# Nehari-Type Ground State Positive Solutions for Superlinear Asymptotically Periodic Schrödinger Equations

Authors: Xiaoyan Lin; X. H. Tang
Journal: Abstract and Applied Analysis
 (2014)

---

## Abstract

We deal with the existence of Nehari-type ground state positive solutions for the nonlinear Schrödinger equation-
Δ
u
+
V
x
u
=
f
x
,
u
,
x
∈
R
N
,
u
∈
H
1
R
N. Under a weaker Nehari condition, we establish some existence criteria to guarantee that the above problem  has Nehari-type ground state solutions by using a more direct method in two cases: the periodic case and the asymptotically periodic case.


---

## Body

## 1. Introduction

Consider the following semilinear Schrödinger equation:(1)
-
Δ
u
+
V
(
x
)
u
=
f
(
x
,
u
)
,
x
∈
R
N
,
u
∈
H
1
(
R
N
)
,
where V
:
R
N
→
R and f
:
R
N
×
R
→
R.The Schrödinger equation has found a great deal of interest last years because not only it is important in applications but also it provides a good model for developing mathematical methods. Many authors have studied the existence of entire solutions of Schrödinger equations under various stipulations (cf., e.g., [1–28] and the references quoted in them).Wheninf
⁡
R
N
V
(
x
)
>
0 and V
(
x
) is periodic, Li et al. [12] made use of a combination of the techniques in [13, 14] with applications of Lions’ concentration compactness principle [26, 29, 30] to establish the following theorem.Theorem 1 (see [12]).
Assume thatV and f satisfy the following assumptions:(V0)
V
∈
C
(
R
N
,
R
) and inf
⁡
R
N
V
(
x
)
>
0;
(V1)
V
(
x
) is 1-periodic in each of x
1
,
x
2
,
…
,
x
N;
(S0)
f
∈
C
1, f
t
′ is a Caratheodory function, and there exists a constant C
>
0 such that
(2)
|
f
t
′
(
x
,
t
)
|
≤
C
(
1
+
|
t
|
2
*
-
2
)
,
lim
⁡
|
t
|
→
∞
|
f
t
′
(
x
,
t
)
|
|
t
|
2
*
-
2
=
0
;
(S1)
f
(
x
,
t
) is 1-periodic in each of x
1
,
x
2
,
…
,
x
N;
(S2)
f
(
x
,
t
)
=
o
(
|
t
|
), as |
t
|
→
0, uniformly in x
∈
R
N;
(S3)
lim
⁡
|
t
|
→
∞
(
|
F
(
x
,
t
)
|
/
|
t
|
2
)
=
∞, uniformly in x
∈
R
N;
(S4)
f
(
x
,
t
)
/
|
t
| is strictly increasing in t on R
∖
{
0
} for every x
∈
R
N.Then problem (1) has a solution u
0
∈
H
1
(
R
N
) such that Φ
(
u
0
)
=
inf
⁡
N
Φ
>
0, where
(3)
Φ
(
u
)
=
1
2
∫
R
N
‍
(
|
∇
u
|
2
+
V
(
x
)
u
2
)
d
x
-
∫
R
N
‍
F
(
x
,
u
)
d
x
,
∀
u
∈
H
1
(
R
N
)
,
(4)
〈
Φ
′
(
u
)
,
v
〉
=
∫
R
N
‍
(
∇
u
·
∇
v
+
V
(
x
)
u
v
)
d
x
-
∫
R
N
‍
f
(
x
,
u
)
v
d
x
,
∀
u
,
v
∈
H
1
(
R
N
)
with F
(
x
,
t
)
≔
∫
0
t
‍
f
(
x
,
s
)
d
s, and
(5)
N
=
{
u
∈
H
1
(
R
N
)
:
〈
Φ
′
(
u
)
,
u
〉
=
0
,
u
≠
0
}
.The setN is the Nehari manifold, which contains infinitely many elements of H
1
(
R
N
). In fact, for any u
∈
H
1
(
R
N
)
∖
{
0
}, there exists t
=
t
(
u
)
>
0 such that t
u
∈
N; see Lemma 9. Since u
0 is a solution at which Φ has least “energy” in set N, we will call it a Nehari-type ground state solution.We must point out that “the least energy solution” (which is sometimes also called the ground state solution in some references) is in fact a nontrivial solutionu
0 which satisfies Φ
(
u
0
)
=
inf
⁡
M
Φ, where
(6)
M
=
{
u
∈
H
1
(
R
N
)
∖
{
0
}
:
Φ
′
(
u
)
=
0
}
is a very small subset of N; it may contain only one element. In general, it is much more difficult to find a solution u
0 for problem (1) with a constraint condition Φ
(
u
0
)
=
inf
⁡
N
Φ than with one Φ
(
u
0
)
=
inf
⁡
M
Φ.To establish the existence of Nehari-type ground state solutions, the so-called Nehari-type condition (S4) seems to be always necessary in the proof of the existence of ground states solutions for problem (1).In recent paper [15, 20], Theorem 1 has been extended to the case where 0 is in the gap of the spectrum σ
(
-
Δ
+
V
), but an additional assumption on the nonlinearity f is assumed in [15].Motivated by [12, 17, 20], in the present paper, we will develop a more direct method to generalize Theorem 1 by relaxing assumptions (V0), (S3), and (S4) in two cases: the periodic case and the asymptotically periodic case.In the periodic case, we establish the following two theorems.Theorem 2.
