# Boundary Value Problems for Fourth Order Nonlinearp-Laplacian Difference Equations

Authors: Qinqin Zhang
Journal: Journal of Applied Mathematics
 (2014)

---

## Abstract

We consider the boundary value problem for a fourth order nonlinearp-Laplacian difference equation containing both advance and retardation. By using Mountain pass lemma and some established inequalities, sufficient conditions of the existence of solutions of the boundary value problem are obtained. And an illustrative example is given in the last part of the paper.


---

## Body

## 1. Introduction

Letℕ, ℤ, and ℝ denote the sets of all natural numbers, integers, and real numbers, respectively. For a
,
b
∈
ℤ, define ℤ
(
a
)
=
{
a
,
a
+
1
,
…
} and ℤ
(
a
,
b
)
=
{
a
,
a
+
1
,
…
,
b
} when a
≤
b.Consider the following fourth order nonlinear difference equation:(1)
Δ
2
(
r
n
-
2
ϕ
p
(
Δ
2
u
n
-
2
)
)
-
f
(
n
,
u
n
+
1
,
u
n
,
u
n
-
1
)
=
0
,
n
∈
ℤ
(
1
,
k
)
,
with boundary value conditions
(2)
u
-
1
=
u
0
=
u
k
+
1
=
u
k
+
2
=
0
,
where k
∈
ℕ, r
j is a positive number for j
∈
ℤ
(
-
1
,
k
), Δ is the forward difference operator defined by Δ
u
n
=
u
n
+
1
-
u
n, Δ
2
u
n
=
Δ
(
Δ
u
n
), and ϕ
p is the p-Laplacian operator; that is, ϕ
p
(
s
)
=
|
s
|
p
-
2
s
(
p
>
1
), f
∈
C
(
ℤ
(
1
,
k
)
×
ℝ
3
,
ℝ
).In the last decade, by using various techniques such as critical point theory, fix point theory, topological degree theory, and coincidence degree theory, a great deal of works have been done on the existence of solutions to boundary value problems of difference equations (see [1–7] and references therein). Among these approaches, the critical point theory seems to be a powerful tool to deal with this problem (see [5, 7–9]). However, compared to the boundary value problems of lower order difference equations ([6, 8, 10–13]), the study of boundary value problems of higher order difference equations is relatively rare (see [9, 14, 15]), especially the works by using the critical point theory [16]. For the background on difference equations, we refer to [17].In this paper, we will consider the existence of solutions of the boundary value problem of (1) with (2). First, we will construct a functional J such that solutions of the boundary value problem (1) with (2) correspond to critical points of J. Then, by using Mountain pass lemma, we obtain the existence of critical points of J. We mention that (1) is a kind of difference equation containing both advance and retardation. This kind of difference equation has many applications both in theory and practice. For example, in [17], Agarwal considered the following difference equation:
(3)
-
ω
2
M
u
(
k
)
+
f
(
-
u
(
k
-
1
)
+
2
u
(
k
)
-
u
(
k
+
1
)
)
=
0
,
with the boundary value conditions
(4)
u
(
0
)
=
u
(
k
+
1
)
=
0
,
as an example. It represents the amplitude of the motion of every particle in the string. And in [7], the authors considered the following second order functional difference equation:
(5)
L
u
n
=
f
(
n
,
u
n
+
1
,
u
n
,
u
n
-
1
)
,
with different boundary value conditions
(6)
Δ
x
0
=
A
,
x
T
+
1
=
B
,
where the operator L is the Jacobi operator given by
(7)
L
u
n
=
a
n
u
n
+
1
+
a
n
-
1
u
n
-
1
+
b
n
u
n
.
In [18], the authors considered the second order p-Laplacian difference equation:
(8)
Δ
(
ϕ
p
(
Δ
u
n
-
1
)
)
+
f
(
n
,
u
n
+
1
,
u
n
,
u
n
-
1
)
=
0
,
n
∈
ℤ
(
1
,
k
)
,
with boundary value conditions
(9)
u
0
=
u
k
+
1
=
0
.
As for the periodic and subharmonic solutions of p-Laplacian difference equations containing both advance and retardation, we refer to [19]. And for the periodic solutions of ϕ-Laplacian difference equations, we refer to [20].Throughout this paper, we assume that there exists a functionF
(
n
,
u
,
v
) which is differentiable in (
u
,
v
) and F
(
n
,
0,0
)
=
0 for each n
∈
ℤ
(
0
,
k
), satisfying
(10)
∂
F
(
n
-
1
,
v
,
w
)
∂
v
+
∂
F
(
n
,
u
,
v
)
∂
v
=
f
(
n
,
u
,
v
,
w
)
,
∀
u
,
v
,
w
∈
ℝ
,
for n
∈
ℤ
(
1
,
k
).


