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Contraction mapping principle with generalized altering distance function in ordered metric spaces and applications to ordinary differential equations

Fixed Point Theory and Applications20142014:227

https://doi.org/10.1186/1687-1812-2014-227

Received: 29 May 2014

Accepted: 8 October 2014

Published: 3 November 2014

Abstract

The aim of this paper is to present the definition of a generalized altering distance function and to extend the results of Yan et al. (Fixed Point Theory Appl. 2012:152, 2012) and some others, and to prove a new fixed point theorem of generalized contraction mappings in a complete metric space endowed with a partial order by using generalized altering distance functions. The results of this paper can be used to investigate a large class of nonlinear problems. As an application, we discuss the existence of a solution for a periodic boundary value problem.

Keywords

contraction mapping principlepartially ordered metric spacesfixed pointgeneralized altering distance functionordinary differential equations

1 Introduction

The Banach contraction mapping principle is a classical and powerful tool in nonlinear analysis. Weak contractions are generalizations of the Banach contraction mapping, which have been studied by several authors. In [18], the authors prove some types of weak contractions in complete metric spaces, respectively. In particular, the existence of a fixed point for weak contractions and generalized contractions was extended to partially ordered metric spaces in [2, 922]. Among them, some involve altering distance functions. Such functions were introduced by Khan et al. in [1], where they present some fixed point theorems with the help of such functions. First, we recall the definition of an altering distance function.

Definition 1.1 An altering distance function is a function ψ : [ 0 , ) [ 0 , ) which satisfies:
  1. (a)

    ψ is continuous and non-decreasing.

     
  2. (b)

    ψ = 0 if and only if t = 0 .

     

Recently, Harjani and Sadarangani proved some fixed point theorems for weak contractions and generalized contractions in partially ordered metric spaces by using the altering distance function in [11, 23], respectively. Their results improve the theorems of [2, 3].

Theorem 1.2 ([11])

Let ( X , ) be a partially ordered set and suppose that there exists a metric d X such that ( X , d ) is a complete metric space. Let f : X X be a continuous and non-decreasing mapping such that
d ( f ( x ) , f ( y ) ) d ( x , y ) ψ ( d ( x , y ) ) , for  x y ,

where ψ : [ 0 , ) [ 0 , ) is continuous and non-decreasing function such that ψ is positive in ( 0 , ) , ψ ( 0 ) = 0 and lim t ψ ( t ) = . If there exists x 0 X with x 0 f ( x 0 ) , then f has a fixed point.

Theorem 1.3 ([23])

Let ( X , ) be a partially ordered set and suppose that there exists a metric d X such that ( X , d ) is a complete metric space. Let f : X X be a continuous and non-decreasing mapping such that
ψ d ( f ( x ) , f ( y ) ) ψ ( d ( x , y ) ) ϕ ( d ( x , y ) ) , for  x y ,

where ψ and ϕ are altering distance functions. If there exists x 0 X with x 0 f ( x 0 ) , then f has a fixed point.

Subsequently, Amini-Harandi and Emami proved another fixed point theorem for contraction type maps in partially ordered metric spaces in [10]. The following class of functions is used in [10].

Let denote the class of those functions β : [ 0 , ) [ 0 , 1 ) which satisfy the condition: β ( t n ) 1 t n 0 .

Theorem 1.4 ([10])

Let ( X , ) be a partially ordered set and suppose that there exists a metric d such that ( X , d ) is a complete metric space. Let f : X X be an increasing mapping such that there exists an element x 0 X with x 0 f ( x 0 ) . Suppose that there exists β such that
d ( f ( x ) , f ( y ) ) β ( d ( x , y ) ) d ( x , y ) for each  x , y X  with  x y .

Assume that either f is continuous or M is such that if an increasing sequence x n x X , then x n x , n. Besides, if for each x , y X there exists z m which is comparable to x and y, then f has a unique fixed point.

In 2012, Yan et al. proved the following result.

Theorem 1.5 ([24])

Let X be a partially ordered set and suppose that there exists a metric d in x such that ( X , d ) is a complete metric space. Let T : X X be a continuous and non-decreasing mapping such that
ψ ( d ( T x , T y ) ) ϕ ( d ( x , y ) ) , x y ,

where ψ is an altering distance function and ϕ : [ 0 , ) [ 0 , ) is a continuous function with the condition ψ ( t ) > ϕ ( t ) for all t > 0 . If there exists x 0 X such that x 0 T x 0 , then T has a fixed point.

