1 Introduction
In the paper [14], d’Avenia and Siciliano have attracted their attention on a new kind of elliptic system, called Schrödinger–Bopp–Podolsky system, which, to the best of our knowledge, was never been considered before in the mathematical literature, although the problem was known among the physicists. The Schrödinger–Bopp–Podolsky system has the following form:
(1.1)
{
-
Δ
u
+
V
(
x
)
u
+
q
2
ϕ
u
=
f
(
u
)
in
ℝ
3
,
-
Δ
ϕ
+
a
2
Δ
2
ϕ
=
4
π
u
2
in
ℝ
3
,
where u,ϕ:ℝ3→ℝ, ω,a>0, q≠0. Such a system appears when we couple a Schrödinger field ψ=ψ(t,x) with its electromagnetic field in the Bopp–Podolsky electromagnetic theory, and, in particular, in the electrostatic case for standing waves ψ(t,x)=eiωtu(x). We refer to the paper of d’Avenia and Siciliano [14] for more details.
The existence of standing waves for scalar fields in dimension 3 has extensively been studied by many authors. The most important scalar fields equation is the Schrödinger equation. In this paper we want to investigate about the existence of nonlinear Schrödinger fields interacting with the electromagnetic field (𝐄,𝐇). Since 𝐄,𝐇 are not assigned, we have to study a system of equations whose unknowns are the Schrödinger field ψ(x,t) and the gauge potentials 𝐀=𝐀(x,t), ϕ=ϕ(x,t) related to the electromagnetic field. In order to construct such a system we shall describe, as usual, the interaction between ψ and 𝐄,𝐇 by using the so called gauge covariant derivatives. The Bopp–Podolsky theory is a second order gauge theory for the electromagnetic field. As the Mie theory [21] and its generalizations given by Born and Infeld [4, 5, 6, 7], it was introduced to solve the so called infinity problem that appears in the classical Maxwell theory. The Bopp–Podolsky theory is developed by Bopp [3] and Podolsky [22], independently. We also shall investigate the case in which 𝐀 and ϕ do not depend on the time t and ψ(x,t) is a standing wave. In this situation we can assume 𝐀=0 and we are reduced to study system (1.1) (see Section 2).
By the well-known Gauss law (or Poisson’s equation), the electrostatic potential ϕ for a given charge distribution whose density is ρ satisfies the equation
(1.2)
-
Δ
ϕ
=
ρ
in
ℝ
3
.
If ρ=4πδx0, with x0∈ℝ3, the fundamental solution of (1.2) is 𝒢(x-x0), where
𝒢
(
x
)
=
1
|
x
|
,
and the electrostatic energy is
ℰ
M
(
𝒢
)
=
1
2
∫
ℝ
3
|
∇
𝒢
|
2
d
x
=
∞
.
Thus, equation (1.2) is replaced by
-
div
(
∇
ϕ
1
-
|
∇
ϕ
|
2
)
=
ρ
in
ℝ
3
in the Born–Infeld theory and by
-
Δ
ϕ
+
a
2
Δ
2
ϕ
=
ρ
in
ℝ
3
in the Bopp–Podolsky one. In both cases, if ρ=4πδx0, we are able to write explicitly the solutions of the respective equations and to see that their energy is finite. In particular, the fundamental solution of the equation
-
Δ
ϕ
+
a
2
Δ
2
ϕ
=
4
π
δ
x
0
is 𝒦(x-x0), where
(1.3)
𝒦
(
x
)
:=
1
-
e
-
|
x
|
a
|
x
|
,
which presents no singularities at x0, since
lim
x
→
x
0
𝒦
(
x
-
x
0
)
=
1
a
.
Furthermore, its energy is
ℰ
BP
(
𝒦
)
=
1
2
∫
ℝ
3
|
∇
𝒦
|
2
d
x
+
a
2
2
∫
ℝ
3
|
Δ
𝒦
|
2
d
x
<
∞
.
We refer to [14] for more details.
Moreover, the Bopp–Podolsky theory may be interpreted as an effective theory for short distances (see [15]) and for large distances it is experimentally indistinguishable from the Maxwell one. Thus, the Bopp–Podolsky parameter a>0, which has dimension of the inverse of mass, can be interpreted as a cut-off distance or can be linked to an effective radius for the electron. For more physical features we refer the interested reader to the recent papers [2, 8, 9, 12, 13] and to the references therein.
In the novel paper [14], d’Avenia and Siciliano deal with the following special form of (1.1):
{
-
Δ
u
+
V
∞
u
+
q
2
ϕ
u
=
|
u
|
p
-
2
u
in
ℝ
3
,
-
Δ
ϕ
+
a
2
Δ
2
ϕ
=
4
π
u
2
in
ℝ
3
,
and study the existence, nonexistence and the behavior of the solution as a→0. Again the solutions converge to the solution of the “limit” problem with a=0. However, d’Avenia and Scilliano in [14] do not cover the critical case. As far as we know this paper is the first attempt to solve this delicate challenging problem in which lack of compactness appears together with the lack of translation invariance. To overcome these difficulties, we have to use a global compactness lemma as well as introduce new inequalities and techniques. In particular, the main results of the present paper extend [14] to the critical case.
This paper is concerned with the existence of ground state solutions for the following Schrödinger–Bopp–Podolsky system with critical Sobolev exponent:
(1.4)
{
-
Δ
u
+
V
(
x
)
u
+
q
2
ϕ
u
=
μ
|
u
|
p
-
1
u
+
|
u
|
4
u
in
ℝ
3
,
-
Δ
ϕ
+
a
2
Δ
2
ϕ
=
4
π
u
2
in
ℝ
3
,
where μ>0 is a parameter, 2<p<5 and the potential V satisfies the following conditions:
V
∈
C
1
(
ℝ
3
)
, (⋅,∇V)∈L∞(ℝ3) and
2
V
(
x
)
+
(
x
,
∇
V
(
x
)
)
≥
0
,
x
∈
ℝ
3
,
where (⋅,⋅) is the usual inner product in ℝ3.
For all x∈ℝ3 it results V(x)≤lim inf|ξ|→∞V(ξ)=V∞∈ℝ+ and the inequality is strict in a subset of positive Lebesgue measure.
There exists a positive number α0 such that
α
0
=
inf
u
∈
H
1
(
ℝ
3
)
∖
{
0
}
∫
ℝ
3
|
∇
u
|
2
+
V
(
x
)
|
u
|
2
d
x
∫
ℝ
3
|
u
|
2
d
x
>
0
.
From now on we assume, without further mentioning, that (V1)–(V3) hold. Then the main results of the paper are stated as follows.
Theorem 1.1.
If p∈(3,5), system (1.4) has a ground state solution for any μ>0, while if p∈(2,3], system (1.4) possesses a ground state solution for μ>0 large enough.
Let us give the main ideas under the proof of Theorem 1.1. The existence of ground state solutions for the Schrödinger–Bopp–Podolsky system (1.4), namely of the couples (u,ϕ) which solve (1.4), is obtained by minimizing the action functional associated to (1.4) among all possible solutions. Motivated by [1, 26, 20, 10], we choose the usual Sobolev space H1(ℝ3) to prove the existence of ground state solutions for the “limit” problem
(1.5)
{
-
Δ
u
+
V
∞
u
+
q
2
ϕ
u
=
μ
|
u
|
p
-
1
u
+
|
u
|
4
u
in
ℝ
3
,
-
Δ
ϕ
+
a
2
Δ
2
ϕ
=
4
π
u
2
in
ℝ
3
.
Then we look for a minimizer of the reduced functional restricted to a suitable manifold ℳμ∞, which was introduced by Ruiz in [24] when a=0. Such a manifold consists of the linear combination of the Pohozaev manifold and the Nehari manifold and is called the Pohozaev–Nehari manifold. It has two perfect characteristics: it is a natural constraint for the reduced functional and it contains every solution of problem (1.5). We shall use the concentration-compactness lemma to establish the following result.
Theorem 1.2.
When p∈(3,5), problem (1.5) has a ground state solution for any μ>0, and when p∈(2,3], problem (1.5) has a ground state solution for μ>0 large enough.
Next, in order to use the monotonicity trick of [17], we introduce a family of functionals ℐλ, which satisfies all the assumptions of Theorem 1.2 and which possesses a bounded (PS)cλ sequence. A global compactness lemma, applied to the functional ℐV,λ and its limit functional ℐλ∞, allows us to prove that the Palais-Smale condition (PS)cλ holds. Finally, choosing a sequence (λn)n approaching 1 at infinity, we show that (uλn)n is a bounded (PS)c1 sequence for ℐV. An application of the global compactness lemma completes the proof of Theorem 1.1.
The paper is organized as follows. In Section 2, we present some preliminaries results. In Section 3, we prove Theorem 1.2. Finally, Section 4 is devoted to the proof of the main Theorem 1.1.
Last, we will mention the very recently paper by Chen and Tang [11]. They also study this type of system, but our method is different with theirs.
2 Variational Setting
We start with some preliminary basic results. Let us consider the nonlinear Schrödinger Lagrangian density
ℒ
Sc
=
i
ℏ
ψ
¯
∂
t
ψ
-
ℏ
2
2
m
|
∇
ψ
|
2
+
2
F
(
ψ
)
,
where ψ:ℝ×ℝ3→ℂ, ℏ,m>0. Let (ϕ,𝐀) be the gauge potential of the electromagnetic field (𝐄,𝐇), namely ϕ:ℝ3→ℝ and 𝐀:ℝ3→ℝ3 satisfy
𝐄
=
-
∇
ϕ
-
1
c
∂
t
𝐀
,
𝐇
=
∇
×
𝐀
.
The coupling of the field ψ with the electromagnetic field (𝐄,𝐇) via the minimal coupling rule, namely the study of the interaction between ψ and its own electromagnetic field, can be obtained replacing in ℒSc the derivatives ∂t and ∇ with the covariant ones
D
t
=
∂
t
+
i
q
ℏ
ϕ
,
𝐃
=
∇
-
i
q
ℏ
c
𝐀
,
respectively. Here q is a coupling constant. This leads to consider
ℒ
CSc
=
i
ℏ
ψ
¯
D
t
ψ
-
ℏ
2
2
m
|
𝐃
ψ
|
2
+
2
F
(
ψ
)
=
i
ℏ
ψ
¯
(
∂
t
+
i
q
ℏ
ϕ
)
ψ
-
ℏ
2
2
m
|
(
∇
-
i
q
ℏ
c
𝐀
)
ψ
|
2
+
2
F
(
ψ
)
.
Now, to get the total Lagrangian density, we have to add to ℒCSc the Lagrangian density of the electromagnetic field. The Bopp–Podolsky Lagrangian density (see [22, formula (3.9)]) is
ℒ
BP
=
1
8
π
{
|
𝐄
|
2
-
|
𝐇
|
2
+
a
2
[
(
div
𝐄
)
2
-
|
∇
×
𝐇
-
1
c
∂
t
𝐄
|
2
]
}
=
1
8
π
{
|
∇
ϕ
+
1
c
∂
t
𝐀
|
2
-
|
∇
×
𝐀
|
2
+
a
2
[
(
Δ
ϕ
+
1
c
div
∂
t
𝐀
)
2
-
|
∇
×
∇
×
𝐀
+
1
c
∂
t
(
∇
ϕ
+
1
c
∂
t
𝐀
)
|
2
]
}
.
Thus the total action is
𝒮
(
ψ
,
ϕ
,
𝐀
)
=
∫
ℝ
3
ℒ
d
x
d
t
,
where ℒ:=ℒCSc+ℒBP is the total Lagrangian density. We refer the interested readers to [14] for a detailed deduction of (1.4).
Thanks to assumptions (V2) and (V3), the Sobolev space H1(ℝ3) can be equipped with the inner product
〈
u
,
v
〉
=
∫
ℝ
3
(
∇
u
∇
v
+
V
(
x
)
u
v
)
d
x
and the corresponding norm
∥
u
∥
=
(
∫
ℝ
3
(
|
∇
u
|
2
+
V
(
x
)
u
2
)
d
x
)
1
2
.
Actually, (V2) and (V3) yield that the above norm is equivalent to the usual norm ∥⋅∥H1. Indeed, from (V3), similar to [17, proof of Lemma 3.4], there exists a constant C>0 such that
∫
ℝ
3
(
|
∇
u
|
2
+
V
(
x
)
u
2
)
d
x
≥
α
0
2
∫
ℝ
3
|
u
|
2
d
x
+
C
∫
ℝ
3
|
∇
u
|
2
d
x
,
while (V2) implies that
∫
ℝ
3
(
|
∇
u
|
2
+
V
(
x
)
u
2
)
d
x
≤
∫
ℝ
3
|
∇
u
|
2
d
x
+
∫
ℝ
3
V
∞
u
2
d
x
.
The above two estimates imply that ∥⋅∥ is an equivalent norm on H1(ℝ3).
It is well known that H1(ℝ3) is continuously embedded into Ls(ℝ3) when 2≤s≤6, and there exists the best constant 𝒮>0 such that
(2.1)
𝒮
=
inf
u
∈
𝒟
1
,
2
∫
ℝ
3
|
∇
u
|
2
d
x
(
∫
ℝ
3
|
u
|
6
d
x
)
1
3
,
where
𝒟
1
,
2
(
ℝ
3
)
:=
{
u
∈
L
6
(
ℝ
3
)
:
∇
u
∈
L
2
(
ℝ
3
)
}
.
Let 𝒟 be the completion of Cc∞(ℝ3) with respect to the norm ∥⋅∥𝒟 induced by the scalar product
〈
φ
,
ψ
〉
𝒟
:=
∫
ℝ
3
(
∇
φ
∇
ψ
+
a
2
Δ
φ
Δ
ψ
)
d
x
.
Then 𝒟 is a Hilbert space continuously embedded into 𝒟1,2(ℝ3) and consequently in L6(ℝ3). In the sequel, we denote by ∥⋅∥p the usual norm of the space Lp(ℝ3), p≥1, the letter ci (i=1,2,…) or Ci (i=1,2,…) denote just positive constants. It is interesting to point out the following properties.
Lemma 2.1 ([14, Lemma 3.1]).
The space D is continuously embedded in L∞(R3).
The next lemma gives a useful characterization of the space 𝒟.
Lemma 2.2 ([14, Lemma 3.2]).
The space Cc∞(R3) is dense in
𝒜
:=
{
ϕ
∈
𝒟
1
,
2
(
ℝ
3
)
:
Δ
ϕ
∈
L
2
(
ℝ
3
)
}
normed by 〈ϕ,ϕ〉D and, therefore, D=A.
For every fixed u∈H1(ℝ3), the Riesz theorem implies that there exists a unique solution ϕu∈𝒟 of the second equation in (1.4). To write explicitly such a solution (see also [22, formula (2.6)]), we take 𝒦 as defined by (1.3). Then the next fundamental properties hold.
Lemma 2.3 ([14, Lemma 3.3]).