## 2.  Preliminaries and Main Results

Lemma 1.
Letp
∈
(
1
,
∞
); then there exist two positive sequences {
c
*
(
n
)
}
n
∈
ℕ and {
c
*
(
n
)
}
n
∈
ℕ such that
(11)
c
*
(
b
-
a
+
1
)
(
∑
j
=
a
b
‍
|
u
j
|
2
)
1
/
2
≤
(
∑
j
=
a
b
‍
|
u
j
|
p
)
1
/
p
≤
c
*
(
b
-
a
+
1
)
(
∑
j
=
a
b
‍
|
u
j
|
2
)
1
/
2
holds for any a
,
b
∈
ℤ with a
<
b, where c
*
(
n
)
=
1, c
*
(
n
)
=
n
(
2
-
p
)
/
2
p for p
∈
(
1,2
] and c
*
(
n
)
=
n
-
(
p
-
2
)
/
2
p, c
*
(
n
)
=
1 for p
∈
(
2
,
∞
).Proof.
If1
<
p
≤
2, by Hölder’s inequality, we have
(12)
∑
j
=
a
b
‍
|
u
j
|
p
≤
(
∑
j
=
a
b
‍
1
2
/
(
2
-
p
)
)
(
2
-
p
)
/
2
(
∑
j
=
a
b
‍
|
u
j
|
2
)
p
/
2
=
(
b
-
a
+
1
)
(
2
-
p
)
/
2
(
∑
j
=
a
b
‍
|
u
j
|
2
)
p
/
2
,
which implies that
(13)
(
∑
j
=
a
b
‍
|
u
j
|
p
)
1
/
p
≤
(
b
-
a
+
1
)
(
2
-
p
)
/
2
p
(
∑
j
=
a
b
‍
|
u
j
|
2
)
1
/
2
,
and c
*
(
n
)
=
1 is obvious. If p
>
2, then we have
(14)
∑
j
=
a
b
‍
|
u
j
|
2
≤
(
∑
j
=
a
b
‍
1
p
/
(
p
-
2
)
)
(
p
-
2
)
/
p
(
∑
j
=
a
b
‍
|
u
j
|
p
)
2
/
p
=
(
b
-
a
+
1
)
(
p
-
2
)
/
p
(
∑
j
=
a
b
‍
|
u
j
|
p
)
2
/
p
,
which implies that
(15)
(
∑
j
=
a
b
‍
|
u
j
|
p
)
1
/
p
≥
(
b
-
a
+
1
)
-
(
p
-
2
)
/
2
p
(
∑
j
=
a
b
‍
|
u
j
|
2
)
1
/
2
,
and c
*
(
n
)
=
1 is obvious. Now the proof is complete.Lemma 2.
There exist two positive sequences{
λ
*
(
n
)
}
n
∈
ℕ and {
λ
*
(
n
)
}
n
∈
ℕ such that
(16)
λ
*
(
b
-
a
+
1
)
∑
j
=
a
b
‍
u
j
2
≤
∑
j
=
a
-
1
b
‍
(
Δ
u
j
)
2
≤
λ
*
(
b
-
a
+
1
)
∑
j
=
a
b
‍
u
j
2
holds for any a
,
b
∈
ℤ with a
<
b and u
a
-
1
=
u
b
+
1
=
0, where
(17)
λ
*
(
n
)
=
4
sin
2
π
2
(
n
+
1
)
,
λ
*
(
n
)
=
4
sin
2
n
π
2
(
n
+
1
)
.Proof.
There is no harm in assuming thata
=
1, b
=
k. Then
(18)
∑
j
=
0
k
‍
(
Δ
u
j
)
2
=
(
u
1
,
u
2
,
…
,
u
k
)
A
(
u
1
,
u
2
,
…
,
u
k
)
tr
,
where (
·
)
tr means the transpose of (
·
), and A is the k
×
k matrix given by
(19)
A
=
(
2
-
1
0
⋯
0
0
-
1
2
-
1
⋯
0
0
0
-
1
2
⋯
0
0
⋯
⋯
⋯
⋯
⋯
⋯
0
0
0
⋯
2
-
1
0
0
0
⋯
-
1
2
)
.
We will calculate the eigenvalues of A. Similar to [21], assume that λ is an eigenvalue of A. Since A
-
r
I is positive-definite for r
<
0 and negative-definite for r
>
4, where I is the k
×
k identity matrix, we see that λ
∈
[
0,4
]. Assume that ξ
=
(
ξ
1
,
ξ
2
,
…
,
ξ
k
)
tr is an eigenvector associated to λ and define the sequence {
y
n
}
n
=
0
k
+
1 as
(20)
y
i
=
ξ
i
,
i
∈
ℤ
(
1
,
k
)
,
y
0
=
y
k
+
1
=
0
.
Then {
y
n
} satisfies
(21)
-
y
n
+
1
+
(
2
-
λ
)
y
n
-
y
n
-
1
=
0
,
y
0
=
y
k
+
1
=
0
,
n
∈
ℤ
(
1
,
k
)
.
Since the roots of the equation -
r
2
+
(
2
-
λ
)
r
-
1
=
0 are
(22)
r
1
=
1
2
(
2
-
λ
+
4
-
(
2
-
λ
)
2
)
i
,
r
2
=
1
2
(
2
-
λ
-
4
-
(
2
-
λ
