# Multiple Solutions for a Class ofN-Laplacian Equations with Critical Growth and Indefinite Weight

Authors: Guoqing Zhang; Ziyan Yao
Journal: Abstract and Applied Analysis
 (2014)

---

## Abstract

Using the suitable Trudinger-Moser inequality and the Mountain Pass Theorem, we prove the existence of multiple solutions for a class ofN-Laplacian equations with critical growth and indefinite weight -
div
∇
u
N
-
2
∇
u
+
V
x
u
N
-
2
u
=
λ
u
N
-
2
u
/
x
β
+
f
x
,
u
/
x
β
+
ɛ
h
x, x
∈
ℝ
N, u
≠
0, x
∈
ℝ
N, where 0
<
β
<
N, V
(
x
) is an indefinite weight, f
:
ℝ
N
×
ℝ
→
ℝ behaves like exp
⁡
α
u
N
/
N
-
1 and does not satisfy the Ambrosetti-Rabinowitz condition, and h
∈
(
W
1
,
N
(
ℝ
N
)
)
*.


---

## Body

## 1. Introduction

In this paper, we consider the existence of multiple solutions for theN-Laplacian elliptic equations with critical growth and singular potentials
(1)
-
div
⁡
(
|
∇
u
|
N
-
2
∇
u
)
+
V
(
x
)
|
u
|
N
-
2
u
=
λ
|
u
|
N
-
2
u
|
x
|
β
+
f
(
x
,
u
)
|
x
|
β
+
ɛ
h
(
x
)
,
x
∈
ℝ
N
,
u
≠
0
,
x
∈
ℝ
N
,
where N
≥
2, 0
<
λ
<
λ
1, λ
1
=
inf
⁡
{
∫
ℝ
N
‍
(
|
∇
u
|
N
+
V
(
x
)
|
u
|
N
)
d
x
:
u
∈
W
1
,
N
(
ℝ
N
)
,
∫
ℝ
N
(
|
u
|
N
/
|
x
|
β
)
‍
d
x
=
1
}, 0
<
β
<
N, h
∈
(
W
1
,
N
(
ℝ
N
)
)
*, Δ
N
u
=
div
⁡
(
|
∇
u
|
N
-
2
∇
u
) is the N-Laplacian, the indefinite weight V
(
x
)
∈
R
(
V
0
), and R
(
V
0
) is the classes of rearrangement of V
0; V
0 satisfies the following conditions:(H1)
V
0
∈
L
q
(
ℝ
N
), ∀
q
≥
1,
(H2)
∥
V
0
-
∥
L
q
(
ℝ
N
)
<
S
N
q
′, or V
0
≥
-
S
N
+
δ, for some δ
>
0,
(H3)
(
1
/
V
0
)
∈
L
1
(
ℝ
N
),
where1
/
q
+
1
/
q
′
=
1, S
r
(
r
=
N
,
N
q
′
) is the best constant
(2)
S
r
∥
u
∥
L
r
(
ℝ
N
)
N
≤
∫
ℝ
N
‍
|
∇
u
|
N
d
x
,
∀
u
∈
W
1
,
N
(
ℝ
N
)
;
that is,(3)
S
r
=
inf
⁡
{
∫
ℝ
N
‍
|
∇
u
|
N
d
x
:
u
∈
W
1
,
N
(
ℝ
N
)
,
∥
u
∥
L
r
(
ℝ
N
)
=
1
}
.Note that ifV is a measurable function which satisfies (H2), there exists δ
0
>
0 such that ∥
V
0
-
∥
L
q
(
ℝ
N
)
≤
(
1
-
δ
0
)
S
N
q
′.Recently,N-Laplacian equations had been studied by many authors. Marcos do Ó [1] studied the existence of nontrivial solutions for the following N-Laplacian equations with critical growth:
(4)
u
∈
W
0
1
,
N
(
Ω
)
,
u
≥
0
,
-
Δ
N
u
=
f
(
x
,
u
)
,
x
∈
Ω
,
where Ω is bounded smooth domain in ℝ
N
(
N
≥
2
). Adimurthi and Sandeep [2] proved that the singular Trudinger-Moser inequality
(5)
sup
⁡
u
∈
W
0
1
,
N
(
Ω
)
∫
Ω
‍
exp
⁡
(
α
|
u
|
N
/
(
N
-
1
)
)
|
x
|
β
d
x
<
+
∞
holds if and only if α
/
α
N
+
β
/
N
≤
1, where α
N
=
N
w
N
-
1
1
/
(
N
-
1
), α
>
0, 0
≤
β
<
N, and ∥
∇
u
∥
L
N
(
Ω
)
≤
1, and studied the corresponding critical exponent problem. For the unbounded domain, Li and Ruf [3] proved that, if we replace the L
N-norm of ∇
u in the supermum by the standard Sobolev norm, the supermum can still be finite. Adimurthi and Yang [4] obtained the following Trudinger-Moser inequality
(6)
∫
ℝ
N
‍
exp
⁡
(
α
|
u
|
N
/
(
N
-
1
)
-
S
N
-
2
(
α
,
u
)
)
|
x
|
β
d
x
<
+
∞
,
where α
>
0, 0
≤
β
<
N, u
∈
W
1
,
N
(
ℝ
N
), and S
N
-
2
(
α
,
u
)
=
∑
k
=
0
N
-
2
(
α
k
/
k
!
)
|
u
|
N
k
/
(
N
-
1
), and studied the existence of nontrivial solution for the corresponding N-Laplacian equations with critical growth. In particular, using inequality (6) and the Mountain Pass Theorem, Lam and Lu [5] studied the following nonuniformly elliptic equations of N-Laplacian type of the form
(7)
-
div
⁡
(
a
(
x
,
∇
u
)
)
+
V
(
x
)
|
u
|
N
-
2
u
=
f
(
x
,
u
)
|
x
|
β
+
ɛ
h
(
x
)
,
div
⁡
(
a
(
x
,
∇
u
)
)
+
V
(
x
)
|
u
|
N
-
2
u
=
ii
f
(
x
,
u
)
|
x
|
β
+
x
∈
ℝ
N
,
where V
(
x
)
>
V
0
>
0, and obtained the existence and multiplicity results of problem (7).On the other hand, some authors have studied the case for the nonlinear term which does not satisfy the Ambrosetti-Rabinowitz condition. Lam and Lu [6, 7] studied the existence of nontrivial solutions for the N-Laplacian equations and systems and polyharmonic equations without Ambrosetti-Rabinowitz conditions, respectively. Miyagaki and Souto [8] discussed a class of superlinear problems for the polynomial case without Ambrosetti-Rabinowitz conditions. Motivated by a suitable Trudinger-Moser inequality, we assume the following growth conditions on the nonlinearity f
(
x
,
u
):(f1)
the functionf
:
ℝ
N
×
ℝ
→
ℝ is continuous, for some constants α
0
,
b
1
,
b
2
>
0 and for all (
x
,
s
)
∈
ℝ
N
×
ℝ,
(8)
|
f
(
x
,
s
)
|
≤
b
1
|
s
|
N
-
1
+
b
2
[
exp
⁡
(
α
0
|
s
|
N
/
(
N
-
1
)
)
-
S
N
-
2
(
α
0
,
s
)
]
;
(f2)
H
(
x
,
t
)
≤
H
(
x
,
s
), for all 0
<
t
<
s, ∀
x
∈
ℝ
N, where
(9)
H
(
x
,
s
)
=
s
f
(
x
,
s
)
-
N
F
(
x
,
s
)
,
F
(
x
,
s
)
=
∫
0
s
‍
f
(
x
,
τ
)
d
τ
;
(f3)
there existsc
>
0 such that for all (
x
,
s
)
∈
ℝ
N
×
ℝ
+,  0
<
F
(
x
,
s
)
≤
c
|
s
|
N
+
c
f
(
x
,
s
);
(f4)
lim
⁡
s
→
∞
(
F
(
x
,
s
)
/
|
s
|
N
)
=
∞, uniformly on x
∈
ℝ
N.
We state our main result in this paper.Theorem 1.
Suppose that (H1)–(H3) and (f1)–(f4) are satisfied and0
<
λ
<
λ
1. Furthermore, assume that (f5)
lim
⁡
sup
⁡
s
→
0
+
(
N
F
(
x
,
s
)
/
|
s
|
N
)
=
0, uniformly on x
∈
ℝ
N,
and there existsr
>
0 such that
(f6)
(10)
lim
⁡
s
→
0
s
f
(
x
,
s
)
exp
⁡
(
-
α
0
|
s
|
N
/
(
N
-
1
)
)
>
2
e
(
α
N
d
(
N
-
β
)
/
N
)

+

C
r
N
-
β

-

(
r
N
-
β
/
(
N
-
β
)
)
×
(
N
-
β
α
0
)
N
-
1
,uniformly on compact subsets of ℝ
N, where d and C are defined in Section 3. Then there exists ε
1
>
0 such that, for each 0
<
ɛ
<
ε
1, problem (1) has at least two nontrivial weak solutions.In this paper, as the functionV
(
x
) is an indefinite weight, we establish a singular Trudinger-Moser inequality (see Lemma 8) and investigate the eigenvalue problem corresponding to problem (1). Using the singular Trudinger-Moser inequality, the eigenvalue problem and the Mountain Pass Theorem, we prove the multiplicity result for problem (1). Furthermore, condition (f2) is used by Lam and Lu [5], and it implies that the function f
(
x
,
u
) does not satisfy the Ambrosetti-Rabinowitz condition.The paper is organized as follows. In Section2, we recall some important lemmas and consider the eigenvalue problem corresponding to problem (1). Section 3 is devote to prove Theorem 1.


