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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

• Accepted: 8 October 2014
• Published:

## 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 principle
• partially ordered metric spaces
• fixed point
• generalized altering distance function
• ordinary 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 $\psi :\left[0,\mathrm{\infty }\right)\to \left[0,\mathrm{\infty }\right)$ which satisfies:
1. (a)

ψ is continuous and non-decreasing.

2. (b)

$\psi =0$ 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 $\left(X,\le \right)$ be a partially ordered set and suppose that there exists a metric $d\in X$ such that $\left(X,d\right)$ is a complete metric space. Let $f:X\to X$ be a continuous and non-decreasing mapping such that

where $\psi :\left[0,\mathrm{\infty }\right)\to \left[0,\mathrm{\infty }\right)$ is continuous and non-decreasing function such that ψ is positive in $\left(0,\mathrm{\infty }\right)$, $\psi \left(0\right)=0$ and ${lim}_{t\to \mathrm{\infty }}\psi \left(t\right)=\mathrm{\infty }$. If there exists ${x}_{0}\in X$ with ${x}_{0}\le f\left({x}_{0}\right)$, then f has a fixed point.

Theorem 1.3 ([23])

Let $\left(X,\le \right)$ be a partially ordered set and suppose that there exists a metric $d\in X$ such that $\left(X,d\right)$ is a complete metric space. Let $f:X\to X$ be a continuous and non-decreasing mapping such that

where ψ and ϕ are altering distance functions. If there exists ${x}_{0}\in X$ with ${x}_{0}\le f\left({x}_{0}\right)$, 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 $\beta :\left[0,\mathrm{\infty }\right)\to \left[0,1\right)$ which satisfy the condition: $\beta \left({t}_{n}\right)\to 1⇒{t}_{n}\to 0$.

Theorem 1.4 ([10])

Let $\left(X,\le \right)$ be a partially ordered set and suppose that there exists a metric d such that $\left(X,d\right)$ is a complete metric space. Let $f:X\to X$ be an increasing mapping such that there exists an element ${x}_{0}\in X$ with ${x}_{0}\le f\left({x}_{0}\right)$. Suppose that there exists $\beta \in \mathrm{\Re }$ such that

Assume that either f is continuous or M is such that if an increasing sequence ${x}_{n}\to x\in X$, then ${x}_{n}\le x$, n. Besides, if for each $x,y\in X$ there exists $z\in m$ 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 $\left(X,d\right)$ is a complete metric space. Let $T:X\to X$ be a continuous and non-decreasing mapping such that
$\psi \left(d\left(Tx,Ty\right)\right)\le \varphi \left(d\left(x,y\right)\right),\phantom{\rule{1em}{0ex}}\mathrm{\forall }x\ge y,$

where ψ is an altering distance function and $\varphi :\left[0,\mathrm{\infty }\right)\to \left[0,\mathrm{\infty }\right)$ is a continuous function with the condition $\psi \left(t\right)>\varphi \left(t\right)$ for all $t>0$. If there exists ${x}_{0}\in X$ such that ${x}_{0}\le T{x}_{0}$, 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 $\psi :\left[0,\mathrm{\infty }\right)\to \left[0,\mathrm{\infty }\right)$ which satisfies:
1. (a)

ψ is non-decreasing;

2. (b)

$\psi =0$ 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 $\left(X,\le \right)$ is a partially ordered set and $T:X\to X$, we say that T is monotone non-decreasing if $x,y\in X$, $x\le y⇒T\left(x\right)\le T\left(y\right)$.

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 $\left(X,d\right)$ is a complete metric space. Let $T:X\to X$ be a continuous and non-decreasing mapping such that
$\psi \left(d\left(Tx,Ty\right)\right)\le \varphi \left(d\left(x,y\right)\right),\phantom{\rule{1em}{0ex}}\mathrm{\forall }x\ge y,$

where ψ is a generalized altering distance function and $\varphi :\left[0,\mathrm{\infty }\right)\to \left[0,\mathrm{\infty }\right)$ is a right upper semi-continuous function with the condition: $\psi \left(t\right)>\varphi \left(t\right)$ for all $t>0$. If there exists ${x}_{0}\in X$ such that ${x}_{0}\le T{x}_{0}$, then T has a fixed point.

