Existence of a solution of integral equations via fixed point theorem

Abstract

In this paper, we establish a solution to the following integral equation:

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

(1)

where $T>0$, $f:[0,T]\times \mathbb{R}\to \mathbb{R}$ and $G:[0,T]\times [0,T]\to [0,\mathrm{\infty })$ are continuous functions. For this purpose, we also obtain some auxiliary fixed point results which generalize, improve and unify some fixed point theorems in the literature.

MSC:47H10, 54H25.

1 Introduction and preliminaries

Fixed point theory is one of the most efficient tools in nonlinear functional analysis to solve the nonlinear differential and integral equations. The existence/uniqueness of a solution of differential/integral equations turns into the existence/uniqueness of a (common) fixed point of the operators which are obtained after suitable substitutions and elementary calculations; see, e.g., [1–14].

In this paper, we first obtain some fixed point theorems to solve the integral equation mentioned above. For the sake of completeness, we recollect some basic definitions and elementary results. Let X be a nonempty set and T be a self-mapping on X. Then, the set of all fixed points of T on X is denoted by $Fix{(T)}_X$. Let Ψ be the set of all functions $\psi :[0,\mathrm{\infty })\times [0,\mathrm{\infty })\to [0,\mathrm{\infty })$ satisfying the following conditions:

• (1) ψ is continuous,

• (2) $\psi (t_1,t_2)=0$ if and only if $t_1=t_2=0$,

• (3) $\psi (t_1,t_2)\le \frac{1}{2}(t_1+t_2)$.

Cyclic mapping and cyclic contraction were introduced by Kirk-Srinavasan-Veeramani to improve the well-known Banach fixed point theorem. Later, various types of cyclic contraction have been investigated by a number of authors; see, e.g., [6, 15–17].

Definition 1.1 [18]

Suppose that $(X,d)$ is a metric space and T is a self-mapping on X. Let m be a natural number and $X_i$, $i=1,\dots ,m$, be nonempty sets. Then $Y={\bigcup }_{i=1}^m{X}_i$ is called a cyclic representation of X with respect to T if

$$T(X_1)\subset X_2,\dots ,T(X_{m-1})\subset X_m,T(X_m)\subset X_{m+1},$$

where $X_{m+1}=X_1$.

Definition 1.2 [17]

Let $T:X\to X$, $r>0$ and $\eta ,\xi :X\to [0,+\mathrm{\infty })$ be two functions. We say that T is r-$(\eta ,\xi )$-admissible if

• (i) $\eta (x)\ge r$ for some $x\in X$ implies $\eta (Tx)\ge r$,

• (ii) $\xi (x)\le r$ for some $x\in X$ implies $\xi (Tx)\le r$.

Definition 1.3 Let $(X,d)$ be a metric space and $T:Y\to Y$ be a self-mapping, where $Y={\bigcup }_{i=1}^m{X}_i$ is a cyclic representation of Y with respect to T. Let $\eta ,\xi :Y\to [0,+\mathrm{\infty })$ be two functions. An operator $T:Y\to Y$ is called:

• a cyclic weak r-$(\eta ,\xi )$-C-contractive mapping of the first kind if

  $$\eta (x)\eta (y)d(Tx,Ty)\le \xi (x)\xi (y)[\frac{1}{2}[d(x,Ty)+d(y,Tx)]-\psi (d(x,Ty),d(y,Tx))]$$

  (2)

  holds for all $x\in X_i$ and $y\in X_{i+1}$, where $\psi \in \mathrm{\Psi }$.

• a cyclic weak r-$(\eta ,\xi )$-C-contractive mapping of the second kind if

  $${[\eta (x)\eta (y)+r]}^{d(Tx,Ty)}\le {[\xi (x)\xi (y)+r]}^{[\frac{1}{2}[d(x,Ty)+d(y,Tx)]-\psi (d(x,Ty),d(y,Tx))]}$$

  (3)

  such that $r^2+r>1$ holds for all $x\in X_i$ and $y\in X_{i+1}$, where $\psi \in \mathrm{\Psi }$.

2 Auxiliary fixed point results

We state the main result of this section as follows.

Theorem 2.1 Let $(X,d)$ be a complete metric space, $m\in \mathbb{N}$, $X_1,X_2,\dots ,X_m$ be nonempty closed subsets of $(X,d)$ and $Y={\bigcup }_{i=1}^m{X}_i$. Suppose that $T:Y\to Y$ is a cyclic weak r-$(\eta ,\xi )$-C-contractive mapping of the first kind such that

• (i) T is r-$(\eta ,\xi )$-admissible;

• (ii) there exists $x_0\in Y$ such that $\eta (x_0)\ge r$ and $\xi (x_0)\le r$;

• (iii) if $\{x_n\}$ is a sequence in Y such that $\eta (x_n)\ge r$ and $\xi (x_n)\le r$ for all $n\in \mathbb{N}$ and $x_n\to x$ as $n\to \mathrm{\infty }$, then $\eta (x)\ge r$ and $\xi (x)\le r$.

Then T has a fixed point $x\in {\bigcap }_{i=1}^n{X}_i$. Moreover, if $\eta (x)\ge r$, $\eta (y)\ge r$, $\xi (x)\le r$, $\xi (y)\le r$ for all $x,y\in Fix{(T)}_Y$, then T has a unique fixed point.

