# Existence and Multiplicity of Fast Homoclinic Solutions for a Class of Damped Vibration Problems with Impulsive Effects

Authors: Qiongfen Zhang
Journal: Abstract and Applied Analysis
 (2014)

---

## Abstract

This paper is concerned with the existence and multiplicity of fast homoclinic solutions for a class of damped vibration problems with impulsive effects. Some new results are obtained under more relaxed conditions by using Mountain Pass Theorem and Symmetric Mountain Pass Theorem in critical point theory. The results obtained in this paper generalize and improve some existing works in the literature.


---

## Body

## 1. Introduction

Consider fast homoclinic solutions of the following problem:(1)
u
′′
(
t
)
+
q
(
t
)
u
′
(
t
)
-
a
(
t
)
u
(
t
)
+
∇
V
(
t
,
u
(
t
)
)
=
0
,
a
.
e
.
t
∈
(
t
j
,
t
j
+
1
)
,
j
∈
Z
,
Δ
u
′
(
t
j
)
=
u
′
(
t
j
+
)
-
u
′
(
t
j
-
)
=
I
(
u
(
t
j
)
)
,
j
∈
Z
,
where V
:
R
×
R
→
R is of class C
1, ∇
V
(
t
,
u
(
t
)
)
=
∂
V
(
t
,
u
(
t
)
)
/
∂
u, I
:
R
→
R, a
∈
C
(
R
,
(
0
,
+
∞
)
), a
(
t
)
→
+
∞ as |
t
|
→
+
∞, Z denotes the sets of integers, and t
j
(
j
∈
Z
) are impulsive points. Moreover, there exist a positive integer p and a positive constant T such that 0
<
t
0
<
t
1
<
⋯
<
t
p
-
1
<
T, t
l
+
k
p
=
t
l
+
k
T, ∀
k
∈
Z, l
=
0,1
,
…
,
p
-
1, u
′
(
t
j
+
)
=
lim
⁡
h
→
0
+
u
′
(
t
j
+
h
) and u
′
(
t
j
-
)
=
lim
⁡
h
→
0
-
u
′
(
t
j
-
h
) represent the right and left limits of u
′
(
t
) at t
=
t
j, respectively, q
:
R
→
R is a continuous function, and Q
(
t
)
=
∫
0
s
‍
q
(
s
)
d
s with
(2)
lim
⁡
|
t
|
→
+
∞
Q
(
t
)
=
+
∞
.
When I
≡
0, problem (1) becomes the following damped vibration problem:
(3)
u
′′
(
t
)
+
q
(
t
)
u
′
(
t
)
-
a
(
t
)
u
(
t
)
+
∇
V
(
t
,
u
(
t
)
)
=
0
,
a
.
e
.
t
∈
R
.Chen et al. [1] investigated problem (3) and obtained some results of fast homoclinic solutions by critical point theory.Whenq
(
t
)
≡
0 and a
(
t
)
≡
0, problem (1) becomes the following problem:
(4)
u
′′
(
t
)
+
∇
V
(
t
,
u
(
t
)
)
=
0
,
a
.
e
.
t
∈
(
t
j
,
t
j
+
1
)
,
j
∈
Z
,
Δ
u
′
(
t
j
)
=
u
′
(
t
j
+
)
-
u
′
(
t
j
-
)
=
I
(
u
(
t
j
)
)
,
j
∈
Z
.
Fang and Duan [2] obtained the following result of homoclinic solutions for (4) by employing Mountain Pass Theorem, a weak convergence argument, and a weak version of Lieb’s methods.Theorem A (see [2]).
Assume that the following conditions hold:(V1)
there exists a positive numberT such that
(5)
∇
V
(
t
+
T
,
x
)
=
∇
V
(
t
,
x
)
,
AA
V
(
t
+
T
,
x
)
=
V
(
t
,
x
)
,
AAAAAAAA
∀
(
t
,
x
)
∈
R
2
;
(V2)
lim
⁡
x
→
0
(
∇
V
(
t
,
x
)
/
x
)
=
0 uniformly for t
∈
R;
(V3)
there exists a constantμ
>
0 such that
(6)
μ
V
(
t
,
x
)
≤
(
∇
V
(
t
,
x
)
,
x
)
,
∀
(
t
,
x
)
∈
R
×
R
∖
{
0
}
;
(V4)
there exist constantsb
0
>
0 and b
>
0 such that
(7)
V
(
t
,
x
)
≥
b
0
|
x
|
μ
,
∀
t
∈
R
,
|
x
|
≥
1
;
V
(
t
,
x
)
≤
b
|
x
|
μ
,
∀
t
∈
R
,
|
x
|
≤
1
;
(I)
there exists a constantb
1 with 0
<
b
1
<
(
μ
-
2
)
/
(
μ
+
2
)
p
T such that
(8)
|
I
(
x
)
|
≤
b
1
|
x
|
,
2
∫
0
x
‍
I
(
t
)
d
t
-
I
(
x
)
x
≤
0
.Then, problem (4) possesses a nontrivial weak homoclinic orbit.ForI
≠
0, problem (1) involves impulsive effects. Impulsive differential equations are suitable for the mathematical simulation of evolutionary processes in which the parameters undergo relatively long periods of smooth variation followed by a short-term rapid change (that is jumps) in their values. Impulsive differential equations are often investigated in various fields of science and technology, for example, many biological phenomena involving thresholds, bursting rhythm models in medicine and biology, optimal control models in economics, pharmacokinetics, and frequency modulated systems, and so on. For more details of impulsive differential equations, we refer the readers to the books [3, 4].In recent years, some researchers have paid attention to the existence and multiplicity of solutions for impulsive differential equations via variational methods. See, for example, [1, 5–17] and references therein. However, there are few papers [2, 18, 19] concerning homoclinic solutions of impulsive differential equations by variational methods. So, it is a novel method to employ variational methods to investigate the existence of homoclinic solutions for impulsive differential equations.Motivated by the above papers, we will establish some new results for (1). In order to introduce the concept of fast homoclinic solutions for (1), we first state some properties of the weighted Sobolev space E on which certain variational functional associated with (1) is defined and the fast homoclinic solutions are the critical points of the certain functional.Let(9)
E
=
{
u
∈
H
1,2
(
R
,
R
)
∣
∫
R
‍
e
Q
(
t
)
[
|
u
′
(
t
)
|
2
+
a
(
t
)
|
u
(
t
)
|
2
]
d
t
<
+
∞
,
{
u
(
t
j
)
}
j
=
-
∞
+
∞
∈
l
2
,
u
(
±
∞
)
=
0
,
u
(
k
T
)
=
0
,
k
∈
Z
∫
R
‍
}
,
where Q
(
t
) is defined in (2) and l
2 denotes the space of sequences whose second powers are summable on Z; that is,
(10)
∑
j
∈
Z
‍
|
a
j
|
2
<
+
∞
,
∀
a
=
{
a
j
}
j
=
-
∞
+
∞
∈
l
2
.
The space l
2 is equipped with the following norm:
(11)
∥
a
∥
l
2
=
(
∑
j
∈
Z
‍
|
a
j
|
2
)
1
/
2
.
For u
,
v
∈
E, let
(12)
〈
u
,
v
〉
=
∫
R
‍
e
Q
(
t
)
[
(
u
′
(
t
)
,
v
′
(
t
)
)
+
(
a
(
t
)
u
(
t
)
,
v
(
t
)
)
]
d
t
.
Similar to [2], it is easy to check that E is a Hilbert space with the norm given by
(13)
∥
u
∥
=
(
∫
R
‍
e
Q
(
t
)
[
|
u
′
(
t
)
|
2
+
a
(
t
)
|
u
(
t
)
|
2
]
d
t
)
1
/
2
.
It is obvious that
(14)
E
⊂
L
2
(
e
Q
(
t
)
)
with the embedding being continuous. Here, L
p
(
e
Q
(
t
)
)
(
2
≤
p
<
+
∞
) denotes the Banach spaces of functions on R with values in R under the norm
(15)
∥
u
∥
p
=
{
∫
R
‍
e
Q
(
t
)
|
u
(
t
)
|
p
d
t
}
1
/
p
.
Similar to [1], we have the following definition of fast homoclinic solutions.Definition 1.
If (2) holds, a solution of (1) is called a fast homoclinic solution if u
∈
E.Here and in subsequence,(
,
) and |
·
| denote the inner product and norm in R, respectively. C
i
(
i
=
0,1
,
…
) denote different positive constants. Now, we state our main results.Theorem 2.
Suppose thatq, a, I, and V satisfy (2), (V1), and the following conditions: (A)
a
∈
C
(
R
,
(
0
,
+
∞
)
) and a
(
t
)
→
+
∞ as |
t
|
→
+
∞,
(V2)′
V
(
t
,
x
)
=
V
1
(
t
,
x
)
-
V
2
(
t
,
x
), V
1
,
V
2
∈
C
1
(
R
×
R
,
R
), and there exists a constant R
>
0 such that
(16)
|
∇
V
(
t
,
x
)
|
=
o
(
|
x
|
)
as
x
⟶
0
uniformly
in
t
∈
(
-
∞
,
-
R
]
∪
[
R
,
+
∞
)
,
(V3)′
there is a constantμ
>
2 such that
(17)
0
<
μ
V
1
(
t
,
x
)
≤
(
∇
V
1
(
t
,
x
)
,
x
)
,
∀
(
t
,
x
)
∈
R
×
R
∖
{
0
}
,
(V5)
V
2
(
t
,
0
)
=
0 and there exists a constant ϱ
∈
(
2
,
μ
) such that
(18)
V
2
(
t
,
x
)
≥
0
,
(
∇
V
2
(
t
,
x
)
,
x
)
≤
ϱ
V
2
(
t
,
x
)
,
AAAAAAAAAAAAAAAA
∀
(
t
,
x
)
∈
R
×
R
,
(I)′
I
∈
C
(
R
,
R
) and there exists a constant c with 0
<
c
<
(
ϱ
-
2
)
e
0
/
(
ϱ
+
2
)
p
T such that
(19)
|
I
(
x
)
|
≤
c
|
x
|
,
(
I
(
x
)
-
I
(
y
)
,
x
-
y
)
≥
0
,
∀
x
,
y
∈
R
,
wheree
0
=
e
min
⁡
{
Q
(
t
)
:
t
∈
R
}.
Then, problem (1) has at least one nontrivial fast homoclinic solution.Theorem 3.
Suppose thatq, a, V, and I satisfy (2), (V1), (A), (V3)′, (I)′, and the following conditions:(V2)′′
V
(
t
,
x
)
=
V
1
(
t
,
x
)
-
V
2
(
t
,
x
), V
1
,
V
2
∈
C
1
(
R
×
R
,
R
), and
(20)
|
∇
V
(
t
,
x
)
|
=
o
(
|
x
|
)
as
x
⟶
0
uniformly
in
t
∈
R
,
(V5)′
V
2
(
t
,
0
)
=
0 and there exists a constant ϱ
∈
(
2
,
μ
) such that
(21)
(
∇
V
2
(
t
,
x
)
,
x
)
≤
ϱ
V
2
(
t
,
x
)
,
∀
(
t
,
x
)
∈
R
×
R
. Then, problem (1) has at least one nontrivial fast homoclinic solution.Theorem 4.
Suppose thatq, a, V, and I satisfy (2), (V1), (A), (V2)′, (V3)′, (V5), (I)′, and the following condition: (V6)
I
(
-
t
)
=
-
I
(
t
)
,
V
(
t
,
-
x
)
=
V
(
t
,
x
)
,
∀
(
t
,
x
)
∈
R
×
R. Then, problem (1) has an unbounded sequence of fast homoclinic solutions.Theorem 5.
Suppose thatq, a, V, and I satisfy (2), (V1), (A), (V2)′′, (V3)′, (V5)′, (I)′, and (V6). Then, problem (1) has an unbounded sequence of fast homoclinic solutions.The rest of this paper is organized as follows. In Section2, some preliminaries are presented. In Section 3, we give the proofs of our results. In Section 4, some examples are given to illustrate our results.