Proof Since T is a non-decreasing function, we obtain by induction that
${x}_{0}\le T{x}_{0}\le {T}^{2}{x}_{0}\le {T}^{3}{x}_{0}\le \cdots \le {T}^{n}{x}_{0}\le {T}^{n+1}{x}_{0}\le \cdots .$
(1)
Put ${x}_{n+1}=T{x}_{n}$. Then, for each integer $n\ge 1$, from (1) and, as the elements ${x}_{n+1}$ and ${x}_{n}$ are comparable, we get
$\psi \left(d\left({x}_{n+1},{x}_{n}\right)\right)=\psi \left(d\left(T{x}_{n},T{x}_{n-1}\right)\right)\le \varphi \left(d\left({x}_{n},{x}_{n-1}\right)\right).$
(2)
Using the condition of Theorem 2.3 we have
$d\left({x}_{n+1},{x}_{n}\right)
(3)
Hence the sequence $\left\{d\left({x}_{n+1},{x}_{n}\right)\right\}$ is decreasing and, consequently, there exists $r\ge 0$ such that
$d\left({x}_{n+1},{x}_{n}\right)\to {r}^{+},$
as $n\to \mathrm{\infty }$. Consider the properties of ψ and ϕ, letting $n\to \mathrm{\infty }$ in (2) we get
$\psi \left(r\right)\le \underset{n\to \mathrm{\infty }}{lim}\psi \left(d\left({x}_{n+1},{x}_{n}\right)\right)\le \underset{n\to \mathrm{\infty }}{lim}\varphi \left(d\left({x}_{n},{x}_{n-1}\right)\right)\le \varphi \left(r\right).$
By using the condition: $\psi \left(t\right)>\varphi \left(t\right)$ for all $t>0$, we have $r=0$, and hence
$d\left({x}_{n+1},{x}_{n}\right)\to 0,$
(4)
as $n\to \mathrm{\infty }$. In what follows, we will show that $\left\{{x}_{n}\right\}$ is a Cauchy sequence. Suppose that $\left\{{x}_{n}\right\}$ is not a Cauchy sequence. Then there exists $\epsilon >0$ for which we can find subsequences $\left\{{x}_{{n}_{k}}\right\}$ with ${n}_{k}>{m}_{k}>k$ such that
$d\left({x}_{{n}_{k}},{x}_{{m}_{k}}\right)\ge \epsilon$
(5)
for all $k\ge 1$. 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
$d\left({x}_{{n}_{k-1}},{x}_{{m}_{k-1}}\right)<\epsilon .$
(6)
From (5) and (6), we have
$\epsilon \le d\left({x}_{{n}_{k}},{x}_{{m}_{k}}\right)\le \left(d\left({x}_{{n}_{k}}\right),{x}_{{n}_{k-1}}\right)+d\left({x}_{{n}_{k-1}},{x}_{{m}_{k}}\right)
Letting $k\to \mathrm{\infty }$ and using (4), we get
$\underset{k\to \mathrm{\infty }}{lim}d\left({x}_{{n}_{k}},{x}_{{m}_{k}}\right)=\epsilon .$
(7)
By using the triangular inequality we have
$\begin{array}{c}d\left({x}_{{n}_{k}},{x}_{{m}_{k}}\right)\le d\left({x}_{{n}_{k}},{x}_{{n}_{k-1}}\right)+d\left({x}_{{n}_{k-1}},{x}_{{m}_{k-1}}\right)+d\left({x}_{{m}_{k-1}},{x}_{{m}_{k}}\right),\hfill \\ d\left({x}_{{n}_{k-1}},{x}_{{m}_{k-1}}\right)\le d\left({x}_{{n}_{k-1}},{x}_{{n}_{k}}\right)+d\left({x}_{{n}_{k}},{x}_{{m}_{k}}\right)+d\left({x}_{{m}_{k}},{x}_{{m}_{k-1}}\right).\hfill \end{array}$
Letting $k\to \mathrm{\infty }$ in the above two inequalities and using (4) and (7), we have
$\underset{k\to \mathrm{\infty }}{lim}d\left({x}_{{n}_{k-1}},{x}_{{m}_{k-1}}\right)=\epsilon .$
(8)
As ${n}_{k}>{m}_{k}$ and ${x}_{{n}_{k-1}}$ and ${x}_{{m}_{k-1}}$ are comparable, using (1) we have
$\psi \left(d\left({x}_{{n}_{k}},{x}_{{m}_{k}}\right)\right)\le \varphi \left(d\left({x}_{{n}_{k-1}},{x}_{{m}_{k-1}}\right)\right).$
Consider the properties of ψ and ϕ, letting $k\to \mathrm{\infty }$ and taking into account (7) and (8), we have
$\psi \left(\epsilon \right)\le \varphi \left(\epsilon \right).$
From the condition $\psi \left(t\right)>\varphi \left(t\right)$ for all $t>0$, we get $\epsilon =0$, which is a contradiction. This shows that $\left\{{x}_{n}\right\}$ is a Cauchy sequence and, since X is a complete metric space, there exists $z\in X$ such that ${x}_{n}\to z$ as $n\to \mathrm{\infty }$. Moreover, the continuity of T implies that
$z=\underset{n\to \mathrm{\infty }}{lim}{x}_{n+1}=\underset{n\to \mathrm{\infty }}{lim}T{x}_{n}=Tz,$

