# Bifurcation and Nodal Solutions for the Half-Linear Problems with Nonasymptotic Nonlinearity at0 and ∞

Authors: Wenguo Shen
Journal: Discrete Dynamics in Nature and Society
 (2016)

---

## Abstract

We study the existence of nodal solutions for the following problem:-
x
″
=
α
x
+
+
β
x
-
+
r
a
(
t
)
f
(
x
), 0
<
t
<
1, x
(
0
)
=
x
(
1
)
=
0, where r
≠
0 is a parameter, a
(
t
)
∈
C
(
[
0,1
]
,
(
0
,
∞
)
) with a
(
t
)
≢
0 on any subinterval of [
0,1
], x
+
=
m
a
x
{
x
,
0
}
,
x
-
=
-
m
i
n
{
x
,
0
}, and α
,
β
∈
C
[
0,1
]; f
∈
C
(
R
,
R
), s
f
(
s
)
>
0 for s
≠
0, and f
0
,
f
∞
∉
(
0
,
∞
), where f
0
=
l
i
m
|
s
|
→
0
f
(
s
)
/
s and f
∞
=
l
i
m
|
s
|
→
+
∞
f
(
s
)
/
s. We use bifurcation techniques to prove our main results.


---

## Body

## 1. Introduction

By applying the bifurcation techniques of Rabinowitz [1, 2], Im et al. [3] studied the existence of positive solutions of the problems and some authors [4–7] studied the existence of nodal solutions of the problems. Recently, Dai and Ma [8] established the unilateral global bifurcation theory for the problems by applying the bifurcation techniques of Dancer [9]. Later, Dai [10] also considered the existence of nodal solutions for the problems with nonasymptotic nonlinearity at 0 or ∞ by applying the unilateral global bifurcation theory of [8]. For the abstract unilateral global bifurcation theory, we refer the reader to [1, 2, 8, 9, 11–13] and the references therein. On the other hand, problems involving nondifferentiable nonlinearity have also been investigated by applying bifurcation techniques; see [13–17]. In particular, Berestycki [14] established the Rabinowitz-type global interval bifurcation result and Ma and Dai [13] established the unilateral global interval bifurcation theorem. Meanwhile, half-linear or half-quasilinear boundary value problems have attracted the attention of many specialists in different equations because of their interesting applications; see [13, 14, 16–18]. Among them, Berestycki [14] (or see [13]) established the spectrum for the following half-linear eigenvalue problem:(1)
-
x
′
′
=
λ
a
t
x
+
α
x
+
+
β
x
-
,
0
<
t
<
1
,
x
0
=
x
1
=
0
,where x
+
=
m
a
x
{
x
,
0
}, x
-
=
-
m
i
n
{
x
,
0
}, α
,
β
∈
C
[
0,1
], and a satisfies the following condition:(
A
1)
a
∈
C
(
[
0,1
]
,
(
0
,
∞
)
) with a
(
t
)
≢
0 on any subinterval of [
0,1
].LetY
=
C
[
0,1
] with the norm x
∞
=
m
a
x
t
∈
[
0,1
]
x
t. Let E
=
{
x
∈
C
1
[
0,1
]
∣
x
(
0
)
=
x
(
1
)
=
0
} with the norm x
=
m
a
x
{
x
∞
,
x
′
∞
}.We define the linear operatorL
:
D
(
L
)
⊂
E
→
Y
(2)
L
x
=
-
x
′
′
,
x
∈
D
Lwith D
(
L
)
=
{
x
∈
C
2
[
0,1
]
∣
x
(
0
)
=
x
(
1
)
=
0
}. Then L is a closed operator and L
-
1
:
Y
→
E is completely continuous.LetE
=
R
×
E under the product topology. Let S
k
+ denote the set of function in E which have exactly k
-
1 interior nodal (i.e., nondegenerate) zeros in (
0,1
) and are positive near t
=
0, set S
k
-
=
-
S
k
+, and S
k
=
S
k
+
∪
S
k
-. They are disjoint and open in E. Finally, let Φ
k
±
=
R
×
S
k
± and Φ
k
=
R
×
S
k.By Theorem 1 of [14], Berestycki [14] obtained the following result.Lemma 1 (see [14, Theorem  2]).
Let(
A
1
) hold. There exist two sequences of simple half eigenvalues for (1)(3)
lim
k
→
∞
⁡
λ
k
+
=
+
∞
,
λ
1
+
<
λ
2
+
<
⋯
<
λ
k
+
<
⋯
,
lim
k
→
∞
⁡
λ
k
-
=
+
∞
,
λ
1
-
<
λ
2
-
<
⋯
<
λ
k
-
<
⋯
.The corresponding half-lines solutions are in {
λ
k
+
}
×
S
k
+ and {
λ
k
-
}
×
S
k
-. Further, aside from these solutions and the trivial ones, there are no other solutions of (1).Furthermore, following Lemma1, Ma and Dai [13] (or see [14]) considered the existence of nodal solutions for the following half-linear eigenvalue problem:(4)
-
x
′
′
=
α
x
+
+
β
x
-
+
r
a
t
f
x
,
0
<
t
<
1
,
x
0
=
x
1
=
0
,where r
≠
0 is a parameter and f
∈
C
(
R
,
R
) satisfies the following assumptions:(A
2)
s
f
(
s
)
>
0 for s
≠
0.
(A
3)
There existf
0
,
f
∞
∈
(
0
,
+
∞
) such that (5)
f
0
=
lim
s
→
0
⁡
f
s
s
,
f
∞
=
lim
s
→
+
∞
⁡
f
s
s
.Firstly, we show that the existence and uniqueness theorem is valid for (4).Lemma 2 (see [13, Lemma  2.2]).
If(
λ
,
x
) is a nontrivial solution of (4) under assumptions (
A
1
)
–
(
A
3
) and x has a double zero, then x
≡
0.Under assumptions (A
1)–(A
3), they obtained the following result.Theorem 3 (see [13, Theorem  4.1]).
