1 Introduction and main results
Over the last three decades, there has been an intensive study on the double phase problems of the form:
(1)
−
div
(
∣
∇
u
∣
p
−
2
∇
u
+
c
(
x
)
∣
∇
u
∣
q
−
2
∇
u
)
=
λ
g
(
x
,
u
)
,
x
∈
Ω
,
u
=
0
,
x
∈
∂
Ω
,
where
λ
>
0
,
Ω
⊂
R
N
is a bounded domain with Lipschitz boundary;
c
:
Ω
¯
→
[
0
,
∞
)
is Lipschitz continuous,
g
:
Ω
×
R
→
R
satisfies the Carathéodory condition; and
p
,
q
, and
N
are constants such that
N
≥
2
,
1
<
p
<
q
<
N
, and
q
p
<
1
+
1
N
.
In [15,16], Zhikov initiated the study of the associated energy functional
I
(
u
)
=
∫
Ω
∣
∇
u
∣
p
+
c
(
x
)
∣
∇
u
∣
q
d
x
, which describes the behavioral change with the point for anisotropic materials. Since then, two directions of study have mainly been conducted. One is to show the regularities of minimizers of (1) (see [1,2,13]) and the other (relatively recent) is to show the existence of solutions under certain conditions on
c
and
g
(see [3,4,9]). We also refer the interested reader to [8] and [14], which provide a discontinuity property for the spectrum of a parametric
(
p
,
q
)
-differential operator and convergence properties of double phase problems, respectively, and [7], which provides an overview of recent results concerning elliptic variational problems with nonstandard growth conditions and related to different kinds of nonuniformly elliptic operators.
However, the one-dimensional double phase problems have rarely been studied as far as the authors know. Thus, we study positive solutions to the following one-dimensional generalized double phase problems:
(2)
−
(
a
(
t
)
φ
p
(
u
′
)
+
b
(
t
)
φ
q
(
u
′
)
)
′
=
λ
h
(
t
)
f
(
u
)
,
t
∈
(
0
,
1
)
,
u
(
0
)
=
0
=
u
(
1
)
,
where
1
<
p
<
q
<
∞
and
φ
m
(
s
)
≔
∣
s
∣
m
−
2
s
. Here
a
,
b
,
h
, and
f
satisfy the following conditions:
a
:
[
0
,
1
]
→
[
0
,
∞
)
and
b
:
[
0
,
1
]
→
[
0
,
∞
)
are continuous such that
a
(
t
)
+
b
(
t
)
>
0
for
t
∈
(
0
,
1
)
,
a
(
t
)
b
(
t
)
>
0
for a.e.
t
∈
[
0
,
1
]
, and
∫
0
1
a
−
1
p
−
1
(
s
)
b
−
1
q
−
1
(
s
)
d
s
<
∞
;
h
:
(
0
,
1
)
→
(
0
,
∞
)
is continuous and integrable;
f
:
[
0
,
∞
)
→
R
is continuous and nondecreasing such that
f
(
s
)
>
0
for
s
≫
1
.
We note that (2) is a generalization of (1) when
N
=
1
since
a
can be zero in
(
0
,
1
)
(if
a
>
0
on
(
0
,
1
)
, then (2) can be transformed into (1)). Furthermore,
(
H
1
)
is essential to define the operator
T
λ
as below whose fixed point is a solution to a modified problem of (2) and to satisfy a comparison principle for
−
(
a
(
⋅
)
φ
p
(
u
′
)
+
b
(
⋅
)
φ
q
(
u
′
)
)
′
under the Dirichlet boundary condition.
The main goal of this article is to investigate the existence, multiplicity, and nonexistence of positive solutions
u
∈
C
1
(
0
,
1
)
∩
C
[
0
,
1
]
to (2) according to the behavior of
f
near 0:
lim
s
→
0
f
(
s
)
ϕ
(
s
)
=
∞
,
lim
s
→
0
f
(
s
)
ϕ
(
s
)
=
0
,
lim
s
→
0
f
(
s
)
ϕ
(
s
)
=
f
0
∈
(
0
,
∞
)
,
lim
s
→
0
f
(
s
)
ϕ
(
s
)
=
−
∞
,
and the behavior of
f
near
∞
:
lim
s
→
∞
f
(
s
)
ϕ
(
s
)
=
∞
,
lim
s
→
∞
f
(
s
)
ϕ
(
s
)
=
0
,
lim
s
→
∞
f
(
s
)
ϕ
(
s
)
=
f
∞
∈
(
0
,
∞
)
,
where
ϕ
(
s
)
≔
φ
p
(
s
)
+
φ
q
(
s
)
. Thus, we shall consider not only positone problems (i.e.,
f
(
0
)
≥
0
) but also semipositone problems (i.e.,
f
(
0
)
<
0
), which are very challenging.
To obtain existence results, we apply the following Krasnoselskii-type fixed point theorem (see Lemma A in [5,6]).
Proposition 1.1
Let
X
be a Banach space and
I
:
X
→
X
be a completely continuous operator. Suppose that there exist a nonzero element
z
∈
X
and positive constants
r
and
R
with
r
≠
R
such that
If
y
∈
X
satisfies
y
=
σ
I
y
for
σ
∈
(
0
,
1
]
, then
‖
y
‖
X
≠
r
;
If
y
∈
X
satisfies
y
=
I
y
+
τ
z
for
τ
≥
0
, then
‖
y
‖
X
≠
R
.
Then
I
has a fixed point
y
∈
X
with
min
{
r
,
R
}
<
‖
y
‖
X
<
max
{
r
,
R
}
.
Such a method was used in [5,6, 11,12] to study the existence results for
m
-Laplacian problems (i.e.,
p
=
q
=
m
and
a
≡
b
≡
1
) and
(
p
,
q
)
-Laplacian problems (i.e.,
a
≡
b
≡
1
). They obtained the results by constructing completely continuous operators consisting of the inverses of the
m
-Laplace operator and the
(
p
,
q
)
-Laplace operator, respectively. However, we cannot directly follow the ideas in this problem due to the presence of the functions
a
and
b
. To overcome this difficulty, let us consider the following modified generalized double phase problem:
−
(
a
(
t
)
φ
p
(
x
′
)
+
b
(
t
)
φ
q
(
x
′
)
)
′
=
λ
h
(
t
)
f
∗
(
y
)
,
t
∈
(
0
,
1
)
,
x
(
0
)
=
0
=
x
(
1
)
,
where
y
∈
C
[
0
,
1
]
and
f
∗
(
y
)
≔
f
(
max
{
0
,
y
}
)
and construct its solution operator
T
λ
:
C
[
0
,
1
]
→
C
[
0
,
1
]
defined by
T
λ
y
(
t
)
≔
∫
0
t
a
1
q
−
p
(
s
)
b
−
1
q
−
p
(
s
)
ϕ
−
1
a
−
q
−
1
q
−
p
(
s
)
b
p
−
1
q
−
p
(
s
)
m
y
+
λ
∫
s
1
h
(
r
)
f
∗
(
y
)
d
r
d
s
,
where
m
y
∈
R
is the constant such that
0
=
∫
0
1
a
1
q
−
p
(
s
)
b
−
1
q
−
p
(
s
)
ϕ
−
1
a
−
q
−
1
q
−
p
(
s
)
b
p
−
1
q
−
p
(
s
)
m
y
+
λ
∫
s
1
h
(
r
)
f
∗
(
y
)
d
r
d
s
.
Then
T
λ
is completely continuous from
C
[
0
,
1
]
to
C
[
0
,
1
]
(see Lemma 2.2). Even though this operator looks complicated, it has useful properties (see Appendix). This is one of the novelties in this article.
It is worth noting that for the cases
(
H
4
a
)
and
(
H
4
b
)
, we can show the existence of at least three positive solutions by employing only the fixed point theorem without the help of other theories (e.g., three-solution theorem in [10]). To obtain the multiplicity results, we assume the ratio
f
ϕ
as follows:
There exist
α
>
0
and
β
>
0
such that
f
(
α
)
ϕ
(
α
)
/
f
(
β
)
ϕ
(
β
)
>
‖
h
‖
1
φ
q
(
16
M
a
b
2
K
a
b
6
)
h
∗
,
where
M
a
b
≔
∫
0
1
a
−
1
p
−
1
(
s
)
b
−
1
q
−
1
(
s
)
d
s
,
K
a
b
≔
max
1
,
‖
a
‖
∞
1
p
−
1
,
‖
b
‖
∞
1
q
−
1
, and
h
∗
≔
min
∫
1
4
1
2
h
(
r
)
d
r
,
∫
1
2
3
4
h
(
r
)
d
r
. If the condition
(
H
5
c
)
is considered with
(
H
4
a
)
or
(
H
4
b
)
, then the multiplicity results are obtained by additionally assuming one of the following conditions to
α
or
β
:
f
(
α
)
ϕ
(
α
)
>
2
f
∞
‖
h
‖
1
φ
q
(
16
M
a
b
2
K
a
b
6
)
h
∗
,
f
(
β
)
ϕ
(
β
)
<
h
∗
f
∞
2
‖
h
‖
1
φ
q
(
16
M
a
b
2
K
a
b
6
)
.
For the case
(
H
4
a
)
, we establish the following results.
Theorem 1.2
Let (
H
1
)–(
H
3
),
(
H
4
a
)
, and
(
H
5
a
)
hold. Then (2) has no positive solution for
λ
≫
1
and has at least two positive solutions
u
1
and
u
2
for
λ
≈
0
such that
‖
u
1
‖
∞
→
0
and
‖
u
2
‖
∞
→
∞
as
λ
→
0
.
Theorem 1.3
Let (
H
1
)–(
H
3
),
(
H
4
a
)
, and
(
H
5
b
)
hold. Then (2) has a positive solution
u
for all
λ
>
0
such that
‖
u
‖
∞
→
0
as
λ
→
0
and
‖
u
‖
∞
→
∞
as
λ
→
∞
. Additionally, if
(
H
6
a
)
is satisfied with
β
<
4
α
M
a
b
K
a
b
2
then (2) has at least three positive solutions
u
1
,
u
2
, and
u
3
for
λ
∈
(
λ
α
,
λ
β
)
such that
‖
u
1
‖
∞
<
β
<
‖
u
2
‖
∞
<
4
α
M
a
b
K
a
b
2
<
‖
u
3
‖
∞
, where
λ
α
≔
φ
q
(
16
M
a
b
K
a
b
4
)
ϕ
(
α
)
h
∗
f
(
α
)
and
λ
β
≔
ϕ
(
β
)
‖
h
‖
1
φ
q
(
M
a
b
K
a
b
2
)
f
(
β
)
.