Assume thatV and f satisfy (V0), (V1), (S1), (S2), and the following assumptions:(S
0
′)
f
∈
C
(
R
N
×
R
,
R
), and there exists a constant C
>
0 such that
(7)
|
f
(
x
,
t
)
|
≤
C
(
1
+
|
t
|
2
*
-
1
)
,
lim
⁡
|
t
|
→
∞
|
f
(
x
,
t
)
|
|
t
|
2
*
-
1
=
0
,
u
n
i
f
o
r
m
l
y
i
n
x
∈
R
N
;
(S
3
′)
lim
⁡
|
t
|
→
∞
(
|
F
(
x
,
t
)
|
/
|
t
|
2
)
=
∞, a
.
e.x
∈
R
N;
(S
4
′)
f
(
x
,
t
)
/
|
t
| is nondecreasing in t on R
∖
{
0
} for every x
∈
R
N.Then problem (1) has a solution u
0
∈
H
1
(
R
N
) such that Φ
(
u
0
)
=
inf
⁡
N
Φ
>
0.Theorem 3.
Assume thatV and f satisfy (V
0), (V
1), (S
0
′), (S
1), (S
2), (S
3
′), and (S
4
′). Then problem (1) has a positive solution u
¯
∈
H
1
(
R
N
) such that Φ
+
(
u
¯
)
=
inf
⁡
N
+
Φ
+
>
0, where
(8)
Φ
+
(
u
)
=
1
2
∫
R
N
‍
(
|
∇
u
|
2
+
V
(
x
)
u
2
)
d
x
-
∫
R
N
‍
F
(
x
,
u
+
)
d
x
,
∀
u
∈
H
1
(
R
N
)
with u
+
(
x
)
=
max
⁡
{
u
(
x
)
,
0
} and u
-
(
x
)
=
max
⁡
{
-
u
(
x
)
,
0
}, and
(9)
N
+
=
{
u
∈
H
1
(
R
N
)
:
〈
Φ
+
′
(
u
)
,
u
〉
=
0
,
u
≠
0
}
.In the asymptotically periodic case,V
(
x
) is allowed to be negative in some bounded domain in R
N. Precisely, we use the following condition instead of (V0).(V
0
′)
V
∈
C
(
R
N
)
∩
L
∞
(
R
N
), V
(
x
)
≤
V
¯
∈
(
0
,
∞
) for all x
∈
R
N and there exists a constant ζ
0
>
0 such that
(10)
∫
R
N
‍
(
|
∇
u
|
2
+
V
(
x
)
u
2
)
d
x
≥
ζ
0
∫
R
N
‍
[
V
¯
-
V
(
x
)
]
|
u
|
2
d
x
,
∀
u
∈
H
1
(
R
N
)
.
(V
0
′)
Was first introduced by Deng et al. [6]; it is satisfied if the following assumption holds (see Lemma 7):
(V
0
′′)
V
∈
C
(
R
N
)
∩
L
∞
(
R
N
), and there exist two constants a
1
,
a
2
>
0 and a bounded measurable set Ω
⊂
R
N such that (
a
1
+
a
2
)
|
Ω
|
2
/
N
≤
S, and
(11)
V
(
x
)
≥
{
-
a
1
,
x
∈
Ω
,
a
2
,
x
∉
Ω
,
whereS
=
inf
⁡
u
∈
D
1,2
(
R
N
)
,
∥
u
∥
2
*
=
1
∥
∇
u
∥
2
2.In this case, we establish the following two theorems.Theorem 4.
Assume thatV and f satisfy (V
0
′), (S
0
′), (S
2), (S
3), (S
4
′), and the following assumptions:(V2)
V
(
x
)
=
V
0
(
x
)
+
V
1
(
x
), V
0
∈
C
(
R
N
,
(
0
,
+
∞
)
), V
0
(
x
) is 1-periodic in each of x
1
,
x
2
,
…
,
x
N, V
1
(
x
)
<
0 for |
x
|
<
1
+
N, and V
1
(
x
)
≤
0 for |
x
|
≥
1
+
N, lim
⁡
|
x
|
→
∞
|
V
1
(
x
)
|
=
0;
(S
1
′)
f
(
x
,
t
)
=
f
0
(
x
,
t
)
+
f
1
(
x
,
t
); f
0
∈
C
(
R
N
×
R
,
R
) satisfies (S1), (S2), and (S
4
′); f
1
∈
C
(
R
N
×
R
,
R
) satisfies that
(12)
0
≤
t
f
1
(
x
,
t
)
≤
a
(
x
)
(
t
2
+
|
t
|
p
0
)
,
∀
(
x
,
t
)
∈
R
N
×
R
,
wherep
0
∈
(
2
,
2
*
) and a
∈
C
(
R
N
,
R
+
) with lim
⁡
|
x
|
→
∞
a
(
x
)
=
0.Then problem (1) has a solution u
0
∈
H
1
(
R
N
) such that Φ
(
u
0
)
=
inf
⁡
N
Φ
>
0.Theorem 5.
