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

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.

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 [1–8], 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, 9–22]. 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 $\psi :[0,\mathrm{\infty })\to [0,\mathrm{\infty })$ which satisfies:

• (a) ψ is continuous and non-decreasing.

• (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 $(X,\le )$ 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 $f:X\to X$ be a continuous and non-decreasing mapping such that

$$d(f(x),f(y))\le d(x,y)-\psi (d(x,y)),\mathit{\text{for }}x\ge y,$$

where $\psi :[0,\mathrm{\infty })\to [0,\mathrm{\infty })$ is continuous and non-decreasing function such that ψ is positive in $(0,\mathrm{\infty })$, $\psi (0)=0$ and ${lim}_{t\to \mathrm{\infty }}\psi (t)=\mathrm{\infty }$. If there exists $x_0\in X$ with $x_0\le f(x_0)$, then f has a fixed point.

Theorem 1.3 ([23])

Let $(X,\le )$ 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 $f:X\to X$ be a continuous and non-decreasing mapping such that

$$\psi d(f(x),f(y))\le \psi (d(x,y))-\varphi (d(x,y)),\mathit{\text{for }}x\ge y,$$

where ψ and ϕ are altering distance functions. If there exists $x_0\in X$ with $x_0\le f(x_0)$, 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 ℜ denote the class of those functions $\beta :[0,\mathrm{\infty })\to [0,1)$ which satisfy the condition: $\beta (t_n)\to 1\Rightarrow t_n\to 0$.

Theorem 1.4 ([10])

Let $(X,\le )$ be a partially ordered set and suppose that there exists a metric d such that $(X,d)$ is a complete metric space. Let $f:X\to X$ be an increasing mapping such that there exists an element $x_0\in X$ with $x_0\le f(x_0)$. Suppose that there exists $\beta \in \mathrm{\Re }$ such that

$$d(f(x),f(y))\le \beta (d(x,y))d(x,y)\mathit{\text{for each }}x,y\in X\mathit{\text{ with }}x\ge y.$$

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 $(X,d)$ is a complete metric space. Let $T:X\to X$ be a continuous and non-decreasing mapping such that

$$\psi (d(Tx,Ty))\le \varphi (d(x,y)),\mathrm{\forall }x\ge y,$$

where ψ is an altering distance function and $\varphi :[0,\mathrm{\infty })\to [0,\mathrm{\infty })$ is a continuous function with the condition $\psi (t)>\varphi (t)$ 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 $\psi :[0,\mathrm{\infty })\to [0,\mathrm{\infty })$ which satisfies:

• (a) ψ is non-decreasing;

• (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 $(X,\le )$ is a partially ordered set and $T:X\to X$, we say that T is monotone non-decreasing if $x,y\in X$, $x\le y\Rightarrow T(x)\le T(y)$.

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 $T:X\to X$ be a continuous and non-decreasing mapping such that

$$\psi (d(Tx,Ty))\le \varphi (d(x,y)),\mathrm{\forall }x\ge y,$$

where ψ is a generalized altering distance function and $\varphi :[0,\mathrm{\infty })\to [0,\mathrm{\infty })$ is a right upper semi-continuous function with the condition: $\psi (t)>\varphi (t)$ 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

$$\psi (d(x_{n+1},x_n))=\psi (d(T{x}_n,T{x}_{n-1}))\le \varphi (d(x_n,x_{n-1})).$$

(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 $r\ge 0$ such that

$$d(x_{n+1},x_n)\to r^{+},$$

as $n\to \mathrm{\infty }$. Consider the properties of ψ and ϕ, letting $n\to \mathrm{\infty }$ in (2) we get

$$\psi (r)\le \underset{n\to \mathrm{\infty }}{lim}\psi (d(x_{n+1},x_n))\le \underset{n\to \mathrm{\infty }}{lim}\varphi (d(x_n,x_{n-1}))\le \varphi (r).$$