## 2. Preliminaries

LetE and ∥
·
∥ be given in Section 1. By a similar argument in [2, 20], we have the following important lemma.Lemma 6.
For anyu
∈
E,
(22)
∥
u
∥
∞
≤
1
2
e
0
a
0
∥
u
∥
=
1
2
e
0
a
0
{
∫
R
‍
e
Q
(
s
)
[
|
u
′
(
s
)
|
2
+
a
(
s
)
|
u
(
s
)
|
2
]
d
s
}
1
/
2
,
(23)
|
u
(
t
)
|
≤
{
∫
t
+
∞
‍
e
-
Q
(
s
)
[
a
(
s
)
]
-
1
/
2
×
[
e
Q
(
s
)
(
|
u
′
(
s
)
|
2
+
a
(
s
)
|
u
(
s
)
|
2
)
]
d
s
∫
t
+
∞
‍
}
1
/
2
,
(24)
|
u
(
t
)
|
≤
{
∫
-
∞
t
‍
e
-
Q
(
s
)
[
a
(
s
)
]
-
1
/
2
×
[
e
Q
(
s
)
(
|
u
′
(
s
)
|
2
+
a
(
s
)
|
u
(
s
)
|
2
)
]
d
s
∫
-
∞
t
‍
}
1
/
2
,
(25)
∑
j
=
-
∞
∞
‍
|
u
(
t
j
)
|
2
≤
p
T
e
0
∥
u
′
∥
2
2
,
where ∥
u
∥
∞
=
ess
sup
⁡
t
∈
R
|
u
(
t
)
|, a
0
=
min
⁡
t
∈
R
{
a
(
t
)
}, and e
0 is the same as that in assumption (I)′.The following two lemmas are Mountain Pass Theorem and Symmetric Mountain Pass Theorem, which are useful in the proofs of our theorems.Lemma 7 (see [21]).
LetE be a real Banach space and φ
∈
C
1
(
E
,
R
) satisfying (PS)-condition. Suppose that φ
(
0
)
=
0 and(i)
there exist constantsρ, α
>
0 such that φ
∂
B
ρ
(
0
)
≥
α;
(ii)
there exists ane
∈
E
∖
B
¯
ρ
(
0
) such that φ
(
e
)
≤
0. Then, φ possesses a critical value c
≥
α which can be characterized as c
=
inf
⁡
h
∈
Φ
max
⁡
s
∈
[
0,1
]
φ
(
h
(
s
)
), where Φ
=
{
h
∈
C
(
[
0,1
]
,
E
)
∣
h
(
0
)
=
0
,
h
(
1
)
=
e
} and B
ρ
(
0
) is an open ball in E of radius ρ centered at 0.Lemma 8 (see [21]).
LetE be a real Banach space and φ
∈
C
1
(
E
,
R
) with I even. Assume that φ
(
0
)
=
0 and φ satisfies (PS)-condition, assumption (i) of Lemma 7, and the following condition:(iii)
for each finite dimensional subspaceE
′
⊂
E, there is r
=
r
(
E
′
)
>
0 such that φ
(
u
)
≤
0, for u
∈
E
′
∖
B
r
(
0
), and B
r
(
0
) is an open ball in E of radius r centered at 0. Then, φ possesses an unbounded sequence of critical values.Remark 9.
Since it is very difficult to check condition (iii) of Lemma8, few results about infinitely many homoclinic solutions can be seen in the literature by using Lemma 8, let alone infinitely many fast homoclinic solutions obtained by this lemma. Motivated by the idea of [22], we will use Lemma 8 to prove that problem (1) has infinitely many homoclinic fast solutions.Lemma 10.
Assume that (V3)′ and (V5) or (V5)′ hold. Then, for every(
t
,
x
)
∈
R
×
R, (i)
s
-
μ
V
1
(
t
,
s
x
) is nondecreasing on (
0
,
+
∞
);
(ii)
s
-
ϱ
V
2
(
t
,
s
x
) is nonincreasing on (
0
,
+
∞
).The proof of Lemma10 is routine and we omit it.The functionalφ corresponding to (1) on E is given by
(26)
φ
(
u
)
=
∫
R
‍
1
2
e
Q
(
t
)
[
|
u
′
(
t
)
|
2
+
a
(
t
)
|
u
(
t
)
|
2
]
d
t
-
∫
R
‍
V
(
t
,
u
(
t
)
)
d
t
+
∑
j
=
-
∞
∞
‍
∫
0
u
(
t
j
)
‍
I
(
s
)
d
s
,
u
∈
E
.
We now show that φ
∈
C
1
(
E
,
R
) and, for u
,
v
∈
E,
(27)
〈
φ
′
(
u
)
,
v
〉
=
∫
R
‍
e
Q
(
t
)
[
(
u
′
(
t
)
,
v
′
(
t
)
)
+
(
a
(
t
)
u
(
t
)
,
v
(
t
)
)
-
(
∇
V
(
t
,
u
(
t
)
)
,
v
(
t
)
)
]
d
t
+
∑
j
=
-
∞
∞
‍
(
I
(
u
(
t
j
)
)
,
v
(
t
j
)
)
.
Firstly, we show that φ
:
E
→
R. By (V2)′, for any given ε
0
>
0, there exists γ
0
>
0 such that
(28)
|
∇
V
(
t
,
x
)
|
≤
2
ɛ
e
0
a
0
|
x
|
,
t
∈
(
-
∞
,
-
R
]
∪
[
R
,
+
∞
)
,
|
x
|
≤
γ
0
,
where a
0
=
min
⁡
t
∈
R
{
a
(
t
)
}, e
0 is the same as that in assumption (I)′. Then, by V
(
t
,
0
)
=
0 and (28), we have
(29)
|
V
(
t
,
x
)
|
=
|
∫
0
1
‍
(
∇
V
(
t
,
s
x
)
,
x
)
d
s
|
≤
ɛ
e
0
a
0
|
x
|
2
,
AAAAA
∀
t
∈
(
-
∞
,
-
R
]
∪
[
R
,
+
∞
)
,
x
∈
R
.
Therefore, from (29), we have
(30)
|
∫
R
∖
(
-
R
,
R
)
‍
V
(
t
,
u
(
t
)
)
d
t
|
≤
∫
R
∖
(
-
R
,
R
)
‍
|
V
(
t
,
u
(
t
)
)
|
d
t
≤
∫
R
∖
(
-
R
,
R
)
‍
ɛ
e
0
a
0
|
u
(
t
)
|
2
d
t
≤
∫
R
∖
(
-
R
,
R
)
‍
