# Infinitely Many Solutions for a Superlinear Fractionalp-Kirchhoff-Type Problem without the (AR) Condition

Authors: Xiangsheng Ren; Jiabin Zuo; Zhenhua Qiao; Lisa Zhu
Journal: Advances in Mathematical Physics
 (2019)

---

## Abstract

In this paper, we investigate the existence of infinitely many solutions to a fractionalp-Kirchhoff-type problem satisfying superlinearity with homogeneous Dirichlet boundary conditions as follows: [
a
+
b
(
∫
R
2
N
u
x
-
u
y
p
K
x
-
y
d
x
d
y
)
]
L
p
s
u
-
λ
|
u
|
p
-
2
u
=
g
x
,
u
,
i
n
Ω
,
u
=
0
,
i
n
R
N
∖
Ω
, where L
p
s is a nonlocal integrodifferential operator with a singular kernel K. We only consider the non-Ambrosetti-Rabinowitz condition to prove our results by using the symmetric mountain pass theorem.


---

## Body

## 1. Introduction

In recent years, the problems with fractional and nonlocal operator have attracted a lot of attention. These types of operators arise in many different contexts. We know that there are population dynamics, stratified materials, minimal surface, water waves, continuum mechanics, and so on. As far as we know, we are able to learn more about their association through referring to [1–6].The problem we are going to deal with also involves fractional and nonlocal operator. Here, we will study thep-Kirchhoff-type problem as follows:(1)
a
+
b
∫
R
2
N
u
x
-
u
y
p
K
x
-
y
d
x
d
y
L
p
s
u
-
λ
u
p
-
2
u
=
g
x
,
u
i
n
Ω
,
u
=
0
i
n
R
N
∖
Ω
.
Ω is an open bounded smooth domain in R
N with Lipschitz boundary ∂
Ω. a
,
b
>
0,1
<
p
<
+
∞; p
s
<
N with s
∈
(
0,1
). λ is a real parameter. g: Ω
×
R
→
R is a Carathéodory function and L
p
s is usually called nonlocal operator. It is defined as follows: (2)
L
p
s
u
x
≔
2
l
i
m
ε
→
0
+
∫
R
N
∖
B
ε
x
u
x
-
u
y
p
-
2
u
x
-
u
y
K
x
-
y
d
y
,for all x
∈
R
N, where B
ε
(
x
)
=
{
z
∣
x
-
z
<
ε
}
. The function K
:
R
N
∖
{
0
}
→
(
0
,
∞
) is measurable. It has the following properties:(3)
l
K
∈
L
1
R
N
w
h
e
r
e
l
x
=
min
⁡
x
p
,
1
,
t
h
e
r
e
e
x
i
s
t
s
γ
>
0
s
u
c
h
t
h
a
t
K
x
≥
γ
x
-
N
+
p
s
f
o
r
a
n
y
x
∈
R
N
∖
0
,
K
x
=
K
-
x
f
o
r
a
n
y
x
∈
R
N
∖
0
.As the singular kernelK satisfies K
(
x
)
=
x
-
(
N
+
p
s
), we call it a typical model. Hence, the fractional p-Laplace operator may be defined as follows: (4)
-
△
p
s
u
x
≔
2
l
i
m
ε
→
0
+
∫
R
N
∖
B
ε
x
u
x
-
u
y
p
-
2
u
x
-
u
y
x
-
y
N
+
p
s
d
y
, for x
∈
R
N. Problem (1) also becomes(5)
a
+
b
∫
R
2
N
u
x
-
u
y
p
x
N
+
p
s
d
x
d
y
-
△
p
s
u
x
-
λ
u
p
-
2
u
=
g
x
,
u
i
n
Ω
,
u
=
0
i
n
R
N
∖
Ω
.Usually, we write the Kirchhoff function asM. Clearly, M
(
t
)
=
a
+
b
t in problem (5). When M
=
1
,
p
=
2
,
λ
=
0, problem (5) becomes the original problem with the following fractional Laplacian form:(6)
-
△
s
u
=
g
x
,
u
i
n
Ω
,
u
=
0
i
n
R
N
∖
Ω
.It is the nonlocality that is a typical characteristic of problem (6). In other words, the value of (
-
△
)
s
u
(
x
) at any point x
∈
Ω relies not only on Ω but actually also on the whole space. We know that the Dirichlet boundary condition was applied to problem (
6
) in [5]. In [7], through the use of the mountain pass theorem, Servadei and Valdinoci obtained the existence of nontrivial weak solutions of problem (6). In [8], Pucci and Saldi studied the Kirchhoff-type eigenvalue problem in whole space. They proved the existence and multiplicity of nontrivial solutions. We also refer to [9] for related problems.On the other hand, the Kirchhoff functionM is transformable. So far, a variety of forms of function M are taken into account in many references on studying Kirchhoff-type problems; see [10–18]. In addition, we notice that more attention has been focused on p-Kirchhoff-type problems.In [19], with the help of the Fountain Theorem, they studied the existence of infinitely many solutions for a fractional p-Kirchhoff equation. In [20], the authors showed the existence and multiplicity of solutions to a degenerate fractional p-Kirchhoff problem. However, we perceive that the A