By using the condition: $\psi (t)>\varphi (t)$ for all $t>0$, we have $r=0$, and hence

$$d(x_{n+1},x_n)\to 0,$$

(4)

as $n\to \mathrm{\infty }$. 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 $\epsilon >0$ for which we can find subsequences $\{x_{n_k}\}$ with $n_k>m_k>k$ such that

$$d(x_{n_k},x_{m_k})\ge \epsilon$$

(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

$$d(x_{n_{k-1}},x_{m_{k-1}})<\epsilon .$$

(6)

From (5) and (6), we have

$$\epsilon \le d(x_{n_k},x_{m_k})\le (d(x_{n_k}),x_{n_{k-1}})+d(x_{n_{k-1}},x_{m_k})<d(x_{n_k},x_{n_{k-1}})+\epsilon .$$

Letting $k\to \mathrm{\infty }$ and using (4), we get

$$\underset{k\to \mathrm{\infty }}{lim}d(x_{n_k},x_{m_k})=\epsilon .$$

(7)

By using the triangular inequality we have

$$\begin{array}{c}d(x_{n_k},x_{m_k})\le d(x_{n_k},x_{n_{k-1}})+d(x_{n_{k-1}},x_{m_{k-1}})+d(x_{m_{k-1}},x_{m_k}),\hfill \\ d(x_{n_{k-1}},x_{m_{k-1}})\le d(x_{n_{k-1}},x_{n_k})+d(x_{n_k},x_{m_k})+d(x_{m_k},x_{m_{k-1}}).\hfill \end{array}$$

Letting $k\to \mathrm{\infty }$ in the above two inequalities and using (4) and (7), we have

$$\underset{k\to \mathrm{\infty }}{lim}d(x_{n_{k-1}},x_{m_{k-1}})=\epsilon .$$

(8)

As $n_k>m_k$ and $x_{n_{k-1}}$ and $x_{m_{k-1}}$ are comparable, using (1) we have

$$\psi (d(x_{n_k},x_{m_k}))\le \varphi (d(x_{n_{k-1}},x_{m_{k-1}})).$$

Consider the properties of ψ and ϕ, letting $k\to \mathrm{\infty }$ and taking into account (7) and (8), we have

$$\psi (\epsilon )\le \varphi (\epsilon ).$$

From the condition $\psi (t)>\varphi (t)$ for all $t>0$, we get $\epsilon =0$, which is a contradiction. This shows that $\{x_n\}$ is a Cauchy sequence and, since X is a complete metric space, there exists $z\in X$ 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:

$$\begin{array}{r}\text{If }(x_n)\text{ is a non-decreasing sequence in }X\text{ such that }{x}_n\to x\\ \text{then }{x}_n\le x\text{ for all }n\in N.\end{array}$$

(9)

Theorem 2.4 Let $(X,\le )$ 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 $T:X\to X$ be a non-decreasing mapping such that

$$\psi (d(Tx,Ty))\le \varphi (d(x,y)),\mathrm{\forall }x\ge y,$$

where ψ is a generalized altering distance functions and ϕ: $[0,\mathrm{\infty })\to [0,\mathrm{\infty })$ is a right upper semi-continuous function with the condition $\psi (t)>\varphi (t)$ 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 $T(z)=z$. As $(x_n)$ 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,

$$\psi \left(d(x_{n+1},T(z))\right)=\psi \left(d(T(x_n),T(z))\right)\le \varphi (d(x_n,z)).$$

Letting $n\to \mathrm{\infty }$ and taking into account that ψ is an altering distance function, we have

$$\psi \left(d(z,T(z))\right)\le \varphi (0).$$

Using condition of theorem we have $\varphi (0)=0$, this implies $\mathrm{\Psi }(d(z,T(z)))=0$. 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 $X=\{(1,0),(0,1)\}\subset R^2$ and consider the usual order $(x,y)\le (z,t)\iff x\le z$, $y\le t$. 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, $(1,0)\le T(1,0)=(1,0)$ 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:

$$\text{for }x,y\in X\text{ there exists a lower bound or an upper bound}.$$

(10)

