d x ≤ ∫ ℝ 3 [ C 5 ( | u n | + | u ¯ | ) + C 6 ( | u n | 5 + | u ¯ | 5 ) ] | u n − u ¯ | d x ≤ C 5 ( ‖ u n ‖ L 2 + ‖ u ¯ ‖ L 2 ) ‖ u n − u ¯ ‖ L 2 + C 6 ( ‖ u n ‖ L 2 5 + ‖ u ¯ ‖ L 2 5 ) ‖ u n − u ¯ ‖ L 2 . (2.21)
By Lemma 2.1, Sobolev inequality and Hölder inequality, it easily gains that:
| 2 ∫ ℝ 3 ( ϕ u n u n − ϕ u u ) ( u n − u ) d x | ≤ | 2 ∫ ℝ 3 ϕ u ( u n − u ) ( u n − u ) d x | + | 2 ∫ ℝ 3 ( ϕ u n − ϕ u ) u n ( u n − u ) d x | ≤ ‖ ϕ u ‖ L 6 ⋅ ‖ u n − u ‖ L 3 ⋅ ‖ u n − u ‖ L 2 + ‖ ϕ u n − ϕ u ‖ L 6 ⋅ ‖ u n − u ‖ L 3 ⋅ ‖ u n ‖ L 2 ≤ ‖ ϕ u ‖ L 6 ⋅ ‖ u n − u ‖ L 3 ⋅ ‖ u n − u ‖ L 2 + C 5 ‖ ϕ u n − ϕ u ‖ D r 1,2 ⋅ ‖ u n − u ‖ L 3 ⋅ ‖ u n ‖ L 2 . (2.22)
From Lemma 2.1 and the boundeness of { u n } , there exists a positive constant C 6 such that:
‖ ϕ u n 2 u n ‖ L 3 2 ≤ ‖ ϕ u n ‖ L 6 2 ‖ u n ‖ L 3 ≤ C 6 . (2.23)
Hence, from (2.23), the sequence { ϕ u n 2 u n } is bounded in L 3 2 ( ℝ 3 ) , so that:
| ∫ ℝ 3 ( ϕ u n 2 u n − ϕ u 2 u ) ( u n − u ) d x | ≤ ‖ ϕ u n 2 u n − ϕ u 2 u ‖ L 3 2 ‖ u n i − u ‖ L 3 ≤ ( ‖ ϕ u n 2 u n ‖ L 3 2 + ‖ ϕ u 2 u ‖ L 3 2 ) ‖ u n − u ‖ L 3 . (2.24)
Since u n → u in L p ( ℝ 3 ) , for any 2 < p < 6 , from (2.21), (2.22) and (2.24), one has:
∫ ℝ 3 [ f ( u n ) − f ( u ) ] ( u n − u ) d x → 0 as n → ∞ . (2.25)
2 ω ∫ ℝ 3 ( ϕ u n u n − ϕ u u ) ( u n − u ) d x → 0, as n → ∞ , (2.26)
∫ ℝ 3 ( ϕ u n 2 u n − ϕ u 2 u ) ( u n − u ) d x → 0, as n → ∞ . (2.27)
From (2.18), (2.19), (2.25), (2.26) and (2.27), we have ‖ u n − u ‖ E → 0 , that is, u n → u in E. £
Lemma 2.4. [
Φ λ ( u ) = A ( u ) − λ B ( u ) ,
with B ( u ) nonnegative and either A ( u ) → + ∞ or B ( u ) → + ∞ , as ‖ u ‖ → + ∞ , and such that Φ λ ( 0 ) = 0 . For any λ ∈ J , we set:
Γ λ = { γ ∈ C ( [ 0 , 1 ] , X ) : γ ( 0 ) = 0 , Φ λ ( γ ( 1 ) ) ≤ 0 } .
If for every λ ∈ J , the set Γ λ is nonempty and
c λ : = inf γ ∈ Γ λ max t ∈ [ 0 , 1 ] Φ λ ( γ ( t ) ) > 0
then for almost every λ ∈ J , there is a sequence { ( u λ ) n } ⊂ X such that:
1) { ( u λ ) n } is bounded;
2) Φ λ ( ( u λ ) n ) → c λ ;
3) Φ ′ λ ( ( u λ ) n ) → 0 in the dual X − 1 of X.