m
b
r
o
s
e
t
t
i
-
R
a
b
i
n
o
w
i
t
z condition was used widely in these papers about p-Laplacian problems. We refer the interested readers to [12, 21–28]. The condition is usually called (AR) condition for short. It is described as follows:(7)
t
h
e
r
e
e
x
i
s
t
t
h
r
e
e
c
o
n
s
t
a
n
t
s
r
>
0
,
μ
>
η
>
1
,
s
u
c
h
t
h
a
t
0
<
μ
G
x
,
t
≤
t
g
x
,
t
f
o
r
a
n
y
x
∈
Ω
,
t
∈
R
a
n
d
t
≥
r
.And G
(
x
,
t
)
=
∫
0
t
g
(
x
,
s
)
d
s.It was introduced for the first time by Ambrosetti and Rabinowitz in [29]. Since then, the (AR) condition has been used far and wide in more and more works involving superlinear elliptic boundary. We know that the importance of (AR) condition is to guarantee the boundedness of familiar (PS) sequences for the energy functional associated with the problem. The nonlinearity function f satisfies superlinear growth under the (AR) condition.Through (7), we can get(8)
G
x
,
t
≥
a
1
t
μ
-
a
2
f
o
r
a
n
y
x
,
t
∈
Ω
×
R
,where μ
>
η, for two constants a
1
,
a
2
>
0.However, there are still lots of functions that dissatisfy the (AR) condition, even though they are superlinear at infinity. We notice another form given by(9)
l
i
m
t
→
∞
G
x
,
t
t
η
=
+
∞
u
n
i
f
o
r
m
l
y
f
o
r
x
∈
Ω
.We find that the nonlinearityg is also superlinear at infinity under condition (9). Obviously, the functional(10)
g
x
,
t
=
t
η
-
1
ln
⁡
1
+
tsatisfies condition (9) and dissatisfies condition (8). So it does not satisfy (7).Motivated by the above works and [20, 24, 25, 30], we study the existence of infinitely many solutions of problem (1) without (AR) condition. Our results are extension of some problems studied by N. Van Thin in [30].Now, we give some assumptions on the functiong: Ω
×
R
→
R.(
g
1
)
There existC
>
0 and q
∈
(
p
,
p
s
∗
) such that |
g
(
x
,
t
)
|
≤
C
(
1
+
|
t
|
q
-
1
) for a.e. x
∈
Ω and all t
∈
R, where p
s
∗
=
N
p
/
N
-
p
s.
(
g
2
)
g
(
x
,
-
t
)
=
-
g
(
x
,
t
) for all x
∈
Ω
,
t
∈
R
.
(
g
3
)
lim
t
→
∞
⁡
G
(
x
,
t
)
/
|
t
|
2
p
=
+
∞ uniformly for x
∈
Ω.
g
4
lim
t
→
0
⁡
g
(
x
,
t
)
/
t
p
-
1
=
0 uniformly for x
∈
Ω.
(
g
5
)
There existst
~
>
0 such that the function t
↦
g
(
x
,
t
)
/
t
2
p
-
1 is decreasing if t
≤
-
t
~
<
0 and increasing if t
≥
t
~
>
0 for all x
∈
Ω
.
(
g
6
)
There existσ
≥
1 and T
∈
L
1
(
Ω
) satisfying T
(
x
)
≥
0 such that G
(
x
,
s
)
≤
σ
G
(
x
,
t
)
+
T
(
x
) for all x
∈
Ω and 0
≤
s
≤
t, where G
(
x
,
t
)
=
1
/
2
p
t
g
(
x
,
t
)
-
G
(
x
,
t
)
.Definition 1.
We claim that a functionu
∈
X
0 is a weak solution of problem (1), if (11)
a
+
b
u
X
0
p
∫
Q
u
x
-
u
y
p
-
2
u
x
-
u
y
ħ
x
-
ħ
y
K
x
-
y
d
x
d
y
-
λ
∫
Ω
u
x
p
-
2
u
x
ħ
x
d
x
-
∫
Ω
g
x
,
u
x
ħ
x
d
x
=
0
,
f
o
r
a
n
y
ħ
∈
X
0
.See Section2 for a detailed description for of X
0.Theorem 2.
LetK
:
R
N
∖
{
0
}
→
(
0
,
∞
) be a function satisfying (3). Let conditions (
g
1
)
-
(
g
5
) hold. Then, for any λ
∈
R, problem (1) has infinitely many nontrivial solutions {
u
k
}
k
∈
N in X
0 with unbounded energy.Corollary 3.
LetK
:
R
N
∖
{
0
}
→
(
0
,
∞
) be a function satisfying (3). Let conditions (
g
1
)
-
(
g
4
) hold. If condition (
g
6
) replaces (
g
5
), then the conclusion of Theorem 2 holds.Remark.Originally, Jeanjean put forward a condition that was similar to (
g
6
) in [31]. It is easy to see that condition (
g
6
) is equivalent to (
g
5
) when σ
=
1. Actually, condition (
g
6
) is weaker than condition (
g
5
). We can find that there are some functions satisfying (
g
1
)
-
(
g
4
)
,
(
g
6
) but dissatisfying (
g
5
). For example, (12)
g
x
,
t
=
2
p
t
2
p
-
2
t
ln
⁡
1
+
t
2
p
+
p
sin
⁡
t
.This paper consists of the following parts. In Section2, we give the definition and some properties for the space X
0 and some preliminary results. Section 3 verifies compactness conditions. In Section 4, we prove Theorem 2 and Corollary 3.