Assume thatV and f satisfy (V
0
′), (V
2), (S
0
′), (S
1
′), (S
2), (S
3), and (S
4
′). Then problem (1) has a positive solution u
¯
∈
H
1
(
R
N
) such that Φ
+
(
u
¯
)
=
inf
⁡
N
+
Φ
+
>
0.In our theorems, we give a new condition (S
4
′) which weakens Nehari-type condition (S3) considerably.


## 2. Preliminaries

Lemma 6 (see [6, Lemma 2.3]).
Assume thatV satisfies (V
0
′). Then there exist two positive constants C
1
,
C
2
>
0 such that
(13)
C
1
∥
u
∥
H
1
(
R
N
)
2
≤
∫
R
N
‍
[
|
∇
u
|
2
+
V
(
x
)
u
2
]
d
x
≤
C
2
∥
u
∥
H
1
(
R
N
)
2
,
∀
u
∈
H
1
(
R
N
)
,
where ∥
u
∥
H
1
(
R
N
) is the usual norm of H
1
(
R
N
)
(14)
∥
u
∥
H
1
(
R
N
)
=
[
∫
R
N
‍
(
|
∇
u
|
2
+
u
2
)
d
x
]
1
/
2
,
u
∈
H
1
(
R
N
)
.By Lemma6, we define an inner product
(15)
(
u
,
v
)
=
∫
R
N
‍
[
(
∇
u
·
∇
v
)
+
V
(
x
)
u
v
]
d
x
,
u
,
v
∈
H
1
(
R
N
)
,
associated with the norm
(16)
∥
u
∥
=
{
∫
R
N
‍
[
|
∇
u
|
2
+
V
(
x
)
u
2
]
d
x
}
1
/
2
,
u
∈
H
1
(
R
N
)
.
Then H
1
(
R
N
) is a Hilbert space with this inner product. Moreover, under assumptions (V
0
′) and (S
0
′), the functional Φ defined by (3) is of class C
1
(
H
1
(
R
N
)
,
R
).Lemma 7.
If (V
0
′′) holds, then (V
0
′) does.Proof.
By virtue of (V
0
′′), the Hölder inequality, and the Sobolev inequality, we have
(17)
∫
R
N
[
|
∇
u
|
2
+
V
(
x
)
u
2
]
‍
d
x
≥
∫
R
N
|
∇
u
|
2
d
x
-
(
a
1
+
a
2
)
∫
Ω
u
2
d
x
+
a
2
∫
R
N
u
2
d
x
≥
∫
R
N
|
∇
u
|
2
d
x
-
(
a
1
+
a
2
)
|
Ω
|
2
/
N
∥
u
∥
2
*
2
+
a
2
∫
R
N
u
2
d
x
≥
∥
∇
u
∥
2
2
-
(
a
1
+
a
2
)
|
Ω
|
2
/
N
S
-
1
∥
∇
u
∥
2
2
+
a
2
∫
R
N
u
2
d
x
≥
a
2
∫
R
N
u
2
d
x
≥
a
2
V
¯
+
a
1
∫
R
N
[
V
¯
-
V
(
x
)
]
u
2
d
x
,
∀
u
∈
H
1
(
R
N
)
.
This shows that (V
0
′) holds with ζ
0
=
a
2
/
(
V
¯
+
a
1
).Lemma 8.
LetX be a Banach space. Let M
0 be a closed subspace of the metric space M and Γ
0
⊂
C
(
M
0
,
X
). Define
(18)
Γ
=
{
γ
∈
C
(
M
,
X
)
:
γ
|
M
0
∈
Γ
0
}
.
If Ψ
∈
C
1
(
X
,
R
) satisfies
(19)
∞
>
c
≔
inf
⁡
γ
∈
Γ
sup
⁡
t
∈
M
Ψ
(
γ
(
t
)
)
>
a
≔
sup
⁡
γ
0
∈
Γ
0
sup
⁡
t
∈
M
0
Ψ
(
γ
0
(
t
)
)
,
then there exists a sequence {
u
n
}
⊂
X satisfying
(20)
Ψ
(
u
n
)
⟶
c
,
∥
Ψ
′
(
u
n
)
∥
(
1
+
∥
u
n
∥
)
⟶
0
.Proof.