)
2
)
i
,
set
(23)
θ
=
arccos
1
2
(
2
-
λ
)
.
Then
(24)
y
n
=
d
1
cos
n
θ
+
d
2
sin
n
θ
for some constants d
1 and d
2. y
0
=
0 implies that d
1
=
0, and y
k
+
1
=
0 implies that sin
(
k
+
1
)
θ
=
0. Therefore, (
k
+
1
)
θ
=
j
π for j
∈
ℤ
(
1
,
k
). By (23), we have λ
=
2
(
1
-
cos
θ
)
=
4
sin
2
θ
/
2 which implies that the eigenvalues of A are
(25)
λ
j
=
4
sin
2
j
π
2
(
k
+
1
)
,
j
∈
ℤ
(
1
,
k
)
.
The maximum eigenvalue of A is λ
k, and the minimal eigenvalue of A is λ
1. Equation (16) follows from (18) and the fact that λ
k
=
λ
*
(
k
) and λ
1
=
λ
*
(
k
).Before we apply the critical point theory, we will establish the corresponding variational framework for (1) with (2).Let(26)
E
=
{
u
=
{
u
n
}
:
ℤ
(
-
1
,
k
+
2
)
⟶
ℝ
∣
u
-
1
=
u
0
=
u
k
+
1
=
u
k
+
2
=
0
}
.
Then E is a k-dimensional Hilbert space.Obviously,E is isomorphic to ℝ
k. In fact, we can find a map I
:
E
→
ℝ
k defined by
(27)
I
:
{
u
n
}
⟶
(
u
1
,
u
2
,
…
,
u
k
)
tr
.Define the inner product onE as
(28)
〈
u
,
v
〉
=
∑
j
=
1
k
‍
u
j
v
j
,
∀
u
,
v
∈
E
.
The corresponding norm ∥
·
∥ can be induced by
(29)
∥
u
∥
=
(
∑
j
=
1
k
‍
|
u
j
|
2
)
1
/
2
,
∀
u
∈
E
.For allu
∈
E, define the functional J
(
u
) on E as follows:
(30)
J
(
u
)
=
1
p
∑
n
=
-
1
k
‍
r
n
|
Δ
2
u
n
|
p
-
∑
n
=
0
k
‍
F
(
n
,
u
n
+
1
,
u
n
)
.
Clearly, J
∈
C
1
(
E
,
ℝ
). We can compute the partial derivative as
(31)
∂
J
(
u
)
∂
u
j
=
Δ
2
(
r
j
-
2
ϕ
p
(
Δ
2
u
j
-
2
)
)
-
f
(
j
,
u
j
+
1
,
u
j
,
u
j
-
1
)
,
for j
∈
ℤ
(
1
,
k
), u
=
{
u
j
}
∈
E. Therefore, u
∈
E is a critical point of J if and only if u is a solution of (1) with (2).Definition 3.
LetE be a real Banach space; the functional J
∈
C
1
(
E
,
ℝ
) is said to satisfy the Palais-Smale (P.S. for short) condition if any sequence {
x
m
} in E such that {
J
(
x
m
)
} is bounded and J
′
(
x
m
)
→
0 as m
→
∞ contains a convergent subsequence.LetB
ρ denote the open ball in E with radius ρ and center 0, and let ∂
B
ρ denote its boundary.In order to obtain the existence of critical points ofJ on E, we need to use the following basic lemma, which is important in the proof of our main results.Lemma 4 (Mountain pass lemma [22]).
LetE be a real Hilbert space and J
∈
C
1
(
E
,
ℝ
) satisfies the P.S. condition, if J
(
0
)
=
0 and the following conditions hold.(
J
1
)
There exist constantsa
>
0 and ρ
>
0 such that J
|
∂
B
ρ
≥
a.
(
J
2
)
There existse
∈
E
∖
B
ρ such that J
(
e
)
<
0.Then J possesses a critical value c
≥
a given by
(32)
c
=
inf
g
∈
Γ
max
s
∈
[
0,1
]
J
(
g
(
s
)
)
,
where
(33)
Γ
=
{
g
∈
C
(
[
0,1
]
,
E
)
∣
g
(
0
)
=
0
,
g
(
1
)
=
e
}
.Let(34)
r
