Báo cáo toán học: " Global Bifurcation Results for General Laplacian Problems" pot

This Provisional PDF corresponds to the article as it appeared upon acceptance. Fully formatted
PDF and full text (HTML) versions will be made available soon.
Global Bifurcation Results for General Laplacian Problems
Fixed Point Theory and Applications 2012, 2012:7 doi:10.1186/1687-1812-2012-7
Eun Kyoung Lee ()
Yong-Hoon Lee ()
Byungjae Son ()
ISSN 1687-1812
Article type Research
Submission date 23 December 2010
Acceptance date 18 January 2012
Publication date 18 January 2012
Article URL />This peer-reviewed article was published immediately upon acceptance. It can be downloaded,
printed and distributed freely for any purposes (see copyright notice below).
For information about publishing your research in Fixed Point Theory and Applications go to
/>For information about other SpringerOpen publications go to

Fixed Point Theory and
Applications
© 2012 Lee et al. ; licensee Springer.
This is an open access article distributed under the terms of the Creative Commons Attribution License ( />which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.
Global bifurcation results for general
Laplacian problems
Eun Kyoung Lee
1
, Yong-Hoon Lee
2∗
and Byungjae Son
2
1
Department of Mathematics Education, Pusan National University,

Busan 609-735, Korea
2
Department of Mathematics, Pusan National University,
Busan 609-735, Korea
*Corresponding author:
Email addresses:
EK LEE:
B SON:
Abstract
In this article, we consider the global bifurcation result and existence of solutions
1
for the following general Laplacian problem,











−(ϕ(u
′
(t)))
′
= λψ(u(t)) + f(t, u, λ), t ∈ (0, 1),
u(0) = u(1) = 0,
(P )

where f : [0, 1] × R × R → R is continuous and ϕ, ψ : R → R are odd increasing
homeomorphisms of R, when ϕ, ψ satisfy the asymptotic homogeneity conditions.
1 Introduction
In this article, we consider the following general Laplacian problem,









−(ϕ(u
′
(t)))
′
= λψ(u(t)) + f(t, u, λ), t ∈ (0, 1),
u(0) = u(1) = 0,
(P )
where f : [0, 1] × R × R → R is continuous with f(t, u, 0) = 0 and ϕ, ψ : R → R are
odd increasing homeomorphisms of R with ϕ(0) = ψ(0) = 0. We consider the following
conditions;
(Φ
1
) lim
t→0
ϕ(σt)
ψ(t)
= σ

p−1
, for all σ ∈ R
+
, for some p > 1.
(Φ
2
) lim
|t|→∞
ϕ(σt)
ψ(t)
= σ
q−1
, for all σ ∈ R
+
, for some q > 1.
(F
1
) f(t, u, λ) = ◦(|ψ(u)|) near zero, uniformly for t and λ in bounded intervals.
(F
2
) f(t, u, λ) = ◦(|ψ(u)|) near infinity, uniformly for t and λ in bounded intervals.
(F
3
) uf(t, u, λ) ≥ 0.
2
We note that ϕ
r
(t) = |t|
r−2
t, r > 1 are special cases of ϕ and ψ. We first prove following

global bifurcation result.
Theorem 1.1. Assume (Φ
1
), (Φ
2
), (F
1
), (F
2
) and (F
3
). Then for any j ∈ N, there exists
a connected component C
j
of the set of nontrivial solutions for (P ) connecting (0, λ
j
(p)) to
(∞, λ
j
(q)) such that (u, λ) ∈ C
j
implies that u has exactly j −1 simple zeros in (0, 1), where
λ
j
(r) is the j-th eigenvalue of (ϕ
r
(u
′
(t)))
′

+ λϕ
r
(u(t)) = 0 and u(0) = u(1) = 0.
By the aid of this theorem, we can prove the following existence result of solutions.
Theorem 1.2. Consider problem









−(ϕ(u
′
(t)))
′
= g(t, u), t ∈ (0, 1),
u(0) = u(1) = 0,
(A)
where g : [0, 1] × R × R → R is continuous and ϕ is odd increasing homeomorphism of R,
3
which satisfy (Φ
1
) and (Φ
2
) with ϕ = ψ. Also ug(t, u) ≥ 0 and there exist positive integers
k, n with k ≤ n such that µ = lim
s→0

g(t,s)
ϕ(s)
< λ
k
(p) ≤ λ
n
(q) < lim
|s|→∞
g(t,s)
ϕ(s)
= ν uniformly
in t ∈ [0, 1]. Then for each integer j with k ≤ j ≤ n, problem (A) has a solution with exactly
j −1 simple zeros in (0, 1). Thus, (A) possesses at least n −k + 1 nontrivial solutions.
In [1], the authors studied the existence of solutions and global bifurcation results for









−(t
N−1
ϕ(u
′
(t)))
′
= t

N−1
λψ(u(t)) + t
N−1
f(t, u, λ), t ∈ (0, R),
u
′
(0) = u(R) = 0.
The main purpose of this article is to derive the same result for N = 1 case with Dirichlet
boundary condition which was not considered in [1].
For p-Laplacian problems, i.e., ϕ = ψ = ϕ
p
, many authors have studied for the existence
and multiplicity of nontrivial solutions [2–6]. In [2, 5, 6], the authors used fixed point theory
or topological degree argument. Also global bifurcation theory was mainly employed in [3, 4].
Moreover, there are some studies related to general Laplacian problems [3, 7, 8], but most
of them are about ϕ = ψ case. In [3], the authors proved some results under several kinds
of boundary conditions and in [7], the authors considered a system of general Laplacian
problems. In [8], the author studied global continuation result for the singular problem.
In this paper, we mainly study the global bifurcation phenomenon for general Laplacian
problem (P ) and prove the existence and multiplicity result for (A).
This article is organized as follows: In Section 2, we set up the equivalent integral operator
of (P ) and compute the degree of this operator. In Section 3, we verify the existence of global
bifurcation having bifurcation points at zero and infinity simultaneously. In Section 4, we
4
introduce an existence result as an application of the previous result and give some examples.
2 Degree estimate
Let us consider problem (P ) with f ≡ 0, i.e.,










