\documentclass[reqno]{amsart} \usepackage{hyperref} \AtBeginDocument{{\noindent\small \emph{Electronic Journal of Differential Equations}, Vol. 2015 (2015), No. 133, pp. 1--15.\newline ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu \newline ftp ejde.math.txstate.edu} \thanks{\copyright 2015 Texas State University - San Marcos.} \vspace{9mm}} \begin{document} \title[\hfilneg EJDE-2015/133\hfil Fixed point results in metric-like spaces] {Fixed point results for generalized $\alpha$-$\psi$-contractions in metric-like spaces and applications} \author[H. Aydi, E. Karapinar \hfil EJDE-2015/133\hfilneg] {Hassen Aydi, Erdal Karapinar} \address{Hassen Aydi \newline Dammam University, Departement of Mathematics, College of Education of Jubail, P.O. 12020, Industrial Jubail 31961, Saudi Arabia} \email{hmaydi@uod.edu.sa} \address{Erdal Karapinar \newline Atilim University, Department of Mathematics, 06836, \.Incek, Ankara, Turkey.\newline Nonlinear Analysis and Applied Mathematics Research Group (NAAM), King Abdulaziz University, 21589, Jeddah, Saudi Arabia} \email{erdalkarapinar@yahoo.com, ekarapinar@atilim.edu.tr} \thanks{Submitted December 2, 2014. Published May 15, 2015.} \subjclass[2010]{47H10, 54H25} \keywords{Metric-like; $\alpha$-admissible mappings; fixed point} \begin{abstract} In this article, we introduce the concept of generalized $\alpha$-$\psi$-con\-traction in the context of metric-like spaces and establish some related fixed point theorems. As consequences, we obtain some known fixed point results in the literature. Some examples and an application on two-point boundary value problems for second order differential equation are also considered. \end{abstract} \maketitle \numberwithin{equation}{section} \newtheorem{theorem}{Theorem}[section] \newtheorem{lemma}[theorem]{Lemma} \newtheorem{corollary}[theorem]{Corollary} \newtheorem{remark}[theorem]{Remark} \newtheorem{definition}[theorem]{Definition} \newtheorem{example}[theorem]{Example} \allowdisplaybreaks \section{Introduction and preliminaries} The notion of metric-like (dislocated) metric spaces was introduced by Hitzler and Seda \cite{HS} in 2000 as a generalization of a metric space. They generalized the Banach Contraction Principle \cite{Ba} in such spaces. Metric-like spaces were discovered by Amini-Harandi \cite{Amini} who established some fixed point results. Very recently, Karapinar and Salimi \cite{Erdal1} established some fixed point theorems for cyclic contractions in the setting of metric-like spaces. Many other (common) fixed point results in the context of metric-like (quasi) spaces have been proved, see for example \cite{Aage,Aage2,Erdal1,Koh,Ren,Zey,Zota}. In the sequel, the letters $\mathbb{R}$, $\mathbb{R}^+_0$ and $\mathbb{N}^{\ast }$ will denote the set of real numbers, the set of nonnegative real numbers and the set of positive integer numbers, respectively. \begin{definition}[\cite{Amini}] \label{def1.1} \rm Let $X$ be a nonempty set. A function $\sigma:X\times X\to \mathbb{R}^+_0$ is said to be a dislocated (metric-like) metric