*
=
min
{
r
n
∣
n
∈
ℤ
(
-
1
,
k
)
}
,
r
*
=
max
{
r
n
∣
n
∈
ℤ
(
-
1
,
k
)
}
,
α
0
=
r
*
p
(
c
*
(
k
+
2
)
c
*
(
k
+
1
)
)
p
(
λ
*
(
k
+
1
)
λ
*
(
k
)
2
)
p
/
2
,
β
0
=
r
*
p
(
c
*
(
k
+
2
)
c
*
(
k
+
1
)
)
p
(
λ
*
(
k
+
1
)
λ
*
(
k
)
2
)
p
/
2
.
Then, for p
∈
(
1,2
],
(35)
α
0
=
r
*
p
(
k
+
1
)
-
(
2
-
p
)
/
2
8
p
/
2
(
sin
π
2
(
k
+
2
)
sin
π
2
(
k
+
1
)
)
p
,
β
0
=
r
*
p
(
k
+
2
)
(
2
-
p
)
/
2
8
p
/
2
(
sin
(
k
+
1
)
π
2
(
k
+
2
)
sin
k
π
2
(
k
+
1
)
)
p
.
For p
∈
(
2
,
∞
),
(36)
α
0
=
r
*
p
(
k
+
2
)
-
(
p
-
2
)
/
2
8
p
/
2
(
sin
π
2
(
k
+
2
)
sin
π
2
(
k
+
1
)
)
p
,
β
0
=
r
*
p
(
k
+
1
)
(
p
-
2
)
/
2
8
p
/
2
(
sin
(
k
+
1
)
π
2
(
k
+
2
)
sin
k
π
2
(
k
+
1
)
)
p
.Now we state our main results.Theorem 5.
Assume thatF
(
n
,
u
,
v
) satisfies the following conditions.(
F
1
)
There exist constantsδ
∈
(
0
,
∞
) and α
∈
(
0
,
α
0
) such that
(37)
F
(
n
,
u
,
v
)
≤
α
(
u
2
+
v
2
)
p
/
2
,
f
o
r
n
∈
ℤ
(
0
,
k
)
,
u
2
+
v
2
≤
δ
2
.
(
F
2
)
There exist constantsβ
∈
(
β
0
,
∞
) and γ
∈
(
0
,
∞
) such that
(38)
F
(
n
,
u
,
v
)
≥
β
(
u
2
+
v
2
)
p
/
2
-
γ
,
f
o
r
n
∈
ℤ
(
0
,
k
)
.Then (1) with (2) possesses at least two nontrivial solutions.Remark 6.
Comparing our results with the results of the boundary value problems of second orderp-Laplacian difference equations in [18], we find that our results are more precisely.In view of (37) and (38), it is easy to obtain the following corollary.Corollary 7.
Assume thatF
(
n
,
u
,
v
) satisfies
(39)
lim
u
2
+
v
2
→
0
F
(
n
,
u
,
v
)
(
u
2
+
v
2
)
p
/
2
=
0
,
∀
n
∈
ℤ
(
0
,
k
)
,
lim
u
2
+
v
2
→
+
∞
F
(
n
,
u
,
v
)
(
u
2
+
v
2
)
p
/
2
=
+
∞
,
∀
n
∈
ℤ
(
0
,
k
)
.
Then (1) with (2) possesses at least two nontrivial solutions.For the case whenp
=
2, we have the following corollary for the boundary value problems of the fourth order nonlinear difference equations.Corollary 8.
Assume thatF
(
n
,
u
,
v
) satisfies the following conditions.(
F
3
)
There exist constantsδ
1
>
0 and 0
<
α
1
<
(
8
r
*
/
p
)
(
sin
(
π
/
2
(
k
+
2
)
)
sin
(
π
/
2
(
k
+
1
)
)
)
2 such that
(40)
F
(
n
,
u
,
v
)
≤
α
1
(
u
2
+
v
2
)
,
f
o
r
n
∈
ℤ
(
0
,
k
)
,
u
2
+
v
2
≤
δ
1
2
.
(
F
4
)
There exist constantsβ
1
>
(
8
r
*
/
p
)
(
sin
(
(
k
+
1
)
π
/
2
(
k
+
2
)
)
sin
(
k
π
/
2
(
k
+
1
)
)
)
2 and γ
1
>
0 such that
(41)
F
(
n
,
u
,
v
)
≥
β
1
(
u
2
+
v
2
)
-
γ
1
,
f
o
r
n
∈
ℤ
(
0
,
k
)
.Then the following fourth order nonlinear difference equation
(42)
Δ
2
(
r
n
-
2
Δ
2
u
n
-
2
)
-
f
(
n
,
u
n
+
1
,
u
n
,
u
n
-
1
)
=
0
,
n
∈
ℤ
(
1
,
k
)
,
with the boundary value conditions (2) possesses at least two nontrivial solutions.