Theorem 1.4
Let (
H
1
)–(
H
3
),
(
H
4
a
)
, and
(
H
5
c
)
hold. Then (2) has no positive solution for
λ
≫
1
and has a positive solution
u
for
λ
<
λ
∞
≔
1
2
f
∞
‖
h
‖
1
φ
q
(
M
a
b
K
a
b
2
)
such that
‖
u
‖
∞
→
0
as
λ
→
0
. Additionally, if
(
H
6
a
)
and
(
H
6
b
)
are satisfied with
β
<
4
α
M
a
b
K
a
b
2
, then (2) has at least three positive solutions
u
1
,
u
2
, and
u
3
for
λ
∈
(
λ
α
,
min
{
λ
β
,
λ
∞
}
)
such that
‖
u
1
‖
∞
<
β
<
‖
u
2
‖
∞
<
4
α
M
a
b
K
a
b
2
<
‖
u
3
‖
∞
.
Next the following results are established for the case
(
H
4
b
)
.
Theorem 1.5
Let (
H
1
)–(
H
3
),
(
H
4
b
)
, and
(
H
5
a
)
hold. Then (2) has a positive solution
u
for all
λ
>
0
such that
‖
u
‖
∞
→
∞
as
λ
→
0
and
‖
u
‖
∞
→
0
as
λ
→
∞
. Additionally, if
(
H
6
a
)
is satisfied with
β
>
4
α
M
a
b
K
a
b
2
, then (2) has at least three positive solutions
u
1
,
u
2
, and
u
3
for
λ
∈
(
λ
α
,
λ
β
)
such that
‖
u
1
‖
∞
<
4
α
M
a
b
K
a
b
2
<
‖
u
2
‖
∞
<
β
<
‖
u
3
‖
∞
.
Theorem 1.6
Let (
H
1
)–(
H
3
),
(
H
4
b
)
, and
(
H
5
b
)
hold. Then (2) has no positive solution for
λ
≈
0
and has at least two positive solutions
u
1
and
u
2
for
λ
≫
1
such that
‖
u
1
‖
∞
→
0
and
‖
u
2
‖
∞
→
∞
as
λ
→
∞
.
Theorem 1.7
Let (
H
1
)–(
H
3
),
(
H
4
b
)
, and
(
H
5
c
)
hold. Then (2) has no positive solution for
λ
≈
0
and has a positive solution
u
for
λ
>
λ
∞
≔
2
φ
q
(
16
M
a
b
K
a
b
4
)
h
∗
f
∞
such that
‖
u
‖
∞
→
0
as
λ
→
∞
. Additionally, if
(
H
6
a
)
and
(
H
6
c
)
are satisfied with
β
>
4
α
M
a
b
K
a
b
2
, then (2) has at least three positive solutions
u
1
,
u
2
and
u
3
for
λ
∈
(
max
{
λ
α
,
λ
∞
}
,
λ
β
)
such that
‖
u
1
‖
∞
<
4
α
M
a
b
K
a
b
2
<
‖
u
2
‖
∞
<
β
<
‖
u
3
‖
∞
.
Remark 1.1
If
f
(
s
)
ϕ
(
s
)
is not a nonincreasing function, then there may exist
α
and
β
satisfying all hypotheses of Theorem 1.3 or Theorem 1.4 (Figure 1). If
f
(
s
)
ϕ
(
s
)
is not a nondecreasing function, then there may exist
α
and
β
satisfying all hypotheses of Theorem 1.5 or Theorem 1.7 (Figure 2).
Remark 1.2
We note that an expected bifurcation curve
λ
versus
‖
u
‖
∞
is
S
-shaped and reversed
S
-shaped if all hypotheses are satisfied in Theorems 1.3 and 1.5, respectively. It is worth noting that the multiplicity result in Theorem 1.5 is new even for the nonautonomous Laplacian problems (i.e.,
p
=
q
=
2
,
a
≡
b
≡
1
and
h
is a nonconstant function) as far as the authors know.
Now we consider the case
(
H
4
c
)
. To show the existence of at least two positive solutions, we assume one of the following conditions:
There exists
γ
>
0
such that
f
(
γ
)
ϕ
(
γ
)
<
h
∗
f
0
2
‖
h
‖
1
φ
q
(
16
M
a
b
2
K
a
b
6
)
,
There exists
δ
>
0
such that
f
(
δ
)
ϕ
(
δ
)
>
2
f
0
‖
h
‖
1
φ
q
(
16
M
a
b
2
K
a
b
6
)
h
∗
,
There exists
ζ
>
0
such that
f
(
ζ
)
ϕ
(
ζ
)
<
h
∗
min
{
f
0
,
f
∞
}
2
‖
h
‖
1
φ
q
(
16
M
a
b
2
K
a
b
6
)
,
There exists
η
>
0
such that
f
(
η
)
ϕ
(
η
)
>
2
‖
h
‖
1
φ
q
(
16
M
a
b
2
K
a
b
6
)
max
{
f
0
,
f
∞
}
h
∗
.
We establish:
Theorem 1.8
Let (
H
1
)–(
H
3
),
(
H
4
c
)
, and
(
H
5
a
)
hold. Then (2) has no positive solution for
λ
≫
1
and has a positive solution
u
for
λ
<
λ
0
≔
1
2
f
0
‖
h
‖
1
φ
q
(
M
a
b
K
a
b
2
)
such that
‖
u
‖
∞
→
∞
as
λ
→
0
. Additionally, if
(
H
7
a
)
is satisfied, then (2) has at least two positive solutions
u
1
and
u
2
for
λ
∈
(
λ
0
,
λ
γ
)
such that
‖
u
1
‖
∞
<
γ
<
‖
u
2
‖
∞
, where
λ
0
≔
2
φ
q
(
16
M
a
b
K
a
b
4
)
h
∗
f
0
and
λ
γ
≔
ϕ
(
γ
)
‖
h
‖
1
φ
q
(
M
a
b
K
a
b
2
)
f
(
γ
)
.
Theorem 1.9
Let (
H
1
)–(
H
3
),
(
H
4
c
)
, and
(
H
5
b
)
hold. Then (2) has no positive solution for
λ
≈
0
and has a positive solution
u
for
λ
>
λ
0
such that
‖
u
‖
∞
→
∞
as
λ
→
∞
. Additionally, if
(
H
7
b
)
is satisfied, then (2) has at least two positive solutions
u
1
and
u
2
for
λ
∈
(
λ
δ
,
λ
0
)
such that
‖
u
1
‖
∞
<
4
δ
M
a
b
K
a
b
2
<
‖
u
2
‖
∞
, where
λ
δ
≔
φ
q
(
16
M
a
b
K
a
b
4
)
ϕ
(
δ
)
h
∗
f
(
δ
)
.
Theorem 1.10
Let (
H
1
)–(
H
3
),
(
H
4
c
)
, and
(
H
5
c
)
hold. Then (2) has no positive solution for
λ
≈
0
and for
λ
≫
1
. If
max
{
f
0
,
f
∞
}
min
{
f
0
,
f
∞
}
>
4
‖
h
‖
1
φ
q
(
16
M
a
b
2
K
a
b
6
)
h
∗
, then (2) has a positive solution for
λ
∈
(
min
{
λ
∞
,
λ
0
}
,
max
{
λ
∞
,
λ
0
}
)
. If
(
H
7
c
)
is satisfied, then (2) has at least two positive solutions
u
1
and
u
2
for
λ
∈
(
max
{
λ
∞
,
λ
0
}
,
λ
ζ
)
such that
‖
u
1
‖
∞
<
ζ
<
‖
u
2
‖
∞
, where
λ
ζ
≔
ϕ
(
ζ
)
‖
h
‖
1
φ
q
(
M
a
b
K
a
b
2
)
f
(
ζ
)
. If
(
H
7
d
)
is satisfied, then (2) has at least two positive solutions
u
1
and
u
2
for
λ
∈
(
λ
η
,
min
{
λ
∞
,
λ
0
}
)
such that
‖
u
1
‖
∞
<
4
η
M
a
b
K
a
b
2
<
‖
u
2
‖
∞
, where
λ
η
≔
φ
q
(
16
M
a
b
K
a
b
4
)
ϕ
(
η
)
h
∗
f
(
η
)
.
Finally, we consider the case
(
H
4
d
)
. Let
M
a
b
∗
≔
2
1
p
−
1
M
a
b
and
ρ
>
0
be the largest constant such that
f
(
s
)
≤
0
for
s
≤
ρ
. To discuss the existence of at least two positive solutions, we assume:
there exist
θ
>
0
and
κ
>
0
such that
f
(
θ
)
ϕ
(
θ
)
/
f
(
κ
)
ϕ
(
κ
)
>
‖
h
‖
1
φ
q
(
32
M
a
b
M
a
b
∗
K
a
b
6
)
h
∗
,
and one of the following conditions will also be imposed on
θ
or
κ
:
θ
≥
2
ρ
and
f
(
θ
)
>
‖
h
‖
1
∣
f
(
0
)
∣
φ
q
(
32
M
a
b
∗
K
a
b
4
)
h
∗
φ
p
(
4
K
a
b
)
,
κ
≥
8
ρ
M
a
b
∗
K
a
b
2
and
f
(
κ
)
>
2
∣
f
(
0
)
∣
.
Furthermore, the following condition is assumed to show an existence result when
(
H
5
c
)
is considered:
there exists
μ
>
0
such that
f
(
μ
)
>
2
∣
f
(
0
)
∣
and
f
(
μ
)
ϕ
(
μ
)
<
h
∗
f
∞
2
‖
h
‖
1
φ
q
(
32
M
a
b
M
a
b
∗
K
a
b
6
)
.
We establish the following results.
Theorem 1.11
Let (
H
1
)–(
H
3
),
(
H
4
d
)
, and
(
H
5
a
)
hold. Then (2) has a positive solution
u
for
λ
≈
0
such that
‖
u
‖
∞
→
∞
as
λ
→
0
. In addition, if
(
H
8
a
)
and
(
H
8
b
)
are satisfied with
κ
>
4
θ
M
a
b
∗
K
a
b
2
, then (2) has at least two positive solutions
u
1
and
u
2
for
λ
∈
(
λ
θ
,
min
{
λ
κ
,
λ
θ
∗
}
)
such that
4
θ
M
a
b
∗
K
a
b
2
<
‖
u
1
‖
∞
<
κ
<
‖
u
2
‖
∞
, where
λ
θ
≔
φ
q
(
32
M
a
b
∗
K
a
b
4
)
ϕ
(
θ
)
h
∗
f
(
θ
)
,
λ
κ
≔
ϕ
(
κ
)
‖
h
‖
1
φ
q
(
M
a
b
K
a
b
2
)
f
(
κ
)
, and
λ
θ
∗
≔
φ
p
(
4
K
a
b
)
ϕ
(
θ
)
‖
h
‖
1
∣
f
(
0
)
∣
.