−(ϕ(u
′
(t)))
′
= λψ(u(t)), t ∈ (0, 1),
u(0) = u(1) = 0.
(P )
We introduce the equivalent integral operator of problem (P ). For this, we consider the
following problem









(ϕ(u
′
(t)))
′
= h(t), a.e., t ∈ (0, 1),
u(0) = u(1) = 0,

(AP )
where h ∈ L
1
(0, 1). Here, a function u is called a solution of (AP ) if u ∈ C
1
0
[0, 1] with ϕ(u
′
)
absolutely continuous which satisfies (AP ). We note that (AP ) is equivalently written as
u(t) = G(h)(t) =

t
0
ϕ
−1

a(h) +

s
0
h(ξ)dξ

ds,
where a : L
1
(0, 1) → R is a continuous function which sends bounded sets of L
1
into bounded
sets of R and satisfying


1
0
ϕ
−1

a(h) +

s
0
h(ξ)dξ

ds = 0. (1)
It is known that G : L
1
(0, 1) → C
1
0
[0, 1] is continuous and maps equi-integrable sets of
L
1
(0, 1) into relatively compact sets of C
1
0
[0, 1]. One may refer Man´asevich-Mawhin [4, 3]
5
and Garcia-Huidobro-Man´asevich-Ward [7] for more details. If we define the operator T
λ
ϕψ
:

C
1
0
[0, 1] → C
1
0
[0, 1] by
T
λ
ϕψ
(u)(t) = G(−λψ(u))(t) =

t
0
ϕ
−1

a(−λψ(u)) +

s
0
−λψ(u(ξ))dξ

ds, (2)
then (P ) is equivalently written as u = T
λ
ϕψ
(u). Now let us consider p-Laplacian problem










−(ϕ
p
(u
′
(t)))
′
= λϕ
p
(u(t)), t ∈ (0, 1),
u(0) = u(1) = 0.
(E
p
)
By the similar argument, we can also get the equivalent integral operator of problem (E
p
),
which is known by Garcia-Huidobro-Man´asevich-Schmitt [1]. Let us define T
λ
p
: C
1
0
[0, 1] →

C
1
0
[0, 1] by
T
λ
p
(u)(t) =

t
0
ϕ
−1
p

a
p
(−λϕ
p
(u)) +

s
0
−λϕ
p
(u(ξ))dξ

ds, (3)
where a
p

: L
1
(0, 1) → R is a continuous function which sends bounded sets of L
1
into
bounded sets of R and satisfying

1
0
ϕ
−1
p

a
p
(h) +

s
0
h(ξ)dξ

ds = 0, for all h ∈ L
1
(0, 1).
Note that a
p
has homogineity property, i.e., a
p
(λt) = λa
p

(t). Problem (E
p
) can be equiva-
lently written as u = T
λ
p
(u). Obviously, T
λ
ϕψ
and T
λ
p
are completely continuous.
The main purpose of this section is to compute the Leray-Schauder degree of I − T
λ
ϕψ
.
Following Lemma is for the property of ϕ and ψ with asymptotic homogeneity condition (Φ
1
)
6
and (Φ
2
), which is very useful for our analysis. The proof can be modified from Proposition
4.1 in [9].
Lemma 2.1. Assume that ϕ, ψ are odd increasing homeomorphisms of R which satisfy (Φ
1
)
and (Φ
2

). Then, we have
(i) lim
t→0
ϕ
−1
(σt)
ψ
−1
(t)
= ϕ
−1
p
(σ), for all σ ∈ R
+
, for some p > 1, (4)
and
(ii) lim
|t|→∞
ϕ
−1
(σt)
ψ
−1
(t)
= ϕ
−1
q
(σ), for all σ ∈ R
+
, for some q > 1. (5)

To compute the degree, we will make use of the following well-known fact [10].
Lemma 2.2. If λ is not an eigenvalue of (E
p
), p > 1 and r > 0, then
deg(I − T
λ
p
, B(0, r), 0) =









1 if λ < λ
1
(p),
(−1)
k
if λ ∈ (λ
k
(p), λ
k+1
(p)).
(6)
Now, let us compute deg(I − T
λ

ϕψ
, B(0, r), 0) when λ is not an eigenvalue of (E
p
).
Theorem 2.3. Assume that ϕ, ψ are odd increasing homeomorphisms of R which satisfy
(Φ
1
) and (Φ
2
). Then,
(i) The Leray-Schauder degree of I − T
λ
ϕψ
is defined for B(0, ε), for all sufficiently small ε.
7
Moreover, we have
deg(I − T
λ
ϕψ
, B(0, ε), 0) =









1 if λ < λ

1
(p),
(−1)
m
if λ ∈ (λ
m
(p), λ
m+1
(p)).
(7)
(ii) The Leray-Schauder degree of I − T
λ
ϕψ
is defined for B(0, M), for all sufficiently large
M, and
deg(I − T
λ
ϕψ
, B(0, M), 0) =









1 if λ < λ
1

(q),
(−1)
l
if λ ∈ (λ
l
(q), λ
l+1
(q)).
(8)
Proof: We give the proof for assertion (i). Proof for the latter case is similar. Define
T
λ
: C
1
0
[0, 1] × [0, 1] → C
1
0
[0, 1] by T
λ
(u, τ ) = τT
λ
ϕψ
(u) + (1 − τ )T
λ
p
(u). We claim that the
Leray-Schauder degree for I −T
λ
(·, τ ) is defined for B(0, ε) in C