Let(
A
1
), (
A
2
), and (
A
3
) hold. For ν
∈
{
+
,
-
}, either λ
ν
/
f
∞
<
r
<
λ
ν
/
f
0 or λ
ν
/
f
0
<
r
<
λ
ν
/
f
∞. Then problem (4) has a solution x
k
ν such that x
k
ν has exactly k
-
1 zeros in (
0,1
) and ν
x
k
ν is positive near 0.Of course, the natural question is that of what would happen iff
0
∉
(
0
,
+
∞
) or f
∞
∉
(
0
,
+
∞
). Obviously, the previous results cannot deal with this case. The purpose of this work is to establish several results similar to those of [13]. The main methods used in this work are global bifurcation techniques and the approximation of connected components. Moreover, we consider the cases of f
0
,
f
∞
∉
(
0
,
∞
), while the authors of [13, 16, 17] only studied the cases of f
0
,
f
∞
∈
(
0
,
∞
).In this paper, we will investigate the existence of nodal solutions for problem (4), where a satisfies condition (
A
1
). Throughout this paper, we assume that f satisfies (
A
2
) and the following assumptions:(H
1)
f
0
∈
(
0
,
∞
) and f
∞
=
0.
(H
2)
f
0
∈
(
0
,
∞
) and f
∞
=
∞.
(H
3)
f
0
=
∞ and f
∞
∈
(
0
,
∞
).
(H
4)
f
0
=
∞ and f
∞
=
0.
(H
5)
f
0
=
∞ and f
∞
=
∞.
(H
6)
f
0
=
0 and f
∞
∈
(
0
,
∞
).
(H
7)
f
0
=
0 and f
∞
=
0.
(H
8)
f
0
=
0 and f
∞
=
∞.The rest of this paper is arranged as follows. In Section2, we gave some preliminaries. In Section 3, we give the interval for the parameter r which ensure the existence of single or multiple nodal solutions for half-linear problem (4) under assumptions (
A
2
) and (H
1)–(H
8) for the nonlinearity f.


## 2. Preliminaries

LetS
k
ν denote the closure of the set of nontrivial solutions of (4) in R
×
E, with S
k
ν denoting the subset of with u
∈
S
k
± and S
k
=
S
k
+
∪
S
k
-.In order to prove our main results, we need the following Sturm type comparison result.Lemma 4.
Letb
2
(
t
)
≥
m
a
x
{
b
1
(
t
)
,
b
1
(
t
)
+
α
(
t
)
+
β
(
t
)
,
b
1
(
t
)
-
α
(
t
)
-
β
(
t
)
} for t
∈
(
0,1
) and b
i
(
t
)
∈
C
(
0,1
), i
=
1,2. Also let u
1, u
2 be solutions of the following differential equations: (6)
u
′
′
+
b
i
t
u
+
α
u
+
+
β
u
-
=
0
,
t
∈
0,1
,
i
=
1,2
,respectively. If (
c
,
d
)
⊆
(
0,1
), and u
1
(
c
)
=
u
1
(
d
)
=
0, u
1
(
t
)
≠
0 in (
c
,
d
), then either there exists τ
∈
(
c
,
d
) such that u
2
(
τ
)
=
0 or b
2
=
b
1 or b
2
=
b
1
+
α
+
β or b
2
=
b
1
-
α
-
β and u
2
(
t
)
=
μ
u
1
(
t
) for some constant μ
≠
0.Proof.
We discuss four cases.
Case  1 (
u
1
(
t
)
>
0
,
u
2
(
t
)
>
0
in
(
c
,
d
)
). By the Picone identity [19] (or (0.4) of [20]), we have(7)
∫
c
d
u
1
u
2
u
2
u
1
′
-
u
1
u
2
′
d
t
=
∫
c
d
u
1
u
2
u
2
L
^
u
1
-
u
1
L
^
u
2
d
t
+
∫
c
d
u
1
′
-
u
1
u
2
u
2
′
2
d
t
,where L
^
x
=
-
L
x
=
x
′
′, x
∈
D
(
L
). It follows that(8)
∫
c
d
u
1
u
2
u
2
u
1
′
-
u
1
u
2
′
d
t
=
∫
c
d
b
2
-
b
1
u
1
2
+
u
1
′
-
u
1
u
2
u
2
′
2
d
t
.
The left-hand side of (8) equals(9)
-
lim
t
→
c
+
⁡
u
1
u
2
u
1
′
-
u
1
u
2
′
u
2
+
lim
t
→
d
-
⁡
u
1
u
2
u
1
′
-
u
1
u
2
′
u
2
=
-
H
c
+
H
d
.
We prove thatH
c
=
0. If u
2
(
c
)
≠
0, then H
c
=
0. If u
2
(
c
)
=
0, noting the conclusion of Lemma 2, then u
2
′
(
c
)
>
0. By L’Hospital rule, we have that(10)
H
c
=
lim
t
→
c
+
⁡
u
1
u
2
u
1
′
-
u
1
u
2
′
u
2
=
l
i
m
t
→
c
+
u
1
′
u
2
u
1
′
-
u
1
u
2
′
+
u
1
u
2
u
1
′
-
u
1
u
2
′
′
u
2
′
=
0
.Similarly, we can obtain that H
d
=
0. Therefore, the left-hand side of (8) equals zero. Hence, the right-hand side of (8) also equals zero. It follows that there exists a constant μ
≠
0 such that u
2
=
μ
u
1 and b
2
=
b
1.
Case  2 (
u
1
(
t
)
>
0
,
u
2
(
t
)
<
0
in
(
c
,
d
)
). Similar to (8), we can get(11)
∫
c
d
u
1
u
2
u
2
u
1
′
-
u
1
u
2
′
d
t
=
∫
c
d
b
2
-
b
1
-
α
-
β
u
1
2
+
u
1
′
-
u
1
u
2
u
2
′
2
d
t
.The above argument is still valid for this case.
Case  3 (
u
1
(
t
)
<
0
,
u
2
(
t
)
<
0
in
(
c
,
d
)
). Similar to Case  1, we can get the result.
Case  4 (
u
1
(
t
)
<
0
,
u
2
(
t
)
>
0
in
(
c
,
d
)
). Similar to (8), we can get(12)
∫
c
d
u
1
u
2
u
2
u
1
′
-
u
1
u
2
′
′
d
t
=
∫
c
d
b
2
-
b
1
+
α
+
β
u
1
2
+
u
1
′
-
u
1
u
2
u
2
′
2
d
t
.
Similar to the proof of Case  1, we can obtain the result.By Lemma4, we obtain the following result that will be used later.Lemma 5.
LetI
=
[
a
,
b
] be such that I
⊂
[
0,1
] and (13)
meas
⁡
I
>
0
.