## 2. Preliminary Results

Firstly, we recall the functional spaceX and X
0 and some lemmas, which will be used in next section for problem (1). We appoint Q
=
R
2
N
∖
Γ where Γ
=
C
(
Ω
)
×
C
(
Ω
)
⊂
R
2
N and C
(
Ω
)
=
R
N
∖
Ω. The spaceX is a linear of Lebesgue measurable functions from R
N to R such that the restriction to Ω of any function u in X belongs to L
p
(
Ω
) and(13)
∫
Q
u
x
-
u
y
p
K
x
-
y
d
x
d
y
<
∞
.
X is endowed with the following norm:(14)
u
X
=
u
L
p
Ω
+
∫
Q
u
x
-
u
y
p
K
x
-
y
d
x
d
y
1
/
p
.In addition, X
0 is endowed with the following norm:(15)
u
X
0
=
∫
R
2
N
u
x
-
u
y
p
K
x
-
y
d
x
d
y
1
/
p
,and (
X
0
,
·
X
0
) is known as the Hilbert space defined by the following scalar product (see [12], Lemma 7).(16)
u
,
ħ
X
0
≔
∫
R
2
N
u
x
-
u
y
p
-
2
u
x
-
u
y
ħ
x
-
ħ
y
K
x
-
y
d
x
d
y
.We denote the usual fractional Sobolev space byW
s
,
p
(
Ω
), which is endowed with norm (the so-called G
a
g
l
i
a
r
d
o
n
o
r
m) as follows:(17)
u
W
s
,
p
Ω
=
u
L
p
Ω
+
∫
Ω
×
Ω
u
x
-
u
y
p
x
-
y
-
N
+
p
s
d
x
d
y
1
/
p
.We observe that the norms (14) and (17) are not the same when K
(
x
)
=
1
/
x
N
+
p
s, since Ω
×
Ω is contained strictly in Q. It makes the space X
0 different from the usual classical fractional Sobolev space. Therefore, from the point of view of the variational method, the classical fractional Sobolev space is insufficient for dealing with our problem.We recall that the spaceX
0 is nonempty due to C
0
∞
⊆
X
0(see [12], Lemma 2.1). The following conclusion is correct if a general kernel K satisfies (3): (18)
X
0
⊂
u
∈
W
s
,
p
R
N
:
u
x
=
0
a
.
e
.
i
n
R
N
∖
Ω
.Particularly, the following characterization holds when K
(
x
)
=
1
/
x
N
+
p
s:(19)
X
0
=
u
∈
W
s
,
p
R
N
:
u
x
=
0
a
.
e
.
i
n
R
N
∖
Ω
.For more details about space X and X
0, we refer to [5, 32].Considering future works, we recall the following eigenvalue problem:(20)
L
p
s
u
=
λ
u
p
-
2
u
i
n
Ω
,
u
=
0
i
n
R
N
∖
Ω
.It has a divergent sequence of positive eigenvalues(21)
0
<
λ
1
≤
λ
2
≤
·
·
·
≤
λ
j
≤
λ
j
+
1
≤
·
·
·whose homologous eigenfunctions are denoted by e
j. From Proposition 9 of [32], we know that e
j
j
∈
N can be chosen in such a way that this sequence provides an orthonormal basis in L
p
(
Ω
) and an orthogonal basis in X
0.Firstly, we define(22)
Ψ
u
=
J
u
-
I
u
-
H
u
, where(23)
J
u
=
a
u
X
0
p
+
b
u
X
0
2
p
,
I
u
=
λ
p
∫
Ω
u
p
d
x
,
H
u
=
∫
Ω
G
x
,
u
d
x
, where G
(
x
,
u
)
≔
∫
0
u
g
(
x
,
s
)
d
s.Clearly, the energy functionalΨ
:
X
0
→
R associated with problem (1) is well defined.For convenience, we writeu
L
p
(
Ω
) as u
p. From Lemma 3.1 of [25], clearly we know that functional J
∈
C
1
(
X
0
,
R
)
. And if (
g
1
) holds, H
∈
C
1
(
X
0
,
R
)
. So, we get that Ψ
∈
C
1
(
X
0
,
R
) and(24)
Ψ
′
u
,
ħ
X
0
=
a
+
b
u
X
0
p
∫
Q
u
x
-
u
y
p
-
2
u
x
-
u
y
ħ
x
-
ħ
y
K
x
-
y
d
x
d
y
-
λ
∫
Ω
u
x
p
-
2
u
x
ħ
x
d
x
-
∫
Ω
g
x
,
u
x
ħ
x
d
xfor all u
,
ħ
∈
X
0. In order to prove the conclusion of problem (1), we need some lemmas.Lemma 4 (see [12]).
Assume that (3) holds. We have the following conclusions:
(1) For anyκ
∈
[
1
,
p
s
∗
), the embedding X
0
↪
L
κ
(
Ω
) is compact when Ω is a bounded domain with continuous boundary.
(2) For allκ
∈
[
1
,
p
s
∗
], the embedding X
0
↪
L
κ
(
Ω
) is continuous.Definition 5.
LetΨ
∈
C
1
(
X
0
,
R
)
. The functional Ψ satisfies (
C
e
)
c at the level c
∈
R, if any sequence {
u
k
}
⊂
X
0, with(25)
Ψ
u
k
→
c
i
n
X
0
,
1
+
u
k
Ψ
′
u
k
→
0
i
n
X
0
′
a
s
k
→
∞
,has a strongly convergent subsequence in X
0. X
0
′ is the dual space of X
0.Theorem 6 (symmetric mountain pass theorem [33]).
Assume that X is an infinite dimensional Banach space.Y
~ is a finite dimensional Banach space and X
=
Y
~
⊕
Z
~
. For any c
>
0, if Ψ
∈
C
1
(
X
,
R
), it satisfies (
C
e
)
c condition, and(a)
Ψ is even and Ψ
(
0
)
=
0
f
o
r
a
l
l
u
∈
X.
(b)
There exist constantsδ
,
ρ
>
0 such that Ψ
∂
B
δ
(
X
)
≥
ρ, where(26)
B
δ
X
=
u
∈
X
:
u
≤
δ
.
(c)
For any finite dimensional subspaceX
¯
⊆
X, there exists R
=
R
(
X
¯
)
>
0 such that Ψ
(
u
)
≤
0 on X
¯
∖
B
R
(
X
¯
).
ThenΨ possesses an unbounded sequence of critical values characterized by a minimax argument.