Lemma 2.5. Γ λ is nonempty, where Γ λ is given by Lemma 2.4.
Proof. From (2.16) and u ∈ E \ { 0 } , we can choose T > 0 such that I λ ( T u ) < 0 . Let γ s ( t ) = T t u , t ∈ [ 0 , 1 ] , such that, γ s ( t ) ∈ C ( [ 0,1 ] , E ) , γ s ( 0 ) = 0 , I λ ( γ s ( 1 ) ) < 0 , and max t ∈ [ 0 , 1 ] I λ ( γ s ( t ) ) < ∞ , for any λ ∈ J . This means that Γ λ is nonempty. £
Lemma 2.6. c λ > 0 , where c λ is given by Lemma 2.4.
Proof. For any γ ∈ Γ λ and any λ ∈ J , we have γ ( 0 ) = 0 and I λ ( γ ( 1 ) ) < 0 . From Lemma 2.2, we get that ‖ γ ( 1 ) ‖ > ζ λ . By continuity, we deduce that there exists t γ ∈ ( 0,1 ) such that ‖ γ ( t γ ) ‖ = ζ λ . From Lemma 2.2, we have I λ ( γ ( t γ ) ) ≥ α λ . Therefore, we have:
∞ > c λ ≥ inf γ ∈ Γ I λ ( γ ( t γ ) ) ≥ α λ > 0.
£
Lemma 2.7. Assume that (V1), (V2) and (F1) - (F3) hold. Then, there exists a sequence { u n } satisfying:
I ( u n ) → c 1 , I ′ ( u n ) → 0. (2.28)
is a bounded ( P S ) c 1 -sequence with c 1 ∈ ( 0, ∞ ) .
Proof. From (2.3) and (2.7), we get that:
I λ ( u ) = 1 2 ∫ ℝ 3 | ∇ u | 2 d x + 1 2 ∫ ℝ 3 V ( x ) u 2 d x − 1 2 ∫ ℝ 3 ω ϕ u u 2 d x + β 16 π ∫ ℝ 3 | ∇ ϕ u | 4 d x − λ ∫ ℝ 3 F ( u ) d x . (2.29)
From Lemma 2.2, we see that I λ has mountain pass geometry. We can define the Mountain Pass level c λ by:
c λ : = inf γ ∈ Γ λ max t ∈ [ 0,1 ] I λ ( γ ( t ) ) (2.30)
where
Γ λ = { γ ∈ C ( [ 0 , 1 ] , E ) : γ ( 0 ) = 0 , γ ( 1 ) = v λ } .
From Lemma 2.6, we have the estimate c 1 ∈ ( 0, ∞ ) , set X = E , J = [ 1 2 , 1 ] , Φ λ = I λ ,
A ( u ) = 1 2 ∫ ℝ 3 | ∇ u | 2 d x + 1 2 ∫ ℝ 3 V ( x ) u 2 d x − 1 2 ∫ ℝ 3 ω ϕ u u 2 d x + β 16 π ∫ ℝ 3 | ∇ ϕ u | 4 d x , B ( u ) = ∫ ℝ 3 F ( u ) d x .
It is easy to know that B ( u ) ≥ 0 for every u ∈ E and A ( u ) → ∞ when ‖ u ‖ → ∞ . Thus, from Lemma 2.4 and Lemma 2.5, for almost every λ ∈ J , there is a sequence { ( u λ ) n } ⊂ E such that:
1) { ( u λ ) n } is bounded in E;
2) I λ ( ( u λ ) n ) → c λ ;
3) I ′ λ ( ( u λ ) n ) → 0 in the dual E − 1 of E.
