# Some Existence Results of Positive Solutions forφ-Laplacian Systems

Authors: Xianghui Xu; Yong-Hoon Lee
Journal: Abstract and Applied Analysis
 (2014)

---

## Abstract

We study the existence of positive solutions for the homogeneous Dirichlet boundary value problem ofφ-Laplacian systems with a singular weight which may not be in L
1.


---

## Body

## 1. Introduction

In this paper, we study nonlinear differential systems of the form(
P
)
-
Φ
(
u
′
)
′
=
h
(
t
)
·
f
(
u
)
,
t
∈
(
0,1
)
,
u
(
0
)
=
0
=
u
(
1
)
,
where Φ
(
u
′
)
=
(
φ
(
u
1
′
)
,
…
,
φ
(
u
N
′
)
) with φ
:
ℝ
→
ℝ an odd increasing homeomorphism, h
(
t
)
=
(
h
1
(
t
)
,
…
,
h
N
(
t
)
) with h
i
:
(
0,1
)
→
ℝ
+, h
i
≢
0 on any subinterval in (
0,1
), and f
(
u
)
=
(
f
1
(
u
)
,
…
,
f
N
(
u
)
) with f
i
:
ℝ
+
N
→
ℝ
+; here we denote ℝ
+
=
[
0
,
+
∞
), ℝ
+
N
=
ℝ
+
×
⋯
×
ℝ
+
︸
N, and x
·
y
=
(
x
1
y
1
,
x
2
y
2
,
…
,
x
N
y
N
) the Hadamard product of x and y in ℝ
N. Thus problem (
P
) can be rewritten as
(1)
i
-
φ
(
u
1
′
)
′
=
h
1
(
t
)
f
1
(
u
)
,
-
φ
(
u
1
′
)
′
i
i
i
⋮
-
φ
(
u
N
′
)
′
=
h
N
(
t
)
f
N
(
u
)
,
t
∈
(
0,1
)
,
iiiiii
u
i
(
0
)
=
0
=
u
i
(
1
)
,
i
=
1
,
…
,
N
.We first give assumptions onφ and h.(A)
There exist an increasing homeomorphismψ of (
0
,
∞
) onto (
0
,
∞
) and a function γ of (
0
,
∞
) into (
0
,
∞
) such that
(2)
ψ
(
σ
)
≤
φ
(
σ
x
)
φ
(
x
)
≤
γ
(
σ
)
,
∀
σ
>
0
,
x
∈
ℝ
.
(H)
h
i
:
(
0,1
)
→
ℝ
+ is locally integrable satisfying
(3)
∫
0
1
/
2
‍
ψ
-
1
(
∫
s
1
/
2
‍
h
i
(
τ
)
d
τ
)
d
s
+
∫
1
/
2
1
‍
ψ
-
1
(
∫
1
/
2
s
‍
h
i
(
τ
)
d
τ
)
d
s
<
∞
,
fori
=
1
,
…
,
N.For convenience, we introduce a new class of weight functions. For a bijectionι
:
ℝ
→
ℝ, define ℋ
ι as a subset of L
loc
⁡
1
(
(
0,1
)
,
ℝ
+
) given by
(4)
ℋ
ι
=
{
g
∣
∫
0
1
/
2
‍
ι
-
1
(
∫
s
1
/
2
‍
g
(
τ
)
d
τ
)
d
s
i
i
i
i
i
i
i
i
i
+
∫
1
/
2
1
‍
ι
-
1
(
∫
1
/
2
s
‍
g
(
τ
)
d
τ
)
d
s
<
∞
}
.
By the notation, condition (
H
) means h
i
∈
ℋ
ψ.The case ofp-Laplace operator, namely, φ
(
x
)
=
φ
p
(
x
)
:
=
|
x
|
p
-
2
x, x
∈
ℝ, p
>
1, satisfies condition (A) with ψ
≡
φ
p
≡
γ. We give one more example of φ and h satisfying conditions (A) and (H).Example 1.
Defineφ
:
ℝ
→
ℝ as an odd function with
(5)
φ
(
x
)
=
x
2
+
x
,
x
≥
0
.
Then φ is obviously an increasing homeomorphism. Define functions ψ and γ given as
(6)
ψ
(
σ
)
=
{
σ
2
,
if
0
<
σ
≤
1
,
σ
,
if
σ
>
1
,
γ
(
σ
)
=
{
1
,
if
0
<
σ
≤
1
,
σ
2
,
if
σ
>
1
.
Then ψ
,
γ
:
(
0
,
∞
)
→
(
0
,
∞
) and ψ is an increasing homeomorphism. This implies that φ satisfies condition (A). Moreover, for h
(
t
)
=
t
-
3
/
2, we can easily calculate to see h
∈
ℋ
ψ.We note thath given in the example above is not integrable near a boundary t
=
0; that is, h
∉
L
1
(
0,1
), and, in this paper, we focus on studying generalized Laplacian systems of condition (
A
) with singular weights which may not be in L
1
(
0,1
). We now give assumptions on f.(F)
f
i
:
ℝ
+
N
→
ℝ
+ is continuous, i
=
1
,
…
,
N.Problems ofp-Laplacian or more generalized ones like problem (
P
) appear in various applications which describe reaction-diffusion systems, nonlinear elasticity, glaciology, population biology, combustion theory, and non-Newtonian fluids (see [1–4]). Recently there is a vast literature related to existence, multiplicity, or nonexistence of positive solutions of problem (
P
) for either p-Laplacian or more generalized Laplacian problems (see [5–11] and the references therein). Specially, for generalized Laplacian problems, one may refer to works of Agarwal et al. (see [12–14]). Let us denote
(7)
f
0
:
=
∑
i
=
1
N
‍
f
0
i
,
f
∞
:
=
∑
i
=
1
N
‍
f
∞
i
,
where
(8)
f
0
i
:
=
lim
⁡
∥
u
∥
→
0
f
i
(
u
)
φ
(
∥
u
∥
)
,
f
∞
i
:
=
lim
⁡
∥
u
∥
→
∞
f
i
(
u
)
φ
(
∥
u
∥
)
,
for all u
∈
ℝ
+
N and i
=
1
,
…
,
N.Among the variety of works mentioned above, we are interested in the following result.Res A. Problem (
P
) has at least one positive solution if either f
0
=
0, f
∞
=
∞ or f
0
=
∞, f
∞
=
0.Wang [10] proved Res A when each h
i
:
[
0,1
]
→
ℝ
+ is continuous and φ satisfies that there exist two increasing homeomorphisms ψ
1 and ψ
2 of (
0
,
∞
) onto (
0
,
∞
) such that
(9)
ψ
1
(
σ
)
φ
(
x
)
≤
φ
(
σ
x
)
≤
ψ
2
(
σ
)
φ
(
x
)
,
for
σ
,
x
>
0
.Do Ó et al. [7] also proved Res A when φ
=
φ
p and each h
i
∈
ℋ
φ
p
(
=
ℋ
ψ
).The aim of this paper is to prove Res A whenφ satisfies condition (A) and each h
i
∈
ℋ
ψ. More precisely, we state our main theorem as follows.Theorem 2.
Assume (A), (H), and (F) hold. Then problem (
P
) satisfies Res A.Extension of results in [10] or [7] to Theorem 2 is not obvious mainly due to the singularity of h
i in comparison with Wang and lack of homogeneity of the general operator φ in comparison with Do et al.For proofs, we introduce a newly developed solution operator for(
P
) motivated by Sim and Lee [15]. And then we make use of the fixed point theorem of a cone for the existence of positive solutions.This paper is organized as follows. In Section2, we introduce a solution operator for problem (
P
) and prove the compactness of the operator. In Section 3, we prove our main theorem.


