# The Neumann Problem for a Degenerate Elliptic System Near Resonance

Authors: Yu-Cheng An; Hong-Min Suo
Journal: Advances in Mathematical Physics
 (2017)

---

## Abstract

This paper studies the following system of degenerate equations-
d
i
v
p
x
∇
u
+
q
x
u
=
α
u
+
β
v
+
g
1
x
,
v
+
h
1
x, x
∈
Ω, -
d
i
v
(
p
(
x
)
∇
v
)
+
q
(
x
)
v
=
β
u
+
α
v
+
g
2
(
x
,
u
)
+
h
2
(
x
), x
∈
Ω, ∂
u
/
∂
ν
=
∂
v
/
∂
ν
=
0, x
∈
∂
Ω
. Here Ω
⊂
R
n is a bounded C
2 domain, and ν is the exterior normal vector on ∂
Ω. The coefficient function p may vanish in Ω
¯, q
∈
L
r
(
Ω
) with r
>
n
s
/
(
2
s
-
n
)
,
s
>
n
/
2. We show that the eigenvalues of the operator -
d
i
v
(
p
(
x
)
∇
u
)
+
q
(
x
)
u are discrete. Secondly, when the linear part is near resonance, we prove the existence of at least two different solutions for the above degenerate system, under suitable conditions on h
1
,
h
2
,
g
1, and g
2.


---

## Body

## 1. Introduction

In recent decades, many kinds of perturbed problems were studied by many scholars, such as [1–11]. Here, we want to say that the authors in [5] studied the following Dirichlet boundary problem:(1)
-
Δ
u
=
μ
u
±
g
x
,
u
+
h
x
,
x
∈
Ω
,
u
=
0
,
x
∈
∂
Ω
.When the parameter μ is close to an eigenvalue of the operator -
Δ, they proved that problem (1) has two different solutions. Moreover, this result was extended to some equations and systems; see [6–10]. In particular, Massa and Rossato [11] studied a nondegenerate elliptic system and two solutions were obtained by using Galerkin techniques. On the other hand, we also mention that many scholars studied some elliptic equations with the Neumann or Robin boundary; see [12–17] and the references therein. Inspired by the above results, we study the following system of degenerate equations:(2)
-
d
i
v
p
x
∇
u
+
q
x
u
=
α
u
+
β
v
+
g
1
x
,
v
+
h
1
x
,
x
∈
Ω
,
-
d
i
v
p
x
∇
v
+
q
x
v
=
β
u
+
α
v
+
g
2
x
,
u
+
h
2
x
,
x
∈
Ω
,
∂
u
∂
ν
=
∂
v
∂
ν
=
0
,
x
∈
∂
Ω
,where Ω
⊂
R
n is a bounded C
2 domain, ν is the exterior normal vector on ∂
Ω, α
,
β
∈
R, and h
1
,
h
2
∈
L
2
(
Ω
). The coefficient p may vanish in Ω
¯, q
∈
L
r
(
Ω
) with r
>
n
s
/
(
2
s
-
n
)
,
s
>
n
/
2; that is, problem (2) may be degenerate; see [18]. As in [11], we will use the critical point theory and Galerkin techniques to obtain the existence of two different solutions for the above degenerate system. Now, we introduce the function set F which consists of functions ω
:
Ω
→
R
+
∪
{
0
} such that(3)
ω
∈
L
loc
⁡
1
Ω
,
ω
-
1
∈
L
loc
⁡
1
Ω
,
ω
-
s
∈
L
1
Ω
with
s
>
n
2
.Throughout the paper, we always assume that there exist ω
∈
F and κ
≥
1 such that(4)
ω
x
κ
≤
p
x
≤
κ
ω
x
,
a.e.
x
∈
Ω
.We also assume that q belongs to L
r
(
Ω
) with r
>
n
s
/
(
2
s
-
n
), and g
i
:
Ω
→
R is a Carathéodory mapping and satisfies the following conditions:(
g
0
) For every M
∈
R
+, there exists l
M
∈
L
2
(
Ω
) such that, for all |
s
|
≤
M
(5)
g
i
x
,
s
≤
l
M
x
,
a.e.
x
∈
Ω
.
(
g
∞
)
lim
|
s
|
→
∞
⁡
g
i
(
x
,
s
)
/
s
=
0, uniformly in x
∈
Ω, i
=
1,2.Although the conditions (
g
0
) and (
g
∞
) were introduced in [10], it is weaker than (
f
1
) in [5] (or (1.2) in [11]). In fact, let g
i
(
x
,
s
)
=
s
-
1
log
⁡
(
1
+
|
s
|
); it is easy to see that g
i satisfies (
g
∞
), but the function g
i does not satisfy the condition (
f
1
) in [5] (or (1.2) in [11]).In Section2, we give some preliminary lemmas and our main results. Meanwhile, we show that the eigenvalues of the operator -
d
i
v
(
p
(
x
)
∇
u
)
+
q
(
x
)
u are discrete under Neumann boundary condition. In Section 3, we prove our main results through Galerkin techniques and saddle point theorem.