The aim of this paper is to present the definition of generalized altering distance function and to extend the results of Yan et al. [24] and some others, and to prove a new fixed point theorem of generalized contraction mappings in a complete metric space endowed with a partial order by using generalized altering distance functions. The results of this paper can be used to investigate a large class of nonlinear problems. As an application, we discuss the existence of a solution for a periodic boundary value problem.

2 Main results

We first give the definition of generalized altering distance function as follows.

Definition 2.1 A generalized altering distance function is a function ψ : [ 0 , ) [ 0 , ) which satisfies:
  1. (a)

    ψ is non-decreasing;

     
  2. (b)

    ψ = 0 if and only if t = 0 .

     

We first recall the following notion of a monotone non-decreasing function in a partially ordered set.

Definition 2.2 If ( X , ) is a partially ordered set and T : X X , we say that T is monotone non-decreasing if x , y X , x y T ( x ) T ( y ) .

This definition coincides with the notion of a non-decreasing function in the case where X = R and ≤ represents the usual total order in R.

In what follows, we prove the following theorem, which is the generalized type of Theorems 1.2-1.5.

Theorem 2.3 Let X be a partially ordered set and suppose that there exists a metric d in x such that ( X , d ) is a complete metric space. Let T : X X be a continuous and non-decreasing mapping such that
ψ ( d ( T x , T y ) ) ϕ ( d ( x , y ) ) , x y ,

where ψ is a generalized altering distance function and ϕ : [ 0 , ) [ 0 , ) is a right upper semi-continuous function with the condition: ψ ( t ) > ϕ ( t ) for all t > 0 . If there exists x 0 X such that x 0 T x 0 , then T has a fixed point.

Proof Since T is a non-decreasing function, we obtain by induction that
x 0 T x 0 T 2 x 0 T 3 x 0 T n x 0 T n + 1 x 0 .
(1)
Put x n + 1 = T x n . Then, for each integer n 1 , from (1) and, as the elements x n + 1 and x n are comparable, we get
ψ ( d ( x n + 1 , x n ) ) = ψ ( d ( T x n , T x n 1 ) ) ϕ ( d ( x n , x n 1 ) ) .
(2)
Using the condition of Theorem 2.3 we have
d ( x n + 1 , x n ) < d ( x n , x n 1 ) .
(3)
Hence the sequence { d ( x n + 1 , x n ) } is decreasing and, consequently, there exists r 0 such that
d ( x n + 1 , x n ) r + ,
as n . Consider the properties of ψ and ϕ, letting n in (2) we get
ψ ( r ) lim n ψ ( d ( x n + 1 , x n ) ) lim n ϕ ( d ( x n , x n 1 ) ) ϕ ( r ) .
By using the condition: ψ ( t ) > ϕ ( t ) for all t > 0 , we have r = 0 , and hence
d ( x n + 1 , x n ) 0 ,
(4)
as n . In what follows, we will show that { x n } is a Cauchy sequence. Suppose that { x n } is not a Cauchy sequence. Then there exists ε > 0 for which we can find subsequences { x n k } with n k > m k > k such that
d ( x n k , x m k ) ε
(5)
for all k 1 . Further, corresponding to m k we can choose n k in such a way that it is the smallest integer with n k > m k and satisfying (5). Then
d ( x n k 1 , x m k 1 ) < ε .
(6)
From (5) and (6), we have
ε d ( x n k , x m k ) ( d ( x n k ) , x n k 1 ) + d ( x n k 1 , x m k ) < d ( x n k , x n k 1 ) + ε .
Letting k and using (4), we get
lim k d ( x n k , x m k ) = ε .
(7)
By using the triangular inequality we have
d ( x n k , x m k ) d ( x n k , x n k 1 ) + d ( x n k 1 , x m k 1 ) + d ( x m k 1 , x m k ) , d ( x n k 1 , x m k 1 ) d ( x n k 1 , x n k ) + d ( x n k , x m k ) + d ( x m k , x m k 1 ) .
Letting k in the above two inequalities and using (4) and (7), we have
lim k d ( x n k 1 , x m k 1 ) = ε .
(8)
As n k > m k and x n k 1 and x m k 1 are comparable, using (1) we have
ψ ( d ( x n k , x m k ) ) ϕ ( d ( x n k 1 , x m k 1 ) ) .
Consider the properties of ψ and ϕ, letting k and taking into account (7) and (8), we have
ψ ( ε ) ϕ ( ε ) .
From the condition ψ ( t ) > ϕ ( t ) for all t > 0 , we get ε = 0 , which is a contradiction. This shows that { x n } is a Cauchy sequence and, since X is a complete metric space, there exists z X such that x n z as n . Moreover, the continuity of T implies that
z = lim n x n + 1 = lim n T x n = T z ,