## 2. A Solution Operator

Let us consider a simple scalar problem of the form(
W
)
-
φ
(
w
′
)
′
=
g
(
t
)
,
t
∈
(
0,1
)
,
(
D
)
w
(
0
)
=
w
(
1
)
=
0
,
where φ satisfies (A) and g
≥
0 with g
∈
ℋ
φ.Sinceg may not be in L
1
(
0,1
) as we see the example in the introduction section, in this case, the solution of (
W
) + (
D
) may not be in C
1
[
0,1
]. So by a solution to this problem, we understand a function w
∈
C
0
[
0,1
]
∩
C
1
(
0,1
) with φ
(
w
′
) absolutely continuous which satisfies (
W
).We first give some remarks for calculations later on.Remark 3.
From condition (A), we get
(10)
σ
x
≤
φ
-
1
[
γ
(
σ
)
φ
(
x
)
]
,
φ
-
1
[
σ
φ
(
x
)
]
≤
ψ
-
1
(
σ
)
x
,
for
σ
,
x
>
0
.Remark 4.
Leth
∈
L
loc
⁡
1
(
(
0,1
)
,
ℝ
+
). Then for any fixed s
∈
(
0,1
/
2
), we know ∫
s
1
/
2
‍
h
(
τ
)
d
τ
<
∞. Applying σ
=
∫
s
1
/
2
‍
h
(
τ
)
d
τ and x
=
φ
-
1
(
1
) in Remark 3, we get
(11)
φ
-
1
(
∫
s
1
/
2
‍
h
(
τ
)
d
τ
)
≤
φ
-
1
(
1
)
ψ
-
1
(
∫
s
1
/
2
‍
h
(
τ
)
d
τ
)
.
This implies ℋ
ψ
⊂
ℋ
φ.Remark 5.
Ifh
∈
ℋ
φ, then, for any fixed σ
∈
(
0,1
),
(12)
φ
-
1
(
∫
s
σ
‍
h
(
τ
)
d
τ
)
∈
L
1
(
0
,
1
2
]
,
φ
-
1
(
∫
σ
s
‍
h
(
τ
)
d
τ
)
∈
L
1
[
1
2
,
1
)
.We need a lemma which guarantees concavity of solutions. The proof is similar to Lemma 2.3 in Wang [10].Lemma 6.
Letw
∈
C
0
[
0,1
]
∩
C
1
(
0,1
) satisfy φ
(
w
′
)
′
≤
0 on (
0,1
). Then w is concave on [
0,1
] and min
⁡
t
∈
[
1
/
4,3
/
4
]
w
(
t
)
≥
(
1
/
4
)
∥
w
∥
∞, where ∥
w
∥
∞ is the supremum norm of w.Letw be a solution of (
W
) + (
D
).Then integrating both sides of(
W
) on the interval [
s
,
1
/
2
] for s
∈
(
0,1
/
2
] and [
1
/
2
,
s
] for s
∈
[
1
/
2,1
), respectively, we find that (
W
) + (
D
) is equivalent to
(13)
w
′
(
s
)
=
φ
-
1
(
a
+
∫
s
1
/
2
‍
g
(
τ
)
d
τ
)
,
w
(
0
)
=
0
,
s
∈
(
0
,
1
2
]
,
w
′
(
s
)
=
φ
-
1
(
-
a
+
∫
1
/
2
s
‍
g
(
τ
)
d
τ
)
,
w
(
1
)
=
0
,
s
∈
[
1
2
,
1
)
,
where a
=
φ
(
w
′
(
1
/
2
)
). We show that φ
-
1
(
a
+
∫
s
1
/
2
‍
g
(
τ
)
d
τ
)
∈
L
1
(
0,1
/
2
]. Indeed, by Lemma 6, solution w has a unique maximal point. That is, there exists a unique σ
w
∈
(
0,1
) such that w
(
σ
w
)
=
max
⁡
t
∈
[
0,1
]
w
(
t
). Since w
′
(
σ
w
)
=
0, we see from (13) that
(14)
φ
-
1
(
a
+
∫
σ
w
1
/
2
‍
g
(
τ
)
d
τ
)
=
0
.
Since φ is an odd homeomorphism, a
+
∫
σ
w
1
/
2
‍
g
(
τ
)
d
τ
=
0, and by Remark 5, we get
(15)
φ
-
1
(
a
+
∫
s
1
/
2
‍
g
(
τ
)
d
τ
)
=
φ
-
1
(
-
∫
σ
w
1
/
2
‍
g
(
τ
)
d
τ
+
∫
s
1
/
2
‍
g
(
τ
)
d
τ
)
=
φ
-
1
(
∫
s
σ
‍
h
(
τ
)
d
τ
)
∈
L
1
(
0
,
1
2
]
.
Similar argument shows that φ
-
1
(
-
a
+
∫
1
/
2
s
‍
g
(
τ
)
d
τ
)
∈
L
1
[
1
/
2,1
). Now we integrate both sides of (13) on the interval [
0
,
t
] for t
∈
[
0,1
/
2
] and on the interval [
t
,
1
] for t
∈
[
1
/
2,1
], respectively. Then we get
(16)
w
(
t
)
=
{
∫
0
t
‍
φ
-
1
(
a
+
∫
s
1
/
2
‍
g
(
τ
)
d
τ
)
d
s
,
t
∈
[
0
,
1
2
]
,
∫
t
1
‍
φ
-
1
(
-
a
+
∫
1
/
2
s
‍
g
(
τ
)
d
τ
)
d
s
,
t
∈
[
1
2
,
1
]
.Let us checkw
(
1
/
2
)
-
=
w
(
1
/
2
)
+. For a
∈
ℝ, define
(17)
G
(
a
)
=
∫
0
1
/
2
‍
φ
-
1
(
a
+
∫
s
1
/
2
‍
g
(
τ
)
d
τ
)
d
s
-
∫
1
/
2
1
‍
φ
-
1
(
-
a
+
∫
1
/
2
s
‍
g
(
τ
)
d
τ
)
d
s
.
Then the function G
:
ℝ
→
ℝ is well-defined. If G has a unique zero, then w
(
1
/
2
)
-
=
w
(
1
/
2
)
+. For this, we give the following lemma. The proof generally follows the lines of proof of Lemma 2.2 in Sim and Lee [15].Lemma 7.
For giveng
∈
ℋ
φ, the function G defined in (17) has a unique zero a
=
a
(
g
) in ℝ.Consequently, ifφ satisfies (A) and g
∈
ℋ
φ, then the solution w of (
W
) + (
D
) can be represented by
(18)
w
(
t
)
=
{
∫
0
t
‍
φ
-
1
(
a
(
g
)
+
∫
s
1
/
2
‍
g
(
τ
)
d
τ
)
d
s
,
t
∈
[
0
,
1
2
]
,
∫
t
1
‍
φ
-
1
(
-
a
(
g
)
+
∫
1
/
2
s
‍
g
(
τ
)
d
τ
)
d
s
,
t
∈
[
1
2
,
1
]
,
where a
(
g
)
∈
ℝ uniquely satisfies
(19)
∫
0
1
/
2
‍
φ
-
1
(
a
(
g
)
+
∫
s
1
/
2
‍
g
(
τ
)
d
τ
)
d
s
=
∫
1
/
2
1
‍
φ
-
1
(
-
a
(
g
)
+
∫
1
/
2
s
‍
g
(
τ
)
d
τ
)
d
s
.On the other hand, it is not hard to see that a functionw defined in (18) satisfies w
∈
C
0
[
0,1
]
∩
C
1
(
0,1
), and φ
(
w
′
) is absolutely continuous on (
0,1
) and w is in turn a solution of (
W
) + (
D
).Now we come back to our main problem(
P
′
)
i
-
φ
(
u
1
′
)
′
=
h
1
(
t
)
f
1
(
u
)
,
iiiiiiiiiiiiiii
⋮
-
φ
(
u
N
′
)
′
=
h
N
(
t
)
f
N
(
u
)
,
t
∈
(
0,1
)
,
i
i
i
i
i
i
u
i
(
0
)
=
0
=
u
i
(
1
)
,
i
=
1
,