Theorem 1.12
Let (
H
1
)–(
H
3
),
(
H
4
d
)
,
(
H
5
b
)
, and
lim
s
→
∞
f
(
s
)
=
∞
hold. Then (2) has no positive solution for
λ
≈
0
and has a positive solution
u
for
λ
≫
1
such that
‖
u
‖
∞
→
∞
as
λ
→
∞
. In addition, if
(
H
8
a
)
and
(
H
8
c
)
are satisfied with
κ
<
4
θ
M
a
b
∗
K
a
b
2
, then (2) has at least two positive solutions
u
1
and
u
2
for
λ
∈
(
λ
θ
,
λ
κ
)
such that
κ
<
‖
u
1
‖
∞
<
4
θ
M
a
b
∗
K
a
b
2
<
‖
u
2
‖
∞
.
Theorem 1.13
Let (
H
1
)–(
H
3
),
(
H
4
d
)
, and
(
H
5
c
)
hold. Then (2) has no positive solution for
λ
≈
0
. In addition, if
(
H
8
d
)
is satisfied, then (2) has a positive solution for
λ
∈
(
λ
̲
,
λ
μ
)
, where
λ
̲
≔
2
φ
q
(
32
M
a
b
∗
K
a
b
4
)
h
∗
f
∞
and
λ
μ
≔
ϕ
(
μ
)
‖
h
‖
1
φ
q
(
M
a
b
K
a
b
2
)
f
(
μ
)
.
Remark 1.3
The other novelty is that we only use one method, the Krasnoselskii fixed point theorem, to obtain all existence results even for
(
H
4
c
)
and
(
H
4
d
)
. Furthermore, the advantage of this method is that it requires fewer assumptions and less information to prove the existence results. For example, the Ambrosetti-Rabinowitz condition and the principal eigenvalue for the
p
-Laplacian with the Dirichlet boundary condition are required in general to obtain the existence results for
p
-Laplacian problems via the variational methods and the methods of sub-supersolutions, respectively. However, we do not require such assumption and information in this method.
2 Preliminaries
In this section, we show the complete continuity of the operator
T
λ
which was defined in Section 1 and lower estimates of solutions to (semi)positone problems. We also discuss the regularities of solutions of (2).
Before going to show the complete continuity of the operator
T
λ
,
we provide the fundamental property of
T
λ
.
Lemma 2.1
Let (
H
1
)–(
H
3
) hold. Then
T
λ
y
∈
C
1
(
0
,
1
)
∩
C
[
0
,
1
]
for
y
∈
C
[
0
,
1
]
.
Proof
If
a
(
t
)
>
0
and
b
(
t
)
>
0
for all
t
∈
(
0
,
1
)
, then it is easy to show
T
λ
y
∈
C
1
(
0
,
1
)
∩
C
[
0
,
1
]
because
(
T
λ
y
)
′
(
t
)
=
a
1
q
−
p
(
t
)
b
−
1
q
−
p
(
t
)
ϕ
−
1
a
−
q
−
1
q
−
p
(
t
)
b
p
−
1
q
−
p
(
t
)
m
y
+
λ
∫
t
1
h
(
r
)
f
∗
(
y
)
d
r
.
Assume that there exists
t
0
∈
(
0
,
1
)
such that
b
(
t
0
)
=
0
. Then
a
(
t
0
)
>
0
, and
a
(
t
)
>
0
and
b
(
t
)
>
0
for
t
≈
t
0
and
t
≠
t
0
. By Proposition 8.2 in Appendix, we have
lim
t
→
t
0
T
λ
y
(
t
)
−
T
λ
y
(
t
0
)
t
−
t
0
=
lim
t
→
t
0
a
1
q
−
p
(
t
)
b
−
1
q
−
p
(
t
)
ϕ
−
1
a
−
q
−
1
q
−
p
(
t
)
b
p
−
1
q
−
p
(
t
)
m
y
+
λ
∫
t
1
h
(
r
)
f
∗
(
y
)
d
r
=
a
−
1
p
−
1
(
t
0
)
φ
p
−
1
m
y
+
λ
∫
t
0
1
h
(
r
)
f
∗
(
y
)
d
r
.
Thus,
T
λ
y
is differentiable at
t
=
t
0
. Furthermore,
(
T
λ
y
)
′
is continuous at
t
=
t
0
since
lim
t
→
t
0
(
T
λ
y
)
′
(
t
)
=
lim
t
→
t
0
a
1
q
−
p
(
t
)
b
−
1
q
−
p
(
t
)
ϕ
−
1
a
−
q
−
1
q
−
p
(
t
)
b
p
−
1
q
−
p
(
t
)
m
y
+
λ
∫
t
1
h
(
r
)
f
∗
(
y
)
d
r
=
a
−
1
p
−
1
(
t
0
)
φ
p
−
1
m
y
+
λ
∫
t
0
1
h
(
r
)
f
∗
(
y
)
d
r
=
(
T
λ
y
)
′
(
t
0
)
.
Hence,
(
T
λ
y
)
′
(
t
0
)
exists and is continuous at
t
=
t
0
.
By similar arguments, if
t
0
∈
(
0
,
1
)
is such that
a
(
t
0
)
=
0
, then we can show that
(
T
λ
y
)
′
(
t
0
)
=
b
−
1
q
−
1
(
t
0
)
φ
q
−
1
m
y
+
λ
∫
t
0
1
h
(
r
)
f
∗
(
y
)
d
r
and
(
T
λ
y
)
′
is continuous at
t
=
t
0
.□
Then
T
λ
y
can be rewritten as
T
λ
y
(
t
)
≔
T
λ
y
(
t
∗
)
+
∫
t
∗
t
a
1
q
−
p
(
s
)
b
−
1
q
−
p
(
s
)
ϕ
−
1
λ
a
−
q
−
1
q
−
p
(
s
)
b
p
−
1
q
−
p
(
s
)
∫
s
t
m
h
(
r
)
f
∗
(
y
)
d
r
d
s
,
0
≤
t
∗
≤
t
≤
t
m
,
T
λ
y
(
t
∗
)
+
∫
t
t
∗
a
1
q
−
p
(
s
)
b
−
1
q
−
p
(
s
)
ϕ
−
1
λ
a
−
q
−
1
q
−
p
(
s
)
b
p
−
1
q
−
p
(
s
)
∫
t
m
s
h
(
r
)
f
∗
(
y
)
d
r
d
s
,
t
m
≤
t
≤
t
∗
≤
1
,
where
t
m
∈
(
0
,
1
)
is such that
‖
T
λ
y
‖
∞
=
∣
T
λ
y
(
t
m
)
∣
. The complete continuity of
T
λ
is essential to apply the fixed point theorem. Thus, we show:
Lemma 2.2
Let (
H
1
)–(
H
3
) hold. Then
T
λ
:
C
[
0
,
1
]
→
C
[
0
,
1
]
is completely continuous.
Proof
We first show that
T
λ
is a compact operator. Let
{
y
n
}
⊂
C
[
0
,
1
]
be a bounded sequence. Let
f
0
∗
(
s
)
≔
max
r
∈
[
0
,
s
]
∣
f
∗
(
s
)
∣
. We note that
{
m
y
n
}
is a bounded sequence. Thus, there exists
M
0
>
0
such that
∣
m
y
n
∣
+
λ
‖
h
‖
1
f
0
∗
(
‖
y
n
‖
∞
)
≤
M
0
for all
n
. By Proposition 8.1 in Appendix, we have
∣
T
λ
y
n
(
t
)
∣
≤
∫
0
t
a
1
q
−
p
(
s
)
b
−
1
q
−
p
(
s
)
ϕ
−
1
a
−
q
−
1
q
−
p
(
s
)
b
p
−
1
q
−
p
(
s
)
(
∣
m
y
n
∣
+
λ
‖
h
‖
1
f
0
∗
(
‖
y
n
‖
∞
)
)
d
s
≤
∫
0
t
a
1
q
−
p
(
s
)
b
−
1
q
−
p
(
s
)
ϕ
−
1
M
0
a
−
q
−
1
q
−
p
(
s
)
b
p
−
1
q
−
p
(
s
)
d
s
≤
K
a
b
ϕ
−
1
(
M
0
)
∫
0
t
a
−
1
p
−
1
(
s
)
b
−
1
q
−
1
(
s
)
d
s
≤
M
a
b
K
a
b
ϕ
−
1
(
M
0
)
.
This implies that
{
T
λ
y
n
}
is uniformly bounded. Furthermore, applying Proposition 8.1 again we obtain
∣
T
λ
y
n
(
t
2
)
−
T
λ
y
n
(
t
1
)
∣
≤
∫
t
1
t
2
a
1
q
−
p
(
s
)
b
−
1
q
−
p
(
s
)
ϕ
−
1
M
0
a
−
q
−
1
q
−
p
(
s
)
b
p
−
1
q
−
p
(
s
)
d
s
≤
K
a
b
ϕ
−
1
(
M
0
)
∫
t
1
t
2
a
−
1
p
−
1
(
s
)
b
−
1
q
−
1
(
s
)
d
s
for
t
1
,
t
2
∈
[
0
,
1
]
. Let
ε
>
0
. By
(
H
1
)
, there exists
δ
0
>
0
such that
K
a
b
ϕ
−
1
(
M
0
)
∫
t
1
t
2
a
−
1
p
−
1
(
s
)
b
−
1
q
−
1
(
s
)
d
s
<
ε
for all
t
1
,
t
2
∈
[
0
,
1
]
such that
∣
t
1
−
t
2
∣
<
δ
0
. This implies that
{
T
λ
y
n
}
is uniformly equicontinuous on
[
0
,
1
]
. Thus,
{
T
λ
y
n
}
has a convergent subsequence in
C
[
0
,
1
]
by the Arzela-Ascoli theorem. Therefore,
T
λ
is a compact operator.
We complete this proof by showing the continuity of
T
λ
. Let
y
n
converge to
y
in
C
[
0
,
1
]
. Then
∣
T
λ
y
n
(
t
)
−
T
λ
y
(
t
)
∣
=
∫
0
t
∣
(
T
λ
y
n
)
′
(
s
)
−
(
T
λ
y
)
′
(
s
)
∣
d
s
≤
∫
0
1
∣
(
T
λ
y
n
)
′
(
s
)
−
(
T
λ
y
)
′
(
s
)
∣
d
s
.
