# Periodic Solutions of Second Order Nonlinear Difference Equations with Singularϕ-Laplacian Operator

Authors: Ruyun Ma; Yanqiong Lu
Journal: Discrete Dynamics in Nature and Society
 (2014)

---

## Abstract

We obtain some existence results of solutions for discrete periodic boundary value problems with singularϕ-Laplacian operator ∇
Δ
u
k
/
1
-
κ
(
Δ
u
k
)
2
+
r
k
u
k
+
m
k
/
(
u
k
)
λ
=
e
k, 2
≤
k
≤
N
-
1, u
1
=
u
N, and Δ
u
1
=
Δ
u
N
-
1 by using the upper and lower solutions method and Brouwer degree theory, where κ
>
0 is a constant, r
=
(
r
2
,
…
,
r
N
-
1
), m
=
(
m
2
,
…
,
m
N
-
1
), e
=
(
e
2
,
…
,
e
N
-
1
)
∈
R
N
-
2, and λ
>
0 is a parameter. We also give some examples with singular nonlinearities to illustrate our main results.


---

## Body

## 1. Introduction

In this paper we present some existence results for the discrete periodic boundary value problems with singularϕ-Laplacian operator
(1)
∇
(
Δ
u
k
1
-
κ
(
Δ
u
k
)
2
)
+
r
k
u
k
+
m
k
(
u
k
)
λ
=
e
k
,
5555555555555555555555
k
∈
[
2
,
N
-
1
]
Z
,
u
1
=
u
N
,
Δ
u
1
=
Δ
u
N
-
1
,
where κ
>
0 is a constant, r
=
(
r
2
,
…
,
r
N
-
1
), m
=
(
m
2
,
…
,
m
N
-
1
), e
=
(
e
2
,
…
,
e
N
-
1
)
∈
R
N
-
2, λ
>
0 is a parameter, and [
2
,
N
-
1
]
Z
≔
{
2,3
,
…
,
N
-
1
} with N
>
4 is an integer.These problems are originated in the study that a particle moves on a straight line, subject to a restoring forceF with relativistic effects, which can be described by the differential equation of motion:
(2)
(
m
0
u
′
1
-
u
′
2
/
c
2
)
′
=
F
,
where m
0 is the rest mass of the particle and c is the speed of light in the vacuum (see [1–3]). Assume that m
0
=
c
=
1. The existence and multiplicity of solutions for (2) subjected to Dirichlet, Robin, periodic, or Neumann boundary conditions have been studied by various methods, such as the method of lower and upper solutions, topological degree theory, and critical point theory; see [4–8] and the references therein.An interesting question is which techniques and theorems regarding the continuous differential equations can be adapted for difference equations (see Kelly and Peterson [9], Agarwal [10], and Bereanu and Mawhin [11, 12]). The purpose of this paper is to show that some known existence and multiplicity results of periodic solution for singular perturbations of the singular ϕ-Laplacian operator also hold for the corresponding difference equation and develop some results for the singular difference equation boundary value problems; see [12–17].In the caseκ
=
1 and r
=
0, Bereanu and Mawhin [12] proved that the discrete periodic problem with an attractive nonlinearity
(3)
∇
(
Δ
u
k
1
-
(
Δ
u
k
)
2
)
+
1
(
u
k
)
λ
=
e
k
,
k
∈
[
2
,
N
-
1
]
Z
,
u
1
=
u
N
,
Δ
u
1
=
Δ
u
N
-
1
has at least one positive solution if and only if e
¯
=
1
/
(
N
-
2
)
∑
s
=
2
N
-
1
e
s
>
0. When λ
≥
1, they also showed that the repulsive singular periodic problem
(4)
∇
(
Δ
u
k
1
-
(
Δ
u
k
)
2
)
-
1
(
u
k
)
λ
=
e
k
,
k
∈
[
2
,
N
-
1
]
Z
,
u
1
=
u
N
,
Δ
u
1
=
Δ
u
N
-
1
has at least one positive solution if and only if e
¯
<
0.In the caseκ
=
0, the problem (1) is the classical discrete periodic problem
(5)
∇
(
Δ
u
k
)
+
r
k
u
k
+
m
k
(
u
k
)
λ
=
e
k
,
k
∈
[
2
,
N
-
1
]
Z
,
(6)
u
1
=
u
N
,
Δ
u
1
=
Δ
u
N
-
1
.
This problem has been studied by Ma and Lu [15–17] where the existence of positive solution needs a necessary condition r
k
<
0 or 0
<
r
k
<
4
sin
2
(
π
/
2
N
) (see [14]). For other results concerning the existence of solutions for singular nonlinear difference equation boundary value problems, see, for example, [9, 10, 18].It is interesting to remark that, in contrast to the classical case, the periodic problem with discrete relativistic acceleration(7)
∇
(
Δ
u
k
1
-
(
Δ
u
k
)
2
)
+
r
u
k
=
e
k
,
k
∈
[
2
,
N
-
1
]
Z
,
u
1
=
u
N
,
Δ
u
1
=
Δ
u
N
-
1
has at least one solution for any r
≠
0 and any forcing term e (see [12, Corollary  2]). Note that, for this type of problems, in some sense, the same situation occurs also if we add a singular nonlinearity.In order to explain the main result, let us introduce some notation. For anyx
∈
R, we write x
+
=
max
⁡
{
x
,
0
} and x
-
=
max
⁡
{
-
x
,
0
}. For e
=
(
e
2
,
…
,
e
N
-
1
)
∈
R
N
-
2, we put
(8)
E
=
∑
k
=
2
N
-
1
‍
e
k
,
E
±
=
∑
k
=
2
N
-
1
‍
e
k
±
and note that E
=
E
+
-
E
-.Motivated by the above results from [12, 14–17], we consider the discrete periodic problem
(9)
∇
(
Δ
u
k
1
-
κ
(
Δ
u
k
)
2
)
+
r
k
u
k
-
1
(
u
k
)
λ
=
e
k
,
5555555555555555555555
k
∈
[
2
,
N
-
1
]
Z
,
u
1
=
u
N
,
Δ
u
1
=
Δ
u
N
-
1
,
where r
=
(
r
2
,
…
,
r
N
-
1
), e
=
(
e
2
,
…
,
e
N
-
1
)
∈
R
N
-
2, and λ
≥
1. If either r
¯
>
0 or r
¯
=
0 and e
¯
<
-
N
R
-
/
2
κ
(
N
-
2
), then we prove that the above problem has at least one solution (see Example 14). In the case r
¯
<
0, we show that the periodic problem with repulsive singular term
(10)
∇
(
Δ
u
k
1
-
κ
(
Δ
u
k
)
2
)
+
r
k
u
k
-
m
k
(
u
k
)
λ
=
e
k
,
5555555555555555555555
k
∈
[
2
,
N
-
1
]
Z
,
u
1
=
u
N
,
Δ
u
1
=
Δ
u
N
-
1
,
with λ
>
0 and m
=
(
m
2
,
…
,
m
N
-
1
)
≥
0, is solvable (see Example 18) provided that
(11)
-
E
>
(
1
+
λ
)
[
|
R
|
λ
M
λ
λ
]
1
/
(
1
+
λ
)
+
N
R
-
2
κ
.
On the other hand, we also consider the periodic problem with attractive singularity
(12)
∇
(
Δ
u
k
1
-
κ
(
Δ
u
k
)
2
)
+
r
k
u
k
+
m
k
(
u
k
)
λ
=
e
k
,
55555555555555555555555
k
∈
[
2
,
N
-
1
]
Z
,
u
1
=
u
N
,
Δ
u
1
=
Δ
u
N
-
1
,
where r
=
(
r
2
,
…
,
r
N
-
1
), m
=
(
m
2
,
…
,
m
N
-
1
), and e
=
(
e
2
,
…
,
e
N
-
1
)
∈
R
N
-
2 with m
≥
0 and λ
>
0. If either r
¯
<
0 or r
=
0 and in both cases E
<
M
(
2
κ
/
N
)
λ
-
(
N
R
-
/
2
κ
), then the problem (12) has at least one solution (see Example 19). Moreover, in the pure attractive case, that is, m
>
0, it follows that (12) is solvable if either r
¯
<
0 or r
¯
=
0 and E
>
N
R
+
/
2
κ (Theorem 21).The rest of the paper is organized as follows. In Section2, we introduce some notations and auxiliary results. In Section 3 we establish the method of non-well-ordered lower and upper solutions and give an application on the discrete periodic problem with the strong repulsive nonlinearities. In addition, we introduce some methods to construct lower and upper solutions. Finally, in Section 4 we give some applications to deal with the singular perturbations periodic problems.