For all y∈R3 the function K(⋅-y) solves
-
Δ
ϕ
+
a
2
Δ
2
ϕ
=
4
π
δ
y
in the sense of distributions. Moreover,
if
g
∈
L
loc
1
(
ℝ
3
)
and the map
ℝ
3
∋
y
↦
g
(
y
)
/
|
x
-
y
|
is summable in
ℝ
3
for a.e.
x
∈
ℝ
3
, then
𝒦
*
g
∈
L
loc
1
(
ℝ
3
)
,
if
g
∈
L
s
(
ℝ
3
)
, with
1
≤
s
<
3
2
, then
𝒦
*
g
∈
L
r
(
ℝ
3
)
for all
r
∈
(
3
s
3
-
2
s
,
∞
]
.
In both cases K*g solves
-
Δ
ϕ
+
a
2
Δ
2
ϕ
=
4
π
g
in the sense of distributions, and has distributional derivatives
∇
(
𝒦
*
g
)
=
(
∇
𝒦
)
*
g
𝑎𝑛𝑑
Δ
(
𝒦
*
g
)
=
(
Δ
𝒦
)
*
g
a.e. in
ℝ
3
.
Fix u∈H1(ℝ3), the unique solution in 𝒟 of the second equation in (1.4) is
ϕ
u
:=
𝒦
*
u
2
.
Furthermore, we define
ψ
u
:=
e
-
|
x
|
a
*
u
2
.
The coming properties will be useful.
Lemma 2.4 ([14, Lemma 3.4]).
For every u∈H1(R3) we have:
ϕ
u
(
⋅
+
y
)
=
ϕ
u
(
⋅
+
y
)
for every
y
∈
ℝ
3
,
ϕ
u
∈
L
s
(
ℝ
3
)
∩
C
(
ℝ
3
)
for every
s
∈
(
3
,
∞
]
,
∇
ϕ
u
=
∇
𝒦
*
u
2
∈
L
s
(
ℝ
3
)
∩
C
(
ℝ
3
)
for every
s
∈
(
3
2
,
∞
]
,
∥
ϕ
u
∥
6
≤
C
∥
u
∥
2
,
ϕ
u
is the unique minimizer
in
𝒟
of the functional
E
(
ϕ
)
=
1
2
∥
∇
ϕ
∥
2
2
+
a
2
2
∥
Δ
ϕ
∥
2
2
-
∫
ℝ
3
ϕ
u
2
d
x
,
ϕ
∈
𝒟
,
∫
ℝ
3
∫
ℝ
3
u
2
(
x
)
u
2
(
y
)
|
x
-
y
|
d
x
d
y
≤
𝒮
2
∥
u
∥
12
5
4
.
Moreover, if vn⇀v in H1(R3), then ϕvn⇀ϕv in D.
In view of [14], under (V1)–(V3), the energy functional of (1.4), defined in H1(ℝ3)×𝒟 by
(2.2)
𝒮
(
u
,
ϕ
)
=
1
2
∫
ℝ
3
[
|
∇
u
|
2
2
+
V
(
x
)
u
2
]
d
x
+
q
2
2
∫
ℝ
3
ϕ
u
2
d
x
-
q
2
16
π
∫
ℝ
3
|
∇
ϕ
|
2
d
x
-
a
2
q
2
16
π
∫
ℝ
3
|
Δ
ϕ
|
2
d
x
-
μ
p
+
1
∫
ℝ
3
|
u
|
p
+
1
d
x
-
1
6
∫
ℝ
3
|
u
|
6
d
x
,
is continuously differentiable and its critical points correspond to the weak solutions of (1.4). Indeed, if (u,ϕ)∈H1(ℝ3)×𝒟 is a critical point of 𝒮, then
0
=
∂
u
𝒮
(
u
,
ϕ
)
[
v
]
=
∫
ℝ
3
[
∇
u
∇
v
+
V
(
x
)
u
v
]
d
x
+
q
2
∫
ℝ
3
ϕ
u
v
d
x
-
μ
∫
ℝ
3
|
u
|
p
-
1
u
v
d
x
-
∫
ℝ
3
|
u
|
4
u
v
d
x
for all v∈H1(ℝ3) and
(2.3)
0
=
∂
ϕ
𝒮
(
u
,
ϕ
)
[
φ
]
=
q
2
2
∫
ℝ
3
u
2
φ
d
x
-
q
2
8
π
∫
ℝ
3
∇
ϕ
∇
φ
d
x
-
a
2
q
2
8
π
∫
ℝ
3
Δ
ϕ
Δ
φ
d
x
for all φ∈𝒟.
In order to avoid the difficulty originated by the strongly indefiniteness of the functional 𝒮, we apply a reduction procedure used in [14]. Since ∂ϕ𝒮 is a C1 functional, if GΦ is the graph of the map Φ defined by H1(ℝ3)∋u↦ϕu∈𝒟, an application of the Implicit Function Theorem gives
G
Φ
=
{
(
u
,
ϕ
)
∈
H
1
(
ℝ
3
)
×
𝒟
:
∂
ϕ
𝒮
(
u
,
ϕ
)
=
0
}
and
Φ
∈
C
1
(
H
1
(
ℝ
3
)
,
𝒟
)
.
Jointly with (2.2) and (2.3), the functional ℐ(u):=𝒮(u,ϕu) has the reduced form
ℐ
(
u
)
=
1
2
∫
ℝ
3
[
|
∇
u
|
2
+
V
(
x
)
u
2
]
d
x
+
q
2
4
∫
ℝ
3
ϕ
u
u
2
d
x
-
μ
p
+
1
∫
ℝ
3
|
u
|
p
+
1
d
x
-
1
6
∫
ℝ
3
|
u
|
6
d
x
,
which is of class C1(H1(ℝ3)) and for all u,v∈H1(ℝ3),
ℐ
′
(
u
)
[
v
]
=
∂
u
𝒮
(
u
,
Φ
(
u
)
)
[
v
]
+
∂
ϕ
𝒮
(
u
,
Φ
(
u
)
)
∘
Φ
′
(
u
)
[
v
]
=
∂
u
𝒮
(
u
,
Φ
(
u
)
)
[
v
]
=
∫
ℝ
3
[
∇
u
∇
v
+
V
(
x
)
u
v
]
d
x
+
q
2
∫
ℝ
3
ϕ
u
u
v
d
x
-
μ
∫
ℝ
3
|
u
|
p
-
1
u
v
d
x
-
∫
ℝ
3
|
u
|
4
u
v
d
x
.
Moreover, the following statements are equivalent:
The pair (u,ϕ)∈H1(ℝ3)×𝒟 is a critical point of 𝒮. i.e. (u,ϕ) is a solution of (1.4).
u is a critical point of ℐ and ϕ=ϕu.
Hence, if u∈H1(ℝ3) is a critical point of ℐ, then the pair (u,ϕu) is a solution of (1.4). For the sake of simplicity, in many cases we just say u∈H1(ℝ3), instead of (u,ϕu)∈H1(ℝ3)×𝒟, is a solution of (1.4).
Let us define the function Ψ:H1(ℝ3)→ℝ by
Ψ
(
u
)
=
∫
ℝ
3
ϕ
u
(
x
)
u
2
(
x
)
d
x
.
It is clear that for all fixed u∈H1(ℝ3) then Ψ(u(⋅+y))=Ψ(u) for any y∈ℝ3 and that Ψ is weakly lower semi-continuous in H1(ℝ3). The next lemma shows that the functional Ψ and its derivative Ψ′ have the B-L splitting property, which is similar to the well-known Brézis–Lieb lemma.
Lemma 2.5.
If un⇀u in H1(R3) and un→u a.e. in R3, then
Ψ
(
u
n
-
u
)
=
Ψ
(
u
n
)
-
Ψ
(
u
)
+
o
(
1
)
,
Ψ
′
(
u
n
-
u
)
=
Ψ
′
(
u
n
)
-
Ψ
′
(
u
)
+
o
(
1
)
in
(
H
1
(
ℝ
3
)
)
′
.
Proof.
(i) This result is proved in [14, Lemma B.2].
(ii) This property is obtained in [25, Lemma 2.2] step by step, thanks to
ϕ
u
:=
1
-
e
-
|
x
|
a
|
x
|
*
u
2
and
1
-
e
-
|
x
|
a
|
x
|
≤
1
|
x
|
.
∎
In the sequel, the Pohozaev identity obtained in [14] will be frequently used.
Proposition 2.6.
Assume that (V1)–(V2) hold. Let f∈C1(R) satisfy for some C>0 and p, with 1≤p≤5,
|
f
(
t
)
|
≤
C
(
|
t
|
+
|
t
|
p
)
,
t
∈
ℝ
.
Let (u,ϕ)∈H1(R3)×D(R3) be a solution of the problem
{
-
Δ
u
+
V
(
x
)
u
+
q
2
ϕ
u
=
f
(
u
)
in
ℝ
3
,
-
Δ
ϕ
+
a
2
Δ
2
ϕ
=
4
π
u
2
in
ℝ
3
.
Then the Pohozaev identity holds true
1
2
∫
ℝ
3
|
∇
u
|
2
d
x
+
1
2
∫
ℝ
3
[
3
V
(
x
)
+
∇
V
(
x
)
⋅
x
]
u
2
d
x
+
5
4
∫
ℝ
3
q
2
ϕ
u
u
2
d
x
+
1
4
∫
ℝ
3
q
2
a
ψ
u
u
2
d
x
=
3
∫
ℝ
3
F
(
u
)
d
x
,
where F(t)=∫0tf(τ)dτ.
The vanishing lemma for Sobolev space is stated as follows.
Lemma 2.7 (Vanishing Lemma, [19]).
Assume that (un)n is bounded sequence in H1(R3) such that
lim
n
→
∞
sup
y
∈
ℝ
3
∫
B
R
(
y
)
|
u
n
(
x
)
|
2
d
x
=
0
for some R>0. Then un→0 in Lr(RN) for every r, with 2<r<6.
The arguments of Ramos, Wang and Willem [23] give sufficient conditions to ensure the convergence to 0 in L6(ℝ3) of a sequence in H1(ℝ3).
Lemma 2.8.
Let R>0 and (un)n be a bounded sequence in H1(R3). If
lim
n
→
∞
sup
y
∈
ℝ
3
∫
B
R
(
y
)
|
u
n
|
6
d
x
=
0
,
then un→0 in Lr(R3) as n→∞ for any r∈(2,6].
The successive concentration-compactness principle is due to P.-L. Lions [19].
Lemma 2.9 ([19, Lemma 1.1]).
Let (ρn)n be a sequence of nonnegative functions in L1(RN) such that for some l>0 fixed ∫RNρndx=l for all n. Then there exists a subsequence, still denoted by (ρn)n, satisfying one of the following three possibilities:
(Compactness) There exists (yn)n⊂ℝN with the property that for any ε>0 there is R>0 such that
lim inf
n
→
∞
∫
B
R
(
y
n
)
ρ
n
(
x
)
d
x
≥
l
-
ε
.
(Vanishing) For any fixed R>0 there holds
lim
n
→
∞
sup
y
∈
ℝ
N
∫
B
R
(
y
)
ρ
n
(
x
)
d
x
=
0
.
(Dichotomy) There exists an α∈(0,l) and (yn)n⊂ℝN with the property that for any ε>0 there is R>0 such that for all r≥R and r′≥R it holds
lim sup
n
→
∞
(
|
α
-
∫
B
r
(
y
n
)
ρ
n
d
x
|
+
|
(
l
-
α
)
-
∫
ℝ
N
∖
B
r
′
(
y
n
)
ρ
n
d
x
|
)
<
ε
.
When V is not a constant, it is more difficult to establish the boundedness of any (PS) sequence. To overcome this difficulty, we use a subtle approach developed by Jeanjean in [16].
Theorem 2.10 ([16, Theorem 1.1]).
Let X be a Banach space and let Λ⊂R+ be an interval. Consider a family {φλ}λ∈Λ of C1(X) functionals, with the form
φ
λ
(
u
)
=
A
(
u
)
-
λ
B
(
u
)
,
λ
∈
Λ
,
where B(u)≥0 for all u∈X, and such that either A(u)→∞ or B(u)→∞ as ∥u∥→∞. If there exist v1, v2∈X such that
c
λ
=
inf
γ
∈
Γ
max
t
∈
[
0
,
1
]
φ
λ
(
γ
(
t
)
)
>
max
{
φ
λ
(
v
1
)
,
φ
λ
(
v
2
)
}
for all
λ
∈
Λ
,
where Γ={γ∈C([0,1],X):γ(0)=v1,γ(1)=v2}, then for a.e. λ∈Λ, there exists a sequence (vn)n in X such that
φ
λ
′
(
v
n
)
→
0
in the dual space
X
′
of
X
.
At last, we give a fundamental inequality we shall use later.
Lemma 2.11.
Let b>0. Then
(2.4)
h
(
t
)
:=
t
3
(
e
-
b
t
-
e
-
b
)
+
1
-
t
3
3
b
e
-
b
≥
0
for all
t
>
0
and
(2.5)
1
-
e
-
b
-
1
3
b
e
-
b
>
0
.
Proof.
A simple calculation shows that h(0+)=be-b3>0 and for all t>0
h
′
(
t
)
=
t
2
[
3
(
e
-
b
t
-
e
-
b
)
+
b
t
e
-
b
t
-
b
e
-
b
]
.
Consequently, h′(0+)=0=h′(1) and t=1 is strict minimum point for h in ℝ0+ so that h(t)>h(1)=0 for all t>0, with t≠1. This proves (2.4). Finally, (2.5) holds actually for all b≥0 by direct computation. ∎
3 The Constant Potential Case
In this section we assume that V is the positive constant V∞ which appears in(V2) and we consider the “limit problem”
(3.1)
{
-
Δ
u
+
V
∞
u
+
q
2
ϕ
u
=
μ
|
u
|
p
-
1
u
+
|
u
|
4
u
in
ℝ
3
,
-
Δ
ϕ
+
a
2
Δ
2
ϕ
=
4
π
u
2
,
in
ℝ
3
,
associated with system (1.4). The norm on the H1(ℝ3) is taken as
∥
u
∥
=
(
∫
ℝ
3
(
|
∇
u
|
2
+
V
∞
u
2
)
d
x
)
1
2
.
The underlying energy functional ℐμ∞:H1(ℝ3)→ℝ, related to (3.1), is defined by
ℐ
μ
∞
(
u
)
=
1
2
∫
ℝ
3
(
|
∇
u
|
2
+
V
∞
u
2
)
d
x
+
q
2
4
∫
ℝ
3
ϕ
u
u
2
d
x
-
μ
p
+
1
∫
ℝ
3
|
u
(
x
)
|
p
+
1
d
x
-
1
6
∫
ℝ
3
|
u
(
x
)
|
6
d
x
.
