# Positive Solutions for a System of Neumann Boundary Value Problems of Second-Order Difference Equations Involving Sign-Changing Nonlinearities

Authors: Jiqiang Jiang; Johnny Henderson; Jiafa Xu; Zhengqing Fu
Journal: Journal of Function Spaces
 (2019)

---

## Abstract

In this paper, we study the existence of positive solutions for the system of second-order difference equations involving Neumann boundary conditions:-
Δ
2
u
1
(
t
-
1
)
=
f
1
(
t
,
u
1
(
t
)
,
u
2
(
t
)
), t
∈
[
1
,
T
]
Z, -
Δ
2
u
2
(
t
-
1
)
=
f
2
(
t
,
u
1
(
t
)
,
u
2
(
t
)
), t
∈
[
1
,
T
]
Z, Δ
u
i
(
0
)
=
Δ
u
i
(
T
)
=
0, i
=
1,2, where T
>
1 is a given positive integer, Δ
u
(
t
)
=
u
(
t
+
1
)
-
u
(
t
), and Δ
2
u
(
t
)
=
Δ
(
Δ
u
(
t
)
). Under some appropriate conditions for our sign-changing nonlinearities, we use the fixed point index to establish our main results.


---

## Body

## 1. Introduction

Fora
,
b
∈
Z with a
<
b, let [
a
,
b
]
Z
=
{
a
,
a
+
1
,
a
+
2
,
…
,
b
-
1
,
b
}. Consider the system of second-order difference equations involving Neumann boundary conditions:(1)
-
Δ
2
u
1
t
-
1
=
f
1
t
,
u
1
t
,
u
2
t
,
t
∈
1
,
T
Z
,
-
Δ
2
u
2
t
-
1
=
f
2
t
,
u
1
t
,
u
2
t
,
t
∈
1
,
T
Z
,
Δ
u
i
0
=
Δ
u
i
T
=
0
,
i
=
1,2
,where f
i
:
[
1
,
T
]
Z
×
R
+
×
R
+
→
R (i
=
1,2) are two continuous functions and there exist h
i
:
[
1
,
T
]
Z
→
R
+ with h
i
(
t
)
≢
0 on [
1
,
T
]
Z(i
=
1,2) and a positive number L
>
0 such that(H0)f
i
(
t
,
u
1
,
u
2
)
+
L
u
i
+
h
i
(
t
)
≥
0, (
t
,
u
1
,
u
2
)
∈
[
1
,
T
]
Z
×
R
+
×
R
+, i=1,2.As known to all, semipositone problems arise in bulking of mechanical systems, design of suspension bridges, chemical reactions, astrophysics, combustion, and management of natural resources; for example, see [1–4]. We note that studying positive solutions for semipositone problems is more difficult than that for positive problems. There are many methods to deal with semipositone (positive) problems, with the usual approaches being variational methods, fixed point theory, subsuper solutions methods, and degree theory; for example, see [3–32] and references therein.In [5], the author used the Guo-Krasnosel’skii fixed point theorem to study the existence of at least one positive solution for the discrete fractional equation:(2)
-
Δ
ν
y
t
=
λ
f
t
+
ν
-
1
,
y
t
+
ν
-
1
,
t
∈
1
,
b
+
1
N
,
y
ν
-
2
=
∑
i
=
1
N
F
i
1
y
t
i
1
,
y
ν
+
b
+
1
=
∑
i
=
1
M
F
i
2
y
t
i
2
,where λ
>
0 is a parameter; the semipositone nonlinear term f satisfies the condition(3)
l
i
m
y
→
+
∞
f
t
,
y
y
=
+
∞
,
l
i
m
y
→
0
+
f
t
,
y
y
=
0
,
u
n
i
f
o
r
m
l
y
f
o
r
t
∈
ν
,
ν
+
b
N
.For semipositone systems, the authors [9, 17, 18] used the similar conditions of (3) to obtain some results for boundary value problems of differential (difference) equations.However, we note that systems of boundary value problems for difference problems have seldom been considered in the literature; we refer to only [8, 9, 33–37] and references therein.In [33], the authors used the Krasnosel’skii-Zabreiko fixed point theorem to investigate the existence and multiplicity of positive solutions for the system of second-order discrete boundary value problems:(4)
Δ
2
u
k
-
1
+
f
k
,
u
k
,
v
k
=
0
,
k
∈
1,2
,
…
,
T
,
Δ
2
v
k
-
1
+
g
k
,
u
k
,
v
k
=
0
,
k
∈
1,2
,
…
,
T
,
u
0
=
u
T
+
1
=
v
0
=
v
T
+
1
=
0
,where f and g are nonnegative continuous functions on {
1,2
,
…
,
T
}
×
R
+
×
R
+.Inspired by the works aforementioned, in this paper we use the fixed point index to consider the existence of positive solutions for (1). The novelty is threefold: (
1
) The nonlinearities may be either bounded or unbounded below; ultimately nonpositive or nonnegative or oscillating, see [6, Page 2]; this improves some conditions for the nonlinearities in [36, 37]. (
2
) Some appropriate nonnegative concave and convex functions are employed to depict the coupling behaviors of nonlinearities. (
3
) Our conditions are better than (3). (
4
) Our a priori estimates for positive solutions are derived by unknown functions u
i
-
w
i,i
=
1,2; see Section 3. This is different from [36, 37].