For anyγ
∈
Γ, define set K
γ
=
{
γ
(
t
)
:
t
∈
M
} in X and the collection K
=
{
K
γ
:
γ
∈
Γ
}. Let A
=
{
γ
0
(
t
)
:
γ
0
∈
Γ
0
,
t
∈
M
0
},
(21)
Λ
=
{
φ
∈
C
(
X
,
X
)
:
φ
-
1
∈
C
(
X
,
X
)
,
both
φ
and
φ
-
1
are
bounded
on
⁡
bounded
sets
(
X
,
X
)
:
φ
-
1
}
,
Λ
(
A
)
=
{
φ
∈
Λ
:
φ
(
u
)
=
u
,
u
∈
A
}
.
For any γ
∈
Γ and φ
∈
Λ
(
A
), let γ
0
=
γ
|
M
0 and γ
~
(
t
)
=
φ
(
γ
(
t
)
)
,
t
∈
M. Then γ
0
∈
Γ
0 and γ
~
∈
C
(
M
,
X
). Hence,
(22)
γ
~
(
t
)
=
φ
(
γ
0
(
t
)
)
=
γ
0
(
t
)
,
∀
t
∈
M
0
;
that is, γ
~
|
M
0
=
γ
0
∈
Γ
0. Therefore,
(23)
φ
(
K
)
∈
K
,
∀
φ
∈
Λ
(
A
)
,
K
∈
K
.
These show that the collection K is a minimax system for A. Since (19) implies
(24)
∞
>
c
≔
inf
⁡
K
∈
K
sup
⁡
K
Ψ
>
a
≔
sup
⁡
A
Ψ
,
it follows from Theorem 2.4 in [18] that the result is true.Lemma 9.
Under assumptions (V
0
′), (S
0
′), (S
2), and (S
3
′), for any u
∈
H
1
(
R
N
)
∖
{
0
}, there exists t
(
u
)
>
0 such that t
(
u
)
u
∈
N.Proof.
Letu
∈
H
1
(
R
N
)
∖
{
0
} be fixed and define the function g
(
t
)
≔
Φ
(
t
u
) on [
0
,
∞
). Clearly we have
(25)
g
′
(
t
)
=
0
⟺
t
u
∈
N
⟺
∥
u
∥
2
=
1
t
∫
R
N
f
(
x
,
t
u
)
u
d
x
.
It is easy to verify, using (S2) and (S
3
′), that g
(
0
)
=
0, g
(
t
)
>
0 for t
>
0 small and g
(
t
)
<
0 for t large. Therefore max
⁡
t
∈
[
0
,
∞
)
g
(
t
) is achieved at a t
0
=
t
(
u
) so that g
′
(
t
0
)
=
0 and t
(
u
)
u
∈
N.Lemma 10.
Under assumptions (V
0
′), (S
0
′), and (S
4
′),
(26)
Φ
(
u
)
≥
Φ
(
t
u
)
+
1
-
t
2
2
〈
Φ
′
(
u
)
,
u
〉
,
∀
u
∈
H
1
(
R
N
)
,
t
≥
0
.Proof.
Forτ
≠
0, (S
4
′) yields
(27)
f
(
x
,
s
)
≤
f
(
x
,
τ
)
|
τ
|
|
s
|
,
s
≤
τ
;
f
(
x
,
s
)
≥
f
(
x
,
τ
)
|
τ
|
|
s
|
,
s
≥
τ
.
It follows that
(28)
1
-
t
2
2
τ
f
(
x
,
τ
)
≥
∫
t
τ
τ
f
(
x
,
s
)
d
s
,
t
≥
0
.
Note that
(29)
〈
Φ
′
(
u
)
,
u
〉
=
∥
u
∥
2
-
∫
R
N
f
(
x
,
u
)
u
d
x
.
Thus, by (3), (28), and (29), one has
(30)
Φ
(
u
)
-
Φ
(
t
u
)
=
1
-
t
2
2
∥
u
∥
2
+
∫
R
N
[
F
(
x
,
t
u
)
-
F
(
x
,
u
)
]
d
x
=
1
-
t
2
2
〈
Φ
′
(
u
)
,
u
〉
+
∫
R
N
[
1
-
t
2
2
f
(
x
,
u
)
u
+
F
(
x
,
t
u
)
-
F
(
x
,
u
)
]
‍
d
x
=
1
-
t
2
2
〈
Φ
′
(
u
)
,
u
〉
+
∫
R
N
[
1
-
t
2
2
f
(
x
,
u
)
u
-
∫
t
u
u
‍
f
(
x
,
s
)
d
s
]
d
x
≥
1
-
t
2
2
〈
Φ
′
(
u
)
,
u
〉
,
t
≥
0
.
This shows that (26) holds.Corollary 11.