Clearly, ℐμ∞∈C1(H1(ℝ3)) and
(
ℐ
μ
∞
)
′
(
u
)
[
φ
]
=
∫
ℝ
3
(
∇
u
∇
φ
+
V
∞
u
φ
)
d
x
+
q
2
∫
ℝ
3
ϕ
u
u
φ
d
x
-
μ
∫
ℝ
3
|
u
|
p
-
1
u
φ
d
x
-
∫
ℝ
3
|
u
|
4
u
φ
d
x
for every φ∈H1(ℝ3). Hence, the critical points of ℐμ∞ in H1(ℝ3) are weak solutions of problem (3.1).
Define 𝒢μ∞:H1(ℝ3)→ℝ as
𝒢
μ
∞
(
u
)
=
2
(
ℐ
μ
∞
)
′
(
u
)
[
u
]
-
𝒫
μ
∞
(
u
)
=
3
2
∫
ℝ
3
|
∇
u
|
2
d
x
+
1
2
∫
ℝ
3
V
∞
|
u
|
2
d
x
+
3
4
∫
ℝ
3
q
2
ϕ
u
u
2
d
x
-
1
4
∫
ℝ
3
q
2
a
ψ
u
u
2
d
x
-
μ
2
p
-
1
p
+
1
∫
ℝ
3
|
u
|
p
+
1
d
x
-
3
2
∫
ℝ
3
|
u
|
6
d
x
,
where 𝒫μ∞ is given by
𝒫
μ
∞
(
u
)
=
1
2
∫
ℝ
3
|
∇
u
|
2
d
x
+
3
2
∫
ℝ
3
V
∞
u
2
d
x
+
5
4
∫
ℝ
3
q
2
ϕ
u
u
2
d
x
+
1
4
∫
ℝ
3
q
2
a
ψ
u
u
2
d
x
-
3
μ
p
+
1
∫
ℝ
3
|
u
|
p
+
1
d
x
-
1
2
∫
ℝ
3
|
u
|
6
d
x
.
See Proposition 2.6.
We study the functional ℐμ∞ restricted on the manifold ℳμ∞ defined as
ℳ
μ
∞
=
{
u
∈
H
1
(
ℝ
3
)
∖
{
0
}
:
𝒢
μ
∞
(
u
)
=
0
}
.
Obviously, if u∈H1(ℝ3) is a nontrivial critical point of ℐμ∞, then u∈ℳμ∞. Hence, if (u,ϕ)∈H1(ℝ3)×𝒟 is a solution of (3.1), then u∈ℳμ∞. The next result describes the properties of the manifold ℳμ∞.
Lemma 3.1.
Let p>2. Then the following properties hold for the manifold Mμ∞:
For any
u
∈
H
1
(
ℝ
3
)
∖
{
0
}
there exists a unique number
θ
0
>
0
such that
u
θ
0
:=
θ
2
u
(
θ
0
⋅
)
is in
ℳ
μ
∞
and
ℐ
μ
∞
(
u
θ
0
)
=
max
θ
≥
0
ℐ
μ
∞
(
u
θ
)
.
ℐ
μ
∞
(
u
)
>
0
for all
u
∈
ℳ
μ
∞
.
(
𝒢
μ
∞
)
′
(
u
)
≠
0
for any
u
∈
ℳ
μ
∞
, that is,
ℳ
μ
∞
is a
C
1
-
manifold.
ℳ
μ
∞
is a natural constraint of
ℐ
μ
∞
, namely every critical point of
ℐ
μ
∞
|
ℳ
μ
∞
is a critical point of
ℐ
μ
∞
.
There exists a positive constant
C
>
0
such that
∥
u
∥
p
+
1
≥
C
for any
u
∈
ℳ
μ
∞
.
Proof.
(1) Fix u∈H1(ℝ3)∖{0} and note that for θ>0,
u
θ
∈
ℳ
μ
∞
⇔
θ
g
′
(
θ
)
=
0
⇔
g
′
(
θ
)
=
0
,
where g is given in ℝ0+ by
g
(
θ
)
=
θ
3
2
∫
|
∇
u
|
2
d
x
+
θ
2
∫
V
∞
|
u
|
2
d
x
+
θ
3
q
2
4
∫
ℝ
3
∫
ℝ
3
1
-
e
-
|
x
-
y
|
θ
a
|
x
-
y
|
u
2
(
x
)
u
2
(
y
)
d
x
d
y
-
μ
θ
2
p
-
1
p
+
1
∫
ℝ
3
|
u
|
p
+
1
d
x
-
θ
9
6
∫
ℝ
3
|
u
|
6
d
x
.
Clearly, g(θ) is positive for small θ>0 and tends to -∞ as θ→∞. Since g′ is continuous in ℝ0+, there exists at least one θ0=θ0(u)>0 such that g′(θ0)=0, which means that uθ0∈ℳμ∞.
To show the uniqueness of θ0, note that g′(θ)=0 and θ>0 imply that
3
2
∫
ℝ
3
|
∇
u
|
2
d
x
=
μ
(
2
p
-
1
)
θ
2
(
p
-
2
)
p
+
1
∫
ℝ
3
|
u
|
p
+
1
d
x
+
3
θ
6
2
∫
ℝ
3
|
u
|
6
d
x
+
q
2
4
a
∫
ℝ
3
∫
ℝ
3
e
-
|
x
-
y
|
θ
a
u
2
(
x
)
u
2
(
y
)
d
x
d
y
-
3
q
2
θ
-
1
4
∫
ℝ
3
∫
ℝ
3
1
-
e
-
|
x
-
y
|
θ
a
|
x
-
y
|
u
2
(
x
)
u
2
(
y
)
d
x
𝑑
y
-
θ
-
2
2
∫
ℝ
3
V
∞
|
u
|
2
d
x
(3.2)
=
:
h
(
θ
)
.
Now the derivative h′ of h is strictly positive in ℝ+, with h(0+)=-∞ and h(∞)=∞. As a consequence, there exists a unique θ0>0 such that (3.2) holds true. The uniqueness of θ0 is verified and (1) is proved.
(2) The Sobolev embedding theorem and (2.5) give
3
2
∫
ℝ
3
|
∇
u
|
2
d
x
+
1
2
∫
ℝ
3
V
∞
u
2
d
x
+
3
q
2
4
∫
ℝ
3
ϕ
u
u
2
d
x
-
q
2
4
a
∫
ℝ
3
ψ
u
u
2
d
x
-
μ
2
p
-
1
p
+
1
∫
ℝ
3
|
u
|
p
+
1
d
x
-
3
2
∫
ℝ
3
|
u
|
6
d
x
=
3
2
∫
ℝ
3
|
∇
u
|
2
d
x
+
1
2
∫
ℝ
3
V
∞
u
2
d
x
+
3
q
2
4
∫
ℝ
3
∫
ℝ
3
1
-
e
-
|
x
-
y
|
a
-
|
x
-
y
|
3
a
e
-
|
x
-
y
|
a
|
x
-
y
|
u
2
(
x
)
u
2
(
y
)
d
x
d
y
-
μ
2
p
-
1
p
+
1
∫
ℝ
3
|
u
|
p
+
1
d
x
-
3
2
∫
ℝ
3
|
u
|
6
d
x
≥
1
2
∥
u
∥
2
-
C
1
∥
u
∥
p
+
1
-
C
2
∥
u
∥
6
,
which is strictly positive for ∥u∥ small. Hence 0∉∂ℳμ∞.
(3) Note that if u∈ℳμ∞ and 2<p<5, then
(3.3)
(
2
p
-
1
)
ℐ
μ
∞
(
u
)
=
(
2
p
-
1
)
ℐ
μ
∞
(
u
)
-
𝒢
μ
∞
(
u
)
=
(
p
-
2
)
∫
ℝ
3
|
∇
u
|
2
d
x
+
(
p
-
1
)
∫
ℝ
3
V
∞
u
2
d
x
+
p
-
2
2
∫
ℝ
3
q
2
ϕ
u
u
2
d
x
+
q
2
4
a
∫
ℝ
3
ψ
u
u
2
d
x
+
(
5
-
p
)
3
∫
ℝ
3
|
u
|
6
d
x
>
0
,
as required.
(4) Suppose by contradiction that (𝒢μ∞)′(u)=0 for some u∈ℳμ∞. Then the equation (𝒢μ∞)′(u)=0 can be written in a weak sense as
(3.4)
-
3
Δ
u
+
V
∞
u
+
3
q
2
ϕ
u
u
-
q
2
a
ψ
u
u
=
μ
(
2
p
-
1
)
|
u
|
p
-
1
u
+
9
|
u
|
4
u
.
Define
a
1
=
1
2
∫
ℝ
3
|
∇
u
|
2
d
x
,
b
1
=
1
2
∫
ℝ
3
V
∞
|
u
|
2
d
x
,
c
1
=
1
4
∫
ℝ
3
q
2
ϕ
u
u
2
d
x
,
d
1
=
1
4
∫
ℝ
3
q
2
a
ψ
u
u
2
d
x
,
e
1
=
μ
p
+
1
∫
ℝ
3
|
u
|
p
+
1
d
x
,
f
1
=
1
6
∫
ℝ
3
|
u
|
6
d
x
.
Then we have
(3.5)
{
3
a
1
+
b
1
+
3
c
1
-
d
1
-
(
2
p
-
1
)
e
1
-
9
f
1
=
0
,
6
a
1
+
2
b
1
+
12
c
1
-
4
d
1
-
(
p
+
1
)
(
2
p
-
1
)
e
1
-
54
f
1
=
0
,
3
a
1
+
3
b
1
+
15
c
1
-
2
d
1
-
3
(
2
p
-
1
)
e
1
-
27
f
1
=
0
,
a
1
+
b
1
+
c
1
-
e
1
-
f
1
=
k
,
where the first equation is u∈ℳμ∞, the second one is (𝒢μ∞)′(u)[u]=0, the third one comes from the Pohozaev identity of (3.4) and the last one is the functional ℐμ∞ on ℳμ∞. From (3.5), fixed f1 and d1 as data, due to the Cramer rule, we get
e
1
=
-
48
f
1
+
2
d
1
+
3
k
4
(
p
-
1
)
(
p
-
2
)
<
0
for any f1,d1,k. This is not possible, since p∈(2,5), f1>0, d1>0 and k>0 when u∈ℳμ∞. Thus, we have (𝒢μ∞)′(u)≠0 for every u∈ℳμ∞ and by the implicit function theorem, ℳμ∞ is a C1-manifold.
(5) Let u be a critical point of the functional ℐμ∞, restricted to the manifold ℳμ∞. By the Lagrange multiplier theorem there exists a ν∈ℝ such that
(
ℐ
μ
∞
)
′
(
u
)
+
ν
(
𝒢
μ
∞
)
′
(
u
)
=
0
.
We claim that ν=0. Evaluating the linear functional above at u∈ℳμ∞, we obtain
(
ℐ
μ
∞
)
′
(
u
)
[
u
]
+
ν
(
𝒢
μ
∞
)
′
(
u
)
[
u
]
=
0
,
which is equivalent to
∫
ℝ
3
(
|
∇
u
|
2
+
V
∞
|
u
|
2
+
ϕ
u
u
2
-
μ
|
u
|
p
+
1
-
|
u
|
6
)
d
x
+
ν
(
3
∫
ℝ
3
|
∇
u
|
2
d
x
+
∫
ℝ
3
V
∞
|
u
|
2
d
x
+
3
q
2
∫
ℝ
3
ϕ
u
u
2
d
x
-
∫
ℝ
3
q
2
a
ψ
u
u
2
d
x
-
(
2
p
-
1
)
∫
ℝ
3
μ
|
u
|
p
+
1
d
x
-
9
∫
ℝ
3
|
u
|
6
d
x
)
=
0
.
The above equality is associated with the equation
-
Δ
u
+
V
∞
u
+
q
2
ϕ
u
u
-
|
u
|
p
-
1
u
-
|
u
|
4
u
+
ν
(
3
(
-
Δ
u
)
+
V
∞
u
+
3
q
2
ϕ
u
u
-
q
2
a
ψ
u
u
-
(
2
p
-
1
)
|
u
|
p
-
1
u
-
9
|
u
|
4
u
)
=
0
,
which can be rewritten as
(
1
+
3
ν
)
(
-
Δ
u
)
+
(
1
+
ν
)
V
∞
u
+
(
1
+
3
ν
)
q
2
ϕ
u
u
=
ν
(
q
2
a
)
ψ
u
u
+
(
1
+
ν
(
2
p
-
1
)
)
|
u
|
p
-
1
u
+
(
1
+
9
ν
)
|
u
|
4
u
.
The solutions of this equation satisfy the following Pohozaev identity
1
+
3
ν
2
∫
ℝ
3
|
∇
u
|
2
d
x
+
3
(
1
+
ν
)
2
∫
ℝ
3
V
∞
|
u
|
2
d
x
+
5
(
1
+
3
ν
)
4
∫
ℝ
3
q
2
ϕ
u
u
2
d
x
+
1
-
2
ν
4
∫
ℝ
3
q
2
a
ψ
u
u
2
d
x
=
3
[
1
+
ν
(
2
p
-
1
]
p
+
1
∫
ℝ
3
|
u
|
p
+
1
d
x
+
1
+
9
ν
2
∫
ℝ
3
|
u
|
6
d
x
.
Using the notations of (3), recalling that u∈ℳμ∞, by multiplying the above equation by u and integrating, and by the Pohozaev identity for the above equation, we get the following linear systems of a1,b1,c1,d1,e1,f1. Namely,
(3.6)
{
a
1
+
b
1
+
c
1
-
e
1
-
f
1
=
k
>
0
,
3
a
1
+
b
1
+
3
c
1
-
d
1
-
(
2
p
-
1
)
e
1
-
9
f
1
=
0
,
2
(
1
+
3
ν
)
a
1
+
2
(
1
+
ν
)
b
1
+
4
(
1
+
3
ν
)
c
1
-
4
ν
d
1
-
(
p
+
1
)
(
1
+
ν
(
2
p
-
1
)
)
e
1
-
6
(
1
+
9
ν
)
f
1
=
0
,
(
1
+
3
ν
)
a
1
+
3
(
1
+
ν
)
b
1
+
5
(
1
+
3
ν
)
c
1
+
(
1
-
2
ν
)
d
1
-
3
(
1
+
ν
(
2
p
-
1
)
)
e
1
-
3
(
1
+
9
ν
)
f
1
=
0
.
Indeed, fixed d1 and f1 as data, we see that the coefficient matrix A of (3.6) is
A
=
(
1
1
1
-
1
3
1
3
-
(
2
p
-
1
)
2
(
1
+
3
ν
)
2
(
1
+
ν
)
4
(
1
+
3
ν
)
-
(
p
+
1
)
(
1
+
ν
(
2
p
-
1
)
)
(
1
+
3
ν
)
3
(
1
+
ν
)
5
(
1
+
3
ν
)
-
3
(
1
+
ν
(
2
p
-
1
)
)
)
.
By computation, the determinant of the coefficient matrix A of (3.6) is
det
(
A
)
=
-
16
ν
(
1
+
3
ν
)
(
p
-
1
)
(
p
-
2
)
.