## 2. Preliminary Results

### 2.1. Key Lemmas

Now, we define the following Sobolev space(11)
E
=
{
u
∈
W
1
,
N
(
ℝ
N
)
:
∫
ℝ
N
‍
|
∇
u
|
N
d
x
+
∫
ℝ
N
‍
V
(
x
)
|
u
|
N
d
x
<
+
∞
}
,
and the corresponding norm,
(12)
∥
u
∥
E
=
(
∫
ℝ
N
‍
(
|
∇
u
|
N
+
V
(
x
)
|
u
|
N
)
d
x
)
1
/
N
.From the Radial Lemma [9, 10], we have
(13)
|
u
(
x
)
|
≤
|
x
|
-
1
(
N
w
N
-
1
)
1
/
N
∥
u
∥
L
N
(
ℝ
N
)
,
∀
x
≠
0
,
for all u
∈
W
1
,
N
(
ℝ
N
) being radially symmetric, where w
N
-
1 is the surface area of the unit sphere in ℝ
N. V
(
x
) is a rearrangement of V
0 if
(14)
|
{
x
∈
ℝ
N
:
V
(
x
)
≥
α
}
|
=
|
{
x
∈
ℝ
N
:
V
0
(
x
)
≥
α
}
|
,
∀
α
∈
ℝ
N
,
where |
·
| denotes the Lebesgue measure.Lemma 2 (see [11]).
LetV
0 satisfy (
H
1
) and (
H
2
). Then there exists δ
0
>
0 such that
(15)
δ
0
∫
ℝ
N
‍
|
∇
u
|
N
d
x
≤
∥
u
∥
E
N
,
∀
V
∈
R
(
V
0
)
.Proof.
Assume that∥
V
0
-
∥
L
N
(
ℝ
N
)
<
S
N
q
′. Since
(16)
∥
u
∥
E
N
≥
∫
ℝ
N
‍
(
|
∇
u
|
N
+
V
-
(
x
)
|
u
|
N
)
d
x
,
∫
ℝ
N
‍
V
-
(
x
)
|
u
|
N
d
x
≤
∥
V
-
(
x
)
∥
L
q
(
ℝ
N
)
∥
u
∥
L
N
q
′
(
ℝ
N
)
N
=
∥
V
0
-
(
x
)
∥
L
q
(
ℝ
N
)
∥
u
∥
L
N
q
′
(
ℝ
N
)
N
.
Then, by (H2), there exists δ
0 such that
(17)
∥
V
0
-
(
x
)
∥
L
q
(
ℝ
N
)
≤
(
1
-
δ
0
)
S
N
q
′
.
Therefore, we have
(18)
∥
u
∥
E
N
≥
δ
0
∫
ℝ
N
‍
|
∇
u
|
N
d
x
.Remark 3.
In this paper, we denoteC as positive (possibly different) constants.Remark 4.
IfV
∈
R
(
V
0
), then V satisfies (H1)–(H3).Lemma 5.
If (H1)–(H3) are satisfied, then(1)
the embeddingE
↪
W
1
,
N
(
ℝ
N
)
↪
L
q
(
ℝ
N
) is continuous, for all 1
≤
q
<
∞;
(2)
the embeddingE
↪
L
q
(
ℝ
N
) is compact, for all q
≥
N.Proof.
(1) From Lemma2 and Sobolev-Poincare inequality, we obtain the conclusion.(2)
Let{
u
k
}
⊂
E satisfy ∥
u
k
∥
E
≤
C for all k, and we assume
(19)
u
k
⇀
u
,
u
k
⟶
u
,
u
k
⟶
u
,
weakly
in
E
,
strongly
in
L
loc
⁡
q
(
ℝ
N
)
,
∀
q
≥
1
,
a
.
e
.
in
ℝ
N
.
In view of (H3), for everyɛ
→
0, there exists R
>
0 such that
(20)
(
∫
|
x
|
>
R
‍
1
V
0
1
/
(
N
-
1
)
d
x
)
1
-
1
/
N
<
ɛ
.
Hence, we have
(21)
∫
|
x
|
>
R
‍
|
u
k
-
u
|
d
x
=
∫
|
x
|
>
R
V
0
1
/
N
V
0
1
/
N
|
u
k
-
u
|
d
x
≤
(
∫
|
x
|
>
R
‍
1
V
0
1
/
(
N
-
1
)
d
x
)
1
-
1
/
N
×
(
∫
|
x
|
>
R
‍
V
0
|
u
k
-
u
|
N
d
x
)
1
/
N
≤
ɛ
∥
u
k
-
u
∥
E
≤
C
ɛ
.
From (19), we have u
k
→
u in L
1
(
B
R
(
0
)
) and B
R
(
0
)
⊂
ℝ
N is the ball centered at 0 with radius R. This together with (21) leads to limsup
⁡
k
→
+
∞
∫
ℝ
N
‍
|
u
k
-
u
|
d
x
≤
C
ɛ. Since ɛ is arbitrary, we have
(22)
lim
⁡
k
→
+
∞
∫
ℝ
N
‍
|
u
k
-
u
|
d
x
=
0
.
Hence, for every q
≥
N, we have
(23)
∫
ℝ
N
‍
|
u
k
-
u
|
q
d
x
≤
∫
ℝ
N
‍
|
u
k
-
u
|
1
/
2
|
u
k
-
u
|
q
-
1
/
2
d
x
≤
(
∫
ℝ
N
|
u
k
-
u
|
d
x
)
1
/
2
×
(
∫
ℝ
N
‍
|
u
k
-
u
|
2
q
-
1
d
x
)
1
/
2
≤
C
(
∫
ℝ
N
‍
|
u
k
-
u
|
d
x
)
1
/
2
⟶
0
.Lemma 6.
E is a reflexive Banach space.Proof.
Suppose that∀
u
1
∈
E, ∀
u
2
∈
E, we have
(24)
∥
u
1
∥
E
≤
1
,
∥
u
2
∥
E
≤
1
,
∥
u
1
-
u
2
∥
E
>
ɛ
,
and there exists δ
=
1
-
1
-
(
ɛ
/
4
)
N
N, using the following inequality
(25)
|
a
+
b
2
|
p
+
|
a
+
b
2
|
p
≤
1
2
(
|
a
|
p
+
|
b
|
p
)
,
|
a
+
b
2
|
p
+
|
a
+
b
2
|
p
≤
4
1
2
∀
a
,
b
,
p
∈
ℝ
,
such that
(26)
∥
u
1
+
u
2
2
∥
E
N
=
∥
u
1
+
u
2
2
∥
E
N
-
∥
u
1
-
u
2
2
∥
E
N
+
∥
u
1
-
u
2
2
∥
E
N
≤
1
2
(
∫
ℝ
N
‍
|
∇
u
1
|
N
d
x
+
∫
ℝ
N
‍
|
∇
u
2
|
N
d
x
)
+
1
2
∫
ℝ
N
‍
V
(
x
)
(
|
u
1
|
N
+
|
u
2
|
N
)
d
x
-
(
ɛ
2
)
N
≤
1
-
(
ɛ
2
)
N
<
(
1
-
δ
)
N
.
Hence, E is uniformly convex. We obtain that E is a reflexive Banach space.
Now, we define the functionalI
:
E
→
ℝ
(27)
I
(
u
)
=
1
N
∫
ℝ
N
‍
|
∇
u
|
N
d
x
+
1
N
∫
ℝ
N
‍
V
(
x
)
|
u
|
N
d
x
-
∫
ℝ
N
‍
F
(
x
,
u
)
|
x
|
β
d
x
-
λ
N
∫
ℝ
N
‍
|
u
|
N
|
x
|
β
d
x
-
ɛ
∫
ℝ
N
‍
h
u
d
x
;
then the functional I
(
u
) is well defined by Lemma 5. Moreover, I
(
u
) is the C
1 functional on E and ∀
u
,
v
∈
E; we have
(28)
D
I
(
u
)
v
=
∫
ℝ
N
‍
|
∇
u
|
N
-
2
∇
u
∇
v
d
x
+
∫
ℝ
N
‍
V
(
x
)
|
u
|
N
-
2
u
v
d
x
-
∫
ℝ
N
‍
f
(
x
,
u
)
v
|
x
|
β
d
x
-
λ
∫
ℝ
N
‍
|
u
|
N
-
2
u
v
|
x
|
β
d
x
-
ɛ
∫
ℝ
N
‍
h
v
d
x
.
Hence, the critical point of the functional I
(
u
) is the weak solution of problem (1).Lemma 7.
Let0
<
α
≤
(
1
-
β
/
N
)
α
N, 0
<
β
<
N, u
∈
E and ∥
u
∥
E
≤
1; then for some q
>
N and α
/
α
N
+
β
/
N
+
1
/
q
≤
1, one has
(29)
∫
ℝ
N
‍
[
exp
⁡
(
α
|
u
|
N
/
(
N
-
1
)
)
-
S
N
-
2
(
α
,
u
)
]
|
u
|
|
x
|
β
d
x
≤
C
∥
u
∥
L
q
(
ℝ
N
)
.Proof.
LetR
(
α
,
u
)
=
exp
⁡
(
α
|
u
|
N
/
(
N
-
1
)
)
-
S
N
-
2
(
α
,
u
); u
* is the Schwarz symmetrization of u; we can conclude that
(30)
∫
ℝ
N
‍
R
(
α
,
u
)
|
u
|
|
x
|
β
d
x
=
∫
ℝ
N
‍
R
(
α
,
u
*
)
|
u
*
|
|
x
|
β
d
x
.
Let u
′
=
u
/
∥
u
∥
E. It is easy to obtain that R
(
α
,
u
) is increasing with respect to |
u
|. If ∥
u
∥
E
≤
1; then there holds
(31)
∫
ℝ
N
‍
exp
⁡
(
α
|
u
|
N
/
(
N
-
1
)
)
-
S
N
-
2
(
α
,
u
)
|
x
|
β
d
x
≤
∫
ℝ
N
‍
exp
⁡
(
α
|
u
′
|
N
/
(
N
-
1
)
)
-
S
N
-
2
(
α
,
u
′
)
|
x
|
β
d
x
.
Now, we prove that there exists a uniform constant C such that, for all radially decreasing symmetric functions u
∈
W
1
,
N
(
ℝ
N
) and ∥
u
∥
E
=
1,
(32)
∫
ℝ
N
‍
exp
⁡
(
α
*
|
u
|
N
/
(
N
-
1
)
)
-
S
N
-
2
(
α
*
,
u
)
|
x
|
β
d
x
≤
C
,
where α
*
=
(
1
-
β
/
N
)
α
N. In the following, assume that u is radially decreasing function in ℝ
N and ∥
u
∥
E
=
1. Take r sufficiently large; that is, r
>
N
1
/
N
w
N
-
1
-
1
/
N
∥
u
*
∥
L
N
(
ℝ
N
). By the radial lemma, for all |
x
|
≥
r, we have u
*
(
x
)
<
1 and
(33)
∫
|
x
|
≥
r
‍
R
(
α
*
,
u
)
|
x
|
β
d
x
=
∫
|
x
|
≥
r
‍
R
(
α
*
,
u
*
)
|
x
|
β
d
x
≤
1
r
β
∫
|
x
|
>
r
‍
(
∑
‍
m
=
N
∞
(
α
*
)
N
-
1
|
u
*
|
N
(
N
-
1
)
!
+
∑
‍
m
=
N
∞
(
α
*
)
m
|
u
*
|
m
N
/
(
N
-
1
)
m
!
)
d
x
≤
∥
u
*
∥
L
N
(
ℝ
N
)
N
r
β
(
∑
‍
m
=
N
-
1
∞
(
α
*
)
m
m
!
)
≤
C
.
Define the set S
=
{
x
∈
𝔹
r
(
0
)
:
|
u
(
x
)
-
u
(
r
)
|
>
2
|
u
(
r
)
|
}. Assume that S is nonempty; then for all x
∈
S and ɛ
>
0 we have
(34)
|
u
(
x
)
|
N
/
(
N
-
1
)
=
|
u
(
x
)
-
u
(
r
)
+
u
(
r
)
|
N
/
(
N
-
1
)
=
|
u
(
x
)
-
u
(
r
)
|
N
/
(
N
-
1
)
(
1
+
u
(
r
)
|
u
(
x
)
-
u
(
r
)
|
)
N
/
(
N
-
1
)
≤
|
u
(
x
)
-
u
(
r
)
|
N
/
(
N
-
1
)
+
N
N
-
1
(
3
2
)
1
/
(
N
-
1
)
×
|
u
(
r
)
|
|
u
(
x
)
-
u
(
r
)
|
1
/
(
N
-
1
)
≤
(
1
+
ɛ
)
|
u
(
x
)
-
u
(
r
)
|
N
/
(
N
-
1
)
+
(
3
2
)
N
/
(
N
-
1
)
2
(
|
u
(
r
)
|
N
(
N
-
1
)
ɛ
)
1
/
N
-
1
.
Since
(35)
∥
u
∥
E
N
=
∥
∇
u
∥
L
N
(
ℝ
N
)
N
+
∫
ℝ
N
‍
V
(
x
)
|
u
|
N
d
x
=
1
,
so we have
(36)
∥
∇
u
∥
L
N
(
ℝ
N
)
N
=
1
-
∫
ℝ
N
‍
V
(
x
)
|
u
|
N
d
x
,
∥
∇
u
∥
L
N
(
ℝ
N
)
N
/
(
N
-
1
)
=
(
1
-
∫
ℝ
N
‍
V
(
x
)
|
u
|
N
d
x
)
1
/
(
N
-
1
)
.
Thus, we obtain
(37)
1
∥
∇
u
∥
L
N
(
ℝ
N
)
N
/
(
N
-
1
)
=
(
1
1
-
∫
ℝ
N
‍
V
(
x
)
|
u
|
N
d
x
)
1
/
(
N
-
1
)
.
From Hardy-Littlewood inequality, we have
(38)
∫
ℝ
N
‍
V
(
x
)
|
u
*
|
N
d
x
≥
∫
ℝ
N
‍
V
0
(
x
)
|
u
|
N
d
x
.
Since V
(
x
)
∈
R
(
V
0
), we have ∫
ℝ
N
‍
V
(
x
)
|
u
*
|
N
d
x
=
∫
ℝ
N
‍
V
(
x
)
|
u
|
N
d
x. Let
(39)
1
+
ɛ
=
1
∥
∇
u
∥
L
N
(
ℝ
N
)
N
/
(
N
-
1
)
=
(
1
1
-
∫
ℝ
N
‍
V
(
x
)
|
u
|
N
d
x
)
1
/
(
N
-
1
)
≥
(
1
1
-
∫
ℝ
N
‍
V
0
(
x
)
|
u
|
N
d
x
)
1
/
(
N
-
1
)
≥
(
1
1
-
(
δ
-
S
N
)
∫
ℝ
N
‍
|
u
|
N
d
x
)
1
/
(
N
-
1
)
.
Applying the mean value theorem to the function ψ
(
t
)
=
t
1
/
(
N
-
1
), we obtain that there exists ξ which satisfies
(40)
1
-
(
δ
-
S
N
)
∥
u
∥
L
N
(
ℝ
N
)
N
≤
ξ
≤
1
,
such that
(41)
1
-
[
1
-
(
δ
-
S
N
)
∥
u
∥
L
N
(
ℝ
N
)
N
]
1
/
(
N
-
1
)
=
(
δ
-
S
N
)
N
-
1
ξ
(
2
-
N
)
/
(
N
-
1
)
∥
u
∥
L
N
(
ℝ
N
)
N
.
So we have
(42)
ɛ
=
∥
u
∥
L
N
(
ℝ
N
)
N
(
N
-
1
)
ξ
(
N
-
2
)
/
(
N
-
1
)
(
1
-
(
δ
-
S
N
)
∥
u
∥
L
N
(
ℝ
N
)
N
)
1
/
(
N
-
1
)
≥
∥
u
∥
L
N
(
ℝ
N
)
N
N
-
1
.
By |
u
(
r
)
|
≤
(
N
/
w
N
-
1
)
1
/
N
∥
u
∥
L
N
(
ℝ
N
)
/
r, we have
(43)
(
3
2
)
N
/
(
N
-
1
)
2
(
|
u
(
r
)
|
N
(
N
-
1
)
ɛ
)
1
/
(
N
-
1
)
≤
C
,
and ∀
x
∈
S,
(44)
|
u
(
x
)
|
N
/
(
N
-
1
)
≤
|
u
(
x
)
-
u
(
r
)
|
N
/
(
N
-
1
)
(
∥
∇
u
∥
L
N
(
ℝ
N
)
N
)
1
/
(
N
-
1
)
+
C
.
Obviously, u
-
u
(
r
)
∈
W
1
,
N
(
𝔹
r
(
0
)
), and
(45)
∫
𝔹
r
(
0
)
‍
|
∇
(
u
-
u
(
r
)
)
|
N
d
x
≤
∫
𝔹
r
(
0
)
‍
|
∇
u
|
N
d
x
≤
1
.
Let u
′
=
(
u
(
x
)
-
u
(
r
)
)
/
∥
∇
(
u
(
x
)
-
u
(
r
)
)
∥
L
N
(
ℝ
N
); we obtain
(46)
∫
𝔹
r
(
0
)
‍
e
α
*
|
u
|
N
/
(
N
-
1
)
|
x
|
β
d
x
=
∫
S
‍
e
α
*
|
u
|
N
/
(
N
-
1
)
|
x
|
β
d
x
+
∫
𝔹
r
(
0
)
∖
S
‍
e
α
*
|
u
|
N
/
(
N
-
1
)
|
x
|
β
d
x
≤
C
∫
𝔹
r
(
0
)
‍
e
α
*
|
u
′
|
N
/
(
N
-
1
)
|
x
|
β
d
x
+
C
≤
C
.
Hence, we obtain that (32) holds. For 0
<
α
≤
(
1
-
β
/
N
)
α
N, ∥
u
∥
E
≤
1, we have
(47)
∫
ℝ
N
‍
exp
⁡
(
α
|
u
|
N
/
(
N
-
1
)
)
-
S
N
-
2
(
α
,
u
)
|
x
|
β
d
x
≤
C
.
Since α
/
α
N
+
β
/
N
+
1
/
q
≤
1 and 1
/
q
+
1
/
q
′
=
1, q
>
N, that is, α
q
′
<
α
N, α
q
′
/
α
N
+
β
q
′
/
N
≤
1, we have
(48)
∫
ℝ
N
‍
R
(
α
,
u
)
|
u
|
|
x
|
β
d
x
≤
∫
ℝ
N
‍
R
(
α
q
′
,
u
)
|
x
|
β
q
′
d
x
∥
u
∥
L
q
(
ℝ
N
)
≤
C
∥
u
∥
L
q
(
ℝ
N
)
.As the proof of Lemma7, we can obtain the following.Lemma 8.
For0
<
α
≤
(
1
-
β
/
N
)
α
N, 0
<
β
<
N, u
∈
E and ∥
u
∥
E
≤
1,  q
>
N, and α
/
α
N
+
β
/
N
+
1
/
q
≤
1, one has
(49)
∫
ℝ
N
‍
[
exp
⁡
(
α
|
u
|
N
/
(
N
-
1
)
)
-
S
N
-
2
(
α
,
u
)
]
|
u
|
q
|
x
|
β
d
x
≤
C
(
N
,
α
)
∥
u
∥
E
q
.


### 2.2. The Eigenvalue Problem

We consider the following eigenvalue problem:(50)
-
div
⁡
(
|
∇
u
|
N
-
2
∇
u
)
+
V
(
x
)
|
u
|
N
-
2
u
=
λ
|
u
|
N
-
2
u
|
x
|
β
,
x
∈
ℝ
N
,
u
≠
0
,
x
∈
ℝ
N
.Now, we denote the setM
=
{
u
∈
E
:
∫
ℝ
N
(
‍
|
u
|
N
/
|
x
|
β
)
d
x
=
1
}, and define
(51)
λ
1
=
inf
⁡
0
≠
u
∈
M
{
I
N
(
u
)
:
u
∈
E
∖
{
0
}
}
>
0
,
0
<
β
<
N
,
where I
N
(
u
)
=
∫
ℝ
N
‍
(
|
∇
u
|
N
+
V
(
x
)
|
u
|
N
)
d
x.Lemma 9 (see [12]).
Letu
>
0, v
>
0 be two continuous functions in Ω differentiable a.e., and
(52)
L
(
u
,
v
)
=
|
∇
u
|
p
+
(
p
-
1
)
u
p
v
p
|
∇
v
|
p
-
p
u
p
-
1
v
p
-
1
|
∇
v
|
p
-
2
∇
v
∇
u
,
R
(
u
,
v
)
=
|
∇
u
|
p
-
|
∇
v
|
p
-
2
∇
(
u
p
v
p
-
1
)
∇
v
.Then(
1
)
L
(
u
,
v
)
=
R
(
u
,
v
)
≥
0, (
2
)
L
(
u
,
v
)
=
0 a.e. in Ω if and only if u
=
k
v for some k
>
0.Proposition 10.
Assume that (H1)–(H3) hold; thenλ
1
>
-
∞ is the lowest eigenvalue of Problem (50) and λ
1 is principal.Proof.
From Lemma2, we have λ
1
>
-
∞. Furthermore, any minimizing sequence {
u
n
} is bounded. Up to a subsequence, there exists u
∈
E such that
(53)
u
n
⇀
u
0
,
in
E
,
u
n
⟶
u
0
,
in
L
N
r
′
(
ℝ
N
)
.
Hence, we have
(54)
I
N
(
u
0
)
≤
lim
⁡
n
→
+
∞
I
N
(
u
n
)
=
λ
1
,
u
0
∈
M
,
and consequently we have
(55)
I
N
(
u
0
)
=
λ
1
.
From Lemma 5, we obtain that M is weakly closed in E. By the Lagrange Multipliers rule, λ
1 is an eigenvalue of problem (50). Moreover I
N
(
|
u
|
)
=
I
N
(
u
) for any u, so that λ
1 possesses a nonnegative eigenfunction. We conclude that the eigenvalue is principal from Harnack inequality in [13].Proposition 11.
The eigenvalueλ
1 is isolated. That is, there exists ɛ
>
0, such that there are no other eigenvalues of problem (50) in the interval (
λ
1
,
λ
1
+
ɛ
).Proof.
Assume by contradiction there exists a sequence of eigenvalueλ
m of problem (50) with 0
<
λ
m
↘
λ
1. Let {
u
m
} be an eigenfunction associated with λ
m. Then {
u
m
} satisfies
(56)
-
Δ
N
u
m
+
V
(
x
)
|
u
m
|
N
-
2
u
m
=
λ
m
|
u
m
|
N
-
2
u
m
|
x
|
β
,
∫
ℝ
N
‍
|
∇
u
m
|
N
d
x
+
∫
ℝ
N
‍
V
(
x
)
|
u
m
|
N
d
x
-
∫
ℝ
N
‍
λ
m
|
x
|
β
|
u
m
|
N
d
x
=
0
.
We define
(57)
v
m
=
u
m
(
∫
ℝ
N
(
|
u
m
|
N
/
|
x
|
β
)
‍
d
x
)
1
/
N
.
The coercivity of the functional I
N
(
u
m
)
=
∫
ℝ
N
‍
|
∇
u
m
|
N
d
x
+
∫
ℝ
N
‍
V
(
x
)
|
u
m
|
N
d
x implies that {
u
m
} is a bounded sequence. Hence {
v
m
} is bounded in E. So there exists a subsequence (still denoted) {
v
m
} and v
∈
E such that
(58)
v
m
⇀
v
,
weakly
in
E
,
v
m
⟶
v
,
strongly
in
L
N
(
ℝ
N
)
,
and ∫
ℝ
N
(
|
u
m
|
N
/
|
x
|
β
)
‍
d
x
=
1. On the other hand, we have
(59)
∫
ℝ
N
‍
|
∇
v
|
N
d
x
+
∫
ℝ
N
‍
V
(
x
)
v
N
d
x
≤
liminf
⁡
n
→
∞
(
∫
ℝ
N
‍
|
∇
v
n
|
N
d
x
+
∫
ℝ
N
‍
V
(
x
)
v
n
N
d
x
)
=
λ
1
,
and ∫
ℝ
N
‍
|
∇
v
|
N
d
x
+
∫
ℝ
N
‍
V
(
x
)
v
N
d
x
=
λ
1
>
0. So we conclude that v is an eigenfunction associated with λ
1 and v
>
0. Then we conclude from the convergence in measure of the sequence {
v
n
} towards v that
(60)
|
Ω
n
-
|
⟶
0
,
where Ω
n
- denotes the negative set of v
n, which contradicts Proposition 11.Proposition 12.
The first eigenvalueλ
1 is simple, in the sense that the eigenfunctions associated with it are merely constant multiples of each other.Proof.
Letφ, ζ be two eigenfunctions associated with λ
1. We assume without restriction that φ
>
0, ζ
>
0; then φ satisfies -
Δ
φ
+
V
(
x
)
φ
N
-
1
=
λ
1
(
N
)
(
φ
N
-
1
/
|
x
|
β
). Testing it with function φ, we get
(61)
∫
ℝ
N
‍
|
∇
φ
|
N
d
x
-
∫
ℝ
N
‍
[
-
V
(
x
)
+
λ
1
|
x
|
β
]
φ
N
d
x
=
0
.
Let ɛ
→
0, from Lemma 9, we have
(62)
0
≤
∫
ℝ
N
‍
L
(
φ
,
ζ
+
ɛ
)
d
x
=
∫
ℝ
N
‍
R
(
φ
,
ζ
+
ɛ
)
d
x
=
∫
ℝ
N
‍
[
-
V
(
x
)
+
λ
1
|
x
|
β
]
φ
N
d
x
-
∫
ℝ
N
‍
|
∇
ζ
|
N
-
2
∇
(
φ
N
(
ζ
+
ɛ
)
N
-
1
)
∇
ζ
d
x
.
The function φ
N
/
(
ζ
+
ɛ
)
N
-
1, where ɛ
>
0, belongs to E and then it is admissible for the weak formulation of -
Δ
N
ζ
+
V
(
x
)
|
ζ
|
N
-
2
ζ
=
λ
1
(
|
ζ
|
N
-
2
ζ
/
|
x
|
β
), a.e., and
(63)
∫
ℝ
N
‍
|
∇
ζ
|
N
-
2
∇
ζ
∇
φ
N
(
ζ
+
ɛ
)
N
-
1
d
x
-
∫
ℝ
N
‍
[
-
V
(
x
)
+
λ
1
|
x
|
β
]
φ
N
ζ
N
-
1
(
ζ
+
ɛ
)
N
-
1
d
x
=
0
.
It follows from (62) and (63) that we have
(64)
0
≤
L
(
φ
,
ζ
+
ɛ
)
=
∫
ℝ
N
‍
λ
1
[
1
-
ζ
N
-
1
(
ζ
+
ɛ
)
N
-
1
]
φ
N
|
x
|
β
d
x
-
∫
ℝ
N
‍
V
(
x
)
φ
N
[
1
-
ζ
N
-
1
(
ζ
+
ɛ
)
N
-
1
]
d
x
.
Let ɛ
→
0; we have L
(
φ
,
ζ
)
=
0. By Lemma 9, there exists k
>
0 such that φ
=
k
ζ.


