Afrika Matematika

, Volume 25, Issue 3, pp 745–756

Weak contractions on chains in a generalized metric space with a partial order

Authors

  • Binayak S. Choudhury
    • Department of MathematicsBengal Engineering and Science University
    • Department of MathematicsBengal Engineering and Science University
Article

DOI: 10.1007/s13370-013-0146-6

Cite this article as:
Choudhury, B.S. & Maity, P. Afr. Mat. (2014) 25: 745. doi:10.1007/s13370-013-0146-6

Abstract

Weak contraction mapping principle is a generalization of the Banach contraction mapping principle. Weakly contractive mappings are intermediate to contraction mappings and nonexpansive mappings. They have been studied in several contexts. Metric fixed point theory in partially ordered spaces have rapidly developed in recent times. In this paper we extend the concept of weak contraction to subset of a partially ordered generalized metric space which are chains by themselves. It is noted that this weak contraction is different from weak contraction on the whole space. We prove here that under certain assumptions the weakly contractive mapping on certain chains will have a fixed point. Two illustrative examples are given.

Keywords

\(G\)-metric spacePartially ordered setWeak contractionFixed pointOrbitMonotone property

Mathematics Subject Classification (2000)

54H25

1 Introduction

\(G\)-metric spaces were introduced by Mustafa and Sims [1, 2]. This is a generalization of metric spaces in which every triplet of elements is assigned to a non-negative real number. Analysis of the structure of this space was done in some detail in [2]. Fixed point theory in this space was initiated in [3]. Particularly, Banach contraction mapping principle was established in this work. After that several fixed point results were proved in this space. Some of these works are noted in ([414]). Several other studies relevant to metric spaces are being extended to \(G\)-metric spaces as, for instances, a best approximation result in these spaces has been established in [15] and the concept of \(\omega \)-distance, which is relevant to minimization problems in metric spaces [16], has been extended to G-metric spaces [10].

In this paper we establish the weak contraction principle in \(G\)-metric spaces. A weak contraction is a generalization of Banach contraction. Banach contraction principle has been generalized by various authors. Over the years, and presently also, it remains an active area of research. Some of the very recent examples from this line of research are noted in [1720]. In particular, weak contraction principle was introduced in Hilbert spaces by [21] and was extended to metric spaces by [22].

The weak contraction mappings are weaker than the contraction mappings but stronger than the nonexpansive mappings. The weak contraction principle states that every weak contraction on a complete metric space necessarily has a unique fixed point. There are several works on weak contractions and weakly contractive type mappings, some of these are noted in [2329].

Fixed point theory in partially ordered metric spaces has rapidly developed in recent times. Some instances of these works are noted in [3035]. One of the reasons of the widespread interest in these problems is that they utilize both analytic and order theoretic aspects of fixed point theory. G-metric spaces with a partial order has been treated in [10].

In this work we have defined the weak contraction on certain subsets of a G-metric spaces with a partial order which are chains by themselves. We have shown that whenever a mapping is a weak contraction on every chain containing a specific orbit, it will have a fixed point. We have given two examples to illustrate our ideas. With the help of one of these examples it is shown that that the weak contraction mentioned above is different from that defined on the whole space.

2 Mathematical preliminaries

Definition 2.1

(\(G\)-metric space [2]) Let \(X\) be a nonempty set and let \(G:X \times X \times X \longrightarrow R^{+}\) be a function satisfying the following properties:
  1. (G1)

    \(G(x, y, z)=0\) if \(x=y=z\);

     
  2. (G2)

    \(0<G(x,x,y)\); for all \(x,y\in X\) with \(x\ne y\);

     
  3. (G3)

    \(G(x,x,y)\le G(x,y,z),\) for all \(x,y,z\in X\) with \(z\ne y\);

     
  4. (G4)

    \(G(x,y,z)=G(x,z,y)=G(y,z,x)=......\),(symmetry in all three variables);

     
  5. (G5)

    \(G(x,y,z)\le G(x,a,a)+G(a,y,z)\), for all \(x,y,z,a\in X\) (rectangle inequality).

     
Then the function \(G\) is called a \(G\)-metric on X and the pair \((X,\,G)\) is called a \(G\)-metric space.

Definition 2.2

[2] Let \((X,\,G)\) be a \(G\)-metric space and let \(\{x_{n}\}\) be a sequence of points of \(X\), a point \(x\in X\) is said to be the limit of the sequence \(\{x_{n}\}\) if \(\underset{n,m\rightarrow \infty }{\lim } G(x,x_{n},x_{m})=0\) and one says that the sequence \(\{x_{n}\}\) is \(G\)-convergent to x.

