# Some Notes on the Existence  of Solution for Ordinary Differential Equations via Fixed Point Theory

Authors: Fei He
Journal: Abstract and Applied Analysis
 (2014)

---

## Abstract

We establish a fixed point theorem withw-distance for nonlinear contractive mappings in complete metric spaces. As applications of our results, we derive the existence and uniqueness of solution for a first-order ordinary differential equation with periodic boundary conditions. Here, we need not assume that the equation has a lower solution.


---

## Body

## 1. Introduction and Preliminaries

The main purpose of this paper is to obtain the existence and uniqueness of solution for a periodic boundary value problem. This topic has been considered in [1–11] and the references therein. A powerful tool to solve the problem is the fixed point theorem in partially ordered metric spaces. On the existing research results, admitting the existence of a lower solution is necessary. In this paper, we establish a fixed point theorem forw-distance contraction type maps in complete metric spaces without partially ordered structure. From this, we obtain some results on the existence and uniqueness for ordinary differential equations. We will see that the assumption of the existence of a lower solution can be removed. In particular, we also show that under assumptions of a recent result in [10] the equation has a unique zero solution.In this note, we consider the following periodic boundary value problem:(1)
u
′
(
t
)
=
f
(
t
,
u
(
t
)
)
,
if
t
∈
I
=
[
0
,
T
]
;
u
(
0
)
=
u
(
T
)
,
where T
>
0 and f
:
I
×
R is a continuous function.Definition 1.
A lower solution for (1) is a function α
∈
C
1
(
I
) such that
(2)
α
′
(
t
)
≤
f
(
t
,
α
(
t
)
)
for
t
∈
I
,
α
(
0
)
≤
α
(
T
)
.
LetX
=
C
(
I
) be the set of all real continuous functions on a closed interval I. We endow X with the norm
(3)
∥
u
∥
=
max
⁡
t
∈
I
|
u
(
t
)
|
,
for all u
∈
X. Obviously, this space is a Banach space and the norm induces a complete metric on X as follows:
(4)
d
(
u
,
v
)
=
∥
u
-
v
∥
=
max
⁡
t
∈
I
|
u
(
t
)
-
v
(
t
)
|
,
for all u
,
v
∈
X.
LetB denote the class of those functions β
:
[
0
,
∞
)
→
[
0,1
) which satisfy the condition
(5)
β
(
t
n
)
⟶
1
implies
t
n
⟶
0
.
LetA denote the class of the functions ϕ
:
[
0
,
∞
)
→
[
0
,
∞
) which have the following properties:(i)
ϕ is increasing;
(ii)
for eachx
>
0, ϕ
(
x
)
<
x;
(iii)
β
(
x
)
=
ϕ
(
x
)
/
x
∈
B.
For example,ϕ
(
t
)
=
μ
t, where 0
≤
μ
<
1, ϕ
(
t
)
=
t
/
(
1
+
t
), and ϕ
(
t
)
=
ln
⁡
(
1
+
t
) are in A.
Recently, Amini-Harandi and Emami [4] proved the following existence theorem.Theorem 2 (see [4, Theorem 3.1]).
Consider problem (1) with f
:
I
×
R
→
R continuous and suppose that there exists λ
>
0 such that for x
,
y
∈
R with y
≥
x,
(6)
0
≤
f
(
t
,
y
)
+
λ
y
-
[
f
(
t
,
x
)
+
λ
x
]
≤
λ
ϕ
(
y
-
x
)
,
where ϕ
∈
A. Then the existence of a lower solution of (1) provides the existence of a unique solution of (1).Then, Caballero et al. [6] also give an existence theorem.Theorem 3 (see [6, Theorem 3.2]).