From the definition of
T
λ
y
n
, it is easy to show that
(
T
λ
y
n
)
′
(
t
)
converges pointwise to
(
T
λ
y
)
′
(
t
)
. Then
∫
0
1
∣
(
T
λ
y
n
)
′
(
s
)
−
(
T
λ
y
)
′
(
s
)
∣
d
s
converges to 0 by the Lebesgue dominated convergence theorem. Hence,
T
λ
y
n
converges to
T
λ
y
in
C
[
0
,
1
]
.□
In the next two lemmas, we observe lower estimates of functions satisfying boundary value problems related to (2). We obtain these results by following and extending the ideas in [6]. Let
v
∈
C
1
(
0
,
1
)
∩
C
[
0
,
1
]
be a function satisfying
(3)
−
(
a
(
t
)
φ
p
(
v
′
)
+
b
(
t
)
φ
q
(
v
′
)
)
′
≥
0
,
t
∈
(
0
,
1
)
,
v
(
0
)
≥
0
,
v
(
1
)
≥
0
.
Then we establish the following result.
Lemma 2.3
Let
v
∈
C
1
(
0
,
1
)
∩
C
[
0
,
1
]
satisfy (3). Let
t
m
∈
(
0
,
1
)
be such that
‖
v
‖
∞
=
∣
v
(
t
m
)
∣
. Then
v
(
t
)
≥
‖
v
‖
∞
M
a
b
K
a
b
2
d
(
t
)
, where
d
(
t
)
≔
min
{
t
,
1
−
t
}
.
Proof
We first show that
v
(
t
m
)
≥
0
. Assume to the contrary that
v
(
t
m
)
<
0
. Integrating (3) from
t
to
t
m
, we have
a
(
t
)
φ
p
(
v
′
(
t
)
)
+
b
(
t
)
φ
q
(
v
′
(
t
)
)
≥
0
for
t
∈
(
0
,
t
m
)
. This implies
v
′
(
t
)
≥
0
for
t
∈
(
0
,
t
m
)
. However, this is a contradiction since
v
(
0
)
≥
0
and
v
(
t
m
)
<
0
. Hence,
v
(
t
m
)
≥
0
.
Now we show that
v
(
t
)
≥
‖
v
‖
∞
M
a
b
K
a
b
2
d
(
t
)
. Let
z
∈
C
1
(
0
,
t
m
]
be the solution of the boundary value problem
−
(
a
(
t
)
φ
p
(
z
′
)
+
b
(
t
)
φ
q
(
z
′
)
)
′
=
0
,
t
∈
(
0
,
t
m
)
,
z
(
0
)
=
0
,
z
(
t
m
)
=
‖
v
‖
∞
.
Then
z
satisfies
z
(
t
)
=
∫
0
t
a
1
q
−
p
(
s
)
b
−
1
q
−
p
(
s
)
ϕ
−
1
m
∗
a
−
q
−
1
q
−
p
(
s
)
b
p
−
1
q
−
p
(
s
)
d
s
,
where
m
∗
>
0
is the constant satisfying
‖
v
‖
∞
=
∫
0
t
m
a
1
q
−
p
(
s
)
b
−
1
q
−
p
(
s
)
ϕ
−
1
m
∗
a
−
q
−
1
q
−
p
(
s
)
b
p
−
1
q
−
p
(
s
)
d
s
.
By Proposition 8.1, we have
‖
v
‖
∞
≤
K
a
b
ϕ
−
1
(
m
∗
)
∫
0
t
m
a
−
1
p
−
1
(
s
)
b
−
1
q
−
1
(
s
)
d
s
≤
M
a
b
K
a
b
ϕ
−
1
(
m
∗
)
.
Applying Proposition 8.1 again, we obtain
z
(
t
)
=
∫
0
t
a
1
q
−
p
(
s
)
b
−
1
q
−
p
(
s
)
ϕ
−
1
m
∗
a
−
q
−
1
q
−
p
(
s
)
b
p
−
1
q
−
p
(
s
)
d
s
≥
∫
0
t
ϕ
−
1
(
m
∗
)
K
a
b
d
s
≥
‖
v
‖
∞
M
a
b
K
a
b
2
d
(
t
)
.
Furthermore,
z
(
t
)
≤
v
(
t
)
by the comparison principle. This implies
v
(
t
)
≥
z
(
t
)
≥
‖
v
‖
∞
M
a
b
K
a
b
2
d
(
t
)
for
t
∈
(
0
,
t
m
)
. By similar arguments, we can also show that
v
(
t
)
≥
‖
v
‖
∞
M
a
b
K
a
b
2
d
(
t
)
for
t
∈
(
t
m
,
1
)
.□
Let
f
(
0
)
<
0
and
v
∈
C
1
(
0
,
1
)
∩
C
[
0
,
1
]
be a function satisfying
(4)
−
(
a
(
t
)
φ
p
(
v
′
)
+
b
(
t
)
φ
q
(
v
′
)
)
′
≥
λ
h
(
t
)
f
(
0
)
,
t
∈
(
0
,
1
)
,
v
(
0
)
≥
0
,
v
(
1
)
≥
0
.
Then
v
satisfies the following property.
Lemma 2.4
Let (
H
1
)–(
H
3
), and
(
H
4
d
)
hold. Let
v
∈
C
1
(
0
,
1
)
∩
C
[
0
,
1
]
satisfy (4), and
t
m
∈
(
0
,
1
)
be such that
‖
v
‖
∞
=
∣
v
(
t
m
)
∣
. If
‖
v
‖
∞
≥
M
a
b
K
a
b
ϕ
−
1
(
2
λ
‖
h
‖
1
∣
f
(
0
)
∣
)
then
v
(
t
)
≥
‖
v
‖
∞
M
a
b
∗
K
a
b
2
d
(
t
)
, where
M
a
b
∗
≔
2
1
p
−
1
M
a
b
.
Proof
We note that we shall repeatedly use Proposition 8.1 in this proof. We first show that
v
(
t
m
)
>
0
. It is clear that
v
(
t
m
)
≠
0
. If
v
(
t
m
)
<
0
, then
v
satisfies
‖
v
‖
∞
≤
−
v
(
0
)
−
∫
0
t
m
a
1
q
−
p
(
s
)
b
−
1
q
−
p
(
s
)
ϕ
−
1
λ
‖
h
‖
1
f
(
0
)
a
−
q
−
1
q
−
p
(
s
)
b
p
−
1
q
−
p
(
s
)
d
s
≤
∫
0
t
m
a
1
q
−
p
(
s
)
b
−
1
q
−
p
(
s
)
ϕ
−
1
λ
‖
h
‖
1
∣
f
(
0
)
∣
a
−
q
−
1
q
−
p
(
s
)
b
p
−
1
q
−
p
(
s
)
d
s
≤
M
a
b
K
a
b
ϕ
−
1
(
λ
‖
h
‖
1
∣
f
(
0
)
∣
)
.
This is a contradiction since
‖
v
‖
∞
≥
M
a
b
K
a
b
ϕ
−
1
(
2
λ
‖
h
‖
1
∣
f
(
0
)
∣
)
. Hence,
v
(
t
m
)
>
0
.
Now we show that
v
(
t
)
≥
‖
v
‖
∞
M
a
b
∗
K
a
b
2
d
(
t
)
. Let
w
∈
C
1
(
0
,
t
m
]
be the solution of the boundary value problem
−
(
a
(
t
)
φ
p
(
w
′
)
+
b
(
t
)
φ
q
(
w
′
)
)
′
=
λ
h
(
t
)
f
(
0
)
,
t
∈
(
0
,
t
m
)
,
w
(
0
)
=
0
,
w
(
t
m
)
=
‖
v
‖
∞
.
Then
w
satisfies
w
(
t
)
=
∫
0
t
a
1
q
−
p
(
s
)
b
−
1
q
−
p
(
s
)
ϕ
−
1
a
−
q
−
1
q
−
p
(
s
)
b
p
−
1
q
−
p
(
s
)
m
0
+
λ
∫
s
t
m
h
(
r
)
f
(
0
)
d
r
d
s
≤
∫
0
t
a
1
q
−
p
(
s
)
b
−
1
q
−
p
(
s
)
ϕ
−
1
m
0
a
−
q
−
1
q
−
p
(
s
)
b
p
−
1
q
−
p
(
s
)
d
s
≤
M
a
b
K
a
b
ϕ
−
1
(
m
0
)
,
where
m
0
≔
a
(
t
m
)
φ
p
(
w
′
(
t
m
)
)
+
b
(
t
m
)
φ
q
(
w
′
(
t
m
)
)
>
0
. Thus,
m
0
≥
ϕ
∣
∣
v
∣
∣
∞
M
a
b
K
a
b
.
Since
‖
v
‖
∞
≥
M
a
b
K
a
b
ϕ
−
1
(
2
λ
‖
h
‖
1
∣
f
(
0
)
∣
)
, we obtain
m
0
≥
2
λ
‖
h
‖
1
∣
f
(
0
)
∣
. Then we have
w
(
t
)
≥
∫
0
t
a
1
q
−
p
(
s
)
b
−
1
q
−
p
(
s
)
ϕ
−
1
a
−
q
−
1
q
−
p
(
s
)
b
p
−
1
q
−
p
(
s
)
(
m
0
−
λ
‖
h
‖
1
∣
f
(
0
)
∣
)
d
s
≥
ϕ
−
1
(
m
0
−
λ
‖
h
‖
1
∣
f
(
0
)
∣
)
K
a
b
d
(
t
)
≥
ϕ
−
1
m
0
2
K
a
b
d
(
t
)
≥
1
K
a
b
ϕ
−
1
1
2
ϕ
‖
v
‖
∞
M
a
b
K
a
b
d
(
t
)
≥
‖
v
‖
∞
M
a
b
∗
K
a
b
2
d
(
t
)
.
Since
w
(
t
)
≤
v
(
t
)
by the comparison principle, we have
v
(
t
)
≥
w
(
t
)
≥
‖
v
‖
∞
M
a
b
∗
K
a
b
2
d
(
t
)
for
t
∈
(
0
,
t
m
)
. By similar arguments, we can show that
v
(
t
)
≥
‖
v
‖
∞
M
a
b
∗
K
a
b
2
d
(
t
)
for
t
∈
(
t
m
,
1
)
.□
We end this section by discussing the regularities of positive solutions of (2).
Lemma 2.5
Let (
H
1
)–(
H
3
) and one of (
H
4
a
)–(
H
4
d
) hold. Let
u
∈
C
1
(
0
,
1
)
∩
C
[
0
,
1
]
be a positive solution of (2). If
a
(
0
)
+
b
(
0
)
>
0
, then
u
∈
C
1
[
0
,
1
)
, and if
a
(
1
)
+
b
(
1
)
>
0
, then
u
∈
C
1
(
0
,
1
]
.
Proof
We first show that if
a
(
0
)
+
b
(
0
)
>
0
then
u
∈
C
1
[
0
,
1
)
. Without loss of generality, we assume
a
(
0
)
>
0
. Let
t
m
∈
(
0
,
1
)
be such that
‖
u
‖
∞
=
u
(
t
m
)
. Then
u
satisfies
(5)
a
(
t
)
φ
p
(
u
′
(
t
)
)
+
b
(
t
)
φ
q
(
u
′
(
t
)
)
=
λ
∫
t
t
m
h
(
s
)
f
(
u
)
d
s
.