## 2. Preliminaries and Main Results

In this section, we first collect some basic facts and then give the properties of the eigenvalues of the operator-
d
i
v
p
x
∇
u
+
q
(
x
)
u. Secondly, we define a new norm and prove it is equivalent to the usual Sobolev norm. At the end of this section, we give the main results of this paper.Firstly, letW
1,2
(
Ω
,
ω
) denote the completion of C
∞
(
Ω
) with respect to the norm (6)
u
ω
=
∫
Ω
ω
x
∇
u
2
+
u
2
d
x
,
∀
ω
∈
F
.The inner product in W
1,2
(
Ω
,
ω
) is denoted by (7)
u
,
v
ω
=
∫
Ω
ω
x
∇
u
∇
v
+
u
v
d
x
,
∀
u
,
v
∈
W
1,2
Ω
,
ω
.From (4), we know that the spaces W
1,2
Ω
,
ω and W
1,2
Ω
,
p are equivalent; see [18]. Let r
′
=
r
/
(
r
-
1
); from r
>
n
s
/
(
2
s
-
n
) with s
>
n
/
2, one has(8)
2
r
′
<
2
s
∗
=
2
n
s
n
s
+
1
-
2
s
.Hence, by the Sobolev embedding theorem of [18], we know that W
1,2
(
Ω
,
ω
) is compactly embedded in L
2
r
′
(
Ω
). Moreover, it follows from Hölder’s inequality that(9)
∫
Ω
q
x
u
2
d
x
≤
q
L
r
Ω
u
L
2
r
′
Ω
2
.Now, we use a similar argument to that of Gasiński and Papageorgiou (see [15]). Let us study the following eigenvalue problem:(10)
-
d
i
v
p
x
∇
u
+
q
x
u
=
μ
u
,
x
∈
Ω
,
∂
u
∂
ν
=
0
,
x
∈
∂
Ω
.Firstly, from (4) and the Sobolev embedding theorem of [18], we know that(11)
W
1,2
Ω
,
p
↪
L
2
r
′
Ω
↪
L
2
Ω
,and the first embedding is compact. Then, for any ε
>
0, we have(12)
u
L
2
r
′
Ω
2
≤
ε
u
p
2
+
c
1
u
L
2
Ω
2
,
∀
u
∈
W
1,2
Ω
,
p
,for some positive constant c
1; see [19].Let us defineσ
:
W
1,2
Ω
,
p
×
W
1,2
(
Ω
,
p
)
→
R
,
∀
u
,
v
∈
W
1,2
Ω
,
p
(13)
σ
u
,
v
=
∫
Ω
p
x
∇
u
∇
v
+
q
x
u
v
d
x
.It follows from (9) and (12) that(14)
1
-
ε
q
L
r
Ω
u
p
2
≤
σ
u
,
u
+
c
1
q
L
r
Ω
+
1
u
L
2
Ω
2
,
∀
u
∈
W
1,2
Ω
,
p
.Choosing ε small enough, then from (14) one gets(15)
u
p
2
≤
c
2
σ
u
,
u
+
u
L
2
Ω
2
,
∀
u
∈
W
1,2
Ω
,
p
,for some positive constants c
2. Hence, by Corollary  7.8 in [20], we conclude that there exists an eigenvalue sequence {
μ
k
} satisfying(16)
-
c
2
<
μ
1
<
μ
2
≤
μ
3
≤
⋯
≤
μ
k
≤
⋯
⟶
+
∞
,as k
→
+
∞ and(17)
μ
1
=
inf
⁡
σ
u
,
u
u
L
2
Ω
2
:
u
∈
W
1,2
Ω
,
p
∖
0
.Let {
φ
k
} be the corresponding eigenfunction sequence; then {
φ
k
} is complete in L
2
(
Ω
) and φ
k
∈
C
1
,
α
(
Ω
) for some α
∈
(
0,1
); see [21].Now, letϑ
=
max
⁡
{
-
μ
1
,
0
}
+
1; since the coefficient p may vanish in Ω
¯, we need to define a new norm:(18)
u
ϑ
=
σ
u
,
u
+
∫
Ω
ϑ
u
2
d
x
,
u
∈
W
1,2
Ω
,
p
,and the corresponding inner product(19)
u
,
v
ϑ
=
σ
u
,
v
+
∫
Ω
ϑ
u
v
d
x
,
∀
u
,
v
∈
W
1,2
Ω
,
p
.Lemma 1.
Letq
∈
L
r
Ω with r
>
n
s
/
(
2
s
-
n
); then the norms ·
ϑ and ·
p are equivalent.Proof.
Firstly, it follows from (17) that(20)
χ
u
=
σ
u
,
u
+
∫
Ω
ϑ
u
2
d
x
≥
0
,
∀
u
∈
W
1,2
Ω
,
p
.We prove that there exists c
3
> such that(21)
χ
u
≥
c
3
u
p
2
,
∀
u
∈
W
1,2
Ω
,
p
.In fact, if (21) is false, by mean of the 2-homogeneity of χ, there exists {
u
n
}
⊂
W
1,2
(
Ω
,
p
) such that u
n
p
=
1 for all n
≥
1 and χ
u
n
→
0
+ as n
→
∞. Without loss of generality, we may assume that(22)
u
n
⇀
u
weakly in
W
1,2
Ω
,
p
,
u
n
⟶
u
strongly in
L
r
′
Ω
.By the sequential weak lower semicontinuity of σ
(
u
,
u
) and the choice of ϑ, we know that(23)
σ
u
,
u
≤
lim
_
n
→
∞
⁡
σ
u
n
,
u
n
≤
μ
1
u
L
2
Ω
2
.By (17), one gets σ
(
u
,
u
)
=
μ
1
u
L
2
(
Ω
)
2 as well as u
=
ϱ
φ
1 for some constant ϱ. If ϱ
=
0, then u
n
→
0 in W
1,2
(
Ω
,
p
), which contradicts u
n
p
=
1, ∀
n
≥
1; if ϱ
≠
0, by (23), one has σ
φ
1
,
φ
1
<
μ
1
φ
1
L
2
Ω
2; this is a contradiction. Hence, (21) is true.
On the flip side, from (9), we have(24)
σ
u
,
u
+
∫
Ω
ϑ
u
2
d
x
≤
∫
Ω
p
x
∇
u
2
+
ϑ
u
2
d
x
+
q
L
r
Ω
u
2
r
′
2
≤
1
+
ϑ
u
p
2
+
K
q
L
r
Ω
u
p
2
≤
max
⁡
1
+
ϑ
,
K
q
L
r
Ω
u
p
2
.Here K is a positive constant. Then, by (21) and (24), one gets(25)
c
3
u
p
2
≤
σ
u
,
u
+
∫
Ω
ϑ
u
2
d
x
≤
max
⁡
1
+
ϑ
,
K
q
L
r
Ω
u
p
2
.This proved the norms ·
ϑ and ·
p are equivalent.From now on, we always assumeW
1,2
(
Ω
,
p
)
=
(
W
1,2
(
Ω
,
p
)
,
·
ϑ
,
〈
·
,
·
〉
ϑ
)
,
φ
k
ϑ
=
1.Lemma 2.
Under the hypotheses of Lemma1, the embedding W
1,2
(
Ω
,
p
)
↪
L
m
(
Ω
) is continuous for 1
,
2
s
∗, compact for 1
,
2
s
∗.Proof.
By Lemma1 and the Compactness Theorem in [18], we directly conclude Lemma 2.
In addition, from Lemma2, there exists l
>
0 such that u
L
m
(
Ω
)
≤
l
u
ϑ. For simplicity, we will assume that l
=
1; that is,(26)
u
L
m
Ω
≤
u
ϑ
,
∀
u
∈
W
1,2
Ω
,
p
.Now, letG
i
(
x
,
s
)
=
∫
0
s
‍
g
i
(
x
,
t
)
d
t and F
=
{
μ
i
}
i
∈
N. For fixed k
≥
1, suppose that μ
k is an eigenvalue of multiplicity τ and denote by E
μ
k the eigenspace associated with the eigenvalue μ
k, Z
±
=
s
p
a
n
φ
,
±
φ
:
φ
∈
E
μ
k. The main results are as follows.Theorem 3.