## 2. Preliminary

For convenience, let(5)
A
=
1
2
L
+
2
+
L
2
+
4
L
,
ρ
=
A
T
-
A
-
T
A
2
-
1
.Then(6)
G
t
,
s
=
1
ρ
A
s
+
A
-
s
+
1
A
t
-
T
+
A
T
-
t
+
1
,
1
≤
s
≤
t
≤
T
+
1
,
A
t
+
A
-
t
+
1
A
s
-
T
+
A
T
-
s
+
1
,
0
≤
t
≤
s
≤
T
,is the Green’s function associated with the linear Neumann boundary value problems(7)
-
Δ
2
u
i
t
-
1
+
L
u
i
t
=
h
i
t
,
t
∈
1
,
T
Z
,
Δ
u
i
0
=
Δ
u
i
T
=
0
,
i
=
1,2
,which is equivalent to(8)
w
i
t
=
∑
s
=
1
T
G
t
,
s
h
i
s
,
i
=
1,2
.Letq
∗
(
t
)
=
1
/
2
A
T
m
i
n
{
A
t
-
T
+
A
T
-
t
+
1
,
A
t
+
A
-
t
+
1
}, t
∈
[
0
,
T
+
1
]
Z. Then 0
<
q
∗
(
t
)
<
1, for t
∈
[
1
,
T
]
Z. Consequently, from [6] we obtain that the Green’s function G has the following properties.Lemma 1.
(i)G
(
t
,
s
)
>
0 for all (
t
,
s
)
∈
[
0
,
T
+
1
]
Z
×
[
1
,
T
]
Z.
(ii)q
∗
(
t
)
G
(
s
,
s
)
≤
G
(
t
,
s
)
≤
G
(
s
,
s
), for all (
t
,
s
)
∈
[
0
,
T
+
1
]
Z
×
[
1
,
T
]
Z.
(iii) Ifφ
(
s
)
=
G
(
s
,
s
), for s
∈
[
1
,
T
]
Z, then we have(9)
∑
t
=
1
T
q
∗
t
φ
t
·
φ
s
≤
∑
t
=
1
T
G
t
,
s
φ
t
≤
∑
t
=
1
T
φ
t
·
φ
s
.These involve direct computations, and so we omit their proofs. For convenience, we denote(10)
κ
1
=
∑
t
=
1
T
q
∗
t
φ
t
,
κ
2
=
∑
t
=
1
T
φ
t
.Define(11)
f
~
i
t
,
u
1
,
u
2
=
f
i
t
,
u
1
,
u
2
+
L
u
i
+
h
i
t
,
t
,
u
1
,
u
2
∈
1
,
T
Z
×
R
+
×
R
+
,
f
i
t
,
0,0
+
h
i
t
,
t
,
u
1
,
u
2
∈
1
,
T
Z
×
R
-
×
R
-
,
i
=
1,2
,
and consider the following modified discrete Neumann boundary value problems(12)
-
Δ
2
u
1
t
-
1
+
L
u
1
t
=
f
~
1
t
,
u
1
t
-
w
1
t
,
u
2
t
-
w
2
t
,
t
∈
1
,
T
Z
,
-
Δ
2
u
2
t
-
1
+
L
u
2
t
=
f
~
2
t
,
u
1
t
-
w
1
t
,
u
2
t
-
w
2
t
,
t
∈
1
,
T
Z
,
Δ
u
i
0
=
Δ
u
i
T
=
0
,
i
=
1,2
.Lemma 2 (see [6]).
(
u
1
,
u
2
) is a positive solution of (1) if and only if (
v
1
,
v
2
)
=
(
u
1
+
w
1
,
u
2
+
w
2
) is a solution of (12) with v
i
(
t
)
≥
w
i
(
t
) for t
∈
[
1
,
T
]
Z.
LetE be the collection of all maps from [
0
,
T
+
1
]
Z to R equipped with the max norm, ·. Then E is a Banach space. Define a set P
⊂
E by P
=
y
∈
E
:
y
(
t
)
≥
0
,
t
∈
[
1
,
T
]
Z
. Then P is a cone on E. Moreover, E
×
E is a Banach space with the norm (
x
,
y
)
≔
m
a
x
{
x
,
y
}, and P
×
P is a cone on E
×
E.
Note that (12) can be expressed in the form(13)
u
1
t
=
∑
s
=
1
T
G
t
,
s
f
~
1
s
,
u
1
s
-
w
1
s
,
u
2
s
-
w
2
s
,
u
2
t
=
∑
s
=
1
T
G
t
,
s
f
~
2
s
,
u
1
s
-
w
1
s
,
u
2
s
-
w
2
s
.
As a result, foru
i
∈
P
(
i
=
1,2
), and t
∈
[
1
,
T
]
Z, we define the operators(14)
B
i
u
1
,
u
2
t
=
∑
s
=
1
T
G
t
,
s
f
~
i
s
,
u
1
s
-
w
1
s
,
u
2
s
-
w
2
s
,and(15)
B
u
1
,
u
2
t
=
B
1
,
B
2
u
1
,
u
2
t
.Then we use the Arzelà-Ascoli theorem in a standard way to establish that B
:
P
×
P
→
P
×
P is a completely continuous operator. It is clear that (
u
1
,
u
2
)
∈
(
P
×
P
)
∖
{
0
} is a positive solution for (12) if and only if (
u
1
,
u
2
)
∈
(
P
×
P
)
∖
{
0
} is a fixed point of B.