This implies
a
(
t
)
φ
p
(
∣
u
′
(
t
)
∣
)
≤
λ
‖
h
‖
1
max
{
∣
f
(
0
)
∣
,
f
(
‖
u
‖
∞
)
}
. Since
a
(
0
)
>
0
,
u
′
(
t
)
is bounded for
t
≈
0
. Then we can show that
lim
t
→
0
u
′
(
t
)
exists from (5). Hence,
u
∈
C
1
[
0
,
1
)
. By similar arguments, we can show that if
a
(
1
)
+
b
(
1
)
>
0
then
u
∈
C
1
(
0
,
1
]
.□
Lemma 2.6
Let (
H
1
)–(
H
3
), and one of (
H
4
a
)–(
H
4
c
) hold. Let
u
∈
C
1
(
0
,
1
)
∩
C
[
0
,
1
]
be a positive solution of (2). If
a
(
0
)
=
b
(
0
)
=
0
, then
u
∉
C
1
[
0
,
1
)
and if
a
(
1
)
=
b
(
1
)
=
0
, then
u
∉
C
1
(
0
,
1
]
.
Proof
We first show that if
a
(
0
)
=
b
(
0
)
=
0
then
u
∉
C
1
[
0
,
1
)
. Assume to the contrary that
u
′
(
0
)
is defined. From (5), we obtain
0
=
a
(
0
)
φ
p
(
u
′
(
0
)
)
+
b
(
0
)
φ
q
(
u
′
(
0
)
)
=
λ
∫
0
t
m
h
(
s
)
f
(
u
)
d
s
>
0
.
This is a contradiction. Hence,
u
∉
C
1
[
0
,
1
)
. By similar arguments, we can show that if
a
(
1
)
=
b
(
1
)
=
0
then
u
∉
C
1
(
0
,
1
]
.□
Remark 2.1
It is worth noting that it is enough to assume that
a
and
b
are continuous and nonnegative instead of
(
H
1
)
to show Lemmas 2.5 and 2.6. We also note that the sufficient conditions of
f
to prove Lemma 2.6 are continuous and nonnegative.
3 Proofs of Theorems 1.2–1.4
In this section, we prove the existence and multiplicity results employing Proposition 1.1. To this end, letting
X
=
C
[
0
,
1
]
,
z
≡
1
, and
I
≡
T
λ
we find constants satisfying conditions
(
a
)
and
(
b
)
in Proposition 1.1.
Proof of Theorem 1.2
We first show the multiplicity result for
λ
≈
0
. To find a constant satisfying
(
a
)
in Proposition 1.1, let
σ
∈
(
0
,
1
]
and
u
∈
C
[
0
,
1
]
be a solution of
u
=
σ
T
λ
u
. By
(
H
1
)
and
(
H
4
a
)
,
u
(
t
)
=
σ
T
λ
u
(
t
)
≥
0
. Thus,
u
has a maximum in
(
0
,
1
)
. Let
t
m
∈
(
0
,
1
)
be such that
‖
u
‖
∞
=
u
(
t
m
)
. Since
‖
u
‖
∞
≤
‖
T
λ
u
‖
∞
and
K
a
b
≥
1
, we have
‖
u
‖
∞
=
∫
0
t
m
a
1
q
−
p
(
s
)
b
−
1
q
−
p
(
s
)
ϕ
−
1
a
−
q
−
1
q
−
p
(
s
)
b
p
−
1
q
−
p
(
s
)
λ
∫
s
t
m
h
(
r
)
f
∗
(
u
)
d
r
d
s
≤
M
a
b
K
a
b
2
ϕ
−
1
(
λ
‖
h
‖
1
f
(
‖
u
‖
∞
)
)
.
Noting that
M
a
b
K
a
b
2
≥
1
, we obtain
λ
‖
h
‖
1
f
(
‖
u
‖
∞
)
≥
ϕ
‖
u
‖
∞
M
a
b
K
a
b
2
≥
ϕ
(
‖
u
‖
∞
)
φ
q
(
M
a
b
K
a
b
2
)
. This implies
(6)
f
(
‖
u
‖
∞
)
ϕ
(
‖
u
‖
∞
)
≥
1
λ
‖
h
‖
1
φ
q
(
M
a
b
K
a
b
2
)
.
Then
‖
u
‖
∞
≠
1
for
λ
≈
0
.
Now we show that there exist two constants satisfying
(
b
)
in Proposition 1.1: one is greater than 1 and one is less than 1. Let
τ
≥
0
and
u
∈
C
[
0
,
1
]
be a solution of
u
=
T
λ
u
+
τ
. Then
u
∈
C
1
(
0
,
1
)
∩
C
[
0
,
1
]
by Lemma 2.1 and
u
satisfies (3). Thus,
u
(
t
)
≥
‖
u
‖
∞
M
a
b
K
a
b
2
d
(
t
)
by Lemma 2.3. Since
u
(
t
)
≥
T
λ
u
(
t
)
≥
0
and
K
a
b
≥
1
, if
t
m
≥
1
2
then
‖
u
‖
∞
≥
∫
0
t
m
a
1
q
−
p
(
s
)
b
−
1
q
−
p
(
s
)
ϕ
−
1
λ
a
−
q
−
1
q
−
p
(
s
)
b
p
−
1
q
−
p
(
s
)
∫
s
t
m
h
(
r
)
f
(
u
)
d
r
d
s
≥
1
K
a
b
∫
0
1
4
ϕ
−
1
λ
∫
1
4
1
2
h
(
r
)
f
(
u
)
d
r
d
s
≥
1
4
K
a
b
2
ϕ
−
1
λ
h
∗
f
‖
u
‖
∞
4
M
a
b
K
a
b
2
,
where
h
∗
is defined in
(
H
6
a
)
. By similar arguments, we can show that if
t
m
<
1
2
then
‖
u
‖
∞
≥
1
4
K
a
b
2
ϕ
−
1
λ
h
∗
f
‖
u
‖
∞
4
M
a
b
K
a
b
2
. Then we have
λ
h
∗
f
‖
u
‖
∞
4
M
a
b
K
a
b
2
≤
ϕ
(
4
K
a
b
2
‖
u
‖
∞
)
≤
φ
q
(
16
M
a
b
K
a
b
4
)
ϕ
‖
u
‖
∞
4
M
a
b
K
a
b
2
.
This implies
(7)
f
‖
u
‖
∞
4
M
a
b
K
a
b
2
ϕ
‖
u
‖
∞
4
M
a
b
K
a
b
2
≤
φ
q
(
16
M
a
b
K
a
b
4
)
λ
h
∗
.
By
(
H
4
a
)
and
(
H
5
a
)
, there exist
r
λ
≈
0
and
R
λ
≫
1
such that
r
λ
<
min
{
1
,
‖
u
‖
∞
}
and
R
λ
>
max
{
1
,
‖
u
‖
∞
}
for
λ
>
0
.
By Proposition 1.1 and Lemma 2.1,
T
λ
has fixed points
v
1
and
v
2
∈
C
1
(
0
,
1
)
∩
C
[
0
,
1
]
for
λ
≈
0
such that
r
λ
<
‖
v
1
‖
∞
<
1
<
‖
v
2
‖
∞
<
R
λ
. Then
v
1
and
v
2
are positive solutions of (2) by Lemma 2.3. Furthermore, we obtain
‖
v
1
‖
∞
→
0
and
‖
v
2
‖
∞
→
∞
as
λ
→
0
from (6).
Next we show the nonexistence result for
λ
≫
1
. Assume to the contrary that (2) has a positive solution
u
for
λ
≫
1
. We note that
inf
s
∈
(
0
,
∞
)
f
(
s
)
ϕ
(
s
)
>
0
by
(
H
4
a
)
and
(
H
5
a
)
. From (7), we obtain
inf
s
∈
(
0
,
∞
)
f
(
s
)
ϕ
(
s
)
≤
φ
q
(
16
M
a
b
K
a
b
4
)
λ
h
∗
. This is a contradiction for
λ
≫
1
.□
Proof of Theorem 1.3
We first show the existence result for
λ
>
0
. For this, we show that there exist a constant satisfying
(
a
)
in Proposition 1.1 and a constant satisfying
(
b
)
in Proposition 1.1. Let
σ
∈
(
0
,
1
]
and
u
∈
C
[
0
,
1
]
be a solution of
u
=
σ
T
λ
u
. Following the arguments in the proof of Theorem 1.2, we can show that
u
satisfies (6). By
(
H
5
b
)
, we can find
R
λ
≫
1
such that
R
λ
>
‖
u
‖
∞
for
λ
>
0
. Let
τ
≥
0
and
u
∈
C
[
0
,
1
]
be a solution of
u
=
T
λ
u
+
τ
. Following the arguments in the proof of Theorem 1.2, we can show that
u
satisfies (7). By
(
H
4
a
)
, we can find
r
λ
≈
0
such that
r
λ
<
min
{
R
λ
,
‖
u
‖
∞
}
for
λ
>
0
. By Proposition 1.1,
T
λ
has a fixed point
v
for
λ
>
0
such that
r
λ
<
‖
v
‖
∞
<
R
λ
. Then
v
is a positive solution of (2) by Lemmas 2.1 and 2.3. Furthermore, we obtain
‖
v
‖
∞
→
0
as
λ
→
0
from (6) and
‖
v
‖
∞
→
∞
as
λ
→
∞
from (7).