## 3. Compactness Conditions

In this section, we are going to give some lemmas about the compactness of functionalΨ and prove them.Lemma 7.
Let(
g
1
) hold. Any bounded sequence u
k
k
∈
N
⊂
X
0, which satisfies 1
+
u
k
Ψ
′
(
u
k
)
→
0 as k
→
∞, possesses a strongly convergent subsequence in X
0Proof.
Suppose that{
u
k
}
k is bounded in X
0. From Lemma 2.4 of [12] and Theorem 1.21 of [34], we know that X
0 is reflexive. Combining with Lemma 4, we have(27)
u
k
⇀
u
i
n
X
0
,
u
k
→
u
i
n
L
κ
Ω
,
1
≤
κ
<
p
s
∗
,
u
k
→
u
a
.
e
.
i
n
R
N
.We just need to prove that u
k
→
u strongly in X
0.
Through the Hölder inequality and(
g
1
), we obtain(28)
∫
Ω
g
x
,
u
k
u
k
-
u
d
x
≤
∫
Ω
C
+
C
u
k
q
-
1
u
k
-
u
d
x
≤
C
Ω
q
-
1
/
q
+
u
k
q
q
-
1
u
k
-
u
q
.By (27), we get(29)
l
i
m
k
→
∞
∫
Ω
g
x
,
u
k
u
k
-
u
d
x
=
0
.We consider the following formula with Ho
¨lder inequality: (30)
∫
Ω
u
k
p
-
2
u
k
u
k
-
u
d
x
≤
u
k
p
p
-
1
u
k
-
u
p
. Hence, by (27), we get(31)
l
i
m
k
→
∞
∫
Ω
u
k
p
-
2
u
k
u
k
-
u
d
x
=
0
.Then, for convenience, we define a new linear functional on X
0 as follows:(32)
B
w
v
≔
∫
R
2
N
w
x
-
w
y
p
-
2
w
x
-
w
y
v
x
-
v
y
K
x
-
y
d
x
d
y
,for all w
,
v
∈
X
0. By means of the Hölder inequality, we have that(33)
B
w
v
≤
w
X
0
p
-
1
·
v
X
0
f
o
r
a
l
l
v
∈
X
0
.Hence, B
w
(
v
) is a continuous functional on X
0. Hence, we obtain that(34)
l
i
m
k
→
∞
B
u
u
k
-
u
d
x
=
0
.Clearly, 〈
Ψ
′
(
u
k
)
,
u
k
-
u
〉
→
0 as k
→
∞, since u
k
⇀
u in X
0 and (
1
+
u
k
X
0
)
Ψ
′
(
u
k
)
→
0 in X
0
′. Hence, by (27), (29), and (31), we have(35)
o
1
=
Ψ
′
u
k
,
u
k
-
u
=
a
+
b
u
k
X
0
p
B
u
k
u
k
-
u
-
λ
∫
Ω
u
k
p
-
2
u
k
u
k
-
u
d
x
+
∫
Ω
g
x
,
u
k
u
k
-
u
d
x
=
a
+
b
u
k
X
0
p
B
u
k
u
k
-
u
+
o
1
a
s
k
→
∞
.Hence, through the boundedness of u
k in X
0 and (34), we have(36)
l
i
m
k
→
∞
B
u
k
u
k
-
u
-
B
u
u
k
-
u
=
0
.Now, we recall the Simon inequalities:(37)
ϑ
-
χ
p
≤
Q
p
ϑ
p
-
2
ϑ
-
χ
p
-
2
χ
·
ϑ
-
χ
p
≥
2
,
(38)
ϑ
-
χ
p
≤
Q
p
~
ϑ
p
-
2
ϑ
-
χ
p
-
2
χ
p
/
2
·
ϑ
p
+
χ
p
2
-
p
/
2
1
<
p
<
2
,for all ϑ
,
χ
∈
R
N, where Q
p
,
Q
p
~
>
0 relying on p. Then if p
≥
2, by (36) and (37), we have (39)
u
k
-
u
X
0
p
≤
Q
p
∫
R
2
N
u
k
x
-
u
k
y
p
-
2
u
k
x
-
u
k
y
-
u
x
-
u
y
p
-
2
u
x
-
u
y
×
u
k
x
-
u
x
-
u
k
y
-
u
y
K
x
-
y
d
x
d
y
=
Q
p
B
u
k
u
k
-
u
-
B
u
u
k
-
u
→
0
, as k
→
∞. When 1
<
p
<
2, by (36), (38), and the boundedness of u
k in X
0, we have (40)
u
k
-
u
X
0
p
≤
Q
p
~
B
u
k
u
k
-
u
-
B
u
u
k
-
u
p
/
2
u
k
X
0
p
+
u
X
0
p
2
-
p
/
2
≤
Q
p
~
B
u
k
u
k
-
u
-
B
u
u
k
-
u
p
/
2
u
k
X
0
p
2
-
p
/
2
+
u
X
0
p
2
-
p
/
2
→
0
, as k
→
∞. Hence, we get u
k
→
u strongly in X
0.Lemma 8.
Let(
g
1
)
,
(
g
3
), and (
g
5
) hold. Then, functional Ψ satisfies the (