Letg
n
(
t
)
∈
C
(
[
0,1
]
,
R
) be such that(14)
lim
n
→
+
∞
⁡
g
n
t
=
+
∞
uniformly
on
I
.
Lety
n
∈
E be a solution of the equation(15)
y
n
′
′
t
+
g
n
t
y
n
+
α
y
n
+
+
β
y
n
-
=
0
,
0
<
t
<
1
.Then y
n must change sign on I as n
→
+
∞.Proof.
Setα
0
≔
m
a
x
t
∈
[
0,1
]
α
t and β
0
≔
m
a
x
t
∈
[
0,1
]
β
t. By simple computation, we can show that (16)
g
n
t
+
α
y
n
+
y
n
+
β
y
n
-
y
n
≥
g
n
t
-
α
0
-
β
0
∀
t
∈
0,1
.After taking a subsequence if necessary, we may assume that (17)
g
n
t
-
α
0
-
β
0
≥
λ
j
∀
t
∈
Ias j
→
+
∞, where λ
j is jth eigenvalue of the following problem:(18)
u
′
′
+
λ
u
=
0
,
0
<
t
<
1
,
u
0
=
u
1
=
0
.Set s
=
(
b
-
a
)
t
+
a, u
(
s
)
=
v
(
(
b
-
a
)
s
+
a
). By some simple computations, we can show(19)
b
-
a
2
v
′
′
+
λ
v
=
0
,
t
∈
a
,
b
,
v
a
=
v
b
=
0
.Let ψ
j be the corresponding eigenfunction of λ
j. Since λ
j
→
∞ as j
→
∞, Lemma 2.1 of [4] implies that ψ
j must change sign on I. Note that the conclusion of Lemma 4 also is valid if α
=
β
=
0. Using these facts and Lemma 4, we can obtain the desired result.Definition 6 (see [21]).
LetX be a Banach space and {
C
n
∣
n
=
1,2
,
…
} be a family of subsets of X. Then the superior limit D of {
C
n
} is defined by(20)
D
≔
limsup
n
→
∞
⁡
C
n
=
x
∈
X
∣
∃
n
i
⊂
N
,
x
n
i
∈
C
n
i
,
s
u
c
h
t
h
a
t
x
n
i
⟶
x
.Lemma 7 (see [22]).
LetX be a Banach space and let {
C
n
∣
n
=
1,2
,
…
} be a family of closed connected subsets of X. Assume that(i)
there existz
n
∈
C
n, n
=
1,2
,
…, and z
∗
∈
X, such that z
n
→
z
∗;
(ii)
r
n
=
s
u
p
{
x
∣
x
∈
C
n
}
=
∞;
(iii)
for allR
>
0, ⋃
n
=
1
∞
C
n
∩
B
R is a relative compact set of X, where(21)
B
R
=
x
∈
X
∣
x
≤
R
.Then there exists an unbounded component C in D and z
∗
∈
C.


## 3. Nodal Solutions for Half-Linear Eigenvalue Problems

In this section, we will study the existence of nodal solutions for problem (4) with f
0
∉
(
0
,
+
∞
) or f
∞
∉
(
0
,
+
∞
).We start this section by studying the following eigenvalue problem:(22)
-
x
′
′
=
α
x
+
+
β
x
-
+
λ
r
a
t
f
x
,
0
<
t
<
1
,
x
0
=
x
1
=
0
,where λ
>
0 is a parameter. Firstly, under the conditions (A
1)–(A
3), let ζ
∈
C
(
R
,
R
) be such that(23)
f
x
=
f
0
x
+
ζ
xwith l
i
m
x
→
0
ζ
(
x
)
/
x
=
0. Let us consider(24)
-
x
′
′
=
λ
r
a
t
f
0
x
+
α
x
+
+
β
x
-
+
λ
r
a
t
ζ
x
,
0
<
t
<
1
,
x
0
=
x
1
=
0
,as a bifurcation problem from the trivial solution x
≡
0. By using Theorem 3 of [13] or Ma and Dai [13, Lemma  4.1] obtained the following Lemma.Lemma 8 (see [13, Lemma  4.1]).
Forν
∈
{
+
,
-
}, (
λ
k
ν
/
r
f
0
,
0
) is a bifurcation point for problem (24). Moreover, there exists an unbounded continuum D
k
ν of solutions of problem (24), such that D
k
ν
⊂
(
Φ
k
ν
∪
{
(
λ
k
ν
/
r
f
0
,
0
)
}
).Remark 9.
Any solution of (22) of the form (
1
,
x
) yields a solution x of (4). In order to prove our main results, one will only show that D
k
ν crosses the hyperplane {
1
}
×
E in R
×
E.
Clearly,(
A
2
) implies f
(
0
)
=
0. Hence, x
=
0 is always the solution of (4). Applying Lemma 8 (or Lemma 4.1 of [13]), we will establish the existence of nodal solutions of (4) as follows.Theorem 10.
Let(
A
1
), (
A
2
), and (
H
1
) hold. For ν
∈
{
+
,
-
}, assume that one of the following conditions holds:(i)
r
∈
(
λ
k
ν
/
f
0
,
+
∞
)
for
λ
k
ν
>
0.
(ii)
r
∈
(
-
∞
,
λ
k
-
/
f
0
)
∪
(
λ
k
+
/
f
0
,
+
∞
)
for
ν
λ
k
ν
>
0.
(iii)
r
∈
(
-
∞
,
λ
k
+
/
f
0
)
∪
(
λ
k
-
/
f
0
,
+
∞
)
for
ν
λ
k
ν
<
0.
(iv)
r
∈
(
-
∞
,
λ
k
ν
/
f
0
)
for
λ
k
ν
<
0.
Then problem (4) possesses two solutions x
k
+ and x
k
- such that x
k
+ has exactly k
-
1 zeros in (
0,1
) and is positive near 0 and x
k
- has exactly k
-
1 zeros in (
0,1
) and is negative near 0.Proof.
We only prove the case of (i) since the proofs of the cases for (ii), (iii), and (iv) can be given similarly.
In view of the proof to prove [13, Theorem 4.1], we only need to show that D
k
ν joins (
λ
k
ν
/
r
f
0
,
0
) to (
∞
,
∞
). To do this, it is enough to prove that [
λ
k
ν
/
r
f
0
,
+
∞
)
⊂
P
r
o
j
R
D
k
ν.