## 2. Some Notations and Auxiliary Results

In this section, we first introduce some notations. Leta
,
b
∈
N with a
<
b; we denote [
a
,
b
]
Z
≔
{
a
,
a
+
1
,
…
,
b
}.Foru
=
(
u
1
,
…
,
u
p
)
∈
R
p, set ∥
u
∥
∞
=
max
⁡
1
≤
k
≤
p
|
u
k
|, ∥
u
∥
1
=
∑
k
=
1
p
‍
|
u
k
|. If α
,
β
∈
R
p, we write α
≤
β (resp., α
<
β) if α
k
≤
β
k (resp., α
k
<
β
k) for all 1
≤
k
≤
p. The following assumption upon ϕ (calledsingular) is made throughout the paper:(H
ϕ) ϕ
:
(
-
a
,
a
)
→
R
(
0
<
a
<
∞
) is an increasing homeomorphism with ϕ
(
0
)
=
0.The model example is(13)
ϕ
(
s
)
=
s
1
-
κ
s
2
,
s
∈
(
-
1
κ
,
1
κ
)
.LetN
∈
N with N
≥
4 be fixed and u
=
(
u
1
,
u
2
,
…
,
u
N
)
∈
R
N. Then we denote
(14)
Δ
u
=
(
Δ
u
1
,
…
,
Δ
u
N
-
1
)
∈
R
N
-
1
,
where Δ
u
k
=
u
k
+
1
-
u
k for k
∈
[
1
,
N
-
1
]
Z and if ∥
Δ
u
∥
∞
≔
max
⁡
k
∈
[
1
,
N
-
1
]
Z
⁡
|
Δ
u
k
|
<
a, define
(15)
∇
[
ϕ
(
Δ
u
)
]
=
(
∇
[
ϕ
(
Δ
u
2
)
]
,
…
,
∇
[
ϕ
(
Δ
u
N
-
1
)
]
)
∈
R
N
-
2
,
where ∇
[
ϕ
(
Δ
u
k
)
]
=
ϕ
(
Δ
u
k
)
-
ϕ
(
Δ
u
k
-
1
) for k
∈
[
2
,
N
-
1
]
Z.Letf
:
[
2
,
N
-
1
]
Z
×
R
2
→
R be a continuous function. Then its Nemytskii operator N
f
(
u
)
:
R
N
→
R
N
-
2 is given by
(16)
N
f
(
u
)
=
(
f
(
2
,
u
2
,
Δ
u
2
)
,
…
,
f
(
N
-
1
,
u
N
-
1
,
Δ
u
N
-
1
)
)
.
It follows that N
f is continuous and takes bounded sets into bounded sets.LetQ be the projectors defined by
(17)
u
¯
=
Q
u
=
1
N
-
2
∑
k
=
2
N
-
1
‍
u
k
∀
u
∈
R
N
.
If u
∈
R
N, we write u
~
=
u
-
u
¯ and we will consider the following closed subspaces of R
N:
(18)
V
N
-
2
=
{
u
∈
R
N
∣
u
1
=
u
N
,
Δ
u
1
=
Δ
u
N
-
1
}
.Let the vector spaceV
N
-
2 be endowed with the orientation of R
N and the norm ∥
u
∥
∞
=
max
⁡
1
≤
k
≤
N
⁡
|
u
k
|. Its elements can be associated with the coordinates (
u
2
,
…
,
u
N
-
1
) and correspond to the elements of R
N of the form
(19)
(
u
2
+
u
N
-
1
2
,
u
2
,
u
3
,
…
,
u
N
-
1
,
u
2
+
u
N
-
1
2
)
.Now, we recall the following technical result given as Lemma 1 from [12].Lemma 1.
For eachh
=
(
h
2
,
…
,
h
N
-
1
), there exists a unique γ
≔
Q
ϕ
(
h
) such that
(20)
2
ϕ
-
1
(
γ
)
+
∑
k
=
3
N
-
1
‍
ϕ
-
1
(
∑
j
=
2
k
-
1
‍
h
j
+
γ
)
=
0
.
Moreover, the function Q
ϕ is continuous.Lemma 2.
LetF
:
R
N
→
R
N
-
2 be a continuous operator which takes bounded sets into bounded sets and consider the abstract discrete periodic problem:
(21)
∇
[
ϕ
(
Δ
u
)
]
=
F
(
u
)
,
u
1
=
u
N
,
Δ
u
1
=
Δ
u
N
-
1
.
A function u is a solution of (21) if and only if u
∈
V
N
-
2 is a fixed point of the continuous operator A
F
:
V
N
-
2
→
V
N
-
2 defined by A
F
(
u
)
=
v, where v
=
(
v
1
,
v
2
,
…
,
v
N
)
∈
V
N
-
2 satisfying
(22)
v
2
=
u
2
+
Q
N
F
(
u
)
,
v
k
=
u
2
+
Q
N
F
(
u
)
+
∑
j
=
3
k
‍
ϕ
-
1
(
∑
l
=
2
j
-
1
‍
F
(
u
k
)
+
Q
ϕ
(
N
F
(
u
)
)
)
,
5555555555555555555555
k
∈
[
3
,
N
-
1
]
Z
.
Furthermore, ∥
Δ
(
A
(
u
)
)
∥
∞
<
a for all u
∈
V
N
-
2 and
(23)
∥
u
~
∥
∞
<
a
(
N
-
2
)
for any solution u of (21).Let us consider the discrete periodic problem(24)
∇
[
ϕ
(
Δ
u
k
)
]
=
f
(
k
,
u
k
,
Δ
u
k
)
,
u
1
=
u
N
,
Δ
u
1
=
Δ
u
N
-
1
.
Obviously, from Lemma 2, the fixed point operator associated with (24) is
(25)
A
f
(
u
)
=
u
.Now, we state the method of upper and lower solutions for discrete periodic problem (24) according to Bereanu and Mawhin [12].Definition 3.
A functionα
=
(
α
1
,
…
,
α
N
) (resp., β
=
(
β
1
,
…
,
β
N
)) is called a lower solution (resp., an upper solution) for (24) if ∥
Δ
α
∥
∞
<
a (resp., ∥
Δ
β
∥
∞
<
a) and
(26)
∇
[
ϕ
(
Δ
α
k
)
]
≥
f
(
k
,
α
k
,
Δ
α
k
)
,
(
resp
.
∇
[
ϕ
(
Δ
β
k
)
]
≤
f
(
k
,
β
k
,
Δ
β
k
)
)
,
55555555555555555
k
∈
[
2
,
N
-
1
]
Z
,
α
1
=
α
N
,
Δ
α
1
≥
Δ
α
N
-
1
,
(
resp
.
β
1
=
β
N
,
Δ
β
1
≤
Δ
β
N
-
1
)
.
Such a lower or an upper solution is called strict if the inequality (26) is strict.Lemma 4 (see [12, Theorem  3]).
If (24) has a lower solution α and an upper solution β such that α
≤
β, then (24) has a solution u such that α
≤
u
≤
β. Moreover, if α and β are strict, then α
<
u
<
β, and
(27)
deg
⁡
[
I
-
A
f
,
Ω
α
,
β
,
0
]
=
-
1
,
where Ω
α
,
β
=
{
u
∈
V
N
-
2
∣
α
<
u
<
β
,
∥
Δ
u
∥
∞
<
a
}.An easy adaption of the proof of [12, Theorem  3] provides the following useful result.Lemma 5.
Assume that (24) has a lower solution α and an upper solution β such that α
<
β, and
(28)
u
≠
A
f
(
u
)
∀
u
∈
∂
Ω
α
,
β
.
Then
(29)
deg
⁡
[
I
-
A
f
,
Ω
α
,
β
,
0
]
=
-
1
.The next result is an elementary estimation of the oscillation of a periodic function.Lemma 6.
Ifu
:
Z
→
R is a N-periodic function, then
(30)
max
⁡
k
∈
[
1
,
N
]
Z
u
k
-
min
⁡
k
∈
[
1
,
N
]
Z
u
k
≤
N
2
∥
Δ
u
∥
∞
.Proof.
Letk
*
∈
[
1
,
N
-
1
]
Z be such that u
k
*
=
min
⁡
k
∈
[
1
,
N
]
Z
⁡
u
k, and let k
*
∈
[
k
*
,
k
*
+
N
]
Z be such that u
k
*
=
max
⁡
k
∈
[
1
,
N
]
Z
⁡
u
k. We have that
(31)
u
k
*
-
u
k
*
=
∑
s
=
k
*
k
*
-
1
‍
Δ
u
s
≤
(
k
*
-
k
*
)
∥
Δ
u
∥
∞
,
u
k
*
-
u
k
*
=
∑
s
=
k
*
k
*
+
N
-
1
‍
(
-
Δ
u
s
)
≤
(
k
*
+
N
-
k
*
)
∥
Δ
u
∥
∞
.
Then, multiplying both inequalities and using that x
y
≤
(
x
+
y
)
2
/
2, for all x
,
y
∈
R, it follows that
(32)
(
u
k
*
-
u
k
*
)
2
≤
[
N
∥
Δ
u
∥
∞
]
2
4
,
and the proof is completed.