1
0
[0, 1] for all small ε. Indeed,
suppose there exist sequences {u
n
}, {τ
n
} and {ε
n
} with ε
n
→ 0 and ∥u
n
∥
0
= ε
n
such that
u
n
= T
λ
(u
n
, τ
n
), i.e.,
u
n
(t) = τ

n

t
0
ϕ
−1

a(−λψ(u
n
)) +

s
0
−λψ(u
n
(ξ))dξ

ds
+ (1 −τ
n
)

t
0
ϕ
−1
p

a
p

(−λϕ
p
(u
n
)) +

s
0
−λϕ
p
(u
n
(ξ))dξ

ds.
Setting v
n
(t) =
u
n
(t)
ε
n
, we have ∥v
n
∥
0
= 1,
v
n

(t) =
τ
n
ε
n

t
0
ϕ
−1

a(−λψ(u
n
)) +

s
0
−λψ(u
n
(ξ))dξ

ds
+ (1 −τ
n
)

t
0
ϕ
−1

p

a
p
(−λϕ
p
(v
n
)) +

s
0
−λϕ
p
(v
n
(ξ))dξ

ds,
8
and
v
′
n
(t) =
τ
n
ε
n
ϕ

−1

a(−λψ(u
n
)) +

t
0
−λψ(u
n
(ξ))dξ

+ (1 −τ
n
)ϕ
−1
p

a
p
(−λϕ
p
(v
n
)) +

t
0
−λϕ
p

(v
n
(ξ))dξ

.
Now, we show that {v
′
n
} is uniformly bounded. Since ∥v
n
∥
0
= 1,

t
0
−λϕ
p
(v
n
(ξ))dξ ≤ λ.
Moreover, there exists C
1
such that a
p
(−λϕ
p
(v
n
)) ≤ C

1
. These results imply the uniform
boundedness of ϕ
−1
p

a
p
(−λϕ
p
(v
n
)) +

t
0
−λϕ
p
(v
n
(ξ))dξ

. Let
q
n
(t) =
1
ε
n
ϕ

−1

a(−λψ(u
n
)) +

t
0
−λψ(u
n
(ξ))dξ

,
and
d
n
(t) =

t
0
λψ(u
n
(ξ))dξ.
Then d
n
∈ C[0, 1], and
∥d
n
∥
0

= max
t∈[0,1]
|

t
0
λψ(u
n
(ξ))dξ| ≤

1
0
λψ(∥u
n
∥
0
)dξ ≤ λψ(ε
n
).
Since

1
0
ϕ
−1

a(−λψ(u
n
)) − d
n

(s)

ds = 0, we have
|a(−λψ(u
n
))| ≤ λψ(ε
n
).
Otherwise,

1
0
ϕ
−1

a(−λψ(u
n
))−d
n
(s)

ds < 0 (or > 0). Now, we show that
1
ε
n
ϕ
−1

2λψ(ε
n

)

is bounded. Indeed, suppose that it is not true, i.e.,
1
ε
n
ϕ
−1

2λψ(ε
n
)

→ ∞ as n → ∞. Then,
for arbitrary A > 0, there exists N
0
∈ N such that
1
ε
n
ϕ
−1

2λψ(ε
n
)

≥ A, for all n > N
0
. This

implies that 2λ ≥
ϕ(Aε
n
)
ψ(ε
n
)
for all n > N
0
. However,
ϕ(Aε
n
)
ψ(ε
n
)
→ ϕ
p
(A) as n → ∞. This is a
9
contradiction. Thus by the above inequality, we get
1
ε
n
ϕ
−1

a(−λψ(u
n
)) +


t
0
−λψ(u
n
(ξ))dξ

≤
1
ε
n
ϕ
−1

2λψ(ε
n
)

≤ C
2
,
for some C
2
> 0. Therefore, {v
′
n
} is uniformly bounded. By the Arzela-Ascoli Theorem, {v
n
}
has a uniformly convergent subsequence in C[0, 1] relabeled as the original sequence so let

lim
n→∞
v
n
= v. Now, we claim that q
n
(t) → q(t), where
q(t) = ϕ
−1
p

a
p
(−λϕ
p
(v)) +

t
0
−λϕ
p
(v(ξ))dξ

.
Clearly,
q
n
(t) =
1
ε

n
ϕ
−1

a(−λψ(u
n
)) +

t
0
−λψ(u
n
(ξ))dξ

=
ϕ
−1

(
a(−λψ(u
n
))
ϕ(ε
n
)
+

t
0
−λ

ψ(v
n
(ξ)ε
n
)
ϕ(ε
n
)
dξ)ϕ(ε
n
)

ψ
−1

ϕ(ε
n
)

ψ
−1

ϕ(ε
n
)

ϕ
−1

ϕ(ε

n
)