Letμ
l
∈
F be the first value above α
-
β and suppose that conditions (
g
0
) and (
g
∞
) hold. Also,(27)
lim
s
→
∞
⁡
G
i
x
,
s
=
+
∞
,
i
=
1,2
,uniformly in x
∈
Ω, and(28)
∫
Ω
h
1
ϕ
+
h
2
ψ
d
x
=
0
,
∀
ϕ
,
ψ
∈
Z
+
.Then for any η
>
0, there exists γ
0
>
0 such that α
+
β
∈
μ
k
-
γ
0
,
μ
k; if d
i
s
t
α
-
β
,
F
>
η, then problem (2) has at least two different solutions.Theorem 4.
If we replace condition (27) of Theorem 3 with(29)
lim
s
→
∞
⁡
G
i
x
,
s
=
-
∞
,
i
=
1,2
,uniformly in x
∈
Ω, then for any η
>
0, there exists γ
1
>
0 such that α
+
β
∈
(
μ
k
,
μ
k
+
γ
1
); if d
i
s
t
(
α
-
β
,
F
)
>
η, then problem (2) has at least two different solutions.Theorem 5.
In addition to conditions(
g
0
), (
g
∞
), and (29), suppose that μ
l
∈
F is the first value above α
+
β and(30)
∫
Ω
h
1
ϕ
+
h
2
ψ
d
x
=
0
,
∀
ϕ
,
ψ
∈
Z
-
.Then for any η
>
0, there exists γ
2
>
0 such that α
-
β
∈
(
μ
k
-
γ
2
,
μ
k
); if d
i
s
t
(
α
+
β
,
F
)
>
η, then problem (2) has at least two different solutions.Theorem 6.
Letμ
l be the first eigenvalue above α
+
β and conditions (
g
0
), (
g
∞
), (27), and (30) hold. Then for any η
>
0, there exists γ
3
>
0 such that α
-
β
∈
(
μ
k
,
μ
k
+
γ
3
); if dist
⁡
α
+
β
,
F
>
η, then problem (2) has at least two different solutions.


## 3. Proof of Main Results

In this section, we firstly prove some preliminary lemmas, and then we prove our main results through variational methods and Galerkin techniques.For the sake of simplicity, letD
W
=
W
1,2
(
Ω
,
p
)
×
W
1,2
(
Ω
,
p
) and D
L
=
L
2
(
Ω
)
×
L
2
(
Ω
), with the norms (31)
z
D
W
=
u
ϑ
2
+
v
ϑ
2
,
z
D
L
=
u
L
2
Ω
2
+
v
L
2
Ω
2
,the inner products ·
,
·
D
W and 〈
·
,
·
〉
D
L, respectively. In addition, we will always use the notation z
=
(
u
,
v
)
,
w
=
(
ϕ
,
ψ
)
∈
D
W, unless otherwise specified.DefineG
:
D
W
→
R and H
:
D
W
→
R
,
∀
z
∈
D
W
(32)
G
z
=
∫
Ω
G
1
x
,
v
+
G
2
x
,
u
d
x
,
H
z
=
∫
Ω
h
1
v
+
h
2
u
d
x
.For every ε
>
0, we claim that there exist positive constants M
ε and C
h such that(33)
G
z
≤
ε
2
z
D
W
2
+
2
M
ε
z
D
W
,
H
z
≤
C
h
z
D
W
,
(34)
G
′
z
,
w
≤
ε
z
D
W
+
2
M
ε
w
D
W
,
H
′
z
,
w
≤
C
h
w
D
W
.In fact, by means of (
g
0
), (
g
∞
), and (26), the arguments of (33) and (34) are quite similar to that of Lemma  3.1 in [11] and so is omitted.Now, we define the functionalJ
(
z
)
:
D
W
→
R, ∀
z
∈
D
W
(35)
J
z
=
1
2
B
z
,
z
-
G
z
-
H
z
,where B
:
D
W
×
D
W
→
R
,
∀
z
,
w
∈
D
W, given by(36)
B
z
,
w
=
∫
Ω
p
x
∇
u
∇
ψ
+
∇
v
∇
ϕ
d
x
+
∫
Ω
q
x
u
ψ
+
v
ϕ
d
x
-
α
∫
Ω
u
ψ
+
v
ϕ
d
x
-
β
∫
Ω
u
ϕ
+
v
ψ
d
x
=
z
,
w
¯
D
W
-
α
+
ϑ
z
,
w
¯
D
L
-
β
z
,
w
D
L
,where w
¯
=
(
ψ
,
ϕ
). By means of (
g
0
) and (
g
∞
), one has J
∈
C
1
(
D
W
,
R
) and (37)
J
′
z
,
w
=
B
z
,
w
-
G
′
z
,
w
-
H
′
z
,
w
,which implies that the critical points of J are exactly weak solutions of problem (2).Next, we need to consider the eigenvalue problem:B
(
z
,
w
)
=
λ
〈
z
,
ϕ
〉
D
W, for λ
∈
R; that is,(38)
B
z
,
w
=
λ
∫
Ω
p
x
∇
u
∇
ϕ
+
∇
v
∇
ψ
+
q
x
+
ϑ
u
ϕ
+
v
ψ
d
x
.Firstly, for any u
,
v
∈
W
1,2
(
Ω
,
p
), one has u
=
∑
k
=
1
∞
‍
a
i
φ
i and v
=
∑
i
=
1
∞
‍
b
i
φ
i for some a
i
,
b
i
∈
R. Now, by φ
i
ϑ
=
1 and using (
ϕ
,
ψ
)
=
(
φ
i
,
0
) and (
ϕ
,
ψ
)
=
(
0
,
φ
i
) in (38), then a straightforward calculation shows that (38) is equivalent to(39)
α
-
μ
i
λ
μ
i
+
ϑ
+
β
λ
μ
i
+
ϑ
+
β
α
-
μ
i
a
i
b
i
=
0
.Obviously, (
a
i
,
b
i
)
≠
0 if and only if (
α
-
μ
i
)
2
-
{
λ
(
ϑ
+
μ
i
)
+
β
}
2
=
0. Hence, we obtain two sequences of eigenvalues(40)
λ
±
i
=
-
β
±
μ
i
-
α
ϑ
+
μ
i
,
∀
i
∈
N
,and the corresponding eigenfunctions(41)