Consider problem (1) with f
:
I
×
R
→
R continuous and suppose that there exist λ
,
α
>
0 with
(7)
α
≤
(
2
λ
(
e
λ
T
-
1
)
T
(
e
λ
T
+
1
)
)
1
/
2
such that for x
,
y
∈
R with y
≥
x
(8)
0
≤
f
(
t
,
y
)
+
λ
y
-
[
f
(
t
,
x
)
+
λ
x
]
≤
α
(
y
-
x
)
ϕ
(
y
-
x
)
,
where ϕ
∈
A. Then the existence of a lower solution of (1) provides the existence of a solution of (1).Very recently, Hussain et al. [10] obtain the following result.Theorem 4 (see [10, Section 3]).
Consider problem (1) with f
:
I
×
R
→
R continuous and suppose that there exists λ
>
0 such that for x
,
y
∈
C
(
I
)
(9)
|
f
(
t
,
x
(
t
)
)
+
λ
x
(
t
)
|
+
|
f
(
t
,
y
(
t
)
)
+
λ
y
(
t
)
|
≤
λ
2
p
-
1
ln
⁡
(
a
2
p
-
1
(
|
x
(
t
)
|
+
|
y
(
t
)
|
)
p
+
1
)
p
,
where 0
≤
a
<
1 and p
>
1. Then the existence of a lower solution of (1) provides the existence of a solution of (1).Remark 5.
If we takex
(
t
)
=
y
(
t
)
=
θ
(
t
)
=
0 for all t
∈
I in the condition (9), then we deduce that
(10)
f
(
t
,
θ
(
t
)
)
=
0
∀
t
∈
I
.
This means thatθ
(
t
) is a solution of (1) (of course, a lower solution of (1)). In Section 3, we will show this fact again; see Remark 17.Now, let us recall the concept ofw-distance, which was introduced by Kada et al. [12].Definition 6.
LetX be a metric space with metric d. Then a function p
:
X
×
X
→
[
0
,
∞
) is called aw-distance on X if the following are satisfied:(w1)
p
(
x
,
z
)
≤
p
(
x
,
y
)
+
p
(
y
,
z
), for any x
,
y
,
z
∈
X;
(w2)
for anyx
∈
X, p
(
x
,
·
)
:
X
→
[
0
,
∞
) is lower semicontinuous; that is, p
(
x
,
y
0
)
≤
liminf
⁡
n
→
∞
p
(
x
,
y
n
), whenever y
n
∈
X and y
n
→
y
0;
(w3)
for anyε
>
0, there exists δ
>
0 such that p
(
z
,
x
)
≤
δ and p
(
z
,
y
)
≤
δ imply d
(
x
,
y
)
≤
ε.Let us give some basic examples ofw-distances (see [12]).Example 7.
Let(
X
,
d
) be a metric space. Then the metric d is aw-distance on X.Example 8.
LetX be a normed space with norm ∥
·
∥. Then the function p
:
X
×
X
→
[
0
,
∞
) defined by
(11)
p
(
x
,
y
)
=
∥
x
∥
+
∥
y
∥
for every x
,
y
∈
X is aw-distance on X.Example 9.
LetX be a normed space with norm ∥
·
∥. Then the function p
:
X
×
X
→
[
0
,
∞
) defined by
(12)
p
(
x
,
y
)
=
∥
y
∥
for every x
,
y
∈
X is aw-distance on X.We also need the following example in Section3.Example 10.
LetX
=
C
(
I
) be endowed with the norm
(13)
∥
u
∥
=
max
⁡
t
∈
I
|
u
(
t
)
|
,
for all u
∈
X. We easily deduce that the function p
:
X
×
X
→
[
0
,
∞
) defined by
(14)
p
(
x
,
y
)
=
max
⁡
t
∈
I
(
|
x
(
t
)
|
+
|
y
(
t
)
|
)
for every x
,
y
∈
X is aw-distance on X.The following lemma has been proved in [12].Lemma 11.