.
Since |a(−λψ(u
n
))| ≤ λψ(ε
n
),
a(−λψ(u
n
))
ϕ(ε
n
)
has a convergent subsequence. Without loss of
generality, we say that the sequence {
a(−λψ(u
n
))
ϕ(ε
n
)
} converges to d. Also by the facts that
ψ(v
n
(ξ)ε
n
)
ϕ(ε

n
)
→ ϕ
p
(v(ξ)) as n → ∞, ϕ(ε
n
) → 0 and (i) of Lemma 2.1, we obtain
q
n
(t) → ϕ
p

d +

t
0
−λϕ
p
(v(ξ))dξ

.
Since

1
0
ϕ
−1

a(−λψ(u
n

)) +

s
0
−λψ(u
n
(ξ))dξ

ds = 0,
1
ε
n

1
0
ϕ
−1

a(−λψ(u
n
)) +

s
0
−λψ(u
n
(ξ))dξ

ds = 0.
10

Thus

1
0
ϕ
−1
p

d +

s
0
−λϕ
p
(v(ξ))dξ

ds = 0 and by the definition of a
p
, d = a
p
(−λϕ
p
(v)).
Therefore, we can easily see that
v(t) =

t
0
ϕ
−1

p

a
p
(−λϕ
p
(v)) +

s
0
−λϕ
p
(v(ξ))dξ

ds,
and

1
0
ϕ
−1
p

a
p
(−λϕ
p
(v)) +

s

0
−λϕ
p
(v(ξ))dξ

ds = 0.
Consequently, v is a solution of (E
p
). Since λ /∈ {λ
n
(p)}, v ≡ 0 and this fact yields a
contradiction. By the properties of the Leray-Schauder degree, we get
deg(I − T
λ
p
, B(0, ε), 0) = deg(I − T
λ
(·, 0), B(0, ε), 0)
= deg(I − T
λ
(·, 1), B(0, ε), 0) = deg(I − T
λ
ϕψ
, B(0, ε), 0)
and the proof is completed by Lemma 2.2. 
3 Existence of unbounded continuum
We begin with this section recalling what we mean by bifurcation at zero and at infinity. Let
X be a Banach space with norm ∥ ·∥, and let F : X × I → X be a completely continuous
operator, where I is some real interval. Consider the equation
u = F(u, λ) (9)

11
Definition 3.1. Suppose that F(0, λ) = 0 for all λ in I, and that
ˆ
λ ∈ I. We say that (0,
ˆ
λ)
is a bifurcation point of (9) at zero if in any neighborhood of (0,
ˆ
λ) in X × I, there is a
nontrivial solution of (9). Or equivalently, if there exist sequences {x
n
̸= 0} and {λ
n
} with
(∥x
n
∥, λ
n
) → (0,
ˆ
λ) and such that (x
n
, λ
n
) satisfies (9) for each n ∈ N.
Definition 3.2. We say that (∞,
ˆ
λ) is a bifurcation point of (9) at infinity if in any neigh-
borhood of (∞,
ˆ

λ) in X ×I, there is a nontrivial solution of (9). Equivalently, if there exist
sequences {x
n
̸= 0} and {λ
n
} with (∥x
n
∥, λ
n
) → (∞,
ˆ
λ) and such that (x
n
, λ
n
) satisfies (9)
for each n ∈ N.
Let u be a solution of problem (P ). Define F(u, λ) by
F(u, λ)(t) =

t
0
ϕ
−1

a(−λψ(u) −f(·, u, λ)) +

s
0
−λψ(u(ξ)) −f(ξ, u, λ)dξ


ds. (10)
We note that (P) is written as u = F(u, λ). It is clear that F : C
1
0
[0, 1] × R → C
1
0
[0, 1] is a
completely continuous operator.
Lemma 3.3. (i) Assume (Φ
1
) and (F
1
). If (0,
ˆ
λ) is a bifurcation point of (P ), then
ˆ
λ = λ
n
(p)
for some p ∈ N.
(ii) Assume (Φ
2
) and (F
2
). If (∞,
ˆ
λ) is a bifurcation point of (P ), then
ˆ

λ = λ
n
(q) for some
q ∈ N.
Proof: We prove assertion (i). Supp ose that (0,
ˆ
λ) is a bifurcation point of (P ). Then there
exists a sequence {(u
n
, λ
n
)} in C
1
0
[0, 1] × R with (u
n
, λ
n
) → (0,
ˆ
λ) and such that (u
n
, λ
n
)
12
satisfies u
n
= F(u
n

, λ
n
) for each n ∈ N. Equivalently, (u
n
, λ
n
) satisfies
u
n
(t) =

t
0
ϕ
−1

a(−λ
n
ψ(u
n
) − f(·, u
n
, λ
n
)) +

s
0
−λ
n

ψ(u
n
(ξ)) −f(ξ, u
n
, λ
n
)dξ

ds
with

1
0
ϕ
−1

a(−λ
n
ψ(u
n
) − f(·, u
n
, λ
n
)) +

s
0
−λ
n

ψ(u
n
(ξ)) −f(ξ, u
n
, λ
n
)dξ

ds = 0.
Let ε
n
= ∥u
n
∥
0
and v
n
(t) =
u
n
(t)
ε
n
. Then
v
n
(t) =
1
ε
n


t
0
ϕ
−1

a(−λ
n
ψ(u
n
) − f(·, u
n
, λ
n
)) +

s
0
−λ
n
ψ(u
n
(ξ)) −f(ξ, u
n
, λ
n
)dξ

ds,
and

v
′
n
(t) =
1
ε
n
ϕ
−1

a(−λ
n
ψ(u
n
) − f(·, u
n
, λ
n
)) +

t
0
−λ
n
ψ(u
n
(ξ)) −f(ξ, u
n
, λ
n

)dξ

.
Now, define d
n
(t) =

t
0
−λ
n
ψ(u
n
(ξ)) − f(ξ, u
n
, λ
n
)dξ. Since f(t, u, λ) = ◦(|ψ(u)|) near zero,
uniformly for t and λ, for some constants K
1
and K
2
.
∥d
n
∥
0
= max
t∈[0,1]
|


t
0
λ
n
ψ(u
n
(ξ)) + f(ξ, u
n
, λ
n
)dξ|
≤ max
t∈[0,1]

t
0
|λ
n
ψ(u
n
(ξ))| + |f(ξ, u
n
, λ
n
)|dξ
≤

1
0

λ
n
ψ(∥u
n
∥
0
) + K
1
ψ(∥u
n
∥
0
)dξ
≤ K
2
ψ(ε
n
).
Since

1
0
ϕ
−1

a(−λ
n
ψ(u
n
) − f(·, u

n
, λ
n
)) − d
n
(s)

ds = 0, we have
|a(−λ
n
ψ(u
n
) − f(·, u
n
, λ
n
))| ≤ K
2
ψ(ε
n
).
13
Otherwise,