Assume on the contrary thats
u
p
{
λ
∣
(
λ
,
u
)
∈
D
k
ν
}
<
+
∞; then there exists a sequence (
μ
n
,
x
n
)
∈
D
k
ν such that(25)
lim
n
→
∞
⁡
x
n
=
+
∞
,
μ
n
≤
c
0
,for some positive constant c
0 not depending on n.
By(
H
1
), let f
¯
(
x
)
=
m
a
x
0
≤
s
≤
x
f
s, then f
¯ is nondecreasing and(26)
lim
x
→
+
∞
⁡
f
¯
x
x
=
0
.We consider the equation(27)
-
x
n
′
′
=
α
x
n
+
+
β
x
n
-
+
μ
n
r
a
t
f
x
n
,
t
∈
0,1
,
x
n
0
=
x
n
1
=
0
.Let y
n
=
x
n
/
x
n, and y
n should be the solutions of problem(28)
-
y
n
′
′
=
α
y
n
+
+
β
y
n
-
+
μ
n
r
a
t
f
x
n
x
n
,
t
∈
0,1
,
y
n
0
=
y
n
1
=
0
.
Sincey
n is bounded in C
2
[
0,1
], choosing a subsequence and relabelling if necessary, we have that y
n
→
y for some y
∈
E and y
=
1.
Furthermore, from (26) and the fact that f
¯ is nondecreasing, we have that(29)
lim
n
→
∞
⁡
f
x
n
x
n
=
0
,since(30)
f
x
n
x
n
≤
f
¯
x
n
x
n
≤
f
¯
x
n
∞
x
n
≤
f
¯
x
n
x
n
⟶
0
,
n
⟶
+
∞
.By (29) and the compactness of L
-
1, we obtain that(31)
-
y
′
′
=
α
y
+
+
β
y
-
,
t
∈
0,1
,
y
0
=
y
1
=
0
. By y
(
0
)
=
y
(
1
), there exists ξ
∈
(
0,1
) such that y
′
(
ξ
)
=
0. Without loss of generality, it follows that (32)
y
t
=
∫
0
t
∫
ξ
s
-
α
y
+
-
β
y
-
d
τ
d
s
≤
M
∫
0
t
y
d
s
,
t
∈
0,1
,
M
=
m
a
x
{
α
0
,
β
0
}.
By the Gronwall-Bellman inequality [3, Lemma 2.1], we get y
t
≡
0, ∀
t
∈
[
0,1
].
This contradictsy
(
t
)
=
1.Theorem 11.
Let(
A
1
), (
A
2
), and (
H
2
) hold. For ν
∈
{
+
,
-
}, assume that one of the following conditions holds:(i)
r
∈
(
0
,
λ
k
ν
/
f
0
)
for
λ
k
ν
>
0.
(ii)
r
∈
(
0
,
λ
k
+
/
f
0
)
∪
(
λ
k
-
/
f
0
,
0
)
for
ν
λ
k
ν
>
0.
(iii)
r
∈
(
0
,
λ
k
-
/
f
0
)
∪
(
λ
k
+
/
f
0
,
0
)
for
ν
λ
k
ν
<
0.
(iv)
r
∈
(
λ
k
ν
/
f
0
,
0
)
for
λ
k
ν
<
0.
Then problem (4) possesses two solutions x
k
+ and x
k
- such that x
k
+ has exactly k
-
1 zeros in (
0,1
) and is positive near 0 and x
k
- has exactly k
-
1 zeros in (
0,1
) and is negative near 0.Proof.
We will only prove the case of (i) since the proofs of the cases for (ii), (iii), and (iv) are completely analogous.
Inspired by the idea of [23] or see [10], we define the cut-off function of f as follows:(33)
f
n
s
≔
f
s
,
s
∈
-
n
,
n
,
2
n
2
-
f
n
n
s
-
n
+
f
n
,
s
∈
n
,
2
n
,
2
n
2
+
f
-
n
n
s
+
n
+
f
-
n
,
s
∈
-
2
n
,
-
n
,
n
s
,
s
∈
-
∞
,
-
2
n
∪
2
n
,
+
∞
.We consider the following problem:(34)
-
x
′
′
=
α
x
+
+
β
x
-
+
λ
r
a
t
f
n
x
,
0
<
t
<
1
,
x
0
=
x
1
=
0
.
Clearly, we can see thatl
i
m
n
→
+
∞
f
[
n
]
(
s
)
=
f
(
s
), (
f
[
n
]
)
0
=
f
0, and (
f
[
n
]
)
∞
=
n.
Similar to the proof of [13, Theorem 4.1], by Lemma 8 and Remark 9, there exists an unbounded continuum D
k
ν
[
n
] of solutions of problem (34) emanating from (
λ
k
ν
/
r
f
0
,
0
), such that D
k
ν
[
n
]
⊂
(
Φ
k
ν
∪
{
(
λ
k
ν
/
r
f
0
,
0
)
}
), and D
k
ν
[
n
] joins (
λ
k
ν
/
r
f
0
,
0
) to (
λ
k
ν
/
r
n
,
∞
).
Takingz
n
=
(
λ
k
ν
/
r
n
,
∞
) and z
∗
=
(
0
,
∞
), we have that z
n
→
z
∗.
So condition (i) in Lemma7 is satisfied with z
∗
=
(
0
,
∞
).
Obviously(35)
r
n
=
sup
⁡
λ
+
x
∣
λ
,
x
∈
D
ν
n
=
∞
,and, accordingly (ii) in Lemma 7 holds. (iii) in Lemma 7 can be deduced directly from the Arzela-Ascoli Theorem and the definition of f
[
n
].
Therefore, by Lemma7, l
i
m
s
u
p
n
→
∞
D
k
ν
[
n
] contains an unbounded connected component D
k
ν with (
0
,
∞
)
∈
D
k
ν.
Froml
i
m
n
→
+
∞
f
[
n
]
(
s
)
=
f
(
s
), (34) can be converted to equivalent equation (22). Since D
k
ν
[
n
]
⊂
Φ
k
ν, we conclude D
k
ν
⊂
Φ
k
ν. Moreover, D
k
ν
⊂
S
k
ν by (4).
In the following, we show that(
λ
k
ν
/
r
f
0
,
0
)
∈
D
k
ν.