Since c λ ∈ ( 0, ∞ ) , there exists u λ ∈ E satisfying:
I ′ λ ( u λ ) = 0 , I λ ( u λ ) = c λ
for almost every λ ∈ J . We can choose a suitable λ n → 1 and u λ n such that:
I ′ λ n ( u λ n ) = 0 , I λ n ( u λ n ) = c λ n → c 1 . (2.31)
We still denote u λ n by u n . From (2.8) and I ′ λ n ( u n ) = 0 , we have:
〈 I ′ λ n ( u n ) , u n 〉 = ∫ ℝ 3 [ | ∇ u n | 2 + V ( x ) u n 2 − ( 2 ω + ϕ u n ) ϕ u n u n 2 ] d x − λ ∫ ℝ 3 f ( u n ) u n d x = 0, (2.32)
and from (2.12), one has that:
G λ n ( u n ) = 〈 I ′ λ n ( u n ) , u n 〉 − 1 2 P λ n ( u n ) = 1 2 ∫ ℝ 3 | ∇ u n | 2 d x − 1 2 ∫ ℝ 3 [ V ( x ) + ( ∇ V ( x ) , x ) − ω ϕ u n ] u n 2 d x − 3 β 16 π ∫ ℝ 3 | ∇ ϕ u n | 4 d x + λ ∫ ℝ 3 [ 3 F ( u n ) − f ( u n ) u n ] d x = 0. (2.33)
Next, we will prove that { u n } is bounded in E.
Case (1): 4 ≤ μ < 6 . By (V1), (F3), (2.1), (2.3), (2.28), (2.29), (2.31), (2.32), − ω ≤ ϕ u ≤ 0 and Hölder inequality, we have:
μ c 1 + o ( 1 ) ≥ μ I λ n ( u n ) − 〈 I ′ λ n ( u n ) , u n 〉 = ( μ 2 − 1 ) ∫ ℝ 3 [ | ∇ u n | 2 + V ( x ) u n 2 ] d x + ( μ 2 + 1 ) ∫ ℝ 3 ϕ u n 2 u n 2 d x + 2 ∫ ℝ 3 ω ϕ u n u n 2 d x + μ 8 π ∫ ℝ 3 | ∇ ϕ u n | 2 d x + 3 β μ 16 π ∫ ℝ 3 | ∇ ϕ u n | 4 d x + λ n ∫ ℝ 3 [ f ( u n ) u n − μ F ( u n ) ] d x
≥ ( μ 2 − 1 ) ∫ ℝ 3 [ | ∇ u n | 2 + V ( x ) u n 2 ] d x + 2 ∫ ℝ 3 ( ϕ u n 2 u n 2 + ω ϕ u n u n 2 ) d x + μ 8 π ∫ ℝ 3 | ∇ ϕ u n | 2 d x + 3 β μ 16 π ∫ ℝ 3 | ∇ ϕ u n | 4 d x + λ n ∫ ℝ 3 [ f ( u n ) u n − μ F ( u n ) ] d x = ( μ 2 − 1 ) ∫ ℝ 3 [ | ∇ u n | 2 + V ( x ) u n 2 ] d x + ( μ 8 π − 1 2 π ) ∫ ℝ 3 | ∇ ϕ u n | 2 d x + ( 3 β μ 16 π − β 2 π ) ∫ ℝ 3 | ∇ ϕ u n | 4 d x + λ n ∫ ℝ 3 [ f ( u n ) u n − μ F ( u n ) ] d x ≥ ( μ 2 − 1 ) ‖ u n ‖ E 2 . (2.34)
When 4 ≤ μ < 6 , from (2.34), we know that { u n } is bounded in E.
Case (2): 2 < μ < 4 . By (V1), (V2), (F3), (2.28), (2.29), (2.31), (2.32), (2.33) and − ω ≤ ϕ u ≤ 0 , we have:
c 1 + o ( 1 ) ≥ I λ n ( u n ) + μ − 4 6 − μ 〈 I ′ λ n ( u n ) , u n 〉 + 2 − μ 6 − μ G λ n ( u n ) = 1 6 − μ ∫ ℝ 3 [ ( μ − 2 ) V ( x ) + 2 ( 3 − μ ) ω ϕ u n + ( 4 − μ ) ϕ u n 2 ] | u n | 2 d x + 2 λ n 6 − μ ∫ ℝ 3 [ f ( u n ) u n − μ F ( u n ) ] d x + μ − 2 2 ( 6 − μ ) ∫ ℝ 3 ( ∇ V ( x ) , x ) | u n | 2 d x + 2 β μ 16 π ( 6 − μ ) ∫ ℝ 3 | ∇ ϕ u n | 4 d x . (2.35)
We will prove the boundedness of ∫ ℝ 3 V ( x ) | u n | 2 d x , to do this, we have two cases to consider.