Proof Let there exist $x_0\in Y$ such that $\eta (x_0)\ge r$ and $\xi (x_0)\le r$. Since T is r-$(\eta ,\xi )$-admissible, then $\eta (T{x}_0)\ge r$ and $\xi (T{x}_0)\le r$. Again, since T is r-$(\eta ,\xi )$-admissible, then $\eta (T^2{x}_0)\ge r$ and $\xi (T^2{x}_0)\le r$. By continuing this process, we get

$$\eta \left(T^n{x}_0\right)\ge r\text{and}\xi \left(T^n{x}_0\right)\le r\text{for all }n\in \mathbb{N}.$$

(4)

On the other hand, since $x_0\in Y$, there exists some $i_0$ such that $x_0\in X_{i_0}$. Now $T(X_{i_0})\subseteq X_{i_0+1}$ implies that $T{x}_0\in X_{i_0+1}$. Thus there exists $x_1$ in $X_{i_0+1}$ such that $T{x}_0=x_1$. Similarly, $T{x}_n=x_{n+1}$, where $x_n\in X_{i_n}$. Hence, for $n\ge 0$, there exists $i_n\in \{1,2,\dots ,m\}$ such that $x_n\in X_{i_n}$ and $x_{n+1}\in X_{i_n+1}$. In case $x_{n_0}=x_{n_0+1}$ for some $n_0=0,1,2,\dots$ , then it is clear that $x_{n_0}$ is a fixed point of T. Now assume that $x_n\ne x_{n+1}$ for all n. Hence, we have $d(x_{n-1},x_n)>0$ for all n. Set $d_n:=d(x_n,x_{n+1})$. We shall show that the sequence $\{d_n\}$ is non-increasing. Due to (2) with $x=x_{n-1}$ and $y=x_n$, we get

$$\begin{array}{r}{r}^{2}d(x_n,x_{n+1})\\ =r^{2}d(T{x}_{n-1},T{x}_n)\le \eta (x_{n-1})\eta (x_n)d(T{x}_{n-1},T{x}_n)\\ \le \xi (x_{n-1})\xi (x_n)[\frac{1}{2}[d(x_{n-1},T{x}_n)+d(x_n,T{x}_{n-1})]-\psi (d(x_{n-1},T{x}_n),d(x_n,T{x}_{n-1}))]\\ =\xi (x_{n-1})\xi (x_n)[\frac{1}{2}d(x_{n-1},x_{n+1})-\psi (d(x_{n-1},x_{n+1}),0)]\\ \le r^2[\frac{1}{2}d(x_{n-1},x_{n+1})-\psi (d(x_{n-1},x_{n+1}),0)],\end{array}$$

which implies

$$\begin{array}{rl}d(x_n,x_{n+1})& \le \frac{1}{2}d(x_{n-1},x_{n+1})-\psi (d(x_{n-1},x_{n+1}),0)\\ \le \frac{1}{2}d(x_{n-1},x_{n+1})\\ \le \frac{1}{2}[d(x_{n-1},x_n)+d(x_n,x_{n+1})],\end{array}$$

(5)

and so $d_n\le d_{n-1}$ for all $n\in \mathbb{N}$. Then there exist $d\ge 0$ such that ${lim}_{n\to \mathrm{\infty }}{d}_n=d$. Suppose, on the contrary, that $d>0$. Also, taking limit as $n\to \mathrm{\infty }$ in (5), we deduce

$$d\le \frac{1}{2}\underset{n\to \mathrm{\infty }}{lim}d(x_{n-1},x_{n+1})\le \frac{1}{2}(d+d)=d,$$

that is,

$$\underset{n\to \mathrm{\infty }}{lim}d(x_{n-1},x_{n+1})=2d.$$

(6)

Taking limit as $n\to \mathrm{\infty }$ in (5) and using (6), we get

$$d\le \frac{1}{2}[2d]-\psi (2d,0).$$

Consequently, we have $\psi (2d,0)=0$, which yields $d=0$. Hence

$$\underset{n\to \mathrm{\infty }}{lim}d(x_n,x_{n+1})=0.$$

(7)

We shall show that $\{x_n\}$ is a Cauchy sequence. To reach this goal, first we prove the following claim:

(K) For every $\epsilon >0$, there exists $n\in \mathbb{N}$ such that if $r,q\ge n$ with $r-q\equiv 1(m)$, then $d(x_r,x_q)<\epsilon$.

Suppose, to the contrary, that there exists $\epsilon >0$ such that for any $n\in \mathbb{N}$, we can find $r_n>q_n\ge n$ with $r_n-q_n\equiv 1(m)$ satisfying

$$d(x_{q_n},x_{r_n})\ge \epsilon .$$

(8)

Now, we take $n>2m$. Then, corresponding to $q_n\ge n$, one can choose $r_n$ in such a way that it is the smallest integer with $r_n>q_n$ satisfying $r_n-q_n\equiv 1(m)$ and $d(x_{q_n},x_{r_n})\ge \epsilon$. Therefore, $d(x_{q_n},x_{r_n-m})<\epsilon$. By using the triangular inequality,

$$\epsilon \le d(x_{q_n},x_{r_n})\le d(x_{q_n},x_{r_n-m})+\sum _{i=1}^{m}d(x_{r_n-i},x_{r_n-i-1})<\epsilon +\sum _{i=1}^{m}d(x_{r_n-i},x_{r_n-i-1}).$$

Letting $n\to \mathrm{\infty }$ in the last inequality, keeping (7) in mind, we derive that

$$\underset{n\to \mathrm{\infty }}{lim}d(x_{q_n},x_{r_n})=\epsilon .$$

(9)

Again,

$$\begin{array}{rcl}\epsilon & \le & d(x_{q_n},x_{r_n})\\ \le & d(x_{q_n},x_{q_n+1})+d(x_{q_n+1},x_{r_n+1})+d(x_{r_n+1},x_{r_n})\\ \le & d(x_{q_n},x_{q_n+1})+d(x_{q_n+1},x_{q_n})+d(x_{q_n},x_{r_n})+d(x_{r_n},x_{r_n+1})+d(x_{r_n+1},x_{r_n}).\end{array}$$

Taking (7) and (9) into account, we get

$$\underset{n\to \mathrm{\infty }}{lim}d(x_{q_n+1},x_{r_n+1})=\epsilon$$

(10)

as $n\to \mathrm{\infty }$ in (9).