Let(
μ
n
,
x
n
)
∈
D
k
ν
∖
{
(
λ
k
ν
/
r
f
0
,
0
)
} satisfy l
i
m
n
→
∞
x
n
=
0.
Letζ
∈
C
(
R
,
R
) be such that(36)
f
s
=
f
0
s
+
ζ
s
.With l
i
m
s
→
0
(
ζ
(
s
)
/
s
)
=
0, let ζ
¯
(
x
)
=
m
a
x
0
≤
s
≤
x
ζ
s, then ζ
¯ is nondecreasing and(37)
lim
x
→
∞
⁡
ζ
¯
x
x
=
0
.
We divide the equation(38)
-
x
n
′
′
=
μ
n
r
a
t
f
0
x
n
+
α
x
n
+
+
β
x
n
-
+
λ
r
a
t
ζ
x
n
,
0
<
t
<
1
,
x
n
0
=
x
n
1
=
0
.
Lety
n
=
x
n
/
x
n, and y
n should be the solutions of problem(39)
-
y
n
′
′
=
μ
n
r
a
t
f
0
y
n
+
α
y
n
+
+
β
y
n
-
+
μ
n
r
a
t
ζ
x
n
x
n
,
t
∈
0,1
,
y
n
0
=
y
n
1
=
0
.
Sincey
n is bounded in C
2
[
0,1
], choosing a subsequence and relabelling if necessary, we have that y
n
→
y for some y
∈
E.
Furthermore, from (37) and the fact that ζ
¯ is nondecreasing, we have that(40)
lim
n
→
∞
⁡
ζ
x
n
x
n
=
0
,since(41)
ζ
x
n
x
n
≤
ζ
¯
x
n
x
n
≤
ζ
¯
x
n
∞
x
n
≤
ζ
¯
x
n
x
n
⟶
0
,
n
⟶
+
∞
.
By (40) and the compactness of L
-
1, we obtain that (42)
L
y
=
μ
r
a
t
f
0
y
+
α
y
+
+
β
y
-
.
It is clear thaty
=
1 and y
∈
D
k
ν
¯
⊆
D
k
ν since D
k
ν is closed in R
×
E. Moreover, by Lemma 1, μ
r
f
0
=
λ
k
ν, so that(43)
μ
=
λ
k
ν
r
f
0
.
ThusD
k
ν joins (
λ
k
ν
/
r
f
0
,
0
) to (
0
,
∞
).Theorem 12.
Let(
A
1
), (
A
2
), and (
H
3
) hold. For ν
∈
{
+
,
-
}, assume that one of the following conditions holds:(i)
r
∈
(
0
,
λ
k
ν
/
f
∞
)
for
λ
k
ν
>
0.
(ii)
r
∈
(
0
,
λ
k
+
/
f
∞
)
∪
(
λ
k
-
/
f
∞
,
0
)
for
ν
λ
k
ν
>
0.
(iii)
r
∈
(
0
,
λ
k
-
/
f
∞
)
∪
(
λ
k
+
/
f
∞
,
0
)
for
ν
λ
k
ν
<
0.
(iv)
r
∈
(
λ
k
ν
/
f
∞
,
0
)
for
λ
k
ν
<
0.
Then problem (4) possesses two solutions x
k
+ and x
k
- such that x
k
+ has exactly k
-
1 zeros in (
0,1
) and is positive near 0 and x
k
- has exactly k
-
1 zeros in (
0,1
) and is negative near 0.Proof.
Define(44)
f
n
s
≔
n
s
,
s
∈
-
1
n
,
1
n
,
f
2
n
-
1
n
s
-
2
+
f
2
n
,
s
∈
1
n
,
2
n
,
-
f
-
2
n
+
1
n
s
+
2
+
f
-
2
n
,
s
∈
-
2
n
,
-
1
n
,
f
s
,
s
∈
-
∞
,
-
2
n
∪
2
n
,
+
∞
.We consider the following problem:(45)
-
x
′
′
=
α
x
+
+
β
x
-
+
λ
r
a
t
f
n
x
,
0
<
t
<
1
,
x
0
=
x
1
=
0
.
Clearly, we can see thatl
i
m
n
→
+
∞
f
[
n
]
(
s
)
=
f
(
s
), (
f
[
n
]
)
0
=
n, and (
f
[
n
]
)
∞
=
f
∞.
Similar to the proof of [13, Theorem 4.1], by Lemma 8 and Remark 9, there exists an unbounded continuum D
k
ν
[
n
] of solutions of problem (45) emanating from (
λ
k
ν
/
r
n
,
0
), such that D
k
ν
[
n
]
⊂
(
Φ
k
ν
∪
{
(
λ
k
ν
/
r
n
,
0
)
}
), and D
k
ν
[
n
] joins (
λ
k
ν
/
r
n
,
0
) to (
λ
k
ν
/
r
f
∞
,
∞
).
Takingz
n
=
(
λ
k
ν
/
r
n
,
0
) and z
∗
=
(
0,0
), we have that z
n
→
z
∗.
So condition (i) in Lemma7 is satisfied with z
∗
=
(
0,0
).
Obviously(46)
r
n
=
sup
⁡
λ
+
x
∣
λ
,
x
∈
D
k
ν
n
=
∞
,and accordingly (ii) in Lemma 7 holds. (iii) in Lemma 7 can be deduced directly from the Arzela-Ascoli Theorem and the definition of f
[
n
].
Therefore, by Lemma7, l
i
m
s
u
p
n
→
∞
D
k
ν
[
n
] contains an unbounded connected component D
k
ν with (
0,0
)
∈
D
k
ν.
Similar to Theorem11, we obtain D
k
ν
⊂
S
k
ν.
Similar to the method of the proof of Case  2 of [13, Theorem 4.1], we can obtain that (
λ
k
ν
/
r
f
∞
,
∞
)
∈
D
k
ν.Theorem 13.
Let(
A
1
), (
A
2
), and (
H
4
) hold. For ν
∈
{
+
,
-
}, assume that one of the following conditions holds:(i)
r
∈
(
0
,
+
∞
)
for
λ
k
ν
>
0.
(ii)
r
∈
(
0
,
+
∞
)
∪
(
-
∞
,
0
)
for
ν
λ
k
ν
>
0
or
ν
λ
k
ν
<
0.
(iii)
r
∈
(
-
∞
,
0
)
for
λ
k
ν
<
0.