## 2.1. Key Lemmas

Now, we define the following Sobolev space(11)
E
=
{
u
∈
W
1
,
N
(
ℝ
N
)
:
∫
ℝ
N
‍
|
∇
u
|
N
d
x
+
∫
ℝ
N
‍
V
(
x
)
|
u
|
N
d
x
<
+
∞
}
,
and the corresponding norm,
(12)
∥
u
∥
E
=
(
∫
ℝ
N
‍
(
|
∇
u
|
N
+
V
(
x
)
|
u
|
N
)
d
x
)
1
/
N
.From the Radial Lemma [9, 10], we have
(13)
|
u
(
x
)
|
≤
|
x
|
-
1
(
N
w
N
-
1
)
1
/
N
∥
u
∥
L
N
(
ℝ
N
)
,
∀
x
≠
0
,
for all u
∈
W
1
,
N
(
ℝ
N
) being radially symmetric, where w
N
-
1 is the surface area of the unit sphere in ℝ
N. V
(
x
) is a rearrangement of V
0 if
(14)
|
{
x
∈
ℝ
N
:
V
(
x
)
≥
α
}
|
=
|
{
x
∈
ℝ
N
:
V
0
(
x
)
≥
α
}
|
,
∀
α
∈
ℝ
N
,
where |
·
| denotes the Lebesgue measure.Lemma 2 (see [11]).
LetV
0 satisfy (
H
1
) and (
H
2
). Then there exists δ
0
>
0 such that
(15)
δ
0
∫
ℝ
N
‍
|
∇
u
|
N
d
x
≤
∥
u
∥
E
N
,
∀
V
∈
R
(
V
0
)
.Proof.
Assume that∥
V
0
-
∥
L
N
(
ℝ
N
)
<
S
N
q
′. Since
(16)
∥
u
∥
E
N
≥
∫
ℝ
N
‍
(
|
∇
u
|
N
+
V
-
(
x
)
|
u
|
N
)
d
x
,
∫
ℝ
N
‍
V
-
(
x
)
|
u
|
N
d
x
≤
∥
V
-
(
x
)
∥
L
q
(
ℝ
N
)
∥
u
∥
L
N
q
′
(
ℝ
N
)
N
=
∥
V
0
-
(
x
)
∥
L
q
(
ℝ
N
)
∥
u
∥
L
N
q
′
(
ℝ
N
)
N
.
Then, by (H2), there exists δ
0 such that
(17)
∥
V
0
-
(
x
)
∥
L
q
(
ℝ
N
)
≤
(
1
-
δ
0
)
S
N
q
′
.
Therefore, we have
(18)
∥
u
∥
E
N
≥
δ
0
∫
ℝ
N
‍
|
∇
u
|
N
d
x
.Remark 3.
In this paper, we denoteC as positive (possibly different) constants.Remark 4.
IfV
∈
R
(
V
0
), then V satisfies (H1)–(H3).Lemma 5.
If (H1)–(H3) are satisfied, then(1)
the embeddingE
↪
W
1
,
N
(
ℝ
N
)
↪
L
q
(
ℝ
N
) is continuous, for all 1
≤
q
<
∞;
(2)
the embeddingE
↪
L
q
(
ℝ
N
) is compact, for all q
≥
N.Proof.
(1) From Lemma2 and Sobolev-Poincare inequality, we obtain the conclusion.(2)
Let{
u
k
}
⊂
E satisfy ∥
u
k
∥
E
≤
C for all k, and we assume
(19)
u
k
⇀
u
,
u
k
⟶
u
,
u
k
⟶
u
,
weakly
in
E
,
strongly
in
L
loc
⁡
q
(
ℝ
N
)
,
∀
q
≥
1
,
a
.
e
.
in
ℝ
N
.
In view of (H3), for everyɛ
→
0, there exists R
>
0 such that
(20)
(
∫
|
x
|
>
R
‍
1
V
0
1
/
(
N
-
1
)
d
x
)
1
-
1
/
N
<
ɛ
.
Hence, we have
(21)
∫
|
x
|
>
R
‍
|
u
k
-
u
|
d
x
=
∫
|
x
|
>
R
V
0
1
/
N
V
0
1
/
N
|
u
k
-
u
|
d
x
≤
(
∫
|
x
|
>
R
‍
1
V
0
1
/
(
N
-
1
)
d
x
)
1
-
1
/
N
×
(
∫
|
x
|
>
R
‍
V
0
|
u
k
-
u
|
N
d
x
)
1
/
N
≤
ɛ
∥
u
k
-
u
∥
E
≤
C
ɛ
.
From (19), we have u
k
→
u in L
1
(
B
R
(
0
)
) and B
R
(
0
)
⊂
ℝ
N is the ball centered at 0 with radius R. This together with (21) leads to limsup
⁡
k
→
+
∞
∫
ℝ
N
‍
|
u
k
-
u
|
d
x
≤
C
ɛ. Since ɛ is arbitrary, we have
(22)
lim
⁡
k
→
+
∞
∫
ℝ
N
‍
|
u
k
-
u
|
d
x
=
0
.
Hence, for every q
≥
N, we have
(23)
∫
ℝ
N
‍
|
u
k
-
u
|
q
d
x
≤
∫
ℝ
N
‍
|
u
k
-
u
|
1
/
2
|
u
k
-
u
|
q
-
1
/
2
d
x
≤
(
∫
ℝ
N
|
u
k
-
u
|
d
x
)
1
/
2
×
(
∫
ℝ
N
‍
|
u
k
-
u
|
2
q
-
1
d
x
)
1
/
2
≤
C
(
∫
ℝ
N
‍
|
u
k
-
u
|
d
x
)
1
/
2
⟶
0
.Lemma 6.
E is a reflexive Banach space.Proof.
Suppose that∀
u
1
∈
E, ∀
u
2
∈
E, we have
(24)
∥
u
1
∥
E
≤
1
,
∥
u
2
∥
E
≤
1
,
∥
u
1
-
u
2
∥
E
>
ɛ
,
and there exists δ
=
1
-
1
-
(
ɛ
/
4
)
N
N, using the following inequality
(25)
|
a
+
b
2
|
p
+
|
a
+
b
2
|
p
≤
1
2
(
|
a
|
p
+
|
b
|
p
)
,
|
a
+
b
2
|
p
+
|
a
+
b
2
|
p
≤
4
1
2
∀
a
,
b
,
p
∈
ℝ
,
such that
(26)
∥
u
1
+
u
2
2
∥
E
N
=
∥
u
1
+
u
2
2
∥
E
N
-
∥
u
1
-
u
2
2
∥
E
N
+
∥
u
1
-
u
2
2
∥
E
N
≤
1
2
(
∫
ℝ
N
‍
|
∇
u
1
|
N
d
x
+
∫
ℝ
N
‍
|
∇
u
2
|
N
d
x
)
+
1
2
∫
ℝ
N
‍
V
(
x
)
(
|
u
1
|
N
+
|
u
2
|
N
)
d
x
-
(
ɛ
2
)
N
≤
1
-
(
ɛ
2
)
N
<
(
1
-
δ
)
N
.
Hence, E is uniformly convex. We obtain that E is a reflexive Banach space.
Now, we define the functionalI
:
E
→
ℝ
(27)
I
(
u
)
=
1
N
∫
ℝ
N
‍
|
∇
u
|
N
d
x
+
1
N
∫
ℝ
N
‍
V
(
x
)
|
u
|
N
d
x
-
∫
ℝ
N
‍
F
(
x
,
u
)
|
x
|
β
d
x
-
λ
N
∫
ℝ
N
‍
|
u
|
N
|
x
|
β
d
x
-
ɛ
∫
ℝ
N
‍
h
u
d
x
;
then the functional I
(
u
) is well defined by Lemma 5. Moreover, I
(
u
) is the C
1 functional on E and ∀
u
,
v
∈
E; we have
(28)
D
I
(
u
)
v
=
∫
ℝ
N
‍
|
∇
u
|
N
-
2
∇
u
∇
v
d
x
+
∫
ℝ
N
‍
V
(
x
)
|
u
|
N
-
2
u
v
d
x
-
∫
ℝ
N
‍
f
(
x
,
u
)
v
|
x
|
β
d
x
-
λ
∫
ℝ
N
‍
|
u
|
N
-
2
u
v
|
x
|
β
d
x
-
ɛ
∫
ℝ
N
‍
h
v
d
x
.
Hence, the critical point of the functional I
(
u
) is the weak solution of problem (1).Lemma 7.
Let0
<
α
≤
(
1
-
β
/
N
)
α
N, 0
<
β
<
N, u
∈
E and ∥
u
∥
E
≤
1; then for some q
>
N and α
/
α
N
+
β
/
N
+
1
/
q
≤
1, one has
(29)
∫
ℝ
N
‍
[
exp
⁡
(
α
|
u
|
N
/
(
N
-
1
)
)
-
S
N
-
2
(
α
,
u
)
]
|
u
|
|
x
|
β
d
x
≤
C
∥
u
∥
L
q
(
ℝ
N
)
.Proof.
LetR
(
α
,
u
)
=
exp
⁡
(
α
|
u
|
N
/
(
N
-
1
)
)
-
S
N
-
2
(
α
,
u
); u
* is the Schwarz symmetrization of u; we can conclude that
(30)
∫
ℝ
N
‍
R
(
α
,
u
)
|
u
|
|
x
|
β
d
x
=
∫
ℝ
N
‍
R
(
α
,
u
*
)
|
u
*
|
|
x
|
β
d
x
.
Let u
′
=
u
/
∥
u
∥
E. It is easy to obtain that R
(
α
,
u
) is increasing with respect to |
u
|. If ∥
u
∥
E
≤
1; then there holds
(31)
∫
ℝ
N
‍
exp
⁡
(
α
|
u
|
N
/
(
N
-
1
)
)
-
S
N
-
2
(
α
,
u
)
|
x
|
β
d
x
≤
∫
ℝ
N
‍
exp
⁡
(
α
|
u
′
|
N
/
(
N
-
1
)
)
-
S
N
-
2
(
α
,
u
′
)
|
x
|
β
d
x
.
Now, we prove that there exists a uniform constant C such that, for all radially decreasing symmetric functions u
∈
W
1
,
N
(
ℝ
N
) and ∥
u
∥
E
=
1,
(32)
∫
ℝ
N
‍
exp
⁡
(
α
*
|
u
|
N
/
(
N
-
1
)
)
-
S
N
-
2
(
α
*
,
u
)
|
x
|
β
d
x
≤
C
,
where α
*
=
(
1
-
β
/
N
)
α
N. In the following, assume that u is radially decreasing function in ℝ
N and ∥
u
∥
E
=
1. Take r sufficiently large; that is, r
>
N
1
/
N
w
N
-
1
-
1
/
N
∥
u
*
∥
L
N
(
ℝ
N
). By the radial lemma, for all |
x
|
≥
r, we have u
*
(
x
)
<
1 and
(33)
∫
|
x
|
≥
r
‍
R
(
α
*
,
u
)
|
x
|
β
d
x
=
∫
|
x
|
≥
r
‍
R
(
α
*
,
u
*
)
|
x
|
β
d
x
≤
1
r
β
∫
|
x
|
>
r
‍
(
∑
‍
m
=
N
∞
(
α
*
)
N
-
1
|
u
*
|
N
(
N
-
1
)
!
+
∑
‍
m
=
N
∞
(
α
*
)
m
|
u
*
|
m
N
/
(
N
-
1
)
m
!
)
d
x
≤
∥
u
*
∥
L
N
(
ℝ
N
)
N
r
β
(
∑
‍
m
=
N
-
1
∞
(
α
*
)
m
m
!
)
≤
C
.
Define the set S
=
{
x
∈
𝔹
r
(
0
)
:
|
u
(
x
)
-
u
(
r
)
|
>
2
|
u
(
r
)
|
}. Assume that S is nonempty; then for all x
∈
S and ɛ
>
0 we have
(34)
|
u
(
x
)
|
N
/
(
N
-
1
)
=
|
u
(
x
)
-
u
(
r
)
+
u
(
r
)
|
N
/
(
N
-
1
)
=
|
u
(
x
)
-
u
(
r
)
|
N
/
(
N
-
1
)
(
1
+
u
(
r
)
|
u
(
x
)
-
u
(
r
)
|
)
N
/
(
N
-
1
)
≤
|
u
(
x
)
-
u
(
r
)
|
N
/
(
N
-
1
)
+
N
N
-
1
(
3
2
)
1
/
(
N
-
1
)
×
|
u
(
r
)
|
|
u
(
x
)
-
u
(
r
)
|
1
/
(
N
-
1
)
≤
(
1
+
ɛ
)
|
u
(
x
)
-
u
(
r
)
|
N
/
(
N
-
1
)
+
(
3
2
)
N
/
(
N
-
1
)
2
(
|
u
(
r
)
|
N
(
N
-
1
)
ɛ
)
1
/
N
-
1
.
Since
(35)
∥
u
∥
E
N
=
∥
∇
u
∥
L
N
(
ℝ
N
)
N
+
∫
ℝ
N
‍
V
(
x
)
|
u
|
N
d
x
=
1
,
so we have
(36)
∥
∇
u
∥
L
N
(
ℝ
N
)
N
=
1
-
∫
ℝ
N
‍
V
(
x
)
|
u
|
N
d
x
,
∥
∇
u
∥
L
N
(
ℝ
N
)
N
/
(
N
-
1
)
=
(
1
-
∫
ℝ
N
‍
V
(
x
)
|
u
|
N
d
x
)
1
/
(
N
-
1
)
.
Thus, we obtain
(37)
1
∥
∇
u
∥
L
N
(
ℝ
N
)
N
/
(
N
-
1
)
=
(
1
1
-
∫
ℝ
N
‍
V
(
x
)
|
u
|
N
d
x
)
1
/
(
N
-
1
)
.
From Hardy-Littlewood inequality, we have
(38)
∫
ℝ
N
‍
V
(
x
)
|
u
*
|
N
d
x
≥
∫
ℝ
N
‍
V
0
(
x
)
|
u
|
N
d
x
.
Since V
(
x
)
∈
R
(
V
0
), we have ∫
ℝ
N
‍
V
(
x
)
|
u
*
|
N
d
x
=
∫
ℝ
N
‍
V
(
x
)
|
u
|
N
d
x. Let
(39)
1
+
ɛ
=
1
∥
∇
u
∥
L
N
(
ℝ
N
)
N
/
(
N
-
1
)
=
(
1
1
-
∫
ℝ
N
‍
V
(
x
)
|
u
|
N
d
x
)
1
/
(
N
-
1
)
≥
(
1
1
-
∫
ℝ
N
‍
V
0
(
x
)
|
u
|
N
d
x
)
1
/
(
N
-
1
)
≥
(
1
1
-
(
δ
-
S
N
)
∫
ℝ
N
‍
|
u
|
N
d
x
)
1
/
(
N
-
1
)
.
Applying the mean value theorem to the function ψ
(
t
)
=
t
1
/
(
N
-
1
), we obtain that there exists ξ which satisfies
(40)
1
-
(
δ
-
S
N
)
∥
u
∥
L
N
(
ℝ
N
)
N
≤
ξ
≤
1
,
such that
(41)
1
-
[
1
-
(
δ
-
S
N
)
∥
u
∥
L
N
(
ℝ
N
)
N
]
1
/
(
N
-
1
)
=
(
δ
-
S
N
)
N
-
1
ξ
(
2
-
N
)
/
(
N
-
1
)
∥
u
∥
L
N
(
ℝ
N
)
N
.
So we have
(42)
ɛ
=
∥
u
∥
L
N
(
ℝ
N
)
N
(
N
-
1
)
ξ
(
N
-
2
)
/
(
N
-
1
)
(
1
-
(
δ
-
S
N
)
∥
u
∥
L
N
(
ℝ
N
)
N
)
1
/
(
N
-
1
)
≥
∥
u
∥
L
N
(
ℝ
N
)
N
N
-
1
.
By |
u
(
r
)
|
≤
(
N
/
w
N
-
1
)
1
/
N
∥
u
∥
L
N
(
ℝ
N
)
/
r, we have
(43)
(
3
2
)
N
/
(
N
-
1
)
2
(
|
u
(
r
)
|
N
(
N
-
1
)
ɛ
)
1
/
(
N
-
1
)
≤
C
,
and ∀
x
∈
S,
(44)
|
u
(
x
)
|
N
/
(
N
-
1
)
≤
|
u
(
x
)
-
u
(
r
)
|
N
/
(
N
-
1
)
(
∥
∇
u
∥
L
N
(
ℝ
N
)
N
)
1
/
(
N
-
1
)
+
C
.
Obviously, u
-
u
(
r
)
∈
W
1
,
N
(
𝔹
r
(
0
)
), and
(45)
∫
𝔹
r
(
0
)
‍
|
∇
(
u
-
u
(
r
)
)
|
N
d
x
≤
∫
𝔹
r
(
0
)
‍
|
∇
u
|
N
d
x
≤
1
.
Let u
′
=
(
u
(
x
)
-
u
(
r
)
)
/
∥
∇
(
u
(
x
)
-
u
(
r
)
)
∥
L
N
(
ℝ
N
); we obtain
(46)
∫
𝔹
r
(
0
)
‍
e
α
*
|
u
|
N
/
(
N
-
1
)
|
x
|
β
d
x
=
∫
S
‍
e
α
*
|
u
|
N
/
(
N
-
1
)
|
x
|
β
d
x
+
∫
𝔹
r
(
0
)
∖
S
‍
e
α
*
|
u
|
N
/
(
N
-
1
)
|
x
|
β
d
x
≤
C
∫
𝔹
r
(
0
)
‍
e
α
*
|
u
′
|
N
/
(
N
-
1
)
|
x
|
β
d
x
+
C
≤
C
.
Hence, we obtain that (32) holds. For 0
<
α
≤
(
1
-
β
/
N
)
α
N, ∥
u
∥
E
≤
1, we have
(47)
∫
ℝ
N
‍
exp
⁡
(
α
|
u
|
N
/
(
N
-
1
)
)
-
S
N
-
2
(
α
,
u
)
|
x
|
β
d
x
≤
C
.
Since α
/
α
N
+
β
/
N
+
1
/
q
≤
1 and 1
/
q
+
1
/
q
′
=
1, q
>
N, that is, α
q
′
<
α
N, α
q
′
/
α
N
+
β
q
′
/
N
≤
1, we have
(48)
∫
ℝ
N
‍
R
(
α
,
u
)
|
u
|
|
x
|
β
d
x
≤
∫
ℝ
N
‍
R
(
α
q
′
,
u
)
|
x
|
β
q
′
d
x
∥
u
∥
L
q
(
ℝ
N
)
≤
C
∥
u
∥
L
q
(
ℝ
N
)
.As the proof of Lemma7, we can obtain the following.Lemma 8.
For0
<
α
≤
(
1
-
β
/
N
)
α
N, 0
<
β
<
N, u
∈
E and ∥
u
∥
E
≤
1,  q
>
N, and α
/
α
N
+
β
/
N
+
1
/
q
≤
1, one has
(49)
∫
ℝ
N
‍
[
exp
⁡
(
α
|
u
|
N
/
(
N
-
1
)
)
-
S
N
-
2
(
α
,
u
)
]
|
u
|
q
|
x
|
β
d
x
≤
C
(
N
,
α
)
∥
u
∥
E
q
.


