# Existence of Nontrivial Solutions for Periodic Schrödinger Equations with New Nonlinearities

Authors: Shaowei Chen; Dawei Zhang
Journal: Abstract and Applied Analysis
 (2014)

---

## Abstract

We study the Schrödinger equation:-
Δ
u
+
V
x
u
+
f
x
,
u
=
0
,
u
∈
H
1
(
R
N
), where V is 1-periodic and f is 1-periodic in the x-variables; 0 is in a gap of the spectrum of the operator -
Δ
+
V. We prove that, under some new assumptions for f, this equation has a nontrivial solution. Our assumptions for the nonlinearity f are very weak and greatly different from the known assumptions in the literature.


---

## Body

## 1. Introduction and Statement of Results

In this paper, we consider the following Schrödinger equation:(1)
-
Δ
u
+
V
(
x
)
u
+
f
(
x
,
u
)
=
0
,
u
∈
H
1
(
R
N
)
,
where N
≥
1. For V and f, we assume the following.(v) V
∈
C
(
R
N
) is 1-periodic in x
j for j
=
1
,
…
,
N, 0 is in a spectral gap (
-
μ
-
1
,
μ
1
) of -
Δ
+
V, and -
μ
-
1 and μ
1 lie in the essential spectrum of -
Δ
+
V.
Denote(2)
μ
0
:
=
min
⁡
{
μ
-
1
,
μ
1
}
.
(f
1) f
∈
C
(
R
N
×
R
) is 1-periodic in x
j for j
=
1
,
…
,
N. And there exist constants C
>
0 and 2
<
p
<
2
* such that
(3)
|
f
(
x
,
t
)
|
≤
C
(
1
+
|
t
|
p
-
1
)
,
∀
(
x
,
t
)
∈
R
N
×
R
,
where(4)
2
*
:
=
{
2
N
(
N
-
2
)
,
N
≥
3
∞
,
N
=
1,2
.
(f
2) The limit lim
⁡
t
→
0
f
(
x
,
t
)
/
t
=
0 holds uniformly for x
∈
R
N. And there exists D
>
0 such that
(5)
inf
⁡
x
∈
R
N
,
|
t
|
≥
D
f
(
x
,
t
)
t
>
max
⁡
R
N
V
-
,
whereV
±
(
x
)
=
max
⁡
{
±
V
(
x
)
,
0
}, ∀
x
∈
R
N.
(f
3) For any (
x
,
t
)
∈
R
N
×
R, F
~
(
x
,
t
)
≥
0, where
(6)
F
~
(
x
,
t
)
:
=
1
2
t
f
(
x
,
t
)
-
F
(
x
,
t
)
,
F
(
x
,
t
)
=
∫
0
t
‍
f
(
x
,
s
)
d
s
.
(f
4) There exist 0
<
κ
<
D and ν
∈
(
0
,
μ
0
) such that, for every (
x
,
t
)
∈
R
N
×
R with |
t
|
<
κ,
(7)
|
f
(
x
,
t
)
|
≤
ν
|
t
|
and, for every(
x
,
t
)
∈
R
N
×
R with κ
≤
|
t
|
≤
D,
(8)
F
~
(
x
,
t
)
>
0
.Remark 1.
By the definitions ofF and F
~, it is easy to verify that, for all (
x
,
t
)
∈
R
N
×
(
R
∖
{
0
}
),
(9)
∂
∂
t
(
F
(
x
,
t
)
t
2
)
=
2
F
~
(
x
,
t
)
t
3
.
Together with f
(
x
,
t
)
=
o
(
t
) as |
t
|
→
0 and (
f
3
), this implies that
(10)
F
(
x
,
t
)
≥
0
∀
(
x
,
t
)
∈
R
N
×
R
.Remark 2.
There are many functions satisfying(
f
1
)
–
(
f
4
). We give several examples here.Example 1.
D
=
1
+
μ
0
/
2
+
e
1
+
max
⁡
R
N
V
-, κ
=
1
+
μ
0
/
2, ν
=
μ
0
/
2, and
(11)
f
(
x
,
t
)
=
{
0
,
|
t
|
≤
1
,
t
ln
⁡
|
t
|
,
|
t
|
>
1
.Example 2.
D
=
3
+
μ
0
/
2
+
2
max
⁡
R
N
V
-, κ
=
3
/
2, ν
=
μ
0
/
2, and
(12)
f
(
x
,
t
)
=
{
0
,
|
t
|
≤
1
,
D
(
t
-
1
)
,
t
>
1
,
D
(
t
+
1
)
,
t
<
-
1
.Example 3.
D
=
μ
0
/
2
+
e
1
+
max
⁡
R
N
V
-, κ
=
ν
=
μ
0
/
2, and f
(
x
,
t
)
=
t
ln
⁡
(
1
+
|
t
|
).A solutionu of (1) is called nontrivial if u
≢
0. Our main results are as follows.Theorem 3.
Suppose(
v
) and (
f
1
)
–
(
f
4
) are satisfied. Then (1) has a nontrivial solution.Note that(
f
2
′
) the limits lim
⁡
t
→
0
f
(
x
,
t
)
/
t
=
0 and lim
⁡
|
t
|
→
∞
(
f
(
x
,
t
)
/
t
)
=
+
∞ hold uniformly for x
∈
R
N.Implying(
f
2
), we have the following corollary.Corollary 4.
Suppose(
v
), (
f
1
), (
f
2
′
), (
f
3
), and (
f
4
) are satisfied. Then (1) has a nontrivial solution.It is easy to verify that the condition(
f
4
′
)  F
~
(
x
,
t
)
>
0, for every (
x
,
t
)
∈
R
N
×
R.
And the assumption thatf
(
x
,
t
)
/
t
→
0 as t
→
0 uniformly for x
∈
R
N imply (
f
3
) and (
f
4
). Therefore, we have the following corollary.Corollary 5.