ϕ
±
i
=
2
2
φ
i
,
±
φ
i
,
∀
i
∈
N
,are the corresponding eigenfunctions.LetZ
0
=
Z
∖
0, for i
,
j
∈
Z
0; a simple calculation yields(42)
ϕ
i
D
W
=
1
,
ϕ
i
,
ϕ
j
D
W
=
δ
i
j
,
μ
i
ϕ
i
,
ϕ
j
D
L
=
δ
i
j
,
B
ϕ
i
,
ϕ
j
=
μ
i
-
1
δ
i
j
,where δ
i
j denotes the Kronecker symbol. Moreover, if z
=
∑
i
∈
Z
0
‍
a
i
ϕ
i, then(43)
z
D
W
=
∑
i
∈
Z
0
a
i
2
,
B
z
,
z
=
∑
i
∈
Z
0
λ
i
a
i
2
,
z
D
L
=
∑
i
∈
Z
0
μ
i
-
1
a
i
2
,In addition, for every γ
>
0, if α
+
β
∈
(
μ
k
-
γ
,
μ
k
), from (40) one gets(44)
0
<
λ
k
=
μ
k
-
α
-
β
ϑ
+
μ
k
<
γ
ϑ
+
μ
k
.Let us fix k
≥
1 and define(45)
I
0
=
i
∈
Z
0
:
λ
i
=
λ
k
,
I
-
=
i
∈
Z
0
:
λ
i
≠
λ
k
,
λ
i
<
0
,
I
+
=
i
∈
Z
0
:
λ
i
≠
λ
k
,
λ
i
>
0
,
Z
=
s
p
a
n
ϕ
i
:
i
∈
I
0
¯
,
V
=
s
p
a
n
ϕ
i
:
i
∈
I
-
¯
,
X
=
s
p
a
n
ϕ
i
:
i
∈
I
+
¯
.Meanwhile, we denote by B
V, B
V
Z, B
X, and B
Z
X the unitary closed balls, with respect to the norm ·
D
W, in the spaces V, V
⊕
Z, X, and Z
⊕
X, respectively, and by S
V, S
V
Z, S
X, and S
Z
X their relative boundaries.Lemma 7.
Suppose thatg
i satisfies (
g
0
), (
g
∞
), α
±
β
∉
F. For fixed k
≥
1, let μ
k and μ
l be the first eigenvalue above α
+
β and α
-
β, respectively.
Ifd
i
s
t
(
α
-
β
,
F
)
>
η
>
0, then we have(46)
J
z
≥
C
α
+
β
η
,
∀
z
∈
Z
⊕
X
,
J
z
<
C
α
+
β
η
,
∀
z
∈
ρ
S
V
,
ρ
≥
ρ
α
+
β
η
,for some constants C
α
+
β
η
∈
R and ρ
α
+
β
η
>
0.
Further, if condition (27) also is satisfied, then there exists a positive constant γ
0 such that, for α
+
β
∈
(
μ
k
-
γ
0
,
μ
k
), there exist D
η, C
α
+
β
η
∈
R, ρ
α
+
β
η
>
R
η
>
0 such that (46) hold and(47)
J
z
≥
D
η
,
∀
z
∈
X
,
(48)
J
z
<
D
η
-
1
,
∀
z
∈
R
η
S
V
Z
,
(49)
J
z
<
D
η
-
1
,
∀
z
∈
V
with
z
D
W
>
R
η
.Here, a value D
η with index δ represents that D
η depend on δ, and other cases are similar.Proof.
Firstly, ifk
≥
2, then μ
k
-
1
<
α
+
β
<
μ
k; if k
=
1, then α
+
β
<
μ
1.
Fork
≥
2, if ϑ
+
α
+
β
≥
0, then the sequence {
λ
i
=
1
-
ϑ
+
α
+
β
/
ϑ
+
μ
i
} is nondecreasing, which implies(50)
λ
i
≥
min
⁡
1
-
ϑ
+
α
+
β
ϑ
+
μ
k
,
-
1
+
ϑ
+
α
+
β
ϑ
+
μ
k
-
1
>
0
,
∀
i
∈
N
.If ϑ
+
α
+
β
<
0, then the sequence {
λ
i
=
1
-
ϑ
+
α
+
β
/
ϑ
+
μ
i
} is nonincreasing, which implies(51)
λ
i
≥
l
i
m
k
→
∞
1
-
ϑ
+
α
+
β
ϑ
+
μ
k
=
1
,
∀
i
∈
N
.
Similarly, fork
=
1, that is, α
+
β
<
μ
1, if ϑ
+
α
+
β
≥
0, then the sequence {
λ
i
=
1
-
ϑ
+
α
+
β
/
ϑ
+
μ
i
} is also nondecreasing, which implies(52)
λ
i
≥
λ
1
=
1
-
ϑ
+
α
+
β
ϑ
+
μ
1
>
0
,
∀
i
∈
N
.If ϑ
+
α
+
β
<
0, then the sequence {
λ
i
=
1
-
ϑ
+
α
+
β
/
ϑ
+
μ
i
} is nonincreasing, which implies λ
i
≥
1 for every i
∈
N.
In a word, for fixedk
≥
1, there exists P
α
+
β
>
0 such that(53)
λ
i
≥
P
α
+
β
,
∀
i
∈
N
.
Secondly, because of the fact thatμ
l is the first eigenvalue above α
-
β and d
i
s
t
(
α
-
β
,
F
)
>
η
>
0, thus α
-
β
<
μ
1
-
δ, if l
=
1; μ
l
-
1
+
δ
<
α
-
β
<
μ
l
-
δ, if l
≥
2. Proceeding as in the proof of the first step, we can also conclude that there exists P
η
>
0 such that(54)
λ
-
i
≥
P
η
,
∀
i
∈
N
.Let P
α
+
β
η
=
m
i
n
{
P
α
+
β
,
P
η
}
>
0; we have by (53) and (54)(55)
λ
±
i
≥
P
α
+
β
η
,
∀
i
∈
N
.Hence, as in the proof of Lemma  4.1 in [11], we know that(56)
B
z
,
z
≤
-
P
α
+
β
η
z
D
W
2
,
∀
z
∈
V
,
(57)
B
z
,
z
≥
P
α
+
β
η
z
D
W
2
,
∀
z
∈
Z
⊕
X
.From (33) and (56), we get(58)
J
z
≤
-
P
α
+
β
η
-
ε
2
z
D
W
2
+
2
M
ε
+
C
h
z
D
W
,
∀
z
∈
V
.From (33) and (57), we get(59)
J
z
≥
P
α
+
β
η
-
ε
2
z
D
W
2
-
2
M
ε
+
C
h
z
D
W
,
∀
z
∈
Z
⊕
X
.By (58) and (59) and choosing ε
<
2
P
α
+
β
η, we conclude that there exist C
α
+
β
η
∈
R and ρ
α
+
β
η
>
0 satisfying (46).