Then problem (4) possesses two solutions x
k
+ and x
k
- such that x
k
+ has exactly k
-
1 zeros in (
0,1
) and is positive near 0 and x
k
- has exactly k
-
1 zeros in (
0,1
) and is negative near 0.Proof.
Define(47)
f
n
s
≔
1
n
s
,
s
∈
-
∞
,
-
2
n
∪
2
n
,
+
∞
.
2
+
f
-
n
n
s
+
n
+
f
-
n
,
s
∈
-
2
n
,
-
n
,
2
-
f
n
n
s
-
n
+
f
n
,
s
∈
n
,
2
n
,
f
s
,
s
∈
-
n
,
-
2
n
∪
2
n
,
n
,
-
f
-
2
n
+
1
n
s
+
2
+
f
-
2
n
,
s
∈
-
2
n
,
-
1
n
,
f
2
n
-
1
n
s
-
2
+
f
2
n
,
s
∈
1
n
,
2
n
,
n
s
,
s
∈
-
1
n
,
1
n
.We consider the following problem:(48)
-
x
′
′
=
α
x
+
+
β
x
-
+
λ
r
a
t
f
n
x
,
0
<
t
<
1
,
x
0
=
x
1
=
0
.
Clearly, we can see thatl
i
m
n
→
+
∞
f
[
n
]
(
s
)
=
f
(
s
), (
f
[
n
]
)
0
=
n, and (
f
[
n
]
)
∞
=
1
/
n.
Applying a similar method used in the proof of Theorem12, we obtain an unbounded connected component D
k
ν
⊂
S
k
ν with (
0,0
)
∈
D
k
ν.
Similar to the method of the proof of Theorem10, we can obtain that (
∞
,
∞
)
∈
D
k
ν. It follows that the result is obtained.Theorem 14.
Let(
A
1
), (
A
2
), and (
H
5
) hold. For ν
∈
{
+
,
-
}, assume that one of the following conditions holds:(i)
There existsλ
ν
k
ν
>
0
for
λ
k
ν
>
0, such that r
∈
(
λ
ν
k
ν
,
+
∞
).
(ii)
There existsν
λ
ν
k
ν
>
0
for
ν
λ
k
ν
>
0, such that r
∈
(
-
∞
,
λ
-
k
-
)
∪
(
λ
+
k
+
,
+
∞
).
(iii)
There existsν
λ
ν
k
ν
<
0
for
ν
λ
k
ν
<
0, such that r
∈
(
-
∞
,
λ
+
k
+
)
∪
(
λ
-
k
-
,
+
∞
).
(iv)
There existsλ
ν
k
ν
<
0
for
λ
k
ν
<
0, such that r
∈
(
-
∞
,
λ
ν
k
ν
).
Then problem (4) possesses two solutions x
k
+ and x
k
- such that x
k
+ has exactly k
-
1 zeros in (
0,1
) and is positive near 0 and x
k
- has exactly k
-
1 zeros in (
0,1
) and is negative near 0.Proof.
Define(49)
f
n
s
≔
n
s
,
s
∈
-
∞
,
-
2
n
∪
2
n
,
+
∞
.
2
n
2
+
f
-
n
n
s
+
n
+
f
-
n
,
s
∈
-
2
n
,
-
n
,
2
n
2
-
f
n
n
s
-
n
+
f
n
,
s
∈
n
,
2
n
,
f
s
,
s
∈
-
n
,
-
2
n
∪
2
n
,
n
,
-
f
-
2
n
+
1
n
s
+
2
+
f
-
2
n
,
s
∈
-
2
n
,
-
1
n
,
f
2
n
-
1
n
s
-
2
+
f
2
n
,
s
∈
1
n
,
2
n
,
n
s
,
s
∈
-
1
n
,
1
n
.We consider the following problem:(50)
-
x
′
′
=
α
x
+
+
β
x
-
+
λ
r
a
t
f
n
x
,
0
<
t
<
1
,
x
0
=
x
1
=
0
.
Clearly, we can see thatl
i
m
n
→
+
∞
f
[
n
]
(
s
)
=
f
(
s
), (
f
[
n
]
)
0
=
n, and (
f
[
n
]
)
∞
=
n.
Applying a similar method used in the proof of Theorem12, we obtain an unbounded connected component D
k
ν
⊂
R
×
Σ
k
ν with (
0,0
)
∈
D
k
ν.
In the following, we can show that(
0
,
∞
)
∈
D
k
ν.
Assume that there exists a sequence(
μ
n
,
x
n
)
∈
D
k
ν such that(51)
lim
n
→
∞
⁡
x
n
+
μ
n
=
+
∞
,
-
x
n
′
′
=
α
x
n
+
+
β
x
n
-
+
μ
n
r
a
t
f
x
n
,
0
<
t
<
1
,
x
n
0
=
x
n
1
=
0
.
We divide the proof into two steps.
Step  1. We show that l
i
m
n
→
∞
x
n
=
+
∞.
If{
x
n
} is bounded, then there exists a constant M
1 not depending on n such that x
n
≤
M
1. It follows that (52)
lim
n
→
∞
⁡
μ
n
=
+
∞
.
Let(53)
0
=
τ
0
,
n
<
τ
1
,
n
<
τ
2
,
n
<
⋯
<
τ
k
-
1
,
n
<
τ
k
,
n
=
1denote the simple zeros of x
n in (
0,1
). Then, after choosing a subsequence and relabelling if necessary, (54)
lim
n
→
+
∞
⁡
τ
l
,
n
=
τ
l
,
∞
,
l
∈
0,1
,
…
,
k
.
We claim that there existsl
0
∈
{
0,1
,
…
,
k
} such that (55)
τ
l
0
,
∞
<
τ
l
0
+
1
,
∞
.
Otherwise, we have(56)
1
=
∑
l
=
0
k
-
1
τ
l
+
1
,
n
-
τ
l
,
n
⟶
∑
l
=
0
k
-
1
τ
l
+
1
,
∞
-
τ
l
,
∞
=
0
.This is a contradiction.