ɛ
e
Q
(
t
)
a
(
t
)
|
u
(
t
)
|
2
d
t
≤
∫
R
‍
ɛ
e
Q
(
t
)
a
(
t
)
|
u
(
t
)
|
2
d
t
≤
ɛ
∥
u
∥
2
,
u
∈
E
.
From (I)′ and Lemma 6, we have
(31)
∑
j
=
-
∞
∞
‍
|
∫
0
u
(
t
j
)
‍
I
(
s
)
d
s
|
≤
∑
j
=
-
∞
∞
‍
∫
min
⁡
⁡
{
0
,
u
(
t
j
)
}
max
⁡
⁡
{
0
,
u
(
t
j
)
}
‍
|
I
(
s
)
|
d
s
≤
1
2
∑
j
=
-
∞
∞
‍
c
|
u
(
t
j
)
|
2
≤
c
p
T
2
e
0
∥
u
′
∥
2
2
≤
c
p
T
2
e
0
∥
u
∥
2
.
It follows from (26), (30), and (31) that φ
:
E
→
R. Next, we prove that φ
∈
C
1
(
E
,
R
). Rewrite φ as the following:
(32)
φ
(
u
)
=
φ
1
(
u
)
-
φ
2
(
u
)
+
φ
3
(
u
)
,
where
(33)
φ
1
(
u
)
:
=
∫
R
‍
1
2
e
Q
(
t
)
[
|
u
′
(
t
)
|
2
+
a
(
t
)
|
u
(
t
)
|
2
]
d
t
,
φ
2
(
u
)
:
=
∫
R
‍
V
(
t
,
u
(
t
)
)
d
t
,
φ
3
(
u
)
:
=
∑
j
=
-
∞
∞
‍
∫
0
u
(
t
j
)
‍
I
(
s
)
d
s
.
It is easy to check that φ
1
,
φ
3
∈
C
1
(
E
,
R
) and
(34)
〈
φ
1
′
(
u
)
,
v
〉
=
∫
R
‍
e
Q
(
t
)
[
(
u
′
(
t
)
,
v
′
(
t
)
)
+
(
a
(
t
)
u
(
t
)
,
v
(
t
)
)
]
d
t
,
〈
φ
3
′
(
u
)
,
v
〉
=
∑
j
=
-
∞
∞
‍
(
I
(
u
(
t
j
)
)
,
v
(
t
j
)
)
,
〈
φ
3
′
(
u
)
,
v
〉
∀
u
,
v
∈
E
.
Next, we prove that φ
2
∈
C
1
(
E
,
R
) and
(35)
〈
φ
2
′
(
u
)
,
v
〉
=
∫
R
‍
(
∇
V
(
t
,
u
(
t
)
)
,
v
(
t
)
)
d
t
,
∀
u
,
v
∈
E
.
Let u
n
→
u in E, without loss of generality, and we can assume that ∥
u
n
∥
≤
γ
0. Since V
∈
C
1
(
R
×
R
,
R
), we have
(36)
|
∇
V
(
t
,
u
n
(
t
)
)
,
v
(
t
)
|
≤
max
⁡
|
x
|
≤
γ
0
/
2
e
0
a
0
|
∇
V
(
t
,
x
)
|
|
v
(
t
)
|
:
=
C
0
|
v
(
t
)
|
,
∀
t
∈
[
-
R
,
R
]
.
By (29) and (36), we have
(37)
|
∇
V
(
t
,
u
n
(
t
)
)
,
v
(
t
)
|
≤
ɛ
e
0
a
0
|
u
n
(
t
)
|
|
v
(
t
)
|
+
C
0
|
v
(
t
)
|
≤
ɛ
e
Q
(
t
)
a
(
t
)
(
|
u
n
(
t
)
-
u
(
t
)
|
+
|
u
(
t
)
|
)
|
v
(
t
)
|
+
C
0
|
v
(
t
)
|
:
=
g
n
(
t
)
,
t
∈
R
.
Since u
n
(
t
)
→
u
(
t
), for almost every t
∈
R, we have
(38)
lim
⁡
n
→
+
∞
g
n
(
t
)
=
ɛ
e
Q
(
t
)
a
(
t
)
|
u
(
t
)
|
|
v
(
t
)
|
+
C
0
|
v
(
t
)
|
:
=
g
(
t
)
,
lim
⁡
n
→
+
∞
∫
R
‍
g
n
(
t
)
d
t
=
lim
⁡
n
→
+
∞
∫
R
‍
[
ɛ
e
Q
(
t
)
a
(
t
)
(
|
u
n
(
t
)
-
u
(
t
)
|
+
|
u
(
t
)
|
)
|
v
(
t
)
|
+
C
0
|
v
(
t
)
|
ɛ
e
Q
(
t
)
a
(
t
)
(
|
u
n
(
t
)
-
u
(
t
)
|
+
|
u
(
t
)
|
)
|
v
(
t
)
|
]
d
t
=
lim
⁡
n
→
+
∞
∫
R
‍
ɛ
e
Q
(
t
)
a
(
t
)
(
|
u
n
(
t
)
-
u
(
t
)
|
)
d
t
+
∫
R
‍
[
ɛ
e
Q
(
t
)
a
(
t
)
|
u
(
t
)
|
|
v
(
t
)
|
+
C
0
|
v
(
t
)
|
]
d
t
=
∫
R
‍
[
ɛ
e
Q
(
t
)
a
(
t
)
|
u
(
t
)
|
|
v
(
t
)
|
+
C
0
|
v
(
t
)
|
]
d
t
:
=
∫
R
‍
g
(
t
)
d
t
<
+
∞
.
Then, by (37), (38), and Lebesgue’s dominated convergence theorem, we have
(39)
lim
⁡
n
→
+
∞
∫
R
‍
(
∇
V
(
t
,
u
n
(
t
)
)
,
v
(
t
)
)
d
t
=
∫
R
‍
(
∇
V
(
t
,
u
(
t
)
)
,
v
(
t
)
)
d
t
.
Therefore, for any u
,
v
∈
E and for any function θ
:
R
→
(
0,1
), from (39), we have
(40)
〈
φ
2
′
(
u
)
,
v
〉
=
lim
⁡
h
→
0
+
φ
2
(
u
+
h
v
)
-
φ
2
(
u
)
h
=
lim
⁡
h
→
0
+
1
h
∫
R
‍
[
V
(
t
,
u
(
t
)
+
h
v
(
t
)
)
-
V
(
t
,
u
(
t
)
)
]
d
t
=
lim
⁡
h
→
0
+
∫
R
‍
(
∇
V
(
t
,
u
(
t
)
+
θ
(
t
)
h
v
(
t
)
)
,
v
(
t
)
)
d
t
=
∫
R
‍
(
∇
V
(
t
,
u
(
t
)
)
,
v
(
t
)
)
d
t
,
∀
u
,
v
∈
E
.
Finally, we prove that φ
2
∈
C
1
(
E
,
R
). From (40), u
n
→
u in E, and V
∈
C
1
(
R
×
R
,
R
), we have
(41)
lim
⁡
n
→
+
∞
|
〈
φ
2
′
(
u
n
)
-
φ
2
′
(
u
)
,
v
〉
|
=
lim
⁡
n
→
+
∞
|
∫
R
‍
(
∇
V
(
t
,
u
n
(
t
)
)
-
∇
V
(
t
,
u
(
t
)
)
,
v
(
t
)
)
d
t
|
≤
lim
⁡
n
→
+
∞
∫
R
‍
|
∇
V
(
t
,
u
n
(
t
)
)
-
∇
V
(
t
,
u
(
t
)
)
|
|
v
(
t
)
|
d
t
=
∫
R
‍
lim
⁡
n
→
+
∞
|
∇
V
(
t
,
u
n
(
t
)
)
-
∇
V
(
t
,
u
(
t
)
)
|
×
|
v
(
t
)
|
d
t
=
0
,
∀
v
∈
E
.
This shows that φ
2
∈
C
1
(
E
,
R
). Therefore, φ
∈
C
1
(
E
,
R
) and (27) holds. Similarly, we can prove φ
∈
C
1
(
E
,
R
) and (27) holds by (V1), (V2)′′, and (I)′. Furthermore, the critical points of φ in E are classical solutions of (1) with u
(
±
∞
)
=
0.