on $X$ if for any $x,y,z\in X$, the following conditions hold: \begin{itemize} \item[(S1)] $\sigma(x,y)=\sigma(x,x)=0\Longrightarrow x=y$; \item[(S2)] $\sigma(x,y)=\sigma(y,x)$; \item[(S3)] $\sigma(x,z)\leq \sigma(x,y)+\sigma(y,z)$. \end{itemize} The pair $(X,\sigma)$ is then called a dislocated (metric-like) metric space. \end{definition} \begin{example} \rm A trivial example of a metric-like space is the pair $(\mathbb{R}^+_0,\sigma)$, where $\sigma:\mathbb{R}^+_0\times \mathbb{R}^+_0\to \mathbb{R}^+_0$ is defined as $\sigma(x,y)=\max \{x,y\}$. Here, $\sigma$ is also a partial metric \cite{Ma}. \label{Ex1} \end{example} \begin{example} \label{examp13} \rm Take $X=\mathbb{R}$ and define the $\sigma$ metric-like as \[ \sigma(x,y)=\frac{|x-y|+|x|+|y|}{2}\quad\text{for all }x,y\in X. \] Notice that $\sigma$ is not a metric. Particularly, if $X=\mathbb{R}^+_0$, we have $\sigma(x,y)=\max\{x,y\}$ and so we return to Example \ref{Ex1}. But, if $X=\mathbb{R}$, we have $\sigma(x,y)\neq \max\{x,y\}$. \end{example} As it is well known, a partial metric \cite{Ma} is a metric-like. The converse is not true. The following example concerns this statement. \begin{example} \label{examp1.4} \rm Take $X=\{1,2,3\}$ and consider the metric-like $\sigma:X \times X\to \mathbb{R}^+_0$ given by \begin{gather*} \sigma(1,1)=0,\quad \sigma(2,2)=1,\quad \sigma(3,3)=\frac{2}{3}, \quad \sigma(1,2)=\sigma(2,1)=\frac{9}{10},\\ \sigma(2,3)=\sigma(3,2)=\frac{4}{5},\quad \sigma(1,3)=\sigma(3,1)=\frac{7}{10}. \end{gather*} Since $\sigma(2,2)\neq 0$, so $\sigma$ is not a metric and since $\sigma(2,2)>\sigma(1,2)$, so $\sigma$ is not a partial metric. \end{example} Each metric-like $\sigma$ on $X$ generates a $T_{0}$ topology $\tau _{\sigma}$ on $X $ which has as a base the family open $\sigma$-balls $\{B_{\sigma}(x,\varepsilon ):x\in X,\varepsilon >0\}$, where $B_{\sigma}(x,\varepsilon )=\{y\in X:|\sigma(x,y)-\sigma(x,x)|<\varepsilon \}$, for all $x\in X$ and $\varepsilon >0$. Observe that a sequence $\{x_{n}\}$ in a metric-like space $(X,\sigma)$ converges to a point $x\in X$, with respect to $\tau _{\sigma}$, if and only if $\sigma(x,x)=\lim_{n\to \infty }\sigma(x,x_{n})$. \begin{definition}[\cite{Amini}] \rm Let $(X,\sigma)$ be a metric-like space. \begin{itemize} \item[(a)] A sequence $\{x_{n}\}$ in $X$ is a Cauchy sequence if $ \lim_{n,m\to \infty }\sigma(x_{n},x_{m})$ exists and is finite. \item[(b)] $(X,\sigma)$ is complete if every Cauchy sequence $\{x_{n}\} $ in $X$ converges with respect to $\tau _{\sigma}$ to a point $x\in X$; that is, \[ \lim_{n\to \infty }\sigma(x,x_{n})=\sigma(x,x) =\lim_{n,m\to \infty }\sigma(x_n,x_{m}). \] \end{itemize} \end{definition} \begin{definition}[\cite{Amini}] \rm Let $(X,\sigma)$ be a metric-like space. A mapping $T:(X,\sigma)\to (X,\sigma)$ is continuous if for any sequence $\{x_n\}$ in $X$ such that $\sigma(x_n,x)\to \sigma(x,x)$ as $n\to \infty$, we have $\sigma(Tx_n,Tx)\to \sigma(Tx,Tx)$ as $n\to \infty$. \end{definition} \begin{lemma}[\cite{Erdal1}] \label{L1} Let $(X,\sigma)$ be a