LetX be a metric space with metric d and let p be a w-distance on X. Let {
x
n
} and {
y
n
} be sequences in X, let {
α
n
} and {
β
n
} be sequences in [
0
,
∞
) converging to 0, and let x
,
y
,
z
∈
X. Then the following hold.(i)
Ifp
(
x
n
,
y
)
≤
α
n and p
(
x
n
,
z
)
≤
β
n for any n
∈
N, then y
=
z. In particular, if p
(
x
,
y
)
=
0 and p
(
x
,
z
)
=
0, then y
=
z.
(ii)
Ifp
(
x
n
,
y
n
)
≤
α
n and p
(
x
n
,
z
)
≤
β
n for any n
∈
N, then {
y
n
} converges to z.
(iii)
Ifp
(
x
n
,
x
m
)
≤
α
n for any n
,
m
∈
N with m
>
n, then {
x
n
} is a Cauchy sequence.


## 2. Fixed Point Results withw-Distance

We are now ready to state and prove our main theorem.Theorem 12.
Let(
X
,
d
) be a complete metric space and let p be aw-distance on X. Let f
:
X
→
X be a map and suppose that there exists β
∈
B such that
(15)
p
(
f
x
,
f
y
)
≤
β
(
p
(
x
,
y
)
)
p
(
x
,
y
)
,
for all x
,
y
∈
X. Then f has a unique fixed point z in X and p
(
z
,
z
)
=
0.Proof.
Starting with an arbitraryx
0
∈
X, we construct by induction a sequence {
f
n
(
x
0
)
} in X. Put x
n
=
f
n
(
x
0
)
,
n
=
1,2
,
….
Applying the condition (15) with x
=
x
n and y
=
x
n
+
1, we have
(16)
p
(
x
n
+
1
,
x
n
+
2
)
=
p
(
f
x
n
,
f
x
n
+
1
)
≤
β
(
p
(
x
n
,
x
n
+
1
)
)
p
(
x
n
,
x
n
+
1
)
≤
p
(
x
n
,
x
n
+
1
)
.
Therefore{
p
(
x
n
,
x
n
+
1
)
} is a decreasing sequence and bounded below and hence lim
⁡
n
→
∞
p
(
x
n
,
x
n
+
1
)
=
r
≥
0.
Let us prove thatr
=
0. Assume that r
>
0. From (15), we obtain
(17)
p
(
x
n
+
1
,
x
n
+
2
)
p
(
x
n
,
x
n
+
1
)
≤
β
(
p
(
x
n
,
x
n
+
1
)
)
<
1
,
n
=
1,2
,
…
.
Taking limit when n
→
∞, the above inequality yields lim
⁡
n
→
∞
β
(
p
(
x
n
,
x
n
+
1
)
)
=
1, and since β
∈
B this implies r
=
0. Thus lim
⁡
n
→
∞
p
(
x
n
,
x
n
+
1
)
=
0. Analogously, it can be proved that lim
⁡
n
→
∞
p
(
x
n
+
1
,
x
n
)
=
0.
Now, we show that(18)
p
(
x
n
,
x
m
)
⟶
0
as
m
>
n
⟶
∞
.
Suppose that this is not true. Then there existsε
0
>
0 for which we can find subsequences {
x
n
k
} and {
x
m
k
} of {
x
n
} with m
k
>
n
k
>
k such that
(19)
p
(
x
n
k
,
x
m
k
)
≥
ε
0
,
k
=
1,2
,
…
.
By the triangle inequality and the condition (15), for every k
∈
N,
(20)
p
(
x
n
k
,
x
m
k
)
≤
p
(
x
n
k
,
x
n
k
+
1
)
+
p
(
x
n
k
+
1
,
x
m
k
+
1
)
+
p
(
x
m
k
+
1
,
x
m
k
)
≤
p
(
x
n
k
,
x
n
k
+
1
)
+
β
(
p
(
x
n
k
,
x
m
k
)
)
p
(
x
n
k
,
x
m
k
)
+
p
(
x
m
k
+
1
,
x
m
k
)
.