Under assumptions (V
0
′), (S
0
′), and (S
4
′), for u
∈
N,
(31)
Φ
(
u
)
≥
Φ
(
t
u
)
,
∀
t
≥
0
.We define(32)
c
1
≔
inf
⁡
N
Φ
,
c
2
≔
inf
⁡
u
∈
E
,
u
≠
0
max
⁡
t
≥
0
Φ
(
t
u
)
,
c
≔
inf
⁡
γ
∈
Γ
sup
⁡
t
∈
[
0,1
]
Φ
(
γ
(
t
)
)
,
where
(33)
Γ
=
{
γ
∈
C
(
[
0,1
]
,
E
)
:
γ
(
0
)
=
0
,
Φ
(
γ
(
1
)
)
<
0
}
.Lemma 12.
Under assumptions (V
0
′), (S
0
′), (S
2), (S
3
′), and (S
4
′), one has that c
1
=
c
2
=
c
>
0 and there exists a sequence {
u
n
}
⊂
H
1
(
R
N
) satisfying
(34)
Φ
(
u
n
)
⟶
c
,
∥
Φ
′
(
u
n
)
∥
(
1
+
∥
u
n
∥
)
⟶
0
.Proof.
(1) Both Lemma9 and Corollary 11 imply that c
1
=
c
2. Next, we prove that c
=
c
1
=
c
2. By the definition of c
2, we choose a sequence {
v
n
}
⊂
E
∖
{
0
} such that
(35)
c
2
≤
max
⁡
t
≥
0
Φ
(
t
v
n
)
<
c
2
+
1
n
,
n
∈
N
.
Since Φ
(
t
u
)
<
0 for u
∈
E
∖
{
0
} and t large, there exist t
n
=
t
(
v
n
)
>
0 and s
n
>
t
n such that
(36)
Φ
(
t
n
v
n
)
=
max
⁡
t
≥
0
Φ
(
t
v
n
)
,
Φ
(
s
n
v
n
)
<
0
,
n
∈
N
.
Let γ
n
(
t
)
=
t
s
n
v
n for t
∈
[
0,1
]. Then γ
n
∈
Γ, and it follows from (35) and (36) that
(37)
sup
⁡
t
∈
[
0,1
]
Φ
(
γ
n
(
t
)
)
=
max
⁡
t
≥
0
Φ
(
t
v
n
)
<
c
2
+
1
n
,
n
∈
N
,
which implies that c
≤
c
2. On the other hand, the manifold N separates H
1
(
R
N
) into two components E
+
=
{
u
∈
H
1
(
R
N
)
:
〈
Φ
′
(
u
)
,
u
〉
>
0
}
∪
{
0
} and E
-
=
{
u
∈
H
1
(
R
N
)
:
〈
Φ
′
(
u
)
,
u
〉
<
0
}. By (S
4
′), one has
(38)
f
(
x
,
t
)
t
≥
2
F
(
x
,
t
)
,
∀
(
x
,
t
)
∈
R
N
×
R
.
It follows that Φ
(
u
)
≥
0 for u
∈
E
+. By (S
0
′) and (S2), E
+ contains a small ball around the origin. Thus every γ
∈
Γ has to cross N, because γ
(
0
)
∈
E
+ and γ
(
1
)
∈
E
-, and so c
1
≤
c.
(2) In order to prove the second part of Lemma12, we apply Lemma 8 with M
=
[
0,1
], M
0
=
{
0,1
}, and
(39)
Γ
0
=
{
γ
0
:
{
0,1
}
⟶
H
1
(
R
N
)
:
γ
0
(
0
)
=
0
,
Φ
(
γ
0
(
1
)
)
<
0
}
.
By (S
0
′) and (S2), there exists r
>
0 such that
(40)
min
⁡
∥
u
∥
≤
r
Φ
(
u
)
=
0
,
inf
⁡
∥
u
∥
=
r
Φ
(
u
)
>
0
.
Hence we obtain
(41)
c
≥
inf
⁡
∥
u
∥
=
r
Φ
(
u
)
>
0
=
sup
⁡
γ
0
∈
Γ
0
sup
⁡
t
∈
M
0
Φ
(
γ
0
(
t
)
)
.
These show that all assumptions of Lemma 8 are satisfied. Therefore there exists a sequence (
u
n
)
⊂
H
1
(
R
N
) satisfying (34).Lemma 13.
Under assumptions (V
0
′), (S
0
′), (S
2), (S
3), and (S
4
′), any sequence {
u
n
}
⊂
H
1
(
R
N
) satisfying (34) is bounded in H
1
(
R
N
).Proof.
To prove the boundedness of{
u
n
}, arguing by contradiction, suppose that ∥
u
n
∥
→
∞. Let v
n
=
u
n
/
∥
u
n
∥. Then ∥
v
n
∥
=
1. By Sobolev embedding theorem, there exists a constant C
3
>
0 such that
(42)
∥
v
n
∥
2
+
∥
v
n
∥
2
*
≤
C
3
.