## 3.  Proof of Theorem5

In order to prove Theorem5, we first establish the following lemma.Lemma 9.
Assume thatF satisfies (
F
2
); then the functional J satisfies the P.S. condition.Proof.
Let{
u
(
s
)
}
s
∈
ℕ be a sequence in E such that {
J
(
u
(
s
)
)
} is bounded and J
′
(
u
(
s
)
)
→
0 as s
→
∞. Then there exists a positive constant C such that |
J
(
u
(
s
)
)
|
≤
C for s
∈
ℕ.
By (11) and (16), we have
(43)
1
p
∑
n
=
-
1
k
‍
r
n
|
Δ
2
u
n
(
s
)
|
p
≤
r
*
p
(
c
*
(
k
+
2
)
)
p
(
∑
n
=
-
1
k
‍
|
Δ
2
u
n
(
s
)
|
2
)
p
/
2
≤
r
*
p
(
c
*
(
k
+
2
)
)
p
(
λ
*
(
k
+
1
)
∑
n
=
0
k
‍
(
Δ
u
n
(
s
)
)
2
)
p
/
2
≤
r
*
p
(
c
*
(
k
+
2
)
)
p
(
λ
*
(
k
+
1
)
λ
*
(
k
)
∑
n
=
1
k
‍
(
u
n
(
s
)
)
2
)
p
/
2
=
r
*
p
(
c
*
(
k
+
2
)
)
p
(
λ
*
(
k
+
1
)
λ
*
(
k
)
)
p
/
2
∥
u
(
s
)
∥
p
=
(
c
*
(
k
+
1
)
)
p
2
p
/
2
β
0
∥
u
(
s
)
∥
p
.
And by (
F
2
), (11), and (16), we have
(44)
∑
n
=
0
k
‍
F
(
n
,
u
n
+
1
(
s
)
,
u
n
(
s
)
)
≥
β
∑
n
=
0
k
‍
(
(
u
n
+
1
(
s
)
)
2
+
(
u
n
(
s
)
)
2
)
p
/
2
-
(
k
+
1
)
γ
≥
β
[
c
*
(
k
+
1
)
(
∑
n
=
0
k
‍
(
(
u
n
+
1
(
s
)
)
2
+
(
u
n
(
s
)
)
2
)
)
1
/
2
]
p
-
(
k
+
1
)
γ
=
(
c
*
(
k
+
1
)
)
p
2
p
/
2
β
∥
u
(
s
)
∥
p
-
(
k
+
1
)
γ
.
Therefore, by (30), we obtain
(45)
J
(
u
(
s
)
)
=
1
p
∑
n
=
-
1
k
‍
r
n
|
Δ
2
u
n
(
s
)
|
p
-
∑
n
=
0
k
‍
F
(
n
,
u
n
+
1
(
s
)
,
u
n
(
s
)
)
J
(
u
(
s
)
)
=
≤
(
c
*
(
k
+
1
)
)
p
2
p
/
2
(
β
0
-
β
)
∥
u
(
s
)
∥
p
J
(
u
(
s
)
)
=
+
(
k
+
1
)
γ
.
Noticing that J
(
u
(
s
)
)
≥
-
C and β
>
β
0, by (45), we have
(46)
∥
u
(
s
)
∥
p
≤
(
k
+
1
)
γ
+
C
(
c
*
(
k
+
1
)
)
p
2
p
/
2
(
β
-
β
0
)
.
Since E is a finite-dimensional space, (46) implies that {
u
(
s
)
} is bounded and has a convergent subsequence. Thus P
.
S. condition is verified.Now we give the proof of Theorem5.Proof.
For anyu
∈
E with ∥
u
∥
≤
δ, according to (11) and (16), we have
(47)
1
p
∑
n
=
-
1
k
‍
r
n
|
Δ
2
u
n
|
p
≥
r
*
p
(
c
*