Also we have the following inequalities:

$$d(x_{q_n},x_{r_n+1})\le d(x_{q_n},x_{r_n})+d(x_{r_n},x_{r_n+1})$$

(11)

and

$$d(x_{q_n},x_{r_n})\le d(x_{q_n},x_{r_n+1})+d(x_{r_n},x_{r_n+1}).$$

(12)

Letting $n\to \mathrm{\infty }$ in (11) and (12), we derive that

$$\underset{n\to \mathrm{\infty }}{lim}d(x_{q_n},x_{r_n+1})=\epsilon .$$

(13)

Again, we have

$$d(x_{r_n},x_{q_n+1})\le d(x_{r_n},x_{r_n+1})+d(x_{r_n+1},x_{q_n+1})$$

(14)

and

$$d(x_{r_n+1},x_{q_n+1})\le d(x_{r_n+1},x_{r_n})+d(x_{r_n},x_{q_n+1}).$$

(15)

Letting $n\to \mathrm{\infty }$ in (14) and (15), we conclude that

$$\underset{n\to \mathrm{\infty }}{lim}d(x_{r_n},x_{q_n+1})=\epsilon .$$

(16)

Since $x_{q_n}$ and $x_{r_n}$ lie in different adjacently labeled sets $X_i$ and $X_{i+1}$ for certain $1\le i\le m$, using the fact that T is a cyclic weak r-$(\eta ,\xi )$-C-contractive mapping of the first kind, we have

$$\begin{array}{r}{r}^{2}d(x_{r_n+1},x_{q_n+1})\\ =r^{2}d(T{x}_{r_n},T{x}_{q_n})\le \eta (x_{r_n})\eta (x_{q_n})d(T{x}_{r_n},T{x}_{q_n})\\ \le \xi (x_{r_n})\xi (x_{q_n})[\frac{1}{2}[d(x_{r_n},T{x}_{q_n})+d(x_{q_n},T{x}_{r_n})]-\psi (d(x_{r_n},T{x}_{q_n}),d(x_{q_n},T{x}_{r_n}))]\\ \le r^2[\frac{1}{2}[d(x_{r_n},x_{q_n+1})+d(x_{q_n},x_{r_n+1})]-\psi (d(x_{r_n},x_{q_n+1}),d(x_{q_n},x_{r_n+1}))],\end{array}$$

(17)

which implies

$$d(x_{r_n+1},x_{q_n+1})\le \frac{1}{2}[d(x_{r_n},x_{q_n+1})+d(x_{q_n},x_{r_n+1})]-\psi (d(x_{r_n},x_{q_n+1}),d(x_{q_n},x_{r_n+1})).$$

Letting $n\to \mathrm{\infty }$ in the inequality above and keeping the expressions (7), (9), (10), (13), (16) in mind, we conclude that

$$\epsilon \le \epsilon -\psi (\epsilon ,\epsilon ).$$

Thus, we have $\psi (\epsilon ,\epsilon )=0$, which yields that $\epsilon =0$. Hence, (K) is satisfied.

We shall show that the sequence $\{x_n\}$ is Cauchy. Fix $\epsilon >0$. By the claim, we find $n_0\in \mathbb{N}$ such that if $r,q\ge n_0$ with $r-q\equiv 1(m)$, then

$$d(x_r,x_q)\le \frac{\epsilon }{2}.$$

(18)

Since ${lim}_{n\to \mathrm{\infty }}d(x_n,x_{n+1})=0$, we also find $n_1\in \mathbb{N}$ such that

$$d(x_n,x_{n+1})\le \frac{\epsilon }{2m}$$

(19)

for any $n\ge n_1$. Suppose that $r,s\ge max\{n_0,n_1\}$ and $s>r$. Then, there exists $k\in \{1,2,\dots ,m\}$ such that $s-r\equiv k(m)$. Therefore, $s-r+\phi \equiv 1(m)$ for $\phi =m-k+1$. So, we have, for $j\in \{1,\dots ,m\}$, $s+j-r\equiv 1(m)$

$$d(x_r,x_s)\le d(x_r,x_{s+j})+d(x_{s+j},x_{s+j-1})+\cdots +d(x_{s+1},x_s).$$

By (18) and (19) and from the last inequality, we get

$$d(x_r,x_s)\le \frac{\epsilon }{2}+j\times \frac{\epsilon }{2m}\le \frac{\epsilon }{2}+m\times \frac{\epsilon }{2m}=\epsilon .$$

This proves that $\{x_n\}$ is a Cauchy sequence. Since Y is closed in $(X,d)$, then $(Y,d)$ is also complete, there exists $x\in Y={\bigcup }_{i=1}^m{X}_i$ such that ${lim}_{n\to \mathrm{\infty }}{x}_n=x$ in $(Y,d)$. In what follows, we prove that x is a fixed point of T. In fact, since ${lim}_{n\to \mathrm{\infty }}{x}_n=x$ and, as $Y={\bigcup }_{i=1}^m{X}_i$ is a cyclic representation of Y with respect to T, the sequence $\{x_n\}$ has infinite terms in each $X_i$ for $i\in \{1,2,\dots ,m\}$. Suppose that $x\in X_i$, $Tx\in X_{i+1}$ and we take a subsequence $x_{n_k}$ of $\{x_n\}$ with $x_{n_k}\in X_{i-1}$. Now from (iii) we have $\eta (x)\ge r$ and $\xi (x)\le r$. By using the contractive condition, we can obtain

$$\begin{array}{rl}{r}^{2}d(Tx,T{x}_{n_k})& \le \eta (x)\eta (x_{n_k})d(Tx,T{x}_{n_k})\\ \le \xi (x)\xi (x_{n_k})[\frac{1}{2}[d(x,T{x}_{n_k})+d(x_{n_k},Tx)]-\psi (d(x,T{x}_{n_k}),d(x_{n_k},Tx))]\\ \le r^2[\frac{1}{2}[d(x,T{x}_{n_k})+d(x_{n_k},Tx)]-\psi (d(x,T{x}_{n_k}),d(x_{n_k},Tx))],\end{array}$$

(20)

which implies

$$d(Tx,x_{n_k+1})\le \frac{1}{2}[d(x,x_{n_k+1})+d(x_{n_k},Tx)]-\psi (d(x,x_{n_k+1}),d(x_{n_k},Tx)).$$