Suppose(
v
), (
f
1
), (
f
2
), and (
f
4
′
) are satisfied. Then (1) has a nontrivial solution.Semilinear Schrödinger equations with periodic coefficients have attracted much attention in recent years due to its numerous applications. One can see [1–24] and the references therein. In [2], the authors used the dual variational method to obtain a nontrivial solution of (1) with f
(
x
,
t
)
=
±
W
(
x
)
|
t
|
p
-
2
t, where W is an asymptotically periodic function. In [20], Troestler and Willem firstly obtained nontrivial solutions for (1) with f being a C
1 function satisfying the Ambrosetti-Rabinowitz condition:(AR)
there existsα
>
2 such that, for every u
≠
0,  0
<
α
G
(
x
,
u
)
≤
g
(
x
,
u
)
u, where g
(
x
,
u
)
=
-
f
(
x
,
u
), G
(
x
,
u
)
=
-
F
(
x
,
u
), and
(13)
|
∂
f
(
x
,
u
)
∂
u
|
≤
C
(
|
u
|
p
-
2
+
|
u
|
q
-
2
)
with2
<
p
<
q
<
2
*. Then, in [9], Kryszewski and Szulkin developed some infinite-dimensional linking theorems. Using these theorems, they improved Troestler and Willem’s results and obtained nontrivial solutions for (1) with f only satisfying (
f
1
) and the (AR) condition. These generalized linking theorems were also used by Li and Szulkin to obtain nontrivial solution for (1) under some asymptotically linear assumptions for f (see [11]). In [13] (see also [14]), existence of nontrivial solutions for (1) under (
f
1
) and the (AR) condition was also obtained by Pankov and Pflüger through approximating (1) by a sequence of equations defined in bounded domains. In the celebrated paper [17], Schechter and Zou combined a generalized linking theorem with the monotonicity methods of Jeanjean (see [8]). They obtained a nontrivial solution of (1) when f exhibits the critical growth. A similar approach was applied by Szulkin and Zou to obtain homoclinic orbits of asymptotically linear Hamiltonian systems (see [19]). Moreover, in [5] (see also [6]), Ding and Lee obtained nontrivial solutions for (1) under some new superlinear assumptions on f different from the classical (AR) conditions.Our assumptions onf are very weak and greatly different from the assumptions mentioned above. In fact, our assumptions (
f
1
)
–
(
f
4
) do not involve the properties of f at infinity. It may be asymptotically linear growth at infinity, that is, limsup
⁡
|
t
|
→
∞
(
f
(
x
,
t
)
/
t
)
<
+
∞, or superlinear growth at infinity as well, that is, liminf
⁡
|
t
|
→
∞
(
f
(
x
,
t
)
/
t
)
=
+
∞. Moreover, the assumptions (
f
1
)
–
(
f
4
) allow f
(
x
,
t
)
≡
0 in a neighborhood of t
=
0 (see Remark 2).In this paper, we use the generalized linking theorem for a class of parameter-dependent functionals (see [17, Theorem 2.1] or Proposition 8 in the present paper) to obtain a sequence of approximate solutions for (1). Then, we prove that these approximate solutions are bounded in L
∞
(
R
N
) and H
1
(
R
N
) (see Lemmas 13 and 14). Finally, using the concentration-compactness principle, we obtain a nontrivial solution of (1).Notation.  B
r
(
a
) denotes the open ball of radius r and center a. For a Banach space E, we denote the dual space of E by E
′ and denote strong and weak convergence in E by → and ⇀, respectively. For φ
∈
C
1
(
E
;
R
), we denote the Fréchet derivative of φ at u by φ
′
(
u
). The Gateaux derivative of φ is denoted by 〈
φ
′
(
u
)
,
v
〉, ∀
u
,
v
∈
E. L
p
(
R
N
) denotes the standard L
p space (
1
≤
p
≤
∞
), and H
1
(
R
N
) denotes the standard Sobolev space with norm ∥
u
∥
H
1
=
(
∫
R
N
‍
(
|
∇
u
|
2
+
u
2
)
d
x
)
1
/
2. We use O
(
h
), o
(
h
) to mean |
O
(
h
)
|
≤
C
|
h
|, o
(
h
)
/
|
h
|
→
0.


## 2. Existence of Approximate Solutions for (1)

Under the assumptions(
v
), (
f
1
), and (
f
2
), the functional
(14)
Φ
(
u
)
=
1
2
∫
R
N
‍
|
∇
u
|
2
d
x
+
1
2
∫
R
N
‍
V
(
x
)
u
2
d
x
+
∫
R
N
‍
F
(
x
,
u
)
d
x
is of class C
1 on X
:
=
H
1
(
R
N
), and the critical points of Φ are weak solutions of (1).There is a standard variational setting for the quadratic form∫
R
N
‍
(
|
∇
u
|
2
+
V
(
x
)
u
2
)
d
x. For the reader’s convenience, we state it here. One can consult [5] or [6] for more details.Assume that(
v
) holds, and let S
=
-
Δ
+
V be the self-adjoint operator acting on L
2
(
R
N
) with domain D
(
S
)
=
H
2
(
R
N
). By virtue of (
v
), we have the orthogonal decomposition
(15)
L
2
=
L
2
(
R
N
)
=
L
+
+
L
-
such that S is negative (resp., positive) in L
- (resp., in L
+). As in [5, Section 2] (see also [6, Chapter 6.2]), let X
=
D
(
|
S
|
1
/
2
) be equipped with the inner product
(16)
(
u
,
v
)
=
(
|
S
|
1
/
2
u
,
|
S
|
1
/
2
v
)
L
2
and norm ∥
u
∥
=
∥
|
S
|
1
/
2
u
∥
L
2, where (
·
,
·
)
L
2 denotes the inner product of L
2. From (
v
),
(17)
X
=
H
1
(
R
N
)
with equivalent norms. Therefore, X continuously embeds in L
q
(
R
N
) for all 2
≤
q
≤
2
N
/
(
N
-
2
) if N
≥
3 and for all q
≥
2 if N
=
1,2. In addition, we have the decomposition
(18)
X
=
X
+
+
X
-
,
where X
±
=
X
∩
L
± is orthogonal with respect to both (
·
,
·
)
L
2 and (
·
,
·
). Therefore, for every u
∈
X, there is a unique decomposition
(19)
u
=
u
+
+
u
-
,
u
±
∈
X
±
with (
u
+
,
u
-
)
=
0 and
(20)
∫
R
N
‍
|
∇
u
|
2
d
x
+
∫
R
N
‍
V
(
x
)
u
2
d
x
=
∥
u
+
∥
2
-
∥
u
-
∥
2
,
u
∈
X
.
Moreover,
(21)
μ
-
1
∥
u
-
∥
L
2
2
≤
∥
u
-
∥
2
,
∀
u
∈
X
,
(22)
μ
1
∥
u
+
∥
L
2
2
≤
∥
u
+
∥
2
,
∀
u
∈
X
.
The functional Φ defined by (14) can be rewritten as
(23)
Φ
(
u
)
=
1
2
(
∥
u
+
∥
2
-
∥
u
-
∥
2
)
+
ψ
(
u
)
,
where
(24)
ψ
(
u
)
=
∫
R
N
‍
F
(
x
,
u
)
d
x
.The above variational setting for the functional (14) is standard. One can consult [5] or [6] for more details.Let{
e
k
±
} be the total orthonormal sequence in X
±. Let P
:
X
→
X
-, Q
:
X
→
X
+ be the orthogonal projections. We define
(25)
|
|
|
u
|
|
|
=
max
⁡
{
∥
P
u
∥
,
∑
k
=
1
∞
‍
1
2
k
+
1
|
(
Q
u
,
e
k
+
)
|
}
on X. The topology generated by |
|
|
·
|
|
| is denoted by τ, and all topological notation related to it will include this symbol.Lemma 6.