Subcase (i): 3 ≤ μ < 4 . In this case, we have:
( 4 − μ ) s 2 + 2 ( 3 − μ ) ω s ≥ 0, ∀ − ω ≤ s ≤ 0. (2.36)
From (2.35), (2.36), (V2) and (F3), we have that ∫ ℝ 3 V ( x ) | u n | 2 d x is bounded.
Subcase (ii): μ ∈ ( 2,3 ) and ω ∈ ( 0, ( μ − 2 ) ( 4 − μ ) V 0 / ( 3 − μ ) ) . For s ∈ [ − ω ,0 ] , by a direct computation, we have that:
( 4 − μ ) s 2 + 2 ( 3 − μ ) ω s + ( μ − 2 ) V 0 ≥ − ( 3 − μ ) 2 4 − μ ω 2 + ( μ − 2 ) V 0 = ( μ − 2 ) ( 4 − μ ) V 0 − ( 3 − μ ) 2 ω 2 4 − μ > 0. (2.37)
Thus, from (2.35), (2.37), (V1), (V2) and (F3), we get:
c n + o ( 1 ) ≥ 1 6 − μ ∫ ℝ 3 [ ( μ − 2 ) V ( x ) + 2 ( 3 − μ ) ω ϕ u n + ( 4 − μ ) ϕ u n 2 ] | u n | 2 d x + μ − 2 2 ( 6 − μ ) ∫ ℝ 3 ( ∇ V ( x ) , x ) | u n | 2 d x + 2 β μ 16 π ( 6 − μ ) ∫ ℝ 3 | ∇ ϕ u n | 4 d x + 2 6 − μ ∫ ℝ 3 [ f ( u n ) u n − μ F ( u n ) ] d x ≥ 1 6 − μ ∫ ℝ 3 [ ( μ − 2 ) V ( x ) + 2 ( 3 − μ ) ω ϕ u n + ( 4 − μ ) ϕ u n 2 ] | u n | 2 d x ≥ 1 6 − μ ∫ ℝ 3 [ ( μ − 2 ) V 0 + 2 ( 3 − μ ) ω ϕ u n + ( 4 − μ ) ϕ u n 2 ] | u n | 2 d x ≥ ( 4 − μ ) ( μ − 2 ) V 0 − ( 3 − μ ) 2 ω 2 ( 6 − μ ) ( 4 − μ ) ∫ ℝ 3 | u n | 2 d x . (2.38)
It follows from (2.38) that ∫ ℝ 3 | u n | 2 d x is bounded when μ ∈ ( 2,3 ) . From Case (1) and Subcase (i), we have that ∫ ℝ 3 | u n | 2 d x is also bounded. Hence, by Lemma 2.1, there exists a positive constant C 7 such that:
| ∫ ℝ 3 ω ϕ u n | u n | 2 d x | ≤ ω 2 ∫ ℝ 3 | u n | 2 d x ≤ C 7 (2.39)
and
| ∫ ℝ 3 ϕ u n 2 | u n | 2 d x | ≤ ω 2 ∫ ℝ 3 | u n | 2 d x ≤ C 7 . (2.40)
From (2.35), (2.39) and (2.40), we know that ∫ ℝ 3 V ( x ) | u n | 2 d x is bounded when μ ∈ ( 2,3 ) and ω ∈ ( 0, ( μ − 2 ) ( 4 − μ ) V 0 / ( 3 − μ ) ) . From Hardy inequality, we have:
∫ ℝ 3 | ∇ u | 2 d x ≥ 1 4 ∫ ℝ 3 u 2 | x | 2 d x , ∀ u ∈ H r 1 ( ℝ 3 ) . (2.41)
By (2.31), (2.32), (2.33), (2.39), (2.40), (2.41) and (V2), we have:
a n + o ( 1 ) ≥ I λ n ( u n ) − 1 3 〈 I ′ λ n ( u n ) , u n 〉 + 1 3 G λ n ( u n ) = 1 3 ∫ ℝ 3 ( ω ϕ u n + ϕ u n 2 ) | u n | 2 d x + 1 3 ∫ ℝ 3 | ∇ u n | 2 d x − 1 6 ∫ ℝ 3 ( ∇ V ( x ) , x ) | u n | 2 d x ≥ 1 − θ 3 ∫ ℝ 3 | ∇ u n | 2 d x − 2 C 7 3 . (2.42)
From (2.42) and θ ∈ [ 0,1 ) , we know that { ∇ u n } is bounded in L 2 ( ℝ 3 ) . Hence, { u n } is bounded in E when μ ∈ ( 2,4 ) . Therefore, { u n } is bounded in E.