On the other hand, letP
0
=
{
y
∈
P
:
y
(
t
)
≥
q
∗
(
t
)
u
2
,
∀
t
∈
[
1
,
T
]
Z
}. Then from Lemma 1(ii) we have(16)
B
i
P
×
P
⊂
P
0
,
i
=
1,2
.Therefore, if we seek a fixed point (
v
1
,
v
2
) of B with (
v
1
,
v
2
)
(
t
)
≥
(
w
1
,
w
2
)
(
t
) (i.e., (
v
1
-
w
1
,
v
2
-
w
2
)
(
t
) is a positive solution for (1)). Then if v
i
∈
P
0
(
i
=
1,2
) we have(17)
v
1
t
-
w
1
t
≥
q
∗
t
v
1
-
∑
s
=
1
T
G
s
,
s
h
1
s
≥
q
0
v
1
-
∑
s
=
1
T
G
s
,
s
h
1
s
,
for
t
∈
1
,
T
Z
,where q
0
=
m
i
n
t
∈
[
1
,
T
]
Z
q
∗
(
t
)
>
0. As a result, v
i
≥
q
0
-
1
∑
s
=
1
T
G
(
s
,
s
)
h
i
(
s
)
≔
M
i
(
i
=
1,2
) implies that (
v
1
,
v
2
)
(
t
)
≥
(
w
1
,
w
2
)
(
t
) for t
∈
[
1
,
T
]
Z.Lemma 3 (see [38]).
LetE be a real Banach space and P a cone on E. Suppose that Ω
⊂
E is a bounded open set and that A
:
Ω
¯
∩
P
→
P is a continuous compact operator. If there exists ω
0
∈
P
∖
{
0
} such that(18)
ω
-
A
ω
≠
λ
ω
0
,
∀
λ
≥
0
,
ω
∈
∂
Ω
∩
P
,then i
(
A
,
Ω
∩
P
,
P
)
=
0, where i denotes the fixed point index on P.Lemma 4 (see [38]).
LetE be a real Banach space and P a cone on E. Suppose that Ω
⊂
E is a bounded open set with 0
∈
Ω and that A
:
Ω
¯
∩
P
→
P is a continuous compact operator. If(19)
ω
-
λ
A
ω
≠
0
,
∀
λ
∈
0,1
,
ω
∈
∂
Ω
∩
P
,then i
(
A
,
Ω
∩
P
,
P
)
=
1.


## 3. Main Results

For convenience, we usec
1
,
c
2
,
… to denote distinct positive constants. Let B
ϱ
≔
{
x
∈
E
:
x
<
ϱ
} for ϱ
>
0. Now, we list our assumptions on f
~
i
(
i
=
1,2
).(H1) There existp
,
q
∈
C
(
R
+
,
R
+
) such that(i)p is concave and strictly increasing on R
+,(ii) there existc
1
>
0
,
d
1
∈
(
0
,
κ
2
-
1
) such that(20)
f
~
1
t
,
x
,
y
≥
d
1
x
+
p
y
-
c
1
,
f
~
2
t
,
x
,
y
≥
q
x
-
c
1
,
∀
t
,
x
,
y
∈
1
,
T
Z
×
R
+
×
R
+
,(iii) there is aγ
1
>
κ
1
-
2
(
1
-
d
1
κ
1
) such that(21)
p
κ
2
q
x
t
≥
κ
2
γ
1
x
t
-
c
1for x
∈
R
+ and t
∈
[
1
,
T
]
Z.(H2) LetM
3
=
m
a
x
{
M
1
,
M
3
}. Then for any (
t
,
x
,
y
)
∈
[
1
,
T
]
Z
×
[
0
,
M
3
]
×
[
0
,
M
3
], we suppose that(22)
f
~
i
t
,
x
,
y
<
κ
2
-
1
M
3
.(H3) There existξ
,
η
∈
C
(
R
+
,
R
+
) such that(i)ξ is convex and strictly increasing on R
+,(ii) there existsd
2
∈
(
0
,
κ
2
-
1
) such that(23)
f
~
1
t
,
x
,
y
≤
d
2
x
+
ξ
y
,
f
~
2
t
,
x
,
y
≤
η
x
,
∀
t
,
x
,
y
∈
1
,
T
Z
×
R
+
×
R
+
,(iii) there arec
2
>
0 and γ
2
<
(
0
,
(
1
-
d
2
κ
2
)
κ
2
-
2
) such that(24)
ξ
κ
2
η
x
t
≤
κ
2
γ
2
x
t
+
c
2for x
∈
R
+ and t
∈
[
1
,
T
]
Z.(H4) For any(
t
,
x
,
y
)
∈
[
1
,
T
]
Z
×
[
0
,
M
3
]
×
[
0
,
M
3
], we suppose that(25)
f
~
i
t
,
x
,
y
>
q
0
-
1
κ
2
-
1
M
3
.We now present a list of remarks and examples in which we discuss how our hypotheses and assumptions are better (weaker) hypotheses and assumptions in some of the closely related papers cited in the reference.Remark 5.
We first provide the growth conditions for the nonlinearities of [36, (H3)(ii) on page 4] given by(26)
lim
sup
e
1
x
+
e
2
y
→
+
∞
⁡
f
~
t
,
x
,
y
e
1
x
+
e
2
y
≤
1
,
lim
sup
e
3
x
+
e
4
y
→
+
∞
⁡
g
~
t
,
x
,
y
e
3
x
+
e
4
y
≤
1
,
u
n
i
f
o
r
m
l
y
o
n
t
∈
ν
-
1
,
b
+
ν
+
1
N
ν
-
1
,where e
i
(
i
=
1,2
,
3,4
) are nonnegative real numbers. However, note that our condition (H3)(i), (ii) of this paper is given by(27)
lim
sup
d
2
x
+
ξ
y
→
+
∞
⁡
f
~
1
t
,
x
,
y
d
2
x
+
ξ
y
≤
1
,
u
n
i
f
o
r
m
l
y
o
n
1
,
T
Z
,and obviously, this includes (26) as a special case.