In [12] it is proved that condition (10) is equivalent to:

$$\text{for }x,y\in X\text{ there exists }z\in X\text{ which is comparable to }x\text{ and }y.$$

(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 $z,y\in X$ 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 $n=0,1,2,\dots$ and

$$\begin{array}{rl}\psi (d(z,y))& =\psi \left(d(T^n(z),T^n(y))\right)\\ \le \varphi \left(d(T^{n-1}(z),T^{n-1}(y))\right)\\ \le \varphi (d(z,y)).\end{array}$$

As we have the condition $\psi (t)>\varphi (t)$ 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 $x\in X$ 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 $n=0,1,2,\dots$ Moreover,

$$\begin{array}{rl}\psi \left(d(z,T^n(x))\right)& =\psi \left(d(T^n(z),T^n(x))\right)\\ \le \varphi \left(d(T^{n-1}(z),T^{n-1}(x))\right)\\ =\varphi \left(d(z,T^{n-1}(x))\right).\end{array}$$

(12)

Hence, ψ is a generalized altering distance function and we have the condition $\psi (t)>\varphi (t)$ 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

$$\underset{n\to \mathrm{\infty }}{lim}d(z,T^n(x))=\gamma .$$

Letting $n\to \mathrm{\infty }$ in (12) and, taking into account the properties of ψ and ϕ, we obtain

$$\psi (\gamma )\le \varphi (\gamma ).$$

This and the condition $\psi (t)>\varphi (t)$ for $t>0$ imply $\gamma =0$. Analogously, it can be proved that

$$\underset{n\to \mathrm{\infty }}{lim}d(y,T^n(x))=0.$$

Finally, as

$$\underset{n\to \mathrm{\infty }}{lim}d(z,T^n(x))=\underset{n\to \mathrm{\infty }}{lim}d(y,T^n(x))=0$$

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 $x\in X$, ${lim}_{n\to \mathrm{\infty }}{T}^n(x)=z$, 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 $\varphi (t)=t-\psi (t)$. Theorem 1.3 is a particular case of our Theorem 2.3 for $\varphi (t)$ being replaced by $\psi (t)-\varphi (t)$. Theorem 1.4 is a particular case of Theorem 2.3 for ψ being the identity function, and $\varphi (t)=\beta (t)t$. 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:

$$\begin{array}{c}{\psi }_1(t)=\{\begin{array}{cc}0,\hfill & t=0,\hfill \\ [t]+1,\hfill & t>0,\hfill \end{array}\hfill \\ {\psi }_2(t)=\{\begin{array}{cc}0,\hfill & t=0,\hfill \\ \lambda ([t]+1),\hfill & t>0,\hfill \end{array}\hfill \end{array}$$

where $\alpha >0$ is a constant.

$${\psi }_3(t)=\{\begin{array}{cc}t,\hfill & 0\le t<1,\hfill \\ \alpha t^2,\hfill & t\ge 1,\hfill \end{array}$$

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

We choose $\psi (t)={\psi }_3(t)$ and

$$\varphi (t)=\{\begin{array}{cc}{t}^2,\hfill & 0\le t<1,\hfill \\ \beta t,\hfill & t\ge 1,\hfill \end{array}$$

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 $(X,d)$ is a complete metric space. Let $T:X\to X$ be a continuous and non-decreasing mapping such that

$$\begin{array}{c}0\le d(Tx,Ty)<1\Rightarrow d(Tx,Ty)\le {(d(x,y))}^2,\hfill \\ d(Tx,Ty)\ge 1\Rightarrow \alpha {(d(Tx,Ty))}^2\le \beta d(x,y)\hfill \end{array}$$

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:

$$\{\begin{array}{cc}{u}^{\prime }(t)=f(t,u(t)),\hfill & t\in [0,T],\hfill \\ u(0)=u(T),\hfill \end{array}$$