Passing to the limit as $k\to \mathrm{\infty }$ in the last inequality, we get

$$\begin{array}{rl}d(x,Tx)& \le \frac{1}{2}d(x,Tx)-\psi (0,d(x,Tx))\\ \le \frac{1}{2}d(x,Tx),\end{array}$$

which implies $d(x,Tx)=0$, i.e., $x=Tx$. Finally, to prove the uniqueness of the fixed point, suppose that $x,y\in Fix{(T)}_Y$ such that $\eta (x)\ge r$, $\eta (y)\ge r$, $\xi (x)\le r$, $\xi (y)\le r$, where $x\ne y$. The cyclic character of T and the fact that $x,y\in X$ are fixed points of T imply that $x,y\in {\bigcap }_{i=1}^m{X}_i$. Suppose that $x\ne y$. That is, $d(x,y)>0$. Using the contractive condition, we obtain

$$\begin{array}{rl}{r}^{2}d(Tx,Ty)& \le \eta (x)\eta (y)d(Tx,Ty)\\ \le \xi (x)\xi (y)[\frac{1}{2}[d(x,Ty)+d(y,Tx)]-\psi (d(x,Ty),d(y,Tx))]\\ \le r^2[\frac{1}{2}[d(x,Ty)+d(y,Tx)]-\psi (d(x,Ty),d(y,Tx))],\end{array}$$

which implies

$$d(x,y)\le d(x,y)-\psi (d(x,y),d(x,y)).$$

Then $\psi (d(x,y),d(x,y))=0$ and so $d(x,y)=0$, i.e., $x=y$, which is a contradiction. This finishes the proof. □

Example 2.2 Let $X=\mathbb{R}$ with the metric $d(x,y)=|x-y|$ for all $x,y\in X$. Suppose $A_1=(-\mathrm{\infty },0]$ and $A_2=[0,\mathrm{\infty })$ and $Y={\bigcup }_{i=1}^2{A}_i$. Define $T:Y\to Y$ and $\eta ,\xi :Y\to [0,\mathrm{\infty })$ by

$$\begin{array}{r}Tx=\{\begin{array}{ll}\frac{x+9}{\sqrt{x^8+1}}& \text{if }x\in (-\mathrm{\infty },-10],\\ \frac{{sin}^{2}x}{x^4}& \text{if }x\in [-10,-5),\\ -3x& \text{if }x\in [-5,-1),\\ 0& \text{if }x\in [-1,1],\\ -5lnx& \text{if }x\in (1,5),\\ \frac{4-x}{(3-x)(2-x)}& \text{if }x\in [5,10),\\ \sqrt[3]{9-x}& \text{if }x\in [10,\mathrm{\infty }),\end{array}\\ \eta (x)=\{\begin{array}{ll}4& \text{if }x\in [-1,1],\\ 0& \text{otherwise,}\end{array}\\ \xi (x)=\{\begin{array}{ll}4& \text{if }x\in [-1,1],\\ 10& \text{otherwise.}\end{array}\end{array}$$

Also, define $\psi :{[0,\mathrm{\infty })}^2\to [0,\mathrm{\infty })$ by $\psi (t,s)=\frac{1}{4}(t+s)$. Clearly, $T{A}_1\subseteq A_2$, $T{A}_2\subseteq A_1$ and $\eta (0)\ge 4$ and $\xi (0)\le 4$. Let $\eta (x)\ge 4$, then $x\in [-1,1]$. On the other hand, $Tw\in [-1,1]$ for all $w\in [-1,1]$, i.e., $\eta (Tx)\ge 1$. Similarly, $\xi (x)\le 4$ implies $\xi (Tx)\le 4$. Therefore, T is an r-$(\eta ,\xi )$-admissible mapping. Let $\{x_n\}$ be a sequence in X such that $\eta (x_n)\ge 1$, $\xi (x_n)\le 1$ and $x_n\to x$ as $n\to \mathrm{\infty }$. Then $x_n\in [-1,1]$. So, $x\in [-1,1]$, i.e., $\eta (x)\ge 1$ and $\xi (x)\le 1$.

Let $x\in A_1$ and $y\in A_2$. Now, if $x\notin [-1,0]$ or $y\notin [0,1]$, then $\eta (x)\eta (y)=0$. Also, if $x\in [-1,0]$ and $y\in [0,1]$, then $d(Tx,Ty)=0$. That is, $\eta (x)\eta (y)d(Tx,Ty)=0$ for all $x\in A_1$ and all $y\in A_2$. Hence,

$$\eta (x)\eta (y)d(Tx,Ty)=0\le \xi (x)\xi (y)[\frac{1}{2}[d(x,Ty)+d(y,Tx)]-\psi (d(x,Ty),d(y,Tx))]$$

for all $x\in A_1$ and $y\in A_2$. Then T is a cyclic weak r-$(\eta ,\xi )$-C-contractive mapping of the first kind. Therefore all the conditions of Theorem 2.1 hold and T has a fixed point in $A_1\cap A_2$. Here, $x=0$ is a fixed point of T.

Theorem 2.3 Let $(X,d)$ be a complete metric space, $m\in \mathbb{N}$, $X_1,X_2,\dots ,X_m$ be nonempty closed subsets of $(X,p)$ and $Y={\bigcup }_{i=1}^m{X}_i$. Suppose that $T:Y\to Y$ is a cyclic weak r-$(\eta ,\xi )$-C-contractive mapping of the second kind such that

• (i) T is r-$(\eta ,\xi )$-admissible;

• (ii) there exists $x_0\in Y$ such that $\eta (x_0)\ge r$ and $\xi (x_0)\le r$;

• (iii) if $\{x_n\}$ is a sequence in Y such that $\eta (x_n)\ge r$ and $\xi (x_n)\le r$ for all $n\in \mathbb{N}$ and $x_n\to x$ as $n\to \mathrm{\infty }$, then $\eta (x)\ge r$ and $\xi (x)\le r$.