Suppose that(
v
) holds. Then (a)
max
⁡
R
N
V
-
≥
μ
-
1, where μ
-
1 is defined in (
v
);
(b)
for anyC
>
μ
-
1, there exists u
0
∈
X
- with ∥
u
0
∥
=
1 such that C
∥
u
0
∥
L
2
>
1.Proof.
(a) We apply an indirect argument, and assume by contradiction that
(26)
max
⁡
R
N
V
-
<
μ
-
1
.
From assumption (
v
), -
μ
-
1 is in the essential spectrum of the operator (with domain D
(
L
)
=
H
2
(
R
N
)):
(27)
L
=
-
Δ
+
V
:
L
2
(
R
N
)
⟶
L
2
(
R
N
)
.
Then, by Weyl’s criterion (see, e.g., [25,  Theorem  VII.12] or [26, Theorem  7.2]), there exists a sequence {
u
n
}
⊂
H
2
(
R
N
) with the properties that ∥
u
n
∥
L
2
=
1, ∀
n and  ∥
-
Δ
u
n
+
V
u
n
+
μ
-
1
u
n
∥
L
2
→
0.
Sinceμ
-
1
>
max
⁡
R
N
⁡
V
_, we deduce that -
V
-
(
x
)
+
μ
-
1
>
0 for all x
∈
R
N. Together with the facts that V is a continuous periodic function and V
=
V
+
-
V
-, this implies
(28)
inf
⁡
x
∈
R
N
⁡
(
V
(
x
)
+
μ
-
1
)
>
0
.
It follows that there exists a constant C
′
>
0 such that
(29)
∫
R
N
‍
(
|
∇
u
|
2
+
(
V
(
x
)
+
μ
-
1
)
u
2
)
d
x
≥
C
′
∥
u
∥
2
,
∀
u
∈
X
.
Note that
(30)
∫
R
N
‍
(
-
Δ
u
n
+
V
(
x
)
u
n
+
μ
-
1
u
n
)
u
n
d
x
=
∫
R
N
‍
(
|
∇
u
n
|
2
+
(
V
(
x
)
+
μ
-
1
)
u
n
2
)
d
x
.
Together with (29) and the fact that ∥
-
Δ
u
n
+
V
u
n
+
μ
-
1
u
n
∥
L
2
→
0 and ∥
u
n
∥
L
2
=
1, this implies ∥
u
n
∥
→
0. It contradicts ∥
u
n
∥
L
2
=
1, ∀
n. Therefore, max
⁡
R
N
V
-
≥
μ
-
1.
(b) It suffices to prove that
(31)
μ
-
1
=
C
-
:
=
inf
⁡
⁡
{
∥
u
∥
2
∣
u
∈
X
-
,
∥
u
∥
L
2
=
1
}
.
From (21), we deduce that μ
-
1
≤
C
-. From assumption (
v
), -
μ
-
1 is in the essential spectrum of L. By Weyl’s criterion, there exists {
u
n
}
⊂
H
2
(
R
N
) such that ∥
u
n
∥
L
2
=
1 and ∥
-
Δ
u
n
+
V
u
n
+
μ
-
1
u
n
∥
L
2
→
0. Multiplying -
Δ
u
n
+
V
u
n
+
μ
-
1
u
n by u
n
+ and then integrating it into R
N, by (20) and (22), we get that
(32)
(
μ
1
+
μ
-
1
)
∥
u
n
+
∥
L
2
2
≤
∫
R
N
‍
(
|
∇
u
n
+
|
2
+
V
(
x
)
(
u
n
+
)
2
+
μ
-
1
(
u
n
+
)
2
)
d
x
=
∫
R
N
‍
(
-
Δ
u
n
+
V
(
x
)
u
n
+
μ
-
1
u
n
)
u
n
+
d
x
⟶
0
.
It follows that ∥
u
n
-
∥
L
2
→
1. Multiplying -
Δ
u
n
+
V
u
n
+
μ
-
1
u
n by u
n
- and then integrating it into R
N, we get that
(33)
-
∥
u
n
-
∥
2
+
μ
-
1
∥
u
n
-
∥
L
2
2
=
∫
R
N
‍
(
|
∇
u
n
-
|
2
+
V
(
x
)
(
u
n
-
)
2
+
μ
-
1
(
u
n
-
)
2
)
d
x
=
∫
R
N
‍
(
-
Δ
u
n
+
V
u
n
+
μ
-
1
u
n
)
u
n
-
d
x
⟶
0
.
It implies that μ
-
1
≥
C
-. This together with μ
-
1
≤
C
- implies μ
-
1
=
C
-.LetR
>
r
>
0 and
(34)
A
:
=
inf
⁡
x
∈
R
N
,
|
t
|
≥
D
f
(
x
,
t
)
t
.
From assumption (5), we have A
>
max
⁡
R
N
V
-. Together with the result (
a
) of Lemma 6, this implies that (
1
/
2
)
(
A
+
μ
-
1
)
>
μ
-
1. Choose
(35)
γ
∈
(
μ
-
1
,
(
A
+
μ
-
1
)
2
)
.
Then by the result (
b
) of Lemma 6, there exists u
0
∈
X
- with ∥
u
0
∥
=
1 such that
(36)
γ
∥
u
0
∥
L
2
>
1
.
Set
(37)
N
=
{
u
∈
X
-
∣
∥
u
∥
=
r
}
,
M
=
{
u
∈
X
+
⊕
R
+
u
0
∣
∥
u
∥
≤
R
}
.
Then, M is a submanifold of X
+
⊕
R
+
u
0 with boundary
(38)
∂
M
=
{
u
∈
X
-
∣
∥
u
∥
≤
R
}
∪
{
u
-
+
t
u
0
∣
u
-
∈
X
-
,
t
>
0
,
∥
u
-
+
t
u
0
∥
=
R
}
.Definition 7.
Letϕ
∈
C
1
(
X
;
R
). A sequence {
u
n
}
⊂
X is called a Palais-Smale sequence at level c ((
P
S
)
c-sequence for short) for ϕ, if ϕ
(
u
n
)
→
c and ∥
ϕ
′
(
u
n
)
∥
X
′
→
0 as n
→
∞.The following proposition is proved in [17] (see [17, Theorem 2.1]).Proposition 8.
Let0
<
K
<
1. The family of C
1-functional {
H
λ
} has the form
(39)
H
λ
(
u
)
=
λ
I
(
u
)
-
J
(
u
)
,
u
∈
X
,
λ
∈
[
K
,
1
]
.