In order to obtain infinitely many solutions of system (1.1), we shall use the following critical point theorem introduced by Bartsch in [
X k = s p a n { e k } , Y k = ⊕ i = 1 k X i , Z k = ⊕ i = k ∞ X i ¯ . (2.43)
Now, we state the following critical points theorem given by Bartsch.
Lemma 2.8. Assume Ψ ∈ C 1 ( X , ℝ ) satisfies the (PS) condition, Ψ ( − u ) = Ψ ( u ) . For every k ∈ ℕ , there exists ρ k > r k > 0 , such that:
1) a k : = max u ∈ Y k , ‖ u ‖ = ρ k Ψ ( u ) ≤ 0 ;
2) b k : = inf u ∈ Z k , ‖ u ‖ = r k Ψ ( u ) → + ∞ as k → ∞ .
Then, Ψ has a sequence of critical values tending to + ∞ .
Lemma 2.9. Assume that (V1) and (F1) - (F4) hold. For every k ∈ ℕ , there exists ρ k > d k > 0 , such that:
1) a k : = max u ∈ Y k , ‖ u ‖ = ρ k I ( u ) ≤ 0 ;
2) b k : = inf u ∈ Z k , ‖ u ‖ = d k I ( u ) → + ∞ as k → ∞ ,
where Y k and Z k are defined by (2.43). Then, I has a sequence of critical values tending to + ∞ .
Proof. From (2.1), (2.4), (2.15) and − ω ≤ ϕ u ≤ 0 , one obtains:
I ( u ) = 1 2 ∫ ℝ 3 [ | ∇ u | 2 + V ( x ) u 2 − ( 2 ω + ϕ u ) ϕ u u 2 ] d x − 1 8 π ∫ ℝ 3 | ∇ ϕ u | 2 d x − β 16 π ∫ ℝ 3 | ∇ ϕ u | 4 d x − ∫ ℝ 3 F ( u ) d x
≤ 1 2 ∫ ℝ 3 [ | ∇ u | 2 + V ( x ) u 2 + 2 ω 2 u 2 ] d x − ∫ ℝ 3 ( C 3 | u | μ − C 4 u 2 ) d x = 1 2 ‖ u ‖ E 2 + r 2 2 ω 2 ‖ u ‖ E 2 − C 3 r μ μ ‖ u ‖ E μ + C 4 r 2 2 ‖ u ‖ E 2 . (2.44)
Since μ > 2 , from (2.44), there exists ρ k > 0 such that:
a k : = max u ∈ Y k , ‖ u ‖ = ρ k I ( u ) ≤ 0.
Subsequently, for any k ∈ ℕ and p ∈ [ 2,6 ) , we set:
β k ( p ) = sup u ∈ Z k , ‖ u ‖ E = 1 ‖ u ‖ p .
Similar to Lemma 2.8 in [
I ( u ) = 1 2 ∫ ℝ 3 [ | ∇ u | 2 + V ( x ) u 2 + ϕ u 2 u 2 ] d x + 1 8 π ∫ ℝ 3 | ∇ ϕ u | 2 d x + 3 β 16 π ∫ ℝ 3 | ∇ ϕ u | 4 d x − ∫ ℝ 3 F ( u ) d x ≥ 1 2 ∫ ℝ 3 [ | ∇ u | 2 + V ( x ) u 2 ] d x − ∫ ℝ 3 F ( u ) d x ≥ 1 2 ‖ u ‖ E 2 − C 1 β k 2 ( 2 ) ‖ u ‖ E 2 − C 2 β k 6 ( 6 ) ‖ u ‖ E 6 ≥ 1 8 ‖ u ‖ E 2 .