Passing to a subsequence, we may assume that v
n
⇀
v in H
1
(
R
N
), v
n
→
v in L
loc
⁡
s
(
R
N
), 2
≤
s
<
2
*, and v
n
→
v a.e. on R
N.
Ifδ
≔
limsup
⁡
n
→
∞
sup
⁡
y
∈
R
N
∫
B
1
(
y
)
‍
|
v
n
|
2
d
x
=
0, then by Lions’ concentration compactness principle [23, Lemma 1.21], v
n
→
0 in L
s
(
R
N
) for 2
<
s
<
2
*. Fix p
∈
(
2
,
2
*
) and R
>
2
c. By (S
0
′) and (S2), for ɛ
=
c
/
4
[
(
R
C
3
)
2
+
(
R
C
3
)
2
*
]
>
0 there exists C
ɛ
>
0 such that
(43)
|
F
(
x
,
u
)
|
≤
ɛ
(
|
u
|
2
+
|
u
|
2
*
)
+
C
ɛ
|
u
|
p
.
It follows that
(44)
limsup
⁡
n
→
∞
∫
R
N
F
(
x
,
R
v
n
)
d
x
≤
ɛ
[
(
R
C
3
)
2
+
(
R
C
3
)
2
*
]
+
R
p
C
ɛ
lim
⁡
n
→
∞
∥
v
n
∥
p
p
=
c
4
.
Hence, by using (34), (44), and Lemma 10, one has
(45)
c
+
o
(
1
)
=
Φ
(
u
n
)
≥
Φ
(
R
v
n
)
+
(
1
2
-
R
2
2
∥
u
n
∥
2
)
〈
Φ
′
(
u
n
)
,
u
n
〉
=
R
2
2
-
∫
R
N
‍
F
(
x
,
R
v
n
)
d
x
+
(
1
2
-
R
2
2
∥
u
n
∥
2
)
〈
Φ
′
(
u
n
)
,
u
n
〉
≥
R
2
2
-
c
4
+
o
(
1
)
>
7
c
4
+
o
(
1
)
,
which is a contradiction. Thus, δ
>
0.
Going if necessary to a subsequence, we may assume the existence ofk
n
∈
Z
N such that ∫
B
1
+
N
(
k
n
)
‍
|
v
n
|
2
d
x
>
δ
/
2. Let w
n
(
x
)
=
v
n
(
x
+
k
n
). Then it follows that
(46)
∫
B
1
+
N
(
0
)
|
w
n
|
2
d
x
>
δ
2
.
Now we define u
~
n
(
x
)
=
u
n
(
x
+
k
n
), then u
~
n
/
∥
u
n
∥
=
w
n, and ∥
w
n
∥
H
1
(
R
N
)
=
∥
u
n
∥
H
1
(
R
N
)
/
∥
u
n
∥
≤
C
4 for some C
4
>
0. Passing to a subsequence, we have w
n
⇀
w in H
1
(
R
N
), w
n
→
w in L
loc
⁡
s
(
R
N
), 2
≤
s
<
2
*, and w
n
→
w a.e. on R
N. Thus, (46) implies that w
≠
0. Hence, it follows from (34), (S3), and Fatou’s lemma that
(47)
0
=
lim
⁡
n
→
∞
c
+
o
(
1
)
∥
u
n
∥
2
=
lim
⁡
n
→
∞
Φ
(
u
n
)
∥
u
n
∥
2
=
lim
⁡
n
→
∞
⁡
[
1
2
-
∫
R
N
F
(
x
+
k
n
,
u
~
n
)
u
~
n
2
w
n
2
d
x
]
≤
1
2
-
liminf
⁡
n
→
∞
∫
R
N
F
(
x
+
k
n
,
u
~
n
)
u
~
n
2
w
n
2
d
x
≤
1
2
-
∫
R
N
liminf
⁡
n
→
∞
F
(
x
+
k
n
,
u
~
n
)
u
~
n
2
w
n
2
d
x
=
-
∞
,
which is a contradiction. Thus {
u
n
} is bounded in H
1
(
R
N
).Remark 14.
In the proof of Lemma13, (S3) is used only in (47). Hence, it can be weakened to (S
3
′) if f
(
x
,
t
) is 1-periodic in each of x
1
,
x
2
,
…
,
x
N.


## 3. The Proofs of Theorems

Proof of Theorem2.
Lemma12 implies the existence of a sequence {
u
n
}
⊂
H
1
(
R
N
) satisfying (34), by a standard argument; we can prove Theorem 2.Proof of Theorem3.