## 3. The Method of Lower and Upper Solutions and Application

In 2008, Bereanu and Mawhin [12] proved that problem (24) has at least one solution if it has a lower solution α and an upper solution β with α
≤
β. In the following result we prove some additional concerning the location of the solution. In particular, we have a posteriori estimations which will be very useful in the sequel (Remark 8).Theorem 7.
Assume that (24) has a lower solution α and an upper solution β such that
(33)
∃
k
⋆
∈
[
1
,
N
]
Z
:
α
k
⋆
>
β
k
⋆
.
Then (24) has at least one solution u such that
(34)
min
⁡
⁡
{
α
k
u
,
β
k
u
}
≤
u
k
u
≤
max
⁡
{
α
k
u
,
β
k
u
}
f
o
r
s
o
m
e
k
u
∈
[
1
,
N
]
Z
.Proof.
Let(35)
u
*
=
∥
α
∥
∞
+
∥
β
∥
∞
+
a
(
N
-
2
)
,
m
=
max
⁡
{
|
f
(
k
,
u
,
v
)
|
+
1
∣
(
k
,
u
,
v
)
∈
[
2
,
N
-
1
]
Z
×
[
-
u
*
-
2
,
u
*
+
2
]
×
[
-
a
,
a
]
}
,
and define the continuous function g
:
[
2
,
N
-
1
]
Z
×
R
2
→
R by
(36)
g
(
k
,
u
,
v
)
=
{
-
m
-
1
,
u
≤
-
u
*
-
1
,
f
(
k
,
u
,
v
)
+
(
u
+
u
*
)
×
(
m
+
1
+
f
(
k
,
u
,
v
)
)
,
-
u
*
-
1
<
u
<
-
u
*
,
f
(
k
,
u
,
v
)
,
-
u
*
≤
u
≤
u
*
,
f
(
k
,
u
,
v
)
+
(
u
-
u
*
)
m
,
u
*
<
u
<
u
*
+
1
,
f
(
k
,
u
,
v
)
+
m
,
u
≥
u
*
+
1
.
Let us consider the modified periodic problem(37)
∇
[
ϕ
(
Δ
u
k
)
]
=
g
(
k
,
u
k
,
Δ
u
k
)
,
u
1
=
u
N
,
Δ
u
1
=
Δ
u
N
-
1
,
and let A
g be the fixed point operator associated with (37).
It is not difficult to verify thatα is a lower solution and β is an upper solution of the problem (37). Moreover, by computation, α
1
=
-
u
*
-
2 is a lower solution of (37) and β
1
=
u
*
+
2 is an upper solution of (37). Notice that
(38)
α
1
<
min
⁡
{
α
,
β
}
≤
max
⁡
{
α
,
β
}
<
β
1
,
which together with (33) imply that
(39)
Ω
α
1
,
β
∪
Ω
α
,
β
1
⊂
Ω
α
1
,
β
1
,
Ω
α
1
,
β
∩
Ω
α
,
β
1
=
∅
.
So, we can consider the open bounded set
(40)
Ω
=
Ω
α
1
,
β
1
∖
[
Ω
α
1
,
β
¯
∪
Ω
α
,
β
1
¯
]
.
It follows that
(41)
Ω
=
{
u
∈
Ω
α
1
,
β
1
∣
u
k
u
>
β
k
u
,
u
s
u
<
α
s
u
for
some
k
u
,
s
u
∈
[
1
,
N
]
Z
}
,
∂
Ω
=
∂
Ω
α
1
,
β
1
∪
∂
Ω
α
1
,
β
∪
∂
Ω
α
,
β
1
.
Clearly, any constant function between β
k
⋆ and α
k
⋆ is contained in Ω, so Ω
≠
∅.
Next, let us consideru
∈
∂
Ω such that A
g
(
u
)
=
u and ∥
u
∥
∞
=
u
*
+
2. Notice that one has ∥
Δ
u
∥
∞
<
a. This implies that there exists k
0
∈
[
1
,
N
]
Z such that u
k
0
=
max
⁡
k
∈
[
1
,
N
]
Z
⁡
u
k
=
u
*
+
2 or u
k
0
=
min
⁡
k
∈
[
1
,
N
]
Z
⁡
u
k
=
-
u
*
-
2. In the first case we can assume that k
0
∈
[
1
,
N
-
1
]
Z. If k
0
∈
[
2
,
N
-
1
]
Z, then Δ
u
k
0
≤
0, Δ
u
k
0
-
1
≥
0. This together with ϕ is an increasing homeomorphism implying ∇
[
ϕ
(
Δ
u
k
0
)
]
≤
0. On the other hand, we have that
(42)
∇
[
ϕ
(
Δ
u
k
0
)
]
=
f
(
k
0
,
u
k
0
,
Δ
u
k
0
)
+
m
>
0
,
which is a contradiction. If k
0
=
1, then from boundary condition u
1
=
u
N, Δ
u
1
=
Δ
u
N
-
1, we can get that Δ
u
1
≤
0 and Δ
u
N
-
1
≥
0, which implies that Δ
u
1
=
Δ
u
N
-
1
=
0. This together with ∇
[
ϕ
(
Δ
u
N
-
1
)
]
=
ϕ
(
Δ
u
N
-
1
)
-
ϕ
(
Δ
u
N
-
2
)
=
f
(
N
-
1
,
u
N
-
1
,
0
)
+
m
>
0 implies that u
N
=
u
N
-
1
<
u
N
-
2; this is a contradiction. Analogously, one can obtain a contradiction in the second case. Consequently,
(43)
[
u
∈
∂
Ω
,
A
g
(
u
)
=
u
]
⟹
∥
u
∥
∞
<
u
*
+
2
.
Now, letu
∈
∂
Ω be such that A
g
(
u
)
=
u. It follows from (43) that ∥
u
∥
∞
<
u
*
+
2
,
∥
Δ
u
∥
∞
<
a and u
∈
∂
Ω
α
1
,
β
∪
∂
Ω
α
,
β
1. We infer that there exists k
0
∈
[
1
,
N
]
Z such that u
k
0
=
α
k
0 or u
k
0
=
β
k
0, implying that |
u
k
0
|
≤
∥
α
∥
∞
+
∥
β
∥
∞. Then,
(44)
|
u
k
|
≤
|
u
k
0
|
+
∑
s
=
2
N
-
1
‍
|
Δ
u
s
|
<
u
*
∀
k
∈
[
1
,
N
]
Z
,
and, consequently,
(45)
[
u
∈
∂
Ω
,
A
g
(
u
)
=
u
]
⟹
∥
u
∥
∞
<
u
*
.
We have divided two cases to discuss.
Case  1. Assume that there exists u
∈
∂
Ω such that A
g
(
u
)
=
u. Using (45), we deduce that ∥
u
∥
∞
<
u
*, implying that u is a solution of (24) and (34) holds. Actually, in this case, there exists k
u
∈
[
1
,
N
]
Z such that u
k
u
=
α
k
u or u
k
u
=
β
k
u.
Case  2. Assume that A
g
(
u
)
≠
u for all u
∈
∂
Ω. Then, from Lemma 5 applied to g, it follows that
(46)
deg
⁡
[
I
-
A
g
,
Ω
α
1
,
β
1
,
0
]
=
deg
⁡
[
I
-
A
g
,
Ω
α
1
,
β
,
0
]
=
deg
⁡
[
I
-
A
g
,
Ω
α
,
β
1
,
0
]
=
-
1
.
This together with the additivity property of the Brouwer degree implies that
(47)
deg
⁡
[
I
-
A
g
,
Ω
,
0
]
=
1
,
which together with the existence property of the Brouwer degree imply that there exists u
∈
Ω such that A
g
(
u
)
=
u. It follows that there exists k
1
,
k
2
∈
[
1
,
N
]
Z such that u
k
1
<
α
k
1 and u
k
2
>
β
k
2. Then, using once again the fact that ∥
Δ
u
∥
∞
<
a, it follows that ∥
u
∥
∞
<
u
* and u is a solution of (24). Moreover, from u
∈
Ω, it follows that (34) is true.Remark 8.
Assume that (24) has a lower solution α and an upper solution β. From Lemma 4 and Theorem 7, we deduce that (24) has at least one solution u satisfying (34). In particular,
(48)
∥
u
∥
∞
<
∥
α
∥
∞
+
∥
β
∥
∞
+
a
(
N
-
2
)
.As an application of Theorem7, we deal with singular strong nonlinearities. Consider the following discrete periodic problem:
(49)
∇
[
ϕ
(
Δ
u
k
)
]
+
h
(
u
k
)
Δ
u
k
=
g
(
u
k
)
+
f
(
k
,