…
,
N
.
We finally introduce the corresponding solution operator for (
P
′
) and prove compactness of the operator. For this purpose, we need a preliminary lemma.Lemma 8.
Ifh
∈
ℋ
ψ, then, for given α
∈
C
[
0,1
], α
h
∈
ℋ
φ.Proof.
Leth
∈
ℋ
ψ and α
∈
C
[
0,1
] be given. Then applying Remark 3 with σ
=
∫
s
1
/
2
‍
h
(
τ
)
d
τ, x
=
φ
-
1
(
∥
α
∥
∞
) and using the fact h
∈
ℋ
ψ, we get
(20)
∫
0
1
/
2
‍
φ
-
1
(
∫
s
1
/
2
‍
α
(
τ
)
h
(
τ
)
d
τ
)
d
s
≤
∫
0
1
/
2
‍
φ
-
1
(
∥
α
∥
∞
∫
s
1
/
2
‍
h
(
τ
)
d
τ
)
d
s
≤
φ
-
1
(
∥
α
∥
∞
)
∫
0
1
/
2
‍
ψ
-
1
(
∫
s
1
/
2
‍
h
(
τ
)
d
τ
)
d
s
<
∞
.
Similarly, we can prove
(21)
∫
1
/
2
1
‍
φ
-
1
(
∫
1
/
2
s
‍
α
(
τ
)
h
(
τ
)
d
τ
)
d
s
<
∞
.This lemma should be more natural if it is valid under assumptionh
∈
ℋ
φ. Even though it is true for the case φ
=
φ
p, the p-Laplace operator, it seems not easy to prove in general mainly caused by lack of homogeneity of φ.To set up the solution operator for(
P
′
), let us define E as the Banach space C
0
[
0,1
]
×
⋯
×
C
0
[
0,1
]
︸
N with norm ∥
u
∥
∞
=
Σ
i
=
1
N
∥
u
i
∥
∞ and define a cone K by taking
(22)
K
=
{
u
∈
E
∣
u
i
is
concave
on
⁡
[
0,1
]
,
i
=
1
,
…
,
N
}
.
Let u
∈
K and h
i
∈
ℋ
ψ, i
=
1
,
…
,
N; then f
i
(
u
)
∈
C
[
0,1
] and by Lemma 8, h
i
f
i
(
u
)
∈
ℋ
φ. Let us apply the solution representation for (
W
) + (
D
) replacing g with h
i
f
i
(
u
); then we get
(23)
u
i
(
t
)
=
{
∫
0
t
‍
φ
-
1
(
∫
1
/
2
s
‍
a
i
(
h
i
f
i
(
u
)
)
∫
0
t
‍
φ
-
1
ii
+
∫
s
1
/
2
‍
h
i
(
τ
)
f
i
(
u
(
τ
)
)
d
τ
)
d
s
,
0
≤
t
≤
1
2
,
∫
t
1
‍
φ
-
1
(
∫
1
/
2
s
‍
-
a
i
(
h
i
f
i
(
u
)
)
∫
0
t
‍
φ
-
1
i
i
+
∫
1
/
2
s
‍
h
i
(
τ
)
f
i
(
u
(
τ
)
)
d
τ
)
d
s
,
1
2
≤
t
≤
1
,
where a
i
(
h
i
f
i
(
u
)
) is a unique zero of
(24)
∫
0
1
/
2
‍
φ
-
1
(
a
i
(
h
i
f
i
(
u
)
)
+
∫
s
1
/
2
‍
h
i
(
τ
)
f
i
(
u
(
τ
)
)
d
τ
)
d
s
=
∫
1
/
2
1
‍
φ
-
1
(
∫
1
/
2
s
‍
-
a
i
(
h
i
f
i
(
u
)
)
+
∫
1
/
2
s
‍
h
i
(
τ
)
f
i
(
u
(
τ
)
)
d
τ
)
d
s
.Now foru
∈
K, let us define
(25)
T
i
(
u
)
(
t
)
=
{
∫
0
t
‍
φ
-
1
(
∫
1
/
2
s
‍
a
i
(
h
i
f
i
(
u
)
)
∫
0
t
‍
φ
-
1
ii
+
∫
s
1
/
2
‍
h
i
(
τ
)
f
i
(
u
(
τ
)
)
d
τ
)
d
s
,
t
∈
[
0
,
1
2
]
,
∫
t
1
‍
φ
-
1
(
∫
1
/
2
s
‍
-
a
i
(
h
i
f
i
(
u
)
)
∫
0
t
‍
φ
-
1
ii
+
∫
1
/
2
s
‍
h
i
(
τ
)
f
i
(
u
(
τ
)
)
d
τ
∫
1
/
2
s
‍
)
d
s
,
t
∈
[
1
2
,
1
]
,
T
(
u
)
=
(
T
1
(
u
)
,
…
,
T
N
(
u
)
)
.
Then by Lemma 6, T
(
K
)
⊂
K and we see that u is a positive solution of (
P
′
) if and only if u
=
T
(
u
) on K.We finally prove the solution operatorT
:
K
→
K is completely continuous. For this, we need a couple of lemmas about the properties of a
i
(
h
i
f
i
(
u
)
). Since h
i and f
i are fixed, we regard a
i
(
h
i
f
i
(
u
)
) as a function of u
∈
K. The proofs of the following two lemmas are mainly induced by the monotonicity of φ and similar to proofs of Lemmas 3.1 and 3.2 in Sim and Lee [15].Lemma 9.
a
i sends bounded sets in K into bounded sets in ℝ for i
=
1
,
…
,
N.Lemma 10.
a
i
:
K
→
ℝ is continuous for i
=
1
,
…
,
N.Lemma 11.
T
:
K
→
K is completely continuous.Proof.
Continuity ofT
i can be done by using the Lebesgue Dominated Convergence Theorem with aid of the continuity of a
i. Let B be a bounded subset of K. Then it is enough to prove T
i
(
B
) is uniformly bounded and equicontinuous. We first prove that T
i
(
B
) is uniformly bounded. Indeed, take M
B
=
sup
⁡
{
∥
f
i
(
u
)
∥
∞
∣
u
∈
B
}, K
i
(
=
K
i
(
h
i
,
M
B
)
)
=
sup
⁡
{
|
a
i
(
h
i
f
i
(
u
)
)
|
∣
u
∈
B
}, and denote simply a
u
i
≜
a
i
(
h
i
f
i
(
u
)
). We compute the bound on the interval (
0,1
/
2
]; the bound on the interval [
1
/
2,1
) can be obtained by the similar way. Consider
(26)
|
T
i
(
u
)
(
t
)
|
≤
∫
0
t
‍
φ
-
1
(
|
a
u
i
|
+
∫
s
1
/
2
‍
h
i
(
τ
)
f
i
(
u
(
τ
)
)
d
τ
)
d
s
≤
∫
0
1
/
2
‍
φ
-
1
(
K
i
+
M
B
∫
s
1
/
2
‍
h
i
(
τ
)
d
τ
)
d
s
.Case
1 (h
i
∈
L
1
(
0,1
/
2
])
(27)
|
T
i
(
u
)
(
t
)
|
≤
∫
0
1
/
2
‍
φ
-
1
(
K
i
+
M
B
∫
0
1
/
2
‍
h
i
(
τ
)
d
τ
)
d
s
|
T
i
(
u
)
(
t
)
|
=
1
2
φ
-
1
(
K
i
+
M
B
∥
h
i
∥
L
1
(
0,1
/
2
]
)
.Case
2 (h
i
∉
L
1
(
0,1
/
2
]). Let H
(
s
)
=
∫
s
1
/
2
‍
h
i
(
τ
)
d
τ; then h
i
∈
L
loc
⁡
1
(
0,1
) implies that H is continuous on (
0,1
/
2
], H
(
s
)
<
∞ for s
∈
(
0,1
/
2
] and H
(
0
+
)
=
∞. Thus we may choose s
*
∈
(
0,1
/
2
) satisfying
(28)
K
i
M
B
=
H
(
s
*
)
(
=
∫
s
*
1
/
2
‍
h
i
(
τ
)
d
τ
)
.
If s