In view of the proofs of Lemmas12 and 13, we can show that the conclusions of Lemmas 12 and 13 still hold if Φ and N are replaced by Φ
+ and N
+, respectively. Hence, there exists a bounded sequence {
u
n
}
⊂
H
1
(
R
N
) satisfying
(48)
Φ
+
(
u
n
)
⟶
c
+
,
∥
Φ
+
′
(
u
n
)
∥
(
1
+
∥
u
n
∥
)
⟶
0
,
where c
+
=
inf
⁡
u
∈
N
+
Φ
+
(
u
). The rest of the proof is standard, so we omit it.To prove Theorems4 and 5, we define functional Φ
0 and Φ
+
,
0 as follows:
(49)
Φ
0
(
u
)
=
1
2
∫
R
N
(
|
∇
u
|
2
+
V
0
(
x
)
u
2
)
d
x
-
∫
R
N
F
0
(
x
,
u
)
d
x
,
u
∈
H
1
(
R
N
)
,
(50)
Φ
+
,
0
(
u
)
=
1
2
∫
R
N
(
|
∇
u
|
2
+
V
0
(
x
)
u
2
)
d
x
-
∫
R
N
F
0
(
x
,
u
+
)
d
x
,
u
∈
H
1
(
R
N
)
,
where F
0
(
x
,
t
)
≔
∫
0
t
‍
f
0
(
x
,
s
)
d
s. Then (V
2), (S
0
′), and (S
1
′) imply that Φ
0
∈
C
1
(
H
1
(
R
N
)
,
R
) and
(51)
〈
Φ
0
′
(
u
)
,
v
〉
=
∫
R
N
(
∇
u
∇
v
+
V
0
(
x
)
u
v
)
d
x
-
∫
R
N
f
0
(
x
,
u
)
v
d
x
,
u
,
v
∈
H
1
(
R
N
)
.Proof of Theorem4.
Lemma12 implies the existence of a sequence {
u
n
}
⊂
H
1
(
R
N
) satisfying (34). By Lemma 13, {
u
n
} is bounded in H
1
(
R
N
). Passing to a subsequence, we have u
n
⇀
u
0 in H
1
(
R
N
). Next, we prove u
0
≠
0.
Arguing by contradiction, suppose thatu
0
=
0; that is, u
n
⇀
0 in H
1
(
R
N
), and so u
n
→
0 in L
loc
⁡
s
(
R
N
), 2
≤
s
<
2
*, and u
n
→
0 a.e. on R
N. For any ɛ
>
0, it follows from (V2) that there exists R
ɛ
>
0 such that |
V
1
(
x
)
|
≤
ɛ for |
x
|
≥
R
ɛ. Hence,
(52)
∫
R
N
|
V
1
(
x
)
|
u
n
2
d
x
=
∫
B
R
ɛ
(
0
)
|
V
1
(
x
)
|
u
n
2
d
x
‍
+
∫
R
N
∖
B
R
ɛ
(
0
)
|
V
1
(
x
)
|
‍
u
n
2
d
x
≤
sup
⁡
R
N
⁡
|
V
1
(
x
)
|
∫
B
R
ɛ
(
0
)
u
n
2
d
x
‍
+
ɛ
∫
R
N
∖
B
R
ɛ
(
0
)
u
n
2
d
x
≤
o
(
1
)
+
ɛ
∥
u
n
∥
2
2
≤
o
(
1
)
+
C
3
ɛ
.
Since ɛ
>
0 is arbitrary, we have
(53)
lim
⁡
n
→
∞
∫
R
N
V
1
(
x
)
u
n
2
d
x
=
0
.
Similarly, by (S
1
′), one has
(54)
lim
⁡
n
→
∞
∫
R
N
F
1
(
x
,
u
n
)
d
x
=
0
,
lim
⁡
n
→
∞
∫
R
N
f
1
(
x
,
u
n
)
u
n
d
x
=
0
.
Note that
(55)
Φ
0
(
u
)
=
Φ
(
u
)
-
1
2
∫
R
N
‍
V
1
(
x
)
u
2
d
x
+
∫
R
N
‍
F
1
(
x
,
u
)
d
x
,
∀
u
∈
H
1
(
R
N
)
,
〈
Φ
0
′
(
u
)
,
v
〉
=
〈
Φ
′
(
u
)
,
v
〉
-
∫
R
N
V
1
(
x
)
u
v
d
x
+
∫
R
N
f
1
(
x
,
u
)
v
d
x
,
∀
u
,
v
∈
H
1
(
R
N
)
.
From (34) and (53)–(55), one has
(56)
Φ
0
(
u
n
)
⟶
c
,
∥
Φ
0
′
(
u
n
)
∥
(
1
+
∥
u
n
∥
)
⟶
0
.
By a standard argument, we may prove that there existsk
n
∈
Z
N, going if necessary to a subsequence, such that
(57)
∫
B
1
+
N
(
k
n
)
|
u
n
|
2
d
x
>
δ
2
>
0
.
Let v
n
(
x
)
=
u
n
(
x
+
k
n
). Then ∥
v
n
∥
H
1
(
R
N
)
=
∥
u
n
∥
H
1
(
R
N
), and
(58)
∫
B
1
+
N
(
0
)
|
v
n
|
2
d
x
>
δ
2
.