## 3. Proofs of Theorems

Proof of Theorem2.
It is clear thatφ
(
0
)
=
0. We first show that the functional φ satisfies the (PS)-condition. Let {
u
n
}
⊂
E satisfying φ
(
u
n
) which is bounded and let φ
′
(
u
n
)
→
0 as n
→
∞. Then, there exists a constant C
1
>
0 such that
(42)
|
φ
(
u
n
)
|
≤
C
1
,
∥
φ
′
(
u
n
)
∥
E
*
≤
μ
C
1
.
From (I)′ and (25), we have
(43)
∑
j
=
-
∞
∞
‍
(
I
(
u
(
t
j
)
)
,
u
(
t
j
)
)
≤
∑
j
=
-
∞
∞
‍
|
I
(
u
(
t
j
)
)
|
|
u
(
t
j
)
|
≤
∑
j
=
-
∞
∞
‍
c
|
u
(
t
j
)
|
2
≤
c
p
T
e
0
∥
u
′
∥
2
2
.
From (26), (27), (31), (42), (43), (V3)′, and (V5), we have
(44)
2
C
1
+
2
C
1
∥
u
n
∥
≥
2
φ
(
u
n
)
-
2
μ
〈
φ
′
(
u
n
)
,
u
n
〉
=
μ
-
2
μ
∥
u
n
∥
2
+
2
∑
j
=
-
∞
∞
‍
∫
0
u
(
t
j
)
‍
I
(
s
)
d
s
-
2
μ
∑
j
=
-
∞
∞
‍
(
I
(
u
n
(
t
j
)
)
,
u
n
(
t
j
)
)
-
2
∫
R
‍
e
Q
(
t
)
[
V
1
(
t
,
u
n
(
t
)
)
-
1
μ
(
∇
V
1
(
t
,
u
n
(
t
)
)
,
u
n
(
t
)
)
]
d
t
+
2
∫
R
‍
e
Q
(
t
)
[
V
2
(
t
,
u
n
(
t
)
)
-
1
μ
(
∇
V
2
(
t
,
u
n
(
t
)
)
,
u
n
(
t
)
)
]
d
t
≥
μ
-
2
μ
∥
u
n
∥
2
-
c
p
T
e
0
∥
u
n
′
∥
2
2
-
2
c
p
T
μ
e
0
∥
u
n
′
∥
2
2
≥
(
μ
-
2
μ
-
(
μ
+
2
)
c
p
T
μ
e
0
)
∥
u
n
∥
2
.
Since μ
>
ϱ
>
2 and 0
<
c
<
(
ϱ
-
2
)
e
0
/
(
ϱ
+
2
)
p
T, the above inequalities imply that there exists a constant C
2
>
0 such that
(45)
∥
u
n
∥
≤
C
2
,
n
∈
N
.
Now, we prove that u
n
→
u
0 in E. Passing to a subsequence if necessary, it can be assumed that u
n
⇀
u
0 in E. For any given number ɛ
>
0, by (V2)′, we can choose ξ
>
0 such that
(46)
|
∇
V
(
t
,
x
)
|
≤
ɛ
a
0
|
x
|
for
|
t
|
≥
R
,
|
x
|
≤
ξ
.
Since Q
(
t
)
→
∞ as |
t
|
→
∞, we can choose R
0
>
R such that
(47)
Q
(
t
)
≥
ln
⁡
(
C
2
ξ
)
for
|
t
|
≥
R
0
.
From (A), we can choose R
0
′
>
R such that
(48)
a
(
t
)
≥
C
2
ξ
for
|
t
|
≥
R
0
′
.
It follows from (23), (45), (47), and (48) that
(49)
|
u
n
(
t
)
|
2
≤
{
∫
t
+
∞
‍
e
-
Q
(
s
)
[
a
(
s
)
]
-
1
/
2
×
[
e
Q
(
s
)
(
|
u
′
(
s
)
|
2
+
a
(
s
)
|
u
(
s
)
|
2
)
]
d
s
∫
t
+
∞
‍
}
1
/
2
≤
ξ
2
C
2
2
∥
u
n
∥
2
≤
ξ
2
for
t
≥
R
1
,
n
∈
N
,
where R
1
=
max
⁡
{
R
0
,
R
0
′
}. Similarly, by (24), (45), (47), and (48), we have
(50)
|
u
n
(
t
)
|
2
≤
ξ
2
for
t
≤
-
R
1
,
n
∈
N
.
Since u
n
⇀
u
0 in E, it is easy to verify that u
n
(
t
) converges to u
0
(
t
) pointwise for all t
∈
R. Hence, it follows from (49) and (50) that
(51)
|
u
0
(
t
)
|
≤
ξ
for
t
∈
(
-
∞
,
-
R
1
]
∪
[
R
1
,
+
∞
)
.
Since e
Q
(
t
)
≥
e
0
>
0 on [
-
R
1
,
R
1
]
=
J, the operator defined by S
:
E
→
H
1
(
J
)
:
u
→
u
|
J is a linear continuous map. So, u
n
→
u
0 in H
1
(
J
). Sobolev theorem implies that u
n
→
u
0 uniformly on J, so there is n
0
∈
N such that
(52)
∫
-
R
1
R
1
‍
e
Q
(
t
)
|
∇
V
(
t
,
u
n
(
t
)
)
-
∇
V
(
t
,
u
0
(
t
)
)
|
×
|
u
n
(
t
)
-
u
0
(
t
)
|
d
t
<
ɛ
for
n
≥
n
0
.
From (45), (46), (49), (50), and (51), we have
(53)
∫
R
∖
[
-
R
1
,
R
1
]
‍
e
Q
(
t
)
|
∇
V
(
t
,
u
n
(
t
)
)
-
∇
V
(
t
,
u
0
(
t
)
)
|
×
|
u
n
(
t
)
-
u
0
(
t
)
|
d
t
≤
∫
R
∖
[
-
R
1
,
R
1
]
‍
e
Q
(
t
)
(
|
∇
V
(
t
,
u
n
(
t
)
)
|
+
|
∇
V
(
t
,
u
0
(
t
)
)
|
)
×
(
|
u
n
(
t
)
|
+
|
u
0
(
t
)
|
)
d
t
≤
ɛ
∫
R
∖
[
-
R
1
,
R
1
]
‍
e
Q
(
t
)
a
0
(
|
u
n
(
t
)
|
+
|
u
0
(
t
)
|
)
×
(
|
u
n
(
t
)
|
+
|
u
0
(
t
)
|
)
d
t
≤
2
ɛ
∫
R
∖
[
-
R
1
,
R
1
]
‍
e
Q
(
t
)
a
0
(
|
u
n
(
t
)
|
2
+
|
u
0
(
t
)
|
2
)
d
t
≤
2
ɛ
∫
R
∖
[
-
R
1
,
R
1
]
‍
e
Q
(
t
)
[
a
(
t
)
|
u
n
(
t
)
|
2
+
a
(
t
)
|
u
0
(
t
)
|
2
]
d
t
≤
2
ɛ
(
∥
u
n
∥
2
+
∥
u
0
∥
2
)
≤
2
ɛ
(
C
2
2
+
∥
u
0
∥
2
)
,
n
∈
N
.
It follows from (52) and (53) that
(54)
∫
R
‍
e
Q
(
t
)
|
∇
V
(
t
,
u
n
(
t
)
)
-
∇
V
(
t
,
u
0
(
t
)
)
|
|
u
n
(
t
)
-
u
0
(
t
)
|
d
t
⟶
0
as
n
⟶
∞
.
From (27) and (I)′, as n
→
∞, we have
(55)
0
⟵
〈
φ
′
(
u
n
)
-
φ
′
(
u
0
)
,
u
n
-
u
0
〉
=
∥
u
n
-
u
0
∥
2
+
∑
j
=
-
∞
∞
‍
(
I
(
u
n
(
t
j
)
)
-
I
(
u
0
(
t
j
)
)
,
u
n
(
t
j
)
-
u
0
(
t
j
)
)
-
∫
R
‍
e
Q
(
t
)
(
∇
V
(
t
,
u
n
(
t
)
)
-
∇
V
(
t
,
u
0
(
t
)
)
,
u
n
(
t
)
-
u
0
(
t
)
)
d
t
≥
∥
u
n
-
u
0
∥
2
-
∫
R
‍
e
Q
(
t
)
(
∇
V
(
t
,
u
n
(
t
)
)
-
∇
V
(
t
,
u
0
(
t
)
)
,
u
n
(
t
)
-
u
0
(
t
)
)
d
t
.
It follows from (54) and (55) that
(56)
∥
u
n
-
u
0
∥
2
⟶
0
as
n
⟶
∞
.
Hence,u
n
→
u
0 in E by (56). This shows that φ satisfies (PS)-condition.
We now show that there exist constantsρ
,
α
>
0 such that assumption (i) of Lemma 7 holds. From (V2)′, there exists δ
∈
(
0,1
) such that
(57)
|
∇
V
(
t
,
x
)
|
≤
a
0
2
|
x
|
for
|
t
|
≥
R
,
|
x
|
≤
δ
.
By V
(
t
,
0
)
=
0 and (57), we have
(58)
|
V
(
t
,
x
)
|
≤
a
0
4
|
x
|
2
for
|
t
|
≥
R
,