1
0
ϕ
−1

a(−λ

n
ψ(u
n
) − f(·, u
n
, λ
n
)) − d
n
(s)

ds > 0 or < 0. Now, let us verify
that
1
ε
n
ϕ
−1

2K
2
ψ(ε
n
)

is bounded. If
1
ε
n
ϕ

−1

2K
2
ψ(ε
n
)

→ ∞ as n → ∞, then for arbitrary
A > 0, there exists N
0
∈ N such that
1
ε
n
ϕ
−1

2K
2
ψ(ε
n
)

≥ A, for all n ≥ N
0
.
This implies that 2K
2
≥

ϕ(Aε
n
)
ψ(ε
n
)
, for all n ≥ N
0
. This is impossible. Thus
1
ε
n
ϕ
−1

a(−λ
n
ψ(u
n
) − f(·, u
n
, λ
n
)) +

t
0
−λ
n
ψ(u

n
(ξ)) −f(ξ, u
n
, λ
n
)dξ

≤ K
3
.
Consequently, {v
′
n
} is uniformly bounded and by the Arzela-Ascoli Theorem, {v
n
} has a
uniformly convergent subsequence in C[0, 1]. Let v
n
→ v in C[0, 1]. Now claim that
v(t) =

t
0
ϕ
−1
p

a
p
(−

ˆ
λϕ
p
(v)) +

s
0
−
ˆ
λϕ
p
(v(ξ))dξ

ds.
Clearly,
v
′
n
(t) =
1
ε
n
ϕ
−1

a(−λ
n
ψ(u
n
) − f(·, u

n
, λ
n
)) +

t
0
−λ
n
ψ(u
n
(ξ)) −f(ξ, u
n
, λ
n
)dξ

=
ϕ
−1

h
n
(t)ψ(ε
n
)

ψ
−1


ψ(ε
n
)

,
where h
n
(t) =
a(−λ
n
ψ(u
n
)−f(·,u
n
,λ
n
))
ψ(ε
n
)
+

t
0
−λ
n
ψ(u
n
(ξ))ϕ(ε
n

)
ϕ(ε
n
)ψ(ε
n
)
−
f(ξ,u
n
,λ
n
)
ψ(ε
n
)
dξ.
Since
a(−λ
n
ψ(u
n
)−f(·,u
n
,λ
n
))
ψ(ε
n
)
is bounded, considering a subsequence if necessary, we may assume

14
that sequence {
a(−λ
n
ψ(u
n
)−f(·,u
n
,λ
n
))
ψ(ε
n
)
} converges to d as n → ∞. This implies that
v
′
n
(t) → ϕ
−1
p

d +

t
0
−
ˆ
λϕ
p

(v(ξ))dξ

as n → ∞,
and thus v(t) =

t
0
ϕ
−1
p

d +

s
0
−
ˆ
λϕ
p
(v(ξ))dξ

ds. Since v
n
(1) = 0 for all n, d = a
p
(−
ˆ
λϕ
p
(v))

and v is a solution of (E
p
). Consequently,
ˆ
λ must be an eigenvalue of the p-Laplacian
operator. 
The converse of first part of Theorem 3.3 is true in our problem.
Lemma 3.4. Assume (Φ
1
) and (F
1
). If µ is an eigenvalue of (E
p
), then (0, µ) is a bifurcation
point.
Proof: Suppose that (0, µ) is not a bifurcation point of (P ). Then there is a neighborhood
of (0, µ) containing no nontrivial solutions of (P ). In particular, we may choose an ε-ball B
ε
such that there are no solutions of (P) on ∂B
ε
×[µ −ε, µ + ε] and µ is the only eigenvalue of
(E
p
) on [µ −ε, µ + ε]. Let Φ(u, λ) = u −F(u, λ). Then deg(Φ(·, λ), B(0, ε), 0) is well-defined
for λ with |λ −µ| ≤ ε. Moreover, from the homotopy invariance theorem,
deg(Φ(·, λ), B(0, ε), 0) ≡ constant, for all λ with |λ −µ| ≤ ε.
Now, we claim that
deg(Φ(·, µ − ε), B(0, ε), 0) = deg(Φ
p
(·, µ −ε), B(0, ε), 0),

15
where Φ
p
(u, µ −ε) = u −T
µ−ε
p
(u). Define H
µ−ε
: C
1
0
[0, 1] ×[0, 1] → C
1
0
[0, 1] by
H
µ−ε
(u, τ )(t) = τ F(u, µ −ε)(t) + (1 − τ)T
µ−ε
p
(u)(t).
We know that F(·, µ−ε) and T
µ−ε
p
are completely continuous. To apply the homotopy invari-
ance theorem, we need to show that 0 /∈ u − H
µ−ε
(u, τ )(∂B
ε
) to guarantee well-definedness

of deg( I −H
µ−ε
(·, τ ), B(0, ε), 0). Suppose that this is not the case, then there exist sequences
{u
n
}, {τ
n
} and {ε
n
} with ε
n
→ 0 and ∥u
n
∥
0
= ε
n
such that u
n
= H
µ−ε
(u
n
, τ
n
), i.e.,
u
n
(t) = τ
n