Thus, if \(x_{n}\rightarrow x\) in a \(G\)-metric space \((X,\,G)\), then for any \(\epsilon >0\), there exists a positive integer \(N\) such that \(G(x,x_{n},x_{m})<\epsilon \), for all \(n,m\ge N\).

It has been shown in [2] that the \(G\)-metric induces a Housdorff topology and the convergence described in the above definition is relative to this topology. The topology being Housdorff, a sequence can converge at most to one point.

Definition 2.3

[2] Let \((X,\,G)\) be a \(G\)-metric space, a sequence \( \{x_{n}\}\) is called \(G\)-cauchy if for every \(\epsilon >0\), there is a positive integer \(N\) such that \(G(x_{n},x_{m},x_{l})<\epsilon \), for all \(n,m\ge N\), that is, if \(G(x_{n},x_{m},x_{l})\rightarrow 0\) as \(n,m\rightarrow \infty \).

We next state the following lemmas.

Lemma 2.4

[2] Let \((X,\,G)\) be a \(G\)-metric space, then the following are equivalent:
  1. (1)

    \(\{x_{n}\}\) is \(G\)-convergent to x.

     
  2. (2)

    \(G(x_{n},x_{n},x)\rightarrow 0\), as \(n\rightarrow \infty .\)

     
  3. (3)

    \(G(x_{n},x,x)\rightarrow 0,\) as \(n\rightarrow \infty .\)

     
  4. (4)

    \(G(x_{m},x_{n},x)\rightarrow 0,\) as \(m,n\rightarrow \infty .\)

     

Lemma 2.5

[1] If \((X,\,G)\) is a \(G\)-metric space, then the following are equivalent:
  1. (1)

    The sequence \(\{x_{n}\}\) is \(G\)-cauchy.

     
  2. (2)

    For every \(\epsilon >0\), there exists a positive integer N such that \(G(x_{n},x_{m},x_{m})<\epsilon \), for all \(n,m\ge N.\)

     

Lemma 2.6

[2] If \((X, G)\) is a \(G\)-metric space then \(G(x,y,y)\le 2G(y,x,x)\) for all \(x,y\in X\).

Lemma 2.7

If \((X,G)\) is a \(G\)-metric space then \(\{x_{n}\}\) is a \(G\)-cauchy sequence if and only if for every \(\epsilon >0\), there exists a positive integer N such that \(G(x_{n},x_{m},x_{m})<\epsilon \), for all \(m>n\ge N.\)

Lemma 2.8

[1] For every \(G\)-metric space \((X, G)\) we can define a metric space \((X, d_{G})\) by \(d_{G}(x,y) = G(x,y,y)+ G(y,x,x)\) ,for all \(x, y\in X \).

Definition 2.9

[2] A \(G\)-metric space \((X,\,G)\) is called symmetric \(G\)-metric space if \(G(x,y,y)=G(y,x,x)\) for all \(x,y\in X\).

Definition 2.10

[2] A \(G\)-metric space \((X,\,G)\) is said to be \(G\)-complete ( or complete \(G\)-metric space) if every \(G\)-cauchy sequence in \((X,\,G)\) is convergent in X.

The following is the Banach contraction principle in the \(G\)-metric space proved by [3].

Theorem 2.11

[3] Let \((X, G)\) be a complete \(G\)-metric space and let \(T:X\rightarrow X\) be a mapping satisfying the following condition:

\(G(T(x),T(y),T(z))\le aG(x,y,z)\) for all \(x,y,z\in X\) where \(0\le a<1\), then \(T\) has a unique fixed point.

In fact it is a special case of the more general result proved by the above mentioned authors (Theorem 2.1 of [3]).

Definition 2.12

[36] Let \((X,d)\) be a metric space and \(T:X\rightarrow X\) be a given function. Given any \(x_{0}\in X\), the orbit of \(x_{0}\) is denoted by \(O(T(x_{0}))\) and defined by the sequence
$$\begin{aligned} \{ x_{0}, x_{1}=T(x_{0}), x_{2}=T^{2}(x_{0}),...............\}, \mathrm{that \ is }, \end{aligned}$$
the sequence \(\{x_{n}\}\) where \(x_{n}=T^{n}(x_{0}) = Tx_{n-1}\).

Definition 2.13

Let \((X,\preceq )\) be a partially ordered set and \(T:X \rightarrow X\) be a mapping. The mapping \(T\) is said to have monotone property if for any \(x_{1},x_{2}\in X,\)\(x_{1}\preceq x_{2}\) implies \(Tx_{1}\preceq Tx_{2}\).

Definition 2.14

A subset \(S\) of a partially ordered set \((X, \preceq )\) is a chain if \( x\preceq y\) or \(y\preceq x\) whenever \(x, y\) in \(S\).

Next we define weak contraction on a chain in a partially ordered \(G\)-metric space.