(13)

where $T>0$ and $f:I\times R\to R$ 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

$$d(x,y)=sup\{|x(t)-y(t)|:t\in I\},\text{for }x,y\in C(I),$$

is a complete metric space. $C(I)$ can also be equipped with a partial order given by

$$x,y\in C(I),x\le y\iff x(t)\le y(t)m,\text{for }t\in I.$$

Clearly, $(C(I),\le )$ satisfies condition (10), since for $x,y\in C(I)$ 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 $(C(I),\le )$ 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 $\alpha \in C^{(1)}(I)$ such that

$$\{\begin{array}{cc}{\alpha }^{\prime }(t)\le f(t,\alpha (t)),\hfill & \text{for }t\in I,\hfill \\ \alpha (0)\le \alpha (T).\hfill \end{array}$$

Theorem 3.2 Consider problem (13) with $f:I\times R\to R$ continuous and suppose that there exist $\lambda ,\alpha >0$ with

$$\alpha \le {\left(\frac{2\lambda (e^{\lambda T}-1)}{T(e^{\lambda T}+1)}\right)}^{\frac{1}{2}}$$

such that for $x,y\in R$ with $x\ge y$

$$0\le f(t,x)+\lambda x-[f(t,y)+\lambda y]\le \alpha \sqrt{g(x-y)},$$

where $g(t):[0,+\mathrm{\infty })\to [0,+\mathrm{\infty })$ is a light upper semi-continuous function with $g(0)=0$, $g(t)<t^2$, $\mathrm{\forall }t>0$. 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

$$\{\begin{array}{cc}{u}^{\prime }(t)+\lambda u(t)=f(t,u(t))+\lambda u(t),\hfill & \text{for }t\in I=[0,T],\hfill \\ u(0)=u(T).\hfill \end{array}$$

This problem is equivalent to the integral equation

$$u(t)={\int }_0^{T}G(t,s)[f(s,u(s))+\lambda u(s)]ds,$$

where $G(t,s)$ is the Green function given by

$$G(t,s)=\{\begin{array}{cc}\frac{e^{\lambda (T+s-t)}}{e^{\lambda T}-1},\hfill & 0\le s<t\le T,\hfill \\ \frac{e^{\lambda (s-t)}}{e^{\lambda T}-1},\hfill & 0\le t<s\le T.\hfill \end{array}$$

Define $F:C(I)\to C(I)$ by

$$(Fu)(t)={\int }_0^{T}G(t,s)[f(s,u(s))+\lambda u(s)]ds.$$

Note that if $u\in C(I)$ is a fixed point of F then $u\in C^1(I)$ 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 $u\ge v$, and using our assumption. We can obtain

$$f(t,u)+\lambda u\ge f(t,v)+\lambda v$$

which implies, since $G(t,s)>0$, that for $t\in I$

$$\begin{array}{rl}(Fu)(t)& ={\int }_0^{T}G(t,s)[f(s,u(s))+\lambda u(s)]ds\\ \ge {\int }_0^{T}G(t,s)[f(s,v(s))+\lambda v(s)]ds=(Fv)(t).\end{array}$$

Besides, for $u\ge v$, we have

$$\begin{array}{rl}d(Fu,Fv)& =\underset{t\in I}{sup}|(Fu)(t)-(Fv)(t)|\\ =\underset{t\in I}{sup}((Fu)(t)-(Fv)(t))\\ =\underset{t\in I}{sup}{\int }_0^{T}G(t,s)[f(s,u(s))+\lambda u(s)-f(s,v(s))-\lambda v(s)]ds\\ \le \underset{t\in I}{sup}{\int }_0^{T}G(t,s)\alpha \sqrt{g(u(s)-v(s))}ds.\end{array}$$

(14)

Using the Cauchy-Schwarz inequality in the last integral we get

$$\begin{array}{r}{\int }_0^{T}G(t,s)\alpha \sqrt{g(u(s)-v(s))}ds\\ \le {\left({\int }_0^{T}G{(t,s)}^{2}ds\right)}^{\frac{1}{2}}{\left({\int }_0^T{\alpha }^{2}g(u(s)-v(s))ds\right)}^{\frac{1}{2}}.\end{array}$$