On the other hand, we note that our growth conditions for nonlinearityf
~
1 depends on two variables x
,
y; however, in [37, (H2)(ii) and (H4)(i)], the corresponding conditions only involve one variable. Finally, our nonlinearities here are allowed to be unbounded from below, which are better than the nonlinearities in [36, 37], which are bounded below due to being semipositone.Remark 6.
Note that (3) is the superlinear condition; i.e., the degree is 1; however, for our conditions, the corresponding degree can be any arbitrary positive number. For example, if we take f
~
2
(
t
,
x
,
y
)
=
q
(
x
)
=
x
γ with γ
>
0 for (
t
,
x
,
y
)
∈
[
1
,
T
]
Z
×
R
+
×
R
+, we see that this function does not satisfy the condition (3) if γ
∈
(
0,1
).Remark 7.
In this paper, we use the functionsp
,
q
,
ξ
,
η (see (H1) and (H3)) to act on u
i
-
w
i and then estimate the norms of u
i; however, in [36, 37] the corresponding parts only involve u
i
(
i
=
1,2
). Moreover, when the nonlinearities in [36] grow sublinearly at +
∞, nonnegative matrices are used to depict the coupling behaviors, yet this is not used in our paper.Example 8.
Letp
(
y
)
=
y
4
/
5, q
(
x
)
=
x
2, x
,
y
∈
R
+. Then l
i
m
x
→
+
∞
p
κ
2
q
(
x
)
/
x
=
l
i
m
x
→
+
∞
κ
2
4
/
5
x
8
/
5
/
x
=
+
∞, and p
,
q satisfy (H1). Moreover, we take f
~
1
,
f
~
2 as follows:(28)
f
~
1
t
,
x
,
y
=
d
1
κ
2
-
1
x
+
1
β
1
+
e
sin
⁡
t
κ
2
-
1
-
d
1
κ
2
-
1
y
,
f
~
2
t
,
x
,
y
=
1
β
2
+
e
cos
⁡
t
κ
2
-
1
M
3
1
-
β
3
x
β
3
,where β
1
,
β
2
>
0
,
β
3
>
2, (
t
,
x
,
y
)
∈
[
1
,
T
]
Z
×
R
+
×
R
+. We next show that f
~
1
,
f
~
2 satisfy (H1)(ii) and (H2). Suppose that (
t
,
x
,
y
)
∈
[
1
,
T
]
Z
×
[
0
,
M
3
]
×
[
0
,
M
3
]; we have(29)
f
~
1
t
,
x
,
y
≤
d
1
κ
2
-
1
M
3
+
1
β
1
+
e
sin
⁡
t
κ
2
-
1
-
d
1
κ
2
-
1
M
3
<
κ
2
-
1
M
3
,
f
~
2
t
,
x
,
y
≤
1
β
2
+
e
cos
⁡
t
κ
2
-
1
M
3
1
-
β
3
M
3
β
3
<
κ
2
-
1
M
3
.On the other hand, we also have(30)
lim
inf
y
→
+
∞
⁡
f
~
1
t
,
x
,
y
d
1
x
+
p
y
=
lim
inf
y
→
+
∞
⁡
d
1
κ
2
-
1
x
+
1
/
β
1
+
e
sin
⁡
t
κ
2
-
1
-
d
1
κ
2
-
1
y
d
1
x
+
y
4
/
5
=
+
∞
,
u
n
i
f
o
r
m
l
y
o
n
t
,
x
∈
1
,
T
Z
×
R
+
,and(31)
lim
inf
x
→
+
∞
⁡
f
~
2
t
,
x
,
y
q
x
=
lim
inf
x
→
+
∞
⁡
1
/
β
2
+
e
cos
⁡
t
κ
2
-
1
M
3
1
-
β
3
x
β
3
x
2
=
+
∞
,
u
n
i
f
o
r
m
l
y
o
n
t
,
y
∈
1
,
T
Z
×
R
+
.And so, (H1)(ii) and (H2) hold.
It follows, from (11), that(32)
f
1
t
,
x
,
y
=
f
~
1
t
,
x
,
y
-
L
x
-
h
1
t
=
d
1
κ
2
-
1
x
+
1
β
1
+
e
sin
⁡
t
κ
2
-
1
-
d
1
κ
2
-
1
y
-
L
x
-
h
1
t
,and this function may be unbounded from below if L and h
1 are large enough. So, this function is not applicable in [36, 37]. Moreover, we also note that f
~
1 is a linear function about x
,
y, and it does not satisfy the condition (3).Example 9.