Definition 2.15

(Weak contraction on a chain) A map \(T:C \rightarrow C\), where \(C\) is a subset of a partially ordered \(G\)-metric space \((X, G)\) such that \(C\) is a chain, is called weakly contractive on the chain \(C\) if for all \(x,y,z\in C\) with \(x\preceq y\preceq z\) and \(z\ne y\),
$$\begin{aligned} G(Tx,Ty,Tz)\le G(x,y,z)-\Psi (G(x,y,z)) \end{aligned}$$
(2.1)
where \(\Psi :[0,\infty )\rightarrow [0,\infty )\) is a continuous and nondecreasing function such that \(\Psi (t)=0\) if and only if t = 0.

3 Main results

Theorem 3.1

Let \((X,\preceq )\) be a partially ordered set and \(G\) be a \(G\)-metric on \(X\) which has the property that whenever a monotone increasing sequence \(\{x_{n}\}\) converges to a point \(p\), it will follow that \(x_{n}\preceq p\). Let \(T:X\rightarrow X\) be a mapping with the monotone property on \(X\). Let \(x_{0}\in X\) be such that \(x_{0}\preceq Tx_{0}\), then the orbit \(O(x_{0})\) is a chain. If \(T\) is weakly contractive on every chain \(C \supseteq O(x_{0})\), then \(T\) has a fixed point.

Proof

Let \(x_{0}\in X\) and \(\{x_{n}\}\) be the sequence defined by \(x_{n+1}=T{x_{n}},n\ge 1\).

Since \(T\) has monotone property on \(X\), we have that
$$\begin{aligned} x_{0}\preceq x_{1}\preceq x_{2}\preceq .....\preceq x_{n}\preceq x_{n+1}........ \end{aligned}$$
(3.1)
If \(x_{n+1}=x_{n}\), then \(T\) has a fixed point. So we assume
$$\begin{aligned} x_{n+1}\ne x_{n}\quad \text{ for } \text{ all } n\ge 0. \end{aligned}$$
(3.2)
Putting \(x=x_{n}\), \(y=x_{n}\) and \(z=x_{n+1}\) in (2.1) we have,
$$\begin{aligned} G(x_{n+1},x_{n+1},x_{n+2})&= G(T{x_{n}},T{x_{n}},T{x_{n+1}})\nonumber \\&\le G(x_{n},x_{n},x_{n+1})-\Psi (G(x_{n},x_{n},x_{n+1}))\,\text{(by } (2.1), (3.1) \text{ and } (3.2)) \nonumber \\&\le G(x_{n},x_{n},x_{n+1}) \end{aligned}$$
(3.3)
Then the sequence \(\{G(x_{n},x_{n},x_{n+1})\}\) is a monotone decreasing sequence of positive numbers and hence possesses a limit \(\rho ^{*}\ge 0\).

Let \(\rho _{n} = G(x_{n},x_{n},x_{n+1}).\)

Suppose \(\rho ^{*}>0\). Since \(\Psi \) is nondecreasing, \(\Psi (\rho _{n})\ge \Psi (\rho ^{*})>0.\) Hence, from (3.3) we have, \(\rho _{n+1}\le \rho _{n}-\Psi (\rho ^{*})\), for all \(n\ge 0\).

Adding over \(n = m, m+1,....N\), we have \(\rho _{N+m}\le \rho _{m}-N\Psi (\rho ^{*})\).

If \(\rho ^{*} \ne 0\), then \(\Psi (\rho ^{*}) \ne 0\), which implies that \(N\Psi (\rho ^{*})\) increases infinitely \(N\) tends to infinity. Also \(\{\rho _{n}\}\) is bounded. Then for large \(N\), the above inequality gives a contradiction. Therefore \(\rho ^{*} = 0\).

Thus we have that
$$\begin{aligned} \underset{n\rightarrow \infty }{\lim } G(x_{n},x_{n},x_{n+1})=0. \end{aligned}$$
(3.4)
Next we prove that \(\{x_{n}\}\) is a Cauchy sequence.

For given \(\epsilon >0\) and a positive integer \(N\), let \(B(N,\epsilon )=\{x:G(x_{N},x_{N},x)<\epsilon \) and \(x_{N}\preceq x \}\).

We next show that for sufficiently large N, \(T\) is a mapping from \(B(N,\epsilon )\) to \(B(N,\epsilon )\).

By virtue of (3.4), there exists \(N\) such that
$$\begin{aligned} G(x_{N},x_{N},x_{N+1})<min\left\{ \frac{\epsilon }{2} ,\Psi (\frac{\epsilon }{2})\right\} . \end{aligned}$$
(3.5)
Let \(x\in B(N,\epsilon )\).