(15)

The first integral gives us

$$\begin{array}{rl}{\int }_0^{T}G{(t,s)}^{2}ds& ={\int }_0^{t}G{(t,s)}^{2}ds+{\int }_t^{T}G{(t,s)}^{2}ds\\ ={\int }_0^t\frac{e^{2\lambda (T+s-t)}}{{(e^{\lambda T}-1)}^2}ds+{\int }_t^T\frac{e^{2\lambda (s-t)}}{{(e^{\lambda T}-1)}^2}ds\\ =\frac{1}{2\lambda {(e^{\lambda T}-1)}^2}{e}^{(2\lambda T-1)}\\ =\frac{e^{\lambda T}+1}{2\lambda (e^{\lambda T}-1)}.\end{array}$$

(16)

The second integral in (15) gives the following estimate:

$$\begin{array}{rl}{\int }_0^T{\alpha }^{2}g(u(s)-v(s))ds& \le {\alpha }^{2}g(\parallel u-v\parallel )\cdot T\\ ={\alpha }^{2}g(d(u,v))\cdot T.\end{array}$$

(17)

Taking into account (14)-(17) we have

$$\begin{array}{rl}d(Fu,Fv)& \le \underset{t\in I}{sup}{\left(\frac{e^{\lambda T}+1}{2\lambda (e^{\lambda T}-1)}\right)}^{\frac{1}{2}}\cdot {({\alpha }^{2}g(d(u,v))\cdot T)}^{\frac{1}{2}}\\ ={\left(\frac{e^{\lambda T}+1}{2\lambda (e^{\lambda T}-1)}\right)}^{\frac{1}{2}}\cdot \alpha \cdot \sqrt{T}\cdot {\left(g(d(u,v))\right)}^{\frac{1}{2}}\end{array}$$

and from the last inequality we obtain

$$d{(Fu,Fv)}^2\le \frac{e^{\lambda T}+1}{2\lambda (e^{\lambda T}-1)}\cdot {\alpha }^2\cdot T\cdot g(d(u,v))$$

or, equivalently.

$$2\lambda (e^{\lambda T}-1)d{(Fu,Fv)}^2\le (e^{\lambda T}+1)\cdot {\alpha }^2\cdot T\cdot g(d(u,v)).$$

By our assumption, as

$$\alpha \le {\left(\frac{2\lambda (e^{\lambda T}-1)}{T(e^{\lambda T}+1)}\right)}^{\frac{1}{2}},$$

the last inequality gives us

$$2\lambda (e^{\lambda T}-1)d{(Fu,Fv)}^2\le 2\lambda (e^{\lambda T}-1)\cdot g(d(u,v)),$$

and, hence,

$$d{(Fu,Fv)}^2\le g(d(u,v)).$$

(18)

Put $\psi (t)=t^2$ and $\varphi (t)=g(t)$. Obviously, ψ is a generalized altering distance function, $\psi (t)$ and $\varphi (t)$ satisfy the condition of $\psi (t)>\varphi (t)$ for $t>0$. From (18), we obtain for $u\ge v$

$$\psi (d(Fu,Fv))\le \varphi (d(u,v)).$$

Finally, let $\alpha (t)$ be a lower solution for (13); we claim that $\alpha \le F(\alpha )$. In fact

$${\alpha }^{\prime }(t)+\lambda \alpha (t)\le f(t,\alpha (t))+\lambda \alpha (t),\text{for }t\in I.$$

We multiply by $e^{\lambda t}$,

$${(\alpha (t)e^{\lambda t})}^{\prime }\le [f(t,\alpha (t))+\lambda \alpha (t)]e^{\lambda t},\text{for }t\in I,$$

and this gives us

$$\alpha (t)e^{\lambda t}\le \alpha (0)+{\int }_0^t[f(s,\alpha (s))+\lambda \alpha (s)]e^{\lambda s}ds,\text{for }t\in I.$$