(
k
+
2
)
)
p
(
∑
n
=
-
1
k
‍
|
Δ
2
u
n
|
2
)
p
/
2
≥
r
*
p
(
c
*
(
k
+
2
)
)
p
(
λ
*
(
k
+
1
)
∑
n
=
0
k
‍
(
Δ
u
n
)
2
)
p
/
2
≥
r
*
p
(
c
*
(
k
+
2
)
)
p
(
λ
*
(
k
+
1
)
λ
*
(
k
)
∑
n
=
1
k
‍
(
u
n
)
2
)
p
/
2
=
r
*
p
(
c
*
(
k
+
2
)
)
p
(
λ
*
(
k
+
1
)
λ
*
(
k
)
)
p
/
2
∥
u
∥
p
.
By (
F
1
), (11), and (16), we have
(48)
∑
n
=
0
k
‍
F
(
n
,
u
n
+
1
,
u
n
)
≤
α
∑
n
=
0
k
‍
(
u
n
+
1
2
+
u
n
2
)
p
/
2
≤
α
[
c
*
(
k
+
1
)
(
∑
n
=
0
k
‍
(
u
n
+
1
2
+
u
n
2
)
)
1
/
2
]
p
=
α
(
c
*
(
k
+
1
)
)
p
2
p
/
2
∥
u
∥
p
.
So, by (30), we get
(49)
J
(
u
)
=
1
p
∑
n
=
-
1
k
‍
r
n
|
Δ
2
u
n
|
p
-
∑
n
=
0
k
‍
F
(
n
,
u
n
+
1
,
u
n
)
≥
[
r
*
p
(
c
*
(
k
+
2
)
)
p
(
λ
*
(
k
+
1
)
λ
*
(
k
)
)
p
/
2
r
*
p
-
α
(
c
*
(
k
+
1
)
)
p
2
p
/
2
]
∥
u
∥
p
=
(
c
*
(
k
+
1
)
)
p
2
p
/
2
(
α
0
-
α
)
∥
u
∥
p
.
Since α
<
α
0, we let a
=
(
c
*
(
k
+
1
)
)
p
2
p
/
2
(
α
0
-
α
)
δ
p and ρ
=
δ. Then by (49),
(50)
J
(
u
)
≥
a
,
∀
u
∈
∂
B
ρ
,
which means that J satisfies the condition (
J
1
) of the Mountain pass lemma.
By our assumptions, it is clear thatJ
(
0
)
=
0. In order to use Mountain pass lemma, it suffices to verify that condition (
J
2
) holds. In fact, similar to the proof of (45), we have
(51)
J
(
u
)
≤
(
c
*
(
k
+
1
)
)
p
2
p
/
2
(
β
0
-
β
)
∥
u
∥
p
+
(
k
+
1
)
γ
,
for any u
∈
E. Since β
0
<
β, it is easy to see that there exists an e
∈
E with ∥
e
∥
>
ρ such that J
(
±
e
)
<
0. Thus (
J
2
) holds.
According to Mountain pass lemma,J possesses a critical value c
≥
a given by
(52)
c
=
inf
g
∈
Γ
1
max
s
∈
[
0,1
]
J
(
g
(
s
)
)
,
where
(53)
Γ
1
=
{
g
∈
C
(
[
0,1
]
,
E
)
∣
g
(
0
)
=
0
,
g
(
1
)
=
e
}
.
Letu
¯
∈
E be a critical point of J corresponding to the critical value c; then u
¯ is nontrivial and J
(
u
¯
)
=
c.
On the other hand, by (51), we have
(54)
lim
∥
u
∥
→
∞
J
(
u
)
=
-
∞
.
Since E is a k-dimensional space, by the continuity of J
(
u
) on u, we see that there exists u