Next we show the multiplicity result for
λ
∈
(
λ
α
,
λ
β
)
, where
λ
α
≔
φ
q
(
16
M
a
b
K
a
b
4
)
ϕ
(
α
)
h
∗
f
(
α
)
and
λ
β
≔
ϕ
(
β
)
‖
h
‖
1
φ
q
(
M
a
b
K
a
b
2
)
f
(
β
)
. To this end, we find two additional constants satisfying the conditions in Proposition 1.1. Let
σ
∈
(
0
,
1
]
and
u
∈
C
[
0
,
1
]
be a solution of
u
=
σ
T
λ
u
. Assume
‖
u
‖
∞
=
β
. From (6), we obtain
f
(
β
)
ϕ
(
β
)
≥
1
λ
‖
h
‖
1
φ
q
(
M
a
b
K
a
b
2
)
. This is a contradiction for
λ
<
λ
β
. Hence,
‖
u
‖
∞
≠
β
for
λ
<
λ
β
. Let
τ
≥
0
and
u
∈
C
[
0
,
1
]
be a solution of
u
=
T
λ
u
+
τ
. Assume
‖
u
‖
∞
=
4
α
M
a
b
K
a
b
2
. From (7), we obtain
f
(
α
)
ϕ
(
α
)
≤
φ
q
(
16
M
a
b
K
a
b
4
)
λ
h
∗
. This is a contradiction for
λ
>
λ
α
. Hence,
‖
u
‖
∞
≠
4
α
M
a
b
K
a
b
2
for
λ
>
λ
α
. We note that
λ
α
<
λ
β
by
(
H
6
a
)
and we can choose
r
λ
≈
0
and
R
λ
≫
1
such that
r
λ
<
β
and
R
λ
>
4
α
M
a
b
K
a
b
2
. By Proposition 1.1,
T
λ
has three fixed points
v
1
,
v
2
, and
v
3
for
λ
∈
(
λ
α
,
λ
β
)
such that
r
λ
<
‖
v
1
‖
∞
<
β
<
‖
v
2
‖
∞
<
4
α
M
a
b
K
a
b
2
<
‖
v
3
‖
∞
<
R
λ
. Following the aforementioned arguments, we can show that
v
1
,
v
2
, and
v
3
are positive solutions of (2) for
λ
∈
(
λ
α
,
λ
β
)
.□
Proof of Theorem 1.4
Noting that
inf
s
∈
(
0
,
∞
)
f
(
s
)
ϕ
(
s
)
>
0
by
(
H
4
a
)
and
(
H
5
c
)
, we can prove the nonexistence result for
λ
≫
1
following the arguments in the proof of Theorem 1.2.
Now we show the existence result for
λ
<
λ
∞
≔
1
2
f
∞
‖
h
‖
1
φ
q
(
M
a
b
K
a
b
2
)
. Let
σ
∈
(
0
,
1
]
and
u
∈
C
[
0
,
1
]
be a solution of
u
=
σ
T
λ
u
. By
(
H
5
c
)
, we can find
R
λ
≫
1
from (6) such that
R
λ
>
‖
u
‖
∞
for
λ
<
λ
∞
. Following the arguments in the proof of Theorem 1.3, we can show that if
u
∈
C
[
0
,
1
]
is a solution of
u
=
T
λ
u
+
τ
then
‖
u
‖
∞
>
r
λ
≈
0
for
λ
>
0
. Then
T
λ
has a fixed point
v
∈
C
[
0
,
1
]
for
λ
<
λ
∞
such that
r
λ
<
‖
v
‖
∞
<
R
λ
by Proposition 1.1, and we can show that
v
is a positive solution of (2) for
λ
<
λ
∞
. Furthermore, we obtain
‖
v
‖
∞
→
0
as
λ
→
0
from (6).
Next we show the multiplicity result for
λ
∈
(
λ
α
,
min
{
λ
β
,
λ
∞
}
)
. By similar arguments in the proof of Theorem 1.3, we can show that if
u
∈
C
[
0
,
1
]
is a solution of
u
=
σ
T
λ
u
, then
‖
u
‖
∞
≠
β
for
λ
<
λ
β
. Furthermore, if
u
∈
C
[
0
,
1
]
is a solution of
u
=
T
λ
u
+
τ
, then
‖
u
‖
∞
≠
4
α
M
a
b
K
a
b
2
for
λ
>
λ
α
. We note that
λ
α
<
min
{
λ
β
,
λ
∞
}
by
(
H
6
a
)
and
(
H
6
b
)
. By Proposition 1.1,
T
λ
has three fixed points
v
1
,
v
2
, and
v
3
for
λ
∈
(
λ
α
,
min
{
λ
β
,
λ
∞
}
)
such that
r
λ
<
‖
v
1
‖
∞
<
β
<
‖
v
2
‖
∞
<
4
α
M
a
b
K
a
b
2
<
‖
v
3
‖
∞
<
R
λ
. Hence, we can show that
v
1
,
v
2
, and
v
3
are positive solutions of (2) for
λ
∈
(
λ
α
,
min
{
λ
β
,
λ
∞
}
)
.□
5 Proofs of Theorems 1.8–1.10
Proof of Theorem 1.8
Noting that
inf
s
∈
(
0
,
∞
)
f
(
s
)
ϕ
(
s
)
>
0
by
(
H
4
c
)
and
(
H
5
a
)
, we can prove the nonexistence result for
λ
≫
1
following the arguments in the proof of Theorem 1.2.
Now we show the existence result for
λ
<
λ
0
≔
1
2
f
0
‖
h
‖
1
φ
q
(
M
a
b
K
a
b
2
)
. Let
σ
∈
(
0
,
1
]
and
u
∈
C
[
0
,
1
]
be a solution of
u
=
σ
T
λ
u
. By
(
H
1
)
and
(
H
4
c
)
,
u
(
t
)
=
σ
T
λ
u
(
t
)
≥
0
. Thus,
u
satisfies (6). Then we can find
r
λ
≈
0
such that
r
λ
<
‖
u
‖
∞
for
λ
<
λ
0
by
(
H
4
c
)
. Following the arguments in the proof of Theorem 1.5, if
u
∈
C
[
0
,
1
]
is a solution of
u
=
T
λ
u
+
τ
, then we can find
R
λ
≫
1
such that
R
λ
>
max
{
r
λ
,
‖
u
‖
∞
}
for
λ
>
0
. Thus, there exists a positive solution
v
of (2) for
λ
<
λ
0
. Furthermore, we obtain
‖
v
‖
∞
→
∞
as
λ
→
0
from (6).
Next we show the multiplicity result for
λ
∈
(
λ
0
,
λ
γ
)
, where
λ
0
≔
2
φ
q
(
16
M
a
b
K
a
b
4
)
h
∗
f
0
and
λ
γ
≔
ϕ
(
γ
)
‖
h
‖
1
φ
q
(
M
a
b
K
a
b
2
)
f
(
γ
)
. Let
σ
∈
(
0
,
1
]
and
u
∈
C
[
0
,
1
]
be a solution of
u
=
σ
T
λ
u
. By
(
H
4
c
)
, we can show that
‖
u
‖
∞
≠
γ
for
λ
<
λ
γ
from (6). Let
τ
≥
0
and
u
∈
C
[
0
,
1
]
be a solution of
u
=
T
λ
u
+
τ
. By
(
H
4
c
)
and
(
H
5
a
)
, we can find
r
λ
≈
0
and
R
λ
≫
1
from (7) such that
r
λ
<
min
{
γ
,
‖
u
‖
∞
}
for
λ
>
λ
0
and
R
λ
>
max
{
γ
,
‖
u
‖
∞
}
for
λ
>
0
. We note that
λ
0
<
λ
γ
by
(
H
7
a
)
. Thus, there exist positive solutions
v
1
and
v
2
of (2) for
λ
∈
(
λ
0
,
λ
γ
)
such that
r
λ
<
‖
v
1
‖
∞
<
γ
<
‖
v
2
‖
∞
<
R
λ
.□
Proof of Theorem 1.9
Noting that
sup
s
∈
(
0
,
∞
)
f
(
s
)
ϕ
(
s
)
<
∞
by
(
H
4
c
)
and
(
H
5
b
)
, we can show the nonexistence result for
λ
≈
0
following the arguments in the proof of Theorem 1.6.
Now we show the existence result for
λ
>
λ
0
. Following the arguments in the proof of Theorem 1.3, if
u
∈
C
[
0
,
1
]
is a solution of
u
=
σ
T
λ
u
then we can find
R
λ
≫
1
from (6) such that
R
λ
>
‖
u
‖
∞
for
λ
>
0
. Let
τ
≥
0
and
u
∈
C
[
0
,
1
]
be a solution of
u
=
T
λ
u
+
τ
. By
(
H
4
c
)
, we can find
r
λ
≈
0
from (7) such that
r
λ
<
min
{
R
λ
,
‖
u
‖
∞
}
for
λ
>
λ
0
. Thus, there exists a positive solution
v
of (2) for
λ
>
λ
0
. Furthermore, we obtain
‖
v
‖
∞
→
∞
as
λ
→
∞
from (7).
Next we show the multiplicity result for
λ
∈
(
λ
δ
,
λ
0
)
, where
λ
δ
≔
φ
q
(
16
M
a
b
K
a
b
4
)
ϕ
(
δ
)
h
∗
f
(
δ
)
. Let
τ
≥
0
and
u
∈
C
[
0
,
1
]
be a solution of
u
=
T
λ
u
+
τ
. By
(
H
4
c
)
, we can show that
‖
u
‖
∞
≠
4
δ
M
a
b
K
a
b
2
for
λ
>
λ
δ
from (7). Let
σ
∈
(
0
,
1
]
and
u
∈
C
[
0
,
1
]
be a solution of
u
=
σ
T
λ
u
. By
(
H
4
c
)
and
(
H
5
b
)
, we can find
r
λ
≈
0
and
R
λ
≫
1
from (6) such that
r
λ
<
min
{
4
δ
M
a
b
K
a
b
2
,
‖
u
‖
∞
}
for
λ
<
λ
0
and
R
λ
>
max
{
4
δ
M
a
b
K
a
b
2
,
‖
u
‖
∞
}
for
λ
>
0
. We note that
λ
δ
<
λ
0
by
(
H
7
b
)
. Thus, there exist positive solutions
v
1
and
v
2
of (2) for
λ
∈
(
λ
δ
,
λ
0
)
such that
r
λ
<
‖
v
1
‖
∞
<
4
δ
M
a
b
K
a
b
2
<
‖
v
2
‖
∞
<
R
λ
.□
Proof of Theorem 1.10
We note that
0
<
inf
s
∈
(
0
,
∞
)
f
(
s
)
ϕ
(
s
)
<
sup
s
∈
(
0
,
∞
)
f
(
s
)
ϕ
(
s
)
<
∞
by
(
H
4
c
)
and
(
H
5
c
)
. Then we can prove the nonexistence results for
λ
≈
0
and
λ
≫
1
following the arguments in the proofs of Theorems 1.6 and 1.2, respectively.
Now we show the existence result for
λ
∈
(
min
{
λ
∞
,
λ
0
}
,
max
{
λ
∞
,
λ
0
}
)
. Without loss of generality, we assume
f
0
≤
f
∞
. Let
σ
∈
(
0
,
1
]
and
u
∈
C
[
0
,
1
]
be a solution of
u
=
σ
T
λ
u
. By
(
H
4
c
)
, we can find
r
λ
≈
0
from (6) such that
r
λ
<
‖
u
‖
∞
for
λ
<
λ
0
. Let
τ
≥
0
and
u
∈
C
[
0
,
1
]
be a solution of
u
=
T
λ
u
+
τ
. By
(
H
5
c
)
, we can find
R
λ
≫
1
from (7) such that
R
λ
>
max
{
r
λ
,
‖
u
‖
∞
}
for
λ
>
λ
∞
. Thus, there exists a positive solution
v
of (2) for
λ
∈
(
λ
∞
,
λ
0
)
.