(19)

As $\alpha (0)\le \alpha (T)$, the last inequality gives us

$$\alpha (0)e^{\lambda t}\le \alpha (T)e^{\lambda T}\le \alpha (0)+{\int }_0^T[f(s,\alpha (s))+\lambda \alpha (s)]e^{\lambda s}ds,$$

and so

$$\alpha (0)\le {\int }_0^T\frac{e^{\lambda s}}{e^{\lambda T}-1}[f(s,\alpha (s))+\lambda \alpha (s)]ds.$$

This and (19) give us

$$\alpha (t)e^{\lambda t}\le {\int }_0^t\frac{e^{\lambda (T+s)}}{e^{\lambda T}-1}[f(s,\alpha (s))+\lambda \alpha (s)]ds+{\int }_t^T\frac{e^{\lambda s}}{e^{\lambda T}-1}[f(s,\alpha (s))+\lambda \alpha (s)]ds$$

and, consequently,

$$\begin{array}{rl}\alpha (t)& \le {\int }_0^t\frac{e^{\lambda (T+s-t)}}{e^{\lambda T}-1}ds+{\int }_0^t\frac{e^{\lambda (s-t)}}{e^{\lambda T}-1}[f(s,\alpha (s))+\lambda \alpha (s)]ds\\ ={\int }_0^{T}G(t,s)[f(s,\alpha (s))+\lambda \alpha (s)]ds\\ =(F\alpha )(t),\text{for }t\in I.\end{array}$$

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) $g_1(t)=ln(t^2+1)$;

• (2)
  $$g_2(t)=\{\begin{array}{cc}{t}^3,\hfill & 0\le t<1,\hfill \\ \frac{1}{2},\hfill & t=1,\hfill \\ t,\hfill & 1<t<+\mathrm{\infty }.\hfill \end{array}$$
• (3)
  $$g_3(t)=\{\begin{array}{cc}{t}^3,\hfill & 0\le t\le \frac{1}{2},\hfill \\ t-\frac{3}{8},\hfill & \frac{1}{2}<t<+\mathrm{\infty }.\hfill \end{array}$$

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:

$$\{\begin{array}{cc}{u}^{\prime }(t)=\frac{sint}{e^t}-\beta x,\hfill & t\in [0,T],\hfill \\ u(0)=u(T).\hfill \end{array}$$

(20)

Let

$$f(t,x)=\frac{sint}{e^t}-\beta x,x\in [0,\mathrm{\infty }),t\in [0,1],$$

then $f(t,x)$ is continuous. Further, for $x\ge y$, we have

$$f(t,x)+\lambda x-[f(t,y)+\lambda y]=2(\lambda -\beta )\frac{x-y}{2}.$$

We chose $\beta \in [0,\lambda ]$ such that

$$2(\lambda -\beta )\le {\left(\frac{2\lambda (e^{\lambda T}-1)}{T(e^{\lambda T}+1)}\right)}^{\frac{1}{2}}.$$

Taking $g(t)={(\frac{t}{2})}^2$ for all $t\in [0,+\mathrm{\infty })$, we have

$$f(t,x)+\lambda x-[f(t,y)+\lambda y]=2(\lambda -\beta )\sqrt{g(x-y)}.$$

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:

$$\{\begin{array}{cc}-\frac{d^{2}x}{d{t}^2}=f(t,x),\hfill & x\in [0,\mathrm{\infty }),t\in [0,1],\hfill \\ x(0)=x(1)=0.\hfill \end{array}$$

(21)

It is well known that $x\in C^2[0,1]$ is a solution of (20) that is equivalent to $x\in C[0,1]$ being a solution of the integral equation

$$x(t)={\int }_0^{1}G(t,s)f(s,x(s))ds,\text{for }t\in [0,1],$$

where $G(t,s)$ is the Green function given by

$$G(t,s)=\{\begin{array}{cc}t(1-s),\hfill & 0\le t\le s\le 1,\hfill \\ s(1-t),\hfill & 0\le s\le t\le 1.\hfill \end{array}$$