If \(x= x_{N}\), then \(G(x_{N},x_{N},Tx)= G(x_{N},x_{N},x_{N+1})<min\{\frac{\epsilon }{2} ,\Psi (\frac{\epsilon }{2})\}< \epsilon .\)

Hence \(Tx\in B(N,\epsilon )\). So we assume that \(x\ne x_{N}\).

Then we have the following two cases.

Case-I\(G(x_{N},x_{N},x)\le \frac{\epsilon }{2}\).

Then
$$\begin{aligned} G(x_{N},x_{N},Tx)&= G(Tx,x_{N},x_{N})\\&\le G(Tx,Tx_{N},Tx_{N})+G(Tx_{N},x_{N},x_{N})\\&= G(Tx_{N},Tx_{N},Tx)+G(x_{N},x_{N},Tx_{N})\\&\le G(x_{N},x_{N},x)-\Psi (G(x_{N},x_{N},x))+G(x_{N},x_{N},x_{N+1})\\&\qquad \qquad \qquad \qquad \qquad \qquad \text{(by } (2.1), \text{ since } x_{N}\prec x)\\&<\frac{\epsilon }{2} +\frac{\epsilon }{2} \text{(by } (3.5))\\&<\epsilon . \end{aligned}$$
Case-II\(\frac{\epsilon }{2}<G(x_{N},x_{N},x)<\epsilon \).
Then
$$\begin{aligned} G(x_{N},x_{N},Tx)&= G(Tx,x_{N},x_{N})\\&\le G(Tx,Tx_{N},Tx_{N})+G(Tx_{N},x_{N},x_{N})\\&= G(Tx_{N},Tx_{N},Tx)+G(x_{N},x_{N},Tx_{N})\\&\le G(x_{N},x_{N},x)-\Psi (G(x_{N},x_{N},x))+G(x_{N},x_{N},x_{N+1})\\&\qquad \qquad \qquad \qquad \qquad \text{(by } \text{(2.1), } \text{ since } x_{N}\prec x \text{) }\\&\le G(x,x_{N},x_{N})- \Psi (\frac{\epsilon }{2})+ \Psi (\frac{\epsilon }{2})\\&\quad (\text{ by } (3.5)) \text{ and } \text{ the } \text{ fact } \text{ that } \Psi \text{ is } \text{ nondecreasing, } \text{ by } \text{ our } \text{ assumption } \text{ that } \\&\quad \frac{\epsilon }{2}<G(x_{N},x_{N},x) )\\&< \epsilon . \end{aligned}$$
Again, \(T\) is monotone increasing and \(x_{N}\preceq x\) for all \(x\in B(N,\epsilon )\).

Therefore, \(Tx_{N}= x_{N+1}\preceq Tx\) for all \(x\in B(N,\epsilon )\).

Then from (3.1), \(x_{N}\preceq x_{N+1}\preceq Tx\).

The above argument thus implies that, \(Tx\in B(N,\epsilon )\), that is, \(T\) is a selfmap of \(B(N,\epsilon )\) for sufficiently large \(N\).

Let \(m>n>N\). Since \(x_{N}\in B(N,\epsilon )\), \( \,\,Tx_{N}= x_{N+1}\in B(N,\epsilon )\Rightarrow Tx_{N+1}=x_{N+2}\in B(N,\epsilon )........x_{N+(n-N)}=x_{n}\in B(N,\epsilon ).......x_{N+(m-N)}=x_{m}\in B(N,\epsilon ).\) So,
$$\begin{aligned} G(x_{n},x_{N},x_{N})<\epsilon \text{ and } G(x_{m},x_{N},x_{N})<\epsilon . \end{aligned}$$
(3.6)
Therefore, for all \( m>n>N \), we have
$$\begin{aligned} G(x_{n},x_{m},x_{m})&\le G(x_{n},x_{N},x_{N})+G(x_{N},x_{m},x_{m})\\&\le G(x_{n},x_{N},x_{N})+ 2G(x_{m},x_{N},x_{N}), \text{( } \text{ by } \text{ Lemma } \text{2.6) }\\&< \epsilon + 2\epsilon ( \text{ by } (3.6))\\&= 3\epsilon . \end{aligned}$$
We conclude, by virtue of Lemma 2.7, that \(\{x_{n}\}\) is a Cauchy sequence and hence is convergent in \(X\).

Let \(\underset{n\rightarrow \infty }{\lim }x_{n}=p\).

Next we show that \(p\) is a fixed point of \(T\).

By our construction, \(\{x_{n}\}\) is a monotone increasing sequence and hence, by our assumption, \(x_{n}\preceq p\) for all \(n\ge 0\). If \(p= x_{m}\) for some \(m\), then it follows that \(p= x_{n}\) for all \(n\ge m\). In particular, we have \(p= x_{m+1}= Tx_{m}= Tp\) , that is, \(p\) is a fixed point of \(T\).