C
e
)
c condition.Proof.
Let(
g
5
) hold. According to the monotonicity of t
↦
g
(
x
,
t
)
/
t
2
p
-
1, we find that there exists a positive constant L
1 such that(41)
G
x
,
s
≤
G
x
,
t
+
L
1
,where G
(
x
,
t
)
=
1
/
2
p
t
g
(
x
,
t
)
-
G
(
x
,
t
), for all x
∈
Ω and 0
≤
|
s
|
≤
|
t
|. Let {
u
k
}
k
∈
N be a Cecrami sequence in X
0. We know that it satisfies(42)
Ψ
u
k
→
c
i
n
X
0
,
(43)
1
+
u
k
X
0
Ψ
′
u
k
→
0
i
n
X
0
′as k
→
∞. By means of Lemma 7, it suffices to prove the boundedness of {
u
k
}. Suppose that u
k
k
∈
N is unbounded in X
0. Then we have(44)
u
k
X
0
→
+
∞
.By (43) and (44), we get (45)
Ψ
′
u
k
→
0
.Hence, we get(46)
u
k
X
0
Ψ
′
u
k
,
u
k
u
k
X
0
→
0
a
s
k
→
∞
. We define ζ
k
=
u
k
/
u
k
X
0. Then ζ
k
X
0
=
1. So ζ
k
k
∈
N is a bounded sequence in X
0. Through Lemma 4, there exists ζ
∞ satisfying(47)
ζ
k
→
ζ
∞
i
n
L
p
R
N
,
ζ
k
→
ζ
∞
i
n
L
q
R
N
,
ζ
k
→
ζ
∞
i
n
R
N
,as k
→
∞. What is more, through Lemma A.1 of [35], there exists a function ε
∈
L
q
(
R
N
) satisfying(48)
ζ
k
x
≤
ε
x
i
n
R
N
.We only need to consider two cases: ζ
∞
=
0 and ζ
∞
≠
0. Firstly, we consider the case ζ
∞
=
0. Refer to [31]; for any k
∈
N, we have l
k
∈
[
0,1
] such that(49)
Ψ
l
k
u
k
=
m
a
x
l
∈
0,1
Ψ
l
u
k
.Because of the unboundedness of {
u
k
}, for any τ
∈
N, we select h
τ
=
(
4
p
/
b
τ
)
1
/
2
p such that(50)
h
τ
u
k
X
0
∈
0,1
,where k is large enough, say k
>
k
¯, with k
¯
=
k
¯
(
τ
). By (47) and ζ
∞
=
0, we get(51)
∫
Ω
h
τ
ζ
k
x
p
d
x
→
0
.Since the function G is continuous, we get that(52)
G
x
,
h
τ
ζ
k
x
→
G
x
,
h
τ
ζ
∞
x
i
n
Ω
,as k
→
∞, for any τ
∈
N. Through (
g
1
), (48), and Ho
¨lder inequality, we obtain that(53)
G
x
,
h
τ
ζ
k
x
≤
C
h
τ
ζ
k
x
+
C
q
h
τ
ζ
k
x
q
≤
C
h
τ
ε
x
+
C
q
h
τ
ε
x
q
∈
L
1
Ω
, for any k
,
τ
∈
N. Hence, we get that(54)
G
·
,
h
τ
ζ
k
·
→
G
·
,
h
τ
ζ
∞
·
i
n
L
1
Ω
,as k
→
∞, for any τ
∈
N, thanks to the Dominated Convergence Theorem and (51) and (52). As a consequence of G
(
x
,
0
)
=
0 for all x
∈
Ω, by (54) and ζ
∞
=
0, we have(55)
∫
Ω
G
x
,
h
τ
ζ
k
x
d
x
→
0
,as k
→
∞, for any τ
∈
N. Therefore, by (50)–(52) and (55), we get(56)
Ψ
l
k
u
k
≥
Ψ
h
τ
u
k
X
0
u
k
=
Ψ
h
τ
ζ
k
=
a
p
h
τ
ζ
k
x
X
0
p
+
b
2
p
h
τ
ζ
k
x
X
0
2
p
-
λ
p
∫
Ω
h
τ
ζ
k
x
p
d
x
-
∫
Ω
G
x
,
h
τ
ζ
k
x
d
x
≥
b
2
p
h
τ
ζ
k
x
X
0
2
p
=
2
τ
, as k
→
∞, for any τ
∈
N. Hence, we infer that(57)
Ψ
l
k
u
k
→
+
∞
a
s
k
→
∞
.Now, we will show that(58)
l
i
m
k
→
∞
s
u
p
Ψ
l
k
u
k
≤
ϖ
,where ϖ
>
0. Because Ω
⊂
R
N is an open bounded smooth domain with Lipschitz boundary, we deduce that there exists a positive constant C
p
,
λ, which depends on λ and p such that(59)
λ
2
p
∫
Ω
u
k
x
p
d
x
<
C
p
,
λ
<
+
∞
.We notice that Ψ
(
0
)