Then
det
(
A
)
=
0
⇔
p
=
1
,
p
=
2
,
ν
=
0
,
ν
=
-
1
3
.
We claim that ν must be equal to zero by excluding the other two possibilities.
(i) If ν≠0, ν≠-13, the linear system (3.6) has a unique solution. Using the Cramer rule, we find that
e
1
=
-
3
k
+
2
d
1
+
48
f
1
4
(
p
-
1
)
(
p
-
2
)
<
0
,
which contradicts the fact that e1>0.
(ii) Assume that ν=-13. In such case, the third equation of (3.6) changes into the following one:
2
b
1
+
2
d
1
+
3
(
p
+
1
)
(
p
-
2
)
e
1
+
18
f
1
=
0
,
which is also impossible, since all b1, d1, e1 and f1 are positive.
In conclusion, ν=0, and as a result, (ℐμ∞)′(u)=0, i.e., u is a critical point of the functional ℐμ∞.
(6) Fix u∈ℳμ∞, so that u∈H1(ℝ3)∖{0} and 𝒢μ∞(u)=0. The Sobolev embedding inequality and (2.5) yield
0
=
3
2
∫
ℝ
3
|
∇
u
|
2
d
x
+
1
2
∫
ℝ
3
V
∞
|
u
|
2
d
x
+
3
q
2
4
∫
ℝ
3
ϕ
u
u
2
d
x
-
1
4
∫
ℝ
3
q
2
a
ψ
u
u
2
d
x
-
2
p
-
1
p
+
1
∫
ℝ
3
|
u
|
p
+
1
d
x
-
3
2
∫
ℝ
3
|
u
|
6
d
x
≥
1
2
∥
u
∥
2
-
2
p
-
1
p
+
1
∫
ℝ
3
|
u
|
p
+
1
d
x
≥
1
2
C
p
∥
u
∥
p
+
1
2
-
2
p
-
1
p
+
1
∥
u
∥
p
+
1
p
+
1
.
Then
∥
u
∥
p
+
1
≥
[
C
p
(
p
+
1
)
2
(
2
p
-
1
)
]
1
p
-
1
:=
C
,
as required. ∎
Property (5) of Lemma 3.1 insures that it is enough to find critical points of ℐμ∞ restricted to ℳμ∞. Set
c
1
∞
=
inf
g
∈
Γ
max
θ
∈
[
0
,
1
]
ℐ
μ
∞
(
g
(
θ
)
)
,
c
2
∞
=
inf
u
≠
0
max
θ
≥
0
ℐ
μ
∞
(
u
θ
)
,
c
3
∞
=
inf
u
∈
ℳ
μ
∞
ℐ
μ
∞
(
u
)
,
where Γ={γ∈C([0,1],H1(ℝ3)):γ(0)=0,ℐμ∞(γ(1))≤0,γ(1)≠0}.
Lemma 3.2.
The following relations hold:
c
∞
:=
c
1
∞
=
c
2
∞
=
c
3
∞
>
0
.
Proof.
When p∈(2,5), we have
(3.7)
ℐ
μ
∞
(
u
θ
)
=
θ
3
2
∫
|
∇
u
|
2
d
x
+
θ
2
∫
V
∞
|
u
|
2
d
x
+
θ
3
q
2
4
∫
ℝ
3
∫
ℝ
3
1
-
e
-
|
x
-
y
|
θ
a
|
x
-
y
|
u
2
(
x
)
u
2
(
y
)
d
x
d
y
-
μ
θ
2
p
-
1
p
+
1
∫
ℝ
3
|
u
|
p
+
1
d
x
-
θ
9
6
∫
ℝ
3
|
u
|
6
d
x
→
-
∞
,
as θ→∞. This implies that c2∞=c3∞.
From (3.7) we see that ℐμ∞(uθ)<0 for u∈H1(ℝ3)∖{0} and θ large enough. Thus, c1∞≤c2∞.
On the other hand, for any γ∈Γ, we claim that γ([0,1])∩ℳμ∞≠∅. Indeed, for every u in
𝒢
:=
{
u
∈
H
1
(
ℝ
3
)
∖
{
0
}
:
𝒢
μ
∞
(
u
)
≥
0
}
∪
{
0
}
the Sobolev embedding theorem and (2.5) give
3
2
∫
ℝ
3
|
∇
u
|
2
d
x
+
1
2
∫
ℝ
3
V
∞
|
u
|
2
d
x
+
3
q
2
4
∫
ℝ
3
ϕ
u
u
2
d
x
-
q
2
4
a
∫
ℝ
3
ψ
u
u
2
d
x
-
2
p
-
1
p
+
1
∫
ℝ
3
|
u
|
p
+
1
d
x
-
3
2
∫
ℝ
3
|
u
|
6
d
x
≥
1
2
∥
u
∥
2
-
C
1
∥
u
∥
p
+
1
-
C
2
∥
u
∥
6
.
This implies that there exists a small neighborhood of 0 such that it is contained in 𝒢. Furthermore, for every u∈𝒢, we have
3
ℐ
μ
∞
(
u
)
=
𝒢
μ
∞
(
u
)
+
∫
ℝ
3
V
∞
u
2
d
x
+
2
(
p
-
2
)
p
+
1
∫
ℝ
3
|
u
|
p
+
1
d
x
+
∫
ℝ
3
|
u
|
6
d
x
+
1
4
∫
ℝ
3
q
2
a
ψ
u
u
2
d
x
≥
0
and ℐμ∞(u)>0 if u≠0. Hence, for any γ∈Γ, satisfying γ(0)=0, ℐμ∞(γ(1))≤0 and γ(1)≠0, the curve γ must across the manifold ℳμ∞. Therefore, c1∞≥c3∞. ∎
Now, we present an upper bound estimate for c∞, which is very important for prove the (PS) condition.
Lemma 3.3.
The following inequalities hold:
0
<
c
∞
<
1
3
𝒮
3
2
,
where S is defined in (2.1), if one of the next conditions is satisfied:
3
<
p
<
5
and
μ
>
0
,
2
<
p
≤
3
and
μ
>
0
large enough.
Proof.
Let us define for all x∈ℝ3,
(3.8)
u
ε
(
x
)
=
ψ
(
x
)
U
ε
(
x
)
,
where
U
ε
(
x
)
=
3
1
4
ε
1
2
(
ε
2
+
|
x
|
2
)
1
2
,
and ψ∈Cc∞(ℝ3) is such that 0≤ψ≤1 in ℝ3, ψ(x)≡1 in Bδ and ψ≡0 in ℝ3∖B2δ, for some δ>0. We know from (2.1) and (3.8) that
(3.9)
∫
ℝ
3
|
∇
u
ε
(
x
)
|
2
d
x
≤
𝒮
3
2
+
O
(
ε
)
,
∫
ℝ
3
|
u
ε
(
x
)
|
6
d
x
≥
𝒮
3
2
+
O
(
ε
3
)
.
By computation, we can deduce that
(3.10)
∫
ℝ
3
|
u
ε
(
x
)
|
p
d
x
=
{
O
(
ε
6
-
p
2
)
,
p
>
3
,
O
(
ε
p
2
|
ln
ε
|
)
,
p
=
3
,
O
(
ε
p
2
)
,
p
<
3
.
From the definition of c∞, it is clear that there holds
(3.11)
0
<
c
∞
≤
max
θ
≥
0
ℐ
μ
∞
(
(
u
ε
)
θ
)
.
Set
g
(
θ
)
=
θ
3
2
∫
ℝ
3
|
∇
u
ε
|
2
d
x
-
θ
9
6
∫
ℝ
3
|
u
ε
|
6
d
x
,
θ
≥
0
.
A direct calculation shows that g attains its maximum at
θ
0
=
(
∫
ℝ
3
|
∇
u
ε
|
2
d
x
∫
ℝ
3
|
u
ε
|
6
d
x
)
1
6
.
Moreover, by (3.9), using the inequality (a+b)p≤ap+p(a+b)p-1b, which holds for any p≥1 and a, b≥0, we deduce that
(3.12)
max
θ
≥
0
g
(
θ
)
=
g
(
θ
0
)
=
1
2
(
∫
ℝ
3
|
∇
u
ε
|
2
d
x
∫
ℝ
3
|
u
ε
|
6
d
x
)
3
6
∫
ℝ
3
|
∇
u
ε
|
2
d
x
-
1
6
(
∫
ℝ
3
|
∇
u
ε
|
2
d
x
∫
ℝ
3
|
u
ε
|
6
d
x
)
3
2
∫
ℝ
3
|
u
ε
|
6
d
x
=
1
3
⋅
∥
∇
u
ε
∥
2
3
∥
u
ε
∥
6
3
≤
1
3
⋅
[
𝒮
3
2
+
O
(
ε
)
]
3
2
[
𝒮
3
2
+
O
(
ε
3
)
]
1
/
2
≤
1
3
𝒮
3
2
+
O
(
ε
)
.
Since ℐμ∞((uε)θ)→-∞ as θ→∞, there exists θε>0 such that
ℐ
μ
∞
(
(
u
ε
)
θ
ε
)
=
max
θ
≥
0
ℐ
μ
∞
(
(
u
ε
)
θ
)
>
0
.
Since 0 is a local minimum of ℐμ∞, there exists a constant C>0, independent of ε, such that ℐμ∞((uε)θε)≥C>0. This implies that θε≥θ1>0, where θ1 is a positive constant. Otherwise, there should exist a sequence (εn)n such that
lim
n
→
∞
ε
n
=
0
=
lim
n
→
∞
θ
ε
n
.
Then Lemma 2.4 and (3.9)–(3.11) imply as n→∞
c
0
<
c
∞
≤
ℐ
μ
∞
(
(
u
ε
n
)
θ
ε
n
)
+
o
(
1
)
=
θ
ε
n
3
2
∫
|
∇
u
ε
n
|
2
d
x
+
θ
ε
n
2
∫
ℝ
3
V
∞
|
u
ε
n
|
2
d
x
+
θ
ε
n
3
4
∫
ℝ
3
ϕ
u
ε
n
u
ε
n
2
(
x
)
d
x
+
o
(
1
)
≤
θ
ε
n
3
2
∫
|
∇
u
ε
n
|
2
d
x
+
θ
ε
n
2
∫
ℝ
3
V
∞
|
u
ε
n
|
2
d
x
+
θ
ε
n
3
4
∫
ℝ
3
∫
ℝ
3
u
ε
n
2
(
x
)
u
ε
n
2
(
y
)
|
x
-
y
|
d
x
d
y
-
μ
θ
ε
n
2
p
-
1
p
+
1
∫
ℝ
3
|
u
ε
n
|
p
+
1
d
x
-
θ
ε
n
9
6
∫
ℝ
3
|
u
ε
n
|
6
d
x
+
o
(
1
)
≤
θ
ε
n
3
2
∫
ℝ
3
|
∇
u
ε
n
|
2
d
x
+
θ
ε
n
2
∫
ℝ
3
V
∞
|
u
ε
n
|
2
d
x
+
θ
ε
n
3
4
(
∫
ℝ
3
|
u
ε
n
|
12
5
d
x
)
5
3
-
μ
θ
ε
n
2
p
-
1
p
+
1
∫
ℝ
3
|
u
ε
n
|
p
+
1
d
x
-
θ
ε
n
9
6
∫
ℝ
3
|
u
ε
n
|
6
d
x
+
o
(
1
)
=
o
(
1
)
.
This is clearly impossible and the claim follows.
On the other hand, from (3.9)–(3.10), for any ε>0,
0
<
C
≤
ℐ
μ
∞
(
(
u
ε
)
θ
ε
)
≤
C
1
θ
ε
3
+
C
2
θ
ε
-
C
3
θ
ε
9
,
which implies that there exists θ2>0 such that θε≤θ2. Thus, 0<θ1≤θε≤θ2 for any ε>0. Now, by (3.12), we deduce that
ℐ
μ
∞
(
(
u
ε
)
θ
ε
)
≤
1
3
𝒮
3
2
+
O
(
ε
)
+
θ
ε
2
∫
ℝ
3
|
u
ε
|
2
d
x
+
θ
ε
3
4
∫
ℝ
3
ϕ
u
ε
u
ε
2
d
x
-
μ
θ
ε
2
p
-
1
p
+
1
∫
ℝ
3
|
u
ε
|
p
+
1
d
x
-
θ
ε
9
6
∫
ℝ
3
|
u
ε
|
6
d
x
≤
1
3
𝒮
3
2
+
O
(
ε
)
+
θ
2
2
∫
ℝ
3
|
u
ε
|
2
d
x
+
θ
2
3
𝒮
4
(
∫
ℝ
3
|
u
ε
|
12
5
d
x
)
5
3
-
μ
θ
1
2
p
-
1
p
+
1
∫
ℝ
3
|
u
ε
|
p
+
1
d
x
.
Next,
ℐ
μ
∞
(
(
u
ε
)
θ
ε
)
≤
1
3
𝒮
3
2
+
O
(
ε
)
+
C
(
∫
ℝ
3
|
u
ε
|
12
5
d
x
)
5
3
-
C
∫
ℝ
3
|
u
ε
|
p
+
1
d
x
.
Observing that p>2, we have
lim
ε
→
0
+
(
∫
ℝ
3
|
u
ε
|
12
5
d
x
)
5
3
ε
≤
lim
ε
→
0
+
O
(
ε
2
)
ε
=
0
,
and noting that 2-12(p+1)<0 if 4<p+1<6, we have
lim
ε
→
0
μ
∫
ℝ
3
|
u
ε
|
p
+
1
d
x
ε
=
{
lim
ε
→
0
μ
O
(
ε
2
-
1
2
(
p
+
1
)
)
=
∞
,
4
<
p
+
1
<
6
,
lim
ε
→
0
μ
O
(
ε
2
-
1
2
(
p
+
1
)
)
,
3
<
p
+
1
<
4
,
lim
ε
→
0
μ
O
(
ε
2
-
1
2
(
p
+
1
)
|
ln
ε
|
)
,
p
+
1
=
3
.
We can choose μ large enough such that the other two limits are equal to ∞, for instance, μ=ε-2.
From the above inequalities, we conclude that
ℐ
μ
∞
(
(
u
ε
)
θ
ε
)
<
1
3
𝒮
3
2
for ε small enough and, combining with (3.11), we complete the proof. ∎
For a minimizing sequence for c∞ on the manifold ℳμ∞ we obtain the following compactness result.
Lemma 3.4.
Let (un)n⊂Mμ∞ be a minimizing sequence for c∞, given in Lemma 3.2. Then there exists a sequence (yn)n⊂R3 such that for any ε>0, there is an R>0 satisfying
∫
ℝ
3
∖
B
R
(
y
n
)
(
|
∇
u
n
|
2
+
V
∞
u
n
2
)
d
x
≤
ε
for all n sufficiently large.
Proof.