So we assume that \(p\ne x_{n}\) for all \(n\).

Also it follows that \(\{x_{n}\}\cup \{p\}\) is a chain and therefore, by a condition of the theorem, \(T\) is a weak contraction on this chain. Putting \(x=x_{n},y=x_{n}\) and \(z=p\) in (2.1), we have,
$$\begin{aligned} G(x_{n+1},x_{n+1},T{p})=G(T{x_{n}},T{x_{n}},T{p})\le G(x_{n},x_{n},p)-\Psi (G(x_{n},x_{n},p)). \end{aligned}$$
Taking \(n\rightarrow \infty \) in the above inequality, we have, by virtue of the facts that \(G\) and \(\Psi \) are continuous, \(G(p,p,T{p})\le 0\), which implies that \(T{p}=p\).

This completes the proof of the theorem.

Example 3.1

Let \(X = \{0,1,2,3....\}\) and \(G:X\times X\times X\rightarrow R^{+}\) be defined as follows.
$$\begin{aligned} G(x,y,z)=\left\{ \begin{array}{ll} x+y+z,&{}\quad \text{ if }\ x,\,y ,\, z \, \text{ are } \text{ all } \text{ distinct } \text{ and } \text{ different } \text{ from } \text{ zero, } \\ x+z,&{}\quad \text{ if }\ x=y\ne z \ \text{ and } \text{ all } \text{ are } \text{ different } \text{ from } \text{ zero, } \\ y+z+1,&{}\quad \,{if}\ x=0, y\ne z \ \text{ and } \, y,z \,\text{ are } \text{ different } \text{ from } \text{ zero, } \\ y+2,&{}\quad \,if\quad x=0,y=z\ne 0, \\ z+1,&{}\quad \ if\quad x=0,y=0,z\ne 0,\\ 0 ,&{}\quad \text{ if }\quad x=y=z.\\ \end{array} \right. \end{aligned}$$
Then \((X,\,G)\) is a complete non-symmetric \(G\)-metric space [8].

Let a partial order \(^{\prime }\preceq ^{\prime }\) on \(X\) be defined as follows:

\(y \preceq z\) whenever \(y\ge z\), for all \(y,z \in X\) and \((y - z)\) is divisible by 2.

Let \(\Psi :[0,\infty )\rightarrow [0,\infty )\) be defined as
$$\begin{aligned} \Psi (t)=\left\{ \begin{array}{ll} 2t^{2},&{}\quad \text{ if }\quad 0<t\le 1,\\ 2,&{}\quad \text{ if }\quad t>1.\\ \end{array} \right. \end{aligned}$$
Let \(T:X\rightarrow X\) be defined as
$$\begin{aligned} Tx=\left\{ \begin{array}{ll} x-2,&{}\quad \text{ if }\quad x\ge 2,\\ 0,&{}\quad \text{ if }\quad x=1,\\ 0,&{}\quad \text{ if }\quad x=0.\\ \end{array} \right. \end{aligned}$$
Let \(x_0 = 2k (k> 0)\). Then \(Tx_0 = 2(k - 1)\) and we have \(x_0 \preceq Tx_0\). The orbit of \(x_0\) is \(O(x_0 = 2k) = \{2k, 2k - 2, ........, 2, 0 \}\). Thus any chain containing \(O(x_0)\) is a subset of the set of all non-negative even integers including zero. Let \(x =2k_1\), \(y = 2k_2\) and \(z = 2k_3\). Then \(x\preceq y\preceq z\) with \(y \ne z\) means that \(k_1 \ge k_2 > k_3\).

It then easily follows that the inequality (2.1) satisfied. Thus all the conditions of the Theorem 3.1 are satisfied. Hence, by Theorem 3.1, a fixed point of \(T\) exists.

Here \(z = 0\) is a fixed point of \(T\).

Remark 3.1

It is noted that in the above example the inequality (2.1) is satisfied on the chain \(0\preceq 2 \preceq 4\preceq ......\), but is not satisfied on the chain \(0 \preceq 1 \preceq 3 \preceq .....\).. This can be seen by assuming \(x=3=y\), \(z=1\), which leads to a violation of (2.1). Hence \(T\) is not weakly contractive over the whole space.

Example 3.2

Let \(X = \{0,2,4\}\) and \(G:X\times X\times X\rightarrow R^{+}\) be defined as follows:

\(G(0,2,2)=3.5\), \(G(0,0,2)=4.5\), \(G(0,4,4)=7.5\), \(G(0,0,4)=7.5\), \(G(0,2,4)=7.5\) and \(G(x,y,z)= \mid x-y \mid + \mid y-z \mid + \mid z-x \mid \), for the rest.