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:
(9)
Theorem 2.4 Let $\left(X,\le \right)$ be a partially ordered set and suppose that there exists a metric d in X such that $\left(X,d\right)$ is a complete metric space. Assume that X satisfies (9). Let $T:X\to X$ be a non-decreasing mapping such that
$\psi \left(d\left(Tx,Ty\right)\right)\le \varphi \left(d\left(x,y\right)\right),\phantom{\rule{1em}{0ex}}\mathrm{\forall }x\ge y,$

where ψ is a generalized altering distance functions and ϕ: $\left[0,\mathrm{\infty }\right)\to \left[0,\mathrm{\infty }\right)$ is a right upper semi-continuous function with the condition $\psi \left(t\right)>\varphi \left(t\right)$ for all $t>0$. If there exists ${x}_{0}\in X$ such that ${x}_{0}\le T{x}_{0}$, then T has a fixed point.

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

Using condition of theorem we have $\varphi \left(0\right)=0$, this implies $\mathrm{\Psi }\left(d\left(z,T\left(z\right)\right)\right)=0$. Thus, $d\left(z,T\left(z\right)\right)=0$ or equivalently, $T\left(z\right)=z$. □

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 $X=\left\{\left(1,0\right),\left(0,1\right)\right\}\subset {R}^{2}$ and consider the usual order $\left(x,y\right)\le \left(z,t\right)⇔x\le z$, $y\le t$. Thus, $\left(x,y\right)$ is a partially ordered set whose different elements are not comparable. Besides $\left(X,{d}_{2}\right)$ is a complete metric space considering ${d}_{2}$ the Euclidean distance. The identity map $T\left(x,y\right)=\left(x,y\right)$ 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, $\left(1,0\right)\le T\left(1,0\right)=\left(1,0\right)$ 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:
(10)
In [12] it is proved that condition (10) is equivalent to:
(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 $z,y\in X$ which are fixed points. We distinguish two cases.

Case 1. If y is comparable to z then ${T}^{n}\left(y\right)=y$ is comparable to ${T}^{n}\left(z\right)=z$ for $n=0,1,2,\dots$ and
$\begin{array}{rl}\psi \left(d\left(z,y\right)\right)& =\psi \left(d\left({T}^{n}\left(z\right),{T}^{n}\left(y\right)\right)\right)\\ \le \varphi \left(d\left({T}^{n-1}\left(z\right),{T}^{n-1}\left(y\right)\right)\right)\\ \le \varphi \left(d\left(z,y\right)\right).\end{array}$

As we have the condition $\psi \left(t\right)>\varphi \left(t\right)$ for $t>0$ we obtain $d\left(z,y\right)=0$ and this implies $z=y$.

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

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 $x\in X$, ${lim}_{n\to \mathrm{\infty }}{T}^{n}\left(x\right)=z$, 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 $\varphi \left(t\right)=t-\psi \left(t\right)$. Theorem 1.3 is a particular case of our Theorem 2.3 for $\varphi \left(t\right)$ being replaced by $\psi \left(t\right)-\varphi \left(t\right)$. Theorem 1.4 is a particular case of Theorem 2.3 for ψ being the identity function, and $\varphi \left(t\right)=\beta \left(t\right)t$. 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:
$\begin{array}{c}{\psi }_{1}\left(t\right)=\left\{\begin{array}{cc}0,\hfill & t=0,\hfill \\ \left[t\right]+1,\hfill & t>0,\hfill \end{array}\hfill \\ {\psi }_{2}\left(t\right)=\left\{\begin{array}{cc}0,\hfill & t=0,\hfill \\ \lambda \left(\left[t\right]+1\right),\hfill & t>0,\hfill \end{array}\hfill \end{array}$
where $\alpha >0$ is a constant.
${\psi }_{3}\left(t\right)=\left\{\begin{array}{cc}t,\hfill & 0\le t<1,\hfill \\ \alpha {t}^{2},\hfill & t\ge 1,\hfill \end{array}$

where $\alpha \ge 1$ is a constant.