=
0. Through (42), (49), and (57), we get that there exists l
k
∈
(
0,1
) such that(60)
d
d
l
l
=
l
k
Ψ
l
u
k
=
0
, for any k
∈
N. Then we have(61)
Ψ
′
l
k
u
k
,
l
k
u
k
=
l
k
d
d
l
l
=
l
k
Ψ
l
u
k
=
0
.Through (41), (59), and l
k
∈
[
0,1
], we have(62)
Ψ
l
k
u
k
=
Ψ
l
k
u
k
-
1
2
p
Ψ
′
l
k
u
k
,
l
k
u
k
=
a
2
p
t
k
u
k
x
X
0
p
-
λ
2
p
∫
Ω
l
k
u
k
x
p
d
x
-
∫
Ω
G
x
,
l
k
u
k
x
d
x
+
∫
Ω
1
2
p
g
x
,
l
k
u
k
x
l
k
u
k
x
d
x
≤
a
2
p
t
k
u
k
x
X
0
p
+
∫
Ω
G
x
,
l
k
u
k
d
x
+
C
p
,
λ
≤
a
2
p
u
k
x
X
0
p
+
∫
Ω
G
x
,
u
k
d
x
+
L
1
Ω
+
C
p
,
λ
=
a
2
p
u
k
x
X
0
p
-
λ
2
p
∫
Ω
u
k
x
p
d
x
-
∫
Ω
G
x
,
u
k
x
d
x
+
∫
Ω
1
2
p
g
x
,
u
k
x
u
k
x
d
x
+
λ
2
p
∫
Ω
u
k
x
p
d
x
+
L
1
Ω
+
C
p
,
λ
≤
Ψ
u
k
-
1
2
p
Ψ
′
u
k
,
u
k
+
2
C
p
,
λ
+
L
1
Ω
=
c
+
2
C
p
,
λ
+
L
1
Ω
<
+
∞
.And k
→
∞. It contradicts with (57). Hence, the sequence {
u
k
}
k
∈
N is bounded in X
0.
Now we consider the caseζ
∞
≠
0. We define Ω
¯
=
{
x
∈
Ω
:
ζ
∞
(
x
)
≠
0
}. Clearly, Ω
¯ is Lebesgue measureable. Through (44), (47), and ζ
∞
≠
0, we have(63)
u
k
x
=
ζ
k
x
·
u
k
X
0
→
+
∞
i
n
Ω
¯
,as k
→
∞. Through (47), (63), and (
g
3
), it holds true that(64)
G
x
,
u
k
x
u
k
X
0
2
p
=
G
x
,
u
k
x
u
k
2
p
·
u
k
2
p
u
k
X
0
2
p
=
G
x
,
u
k
x
u
k
x
2
p
ζ
k
x
2
p
→
+
∞
,
i
n
Ω
¯, as k
→
∞. Through the Fatou Lemma and (64), we have(65)
∫
Ω
¯
G
x
,
u
k
x
u
k
X
0
2
p
d
x
→
+
∞
.Next, we discusses the case in Ω
∖
Ω
¯. By (
g
3
), we get(66)
l
i
m
t
→
∞
G
x
,
t
=
+
∞
,
x
∈
Ω
.Therefore, there exist constants t
1
,
A
>
0 such that(67)
G
x
,
t
≥
A
,for all x
∈
Ω and t
>
t
1. Moreover, by means of the continuity of G in Ω
×
R, we get(68)
G
x
,
t
≥
m
i
n
x
,
t
∈
Ω
×
-
t
1
,
t
1
G
x
,
t
,for all t
≤
t
1. Hence, we obtain that(69)
G
x
,
t
≥
m
i
n
A
,
m
i
n
x
,
t
∈
Ω
×
-
t
1
,
t
1
G
x
,
t
≔
A
~
,for any (
x
,
t
)
∈
Ω
×
R, thanks to (67) and (68). So we get(70)
l
i
m
t
→
∞
∫
Ω
∖
Ω
¯
G
x
,
u
k
x
u
k
X
0
2
p
d
x
≥
0
.By (42) and (44), we get(71)
o
1
=
Ψ
u
k
u
k
X
0
2
p
=
a
p
u
k
X
0
2
p
+
b
2
p
-
λ
p
∫
Ω
u
k
x
p
u
k
X
0
2
p
d
x
-
∫
Ω
¯
G
x
,
u
k
x
u
k
X
0
2
p
d
x
-
∫
Ω
∖
Ω
¯
G
x
,
u
k
x
u
k
X
0
2
p
d
x
, as k
→
∞. Consider (63), (65), (70), and the variational characterization of λ
j defined as follows:(72)
λ
j
=
m
i
n
u
∈
X
0
∖
0
∫
R
2
N
u
x
-
u
y
p
K
x
-
y
d
x
d
y
∫
Ω
u
x
p
d
x
.We have(73)
o
1
=
a
p
u
k
X
0
2
p
+
b
2
p
-
λ
p
∫
Ω
u
k
x
p
u
k
X
0
2
p
d
x
-
∫
Ω
¯
G
x
,
u
k
u
k
X
0
2
p
d
x
-
∫
Ω
∖
Ω
¯
G
x
,
u
k
u
k
X
0
2
p
d
x
≤
b
2
p
+
m
a
x
0
,
-
λ
p
λ
1
∫
Ω
u
k
X
0
p
u
k
X
0
2
p
d
x
-
∫
Ω
¯
G
x
,
u
k
u
k
X
0
2
p
d
x
-
∫
Ω
∖
Ω
¯
G
x
,
u
k
u
k
X
0
2
p
d
x
≤
-
∞
.We find that (73) is a contradictory result. Thus, the sequence {
u
k
}
k
∈
N is bounded in X
0. This ends the proof of Lemma 8.