≤
s
*, then
(29)
∫
0
s
*
‍
φ
-
1
(
K
i
+
M
B
∫
s
1
/
2
‍
h
i
(
τ
)
d
τ
)
d
s
=
∫
0
s
*
‍
φ
-
1
(
M
B
(
∫
s
*
1
/
2
‍
h
i
(
τ
)
d
τ
+
∫
s
1
/
2
‍
h
i
(
τ
)
d
τ
)
)
d
s
≤
∫
0
s
*
‍
φ
-
1
(
2
M
B
∫
s
1
/
2
‍
h
i
(
τ
)
d
τ
)
d
s
.
On the other hand, if s
>
s
*, then
(30)
∫
s
*
1
/
2
‍
φ
-
1
(
K
i
+
M
B
∫
s
1
/
2
‍
h
i
(
τ
)
d
τ
)
d
s
≤
∫
s
*
1
/
2
‍
φ
-
1
(
K
i
+
M
B
∫
s
*
1
/
2
‍
h
i
(
τ
)
d
τ
)
d
s
≤
∫
s
*
1
/
2
‍
φ
-
1
(
2
M
B
∫
s
*
1
/
2
‍
h
i
(
τ
)
d
τ
)
d
s
.
Applying Remark 3 with σ
=
∫
s
1
/
2
‍
h
i
(
τ
)
d
τ and x
=
φ
-
1
(
2
M
B
), we get
(31)
|
T
i
(
u
)
(
t
)
|
≤
∫
0
s
*
‍
φ
-
1
(
K
i
+
M
B
∫
s
1
/
2
‍
h
i
(
τ
)
d
τ
)
d
s
+
∫
s
*
1
/
2
‍
φ
-
1
(
K
i
+
M
B
∫
s
1
/
2
‍
h
i
(
τ
)
d
τ
)
d
s
≤
∫
0
s
*
‍
φ
-
1
(
2
M
B
∫
s
1
/
2
‍
h
i
(
τ
)
d
τ
)
d
s
+
∫
s
*
1
/
2
‍
φ
-
1
(
2
M
B
∫
s
*
1
/
2
‍
h
i
(
τ
)
d
τ
)
d
s
≤
φ
-
1
(
2
M
B
)
∫
0
s
*
‍
ψ
-
1
(
∫
s
1
/
2
‍
h
i
(
τ
)
d
τ
)
d
s
+
(
1
2
-
s
*
)
φ
-
1
(
2
M
B
∥
h
i
∥
L
1
(
s
*
,
1
/
2
]
)
.
By the fact h
i
∈
ℋ
ψ, all bounds above are finite and independent on u
∈
B and t
∈
[
0,1
/
2
]. Thus T
i
(
B
) is uniformly bounded.We finally prove the equicontinuity ofT
i
(
B
). Assume t
1
<
t
2.Case
1 (t
1
,
t
2
∈
[
0,1
/
2
])
(32)
|
T
i
(
u
)
(
t
1
)
-
T
i
(
u
)
(
t
2
)
|
≤
∫
t
1
t
2
‍
φ
-
1
(
|
a
u
i
|
+
∫
s
1
/
2
‍
h
i
(
τ
)
f
i
(
u
(
τ
)
)
d
τ
)
d
s
≤
∫
t
1
t
2
‍
φ
-
1
(
K
i
+
M
B
∫
s
1
/
2
‍
h
i
(
τ
)
d
τ
)
d
s
.
Let h
i
∈
L
1
(
0,1
/
2
]; then we can easily see
(33)
|
T
i
(
u
)
(
t
1
)
-
T
i
(
u
)
(
t
2
)
|
≤
φ
-
1
(
K
i
+
M
B
∥
h
i
∥
L
1
(
0,1
/
2
]
)
|
t
1
-
t
2
|
.
Let h
i
∉
L
1
(
0,1
/
2
]; then, for s
*
∈
(
0,1
/
2
) defined in (28),
(34)
|
T
i
(
u
)
(
t
1
)
-
T
i
(
u
)
(
t
2
)
|
≤
∫
t
1
t
2
φ
-
1
(
M
B
(
∫
s
*
1
/
2
‍
h
i
(
τ
)
d
τ
+
∫
s
1
/
2
‍
h
i
(
τ
)
d
τ
)
)
d
s
.Subcase
1 (0
≤
t
1
<
t
2
≤
s
*). Applying Remark 3 with σ
=
∫
s
1
/
2
‍
h
i
(
τ
)
d
τ and x
=
φ
-
1
(
2
M
B
), we get
(35)
|
T
i
(
u
)
(
t
1
)
-
T
i
(
u
)
(
t
2
)
|
≤
∫
t
1
t
2
φ
-
1
(
2
M
B
∫
s
1
/
2
‍
h
i
(
τ
)
d
τ
)
d
s
≤
φ
-
1
(
2
M
B
)
∫
t
1
t
2
‍
ψ
-
1
(
∫
s
1
/
2
‍
h
i
(
τ
)
d
τ
)
d
s
.Subcase
2 (s
*
≤
t
1
<
t
2)
(36)
|
T
i
(
u
)
(
t
1
)
-
T
i
(
u
)
(
t
2
)
|
≤
∫
t
1
t
2
‍
φ
-
1
(
2
M
B
∫
s
*
1
/
2
‍
h
i
(
τ
)
d
τ
)
d
s
≤
φ
-
1
(
2
M
B
∥
h
i
∥
L
1
[
s
*
,
1
/
2
]
)
|
t
1
-
t
2
|
.Subcase
3 (0
≤
t
1
≤
s
*
<
t
2). Consider
(37)
|
T
i
(
u
)
(
t
1
)
-
T
i
(
u
)
(
t
2
)
|
≤
∫
t
1
s
*
‍
φ
-
1
(
M
B
(
∫
s
*
1
/
2
‍
h
i
(
τ
)
d
τ
+
∫
s
1
/
2
‍
h
i
(
τ
)
d
τ
)
)
d
s
+
∫
s
*
t
2
‍
φ
-
1
(
M
B
(
∫
s
*
1
/
2
‍
h
i
(
τ
)
d
τ
+
∫
s
1
/
2
‍
h
i
(
τ
)
d
τ
)
)
d
s
≤
∫
t
1
s
*
‍
φ
-
1
(
2
M
B
∫
s
1
/
2
‍
h
i
(
τ
)
d
τ
)
d
s
+
∫
s
*
t
2
‍
φ
-
1
(
2
M
B
∫
s
*
1
/
2
‍
h
i
(
τ
)
d
τ
)
d
s
≤
φ
-
1
(
2
M
B
)
∫
t
1
t
2
‍
ψ
-
1
(
∫
s
1
/
2
‍
h
i
(
τ
)
d
τ
)
d
s
+
φ
-
1
(
2
M
B
∥
h
i
∥
L
1
[
s
*
,
1
/
2
]
)
|
t
1
-
t
2
|
.
Bounds of all cases above are independent of u
∈
B and by the fact h
i
∈
ℋ
ψ, we see that each bound converges to 0 as |
t
1
-
t
2
|
→
0.Case
2 (t
1
,
t
2
∈
[
1
/
2,1
]). Proof can be done by the same argument as Case 1.Case
3 (0
<
t
1
≤
1
/
2
<
t
2
<
1). Without loss of generality, we assume 1
/
4
≤
t
1
≤
1
/
2
<
t
2
≤
3
/
4. Then, by using the definition of a
u
i, we obtain
(38)
|
T
i
(
u
)
(
t
1
)
-
T
i
(
u
)
(
t
2
)
|
=
|
∫
0
t
1
‍
φ
-
1
(
a
u
i
+
∫
s
1
/
2
‍
h
i
(
τ
)
f
i
(
u
(
τ
)
)
d
τ
)
d
s
-
∫
t
2
1
‍
φ
-
1
(
-
a
u
i
+
∫
1
/
2
s
‍
h
i
(
τ
)
f
i
(
u
(
τ
)
)
d
τ
)
d
s
|
=
|
∫
0
t
1
‍
φ
-
1
(
a
u
i
+
∫
s
1
/
2
‍
h
i
(
τ
)
f
i
(
u
(
τ
)
)
d
τ
)
d
s
-
∫
0
1
/
2
‍
φ
-
1
(
a
u
i
+
∫
s
1
/
2
‍
h
i
(
τ
)
f
i
(
u
(
τ
)
)
d
τ
)
d
s
+
∫
1
/
2
1
‍
φ
-
1
(
-
a
u
i
+
∫
1
/
2
s
‍
h
i
(
τ
)
f
i
(
u
(
τ
)
)
d
τ
)
d
s
-
∫
t
2
1
‍
φ
-
1
(
-
a
u
i
+
∫
1
/
2
s
‍
h
i
(
τ
)
f
i
(
u
(
τ
)
)
d
τ
)
d
s
|
≤
∫
t
1
1
/
2
‍
φ
-
1
(
K
i
+
M
B
∫
s
1
/
2
‍
h
i
(
τ
)
d
τ
)
d
s
+
∫
1
/
2
t
2
‍
φ
-
1
(
K
i
+
M
B
∫
1
/
2
s
‍
h
i
(
τ
)
d
τ
)
d
s
≤
∫
t
1
1
/
2
‍
φ
-
1
(
K
i
+
M
B
∫
1
/
4
1
/
2
‍
h
i
(
τ
)
d
τ
)
d
s
+
∫
1
/
2
t
2
‍
φ
-
1
(
K
i
+
M
B
∫
1
/
2
3
/
4
‍
h
i
(
τ
)
d
τ
)
d
s
=
φ
-
1
(
K
i
+
M
B
∥
h
i
∥
L
1
[
1
/
4,1
/
2
]
)
|
t
1
-
1
2
|
+
φ
-
1
(
K
i
+
M
B
∥
h
i
∥
L
1
[
1
/
2
,
3
/
4
]
)
|
t
2
-
1
2
|
≤
2
φ
-
1
(
K
i
+
M
B
∥
h
i
∥
L
1
[
1
/
4
,
3
/
4
]
)
|
t
1
-
t
2
|
.
Conclusion is the same as Case 1 and it completes the proof of equicontinuity.