metric-like space. Let $\{x_n\}$ be a sequence in $X$ such that $x_n\to x$ where $x\in X$ and $\sigma(x,x)=0$. Then, for all $y\in X$, we have \[ \lim_{n\to \infty} \sigma(x_n,y)=\sigma(x,y). \] \end{lemma} Let $\Psi $ be the family of functions $\psi :[0,\infty )\to [0,\infty )$ satisfying the following conditions: \begin{itemize} \item[(i)] $\psi $ is nondecreasing; \item[(ii)] $\sum_{n=1}^{+\infty} \psi ^{n}(t)< \infty$ for all $t>0$. \end{itemize} Note that if $\psi\in\Psi$, we have $\psi(t)0$. In 2012, Samet et al \cite{Samet1} introduced the class of $\alpha$-admissible mappings. \begin{definition} \cite{Samet1} For a nonempty set $X$, let $T: X\to X$ and $\alpha: X\times X\to [0,\infty)$ be given mappings. We say that $T$ is $\alpha$-admissible if for all $x,y\in X$, we have \begin{equation} \label{eqA} \alpha(x,y)\geq 1 \Longrightarrow \alpha(Tx,Ty)\geq 1. \end{equation} \end{definition} The notion of $\alpha-\psi$-contractive mappings is also defined in the following way. \begin{definition}[\cite{Samet1}] \rm Let $(X,d)$ be a metric space and $T: X\to X$ be a given mapping. We say that $T$ is a $\alpha-\psi$ contractive mapping if there exist two functions $\alpha: X\times X \to [0, \infty)$ and $\psi\in\Psi$ such that \begin{equation} \label{eq1x} \alpha(x,y)d(Tx,Ty)\leq\psi(d(x,y)), \text{ for all } x,y\in X. \end{equation} \end{definition} Many authors have proved fixed point results for generalized contractions using the function $\alpha$, see for instance \cite{AK,AJK,AJK2,KS}. Now, we state in the following definition a generalization of the notion of $\alpha-\psi$ contractive mappings in the context of a metric-like space. \begin{definition} \rm Let $(X,\sigma)$ be a metric-like space and $T: X\to X$ be a given mapping. We say that $T$ is a generalized $\alpha-\psi$ contractive mapping of type A if there exist two functions $\alpha: X\times X \to [0, \infty)$ and $\psi\in\Psi$ such that \begin{equation} \label{e1g} \alpha(x,y)\sigma(Tx,Ty)\leq\psi(M(x,y)), \quad \text{ for all } x,y\in X, \end{equation} where \begin{equation} \label{Mxy} M(x,y)=\max\{\sigma(x,y),\sigma(x,Tx),\sigma(y,Ty),\frac{\sigma(x,Ty)+\sigma(y,Tx)}{4}\}. \end{equation} \end{definition} Our aim in this article is to provide some fixed point results for variant generalized $\alpha-\psi$ contractive mappings in the setting of metric-like spaces. We support our obtained theorems by some concrete examples and an application. \section{Main results} Our first fixed point result read as follows. \begin{theorem} \label{thm1} Let $(X,\sigma)$ be a complete metric-like space and $T: X\to X$ be a generalized $\alpha-\psi$ contractive mapping of type A. Suppose that \begin{itemize} \item[(i)] $T$ is $\alpha$-admissible; \item[(ii)] there exists $x_{0}\in X$ such that $\alpha(x_{0},Tx_{0})\geq 1$; \item[(iii)] $T$ is continuous. \end{itemize} Then there exists a $u\in X$ such that $\sigma(u,u)=0$. Assume in addition that \begin{itemize} \item[(H1)] If $\sigma(x,x)=0$ for some $x\in X$, then $\alpha(x,x)\geq 1$. \end{itemize} Then such $u$ is a fixed point of $T$, that is, $Tu=u$. \end{theorem} \begin{proof} By assumption (ii), there exists a point $x_0\in X$ such that $\alpha(x_{0},Tx_{0})\geq 1$. We define a sequence $\{x_n\}$ in $X$ by $x_{n+1}=Tx_{n}=T^{n+1} x_0$ for all $n\geq 0$. Suppose that $x_{n_0}=x_{n_0+1}$ for some $n_0$. So the proof is completed since $u=x_{n_0}=x_{n_0+1}=Tx_{n_0}=Tu$. Consequently, throughout the proof, we assume that \begin{equation}\label{e2a} x_{n}\neq x_{n+1} \quad \text{for all } n. \end{equation} Observe that \[ \alpha(x_{0},x_{1})=\alpha(x_{0},Tx_{0})\geq 1\Rightarrow \alpha(Tx_{0},Tx_{1})=\alpha(x_{1},x_{2})\geq 1, \] since $T$ is $\alpha$-admissible. By repeating the process above, we derive that \begin{equation}\label{e2} \alpha(x_{n},x_{n+1})\geq 1, \quad \text{for all } n=0,1,\ldots. \end{equation} \smallskip \noindent\textbf{Step 1:} We shall prove that \begin{equation}\label{e7EKC1} \lim_{n\to \infty} \sigma(x_{n},x_{n+1})=0. \end{equation} Combining \eqref{e1g} and \ref{e2}, we find that \begin{equation}\label{e3C1} \sigma(x_{n},x_{n+1})=\sigma(Tx_{n-1},Tx_{n}) \leq \alpha(x_{n-1},x_{n})\sigma(Tx_{n-1},Tx_{n})\leq\psi(M(x_{n-1},x_{n})), \end{equation} for all $n\geq 1$, where \begin{equation}\label{g1} \begin{aligned} &M(x_{n-1},x_{n})\\ &=\max\{\sigma(x_{n-1},x_{n}),\sigma(x_{n-1},Tx_{n-1}),\sigma(x_{n},Tx_{n}), \frac{\sigma(x_{n-1},Tx_{n})+\sigma(x_{n},Tx_{n-1})}{4}\}\\ &\\ &=\max\{\sigma(x_{n-1},x_{n}),\sigma(x_{n-1},x_{n}),\sigma(x_{n},x_{n+1}), \frac{\sigma(x_{n-1},x_{n+1})+\sigma(x_{n},x_{n})}{4}\}\\ &\\ &\leq \max\{\sigma(x_{n-1},x_{n}),\sigma(x_{n},x_{n+1}), \frac{\sigma(x_{n-1},x_{n})+3\sigma(x_{n},x_{n+1})}{4}\}\\ &= \max\{\sigma(x_{n-1},x_{n}),\sigma(x_{n},x_{n+1})\}. \end{aligned} \end{equation} If for some $n$, $\max\{\sigma(x_{n-1},x_{n}),\sigma(x_{n},x_{n+1})\} =\sigma(x_{n},x_{n+1}) (\neq 0)$, then \eqref{e3C1} and \eqref{g1} turn into \[ \sigma(x_{n},x_{n+1})\leq\psi(M(x_{n-1},x_{n})) \leq \psi(\sigma(x_{n},x_{n+1}))<\sigma(x_{n},x_{n+1}), \] which is a contradiction. Hence, $\max\{\sigma(x_{n-1},x_{n}),\sigma(x_{n},x_{n+1})\}=\sigma(x_{n-1},x_{n})$ for all $n\in \mathbb{N}^*$ and \eqref{e3C1} becomes \begin{equation}\label{e3C1+} \sigma(x_{n},x_{n+1})\leq\psi(\sigma(x_{n-1},x_{n}))\quad\text{for all } n\geq 1. \end{equation} This yields \begin{equation} \label{aa1} \sigma(x_{n},x_{n+1})<\sigma(x_{n-1},x_{n})\quad\text{for all } n\geq 1. \end{equation} By \eqref{e3C1+}, we find that \begin{equation}\label{e7C1} \sigma(x_{n},x_{n+1})\leq \psi^n (\sigma(x_0,x_{1})), \quad\text{for all } n\in \mathbb{N}. \end{equation} By the properties of $\psi$, we have \[ \lim_{n\to \infty} \sigma(x_{n},x_{n+1})=0. \] \smallskip \noindent\textbf{Step 2:} We shall prove that $\{x_n\}$ is a Cauchy sequence. First, by using (S3) and \eqref{e7C1} \begin{equation}\label{C1EK} \begin{aligned} \sigma(x_n, x_{n+k}) &\leq \sigma(x_{n},x_{n+1})+ \sigma(x_{n+1},x_{n+2})+\ldots+ \sigma(x_{n+k-1},x_{n+k})\\ &\leq \sum_{p=n}^{n+k-1}\psi^p (\sigma(x_0,x_{1}))\\ &\leq \sum_{p=n}^{+\infty}\psi^p (\sigma(x_0,x_{1})) \to 