t
0
ϕ
−1

a(−(µ − ε)ψ(u
n
) − f(·, u
n
, µ −ε))
+

s
0
−(µ − ε)ψ(u
n
(ξ)) −f(ξ, u
n
, µ −ε)dξ

ds
+ (1 −τ
n
)

t
0
ϕ
−1

p

a
p
(−(µ − ε)ϕ
p
(u
n
)) +

s
0
−(µ − ε)ϕ
p
(u
n
(ξ))dξ

ds.
Setting v
n
(t) =
u
n
(t)
ε
n
, we have that ∥v
n
∥

0
= 1 and
v
n
(t) =
τ
n
ε
n

t
0
ϕ
−1

a(−(µ − ε)ψ(u
n
) − f(·, u
n
, µ −ε))
+

s
0
−(µ − ε)ψ(u
n
(ξ)) −f(ξ, u
n
, µ −ε)dξ


ds
+ (1 −τ
n
)

t
0
ϕ
−1
p

a
p
(−(µ − ε)ϕ
p
(v
n
)) +

s
0
−(µ − ε)ϕ
p
(v
n
(ξ))dξ

ds.
16
Hence, we obtain that

v
′
n
(t) =
τ
n
ε
n
ϕ
−1

a(−(µ − ε)ψ(u
n
) − f(·, u
n
, µ −ε))
+

t
0
−(µ − ε)ψ(u
n
(ξ)) −f(ξ, u
n
, µ −ε)dξ

+ (1 −τ
n
)ϕ
−1

p

a
p
(−(µ − ε)ϕ
p
(v
n
)) +

t
0
−(µ − ε)ϕ
p
(v
n
(ξ))dξ

,
and we see that {v
′
n
} is uniformly bounded. Therefore, by the Arzela-Ascoli Theorem, {v
n
}
has a uniformly convergent subsequence in C[0, 1]. Without loss of generality, let v
n
→ v.
Moreover, using the fact that
1

ε
n
ϕ
−1

a(−(µ − ε)ψ(u
n
) − f(·, u
n
, µ − ε)) +

t
0
−(µ − ε)ψ(u
n
(ξ)) − f (ξ, u
n
, µ − ε)dξ

→ ϕ
−1
p

a
p
(−(µ − ε)ϕ
p
(v)) +

t

0
−(µ − ε)ϕ
p
(v(ξ))dξ

,
we can obtain that
v(t) =

t
0
ϕ
−1
p

a
p
(−(µ − ε)ϕ
p
(v)) +

s
0
−(µ − ε)ϕ
p
(v(ξ))dξ

ds.
This implies v ≡ 0 and this is a contradiction. Consequently, deg (I − H
µ−ε

(·, τ ), B(0, ε), 0)
is well defined. Therefore, by the homotopy invariance theorem,
deg(Φ(·, µ − ε), B(0, ε), 0) = deg(Φ
p
(·, µ −ε), B(0, ε), 0).
Similarly,
deg(Φ(·, µ + ε), B(0, ε), 0) = deg(Φ
p
(·, µ + ε), B(0, ε), 0).
17
Let µ is k-th eigenvalue of (E
p
). Then by Lemma 2.2, we get
deg(Φ(·, µ − ε), B(0, ε), 0) = (−1)
k−1
and deg(Φ(·, µ + ε), B(0, ε), 0) = (−1)
k
.
This is a contradiction to the fact deg(Φ(·, µ − ε), B(0, ε), 0) = deg(Φ(·, µ + ε), B(0, ε), 0).
Thus (0, µ) is a bifurcation point of (P ). 
Now, we shall adopt Rabinowitz’s standard arguement [11]. Let S denote the closure of
the set of nontrivial solutions of (P ) and S
+
k
denote the set of u ∈ C
1
0
[0, 1] such that u has
exactly k − 1 simple zeros in (0,1), u > 0 near 0, and all zeros of u in [0,1] are simple. Let
S

−
k
= −S
+
k
and S
k
= S
+
k
∪ S
−
k
. We note that the sets S
+
k
, S
−
k
and S
k
are open in C
1
0
[0, 1].
Moreover, let C
k
denote the component of S which meets (0, µ
k
), where µ

k
= λ
k
(p). By
the similar argument of Theorem 1.10 in [11], we can show the existence of two types of
components C emanating from (0, µ) contained in S, when µ is an eigenvalue of (E
p
); either
it is unbounded or it contains (0, ˆµ), where ˆµ(̸= µ) is an eigenvalue of (E
p
). The existence
of a neighborhood O
k
of (0, µ
k
) such that (u, λ) ∈ S ∩ O
k
and u ̸≡ 0 imply u ∈ S
k
is also
proved in [11]. Actually, only the first alternative is possible as shall be shown next.
Lemma 3.5. Assume (Φ
1
), (Φ
2
), and (F
1
). Then, C
k
is unbounded in S

k
× R.
Proof: Suppose C
k
⊂ (S
k
×R) ∪{(0, µ
k
)}. Then since S
k
∩S
j
= ∅ for j ̸= k, it follows from
the above facts, C
k
must be unbounded in S
k
×R. Hence, Lemma 3.5 will be established once
we show C
k
̸⊂ (S
k
×R)∪{(0, µ
k
)} is impossible. It is clear that C
k
∩O
k
⊂ (S
k

×R)∪{(0, µ
k
)}.
Hence if C
k
̸⊂ (S
k
× R)∪{(0, µ
k
)}, then there exists (u, λ) ∈ C
k
∩( ∂S
k
×R) with (u, λ) ̸= (0, µ
k
)
and (u, λ) = lim
n→∞
(u
n
, λ
n
), u
n
∈ S
k
. If u ∈ ∂S
k
, u ≡ 0 because u dose not have double
zero. Henceforth λ = µ

j
, j ̸= k. But then, (u
n
, λ
n
) ∈ (S
k
× R) ∩ O
j
for large n which is
18
impossible by the fact that (u
n
, λ
n
) ∈ S ∩O
j
implies u
n
∈ S
j
. The proof is complete. 
Lemma 3.6. Assume (Φ
1
), (Φ
2
), (F
1
), and (F
3