Letξ
(
y
)
=
y
2, η
(
x
)
=
ln
⁡
(
x
+
1
), and x
,
y
∈
R
+. Then l
i
m
x
→
+
∞
ξ
κ
2
η
(
x
)
/
x
=
l
i
m
x
→
+
∞
κ
2
2
l
n
2
(
x
+
1
)
/
x
=
0, and ξ
,
η satisfy (H3). Moreover, we chose f
~
1
,
f
~
2 as follows:(33)
f
~
1
t
,
x
,
y
=
d
2
x
+
q
0
-
1
κ
2
-
1
M
3
e
M
3
+
β
4
+
sin
⁡
t
e
-
y
,
f
~
2
t
,
x
,
y
=
q
0
-
1
κ
2
-
1
M
3
e
M
3
+
β
5
+
cos
⁡
t
y
e
-
x
,where β
4
,
β
5
>
0, (
t
,
x
,
y
)
∈
[
1
,
T
]
Z
×
R
+
×
R
+. In what follows, we prove that f
~
1
,
f
~
2 satisfy (H3)(ii) and (H4). Indeed, if (
t
,
x
,
y
)
∈
[
1
,
T
]
Z
×
0
,
M
3
×
0
,
M
3, we have(34)
f
~
1
t
,
x
,
y
≥
q
0
-
1
κ
2
-
1
M
3
e
M
3
+
β
4
+
sin
⁡
t
e
-
y
>
q
0
-
1
κ
2
-
1
M
3
e
M
3
e
-
M
3
=
q
0
-
1
κ
2
-
1
M
3
,and(35)
f
~
2
t
,
x
,
y
>
q
0
-
1
κ
2
-
1
M
3
e
M
3
e
-
x
≥
q
0
-
1
κ
2
-
1
M
3
e
M
3
e
-
M
3
=
q
0
-
1
κ
2
-
1
M
3
.On the other hand, we obtain(36)
lim
sup
y
→
+
∞
⁡
f
~
1
t
,
x
,
y
d
2
x
+
ξ
y
=
lim
sup
y
→
+
∞
⁡
d
2
x
+
q
0
-
1
κ
2
-
1
M
3
e
M
3
+
β
4
+
sin
⁡
t
e
-
y
d
2
x
+
y
2
=
0
,
u
n
i
f
o
r
m
l
y
o
n
t
,
x
∈
1
,
T
Z
×
R
+
,and(37)
lim
sup
x
→
+
∞
⁡
f
~
2
t
,
x
,
y
η
x
=
lim
sup
x
→
+
∞
⁡
q
0
-
1
κ
2
-
1
M
3
e
M
3
+
β
5
+
cos
⁡
t
y
e
-
x
ln
⁡
x
+
1
=
0
,
u
n
i
f
o
r
m
l
y
o
n
t
,
y
∈
1
,
T
Z
×
R
+
.Consequently, (H3)(ii) and (H4) hold.
Next, from (11) we see that(38)
f
2
t
,
x
,
y
=
f
~
2
t
,
x
,
y
-
L
y
-
h
2
t
=
q
0
-
1
κ
2
-
1
M
3
e
M
3
+
β
5
+
cos
⁡
t
y
e
-
x
-
L
y
-
h
2
t
,and this function may be unbounded from below if L and h
2 are large enough. So, this function is not applicable in [36, 37]. Moreover, we also note that f
~
2 is a sublinear function about x, and it does not satisfy the condition (3).Theorem 10.
Suppose that (H0)-(H2) hold. Then (1) has at least one positive solution.Proof.
There exists a sufficiently largeR
>
max
⁡
{
M
1
,
M
2
}
=
M
3, for which we will prove that(39)
u
1
,
u
2
≠
B
u
1
,
u
2
+
λ
x
0
,
y
0
,
∀
u
1
,
u
2
∈
∂
B
R
×
B
R
∩
P
×
P
,
λ
≥
0
,where x
0
,
y
0
∈
P
0 are two given functions. Indeed, if not, there exist (
u
1
,
u
2
)
∈
∂
(
B
R
×
B
R
)
∩
(
P
×
P
) and λ
≥
0 such that (
u
1
,
u
2
)
=
B
(
u
1
,
u
2
)
+
λ
(
x
0
,
y
0
), and then(40)
u
1
t
=
B
1
u
1
,
u
2
t
+
λ
x
0
t
,
u
2
t
=
B
2
u
1
,
u
2
t
+
λ
y
0
t
,
f
o
r
t
∈
1
,
T
Z
.This implies u
1
(
t
)
≥
B
1
(
u
1
,
u
2
)
(
t
), and u
2
(
t
)
≥
B
2
(
u
1
,
u
2
)
(
t
) for t
∈
[
1
,
T
]
Z. Note that (
u
1
,
u
2
)
∈
∂
(
B
R
×
B
R
)
∩
(
P
×
P
) with R
>
max
⁡
{
M
1
,
M
2
}; this implies u
i
(
t
)
≥
w
i
(
t
) for t
∈
[
1
,
T
]
Z
,
i
=
1,2. From (H1) we have(41)
u
1
t
-
w
1
t
≥
B
1
u
1
,
u
2
t
-
w
1
t
≥
∑
s
=
1
T
G
t
,
s
d
1
u
1
s
-
w
1
s
+
p
u
2
s
-
w
2
s
-
c
1
-
w
1
t
≥
∑
s
=
1
T
G
t
,
s
d
1
u
1
s
-
w
1
s
+
p
u
2
s
-
w
2
s
-
c
3
,
f
o
r
t
∈
1
,
T
Z
,and(42)
u
2
t
-
w
2
t
≥
B
2
u
1
,
u
2
t
-
w
2
t
≥
∑
s
=
1
T
G
t
,
s
q
u
1
s
-
w