We choose $\psi \left(t\right)={\psi }_{3}\left(t\right)$ and
$\varphi \left(t\right)=\left\{\begin{array}{cc}{t}^{2},\hfill & 0\le t<1,\hfill \\ \beta t,\hfill & t\ge 1,\hfill \end{array}$

where $0<\beta <\alpha$ 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 $\left(X,d\right)$ is a complete metric space. Let $T:X\to X$ be a continuous and non-decreasing mapping such that
$\begin{array}{c}0\le d\left(Tx,Ty\right)<1\phantom{\rule{1em}{0ex}}⇒\phantom{\rule{1em}{0ex}}d\left(Tx,Ty\right)\le {\left(d\left(x,y\right)\right)}^{2},\hfill \\ d\left(Tx,Ty\right)\ge 1\phantom{\rule{1em}{0ex}}⇒\phantom{\rule{1em}{0ex}}\alpha {\left(d\left(Tx,Ty\right)\right)}^{2}\le \beta d\left(x,y\right)\hfill \end{array}$

for any $x,y\in X$. If there exists ${x}_{0}\in X$ such that ${x}_{0}\le T{x}_{0}$, 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:
$\left\{\begin{array}{cc}{u}^{\prime }\left(t\right)=f\left(t,u\left(t\right)\right),\hfill & t\in \left[0,T\right],\hfill \\ u\left(0\right)=u\left(T\right),\hfill \end{array}$
(13)
where $T>0$ and $f:I×R\to R$ is a continuous function. Previously, we considered the space $C\left(I\right)$ ($I=\left[0,T\right]$) of continuous functions defined on I. Obviously, this space with the metric given by
is a complete metric space. $C\left(I\right)$ can also be equipped with a partial order given by

Clearly, $\left(C\left(I\right),\le \right)$ satisfies condition (10), since for $x,y\in C\left(I\right)$ the functions $max\left\{x,y\right\}$ and $min\left\{x,y\right\}$ are least upper and greatest lower bounds of x and y, respectively. Moreover, in [17] it is proved that $\left(C\left(I\right),\le \right)$ 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 $\alpha \in {C}^{\left(1\right)}\left(I\right)$ such that
Theorem 3.2 Consider problem (13) with $f:I×R\to R$ continuous and suppose that there exist $\lambda ,\alpha >0$ with
$\alpha \le {\left(\frac{2\lambda \left({e}^{\lambda T}-1\right)}{T\left({e}^{\lambda T}+1\right)}\right)}^{\frac{1}{2}}$
such that for $x,y\in R$ with $x\ge y$
$0\le f\left(t,x\right)+\lambda x-\left[f\left(t,y\right)+\lambda y\right]\le \alpha \sqrt{g\left(x-y\right)},$