Then T has a fixed point $x\in {\bigcap }_{i=1}^n{X}_i$. Moreover, if $\eta (x)\ge r$, $\eta (y)\ge r$, $\xi (x)\le r$, $\xi (y)\le r$ for all $x,y\in Fix{(T)}_Y$, then T has a unique fixed point.

Proof By a similar method as in the proof of Theorem 2.1, we have

$$x_{n+1}=T{x}_n,\eta (x_n)\ge r\text{and}\xi (x_n)\le r\text{for all }n\in \mathbb{N}.$$

(21)

We shall show that the sequence $\{d_n:=d(x_n,x_{n+1})\}$ is non-increasing. Due to (3) with $x=x_{n-1}$ and $y=x_n$, we get

$$\begin{array}{rl}{(r^2+r)}^{d(x_n,x_{n+1})}& ={(r^2+r)}^{d(T{x}_{n-1},T{x}_n)}\le {(\eta (x_{n-1})\eta (x_n)+r)}^{d(T{x}_{n-1},T{x}_n)}\\ \le {(\xi (x_{n-1})\xi (x_n)+r)}^{[\frac{1}{2}[d(x_{n-1},T{x}_n)+d(x_n,T{x}_{n-1})]-\psi (d(x_{n-1},T{x}_n),d(x_n,T{x}_{n-1}))]}\\ ={(\xi (x_{n-1})\xi (x_n)+r)}^{[\frac{1}{2}d(x_{n-1},x_{n+1})-\psi (d(x_{n-1},x_{n+1}),0)]}\\ \le {(r^2+r)}^{[\frac{1}{2}d(x_{n-1},x_{n+1})-\psi (d(x_{n-1},x_{n+1}),0)]},\end{array}$$

which implies

$$\begin{array}{rl}d(x_n,x_{n+1})& \le \frac{1}{2}d(x_{n-1},x_{n+1})-\psi (d(x_{n-1},x_{n+1}),0)\\ \le \frac{1}{2}d(x_{n-1},x_{n+1})\\ \le \frac{1}{2}[d(x_{n-1},x_n)+d(x_n,x_{n+1})],\end{array}$$

(22)

and so $d_n\le d_{n-1}$ for all $n\in \mathbb{N}$. Then there exists $d\ge 0$ such that ${lim}_{n\to \mathrm{\infty }}{d}_n=d$. We shall show that $d=0$ by the method of reductio ad absurdum. Suppose that $d>0$. By letting $n\to \mathrm{\infty }$ in (22), we deduce

$$d\le \frac{1}{2}\underset{n\to \mathrm{\infty }}{lim}d(x_{n-1},x_{n+1})\le \frac{1}{2}(d+d)=d,$$

that is,

$$\underset{n\to \mathrm{\infty }}{lim}d(x_{n-1},x_{n+1})=2d.$$

(23)

Taking limit as $n\to \mathrm{\infty }$ in (22) and using (23), we get

$$d\le \frac{1}{2}[2d]-\psi (2d,0).$$

Thus, we have $\psi (2d,0)=0$ and hence $d=0$, which is a contradiction. Consequently, we have

$$\underset{n\to \mathrm{\infty }}{lim}{d}_n=\underset{n\to \mathrm{\infty }}{lim}d(x_n,x_{n+1})=0.$$

(24)

We shall show that $\{x_n\}$ is a Cauchy sequence. To reach this goal, first we prove the following claim:

(K) For every $\epsilon >0$, there exists $n\in \mathbb{N}$ such that if $r,q\ge n$ with $r-q\equiv 1(m)$, then $d(x_r,x_q)<\epsilon$.

Suppose, to the contrary, that there exists $\epsilon >0$ such that for any $n\in \mathbb{N}$ we can find $r_n>q_n\ge n$ with $r_n-q_n\equiv 1(m)$ satisfying

$$d(x_{q_n},x_{r_n})\ge \epsilon .$$

(25)

Following the related lines in Theorem 2.1, we deduce

$$\underset{n\to \mathrm{\infty }}{lim}d(x_{q_n},x_{r_n})=\epsilon ,$$

(26)

$$\underset{n\to \mathrm{\infty }}{lim}d(x_{q_n+1},x_{r_n+1})=\epsilon ,$$

(27)

$$\underset{n\to \mathrm{\infty }}{lim}d(x_{q_n},x_{r_n+1})=\epsilon$$

(28)

and

$$\underset{n\to \mathrm{\infty }}{lim}d(x_{r_n},x_{q_n+1})=\epsilon .$$

(29)

Since $x_{q_n}$ and $x_{r_n}$ lie in different adjacently labeled sets $X_i$ and $X_{i+1}$ for certain $1\le i\le m$, using the fact that a cyclic weak r-$(\eta ,\xi )$-C-contractive mapping of the second kind, we have

$$\begin{array}{rl}{(r^2+r)}^{d(x_{r_n+1},x_{q_n+1})}& ={(r^2+r)}^{d(T{x}_{r_n},T{x}_{q_n})}\le {(\eta (x_{r_n})\eta (x_{q_n})+r)}^{d(T{x}_{r_n},T{x}_{q_n})}\\ \le {(\xi (x_{r_n})\xi (x_{q_n})+r)}^{[\frac{1}{2}[d(x_{r_n},T{x}_{q_n})+d(x_{q_n},T{x}_{r_n})]-\psi (d(x_{r_n},T{x}_{q_n}),d(x_{q_n},T{x}_{r_n}))]}\\ \le {(r^2+r)}^{[\frac{1}{2}[d(x_{r_n},x_{q_n+1})+d(x_{q_n},x_{r_n+1})]-\psi (d(x_{r_n},x_{q_n+1}),d(x_{q_n},T{x}_{r_n+1}))],}\end{array}$$

which implies

$$d(x_{r_n+1},x_{q_n+1})\le \frac{1}{2}[d(x_{r_n},x_{q_n+1})+d(x_{q_n},x_{r_n+1})]-\psi (d(x_{r_n},x_{q_n+1}),d(x_{q_n},T{x}_{r_n+1})).$$