From this, we obtain(21)
1
-
β
(
p
(
x
n
k
,
x
m
k
)
)
≤
p
(
x
n
k
,
x
m
k
)
-
1
·
[
p
(
x
n
k
,
x
n
k
+
1
)
+
p
(
x
m
k
+
1
,
x
m
k
)
]
≤
ε
0
-
1
·
[
p
(
x
n
k
,
x
n
k
+
1
)
+
p
(
x
m
k
+
1
,
x
m
k
)
]
.
Lettingk
→
∞ in the above inequality, using lim
⁡
n
→
∞
p
(
x
n
,
x
n
+
1
)
=
0 and lim
⁡
n
→
∞
p
(
x
n
+
1
,
x
n
)
=
0, we deduce that
(22)
lim
⁡
k
→
∞
β
(
p
(
x
n
k
,
x
m
k
)
)
=
1
.
But sinceβ
∈
B we get lim
⁡
k
→
∞
p
(
x
n
k
,
x
m
k
)
=
0. This is a contradiction and hence we have shown that (18) holds.
From (18) and Lemma 11(iii), it follows that {
x
n
} is a Cauchy sequence in X. Since (
X
,
d
) is a complete metric space, there exists a z
∈
X such that lim
⁡
n
→
∞
x
n
=
z.
Next, we prove thatz is a fixed point of f. To the end, we prove that lim
⁡
n
→
∞
p
(
x
n
,
z
)
=
0 and lim
⁡
n
→
∞
p
(
x
n
,
f
z
)
=
0. Let ε
>
0 be given. Using (18), there exists n
0
∈
N such that p
(
x
n
,
x
m
)
<
ε for all m
>
n
≥
n
0. Letting m
→
∞, by (w2) we have p
(
x
n
,
z
)
≤
liminf
⁡
m
→
∞
p
(
x
n
,
x
m
)
≤
ε for all n
≥
n
0. This means that lim
⁡
n
→
∞
p
(
x
n
,
z
)
=
0. Applying the condition (15) with x
=
x
n and y
=
z, we have
(23)
p
(
x
n
+
1
,
f
z
)
=
p
(
f
x
n
,
f
z
)
≤
β
(
p
(
x
n
,
z
)
)
p
(
x
n
,
z
)
≤
p
(
x
n
,
z
)
.
Due tolim
⁡
n
→
∞
p
(
x
n
,
z
)
=
0, it follows that lim
⁡
n
→
∞
p
(
x
n
,
f
z
)
=
0. According to Lemma 11(i), we obtain f
z
=
z; that is, z is a fixed point of f.
To prove the uniqueness of the fixed point off, let us suppose that y is another fixed point of f. Using the condition (15), we get
(24)
p
(
x
n
+
1
,
y
)
=
p
(
f
x
n
,
f
y
)
≤
β
(
p
(
x
n
,
y
)
)
p
(
x
n
,
y
)
≤
p
(
x
n
,
y
)
.
Therefore{
p
(
x
n
,
y
)
} is a decreasing sequence and bounded below and hence lim
⁡
n
→
∞
p
(
x
n
,
y
)
=
s
≥
0. Suppose that s
>
0. Using the condition (15), we have
(25)
p
(
x
n
+
1
,
y
)
p
(
x
n
,
y
)
≤
β
(
p
(
x
n
,
y
)
)
<
1
,
n
=
1,2
,
…
.
Taking limit whenn
→
∞, the above inequality yields lim
⁡
n
→
∞
β
(
p
(
x
n
,
y
)
)
=
1, and since β
∈
B this implies s
=
0. Then lim
⁡
n
→
∞
p
(
x
n
,
y
)
=
0. Combining this and lim
⁡
n
→
∞
p
(
x
n
,
z
)
=
0, we get by Lemma 11(i) that y
=
z.