^
∈
E such that
(55)
J
(
u
^
)
=
max
{
J
(
u
)
∣
u
∈
E
}
.
Clearly, u
^ is a nonzero critical point of J, and J
(
u
^
)
≥
c
≥
a
>
0.
Ifu
^
≠
u
¯, then the proof is finished. Otherwise, u
^
=
u
¯. Since ∥
-
e
∥
>
ρ and J
(
-
e
)
<
0, then by Mountain pass lemma again, J possesses a critical value c
~
≥
a given by
(56)
c
~
=
inf
g
∈
Γ
2
max
s
∈
[
0,1
]
J
(
g
(
s
)
)
,
where
(57)
Γ
2
=
{
g
∈
C
(
[
0,1
]
,
E
)
∣
g
(
0
)
=
0
,
g
(
1
)
=
-
e
}
.
Letu
~
∈
E be a critical point of J corresponding to the critical value c
~. If u
~
≠
u
^, then the proof is finished. Otherwise u
~
=
u
^
=
u
¯. Let g
1
(
s
)
=
s
e for s
∈
[
0,1
]; then g
1
∈
Γ
1. By the definition of c, we see that there exists s
1
∈
(
0,1
) such that J
(
s
1
e
)
=
J
(
u
^
). Thus s
1
e is a critical point of J. Similar, let g
2
(
s
)
=
-
s
e for s
∈
[
0,1
]; then g
2
∈
Γ
2. By the definition of c
~, we see that there exists s
2
∈
(
0,1
) such that J
(
-
s
2
e
)
=
J
(
u
^
). And -
s
2
e is a critical point of J. Clearly s
1
e
≠
-
s
2
e. The proof is now completed.In the last part of this paper, we give an example to illustrate our results.Example 10.
Consider (1) with (2), where f is defined by
(58)
f
(
n
,
u
,
v
,
w
)
=
2
μ
n
v
(
u
2
+
v
2
)
p
/
2
1
+
u
2
+
v
2
+
μ
n
p
v
ln
(
1
+
u
2
+
v
2
)
(
u
2
+
v
2
)
(
p
/
2
)
-
1
+
2
μ
n
-
1
v
(
v
2
+
w
2
)
p
/
2
1
+
v
2
+
w
2
+
μ
n
-
1
p
v
ln
(
1
+
v
2
+
w
2
)
(
v
2
+
w
2
)
(
p
/
2
)
-
1
,
for n
∈
ℤ
(
1
,
k
). Here μ
n
>
0 for n
∈
ℤ
(
0
,
k
). Define
(59)
F
(
n
,
u
,
v
)
=
μ
n
ln
(
1
+
u
2
+
v
2
)
(
u
2
+
v
2
)
p
/
2
,
n
∈
ℤ
(
0
,
k
)
.
Then F
(
n
,
0,0
)
=
0 for n
∈
ℤ
(
0
,
k
) and (10) holds. Moreover, it is easy to see that F
(
n
,
u
,
v
) satisfies (39) for n
∈
ℤ
(
0
,
k
). By Corollary 7, we see that (1) with (2) when f is defined by (58) has at least two nontrivial solutions.



---

*Source: 343129-2014-01-30.xml*