1
s
-
c
1
-
w
2
t
≥
∑
s
=
1
T
G
t
,
s
q
u
1
s
-
w
1
s
-
c
3
,
f
o
r
t
∈
1
,
T
Z
.As a result, for t
∈
[
1
,
T
]
Z, we have(43)
p
u
2
t
-
w
2
t
+
p
c
3
≥
p
u
2
t
-
w
2
t
+
c
3
≥
p
∑
s
=
1
T
G
t
,
s
q
u
1
s
-
w
1
s
=
p
∑
s
=
1
T
G
t
,
s
κ
2
-
1
κ
2
q
u
1
s
-
w
1
s
≥
∑
s
=
1
T
G
t
,
s
κ
2
-
1
p
κ
2
q
u
1
s
-
w
1
s
≥
∑
s
=
1
T
G
t
,
s
κ
2
-
1
κ
2
γ
1
u
1
s
-
w
1
s
-
c
1
≥
γ
1
∑
s
=
1
T
G
t
,
s
u
1
s
-
w
1
s
-
c
4
.This means(44)
p
u
2
t
-
w
2
t
≥
γ
1
∑
s
=
1
T
G
t
,
s
u
1
s
-
w
1
s
-
c
5
.Therefore,(45)
u
1
t
-
w
1
t
≥
∑
s
=
1
T
G
t
,
s
d
1
u
1
s
-
w
1
s
+
p
u
2
s
-
w
2
s
-
c
3
≥
d
1
∑
s
=
1
T
G
t
,
s
u
1
s
-
w
1
s
+
∑
s
=
1
T
G
t
,
s
γ
1
∑
τ
=
1
T
G
s
,
τ
u
1
τ
-
w
1
τ
-
c
5
-
c
3
≥
d
1
∑
s
=
1
T
G
t
,
s
u
1
s
-
w
1
s
+
γ
1
∑
s
=
1
T
G
t
,
s
∑
τ
=
1
T
G
s
,
τ
u
1
τ
-
w
1
τ
-
c
6
,
f
o
r
t
∈
1
,
T
Z
.Multiply both sides of the above inequality by φ
(
t
) and sum from 1 to T. Then together with Lemma 1(iii) we obtain(46)
∑
t
=
1
T
u
1
t
-
w
1
t
φ
t
≥
∑
t
=
1
T
φ
t
d
1
∑
s
=
1
T
G
t
,
s
u
1
s
-
w
1
s
+
γ
1
∑
s
=
1
T
G
t
,
s
∑
τ
=
1
T
G
s
,
τ
u
1
τ
-
w
1
τ
-
c
6
≥
κ
1
d
1
∑
t
=
1
T
u
1
t
-
w
1
t
φ
t
+
γ
1
κ
1
2
∑
t
=
1
T
u
1
t
-
w
1
t
φ
t
-
c
7
.Hence, we have(47)
∑
t
=
1
T
u
1
t
-
w
1
t
φ
t
≤
c
7
γ
1
κ
1
2
+
d
1
κ
1
-
1
,
∑
t
=
1
T
u
1
t
φ
t
≤
c
7
γ
1
κ
1
2
+
d
1
κ
1
-
1
+
∑
t
=
1
T
w
1
t
φ
t
≔
N
1
.From (16), (40), and x
0
∈
P
0 we have u
1
∈
P
0. This implies(48)
κ
1
u
1
=
u
1
∑
t
=
1
T
q
∗
t
φ
t
≤
∑
t
=
1
T
u
1
t
φ
t
≤
N
1
,
u
1
≤
κ
1
-
1
N
1
.Note that from (16), (40), and y
0
∈
P
0, we find u
2
∈
P
0. From the definition of w
2, we know w
2
∈
P
0, and this implies u
2
-
w
2
∈
P
0. Moreover, we may assume u
2
(
t
)
-
w
2
(
t
)
≢
0, for t
∈
[
1
,
T
]
Z. Then u
2
-
w
2
>
0 and p
(
u
2
-
w
2
)
>
0. Thus, from the concavity of p, we have(49)
κ
1
u
2
-
w
2
=
u
2
-
w
2
∑
t
=
1
T
q
∗
t
φ
t
≤
∑
t
=
1
T
u
2
t
-
w
2
t
φ
t
=
u
2
-
w
2
p
u
2
-
w
2
∑
t
=
1
T
u
2
t
-
w
2
t
u
2
-
w
2
p
u
2
-
w
2
φ
t
≤
u
2
-
w
2
p
u
2
-
w
2
∑
t
=
1
T
p
u
2
t
-
w
2
t
φ
t
.This implies that(50)
p
u
2
-
w
2
≤
κ
1
-
1
∑
t
=
1
T
p
u
2
t
-
w
2
t
φ
t
.
On the other hand, from (41) and Lemma 1(ii) we obtain(51)
u
1
t
+
c
3
≥
∑
s
=
1
T
G
t
,
s
p
u
2
s
-
w
2
s
≥
∑
s
=
1
T
q
∗
t
φ
s
p
u
2
s
-
w
2
s
≥
q
0
∑
s
=
1
T
p
u
2
t
-
w
2
t
φ
t
.Combining the above two inequalities, we get(52)
p
u
2
-
w
2
≤
κ
1
q
0
-
1
u
1
t
+
c
3
≤
κ
1
q
0
-
1
κ
1
-
1
N
1
+
c
3
.Note that triangular inequality p
(
u
2
)
≤
p
(
u
2
-
w
2
)
+
p
(
w
2
) and from (H1), l
i
m
z
→
+
∞
p
(
z
)
=
+
∞, and thus there exists N
2
>
0 such that u
2
≤
N
2.
Consequently, we obtain thatu
1
≤
κ
1
-
1
N
1 and u
2
≤
N
2. As a result, we can choose R
>
m
a
x
{
M
3
,
κ
1
-
1
N
1
,
N
2
} and thus (39) holds true. Consequently, Lemma 3 indicates that(53)
i
B
,
B
R
×
B
R
∩
P
×
P
,
P
×
P
=
0
.