Assume(a)
J
(
u
)
≥
0, ∀
u
∈
X;
(b)
|
I
(
u
)
|
+
J
(
u
)
→
+
∞
as
∥
u
∥
→
+
∞;
(c)
for all
λ
∈
[
K
,
1
], H
λ
is
τ
-sequentially upper semicontinuous; that is, if
|
|
|
u
n
-
u
|
|
|
→
0,then
(40)
limsup
⁡
n
→
∞
H
λ
(
u
n
)
≤
H
λ
(
u
)
,
and
H
λ
′
is weakly sequentially continuous. Moreover,
H
λ
maps bounded sets to bounded sets;
(d)
there exist
u
0
∈
X
-
∖
{
0
}
with
∥
u
0
∥
=
1
and
R
>
r
>
0
such that, for all
λ
∈
[
K
,
1
]
,
(41)
inf
⁡
N
H
λ
>
sup
⁡
∂
M
H
λ
. Then there exists E
⊂
[
K
,
1
] such that the Lebesgue measure of [
K
,
1
]
∖
E is zero and, for every λ
∈
E, there exist c
λ and a bounded (
P
S
)
c
λ-sequence for H
λ, where c
λ satisfies
(42)
sup
⁡
M
H
λ
≥
c
λ
≥
inf
⁡
N
H
λ
.For0
<
K
<
1 and λ
∈
[
K
,
1
], let
(43)
Ψ
λ
(
u
)
=
λ
2
∫
R
N
‍
V
-
(
x
)
u
2
d
x
-
(
1
2
∫
R
N
‍
(
|
∇
u
|
2
+
V
+
(
x
)
u
2
)
d
x
+
ψ
(
u
)
)
,
u
∈
X
.
Then
(44)
Ψ
1
=
-
Φ
and it is easy to verify that a critical point u of Ψ
λ is a weak solution of
(45)
-
Δ
u
+
V
λ
(
x
)
u
+
f
(
x
,
u
)
=
0
,
u
∈
X
,
where
(46)
V
λ
=
V
+
-
λ
V
-
.Lemma 9.
Suppose that(
v
) and (
f
1
)
–
(
f
3
) hold. Then, there exist 0
<
K
*
<
1 and E
⊂
[
K
*
,
1
] such that the Lebesgue measure of [
K
*
,
1
]
∖
E is zero and, for every λ
∈
E, there exist c
λ and a bounded (
P
S
)
c
λ-sequence for Ψ
λ, where c
λ satisfies
(47)
+
∞
>
sup
⁡
λ
∈
E
c
λ
≥
inf
⁡
λ
∈
E
c
λ
>
0
.Proof.
Foru
∈
X, let
(48)
I
(
u
)
=
1
2
∫
R
N
‍
V
-
(
x
)
u
2
d
x
,
J
(
u
)
=
1
2
∫
R
N
‍
(
|
∇
u
|
2
+
V
+
(
x
)
u
2
)
d
x
+
ψ
(
u
)
.
Then, I and J satisfy assumptions (a) and (b) in Proposition 8, and, by (43), Ψ
λ
(
u
)
=
λ
I
(
u
)
-
J
(
u
).
From (43) and (20), for any u
∈
X and λ
∈
[
K
,
1
], we have
(49)
Ψ
λ
(
u
)
=
λ
-
1
2
∫
R
N
‍
V
-
(
x
)
u
2
d
x
-
(
1
2
∫
R
N
‍
(
|
∇
u
|
2
+
V
(
x
)
u
2
)
d
x
+
∫
R
N
‍
F
(
x
,
u
)
d
x
)
=
1
2
∥
u
-
∥
2
-
1
2
∥
u
+
∥
2
-
1
-
λ
2
∫
R
N
‍
V
-
(
x
)
u
2
d
x
-
∫
R
N
‍
F
(
x
,
u
)
d
x
.
Let u
*
∈
X and {
u
n
}
⊂
X be such that |
|
|
u
n
-
u
*
|
|
|
→
0. It follows that u
n
-
→
u
*
-, u
n
+
⇀
u
*
+, and u
n
⇀
u
*. In addition, up to a subsequence, we can assume that u
n
→
u
* a.e. in R
N. Then, we have
(50)
∥
u
n
-
∥
2
⟶
∥
u
*
-
∥
2
,
(51)
liminf
⁡
n
→
∞
∫
R
N
‍
V
-
(
x
)
u
n
2
d
x
≥
∫
R
N
‍
V
-
(
x
)
u
*
2
d
x
(
by Fatou
,
s lemma
)
,
liminf
⁡
n
→
∞
∥
u
n
+
∥
2
≥
∥
u
*
+
∥
2
.
By Remark 1, F
(
x
,
t
)
≥
0 for all x and t. This together with Fatou’s lemma implies
(52)
liminf
⁡
n
→
∞
∫
R
N
‍
F
(
x
,
u
n
)
d
x
≥
∫
R
N
‍
F
(
x
,
u
*
)
d
x
.
Then, by (49), we obtain
(53)
limsup
⁡
n
→
∞
Ψ
λ
(
u
n
)
≤
Ψ
λ
(
u
*
)
.
This implies that Ψ
λ is τ-sequentially upper semicontinuous.
Ifu
n
⇀
u
* in X, then, for any fixed φ
∈
X, as n
→
∞,
(54)
〈
-
Ψ
λ
′
(
u
n
)
,
φ
〉
=
∫
R
N
‍
(
∇
u
n
∇
φ
+
V
λ
u
n
φ
)
d
x
+
∫
R
N
‍
f
(
x
,
u
n
)
φ
d
x
⟶
∫
R
N
‍
(
∇
u
*
∇
φ
+
V
λ
u
*
φ
)
d
x
+
∫
R
N
‍
f
(
x
,
u
*
)
φ
d
x
=
〈
-
Ψ
λ
′
(
u
*
)
,
φ
〉
.
This implies that Ψ
λ
′ is weakly sequentially continuous. Moreover, it is easy to see that Ψ
λ maps bounded sets to bounded sets. Therefore, Ψ
λ satisfies assumption (c) in Proposition 8.
Finally, we will verify assumption (d) in Proposition8 for Ψ
λ.
From assumption(
f
1
) and f
(
x
,
t
)
/
t
→
0 as t
→
0 uniformly for x
∈
R
N, we deduce that, for any ϵ
>
0, there exists C
ϵ
>
0 such that
(55)
F
(
x
,
t
)
≤
ϵ
t
2
+
C
ϵ
|
t
|
p
,
∀
(
x
,
t
)
∈
R
N
×
R
.
From (49) and (55), we have, for u
∈
N,
(56)
Ψ
λ
(
u
)
≥
1
2
∥
u
∥
2
-
1
-
λ
2
∫
R
N
‍
V
-
(
x
)
u
2
d
x
-
ϵ
∫
R
N
‍
u
2
d
x
-
C
ϵ
∫
R
N
‍
|
u
|
p
d
x
.