Finally, we provep
(
z
,
z
)
=
0 when z is a fixed point of f. Applying the condition (15) with x
=
z and y
=
z, we obtain
(26)
p
(
z
,
z
)
=
p
(
f
z
,
f
z
)
≤
β
(
p
(
z
,
z
)
)
p
(
z
,
z
)
and hence (
1
-
β
(
p
(
z
,
z
)
)
)
·
p
(
z
,
z
)
=
0. Now that β
∈
B, it follows that p
(
z
,
z
)
=
0.The following corollary is immediate result from Theorem12 and Example 7.Corollary 13.
Let(
X
,
d
) be a complete metric space. Let f
:
X
→
X be a map and suppose that there exists β
∈
B such that
(27)
d
(
f
x
,
f
y
)
≤
β
(
d
(
x
,
y
)
)
d
(
x
,
y
)
for all x
,
y
∈
X. Then f has a unique fixed point z in X.


## 3. Application to Ordinary Differential Equations

Now, we prove some results on the existence of solution for problem (1).Theorem 14.
Consider problem (1) with f continuous and suppose that there exists λ
>
0 such that for any x
,
y
∈
R
(28)
|
f
(
t
,
y
)
+
λ
y
-
[
f
(
t
,
x
)
+
λ
x
]
|
≤
λ
ϕ
(
|
y
-
x
|
)
,
where ϕ
∈
A. Then there exists a unique solution for (1).Proof.
Problem (1) can be rewritten as
(29)
u
′
(
t
)
+
λ
u
(
t
)
=
f
(
t
,
u
(
t
)
)
+
λ
u
(
t
)
,
if
t
∈
I
=
[
0
,
T
]
;
u
(
0
)
=
u
(
T
)
.
Using variation of parameters formula, we can easily deduce that problem (1) is equivalent to the integral equation
(30)
u
(
t
)
=
∫
0
T
‍
G
(
t
,
s
)
[
f
(
s
,
u
(
s
)
)
+
λ
u
(
s
)
]
d
s
,
where
(31)
G
(
t
,
s
)
=
{
e
λ
(
T
+
s
-
t
)
e
λ
T
-
1
,
0
≤
s
<
t
≤
T
,
e
λ
(
s
-
t
)
e
λ
T
-
1
,
0
≤
t
<
s
≤
T
.
Define F
:
C
(
I
)
→
C
(
I
) by
(32)
(
F
u
)
(
t
)
=
∫
0
T
‍
G
(
t
,
s
)
[
f
(
s
,
u
(
s
)
)
+
λ
u
(
s
)
]
d
s
.
Note that if u
∈
C
(
I
) is a fixed point of F then u
∈
C
1
(
I
) is a solution of (1).
Now, we check that hypotheses of Corollary13. Consider the space X
=
C
(
I
) with the metric
(33)
d
(
x
,
y
)
=
∥
x
-
y
∥
=
max
⁡
t
∈
I
|
x
(
t
)
-
y
(
t
)
|
.
Then by (28) for any u
,
v
∈
X,
(34)
d
(
F
u
,
F
v
)
=
max
⁡
t
∈
I
|
(
F
u
)
(
t
)
-
(
F
v
)
(
t
)
|
≤
max
⁡
t
∈
I
∫
0
T
‍
G
(
t
,
s
)
|
f
(
t
,
u
(
s
)
)
+
λ
u
(
s
)
gggggggggggggggg
-
[
f
(
t
,
v
(
s
)
)
+
λ
v
(
s
)
]
|
d
s
≤
max
⁡
t
∈
I
∫
0
T
‍
G
(
t
,
s
)
λ
ϕ
(
|
u
(
s
)
-
v
(
s
)
|
)
d
s
.