Then \(G\) is a non symmetric \(G\)-metric space.

Let a partial order \(^{\prime }\preceq ^{\prime }\) on \(X\) be defined as follows: \(4 \preceq 2 \preceq 0\).

Let \(\Psi :[0,\infty )\rightarrow [0,\infty )\) be defined as
$$\begin{aligned} \Psi (t)=\left\{ \begin{array}{ll} 0.1t^{2},&{}\quad \text{ if }\quad 0<t\le 1,\\ 0.1,&{}\quad \text{ if }\quad t>1.\\ \end{array} \right. \end{aligned}$$
Let \(T:X\rightarrow X\) be defined as
$$\begin{aligned} Tx=\left\{ \begin{array}{ll} 2,\quad \text{ if }\quad x=4, \\ 0,\quad \text{ if }\quad x=2,\\ 0,\quad \text{ if }\quad x=0.\\ \end{array} \right. \end{aligned}$$
Then all the conditions of Theorem 3.1 are satisfied.

Remark 3.2

The condition \(x\preceq y\preceq z\) with \(y\ne z\) is essential in the definition of the weak contraction on a chain in a partially ordered \(G\)-metric space in order to make the definition meaningful. Otherwise, the inequality is reducible to a weak contraction inequality in \((X,d_{G})\) where \(d_{G}\) is the metric induced by the \(G\)-metric as described in Lemma 2.8 and the result of the theorem can be obtained by an application of a result in metric space. In Example 3.2, the inequality (2.1) is not valid with \(z=y= 2\), \(x=4\). It is worth mentioning that several fixed point problems in \(G\)-metric spaces are reducible to metric fixed point problems. Our case is outside this category.

4 An application to a boundary value problem

In this section we consider an application of our theorem to a problem of differential equation, namely the following periodic boundary value problem.
$$\begin{aligned}&u^{^{\prime }}(t) = f(t, x(t))\end{aligned}$$
(4.1)
$$\begin{aligned}&u(0) = u(T) \end{aligned}$$
(4.2)
where \(T > 0\) and \(f\) is continuous.

The above problem, with some conditions imposed on \(f\), was considered by [33] and later by [32].

The problem can be restated as
$$\begin{aligned}&x^{^{\prime }}(t) + \lambda x(t) = f(t, x(t)) + \lambda x(t), t\in [0, T]\end{aligned}$$
(4.3)
$$\begin{aligned}&x(0) = x(T), \end{aligned}$$
(4.4)
where \(T > 0\) and \(f\) is continuous.
Further the problem can be restated in the following equivalent integral equation form [33]
$$\begin{aligned} x(t) = \displaystyle \int \limits _{0}^{T} K(t, s)[f(s, x(s)) + \lambda x(s)] \end{aligned}$$
(4.5)
where the Green’s function \(K(t, s)\) is,
$$\begin{aligned} K(t, s)=\left\{ \begin{array}{ll} \frac{e^ {\lambda (T + s -t )}}{e^ {\lambda T} - 1},\quad \quad \quad 0\le s < t\le T,\\ \frac{e^{ \lambda (s - t)}}{e^ {\lambda T} - 1},\quad \quad \quad \quad 0\le t < s\le T.\\ \end{array} \right. \end{aligned}$$
(4.6)
We consider the complete \(G\)-metric space \(C[0,T]\), the set of all real continuous functions on \([0,T]\), with the \(G\)-metric
$$\begin{aligned} G(x,y,z)\!=\!\underset{t\in [0,T] }{ Sup } \mid x(t)\!-\!y(t)\mid \!+\! \underset{t\in [0,T] }{ Sup } \mid y(t)\!-\!z(t)\mid \!+\! \underset{t\in [0,T] }{ Sup } \mid z(t)\!-\!x(t)\mid . \end{aligned}$$
(4.7)
We consider the partial order in \(C[0,T]\) defined by \(x\preceq y\) whenever \(x(t)\le y(t)\) for all \(t\in [0,1]\). A lower solution of (4.1)–(4.2) is \(x_{0}\in C^{\prime }[0,1]\) such that \(x^{^{\prime }}_{0}(t)\le f(t, x_{0}(t))\) and \(x_{0}(0)\le x_{0}(T)\) [33].
We consider in the following theorem a continuous and monotonically increasing function \(\phi : [0, \infty )\rightarrow [0, \infty )\) with \(\phi (t)=0\) if and only if \(t=0\) for which the function \(\Phi : [0, \infty )\rightarrow [0, \infty )\) defined as
$$\begin{aligned} \Phi (s)= \underset{{s_1+s_2+s_3=s},~{s_1,~ s_2, ~s_3\ge 0}}{Sup}(\phi (s_1)+\phi (s_2)+\phi (s_3))\end{aligned}$$
(4.8)
$$\begin{aligned} \text{ is } \text{ such } \text{ that } \ {s- \Phi (s)} \text{ is } \text{ monotonic } \text{ increasing } \text{ for } \text{ all } \ {s> 0}. \end{aligned}$$
(4.9)
From its definition (4.8), since the set\(\{(s_1,~ s_2, ~s_3):s_1+s_2+s_3=s,~s_1,~ s_2, ~s_3\ge 0\}\) is compact for fixed \(s\ge 0\), it follows that \(\Phi \) is monotone increasing, continuous and \(\Phi (s)=0\) if and only if \(s=0\).