In addition, ifα
+
β is near enough to μ
k, in particular, if α
+
β
>
0 and d
i
s
t
(
α
+
β
,
F
∖
{
μ
k
}
)
>
d
>
0, we claim that there exists Q
d
η
>
0 such that(60)
B
z
,
z
≤
-
Q
d
η
z
D
W
2
,
∀
z
∈
V
,
(61)
B
z
,
z
≥
Q
d
η
z
D
W
2
,
∀
z
∈
X
.In fact, if α
+
β
>
0 and d
i
s
t
(
α
+
β
,
F
∖
{
μ
k
}
)
>
d
>
0, for k
≥
2, we have(62)
min
i
∈
N
∖
k
,
…
,
k
+
τ
-
1
⁡
λ
i
=
min
⁡
1
-
ϑ
+
α
+
β
ϑ
+
μ
k
+
τ
,
-
1
+
ϑ
+
α
+
β
ϑ
+
μ
k
-
1
>
min
⁡
d
ϑ
+
μ
k
+
τ
,
d
ϑ
+
μ
k
-
1
>
0
.And, for k
=
1, we have(63)
inf
i
∈
N
∖
1
⁡
λ
i
=
μ
2
-
α
-
β
ϑ
+
μ
2
≥
d
ϑ
+
μ
2
>
0
.Hence, for fixed k
≥
1, there exists Q
d
η
>
0 such that(64)
λ
i
≥
Q
d
η
,
∀
i
∈
Z
0
∖
k
,
…
,
k
+
τ
-
1
.From this we easily get the estimates (60) and (61).
Next, we prove (47), (48), and (49). Let d
=
1
/
2
dist
⁡
(
μ
k
,
F
∖
{
μ
k
}
) and γ
0
∈
(
0
,
d
). If α
+
β
∈
(
μ
k
-
γ
0
,
μ
k
), then d
i
s
t
(
α
+
β
,
F
∖
{
μ
k
}
)
>
d.
Ifz
∈
X, it follows from (33) and (61) that(65)
J
z
≥
Q
d
η
-
ε
2
z
D
W
2
-
2
M
ε
+
C
h
z
D
W
.By choosing ε
=
Q
d
η, then there exists D
η
∈
R satisfying (47).
Ifz
∈
V, by (33) and (60), we obtain(66)
J
z
≤
-
Q
d
η
-
ε
2
z
D
W
2
+
2
M
ε
+
C
h
z
D
W
.Let us choose ε small enough; then there exists R
~
>
0 satisfying (49) for R
η
>
R
~.
Now, we prove the estimate (48). If (48) is not true, then, for any sequences γ
n
→
0
+ and R
n
>
R
~, there exist z
n
∈
R
n
S
V
Z and α
n
,
β
n
∈
R such that(67)
α
n
+
β
n
∈
μ
k
-
γ
n
,
μ
k
,
dist
⁡
α
n
-
β
n
,
F
∖
μ
k
>
η
,
J
n
z
n
≥
D
η
-
1
.Here B
n (or J
n) denotes the form B (or the functional J) with α
=
α
n and β
=
β
n. And, {
λ
i
n
}
i
∈
Z
0 denotes the eigenvalues of the bilinear form B
n.
Letz
n
=
ϖ
n
+
τ
n
∈
V
⊕
Z, and suppose R
n
→
∞ and γ
n
R
n
2
→
0 as n
→
∞. By (44), one has 0
<
λ
k
n
<
γ
n
/
ϑ
+
μ
k, which implies B
n
(
τ
n
,
τ
n
)
≤
γ
n
/
ϑ
+
μ
k
τ
n
D
W
2. So, by (60), we get(68)
B
n
z
n
,
z
n
≤
B
n
ϖ
n
,
ϖ
n
+
B
n
τ
n
,
τ
n
≤
γ
n
ϑ
+
μ
k
τ
n
D
W
2
-
Q
d
η
ϖ
n
D
W
2
,for all positive integers n; we get by (68)(69)
D
η
-
1
≤
J
n
z
n
≤
γ
n
ϑ
+
μ
k
τ
n
D
W
2
-
Q
d
η
ϖ
n
D
W
2
-
G
z
n
-
H
z
n
.It follows from (33) and (69) that(70)
Q
d
η
-
ε
2
ϖ
n
D
W
2
R
n
2
≤
γ
n
ϑ
+
μ
k
+
ε
2
τ
n
D
W
2
R
n
2
-
D
η
-
1
R
n
2
+
2
M
ε
+
C
h
z
D
W
R
n
2
.We note that τ
n
D
W
≤
z
n
D
W
=
R
n; then from (70) we obtain(71)
lim
⁡
s
u
p
n
→
∞
ϖ
n
D
W
2
R
n
2
≤
ε
′
,where ε
′
=
ε
/
(
2
Q
d
η
-
ε
). Note that dim
⁡
Z
<
∞
(72)
ϖ
n
D
W
2
R
n
2
+
τ
n
D
W
2
R
n
2
=
1
,
τ
n
D
W
R
n
∈
Z
.Then, we get by (71)(73)
l
i
m
n
→
∞
τ
n
D
W
R
n
≥
1
-
ε
′
,
l
i
m
n
→
∞
ϖ
n
D
W
R
n
≤
ε
′
.Let z
~
n
=
ϖ
~
n
+
τ
~
n, where ϖ
~
n
=
ϖ
n
/
R
n, τ
~
n
=
τ
n
/
R
n. Then, by (73), there exists z
~
0
=
ϖ
~
0
+
τ
~
0
∈
D
W with z
~
0
D
W
=
1, such that(74)
l
i
m
n
→
∞
τ
~
n
D
W
=
τ
~
0
D
W
≥
1
-
ε
′
,
l
i
m
n
→
∞
ϖ
~
n
D
W
=
ϖ
~
0
D
W
≤
ε
′
.Let z
=
(
P
1
z
,
P
2
z
)
∈
D
W. Then, we have (75)
z
~
n
D
W
2
=
P
1
z
~
n
,
P
2
z
~
n
D
W
2
=
P
1
ϖ
~
n
+
P
1
τ
~
n
,
P
2
ϖ
~
n
+
P
2
τ
~
n
D
W
2
=
P
1
ϖ
~
n
ϑ
2
+
P
1
τ
~
n
ϑ
2
+
P
2
ϖ
~
n
ϑ
2
+
P
2
τ
~
n
ϑ
2
=
1
.From this and (74), without loss of generality, we assume(76)
P
1
τ
~
0
ϑ
≥
1
-
ε
′
2
,
P
1
ϖ
~
0
ϑ
≤
ε
′
.By dim
⁡
Z
<
∞, one has P
1
τ
~
0
L
1
(
Ω
)
>
C
1
-
ε
′
/
2 for some positive constant C. Besides, from (26) and the second inequality of (76), we obtain P
1
ϖ
~
0
L
1
(
Ω
)
<
ε
′. Thus, by choosing ε small enough, there exists ζ
>
0 such that(77)
P
1
τ
~
0
L
1
Ω
>
3
ζ
,
P
1
ϖ
~
0