Following the idea of [18], let (un)n⊂ℳμ∞ be such that
(3.13)
0
<
lim
n
→
∞
ℐ
μ
∞
(
u
n
)
=
c
∞
<
1
3
𝒮
3
2
.
Since (un)n⊂ℳμ∞, by (3.3) we see that
(3.14)
ℐ
μ
∞
(
u
n
)
=
1
2
∫
ℝ
3
(
|
∇
u
n
|
2
+
V
∞
|
u
n
|
2
)
d
x
+
1
4
∫
ϕ
u
n
u
n
2
d
x
-
μ
p
+
1
∫
|
u
n
|
p
+
1
d
x
-
1
6
∫
ℝ
3
|
u
n
|
6
d
x
=
p
-
2
2
p
-
1
∫
ℝ
3
|
∇
u
n
|
2
d
x
+
p
-
1
2
p
-
1
∫
ℝ
3
V
∞
|
u
n
|
2
d
x
+
q
2
(
p
-
2
)
2
(
2
p
-
1
)
∫
ℝ
3
ϕ
u
n
u
n
2
d
x
+
q
2
4
a
(
2
p
-
1
)
∫
ℝ
3
ψ
u
u
2
d
x
+
5
-
p
3
(
2
p
-
1
)
∫
ℝ
3
|
u
n
|
6
d
x
≜
J
(
u
n
)
≥
0
.
From (3.13), it follows that (un)n is bounded in H1(ℝ3). Next, we use Lemma 2.9 to conclude the compactness of the sequence (un)n. Let
ρ
n
=
p
-
2
2
p
-
1
|
∇
u
n
|
2
+
p
-
1
2
p
-
1
V
∞
|
u
n
|
2
+
q
2
(
p
-
2
)
2
(
2
p
-
1
)
ϕ
u
n
u
n
2
+
q
2
4
a
(
2
p
-
1
)
ψ
u
u
2
+
5
-
p
3
(
2
p
-
1
)
|
u
n
|
6
.
Then (ρn)n is a sequence of nonnegative L1 functions on ℝ3 by (3.14), and as n→∞
(3.15)
∫
ℝ
3
ρ
n
d
x
→
c
∞
,
thanks to (3.13).
(i) Vanishing does not occur. Suppose by contradiction that for all R>0,
lim
n
→
∞
sup
y
∈
ℝ
3
∫
B
R
(
y
)
ρ
n
(
x
)
d
x
=
0
.
Hence
lim
n
→
∞
sup
y
∈
ℝ
3
∫
B
R
(
y
)
|
u
n
|
2
d
x
=
0
and
lim
n
→
∞
sup
y
∈
ℝ
3
∫
B
R
(
y
)
|
u
n
|
6
d
x
=
0
.
The Vanishing Lemma 2.7 and Lemma 2.8 yield that
(3.16)
u
n
→
0
in
L
t
(
ℝ
3
)
for any
t
∈
(
2
,
6
]
.
A consequence of Lemma 2.4 gives as n→∞
(3.17)
∫
ℝ
3
ϕ
u
n
u
n
2
d
x
→
0
and
∫
ℝ
3
ψ
u
n
u
n
2
d
x
→
0
.
Since un∈ℳμ∞, properties (3.14), (3.16) and (3.17) imply that
lim
n
→
∞
ℐ
μ
∞
(
u
n
)
=
0
.
This contradicts either (3.13) or (3.15).
(ii) Dichotomy does not occur. Suppose by contradiction that there exist an α∈(0,c∞) and (yn)n⊂ℝ3 such that for all ε>0 there is (Rn)n⊂ℝ+, with Rn→∞, satisfying
(3.18)
lim sup
n
→
∞
(
|
α
-
∫
B
R
n
(
y
n
)
ρ
n
d
x
|
+
|
c
∞
-
α
-
∫
ℝ
3
∖
B
2
R
n
(
y
n
)
ρ
n
d
x
|
)
<
ε
.
Let ξ:ℝ+∪{0}→ℝ+ be a cut-off function such that 0≤ξ≤1, ξ(t)=1 for t≤1,ξ(t)=0 for t≥2 and |ξ′(t)|≤2. Set
v
n
(
x
)
=
ξ
(
|
x
-
y
n
|
R
n
)
u
n
(
x
)
,
w
n
(
x
)
=
(
1
-
ξ
(
|
x
-
y
n
|
R
n
)
)
u
n
(
x
)
.
Then by (3.14) and (3.18), we see that
lim inf
n
→
∞
∫
ℝ
3
J
(
v
n
)
d
x
≥
α
,
lim inf
n
→
∞
∫
ℝ
3
J
(
w
n
)
d
x
≥
c
∞
-
α
.
Put Ωn=B2Rn(yn)∖BRn(yn). The above inequalities and (3.14) give
∫
Ω
n
(
|
∇
u
n
|
2
+
V
∞
u
n
2
)
d
x
→
0
,
∫
Ω
n
ϕ
u
n
u
n
2
d
x
→
0
,
∫
Ω
n
|
u
n
|
6
d
x
→
0
,
as n→∞. By direct computation we have as n→∞
(3.19)
∫
ℝ
3
|
∇
u
n
|
2
d
x
=
∫
ℝ
3
|
∇
v
n
|
2
d
x
+
∫
ℝ
3
|
∇
w
n
|
2
d
x
+
o
n
(
1
)
,
(3.20)
∫
ℝ
3
u
n
2
(
x
)
d
x
=
∫
ℝ
3
v
n
2
(
x
)
d
x
+
∫
ℝ
3
w
n
2
(
x
)
d
x
+
o
n
(
1
)
,
(3.21)
∫
ℝ
3
|
u
n
|
p
+
1
d
x
=
∫
ℝ
3
|
v
n
|
p
+
1
d
x
+
∫
ℝ
3
|
w
n
|
p
+
1
d
x
+
o
n
(
1
)
,
(3.22)
∫
ℝ
3
|
u
n
|
6
d
x
=
∫
ℝ
3
|
v
n
|
6
d
x
+
∫
ℝ
3
|
w
n
|
6
d
x
+
o
n
(
1
)
,
(3.23)
∫
ℝ
3
ϕ
u
n
u
n
2
d
x
≥
∫
ϕ
v
n
v
n
2
d
x
+
∫
ℝ
3
ϕ
w
n
w
n
2
d
x
+
o
n
(
1
)
,
(3.24)
∫
ℝ
3
ψ
u
n
u
n
2
d
x
≥
∫
ψ
v
n
v
n
2
d
x
+
∫
ℝ
3
ψ
w
n
w
n
2
d
x
+
o
n
(
1
)
.
Hence, by (3.19), (3.20) and (3.22)–(3.24), we get
J
(
u
n
)
≥
J
(
v
n
)
+
J
(
w
n
)
+
o
n
(
1
)
n
→
∞
. Then
c
∞
=
lim
n
→
∞
J
(
u
n
)
≥
lim inf
n
→
∞
J
(
v
n
)
+
lim inf
n
→
∞
J
(
w
n
)
≥
α
+
c
∞
-
α
=
c
∞
.
Therefore,
(3.25)
lim
n
→
∞
J
(
v
n
)
=
α
,
lim
n
→
∞
J
(
w
n
)
=
c
∞
-
α
.
Now, 𝒢μ∞(un)=0, since un∈ℳμ∞. By (3.19)–(3.24) we have
(3.26)
0
=
𝒢
μ
∞
(
u
n
)
≥
𝒢
μ
∞
(
v
n
)
+
𝒢
μ
∞
(
w
n
)
+
o
n
(
1
)
.
We distinguish the following two cases.
Case 1. Up to a subsequence, we may assume that either 𝒢μ∞(vn)≤0 or 𝒢μ∞(wn)≤0. Without loss of generality, we suppose that 𝒢μ∞(vn)≤0. Then
3
2
∫
ℝ
3
|
∇
v
n
|
2
d
x
+
1
2
∫
ℝ
3
V
∞
|
v
n
|
2
d
x
+
3
q
2
4
∫
ℝ
3
ϕ
v
n
v
n
2
d
x
-
q
2
4
a
∫
ℝ
3
ψ
v
n
v
n
2
d
x
-
μ
2
p
-
1
p
+
1
∫
ℝ
3
|
v
n
|
p
+
1
d
x
-
3
2
∫
ℝ
3
|
v
n
|
6
d
x
≤
0
.
By Lemma 3.1 for any n there exists θn>0 such that θn2vn(θx)∈ℳμ∞ and so 𝒢μ∞(θn2vn(θx))=0. Thus, (2.5) yields that
3
2
(
θ
n
2
(
p
-
1
)
-
θ
n
2
)
∫
ℝ
3
|
∇
v
n
|
2
d
x
+
1
2
(
θ
n
2
(
p
-
1
)
-
1
)
∫
ℝ
3
V
∞
|
v
n
|
2
d
x
+
3
q
2
4
(
θ
n
2
(
p
-
1
)
-
θ
n
2
)
∫
ℝ
3
∫
ℝ
3
1
-
e
-
|
x
-
y
|
θ
n
a
|
x
-
y
|
u
2
(
x
)
u
2
(
y
)
d
x
d
y
-
q
2
4
a
(
θ
n
2
(
p
-
1
)
-
θ
n
2
)
∫
ℝ
3
∫
ℝ
3
e
-
|
x
-
y
|
θ
n
a
u
2
(
x
)
u
2
(
y
)
d
x
d
y
+
3
2
(
θ
n
8
-
θ
n
2
(
p
-
1
)
)
∫
ℝ
3
|
v
n
|
2
s
*
d
x
=
3
2
(
θ
n
2
(
p
-
1
)
-
θ
n
2
)
∫
ℝ
3
|
∇
v
n
|
2
d
x
+
1
2
(
θ
n
2
(
p
-
1
)
-
1
)
∫
ℝ
3
V
∞
|
v
n
|
2
d
x
+
q
2
4
(
θ
n
2
(
p
-
1
)
-
θ
n
2
)
∫
ℝ
3
∫
ℝ
3
3
(
1
-
e
-
|
x
-
y
|
θ
n
a
)
-
|
x
-
y
|
θ
n
a
e
-
|
x
-
y
|
θ
a
|
x
-
y
|
u
2
(
x
)
u
2
(
y
)
d
x
d
y
+
3
2
(
θ
n
8
-
θ
n
2
(
p
-
1
)
)
∫
ℝ
3
|
v
n
|
2
s
*
d
x
≤
0
,
which implies that θn≤1. Then by (3.25), we have
(3.27)
c
∞
≤
ℐ
μ
∞
(
θ
n
2
v
n
(
θ
x
)
)
=
J
(
θ
n
2
v
n
(
θ
x
)
)
≤
J
(
v
n
)
→
α
<
c
∞
,
which is the desired contradiction.
Case 2. Up to a subsequence, we may assume that 𝒢μ∞(vn)>0 and 𝒢μ∞(wn)>0. By formula (3.26), we see that 𝒢μ∞(vn)=on(1) and 𝒢μ∞(wn)=on(1). Repeating the argument of Case 1, we obtain a contradiction of type (3.27). Thus we suppose that
lim
n
→
∞
θ
n
=
θ
0
>
1
.
Now, as n→∞
o
n
(
1
)
=
𝒢
μ
∞
(
v
n
)
=
3
2
∫
ℝ
3
|
∇
v
n
|
2
d
x
+
1
2
∫
ℝ
3
V
∞
|
v
n
|
2
d
x
+
3
q
2
4
∫
ℝ
3
ϕ
v
n
v
n
2
d
x
-
q
2
4
a
∫
ℝ
3
ψ
v
n
v
n
2
d
x
-
μ
2
p
-
1
p
+
1
∫
ℝ
3
|
v
n
|
p
+
1
d
x
-
3
2
∫
ℝ
3
|
v
n
|
6
d
x
≤
3
2
(
1
-
1
θ
n
2
p
-
4
)
∫
ℝ
3
|
∇
v
n
|
2
d
x
+
1
2
(
1
-
1
θ
n
2
(
p
-
1
)
)
∫
ℝ
3
V
∞
|
v
n
|
2
d
x
+
3
q
2
4
(
1
-
1
θ
n
2
p
-
4
)
∫
ℝ
3
∫
ℝ
3
1
-
e
-
|
x
-
y
|
θ
n
a
|
x
-
y
|
u
2
(
x
)
u
2
(
y
)
d
x
d
y
-
q
2
4
a
(
1
-
1
θ
n
2
p
-
4
)
∫
ℝ
3
∫
ℝ
3
e
-
|
x
-
y
|
θ
n
a
u
2
(
x
)
u
2
(
y
)
d
x
d
y
+
3
2
(
θ
n
2
(
5
-
p
)
-
1
)
∫
ℝ
3
|
v
n
|
6
d
x
.
This implies that vn→0 in H1(ℝ3), which is impossible by (3.25). Hence, we conclude that dichotomy cannot occur.
As a result, compactness holds for the sequence (ρn)n by Lemma 2.9, i.e., there exists (yn)n⊂ℝ3 such that for any ε>0 there is an R>0 satisfying
lim inf
n
→
∞
∫
B
R
(
y
n
)
ρ
n
(
x
)
d
x
≥
c
∞
-
ε
.
Since limn→∞ℐμ∞(un)=limn→∞J(un)=c∞, it follows that
ε
>
c
∞
-
(
c
∞
-
ε
)
≥
lim
n
→
∞
J
(
u
n
)
-
lim inf
n
→
∞
∫
B
R
(
y
n
)
ρ
n
(
x
)
d
x
=
lim inf
n
→
∞
∫
ℝ
3
∖
B
R
(
y
n
)
ρ
n
(
x
)
d
x
,
which implies that the conclusion holds true for all n sufficiently large. ∎
Proof of Theorem 1.2.
Let (un)n⊂ℳμ∞ be a minimizing sequence for c∞. By Lemma 3.4 there exists (yn)n⊂ℝ3 such that for any ε>0 there exists an R>0 satisfying
(3.28)
∫
ℝ
3
∖
B
R
(
y
n
)
(
|
∇
u
n
|
2
+
V
∞
u
n
2
)
d
x
≤
ε
.
Define un~(x)=un(x-yn)∈H1(ℝ3); then we have ϕun~=ϕun(⋅-yn) by Lemma 2.4 and thus un~∈ℳμ∞ and also ℐμ∞(un)=ℐμ∞(un~). This means that (un~)n is also a minimizing sequence for c∞. Hence, by (3.28) for any ε>0 there exists an R>0 such that
(3.29)
∫
ℝ
3
∖
B
R
(
0
)
(
|
∇
u
n
~
|
2
+
V
∞
u
n
~
2
)
d
x
≤
ε
.
Since (un~)n is bounded in H1(ℝ3), up to a subsequence, we may assume that there exists u~∈H1(ℝ3) such that
(3.30)
{
u
n
~
⇀
u
~
in
H
1
(
ℝ
3
)
,
u
n
~
→
u
~
in
L
loc
r
(
ℝ
3
)
, with
1
≤
r
<
6
,
u
n
~
→
u
~
a.e. in
ℝ
3
.