u
k
,
Δ
u
k
)
,
u
1
=
u
N
,
Δ
u
1
=
Δ
u
N
-
1
,
where f
:
[
2
,
N
-
1
]
Z
×
R
2
→
R and h
,
g
:
(
0
,
∞
)
→
R are continuous functions such that
(50)
lim
⁡
u
→
0
+
g
(
u
)
=
+
∞
and h
≥
0. Under those assumptions we have the following theorem.Theorem 9.
Assume that (49) has a lower solution α
>
0 and an upper solution β
>
0. Then (49) has at least one solution u which satisfies (34).Proof.
First, we define some notations as follows:(51)
δ
=
min
⁡
[
1
,
N
]
Z
min
⁡
⁡
{
α
,
β
}
,
B
=
∥
α
∥
∞
+
∥
β
∥
∞
+
a
(
N
-
2
)
,
m
=
min
⁡
[
1
,
N
]
Z
×
[
-
B
,
B
]
×
[
-
a
,
a
]
f
.
From (50), there exists ɛ
∈
(
0
,
δ
) such that
(52)
g
(
u
k
)
>
h
(
ɛ
)
∑
s
∈
I
1
‍
Δ
u
s
+
∑
s
∈
I
2
‍
h
(
u
s
)
Δ
u
s
-
(
N
-
2
)
m
-
∑
s
∈
I
2
‍
g
(
u
s
)
,
u
k
∈
(
0
,
ɛ
]
,
where I
1
=
{
k
∣
0
<
u
k
≤
ɛ
}, I
2
=
{
k
∣
u
k
>
ɛ
}.
Letg
^, h
^
:
R
→
R be the continuous functions given by
(53)
g
^
(
u
)
=
{
g
(
u
)
,
u
≥
ɛ
,
g
(
ɛ
)
,
u
≤
ɛ
,
h
^
(
u
)
=
{
h
(
u
)
,
u
≥
ɛ
,
h
(
ɛ
)
,
u
≤
ɛ
,
and consider the auxiliary periodic problem
(54)
∇
[
ϕ
(
Δ
u
k
)
]
+
h
^
(
u
k
)
Δ
u
k
=
g
^
(
u
k
)
+
f
(
k
,
u
k
,
Δ
u
k
)
,
u
1
=
u
N
,
Δ
u
1
=
Δ
u
N
-
1
.
From ɛ
<
δ, it follows that α and β are lower and upper solutions of (54), respectively.
Ifα
≤
β, then (54) has a solution u satisfying α
≤
u
≤
β from Lemma 4 and [12, Remark  3] (without any additional assumption). If condition (33) holds, then (54) has a solution u satisfying (34). Obviously, the solution u satisfies
(55)
-
B
≤
u
≤
B
,
∥
Δ
u
∥
∞
<
a
.
Next, we will prove that u
>
ɛ. From (55), there exists k
0
∈
[
1
,
N
]
Z such that u
k
0
=
min
⁡
k
∈
[
1
,
N
]
Z
⁡
u
k. Suppose on the contrary that u
k
0
≤
ɛ, summing from s
=
2 to s
=
N
-
1 for (54); then we have
(56)
∑
s
=
2
N
-
1
‍
h
^
(
u
s
)
Δ
u
s
=
∑
s
=
2
N
-
1
‍
g
^
(
u
s
)
+
∑
s
=
2
N
-
1
‍
f
(
s
,
u
s
,
Δ
u
s
)
.
This together with (52) implies that
(57)
0
=
∑
s
=
2
N
-
1
g
^
(
u
s
)
-
∑
s
=
2
N
-
1
h
^
(
u
s
)
Δ
u
s
+
∑
s
=
2
N
-
1
f
(
s
,
u
s
,
Δ
u
s
)
=
∑
s
∈
I
1
g
^
(
u
s
)
+
∑
s
∈
I
2
g
^
(
u
s
)
-
∑
s
=
2
N
-
1
h
^
(
u
s
)
Δ
u
s
+
∑
s
=
2
N
-
1
f
(
s
,
u
s
,
Δ
u
s
)
=
g
(
ɛ
)
∑
s
∈
I
1
1
+
∑
s
∈
I
2
g
(
u
s
)
-
h
(
ɛ
)
∑
s
∈
I
1
Δ
u
s
-
∑
s
∈
I
2
h
(
u
s
)
Δ
u
s
‍
+
∑
s
=
2
N
-
1
f
(
s
,
u
s
,
Δ
u
s
)
≥
g
(
ɛ
)
+
∑
s
∈
I
2
g
(
u
s
)
-
h
(
ɛ
)
∑
s
∈
I
1
Δ
u
s
-
∑
s
∈
I
2
h
(
u
s
)
Δ
u
s
‍
+
∑
s
=
2
N
-
1
f
(
s
,
u
s
,
Δ
u
s
)
>
0
,
which is a contradiction. Hence, u
>
ɛ, implying that u is also a solution of (49).Now, we give a method to construct the lower solution and the upper solution of the following discrete periodic problem:(58)
∇
[
ϕ
(
Δ
u
k
)
]
=
g
0
(
k
,
u
k
)
+
e
k
,
k
∈
[
2
,
N
-
1
]
Z
,
u
1
=
u
N
,
Δ
u
1
=
Δ
u
N
-
1
,
where g
0
:
[
2
,
N
-
1
]
Z
×
(
0
,
∞
)
→
R is a continuous singular nonlinearity and e
=
(
e
2
,
…
,
e
N
-
1
)
∈
R
N
-
2.The following result gives a method to construct a lower solution to (58), getting also control on its localization.Theorem 10.
Suppose that there existu
1
>
0 and c
=
(
c
2
,
…
,
c
N
-
1
)
∈
R
N
-
2 such that
(59)
g
0
(
k
,
u
)
≤
c
k
,
∀
(
k
,
u
)
∈
[
2
,
N
-
1
]
Z
×
[
u
1
,
u
1
+
a
N
2
]
.
If
(60)
c
¯
+
e
¯
≤
0
,
then (58) has a lower solution α such that
(61)
u
1
≤
α
<
u
1
+
a
N
2
.Proof.
Consider the functionψ
=
c
+
e. We have two cases.
Case  1. Assume that Ψ
+
=
0. Taking α
≡
u
1 and using that c
+
e
≤
0, it follows from (59) that α is a lower solution of (58).
Case  2. Assume that Ψ
+
>
0. Let h
k
=
ψ
k
+
Ψ
-
-
ψ
k
-
Ψ
+. Then using
(62)
∑
k
=
2
N
-
1
h
k
=
∑
k
=
2
N
-
1
[
ψ
k
+
Ψ
-
-
ψ
k
-
Ψ
+
]
=
0
and [12, Proposition  3], it follows that there exists w
∈
V
N
-
2 such that
(63)
∇
[
ϕ
(
Δ
w
k
)
]
=
h
k
,
w
1
=
w
N
,
Δ
w
1
=
Δ
w
N
-
1
.
Let us take u
0
=
1
/
Ψ
+ and ϖ
j
=
min
⁡
{
0
,
ϕ
-
1
(
∑
l
=
2
j
-
1
u
0
h
l
+
Q
ϕ
(
u
0
h
)
)
} for j
=
3
,
…
,
N
-
1. Then we define
(64)
α
2
=
u
1
-
min
⁡
k
∈
[
3
,
N
-
1
]
Z
⁡
∑
j
=
3
k
ϖ
j
,
α
k
=
u
1
+
∑
j
=
3
k
ϕ
-
1
(
∑
l
=
2
j
-
1
u
0
h
l
+
Q
ϕ
(
u
0
h
)
)
-
min
⁡
k
∈
[
3
,
N
-
1
]
Z
⁡
∑
j
=
3
k
ϖ
j
,
k
∈
[
3
,
N
-
1
]
Z
.
Let α
1
=
α
N
=
(
α
2
+
α
N
-
1
)
/
2; then Δ
α
1
=
Δ
α
N
-
1. On the other hand, we have that
(65)
Δ
α
k
=
ϕ
-
1
(
∑
l
=
2
k
‍
u
0
h
l
+
Q
ϕ
(
u
0
h
)
)
,
2
≤
k
≤
N
-
1
.
Since min
⁡
k
∈
[
3
,
N
-
1
]
Z
⁡
∑
j
=
3
k
ϖ
j
≤
0, Lemma 6 implies (61). Now, we will show that α is the lower solution of (58). By using (60), it follows that Ψ
+
≤
Ψ
-; this together with the definitions of α, u
0, and h
k implies that
(66)
∇
[
ϕ
(
Δ
α
k
)
]
=
u
0
h
k
=
1
Ψ
+
[
ψ
k
+
Ψ
-
-
ψ
k
-
Ψ
+
]
≥
ψ
k
+
-
ψ
k
-
=
ψ
k
,
k
∈
[
2
,
N
-
1
]
Z
.
From (59) and (61), we deduce that
(67)
g
0
(
k
,
α
k
)
+
e
k
≤
ψ
k
,
∀
k
∈
[
2
,
N
-
1
]
Z
.
Consequently,
(68)
∇
[
ϕ
(
Δ
α
k
)
]
≥
g
0
(
k
,
α
k
)
+
e
k
,
∀
k
∈
[
2
,
N
-
1
]
Z
.By a similar argument, it is easy to prove the following theorem.Theorem 11.
Suppose that there existu
2
>
0 and d
=
(
d
2
,
…
,
d
N
-
1
)
∈
R
N
-
2 such that
(69)
g
0
(
k
,
u
)
≥
d
k
,
f
o
r
a
n
y
(
k
,
u
)
∈
[
2
,
N
-
1
]
Z
×
[
u
2
,
u
2
+
a
N
2
]
.
If
(70)
d
¯
+
e
¯
≥
0
,
then (58) has an upper solution β such that
(71)
u
2
≤
β
<
u
2
+
a
N
2
.