## 2.2. The Eigenvalue Problem

We consider the following eigenvalue problem:(50)
-
div
⁡
(
|
∇
u
|
N
-
2
∇
u
)
+
V
(
x
)
|
u
|
N
-
2
u
=
λ
|
u
|
N
-
2
u
|
x
|
β
,
x
∈
ℝ
N
,
u
≠
0
,
x
∈
ℝ
N
.Now, we denote the setM
=
{
u
∈
E
:
∫
ℝ
N
(
‍
|
u
|
N
/
|
x
|
β
)
d
x
=
1
}, and define
(51)
λ
1
=
inf
⁡
0
≠
u
∈
M
{
I
N
(
u
)
:
u
∈
E
∖
{
0
}
}
>
0
,
0
<
β
<
N
,
where I
N
(
u
)
=
∫
ℝ
N
‍
(
|
∇
u
|
N
+
V
(
x
)
|
u
|
N
)
d
x.Lemma 9 (see [12]).
Letu
>
0, v
>
0 be two continuous functions in Ω differentiable a.e., and
(52)
L
(
u
,
v
)
=
|
∇
u
|
p
+
(
p
-
1
)
u
p
v
p
|
∇
v
|
p
-
p
u
p
-
1
v
p
-
1
|
∇
v
|
p
-
2
∇
v
∇
u
,
R
(
u
,
v
)
=
|
∇
u
|
p
-
|
∇
v
|
p
-
2
∇
(
u
p
v
p
-
1
)
∇
v
.Then(
1
)
L
(
u
,
v
)
=
R
(
u
,
v
)
≥
0, (
2
)
L
(
u
,
v
)
=
0 a.e. in Ω if and only if u
=
k
v for some k
>
0.Proposition 10.
Assume that (H1)–(H3) hold; thenλ
1
>
-
∞ is the lowest eigenvalue of Problem (50) and λ
1 is principal.Proof.
From Lemma2, we have λ
1
>
-
∞. Furthermore, any minimizing sequence {
u
n
} is bounded. Up to a subsequence, there exists u
∈
E such that
(53)
u
n
⇀
u
0
,
in
E
,
u
n
⟶
u
0
,
in
L
N
r
′
(
ℝ
N
)
.
Hence, we have
(54)
I
N
(
u
0
)
≤
lim
⁡
n
→
+
∞
I
N
(
u
n
)
=
λ
1
,
u
0
∈
M
,
and consequently we have
(55)
I
N
(
u
0
)
=
λ
1
.
From Lemma 5, we obtain that M is weakly closed in E. By the Lagrange Multipliers rule, λ
1 is an eigenvalue of problem (50). Moreover I
N
(
|
u
|
)
=
I
N
(
u
) for any u, so that λ
1 possesses a nonnegative eigenfunction. We conclude that the eigenvalue is principal from Harnack inequality in [13].Proposition 11.
The eigenvalueλ
1 is isolated. That is, there exists ɛ
>
0, such that there are no other eigenvalues of problem (50) in the interval (
λ
1
,
λ
1
+
ɛ
).Proof.
Assume by contradiction there exists a sequence of eigenvalueλ
m of problem (50) with 0
<
λ
m
↘
λ
1. Let {
u
m
} be an eigenfunction associated with λ
m. Then {
u
m
} satisfies
(56)
-
Δ
N
u
m
+
V
(
x
)
|
u
m
|
N
-
2
u
m
=
λ
m
|
u
m
|
N
-
2
u
m
|
x
|
β
,
∫
ℝ
N
‍
|
∇
u
m
|
N
d
x
+
∫
ℝ
N
‍
V
(
x
)
|
u
m
|
N
d
x
-
∫
ℝ
N
‍
λ
m
|
x
|
β
|
u
m
|
N
d
x
=
0
.
We define
(57)
v
m
=
u
m
(
∫
ℝ
N
(
|
u
m
|
N
/
|
x
|
β
)
‍
d
x
)
1
/
N
.
The coercivity of the functional I
N
(
u
m
)
=
∫
ℝ
N
‍
|
∇
u
m
|
N
d
x
+
∫
ℝ
N
‍
V
(
x
)
|
u
m
|
N
d
x implies that {
u
m
} is a bounded sequence. Hence {
v
m
} is bounded in E. So there exists a subsequence (still denoted) {
v
m
} and v
∈
E such that
(58)
v
m
⇀
v
,
weakly
in
E
,
v
m
⟶
v
,
strongly
in
L
N
(
ℝ
N
)
,
and ∫
ℝ
N
(
|
u
m
|
N
/
|
x
|
β
)
‍
d
x
=
1. On the other hand, we have
(59)
∫
ℝ
N
‍
|
∇
v
|
N
d
x
+
∫
ℝ
N
‍
V
(
x
)
v
N
d
x
≤
liminf
⁡
n
→
∞
(
∫
ℝ
N
‍
|
∇
v
n
|
N
d
x
+
∫
ℝ
N
‍
V
(
x
)
v
n
N
d
x
)
=
λ
1
,
and ∫
ℝ
N
‍
|
∇
v
|
N
d
x
+
∫
ℝ
N
‍
V
(
x
)
v
N
d
x
=
λ
1
>
0. So we conclude that v is an eigenfunction associated with λ
1 and v
>
0. Then we conclude from the convergence in measure of the sequence {
v
n
} towards v that
(60)
|
Ω
n
-
|
⟶
0
,
where Ω
n
- denotes the negative set of v
n, which contradicts Proposition 11.Proposition 12.
The first eigenvalueλ
1 is simple, in the sense that the eigenfunctions associated with it are merely constant multiples of each other.Proof.
Letφ, ζ be two eigenfunctions associated with λ
1. We assume without restriction that φ
>
0, ζ
>
0; then φ satisfies -
Δ
φ
+
V
(
x
)
φ
N
-
1
=
λ
1
(
N
)
(
φ
N
-
1
/
|
x
|
β
). Testing it with function φ, we get
(61)
∫
ℝ
N
‍
|
∇
φ
|
N
d
x
-
∫
ℝ
N
‍
[
-
V
(
x
)
+
λ
1
|
x
|
β
]
φ
N
d
x
=
0
.
Let ɛ
→
0, from Lemma 9, we have
(62)
0
≤
∫
ℝ
N
‍
L
(
φ
,
ζ
+
ɛ
)
d
x
=
∫
ℝ
N
‍
R
(
φ
,
ζ
+
ɛ
)
d
x
=
∫
ℝ
N
‍
[
-
V
(
x
)
+
λ
1
|
x
|
β
]
φ
N
d
x
-
∫
ℝ
N
‍
|
∇
ζ
|
N
-
2
∇
(
φ
N
(
ζ
+
ɛ
)
N
-
1
)
∇
ζ
d
x
.
The function φ
N
/
(
ζ
+
ɛ
)
N
-
1, where ɛ
>
0, belongs to E and then it is admissible for the weak formulation of -
Δ
N
ζ
+
V
(
x
)
|
ζ
|
N
-
2
ζ
=
λ
1
(
|
ζ
|
N
-
2
ζ
/
|
x
|
β
), a.e., and
(63)
∫
ℝ
N
‍
|
∇
ζ
|
N
-
2
∇
ζ
∇
φ
N
(
ζ
+
ɛ
)
N
-
1
d
x
-
∫
ℝ
N
‍
[
-
V
(
x
)
+
λ
1
|
x
|
β
]
φ
N
ζ
N
-
1
(
ζ
+
ɛ
)
N
-
1
d
x
=
0
.
It follows from (62) and (63) that we have
(64)
0
≤
L
(
φ
,
ζ
+
ɛ
)
=
∫
ℝ
N
‍
λ
1
[
1
-
ζ
N
-
1
(
ζ
+
ɛ
)
N
-
1
]
φ
N
|
x
|
β
d
x
-
∫
ℝ
N
‍
V
(
x
)
φ
N
[
1
-
ζ
N
-
1
(
ζ
+
ɛ
)
N
-
1
]
d
x
.
Let ɛ
→
0; we have L
(
φ
,
ζ
)
=
0. By Lemma 9, there exists k
>
0 such that φ
=
k
ζ.