and this proves that z is a fixed point. This completes the proof. □

In what follows, we prove that Theorem 2.3 is still valid for T being not necessarily continuous, assuming the following hypothesis in X:
If  ( x n )  is a non-decreasing sequence in  X  such that  x n x then  x n x  for all  n N .
(9)
Theorem 2.4 Let ( X , ) be a partially ordered set and suppose that there exists a metric d in X such that ( X , d ) is a complete metric space. Assume that X satisfies (9). Let T : X X be a non-decreasing mapping such that
ψ ( d ( T x , T y ) ) ϕ ( d ( x , y ) ) , x y ,

where ψ is a generalized altering distance functions and ϕ: [ 0 , ) [ 0 , ) is a right upper semi-continuous function with the condition ψ ( t ) > ϕ ( t ) for all t > 0 . If there exists x 0 X such that x 0 T x 0 , then T has a fixed point.

Proof Following the proof of Theorem 2.3 we only have to check that T ( z ) = z . As ( x n ) is a non-decreasing sequence in X and lim n x n = z the condition (9) gives us that x n z for every n N and consequently,
ψ ( d ( x n + 1 , T ( z ) ) ) = ψ ( d ( T ( x n ) , T ( z ) ) ) ϕ ( d ( x n , z ) ) .
Letting n and taking into account that ψ is an altering distance function, we have
ψ ( d ( z , T ( z ) ) ) ϕ ( 0 ) .

Using condition of theorem we have ϕ ( 0 ) = 0 , this implies Ψ ( d ( z , T ( z ) ) ) = 0 . Thus, d ( z , T ( z ) ) = 0 or equivalently, T ( z ) = z . □

Now, we present an example where it can be appreciated that the hypotheses in Theorems 2.3 and Theorems 2.4 do not guarantee uniqueness of the fixed point. An example appears in [12].

Let X = { ( 1 , 0 ) , ( 0 , 1 ) } R 2 and consider the usual order ( x , y ) ( z , t ) x z , y t . Thus, ( x , y ) is a partially ordered set whose different elements are not comparable. Besides ( X , d 2 ) is a complete metric space considering d 2 the Euclidean distance. The identity map T ( x , y ) = ( x , y ) is trivially continuous and non-decreasing and condition (1) of Theorem 2.4 is satisfied, since the elements in X are only comparable to themselves. Moreover, ( 1 , 0 ) T ( 1 , 0 ) = ( 1 , 0 ) and T has two fixed points in X.

In what follows, we give a sufficient condition for the uniqueness of the point in Theorems 2.3 and 2.4. This condition is:
for  x , y X  there exists a lower bound or an upper bound .
(10)
In [12] it is proved that condition (10) is equivalent to:
for  x , y X  there exists  z X  which is comparable to  x  and  y .
(11)

Theorem 2.5 Adding condition (11) to the hypotheses of Theorem  2.3 (resp. Theorem  2.4) we obtain the uniqueness of the fixed point of T.

Proof Suppose that there exist z , y X which are fixed points. We distinguish two cases.

Case 1. If y is comparable to z then T n ( y ) = y is comparable to T n ( z ) = z for n = 0 , 1 , 2 , and
ψ ( d ( z , y ) ) = ψ ( d ( T n ( z ) , T n ( y ) ) ) ϕ ( d ( T n 1 ( z ) , T n 1 ( y ) ) ) ϕ ( d ( z , y ) ) .

As we have the condition ψ ( t ) > ϕ ( t ) for t > 0 we obtain d ( z , y ) = 0 and this implies z = y .