Next we show the multiplicity result for
λ
∈
(
max
{
λ
∞
,
λ
0
}
,
λ
ζ
)
, where
λ
ζ
≔
ϕ
(
ζ
)
‖
h
‖
1
φ
q
(
M
a
b
K
a
b
2
)
f
(
ζ
)
. Let
σ
∈
(
0
,
1
]
and
u
∈
C
[
0
,
1
]
be a solution of
u
=
σ
T
λ
u
. From (6), we have
‖
u
‖
∞
≠
ζ
for
λ
<
λ
ζ
. Let
τ
≥
0
and
u
∈
C
[
0
,
1
]
be a solution of
u
=
T
λ
u
+
τ
. By
(
H
4
c
)
and
(
H
5
c
)
, we can find
r
λ
≈
0
and
R
λ
≫
1
from (7) such that
r
λ
<
min
{
ζ
,
‖
u
‖
∞
}
for
λ
>
λ
0
and
R
λ
>
max
{
ζ
,
‖
u
‖
∞
}
for
λ
>
λ
∞
. We note that
max
{
λ
∞
,
λ
0
}
<
λ
ζ
by
(
H
7
c
)
. Thus, there exist positive solutions
v
1
and
v
2
of (2) for
λ
∈
(
max
{
λ
∞
,
λ
0
}
,
λ
ζ
)
such that
r
λ
<
‖
v
1
‖
∞
<
ζ
<
‖
v
2
‖
∞
<
R
λ
.
Finally, we show the multiplicity result for
λ
∈
(
λ
η
,
min
{
λ
∞
,
λ
0
}
)
, where
λ
η
≔
φ
q
(
16
M
a
b
K
a
b
4
)
ϕ
(
η
)
h
∗
f
(
η
)
. Let
τ
≥
0
and
u
∈
C
[
0
,
1
]
be a solution of
u
=
T
λ
u
+
τ
. From (7), we have
‖
u
‖
∞
≠
4
η
M
a
b
K
a
b
2
for
λ
>
λ
η
. Let
σ
∈
(
0
,
1
]
and
u
∈
C
[
0
,
1
]
be a solution of
u
=
σ
T
λ
u
. By
(
H
4
c
)
and
(
H
5
c
)
, we can find
r
λ
≈
0
and
R
λ
≫
1
from (6) such that
r
λ
<
min
{
4
η
M
a
b
K
a
b
2
,
‖
u
‖
∞
}
for
λ
<
λ
0
and
R
λ
>
max
{
4
η
M
a
b
K
a
b
2
,
‖
u
‖
∞
}
for
λ
<
λ
∞
. We note that
λ
η
<
min
{
λ
∞
,
λ
0
}
by
(
H
7
d
)
. Thus, there exist positive solutions
v
1
and
v
2
of (2) for
λ
∈
(
λ
η
,
min
{
λ
∞
,
λ
0
}
)
such that
r
λ
<
‖
v
1
‖
∞
<
4
η
M
a
b
K
a
b
2
<
‖
v
2
‖
∞
<
R
λ
.□
6 Proofs of Theorems 1.11–1.13
In this section, we make use of the fact that
M
a
b
∗
K
a
b
>
M
a
b
K
a
b
≥
1
in several places.
Proof of Theorem 1.11
We first show the existence result for
λ
≈
0
. Let
σ
∈
(
0
,
1
]
and
u
∈
C
[
0
,
1
]
be a solution of
u
=
σ
T
λ
u
. Assume
‖
u
‖
∞
=
K
λ
≔
max
{
ρ
,
M
a
b
K
a
b
ϕ
−
1
(
2
λ
‖
h
‖
1
∣
f
(
0
)
∣
)
}
. Since
‖
T
λ
u
‖
∞
≥
‖
u
‖
∞
, we obtain
T
λ
u
(
t
)
≥
0
by Lemma 2.4. This implies
u
(
t
)
≥
0
. Then we can show that
u
satisfies (6) following the arguments in the proof of Theorem 1.2. However, this is a contradiction for
λ
≈
0
since
f
(
K
λ
)
=
0
for
λ
≈
0
. Hence,
‖
u
‖
∞
≠
K
λ
for
λ
≈
0
.
Let
τ
≥
0
and
u
∈
C
[
0
,
1
]
be a solution of
u
=
T
λ
u
+
τ
. We show that there exists a constant that is greater than
K
λ
and satisfies
(
b
)
in Proposition 1.1. Assume
‖
u
‖
∞
≥
K
λ
∗
≔
max
{
K
λ
,
8
ρ
M
a
b
∗
K
a
b
2
}
. It is easy to show that
u
satisfies (4) by
(
H
3
)
. Thus,
u
(
t
)
≥
ρ
for
t
∈
1
8
,
7
8
by Lemma 2.4. If
t
m
≥
1
2
, then
‖
u
‖
∞
=
u
1
8
+
∫
1
8
t
m
a
1
q
−
p
(
s
)
b
−
1
q
−
p
(
s
)
ϕ
−
1
λ
a
−
q
−
1
q
−
p
(
s
)
b
p
−
1
q
−
p
(
s
)
∫
s
t
m
h
(
r
)
f
(
u
)
d
r
d
s
≥
∫
1
8
1
4
a
1
q
−
p
(
s
)
b
−
1
q
−
p
(
s
)
ϕ
−
1
λ
a
−
q
−
1
q
−
p
(
s
)
b
p
−
1
q
−
p
(
s
)
∫
1
4
1
2
h
(
r
)
f
(
u
)
d
r
d
s
≥
1
8
K
a
b
2
ϕ
−
1
λ
h
∗
f
‖
u
‖
∞
4
M
a
b
∗
K
a
b
2
.
By similar arguments, we can show that if
t
m
<
1
2
then
‖
u
‖
∞
≥
1
8
K
a
b
2
ϕ
−
1
λ
h
∗
f
‖
u
‖
∞
4
M
a
b
∗
K
a
b
2
. Then
λ
h
∗
f
‖
u
‖
∞
4
M
a
b
∗
K
a
b
2
≤
ϕ
(
8
K
a
b
2
‖
u
‖
∞
)
≤
φ
q
(
32
M
a
b
∗
K
a
b
4
)
ϕ
‖
u
‖
∞
4
M
a
b
∗
K
a
b
2
. This implies
(8)
f
‖
u
‖
∞
4
M
a
b
∗
K
a
b
2
ϕ
‖
u
‖
∞
4
M
a
b
∗
K
a
b
2
≤
φ
q
(
32
M
a
b
∗
K
a
b
4
)
λ
h
∗
.
By
(
H
5
a
)
, there exists
R
λ
≫
1
such that
R
λ
>
max
{
K
λ
∗
,
‖
u
‖
∞
}
for
λ
>
0
.
By Proposition 1.1 and Lemma 2.1,
T
λ
has a fixed point
v
∈
C
1
(
0
,
1
)
∩
C
[
0
,
1
]
for
λ
≈
0
such that
K
λ
<
‖
v
‖
∞
<
R
λ
. Then
v
(
t
)
>
0
on
(
0
,
1
)
by Lemma 2.4. Thus,
v
is a positive solution of (2). Furthermore, we obtain
‖
v
‖
∞
→
∞
as
λ
→
0
from (6).
Next we show the multiplicity result for
λ
∈
(
λ
θ
,
min
{
λ
κ
,
λ
θ
∗
}
)
, where
λ
θ
≔
φ
q
(
32
M
a
b
∗
K
a
b
4
)
ϕ
(
θ
)
h
∗
f
(
θ
)
,
λ
κ
≔
ϕ
(
κ
)
‖
h
‖
1
φ
q
(
M
a
b
K
a
b
2
)
f
(
κ
)
and
λ
θ
∗
≔
φ
p
(
4
K
a
b
)
ϕ
(
θ
)
‖
h
‖
1
∣
f
(
0
)
∣
. We note that
λ
θ
<
min
{
λ
κ
,
λ
θ
∗
}
by
(
H
8
a
)
and
(
H
8
b
)
. Furthermore, by the definition of
λ
θ
∗
, if
λ
<
λ
θ
∗
, then we have
λ
<
φ
p
(
4
K
a
b
)
ϕ
(
θ
)
‖
h
‖
1
∣
f
(
0
)
∣
≤
φ
p
2
1
p
−
1
4
θ
K
a
b
+
φ
q
2
1
p
−
1
4
θ
K
a
b
2
‖
h
‖
1
∣
f
(
0
)
∣
=
ϕ
2
1
p
−
1
4
θ
K
a
b
2
‖
h
‖
1
∣
f
(
0
)
∣
.
This implies
4
θ
M
a
b
∗
K
a
b
2
>
M
a
b
K
a
b
ϕ
−
1
(
2
λ
‖
h
‖
1
∣
f
(
0
)
∣
)
for
λ
<
λ
θ
∗
. Since
κ
>
4
θ
M
a
b
∗
K
a
b
2
and
θ
≥
2
ρ
by
(
H
8
b
)
,
we obtain
κ
>
4
θ
M
a
b
∗
K
a
b
2
≥
K
λ
∗
≥
K
λ
for
λ
<
λ
θ
∗
.
Let
σ
∈
(
0
,
1
]
and
u
∈
C
[
0
,
1
]
be a solution of
u
=
σ
T
λ
u
. Assume
‖
u
‖
∞
=
κ
for
λ
<
min
{
λ
κ
,
λ
θ
∗
}
. Since
κ
>
K
λ
for
λ
<
λ
θ
∗
,
u
(
t
)
≥
0
by Lemma 2.4. Thus,
u
satisfies (6) and we have
f
(
κ
)
ϕ
(
κ
)
≥
1
λ
‖
h
‖
1
φ
q
(
M
a
b
K
a
b
2
)
. However, this is a contradiction since
λ
<
λ
κ
. Hence,
‖
u
‖
∞
≠
κ
for
λ
<
min
{
λ
κ
,
λ
θ
∗
}
. Let
τ
≥
0
and
u
∈
C
[
0
,
1
]
be a solution of
u
=
T
λ
u
+
τ
. Assume
‖
u
‖
∞
=
4
θ
M
a
b
∗
K
a
b
2
for
λ
∈
(
λ
θ
,
λ
θ
∗
)
. Since
4
θ
M
a
b
∗
K
a
b
2
≥
K
λ
for
λ
<
λ
θ
∗
,
u
(
t
)
≥
0
by Lemma 2.4. Thus,
u
satisfies (8) and we have
f
(
θ
)
ϕ
(
θ
)
≤
φ
q
(
32
M
a
b
∗
K
a
b
4
)
λ
h
∗
. However, this is a contradiction since
λ
>
λ
θ
. Hence,
‖
u
‖
∞
≠
4
θ
M
a
b
∗
K
a
b
2
for
λ
∈
(
λ
θ
,
λ
θ
∗
)
. We note that we can choose
R
λ
≫
1
such that
R
λ
>
κ
. By Proposition 1.1,
T
λ
has two fixed points
v
1
and
v
2
for
λ
∈
(
λ
θ
,
min
{
λ
κ
,
λ
θ
∗
}
)
such that
4
θ
M
a
b
∗
K
a
b
2
<
‖
v
1
‖
∞
<
κ
<
‖
v
2
‖
∞
<
R
λ
. Since
4
θ
M
a
b
∗
K
a
b
2
>
K
λ
,
v
1
(
t
)
>
0
, and
v
2
(
t
)
>
0
on
(
0
,
1
)
by Lemma 2.4. Hence,
v
1
and
v
2
are positive solutions of (2) for
λ
∈
(
λ
θ
,
min
{
λ
κ
,
λ
θ
∗
}
)
.□
Proof of Theorem 1.12
Noting that
sup
s
∈
(
0
,
∞
)
f
(
s
)
ϕ
(
s
)
<
∞
by
(
H
4
d
)
and
(
H
5
b
)
, we can show the nonexistence result for
λ
≈
0
following the arguments in the proof of Theorem 1.6.