Then by the Sobolev inequality ∥
u
∥
L
p
(
R
N
)
≤
C
∥
u
∥ and ∥
u
∥
L
2
≤
C
∥
u
∥ (by (21) and (22)), we deduce that there exists a constant C
>
0 such that
(57)
Ψ
λ
(
u
)
≥
1
2
∥
u
∥
2
-
C
(
1
-
λ
)
max
⁡
R
N
V
-
(
x
)
∥
u
∥
2
-
ϵ
C
∥
u
∥
2
-
C
C
ϵ
∥
u
∥
p
.
Choose 0
<
K
*
<
1 and ϵ
>
0 such that C
(
1
-
K
*
)
max
⁡
R
N
V
-
(
x
)
<
1
/
4 and C
ϵ
=
1
/
8. Then, for every λ
∈
[
K
*
,
1
], we have
(58)
Ψ
λ
(
u
)
≥
1
8
∥
u
∥
2
-
C
C
ϵ
∥
u
∥
p
.
Let r
>
0 be such that r
p
-
2
C
C
ϵ
=
1
/
16 and β
=
r
2
/
16. Then, from (58), we deduce that, for N
=
{
u
∈
X
-
∣
∥
u
∥
=
r
},
(59)
inf
⁡
N
Ψ
λ
≥
β
,
∀
λ
∈
[
K
*
,
1
]
.
We will prove thatsup
⁡
K
*
≤
λ
≤
1
Ψ
λ
(
u
)
→
-
∞ as ∥
u
∥
→
∞ and u
∈
X
+
⊕
R
+
u
0. Arguing indirectly, assume that, for some sequences λ
n
∈
[
K
*
,
1
] and u
n
∈
X
+
⊕
R
+
u
0 with ∥
u
n
∥
→
+
∞, there is L
>
0 such that Ψ
λ
n
(
u
n
)
≥
-
L for all n. Then, setting w
n
=
u
n
/
∥
u
n
∥, we have ∥
w
n
∥
=
1, and, up to a subsequence, w
n
⇀
w, w
n
-
→
w
-
∈
X
- and w
n
+
⇀
w
+
∈
X
+.
First, we consider the casew
≠
0. Dividing both sides of (49) by ∥
u
n
∥
2, we get that
(60)
-
L
∥
u
n
∥
2
≤
Ψ
λ
n
(
u
n
)
∥
u
n
∥
2
=
1
2
∥
w
n
-
∥
2
-
1
2
∥
w
n
+
∥
2
-
1
-
λ
n
2
∫
R
N
‍
V
-
(
x
)
w
n
2
d
x
-
∫
R
N
F
(
x
,
u
n
)
∥
u
n
∥
2
d
x
.
From (5) and the result (
a
) of Lemma 6, we deduce that
(61)
liminf
⁡
|
t
|
→
∞
F
(
x
,
t
)
t
2
≥
A
2
>
1
2
max
⁡
R
N
V
-
≥
1
2
μ
-
1
,
where A is defined by (34). Note that, for x
∈
{
x
∈
R
N
∣
w
≠
0
}, we have |
u
n
(
x
)
|
→
+
∞. This implies that, when n is large enough,
(62)
∫
{
x
∈
R
N
∣
w
≠
0
}
‍
F
(
x
,
u
n
)
u
n
2
w
n
2
d
x
≥
A
+
μ
-
1
4
∫
{
x
∈
R
N
∣
w
≠
0
}
‍
w
n
2
d
x
.
By (10), we have, when n is large enough,
(63)
∫
R
N
‍
F
(
x
,
u
n
)
∥
u
n
∥
2
d
x
=
∫
R
N
‍
F
(
x
,
u
n
)
u
n
2
w
n
2
d
x
≥
∫
{
x
∈
R
N
∣
w
≠
0
}
‍
F
(
x
,
u
n
)
u
n
2
w
n
2
d
x
.
Combining the above two inequalities yields
(64)
liminf
⁡
n
→
∞
(
∫
R
N
‍
F
(
x
,
u
n
)
∥
u
n
∥
2
1
2
∥
w
n
-
∥
2
-
1
2
∥
w
n
+
∥
2
-
1
-
λ
n
2
∫
R
N
‍
V
-
(
x
)
w
n
2
d
x
-
∫
R
N
‍
F
(
x
,
u
n
)
∥
u
n
∥
2
d
x
)
≤
liminf
⁡
n
→
∞
(
∫
R
N
‍
1
2
∥
w
n
-
∥
2
-
1
2
∥
w
n
+
∥
2
-
A
+
μ
-
1
4
∫
{
x
∈
R
N
∣
w
≠
0
}
‍
w
n
2
d
x
∫
R
N
‍
)
≤
1
2
∥
w
-
∥
2
-
1
2
∥
w
+
∥
2
-
A
+
μ
-
1
4
∫
R
N
‍
w
2
d
x
≤
1
2
∥
w
-
∥
2
-
1
2
∥
w
+
∥
2
-
A
+
μ
-
1
4
∥
w
-
∥
L
2
2
.
We used the inequalities
(65)
lim
⁡
n
→
∞
∥
w
n
-
∥
2
=
∥
w
-
∥
2
,
liminf
⁡
n
→
∞
∥
w
n
+
∥
2
≥
∥
w
+
∥
2
,
liminf
⁡
n
→
∞
∫
{
x
∈
R
N
∣
w
≠
0
}
‍
w
n
2
d
x
≥
∫
R
N
‍
w
2
d
x
in the second inequality of (64).
Sincew
-
=
t
u
0 for some t
∈
R, by (36), we get that
(66)
A
+
μ
-
1
4
∥
w
-
∥
L
2
2
≥
A
+
μ
-
1
4
γ
∥
w
-
∥
2
.
Note that, by the choice of γ (see (35)), we have (
(
A
+
μ
-
1
)
/
4
γ
)
>
1
/
2. Then by (64) and the fact that w
≠
0, we have that
(67)
liminf
⁡
n
→
∞
(
F
(
x
,
u
n
)
∥
u
n
∥
2
d
1
2
∥
w
n
-
∥
2
-
1
2
∥
w
n
+
∥
2
-
1
-
λ
n
2
∫
R
N
‍
V
-
(
x
)
w
n
2
d
x
-
∫
R
N
‍
F
(
x
,
u
n
)
∥
u
n
∥
2
d
x
)
≤
-
(
A
+
μ
-
1
4
γ
-
1
2
)
∥
w
-
∥
2
-
1
2
∥
w
+
∥
2
<
0
.
It contradicts (60), since -
L
/
∥
u
n
∥
2
→
0 as n
→
∞.