Case 2. If y is not comparable to z then there exists x X comparable to y and z. Monotonicity of T implies that T n ( x ) is comparable to T n ( y ) and to T n ( z ) = z , for n = 0 , 1 , 2 , Moreover,
ψ ( d ( z , T n ( x ) ) ) = ψ ( d ( T n ( z ) , T n ( x ) ) ) ϕ ( d ( T n 1 ( z ) , T n 1 ( x ) ) ) = ϕ ( d ( z , T n 1 ( x ) ) ) .
(12)
Hence, ψ is a generalized altering distance function and we have the condition ψ ( t ) > ϕ ( t ) for t > 0 , this gives us that { d ( z , f n ( x ) ) } is a non-negative decreasing sequence and, consequently, there exists γ such that
lim n d ( z , T n ( x ) ) = γ .
Letting n in (12) and, taking into account the properties of ψ and ϕ, we obtain
ψ ( γ ) ϕ ( γ ) .
This and the condition ψ ( t ) > ϕ ( t ) for t > 0 imply γ = 0 . Analogously, it can be proved that
lim n d ( y , T n ( x ) ) = 0 .
Finally, as
lim n d ( z , T n ( x ) ) = lim n d ( y , T n ( x ) ) = 0

the uniqueness of the limit gives us y = z . This finishes the proof. □

Remark 2.6 Under the assumption of Theorem 2.3, it can be proved that for every x X , lim n T n ( x ) = z , where z is the fixed point (i.e. the operator f is Picard).

Remark 2.7 Theorem 1.2 is a particular case of Theorem 2.3 for ψ being the identity function, and ϕ ( t ) = t ψ ( t ) . Theorem 1.3 is a particular case of our Theorem 2.3 for ϕ ( t ) being replaced by ψ ( t ) ϕ ( t ) . Theorem 1.4 is a particular case of Theorem 2.3 for ψ being the identity function, and ϕ ( t ) = β ( t ) t . Theorem 1.5 is also a particular case of Theorem 2.3 for ψ and ϕ being continuous.

Example 2.8 The following are some generalized altering distance functions:
ψ 1 ( t ) = { 0 , t = 0 , [ t ] + 1 , t > 0 , ψ 2 ( t ) = { 0 , t = 0 , λ ( [ t ] + 1 ) , t > 0 ,
where α > 0 is a constant.
ψ 3 ( t ) = { t , 0 t < 1 , α t 2 , t 1 ,

where α 1 is a constant.

We choose ψ ( t ) = ψ 3 ( t ) and
ϕ ( t ) = { t 2 , 0 t < 1 , β t , t 1 ,

where 0 < β < α is a constant. By using Theorem 2.3, we can get the following result.

Theorem 2.9 Let X be a partially ordered set and suppose that there exists a metric d in x such that ( X , d ) is a complete metric space. Let T : X X be a continuous and non-decreasing mapping such that
0 d ( T x , T y ) < 1 d ( T x , T y ) ( d ( x , y ) ) 2 , d ( T x , T y ) 1 α ( d ( T x , T y ) ) 2 β d ( x , y )

for any x , y X . If there exists x 0 X such that x 0 T x 0 , then T has a fixed point.

3 Application to ordinary differential equations

In this section we present two examples where our Theorems 2.3 and 2.4 can be applied. The first example is inspired by [17]. We study the existence of a solution for the following first-order periodic problem:
{ u ( t ) = f ( t , u ( t ) ) , t [ 0 , T ] , u ( 0 ) = u ( T ) ,
(13)
where T > 0 and f : I × R R is a continuous function. Previously, we considered the space C ( I ) ( I = [ 0 , T ] ) of continuous functions defined on I. Obviously, this space with the metric given by
d ( x , y ) = sup { | x ( t ) y ( t ) | : t I } , for  x , y C ( I ) ,
is a complete metric space. C ( I ) can also be equipped with a partial order given by
x , y C ( I ) , x y x ( t ) y ( t ) m , for  t I .

Clearly, ( C ( I ) , ) satisfies condition (10), since for x , y C ( I ) the functions max { x , y } and min { x , y } are least upper and greatest lower bounds of x and y, respectively. Moreover, in [17] it is proved that ( C ( I ) , ) with the above mentioned metric satisfies condition (9).

Now we give the following definition.