|
x
|
≤
δ
.
Let
(59)
C
3
=
sup
⁡
{
V
1
(
t
,
x
)
a
0
∣
t
∈
[
-
R
,
R
]
,
x
∈
R
,
|
x
|
=
1
}
.
Setting σ
=
min
⁡
{
1
/
(
4
C
3
+
1
)
1
/
(
μ
-
2
)
,
δ
} and ∥
u
∥
=
2
e
0
a
0
σ
:
=
ρ, it follows from Lemma 6 that |
u
(
t
)
|
≤
σ
≤
δ
<
1 for t
∈
R. From Lemma 10 (i) and (59), we have
(60)
∫
-
R
R
‍
e
Q
(
t
)
V
1
(
t
,
u
(
t
)
)
d
t
≤
∫
{
t
∈
[
-
R
,
R
]
:
u
(
t
)
≠
0
}
‍
e
Q
(
t
)
V
1
(
t
,
u
(
t
)
|
u
(
t
)
|
)
|
u
(
t
)
|
μ
d
t
≤
C
3
∫
-
R
R
‍
e
Q
(
t
)
a
0
|
u
(
t
)
|
μ
d
t
≤
C
3
σ
μ
-
2
∫
-
R
R
‍
e
Q
(
t
)
a
0
|
u
(
t
)
|
2
d
t
≤
C
3
σ
μ
-
2
∫
-
R
R
‍
e
Q
(
t
)
a
(
t
)
|
u
(
t
)
|
2
d
t
≤
1
4
∫
-
R
R
‍
e
Q
(
t
)
a
(
t
)
|
u
(
t
)
|
2
d
t
.
By (V5), (31), (58), and (60), we have
(61)
φ
(
u
)
=
1
2
∥
u
∥
2
+
∑
j
=
-
∞
∞
‍
∫
0
u
(
t
j
)
‍
I
(
s
)
d
s
-
∫
R
‍
e
Q
(
t
)
V
(
t
,
u
(
t
)
)
d
t
≥
1
2
∥
u
∥
2
+
∑
j
=
-
∞
∞
‍
∫
0
u
(
t
j
)
‍
I
(
s
)
d
s
-
∫
R
∖
[
-
R
,
R
]
‍
e
Q
(
t
)
V
(
t
,
u
(
t
)
)
d
t
-
∫
-
R
R
‍
e
Q
(
t
)
V
1
(
t
,
u
(
t
)
)
d
t
≥
1
2
∥
u
∥
2
-
c
p
T
2
e
0
∥
u
′
∥
2
2
-
1
4
∫
R
∖
[
-
R
,
R
]
‍
a
0
e
Q
(
t
)
|
u
(
t
)
|
2
d
t
-
1
4
∫
-
R
R
‍
e
Q
(
t
)
a
(
t
)
|
u
(
t
)
|
2
d
t
≥
1
2
∥
u
∥
2
-
c
p
T
2
e
0
∥
u
′
∥
2
2
-
1
4
∫
R
∖
[
-
R
,
R
]
‍
e
Q
(
t
)
a
(
t
)
|
u
(
t
)
|
2
d
t
-
1
4
∫
-
R
R
‍
e
Q
(
t
)
a
(
t
)
|
u
(
t
)
|
2
d
t
=
1
2
∥
u
∥
2
-
c
p
T
2
e
0
∥
u
′
∥
2
2
-
1
4
∫
R
‍
e
Q
(
t
)
a
(
t
)
|
u
(
t
)
|
2
d
t
≥
min
⁡
{
1
2
-
c
p
T
2
e
0
,
1
4
}
∥
u
∥
2
.
Therefore, we can choose a constant α
>
0 depending on ρ such that φ
(
u
)
≥
α for any u
∈
E with ∥
u
∥
=
ρ, which shows that φ satisfies assumption (i) of Lemma 7.
Finally, it remains to show thatφ satisfies assumption (ii) of Lemma 7. From Lemma 10 (ii) and (22), we have for any u
∈
E
(62)
∫
-
3
3
‍
e
Q
(
t
)
V
2
(
t
,
u
(
t
)
)
d
t
=
∫
{
t
∈
[
-
3,3
]
:
|
u
(
t
)
|
>
1
}
‍
e
Q
(
t
)
V
2
(
t
,
u
(
t
)
)
d
t
+
∫
{
t
∈
[
-
3,3
]
:
|
u
(
t
)
|
≤
1
}
‍
e
Q
(
t
)
V
2
(
t
,
u
(
t
)
)
d
t
≤
∫
{
t
∈
[
-
3,3
]
:
|
u
(
t
)
|
>
1
}
‍
e
Q
(
t
)
V
2
(
t
,
u
(
t
)
|
u
(
t
)
|
)
|
u
(
t
)
|
ϱ
d
t
+
∫
-
3
3
‍
e
Q
(
t
)
max
⁡
|
x
|
≤
1
V
2
(
t
,
x
)
d
t
≤
∥
u
∥
∞
ϱ
∫
-
3
3
‍
e
Q
(
t
)
max
⁡
|
x
|
=
1
V
2
(
t
,
x
)
d
t
+
∫
-
3
3
‍
e
Q
(
t
)
max
⁡
|
x
|
≤
1
V
2
(
t
,
x
)
d
t
≤
(
1
2
e
0
a
0
)
ϱ
∥
u
∥
ϱ
∫
-
3
3
‍
e
Q
(
t
)
max
⁡
|
x
|
=
1
V
2
(
t
,
x
)
d
t
+
∫
-
3
3
‍
e
Q
(
t
)
max
⁡
|
x
|
≤
1
V
2
(
t
,
x
)
d
t
=
C
4
∥
u
∥
ϱ
+
C
5
,
where C
4
=
(
1
/
2
e
0
a
0
)
ϱ
∫
-
3
3
‍
e
Q
(
t
)
max
⁡
|
x
|
=
1
V
2
(
t
,
x
)
d
t, C
5
=
∫
-
3
3
‍
e
Q
(
t
)
max
⁡
|
x
|
≤
1
V
2
(
t
,
x
)
d
t. Take ω
∈
E such that
(63)
|
ω
(
t
)
|
=
{
1
,
for
|
t
|
≤
1
,
0
,
for
|
t
|
≥
3
,
and |
ω
(
t
)
|
≤
1 for |
t
|
∈
(
1,3
]. For s
>
1, from Lemma 10 (i) and (63), we get
(64)
∫
-
1
1
‍
e
Q
(
t
)
V
1
(
t
,
s
ω
(
t
)
)
d
t
≥
s
μ
∫
-
1
1
‍
e
Q
(
t
)
V
1
(
t
,
ω
(
t
)
)
d
t
=
C
6
s
μ
,
where C
6
=
∫
-
1
1
‍
e
Q
(
t
)
V
1
(
t
,
ω
(
t
)
)
d
t
>
0. From (26), (31), (62), (63), and (64), we get for s
>
1
(65)
φ
(
s
ω
)
=
s
2
2
∥
ω
∥
2
+
∑
j
=
-
∞
∞
‍
∫
0
s
ω
(
t
)
‍
I
(
t
)
d
t
+
∫
R
‍
e
Q
(
t
)
[
V
2
(
t
,
s
ω
(
t
)
)
-
V
1
(
t
,
s
ω
(
t
)
)
]
d
t
≤
s
2
2
∥
ω
∥
2
+
c
p
T
s
2
2
e
0
∥
ω
′
∥
2
2
+
∫
-
3
3
‍
e
Q
(
t
)
V
2
(
t
,
s
ω
(
t
)
)
d
t
-
∫
-
1
1
‍
e
Q
(
t
)
V
1
(
t
,
s
ω
(
t
)
)
d
t
≤
(
e
0
+
c
p
T
2
e
0
)
s
2
∥
ω
∥
2
+
C
4
s
ϱ
∥
ω
∥
ϱ
+
C
5
-
C
6
s
μ
.
Since μ
>
ϱ
>
2 and C
6
>
0, it follows from (65) that there exists s
1
>
1 such that ∥
s
1
ω
∥
>
ρ and φ
(
s
1
ω
)
<
0. Set e
=
s
1
ω
(
t
), and then e
∈
E, ∥
e
∥
=
∥
s
1
ω
∥
>
ρ, and φ
(
e
)
=
φ
(
s
1
ω
)
<
0. By Lemma 7, φ has a critical value c
>
α given by
(66)
c
=
inf
⁡
g
∈
Φ
max
⁡
s
∈
[
0,1
]
φ
(
g
(
s
)
)
,
where
(67)
Φ
=
{
g
∈
C
(
[
0,1
]
,
E
)
:
g
(
0
)
=
0
,
g
(
1
)
=
e
}
.
Hence, there exists u
*
∈
E such that
(68)
φ
(
u
*
)
=
c
,
φ
′
(
u
*
)
=
0
.
The function u
* is a desired solution of problem (1). Since c
>
0, u
* is a nontrivial fast homoclinic solution. The proof is complete.Proof of Theorem3.
In the proof of Theorem2, the condition V
2
(
t
,
x
)
≥
0 in (V5) is only used in the proofs of (45) and assumption (i) of Lemma 7. Therefore, we only need to prove that (45) and assumption (i) of Lemma 7 still hold if we use (V2)′′ and (V5)′ instead of (V2)′ and (V5), respectively. We first prove that (45) holds. From (V3)′, (V5)′, (26), (27), (31), (42), and (43), we have
(69)
2
C
1
+
2
C
1
μ
ϱ
∥
u
n
∥
≥
2
φ
(
u
n
)
-
2
ϱ
〈
φ
′
(
u
n
)
,
u
n
〉
=
(
ϱ
-
2
)
ϱ
∥
u
n
∥
2
+
2
∫