where $g\left(t\right):\left[0,+\mathrm{\infty }\right)\to \left[0,+\mathrm{\infty }\right)$ is a light upper semi-continuous function with $g\left(0\right)=0$, $g\left(t\right)<{t}^{2}$, $\mathrm{\forall }t>0$. 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
This problem is equivalent to the integral equation
$u\left(t\right)={\int }_{0}^{T}G\left(t,s\right)\left[f\left(s,u\left(s\right)\right)+\lambda u\left(s\right)\right]\phantom{\rule{0.2em}{0ex}}ds,$
where $G\left(t,s\right)$ is the Green function given by
$G\left(t,s\right)=\left\{\begin{array}{cc}\frac{{e}^{\lambda \left(T+s-t\right)}}{{e}^{\lambda T}-1},\hfill & 0\le s
Define $F:C\left(I\right)\to C\left(I\right)$ by
$\left(Fu\right)\left(t\right)={\int }_{0}^{T}G\left(t,s\right)\left[f\left(s,u\left(s\right)\right)+\lambda u\left(s\right)\right]\phantom{\rule{0.2em}{0ex}}ds.$
Note that if $u\in C\left(I\right)$ is a fixed point of F then $u\in {C}^{1}\left(I\right)$ 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 $u\ge v$, and using our assumption. We can obtain
$f\left(t,u\right)+\lambda u\ge f\left(t,v\right)+\lambda v$
which implies, since $G\left(t,s\right)>0$, that for $t\in I$
$\begin{array}{rl}\left(Fu\right)\left(t\right)& ={\int }_{0}^{T}G\left(t,s\right)\left[f\left(s,u\left(s\right)\right)+\lambda u\left(s\right)\right]\phantom{\rule{0.2em}{0ex}}ds\\ \ge {\int }_{0}^{T}G\left(t,s\right)\left[f\left(s,v\left(s\right)\right)+\lambda v\left(s\right)\right]\phantom{\rule{0.2em}{0ex}}ds=\left(Fv\right)\left(t\right).\end{array}$
Besides, for $u\ge v$, we have
$\begin{array}{rl}d\left(Fu,Fv\right)& =\underset{t\in I}{sup}|\left(Fu\right)\left(t\right)-\left(Fv\right)\left(t\right)|\\ =\underset{t\in I}{sup}\left(\left(Fu\right)\left(t\right)-\left(Fv\right)\left(t\right)\right)\\ =\underset{t\in I}{sup}{\int }_{0}^{T}G\left(t,s\right)\left[f\left(s,u\left(s\right)\right)+\lambda u\left(s\right)-f\left(s,v\left(s\right)\right)-\lambda v\left(s\right)\right]\phantom{\rule{0.2em}{0ex}}ds\\ \le \underset{t\in I}{sup}{\int }_{0}^{T}G\left(t,s\right)\alpha \sqrt{g\left(u\left(s\right)-v\left(s\right)\right)}\phantom{\rule{0.2em}{0ex}}ds.\end{array}$
(14)
Using the Cauchy-Schwarz inequality in the last integral we get
$\begin{array}{r}{\int }_{0}^{T}G\left(t,s\right)\alpha \sqrt{g\left(u\left(s\right)-v\left(s\right)\right)}\phantom{\rule{0.2em}{0ex}}ds\\ \phantom{\rule{1em}{0ex}}\le {\left({\int }_{0}^{T}G{\left(t,s\right)}^{2}\phantom{\rule{0.2em}{0ex}}ds\right)}^{\frac{1}{2}}{\left({\int }_{0}^{T}{\alpha }^{2}g\left(u\left(s\right)-v\left(s\right)\right)\phantom{\rule{0.2em}{0ex}}ds\right)}^{\frac{1}{2}}.\end{array}$
(15)
The first integral gives us
$\begin{array}{rl}{\int }_{0}^{T}G{\left(t,s\right)}^{2}\phantom{\rule{0.2em}{0ex}}ds& ={\int }_{0}^{t}G{\left(t,s\right)}^{2}\phantom{\rule{0.2em}{0ex}}ds+{\int }_{t}^{T}G{\left(t,s\right)}^{2}\phantom{\rule{0.2em}{0ex}}ds\\ ={\int }_{0}^{t}\frac{{e}^{2\lambda \left(T+s-t\right)}}{{\left({e}^{\lambda T}-1\right)}^{2}}\phantom{\rule{0.2em}{0ex}}ds+{\int }_{t}^{T}\frac{{e}^{2\lambda \left(s-t\right)}}{{\left({e}^{\lambda T}-1\right)}^{2}}\phantom{\rule{0.2em}{0ex}}ds\\ =\frac{1}{2\lambda {\left({e}^{\lambda T}-1\right)}^{2}}{e}^{\left(2\lambda T-1\right)}\\ =\frac{{e}^{\lambda T}+1}{2\lambda \left({e}^{\lambda T}-1\right)}.\end{array}$