L
1
Ω
<
ζ
.From this and (74), there exist η
>
0 and n
0
>
0 such that(78)
Ω
P
1
z
~
n
=
x
∈
Ω
:
P
1
z
~
n
x
>
η
>
η
for
n
>
n
0
.By (
g
0
) and (27), proceeding as in the proof of Proposition  4.6 in [11], we have(79)
lim
R
→
∞
⁡
inf
n
>
n
0
⁡
∫
Ω
G
2
x
,
R
P
1
z
~
n
d
x
=
+
∞
.Further, because of the fact that ∫
Ω
‍
G
1
(
x
,
R
P
2
z
~
n
)
d
x is bounded from below, we get(80)
lim
n
→
∞
⁡
G
z
n
=
lim
n
→
∞
⁡
G
R
n
z
~
n
=
+
∞
.In addition, for fixed d and η, from (54) and (64), we can choose γ small enough such that μ
i
=
μ
k
,
i
=
k
,
…
,
k
+
τ
-
1; then we get Z
=
Z
+. Hence, by (28), it is easy to see that Q
d
η
ϖ
n
D
W
2
≥
-
δ
2 for some positive constant δ
2. Moreover, it follows from (69) that (81)
G
z
n
≤
γ
n
ϑ
+
μ
k
τ
n
D
W
2
-
D
η
+
1
+
δ
2
.Recall that τ
n
D
W
≤
R
n and γ
n
R
n
2
→
0 as n
→
∞, which contradicts with (80). Hence, there exist R
η
>
R
~
>
0 and γ
0
∈
(
0
,
d
) which satisfy (48).
Finally, by the process of the above proof, we easily know that all the constants of the estimates above are not contradictory; then we finished the proof of Lemma7.Remark 8.
In fact, from (46), we can get a solution of problem (2). And, from (47), (48), and (49), we can also get a solution of problem (2). But we do not know whether they are different. So we will use Galerkin techniques to show the existence of two different solutions.
Now, we assume thatn
>
k
+
τ
(82)
D
W
n
=
span
⁡
ϕ
-
n
,
…
,
ϕ
n
,
V
n
=
V
∩
D
W
n
,
X
n
=
X
∩
D
W
n
.It follows from this that Z
⊂
D
W
n. As before, let B
V
n
=
B
1
∩
V
n, B
V
Z
n
=
B
1
∩
(
V
n
⊕
Z
), B
Z
X
n
=
B
1
∩
(
Z
⊕
X
n
), S
V
n, S
V
Z
n, and S
Z
X
n represent their relative boundary. In addition, we know that all the estimates of Lemma 7 are true when the conditions of Theorem 3 are satisfied. Further, if d
i
s
t
(
α
-
β
,
F
)
>
η
>
0 and α
+
β
∈
(
μ
k
-
γ
0
,
μ
k
), we can also check that the similar estimates of Lemma 7, with respect to the functional J
n, hold on the spaces V
n, Z, and X
n.Lemma 9.
Fixn
>
k
+
τ, assume that α
±
β
∉
F, h
1
,
h
2
∈
L
2
(
Ω
), and g
1 and g
2 satisfy (
g
0
) and (
g
∞
). In addition, let {
z
i
}
⊂
D
W
n satisfy(83)
J
′
z
i
,
w
<
δ
i
w
D
W
,
∀
w
=
ϕ
,
ψ
∈
D
W
n
,where δ
i
→
0 as i
→
∞. Then, there exists a subsequence {
z
i
k
}
⊂
{
z
i
} such that {
z
i
k
} converges in D
W
n.Proof.
Ifi
∈
Z
0, then λ
i
≠
0. Hence, we may divide the space D
W in the two orthogonal components(84)
D
W
-
=
s
p
a
n
ϕ
i
:
i
∈
Z
0
,
λ
i
<
0
¯
,
D
W
+
=
D
W
∖
D
W
-
.For j
>
k
+
τ, let D
W
j
+
=
D
W
+
∩
D
W
j
,
D
W
j
-
=
D
W
-
∩
D
W
j. In addition, because of λ
i
→
±
1 as i
→
±
∞, then we may set λ
0
=
i
n
f
|
λ
i
|
:
i
∈
Z
0
>
0, and then(85)
B
z
,
z
≤
-
λ
0
z
D
W
2
,
∀
z
∈
D
W
-
,
(86)
B
z
,
z
≥
λ
0
z
D
W
2
,
∀
z
∈
D
W
+
.Setting z
i
=
z
i
-
+
z
i
+
∈
D
W
-
⊕
D
W
+, by testing (83) with w
=
z
i
- we have(87)
B
z
i
,
z
i
-
-
G
′
z
i
,
z
i
-
-
H
′
z
i
,
z
i
-
≤
δ
i
z
i
-
D
W
.From this, (34), (42), and (85), we get(88)
λ
0
z
i
-
D
W
≤
δ
i
+
ε
z
i
D
W
+
2
M
ε
+
C
h
.In the same way, by testing (83) with w
=
z
i
+ we have(89)
λ
0
z
i
+
D
W
≤
δ
i
+
ε
z
i
D
W
+
2
M
ε
+
C
h
.From (88) and (89), we have(90)
λ
0
-
2
ε
z
i
D
W
≤
2
δ
i
+
4
M
ε
+
2
C
h
.Let us choose ε
<
λ
0
/
2; then {
z
i
} is bounded in D
W
n. By dim
⁡
D
W
n
<
∞, then there exists {
z
i
k
}
⊂
{
z
i
} such that {
z
i
k
} converges in D
W
n.From the proof of Lemma9, we can also get Lemma 10 below.Lemma 10.