As the function ϕ
(
x
) is increasing, we obtain
(35)
d
(
F
u
,
F
v
)
≤
max
⁡
t
∈
I
∫
0
T
‍
G
(
t
,
s
)
λ
ϕ
(
|
u
(
s
)
-
v
(
s
)
|
)
d
s
≤
λ
ϕ
(
d
(
u
,
v
)
)
·
max
⁡
t
∈
I
∫
0
T
‍
G
(
t
,
s
)
d
s
=
λ
ϕ
(
d
(
u
,
v
)
)
·
max
⁡
t
∈
I
1
e
λ
T
-
1
×
(
1
λ
e
λ
(
T
+
s
-
t
)
]
0
t
+
1
λ
e
λ
(
s
-
t
)
]
t
T
)
=
λ
ϕ
(
d
(
u
,
v
)
)
·
1
e
λ
T
-
1
·
1
λ
(
e
λ
T
-
1
)
=
ϕ
(
d
(
u
,
v
)
)
=
ϕ
(
d
(
u
,
v
)
)
d
(
u
,
v
)
·
d
(
u
,
v
)
=
β
(
d
(
u
,
v
)
)
d
(
u
,
v
)
.
From Corollary 13, we see that F has a unique fixed point.Remark 15.
Clearly, we see that the condition (6) of Theorem 2 implies the condition (28) of Theorem 14. In addition, Theorem 14 need not assume the existence of a lower solution for (1). Hence this result is an improvement of Theorem 2.Theorem 16.
Consider problem (1) with f continuous and suppose that there exists λ
>
0 such that for any x
,
y
∈
R
(36)
|
f
(
t
,
x
)
+
λ
x
|
+
|
f
(
t
,
y
)
+
λ
y
|
≤
λ
ϕ
(
|
x
|
+
|
y
|
)
,
where ϕ
∈
A. Then there exists a unique solution for (1).Proof.
Problem (1) is equivalent to the integral equation
(37)
u
(
t
)
=
∫
0
T
‍
G
(
t
,
s
)
[
f
(
s
,
u
(
s
)
)
+
λ
u
(
s
)
]
d
s
,
where G
(
t
,
s
) is the same as (31). Define F
:
C
(
I
)
→
C
(
I
) by
(38)
(
F
u
)
(
t
)
=
∫
0
T
‍
G
(
t
,
s
)
[
f
(
s
,
u
(
s
)
)
+
λ
u
(
s
)
]
d
s
.
Notice that if u
∈
C
(
I
) is a fixed point of F then u
∈
C
1
(
I
) is a solution of (1).
Now, we check that hypotheses of Theorem12. Consider the space X
=
C
(
I
) with the metric
(39)
d
(
x
,
y
)
=
∥
x
-
y
∥
=
max
⁡
t
∈
I
|
x
(
t
)
-
y
(
t
)
|
.
We take aw-distance on X
=
C
(
I
) by p
(
x
,
y
)
=
max
⁡
t
∈
I
(
|
x
(
t
)
|
+
|
y
(
t
)
|
); see Example 10. Using the monotene property of the function ϕ
(
x
) and condition (36), for any u
,
v
∈
X,
(40)
p
(
F
u
,
F
v
)
=
max
⁡
t
∈
I
(
|
(
F
u
)
(
t
)
|
+
|
(
F
v
)
(
t
)
|
)
≤
max
⁡
t
∈
I
(
∫
0
T
‍
G
(
t
,
s
)
(
|
f
(
t
,
u
(
s
)
)
+
λ
u
(
s
)
|
n
n
n
n
n
n
n
n
n
n
n
n
n
n
n
n
+
|
f
(
t
,
v
(
s
)
)
+
λ
v
(
s
)
|
)
d
s
∫
0
T
‍
)
≤
max
⁡
t
∈
I
∫
0
T
‍
G
(
t
,
s
)
λ
ϕ
(
|
u
(
s
)
|
+
|
v
(
s
)
|
)
d
s
≤
λ
ϕ
(
p
(
u
,
v
)
)
·
max
⁡
t
∈
I
∫
0
T
‍
G
(
t
,
s
)
d
s
=
λ
ϕ
(
p
(
u
,
v
)
)
·
1
λ
=
ϕ
(
p
(
u
,
v
)
)
=
ϕ
(
p
(
u
,
v
)
)
p
(
u
,
v
)
·
p
(
u
,
v
)
=
β
(
p
(
u
,
v
)
)
p
(
u
,
v
)
.