(16)
The second integral in (15) gives the following estimate:
$\begin{array}{rl}{\int }_{0}^{T}{\alpha }^{2}g\left(u\left(s\right)-v\left(s\right)\right)\phantom{\rule{0.2em}{0ex}}ds& \le {\alpha }^{2}g\left(\parallel u-v\parallel \right)\cdot T\\ ={\alpha }^{2}g\left(d\left(u,v\right)\right)\cdot T.\end{array}$
(17)
Taking into account (14)-(17) we have
$\begin{array}{rl}d\left(Fu,Fv\right)& \le \underset{t\in I}{sup}{\left(\frac{{e}^{\lambda T}+1}{2\lambda \left({e}^{\lambda T}-1\right)}\right)}^{\frac{1}{2}}\cdot {\left({\alpha }^{2}g\left(d\left(u,v\right)\right)\cdot T\right)}^{\frac{1}{2}}\\ ={\left(\frac{{e}^{\lambda T}+1}{2\lambda \left({e}^{\lambda T}-1\right)}\right)}^{\frac{1}{2}}\cdot \alpha \cdot \sqrt{T}\cdot {\left(g\left(d\left(u,v\right)\right)\right)}^{\frac{1}{2}}\end{array}$
and from the last inequality we obtain
$d{\left(Fu,Fv\right)}^{2}\le \frac{{e}^{\lambda T}+1}{2\lambda \left({e}^{\lambda T}-1\right)}\cdot {\alpha }^{2}\cdot T\cdot g\left(d\left(u,v\right)\right)$
or, equivalently.
$2\lambda \left({e}^{\lambda T}-1\right)d{\left(Fu,Fv\right)}^{2}\le \left({e}^{\lambda T}+1\right)\cdot {\alpha }^{2}\cdot T\cdot g\left(d\left(u,v\right)\right).$
By our assumption, as
$\alpha \le {\left(\frac{2\lambda \left({e}^{\lambda T}-1\right)}{T\left({e}^{\lambda T}+1\right)}\right)}^{\frac{1}{2}},$
the last inequality gives us
$2\lambda \left({e}^{\lambda T}-1\right)d{\left(Fu,Fv\right)}^{2}\le 2\lambda \left({e}^{\lambda T}-1\right)\cdot g\left(d\left(u,v\right)\right),$
and, hence,
$d{\left(Fu,Fv\right)}^{2}\le g\left(d\left(u,v\right)\right).$
(18)
Put $\psi \left(t\right)={t}^{2}$ and $\varphi \left(t\right)=g\left(t\right)$. Obviously, ψ is a generalized altering distance function, $\psi \left(t\right)$ and $\varphi \left(t\right)$ satisfy the condition of $\psi \left(t\right)>\varphi \left(t\right)$ for $t>0$. From (18), we obtain for $u\ge v$
$\psi \left(d\left(Fu,Fv\right)\right)\le \varphi \left(d\left(u,v\right)\right).$
Finally, let $\alpha \left(t\right)$ be a lower solution for (13); we claim that $\alpha \le F\left(\alpha \right)$. In fact
We multiply by ${e}^{\lambda t}$,
and this gives us
(19)
As $\alpha \left(0\right)\le \alpha \left(T\right)$, the last inequality gives us
$\alpha \left(0\right){e}^{\lambda t}\le \alpha \left(T\right){e}^{\lambda T}\le \alpha \left(0\right)+{\int }_{0}^{T}\left[f\left(s,\alpha \left(s\right)\right)+\lambda \alpha \left(s\right)\right]{e}^{\lambda s}\phantom{\rule{0.2em}{0ex}}ds,$
and so
$\alpha \left(0\right)\le {\int }_{0}^{T}\frac{{e}^{\lambda s}}{{e}^{\lambda T}-1}\left[f\left(s,\alpha \left(s\right)\right)+\lambda \alpha \left(s\right)\right]\phantom{\rule{0.2em}{0ex}}ds.$
This and (19) give us
$\alpha \left(t\right){e}^{\lambda t}\le {\int }_{0}^{t}\frac{{e}^{\lambda \left(T+s\right)}}{{e}^{\lambda T}-1}\left[f\left(s,\alpha \left(s\right)\right)+\lambda \alpha \left(s\right)\right]\phantom{\rule{0.2em}{0ex}}ds+{\int }_{t}^{T}\frac{{e}^{\lambda s}}{{e}^{\lambda T}-1}\left[f\left(s,\alpha \left(s\right)\right)+\lambda \alpha \left(s\right)\right]\phantom{\rule{0.2em}{0ex}}ds$
and, consequently,