Assume thatα
±
β
∉
F, h
1
,
h
2
∈
L
2
(
Ω
), and g
1 and g
2 satisfy (
g
0
) and (
g
∞
). In addition, let {
z
i
}
⊂
D
W
i satisfy(91)
J
′
z
i
,
w
<
δ
i
w
D
W
for every
w
∈
D
W
i
,where δ
i
→
0 as i
→
∞. Then {
z
i
} is bounded in D
W.
Clearly, by using the Saddle Point Theorem, Lemma9, (46), (47), and (48), there exist z
n
=
(
u
n
,
v
n
) and w
n in D
W
n such that J
n
(
z
n
)
=
c, J
n
′
(
z
n
)
=
0, and J
n
(
w
n
)
=
s
n, J
n
′
(
w
n
)
=
0, where(92)
c
n
=
inf
γ
∈
Γ
V
n
⁡
sup
z
∈
ρ
α
+
β
η
B
V
n
⁡
J
n
γ
z
≥
C
α
+
β
η
,
s
n
=
inf
γ
∈
Γ
VZ
n
⁡
sup
w
∈
R
η
B
VZ
n
⁡
J
n
γ
z
≥
D
η
,
Γ
V
n
=
γ
∈
C
ρ
α
+
β
η
B
V
n
,
D
W
n
:
γ
z
=
z
,
u
D
W
=
ρ
α
+
β
η
,
Γ
V
Z
n
=
γ
∈
C
R
η
B
V
Z
n
,
D
W
n
:
γ
w
=
w
,
w
D
W
=
R
η
.Lemma 11.
Assume that the conditions are the same as Lemma7, α
+
β
∈
(
μ
k
-
γ
0
,
μ
k
), and d
i
s
t
(
α
-
β
,
F
)
>
η
>
0. Then, for all n
>
k
+
τ, there exists T
η
>
0 such that c
n
∈
[
C
α
+
β
η
,
D
η
-
1
] and s
n
∈
[
D
η
,
T
η
].Proof.
Firstly, by (92), one gets c
n
≥
D
α
+
β and s
n
≥
D
η. Define(93)
γ
0
p
=
p
,
R
η
≤
p
D
W
≤
ρ
α
+
β
η
,
p
+
R
η
2
-
p
D
W
2
ϕ
k
,
p
D
W
≤
R
η
.So it follows from (48) and (49) that (94)
s
u
p
z
∈
ρ
α
+
β
η
B
V
n
J
n
γ
0
z
<
D
η
-
1
,which implies c
n
<
D
η
-
1. In addition, by I
d
|
R
η
B
V
Z
n
∈
Γ
V
Z
n, one gets s
n
<
s
u
p
w
∈
R
η
B
V
Z
n
J
(
w
).
Forz
∈
V
⊕
Z, it follows from α
+
β
∈
(
μ
k
-
γ
0
,
μ
k
) that(95)
-
P
α
+
β
η
<
0
<
λ
k
≤
γ
0
ϑ
+
μ
k
.From this and (33), we get for any z
∈
R
η
B
V
Z
n
(96)
J
z
≤
γ
0
ϑ
+
μ
k
+
ε
z
D
W
2
+
2
M
ε
+
C
h
z
D
W
.Hence, there exists T
η
>
0 such that s
n
≤
T
η. This finishes the proof of Lemma 11.Now, we start to prove our main theorems.Proof of Theorem3.
Firstly, by Lemma11 and without loss of generality, there exist c
∈
[
C
α
+
β
η
,
D
η
-
1
] and s
∈
[
D
η
,
T
η
] such that c
n
→
c and s
n
→
s as n
→
+
∞.
Next, we prove that there existsz
∈
D
W such that J
(
z
)
=
c and J
′
(
z
)
=
0. In fact, for all n
>
k
+
τ, we have(97)
J
n
z
n
=
c
n
,
(98)
J
n
′
z
n
,
w
=
0
,
∀
w
∈
D
W
n
.Then, it follows from Lemma 10 that {
z
n
} is bounded in D
W. Hence, without loss of generality, there exists z
∈
D
W such that(99)
z
n
⇀
z
in
DW
,
z
n
⟶
z
in
D
L
.For l
>
k
+
τ, let w
=
(
ϕ
,
0
)
∈
D
W
l and w
=
(
0
,
ψ
)
∈
D
W
l in (98), respectively. From a direct calculation, we have for all n
>
l
(100)
u
n
,
ψ
ϑ
=
α
+
ϑ
∫
Ω
u
n
ψ
d
x
+
β
∫
Ω
v
n
ψ
d
x
+
∫
Ω
g
1
x
,
v
n
ψ
d
x
+
∫
Ω
h
1
ψ
d
x
,
v
n
,
ϕ
ϑ
=
α
+
ϑ
∫
Ω
v
n
ϕ
d
x
+
β
∫
Ω
u
n
ϕ
d
x
+
∫
Ω
g
2
x
,
u
n
ϕ
d
x
+
∫
Ω
h
2
ϕ
d
x
.Let n
→
∞; we get 〈
J
′
(
z
)
,
w
〉
=
0 for any w
∈
D
W
l. It follows from D
W
=
∪
l
∈
N
D
W
l
¯ that J
′
(
z
)
=
0. That is, z
=
(
u
,
v
) is a critical point of the functional J.