Definition 3.1 A lower solution for (13) is a function α C ( 1 ) ( I ) such that
{ α ( t ) f ( t , α ( t ) ) , for  t I , α ( 0 ) α ( T ) .
Theorem 3.2 Consider problem (13) with f : I × R R continuous and suppose that there exist λ , α > 0 with
α ( 2 λ ( e λ T 1 ) T ( e λ T + 1 ) ) 1 2
such that for x , y R with x y
0 f ( t , x ) + λ x [ f ( t , y ) + λ y ] α g ( x y ) ,

where g ( t ) : [ 0 , + ) [ 0 , + ) is a light upper semi-continuous function with g ( 0 ) = 0 , g ( t ) < t 2 , t > 0 . Then the existence of a lower solution for (13) provides the existence of an unique solution of (13).

Proof Problem (13) can be written as
{ u ( t ) + λ u ( t ) = f ( t , u ( t ) ) + λ u ( t ) , for  t I = [ 0 , T ] , u ( 0 ) = u ( T ) .
This problem is equivalent to the integral equation
u ( t ) = 0 T G ( t , s ) [ f ( s , u ( s ) ) + λ u ( s ) ] d s ,
where G ( t , s ) is the Green function given by
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
( 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 (13). In what follows, we check that the hypotheses in Theorems 2.3 and 2.4 are satisfied. The mapping F is non-decreasing, since we have u v , and using our assumption. We can obtain
f ( t , u ) + λ u f ( t , v ) + λ v
which implies, since G ( t , s ) > 0 , that for t I
( F u ) ( t ) = 0 T G ( t , s ) [ f ( s , u ( s ) ) + λ u ( s ) ] d s 0 T G ( t , s ) [ f ( s , v ( s ) ) + λ v ( s ) ] d s = ( F v ) ( t ) .
Besides, for u v , we have
d ( F u , F v ) = sup t I | ( F u ) ( t ) ( F v ) ( t ) | = sup t I ( ( F u ) ( t ) ( F v ) ( t ) ) = sup t I 0 T G ( t , s ) [ f ( s , u ( s ) ) + λ u ( s ) f ( s , v ( s ) ) λ v ( s ) ] d s sup t I 0 T G ( t , s ) α g ( u ( s ) v ( s ) ) d s .
(14)
Using the Cauchy-Schwarz inequality in the last integral we get
0 T G ( t , s ) α g ( u ( s ) v ( s ) ) d s ( 0 T G ( t , s ) 2 d s ) 1 2 ( 0 T α 2 g ( u ( s ) v ( s ) ) d s ) 1 2 .
(15)
The first integral gives us
0 T G ( t , s ) 2 d s = 0 t G ( t , s ) 2 d s + t T G ( t , s ) 2 d s = 0 t e 2 λ ( T + s t ) ( e λ T 1 ) 2 d s + t T e 2 λ ( s t ) ( e λ T 1 ) 2 d s = 1 2 λ ( e λ T 1 ) 2 e ( 2 λ T 1 ) = e λ T + 1 2 λ ( e λ T 1 ) .
(16)
The second integral in (15) gives the following estimate:
0 T α 2 g ( u ( s ) v ( s ) ) d s α 2 g ( u v ) T = α 2 g ( d ( u , v ) ) T .
(17)
Taking into account (14)-(17) we have
d ( F u , F v ) sup t I ( e λ T + 1 2 λ ( e λ T 1 ) ) 1 2 ( α 2 g ( d ( u , v ) ) T ) 1 2 = ( e λ T + 1 2 λ ( e λ T 1 ) ) 1 2 α T ( g ( d ( u , v ) ) ) 1 2
and from the last inequality we obtain
d ( F u , F v ) 2 e λ T + 1 2 λ ( e λ T 1 ) α 2 T g ( d ( u , v ) )
or, equivalently.
2 λ ( e λ T 1 ) d ( F u , F v ) 2 ( e λ T + 1 ) α 2 T g ( d ( u , v ) ) .
By our assumption, as
α ( 2 λ ( e λ T 1 ) T ( e λ T + 1 ) ) 1 2 ,
the last inequality gives us
2 λ ( e λ T 1 ) d ( F u , F v ) 2 2 λ ( e λ T 1 ) g ( d ( u , v ) ) ,
and, hence,
d ( F u , F v ) 2 g ( d ( u , v ) ) .
(18)
Put ψ ( t ) = t 2 and ϕ ( t ) = g ( t ) . Obviously, ψ is a generalized altering distance function, ψ ( t ) and ϕ ( t ) satisfy the condition of ψ ( t ) > ϕ ( t ) for t > 0 . From (18), we obtain for u v
ψ ( d ( F u , F v ) ) ϕ ( d ( u , v ) ) .
Finally, let α ( t ) be a lower solution for (13); we claim that α F ( α ) . In fact
α ( t ) + λ α ( t ) f ( t , α ( t ) ) + λ α ( t ) , for  t I .
We multiply by e λ t ,
( α ( t ) e λ t ) [ f ( t , α ( t ) ) + λ α ( t ) ] e λ t , for  t I ,
and this gives us
α ( t ) e λ t α ( 0 ) + 0 t [ f ( s , α ( s ) ) + λ α ( s ) ] e λ s d s , for  t I .
(19)
As α ( 0 ) α ( T ) , the last inequality gives us
α ( 0 ) e λ t α ( T ) e λ T α ( 0 ) + 0 T [ f ( s , α ( s ) ) + λ α ( s ) ] e λ s d s ,
and so
α ( 0 ) 0 T e λ s e λ T 1 [ f ( s , α ( s ) ) + λ α ( s ) ] d s .
This and (19) give us
α ( t ) e λ t 0 t e λ ( T + s ) e λ T 1 [ f ( s , α ( s ) ) + λ α ( s ) ] d s + t T e λ s e λ T 1 [ f ( s , α ( s ) ) + λ α ( s ) ] d s
and, consequently,
α ( t ) 0 t e λ ( T + s t ) e λ T 1 d s + 0 t e λ ( s t ) e λ T 1 [ f ( s , α ( s ) ) + λ α ( s ) ] d s = 0 T G ( t , s ) [ f ( s , α ( s ) ) + λ α ( s ) ] d s = ( F α ) ( t ) , for  t I .