Letting $n\to \mathrm{\infty }$ in the inequality above and by applying (24) (26), (27), (28), (29), we deduce that

$$\epsilon \le \epsilon -\psi (\epsilon ,\epsilon ).$$

Consequently, we have $\psi (\epsilon ,\epsilon )=0$, and hence $\epsilon =0$. As a result, we conclude that (K) is satisfied. We assert that the sequence $\{x_n\}$ is Cauchy. Fix $\epsilon >0$. By the claim, we find $n_0\in \mathbb{N}$ such that if $r,q\ge n_0$ with $r-q\equiv 1(m)$, then

$$d(x_r,x_q)\le \frac{\epsilon }{2}.$$

(30)

Since ${lim}_{n\to \mathrm{\infty }}d(x_n,x_{n+1})=0$, we also find $n_1\in \mathbb{N}$ such that

$$d(x_n,x_{n+1})\le \frac{\epsilon }{2m}$$

(31)

for any $n\ge n_1$. Suppose that $r,s\ge max\{n_0,n_1\}$ and $s>r$. Then there exists $k\in \{1,2,\dots ,m\}$ such that $s-r\equiv k(m)$. Therefore, $s-r+\phi \equiv 1(m)$ for $\phi =m-k+1$. So, we have, for $j\in \{1,\dots ,m\}$, $s+j-r\equiv 1(m)$,

$$d(x_r,x_s)\le d(x_r,x_{s+j})+d(x_{s+j},x_{s+j-1})+\cdots +d(x_{s+1},x_s).$$

By (30) and (31) and from the last inequality, we get

$$\begin{array}{rcl}d(x_r,x_s)& \le & \frac{\epsilon }{2}+j\times \frac{\epsilon }{2m}\\ \le & \frac{\epsilon }{2}+m\times \frac{\epsilon }{2m}=\epsilon .\end{array}$$

This proves that $\{x_n\}$ is a Cauchy sequence. Since Y is closed in $(X,d)$, then $(Y,d)$ is also complete, there exists $x\in Y={\bigcup }_{i=1}^m{X}_i$ such that ${lim}_{n\to \mathrm{\infty }}{x}_n=x$ in $(Y,d)$. In what follows, we prove that x is a fixed point of T. In fact, since ${lim}_{n\to \mathrm{\infty }}{x}_n=x$ and, as $Y={\bigcup }_{i=1}^m{X}_i$ is a cyclic representation of Y with respect to T, the sequence $\{x_n\}$ has infinite terms in each $X_i$ for $i\in \{1,2,\dots ,m\}$. Suppose that $x\in X_i$, $Tx\in X_{i+1}$ and we take a subsequence $x_{n_k}$ of $\{x_n\}$ with $x_{n_k}\in X_{i-1}$. Now from (iii) we have $\eta (x)\ge r$ and $\xi (x)\le r$. By using the contractive condition, we can obtain

$$\begin{array}{rl}{(r^2+r)}^{d(Tx,T{x}_{n_k})}& \le {(\eta (x)\eta (x_{n_k})+r)}^{d(Tx,T{x}_{n_k})}\\ \le {(\xi (x)\xi (x_{n_k})+r)}^{[\frac{1}{2}[d(x,T{x}_{n_k})+d(x_{n_k},Tx)]-\psi (d(x,T{x}_{n_k}),d(x_{n_k},Tx))]}\\ \le {(r^2+r)}^{[\frac{1}{2}[d(x,T{x}_{n_k})+d(x_{n_k},Tx)]-\psi (d(x,T{x}_{n_k}),d(x_{n_k},Tx))]},\end{array}$$

(32)

which implies

$$d(Tx,x_{n_k+1})\le \frac{1}{2}[d(x,x_{n_k+1})+d(x_{n_k},Tx)]-\psi (d(x,x_{n_k+1}),d(x_{n_k},Tx)).$$

Passing to the limit as $k\to \mathrm{\infty }$ in the last inequality, we get

$$d(x,Tx)\le \frac{1}{2}d(x,Tx)-\psi (0,d(x,Tx))\le \frac{1}{2}d(x,Tx),$$

which implies $d(x,Tx)=0$, i.e., $x=Tx$. Finally, to prove the uniqueness of the fixed point, suppose that $x,y\in Fix{(T)}_Y$ such that $\eta (x)\ge r$, $\eta (y)\ge r$, $\xi (x)\le r$, $\xi (y)\le r$, where $x\ne y$. The cyclic character of T and the fact that $x,y\in X$ are fixed points of T imply that $x,y\in {\bigcap }_{i=1}^m{X}_i$. Suppose that $x\ne y$. That is, $d(x,y)>0$. Using the contractive condition, we obtain

$$\begin{array}{rl}{(r^2+r)}^{d(Tx,Ty)}& \le {(\eta (x)\eta (y)+r)}^{d(Tx,Ty)}\\ \le {(\xi (x)\xi (y)+r)}^{[\frac{1}{2}[d(x,Ty)+d(y,Tx)]-\psi (d(x,Ty),d(y,Tx))]}\\ \le {(r^2+r)}^{[\frac{1}{2}[d(x,Ty)+d(y,Tx)]-\psi (d(x,Ty),d(y,Tx))],}\end{array}$$

which implies

$$d(x,y)\le d(x,y)-\psi (d(x,y),d(x,y)).$$

Hence, we obtain $\psi (d(x,y),d(x,y))=0$, which implies $d(x,y)=0$, that is, $x=y$ a contradiction. □

3 Existence of solutions of an integral equation

For $T>0$, we denote by $X=C([0,T])$ the set of real continuous functions on $[0,T]$. We endow X with the metric

$$d_{\mathrm{\infty }}(u,v)={\parallel u-v\parallel }_{\mathrm{\infty }}\text{for all }u,v\in X.$$

It is evident that $(X,d_{\mathrm{\infty }})$ is a complete metric space.

Consider the integral equation

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

(33)

• (1) $f:[0,T]\times \mathbb{R}\to \mathbb{R}$ and $G:[0,T]\times [0,T]\to [0,\mathrm{\infty })$ are continuous functions.