Then we show that(54)
u
1
,
u
2
≠
λ
B
u
1
,
u
2
,
∀
u
1
,
u
2
∈
∂
B
M
3
×
B
M
3
∩
P
×
P
,
λ
∈
0,1
.Indeed, if not, there exist (
u
1
,
u
2
)
∈
∂
(
B
M
3
×
B
M
3
)
∩
(
P
×
P
) and λ
0
∈
[
0,1
] such that (
u
1
,
u
2
)
=
λ
0
B
(
u
1
,
u
2
). This implies that(55)
u
1
t
≤
B
1
u
1
,
u
2
t
,
u
2
t
≤
B
2
u
1
,
u
2
t
,
f
o
r
t
∈
1
,
T
Z
.Hence, u
1
≤
B
1
(
u
1
,
u
2
) and u
2
≤
B
2
(
u
1
,
u
2
). However, from (H2) we have(56)
B
1
u
1
,
u
2
t
=
∑
s
=
1
T
G
t
,
s
f
~
s
,
u
1
s
-
w
1
s
,
u
2
s
-
w
2
s
<
∑
s
=
1
T
φ
s
κ
2
-
1
M
3
=
M
3
,for all t
∈
[
1
,
T
]
Z. Similarly, B
2
(
u
1
,
u
2
)
<
M
3. Note that(57)
u
1
,
u
2
=
max
⁡
u
1
,
u
2
≤
max
⁡
B
1
u
1
,
u
2
,
B
2
u
1
,
u
2
<
M
3
=
u
1
,
u
2with (
u
1
,
u
2
)
∈
∂
(
B
M
3
×
B
M
3
)
∩
(
P
×
P
). This is a contradiction. So (54) is true. It follows from Lemma 4 that(58)
i
B
,
B
M
3
×
B
M
3
∩
P
×
P
,
P
×
P
=
1
.
From (53) and (58) we have(59)
i
B
,
B
R
×
B
R
∖
B
¯
M
3
×
B
¯
M
3
∩
P
×
P
,
P
×
P
=
0
-
1
=
-
1
.Therefore B has at least one fixed point (
u
1
,
u
2
) in (
(
B
R
×
B
R
)
∖
(
B
¯
M
3
×
B
¯
M
3
)
)
∩
(
P
×
P
) with u
1
≥
M
1
,
u
2
≥
M
2 (note that u
i
=
M
3
≥
M
i
,
i
=
1,2), and then (
u
1
-
w
1
,
u
2
-
w
2
) is a positive solution for (1). This completes the proof.Theorem 11.
Suppose that (H0), (H3), and (H4) hold. Then (1) has at least one positive solution.Proof.
There exists a sufficiently largeR
>
m
a
x
{
M
1
,
M
2
}, for which we shall prove that(60)
u
1
,
u
2
≠
λ
B
u
1
,
u
2
,
∀
u
1
,
u
2
∈
∂
B
R
×
B
R
∩
P
×
P
,
λ
∈
0,1
.Indeed, if not, there exist (
u
1
,
u
2
)
∈
∂
(
B
R
×
B
R
)
∩
(
P
×
P
) and λ
0
∈
[
0,1
] such that (
u
1
,
u
2
)
=
λ
0
B
(
u
1
,
u
2
). This indicates that(61)
u
1
t
≤
B
1
u
1
,
u
2
t
,
u
2
t
≤
B
2
u
1
,
u
2
t
,
f
o
r
t
∈
1
,
T
Z
.Note that (
u
1
,
u
2
)
∈
∂
(
B
R
×
B
R
)
∩
(
P
×
P
) with R
>
m
a
x
{
M
1
,
M
2
}; this implies u
i
(
t
)
≥
w
i
(
t
) for t
∈
[
1
,
T
]
Z
,
i
=
1,2. From (H3) we have(62)
u
1
t
-
w
1
t
≤
∑
s
=
1
T
G
t
,
s
d
2
u
1
s
-
w
1
s
+
ξ
u
2
s
-
w
2
s
-
w
1
t
≤
∑
s
=
1
T
G
t
,
s
d
2
u
1
s
-
w
1
s
+
ξ
u
2
s
-
w
2
s
,and(63)
u
2
t
-
w
2
t
≤
∑
s
=
1
T
G
t
,
s
η
u
1
s
-
w
1
s
-
w
2
t
≤
∑
s
=
1
T
G
t
,
s
η
u
1
s
-
w
1
s
,for t
∈
[
1
,
T
]
Z. Consequently, for all t
∈
[
1
,
T
]
Z, we have(64)
ξ
u
2
t
-
w
2
t
≤
ξ
∑
s
=
1
T
G
t
,
s
η
u
1
s
-
w
1
s
=
ξ
∑
s
=
1
T
G
t
,
s
κ
2
-
1
κ
2
η
u
1
s
-
w
1
s
≤
∑
s
=
1
T
G
t
,
s
κ
2
-
1
ξ
κ
2
η
u
1
s
-
w
1
s
≤
∑
s
=
1
T
G
t
,
s
κ
2
-
1
κ
2
γ
2
u
1
s
-
w
1
s
+
c
2
≤
γ
2
∑
s
=
1
T
G
t
,
s
u
1
s
-
w
1
s
+
c
8
.This, together with (62), implies that(65)
u
1
t
-
w
1
t
≤
∑
s
=
1
T
G
t
,
s
d
2
u
1
s
-
w
1
s
+
ξ
u
2
s
-
w
2
s
≤
d
2
∑
s
=
1
T
G
t
,
s
u
1
s
-
w
1
s
+
γ
2
∑
s
=
1
T
G
t
,
s
∑
τ
=
1
T
G
s
,
τ
u
1
τ
-
w
1
τ
+
c
9
.Multiply both sides of the above inequality by φ