From Theorem 12, we deduce that F has a unique fixed point.Remark 17.
Let0
≤
a
<
1 and p
>
1. If we denote
(41)
ϕ
(
t
)
=
1
2
p
-
1
ln
⁡
(
a
2
p
-
1
t
p
+
1
)
p
,
then ϕ
∈
A and ϕ
(
t
)
≤
ln
⁡
(
t
+
1
). Thus Theorem 4 is a sepical case of Theorem 16. However, from Theorem 12 we get p
(
u
,
u
)
=
max
⁡
t
∈
I
(
2
|
u
(
t
)
|
)
=
2
∥
u
∥
=
0 if u is a solution of (1). This means u
(
t
)
=
0 for all t
∈
I. In other words, the unique solution for (1) from Theorem 16 (in particular, from Theorem 4) can only be a zero function.Theorem 18.
Consider problem (1) with f continuous and suppose that there exist λ
,
α
>
0 and p
>
1 with
(42)
α
≤
(
p
λ
(
e
λ
T
-
1
)
p
T
p
-
1
(
e
p
λ
T
-
1
)
)
1
/
p
such that for x
,
y
∈
R
(43)
|
f
(
t
,
y
)
+
λ
y
-
[
f
(
t
,
x
)
+
λ
x
]
|
≤
α
ϕ
(
|
y
-
x
|
)
,
where ϕ
∈
A. Then there exists a unique solution for (1).Proof.
Problem (1) is equivalent to the integral equation
(44)
u
(
t
)
=
∫
0
T
‍
G
(
t
,
s
)
[
f
(
s
,
u
(
s
)
)
+
λ
u
(
s
)
]
d
s
,
where G
(
t
,
s
) is the same as (31). Define F
:
C
(
I
)
→
C
(
I
) by
(45)
(
F
u
)
(
t
)
=
∫
0
T
‍
G
(
t
,
s
)
[
f
(
s
,
u
(
s
)
)
+
λ
u
(
s
)
]
d
s
.
Notice that if u
∈
C
(
I
) is a fixed point of F then u
∈
C
1
(
I
) is a solution of (1).
Now, we check that hypotheses of Corollary13. By (43), for any u
,
v
∈
X,
(46)
d
(
F
u
,
F
v
)
=
max
⁡
t
∈
I
|
(
F
u
)
(
t
)
-
(
F
v
)
(
t
)
|
≤
max
⁡
t
∈
I
∫
0
T
‍
G
(
t
,
s
)
|
f
(
t
,
u
(
s
)
)
+
λ
u
(
s
)
g
g
g
g
g
g
g
g
g
g
g
i
g
g
-
[
f
(
t
,
v
(
s
)
)
+
λ
v
(
s
)
]
|
d
s
≤
max
⁡
t
∈
I
∫
0
T
‍
G
(
t
,
s
)
α
ϕ
(
|
u
(
s
)
-
v
(
s
)
|
)
d
s
=
α
max
⁡
t
∈
I
∫
0
T
‍
G
(
t
,
s
)
ϕ
(
|
u
(
s
)
-
v
(
s
)
|
)
d
s
.