## 4. Proof of Theorem2 and Corollary 3

We know thatX
0 is a Hilbert space. Hence, we divide X
0 into two parts. Let X
0 = ⨁
j
≥
1
X
j, where X
j = span{
e
j
}
j
∈
N. Now we define(74)
Y
m
=
⨁
j
=
1
m
X
j
,
Z
m
=
⨁
j
=
m
∞
X
j
¯
.Obviously, Y
m is a finite-dimensional space.Lemma 9.
Letν
∈
[
1
,
p
s
∗
). We have (75)
ξ
m
ν
≔
s
u
p
u
ν
:
u
∈
Z
m
,
u
X
0
=
1
→
0
,
m
→
∞
.Proof.
Clearlyξ
m
+
1
≥
ξ
m
≥
0, so that ξ
m
→
ξ
≥
0, as m
→
∞. For every m
∈
N, there exists u
m
∈
Z
m such that u
m
ν
>
ξ
m
/
2 and u
m
X
0
=
1. Through the definition of Z
m, we can get u
m
⇀
0 in X
0. According to Lemma 4, the embedding X
0
↪
↪
L
κ
(
Ω
)
(
1
≤
κ
<
p
s
∗
) implies that u
m
→
0 in L
κ
(
Ω
). Therefore, we get ξ
=
0, as m
→
∞. This implies that the proof is complete.Proof of Theorem2.
Obviously, all norms are equivalent inY
m. Hence, there exist two constants C
1
,
C
2
>
0 such that(76)
C
1
u
X
0
≤
u
ν
≤
C
2
u
X
0
f
o
r
a
n
y
ν
∈
1
,
p
s
∗
.According to Theorem 6 and Lemma 8, we just need to prove (
a
)
,
(
b
), and (
c
) of Theorem 6. Let (
g
2
) hold. Clearly, Ψ
(
0
)
=
0. Hence, condition (
a
) of Theorem 6 is true. Then, we take into account the range of values of λ. According to (72), if λ
/
a
>
λ
1, we can find j
∈
N and j
>
2 such that λ
/
a
∈
[
λ
j
-
1
,
λ
j
). So, for all j, there exists j
∈
N and j
≥
1 such that λ
/
a
<
λ
j for any λ
∈
R.
By(
g
1
) and (
g
4
), for any ϱ
>
0, there exists C
ϱ
>
0 such that(77)
G
x
,
t
≤
ϱ
p
t
p
+
C
ϱ
q
t
q
,for any (
x
,
t
)
∈
Ω
×
R. By Lemma 9, for any fixed ϱ
>
0, choose an integer m
~
≥
1 such that(78)
u
p
p
≤
min
⁡
a
,
a
-
λ
/
λ
j
2
ϱ
u
X
0
p
,
u
q
q
≤
q
min
⁡
a
,
a
-
λ
/
λ
j
2
p
C
ϱ
u
X
0
q
,for any u
∈
Z
m
~
. Choose δ
≔
u
X
0
=
1
/
2, for all u
∈
Z
m
~. Note that q
>
p; from (77) and (78), we have(79)
Ψ
u
=
a
p
u
X
0
p
+
b
2
p
u
X
0
2
p
-
λ
p
∫
Ω
u
x
p
d
x
-
∫
Ω
G
x
,
u
x
d
x
≥
a
p
u
X
0
p
+