R
‍
e
Q
(
t
)
[
V
2
(
t
,
u
n
(
t
)
)
-
1
ϱ
(
∇
V
2
(
t
,
u
n
(
t
)
)
,
u
n
(
t
)
)
]
d
t
-
2
∫
R
‍
e
Q
(
t
)
[
V
1
(
t
,
u
n
(
t
)
)
-
1
ϱ
(
∇
V
1
(
t
,
u
n
(
t
)
)
,
u
n
(
t
)
)
]
d
t
+
2
∑
j
=
-
∞
∞
‍
∫
0
u
(
t
j
)
‍
I
(
t
)
d
t
-
2
ϱ
∑
j
=
-
∞
∞
‍
(
I
(
u
(
t
j
)
)
,
u
(
t
j
)
)
≥
(
ϱ
-
2
)
ϱ
∥
u
n
∥
2
-
c
p
T
e
0
∥
u
n
′
∥
2
2
-
2
c
p
T
ϱ
e
0
∥
u
n
′
∥
2
2
≥
(
ϱ
-
2
)
ϱ
∥
u
n
∥
2
-
c
p
T
e
0
∥
u
n
∥
2
-
2
c
p
T
ϱ
e
0
∥
u
n
∥
2
=
(
ϱ
-
2
ϱ
-
(
ϱ
+
2
)
c
p
T
ϱ
e
0
)
∥
u
n
∥
2
,
which implies that there exists a constant C
2
>
0 such that (45) holds. Next, we prove that assumption (i) of Lemma 7 still holds. From (V2)′′, there exists δ
∈
(
0,1
) such that
(70)
|
∇
V
(
t
,
x
)
|
≤
a
0
2
|
x
|
for
t
∈
R
,
|
x
|
≤
δ
.
By V
(
t
,
0
)
=
0 and (70), we have
(71)
|
V
(
t
,
x
)
|
≤
a
0
4
|
x
|
2
for
t
∈
R
,
|
x
|
≤
δ
.
Let ∥
u
∥
=
2
e
0
a
0
δ
:
=
ρ, and it follows from Lemma 6 that |
u
(
t
)
|
≤
δ. It follows from (31) and (71) that
(72)
φ
(
u
)
=
1
2
∥
u
∥
2
+
∑
j
=
-
∞
∞
‍
∫
0
u
(
t
j
)
‍
I
(
t
)
d
t
-
∫
R
‍
e
Q
(
t
)
V
(
t
,
u
(
t
)
)
d
t
≥
1
2
∥
u
∥
2
-
c
p
T
2
e
0
∥
u
′
∥
2
2
-
1
4
∫
R
‍
a
0
e
Q
(
t
)
|
u
(
t
)
|
2
d
t
≥
1
2
∥
u
∥
2
-
c
p
T
2
e
0
∥
u
′
∥
2
2
-
1
4
∫
R
‍
a
(
t
)
e
Q
(
t
)
|
u
(
t
)
|
2
d
t
≥
min
⁡
{
1
2
-
c
p
T
2
e
0
,
1
4
}
∥
u
∥
2
.
Therefore, we can choose a constant α
>
0 depending on ρ such that φ
(
u
)
≥
α for any u
∈
E with ∥
u
∥
=
ρ. The proof of Theorem 3 is complete.Proof of Theorem4.
Condition (V6) shows thatφ is even. In view of the proof of Theorem 2, we know that φ
∈
C
1
(
E
,
R
) and satisfies (PS)-condition and assumption (i) of Lemma 7. Now, we prove that (iii) of Lemma 8 holds. Let E
′ be a finite dimensional subspace of E. Since all norms of a finite dimensional space are equivalent, there exists d
>
0 such that
(73)
∥
u
∥
≤
d
∥
u
∥
∞
,
u
∈
E
′
.
Assume that dimE
′
=
m and {
u
1
,
u
2
,
…
,
u
m
} is a base of E
′ such that
(74)
∥
u
i
∥
=
d
,
i
=
1,2
,
…
,
m
.
For any u
∈
E
′, there exist λ
i
∈
R, i
=
1,2
,
…
,
m such that
(75)
u
(
t
)
=
∑
i
=
1
m
‍
λ
i
u
i
(
t
)
for
t
∈
R
.
Let
(76)
∥
u
∥
*
=
∑
i
=
1
m
‍
|
λ
i
|
∥
u
i
∥
.
It is easy to see that ∥
·
∥
* is a norm of E
′. Hence, there exists a constant d
′
>
0 such that d
′
∥
u
∥
*
≤
∥
u
∥. Since u
i
∈
E, by Lemma 6, we can choose R
2
>
R such that
(77)
|
u
i
(
t
)
|
<
d
′
δ
m
+
d
′
,
|
t
|
>
R
2
,
i
=
1,2
,
…
,
m
,
where δ is given in (71). Let
(78)
Θ
=
{
∑
i
=
1
m
‍
λ
i
u
i
(
t
)
:
λ
i
∈
R
,
i
=
1,2
,
…
,
m
;
∑
i
=
1
m
‍
|
λ
i
|
=
1
}
=
{
u
∈
E
′
:
∥
u
∥
*
=
d
}
.
Hence, for u
∈
Θ, let t
0
=
t
0
(
u
)
∈
R such that
(79)
|
u
(
t
0
)
|
=
∥
u
∥
∞
.
Then, by (73)–(76), (78), and (79), we have
(80)
d
d
′
=
d
d
′
∑
i
=
1
m
‍
|
λ
i
|
=
d
′
∑
i
=
1
m
‍
|
λ
i
|
∥
u
i
∥
=
d
′
∥
u
∥
*
≤
∥
u
∥
≤
d
∥
u
∥
∞
=
d
|
u
(
t
0
)
|
=
d
|
∑
i
=
1
m
‍
λ
i
u
i
(
t
0
)
|
≤
d
∑
i
=
1
m
‍
|
λ
i
|
|
u
i
(
t
0
)
|
,
u
∈
Θ
.
This shows that |
u
(
t
0
)
|
≥
d
′ and there exists i
0
∈
{
1,2
,
…
,
m
} such that |
u
i
0
(
t
0
)
|
≥
d
′
/
m, which together with (77) implies that |
t
0
|
≤
R
2. Let R
3
=
R
2
+
1 and
(81)
γ
=
min
⁡
{
d
2
e
0
a
0
e
Q
(
t
)
V
1
(
t
,
x
)
:
-
R
3
≤
t
≤
R
3
,
d
′
2
≤
|
x
|
≤
d
2
e
0
a
0
}
.
Since V
1
(
t
,
x
)
>
0 for all t
∈
R and x
∈
R
∖
{
0
} and V
1
∈
C
1
(
R
×
R
,
R
), it follows that γ
>
0. For any u
∈
E, from Lemmas 6 and 10 (i), we have
(82)
∫
-
R
3
R
3
‍
e
Q
(
t
)
V
2
(
t
,
u
(
t
)
)
d
t
=
∫
{
t
∈
[
-
R
3
,
R
3
]
:
|
u
(
t
)
|
>
1
}
‍
e
Q
(
t
)
V
2
(
t
,
u
(
t
)
)
d
t
+
∫
{
t
∈
[
-
R
3
,
R
3
]
:
|
u
(
t
)
|
≤
1
}
‍
e
Q
(
t
)
V
2
(
t
,
u
(
t
)
)
d
t
≤
∫
{
t
∈
[
-
R
3
,
R
3
]
:
|
u
(
t
)
|
>
1
}
‍
e
Q
(
t
)
V
2
(
t
,
u
(
t
)
|
u
(
t
)
|
)
|
u
(
t
)
|
ϱ
d
t
+
∫
-
R
3
R
3
‍
e
Q
(
t
)
max
⁡
|
x
|
≤
1
V
2
(
t
,
x
)
d
t
≤
∥
u
∥
∞
ϱ
∫
-
R
3
R
3
‍
e
Q
(
t
)
max
⁡
|
x
|
=
1
V
2
(
t
,
x
)
d
t
+
∫
-
R
3
R
3
‍
e
Q
(
t
)
max
⁡
|
x
|
≤
1
V
2
(
t
,
x
)
d
t
≤
(
1
2
e
0
a
0
)
ϱ
∥
u
∥
ϱ
∫
-
R
3
R
3
‍
e
Q
(
t
)
max
⁡
|
x
|
=
1
V
2
(
t
,
x
)
d
t
+
∫
-
R
3
R
3
‍
e
Q
(
t
)
max
⁡
|
x
|
≤
1
V
2
(
t
,
x
)
d
t
=
C
7
∥
u
∥
ϱ
+
C
8
,
where C
7
=
(
1
/
2
e
0
a
0
)
ϱ
∫
-
R
3
R
3
‍
e
Q
(
t
)
max
⁡
|
x
|
=
1
V
2
(
t
,
x
)
d
t and C
8
=
∫
-
R
3
R
3
‍
e
Q
(
t
)
max
⁡
|
x
|
≤
1
V
2
(
t
,
x
)
d
t. Since u
i
′
∈
L
2
(
e
Q
(
t
)
), i
=
1,2
,
…
,
m, it follows that there exists ε
1
∈
(
0
,
(
d
′
)
2
e
0
/
32
m
2
d
2
) such that
(83)
∫
t
-
ε
1
t
+
ε
1
‍
|
u
i
′
(
s
)
|
d
s
=
∫
t
-
ε
1
t
+
ε
1
‍
e
-
Q
(
s
)
/