## 3. Proof of Theorem2

In this section, we prove our main theorem. Basic tool for the proof is the following well-known fixed point theorem (see [16, 17]).Lemma 12.
LetE be a Banach space and let K be a cone in E. Assume that Ω
1 and Ω
2 are open subsets of E with 0
∈
Ω
1, Ω
1
¯
⊂
Ω
2. Assume that T
:
K
∩
Ω
2
¯
∖
Ω
1
→
K is completely continuous such that either
(39)
or
∥
T
u
∥
≤
∥
u
∥
,
f
o
r
u
∈
K
∩
∂
Ω
1
;
or
∥
T
u
∥
≥
∥
u
∥
,
f
o
r
u
∈
K
∩
∂
Ω
2
;
o
r
∥
T
u
∥
≥
∥
u
∥
,
f
o
r
u
∈
K
∩
∂
Ω
1
;
or
∥
T
u
∥
≤
∥
u
∥
,
f
o
r
u
∈
K
∩
∂
Ω
2
.
Then T has a fixed point in K
∩
Ω
2
¯
∖
Ω
1.Proof of Theorem2.
(1) Letf
0
=
0; then f
0
i
=
0, i
=
1
,
…
,
N. For convenience, we denote
(40)
H
0
i
≜
∫
0
1
/
2
‍
ψ
-
1
(
∫
s
1
/
2
‍
h
i
(
τ
)
d
τ
)
d
s
,
H
1
i
≜
∫
1
/
2
1
‍
ψ
-
1
(
∫
1
/
2
s
‍
h
i
(
τ
)
d
τ
)
d
s
,wherei
=
1
,
…
,
N. Then h
i
∈
ℋ
ψ implies H
0
i, H
1
i
<
∞. Choose ϵ
>
0 sufficiently small so that
(41)
ψ
-
1
(
ϵ
)
max
⁡
{
H
0
i
,
H
1
i
∣
i
=
1
,
…
,
N
}
≤
1
N
.
Then we see that
(42)
ψ
-
1
(
ϵ
)
max
⁡
{
H
0
i
,
H
1
i
}
≤
1
N
,
for
i
=
1
,
…
,
N
.
Since f
0
i
=
0, there exists r
1
i
(
=
r
1
i
(
ϵ
)
)
>
0 such that, for x
∈
ℝ
+
N with ∥
x
∥
≤
r
1
i,
(43)
f
i
(
x
)
≤
ε
φ
(
∥
x
∥
)
,
for
i
=
1
,
…
,
N
.
Denote K
a
=
{
u
∈
K
∣
∥
u
∥
∞
<
a
} for a
>
0 and take r
1
=
min
⁡
{
r
1
i
∣
i
=
1
,
…
,
N
}. Then since T
(
u
)
∈
K for u
∈
∂
K
r
1, there exists unique σ
i
∈
(
0,1
) such that T
i
(
u
)
(
σ
i
)
=
max
⁡
t
∈
[
0,1
]
T
i
(
u
)
(
t
) and T
i
(
u
)
′
(
σ
i
)
=
0. We first consider the case σ
i
∈
(
0,1
/
2
]. Consider
(44)
0
=
T
i
(
u
)
′
(
σ
i
)
=
φ
-
1
(
a
u
i
+
∫
σ
i
1
/
2
‍
h
i
(
τ
)
f
i
(
u
(
τ
)
)
d
τ
)
.
Since φ is an odd homeomorphism, a
u
i
=
-
∫
σ
i
1
/
2
‍
h
i
(
τ
)
f
i
(
u
(
τ
)
)
d
τ. Using (43) and applying Remark 3 with σ
=
ϵ, x
=
φ
-
1
(
φ
(
r
1
)
∫
s
1
/
2
‍
h
i
(
τ
)
d
τ
), and then σ
=
∫
s
1
/
2
‍
h
i
(
τ
)
d
τ, x
=
r
1 consecutively, we obtain
(45)
∥
T
i
(
u
)
∥
∞
=
T
i
(
u
)
(
σ
i
)
=
∫
0
σ
i
‍
φ
-
1
(
a
u
i
+
∫
s
1
/
2
‍
h
i
(
τ
)
f
i
(
u
(
τ
)
)
d
τ
)
d
s
=
∫
0
σ
i
‍
φ
-
1
(
-
∫
σ
i
1
/
2
‍
h
i
(
τ
)
f
i
(
u
(
τ
)
)
d
τ
iiiiiiiiiii
+
∫
s
1
/
2
‍
h
i
(
τ
)
f
i
(
u
(
τ
)
)
d
τ
)
d
s
=
∫
0
σ
i
‍
φ
-
1
(
∫
s
σ
i
‍
h
i
(
τ
)
f
i
(
u
(
τ
)
)
d
τ
)
d
s
≤
∫
0
1
/
2
‍
φ
-
1
(
∫
s
1
/
2
‍
h
i
(
τ
)
f
i
(
u
(
τ
)
)
d
τ
)
d
s
≤
∫
0
1
/
2
‍
φ
-
1
(
ϵ
φ
(
r
1
)
∫
s
1
/
2
‍
h
i
(
τ
)
d
τ
)
d
s
≤
ψ
-
1
(
ϵ
)
∫
0
1
/
2
‍
φ
-
1
(
φ
(
r
1
)
∫
s
1
/
2
‍
h
i
(
τ
)
d
τ
)
d
s
≤
ψ
-
1
(
ϵ
)
[
∫
0
1
/
2
‍
ψ
-
1
(
∫
s
1
/
2
‍
h
i
(
τ
)
d
τ
)
d
s
]
r
1
=
ψ
-
1
(
ϵ
)
H
0
i
r
1
.
Similarly for the case σ
i
∈
[
1
/
2,1
), we get
(46)
∥
T
i
(
u
)
∥
∞
≤
ψ
-
1
(
ϵ
)
H
1
i
r
1
.
Therefore combining the above two inequalities and using the definition of ϵ, we get
(47)
∥
T
i
(
u
)
∥
∞
≤
ψ
-
1
(
ϵ
)
max
⁡
{
H
0
i
,
H
1
i
}
r
1
≤
r
1
N
,
iiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiii
for
i
=
1
,
…
,
N
,
and thus
(48)
∥
T
(
u
)
∥
∞
=
∑
i
=
1
N
‍
∥
T
i
(
u
)
∥
∞
≤