Next we show the existence result for
λ
≫
1
. Let
τ
≥
0
and
u
∈
C
[
0
,
1
]
be a solution of
u
=
T
λ
u
+
τ
. If
‖
u
‖
∞
=
K
¯
λ
≔
4
M
a
b
∗
K
a
b
2
ϕ
−
1
(
2
λ
‖
h
‖
1
∣
f
(
0
)
∣
)
then
‖
u
‖
∞
≥
K
λ
∗
for
λ
≫
1
. Thus,
u
satisfies (8). Then we obtain
f
(
ϕ
−
1
(
2
λ
‖
h
‖
1
∣
f
(
0
)
∣
)
)
≤
2
‖
h
‖
1
∣
f
(
0
)
∣
φ
q
(
32
M
a
b
∗
K
a
b
4
)
h
∗
. However, this is a contradiction for
λ
≫
1
since
lim
s
→
∞
f
(
s
)
=
∞
. Thus,
‖
u
‖
∞
≠
K
¯
λ
for
λ
≫
1
. Let
σ
∈
(
0
,
1
]
and
u
∈
C
[
0
,
1
]
be a solution of
u
=
σ
T
λ
u
. Assume
‖
u
‖
∞
≥
K
¯
λ
. Then
u
(
t
)
≥
0
by Lemma 2.4. Thus,
u
satisfies (6) and we can find
R
λ
≫
1
such that
R
λ
>
max
{
K
¯
λ
,
‖
u
‖
∞
}
for
λ
>
0
by
(
H
5
b
)
. Hence, (2) has a positive solution
v
for
λ
≫
1
such that
K
¯
λ
<
‖
v
‖
∞
<
R
λ
by Proposition 1.1, Lemmas 2.1, and 2.4. Furthermore, we obtain
‖
v
‖
∞
→
∞
as
λ
→
∞
from (8).
Now we show the multiplicity result for
λ
∈
(
λ
θ
,
λ
κ
)
. We note that
λ
θ
<
λ
κ
by
(
H
8
a
)
. Furthermore, since
f
(
κ
)
>
2
∣
f
(
0
)
∣
by
(
H
8
c
)
, if
λ
<
λ
κ
, then
(9)
λ
<
ϕ
(
κ
)
‖
h
‖
1
φ
q
(
M
a
b
K
a
b
2
)
f
(
κ
)
<
ϕ
(
κ
)
2
‖
h
‖
1
φ
q
(
M
a
b
K
a
b
2
)
∣
f
(
0
)
∣
≤
ϕ
(
κ
M
a
b
K
a
b
)
2
‖
h
‖
1
∣
f
(
0
)
∣
.
This implies
κ
>
M
a
b
K
a
b
ϕ
−
1
(
2
λ
‖
h
‖
1
∣
f
(
0
)
∣
)
for
λ
<
λ
κ
.
Since
4
θ
M
a
b
∗
K
a
b
2
>
κ
and
κ
≥
8
ρ
M
a
b
∗
K
a
b
2
, we have
4
θ
M
a
b
∗
K
a
b
2
>
κ
≥
K
λ
∗
≥
K
λ
for
λ
<
λ
κ
.
Let
σ
∈
(
0
,
1
]
and
u
∈
C
[
0
,
1
]
be a solution of
u
=
σ
T
λ
u
. Assume
‖
u
‖
∞
=
κ
for
λ
<
λ
κ
. Since
κ
>
K
λ
,
u
(
t
)
≥
0
and
u
satisfies (6). Thus, we obtain
f
(
κ
)
ϕ
(
κ
)
≥
1
λ
‖
h
‖
1
φ
q
(
M
a
b
K
a
b
2
)
. However, this is a contradiction since
λ
<
λ
κ
. Hence,
‖
u
‖
∞
≠
κ
for
λ
<
λ
κ
. Let
τ
≥
0
and
u
∈
C
[
0
,
1
]
be a solution of
u
=
T
λ
u
+
τ
. Assume
‖
u
‖
∞
=
4
θ
M
a
b
∗
K
a
b
2
for
λ
<
λ
κ
. Then we obtain
f
(
θ
)
ϕ
(
θ
)
≤
φ
q
(
32
M
a
b
∗
K
a
b
2
)
λ
h
∗
from (8). However, this is a contradiction since
λ
>
λ
θ
. Hence,
‖
u
‖
∞
≠
4
θ
M
a
b
∗
K
a
b
2
for
λ
∈
(
λ
θ
,
λ
κ
)
. We note that we can choose
R
λ
≫
1
such that
R
λ
>
4
θ
M
a
b
∗
K
a
b
2
. Thus, (2) has positive solutions
v
1
and
v
2
for
λ
∈
(
λ
θ
,
λ
κ
)
such that
κ
<
‖
v
1
‖
∞
<
4
θ
M
a
b
∗
K
a
b
2
<
‖
v
2
‖
∞
<
R
λ
.□
Proof of Theorem 1.13
Noting that
sup
s
∈
(
0
,
∞
)
f
(
s
)
ϕ
(
s
)
<
∞
by
(
H
4
d
)
and
(
H
5
c
)
, we can show the nonexistence result for
λ
≈
0
following the arguments in the proof of Theorem 1.6.
Next we show the existence result for
λ
∈
(
λ
̲
,
λ
μ
)
, where
λ
̲
≔
2
φ
q
(
32
M
a
b
∗
K
a
b
4
)
h
∗
f
∞
and
λ
μ
≔
ϕ
(
μ
)
‖
h
‖
1
φ
q
(
M
a
b
K
a
b
2
)
f
(
μ
)
. We note that
λ
̲
<
λ
μ
by
(
H
8
d
)
and
μ
≥
K
λ
for
λ
<
λ
μ
by similar arguments in (9) to
(
H
8
d
)
. Let
σ
∈
(
0
,
1
]
and
u
∈
C
[
0
,
1
]
be a solution of
u
=
σ
T
λ
u
. If
‖
u
‖
∞
=
μ
for
λ
<
λ
μ
, then
u
satisfies (6). Thus, we have
f
(
μ
)
ϕ
(
μ
)
≥
1
λ
‖
h
‖
1
φ
q
(
M
a
b
K
a
b
2
)
. However, this is a contradiction since
λ
<
λ
μ
. Hence,
‖
u
‖
∞
≠
μ
for
λ
<
λ
μ
. Let
τ
≥
0
and
u
∈
C
[
0
,
1
]
be a solution of
u
=
T
λ
u
+
τ
. Assume
‖
u
‖
∞
≥
K
λ
∗
. Then
u
satisfies (8) and we can find
R
λ
≫
1
such that
R
λ
>
max
{
μ
,
‖
u
‖
∞
}
for
λ
>
λ
̲
by
(
H
5
c
)
. Hence, (2) has a positive solution for
λ
∈
(
λ
̲
,
λ
μ
)
.□
7 Application: Generalized double phase problems on the exterior of a ball
In this section, we consider the generalized double phase problems on the exterior of a ball:
(10)
−
div
(
a
ˆ
(
∣
x
∣
)
∣
∇
u
∣
p
−
2
∇
u
+
b
ˆ
(
∣
x
∣
)
∣
∇
u
∣
q
−
2
∇
u
)
=
λ
K
(
∣
x
∣
)
f
(
u
)
,
x
∈
Ω
,
u
=
0
,
∣
x
∣
=
r
0
,
u
→
0
,
∣
x
∣
→
∞
,
where
λ
>
0
,
1
<
p
<
q
<
N
,
Ω
≔
{
x
∈
R
N
∣
∣
x
∣
>
r
0
>
0
}
,
f
satisfies
(
H
3
)
,
K
:
(
r
0
,
∞
)
→
(
0
,
∞
)
is continuous such that
K
(
r
)
≤
1
r
N
+
ν
for some
ν
>
0
, and
a
ˆ
,
b
ˆ
:
[
r
0
,
∞
)
→
[
0
,
∞
)
are continuous such that
a
ˆ
(
r
)
+
b
(
r
)
ˆ
>
0
for
r
∈
(
r
0
,
∞
)
,
a
ˆ
(
r
)
b
(
r
)
ˆ
>
0
for a.e.
r
∈
[
r
0
,
∞
)
and
∫
r
0
∞
r
−
N
−
1
q
−
1
a
ˆ
(
r
)
−
1
p
−
1
b
(
r
)
ˆ
−
1
q
−
1
d
r
<
∞
.
The Kelvin-type transformation, namely
r
=
∣
x
∣
and
t
=
r
r
0
N
−
p
1
−
p
reduces the study of positive radial solutions to (10) to the study of positive solutions to the boundary value problem:
(11)
−
(
a
∗
(
t
)
φ
p
(
u
′
)
+
b
∗
(
t
)
φ
q
(
u
′
)
)
′
=
λ
k
(
t
)
f
(
u
)
,
t
∈
(
0
,
1
)
,
u
(
0
)
=
0
=
u
(
1
)
,
where
a
∗
(
t
)
≔
a
ˆ
r
0
t
1
−
p
N
−
p
,
b
∗
(
t
)
≔
N
−
p
r
0
(
p
−
1
)
q
−
p
t
(
N
−
1
)
(
q
−
p
)
N
−
p
b
ˆ
r
0
t
1
−
p
N
−
p
, and
k
(
t
)
≔
r
0
(
p
−
1
)
N
−
p
p
t
p
(
1
−
N
)
N
−
p
K
r
0
t
1
−
p
N
−
p
. Then
a
∗
and
b
∗
satisfy
(
H
1
)
and
k
satisfies
(
H
2
)
. Thus, if
f
satisfies the assumptions of each theorem, then Theorems 1.2–1.13 for (11) are valid. Hence, we obtain various existence results for positive radial solutions for (10).