By Fatou’s lemma and (3.29) we get
(3.31)
∫
ℝ
3
∖
B
R
(
0
)
(
|
∇
u
~
|
2
+
V
∞
u
~
2
)
d
x
≤
ε
.
By (3.29)–(3.31), and the Sobolev embedding theorem, we have that for any r∈[2,6) and any ε>0 there exists a C>0 such that
∫
ℝ
3
|
u
n
~
-
u
~
|
r
d
x
=
∫
B
R
(
0
)
|
u
n
~
-
u
~
|
r
d
x
+
∫
ℝ
3
∖
B
R
(
0
)
|
u
n
~
-
u
~
|
r
d
x
≤
ε
+
C
(
∥
u
n
~
∥
H
1
(
ℝ
3
∖
B
R
(
0
)
)
+
∥
u
~
∥
H
1
(
ℝ
3
∖
B
R
(
0
)
)
)
≤
(
1
+
2
C
)
ε
,
where C>0 is the constant of the embedding H1(BR(0))↪Lr(BR(0)). Hence, we have proved that
(3.32)
u
n
~
→
u
~
in
L
r
(
ℝ
3
)
for any
r
∈
[
2
,
6
)
.
Since un~∈ℳμ∞, Lemma 3.1 yields that ∥un~∥p+1≥C>0. Hence ∥u~∥p+1≥C>0, and as a result u~≠0.
Finally, we show that un~→u~ in H1(ℝ3). From Lemma 2.4 and (3.32) we deduce that
ϕ
u
n
~
→
ϕ
u
~
in
𝒟
(
ℝ
3
)
,
and thus
(3.33)
∫
ℝ
3
ϕ
u
n
~
u
n
~
2
d
x
→
∫
ℝ
3
ϕ
u
~
u
~
2
d
x
.
Set vn~=un~-u~. By (3.30), we have as n→∞
∥
u
~
n
∥
2
-
∥
u
~
∥
2
=
∥
v
n
~
∥
2
+
o
n
(
1
)
,
which implies that
(3.34)
∥
∇
u
n
~
∥
2
2
-
∥
∇
u
~
∥
2
2
=
∥
∇
v
~
n
∥
2
2
+
o
n
(
1
)
.
The Brézis–Lieb lemma and (3.30) give as n→∞
(3.35)
∥
u
~
n
∥
6
6
-
∥
u
~
∥
6
6
=
∥
v
~
n
∥
6
6
+
o
n
(
1
)
.
Hence, from (3.30), (3.33), (3.34) and (3.35) it follows as n→∞ that
(3.36)
c
∞
-
ℐ
μ
∞
(
u
~
)
=
ℐ
μ
∞
(
u
n
~
)
-
ℐ
μ
∞
(
u
~
)
+
o
n
(
1
)
=
1
2
∥
∇
v
~
n
∥
2
2
-
1
6
∥
v
n
~
∥
6
6
+
o
n
(
1
)
.
Next, we claim that ℐμ∞(u~)≥0. We first prove that 𝒢μ∞(u~)≥0. Suppose by contradiction that 𝒢μ∞(u~)<0 and so there exists θ∈(0,1) such that u~θ∈ℳμ∞. By (3.3) and (3.30), we deduce that
ℐ
μ
∞
(
u
~
θ
)
=
p
-
2
2
p
-
1
∫
ℝ
3
|
∇
u
~
θ
|
2
d
x
+
p
-
1
2
p
-
1
∫
ℝ
3
V
∞
|
u
~
θ
|
2
d
x
+
p
-
2
4
p
-
2
∫
ℝ
3
q
2
ϕ
u
~
θ
|
u
~
θ
|
2
d
x
+
q
2
4
a
(
2
p
-
1
)
∫
ℝ
3
∫
ℝ
3
e
-
|
x
-
y
|
a
u
~
θ
2
(
x
)
u
~
θ
2
(
y
)
d
x
d
y
+
5
-
p
6
p
-
3
∫
ℝ
3
|
u
~
θ
|
6
d
x
=
p
-
2
2
p
-
1
θ
3
∫
ℝ
3
|
∇
u
~
|
2
d
x
+
p
-
1
2
p
-
1
θ
∫
ℝ
3
V
∞
|
u
~
|
2
d
x
+
p
-
2
4
p
-
2
θ
3
∫
ℝ
3
q
2
ϕ
u
¯
|
u
~
|
2
d
x
+
θ
2
q
2
4
a
(
2
p
-
1
)
∫
ℝ
3
∫
ℝ
3
e
-
|
x
-
y
|
θ
a
u
~
2
(
x
)
u
~
2
(
y
)
d
x
d
y
+
5
-
p
6
p
-
3
θ
9
∫
ℝ
3
|
u
~
|
6
d
x
<
p
-
2
2
p
-
1
∫
ℝ
3
|
∇
u
~
|
2
d
x
+
p
-
1
2
p
-
1
∫
ℝ
3
V
∞
|
u
~
|
2
d
x
+
p
-
2
4
p
-
2
∫
ℝ
3
ϕ
u
~
|
u
~
|
2
d
x
+
q
2
4
a
(
2
p
-
1
)
∫
ℝ
3
∫
ℝ
3
e
-
|
x
-
y
|
a
u
~
2
(
x
)
u
~
2
(
y
)
d
x
d
y
+
5
-
p
6
p
-
3
∫
ℝ
3
|
u
~
|
6
d
x
≤
lim inf
n
→
∞
[
p
-
2
2
p
-
1
∫
ℝ
3
|
∇
u
n
~
|
2
d
x
+
p
-
1
2
p
-
1
∫
ℝ
3
V
∞
|
u
n
~
|
2
d
x
+
p
-
2
4
p
-
2
∫
ℝ
3
q
2
ϕ
u
~
n
|
u
~
|
2
d
x
+
q
2
4
a
(
2
p
-
1
)
∫
ℝ
3
∫
ℝ
3
e
-
|
x
-
y
|
a
u
~
n
2
(
x
)
u
~
n
2
(
y
)
d
x
d
y
+
5
-
p
6
p
-
3
∫
ℝ
3
|
u
~
n
|
6
d
x
]
≤
lim inf
n
→
∞
ℐ
μ
∞
(
u
n
~
)
=
c
∞
,
which contradicts the fact that ℐμ∞(u~θ)≥c∞. Therefore, the claim is proved and we can infer that
(
2
p
-
1
)
ℐ
μ
∞
(
u
~
)
≥
(
2
p
-
1
)
ℐ
μ
∞
(
u
~
)
-
𝒢
μ
∞
(
u
~
)
=
(
p
-
2
)
∫
ℝ
3
|
∇
u
~
|
2
d
x
+
(
p
-
1
)
∫
ℝ
3
V
∞
|
u
~
|
2
d
x
+
p
-
2
2
∫
ℝ
3
q
2
ϕ
u
~
|
u
~
|
2
d
x
+
5
-
p
3
∫
ℝ
3
|
u
~
|
6
d
x
+
q
2
4
a
∫
ℝ
3
∫
ℝ
3
e
-
|
x
-
y
|
a
u
~
2
(
x
)
u
~
2
(
y
)
d
x
d
y
>
0
.
Hence, Lemma 3.3 and (3.36) yield as n→∞
(3.37)
1
2
∥
∇
v
~
n
∥
2
2
-
1
6
∥
v
~
n
∥
6
6
+
o
n
(
1
)
=
c
∞
-
ℐ
μ
∞
(
u
~
)
<
1
3
𝒮
3
2
.
On the other hand, it follows from (3.30) that
(
ℐ
μ
∞
)
′
(
v
~
n
)
[
φ
]
=
∫
ℝ
3
(
∇
v
~
n
∇
φ
+
V
∞
v
~
n
φ
)
d
x
+
∫
ℝ
3
ϕ
v
~
n
v
~
n
φ
d
x
-
μ
∫
ℝ
3
|
v
~
n
|
p
-
1
v
n
φ
d
x
-
∫
ℝ
3
|
v
n
~
|
4
v
n
φ
d
x
→
0
for any φ∈H1(ℝ3). We take φ=v~n, using (3.30) and the boundedness of (vn~)n, we have as n→∞
(3.38)
∥
∇
v
n
~
∥
2
2
-
∥
v
n
~
∥
6
6
=
o
n
(
1
)
.
We may assume that limn→∞∥∇v~n∥22=l≥0. Thus, (3.38) gives that limn→∞∥vn~∥66=l. If l>0, from the definition of 𝒮, we have that
𝒮
≤
∥
∇
v
~
n
∥
2
2
∥
v
n
~
∥
6
6
,
which implies that l≥𝒮32. Therefore, we get
lim
n
→
∞
[
1
2
∥
∇
v
n
~
∥
2
2
-
1
6
∥
v
n
~
∥
6
6
]
=
1
3
l
≥
1
3
𝒮
3
2
.
This contradicts (3.37). Hence, we have l=0, that is un~→u~ in H1(ℝ3) and so we conclude that u~∈ℳμ∞ and ℐμ∞(u~)=c∞. ∎
4 The Nonconstant Potential Case
Let Λ=[12,1]. We consider a family of functionals ℐV,λ:H1(ℝ3)→ℝ defined by
ℐ
V
,
λ
(
u
)
=
1
2
∫
ℝ
3
(
|
∇
u
|
2
+
V
(
x
)
u
2
)
d
x
+
q
2
4
∫
ℝ
3
ϕ
u
u
2
d
x
-
μ
λ
p
+
1
∫
ℝ
3
|
u
|
p
+
1
d
x
-
λ
6
∫
ℝ
3
|
u
|
6
d
x
.
Let ℐV,λ(u)=A(u)-λB(u), where
A
(
u
)
=
1
2
∫
ℝ
3
(
|
∇
u
|
2
+
V
(
x
)
u
2
)
d
x
+
q
2
4
∫
ℝ
3
ϕ
u
u
2
d
x
→
∞
as
∥
u
∥
→
∞
and
B
(
u
)
=
μ
p
+
1
∫
ℝ
3
|
u
|
p
+
1
d
x
+
1
6
∫
ℝ
3
|
u
|
6
d
x
.
Let us first show that the family {ℐV,λ}λ∈Λ verifies all the assumptions of Theorem 2.10.
Lemma 4.1.
Suppose that (V1) and (V2) hold and that 2<p<5. Then:
There exists a
v
0
∈
H
1
(
ℝ
3
)
∖
{
0
}
such that
ℐ
V
,
λ
(
v
0
)
<
0
for any
λ
∈
Λ
.
c
λ
=
inf
γ
∈
Γ
max
t
∈
[
0
,
1
]
ℐ
V
,
λ
(
γ
(
t
)
)
>
max
{
ℐ
V
,
λ
(
0
)
,
ℐ
V
,
λ
(
v
0
)
}
=
0
for all
λ
∈
Λ
, where
Γ
=
{
γ
∈
C
(
[
0
,
1
]
,
H
1
(
ℝ
3
)
)
:
γ
(
0
)
=
0
,
γ
(
1
)
=
v
0
}
.
Proof.
(i) Fix v∈H1(ℝ3)∖{0} and λ∈Λ. Then
ℐ
V
,
λ
(
v
)
≤
ℐ
λ
∞
(
v
)
=
1
2
∫
ℝ
3
(
|
∇
v
|
2
+
V
∞
v
2
)
d
x
+
q
2
4
∫
ℝ
3
ϕ
v
v
2
d
x
-
μ
λ
p
+
1
∫
ℝ
3
|
v
|
p
+
1
d
x
-
λ
6
∫
ℝ
3
|
v
|
6
d
x
.
Set vθ=θ2v(θx) for all θ>0. Lemma 3.2 gives
ℐ
λ
∞
(
v
θ
)
→
-
∞
as
θ
→
∞
.
Hence, take v0=vθ for θ large, so that ℐV,λ(v0)≤ℐλ∞(v0)<0.
(ii) The Sobolev embedding theorem and Lemma 2.4 yield
ℐ
V
,
λ
(
v
)
≥
1
2
∥
v
∥
2
-
μ
λ
p
+
1
∫
ℝ
3
|
v
|
p
+
1
d
x
-
λ
6
∫
ℝ
3
|
v
|
6
d
x
≥
1
2
∥
v
∥
2
-
C
1
∥
v
∥
p
+
1
-
C
2
∥
v
∥
6
.
Since p>2, we see that there exist β>0 and r0>0 such that
ℐ
V
,
λ
(
v
)
≥
β
>
0
for all
v
, with
∥
v
∥
=
r
0
and for any
λ
∈
Λ
.
Therefore, for any γ∈Γ, there exists a t0∈(0,1) such that ∥γ(t0)∥=r0 and so
max
t
∈
[
0
,
1
]
ℐ
V
,
λ
(
γ
(
t
)
)
≥
ℐ
V
,
λ
(
γ
(
t
0
)
)
≥
β
>
max
{
ℐ
V
,
λ
(
0
)
,
ℐ
V
,
λ
(
v
0
)
}
=
0
,
which implies that cλ>0. ∎
Thanks to Lemma 4.1 we can apply Theorem 2.10 and get that for a.e. λ∈Λ, there exists a bounded sequence (un)n⊂H1(ℝ3), which, for simplicity, we denote by (un)n instead of (un(λ))n, such that
ℐ
V
,
λ
(
u
n
)
→
c
λ
,
ℐ
V
,
λ
′
(
u
n
)
→
0
as n→∞. Next, we aim to prove the strong convergence of the above sequence (un)n in H1(ℝ3).
Now, for any λ∈Λ the limit problem corresponding to (1.4) is
{
-
Δ
u
+
V
∞
u
+
ϕ
(
x
)
u
=
μ
λ
|
u
|
p
-
1
u
+
λ
|
u
|
4
u
in
ℝ
3
,
-
Δ
ϕ
+
a
2
Δ
2
ϕ
=
4
π
u
2
in
ℝ
3
,
with 2<p<5, and has a ground state solution in H1(ℝ3) by Theorem 1.2, i.e. for any λ∈Λ,
c
λ
∞
=
inf
u
∈
ℳ
λ
∞
ℐ
λ
∞
(
u
)
can be achieved at some uλ∞∈ℳλ∞ and (ℐλ∞)′(uλ∞)=0, where
ℐ
λ
∞
(
u
)
=
1
2
∫
ℝ
3
(
|
∇
u
|
2
+
V
∞
u
2
)
d
x
+
q
2
4
∫
ℝ
3
ϕ
u
u
2
d
x
-
μ
λ
p
+
1
∫
ℝ
3
|
u
|
p
+
1
d
x
-
λ
6
∫
ℝ
3
|
u
|
6
d
x
,
ℳ
λ
∞
=
{
u
∈
H
1
(
ℝ
3
)
∖
{
0
}
:
𝒢
λ
∞
(
u
)
=
0
}
and
𝒢
λ
∞
(
u
)
=
3
2
∫
ℝ
3
|
∇
u
|
2
d
x
+
1
2
∫
ℝ
3
V
∞
|
u
|
2
d
x
+
3
q
2
4
∫
ℝ
3
ϕ
u
u
2
d
x
-
1
4
∫
ℝ
3
q
2
a
ψ
u
u
2
d
x
-
λ
μ
2
p
-
1
p
+
1
∫
ℝ
3
|
u
|
p
+
1
d
x
-
3
λ
2
∫
ℝ
3
|
u
|
6
d
x
.