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 $g\left(t\right)$ as follows:
1. (1)

${g}_{1}\left(t\right)=ln\left({t}^{2}+1\right)$;

2. (2)
${g}_{2}\left(t\right)=\left\{\begin{array}{cc}{t}^{3},\hfill & 0\le t<1,\hfill \\ \frac{1}{2},\hfill & t=1,\hfill \\ t,\hfill & 1

3. (3)
${g}_{3}\left(t\right)=\left\{\begin{array}{cc}{t}^{3},\hfill & 0\le t\le \frac{1}{2},\hfill \\ t-\frac{3}{8},\hfill & \frac{1}{2}

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

Example 3.4 Consider the following first-order periodic problem:
$\left\{\begin{array}{cc}{u}^{\prime }\left(t\right)=\frac{sint}{{e}^{t}}-\beta x,\hfill & t\in \left[0,T\right],\hfill \\ u\left(0\right)=u\left(T\right).\hfill \end{array}$
(20)
Let
$f\left(t,x\right)=\frac{sint}{{e}^{t}}-\beta x,\phantom{\rule{1em}{0ex}}x\in \left[0,\mathrm{\infty }\right),t\in \left[0,1\right],$
then $f\left(t,x\right)$ is continuous. Further, for $x\ge y$, we have
$f\left(t,x\right)+\lambda x-\left[f\left(t,y\right)+\lambda y\right]=2\left(\lambda -\beta \right)\frac{x-y}{2}.$
We chose $\beta \in \left[0,\lambda \right]$ such that
$2\left(\lambda -\beta \right)\le {\left(\frac{2\lambda \left({e}^{\lambda T}-1\right)}{T\left({e}^{\lambda T}+1\right)}\right)}^{\frac{1}{2}}.$
Taking $g\left(t\right)={\left(\frac{t}{2}\right)}^{2}$ for all $t\in \left[0,+\mathrm{\infty }\right)$, we have
$f\left(t,x\right)+\lambda x-\left[f\left(t,y\right)+\lambda y\right]=2\left(\lambda -\beta \right)\sqrt{g\left(x-y\right)}.$

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:
$\left\{\begin{array}{cc}-\frac{{d}^{2}x}{d{t}^{2}}=f\left(t,x\right),\hfill & x\in \left[0,\mathrm{\infty }\right),t\in \left[0,1\right],\hfill \\ x\left(0\right)=x\left(1\right)=0.\hfill \end{array}$
(21)
It is well known that $x\in {C}^{2}\left[0,1\right]$ is a solution of (20) that is equivalent to $x\in C\left[0,1\right]$ being a solution of the integral equation
where $G\left(t,s\right)$ is the Green function given by
$G\left(t,s\right)=\left\{\begin{array}{cc}t\left(1-s\right),\hfill & 0\le t\le s\le 1,\hfill \\ s\left(1-t\right),\hfill & 0\le s\le t\le 1.\hfill \end{array}$
(22)
Theorem 3.5 Consider problem (21) with $f:I×R\to \left[0,\mathrm{\infty }\right)$ continuous and non-decreasing with respect to the second variable and suppose that there exists $0\le \alpha \le 8$ such that for $x,y\in R$ with $x\ge y$
$f\left(t,x\right)-f\left(t,y\right)\le \alpha \sqrt{g\left(x-y\right)},$
(23)

where $g\left(t\right):\left[0,+\mathrm{\infty }\right)\to \left[0,+\mathrm{\infty }\right)$ is a light upper semi-continuous function with $g\left(0\right)=0$, $g\left(t\right)<{t}^{2}$, $\mathrm{\forall }t>0$. Then our problem (21) has a unique non-negative solution.

Proof Consider the cone
$P=\left\{x\in C\left[0,1\right]:x\left(t\right)\ge 0\right\}.$
Obviously, $\left(P,d\right)$ with $d\left(x,y\right)=sup\left\{|x\left(t\right)-y\left(t\right)|:t\in \left[0,1\right]\right\}$ is a complete metric space. Consider the operator given by

where $G\left(t,s\right)$ is the Green function appearing in (22).

As f is non-decreasing with respect to the second variable, for $x,y\in P$ with $y\ge x$ and $t\in \left[0,1\right]$, we have
$\left(Ty\right)\left(t\right)={\int }_{0}^{1}G\left(t,s\right)f\left(s,y\left(s\right)\right)\phantom{\rule{0.2em}{0ex}}ds\ge {\int }_{0}^{1}G\left(t,s\right)f\left(s,x\left(s\right)\right)\phantom{\rule{0.2em}{0ex}}ds\ge \left(Tx\right)\left(t\right),$