Second, we consider the casew
=
0. In this case, lim
⁡
n
→
∞
∥
w
n
-
∥
=
0. It follows that
(68)
liminf
⁡
n
→
∞
∥
w
n
+
∥
≥
1
,
since ∥
w
n
∥
=
1 and w
n
=
w
n
+
+
w
n
-. Therefore, the right hand side of (60) is less than -
1
/
4 when n is large enough. However, as n
→
∞, the left hand side of (60) converges to zero. It induces a contradiction.
Therefore, there existsR
>
r such that
(69)
sup
⁡
λ
∈
[
K
*
,
1
]
sup
⁡
∂
M
Ψ
λ
≤
0
.
This implies that Ψ
λ satisfies assumption (d) in Proposition 8 if λ
∈
[
K
*
,
1
]. Finally, it is easy to see that
(70)
sup
⁡
λ
∈
[
K
*
,
1
]
sup
⁡
M
Ψ
λ
<
+
∞
.
Then, the results of this lemma follow immediately from Proposition 8.Lemma 10.
Suppose that(
v
) and (
f
1
)
–
(
f
3
) are satisfied. Let λ
∈
[
K
*
,
1
] be fixed, where K
* is the constant in Lemma 9. If {
v
n
} is a bounded (
P
S
)
c-sequence for Ψ
λ with c
≠
0, then, for every n
∈
N, there exists a
n
∈
Z
N such that, up to a subsequence, u
n
:
=
v
n
(
·
+
a
n
) satisfies
(71)
u
n
⇀
u
λ
≠
0
,
Ψ
λ
(
u
λ
)
≤
c
,
Ψ
λ
′
(
u
λ
)
=
0
.Proof.
The proof of this lemma is inspired by the proof of Lemma 3.7 in [19]. Because {
v
n
} is a bounded sequence in X, up to a subsequence, either (a)
lim
⁡
n
→
∞
sup
⁡
y
∈
R
N
∫
B
1
(
y
)
‍
|
v
n
|
2
d
x
=
0 or
(b)
there existϱ
>
0 and a
n
∈
Z
N such that ∫
B
1
(
a
n
)
‍
|
v
n
|
2
d
x
≥
ϱ.
If (a) occurs, using the Lions lemma (see, e.g., [21, Lemma 1.21]), a similar argument as for the proof of [19, Lemma 3.6] shows that
(72)
lim
⁡
n
→
∞
∫
R
N
‍
F
(
x
,
v
n
)
d
x
=
0
,
lim
⁡
n
→
∞
∫
R
N
‍
f
(
x
,
v
n
)
v
n
±
d
x
=
0
.
It follows that
(73)
lim
⁡
n
→
∞
∫
R
N
‍
(
2
F
(
x
,
v
n
)
-
f
(
x
,
v
n
)
v
n
)
d
x
=
0
.
On the other hand, as {
v
n
} is a (
P
S
)
c-sequence of Ψ
λ, we have 〈
Ψ
λ
′
(
v
n
)
,
v
n
〉
→
0 and Ψ
λ
(
v
n
)
→
c
≠
0. It follows that
(74)
∫
R
N
‍
(
f
(
x
,
v
n
)
v
n
-
2
F
(
x
,
v
n
)
)
d
x
=
2
Ψ
λ
(
v
n
)
-
〈
Ψ
λ
′
(
v
n
)
,
v
n
〉
⟶
2
c
≠
0
,
n
⟶
∞
.
This contradicts (73). Therefore, case (a) cannot occur.
If case (b) occurs, letu
n
=
v
n
(
·
+
a
n
). For every n,
(75)
∫
B
1
(
0
)
‍
|
u
n
|
2
d
x
≥
ϱ
.
Because V and F
(
x
,
t
) are 1-periodic in every x
j, {
u
n
} is still bounded in X,
(76)
lim
⁡
n
→
∞
Ψ
λ
(
u
n
)
≤
c
,
Ψ
λ
′
(
u
n
)
⇀
0
,
n
⟶
∞
.
Up to a subsequence, we assume that u
n
⇀
u
λ in X as n
→
∞. Since u
n
→
u
λ in L
loc
⁡
2
(
R
N
), it follows from (75) that u
λ
≠
0. Recall that Ψ
λ
′
(
u
n
) is weakly sequentially continuous. Therefore, Ψ
λ
′
(
u
n
)
⇀
Ψ
λ
′
(
u
λ
) and, by (76), Ψ
λ
′
(
u
λ
)
=
0.
Finally, by(
f
3
) and Fatou’s lemma
(77)
c
=
lim
⁡
n
→
∞
(
Ψ
λ
(
u
n
)
-
1
2
〈
Ψ
λ
′
(
u
n
)
,
u
n
〉
)
=
lim
⁡
n
→
∞
∫
R
N
‍
F
~
(
x
,
u
n
)
≥
∫
R
N
‍
F
~
(
x
,
u
λ
)
=
Ψ
λ
(
u
λ
)
.Lemma 11.
There exist0
<
K
*
*
<
1 and η
>
0 such that, for any λ
∈
[
K
*
*
,
1
], if u
≠
0 satisfies Ψ
λ
′
(
u
)
=
0, then ∥
u
∥
≥
η.Proof.
We adapt the arguments of Yang [23, p. 2626] and Liu [12, Lemma 2.2]. Note that, by (
f
1
) and (
f
2
), for any ϵ
>
0, there exists C
ϵ
>
0 such that
(78)
|
f
(
x
,
t
)
|
≤
ϵ
|
t
|
+
C
ϵ
|
t
|
p
-
1
.
Let u
≠
0 be a critical point of Ψ
λ. Then u is a solution of
(79)
-
Δ
u
+
V
λ
u
+
f
(
x
,
u
)
=
0
,
u
∈
X
.
Multiplying both sides of this equation by u
±, respectively, and then integrating into R
N, we get that
(80)
0
=
±
∥
u
±
∥
2
+
(
1
-
λ
)
∫
R
N
‍
V
-
(
x
)
u
n
u
±
d
x
+
∫
R
N
‍
f
(
x
,
u
)
u
±
d
x
.
It follows that
(81)
∥
u
±
∥
2
=
∓
(
1
-
λ
)
∫
R
N
‍
V
-
(
x
)
u
u
±
d
x
∓
∫
R
N
‍
f
(
x
,
u
)
u
±
d
x
≤
(
1
-
λ
)
sup
⁡
R
N
V
-
∫
R
N
‍
|
u
|
·
|
u
±
|
d
x
+
ϵ
∫
R
N
‍
|
u
|
·
|
u
±
|
d
x
+
C
ϵ
∫
R
N
‍
|
u
|
p
-
1
|
u
±
|
d
x
≤
C
1
(
(
1
-
λ
)
+
ϵ
)
∥
u
∥
·
∥
u
±
∥
+
C
2
∥
u
∥
p
-
1
∥
u
±
∥
,
where C
1 and C
2 are positive constants related to the Sobolev inequalities and sup
⁡
R
N
V
-. From the above two inequalities, we obtain
(82)
∥
u
∥
2
=
∥
u
+
∥
2
+
∥
u
-
∥
2
≤
2
C
1
(
(
1
-
λ
)
+
ϵ
)
∥
u
∥
2
+
2
C
2
∥
u
∥
p
.