## 3. The Proof of Theorem1

### 3.1. Palais-Smale Sequence

Now, we check that the functionalI satisfies the geometric conditions of the Mountain Pass Theorem.Lemma 13.
Suppose that (H1)–(H3) and (f1)–(f5) hold. Then there existsε
2 such that, for 0
<
ɛ
<
ε
2, there exists ρ
ɛ
>
0 such that I
(
u
)
>
0 if ∥
u
∥
E
=
ρ
ɛ. Furthermore, ρ
ɛ can be chosen such that ρ
ɛ
→
0, as ɛ
→
0.Proof.
From (f5), for everyɛ
>
0, there exists σ
>
0 such that |
u
|
≤
σ implies
(65)
F
(
x
,
u
)
≤
ɛ
N
|
u
|
N
,
∀
x
∈
ℝ
N
.
Moreover, using (f1), for each q
>
N and k
/
α
N
+
β
/
N
+
1
/
q
≤
1, we find a constant C such that
(66)
F
(
x
,
u
)
≤
C
|
u
|
q
[
exp
⁡
(
k
|
u
|
N
/
(
N
-
1
)
)
-
S
N
-
2
(
k
,
u
)
]
,
F
(
x
,
u
)
≤
F
(
x
,
u
)
≤
|
u
|
q
C
|
u
|
q
∀
|
u
|
≥
σ
,
x
∈
ℝ
N
.
Combining (65) and (66), we have
(67)
F
(
x
,
u
)
≤
ɛ
N
|
u
|
N
+
C
|
u
|
q
×
[
exp
⁡
(
k
|
u
|
N
/
(
N
-
1
)
)
-
S
N
-
2
(
k
,
u
)
]
,
∀
(
x
,
u
)
∈
ℝ
N
×
ℝ
.
Since the embedding E
↪
L
N
(
ℝ
N
) is continuous, we obtain
(68)
I
(
u
)
≥
1
N
∥
u
∥
E
N
-
ɛ
+
λ
N
∫
ℝ
N
‍
|
u
|
N
|
x
|
β
d
x
-
C
∥
u
∥
E
q
-
ɛ
∥
h
∥
*
∥
u
∥
E
≥
1
N
(
1
-
λ
+
ɛ
λ
1
)
∥
u
∥
E
N
-
C
∥
u
∥
E
q
-
ɛ
∥
h
∥
*
∥
u
∥
E
.
Thus, we have
(69)
I
(
u
)
≥
∥
u
∥
E
[
1
N
(
1
-
λ
+
ɛ
λ
1
)
∥
u
∥
E
N
-
1
-
C
∥
u
∥
E
q
-
1
-
ɛ
∥
h
∥
*
(
1
-
λ
+
ɛ
λ
1
)
]
.
Since q
>
N and 0
<
λ
<
λ
1 and letting ɛ
<
λ
1
-
λ, we choose ρ
>
0 such that (
1
/
N
)
(
1
-
(
λ
+
ɛ
)
/
λ
1
)
ρ
N
-
1
-
C
ρ
q
-
1
-
ɛ
∥
h
∥
*
>
0. Thus, if ɛ is sufficiently small, we find some ρ
ɛ
>
0 such that I
(
u
)
>
0 if ∥
u
∥
E
=
ρ
ɛ and even ρ
ɛ
→
0 as ɛ
→
0.Lemma 14.
If0
<
λ
<
λ
1, there exists e
∈
E, with ∥
e
∥
E
>
ρ
ɛ, such that I
(
e
)
<
inf
⁡
∥
u
∥
E
=
ρ
ɛ
I
(
u
).Proof.
Letu
∈
E
∖
{
0
}, u
>
0 with compact support Ω
=
supp
⁡
(
u
). By (f4), we obtain that for p
>
N, there exists a positive constant C
>
0 such that for every M
>
0,
(70)
∀
|
u
|
>
C
,
∀
x
∈
Ω
,
F
(
x
,
u
)
≥
M
|
u
|
N
.
Then, we have
(71)
I
(
t
u
)
≤
t
N
N
∥
u
∥
E
N
-
M
t
N
∫
Ω
‍
|
u
|
N
|
x
|
β
d
x
-
ɛ
t
∫
Ω
‍
h
u
d
x
-
λ
N
∫
Ω
‍
|
t
u
|
N
|
x
|
β
d
x
≤
t
N
N
∥
u
∥
E
N
-
t
N
(
M
+
λ
N
)
∫
Ω
‍
|
u
|
N
|
x
|
β
d
x
-
ɛ
t
∫
Ω
‍
h
u
d
x
≤
t
N
(
∥
u
∥
E
N
N
-
(
M
+
λ
N
)
∫
|
x
|
<
R
‍
|
u
|
N
|
x
|
β
d
x
)
.
Choose R
>
0 and B
R
(
0
)
⊂
Ω
¯ and let
(72)
M
>
R
β
∥
u
∥
E
N
N
∥
u
∥
L
N
(
ℝ
N
)
N
-
λ
N
>
R
β
∥
u
∥
E
N
N
∥
u
∥
L
N
(
ℝ
N
)
N
-
λ
1
N
;
we have I
(
t
u
)
→
-
∞ as t
→
∞. Setting e
=
t
u with t being sufficient large, we obtain the conclusion.It is well known that the failure of the (PS) compactness condition creates some difficulties in studying the class of elliptic problems involving critical growth. In Lemma15, instead of (PS) sequence, we analyze the compactness of Cerami sequences of the functional I.Lemma 15.
Let(
u
n
)
⊂
E be a Cerami sequence of I; that is,
(73)
I
(
u
n
)
→
C
M
,
(
1
+
∥
u
n
∥
E
)
∥
D
I
(
u
n
)
∥
E
′
⟶
0
,
I
(
u
n
)
⟶
C
M
,
I
6
→
(
1
+
∥
u
n
∥
E
)
a
s
n
⟶
∞
.
Then there exists a subsequence of (
u
n
) (still denoted by (
u
n
)) and u
∈
E such that
(74)
f
(
x
,
u
n
)
|
x
|
β
⟶
f
(
x
,
u
)
|
x
|
β
,
i
n
L
loc
⁡
1
(
ℝ
N
)
,
∇
u
n
⟶
∇
u
,
a
.
e
.
i
n
ℝ
N
,
|
∇
u
n
|
N
-
2
|
∇
u
n
|
⇀
|
∇
u
|
N
-
2
|
∇
u
|
,
(
L
loc
⁡
N
/
(
N
-
1
)
(
ℝ
N
)
)
N
,
u
n
⇀
u
,
i
n
E
,
where C
M
∈
(
0
,
(
1
/
N
)
(
1
-
β
/
N
)
(
α
N
/
α
0
)
). Furthermore, u is a nontrivial weak solution of problem (1).Proof.
Letu
n
∈
E, v
∈
E, as n
→
∞; we have
(75)
1
N
∫
ℝ
N
‍
|
∇
u
n
|
N
d
x
+
1
N
∫
ℝ
N
‍
V
(
x
)
|
u
n
|
N
d
x
-
∫
ℝ
N
‍
F
(
x
,
u
n
)
|
x
|
β
d
x
-
λ
N
∫
ℝ
N
‍
|
u
n
|
N
|
x
|
β
d
x
-
ɛ
∫
ℝ
N
‍
h
u
n
⟶
C
M
,
(76)
|
D
I
(
u
n
)
v
|
=
|
|
u
n
|
N
-
2
u
n
v
|
x
|
β
∫
ℝ
N
|
∇
u
n
|
N
-
2
∇
u
n
∇
v
d
x
+
∫
ℝ
N
‍
V
(
x
)
|
u
n
|
N
-
2
u
n
v
d
x
-
∫
ℝ
N
‍
f
(
x
,
u
n
)
v
|
x
|
β
d
x
-
λ
∫
ℝ
N
|
u
n
|
N
-
2
u
n
v
|
x
|
β
d
x
-
ε
∫
ℝ
N
h
v
‍
d
x
|
≤
τ
n
∥
v
∥
E
(
1
+
∥
u
n
∥
E
)
,
where τ
n
→
0 as n
→
∞. Let v
=
u
n in (76); we have
(77)
-
∫
ℝ
N
‍
|
∇
u
n
|
N
d
x
-
∫
ℝ
N
‍
V
(
x
)
|
u
n
|
N
d
x
+
∫
ℝ
N
‍
f
(
x
,
u
n
)
u
n
|
x
|
β
d
x
+
λ
∫
ℝ
N
‍
|
u
n
|
N
|
x
|
β
d
x
+
ɛ
∫
ℝ
N
‍
h
u
n
d
x
≤
τ
n
∥
u
n
∥
E
(
1
+
∥
u
n
∥
E
)
⟶
0
,
as
n
⟶
∞
.
Suppose that
(78)
∥
u
n
∥
E
⟶
∞
.
Set
(79)
v
n
=
u
n
∥
u
n
∥
E
;
we have ∥
v
n
∥
E
=
1. From [5], we have
(80)
∫
ℝ
N
‍
liminf
⁡
n
→
+
∞
F
(
x
,
u
n
)
|
x
|
β
|
u
n
+
(
x
)
|
N
|
v
n
+
|
N
d
x
=
+
∞
.
However, since {
u
n
} is the Cerami sequence at the level C
M, we have that
(81)
∥
u
n
∥
E
N
=
N
I
(
u
n
)
+
N
∫
ℝ
N
‍
F
(
x
,
u
n
)
|
x
|
β
d
x
+
N
ɛ
∫
ℝ
N
‍
h
u
n
d
x
+
λ
∫
ℝ
N
‍
|
u
n
|
N
|
x
|
β
d
x
+
o
(
1
)
.
Then there exists some constant C such that
(82)
∥
u
n
∥
E
[
1
-
λ
λ
1
-
N
ɛ
∥
h
∥
*
]
≤
N
C
M
+
N
∫
ℝ
N
‍
F
(
x
,
u
n
)
|
x
|
β
d
x
+
o
(
1
)
,
as
ɛ
⟶
0
,
which implies that
(83)
∫
ℝ
N
‍
F
(
x
,
u
n
)
|
x
|
β
d
x
⟶
∞
,
liminf
⁡
n
→
∞
∫
ℝ
N
‍
F
(
x
,
u
n
)
|
x
|
β
|
u
n
+
|
N
|
v
n
+
|
N
d
x
=
liminf
⁡
n
→
∞
∫
ℝ
N
‍
F
(
x
,
u
n
)
|
x
|
β
∥
u
n
+
∥
E
N
d
x
.
Let Ψ
=
∫
ℝ
N
‍
(
F
(
x
,
u
n
)
/
|
x
|
β
)
d
x; then we have
(84)
liminf
⁡
n
→
∞
∫
ℝ
N
‍
F
(
x
,
u
n
)
|
x
|
β
|
u
n
+
|
N
|
v
n
+
|
N
d
x
≤
liminf
⁡
n
→
∞
Ψ
N
C
M
+
N
Ψ
+
N
ɛ
∫
ℝ
N
‍
h
u
n
+
d
x
+
o
(
1
)
.
So we can conclude that
(85)
liminf
⁡
n
→
∞
∫
ℝ
N
‍
F
(
x
,
u
n
)
|
x
|
β
|
u
n
+
|
N
|
v
n
+
|
N
d
x
≤
liminf
⁡
n
→
∞
Ψ
N
C
M
+
N
Ψ
+
N
ɛ
∫
ℝ
N
‍
h
u
n
+
d
x
+
o
(
1
)
=
1
N
.
Note that F
(
x
,
u
n
)
≥
0; by Fatou Lemma, (80), and (85), we get a contradiction. So v
≤
0 which means that v
k
+
⇀
0 in E.
Lett
n
∈
[
0,1
], such that
(86)
I
(
t
n
u
n
)
=
max
⁡
t
∈
[
0,1
]
I
(
t
u
n
)
.
For any given R
∈
(
0
,
(
1
-
β
/
N
)
α
N
/
α
0
)
(
N
-
1
)
/
N, let ɛ
=
(
1
-
β
/
N
)
α
N
/
R
N
/
(
N
-
1
)
-
α
0
>
0, by (f1); there exists C
=
C
(
R
)
>
0 such that
(87)
F
(
x
,
u
n
)
≤
C
|
u
n
|
N
+
|
(
1
-
β
/
N
)
α
N
R
N
/
(
N
-
1
)
-
α
0
|
R
(
α
0
+
ɛ
,
u
n
)
,
(
x
,
u
n
)
≤
|
u
n
|
N
+
|
(
-
β
/
N
)
α
N
R
N
/
(
N
-
1
)
-
α
0
|
∀
(
x
,
u
n
)
∈
Ω
¯
×
ℝ
N
.
Since ∥
u
n
∥
E
→
∞, we have
(88)
I
(
t
n
u
n
)
≥
I
(
R
u
n
∥
u
n
∥
E
)
=
I
(
R
v
n
)
,
and by (f3), ∥
v
n
∥
E
=
1, and ∫
ℝ
N
‍
(
F
(
x
,
v
n
+
)
/
|
x
|
β
)
d
x
=
∫
ℝ
N
‍
(
F
(
x
,
v
n
)
/
|
x
|
β
)
d
x, we have
(89)
N
I
(
R
v
n
)
≥
R
N
-
(
N
C
R
N
+
λ
R
N
)
∫
ℝ
N
‍
|
v
n
+
|
N
|
x
|
β
d
x
-
N
|
(
1
-
β
/
N
)
α
N
R
N
/
(
N
-
1
)
-
α
0
|
∫
ℝ
N
‍
R
(
α
0
+
ɛ
,
R
|
v
n
+
|
)
|
x
|
β
d
x
-
N
R
|
(
1
-
β
/
N
)
α
N
R
N
/
(
N
-
1
)
-
α
0
|
∫
ℝ
N
‍
h
|
v
n
+
|
d
x
≥
R
N
-
(
N
C
R
N
+
λ
R
N
)
∫
ℝ
N
‍
|
v
n
+
|
N
|
x
|
β
d
x
-
N
|
(
1
-
β
/
N
)
α
N
R
N
/
(
N
-
1
)
-
α
0
|
×
∫
ℝ
N
‍
R
(
(
α
0
+
ɛ
)
R
N
/
(
N
-
1
)
,
|
v
n
+
|
)
|
x
|
β
d
x
-
N
R
|
(
1
-
β
/
N
)
α
N
R
N
/
(
N
-
1
)
-
α
0
|
∫
ℝ
N
‍
h
|
v
n
+
|
d
x
≥
R
N
-
(
N
C
R
N
+
λ
R
N
)
∫
ℝ
N
‍
|
v
n
+
|
N
|
x
|
β
d
x
-
N
|
(
1
-
β
/
N
)
α
N
R
N
/
(
N
-
1
)
-
α
0
|
×
∫
ℝ
N
‍
R
(
(
1
-
β
/
N
)
α
N
,
|
v
n
+
|
)
|
x
|
β
d
x
-
N
R
|
(
1
-
β
/
N
)
α
N
R
N
/
(
N
-
1
)
-
α
0
|
∫
ℝ
N
‍
h
|
v
n
+
|
d
x
.
Since v
n
+
⇀
0 in E and the embedding E
↪
L
N
(
ℝ
N
) is compact from and the Hölder inequality, we have ∫
ℝ
N
‍
(
|
v
n
+
|
N
/
|
x
|
β
)
d
x
→
0
(
n
→
∞
). By Lemma 7, we have ∫
ℝ
N
‍
(
R
(
(
1
-
β
/
N
)
α
N
,
|
v
n
+
|
)
/
|
x
|
β
)
d
x
≤
C.
Letn
→
∞ in (89) and R
→
[
(
1
-
β
/
N
)
α
N
/
α
0
]
(
N
-
1
)
/
N; we get
(90)
liminf
⁡
n
→
∞
I
(
t
n
u
n
)
≥
1
N
[
(
1
-
β
/
N
)
α
N
α
0
]
N
-
1
>
C
M
.
Note that I
(
0
)
=
0 and I
(
u
n
)
→
C
M; we suppose that t
n
∈
(
0,1
). Since D
I
(
t
n
u
n
)
t
n
u
n
=
0, we have
(91)
t
n
N
∥
u
n
∥
E
N
=
∫
ℝ
N
‍
f
(
x
,
t
n
u
n
)
t
n
u
n
|
x
|
β
d
x
+
ɛ
t
n
∫
ℝ
N
‍
h
u
n
d
x
+
λ
t
n
N
∫
ℝ
N
‍
|
u
n
|
N
|
x
|
β
d
x
.
By (f2) and ε
n
→
0, we have
(92)
N
I
(
t
n
u
n
)
=
∫
ℝ
N
‍
f
(
x
,
t
n
u
n
)
t
n
u
n
|
x
|
β
d
x
-
N
∫
ℝ
N
‍
F
(
x
,
t
n
u
n
)
|
x
|
β
d
x
+
o
(
1
)
.
Moreover, we have
(93)
∫
ℝ
N
‍
f
(
x
,
u
n
)
u
n
|
x
|
β
d
x
-
N
∫
ℝ
N
‍
F
(
x
,
u
n
)
|
x
|
β
d
x
=
∥
u
n
∥
E
N
+
N
C
M
-
∥
u
n
∥
E
N
+
o
(
1
)
,
which is a contraction to (75); this proves that {
u
n
} is bounded in E. Thus, we have
(94)
∫
ℝ
N
‍
f
(
x
,
u
n
)
u
n
|
x
|
β
d
x
≤
C
,
∫
ℝ
N
‍
F
(
x
,
u
n
)
|
x
|
β
d
x
≤
C
.
From Lemma 5, the embedding E
↪
L
q
(
ℝ
N
) is compact for all q
≥
N. If {
u
n
}
∈
E, we get
(95)
u
n
⇀
u
,
in
E
,
u
n
⟶
u
,
in
L
loc
⁡
q
(
ℝ
N
)
,
u
n
⟶
u
,
a
.
e
.
in
ℝ
N
.
From (f1), the Trudinger-Moser inequality, and the Hölder inequality, we have f
(
x
,
u
n
)
/
|
x
|
β
∈
L
loc
⁡
1
(
ℝ
N
). From Lemma 2.1 in [14], we have
(96)
f
(
x
,
u
n
)
|
x
|
β
⟶
f
(
x
,
u
)
|
x
|
β
,
in
L
loc
⁡
1
(
ℝ
N
)
.
For any fixed δ
>
0, set
(97)
Σ
δ
=
{
x
∈
ℝ
N
:
lim
⁡
r
→
0
lim
⁡
n
→
∞
∫
𝔹
r
(
x
)
(
|
∇
u
n
|
N
+
|
u
n
|
N
)
‍
d
x
≥
δ
∫
𝔹
r
(
x
)
(
|
∇
u
n
|
N
+
|
u
n
|
N
)
‍
}
.
Because {
u
n
} is bounded,Σ
δ is a finite set. From Lemma 4.4 in ([4]), for any compact set K
⊂
⊂
ℝ
N
∖
Σ
δ, we have
(98)
lim
⁡
n
→
∞
∫
K
‍
|
f
(
x
,
u
n
)
u
n
-
f
(
x
,
u
)
u
|
|
x
|
β
d
x
=
0
.
Now, we prove that
(99)
lim
⁡
n
→
∞
∫
K
‍
|
∇
u
n
-
∇
u
|
N
d
x
=
0
.
It is enough to prove that for any x
′
∈
ℝ
N
∖
Σ
δ and B
r
(
x
′
)
⊂
ℝ
N
∖
Σ
δ there holds
(100)
lim
⁡
n
→
∞
∫
𝔹
r
/
2
(
x
′
)
‍
|
∇
u
n
-
∇
u
|
N
d
x
=
0
.
We take ϕ
∈
C
0
∞
(
B
r
(
x
′
)
) with 0
≤
ϕ
≤
1 and ϕ
=
1 on 𝔹
r
/
2
(
x
′
). Then {
ϕ
u
n
} is a bounded sequence. Choosing v
n
=
ϕ
u
n and v
=
ϕ
u in (76), we have
(101)
∫
B
r
(
x
′
)
‍
ϕ
(
|
∇
u
n
|
N
-
2
∇
u
n
-
|
∇
u
|
N
-
2
∇
u
)
(
∇
u
n
-
∇
u
)
d
x
≤
∫
B
r
(
x
′
)
‍
|
∇
u
n
|
N
-
2
∇
u
n
∇
ϕ
(
u
-
u
n
)
d
x
+
∫
B
r
(
x
′
)
‍
ϕ
|
∇
u
|
N
-
2
∇
u
(
∇
u
-
∇
u
n
)
d
x
+
∫
B
r
(
x
′
)
‍
ϕ
(
u
n
-
u
)
f
(
x
,
u
n
)
|
x
|
β
d
x
+
τ
n
∥
ϕ
u
n
∥
E
+
τ
n
∥
ϕ
u
∥
E
+
ɛ
∫
B
r
(
x
′
)
‍
ϕ
h
(
u
n
-
u
)
d
x
+
λ
∫
B
r
(
x
′
)
‍
|
u
n
|
u
n
ϕ
|
x
|
β
(
u
n
-
u
)
d
x
.
Adapting an argument similar to [4], we have
(102)
lim
⁡
n
→
∞
∫
K
‍
|
∇
u
n
-
∇
u
|
N
d
x
=
0
.
Since Σ
δ is finite, it follows that ∇
u
n
→
∇
u a.e. This implies, up to a subsequence that |
∇
u
n
|
N
-
2
∇
u
n
⇀
|
∇
u
|
N
-
2
∇
u in (
L
loc
⁡
N
/
(
N
-
1
)
(
ℝ
N
)
)
N. Let n
→
∞ in (76), and f
(
x
,
u
n
)
/
|
x
|
β
→
f
(
x
,
u
)
/
|
x
|
β in L
loc
⁡
1
(
ℝ
N
); we obtain
(103)
〈
D
I
(
u
)
,
v
〉
=
0
,
∀
v
∈
C
0
∞
(
ℝ
N
)
.Remark 16.
The idea and proof of Lemma15 follow as in Lemma 4.1 in [5].


### 3.2. Min-Max Value

In order to get a more precise information about the minimax level obtained by the Mountain Pass Theorem, we consider the following sequence of scale which is called the Moser function:(104)
M
l
(
x
,
r
)
~
=
1
w
N
-
1
1
/
N
{
(
log
⁡
l
)
(
N
-
1
)
/
N
,
if
|
x
|
≤
r
l
,
log
⁡
⁡
(
r
/
|
x
|
)
(
log
⁡
l
)
1
/
N
,
if
r
l
≤
|
x
|
≤
r
,
0
,
if
|
x
|
≥
r
.
Hence, we have M
l
(
x
,
r
)
~
∈
W
1
,
N
(
ℝ
N
), the support of M
l
(
x
,
r
)
~ is the ball B
r
(
0
), and
(105)
∫
ℝ
N
‍
|
∇
M
l
(
x
,
r
)
~
|
N
d
x
=
1
,
∫
ℝ
N
‍
|
M
l
(
x
,
r
)
~
|
N
d
x
=
o
(
1
log
⁡
l
)
.LetM
l
(
x
,
r
)
=
M
l
(
x
,
r
)
~
/
∥
M
l
(
x
,
r
)
~
∥
E
N; we have
(106)
M
l
N
/
(
N
-
1
)
(
x
,
r
)
=
w
N
-
1
-
1
/
(
N
-
1
)
log
⁡
l
+
d
l
,
for
|
x
|
≤
r
l
,
where
(107)
d
l
=
w
N
-
1
-
1
/
(
N
-
1
)
log
⁡
l
(
∥
M
l
(
x
,
r
)
~
∥
E
-
1
/
(
N
-
1
)
-
1
)
.
From (105), we conclude that ∥
M
l
(
x
,
r
)
∥
E
→
1, as l
→
∞. Consequently, we have
(108)
d
l
log
⁡
l
⟶
0
,
as
l
⟶
∞
.Lemma 17.
Suppose that (H1)–(H3) and (f1)–(f6) hold. Then there existsk
∈
ℕ such that
(109)
max
⁡
t
≥
0
{
t
N
N
-
∫
ℝ
N
‍
F
(
x
,
t
M
k
)
|
x
|
β
d
x
-
λ
N
∫
ℝ
N
‍
|
t
M
k
|
N
|
x
|
β
d
x
}
<
1
N
(
N
-
β
N
α
N
α
0
)
N
-
1
.Proof.
Chooser
>
0 as in (
f
6
) and β
0
>
0 such that
(110)
lim
⁡
s
→
∞
s
f
(
x
,
s
)
exp
⁡
(
-
α
0
|
s
|
N
/
(
N
-
1
)
)
≥
β
0
>
2
e
(
α
N
d
(
N
-
β
)
/
N
)
+
C
r
N
-
β
-
r
N
-
β
/
(
N
-
β
)
×
(
N
-
β
α
0
)
N
-
1
,
where
(111)
C
=
lim
⁡
k
→
∞
ξ
k
log
⁡
k
∫
0
ξ
k
‍
exp
⁡
[
s
N
/
(
N
-
1
)
(
N
-
β
)
×
log
⁡
k
(
s
N
/
(
N
-
1
)
-
ξ
k
s
)
]
d
s
>
0
,
ξ
k
=
∥
M
k
~
∥
E
,
C
≥
1
-
e
-
(
N
-
β
)
log
⁡
n
N
-
β
.
Suppose, by contradiction, that, for all k, we get
(112)
max
⁡
t
≥
0
{
t
N
N
-
∫
ℝ
N
‍
F
(
x
,
t
M
k
)
|
x
|
β
d
x
-
λ
N
∫
ℝ
N
‍
|
t
M
k
|
N
|
x
|
β
d
x
}
≥
1
N
(
(
N
-
β
)
N
α
N
α
0
)
N
-
1
,
where M
k
(
x
)
=
M
k
(
x
,
r
). For each k, there exists t
k
>
0 such that
(113)
t
k
N
N
-
∫
ℝ
N
‍
F
(
x
,
t
k
M
k
)
|
x
|
β
d
x
-
λ
N
∫
ℝ
N
‍
|
t
k
M
k
|
N
|
x
|
β
d
x
=
max
⁡
t
≥
0
{
t
N
N
-
∫
ℝ
N
‍
F
(
x
,
t
M
k
)
|
x
|
β
d
x
-
λ
N
∫
ℝ
N
‍
|
t
M
k
|
N
|
x
|
β
d
x
}
.
Thus; we have
(114)
t
k
N
N
-
∫
ℝ
N
‍
F
(
x
,
t
k
M
k
)
|
x
|
β
d
x
-
λ
N
∫
ℝ
N
‍
|
t
k
M
k
|
N
|
x
|
β
d
x
≥
1
N
(
(
N
-
β
)
N
α
N
α
0
)
N
-
1
.
From F
(
x
,
u
)
≥
0, 0
<
λ
<
λ
1, we obtain
(115)
t
k
N
≥
(
N
-
β
N
α
N
α
0
)
N
-
1
.
Let t
=
t
k, we have
(116)
t
k
N
=
∫
ℝ
N
(
t
k
M
k
f
(
x
,
t
k
M
k
)
/
|
x
|
β
)
‍
d
x
(
1
-
λ
∫
|
x
|
≤
r
‍
(
|
M
k
|
N
/
|
x
|
β
)
d
x
)
≥
∫
|
x
|
≤
r
‍
(
t
k
M
k
f
(
x
,
t
k
M
k
)
/
|
x
|
β
)
d
x
1
+
λ
/
λ
1
.
By (f6), given that τ
>
0, there exist R
τ
>
0 and |
x
|
≤
r; we have
(117)
u
f
(
x
,
u
)
≥
(
β
0
-
τ
)
exp
⁡
(
α
0
|
u
|
N
/
(
N
-
1
)
)
.
From (116) and (117), for large k, we obtain
(118)
t
k
N
≥
λ
1
(
β
0
-
τ
)
∫
|
x
|
≤
r
/
k
‍
(
exp
⁡
⁡
(
α
0
|
t
k
M
k
|
N
/
(
N
-
1
)
)
/
|
x
|
β
)
d
x
(
λ
1
+
λ
)
≥
(
β
0
-
τ
)
k
-
N
+
β
w
N
-
1
r
N
-
β
2
(
N
-
β
)
×
exp
⁡
(
α
0
t
k
N
/
(
N
-
1
)
w
N
-
1
-
1
/
(
N
-
1
)
log
⁡
k
+
α
0
t
k
N
/
(
N
-
1
)
d
k
)
.
Let
(119)
L
k
=
α
0
N
log
⁡
k
α
N
t
k
N
/
(
N
-
1
)
+
α
0
t
k
N
/
(
N
-
1
)
d
k
-
(
N
-
β
)
log
⁡
k
;
we have
(120)
1
≥
(
β
0
-
τ
)
r
N
-
β
w
N
-
1
2
(
N
-
β
)
exp
⁡
L
(
k
)
.
Hence, the sequence {
t
k
} is bounded. Otherwise, up to subsequences, we have lim
⁡
k
→
∞
L
(
k
)
=
∞, which leads to a contradiction. From (108) and (115) and
(121)
t
k
N
≥
(
β
0
-
τ
)
r
N
-
β
w
N
-
1
2
(
N
-
β
)
×
exp
⁡
[
(
N
α
0
t
k
N
/
(
N
-
1
)
α
N
-
(
N
-
β
)
)
log
⁡
k
+
α
0
t
k
N
/
(
N
-
1
)
d
k
(
N
α
0
t
k
N
/
(
N
-
1
)
α
N
-
(
N
-
β
)
)
]
,
it follows that
(122)
t
k
N
⟶
(
(
N
-
β
)
N
α
N
α
0
)
N
-
1
.
From [4], we have
(123)
∫
|
x
|
≤
r
‍
exp
⁡
⁡
(
α
0
|
t
k
M
k
|
N
/
(
N
-
1
)
)
|
x
|
β
d
x
≥
∫
|
x
|
≤
r
/
k
‍
exp
⁡
(
α
N
|
M
k
|
N
/
(
N
-
1
)
(
N
-
β
)
/
N
)
|
x
|
β
d
x
+
∫
r
/
k
≤
|
x
|
≤
r
‍
exp
⁡
(
α
N
|
M
k
|
N
/
(
N
-
1
)
(
N
-
β
)
/
N
)
|
x
|
β
d
x
,
∫
|
x
|
≤
r
/
k
‍
exp
⁡
(
α
N
|
M
k
|
N
/
(
N
-
1
)
(
N
-
β
)
/
N
)
|
x
|
β
d
x
=
w
N
-
1
N
-
β
r
N
-
β
k
[
N
-
β
+
(
d
k
α
N
/
log
⁡
k
)
(
N
-
β
)
/
N
]
.
Now, using the change of variable
(124)
s
=
log
⁡
⁡
(
r
/
|
x
|
)
ζ
k
log
⁡
k
with
ζ
k
=
∥
M
k
∥
E
,
by straight forward computation, we have
(125)
∫
r
/
k
≤
|
x
|
≤
r
‍
exp
⁡
(
α
N
|
M
k
|
N
/
(
N
-
1
)
(
N
-
β
)
/
N
)
|
x
|
β
d
x
=
w
N
-
1
r
N
-
β
ζ
k
log
⁡
k
×
∫
0
ζ
k
-
1
‍
exp
⁡
[
(
N
-
β
)
log
⁡
k
(
s
N
/
(
N
-
1
)
-
ζ
k
s
)
]
d
s
,
which converges to C
w
N
-
1
r
N
-
β as k
→
∞, where
(126)
C
=
lim
⁡
k
→
∞
ζ
k
log
⁡
k
∫
0
ξ
k
‍
exp
⁡
[
(
s
N
/
(
N
-
1
)
-
ξ
k
s
)
(
N
-
β
)
×
log
⁡
k
(
s
N
/
(
N
-
1
)
-
ξ
k
s
)
]
d
s