Let[
c
,
d
]
⊂
(
τ
(
l
0
,
∞
)
,
τ
(
l
0
+
1
,
∞
)
), with c
<
d. For all n sufficiently large, we have [
c
,
d
]
⊂
(
τ
(
l
0
,
n
)
,
τ
(
l
0
+
1
,
n
)
). So x
n
(
t
) does not change its sign in [
c
,
d
]. Without loss of generality, we assume that x
n
(
t
)
>
0, t
∈
[
c
,
d
]. Moreover, we have m
a
x
t
∈
[
c
,
d
]
x
n
(
t
)
≤
M
1. Combining this with the fact (57)
f
x
n
x
n
t
≥
inf
⁡
f
s
s
∣
0
<
s
≤
M
1
>
0
,
t
∈
c
,
dand with the relation(58)
-
x
n
′
′
=
α
x
n
+
+
β
x
n
-
+
μ
n
r
a
t
f
x
n
x
n
t
x
n
t
,
t
∈
c
,
d
.Furthermore, we obtain that l
i
m
n
→
+
∞
μ
n
r
a
(
t
)
(
f
(
x
n
(
t
)
)
/
x
n
(
t
)
)
=
±
∞ for any t
∈
[
c
,
d
]. Let φ
1 be the corresponding eigenfunction of λ
1
+ of (1). If l
i
m
n
→
+
∞
μ
n
r
a
(
t
)
(
f
(
x
n
(
t
)
)
/
x
n
(
t
)
)
=
-
∞ for any t
∈
[
c
,
d
], applying Lemma 4 to y
n and φ
1 on [
c
,
d
], we can get that φ
1 must change its sign in [
c
,
d
] for n large enough. While, this is impossible. So we have that l
i
m
n
→
+
∞
μ
n
r
a
(
t
)
(
f
(
x
n
(
t
)
)
/
x
n
(
t
)
)
=
+
∞ for any t
∈
[
c
,
d
]. Applying Lemma 5 to x
n
(
t
) on [
c
,
d
], we can get that x
n
(
t
) must change its sign in [
c
,
d
] for n large enough. This is a contradiction.
Step  2. We show that l
i
m
n
→
∞
μ
n
=
0.
Assume on the contrary thatμ
n
>
c
0. We consider the problem (59)
-
x
n
′
′
=
α
x
n
t
+
μ
n
r
a
t
f
x
n
x
n
t
x
n
t
,
t
∈
c
,
d
.
By Lemma 2.1 of [7], we have(60)
l
i
m
n
→
∞
min
⁡
x
n
t
∣
t
∈
c
,
d
=
∞
.
We have from (60) that we may assume, after taking a subsequence if necessary, that(61)
l
i
m
n
→
∞
min
⁡
x
n
t
∣
t
∈
c
,
d
=
+
∞
x
n
t
>
0
or
l
i
m
n
→
∞
max
⁡
x
n
t
∣
t
∈
c
,
d
=
-
∞
x
n
t
<
0
.From (
A
1
) and (61) and the condition f
∞
=
∞, we have that (62)
lim
n
→
+
∞
⁡
μ
n
r
a
t
f
x
n
t
x
n
t
=
±
∞
,
uniformly  for
t
∈
c
,
d
.Similar to the proof of Step  1, we can get a contradiction.Theorem 15.
Let(
A
1
), (
A
2
), and (
H
6
) hold. For ν
∈
{
+
,
-
}, assume that one of the following conditions holds:(i)
r
∈
(
λ
k
ν
/
f
∞
,
+
∞
)
for
λ
k
ν
>
0.
(ii)
r
∈
(
λ
k
+
/
f
∞
,
+
∞
)
∪
(
-
∞
,
λ
k
-
/
f
∞
)
for
ν
λ
k
ν
>
0.
(iii)
r
∈
(
λ
k
-
/
f
∞
,
+
∞
)
∪
(
-
∞
,
λ
k
+
/
f
∞
)
for
ν
λ
k
ν
<
0.
(iv)
r
∈
(
-
∞
,
λ
k
ν
/
f
∞
)
for
λ
k
ν
<
0.
Then problem (4) possesses two solutions x
k
+ and x
k
- such that x
k
+ has exactly k
-
1 zeros in (
0,1
) and is positive near 0 and x
k
- has exactly k
-
1 zeros in (
0,1
) and is negative near 0.Proof.
Define(63)
f
n
s
≔
1
n
s
,
s
∈
-
1
n
,
1
n
,
f
2
n
-
1
n
2
n
s
-
2
+
f
2
n
,
s
∈
1
n
,
2
n
,
-
f
-
2
n
+
1
n
2
n
s
+
2
+
f
-
2
n
,
s
∈
-
2
n
,
-
1
n
,
f
s
,
s
∈
-
∞
,
-
2
n
∪
2
n
,
+
∞
.We consider the following problem:(64)
-
x
′
′
=
α
x
+
+
β
x
-
+
λ
r
a
t
f
n
x
,
0
<
t
<
1
,
x
0
=
x
1
=
0
.
Clearly, we can see thatl
i
m
n
→
+
∞
f
[
n
]
(
s
)
=
f
(
s
), (
f
[
n
]
)
0
=
1
/
n, and (
f
[
n
]
)
∞
=
f
∞.
Similar to the proof of [13, Theorem 4.1], by Lemma 8 and Remark 9, there exists an unbounded continuum D
k
ν
[
n
] of solutions of problem (64) emanating from (
n
λ
ν
/
r
,
0
), such that D
k
ν
[
n
]
⊂
(
Φ
k
ν
∪
{
(
n
λ
ν
/
r
,
0
)
}
), and D
k
ν
[
n
] joins (
n
λ
ν
/
r
,
0
) to (
λ
ν
/
r
f
∞
,
∞
).
Takingz
n
=
(
n
λ
ν
/
r
,
0
) and z
∗
=
(
∞
,
0
), we have that z
n
→
z
∗.
So condition (i) in Lemma7 is satisfied with z
∗
=
(
∞
,
0
).
Obviously(65)
r
n
=
sup
⁡
λ
+
x
∣
λ
,
x
∈
D
k
ν
n
=
∞
,and accordingly (ii) in Lemma 7 holds. (iii) in Lemma 7 can be deduced directly from the Arzela-Ascoli Theorem and the definition of f
[
n
].
Therefore, by Lemma7, l
i
m
s
u
p
n
→
∞
D
k
ν
[
n
] contains an unbounded connected component D
k
ν with (
∞
,
0
)
∈
D
k
v.
Similar to Theorem11, we obtain D
k
ν
⊂
S
k
ν.
Similar to the method of the proof of Case  2 of [13, Theorem 4.1], we can obtain that (
λ
k
ν
/
r
f
∞
,
∞
)
∈
D
k
ν.Theorem 16.