Particular choices of \(\phi \) satisfying the required conditions are \(\phi (t) = kt, \, 0<k<1\) and \(\phi (t) = log (1 + t)\).

Theorem 4.1

If in the problem described in (4.1)–(4.2) the function \(f\) is such that there exists \(\lambda > 0\) for which
$$\begin{aligned} 0\le (f(t,s_{1})+ \lambda s_{1})- (f(t,s_{2})+ \lambda s_{2})\le \lambda \phi ( s_{1}-s_{2}) \end{aligned}$$
(4.10)
where \(s_{1}\ge s_{2}\), \(\phi \) is a function defined as in the above and there exists a lower solution of (4.1)–(4.2), then there exists a solution of (4.1)–(4.2).

Proof

We consider the equivalent integral equation form of (4.1)–(4.2), that is, (4.5)–(4.6) with \(\lambda >0\) taken to be the same as in (4.10). We construct the function
$$\begin{aligned} Sx(t)= \int \limits ^{T}_{0} K(t,s)(f(s,x(s))+ \lambda x(s))ds \end{aligned}$$
(4.11)
If we can show that \(Sx\) has a fixed point, then we have solved the problem.

Since \(K(t,s)\ge 0\), (4.10) implies that \(Sx(t)\ge Sy(t)\) for all \(t\in [0,1]\) whenever \(x(t)\ge y(t)\) for all \(t\in [0,1]\), that is, \(Sy\preceq Sx\) whenever \(y\preceq x\). This proves that \(S\) is monotone.

For all \(x\preceq y\preceq z\) with \(z\ne y\), we have
$$\begin{aligned} G(Sx,Sy,Sz)&= \underset{t\in [0,T] }{ Sup }\mid \int \limits ^{T}_{0}K(t,s)[(f(s,x(s))\!+\! \lambda x(s))\!-\!(f(s,y(s))\!+\! \lambda y(s))]ds\nonumber \\&+\! \underset{t\in [0,T] }{ Sup }\mid \int \limits ^{T}_{0}K(t,s)[(f(s,y(s))\!+\! \lambda y(s))\!-\!(f(s,z(s))\!+\! \lambda z(s))] ds\nonumber \\&+\underset{t\in [0,T] }{ Sup }\mid \int \limits ^{T}_{0}K(t,s)[(f(s,z(s))\!+\! \lambda z(s))\!-\!(f(s,x(s))\!+\! \lambda x(s))]ds\nonumber \\&\le \underset{t\in [0,T] }{ Sup } \int \limits ^{T}_{0}K(t,s)[\lambda \phi (\mid x(s)\!-\! y(s)\mid )\!+\! \lambda \phi (\mid y(s)\!-\! z(s)\mid )\nonumber \\&+\, \lambda \phi ( \mid z(s)\!-\! x(s)\mid )]ds\nonumber \\&\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad (\text{ by } (4.7))\nonumber \\&\le \lambda ( \underset{s\in [0,T] }{ Sup }( \phi (\mid x(s)\!-\! y(s)\mid )\!+\! \underset{s\in [0,T] }{ Sup }\phi (\mid y(s)\!-\! z(s)\mid )\nonumber \\&+ \underset{s\in [0,T] }{ Sup }\phi ( \mid z(s)\!-\! x(s)\mid ))(\underset{t\in [0,T] }{ Sup } \int \limits ^{T}_{0}K(t,s)ds).\nonumber \\&= \lambda (\phi ( \underset{s\in [0,T] }{ Sup }(\mid x(s)\!-\! y(s)\mid ))\!+\! \phi ( \underset{s\in [0,T] }{ Sup } (\mid y(s)\!-\! z(s)\mid ))\nonumber \\&+\, \phi ( \underset{s\in [0,T] }{ Sup }( \mid z(s)\!-\! x(s)\mid )) (\underset{t\in [0,T] }{ Sup } \int \limits ^{T}_{0}K(t,s)ds). \end{aligned}$$
(4.12)
(Since \(\phi \) is continuous and monotone)
Now,
$$\begin{aligned} \int \limits ^{T}_{0}K(t,s) ds&= \int \limits ^{t}_{0}\frac{e^{\lambda (T+s-t)}}{e^{\lambda }-1}ds+ \int \limits ^{T}_{t}\frac{e^{\lambda (s-t)}}{e^{\lambda }-1}ds\nonumber \\&= \frac{1}{e^{\lambda T} - 1}\left( \frac{1}{\lambda }e^{\lambda (T +s -t)}]_{0}^{t }+ \frac{1}{\lambda }e^{\lambda (s - t)}]_{t}^{T}\right) \nonumber \\&= \frac{1}{\lambda (e^{\lambda T} - 1)}(e^{\lambda T}-1)\nonumber \\&= \frac{1}{\lambda } \end{aligned}$$
(4.13)
Then, from (4.8), (4.12) and (4.13),
$$\begin{aligned} G(Sx,Sy,Sz)&\le \lambda \frac{1}{\lambda } \Phi (\underset{s\in [0,1] }{ Sup }\mid x(s)\!-\! y(s)\mid \!+\! \underset{s\in [0,1] }{ Sup }\mid y(s)\!-\! z(s)\mid \!+\! \underset{s\in [0,1] }{ Sup }\mid z(s)\!-\! x(s)\mid )\\&= G(x,y,z)\!-\!(G(x,y,z)\!-\!\Phi (G(x,y,z)))\\&= G(x,y,z)\!-\!\psi (G(x,y,z)) \end{aligned}$$
where \(\psi (t)= t- \Phi (t)\).