>

0
.
Finally, let k
→
∞ in (118); from (108) and (115), we have
(127)
(
(
N
-
β
)
N
α
N
α
0
)
N
-
1
≥
(
β
0
-
τ
)
2
×
[
r
N
-
β
w
N
-
1
(
N
-
β
)
e
(
α
N
d
(
N
-
β
)
/
N
)
,
+
C
r
N
-
β
w
N
-
1
-
r
N
-
β
w
N
-
1
N
-
β
]
which implies that
(128)
β
0
≤
2
e
(
α
N
d
(
N
-
β
)
/
N
)
+
C
r
N
-
β
-
r
N
-
β
/
(
N
-
β
)
×
(
N
-
β
α
0
)
N
-
1
.Remark 18.
The idea and the proof of Lemma17 come from Lemma 3.6 in [5].Lemma 19.
There existτ
>
0 and v
∈
E with ∥
v
∥
E
=
1 such that I
(
t
v
)
<
0 for all 0
<
t
<
ς. In particular, inf
⁡
∥
v
∥
E
≤
ς
I
(
u
)
<
0.Proof.
See Lemma 3.3 in [10].Corollary 20.
Under the conditions (H1)–(H3) and (f1)–(f4), ifɛ
→
0, then one has
(129)
max
⁡
t
≥
0
I
(
t
M
k
)
<
1
N
(
(
N
-
β
)
N
α
N
α
0
)
N
-
1
.From Lemmas13 and 19, we conclude that
(130)
∞
<
C
0
=
inf
⁡
∥
v
∥
E
≤
ς
I
(
u
)
<
0
.Corollary 21.
There existε
2
∈
(
0
,
ε
1
] and u
∈
W
1
,
N
(
ℝ
N
) with compact support such that, for all 0
<
ɛ
<
ε
2,
(131)
I
(
t
u
)
<
C
0
+
1
N
(
(
N
-
β
)
N
α
N
α
0
)
N
-
1
,
∀
t
≥
0
.Lemma 22.
If{
u
k
} is a Cerami sequence for I
(
u
) at any level with
(132)
liminf
⁡
n
→
∞
∥
u
k
∥
E
<
(
(
N
-
β
)
N
α
N
α
0
)
N
-
1
,
then {
u
k
} possesses a subsequence which converges strongly to a solution u
0 of problem (1).Proof.
See Lemma 4.6 in [4].
In conclusion, we have(133)
0
<
C
M
<
C
0
+
1
N
(
(
N
-
β
)
N
α
N
α
0
)
N
-
1
.


### 3.3. Multiplicity Results

In order to prove the existence of the second solution of problem (1) follows by the minimum argument and Ekeland's Variational Principle.Proposition 23.
Under the conditions (H1)–(H3) and (f1)–(f6), there existsε
1
>
0 such that, for each ɛ with 0
<
ɛ
<
ε
1, problem (1) has a solution u
M via Mountain Pass Theorem.Proof.
See Proposition 4.1 in [5].Proposition 24.
There existsε
2
>
0 such that, for each ɛ with 0
<
ɛ
<
ε
2, (1) has a minimum type solution u
0 with I
(
u
0
)
=
C
0
<
0, where C
0 is defined in (130).Proof.
See Proposition 5.1 in [5].Proposition 25.
Ifε
2
>
0 is sufficiently small, then the solutions of problem (1) obtained in Propositions 23 and 24 are distinct.Proof.
By Propositions23 and 24, there exist sequences {
u
n
}, {
v
n
} in E such that
(134)
u
n
⟶
u
0
,
I
(
u
n
)
⟶
C
0
,
D
I
(
u
n
)
u
n
⟶
0
,
v
n
⇀
u
M
,
I
(
v
n
)
⟶
C
M
>
0
,
D
I
(
v
n
)
v
n
⟶
0
,
∇
v
n
⟶
∇
u
M
a
.
e
.
in
ℝ
N
.
Suppose by contradiction that u
0
=
u
M. As in the proof of Lemma 15, we have
(135)
f
(
x
,
v
n
)
|
x
|
β
⟶
f
(
x
,
u
0
)
|
x
|
β
in
L
loc
⁡
1
(
ℝ
N
)
,
as
n
⟶
∞
.
Hence, by (f2) and (f3) and Generalized Lebesgue's Dominated Convergence Theorem, we obtain that there exists R
>
0 such that
(136)
F
(
x
,
v
n
)
|
x
|
β
⟶
F
(
x
,
u
0
)
|
x
|
β
in
L
loc
⁡
1
(
B
R
)
,
as
n
⟶
∞
.
Claim  1. ∫
ℝ
N
‍
(
F
(
x
,
v
n
)
/
|
x
|
β
)
d
x
→
∫
ℝ
N
(
F
(
x
,
u
0
)
/
|
x
|
β
)
‍
d
x, as n
→
∞. Indeed, by (f2) and (f3), we have
(137)
F
(
x
,
s
)
≤
C
|
s
|
N
+
C
f
(
x
,
s
)
≤
C
|
s
|
N
+
C
R
(
α
0
,
s
)
s
,
∫
ℝ
N
‍
f
(
x
,
v
n
)
v
n
|
x
|
β
d
x
≤
C
,
∫
ℝ
N
‍
F
(
x
,
v
n
)
|
x
|
β
d
x
≤
C
.
Hence, on the domain {
|
x
|
>
R and |
v
n
|
>
A
}, we have
(138)
∫
{
|
x
|
>
R
,
|
v
n
|
>
A
}
‍
F
(
x
,
v
n
)
|
x
|
β
d
x
≤
C
∫
|
x
|
>
R
‍
|
v
n
|
N
|
x
|
β
d
x
+
C
∫
{
|
x
|
>
R
,
|
v
n
|
>
A
}
‍
f
(
x
,
v
n
)
|
x
|
β
d
x
≤
C
R
β
∥
v
n
∥
E
N
+
C
A
∫
ℝ
N
‍
f
(
x
,
v
n
)
v
n
|
x
|
β
d
x
.
Since ∥
v
n
∥
E
N is bounded, and using (137), we have
(139)
∫
{
|
x
|
>
R
,
|
v
n
|
>
A
}
‍
F
(
x
,
v
n
)
|
x
|
β
d
x
≤
2
δ
.
For |
s
|
≤
A, we have
(140)
|
F
(
x
,
s
)
|
≤
C
|
s
|
N
+
C
R
(
α
0
,
s
)
s
≤
|
s
|
N
[
C
+
C
∑
j
=
N
-
1
∞
‍
α
0
j
j
!
A
(
N
j
/
(
N
-
1
)
+
1
-
N
)
]
≤
C
(
α
0
,
A
)
|
s
|
N
.
Since ∥
v
n
∥
E
N is bounded, we have
(141)
∫
{
|
x
|
>
R
,
|
v
n
|
≤
A
}
‍
F
(
x
,
v
n
)
|
x
|
β
d
x
≤
δ
.
Combining (139) and (141), we have
(142)
∫
|
x
|
>
R
‍
F
(
x
,
v
n
)
|
x
|
β
d
x
≤
3
δ
.
Similarly, we also have
(143)
∫
|
x
|
>
R
‍
F
(
x
,
u
0
)
|
x
|
β
d
x
≤
3
δ
.
Combining (136), (142), and (143), we obtain
(144)
|
∫
ℝ
N
‍
F
(
x
,
v
n
)
|
x
|
β
d
x
-
∫
ℝ
N
‍
F
(
x
,
u
0
)
|
x
|
β
d
x
|
≤
|
∫
B
R
‍
F
(
x
,
v
n
)
|
x
|
β
d
x
-
∫
B
R
‍
F
(
x
,
u
0
)
|
x
|
β
d
x
|
+
|
∫
|
x
|
>
R
‍
F
(
x
,
v
n
)
|
x
|
β
d
x
-
∫
|
x
|
>
R
‍
F
(
x
,
u
0
)
|
x
|
β
d
x
|
≤
C
δ
.
Hence, the claim is proved.
Claim  2. I
(
v
n
)
→
I
(
u
0
)
=
C
0
<
0. Indeed, we have
(145)
lim
⁡
n
→
∞
∥
∇
v
n
∥
L
N
(
ℝ
N
)
N
=
N
C
M
-
lim
⁡
n
→
∞
∫
ℝ
N
‍
V
(
x
)
|
v
n
|
N
d
x
+
N
∫
ℝ
N
‍
F
(
x
,
u
0
)
|
x
|
β
d
x
+
λ
∫
ℝ
N
‍
|
v
n
|
N
|
x
|
β
d
x
.
From [10], we have
(146)
|
∫
ℝ
N
‍
f
(
x
,
v
n
)
(
v
n
-
u
0
)
|
x
|
β
d
x
|
⟶
0
,
∫
ℝ
N
|
∇
v
n
|
N
-
2
∇
v
n
(
∇
v
n
-
∇
u
0
)
d
x
+
∫
ℝ
N
‍
V
(
x
)
|
v
n
|
N
-
2
v
n
(
v
n
-
u
0
)
d
x
⟶
0
.
On the other hand, since v
n
→
u
0, we have
(147)
∫
ℝ
N
‍
|
∇
u
0
|
N
-
2
∇
u
0
(
∇
v
n
-
∇
u
0
)
d
x
⟶
0
,
∫
ℝ
N
‍
V
(
x
)
|
u
0
|
N
-
2
u
0
(
v
n
-
u
0
)
d
x
⟶
0
.
By the inequality (
|
x
|
N
-
2
x
-
|
y
|
N
-
2
y
)
(
x
-
y
)
≥
2
2
-
N
|
x
-
y
|
N, we have
(148)
∫
ℝ
N
‍
|
∇
v
n
-
∇
u
0
|
N
d
x
+
∫
ℝ
N
‍
V
(
x
)
|
v
n
-
u
0
|
N
d
x
≤
C
1
∫
ℝ
N
‍
(
|
∇
v
n
|
N
-
2
∇
v
n
-
|
∇
u
0
|
N
-
2
∇
u
0
)
(
v
n
-
u
0
)
d
x
+
C
1
∫
ℝ
N
‍
V
(
x
)
(
|
v
n
|
N
-
2
v
n
-
|
u
0
N
-
2
|
u
0
(
v
n
-
u
0
)
)
d
x
.
Hence, we have v
n
→
u
0 in E and I
(
v
n
)
→
I
(
u
0
)
=
C
0
<
0. It is a contradiction. The proof is complete.