Because p
>
2, this implies that ∥
u
∥
≥
η for some η
>
0 if ϵ
>
0 and 1
-
K
*
*
>
0 are small enough and λ
∈
[
K
*
*
,
1
]. The desired result follows.LetK
=
max
⁡
{
K
*
,
K
*
*
}, where K
* and K
*
* are the constants that appeared in Lemmas 9 and 11, respectively. Combining Lemmas 9–11, we obtain the following lemma.Lemma 12.
Suppose(
v
) and (
f
1
)
–
(
f
3
) are satisfied. Then, there exist η
>
0,  {
λ
n
}
⊂
[
K
,
1
], and {
u
n
}
⊂
X such that λ
n
→
1,
(83)
sup
⁡
n
Ψ
λ
n
(
u
n
)
<
+
∞
,
∥
u
n
∥
≥
η
,
Ψ
λ
n
′
(
u
n
)
=
0
.


## 3. A Priori Bound of Approximate Solutions and Proof of the Main Theorem

In this section, we give a priori bound for the sequence of approximate solutions{
u
n
} obtained in Lemma 12. We then give the proofs of Theorem 3.Lemma 13.
Suppose(
v
) and (
f
1
)
–
(
f
3
) are satisfied. Let {
u
n
} be the sequence obtained in Lemma 12. Then, {
u
n
}
⊂
L
∞
(
R
N
) and
(84)
sup
⁡
n
∥
u
n
∥
L
∞
(
R
N
)
≤
D
.Proof.
FromΨ
λ
n
′
(
u
n
)
=
0, we deduce that u
n is a weak solution of (45) with λ
=
λ
n; that is,
(85)
-
Δ
u
n
+
V
λ
n
(
x
)
u
n
+
f
(
x
,
u
n
)
=
0
in
R
N
.
By assumption (
f
1
) and the bootstrap argument of elliptic equations, we deduce that u
n
∈
L
∞
(
R
N
).
Multiplying both sides of (85) by v
n
=
(
u
n
-
D
)
+
:
=
max
⁡
{
u
n
-
D
,
0
} and integrating into R
N, we get that
(86)
∫
R
N
‍
|
∇
v
n
|
2
d
x
+
∫
u
n
≥
D
‍
(
V
λ
n
(
x
)
u
n
+
f
(
x
,
u
n
)
)
v
n
d
x
=
0
.
Recall that V
λ
n
=
V
+
-
λ
n
V
- and λ
n
≤
1. Then by (5), we get that
(87)
∫
u
n
≥
D
‍
(
V
λ
n
(
x
)
u
n
+
f
(
x
,
u
n
)
)
v
n
d
x
=
∫
u
n
≥
D
‍
(
V
λ
n
(
x
)
+
f
(
x
,
u
n
)
u
n
)
u
n
v
n
d
x
≥
0
.
This together with (86) yields v
n
=
0. It follows that u
n
(
x
)
≤
D on R
N.
Similarly, multiplying both sides of (85) by w
n
=
(
u
n
+
D
)
-
:
=
max
⁡
{
-
(
u
n
+
D
)
,
0
} and integrating into R
N, we can get that u
n
≥
-
D on R
N. Therefore, for all n
,
∥
u
n
∥
L
∞
(
R
N
)
≤
D.Lemma 14.
Suppose that(
v
), (
f
1
), (
f
2
), (
f
3
), and (
f
4
) are satisfied. Let {
u
n
} be the sequence obtained in Lemma 12. Then
(88)
0
<
inf
⁡
n
∥
u
n
∥
≤
sup
⁡
n
∥
u
n
∥
<
+
∞
.Proof.
AsΨ
λ
n
′
(
u
n
)
=
0 and u
n
≠
0, Lemma 11 implies that inf
⁡
n
∥
u
n
∥
>
0.
To provesup
⁡
n
∥
u
n
∥
<
+
∞, we apply an indirect argument and assume by contradiction that ∥
u
n
∥
→
+
∞.
SinceΨ
λ
n
′
(
u
n
)
=
0, by (81), we get that
(89)
∥
u
n
±
∥
2
=
∓
(
1
-
λ
n
)
∫
R
N
‍
V
-
(
x
)
u
n
u
n
±
d
x
∓
∫
R
N
‍
f
(
x
,
u
n
)
u
n
±
d
x
=
∓
∫
R
N
‍
f
(
x
,
u
n
)
u
n
±
d
x
+
(
1
-
λ
n
)
O
(
∥
u
n
∥
2
)
.
It follows that
(90)
∥
u
n
∥
2
+
∫
R
N
‍
f
(
x
,
u
n
)
(
u
n
+
-
u
n
-
)
d
x
=
∥
u
n
+
∥
2
+
∥
u
n
-
∥
2
+
∫
R
N
‍
f
(
x
,
u
n
)
(
u
n
+
-
u
n
-
)
d
x
=
(
1
-
λ
n
)
O
(
∥
u
n
∥
2
)
.
Set w
n
=
u
n
/
∥
u
n
∥. Then, by (90),
(91)
∥
u
n
∥
2
(
1
+
∫
R
N
‍
f
(
x
,
u
n
)
u
n
(
w
n
+
-
w
n
-
)
w
n
d
x
)
=
(
1
-
λ
n
)
O
(
∥
u
n
∥
2
)
.
Then, by λ
n
→
1 as n
→
∞, we have that
(92)
∫
R
N
‍
f
(
x
,
u
n
)
u
n
(
w
n
+
-
w
n
-
)
w
n
d
x
⟶
-
1
,
n
⟶
∞
.
From Lemma12,
(93)
C
0
:
=
sup
⁡
n
Ψ
λ
n
(
u
n
)
<
+
∞
.
Then, by Ψ
λ
n
′
(
u
n
)
=
0, we obtain
(94)
2
C
0
≥
2
Ψ
λ
n
(
u
n
)
-
〈
Ψ
λ
n
′
(
u
n
)
,
u
n
〉
=
2
∫
R
N
‍
F
~
(
x
,
u
n
)
d
x
.