• (2) Let $(\alpha ,\beta )\in X^2$, $({\alpha }_0,{\beta }_0)\in {\mathbb{R}}^2$ such that

  $${\alpha }_0\le \alpha (t)\le \beta (t)\le {\beta }_0\text{for all }t\in [0,T].$$

  (34)

  Assume that for all $t\in [0,T]$, we have

  $$\alpha (t)\le {\int }_0^{T}G(t,s)f(s,\beta (s))ds$$

  (35)

  and

  $$\beta (t)\ge {\int }_0^{T}G(t,s)f(s,\alpha (s))ds.$$

  (36)

  Let for all $s\in [0,T]$, $f(s,\cdot )$ be a decreasing function, that is,

  $$x,y\in \mathbb{R},x\ge y⟹f(s,x)\le f(s,y).$$

  (37)

  Let $Z:=\{u\in X:u\le \beta \}\cup \{u\in X:u\ge \alpha \}$. There exist $0\le r<1$ and $\theta ,\pi :Z\to \mathbb{R}$ such that if $\theta (x)\ge 0$ and $\theta (y)\ge 0$ with ($x\le {\beta }_0$ and $y\ge {\alpha }_0$) or ($x\ge {\alpha }_0$ and $y\le {\beta }_0$), then for every $s\in [0,T]$, we have

  $$|f(s,x(s))-f(s,y(s))|\le \frac{r|\pi (y)|}{2}\left(\right|x(s)-Ty(s)|+|y(s)-Tx(s)\left|\right).$$

  (38)

• (3) Assume that

  $${\parallel {\int }_0^T|\pi (y)|G(t,s)ds\parallel }_{\mathrm{\infty }}\le 1$$

  (39)

  for all $x\in Z$, where $\theta (x)\ge 0$. Suppose that

  $$\theta (x)\ge 0⟹\theta (Tx)\ge 0\text{for }x\in \{u\in X:u\le \beta \}\cup \{u\in X:u\ge \alpha \}.$$

  (40)

• (4) If $\{x_n\}$ is a sequence in $\{u\in X:u\le \beta \}\cup \{u\in X:u\ge \alpha \}$ such that $\theta (x_n)\ge 0$ for all $n\in \mathbb{N}$ and $x_n\to x$ as $n\to \mathrm{\infty }$, then $\theta (x)\ge 0$.

• (5) There exists $x_0\in \{u\in X:u\le \beta \}\cup \{u\in X:u\ge \alpha \}$ such that $\theta (x_0)\ge 0$.

Theorem 3.1 Under assumptions (1)-(5), integral equation (33) has a solution in $\{u\in C([0,T]):\alpha (t)\le u(t)\le \beta (t)\mathit{\text{for all}}t\in [0,T]\}$.

Proof Define the closed subsets of X, $A_1$ and $A_2$ by

$$A_1=\{u\in X:u\le \beta \}$$

and

$$A_2=\{u\in X:u\ge \alpha \}.$$

Also define the mapping $T:X\to X$ by

$$Tu(t)={\int }_0^{T}G(t,s)f(s,u(s))ds\text{for all }t\in [0,T].$$

Let us prove that

$$T(A_1)\subseteq A_2\text{and}T(A_2)\subseteq A_1.$$

(41)

Suppose $u\in A_1$, that is,

$$u(s)\le \beta (s)\text{for all }s\in [0,T].$$

Applying condition (37), since $G(t,s)\ge 0$ for all $t,s\in [0,T]$, we obtain that

$$G(t,s)f(s,u(s))\ge G(t,s)f(s,\beta (s))\text{for all }t,s\in [0,T].$$

The above inequality with condition (35) imply that

$${\int }_0^{T}G(t,s)f(s,u(s))ds\ge {\int }_0^{T}G(t,s)f(s,\beta (s))ds\ge \alpha (t)$$

for all $t\in [0,T]$. Then we have $Tu\in A_2$.

Similarly, let $u\in A_2$, that is,

$$u(s)\ge \alpha (s)\text{for all }s\in [0,T].$$

Using condition (37), since $G(t,s)\ge 0$ for all $t,s\in [0,T]$, we obtain that

$$G(t,s)f(s,u(s))\le G(t,s)f(s,\alpha (s))\text{for all }t,s\in [0,T].$$

The above inequality with condition (36) imply that

$${\int }_0^{T}G(t,s)f(s,u(s))ds\le {\int }_0^{T}G(t,s)f(s,\alpha (s))ds\le \beta (t)$$

for all $t\in [0,T]$. Then we have $Tu\in A_1$. Also, we deduce that (41) holds.

Now, let $(u,v)\in A_1\times A_2$, that is, for all $t\in [0,T]$,

$$u(t)\le \beta (t),v(t)\ge \alpha (t).$$

This implies from condition (34) that for all $t\in [0,T]$,

$$u(t)\le {\beta }_0,v(t)\ge {\alpha }_0.$$

Now, by conditions (39) and (38), we have, for all $s\in [0,T]$,

$$\begin{array}{rl}|Tu(t)-Tv(t)|& =|{\int }_0^{T}G(t,s)[f(s,u(s))-f(s,v(s))]ds|\\ \le {\int }_0^{T}G(t,s)|f(s,u(s))-f(s,v(s))|ds\\ \le {\int }_0^{T}G(t,s)\frac{r|\pi (y)|}{2}\left(\right|u(s)-Tv(s)|+|v(s)-Tu(s)\left|\right)ds\\ \le \frac{r}{2}({\parallel u-Tv\parallel }_{\mathrm{\infty }}+{\parallel v-Tu\parallel }_{\mathrm{\infty }}){\parallel {\int }_0^T|\pi (v)|G(t,s)ds\parallel }_{\mathrm{\infty }}\\ \le \frac{r}{2}({\parallel u-Tv\parallel }_{\mathrm{\infty }}+{\parallel v-Tu\parallel }_{\mathrm{\infty }}),\end{array}$$

which implies

$${\parallel Tu-Tv\parallel }_{\mathrm{\infty }}\le \frac{r}{2}({\parallel u-Tv\parallel }_{\mathrm{\infty }}+{\parallel v-Tu\parallel }_{\mathrm{\infty }}).$$

Define $\eta ,\xi :Z\to [0,\mathrm{\infty })$ by $\eta (u)=\{\begin{array}{ll}1,& \theta (u)\ge 0,\\ 0,& \text{otherwise}\end{array}$ and $\xi (u)=1$. Further, $\psi (t_1,t_2)=\frac{(1-r)}{2}(t_1+t_2)$.