## 4. Some Applications for Singular Perturbations Problems

### 4.1. Strong Repulsive Perturbations

Consider the discrete periodic problem(72)
∇
[
ϕ
(
Δ
u
k
)
]
+
r
k
u
k
-
g
(
u
k
)
=
e
k
,
u
1
=
u
N
,
Δ
u
1
=
Δ
u
N
-
1
,
where r
=
(
r
2
,
…
,
r
N
-
1
), e
=
(
e
2
,
…
,
e
N
-
1
)
∈
R
N
-
2, and g
:
(
0
,
∞
)
→
R is continuous and satisfies
(73)
lim
⁡
x
→
0
g
(
x
)
=
+
∞
,
lim
⁡
x
→
∞
g
(
x
)
=
0
.The main result of this subsection is the following theorem.Theorem 12.
Assume that (73) holds. If either
(74)
r
¯
>
0
o
r
r
¯
=
0
,
e
¯
<
-
a
N
R
-
2
(
N
-
2
)
,
then problem (72) has at least one solution.Proof.
Notice that from (73) it follows that there exists a constant β sufficiently small such that
(75)
0
≤
g
(
β
)
+
e
k
-
r
k
β
,
which means that β is an upper solution of (72).
Now we construct a lower solution of (72) by applying Theorem 10. Consider the continuous function g
0
:
[
2
,
N
-
1
]
Z
×
(
0
,
∞
)
→
R defined by
(76)
g
0
(
k
,
u
k
)
=
-
r
k
u
k
+
g
(
u
k
)
,
g
*
:
(
0
,
∞
)
→
R given by
(77)
g
*
(
u
)
=
max
⁡
x
∈
[
u
,
u
+
a
N
/
2
]
g
(
x
)
,
and γ
*
:
(
0
,
∞
)
→
R defined by
(78)
γ
*
(
u
)
=
-
R
u
+
a
N
2
R
-
+
(
N
-
2
)
g
*
(
u
)
.
Case  1. Assume that r
¯
>
0. This together with (73) implies that
(79)
lim
⁡
u
→
∞
γ
*
(
u
)
=
-
∞
,
so there exists u
1
>
0 such that γ
*
(
u
1
)
≤
-
E. In order to apply Theorem 10, define
(80)
c
k
=
r
k
-
(
u
1
+
a
N
2
)
-
r
k
+
u
1
+
g
*
(
u
1
)
,
k
∈
[
2
,
N
-
1
]
Z
.
It follows that C
=
γ
*
(
u
1
) and C
+
E
≤
0, meaning that condition (60) is fulfilled. One has that
(81)
g
0
(
k
,
u
k
)
=
r
k
-
u
k
-
r
k
+
u
k
+
g
(
u
k
)
≤
r
k
-
(
u
1
+
a
N
2
)
-
r
k
+
u
1
+
g
*
(
u
1
)
,
for all (
k
,
u
k
)
∈
[
2
,
N
-
1
]
Z
×
[
u
1
,
u
1
+
a
N
/
2
]. So, condition (59) holds. Then, from Theorem 10 we infer that (72) has a lower solution α. Therefore, from Theorem 9, we can obtain the result.
Case  2. Assume that r
¯
=
0 and e
¯
<
-
a
N
R
-
/
2
(
N
-
2
). It follows that
(82)
γ
*
(
u
)
=
a
N
2
R
-
+
(
N
-
2
)
g
*
(
u
)
,
lim
⁡
u
→
∞
γ
*
(
u
)
=
a
N
2
R
-
.
Then, there exists u
1
>
0 such that γ
*
(
u
1
)
≤
-
E, since e
¯
<
-
a
N
R
-
/
2
(
N
-
2
). The result follows by a similar argument to that used in Case  1.Remark 13.
Theorem  9 in [12] follows from Theorem 12 just taking r
=
0.Example 14.
Consider the repulsive singular periodic problem(83)
∇
(
Δ
u
k
1
-
κ
(
Δ
u
k
)
2
)
+
r
k
u
k
-
1
(
u
k
)
λ
=
e
k
,
5555555555555555555555
k
∈
[
2
,
N
-
1
]
Z
,
u
1
=
u
N
,
Δ
u
1
=
Δ
u
N
-
1
,
where r
=
(
r
2
,
…
,
r
N
-
1
), e
=
(
e
2
,
…
,
e
N
-
1
)
∈
R
N
-
2, and λ
≥
1. If either r
¯
>
0 or r
¯
=
0, e
¯
<
-
N
R
-
/
2
κ
(
N
-
2
), then (83) has at least one solution.In the caser
<
0, there exists s
0
<
0 such that (83) has at least two solutions provided that e
≤
s
0 holds true. In fact, in this case, problem (83) has two strict upper solutions β
1, β
2
>
0 and a strict lower solution α
>
0 such that β
1
<
α
<
β
2. Thus, the result follows from Lemma 4 and Theorem 9.