Since V
0
(
x
) and f
0
(
x
,
u
) are periodic, we have
(59)
Φ
0
(
v
n
)
⟶
c
,
∥
Φ
0
′
(
v
n
)
∥
(
1
+
∥
v
n
∥
)
⟶
0
.
Since {
v
n
} is bounded in H
1
(
R
N
), passing to a subsequence, we have v
n
⇀
v
¯ in H
1
(
R
N
), v
n
→
v
¯ in L
loc
⁡
s
(
R
N
), 2
≤
s
<
2
*, and v
n
→
v
¯ a.e. on R
N. Obviously, (58) implies that v
¯
(
x
)
≢
0 for x
∈
B
1
+
N
(
0
). By a standard argument, we can prove that Φ
0
′
(
v
¯
)
=
0 and Φ
0
(
v
¯
)
≤
c by using (59).
Sincev
¯
≠
0, it follows from Lemma 9 that there exists t
0
=
t
(
v
¯
) such that t
0
v
¯
∈
N, and so Φ
(
t
0
v
¯
)
≥
c. On the other hand, from (49), (51), (V2), (S
1
′), and (S
4
′), we have
(60)
c
≥
Φ
0
(
v
¯
)
=
Φ
0
(
t
0
v
¯
)
+
∫
R
N
[
1
-
t
0
2
2
f
0
(
x
,
v
¯
)
v
¯
+
F
0
(
x
,
t
0
v
¯
)
-
F
0
(
x
,
v
¯
)
]
d
x
≥
Φ
0
(
t
0
v
¯
)
=
Φ
(
t
0
v
¯
)
-
t
0
2
2
∫
R
N
V
1
(
x
)
v
¯
2
d
x
+
∫
R
N
F
1
(
x
,
t
0
v
¯
)
d
x
>
Φ
(
t
0
v
¯
)
≥
c
.
This contradiction implies that u
0
≠
0. By a standard argument, we can prove that Φ
′
(
u
0
)
=
0 and Φ
(
u
0
)
=
c
=
inf
⁡
N
Φ. This shows that u
0
∈
H
1
(
R
N
) is a solution for problem (1) with Φ
(
u
0
)
=
inf
⁡
N
Φ
>
0.Proof of Theorem5.
Similar to the proof of Theorem3, there exists a bounded sequence {
u
n
}
⊂
H
1
(
R
N
) satisfying (48). Passing to an appropriate subsequence, we have that u
n
⇀
u
¯ in H
1
(
R
N
). Next, we prove u
¯
≠
0.
Arguing by contradiction, suppose thatu
¯
=
0; that is, u
n
⇀
0 in H
1
(
R
N
). Then, u
n
→
0 in L
loc
⁡
s
(
R
N
), 2
≤
s
<
2
*, and u
n
→
0 a.e. on R
N. Analogous to the proof of Theorem 4, we can demonstrate that there exists a v
¯
∈
H
1
(
R
N
)
∖
{
0
} with v
¯
(
x
)
≢
0 for x
∈
B
1
+
N
(
0
) such that Φ
+
,
0
′
(
v
¯
)
=
0 and Φ
+
,
0
(
v
¯
)
≤
c. By a standard argument, we can show that v
¯
≥
0.
Sincev
¯
≥
(
≢
)
0, it follows from Lemma 9 that there exists t
0
=
t
(
v
¯
) such that t
0
v
¯
∈
N
+, and so Φ
+
(
t
0
v
¯
)
≥
c. On the other hand, from (49), (51), (V2), (S
1
′), and (S
4
′), we have
(61)
c
≥
Φ
+
,
0
(
v
¯
)
=
Φ
+
,
0
(
t
0
v
¯
)
+
∫
R
N
[
1
-
t
0
2
2
f
0
(
x
,
v
¯
)
v
¯
+
F
0
(
x
,
t
0
v
¯
)
-
F
0
(
x
,
v
¯
)
1
-
t
0
2
2
]
d
x
≥
Φ
+
,
0
(
t
0
v
¯
)
=
Φ
+
(
t
0
v
¯
)
-
t
0
2
2
∫
R
N
V
1
(
x
)
v
¯
2
d
x
+
∫
R
N
F
1
(
x
,
t
0
v
¯
)
d
x
>
Φ
+
(
t
0
v
¯
)
≥
c
.
This contradiction shows that u
¯
≠
0. In the same way as the last part of the proof of Theorem 1, we can deduce that Φ
+
′
(
u
¯
)
=
0 and Φ
+
(
u
¯
)
=
c
=
inf
⁡
N
+
Φ
+. By a standard argument, we can demonstrate that u
¯
≥
0. Therefore, u
¯
∈
H
1
(
R
N
) is a positive solution for problem (1) with Φ
+
(
u
¯
)
=
inf
⁡
N
+
Φ
+
>
0.



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*Source: 607078-2014-06-29.xml*