Thus, we obtain b k : = inf u ∈ Z k , ‖ u ‖ = d k I ( u ) ≥ 1 8 d k 2 → + ∞ as → ∞ . □
Proof of Theorem 1.1. It follows from Lemma 2.3 and Lemma 2.7 that I satisfies the (PS) condition. By (F1), (F2) and (F4), it is easy to see that I ( 0 ) = 0 and I ( − u ) = I ( u ) . By Lemma 2.9, the functional I satisfies the geometric conditions of Lemma 2.8. Hence, problem (1.1) has infinitely many nontrivial solutions ( u n , ϕ n ) ∈ E × D r 1,2 ( ℝ 3 ) . This completes the proof. □
Proof of Theorem 1.2. First, we show that the set S ≠ 0 . Similar to Lemma 2.7, we can prove that, there exists a sequence { w n } bounded in H r 1 ( ℝ 3 ) and I ( w n ) = c 1 , I ′ ( w n ) = 0 . We claim that:
lim sup n → ∞ sup y ∈ ℝ 3 ∫ B 1 ( y ) | w n | 2 d x > 0. (3.1)
If not, from Lion’s concentration compactness principle [
∫ ℝ 3 [ 1 2 f ( w n ) w n − F ( w n ) ] d x ≤ C 8 ‖ w n ‖ L 2 2 + C 9 ‖ w n ‖ L s s + C 10 ‖ w n ‖ L 6 6 ≤ c 1 2 + o ( 1 ) .
Thus,
c 1 + o ( 1 ) ≤ I ( w n ) − 1 2 〈 I ′ ( w n ) , w n 〉 = − 1 8 π ∫ ℝ 3 | ∇ ϕ w n | 2 d x − β 16 π ∫ ℝ 3 | ∇ ϕ w n | 4 d x + ∫ ℝ 3 [ 1 2 f ( w n ) w n − F ( w n ) ] d x ≤ c 1 2 + o ( 1 ) .
This contradiction shows that (3.1) holds, and so there exist δ > 0 and { y n } ⊂ ℝ 3 such that:
∫ B 2 ( y n ) | w n | 2 d x > δ > 0.
Let w ¯ n = w n ( x + y n ) , thus ‖ w ¯ n ‖ = ‖ w n ‖ , I ( w ¯ n ) = c 1 , I ′ ( w ¯ n ) = 0 and
∫ B 2 ( 0 ) | w ¯ n | 2 d x > δ > 0
which implies
w ¯ n ⇀ w ¯ ≡ 0 in H r 1 ( ℝ 3 ) .
By a standard argument, we can show that I ′ ( w ¯ ) = 0 , and so w ¯ ∈ S .