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

Besides, for $y\ge x$ and taking into account (23), we obtain
$\begin{array}{rl}d\left(Ty,Tx\right)& =\underset{t\in \left[0,1\right]}{sup}|\left(Tx\right)\left(t\right)-\left(Ty\right)\left(t\right)|\\ =\underset{t\in \left[0,1\right]}{sup}\left(\left(Tx\right)\left(t\right)-\left(Ty\right)\left(t\right)\right)\\ =\underset{t\in \left[0,1\right]}{sup}{\int }_{0}^{1}G\left(t,s\right)\left(f\left(s,x\left(s\right)\right)-f\left(s,y\left(s\right)\right)\right)\phantom{\rule{0.2em}{0ex}}ds\\ \le \underset{t\in \left[0,1\right]}{sup}{\int }_{0}^{1}G\left(t,s\right)\alpha \sqrt{g\left(x\left(s\right)-y\left(s\right)\right)}\\ \le \underset{t\in \left[0,1\right]}{sup}{\int }_{0}^{1}G\left(t,s\right)\alpha \sqrt{g\left(x\left(s\right)-y\left(s\right)\right)}\phantom{\rule{0.2em}{0ex}}ds\\ =\alpha \sqrt{ln\left[{\parallel y-x\parallel }^{2}+1\right]}\underset{t\in \left[0,1\right]}{sup}{\int }_{0}^{1}G\left(t,s\right)\phantom{\rule{0.2em}{0ex}}ds.\end{array}$
(24)
It is easy to verify that
${\int }_{0}^{1}G\left(t,s\right)\phantom{\rule{0.2em}{0ex}}ds=\frac{-{t}^{2}}{2}+\frac{t}{2}$
and that
$\underset{t\in \left[0,1\right]}{sup}{\int }_{0}^{1}G\left(t,s\right)\phantom{\rule{0.2em}{0ex}}ds=\frac{1}{8}.$
These facts, the inequality (24), and the hypothesis $0<\alpha \le 8$ give us
$\begin{array}{rl}d\left(Tx,Ty\right)& \le \frac{\alpha }{8}\sqrt{g\left(x-y\right)}\\ \le \sqrt{g\left(\parallel x-y\parallel \right)}=\sqrt{g\left(d\left(x,y\right)\right)}.\end{array}$
Hence
$d{\left(Ty,Tx\right)}^{2}\le g\left(d\left(x,y\right)\right).$
Put $\psi \left(t\right)={t}^{2}$, $\varphi \left(t\right)=g\left(t\right)$, obviously ψ is an altering distance function, ψ and ϕ satisfy the condition of $\psi \left(t\right)>\varphi \left(t\right)$, for $t>0$. From the last inequality, we have
$\psi \left(d\left(Tx,Ty\right)\right)\le \varphi \left(d\left(x,y\right)\right).$
Finally, as f and G are non-negative functions
$T0={\int }_{0}^{1}G\left(t,s\right)f\left(s,0\right)\phantom{\rule{0.2em}{0ex}}ds\ge 0$

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 $g\left(t\right)$ as ${g}_{1}\left(1\right)$, ${g}_{2}\left(t\right)$, and ${g}_{3}\left(t\right)$ as well as in Theorem 3.2.

Example 3.7 Consider the following two-point boundary value problem of the second order differential equation:
$\left\{\begin{array}{cc}-\frac{{d}^{2}x}{d{t}^{2}}=\frac{sint}{{e}^{t}}+\frac{x}{1+cost\pi },\hfill & x\in \left[0,\mathrm{\infty }\right),t\in \left[0,1\right],\hfill \\ x\left(0\right)=x\left(1\right)=0.\hfill \end{array}$
(25)
Let
$f\left(t,x\right)=\frac{sint}{{e}^{t}}+\frac{x}{1+cost\pi },\phantom{\rule{1em}{0ex}}x\in \left[0,\mathrm{\infty }\right),t\in \left[0,1\right],$
then $f\left(t,x\right)$ is continuous and non-decreasing with respect to the second variable. Further, for $x\ge y$, we have
$f\left(t,x\right)-f\left(t,y\right)=\frac{x}{1+cost\pi }-\frac{y}{1+cost\pi }\le \sqrt{{\left(\frac{x-y}{2}\right)}^{2}}.$

Taking $g\left(t\right)=\frac{t}{2}$ for all $t\in \left[0,+\mathrm{\infty }\right)$. By using Theorem 3.2, we know that the two-point boundary value problem (25) has a unique non-negative solution.

## Authors’ Affiliations

(1)
Department of Mathematics, Tianjin Polytechnic University, Tianjin, 300387, P.R. China

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