### 4.2. Mixed Singularities

Consider the discrete periodic problem(84)
∇
[
ϕ
(
Δ
u
k
)
]
+
r
k
u
k
+
g
(
k
,
u
k
)
=
e
k
,
u
1
=
u
N
,
Δ
u
1
=
Δ
u
N
-
1
,
where r
=
(
r
2
,
…
,
r
N
-
1
), e
=
(
e
2
,
…
,
e
N
-
1
)
∈
R
N
-
2, and g
:
[
2
,
N
-
1
]
Z
×
(
0
,
∞
)
→
R is continuous and satisfies
(85)
lim
⁡
u
→
∞
g
(
k
,
u
)
=
0
,
uniformly
with
k
∈
[
2
,
N
-
1
]
Z
.
Let the continuous function g
0
:
[
2
,
N
-
1
]
Z
×
(
0
,
∞
)
→
R defined by
(86)
g
0
(
k
,
u
k
)
=
-
r
k
u
k
-
g
(
k
,
u
k
)
,
g
*, g
*
:
[
2
,
N
-
1
]
Z
×
(
0
,
∞
)
→
R given by
(87)
g
*
(
k
,
u
)
=
min
⁡
x
∈
[
u
,
u
+
aN
/
2
]
⁡
g
(
k
,
x
)
,
g
*
(
k
,
u
)
=
max
⁡
x
∈
[
u
,
u
+
aN
/
2
]
⁡
g
(
k
,
x
)
,
and γ
*
:
(
a
N
/
2
,
∞
)
→
R, γ
*
:
(
0
,
∞
)
→
R, defined by
(88)
γ
*
(
u
)
=
-
R
u
+
a
N
2
R
+
-
∑
j
=
2
N
-
1
‍
g
*
(
j
,
u
-
a
N
2
)
,
γ
*
(
u
)
=
-
R
u
-
a
N
2
R
+
-
∑
j
=
2
N
-
1
‍
g
*
(
j
,
u
)
.The following lemma plays a key role to prove the main result in this subsection.Lemma 15.
Let (85) hold and γ
*
m
≔
inf
⁡
γ
*. If r
¯
<
0 and -
E
>
γ
*
m, then (84) has at least one solution.Proof.
Since-
E
>
γ
*
m, there exists z
>
a
N
/
2 such that γ
*
(
z
)
≤
-
E. Let us take u
1
=
z
-
a
N
/
2
>
0 and c
=
(
c
2
,
…
,
c
N
-
1
)
∈
R
N
-
2 by
(89)
c
k
=
r
k
-
(
u
1
+
a
N
2
)
-
r
k
+
u
1
-
g
*
(
k
,
u
1
)
.
Then it follows that conditions (59) and (60) hold. Thus, from Theorem 10 we infer that (84) has a lower solution α such that u
1
≤
α
<
u
1
+
a
N
/
2.
On the other hand, using the fact thatr
¯
<
0, there exists u
2
≥
z such that γ
*
(
u
2
)
≥
-
E. Consider d
=
(
d
2
,
…
,
d
N
-
1
)
∈
R
N
-
2 by
(90)
d
k
=
r
k
-
u
2
-
r
k
+
(
u
2
+
a
N
2
)
-
g
*
(
k
,
u
2
)
.
Then, it follows that conditions (69) and (70) hold. Therefore, from Theorem 11 we can get that (84) has an upper solution β such that α
≤
β. The result follows from Lemma 4.Remark 16.
From Lemma15, the solution u of (84) is a positive solution since 0
<
u
1
≤
α
≤
u
≤
β
<
u
2
+
a
N
/
2.Let us consider the discrete periodic problem(91)
∇
[
ϕ
(
Δ
u
k
)
]
+
r
k
u
k
+
m
k
(
u
k
)
λ
=
e
k
,
k
∈
[
2
,
N
-
1
]
Z
,
u
1
=
u
N
,
Δ
u
1
=
Δ
u
N
-
1
,
where r
=
(
r
2
,
…
,
r
N
-
1
), m
=
(
m
2
,
…
,
m
N
-
1
), e
=
(
e
2
,
…
,
e
N
-
1
)
∈
R
N
-
2, and λ
>
0. We have the following theorem.Theorem 17.
Ifr
¯
<
0 and
(92)
-
E
>
(
1
+
λ
)
[
|
R
|
λ
M
-
λ
λ
]
1
/
(
1
+
λ
)
+
a
N
2
R
-
-
M
+
[
a
N
2
+
(
λ
M
-
|
R
|
)
1
/
(
1
+
λ
)
]
-
λ
,
then (91) has at least one solution.Proof.
We have divided two cases.
Case  1. Assume that M
-
=
0. In this case one has that
(93)
γ
*
(
u
)
=
-
R
u
+
a
N
2
R
+
-
M
+
u
λ
,
implying that γ
*
m
=
γ
*
(
a
N
/
2
). So, (92) becomes -
E
>
γ
*
m, and the result follows from Lemma 15.
Case  2. Assume that M
-
>
0. Notice that the minimum of u
↦
-
R
u
+
(
a
N
/
2
)
R
+
+
M
-
/
(
u
-
a
N
/
2
)
λ is attained in u
0
=
a
N
/
2
+
[
λ
M
-
/
|
R
|
]
1
/
(
λ
+
1
) and
(94)
γ
*
(
u
0
)
≤
-
R
u
0
+
a
N
2
R
+
+
M
-
(
u
0
-
a
N
/
2
)
λ
-
M
+
(
u
0
)
λ
.
It is not difficult to verify that γ
*
(
u
0
)
<
-
E by using (92). Hence, -
E
>
r
*
m, and the result follows from Lemma 15.Example 18.
Consider the discrete periodic problem with repulsive singularity:(95)
∇
(
Δ
u
k
1
-
κ
(
Δ
u
k
)
2
)
+
r
k
u
k
-
m
k
(
u
k
)
λ
=
e
k
,
5555555555555555555555
k
∈
[
2
,
N
-
1
]
Z
,
u
1
=
u
N
,
Δ
u
1
=
Δ
u
N
-
1
,
where r
=
(
r
2
,
…
,
r
N
-
1
), m
=
(
m
2
,
…
,
m
N
-
1
), and e
=
(
e
2
,
…
,
e
N
-
1
)
∈
R
N
-
2 with m
≥
0 and λ
>
0. If r
¯
<
0 and
(96)
-
E
>
(
1
+
λ
)
[
|
R
|
λ
M
λ
λ
]
1
/
(
1
+
λ
)
+
N
R
-
2
κ
,
then the above problem has at least one solution.Example 19.
Consider the periodic problem with attractive singularity(97)
∇
(
Δ
u
k
1
-
κ
(
Δ
u
k
)
2
)
+
r
k
u
k
+
m
k
(
u
k
)
λ
=
e
k
,
5555555555555555555555
k
∈
[
2
,
N
-
1
]
Z
,
u
1
=
u
N
,
Δ
u
1
=
Δ
u
N
-
1
,
where r
=
(
r
2
,
…
,
r
N
-
1
), m
=
(
m
2
,
…
,
m
N
-
1
), and e
=
(
e
2
,
…
,
e
N
-
1
)
∈
R
N
-
2 with m
≥
0 and λ
>
0. If r
¯
<
0 and
(98)
E
<
M
(
2
κ
N
)
λ
-
N
R
-
2
κ
,
then the above problem has at least one solution.In connection with Example19, if r
=
0, then we have the following theorem.Theorem 20.
Consider the discrete periodic problem with attractive singularity(99)
∇
[
ϕ
(
Δ
u
k
)
]
+
m
k
(
u
k
)
λ
=
e
k
,
k
∈
[
2
,
N
-
1
]
Z
,
u
1
=
u
N
,
Δ
u
1
=
Δ
u
N
-
1
,
where m
=
(
m
2
,
…
,
m
N
-
1
), e
=
(
e
2
,
…
,
e
N
-
1
)
∈
R
N
-
2 with m
≥
0 and λ
>
0. If
(100)
0
<
E
<
M
(
2
a
N
)
λ
,
then (99) has at least one solution.Proof.
We will use the same strategy as in the proof of Lemma15. In this case one has that g
0
(
k
,
u
)
=
-
m
k
/
u
λ.
Clearly, it follows from (100) that there exists z
>
a
N
/
2 such that E
<
M
z
-
λ. Let us define u
1
=
z
-
a
N
/
2
>
0 and c
=
(
c
2
,
…
,
c
N
-
1
)
∈
R
N
-
2 by
(101)
c
k
=
-
m
k
(
u
1
+
a
N
2
)
-
λ
.
Then, it follows that conditions (59) and (60) hold. Therefore, from Theorem 10 we infer that (99) has a lower solution α such that u
1
≤
α
<
u
1
+
a
N
/
2.
Using again (100), it follows that there exists u
2
>
z such that E
≥
M
(
u
2
)
-
λ. Let us define d
=
(
d
2
,
…
,
d
N
-
1
)
∈
R
N
-
2 by d
k
=
-
m
k
(
u
2
)
-
λ. Then it follows that conditions (69) and (70) are true. Hence, from Theorem 11 we infer that (99) has an upper solution β such that u
2
≤
β
<
u
2
+
a
N
/
2.
Consequently, (99) has a lower solution α and an upper solution β such that α
≤
β. The result follows from Lemma 4.In the “pure” attractive case we have the following result concerning (84).Theorem 21.
Assume that (85) and
(102)
lim
⁡
u
→
0
g
(
k
,
u
)
=
+
∞
,
u
n
i
f
o
r
m
l
y
w
i
t
h
k
∈
[
2
,
N
-
1
]
Z
hold. Then (84) has at least one solution provided that either
(103)
r
¯
<
0
o
r
r
¯
=
0
,
E
>
a
N
2
R
+
.Proof.
Notice that from (102) it follows that any sufficiently small positive constant α is a lower solution of (84). The construction of an upper solution β
≥
α for (84) is similar as in Lemma 15. Then the result follows from Lemma 4.Remark 22.
Theorem21 follows from [12] taking r
=
0 in Theorem 21.Example 23.
Let us consider again problem (97), with the condition m
>
0. If either r
¯
<
0 or r
¯
=
0 and E
>
N
R
+
/
2
κ, then (97) has at least one solution.Example 24.
Letλ
>
0, μ
≥
1 with μ
>
λ and consider the discrete periodic problem
(104)
∇
(
Δ
u
k
1
-
(
Δ
u
k
)
2
)
+
c
Δ
u
k
(
u
k
)
4
/
5
+
1
(
u
k
)
λ
-
1
(
u
k
)
μ
=
e
k
(
u
k
)
λ
,
u
1
=
u
N
,
Δ
u
0
=
Δ
u
N
-
1
,
where e
=
(
e
2
,
…
,
e
N
-
1
)
∈
R
N
-
2 with e
≤
0. It is not difficult to prove that the above problem has at least one solution for any c
∈
R by using Theorem 9.