Hence,

$$\eta (u)\eta (v)d_{\mathrm{\infty }}(Tu,Tv)\le \frac{r}{2}(d_{\mathrm{\infty }}(u,Tv)+d_{\mathrm{\infty }}(v,Tu))$$

for all $(u,v)\in A_1\times A_2$. By a similar method, we can show that the above inequality holds if $(u,v)\in A_2\times A_1$. Now, all the conditions of Theorem 2.1 hold and T has a fixed point $z^{\ast }$ in

$$A_1\cap A_2=\{u\in C([0,T]):\alpha \le u(t)\le \beta \text{ for all }t\in [0,T]\}.$$

That is, $z^{\ast }\in A_1\cap A_2$ is the solution to (33). □

Example 3.2 In this example, we denote by $X=C([0,1])$ the set of real continuous functions on $[0,1]$. We endow X with the metric

$$d_{\mathrm{\infty }}(u,v)={\parallel u-v\parallel }_{\mathrm{\infty }}\text{for all }u,v\in X.$$

Consider the following continuous functions:

$$f(t,x)=\{\begin{array}{ll}{x}^3& \text{if }x\in (-\mathrm{\infty },0),\\ 0& \text{if }x\in [0,1],\\ x^2-1& \text{if }x\in (1,4),\\ 15& \text{if }x\in [4,\sqrt{e^{16}-1}],\\ x^2+16-e^{16}& \text{if }x\in (\sqrt{e^{16}-1},\mathrm{\infty })\end{array}\text{for all }t\in [0,1]$$

and

$$G(t,s)=\frac{t}{1+t}{e}^s\text{for all }s,t\in [0,1]\times [0,1].$$

Let $\alpha (t)=0$ and $\beta (t)=1$. Then, for $({\alpha }_0,{\beta }_0)=(0,1)\in {\mathbb{R}}^2$, we have

$$\begin{array}{c}{\alpha }_0\le \alpha (t)\le \beta (t)\le {\beta }_0;\hfill \\ \alpha (t)=0\le {\int }_0^{1}G(t,s)f(s,\beta (s))ds=0\hfill \end{array}$$

and

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

Also, $Z:=\{u\in X:u\le \beta \}\cup \{u\in X:u\ge \alpha \}=X$. Define $\theta ,\pi :Z\to \mathbb{R}$ by

$$\theta (x(t))=\{\begin{array}{ll}0& \text{if }0\le x(t)\le 1\text{ for all }t\in [0,1]\\ -1,& \text{otherwise}\end{array}\text{and}\pi (x)=\frac{1}{e-1}.$$

Clearly, $\theta (0)\ge 0$. Also, if $\theta (x(t))\ge 0$, then $0\le x(t)\le 1$. On the other hand,

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

for all $0\le u(t)\le 1$. That is, $\theta (Tx(t))\ge 0$. Hence, $\theta (x)\ge 0$ implies $\theta (Tx)\ge 0$.

Assume $\theta (x(s))\ge 0$ and $\theta (y(s))\ge 0$ with ($x\le {\beta }_0$ and $y\ge {\alpha }_0$) or ($x\ge {\alpha }_0$ and $y\le {\beta }_0$). Thus, $0\le x(s)\le 1$ and $0\le y(s)\le 1$, which implies $f(s,x(s))=f(s,y(s))=0$. That is,

$$|f(s,x(s))-f(s,y(s))|=0\le \frac{r|\pi (y)|}{2}\left(\right|x(s)-Ty(s)|+|y(s)-Tx(s)\left|\right)$$

for all $s\in [0,1]$, where $0\le r<1$. Further,

$${\int }_0^1|\pi (y)|G(t,s)ds={\int }_0^1\frac{1}{e-1}\frac{t}{1+t}{e}^{s}ds=\frac{t}{1+t}\le 1,$$

and so

$${\parallel {\int }_0^T|\pi (y)|G(t,s)ds\parallel }_{\mathrm{\infty }}\le 1.$$

Assume that $\{x_n\}$ is a sequence in X such that $\theta (x_n)\ge 0$ for all $n\in \mathbb{N}$ and $x_n\to x$ as $n\to \mathrm{\infty }$. Then $0\le x_n\le 1$. So, $0\le x\le 1$. That is, $\theta (x)\ge 0$.

Therefore, all of the conditions of Theorem 3.1 are satisfied. Then the integral equation

$$u(t)=\frac{t}{1+t}{\int }_0^1{e}^{s}f(s,u(s))ds$$

has a solution in $\{u\in C([0,1]):0\le u(t)\le 1\text{ for all }t\in [0,1]\}$. Here, $u(t)=0$ is a solution.

But if we chose $x_0(t)=0$ and $y_0(t)=\sqrt{e^{16}-1}$, then $f(s,x_0(s))=0$ and $f(s,y_0(s))=15$. That is,

$$|f(s,x_0(s))-f(s,y_0(s))|=15.$$

Also,

$$\sqrt{ln\left(\right|x_0(s)-y_0(s){|}^2+1)}=\sqrt{ln\left(\right|0-\sqrt{e^{16}-1}{|}^2+1)}=\sqrt{ln{e}^{16}}=4,$$

and so

$$|f(s,x_0(s))-f(s,y_0(s))|=15>4=\sqrt{ln\left(\right|x_0(s)-y_0(s){|}^2+1)}.$$

That is, Theorem 3.1 of [6] cannot be applied to this example.