Next we show that the lower solution \(x_0\) satisfies \(x_0\preceq Sx_0\). We follow the argument of [33] in the following.

By (4.9) and the properties of \(\Phi \), we have that \(\psi \) is continuous, monotonically increasing and \(\psi (t) = 0\) if and only if \(t =0\).

Also, from the existence of the lower solution \(x_0(t)\)
$$\begin{aligned} x_0^{\prime }(t)+ \lambda x_0(t)\le f(t,x_0(t))+\lambda x_0(t),~ t\in [0,T]. \end{aligned}$$
Multiplying by \(e^{\lambda t}\), we obtain that
$$\begin{aligned} (x_0(t)e^{\lambda t})^{\prime }\le [f(t,x_0(t))+\lambda x_0(t)]e^{\lambda t},~ t\in [0,T], \end{aligned}$$
that is,
$$\begin{aligned} x_0(t)e^{\lambda t}\le x_0(0)+\int \limits _{0}^{t}[f(s,x_0(s))+\lambda x_0(s)]e^{\lambda s}ds,~ t\in [0,T], \end{aligned}$$
(4.14)
which implies that
$$\begin{aligned} x_0(0)e^{\lambda T}\le x_0(T)e^{\lambda T} \le x_0(0)+\int \limits _{0}^{T}[f(s,x_0(s))+\lambda x_0(s)]e^{\lambda s}ds, \end{aligned}$$
(Since \(x_0(0)\le x_0(T)\))
so that
$$\begin{aligned} x_0(0)\le \int \limits _{0}^{T}\frac{e^{\lambda s}}{e^{\lambda T}-1}[f(s,x_0(s))+\lambda x_0(s)]ds. \end{aligned}$$
From this inequality and (4.14), we obtain
$$\begin{aligned} x_0(t)e^{\lambda t}&\le \int \limits _{0}^{t}\frac{e^{\lambda (T+s)}}{e^{\lambda T}-1}[f(s,x_0(s))+\lambda x_0(s)]ds+ \int \limits _{t}^{T}\frac{e^{\lambda s}}{e^{\lambda T}-1}[f(s,x_0(s))\\&+\lambda x_0(s)]ds,~ t\in [0,T], \end{aligned}$$
and hence
$$\begin{aligned} x_0(t)\le \int \limits _{0}^{T}K(t, s )[f(s,x_0(s))+\lambda x_0(s)]ds=Sx_0(t),~ t\in [0,T]. \end{aligned}$$
Thus we have \(x_0 \preceq S x_0\).

Thus all the conditions of the Theorem 3.1 are satisfied by \(S\). Hence \(S\) must have a fixed point, that is, there exists \(x(t)\) such that \(x(t)= Sx(t)= \int ^{T}_{0}G(t,s)[f(s,x(s))+ \lambda x(s)]\).

This proves that the system of differential Eqs. (4.1)–(4.2) has a solution.

Acknowledgments

The work is supported by the Council of Scientific and Industrial Research, Government of India, under Research Project No - 25(0168)/09/EMR-II. The support is gratefully acknowledged. The authors acknowledge the suggestions of the learned referees.

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© African Mathematical Union and Springer-Verlag Berlin Heidelberg 2013