2
e
Q
(
s
)
/
2
|
u
i
′
(
s
)
|
d
s
≤
1
e
0
∫
t
-
ε
1
t
+
ε
1
‍
e
Q
(
s
)
/
2
|
u
i
′
(
s
)
|
d
s
≤
1
e
0
(
2
ε
1
)
1
/
2
(
∫
t
-
ɛ
t
+
ε
1
‍
e
Q
(
s
)
|
u
i
′
(
s
)
|
2
d
s
)
1
/
2
≤
(
2
ε
1
e
0
)
1
/
2
∥
u
i
′
∥
2
≤
d
′
4
m
for
t
∈
R
,
i
=
1,2
,
…
,
m
.
Then, for u
∈
Θ with |
u
(
t
0
)
|
=
∥
u
∥
∞ and t
∈
[
t
0
-
ε
1
,
t
0
+
ε
1
], it follows from (75), (78), (79), (80), and (83) that
(84)
|
u
(
t
)
|
2
=
|
u
(
t
0
)
|
2
+
2
∫
t
0
t
‍
(
u
′
(
s
)
,
u
(
s
)
)
d
s
≥
|
u
(
t
0
)
|
2
-
2
∫
t
0
-
ε
1
t
0
+
ε
1
‍
|
u
(
s
)
|
|
u
′
(
s
)
|
d
s
≥
|
u
(
t
0
)
|
2
-
2
|
u
(
t
0
)
|
∫
t
0
-
ε
1
t
0
+
ε
1
‍
|
u
′
(
s
)
|
d
s
≥
|
u
(
t
0
)
|
2
-
2
|
u
(
t
0
)
|
∑
i
=
1
m
‍
|
λ
i
|
∫
t
0
-
ε
1
t
0
+
ε
1
‍
|
u
i
′
(
s
)
|
d
s
≥
(
d
′
)
2
2
.
On the other hand, since ∥
u
∥
≤
d for u
∈
Θ, then
(85)
|
u
(
t
)
|
≤
∥
u
∥
∞
≤
d
2
e
0
a
0
,
t
∈
R
,
u
∈
Θ
.
Therefore, from (81), (84), and (85), we have
(86)
∫
-
R
3
R
3
‍
e
Q
(
t
)
V
1
(
t
,
u
(
t
)
)
d
t
≥
∫
t
0
-
ε
1
t
0
+
ε
1
‍
e
Q
(
t
)
V
1
(
t
,
u
(
t
)
)
d
t
≥
2
ε
1
γ
for
u
∈
Θ
.
By (77) and (78), we have
(87)
|
u
(
t
)
|
≤
∑
i
=
1
m
‍
|
λ
i
|
|
u
i
(
t
)
|
≤
δ
for
|
t
|
≥
R
2
,
u
∈
Θ
.
By (26), (31), (58), (82), (86), (87), and Lemma 10, we have for u
∈
Θ and r
>
1
(88)
φ
(
r
u
)
=
r
2
2
∥
u
∥
2
+
∑
j
=
-
∞
∞
‍
∫
0
r
u
(
t
j
)
‍
I
(
t
)
d
t
+
∫
R
‍
e
Q
(
t
)
[
V
2
(
t
,
r
u
(
t
)
)
-
V
1
(
t
,
r
u
(
t
)
)
]
d
t
≤
r
2
2
∥
u
∥
2
+
c
p
T
r
2
2
e
0
∥
u
′
∥
2
2
+
r
ϱ
∫
R
‍
e
Q
(
t
)
V
2
(
t
,
u
(
t
)
)
d
t
-
r
μ
∫
R
‍
e
Q
(
t
)
V
1
(
t
,
u
(
t
)
)
d
t
=
r
2
2
∥
u
∥
2
+
c
p
T
r
2
2
e
0
∥
u
′
∥
2
2
+
r
ϱ
∫
R
∖
(
-
R
3
,
R
3
)
‍
e
Q
(
t
)
V
2
(
t
,
u
(
t
)
)
d
t
-
r
μ
∫
-
R
3
R
3
‍
e
Q
(
t
)
V
1
(
t
,
u
(
t
)
)
d
t
-
r
μ
∫
R
∖
(
-
R
3
,
R
3
)
‍
e
Q
(
t
)
V
1
(
t
,
u
(
t
)
)
d
t
+
r
ϱ
∫
-
R
3
R
3
‍
e
Q
(
t
)
V
2
(
t
,
u
(
t
)
)
d
t
≤
r
2
2
∥
u
∥
2
+
c
p
T
r
2
2
e
0
∥
u
′
∥
2
2
-
r
ϱ
∫
R
∖
(
-
R
3
,
R
3
)
‍
e
Q
(
t
)
V
(
t
,
u
(
t
)
)
d
t
-
r
μ
∫
-
R
3
R
3
‍
e
Q
(
t
)
V
1
(
t
,
u
(
t
)
)
d
t
+
r
ϱ
∫
-
R
3
R
3
‍
e
Q
(
t
)
V
2
(
t
,
u
(
t
)
)
d
t
≤
r
2
2
∥
u
∥
2
+
c
p
T
r
2
2
e
0
∥
u
′
∥
2
2
+
r
ϱ
4
∫
R
∖
(
-
R
3
,
R
3
)
‍
a
0
e
Q
(
t
)
|
u
(
t
)
|
2
d
t
+
r
ϱ
(
C
7
∥
u
∥
ϱ
+
C
8
)
-
2
ε
1
γ
r
μ
≤
r
2
2
∥
u
∥
2
+
c
p
T
r
2
2
e
0
∥
u
′
∥
2
2
+
r
ϱ
4
∫
R
∖
(
-
R
3
,
R
3
)
‍
a
(
t
)
e
Q
(
t
)
|
u
(
t
)
|
2
d
t
+
r
ϱ
(
C
7
∥
u
∥
ϱ
+
C
8
)
-
2
ε
1
γ
r
μ
≤
(
1
2
+
c
p
T
2
e
0
)
r
2
∥
u
∥
2
+
r
ϱ
4
∥
u
∥
2
+
r
ϱ
(
C
7
∥
u
∥
ϱ
+
C
8
)
-
2
ε
1
γ
r
μ
≤
(
1
2
+
c
p
T
2
e
0
)
r
2
d
2
+
r
ϱ
4
d
2
+
C
7
(
r
d
)
ϱ
+
C
8
r
ϱ
-
2
ε
1
γ
r
μ
.
Since μ
>
ϱ
>
2, there exists r
0
=
r
0
(
c
,
p
,
T
,
e
0
,
d
,
d
′
,
C
7
,
C
8
,
R
2
,
R
3
,
ε
1
,
γ
)
=
r
0
(
E
′
)
>
1 such that
(89)
φ
(
r
u
)
<
0
for
u
∈
Θ
,
r
≥
r
0
.
It follows that
(90)
φ
(
u
)
<
0
for
u
∈
E
′
,
∥
u
∥
≥
d
r
0
,
which shows that (iii) of Lemma 8 holds. By Lemma 8, φ possesses an unbounded sequence {
c
n
}
n
=
1
∞ of critical values with c
n
=
φ
(
u
n
), where u
n is such that φ
′
(
u
n
)
=
0 for n
=
1,2
,
… . If {
∥
u
n
∥
} is bounded, then there exists C
9
>
0 such that
(91)
∥
u
n
∥
≤
C
9
for
n
∈
N
.
By a similar fashion for the proof of (49) and (50), for the given δ in (58), there exists R
4
>
R such that
(92)
|
u
n
(
t
)
|
≤
δ
for
|
t
|
≥
R
4
,
n
∈
N
.
Hence, by (22), (26), (31), (58), (91), and (92), we have
(93)
1
2
∥
u
n
∥
2
=
c
n
+
∫
R
‍
e
Q
(
t
)
V
(
t
,
u
n
(
t
)
)
d
t
-
∑
j
=
-
∞
∞
‍
∫
0
u
n
(
t
j
)
‍
I
(
t
)
d
t
=
c
n
+
∫
R
∖
(
-
R
4
,
R
4
)
‍
e
Q
(
t
)
V
(
t
,
u
n
(
t
)
)
d
t
+
∫
-
R
4
R
4
‍
e
Q
(
t
)
V
(
t
,
u
n
(
t
)
)
d
t
-
∑
j
=
-
∞
∞
‍
∫
0
u
n
(
t
j
)
‍
I
(
t
)
d
t
≥
c
n
-
1
4
∫
R
∖
(
-
R
4
,
R
4
)
‍
a
0
e
Q
(
t
)
|
u
n
(
t
)
|
2
d
t
-
∫
-
R
4
R
4
‍
e
Q
(
t
)
|
V
(
t
,
u
n
(
t
)
)
|
d
t
-
c
p
T
2
e
0
∥
u
n
′
∥
2
2
≥
c
n
-
1
4
∫
R
∖
(
-
R
4
,
R
4
)
‍
e
Q
(
t
)
a
(
t
)
|
u
n
(
t
)
|
2
d
t
-
∫
-
R
4
R
4
‍
e
Q
(
t
)
|
V
(
t
,
u
n
(
t
)
)
|
d
t
-
c
p
T
2
e
0
∥
u
n
′
∥
2
2
≥
c
n
-
1
4
∥
u
n
∥
2
-
∫
-
R
4
R
4
‍
e
Q
(
t
)
max
⁡
|
x
|
≤
2
e
0
a
0
C
9
|
V
(
t
,
x
)
|
d
t
-
c
p
T
2
e
0
∥
u
n
∥
2
.
It follows from (93) that
(94)
c
n
≤
(
3
4
+
c
p
T
2
e
0
)
∥
u
n
∥
2
+
∫
-
R
3
R
3
‍
e
Q
(
t
)
max
⁡
|
x
|
≤
2
e
0
a
0
C
9
|
V
(
t
,
x
)
|
d
t
<
+
∞
.
This contradicts the fact that {
c
n
}
n
=
1
∞ is unbounded, and so {
∥
u
n
∥
} is unbounded. The proof is complete.Proof of Theorem5.
In view of the proofs of Theorems3 and 4, the conclusion of Theorem 5 holds. The proof is complete.