(22)

Theorem 3.5 Consider problem (21) with $f:I\times R\to [0,\mathrm{\infty })$ 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 $x\ge y$

$$f(t,x)-f(t,y)\le \alpha \sqrt{g(x-y)},$$

(23)

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

Proof Consider the cone

$$P=\{x\in C[0,1]:x(t)\ge 0\}.$$

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

$$(Tx)(t)={\int }_0^{1}G(t,s)f(s,x(s))ds,\text{for }x\in P,$$

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

As f is non-decreasing with respect to the second variable, for $x,y\in P$ with $y\ge x$ and $t\in [0,1]$, we have

$$(Ty)(t)={\int }_0^{1}G(t,s)f(s,y(s))ds\ge {\int }_0^{1}G(t,s)f(s,x(s))ds\ge (Tx)(t),$$

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

Besides, for $y\ge x$ and taking into account (23), we obtain

$$\begin{array}{rl}d(Ty,Tx)& =\underset{t\in [0,1]}{sup}|(Tx)(t)-(Ty)(t)|\\ =\underset{t\in [0,1]}{sup}((Tx)(t)-(Ty)(t))\\ =\underset{t\in [0,1]}{sup}{\int }_0^{1}G(t,s)(f(s,x(s))-f(s,y(s)))ds\\ \le \underset{t\in [0,1]}{sup}{\int }_0^{1}G(t,s)\alpha \sqrt{g(x(s)-y(s))}\\ \le \underset{t\in [0,1]}{sup}{\int }_0^{1}G(t,s)\alpha \sqrt{g(x(s)-y(s))}ds\\ =\alpha \sqrt{ln[{\parallel y-x\parallel }^2+1]}\underset{t\in [0,1]}{sup}{\int }_0^{1}G(t,s)ds.\end{array}$$

(24)

It is easy to verify that

$${\int }_0^{1}G(t,s)ds=\frac{-t^2}{2}+\frac{t}{2}$$

and that

$$\underset{t\in [0,1]}{sup}{\int }_0^{1}G(t,s)ds=\frac{1}{8}.$$

These facts, the inequality (24), and the hypothesis $0<\alpha \le 8$ give us

$$\begin{array}{rl}d(Tx,Ty)& \le \frac{\alpha }{8}\sqrt{g(x-y)}\\ \le \sqrt{g(\parallel x-y\parallel )}=\sqrt{g(d(x,y))}.\end{array}$$

Hence

$$d{(Ty,Tx)}^2\le g(d(x,y)).$$

Put $\psi (t)=t^2$, $\varphi (t)=g(t)$, obviously ψ is an altering distance function, ψ and ϕ satisfy the condition of $\psi (t)>\varphi (t)$, for $t>0$. From the last inequality, we have

$$\psi (d(Tx,Ty))\le \varphi (d(x,y)).$$

Finally, as f and G are non-negative functions

$$T0={\int }_0^{1}G(t,s)f(s,0)ds\ge 0$$

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 $g(t)$ 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:

$$\{\begin{array}{cc}-\frac{d^{2}x}{d{t}^2}=\frac{sint}{e^t}+\frac{x}{1+cost\pi },\hfill & x\in [0,\mathrm{\infty }),t\in [0,1],\hfill \\ x(0)=x(1)=0.\hfill \end{array}$$

(25)

Let

$$f(t,x)=\frac{sint}{e^t}+\frac{x}{1+cost\pi },x\in [0,\mathrm{\infty }),t\in [0,1],$$

then $f(t,x)$ is continuous and non-decreasing with respect to the second variable. Further, for $x\ge y$, we have

$$f(t,x)-f(t,y)=\frac{x}{1+cost\pi }-\frac{y}{1+cost\pi }\le \sqrt{{\left(\frac{x-y}{2}\right)}^2}.$$

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