). Then for each k ∈ N, there exists a
constant M
k
∈ (0, ∞) such that λ ≤ M
k
for every λ with (u, λ) ∈ C
k
.
Proof: Suppose it is not true, then there exists a sequence {(u
n
, λ
n
)} ⊂ C
k
such that
λ
n
→ ∞. Let ρ
j
n
be the jth zero of u
n
. Then there exists j ∈ {1, . . . , k − 1} such that
|ρ
(j+1)
n
− ρ
j
n
| ≥

1
k
. Thus for each n, there exists σ
j
n
∈ (ρ
j
n
, ρ
(j+1)
n
) such that u
′
n
(σ
j
n
) = 0.
Let u
n
(t) > 0 for all t ∈ (ρ
j
n
, ρ
(j+1)
n
). Suppose σ
j
n
∈ (ρ

j
n
,
ρ
j
n
+3ρ
(j+1)
n
4
]. Then by integrating
the equation in (P ) from σ
j
n
to t ∈ [σ
j
n
, ρ
(j+1)
n
], we see that u
n
satisfies
u
n
(t) =

ρ
(j+1)
n

t
ϕ
−1


s
σ
j
n
λ
n
ψ(u
n
(ξ)) + f(ξ, u
n
, λ
n
)dξ

ds.
For t ∈ [
ρ
j
n
+4ρ
(j+1)
n
5
,
ρ

j
n
+5ρ
(j+1)
n
6
],
u
n
(t) ≥

ρ
(j+1)
n
ρ
j
n
+5ρ
(j+1)
n
6
ϕ
−1


t
σ
j
n
λ

n
ψ(u
n
(t))dξ

ds
≥

ρ
(j+1)
n
ρ
j
n
+5ρ
(j+1)
n
6
ϕ
−1


ρ
j
n
+4ρ
(j+1)
n
5
ρ

j
n
+3ρ
(j+1)
n
4
λ
n
ψ(u
n
(t))dξ

ds
=
ρ
(j+1)
n
− ρ
j
n
6
ϕ
−1

ρ
(j+1)
n
− ρ
j
n

20
λ
n
ψ(u
n
(t))

≥
1
6k
ϕ
−1

1
20k
λ
n
ψ(u
n
(t))

.
Thus
ϕ(6ku
n
(t))
ψ(u
n
(t))
≥

λ
n
20k
. (11)
The left side of (11) is bounded and independent on n, but the right side goes to ∞ as
19
n → ∞. This is impossible. Now, if σ
j
n
∈ (
ρ
j
n
+3ρ
(j+1)
n
4
, ρ
(j+1)
n
), then by integrating the
equation in (P ) from t ∈ [ρ
j
n
, σ
j
n
] to σ
j
n

, we see that u
n
satisfies
u
n
(t) =

t
ρ
j
n
ϕ
−1


σ
j
n
s
λ
n
ψ(u
n
(t)) + f(ξ, u
n
, λ
n
)dξ

ds.

For t ∈ [
ρ
j
n
+ρ
(j+1)
n
2
,
ρ
j
n
+2ρ
(j+1)
n
3
],
u
n
(t) ≥

t
ρ
j
n
ϕ
−1


σ

j
n
t
λ
n
ψ(u
n
(t))dξ

ds
≥

ρ
j
n
+ρ
(j+1)
n
2
σ
j
n
ϕ
−1


ρ
j
n
+3ρ

(j+1)
n
4
ρ
j
n
+2ρ
(j+1)
n
3
λ
n
ψ(u
n
(t))dξ

ds
=
ρ
(j+1)
n
− ρ
j
n
2
ϕ
−1

ρ
(j+1)

n
− ρ
j
n
12
λ
n
ψ(u
n
(t))

≥
1
2k
ϕ
−1

1
12k
λ
n
ψ(u
n
(t))

.
From the above argument, we get
ϕ(2ku
n
(t))

ψ(u
n
(t))
≥
λ
n
12k
. (12)
This is impossible because the left side is bounded and independent on n, but the right side
goes to infinity as n goes to infinity. We can get similar results when u
n
(t) < 0. Indeed, if
σ
j
n
∈ (ρ
j
n
,
ρ
j
n
+3ρ
(j+1)
n
4
], then we have
ϕ(6k|u
n
(t)|)

ψ(|u
n
(t)|)
≥
λ
n
20k
. (13)
20
Also if σ
j
n
∈ (
ρ
j
n
+3ρ
(j+1)
n
4
, ρ
(j+1)
n
), then we have
ϕ(2k|u
n
(t)|)
ψ(|u
n
(t)|)

≥
λ
n
12k
. (14)
Since both (13) and (14) are impossible, there is no sequence {(u
n
, λ
n
)} ⊂ C
k
satisfying
λ
n
→ ∞. Consequently, there exists an M
k
∈ (0, ∞) such that λ ≤ M
k
. 
Proof of Theorem 1.1
By Lemmas 3.3, 3.4, and 3.5, for any j ∈ N, there exists an unbounded connected component
C
j
of the set of nontrivial solutions emanating from (0, λ
j
(p)) such that (u, λ) ∈ C
j
implies u
has exactly j −1 simple zeros in (0,1). From Lemma 3.6, there is an M
j

such that (u, λ) ∈ C
j
implies that λ ≤ M
j
, and there are no nontrivial solutions of (P ) for λ = 0, it follows that
for any M > 0, there is (u, λ) ∈ C
j
such that ∥u∥
1
> M. Hence, we can choose subsequence
{(u
n
, λ
n
)} ⊂ C
j
such that λ
n
→
ˆ
λ and ∥u
n
∥
1
→ ∞. Thus, (∞,
ˆ
λ) is a bifurcation point and
ˆ
λ = λ
j