∥
u
∥
∞
,
for
u
∈
∂
K
r
1
.
We now use the assumption f
∞
=
∞. In this case, we may choose an index i
0 satisfying f
∞
i
0
=
∞. Take
(49)
M
=
γ
(
32
)
min
⁡
{
∫
1
/
4
1
/
2
‍
h
i
0
(
τ
)
d
τ
,
∫
1
/
2
3
/
4
‍
h
i
0
(
τ
)
d
τ
}
>
0
,
where γ is the function appeared in condition (
A
). Then there exists R
M
>
0 such that, for x
∈
ℝ
+
N with ∥
x
∥
≥
R
M, we have
(50)
f
i
0
(
x
)
≥
M
φ
(
∥
x
∥
)
.
If u
∈
K with ∥
u
∥
∞
≥
4
R
M, then by Lemma 6, for t
∈
[
1
/
4,3
/
4
],
(51)
∥
u
(
t
)
∥
=
∑
i
=
1
N
‍
u
i
(
t
)
≥
min
⁡
t
∈
[
1
/
4
,
3
/
4
]
∑
i
=
1
N
‍
u
i
(
t
)
≥
1
4
∥
u
∥
∞
≥
R
M
,
(52)
f
i
0
(
u
(
t
)
)
≥
M
φ
(
∥
u
(
t
)
∥
)
≥
M
φ
(
1
4
∥
u
∥
∞
)
.
Take r
2
>
max
⁡
{
r
1
,
4
R
M
}. Then for u
∈
∂
K
r
2, we get
(53)
2
T
i
0
(
u
)
(
1
2
)
=
∫
0
1
/
2
‍
φ
-
1
(
a
u
i
0
+
∫
s
1
/
2
‍
h
i
0
(
τ
)
f
i
0
(
u
(
τ
)
)
d
τ
)
d
s
+
∫
1
/
2
1
‍
φ
-
1
(
-
a
u
i
0
+
∫
1
/
2
s
‍
h
i
0
(
τ
)
f
i
0
(
u
(
τ
)
)
d
τ
)
d
s
.
If a
u
i
0
≥
0, then
(54)
∫
0
1
/
2
‍
φ
-
1
(
a
u
i
0
+
∫
s
1
/
2
‍
h
i
0
(
τ
)
f
i
0
(
u
(
τ
)
)
d
τ
)
d
s
≥
∫
0
1
/
2
‍
φ
-
1
(
∫
s
1
/
2
‍
h
i
0
(
τ
)
f
i
0
(
u
(
τ
)
)
d
τ
)
d
s
,
and by the definition of a
u
i
0,
(55)
∫
1
/
2
1
‍
φ
-
1
(
-
a
u
i
0
+
∫
1
/
2
s
‍
h
i
0
(
τ
)
f
i
0
(
u
(
τ
)
)
d
τ
)
d
s
=
∫
0
1
/
2
‍
φ
-
1
(
a
u
i
0
+
∫
s
1
/
2
‍
h
i
0
(
τ
)
f
i
0
(
u
(
τ
)
)
d
τ
)
d
s
≥
0
.
Thus
(56)
2
T
i
0
(
u
)
(
1
2
)
≥
∫
0
1
/
2
‍
φ
-
1
(
∫
s
1
/
2
‍
h
i
0
(
τ
)
f
i
0
(
u
(
τ
)
)
d
τ
)
d
s
.
If a
u
i
0
<
0, then -
a
u
i
0
>
0 and
(57)
∫
1
/
2
1
‍
φ
-
1
(
-
a
u
i
0
+
∫
1
/
2
s
‍
h
i
0
(
τ
)
f
i
0
(
u
(
τ
)
)
d
τ
)
d
s
≥
∫
1
/
2
1
‍
φ
-
1
(
∫
1
/
2
s
‍
h
i
0
(
τ
)
f
i
0
(
u
(
τ
)
)
d
τ
)
d
s
,
and by the same argument, we get
(58)
2
T
i
0
(
u
)
(
1
2
)
≥
∫
1
/
2
1
‍
φ
-
1
(
∫
1
/
2
s
‍
h
i
0
(
τ
)
f
i
0
(
u
(
τ
)
)
d
τ
)
d
s
.
Thus by using (52), we get
(59)
2
∥
T
i
0
(
u
)
∥
∞
≥
2
T
i
0
(
u
)
(
1
2
)
≥
min
⁡
{
∫
0
1
/
2
‍
φ
-
1
(
∫
s
1
/
2
‍
h
i
0
(
τ
)
f
i
0
(
u
(
τ
)
)
d
τ
)
d
s
,
iiiii
∫
1
/
2
1
‍
φ
-
1
(
∫
1
/
2
s
‍
h
i
0
(
τ
)
f
i
0
(
u
(
τ
)
)
d
τ
)
d
s
}
≥
min
⁡
{
∫
0
1
/
4
‍
φ
-
1
(
∫
s
1
/
2
‍
h
i
0
(
τ
)
f
i
0
(
u
(
τ
)
)
d
τ
)
d
s
,
iiiii
∫
3
/
4
1
‍
φ
-
1
(
∫
1
/
2
s
‍
h
i
0
(
τ
)
f
i
0
(
u
(
τ
)
)
d
τ
)
d
s
}
≥
min
⁡
{
∫
0
1
/
4
‍
φ
-
1
(
∫
1
/
4
1
/
2
‍
h
i
0
(
τ
)
f
i
0
(
u
(
τ
)
)
d
τ
)
d
s
,
iiiii
∫
3
/
4
1
‍
φ
-
1
(
∫
1
/
2
3
/
4
‍
h
i
0
(
τ
)
f
i
0
(
u
(
τ
)
)
d
τ
)
d
s
}
≥
min
⁡
{
∫
0
1
/
4
‍
φ
-
1
(
M
φ
(
1
4
∥
u
∥
∞
)
∫
1
/
4
1
/
2
‍
h
i
0
(
τ
)
d
τ
)
d
s
,
iiiii
∫
3
/
4
1
‍
φ
-
1
(
M
φ
(
1
4
∥
u
∥
∞
)
∫
1
/
2
3
/
4
‍
h
i
0
(
τ
)
d
τ
)
d
s
}
=
1
4
φ
-
1
(
∫
1
/
4
1
/
2
‍
M
φ
(
1
4
∥
u
∥
∞
)
iiiiiiiii
×
min
⁡
{
∫
1
/
4
1
/
2
‍
h
i
0
(
τ
)
d
τ
,
∫
1
/
2
3
/
4
‍
h
i
0
(
τ
)
d
τ
}
)
.
By the definition of M, we get
(60)
2
∥
T
i
0
(
u
)
∥
∞
≥
1
4
φ
-
1
(
γ
(
32
)
φ
(
1
4
∥
u
∥
∞
)
)
.
Applying Remark 3 with σ
=
32 and x
=
(
1
/
4
)
∥
u
∥
∞, we get
(61)
2
∥
T
i
0
(
u
)
∥
∞
≥
1
4
·
32
·
1
4
∥
u
∥
∞
=
2
∥
u
∥
∞
.
Thus
(62)
∥
T
(
u