## 4.1. Strong Repulsive Perturbations

Consider the discrete periodic problem(72)
∇
[
ϕ
(
Δ
u
k
)
]
+
r
k
u
k
-
g
(
u
k
)
=
e
k
,
u
1
=
u
N
,
Δ
u
1
=
Δ
u
N
-
1
,
where r
=
(
r
2
,
…
,
r
N
-
1
), e
=
(
e
2
,
…
,
e
N
-
1
)
∈
R
N
-
2, and g
:
(
0
,
∞
)
→
R is continuous and satisfies
(73)
lim
⁡
x
→
0
g
(
x
)
=
+
∞
,
lim
⁡
x
→
∞
g
(
x
)
=
0
.The main result of this subsection is the following theorem.Theorem 12.
Assume that (73) holds. If either
(74)
r
¯
>
0
o
r
r
¯
=
0
,
e
¯
<
-
a
N
R
-
2
(
N
-
2
)
,
then problem (72) has at least one solution.Proof.
Notice that from (73) it follows that there exists a constant β sufficiently small such that
(75)
0
≤
g
(
β
)
+
e
k
-
r
k
β
,
which means that β is an upper solution of (72).
Now we construct a lower solution of (72) by applying Theorem 10. Consider the continuous function g
0
:
[
2
,
N
-
1
]
Z
×
(
0
,
∞
)
→
R defined by
(76)
g
0
(
k
,
u
k
)
=
-
r
k
u
k
+
g
(
u
k
)
,
g
*
:
(
0
,
∞
)
→
R given by
(77)
g
*
(
u
)
=
max
⁡
x
∈
[
u
,
u
+
a
N
/
2
]
g
(
x
)
,
and γ
*
:
(
0
,
∞
)
→
R defined by
(78)
γ
*
(
u
)
=
-
R
u
+
a
N
2
R
-
+
(
N
-
2
)
g
*
(
u
)
.
Case  1. Assume that r
¯
>
0. This together with (73) implies that
(79)
lim
⁡
u
→
∞
γ
*
(
u
)
=
-
∞
,
so there exists u
1
>
0 such that γ
*
(
u
1
)
≤
-
E. In order to apply Theorem 10, define
(80)
c
k
=
r
k
-
(
u
1
+
a
N
2
)
-
r
k
+
u
1
+
g
*
(
u
1
)
,
k
∈
[
2
,
N
-
1
]
Z
.
It follows that C
=
γ
*
(
u
1
) and C
+
E
≤
0, meaning that condition (60) is fulfilled. One has that
(81)
g
0
(
k
,
u
k
)
=
r
k
-
u
k
-
r
k
+
u
k
+
g
(
u
k
)
≤
r
k
-
(
u
1
+
a
N
2
)
-
r
k
+
u
1
+
g
*
(
u
1
)
,
for all (
k
,
u
k
)
∈
[
2
,
N
-
1
]
Z
×
[
u
1
,
u
1
+
a
N
/
2
]. So, condition (59) holds. Then, from Theorem 10 we infer that (72) has a lower solution α. Therefore, from Theorem 9, we can obtain the result.
Case  2. Assume that r
¯
=
0 and e
¯
<
-
a
N
R
-
/
2
(
N
-
2
). It follows that
(82)
γ
*
(
u
)
=
a
N
2
R
-
+
(
N
-
2
)
g
*
(
u
)
,
lim
⁡
u
→
∞
γ
*
(
u
)
=
a
N
2
R
-
.
Then, there exists u
1
>
0 such that γ
*
(
u
1
)
≤
-
E, since e
¯
<
-
a
N
R
-
/
2
(
N
-
2
). The result follows by a similar argument to that used in Case  1.Remark 13.
Theorem  9 in [12] follows from Theorem 12 just taking r
=
0.Example 14.
Consider the repulsive singular periodic problem(83)
∇
(
Δ
u
k
1
-
κ
(
Δ
u
k
)
2
)
+
r
k
u
k
-
1
(
u
k
)
λ
=
e
k
,
5555555555555555555555
k
∈
[
2
,
N
-
1
]
Z
,
u
1
=
u
N
,
Δ
u
1
=
Δ
u
N
-
1
,
where r
=
(
r
2
,
…
,
r
N
-
1
), e
=
(
e
2
,
…
,
e
N
-
1
)
∈
R
N
-
2, and λ
≥
1. If either r
¯
>
0 or r
¯
=
0, e
¯
<
-
N
R
-
/
2
κ
(
N
-
2
), then (83) has at least one solution.In the caser
<
0, there exists s
0
<
0 such that (83) has at least two solutions provided that e
≤
s
0 holds true. In fact, in this case, problem (83) has two strict upper solutions β
1, β
2
>
0 and a strict lower solution α
>
0 such that β
1
<
α
<
β
2. Thus, the result follows from Lemma 4 and Theorem 9.