From (
f
3
), we have
(95)
2
C
0
≥
2
∫
R
N
‍
F
~
(
x
,
u
n
)
d
x
≥
2
∫
{
x
∣
D
≥
|
u
n
(
x
)
|
≥
κ
}
‍
F
~
(
x
,
u
n
)
d
x
,
where κ is the constant in (
f
4
). As the continuous function F
~ is 1-periodic in every x
j variable, we deduce from (8) that there exists a constant C
′
>
0 such that
(96)
F
~
(
x
,
t
)
≥
C
′
t
2
,
for
every
(
x
,
t
)
∈
R
N
×
R
with
κ
≤
|
t
|
≤
D
.
Combining (95) and (96) leads to
(97)
C
0
≥
C
′
∫
{
x
∣
D
≥
|
u
n
(
x
)
|
≥
κ
}
‍
u
n
2
d
x
.
Dividing both sides of this inequality by ∥
u
n
∥
2 and sending n
→
∞, we obtain
(98)
lim
⁡
n
→
∞
∫
{
x
∣
D
≥
|
u
n
(
x
)
|
≥
κ
}
‍
w
n
2
d
x
=
0
.
From (7), (21), and (22), we have that
(99)
∫
{
x
∣
|
u
n
(
x
)
|
<
κ
}
‍
|
f
(
x
,
u
n
)
u
n
(
w
n
+
-
w
n
-
)
w
n
|
d
x
≤
ν
∫
{
x
∣
|
u
n
(
x
)
|
<
κ
}
‍
|
(
w
n
+
-
w
n
-
)
w
n
|
d
x
≤
ν
∫
R
N
‍
|
(
w
n
+
-
w
n
-
)
w
n
|
d
x
≤
ν
∥
w
n
∥
L
2
2
≤
ν
μ
0
∥
w
n
∥
2
=
ν
μ
0
<
1
,
where μ
0 is the constant defined in (
v
).
Sincef
∈
C
(
R
N
×
R
) and lim
⁡
t
→
0
f
(
x
,
t
)
/
t
=
0, we deduce that there exists C
>
0 such that, for every (
x
,
t
)
∈
R
N
×
R with |
t
|
≤
D,
(100)
|
f
(
x
,
t
)
|
≤
C
|
t
|
.
This together with (98) gives
(101)
∫
{
x
∣
D
≥
|
u
n
(
x
)
|
≥
κ
}
‍
|
f
(
x
,
u
n
)
u
n
(
w
n
+
-
w
n
-
)
w
n
|
d
x
≤
C
∫
{
x
∣
D
≥
|
u
n
(
x
)
|
≥
κ
}
‍
|
(
w
n
+
-
w
n
-
)
w
n
|
d
x
≤
C
∥
w
n
+
-
w
n
-
∥
L
2
(
∫
{
x
∣
D
≥
|
u
n
(
x
)
|
≥
κ
}
‍
w
n
2
d
x
)
1
/
2
≤
2
C
∥
w
n
∥
L
2
(
∫
{
x
∣
D
≥
|
u
n
(
x
)
|
≥
κ
}
‍
w
n
2
d
x
)
1
/
2
⟶
0
,
n
⟶
∞
.
Combining (99) and (101) yields
(102)
limsup
⁡
n
→
∞
∫
R
N
‍
|
f
(
x
,
u
n
)
u
n
(
w
n
+
-
w
n
-
)
w
n
|
d
x
≤
limsup
⁡
n
→
∞
∫
{
x
∣
|
u
n
(
x
)
|
<
κ
}
‍
|
f
(
x
,
u
n
)
u
n
(
w
n
+
-
w
n
-
)
w
n
|
d
x
+
limsup
⁡
n
→
∞
∫
{
x
∣
D
≥
|
u
n
(
x
)
|
≥
κ
}
‍
|
f
(
x
,
u
n
)
u
n
(
w
n
+
-
w
n
-
)
w
n
|
d
x
<
1
.
This contradicts (92). Therefore, {
u
n
} is bounded in X.Proof of Theorem3.
Let{
u
n
} be the sequence obtained in Lemma 12. From Lemma 14, {
u
n
} is bounded in X. Therefore, up to a subsequence, either (a)
lim
⁡
n
→
∞
sup
⁡
y
∈
R
N
∫
B
1
(
y
)
‍
|
u
n
|
2
d
x
=
0 or
(b)
there existϱ
>
0 and y
n
∈
Z
N such that ∫
B
1
(
y
n
)
‍
|
u
n
|
2
d
x
≥
ϱ. According to (72), if case (a) occurs,
(103)
lim
⁡
n
→
∞
∫
R
N
‍
f
(
x
,
u
n
)
u
n
±
d
x
=
0
.
Then, by (81) and λ
n
→
1, we have
(104)
∥
u
n
±
∥
2
=
∓
(
1
-
λ
n
)
∫
R
N
‍
V
-
(
x
)
u
n
u
n
±
d
x
∓
∫
R
N
‍
f
(
x
,
u
n
)
u
n
±
d
x
≤
C
(
1
-
λ
n
)
∥
u
n
∥
L
2
2
+
|
∫
R
N
‍
f
(
x
,
u
n
)
u
n
±
d
x
|
⟶
0
.
This contradicts inf
⁡
n
∥
u
n
∥
>
0 (see (88)). Therefore, case (a) cannot occur. As case (b) therefore occurs, w
n
=
u
n
(
·
+
y
n
) satisfies w
n
⇀
u
0
≠
0. From (14) and (43), we have that
(105)
Ψ
λ
(
u
)
=
-
Φ
(
u
)
+
λ
-
1
2
∫
R
N
‍
V
-
u
2
d
x
,
∀
u
∈
X
.
It follows that
(106)
〈
Ψ
λ
′
(
u
)
,
φ
〉
=
-
〈
Φ
′
(
u
)
,
φ
〉
+
(
λ
-
1
)
∫
R
N
‍
V
-
u
φ
d
x
,
iiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiii
∀
u
,
φ
∈
X
.
By Ψ
λ
n
′
(
u
n
)
=
0 (see Lemma 12), we have Ψ
′
λ
n
(
w
n
)
=
0. From (106), we have that, for any φ
∈
X,
(107)
〈
Ψ
′
λ
n
(
w
n
)
,
φ
〉
=
-
〈
Φ
′
(
w
n
)
,
φ
〉
+
(
λ
n
-
1
)
×
∫
R
N
‍
V
-
(
x
)
w
n
φ
d
x
.
Together with Ψ
λ
n
′
(
w
n
)
=
0 and λ
n
→
1, this yields
(108)
〈
Φ
′
(
w
n
)
,
φ
〉
⟶
0
,
∀
φ
∈
X
.
Finally, by w
n
⇀
u
0
≠
0 and the weakly sequential continuity of Φ
′, we have that Φ
′
(
u
0
)
=
0. Therefore, u
0 is a nontrivial solution of (1). This completes the proof.



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*Source: 539639-2014-06-15.xml*