## 3.1. Palais-Smale Sequence

Now, we check that the functionalI satisfies the geometric conditions of the Mountain Pass Theorem.Lemma 13.
Suppose that (H1)–(H3) and (f1)–(f5) hold. Then there existsε
2 such that, for 0
<
ɛ
<
ε
2, there exists ρ
ɛ
>
0 such that I
(
u
)
>
0 if ∥
u
∥
E
=
ρ
ɛ. Furthermore, ρ
ɛ can be chosen such that ρ
ɛ
→
0, as ɛ
→
0.Proof.
From (f5), for everyɛ
>
0, there exists σ
>
0 such that |
u
|
≤
σ implies
(65)
F
(
x
,
u
)
≤
ɛ
N
|
u
|
N
,
∀
x
∈
ℝ
N
.
Moreover, using (f1), for each q
>
N and k
/
α
N
+
β
/
N
+
1
/
q
≤
1, we find a constant C such that
(66)
F
(
x
,
u
)
≤
C
|
u
|
q
[
exp
⁡
(
k
|
u
|
N
/
(
N
-
1
)
)
-
S
N
-
2
(
k
,
u
)
]
,
F
(
x
,
u
)
≤
F
(
x
,
u
)
≤
|
u
|
q
C
|
u
|
q
∀
|
u
|
≥
σ
,
x
∈
ℝ
N
.
Combining (65) and (66), we have
(67)
F
(
x
,
u
)
≤
ɛ
N
|
u
|
N
+
C
|
u
|
q
×
[
exp
⁡
(
k
|
u
|
N
/
(
N
-
1
)
)
-
S
N
-
2
(
k
,
u
)
]
,
∀
(
x
,
u
)
∈
ℝ
N
×
ℝ
.
Since the embedding E
↪
L
N
(
ℝ
N
) is continuous, we obtain
(68)
I
(
u
)
≥
1
N
∥
u
∥
E
N
-
ɛ
+
λ
N
∫
ℝ
N
‍
|
u
|
N
|
x
|
β
d
x
-
C
∥
u
∥
E
q
-
ɛ
∥
h
∥
*
∥
u
∥
E
≥
1
N
(
1
-
λ
+
ɛ
λ
1
)
∥
u
∥
E
N
-
C
∥
u
∥
E
q
-
ɛ
∥
h
∥
*
∥
u
∥
E
.
Thus, we have
(69)
I
(
u
)
≥
∥
u
∥
E
[
1
N
(
1
-
λ
+
ɛ
λ
1
)
∥
u
∥
E
N
-
1
-
C
∥
u
∥
E
q
-
1
-
ɛ
∥
h
∥
*
(
1
-
λ
+
ɛ
λ
1
)
]
.
Since q
>
N and 0
<
λ
<
λ
1 and letting ɛ
<
λ
1
-
λ, we choose ρ
>
0 such that (
1
/
N
)
(
1
-
(
λ
+
ɛ
)
/
λ
1
)
ρ
N
-
1
-
C
ρ
q
-
1
-
ɛ
∥
h
∥
*
>
0. Thus, if ɛ is sufficiently small, we find some ρ
ɛ
>
0 such that I
(
u
)
>
0 if ∥
u
∥
E
=
ρ
ɛ and even ρ
ɛ
→
0 as ɛ
→
0.Lemma 14.
If0
<
λ
<
λ
1, there exists e
∈
E, with ∥
e
∥
E
>
ρ
ɛ, such that I
(
e
)
<
inf
⁡
∥
u
∥
E
=
ρ
ɛ
I
(
u
).Proof.
Letu
∈
E
∖
{
0
}, u
>
0 with compact support Ω
=
supp
⁡
(
u
). By (f4), we obtain that for p
>
N, there exists a positive constant C
>
0 such that for every M
>
0,
(70)
∀
|
u
|
>
C
,
∀
x
∈
Ω
,
F
(
x
,
u
)
≥
M
|
u
|
N
.
Then, we have
(71)
I
(
t
u
)
≤
t
N
N
∥
u
∥
E
N
-
M
t
N
∫
Ω
‍
|
u
|
N
|
x
|
β
d
x
-
ɛ
t
∫
Ω
‍
h
u
d
x
-
λ
N
∫
Ω
‍
|
t
u
|
N
|
x
|
β
d
x
≤
t
N
N
∥
u
∥
E
N
-
t
N
(
M
+
λ
N
)
∫
Ω
‍
|
u
|
N
|
x
|
β
d
x
-
ɛ
t
∫
Ω
‍
h
u
d
x
≤
t
N
(
∥
u
∥
E
N
N
-
(
M
+
λ
N
)
∫
|
x
|
<
R
‍
|
u
|
N
|
x
|
β
d
x
)
.
Choose R
>
0 and B
R
(
0
)
⊂
Ω
¯ and let
(72)
M
>
R
β
∥
u
∥
E
N
N
∥
u
∥
L
N
(
ℝ
N
)
N
-
λ
N
>
R
β
∥
u
∥
E
N
N
∥
u
∥
L
N
(
ℝ
N
)
N
-
λ
1
N
;
we have I
(
t
u
)
→
-
∞ as t
→
∞. Setting e
=
t
u with t being sufficient large, we obtain the conclusion.It is well known that the failure of the (PS) compactness condition creates some difficulties in studying the class of elliptic problems involving critical growth. In Lemma15, instead of (PS) sequence, we analyze the compactness of Cerami sequences of the functional I.Lemma 15.
Let(
u
n
)
⊂
E be a Cerami sequence of I; that is,
(73)
I
(
u
n
)
→
C
M
,
(
1
+
∥
u
n
∥
E
)
∥
D
I
(
u
n
)
∥
E
′
⟶
0
,
I
(
u
n
)
⟶
C
M
,
I
6
→
(
1
+
∥
u
n
∥
E
)
a
s
n
⟶
∞
.
Then there exists a subsequence of (
u
n
) (still denoted by (
u
n
)) and u
∈
E such that
(74)
f
(
x
,
u
n
)
|
x
|
β
⟶
f
(
x
,
u
)
|
x
|
β
,
i
n
L
loc
⁡
1
(
ℝ
N
)
,
∇
u
n
⟶
∇
u
,
a
.
e
.
i
n
ℝ
N
,
|
∇
u
n
|
N
-
2
|
∇
u
n
|
⇀
|
∇
u
|
N
-
2
|
∇
u
|
,
(
L
loc
⁡
N
/
(
N
-
1
)
(
ℝ
N
)
)
N
,
u
n
⇀
u
,
i
n
E
,
where C
M
∈
(
0
,
(
1
/
N
)
(
1
-
β
/
N
)
(
α
N
/
α
0
)
). Furthermore, u is a nontrivial weak solution of problem (1).Proof.
Letu
n
∈
E, v
∈
E, as n
→
∞; we have
(75)
1
N
∫
ℝ
N
‍
|
∇
u
n
|
N
d
x
+
1
N
∫
ℝ
N
‍
V
(
x
)
|
u
n
|
N
d
x
-
∫
ℝ
N
‍
F
(
x
,
u
n
)
|
x
|
β
d
x
-
λ
N
∫
ℝ
N
‍
|
u
n
|
N
|
x
|
β
d
x
-
ɛ
∫
ℝ
N
‍
h
u
n
⟶
C
M
,
(76)
|
D
I
(
u
n
)
v
|
=
|
|
u
n
|
N
-
2
u
n
v
|
x
|
β
∫
ℝ
N
|
∇
u
n
|
N
-
2
∇
u
n
∇
v
d
x
+
∫
ℝ
N
‍
V
(
x
)
|
u
n
|
N
-
2
u
n
v
d
x
-
∫
ℝ
N
‍
f
(
x
,
u
n
)
v
|
x
|
β
d
x
-
λ
∫
ℝ
N
|
u
n
|
N
-
2
u
n
v
|
x
|
β
d
x
-
ε
∫
ℝ
N
h
v
‍
d
x
|
≤
τ
n
∥
v
∥
E
(
1
+
∥
u
n
∥
E
)
,
where τ
n
→
0 as n
→
∞. Let v
=
u
n in (76); we have
(77)
-
∫
ℝ
N
‍
|
∇
u
n
|
N
d
x
-
∫
ℝ
N
‍
V
(
x
)
|
u
n
|
N
d
x
+
∫
ℝ
N
‍
f
(
x
,
u
n
)
u
n
|
x
|
β
d
x
+
λ
∫
ℝ
N
‍
|
u
n
|
N
|
x
|
β
d
x
+
ɛ
∫
ℝ
N
‍
h
u
n
d
x
≤
τ
n
∥
u
n
∥
E
(
1
+
∥
u
n
∥
E
)
⟶
0
,
as
n
⟶
∞
.
Suppose that
(78)
∥
u
n
∥
E
⟶
∞
.
Set
(79)
v
n
=
u
n
∥
u
n
∥
E
;
we have ∥
v
n
∥
E
=
1. From [5], we have
(80)
∫
ℝ
N
‍
liminf
⁡
n
→
+
∞
F
(
x
,
u
n
)
|
x
|
β
|
u
n
+
(
x
)
|
N
|
v
n
+
|
N
d
x
=
+
∞
.
However, since {
u
n
} is the Cerami sequence at the level C
M, we have that
(81)
∥
u
n
∥
E
N
=
N
I
(
u
n
)
+
N
∫
ℝ
N
‍
F
(
x
,
u
n
)
|
x
|
β
d
x
+
N
ɛ
∫
ℝ
N
‍
h
u
n
d
x
+
λ
∫
ℝ
N
‍
|
u
n
|
N
|
x
|
β
d
x
+
o
(
1
)
.
Then there exists some constant C such that
(82)
∥
u
n
∥
E
[
1
-
λ
λ
1
-
N
ɛ
∥
h
∥
*
]
≤
N
C
M
+
N
∫
ℝ
N
‍
F
(
x
,
u
n
)
|
x
|
β
d
x
+
o
(
1
)
,
as
ɛ
⟶
0
,
which implies that
(83)
∫
ℝ
N
‍
F
(
x
,
u
n
)
|
x
|
β
d
x
⟶
∞
,
liminf
⁡
n
→
∞
∫
ℝ
N
‍
F
(
x
,
u
n
)
|
x
|
β
|
u
n
+
|
N
|
v
n
+
|
N
d
x
=
liminf
⁡
n
→
∞
∫
ℝ
N
‍
F
(
x
,
u
n
)
|
x
|
β
∥
u
n
+
∥
E
N
d
x
.
Let Ψ
=
∫
ℝ
N
‍
(
F
(
x
,
u
n
)
/
|
x
|
β
)
d
x; then we have
(84)
liminf
⁡
n
→
∞
∫
ℝ
N
‍
F
(
x
,
u
n
)
|
x
|
β
|
u
n
+
|
N
|
v
n
+
|
N
d
x
≤
liminf
⁡
n
→
∞
Ψ
N
C
M
+
N
Ψ
+
N
ɛ
∫
ℝ
N
‍
h
u
n
+
d
x
+
o
(
1
)
.
So we can conclude that
(85)
liminf
⁡
n
→
∞
∫
ℝ
N
‍
F
(
x
,
u
n
)
|
x
|
β
|
u
n
+
|
N
|
v
n
+
|
N
d
x
≤
liminf
⁡
n
→
∞
Ψ
N
C
M
+
N
Ψ
+
N
ɛ
∫
ℝ
N
‍
h
u
n
+
d
x
+
o
(
1
)
=
1
N
.
Note that F
(
x
,
u
n
)
≥
0; by Fatou Lemma, (80), and (85), we get a contradiction. So v
≤
0 which means that v
k
+
⇀
0 in E.
Lett
n
∈
[
0,1
], such that
(86)
I
(
t
n
u
n
)
=
max
⁡
t
∈
[
0,1
]
I
(
t
u
n
)
.
For any given R
∈
(
0
,
(
1
-
β
/
N
)
α
N
/
α
0
)
(
N
-
1
)
/
N, let ɛ
=
(
1
-
β
/
N
)
α
N
/
R
N
/
(
N
-
1
)
-
α
0
>
0, by (f1); there exists C
=
C
(
R
)
>
0 such that
(87)
F
(
x
,
u
n
)
≤
C
|
u
n
|
N
+
|
(
1
-
β
/
N
)
α
N
R
N
/
(
N
-
1
)
-
α
0
|
R
(
α
0
+
ɛ
,
u
n
)
,
(
x
,
u
n
)
≤
|
u
n
|
N
+
|
(
-
β
/
N
)
α
N
R
N
/
(
N
-
1
)
-
α
0
|
∀
(
x
,
u
n
)
∈
Ω
¯
×
ℝ
N
.
Since ∥
u
n
∥
E
→
∞, we have
(88)
I
(
t
n
u
n
)
≥
I
(
R
u
n
∥
u
n
∥
E
)
=
I
(
R
v
n
)
,
and by (f3), ∥
v
n
∥
E
=
1, and ∫
ℝ
N
‍
(
F
(
x
,
v
n
+
)
/
|
x
|
β
)
d
x
=
∫
ℝ
N
‍
(
F
(
x
,
v
n
)
/
|
x
|
β
)
d
x, we have
(89)
N
I
(
R
v
n
)
≥
R
N
-
(
N
C
R
N
+
λ
R
N
)
∫
ℝ
N
‍
|
v
n
+
|
N
|
x
|
β
d
x
-
N
|
(
1
-
β
/
N
)
α
N
R
N
/
(
N
-
1
)
-
α
0
|
∫
ℝ
N
‍
R
(
α
0
+
ɛ
,
R
|
v
n
+
|
)
|
x
|
β
d
x
-
N
R
|
(
1
-
β
/
N
)
α
N
R
N
/
(
N
-
1
)
-
α
0
|
∫
ℝ
N
‍
h
|
v
n
+
|
d
x
≥
R
N
-
(
N
C
R
N
+
λ
R
N
)
∫
ℝ
N
‍
|
v
n
+
|
N
|
x
|
β
d
x
-
N
|
(
1
-
β
/
N
)
α
N
R
N
/
(
N
-
1
)
-
α
0
|
×
∫
ℝ
N
‍
R
(
(
α
0
+
ɛ
)
R
N
/
(
N
-
1
)
,
|
v
n
+
|
)
|
x
|
β
d
x
-
N
R
|
(
1
-
β
/
N
)
α
N
R
N
/
(
N
-
1
)
-
α
0
|
∫
ℝ
N
‍
h
|
v
n
+
|
d
x
≥
R
N
-
(
N
C
R
N
+
λ
R
N
)
∫
ℝ
N
‍
|
v
n
+
|
N
|
x
|
β
d
x
-
N
|
(
1
-
β
/
N
)
α
N
R
N
/
(
N
-
1
)
-
α
0
|
×
∫
ℝ
N
‍
R
(
(
1
-
β
/
N
)
α
N
,
|
v
n
+
|
)
|
x
|
β
d
x
-
N
R
|
(
1
-
β
/
N
)
α
N
R
N
/
(
N
-
1
)
-
α
0
|
∫
ℝ
N
‍
h
|
v
n
+
|
d
x
.
Since v
n
+
⇀
0 in E and the embedding E
↪
L
N
(
ℝ
N
) is compact from and the Hölder inequality, we have ∫
ℝ
N
‍
(
|
v
n
+
|
N
/
|
x
|
β
)
d
x
→
0
(
n
→
∞
). By Lemma 7, we have ∫
ℝ
N
‍
(
R
(
(
1
-
β
/
N
)
α
N
,
|
v
n
+
|
)
/
|
x
|
β
)
d
x
≤
C.
Letn
→
∞ in (89) and R
→
[
(
1
-
β
/
N
)
α
N
/
α
0
]
(
N
-
1
)
/
N; we get
(90)
liminf
⁡
n
→
∞
I
(
t
n
u
n
)
≥
1
N
[
(
1
-
β
/
N
)
α
N
α
0
]
N
-
1
>
C
M
.
Note that I
(
0
)
=
0 and I
(
u
n
)
→
C
M; we suppose that t
n
∈
(
0,1
). Since D
I
(
t
n
u
n
)
t
n
u
n
=
0, we have
(91)
t
n
N
∥
u
n
∥
E
N
=
∫
ℝ
N
‍
f
(
x
,
t
n
u
n
)
t
n
u
n
|
x
|
β
d
x
+
ɛ
t
n
∫
ℝ
N
‍
h
u
n
d
x
+
λ
t
n
N
∫
ℝ
N
‍
|
u
n
|
N
|
x
|
β
d
x
.
By (f2) and ε
n
→
0, we have
(92)
N
I
(
t
n
u
n
)
=
∫
ℝ
N
‍
f
(
x
,
t
n
u
n
)
t
n
u
n
|
x
|
β
d
x
-
N
∫
ℝ
N
‍
F
(
x
,
t
n
u
n
)
|
x
|
β
d
x
+
o
(
1
)
.
Moreover, we have
(93)
∫
ℝ
N
‍
f
(
x
,
u
n
)
u
n
|
x
|
β
d
x
-
N
∫
ℝ
N
‍
F
(
x
,
u
n
)
|
x
|
β
d
x
=
∥
u
n
∥
E
N
+
N
C
M
-
∥
u
n
∥
E
N
+
o
(
1
)
,
which is a contraction to (75); this proves that {
u
n
} is bounded in E. Thus, we have
(94)
∫
ℝ
N
‍
f
(
x
,
u
n
)
u
n
|
x
|
β
d
x
≤
C
,
∫
ℝ
N
‍
F
(
x
,
u
n
)
|
x
|
β
d
x
≤
C
.
From Lemma 5, the embedding E
↪
L
q
(
ℝ
N
) is compact for all q
≥
N. If {
u
n
}
∈
E, we get
(95)
u
n
⇀
u
,
in
E
,
u
n
⟶
u
,
in
L
loc
⁡
q
(
ℝ
N
)
,
u
n
⟶
u
,
a
.
e
.
in
ℝ
N
.
From (f1), the Trudinger-Moser inequality, and the Hölder inequality, we have f
(
x
,
u
n
)
/
|
x
|
β
∈
L
loc
⁡
1
(
ℝ
N
). From Lemma 2.1 in [14], we have
(96)
f
(
x
,
u
n
)
|
x
|
β
⟶
f
(
x
,
u
)
|
x
|
β
,
in
L
loc
⁡
1
(
ℝ
N
)
.
For any fixed δ
>
0, set
(97)
Σ
δ
=
{
x
∈
ℝ
N
:
lim
⁡
r
→
0
lim
⁡
n
→
∞
∫
𝔹
r
(
x
)
(
|
∇
u
n
|
N
+
|
u
n
|
N
)
‍
d
x
≥
δ
∫
𝔹
r
(
x
)
(
|
∇
u
n
|
N
+
|
u
n
|
N
)
‍
}
.
Because {
u
n
} is bounded,Σ
δ is a finite set. From Lemma 4.4 in ([4]), for any compact set K
⊂
⊂
ℝ
N
∖
Σ
δ, we have
(98)
lim
⁡
n
→
∞
∫
K
‍
|
f
(
x
,
u
n
)
u
n
-
f
(
x
,
u
)
u
|
|
x
|
β
d
x
=
0
.
Now, we prove that
(99)
lim
⁡
n
→
∞
∫
K
‍
|
∇
u
n
-
∇
u
|
N
d
x
=
0
.
It is enough to prove that for any x
′
∈
ℝ
N
∖
Σ
δ and B
r
(
x
′
)
⊂
ℝ
N
∖
Σ
δ there holds
(100)
lim
⁡
n
→
∞
∫
𝔹
r
/
2
(
x
′
)
‍
|
∇
u
n
-
∇
u
|
N
d
x
=
0
.
We take ϕ
∈
C
0
∞
(
B
r
(
x
′
)
) with 0
≤
ϕ
≤
1 and ϕ
=
1 on 𝔹
r
/
2
(
x
′
). Then {
ϕ
u
n
} is a bounded sequence. Choosing v
n
=
ϕ
u
n and v
=
ϕ
u in (76), we have
(101)
∫
B
r
(
x
′
)
‍
ϕ
(
|
∇
u
n
|
N
-
2
∇
u
n
-
|
∇
u
|
N
-
2
∇
u
)
(
∇
u
n
-
∇
u
)
d
x
≤
∫
B
r
(
x
′
)
‍
|
∇
u
n
|
N
-
2
∇
u
n
∇
ϕ
(
u
-
u
n
)
d
x
+
∫
B
r
(
x
′
)
‍
ϕ
|
∇
u
|
N
-
2
∇
u
(
∇
u
-
∇
u
n
)
d
x
+
∫
B
r
(
x
′
)
‍
ϕ
(
u
n
-
u
)
f
(
x
,
u
n
)
|
x
|
β
d
x
+
τ
n
∥
ϕ
u
n
∥
E
+
τ
n
∥
ϕ
u
∥
E
+
ɛ
∫
B
r
(
x
′
)
‍
ϕ
h
(
u
n
-
u
)
d
x
+
λ
∫
B
r
(
x
′
)
‍
|
u
n
|
u
n
ϕ
|
x
|
β
(
u
n
-
u
)
d
x
.
Adapting an argument similar to [4], we have
(102)
lim
⁡
n
→
∞
∫
K
‍
|
∇
u
n
-
∇
u
|
N
d
x
=
0
.
Since Σ
δ is finite, it follows that ∇
u
n
→
∇
u a.e. This implies, up to a subsequence that |
∇
u
n
|
N
-
2
∇
u
n
⇀
|
∇
u
|
N
-
2
∇
u in (
L
loc
⁡
N
/
(
N
-
1
)
(
ℝ
N
)
)
N. Let n
→
∞ in (76), and f
(
x
,
u
n
)
/
|
x
|
β
→
f
(
x
,
u
)
/
|
x
|
β in L
loc
⁡
1
(
ℝ
N
); we obtain
(103)
〈
D
I
(
u
)
,
v
〉
=
0
,
∀
v
∈
C
0
∞
(
ℝ
N
)
.Remark 16.
The idea and proof of Lemma15 follow as in Lemma 4.1 in [5].