)
∥
∞
≥
∥
T
i
0
(
u
)
∥
∞
≥
∥
u
∥
∞
,
for
u
∈
∂
K
r
2
.
Combining (48) and (62), we conclude that problem (
P
) has at least one positive solution u with r
1
≤
∥
u
∥
∞
≤
r
2.
(2) We now prove the second result of Theorem2. Let f
0
=
∞; then there exists an index i
0 satisfying f
0
i
0
=
∞. Take
(63)
M
=
γ
(
32
)
min
⁡
{
∫
1
/
4
1
/
2
‍
h
i
0
(
τ
)
d
τ
,
∫
1
/
2
3
/
4
‍
h
i
0
(
τ
)
d
τ
}
>
0
.
Then there exists r
M
>
0 such that, for x
∈
ℝ
+
N with ∥
x
∥
≤
r
M, we have
(64)
f
i
0
(
x
)
≥
M
φ
(
∥
x
∥
)
.
If u
∈
K with ∥
u
∥
∞
≤
r
M, then by Lemma 6, for t
∈
[
1
/
4,3
/
4
],
(65)
∥
u
(
t
)
∥
≤
∥
u
∥
∞
≤
r
M
,
(66)
f
i
0
(
u
(
t
)
)
≥
M
φ
(
∥
u
(
t
)
∥
)
≥
M
φ
(
1
4
∥
u
∥
∞
)
.
Take r
1
=
r
M and let u
∈
∂
K
r
1. Then
(67)
2
T
i
0
(
u
)
(
1
2
)
=
∫
0
1
/
2
‍
φ
-
1
(
a
u
i
0
+
∫
s
1
/
2
‍
h
i
0
(
τ
)
f
i
0
(
u
(
τ
)
)
d
τ
)
d
s
+
∫
1
/
2
1
‍
φ
-
1
(
-
a
u
i
0
+
∫
1
/
2
s
‍
h
i
0
(
τ
)
f
i
0
(
u
(
τ
)
)
d
τ
)
d
s
.
We also consider two cases a
u
i
0
≥
0 and a
u
i
0
<
0. Applying the same argument in (1) with aid of (66), we get
(68)
2
∥
T
i
0
(
u
)
∥
∞
≥
2
T
i
0
(
u
)
(
1
2
)
=
1
4
φ
-
1
(
∫
1
/
4
1
/
2
‍
M
φ
(
1
4
∥
u
∥
∞
)
iiiiiiii
×
min
⁡
{
∫
1
/
4
1
/
2
‍
h
i
0
(
τ
)
d
τ
,
∫
1
/
2
3
/
4
‍
h
i
0
(
τ
)
d
τ
}
)
.
By the definition of M, we get
(69)
2
∥
T
i
0
(
u
)
∥
∞
≥
1
4
φ
-
1
(
γ
(
32
)
φ
(
1
4
∥
u
∥
∞
)
)
.
Applying Remark 3 with σ
=
32 and x
=
(
1
/
4
)
∥
u
∥
∞, we get
(70)
2
∥
T
i
0
(
u
)
∥
∞
≥
1
4
·
32
·
1
4
∥
u
∥
∞
=
2
∥
u
∥
∞
.
Thus
(71)
∥
T
(
u
)
∥
∞
≥
∥
T
i
0
(
u
)
∥
∞
≥
∥
u
∥
∞
,
for
u
∈
∂
K
r
1
.
Letf
∞
=
0; then f
∞
i
=
0, i
=
1
,
…
,
N. Define a function f
^
i
(
t
)
:
ℝ
+
→
ℝ
+ by
(72)
f
^
i
(
t
)
=
max
⁡
{
f
i
(
x
)
∣
x
∈
ℝ
+
N
,
∥
x
∥
≤
t
}
.
By Lemma 2.8 in Wang [10], we have
(73)
f
^
∞
i
=
lim
⁡
t
→
∞
f
^
i
(
t
)
φ
(
t
)
=
f
∞
i
=
0
.
Choose ϵ
>
0 sufficiently small so that
(74)
ψ
-
1
(
ϵ
)
max
⁡
{
H
0
i
,
H
1
i
∣
i
=
1
,
…
,
N
}
≤
1
N
,
where H
0
i and H
1
i are defined as in part (1). Then we see that
(75)
ψ
-
1
(
ϵ
)
max
⁡
{
H
0
i
,
H
1
i
}
≤
1
N
,
for
i
=
1
,
…
,
N
.
Since f
^
∞
i
=
0, there exists r
2
i
(
=
r
2
i
(
ϵ
)
)
> such that, for t
∈
ℝ
+ with t
≥
r
2
i,
(76)
f
^
i
(
t
)
≤
ε
φ
(
t
)
,
for
i
=
1
,
…
,
N
.
Take r
2
>
max
⁡
{
r
1
,
max
⁡
{
r
2
i
∣
i
=
1
,
…
,
N
}
}. Then for u
∈
∂
K
r
2, we get
(77)
f
i
(
u
(
t
)
)
≤
f
^
i
(
r
2
)
≤
ε
φ
(
r
2
)
,
for
i
=
1
,
…
,
N
.
Since T
(
u
)
∈
K, there exists unique σ
i
∈
(
0,1
) such that T
i
(
u
)
(
σ
i
)
=
max
⁡
t
∈
[
0,1
]
T
i
(
u
)
(
t
) and T
i
(
u
)
′
(
σ
i
)
=
0. Considering two cases σ
i
∈
(
0,1
/
2
] and σ
i
∈
[
1
/
2,1
) with the same argument in (1) and using (77), we get
(78)
∥
T
i
(
u
)
∥
∞
≤
ψ
-
1
(
ϵ
)
max
⁡
{
H
0
i
,
H
1
i
}
r
2
,
for
i
=
1
,
…
,
N
,
(79)
∥
T
(
u
)
∥
∞
=
∑
i
=
1
N
‍
∥
T
i
(
u
)
∥
∞
≤
∥
u
∥
∞
,
for
u
∈
∂
K
r
2
.
Combining (71) and (79), we conclude that problem (
P
) has at least one positive solution u with r
1
≤
∥
u
∥
∞
≤
r
2 and the proof is complete.