m
i
n
0
,
-
λ
p
λ
j
u
X
0
p
-
ϱ
p
u
p
p
-
C
ϱ
q
u
q
q
≥
1
p
m
i
n
a
,
a
-
λ
λ
j
u
X
0
p
-
1
2
p
m
i
n
a
,
a
-
λ
λ
j
u
X
0
p
-
1
2
p
m
i
n
a
,
a
-
λ
λ
j
u
X
0
q
≥
1
2
p
m
i
n
a
,
a
-
λ
λ
j
u
X
0
p
-
u
X
0
q
=
1
2
p
m
i
n
a
,
a
-
λ
λ
j
1
2
p
-
1
2
q
≔
ρ
>
0
.Then, condition (b) of Theorem 6 is true.
In the end, we demonstrate condition (c) of Theorem6. In view of (
g
3
), there exist constants α
>
b
/
2
p
C
1
2
p
,
β
>
0 such that(80)
G
x
,
t
≥
α
t
2
p
,for any x
∈
Ω and |
t
|
>
β. Considering condition (
g
1
), we have(81)
G
x
,
t
≤
C
1
+
β
q
-
1
t
,for any x
∈
Ω and |
t
|
≤
β. Take C
∗
=
C
(
1
+
β
q
-
1
), where C
∗
>
0. We obtain(82)
G
x
,
t
≥
α
t
2
p
-
C
∗
t
,for any (
x
,
t
)
∈
Ω
×
R. Then, by (76) and (82), we have(83)
Ψ
u
=
a
p
u
X
0
p
+
b
2
p
u
X
0
2
p
-
λ
p
∫
Ω
u
x
p
d
x
-
∫
Ω
G
x
,
u
x
d
x
≤
a
p
u
X
0
p
+
b
2
p
u
X
0
2
p
+
m
a
x
0
,
-
λ
p
λ
j
u
X
0
p
-
α
u
2
p
2
p
-
C
∗
u
1
≤
1
p
max
⁡
a
,
a
-
λ
λ
j
u
X
0
p
+
b
2
p
-
C
1
2
p
α
u
X
0
2
p
-
C
1
C
∗
u
X
0
.Let R
=
R
(
X
¯
)
>
0 be large enough. When u
X
0
>
R, Ψ
(
u
)
≤
0. Hence, condition (c) of Theorem 6 is true. In conclusion, Theorem 2 is proven.Proof of Corollary3.
Let(
g
6
) hold. Similar to Theorem 2, we only need to prove inequality (62).(84)
1
σ
Ψ
l
k
u
k
=
1
σ
Ψ
l
k
u
k
-
1
2
p
Ψ
′
l
k
u
k
,
l
k
u
k
=
1
σ
a
2
p
l
k
u
k
x
X
0
p
-
λ
2
p
∫
Ω
l
k
u
k
x
p
d
x
-
∫
Ω
G
x
,
l
k
u
k
x
d
x
+
∫
Ω
1
2
p
g
x
,
l
k
u
k
x
l
k
u
k
x
d
x
≤
1
σ
a
2
p
l
k
u
k
x
X
0
p
+
∫
Ω
G
x
,
l
k
u
k
d
x
+
C
p
,
λ
≤
a
2
p
u
k
x
X
0
p
+
∫
Ω
G
x
,
u
k
d
x
+
1
σ
∫
Ω
T
x
d
x
+
C
p
,
λ
=
a
2
p
u
k
x
X
0
p
-
λ
2
p
∫
Ω
u
k
x
p
d
x
-
∫
Ω
G
x
,
u
k
x
d
x
+
∫
Ω
1
2
p
g
x
,
u
k
x
u
k
x
d
x
+
λ
2
p
∫
Ω
u
k
x
p
d
x
+
1
σ
∫
Ω
T
x
d
x
+
C
p
,
λ
≤
Ψ
u
k
-
1
2
p
Ψ
′
u
k
,
u
k
+
2
C
p
,
λ
+
1
σ
∫
Ω
T
x
d
x
≤
c
+
2
C
p
,
λ
+
1
σ
∫
Ω
T
x
d
x
<
+
∞
.The proof is completed.



---

*Source: 1353961-2019-04-07.xml*