## 4.2. Mixed Singularities

Consider the discrete periodic problem(84)
∇
[
ϕ
(
Δ
u
k
)
]
+
r
k
u
k
+
g
(
k
,
u
k
)
=
e
k
,
u
1
=
u
N
,
Δ
u
1
=
Δ
u
N
-
1
,
where r
=
(
r
2
,
…
,
r
N
-
1
), e
=
(
e
2
,
…
,
e
N
-
1
)
∈
R
N
-
2, and g
:
[
2
,
N
-
1
]
Z
×
(
0
,
∞
)
→
R is continuous and satisfies
(85)
lim
⁡
u
→
∞
g
(
k
,
u
)
=
0
,
uniformly
with
k
∈
[
2
,
N
-
1
]
Z
.
Let the continuous function g
0
:
[
2
,
N
-
1
]
Z
×
(
0
,
∞
)
→
R defined by
(86)
g
0
(
k
,
u
k
)
=
-
r
k
u
k
-
g
(
k
,
u
k
)
,
g
*, g
*
:
[
2
,
N
-
1
]
Z
×
(
0
,
∞
)
→
R given by
(87)
g
*
(
k
,
u
)
=
min
⁡
x
∈
[
u
,
u
+
aN
/
2
]
⁡
g
(
k
,
x
)
,
g
*
(
k
,
u
)
=
max
⁡
x
∈
[
u
,
u
+
aN
/
2
]
⁡
g
(
k
,
x
)
,
and γ
*
:
(
a
N
/
2
,
∞
)
→
R, γ
*
:
(
0
,
∞
)
→
R, defined by
(88)
γ
*
(
u
)
=
-
R
u
+
a
N
2
R
+
-
∑
j
=
2
N
-
1
‍
g
*
(
j
,
u
-
a
N
2
)
,
γ
*
(
u
)
=
-
R
u
-
a
N
2
R
+
-
∑
j
=
2
N
-
1
‍
g
*
(
j
,
u
)
.The following lemma plays a key role to prove the main result in this subsection.Lemma 15.
Let (85) hold and γ
*
m
≔
inf
⁡
γ
*. If r
¯
<
0 and -
E
>
γ
*
m, then (84) has at least one solution.Proof.
Since-
E
>
γ
*
m, there exists z
>
a
N
/
2 such that γ
*
(
z
)
≤
-
E. Let us take u
1
=
z
-
a
N
/
2
>
0 and c
=
(
c
2
,
…
,
c
N
-
1
)
∈
R
N
-
2 by
(89)
c
k
=
r
k
-
(
u
1
+
a
N
2
)
-
r
k
+
u
1
-
g
*
(
k
,
u
1
)
.
Then it follows that conditions (59) and (60) hold. Thus, from Theorem 10 we infer that (84) has a lower solution α such that u
1
≤
α
<
u
1
+
a
N
/
2.
On the other hand, using the fact thatr
¯
<
0, there exists u
2
≥
z such that γ
*
(
u
2
)
≥
-
E. Consider d
=
(
d
2
,
…
,
d
N
-
1
)
∈
R
N
-
2 by
(90)
d
k
=
r
k
-
u
2
-
r
k
+
(
u
2
+
a
N
2
)
-
g
*
(
k
,
u
2
)
.
Then, it follows that conditions (69) and (70) hold. Therefore, from Theorem 11 we can get that (84) has an upper solution β such that α
≤
β. The result follows from Lemma 4.Remark 16.
From Lemma15, the solution u of (84) is a positive solution since 0
<
u
1
≤
α
≤
u
≤
β
<
u
2
+
a
N
/
2.Let us consider the discrete periodic problem(91)
∇
[
ϕ
(
Δ
u
k
)
]
+
r
k
u
k
+
m
k
(
u
k
)
λ
=
e
k
,
k
∈
[
2
,
N
-
1
]
Z
,
u
1
=
u
N
,
Δ
u
1
=
Δ
u
N
-
1
,
where r
=
(
r
2
,
…
,
r
N
-
1
), m
=
(
m
2
,
…
,
m
N
-
1
), e
=
(
e
2
,
…
,
e
N
-
1
)
∈
R
N
-
2, and λ
>
0. We have the following theorem.Theorem 17.
Ifr
¯
<
0 and
(92)
-
E
>
(
1
+
λ
)
[
|
R
|
λ
M
-
λ
λ
]
1
/
(
1
+
λ
)
+
a
N
2
R
-
-
M
+
[
a
N
2
+
(
λ
M
-
|
R
|
)
1
/
(
1
+
λ
)
]
-
λ
,
then (91) has at least one solution.Proof.
We have divided two cases.
Case  1. Assume that M
-
=
0. In this case one has that
(93)
γ
*
(
u
)
=
-
R
u
+
a
N
2
R
+
-
M
+
u
λ
,
implying that γ
*
m
=
γ
*
(
a
N
/
2
). So, (92) becomes -
E
>
γ
*
m, and the result follows from Lemma 15.
Case  2. Assume that M
-
>
0. Notice that the minimum of u
↦
-
R
u
+
(
a
N
/
2
)
R
+
+
M
-
/
(
u
-
a
N
/
2
)
λ is attained in u
0
=
a
N
/
2
+
[
λ
M
-
/
|
R
|
]
1
/
(
λ
+
1
) and
(94)
γ
*
(
u
0
)
≤
-
R
u
0
+
a
N
2
R
+
+
M
-
(
u
0
-
a
N
/
2
)
λ
-
M
+
(
u
0
)
λ
.
It is not difficult to verify that γ
*
(
u
0
)
<
-
E by using (92). Hence, -
E
>
r
*
m, and the result follows from Lemma 15.Example 18.
Consider the discrete periodic problem with repulsive singularity:(95)
∇
(
Δ
u
k
1
-
κ
(
Δ
u
k
)
2
)
+
r
k
u
k
-
m
k
(
u
k
)
λ
=
e
k
,
5555555555555555555555
k
∈
[
2
,
N
-
1
]
Z
,
u
1
=
u
N
,
Δ
u
1
=
Δ
u
N
-
1
,
where r
=
(
r
2
,
…
,
r
N
-
1
), m
=
(
m
2
,
…
,
m
N
-
1
), and e
=
(
e
2
,
…
,
e
N
-
1
)
∈
R
N
-
2 with m
≥
0 and λ
>
0. If r
¯
<
0 and
(96)
-
E
>
(
1
+
λ
)
[
|
R
|
λ
M
λ
λ
]
1
/
(
1
+
λ
)
+
N
R
-
2
κ
,
then the above problem has at least one solution.Example 19.
Consider the periodic problem with attractive singularity(97)
∇
(
Δ
u
k
1
-
κ
(
Δ
u
k
)
2
)
+
r
k
u
k
+
m
k
(
u
k
)
λ
=
e
k
,
5555555555555555555555
k
∈
[
2
,
N
-
1
]
Z
,
u
1
=
u
N
,
Δ
u
1
=
Δ
u
N
-
1
,
where r
=
(
r
2
,
…
,
r
N
-
1
), m
=
(
m
2
,
…
,
m
N
-
1
), and e
=
(
e
2
,
…
,
e
N
-
1
)
∈
R
N
-
2 with m
≥
0 and λ
>
0. If r
¯
<
0 and
(98)
E
<
M
(
2
κ
N
)
λ
-
N
R
-
2
κ
,
then the above problem has at least one solution.In connection with Example19, if r
=
0, then we have the following theorem.Theorem 20.
Consider the discrete periodic problem with attractive singularity(99)
∇
[
ϕ
(
Δ
u
k
)
]
+
m
k
(
u
k
)
λ
=
e
k
,
k
∈
[
2
,
N
-
1
]
Z
,
u
1
=
u
N
,
Δ
u
1
=
Δ
u
N
-
1
,
where m
=
(
m
2
,
…
,
m
N
-
1
), e
=
(
e
2
,
…
,
e
N
-
1
)
∈
R
N
-
2 with m
≥
0 and λ
>
0. If
(100)
0
<
E
<
M
(
2
a
N
)
λ
,
then (99) has at least one solution.Proof.
We will use the same strategy as in the proof of Lemma15. In this case one has that g
0
(
k
,
u
)
=
-
m
k
/
u
λ.
Clearly, it follows from (100) that there exists z
>
a
N
/
2 such that E
<
M
z
-
λ. Let us define u
1
=
z
-
a
N
/
2
>
0 and c
=
(
c
2
,
…
,
c
N
-
1
)
∈
R
N
-
2 by
(101)
c
k
=
-
m
k
(
u
1
+
a
N
2
)
-
λ
.
Then, it follows that conditions (59) and (60) hold. Therefore, from Theorem 10 we infer that (99) has a lower solution α such that u
1
≤
α
<
u
1
+
a
N
/
2.
Using again (100), it follows that there exists u
2
>
z such that E
≥
M
(
u
2
)
-
λ. Let us define d
=
(
d
2
,
…
,
d
N
-
1
)
∈
R
N
-
2 by d
k
=
-
m
k
(
u
2
)
-
λ. Then it follows that conditions (69) and (70) are true. Hence, from Theorem 11 we infer that (99) has an upper solution β such that u
2
≤
β
<
u
2
+
a
N
/
2.
Consequently, (99) has a lower solution α and an upper solution β such that α
≤
β. The result follows from Lemma 4.In the “pure” attractive case we have the following result concerning (84).Theorem 21.
Assume that (85) and
(102)
lim
⁡
u
→
0
g
(
k
,
u
)
=
+
∞
,
u
n
i
f
o
r
m
l
y
w
i
t
h
k
∈
[
2
,
N
-
1
]
Z
hold. Then (84) has at least one solution provided that either
(103)
r
¯
<
0
o
r
r
¯
=
0
,
E
>
a
N
2
R
+
.Proof.
Notice that from (102) it follows that any sufficiently small positive constant α is a lower solution of (84). The construction of an upper solution β
≥
α for (84) is similar as in Lemma 15. Then the result follows from Lemma 4.Remark 22.
Theorem21 follows from [12] taking r
=
0 in Theorem 21.Example 23.
Let us consider again problem (97), with the condition m
>
0. If either r
¯
<
0 or r
¯
=
0 and E
>
N
R
+
/
2
κ, then (97) has at least one solution.Example 24.
Letλ
>
0, μ
≥
1 with μ
>
λ and consider the discrete periodic problem
(104)
∇
(
Δ
u
k
1
-
(
Δ
u
k
)
2
)
+
c
Δ
u
k
(
u
k
)
4
/
5
+
1
(
u
k
)
λ
-
1
(
u
k
)
μ
=
e
k
(
u
k
)
λ
,
u
1
=
u
N
,
Δ
u
0
=
Δ
u
N
-
1
,
where e
=
(
e
2
,
…
,
e
N
-
1
)
∈
R
N
-
2 with e
≤
0. It is not difficult to prove that the above problem has at least one solution for any c
∈
R by using Theorem 9.



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*Source: 637242-2014-07-10.xml*