Let(
A
1
), (
A
2
), and (
H
7
) hold. For ν
∈
{
+
,
-
}, assume that one of the following conditions holds:(i)
There existsλ
ν
k
ν
>
0
for
λ
k
ν
>
0, such that r
∈
(
λ
ν
k
ν
,
+
∞
).
(ii)
There existsν
λ
ν
k
ν
>
0
for
ν
λ
k
ν
>
0, such that r
∈
(
-
∞
,
λ
-
k
-
)
∪
(
λ
+
k
+
,
+
∞
).
(iii)
There existsν
λ
ν
k
ν
<
0
for
ν
λ
k
ν
<
0, such that r
∈
(
-
∞
,
λ
+
k
+
)
∪
(
λ
-
k
-
,
+
∞
).
(iv)
There existsλ
ν
k
ν
<
0
for
λ
k
ν
<
0, such that r
∈
(
-
∞
,
λ
ν
k
ν
).
Then problem (4) possesses two solutions x
k
+ and x
k
- such that x
k
+ has exactly k
-
1 zeros in (
0,1
) and is positive near 0 and x
k
- has exactly k
-
1 zeros in (
0,1
) and is negative near 0.Proof.
Define(66)
f
n
s
≔
1
n
s
,
s
∈
-
∞
,
-
2
n
∪
2
n
,
+
∞
.
2
+
f
-
n
n
s
+
n
+
f
-
n
,
s
∈
-
2
n
,
-
n
,
2
-
f
n
n
s
-
n
+
f
n
,
s
∈
n
,
2
n
,
f
s
,
s
∈
-
n
,
-
2
n
∪
2
n
,
n
,
-
f
-
2
n
+
1
n
2
n
s
+
2
+
f
-
2
n
,
s
∈
-
2
n
,
-
1
n
,
f
2
n
-
1
n
2
n
s
-
2
+
f
2
n
,
s
∈
1
n
,
2
n
,
1
n
s
,
s
∈
-
1
n
,
1
n
.We consider the following problem:(67)
-
x
′
′
=
α
x
+
+
β
x
-
+
λ
r
a
t
f
n
x
,
0
<
t
<
1
,
x
0
=
x
1
=
0
.
Clearly, we can see thatl
i
m
n
→
+
∞
f
[
n
]
(
s
)
=
f
(
s
), (
f
[
n
]
)
0
=
1
/
n, and (
f
[
n
]
)
∞
=
1
/
n.
Applying a similar method used in the proof of Theorem15, we obtain an unbounded connected component D
k
ν
⊂
S
k
ν with (
∞
,
0
)
∈
D
k
ν.
Similar to the proof of Theorem10, we can show that (
∞
,
∞
)
∈
D
k
ν.Theorem 17.
Let(
A
1
), (
A
2
), and (
H
8
) hold. For ν
∈
{
+
,
-
}, assume that one of the following conditions holds:(i)
r
∈
(
0
,
+
∞
)
for
λ
k
ν
>
0.
(ii)
r
∈
(
0
,
+
∞
)
∪
(
-
∞
,
0
)
for
ν
λ
k
ν
>
0
or
ν
λ
k
ν
<
0.
(iii)
r
∈
(
-
∞
,
0
)
for
λ
k
ν
<
0.
Then problem (4) possesses two solutions x
k
+ and x
k
- such that x
k
+ has exactly k
-
1 zeros in (
0,1
) and is positive near 0 and x
k
- has exactly k
-
1 zeros in (
0,1
) and is negative near 0.Proof.
Define(68)
f
n
s
≔
n
s
,
s
∈
-
∞
,
-
2
n
∪
2
n
,
+
∞
.
2
n
2
+
f
-
n
n
s
+
n
+
f
-
n
,
s
∈
-
2
n
,
-
n
,
2
n
2
-
f
n
n
s
-
n
+
f
n
,
s
∈
n
,
2
n
,
f
s
,
s
∈
-
n
,
-
2
n
∪
2
n
,
n
,
-
f
-
2
n
+
1
n
2
n
s
+
2
+
f
-
2
n
,
s
∈
-
2
n
,
-
1
n
,
f
2
n
-
1
n
2
n
s
-
2
+
f
2
n
,
s
∈
1
n
,
2
n
,
1
n
s
,
s
∈
-
1
n
,
1
n
.We consider the following problem:(69)
-
x
′
′
=
α
x
+
+
β
x
-
+
λ
r
a
t
f
n
x
,
0
<
t
<
1
,
x
0
=
x
1
=
0
.
Clearly, we can see thatl
i
m
n
→
+
∞
f
[
n
]
(
s
)
=
f
(
s
), (
f
[
n
]
)
0
=
n, and (
f
[
n
]
)
∞
=
1
/
n.
Applying a similar method used in the proof of Theorem15, we obtain an unbounded connected component D
k
ν
⊂
S
k
ν with (
∞
,
0
)
∈
D
k
ν.
Similar to the proof of Theorem14, we can show that (
0
,
∞
)
∈
D
k
ν.Remark 18.
Clearly, ifα
=
β
=
0, [6, 7] studied the problem (4) under conditions (
A
2
) and (
H
4
) or (
H
8
). Hence, the results of Theorems 13 and 17 improve on Theorems 1.1 and 1.2 of [6] or [7].Remark 19.
Note that ifα
=
β
≡
0 the result of Theorems 10–17 is equivalent to those of Theorems 2.1–2.8 of [10] (when p
=
2), respectively. Hence, Theorems 10–17 extend the corresponding results of [10] (when p
=
2).Remark 20.
The nonlinear term of (4) is not necessarily homogeneous linearizable at the origin and infinity because of the influence of the term α
x
+
+
β
x
-. Clearly, so the bifurcation results of [1, 2, 8–12] cannot be applied directly to obtain our results.Remark 21.
We consider the cases off
0
,
f
∞
∉
(
0
,
∞
), while the authors of [13] only studied the cases of f
0
,
f
∞
∈
(
0
,
∞
). Hence, Theorems 10–17 extend Theorem 4.1 of [13].Remark 22.
Previous cases have been investigated by several authors by applying also different methods. In particular, under the conditionsα
=
β
=
0 and (
H
4
) or (
H
8
), we can see [24–29].



---

*Source: 2386287-2016-01-14.xml*