## 3.2. Min-Max Value

In order to get a more precise information about the minimax level obtained by the Mountain Pass Theorem, we consider the following sequence of scale which is called the Moser function:(104)
M
l
(
x
,
r
)
~
=
1
w
N
-
1
1
/
N
{
(
log
⁡
l
)
(
N
-
1
)
/
N
,
if
|
x
|
≤
r
l
,
log
⁡
⁡
(
r
/
|
x
|
)
(
log
⁡
l
)
1
/
N
,
if
r
l
≤
|
x
|
≤
r
,
0
,
if
|
x
|
≥
r
.
Hence, we have M
l
(
x
,
r
)
~
∈
W
1
,
N
(
ℝ
N
), the support of M
l
(
x
,
r
)
~ is the ball B
r
(
0
), and
(105)
∫
ℝ
N
‍
|
∇
M
l
(
x
,
r
)
~
|
N
d
x
=
1
,
∫
ℝ
N
‍
|
M
l
(
x
,
r
)
~
|
N
d
x
=
o
(
1
log
⁡
l
)
.LetM
l
(
x
,
r
)
=
M
l
(
x
,
r
)
~
/
∥
M
l
(
x
,
r
)
~
∥
E
N; we have
(106)
M
l
N
/
(
N
-
1
)
(
x
,
r
)
=
w
N
-
1
-
1
/
(
N
-
1
)
log
⁡
l
+
d
l
,
for
|
x
|
≤
r
l
,
where
(107)
d
l
=
w
N
-
1
-
1
/
(
N
-
1
)
log
⁡
l
(
∥
M
l
(
x
,
r
)
~
∥
E
-
1
/
(
N
-
1
)
-
1
)
.
From (105), we conclude that ∥
M
l
(
x
,
r
)
∥
E
→
1, as l
→
∞. Consequently, we have
(108)
d
l
log
⁡
l
⟶
0
,
as
l
⟶
∞
.Lemma 17.
Suppose that (H1)–(H3) and (f1)–(f6) hold. Then there existsk
∈
ℕ such that
(109)
max
⁡
t
≥
0
{
t
N
N
-
∫
ℝ
N
‍
F
(
x
,
t
M
k
)
|
x
|
β
d
x
-
λ
N
∫
ℝ
N
‍
|
t
M
k
|
N
|
x
|
β
d
x
}
<
1
N
(
N
-
β
N
α
N
α
0
)
N
-
1
.Proof.
Chooser
>
0 as in (
f
6
) and β
0
>
0 such that
(110)
lim
⁡
s
→
∞
s
f
(
x
,
s
)
exp
⁡
(
-
α
0
|
s
|
N
/
(
N
-
1
)
)
≥
β
0
>
2
e
(
α
N
d
(
N
-
β
)
/
N
)
+
C
r
N
-
β
-
r
N
-
β
/
(
N
-
β
)
×
(
N
-
β
α
0
)
N
-
1
,
where
(111)
C
=
lim
⁡
k
→
∞
ξ
k
log
⁡
k
∫
0
ξ
k
‍
exp
⁡
[
s
N
/
(
N
-
1
)
(
N
-
β
)
×
log
⁡
k
(
s
N
/
(
N
-
1
)
-
ξ
k
s
)
]
d
s
>
0
,
ξ
k
=
∥
M
k
~
∥
E
,
C
≥
1
-
e
-
(
N
-
β
)
log
⁡
n
N
-
β
.
Suppose, by contradiction, that, for all k, we get
(112)
max
⁡
t
≥
0
{
t
N
N
-
∫
ℝ
N
‍
F
(
x
,
t
M
k
)
|
x
|
β
d
x
-
λ
N
∫
ℝ
N
‍
|
t
M
k
|
N
|
x
|
β
d
x
}
≥
1
N
(
(
N
-
β
)
N
α
N
α
0
)
N
-
1
,
where M
k
(
x
)
=
M
k
(
x
,
r
). For each k, there exists t
k
>
0 such that
(113)
t
k
N
N
-
∫
ℝ
N
‍
F
(
x
,
t
k
M
k
)
|
x
|
β
d
x
-
λ
N
∫
ℝ
N
‍
|
t
k
M
k
|
N
|
x
|
β
d
x
=
max
⁡
t
≥
0
{
t
N
N
-
∫
ℝ
N
‍
F
(
x
,
t
M
k
)
|
x
|
β
d
x
-
λ
N
∫
ℝ
N
‍
|
t
M
k
|
N
|
x
|
β
d
x
}
.
Thus; we have
(114)
t
k
N
N
-
∫
ℝ
N
‍
F
(
x
,
t
k
M
k
)
|
x
|
β
d
x
-
λ
N
∫
ℝ
N
‍
|
t
k
M
k
|
N
|
x
|
β
d
x
≥
1
N
(
(
N
-
β
)
N
α
N
α
0
)
N
-
1
.
From F
(
x
,
u
)
≥
0, 0
<
λ
<
λ
1, we obtain
(115)
t
k
N
≥
(
N
-
β
N
α
N
α
0
)
N
-
1
.
Let t
=
t
k, we have
(116)
t
k
N
=
∫
ℝ
N
(
t
k
M
k
f
(
x
,
t
k
M
k
)
/
|
x
|
β
)
‍
d
x
(
1
-
λ
∫
|
x
|
≤
r
‍
(
|
M
k
|
N
/
|
x
|
β
)
d
x
)
≥
∫
|
x
|
≤
r
‍
(
t
k
M
k
f
(
x
,
t
k
M
k
)
/
|
x
|
β
)
d
x
1
+
λ
/
λ
1
.
By (f6), given that τ
>
0, there exist R
τ
>
0 and |
x
|
≤
r; we have
(117)
u
f
(
x
,
u
)
≥
(
β
0
-
τ
)
exp
⁡
(
α
0
|
u
|
N
/
(
N
-
1
)
)
.
From (116) and (117), for large k, we obtain
(118)
t
k
N
≥
λ
1
(
β
0
-
τ
)
∫
|
x
|
≤
r
/
k
‍
(
exp
⁡
⁡
(
α
0
|
t
k
M
k
|
N
/
(
N
-
1
)
)
/
|
x
|
β
)
d
x
(
λ
1
+
λ
)
≥
(
β
0
-
τ
)
k
-
N
+
β
w
N
-
1
r
N
-
β
2
(
N
-
β
)
×
exp
⁡
(
α
0
t
k
N
/
(
N
-
1
)
w
N
-
1
-
1
/
(
N
-
1
)
log
⁡
k
+
α
0
t
k
N
/
(
N
-
1
)
d
k
)
.
Let
(119)
L
k
=
α
0
N
log
⁡
k
α
N
t
k
N
/
(
N
-
1
)
+
α
0
t
k
N
/
(
N
-
1
)
d
k
-
(
N
-
β
)
log
⁡
k
;
we have
(120)
1
≥
(
β
0
-
τ
)
r
N
-
β
w
N
-
1
2
(
N
-
β
)
exp
⁡
L
(
k
)
.
Hence, the sequence {
t
k
} is bounded. Otherwise, up to subsequences, we have lim
⁡
k
→
∞
L
(
k
)
=
∞, which leads to a contradiction. From (108) and (115) and
(121)
t
k
N
≥
(
β
0
-
τ
)
r
N
-
β
w
N
-
1
2
(
N
-
β
)
×
exp
⁡
[
(
N
α
0
t
k
N
/
(
N
-
1
)
α
N
-
(
N
-
β
)
)
log
⁡
k
+
α
0
t
k
N
/
(
N
-
1
)
d
k
(
N
α
0
t
k
N
/
(
N
-
1
)
α
N
-
(
N
-
β
)
)
]
,
it follows that
(122)
t
k
N
⟶
(
(
N
-
β
)
N
α
N
α
0
)
N
-
1
.
From [4], we have
(123)
∫
|
x
|
≤
r
‍
exp
⁡
⁡
(
α
0
|
t
k
M
k
|
N
/
(
N
-
1
)
)
|
x
|
β
d
x
≥
∫
|
x
|
≤
r
/
k
‍
exp
⁡
(
α
N
|
M
k
|
N
/
(
N
-
1
)
(
N
-
β
)
/
N
)
|
x
|
β
d
x
+
∫
r
/
k
≤
|
x
|
≤
r
‍
exp
⁡
(
α
N
|
M
k
|
N
/
(
N
-
1
)
(
N
-
β
)
/
N
)
|
x
|
β
d
x
,
∫
|
x
|
≤
r
/
k
‍
exp
⁡
(
α
N
|
M
k
|
N
/
(
N
-
1
)
(
N
-
β
)
/
N
)
|
x
|
β
d
x
=
w
N
-
1
N
-
β
r
N
-
β
k
[
N
-
β
+
(
d
k
α
N
/
log
⁡
k
)
(
N
-
β
)
/
N
]
.
Now, using the change of variable
(124)
s
=
log
⁡
⁡
(
r
/
|
x
|
)
ζ
k
log
⁡
k
with
ζ
k
=
∥
M
k
∥
E
,
by straight forward computation, we have
(125)
∫
r
/
k
≤
|
x
|
≤
r
‍
exp
⁡
(
α
N
|
M
k
|
N
/
(
N
-
1
)
(
N
-
β
)
/
N
)
|
x
|
β
d
x
=
w
N
-
1
r
N
-
β
ζ
k
log
⁡
k
×
∫
0
ζ
k
-
1
‍
exp
⁡
[
(
N
-
β
)
log
⁡
k
(
s
N
/
(
N
-
1
)
-
ζ
k
s
)
]
d
s
,
which converges to C
w
N
-
1
r
N
-
β as k
→
∞, where
(126)
C
=
lim
⁡
k
→
∞
ζ
k
log
⁡
k
∫
0
ξ
k
‍
exp
⁡
[
(
s
N
/
(
N
-
1
)
-
ξ
k
s
)
(
N
-
β
)
×
log
⁡
k
(
s
N
/
(
N
-
1
)
-
ξ
k
s
)
]
d
s

>

0
.
Finally, let k
→
∞ in (118); from (108) and (115), we have
(127)
(
(
N
-
β
)
N
α
N
α
0
)
N
-
1
≥
(
β
0
-
τ
)
2
×
[
r
N
-
β
w
N
-
1
(
N
-
β
)
e
(
α
N
d
(
N
-
β
)
/
N
)
,
+
C
r
N
-
β
w
N
-
1
-
r
N
-
β
w
N
-
1
N
-
β
]
which implies that
(128)
β
0
≤
2
e
(
α
N
d
(
N
-
β
)
/
N
)
+
C
r
N
-
β
-
r
N
-
β
/
(
N
-
β
)
×
(
N
-
β
α
0
)
N
-
1
.Remark 18.
The idea and the proof of Lemma17 come from Lemma 3.6 in [5].Lemma 19.
There existτ
>
0 and v
∈
E with ∥
v
∥
E
=
1 such that I
(
t
v
)
<
0 for all 0
<
t
<
ς. In particular, inf
⁡
∥
v
∥
E
≤
ς
I
(
u
)
<
0.Proof.
See Lemma 3.3 in [10].Corollary 20.
Under the conditions (H1)–(H3) and (f1)–(f4), ifɛ
→
0, then one has
(129)
max
⁡
t
≥
0
I
(
t
M
k
)
<
1
N
(
(
N
-
β
)
N
α
N
α
0
)
N
-
1
.From Lemmas13 and 19, we conclude that
(130)
∞
<
C
0
=
inf
⁡
∥
v
∥
E
≤
ς
I
(
u
)
<
0
.Corollary 21.
There existε
2
∈
(
0
,
ε
1
] and u
∈
W
1
,
N
(
ℝ
N
) with compact support such that, for all 0
<
ɛ
<
ε
2,
(131)
I
(
t
u
)
<
C
0
+
1
N
(
(
N
-
β
)
N
α
N
α
0
)
N
-
1
,
∀
t
≥
0
.Lemma 22.
If{
u
k
} is a Cerami sequence for I
(
u
) at any level with
(132)
liminf
⁡
n
→
∞
∥
u
k
∥
E
<
(
(
N
-
β
)
N
α
N
α
0
)
N
-
1
,
then {
u
k
} possesses a subsequence which converges strongly to a solution u
0 of problem (1).Proof.
See Lemma 4.6 in [4].
In conclusion, we have(133)
0
<
C
M
<
C
0
+
1
N
(
(
N
-
β
)
N
α
N
α
0
)
N
-
1
.


## 3.3. Multiplicity Results

In order to prove the existence of the second solution of problem (1) follows by the minimum argument and Ekeland's Variational Principle.Proposition 23.
Under the conditions (H1)–(H3) and (f1)–(f6), there existsε
1
>
0 such that, for each ɛ with 0
<
ɛ
<
ε
1, problem (1) has a solution u
M via Mountain Pass Theorem.Proof.
See Proposition 4.1 in [5].Proposition 24.
There existsε
2
>
0 such that, for each ɛ with 0
<
ɛ
<
ε
2, (1) has a minimum type solution u
0 with I
(
u
0
)
=
C
0
<
0, where C
0 is defined in (130).Proof.
See Proposition 5.1 in [5].Proposition 25.
Ifε
2
>
0 is sufficiently small, then the solutions of problem (1) obtained in Propositions 23 and 24 are distinct.Proof.
By Propositions23 and 24, there exist sequences {
u
n
}, {
v
n
} in E such that
(134)
u
n
⟶
u
0
,
I
(
u
n
)
⟶
C
0
,
D
I
(
u
n
)
u
n
⟶
0
,
v
n
⇀
u
M
,
I
(
v
n
)
⟶
C
M
>
0
,
D
I
(
v
n
)
v
n
⟶
0
,
∇
v
n
⟶
∇
u
M
a
.
e
.
in
ℝ
N
.
Suppose by contradiction that u
0
=
u
M. As in the proof of Lemma 15, we have
(135)
f
(
x
,
v
n
)
|
x
|
β
⟶
f
(
x
,
u
0
)
|
x
|
β
in
L
loc
⁡
1
(
ℝ
N
)
,
as
n
⟶
∞
.
Hence, by (f2) and (f3) and Generalized Lebesgue's Dominated Convergence Theorem, we obtain that there exists R
>
0 such that
(136)
F
(
x
,
v
n
)
|
x
|
β
⟶
F
(
x
,
u
0
)
|
x
|
β
in
L
loc
⁡
1
(
B
R
)
,
as
n
⟶
∞
.
Claim  1. ∫
ℝ
N
‍
(
F
(
x
,
v
n
)
/
|
x
|
β
)
d
x
→
∫
ℝ
N
(
F
(
x
,
u
0
)
/
|
x
|
β
)
‍
d
x, as n
→
∞. Indeed, by (f2) and (f3), we have
(137)
F
(
x
,
s
)
≤
C
|
s
|
N
+
C
f
(
x
,
s
)
≤
C
|
s
|
N
+
C
R
(
α
0
,
s
)
s
,
∫
ℝ
N
‍
f
(
x
,
v
n
)
v
n
|
x
|
β
d
x
≤
C
,
∫
ℝ
N
‍
F
(
x
,
v
n
)
|
x
|
β
d
x
≤
C
.
Hence, on the domain {
|
x
|
>
R and |
v
n
|
>
A
}, we have
(138)
∫
{
|
x
|
>
R
,
|
v
n
|
>
A
}
‍
F
(
x
,
v
n
)
|
x
|
β
d
x
≤
C
∫
|
x
|
>
R
‍
|
v
n
|
N
|
x
|
β
d
x
+
C
∫
{
|
x
|
>
R
,
|
v
n
|
>
A
}
‍
f
(
x
,
v
n
)
|
x
|
β
d
x
≤
C
R
β
∥
v
n
∥
E
N
+
C
A
∫
ℝ
N
‍
f
(
x
,
v
n
)
v
n
|
x
|
β
d
x
.
Since ∥
v
n
∥
E
N is bounded, and using (137), we have
(139)
∫
{
|
x
|
>
R
,
|
v
n
|
>
A
}
‍
F
(
x
,
v
n
)
|
x
|
β
d
x
≤
2
δ
.
For |
s
|
≤
A, we have
(140)
|
F
(
x
,
s
)
|
≤
C
|
s
|
N
+
C
R
(
α
0
,
s
)
s
≤
|
s
|
N
[
C
+
C
∑
j
=
N
-
1
∞
‍
α
0
j
j
!
A
(
N
j
/
(
N
-
1
)
+
1
-
N
)
]
≤
C
(
α
0
,
A
)
|
s
|
N
.
Since ∥
v
n
∥
E
N is bounded, we have
(141)
∫
{
|
x
|
>
R
,
|
v
n
|
≤
A
}
‍
F
(
x
,
v
n
)
|
x
|
β
d
x
≤
δ
.
Combining (139) and (141), we have
(142)
∫
|
x
|
>
R
‍
F
(
x
,
v
n
)
|
x
|
β
d
x
≤
3
δ
.
Similarly, we also have
(143)
∫
|
x
|
>
R
‍
F
(
x
,
u
0
)
|
x
|
β
d
x
≤
3
δ
.
Combining (136), (142), and (143), we obtain
(144)
|
∫
ℝ
N
‍
F
(
x
,
v
n
)
|
x
|
β
d
x
-
∫
ℝ
N
‍
F
(
x
,
u
0
)
|
x
|
β
d
x
|
≤
|
∫
B
R
‍
F
(
x
,
v
n
)
|
x
|
β
d
x
-
∫
B
R
‍
F
(
x
,
u
0
)
|
x
|
β
d
x
|
+
|
∫
|
x
|
>
R
‍
F
(
x
,
v
n
)
|
x
|
β
d
x
-
∫
|
x
|
>
R
‍
F
(
x
,
u
0
)
|
x
|
β
d
x
|
≤
C
δ
.
Hence, the claim is proved.
Claim  2. I
(
v
n
)
→
I
(
u
0
)
=
C
0
<
0. Indeed, we have
(145)
lim
⁡
n
→
∞
∥
∇
v
n
∥
L
N
(
ℝ
N
)
N
=
N
C
M
-
lim
⁡
n
→
∞
∫
ℝ
N
‍
V
(
x
)
|
v
n
|
N
d
x
+
N
∫
ℝ
N
‍
F
(
x
,
u
0
)
|
x
|
β
d
x
+
λ
∫
ℝ
N
‍
|
v
n
|
N
|
x
|
β
d
x
.
From [10], we have
(146)
|
∫
ℝ
N
‍
f
(
x
,
v
n
)
(
v
n
-
u
0
)
|
x
|
β
d
x
|
⟶
0
,
∫
ℝ
N
|
∇
v
n
|
N
-
2
∇
v
n
(
∇
v
n
-
∇
u
0
)
d
x
+
∫
ℝ
N
‍
V
(
x
)
|
v
n
|
N
-
2
v
n
(
v
n
-
u
0
)
d
x
⟶
0
.
On the other hand, since v
n
→
u
0, we have
(147)
∫
ℝ
N
‍
|
∇
u
0
|
N
-
2
∇
u
0
(
∇
v
n
-
∇
u
0
)
d
x
⟶
0
,
∫
ℝ
N
‍
V
(
x
)
|
u
0
|
N
-
2
u
0
(
v
n
-
u
0
)
d
x
⟶
0
.
By the inequality (
|
x
|
N
-
2
x
-
|
y
|
N
-
2
y
)
(
x
-
y
)
≥
2
2
-
N
|
x
-
y
|
N, we have
(148)
∫
ℝ
N
‍
|
∇
v
n
-
∇
u
0
|
N
d
x
+
∫
ℝ
N
‍
V
(
x
)
|
v
n
-
u
0
|
N
d
x
≤
C
1
∫
ℝ
N
‍
(
|
∇
v
n
|
N
-
2
∇
v
n
-
|
∇
u
0
|
N
-
2
∇
u
0
)
(
v
n
-
u
0
)
d
x
+
C
1
∫
ℝ
N
‍
V
(
x
)
(
|
v
n
|
N
-
2
v
n
-
|
u
0
N
-
2
|
u
0
(
v
n
-
u
0
)
)
d
x
.
Hence, we have v
n
→
u
0 in E and I
(
v
n
)
→
I
(
u
0
)
=
C
0
<
0. It is a contradiction. The proof is complete.



---

*Source: 942092-2014-01-23.xml*