Finally, Theorems 2.3 and 2.4 show that F has an unique fixed point. □

Example 3.3 In Theorem 3.2, we can choose the function g ( t ) as follows:
  1. (1)

    g 1 ( t ) = ln ( t 2 + 1 ) ;

     
  2. (2)
    g 2 ( t ) = { t 3 , 0 t < 1 , 1 2 , t = 1 , t , 1 < t < + .
     
  3. (3)
    g 3 ( t ) = { t 3 , 0 t 1 2 , t 3 8 , 1 2 < t < + .
     

The functions g 1 ( t ) , g 2 ( t ) are continuous and non-decreasing. The function g 3 ( t ) is right upper semi-continuous. If we choose g ( t ) = g 1 ( t ) in Theorem 3.2, we obtain the result of [5].

Example 3.4 Consider the following first-order periodic problem:
{ u ( t ) = sin t e t β x , t [ 0 , T ] , u ( 0 ) = u ( T ) .
(20)
Let
f ( t , x ) = sin t e t β x , x [ 0 , ) , t [ 0 , 1 ] ,
then f ( t , x ) is continuous. Further, for x y , we have
f ( t , x ) + λ x [ f ( t , y ) + λ y ] = 2 ( λ β ) x y 2 .
We chose β [ 0 , λ ] such that
2 ( λ β ) ( 2 λ ( e λ T 1 ) T ( e λ T + 1 ) ) 1 2 .
Taking g ( t ) = ( t 2 ) 2 for all t [ 0 , + ) , we have
f ( t , x ) + λ x [ f ( t , y ) + λ y ] = 2 ( λ β ) g ( x y ) .

By using Theorem 3.2, we know that the first-order periodic problem (20) has a unique solution.

A second example where our results can be applied is the following two-point boundary value problem of the second order differential equation:
{ d 2 x d t 2 = f ( t , x ) , x [ 0 , ) , t [ 0 , 1 ] , x ( 0 ) = x ( 1 ) = 0 .
(21)
It is well known that x C 2 [ 0 , 1 ] is a solution of (20) that is equivalent to x C [ 0 , 1 ] being a solution of the integral equation
x ( t ) = 0 1 G ( t , s ) f ( s , x ( s ) ) d s , for  t [ 0 , 1 ] ,
where G ( t , s ) is the Green function given by
G ( t , s ) = { t ( 1 s ) , 0 t s 1 , s ( 1 t ) , 0 s t 1 .
(22)
Theorem 3.5 Consider problem (21) with f : I × R [ 0 , ) continuous and non-decreasing with respect to the second variable and suppose that there exists 0 α 8 such that for x , y R with x y
f ( t , x ) f ( t , y ) α g ( x y ) ,
(23)

where g ( t ) : [ 0 , + ) [ 0 , + ) is a light upper semi-continuous function with g ( 0 ) = 0 , g ( t ) < t 2 , t > 0 . Then our problem (21) has a unique non-negative solution.