## 4. Examples

Example 1.
Consider the following system:(95)
u
′′
(
t
)
+
t
3
u
′
(
t
)
-
a
(
t
)
u
(
t
)
+
∇
V
(
t
,
u
(
t
)
)
=
0
,
a
.
e
.
t
∈
(
t
j
,
t
j
+
1
)
,
j
∈
Z
,
Δ
u
′
(
t
j
)
=
u
′
(
t
j
+
)
-
u
′
(
t
j
-
)
=
I
(
u
(
t
j
)
)
,
j
∈
Z
,
where q
(
t
)
=
t
3, t
∈
R, u
∈
R, a
∈
C
(
R
,
(
0
,
+
∞
)
), and a
(
t
)
→
+
∞ as |
t
|
→
+
∞. Let
(96)
V
(
t
,
x
)
=
∑
i
=
1
m
‍
(
a
i
+
1
+
sin
t
)
|
x
|
μ
i
-
∑
j
=
1
n
‍
(
b
j
+
1
+
cos
⁡
t
)
|
x
|
ϱ
j
,
I
(
x
)
=
(
ϱ
1
-
2
)
x
4
(
ϱ
1
+
2
)
p
π
,
where μ
1
>
μ
2
>
⋯
>
μ
m
>
ϱ
1
>
ϱ
2
>
⋯
>
ϱ
n
>
2, a
i, b
j
>
0, i
=
1
,
…
,
m, and j
=
1
,
…
,
n. Let
(97)
V
1
(
t
,
x
)
=
∑
i
=
1
m
‍
(
a
i
+
1
+
sin
t
)
|
x
|
μ
i
,
V
2
(
t
,
x
)
=
∑
j
=
1
n
‍
(
b
j
+
1
+
cos
⁡
t
)
|
x
|
ϱ
j
.
Then, it is easy to check that all the conditions of Theorem 4 are satisfied with μ
=
μ
m and ϱ
=
ϱ
1. Hence, problem (95) has an unbounded sequence of fast homoclinic solutions.Example 2.
Consider the following system:(98)
u
′′
(
t
)
+
(
t
+
t
3
)
u
′
(
t
)
-
a
(
t
)
u
(
t
)
+
∇
V
(
t
,
u
(
t
)
)
=
0
,
a
.
e
.
t
∈
(
t
j
,
t
j
+
1
)
,
j
∈
Z
,
Δ
u
′
(
t
j
)
=
u
′
(
t
j
+
)
-
u
′
(
t
j
-
)
=
I
(
u
(
t
j
)
)
,
j
∈
Z
,
where q
(
t
)
=
t
+
t
3, t
∈
R, u
∈
R, a
∈
C
(
R
,
(
0
,
+
∞
)
), and a
(
t
)
→
+
∞ as |
t
|
→
+
∞. Let
(99)
V
(
t
,
x
)
=
(
a
1
+
1
+
sin
t
)
|
x
|
μ
1
+
(
a
2
+
2
+
sin
t
)
|
x
|
μ
2
-
b
1
(
cos
⁡
t
)
|
x
|
ϱ
1
-
(
b
2
+
1
+
cos
⁡
t
)
|
x
|
ϱ
2
,
where μ
1
>
μ
2
>
ϱ
1
>
ϱ
2
>
2, a
1, a
2
>
0, b
1, and b
2
>
0. Let
(100)
V
1
(
t
,
x
)
=
(
a
1
+
1
+
sin
t
)
|
x
|
μ
1
+
(
a
2
+
2
+
sin
t
)
|
x
|
μ
2
,
V
2
(
t
,
x
)
=
b
1
(
cos
⁡
t
)
|
x
|
ϱ
1
+
(
b
2
+
1
+
cos
⁡
t
)
|
x
|
ϱ
2
,
I
(
x
)
=
(
ϱ
1
-
2
)
x
6
(
ϱ
1
+
2
)
p
π
.
Then, it is easy to check that all the conditions of Theorem 5 are satisfied with μ
=
μ
2 and ϱ
=
ϱ
1. Hence, by Theorem 5, problem (98) has an unbounded sequence of fast homoclinic solutions.



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*Source: 960276-2014-06-15.xml*