(
t
) and sum from 1 to T. Then together with Lemma 1(iii) we obtain(66)
∑
t
=
1
T
u
1
t
-
w
1
t
φ
t
≤
∑
t
=
1
T
φ
t
d
2
∑
s
=
1
T
G
t
,
s
u
1
s
-
w
1
s
+
γ
2
∑
s
=
1
T
G
t
,
s
∑
τ
=
1
T
G
s
,
τ
u
1
τ
-
w
1
τ
+
c
9
≤
d
2
κ
2
∑
t
=
1
T
u
1
t
-
w
1
t
φ
t
+
γ
2
κ
2
2
∑
t
=
1
T
u
1
t
-
w
1
t
φ
t
+
c
9
∑
t
=
1
T
φ
t
.Hence, we have(67)
∑
t
=
1
T
u
1
t
-
w
1
t
φ
t
≤
κ
2
c
9
1
-
d
2
κ
2
-
γ
2
κ
2
2
,
∑
t
=
1
T
u
1
t
φ
t
≤
κ
2
c
9
1
-
d
2
κ
2
-
γ
2
κ
2
2
+
∑
t
=
1
T
w
1
t
φ
t
≔
N
3
.Note that u
1
∈
P
0 from the fact that B
1
(
P
×
P
)
⊂
P
0. This implies(68)
u
1
∑
t
=
1
T
q
∗
t
φ
t
≤
∑
t
=
1
T
u
1
t
φ
t
≤
N
3
,
u
1
≤
κ
1
-
1
N
3
.On the other hand, for t
∈
[
1
,
T
]
Z, by (64) and Lemma 1(ii) we obtain(69)
ξ
u
2
t
-
w
2
t
≤
γ
2
∑
s
=
1
T
φ
s
u
1
s
+
c
8
≤
γ
2
N
3
+
c
8
.Note that u
2
∈
P
0 for B
2
(
P
×
P
)
⊂
P
0, and from the definition of w
2, we also have w
2
∈
P
0. This, for t
∈
[
1
,
T
]
Z, implies that(70)
ξ
q
0
u
2
-
w
2
≤
ξ
u
2
t
-
w
2
t
≤
γ
2
N
3
+
c
8
.As a result, combining with triangular inequality of norm, there exists N
4
>
0 such that u
2
≤
N
4.
Consequently, we can conclude thatu
1
≤
κ
1
-
1
N
3 and u
2
≤
N
4. Therefore, we choose R
>
max
⁡
M
3
,
κ
1
-
1
N
3
,
N
4 such that (60) holds true. Lemma 4 implies that(71)
i
B
,
B
R
×
B
R
∩
P
×
P
,
P
×
P
=
1
.
Note that the definition ofM
3. Secondly, we prove that(72)
u
1
,
u
2
≠
B
u
1
,
u
2
+
λ
x
0
,
y
0
,
∀
u
1
,
u
2
∈
∂
B
M
3
×
B
M
3
∩
P
×
P
,
λ
≥
0
,where x
0
,
y
0
∈
P are two fixed functions. Indeed, if not, there exist (
u
1
,
u
2
)
∈
∂
(
B
M
3
×
B
M
3
)
∩
(
P
×
P
) and λ
0
≥
0 such that (
u
1
,
u
2
)
=
B
(
u
1
,
u
2
)
+
λ
0
(
x
0
,
y
0
). This implies that(73)
u
1
t
≥
B
1
u
1
,
u
2
t
,
u
2
t
≥
B
2
u
1
,
u
2
t
,
f
o
r
t
∈
1
,
T
Z
.Hence, u
1
≥
B
1
(
u
1
,
u
2
) and u
2
≥
B
2
(
u
1
,
u
2
). However, from (H4) we have(74)
B
1
u
1
,
u
2
t
=
∑
s
=
1
T
G
t
,
s
f
~
s
,
u
1
s
-
w
1
s
,
u
2
s
-
w
2
s
>
∑
s
=
1
T
q
∗
t
φ
s
q
0
-
1
κ
2
-
1
M
3
≥
M
3
,for all t
∈
[
1
,
T
]
Z. This implies B
1
(
u
1
,
u
2
)
>
M
3. Similarly, B
2
(
u
1
,
u
2
)
>
M
3. This implies(75)
u
1
,
u
2
=
max
⁡
u
1
,
u
2
≥
max
⁡
B
1
u
1
,
u
2
,
B
2
u
1
,
u
2
>
M
3
=
u
1
,
u
2with (
u
1
,
u
2
)
∈
∂
(
B
M
3
×
B
M
3
)
∩
(
P
×
P
). This is a contradiction. Hence, (72) is true. Lemma 3 yields that(76)
i
B
,
B
M
3
×
B
M
3
∩
P
×
P
,
P
×
P
=
0
.
From (71) and (76) we have(77)
i
(
B
,
B
R
×
B
R
∖
B
¯
M
3
×
B
¯
M
3
∩
P
×
P
,
P
×
P
=
1
-
0
=
1
.Hence B has at least one fixed point (
(
B
R
×
B
R
)
∖
(
B
¯
M
3
×
B
¯
M
3
)
∩
(
P
×
P
) with u
1
≥
M
1, u
2
≥
M
2(note that u
i
=
M
3
≥
M
i
,
i
=
1,2), and thus (
u
1
-
w
1
,
u
2
-
w
2
) is a positive solution for (1). This completes the proof.



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*Source: 3203401-2019-02-03.xml*