Proof Consider the cone
P = { x C [ 0 , 1 ] : x ( t ) 0 } .
Obviously, ( P , d ) with d ( x , y ) = sup { | x ( t ) y ( t ) | : t [ 0 , 1 ] } is a complete metric space. Consider the operator given by
( T x ) ( t ) = 0 1 G ( t , s ) f ( s , x ( s ) ) d s , for  x P ,

where G ( t , s ) is the Green function appearing in (22).

As f is non-decreasing with respect to the second variable, for x , y P with y x and t [ 0 , 1 ] , we have
( T y ) ( t ) = 0 1 G ( t , s ) f ( s , y ( s ) ) d s 0 1 G ( t , s ) f ( s , x ( s ) ) d s ( T x ) ( t ) ,

and this proves that T is a non-decreasing operator.

Besides, for y x and taking into account (23), we obtain
d ( T y , T x ) = sup t [ 0 , 1 ] | ( T x ) ( t ) ( T y ) ( t ) | = sup t [ 0 , 1 ] ( ( T x ) ( t ) ( T y ) ( t ) ) = sup t [ 0 , 1 ] 0 1 G ( t , s ) ( f ( s , x ( s ) ) f ( s , y ( s ) ) ) d s sup t [ 0 , 1 ] 0 1 G ( t , s ) α g ( x ( s ) y ( s ) ) sup t [ 0 , 1 ] 0 1 G ( t , s ) α g ( x ( s ) y ( s ) ) d s = α ln [ y x 2 + 1 ] sup t [ 0 , 1 ] 0 1 G ( t , s ) d s .
(24)
It is easy to verify that
0 1 G ( t , s ) d s = t 2 2 + t 2
and that
sup t [ 0 , 1 ] 0 1 G ( t , s ) d s = 1 8 .
These facts, the inequality (24), and the hypothesis 0 < α 8 give us
d ( T x , T y ) α 8 g ( x y ) g ( x y ) = g ( d ( x , y ) ) .
Hence
d ( T y , T x ) 2 g ( d ( x , y ) ) .
Put ψ ( t ) = t 2 , ϕ ( t ) = g ( t ) , obviously ψ is an altering distance function, ψ and ϕ satisfy the condition of ψ ( t ) > ϕ ( t ) , for t > 0 . From the last inequality, we have
ψ ( d ( T x , T y ) ) ϕ ( d ( x , y ) ) .
Finally, as f and G are non-negative functions
T 0 = 0 1 G ( t , s ) f ( s , 0 ) d s 0

and Theorems 2.3 and 2.4 tell us that F has a unique non-negative solution. □

Remark 3.6 In Theorem 3.5, we can choose g ( t ) as g 1 ( 1 ) , g 2 ( t ) , and g 3 ( t ) as well as in Theorem 3.2.

Example 3.7 Consider the following two-point boundary value problem of the second order differential equation:
{ d 2 x d t 2 = sin t e t + x 1 + cos t π , x [ 0 , ) , t [ 0 , 1 ] , x ( 0 ) = x ( 1 ) = 0 .
(25)
Let
f ( t , x ) = sin t e t + x 1 + cos t π , x [ 0 , ) , t [ 0 , 1 ] ,
then f ( t , x ) is continuous and non-decreasing with respect to the second variable. Further, for x y , we have
f ( t , x ) f ( t , y ) = x 1 + cos t π y 1 + cos t π ( x y 2 ) 2 .

Taking g ( t ) = t 2 for all t [ 0 , + ) . By using Theorem 3.2, we know that the two-point boundary value problem (25) has a unique non-negative solution.

Declarations

Authors’ Affiliations

(1)
Department of Mathematics, Tianjin Polytechnic University

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© Su; licensee Springer. 2014

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