Lemma 4.2.
Let (un)n be a bounded sequence in H1(R3) such that un→u in Llocr(R3) for r∈[1,6) and un→u a.e. in R3. Then, for any p∈[1,5], there holds
∫
ℝ
3
(
|
u
n
|
p
-
1
u
n
-
|
u
|
p
-
1
u
)
φ
d
x
→
0
for all
φ
∈
C
0
∞
(
ℝ
3
)
.
Proof.
By Hölder’s inequality and, for any m≥1, by the elementary inequality |am-bm|≤L|a-b|m valid for all a, b≥0, and some appropriate L=L(m)≥1, we have for any p∈[1,5],
|
∫
ℝ
3
(
|
u
n
|
p
-
1
u
n
-
|
u
|
p
-
1
u
)
φ
d
x
|
≤
∫
ℝ
3
|
u
n
|
p
-
1
|
u
n
-
u
|
|
φ
|
d
x
+
∫
ℝ
3
|
|
u
n
|
p
-
1
-
|
u
|
p
-
1
|
|
u
φ
|
d
x
≤
∥
φ
∥
∞
∥
u
n
∥
p
p
-
1
(
∫
supp
φ
|
u
n
-
u
|
p
d
x
)
1
p
+
∥
φ
∥
∞
∥
u
∥
p
(
∫
supp
φ
|
|
u
n
|
p
-
1
-
|
u
|
p
-
1
|
p
p
-
1
d
x
)
p
-
1
p
≤
∥
φ
∥
∞
∥
u
n
∥
p
p
-
1
(
∫
supp
φ
|
u
n
-
u
|
p
d
x
)
1
p
+
C
∥
φ
∥
∞
∥
u
∥
p
(
∫
supp
φ
|
u
n
-
u
|
p
d
x
)
p
-
1
p
→
0
.
∎
Lemma 4.3 (Global Compactness Lemma).
Assume that (V1)–(V3) hold and let (un)n be a bounded (PS)cλ sequence for the functional IV,λ, with cλ<S3/23λ for any λ∈Λ. Then there exist a subsequence of (un)n, still denoted by (un)n, an integer l∈N∪{0}, sequences (ynk)⊂R3, functions wk∈H1(R3) for 1≤k≤l such that
u
n
⇀
u
0
and
ℐ
V
,
λ
′
(
u
0
)
=
0
,
|
y
n
k
|
→
∞
and
|
y
n
k
-
y
n
k
′
|
→
∞
for
k
≠
k
′
,
w
k
≠
0
and
(
ℐ
λ
∞
)
′
(
w
k
)
=
0
for
1
≤
k
≤
l
,
∥
u
n
-
u
0
-
∑
k
=
1
l
w
k
(
⋅
-
y
n
k
)
∥
→
0
,
ℐ
V
,
λ
(
u
n
)
→
ℐ
V
,
λ
(
u
0
)
+
∑
k
=
1
l
ℐ
λ
∞
(
w
k
)
.
Here we agree that in the case l=0 the above properties (iv) and (v) hold without wk and ynk.
Proof.
We divide the proof into three steps.
Step 1. Since (un)n is bounded in H1(ℝ3), we may assume that, up to a subsequence, un⇀u0 weakly in H1(ℝ3), un→u0 in Llocr(ℝ3) for 1≤r<6 and un→u0 a.e. in ℝ3. Let us prove that ℐV,λ′(u0)=0. Noting that C0∞(ℝ3) is dense in H1(ℝ3), it is sufficient to check that ℐV,λ′(u0)[φ]=0 for all φ∈C0∞(ℝ3). Observe that
(4.1)
ℐ
V
,
λ
′
(
u
n
)
[
φ
]
-
ℐ
V
,
λ
′
(
u
0
)
[
φ
]
=
∫
ℝ
3
(
∇
(
u
n
-
u
0
)
∇
φ
+
V
(
x
)
(
u
n
-
u
0
)
φ
)
d
x
+
∫
ℝ
3
(
ϕ
u
n
u
n
-
ϕ
u
0
u
0
)
φ
d
x
-
μ
λ
∫
ℝ
3
(
|
u
n
|
p
-
1
u
n
-
|
u
|
p
-
1
u
)
φ
d
x
-
λ
∫
ℝ
3
(
|
u
n
|
4
u
n
-
|
u
0
|
4
u
0
)
φ
d
x
.
Since un⇀u0 weakly in H1(ℝ3), we have 〈un-u0,φ〉→0.
By (ii) of Lemma 2.5 and Hölder’s inequality we deduce that
∫
ℝ
3
ϕ
u
n
u
n
φ
d
x
-
∫
ℝ
3
ϕ
u
0
u
0
φ
d
x
→
0
.
The definition of weak convergence, Lemma 4.2 and (4.1) give that
ℐ
V
,
λ
′
(
u
n
)
[
φ
]
-
ℐ
V
,
λ
′
(
u
0
)
[
φ
]
→
0
.
This implies that ℐV,λ′(u0)=0. On the other hand, ℐV,λ(u0)≥0. Indeed, put
a
0
=
1
2
∫
ℝ
3
|
∇
u
0
|
2
d
x
,
b
0
=
1
2
∫
ℝ
3
V
(
x
)
|
u
0
|
2
d
x
,
c
0
=
q
2
4
∫
ℝ
3
ϕ
u
0
u
0
2
d
x
,
d
0
=
1
4
∫
ℝ
3
q
2
a
ψ
u
0
u
0
2
d
x
,
e
0
=
λ
μ
p
+
1
∫
ℝ
3
|
u
0
|
p
+
1
d
x
,
f
0
=
λ
6
∫
ℝ
3
|
u
0
|
6
d
x
,
g
0
=
1
2
∫
ℝ
3
(
x
,
∇
V
(
x
)
)
|
u
0
|
2
d
x
.
Then we get the following linear system in a0, b0, c0, d0, e0, f0, g0:
(4.2)
{
a
0
+
b
0
+
c
0
-
e
0
-
f
0
=
ℐ
V
,
λ
(
u
0
)
,
2
a
0
+
2
b
0
+
4
c
0
-
(
p
+
1
)
e
0
-
6
f
0
=
0
,
a
0
+
3
b
0
+
5
c
0
+
d
0
-
3
e
0
-
3
f
0
+
g
0
=
0
,
where the first equation comes from the definition of ℐV,λ(u0), the second is (ℐV,λ)′(u0)[u0]=0, and the last comes from the Pohozaev identity in Proposition 2.6. System (4.2) and assumption (V1) yield that
3
ℐ
V
,
λ
(
u
0
)
=
(
2
b
0
+
g
0
)
+
d
0
+
2
(
p
-
2
)
e
0
+
6
f
0
≥
0
.
Step 2. Set vn1=un-u0; then vn1⇀0 weakly in H1(ℝ3). By (i) of Lemma 2.4, we have
(4.3)
∫
ℝ
3
ϕ
u
n
u
n
2
d
x
-
∫
ℝ
3
ϕ
u
0
u
0
2
d
x
-
∫
ℝ
3
ϕ
u
n
-
u
0
(
u
n
-
u
0
)
2
d
x
→
0
as n→∞. Moreover, it follows from the Brézis–Lieb lemma that
(4.4)
{
∥
v
n
1
∥
2
=
∥
u
n
∥
2
-
∥
u
0
∥
2
+
o
n
(
1
)
,
∥
v
n
1
∥
p
+
1
p
+
1
=
∥
u
n
∥
p
+
1
p
+
1
-
∥
u
0
∥
p
+
1
p
+
1
+
o
n
(
1
)
,
∥
v
n
1
∥
6
6
=
∥
u
n
∥
6
6
-
∥
u
0
∥
6
6
+
o
n
(
1
)
.
By (4.3) and (4.4), it is easy to check that
(4.5)
ℐ
V
,
λ
(
v
n
1
)
=
ℐ
V
,
λ
(
u
n
)
-
ℐ
V
,
λ
(
u
0
)
+
o
n
(
1
)
and as n→∞
(4.6)
ℐ
V
,
λ
′
(
v
n
1
)
[
(
v
n
)
1
]
=
ℐ
V
,
λ
′
(
u
n
)
,
[
u
n
]
-
ℐ
V
,
λ
′
(
u
0
)
[
u
0
]
+
o
n
(
1
)
=
o
n
(
1
)
.
Recalling that ℐV,λ(u0)≥0, we have that
ℐ
V
,
λ
(
v
n
1
)
≤
ℐ
V
,
λ
(
u
n
)
<
1
3
λ
𝒮
3
2
.
Let us introduce
δ
:=
lim sup
n
→
∞
sup
y
∈
ℝ
3
∫
B
1
(
y
)
|
v
n
1
|
2
d
x
.
Case 1: δ=0. Namely,
sup
y
∈
ℝ
3
∫
B
1
(
y
)
|
v
n
1
|
2
d
x
→
0
.
Using the Vanishing Lemma 2.7, we get vn1→0 in Lr(ℝ3) for r, with 2<r<2s*. Lemma 2.4 yields
(4.7)
∫
ℝ
3
ϕ
v
n
1
(
v
n
1
)
2
d
x
≤
C
∥
v
n
1
∥
12
5
4
→
0
.
Hence, (4.6) and (4.7) give as n→∞
∥
v
n
1
∥
2
=
λ
∥
v
n
1
∥
6
6
+
o
n
(
1
)
.
Up to a subsequence, we may assume that ∥vn1∥2→η and λ∥vn1∥66→η, with η≥0. Suppose that η>0. By (4.5) and (4.7), we have
ℐ
V
,
λ
(
v
n
1
)
=
c
λ
-
ℐ
V
,
λ
(
u
0
)
+
o
n
(
1
)
→
η
3
and thus
(4.8)
η
=
3
(
c
λ
-
ℐ
V
,
λ
(
u
0
)
)
<
1
λ
𝒮
3
2
.
But, by (2.1)
∥
v
n
1
∥
2
≥
∫
ℝ
3
|
∇
v
n
1
|
2
d
x
≥
𝒮
(
∫
ℝ
3
|
v
n
1
|
6
d
x
)
1
3
.
This implies that
η
≥
1
λ
𝒮
3
2
for all
λ
∈
Λ
.
This contradicts (4.8). Hence, limn→∞∥vn1∥=0.
Case 2: δ>0. We may assume that there exists yn1∈ℝ3 such that
∫
B
1
(
y
n
1
)
|
v
n
1
|
2
d
x
>
δ
2
>
0
,
Let us define vn1~(⋅):=vn1(⋅+yn1). Then (vn1~)n is bounded in H1(ℝ3) and we may assume that vn1~⇀w1 in H1(ℝ3) and vn1~→w1 in Llocr(ℝ3) for 1≤r<6 and vn1~→w1 a.e. in ℝ3. Since
∫
B
1
(
0
)
|
v
n
1
~
|
2
d
x
>
δ
2
,
we have
∫
B
1
(
0
)
|
w
1
|
2
d
x
>
δ
2
,
and w1≠0. But, since vn1⇀0 in H1(ℝ3), it follows that (yn1)n must be unbounded. Up to a subsequence, we suppose that |yn1|→∞.
We claim that (ℐλ∞)′(w1)=0. Similar to the proof of (4.1), we see that
(
ℐ
λ
∞
)
′
(
v
n
1
~
)
[
φ
]
-
(
ℐ
λ
∞
)
′
(
w
1
)
[
φ
]
→
0
for any fixed φ∈C0∞(ℝ3). Since vn1⇀0 in H1(ℝ3), similar as the proof of (4.1), we obtain that
ℐ
V
,
λ
′
(
v
n
1
)
,
[
φ
(
⋅
-
y
n
1
)
]
-
ℐ
V
,
λ
′
(
0
)
[
φ
(
⋅
-
y
n
1
)
]
→
0
,
which implies that
(4.9)
ℐ
V
,
λ
′
(
v
n
1
)
[
φ
(
⋅
-
y
n
1
)
]
→
0
.
By (V2), for n large enough, we have
(4.10)
∫
ℝ
3
V
(
x
+
y
n
1
)
v
n
1
(
x
)
φ
(
x
)
d
x
-
∫
ℝ
3
V
∞
v
~
n
1
(
x
)
φ
(
x
)
d
x
→
0
.
Thus, we use (4.9) minus (ℐλ∞)′(vn1~)[φ] and (4.10) to deduce that
(
ℐ
λ
∞
)
′
(
v
n
1
~
)
[
φ
]
→
0
.
In view of (V2) and the locally compactness of Sobolev embedding, we have
∫
ℝ
3
(
V
(
x
)
-
V
∞
)
(
u
n
-
u
0
)
2
d
x
→
0
.
Thus, by (4.3), (4.4) and (4.5), we conclude that
(4.11)
ℐ
V
,
λ
(
u
n
)
-
ℐ
V
,
λ
(
u
0
)
-
ℐ
λ
∞
(
v
n
1
)
→
0
.
Step 3. Let us set vn2(⋅):=vn1(⋅)-w1(⋅-yn1); then vn2→0 in Hs(ℝ3). From the Brézis–Lieb lemma and Lemma 2.5 we get as n→∞
(4.12)
∥
v
n
2
∥
2
=
∥
u
n
∥
2
-
∥
u
0
∥
2
-
∥
w
1
(
⋅
-
y
n
1
)
∥
2
+
o
n
(
1
)
,
(4.13)
∥
v
n
2
∥
6
6
=
∥
u
n
∥
6
6
-
∥
u
0
∥
6
6
-
∥
w
1
∥
6
6
+
o
n
(
1
)
,
(4.14)
∥
v
n
2
∥
p
+
1
p
+
1
=
∥
u
n
∥
p
+
1
p
+
1
-
∥
u
0
∥
p
+
1
p
+
1
-
∥
w
1
∥
p
+
1
p
+
1
+
o
n
(
1
)
,
(4.15)
∫
ℝ
3
ϕ
v
n
2
(
v
n
2
)
2
d
x
=
∫
ℝ
3
ϕ
u
n
u
n
2
d
x
-
∫
ℝ
3
ϕ
u
0
u
0
2
d
x
-
∫
ℝ
3
ϕ
w
1
(
w
1
)
2
d
x
+
o
n
(
1
)
,
(4.16)
∫
ℝ
3
ϕ
v
n
2
v
n
2
φ
d
x
=
∫
ℝ
3
ϕ
u
n
u
n
φ
d
x
-
∫
ℝ
3
ϕ
u
0
u
0
φ
d
x
-
∫
ℝ
3
ϕ
w
1
(
x
-
y
n
1
)
w
1
(
x
-
y
n
1
)
φ
d
x
+
o
n
(
1
)
,
for any φ∈(H1(ℝ3))′ and
(4.17)
∫
V
(
x
)
|
v
n
2
|
2
d
x
=
∫
V
(
x
)
|
u
n
|
2
d
x
-
∫
V
(
x
)
|
u
0
|
2
d
x
-
∫
ℝ
3
V
(
x
)
|
w
1
(
x
-
y
n
1
)
|
2
d
x
+
o
n
(
1
)
.