Next, we will prove 0 < k : = inf u ∈ S I ( u ) is achieved. Let { w n } ⊂ S be such that I ( w n ) → k and I ′ ( w n ) = G ( w n ) = 0 as n → ∞ . Arguing as before, we can prove that there exists w ˜ ∈ S such that I ( w ˜ ) ≥ k and I ′ ( w ˜ ) = G ( w ˜ ) = 0 . If μ ∈ [ 4,6 ) , from (V2), (F3), Lemma 2.1, (2.5), (2.6) and Fatou’s Lemma, we have:
k = l i m n → ∞ [ I ( w n ) − 1 μ 〈 I ′ ( w n ) , w n 〉 ] = l i m n → ∞ { ( 1 2 − 1 μ ) ∫ ℝ 3 [ | ∇ w n | 2 + V ( x ) | w n | 2 ] d x + β 16 π ∫ ℝ 3 | ∇ ϕ w n | 4 d x + ∫ ℝ 3 [ 1 μ ϕ w n 2 − ( 1 2 − 2 μ ) ω ϕ w n ] | w n | 2 d x + ∫ ℝ 3 [ 1 μ f ( w n ) w n − F ( w n ) ] d x } ≥ ( 1 2 − 1 μ ) ∫ ℝ 3 [ | ∇ w ˜ | 2 + V ( x ) | w ˜ | 2 ] d x + β 16 π ∫ ℝ 3 | ∇ ϕ w ˜ | 4 d x + ∫ ℝ 3 [ 1 μ ϕ w ˜ 2 − ( 1 2 − 2 μ ) ω ϕ w ˜ ] | w ˜ | 2 d x + ∫ ℝ 3 [ 1 μ f ( w ˜ ) w ˜ − F ( w ˜ ) ] d x = I ( w ˜ ) − 1 μ 〈 I ( w ˜ ) , w ˜ 〉 . (3.2)
If μ ∈ ( 2,4 ) , from Lemma 2.1, (2.5), (2.6), (2.36), (2.37) and Fatou’s Lemma, we obtain:
k = l i m n → ∞ [ I ( w n ) + μ − 4 6 − μ 〈 I ′ ( w n ) , w n 〉 + 2 − μ 6 − μ G ( w n ) ] = l i m n → ∞ { μ − 2 6 − μ ∫ ℝ 3 V ( x ) | w n | 2 d x + μ − 2 2 ( 6 − μ ) ∫ ℝ 3 ( ∇ V ( x ) , x ) | w n | 2 d x
+ 1 6 − μ ∫ ℝ 3 [ 2 ( 3 − μ ) ω ϕ w n + ( 4 − μ ) ϕ w n 2 ] | w n | 2 d x + 2 6 − μ ∫ ℝ 3 [ f ( w n ) w n − μ F ( w n ) ] d x + 2 β μ 16 π ∫ ℝ 3 | ∇ ϕ w n | 4 d x }
≥ μ − 2 6 − μ ∫ ℝ 3 V ( x ) | w ˜ | 2 d x + μ − 2 2 ( 6 − μ ) ∫ ℝ 3 ( ∇ V ( x ) , x ) | w ˜ | 2 d x + 1 6 − μ ∫ ℝ 3 [ 2 ( 3 − μ ) ω ϕ w ˜ + ( 4 − μ ) ϕ w ˜ 2 ] | w ˜ | 2 d x + 2 6 − μ ∫ ℝ 3 [ f ( w ˜ ) w ˜ − μ F ( w ˜ ) ] d x + 2 β μ 16 π ∫ ℝ 3 | ∇ ϕ w ˜ | 4 d x = I ( w ˜ ) + μ − 4 6 − μ 〈 I ′ ( w ˜ ) , w ˜ 〉 + 2 − μ 6 − μ G ( w ˜ ) . (3.3)
It follows from (3.2), (3.3) and I ( w ˜ ) ≥ k that w ˜ ∈ S and I ( w ˜ ) = k = inf u ∈ S I ( u ) . Hence, it follows from (2.2) and (2.5) that
J ( u , ϕ u ) = I ( u ) ≥ I ( w ˜ ) = J ( w ˜ , ϕ w ˜ ) = k , ∀ u ∈ S .
This proves that ( w ˜ , ϕ w ˜ ) is a ground-state solution for system (1.1). □
In this paper, we used Pohožaev identity of (1.1) to certify the boundedness of Palais-Smale sequence of energy functional of problem (1.1) at level c 1 , and then certified the boundedness of Palais-Smale sequence. By using critical point theory and the method of Nehari manifold, we obtained two existing results of infinitely many high-energy radial solutions and a ground-state solution for the system (1.1).
Supported by National Nature Science Foundation of China (No. 11961014) and Guangxi Natural Science Foundation (No. 2021GXNSFAA196040).
The authors declare no conflicts of interest regarding the publication of this paper.
Huang, F.F. and Zhang, Q.F. (2024) Infinitely Many Solutions and a Ground-State Solution for Klein-Gordon Equation Coupled with Born-Infeld Theory. Journal of Applied Mathematics and Physics, 12, 1441-1458. https://doi.org/10.4236/jamp.2024.124089