## 4. Examples

In this section, we give some examples applicable to our main results.Example 13.
Consider the followingφ-Laplacian system:
(
E
1
)
φ
(
u
′
)
′
+
t
-
α
[
(
u
+
v
)
p
-
q
+
1
]
=
0
,
φ
(
v
′
)
′
+
t
-
β
u
q
-
1
(
1
-
e
-
v
)
=
0
,
t
∈
(
0,1
)
,
u
(
0
)
=
v
(
0
)
=
u
(
1
)
=
v
(
1
)
=
0
,
where φ
(
x
)
=
|
x
|
p
-
2
x
+
|
x
|
q
-
2
x, x
∈
ℝ, 1
<
q
<
p, 1
<
α
,
β
<
min
⁡
{
2
,
q
}. We note that both h
(
t
)
=
t
-
α and h
(
t
)
=
t
-
β are not in L
1
(
0,1
). It is easy to see that φ is an odd increasing homeomorphism. Define functions ψ and γ given as
(80)
ψ
(
σ
)
=
{
σ
p
-
1
,
if
0
<
σ
≤
1
,
σ
q
-
1
,
if
σ
>
1
,
γ
(
σ
)
=
{
1
,
if
0
<
σ
≤
1
,
σ
p
-
1
,
if
σ
>
1
.
Then ψ
,
γ
:
(
0
,
∞
)
→
(
0
,
∞
) and ψ is an increasing homeomorphism with
(81)
ψ
-
1
(
σ
)
=
{
σ
1
/
(
p
-
1
)
,
if
0
<
σ
≤
1
,
σ
1
/
(
q
-
1
)
,
if
σ
>
1
.
If 0
<
σ
≤
1, then σ
-
(
p
-
q
)
≥
1 and
(82)
φ
(
σ
x
)
φ
(
x
)
=
σ
p
-
1
[
|
x
|
p
-
2
x
+
σ
-
(
p
-
q
)
|
x
|
q
-
2
x
]
|
x
|
p
-
2
x
+
|
x
|
q
-
2
x
≥
σ
p
-
1
=
ψ
(
σ
)
.
If σ
>
1, then σ
p
-
q
>
1 and
(83)
φ
(
σ
x
)
φ
(
x
)
=
σ
q
-
1
[
σ
p
-
q
|
x
|
p
-
2
x
+
|
x
|
q
-
2
x
]
|
x
|
p
-
2
x
+
|
x
|
q
-
2
x
≥
σ
q
-
1
=
ψ
(
σ
)
.
If 0
<
σ
≤
1, then σ
p
-
q
≤
1 and
(84)
φ
(
σ
x
)
φ
(
x
)
=
σ
q
-
1
[
σ
p
-
q
|
x
|
p
-
2
x
+
|
x
|
q
-
2
x
]
|
x
|
p
-
2
x
+
|
x
|
q
-
2
x
≤
σ
q
-
1
≤
1
=
γ
(
σ
)
.
If σ
>
1, then σ
-
(
p
-
q
)
<
1 and
(85)
φ
(
σ
x
)
φ
(
x
)
=
σ
p
-
1
[
|
x
|
p
-
2
x
+
σ
-
(
p
-
q
)
|
x
|
q
-
2
x
]
|
x
|
p
-
2
x
+
|
x
|
q
-
2
x
≤
σ
p
-
1
=
γ
(
σ
)
.
Thus, it follows that
(86)
ψ
(
σ
)
≤
φ
(
σ
x
)
φ
(
σ
)
≤
γ
(
σ
)
,
∀
σ
>
0
,
x
∈
ℝ
.
Next, we show that h
(
t
)
=
t
-
α
∈
ℋ
ψ. Consider
(87)
∫
s
1
/
2
‍
τ
-
α
d
τ
=
-
1
α
-
1
τ
-
(
α
-
1
)
|
s
1
/
2
=
-
1
α
-
1
[
(
1
2
)
-
(
α
-
1
)
-
s
-
(
α
-
1
)
]
=
1
α
-
1
[
s
-
(
α
-
1
)
-
2
α
-
1
]
≤
1
α
-
1
s
-
(
α
-
1
)
.
Since 1
<
α
<
min
⁡
{
2
,
q
}, then (
1
/
(
α
-
1
)
)
1
/
(
α
-
1
)
>
1 and (
1
/
(
α
-
1
)
)
1
/
(
α
-
1
)
>
s, for s
∈
(
0,1
). Thus, 1
/
(
α
-
1
)
>
s
α
-
1, (
1
/
(
α
-
1
)
)
s
-
(
α
-
1
)
>
1, and
(88)
∫
0
1
/
2
‍
ψ
-
1
(
∫
s
1
/
2
‍
τ
-
α
d
τ
)
d
s
≤
∫
0
1
/
2
‍
ψ
-
1
(
1
α
-
1
s
-
(
α
-
1
)
)
d
s
=
∫
0
1
/
2
‍
(
s
-
(
α
-
1
)
α
-
1
)
1
/
(
q
-
1
)
d
s
=
q
-
1
(
α
-
1
)
1
/
(
q
-
1
)
(
q
-
α
)
s
(
q
-
α
)
/
(
q
-
1
)
|
0
1
/
2
<
∞
,
since q
-
α
>
0 and q
-
1
>
0. The continuity of h
(
t
)
=
t
-
α on [
1
/
2,1
] obviously implies that ∫
1
/
2
1
‍
ψ
-
1
(
∫
1
/
2
s
‍
τ
-
α
d
τ
)
d
s
<
∞. Similarly, we can show that h
(
t
)
=
t
-
β
∈
ℋ
ψ. We now check the conditions on the nonlinear terms. Both f
1
(
u
,
v
)
=
(
u
+
v
)
p
-
q
+
1 and f
2
(
u
,
v
)
=
u
q
-
1
(
1
-
e
-
v
) satisfy (F) and
(89)
f
0
1
=
lim
⁡
∥
(
u
,
v
)
∥
→
0
f
1
(
u
,
v
)
φ
(
∥
(
u
,
v
)
∥
)
f
0
1
=
lim
⁡
∥
(
u
,
v
)
∥
→
0
(
u
+
v
)
p
-
q
+
1
(
u
+
v
)
q
-
1
[
(
u
+
v
)
p
-
q
+
1
]
=
∞
,
f
∞
1
=
lim
⁡
∥
(
u
,
v
)
∥
→
∞
f
1
(
u
,
v
)
φ
(
∥
(
u
,
v
)
∥
)
f
0
1
=
lim
⁡
∥
(
u
,
v
)
∥
→
∞
1
(
u
+
v
)
q
-
1
=
0
,
f
0
2
=
lim
⁡
∥
(
u
,
v
)
∥
→
0
(
1
-
e
-
v
)
·
u
q
-
1
(
u
+
v
)
p
-
1
+
(
u
+
v
)
q
-
1
f
0
1
≤
lim
⁡
∥
(
u
,
v
)
∥
→
0
(
1
-
e
-
v
)
=
0
,
f
∞
2
=
lim
⁡
∥
(
u
,
v
)
∥
→
∞
(
1
-
e
-
v
)
·
u
q
-
1
(
u
+
v
)
p
-
1
+
(
u
+
v
)
q
-
1
f
0
1
≤
lim
⁡
∥
(
u
,
v
)
∥
→
∞
(
1
-
e
-
v
)
·
(
u
+
v
)
q
-
1
(
u
+
v
)
p
-
1
+
(
u
+
v
)
q
-
1
f
0
1
≤
lim
⁡
∥
(
u
,
v
)
∥
→
∞
1
(
u
+
v
)
p
-
q
+
1
=
0
.
Thus, f
0
=
f
0
1
+
f
0
2
=
∞, f
∞
=
f
∞
1
+
f
∞
2
=
0. Consequently, by Theorem 2, we see that problem (
E
1
) has at least one positive solution.Example 14.
Consider the followingφ-Laplacian system:
(
E
2
)
φ
(
u
′
)
′
+
t
-
5
/
4
(
u
+
v
)
1
/
2
=
0
,
φ
(
v
′
)
′
+
t
-
6
/
5
(
1
-
e
-
(
u
+
v
)
)
(
u
+
v
)
1
/
3
=
0
,
t
∈
(
0,1
)
,
u
(
0
)
=
v
(
0
)
=
u
(
1
)
=
v
(
1
)
=
0
,
where φ
(
x
)
=
x
1
/
3, x
∈
ℝ, is an odd increasing homeomorphism. By the homogeneity of φ, taking ψ
(
σ
)
=
γ
(
σ
)
≡
φ
(
σ
), we see that condition (A) is satisfied. Consider
(90)
∫
0
1
/
2
‍
ψ
-
1
(
∫
s
1
/
2
‍
τ
-
5
/
4
d
τ
)
d
s
=
∫
0
1
/
2
‍
ψ
-
1
(
4
(
s
-
1
/
4
-
2
1
/
4
)
)
d
s
=
∫
0
1
/
2
‍
(
4
(
s
-
1
/
4
-
2
1
/
4
)
)
3
d
s
≤
64
∫
0
1
/
2
‍
s
-
3
/
4
d
s
=
256
s
1
/
4
|
0
1
/
2
<
∞
,
and the continuity of h
(
t
)
=
t
-
5
/
4 on [
1
/
2,1
] implies that h
(
t
)
=
t
-
5
/
4
∈
ℋ
ψ. Similarly, we can show that h
(
t
)
=
t
-
6
/
5
∈
ℋ
ψ. For the nonlinear terms, both f
1
(
u
,
v
)
=
(
u
+
v
)
1
/
2 and f
2
(
u
,
v
)
=
(
1
-
e
-
(
u
+
v
)
)
(
u
+
v
)
1
/
3 satisfy condition (F) and
(91)
f
0
1
=
lim
⁡
∥
(
u
,
v
)
∥
→
0
f
1
(
u
,
v
)
φ
(
∥
(
u
,
v
)
∥
)
=
lim
⁡
∥
(
u
,
v
)
∥
→
0
(
u
+
v
)
1
/
2
(
u
+
v
)
1
/
3
=
lim
⁡
∥
(
u
,
v
)
∥
→
0
(
u
+
v
)
1
/
6
=
0
,
f
∞
1
=
lim
⁡
∥
(
u
,
v
)
∥
→
∞
f
1
(
u
,
v
)
φ
(
∥
(
u
,
v
)
∥
)
=
lim
⁡
∥
(
u
,
v
)
∥
→
∞
(
u
+
v
)
1
/
6
=
∞
,
f
0
2
=
lim
⁡
∥
(
u
,
v
)
∥
→
0
(
1
-
e
-
(
u
+
v
)
)
·
(
u
+
v
)
1
/
3
(
u
+
v
)
1
/
3
=
lim
⁡
∥
(
u
,
v
)
∥
→
0
(
1
-
e
-
(
u
+
v
)
)
=
0
,
f
∞
2
=
lim
⁡
∥
(
u
,
v
)
∥
→
∞
(
1
-
e
-
(
u
+
v
)
)
=
1
.
Thus, f
0
=
f
0
1
+
f
0
2
=
0, f
∞
=
f
∞
1
+
f
∞
2
=
∞. Consequently, by Theorem 2, we see that problem (
E
2
) has at least one positive solution.



---

*Source: 814312-2014-05-04.xml*