By (4.12)–(4.17), we can similarly check that
(4.18)
{
ℐ
V
,
λ
(
v
n
2
)
=
ℐ
V
,
λ
(
u
n
)
-
ℐ
V
,
λ
(
u
0
)
-
ℐ
λ
∞
(
w
1
)
+
o
n
(
1
)
,
ℐ
λ
∞
(
v
n
2
)
=
ℐ
V
,
λ
(
v
n
1
)
-
ℐ
λ
∞
(
w
1
)
+
o
n
(
1
)
,
ℐ
V
,
λ
′
(
v
n
2
)
[
φ
]
=
ℐ
V
,
λ
′
(
u
n
)
[
φ
]
-
ℐ
V
,
λ
′
(
u
0
)
[
φ
]
-
(
ℐ
λ
∞
)
′
(
w
1
)
[
φ
]
+
o
n
(
1
)
=
o
n
(
1
)
.
Hence, (4.11) gives as n→∞
ℐ
V
,
λ
(
u
n
)
=
ℐ
V
,
λ
(
u
0
)
+
ℐ
λ
∞
(
v
n
1
)
+
o
n
(
1
)
=
ℐ
V
,
λ
(
u
0
)
+
ℐ
λ
∞
(
w
1
)
+
ℐ
λ
∞
(
v
n
2
)
+
o
n
(
1
)
.
Recalling that any critical point of ℐλ∞ is at nonnegative level, then ℐλ∞(w1)≥0, and from Step 1, we know that ℐV,λ(u0)≥0. Consequently,
ℐ
V
,
λ
(
v
n
2
)
=
ℐ
V
,
λ
(
u
n
)
-
ℐ
V
,
λ
(
u
0
)
-
ℐ
λ
∞
(
w
1
)
+
o
n
(
1
)
≤
c
λ
<
1
3
λ
𝒮
3
2
.
Similar to the arguments in Step 2, let
δ
1
=
lim sup
n
→
∞
sup
y
∈
ℝ
3
∫
B
1
(
y
)
|
v
n
2
|
2
d
x
.
If δ1=0, then ∥vn2∥→0, i.e. ∥un-u0-w1(⋅-yn1)∥→0 and so Lemma 4.3 applies, with k=1. If δ1>0, then there exists a sequence (yn2)n⊂ℝ3 and w2∈H1(ℝ3) such that the sequence vn2~(x):=vn2(x+yn2)→w2 in H1(ℝ3). Hence, (ℐλ∞)′(w2)=0 by (4.18). Furthermore, vn2⇀0 in H1(ℝ3) implies that |yn2|→∞ and |yn1-yn2|→∞. By iterating this procedure we obtain sequences of points (ynk)n⊂ℝ3 such that |ynk|→∞ and |ynk-ynk′|→∞ for k≠k′ and vnk=vnk-1-wk-1(⋅-ynk-1), with k≥2, such that
v
n
k
→
0
in
H
1
(
ℝ
3
)
and
(
ℐ
λ
∞
)
′
(
w
k
)
=
0
and
(4.19)
{
∥
u
n
∥
2
-
∥
u
0
∥
2
-
∑
j
=
1
k
-
1
∥
w
j
(
⋅
-
y
n
j
)
∥
2
=
∥
u
n
-
u
0
-
∑
j
=
1
k
-
1
w
j
(
⋅
-
y
n
j
)
∥
2
,
ℐ
V
,
λ
(
u
n
)
-
ℐ
V
,
λ
(
u
0
)
-
∑
j
=
1
k
-
1
ℐ
λ
∞
(
w
j
)
-
ℐ
λ
∞
(
v
n
k
)
=
o
n
(
1
)
.
Since (un)n is bounded in H1(ℝ3), system (4.19) implies that the iteration stops at some finite index l+1. Therefore, vnl+1→0 in H1(ℝ3) by (4.19), and it is easy to verify that conclusions (iv) and (v) hold. The proof is completed. ∎
Based on Global Compactness Lemma 4.3, we can prove that the functional ℐV,λ verifies the (PS)cλ condition. That is, we have the following result.
Lemma 4.4.
Assume that (V1)–(V3) hold and 2<p<5, and let (un)n be a bounded (PS)cλ sequence for IV,λ. Then, up to a subsequence, (un)n converges to a nontrivial critical point uλ of IV,λ, with IV,λ(uλ)=cλ for any λ∈Λ.
Proof.
Fix any λ∈Λ. Let uλ∞ be the minimizer of cλ∞, so that ℐλ∞(uλ∞)=maxθ≥0ℐλ∞(θ2uλ∞(θx)) by Lemma 3.1. Then choosing v(x)=θ2uλ∞(θx) for θ large enough in Lemma 4.1, by (V2) we have
(4.20)
c
λ
≤
max
θ
≥
0
ℐ
V
,
λ
(
θ
2
u
λ
∞
(
θ
x
)
)
<
max
θ
≥
0
ℐ
λ
∞
(
θ
2
u
λ
∞
(
θ
x
)
)
=
ℐ
λ
∞
(
u
λ
∞
)
=
c
λ
∞
<
1
3
λ
𝒮
3
2
.
By Lemma 4.3 there exist l∈ℕ∪{0} and (ynk)n⊂ℝ3, with |ynk|→∞ for each 1≤k≤l, and uλ∈H1(ℝ3), wk∈H1(ℝ3) such that
ℐ
V
,
λ
′
(
u
λ
)
=
0
,
u
n
⇀
u
λ
,
ℐ
V
,
λ
(
u
n
)
→
ℐ
V
,
λ
(
u
λ
)
+
∑
k
=
1
l
ℐ
λ
∞
(
w
k
)
,
where wj is a critical point of ℐλ∞ for 1≤k≤l. Set
a
λ
=
1
2
∫
ℝ
3
|
∇
u
λ
|
2
d
x
,
b
λ
=
1
2
∫
ℝ
3
V
(
x
)
|
u
λ
|
2
d
x
,
c
λ
=
q
2
4
∫
ℝ
3
ϕ
u
λ
u
λ
2
d
x
,
d
λ
=
q
2
4
a
∫
ℝ
3
ψ
u
λ
u
λ
2
d
x
,
e
λ
=
μ
λ
p
+
1
∫
ℝ
3
|
u
λ
|
p
+
1
d
x
,
f
λ
=
λ
6
∫
ℝ
3
|
u
λ
|
6
d
x
,
g
λ
=
1
2
∫
ℝ
3
(
x
,
∇
V
(
x
)
)
|
u
λ
|
2
d
x
.
Then
{
a
λ
+
b
λ
+
c
λ
-
e
λ
-
f
λ
=
ℐ
V
,
λ
(
u
λ
)
,
2
a
λ
+
2
b
λ
+
4
c
λ
-
(
p
+
1
)
e
λ
-
6
f
λ
=
0
,
a
λ
+
3
b
λ
+
5
c
λ
+
d
λ
-
3
e
λ
-
3
f
λ
+
g
λ
=
0
.
Similarly to the arguments used for handling system (4.2), we get
3
ℐ
V
,
λ
(
u
λ
)
=
(
2
b
λ
+
g
λ
)
+
d
λ
+
(
2
p
-
4
)
e
λ
+
6
f
λ
≥
0
.
Thus, if l≠0, we have
c
λ
=
lim
n
→
∞
ℐ
V
,
λ
(
u
n
)
=
ℐ
V
,
λ
(
u
λ
)
+
∑
k
=
1
l
ℐ
λ
∞
(
w
k
)
≥
c
λ
∞
,
which contradicts (4.20). Hence, l=0 and Lemma 4.3 yields that un→uλ and cλ=ℐV,λ(uλ). ∎
Proof of Theorem 1.1.
From Lemma 4.1, it follows that for a.e. λ∈Λ there exists a nontrivial critical point uλ∈H1(ℝ3) for ℐV,λ and ℐV,λ(uλ)=cλ. Let us choose a sequence λn∈[12,1], with λn→1. Then there exists a sequence of nontrivial critical points uλn)n of ℐV,λn and ℐV,λn(uλn)=cλn. Next we claim that {uλn} is bounded in H1(ℝ3). Set
a
λ
n
=
1
2
∫
ℝ
3
|
∇
u
λ
n
|
2
d
x
,
b
λ
n
=
1
2
∫
ℝ
3
V
(
x
)
|
u
λ
n
|
2
d
x
,
c
λ
n
=
q
2
4
∫
ℝ
3
ϕ
u
λ
n
u
λ
n
2
d
x
,
d
λ
n
=
q
2
4
a
∫
ℝ
3
ψ
u
λ
n
u
λ
n
2
d
x
,
e
λ
n
=
λ
n
p
+
1
∫
ℝ
3
|
u
λ
n
|
p
+
1
d
x
,
f
λ
n
=
λ
n
6
∫
ℝ
3
|
u
λ
n
|
6
d
x
,
g
λ
n
=
∫
ℝ
3
(
x
,
∇
V
(
x
)
)
|
u
λ
n
|
2
d
x
.
Then
(4.21)
{
a
λ
n
+
b
λ
n
+
c
λ
n
-
e
λ
n
-
f
λ
n
=
ℐ
V
,
λ
n
(
u
λ
n
)
,
2
a
λ
n
+
2
b
λ
n
+
4
c
λ
n
-
(
p
+
1
)
e
λ
n
-
6
f
λ
n
=
0
,
a
λ
n
+
3
b
λ
n
+
5
c
λ
n
+
d
λ
n
-
3
e
λ
n
-
3
f
λ
n
+
g
λ
n
=
0
.
Similarly to the arguments used for (4.2), we get
(
2
b
λ
n
+
g
λ
n
)
+
d
λ
n
+
2
(
p
-
2
)
e
λ
n
+
6
f
λ
n
=
3
ℐ
V
,
λ
n
(
u
λ
n
)
≤
3
c
1
2
and
2
(
a
λ
n
+
b
λ
n
)
+
(
p
-
3
)
e
λ
n
+
2
f
λ
n
=
4
c
λ
n
≤
4
c
1
2
.
In view of (V1) we deduce that (aλn+bλn)n is bounded, that is, (uλn)n is bounded in H1(ℝ3). Therefore, using the fact that the map λ→cλ is left continuous, we have
lim
n
→
∞
ℐ
V
,
1
(
u
λ
n
)
=
lim
n
→
∞
(
ℐ
V
,
λ
n
(
u
λ
n
)
+
(
λ
n
-
1
)
∫
ℝ
3
(
μ
p
+
1
|
u
λ
n
|
p
+
1
+
1
6
|
u
λ
n
|
6
)
d
x
)
=
lim
n
→
∞
c
λ
n
=
c
1
and
lim
n
→
∞
ℐ
V
,
1
′
(
u
λ
n
)
[
φ
]
=
lim
n
→
∞
(
ℐ
V
,
λ
n
′
(
u
λ
n
)
[
φ
]
+
(
λ
n
-
1
)
∫
ℝ
3
(
μ
|
u
λ
n
|
p
-
1
u
λ
n
+
|
u
λ
n
|
4
u
λ
n
)
φ
d
x
)
=
0
.
These show that (uλn)n is a bounded (PS)c1 sequence for ℐV:=ℐV,1. Then by Lemma 4.4 there exists a nontrivial critical point u∈H1(ℝ3) for ℐV and ℐV(u)=c1.
Finally, we prove the existence of a ground state solution for problem (1.4). Set
m
=
inf
{
ℐ
V
(
u
)
:
u
≠
0
,
ℐ
V
′
(
u
)
=
0
}
.
Then 0≤m≤ℐV(u)=c1<∞. We rule out the case m=0. By contradiction, let (un)n be a (PS)0 sequence for ℐV. Hence,
0
=
ℐ
V
′
(
u
n
)
[
u
n
]
≥
1
2
∥
u
n
∥
2
-
μ
p
+
1
∫
ℝ
3
|
u
n
|
p
+
1
d
x
-
1
6
∫
ℝ
3
|
u
n
|
6
d
x
,
which implies that
(4.22)
∥
u
n
∥
≥
C
>
0
for all
n
∈
ℕ
.
Since ℐV′(un)=0 for any n∈ℕ, Proposition 2.6 and (V1) give
3
ℐ
V
(
u
n
)
=
3
ℐ
V
(
u
n
)
-
𝒢
V
(
u
n
)
=
μ
2
(
p
-
2
)
p
+
1
∫
ℝ
3
|
u
n
|
p
+
1
d
x
+
∫
ℝ
3
|
u
n
|
6
d
x
+
1
4
∫
ℝ
3
q
2
a
ψ
u
n
u
n
2
d
x
+
1
2
∫
ℝ
3
(
2
V
(
x
)
+
(
x
,
∇
V
(
x
)
)
u
n
2
d
x
≥
μ
2
(
p
-
2
)
p
+
1
∫
ℝ
3
|
u
n
|
p
+
1
d
x
+
∫
ℝ
3
|
u
n
|
6
d
x
,
where 𝒢V is defined by
𝒢
V
(
u
)
=
3
2
∫
ℝ
3
|
∇
u
|
2
d
x
+
3
2
∫
ℝ
3
V
(
x
)
u
2
d
x
+
1
2
∫
ℝ
3
u
2
(
x
⋅
∇
V
(
x
)
)
d
x
+
3
q
2
4
∫
ℝ
3
ϕ
u
u
2
d
x
-
q
2
4
a
∫
ℝ
3
ψ
u
u
2
d
x
-
μ
2
p
-
1
p
+
1
∫
ℝ
3
|
u
|
p
+
1
d
x
-
3
2
∫
ℝ
3
|
u
|
6
d
x
.
Therefore, limn→∞∥un∥p+1=0 and limn→∞∥un∥6=0. Combining them with ℐV′(un)[un]=0, it is easy to verify that limn→∞∥un∥=0. This contradicts (4.22).
In order to complete the proof, it suffices to prove that ℐV can be achieved in H1(ℝ3). Let (un)n be a sequence of nontrivial critical points of ℐV satisfying ℐV′(un)=0 and ℐV(un)→m<13𝒮32. Since (ℐV(un))n is bounded, by similar arguments used as in (4.21), we conclude that (un)n is bounded in H1(ℝ3) and so (un)n is a (PS)m sequence of ℐV. Arguing as the proof of Lemma 4.4, we show that there exists a nontrivial critical point u∈H1(ℝ3) of ℐV, with ℐV(u)=m. ∎
,
Patrizia Pucci
and
Xianhua Tang