(q). 
4 Application and some examples
Proof of Theorem 1.2
Let us consider the bifurcation problem









−(ϕ(u
′
(t)))
′
= λϕ(u(t)) + g(t, u), t ∈ (0, 1),
u(0) = u(1) = 0
(A
g
)
21
Put f(t, u, λ) = −µϕ(u) + g(t, u). We can easily see that f(t, u, λ) = ◦(|ϕ(u)|) near zero
uniformly for t and λ in bounded intervals. The equation in (A
g
) can be equivalently changed
into the following equation










−(ϕ(u
′
(t)))
′
= (λ + µ)ϕ(u(t)) + f(t, u, λ), t ∈ (0, 1),
u(0) = u(1) = 0.
(A
f
)
By the similar argument in the proof of Theorem 1.1, for each k ≤ j ≤ n, there is a
connected branch C
j
of solutions to (A
f
) emanating from (0, λ
j
(p) −µ) which is unbounded
in C
1
0
[0, 1]×R and such that (u, λ) ∈ C
j
implies that u has exactly j −1 simple zeros in (0,1).

From the fact ug(t, u) ≥ 0, it can be proved that there is an M
j
> 0 such that (u, λ) ∈ C
j
implies that λ ≤ M
j
, by the same argument as in the proof of Lemma 3.6. Since there is
a constant K
g
> 0 such that g(t, s) ≤ K
g
ϕ(s) for all (t, s) ∈ [0, 1] × R, if (u, λ) ∈ C
j
, then
λ > −K
g
. Hence C
j
will bifurcate from infinity also, which can only happen for λ = λ
j
(q)−ν.
Since λ
j
(q) −ν < 0 < λ
j
(p) −µ and C
j
is connected, there exists u ̸= 0 such that (u, 0) ∈ C
j
.

This u is a solution of (A). Since this is true for every such j, (A) has at least n − k + 1
nontrivial solutions. 
Finally, we illustrate several examples of Theorems 1.1 and 1.2.
Example 4.1. Define ϕ, ψ, f by
ϕ(u) =







u + u
2
+ u
3
, if u ≥ 0,
u − u
2
+ u
3
, if u < 0,
ψ(u) =








u + 2u
2
+ u
3
, if u ≥ 0,
u − 2u
2
+ u
3
, if u < 0,
22
f(t, u, λ) =







λu
2
, if u ≥ 0,
−λu
2
, if u < 0.
Then ϕ and ψ are odd increasing homeomorphisms of R and
lim
u→0
ϕ(σu)
ψ(u)

= σ = ϕ
2
(σ), lim
|u|→∞
ϕ(σu)
ψ(u)
= σ
3
= ϕ
4
(σ).
Moreover, f satisfies f(t, u, λ) = ◦(|ψ(u)|) near zero and infinity, uniformly in t and λ, and
uf(t, u, λ) ≥ 0. Therefore, all hypotheses of Theorem 1.1 are satisfied.
Example 4.2. Define ϕ, g by
ϕ(u) =







u + u
2
, if u ≥ 0,
u − u
2
, if u < 0,
g(t, u) =








πu + π
3
u
2
, if u ≥ 0,
πu −π
3
u
2
, if u < 0.
Then ϕ is odd increasing homeomorphism of R and
lim
u→0
ϕ(σu)
ϕ(u)
= σ = ϕ
2
(σ), lim
|u|→∞
ϕ(σu)
ϕ(u)
= σ
2
= ϕ

3
(σ).
Moreover, ug(t, u) ≥ 0 and
lim
u→0
g(t, u)
ϕ(u)
= π, lim
|u|→∞
g(t, u)
ϕ(u)
= π
3
.
Thus we can check on the fact that
lim
u→0
g(t, u)
ϕ(u)
< λ
1
(2) = π
2
< λ
1
(3) =

4
√
3π

9

3
< lim
|u|→∞
g(t, u)
ϕ(u)
23
All hypotheses of Theorem 1.2 for k = n = 1 are satisfied so that (A) possesses at least one
nontrivial solution.
Example 4.3. Define ϕ, g by
ϕ(u) =







u
2
ln(u + 1), if u ≥ 0,
−u
2
ln(−u + 1), if u < 0,
g(t, u) =








2
10
u
2
ln(u + 1) tan
−1
u, if u ≥ 0,
2
10
u
2
ln(−u + 1) tan
−1
u, if u < 0.
Then ϕ is odd increasing homeomorphism of R and
lim
u→0
ϕ(σu)
ϕ(u)
= σ
3
= ϕ
4
(σ), lim
|u|→∞
ϕ(σu)
ϕ(u)

= σ
2
= ϕ
3
(σ).
Moreover, ug(t, u) ≥ 0 and
lim
u→0
g(t, u)
ϕ(u)
= 0, lim
|u|→∞
g(t, u)
ϕ(u)
= 2
9
π.
Thus we can check on the fact that
lim
u→0
g(t, u)
ϕ(u)
< λ
1
(4) =

√
2π
2


4
< λ
3
(3) =

4
√
3π
3

3
< lim
|u|→∞
g(t, u)
ϕ(u)
All hypotheses of Theorem 1.2 for k = 1 and n = 3 are satisfied so that (A) possesses at
least three nontrivial solutions.
Competing interests
The authors declare that they have no competing interests.
24