Using the Hölder inequality in the last integral, we get(47)
∫
0
T
‍
G
(
t
,
s
)
ϕ
(
|
u
(
s
)
-
v
(
s
)
|
)
d
s
≤
(
∫
0
T
‍
G
(
t
,
s
)
p
d
s
)
1
/
p
(
∫
0
T
‍
ϕ
(
|
u
(
s
)
-
v
(
s
)
|
)
q
d
s
)
1
/
q
,
where q
>
0 with 1
/
p
+
1
/
q
=
1. The first integral gives us
(48)
∫
0
T
‍
G
(
t
,
s
)
p
d
s
=
∫
0
t
‍
G
(
t
,
s
)
p
d
s
+
∫
t
T
‍
G
(
t
,
s
)
p
d
s
=
∫
0
t
‍
e
p
λ
(
T
+
s
-
t
)
(
e
λ
T
-
1
)
p
d
s
+
∫
t
T
‍
e
p
λ
(
s
-
t
)
(
e
λ
T
-
1
)
p
d
s
=
e
p
λ
T
-
1
p
λ
(
e
λ
T
-
1
)
p
.
As the function ϕ
(
x
) is increasing and α
≤
(
p
λ
(
e
λ
T
-
1
)
p
/
T
p
-
1
(
e
p
λ
T
-
1
)
)
1
/
p, from (47) we obtain
(49)
d
(
F
u
,
F
v
)
≤
α
max
⁡
t
∈
I
[
(
∫
0
T
‍
G
(
t
,
s
)
p
d
s
)
1
/
p
g
g
g
g
g
g
g
×
(
∫
0
T
‍
ϕ
(
|
u
(
s
)
-
v
(
s
)
|
)
q
d
s
)
1
/
q
]
≤
α
(
e
p
λ
T
-
1
p
λ
(
e
λ
T
-
1
)
p
)
1
/
p
(
∫
0
T
‍
ϕ
(
∥
u
-
v
∥
)
q
d
s
)
1
/
q
=
α
·
(
e
p
λ
T
-
1
p
λ
(
e
λ
T
-
1
)
p
)
1
/
p
·
T
1
/
q
·
ϕ
(
∥
u
-
v
∥
)
≤
(
p
λ
(
e
λ
T
-
1
)
p
T
p
-
1
(
e
p
λ
T
-
1
)
)
1
/
p
·
(
e
p
λ
T
-
1
p
λ
(
e
λ
T
-
1
)
p
)
1
/
p
·
T
1
/
q
·
ϕ
(
d
(
u
,
v
)
)
=
ϕ
(
d
(
u
,
v
)
)
=
ϕ
(
d
(
u
,
v
)
)
d
(
u
,
v
)
·
d
(
u
,
v
)
=
β
(
d
(
u
,
v
)
)
d
(
u
,
v
)
.
From Corollary 13, we deduce that F has a unique fixed point.From Theorem18, we obtain the following result when we take p
=
2.Corollary 19.
Consider problem (1) with f continuous and suppose that there exist λ
,
α
>
0 with
(50)
α
≤
(
2
λ
(
e
λ
T
-
1
)
T
(
e
λ
T
+
1
)
)
1
/
2
such that for x
,
y
∈
R
(51)
|
f
(
t
,
y
)
+
λ
y
-
[
f
(
t
,
x
)
+
λ
x
]
|
≤
α
ϕ
(
|
y
-
x
|
)
,
where ϕ
∈
A. Then there exists a unique solution for (1).Remark 20.
From [6, Remark 3.3], we see that if ϕ
∈
A, then φ
(
x
)
=
x
ϕ
(
x
)
∈
A. The condition (8) of Theorem 3 implies the condition (51) of Corollary 19. In addition, Corollary 19 need not assume the existence of a lower solution for (1). Thus Corollary 19 is an improvement of Theorem 3.Remark 21.
We easily check thatϕ
(
x
)
=
ln
⁡
(
x
2
+
1
) and ϕ
(
x
)
=
x
2
/
(
x
2
+
1
) are in A. Therefore the condition of [8, Theorem 3.1], [11, Theorem 25], or [7, Theorem 3.2] implies the condition (51) of Corollary 19. In addition, Corollary 19 need not assume the existence of a lower solution for (1). Thus Corollary 19 is an improvement of [8, Theorem 3.1], [11, Theorem 25], and [7, Theorem 3.2].



---

*Source: 309613-2014-10-14.xml*