Next, we proveJ
(
z
)
=
c. Firstly, let us define the orthogonal projection (101)
P
n
:
W
1,2
Ω
,
p
⟶
span
⁡
φ
1
,
…
,
φ
n
,which implies P
n
u
→
u, P
n
v
→
v in W
1,2
(
Ω
,
p
). Then(102)
l
i
m
n
→
∞
u
n
-
P
n
u
L
2
Ω
+
v
n
-
P
n
v
L
2
Ω
=
0
,
u
n
L
2
Ω
+
v
n
L
2
Ω
<
∞
.It follows from (102) that(103)
α
∫
Ω
u
n
u
n
-
P
n
u
d
x
+
β
∫
Ω
v
n
u
n
-
P
n
u
d
x
+
∫
Ω
g
1
x
,
v
n
u
n
-
P
n
u
d
x
⟶
0
,
α
∫
Ω
v
n
v
n
-
P
n
v
d
x
+
β
∫
Ω
u
n
v
n
-
P
n
v
d
x
+
∫
Ω
g
2
x
,
u
n
v
n
-
P
n
v
d
x
⟶
0
,
∫
Ω
h
1
u
n
-
P
n
u
d
x
+
∫
Ω
h
2
v
n
-
P
n
v
d
x
⟶
0
.Let w
=
(
v
n
-
P
n
v
,
u
n
-
P
n
u
) in (98); one gets(104)
J
n
′
z
n
,
v
n
-
P
n
v
,
u
n
-
P
n
u
=
0
,
∀
n
>
k
+
τ
.From this, (103), we obtain(105)
∫
Ω
p
x
∇
u
n
∇
u
n
-
P
n
u
+
q
x
u
n
u
n
-
P
n
u
d
x
+
∫
Ω
p
x
∇
v
n
∇
v
n
-
P
n
v
+
q
x
v
n
v
n
-
P
n
v
d
x
⟶
0
,which implies(106)
∫
Ω
p
x
∇
u
n
2
+
∇
v
n
2
d
x
+
q
x
u
n
2
+
v
n
2
d
x
-
∫
Ω
p
x
∇
u
n
∇
u
+
∇
v
n
∇
v
+
q
x
u
n
u
+
v
n
v
d
x
+
∫
Ω
p
x
∇
u
n
∇
u
-
P
n
u
+
∇
v
n
∇
v
-
P
n
v
d
x
+
∫
Ω
q
x
u
n
u
-
P
n
u
+
v
n
v
-
P
n
v
d
x
⟶
0
.By (
P
n
u
,
P
n
v
)
-
(
u
,
v
)
D
W
→
0 and (106), one has z
n
D
W
→
z
D
W. From the uniform convexity of D
W, we have z
n
-
z
D
W
→
0 as n
→
∞, which implies that J
(
z
)
=
c.
In the same way, we can also prove that there existsw
∈
D
W such that w
n
-
w
D
W
→
0, J
′
(
w
)
=
0, and J
(
w
)
=
s. Note that (107)
c
=
J
z
≤
D
η
-
1
<
D
η
≤
J
ψ
=
s
.Hence, we get z
≠
w, which finished the proof of Theorem 3.Proof of Theorem4.
LetB
~
(
z
,
w
)
=
-
B
(
z
,
w
), and we use the same approach as before; we also get eigenvalues of B
~:(108)
λ
~
±
i
=
-
λ
±
i
=
-
-
β
±
μ
i
-
α
ϑ
+
μ
i
,
∀
i
∈
N
,and the same eigenfunctions(109)
ϕ
±
i
=
2
2
φ
i
,
±
φ
i
,
i
∈
N
.In the same way, we can also define the subspaces V
~, Z
~, and X
~. If α
+
β
∈
(
μ
k
,
μ
k
+
γ
) for some γ
>
0, then(110)
0
<
λ
~
k
=
α
+
β
-
μ
k
ϑ
+
μ
k
<
γ
ϑ
+
μ
k
.Next, we prove that (56), (57), (60), and (61), with respect to B
~, hold on the new subspaces V
~, Z
~, and X
~. Firstly, we prove the following claim.
Assume thatμ
k and μ
l are the first eigenvalue above α
+
β and α
-
β, respectively. If α
±
β
∉
F and d
i
s
t
(
α
-
β
,
F
)
>
η
>
0, then there exists P
α
+
β
η
>
0 such that(111)
B
~
z
,
z
≤
-
P
α
+
β
η
z
D
W
2
,
∀
z
∈
V
~
,
B
~
z
,
z
≥
P
α
+
β
η
z
D
W
2
,
∀
z
∈
Z
~
⊕
X
~
.Besides, if α
+
β is close to λ
k, or d
i
s
t
(
α
+
β
,
F
∖
{
μ
k
}
)
>
d
>
0, then(112)
B
~
z
,
z
≤
-
Q
d
η
z
D
W
2
,
∀
z
∈
V
~
,
B
~
z
,
z
≥
Q
d
η
z
D
W
2
,
∀
z
∈
X
~
,for some positive constant Q
d
η.
Actually, from (108) one can prove that an estimate like (55) holds for these new eigenvalues λ
~
±
i, and then we get (111). Moreover, when α
+
β is close to μ
k, we can also check that, as in the proof of (60) and (61), there exists the positive constant Q
d
η satisfying (112).
Similarly, we can also chooseγ small enough such that Z
~
=
Z
+. By (111) and (112), we can prove a similar result of Lemma 7. In other words, if the conditions of Theorem 4 are satisfied, we can conclude that there exist positive constants C
α
+
β
η and ρ
α
+
β
η such that the functional J satisfies (46) on the new subspaces V
~, Z
~, and X
~. Further, there exists γ
1
>
0 such that if α
+
β
∈
(
λ
k
,
λ
k
+
γ
1
), then the functional J satisfies (46), (47), (48), and (49). Hence, we can also obtain two critical point sequences {
z
~
n
} and {
ψ
~
n
} of J
n
=
J
|
D
W
n, at critical levels c
~
n, s
~
n, similar to (92). Next, the remainder of the argument is similar to the proof of Theorem 3.Proof of Theorem5 (or Theorem 6).
In order to take advantage of the conclusion of Theorem3 (or Theorem 4), we consider the following degenerate system:(113)
-
div
⁡
p
x
∇
u
+
q
x
u
=
α
u
+
β
~
v
+
g
~
1
x
,
v
+
h
1
x
,
x
∈
Ω
,
-
div
⁡
p
x
∇
v
+
q
x
v
=
β
~
u
+
α
v
+
g
~
2
x
,
u
+
h
~
2
x
,
x
∈
Ω
,
∂
u
∂
ν
=
∂
v
∂
ν
=
0
,
x
∈
∂
Ω
,where β
~
=
-
β, g
~
1
(
x
,
v
)
=
g
1
(
x
,
-
v
), g
~
2
(
x
,
u
)
=
-
g
2
(
x
,
u
), and h
~
2
=
-
h
2. Obviously, under the hypotheses of Theorem 5 (or Theorem 6), the similar hypotheses of Theorem 3 (or Theorem 4) are satisfied for problem (113). Meanwhile, if (
u
,
v
) is a solution of (113), then (
u
,
-
v
) is a solution of problem (2). Hence, by the proof of Theorem 3 (or Theorem 4), we know that Theorem 5 (or Theorem 6) is true.



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*Source: 2925065-2017-01-11.xml*