0 \quad \text{as } n\to \infty. \end{aligned} \end{equation} Thus, by the symmetry of $\sigma$, we obtain \begin{equation} \label{b9} \lim_{n,m\to \infty} \sigma(x_n, x_{m})=0. \end{equation} We conclude that $\{x_n\}$ is a Cauchy sequence in $(X,\sigma)$. Since $(X,\sigma)$ is complete, there exists $u\in X$ such that \begin{equation}\label{e8} \lim_{n\to \infty} \sigma(x_n,u)=\sigma(u,u)= \lim_{n,m\to \infty} \sigma(x_n,x_m)=0. \end{equation} Since $T$ is continuous, from \eqref{e8} we obtain that \begin{equation}\label{e9} \lim_{n\to \infty} \sigma(x_{n+1},Tu)=\lim_{n\to \infty} \sigma(Tx_{n},Tu)=\sigma(Tu,Tu), \end{equation} On the other hand, by \eqref{e8} and Lemma \ref{L1} \begin{equation}\label{e10} \lim_{n\to \infty} \sigma(x_{n+1},Tu)=\sigma(u,Tu). \end{equation} Comparing \eqref{e9} and \eqref{e10}, we get $\sigma(u,Tu)=\sigma(Tu,Tu)$. By \eqref{e1g}, \[ \alpha(u,u)\sigma(Tu,Tu)\leq\psi(M(u,u)), \] where \begin{align*} M(u,u) &=\max\{\sigma(u,u),\sigma(u,Tu),\sigma(u,Tu),\frac{\sigma(u,Tu)+\sigma(u,Tu)}{4}\}\\ &=\max\{0,\sigma(u,Tu)\}=\sigma(u,Tu). \end{align*} From hypothesis (H1) and the fact that $\sigma(u,u)=0$, we have $\alpha(u,u)\geq 1$. Therefore, by \eqref{e1g} \[ \sigma(u,Tu)\leq \alpha(u,u)\sigma(u,Tu)\leq \psi(\sigma(u,Tu)), \] which holds unless $\sigma(u,Tu)=0$, that is $Tu=u$. So $u$ is a fixed point of $T$. \end{proof} Theorem \ref{thm1} remains true if we replace the continuity hypothesis by the following property: \begin{quote} If $\{x_n\}$ is a sequence in $X$ such that $\alpha(x_n,x_{n+1})\geq 1$ for all $n$ and $x_n \to x\in X$ as $n\to \infty$, then there exists a subsequence $\{x_{n(k)}\}$ of $\{x_n\}$ such that $\alpha(x_{n(k)},x)\geq 1$ for all $k$. \end{quote} This statement is given as follows. \begin{theorem}\label{thm2} Let $(X, d)$ be a complete metric-like space and $T: X\to X$ be a generalized $\alpha-\psi$ contractive mapping of type A. Suppose that \begin{itemize} \item[(i)] $T$ is $\alpha$-admissible; \item[(ii)] there exists $x_{0}\in X$ such that $\alpha(x_{0},Tx_{0})\geq 1$; \item[(iii)] if $\{x_n\}$ is a sequence in $X$ such that $\alpha(x_n,x_{n+1})\geq 1$ for all $n$ and $x_n \to x\in X$ as $n\to \infty$, then there exists a subsequence $\{x_{n(k)}\}$ of $\{x_n\}$ such that $\alpha(x_{n(k)},x)\geq 1$ for all $k$. \end{itemize} Then, there exists $u\in X$ such that $Tu=u$. \end{theorem} \begin{proof} Following the proof of Theorem \ref{thm1}, we know that the sequence $\{x_n\}$ defined by $x_{n+1}=Tx_n$ for all $n\geq 0$ is Cauchy in $(X,\sigma)$ and converges to some $u\in X$. Also, \eqref{e8}) holds, so \begin{equation} \lim_{k\to \infty} \sigma(x_{n(k)+1},Tu)=\sigma(u,Tu). \label{b10} \end{equation} We shall show that $Tu=u$. Suppose, on the contrary, that $Tu\neq u$, i.e, $\sigma(Tu,u)>0$. From \eqref{e2} and condition (iii), there exists a subsequence $\{x_{n(k)}\}$ of $\{x_n\}$ such that $\alpha(x_{n(k)},u)\geq 1$ for all $k$. By applying \eqref{e1g}, we obtain \begin{equation}\label{h1} \sigma(x_{n(k)+1},Tu)\leq \alpha(x_{n(k)},u)\sigma(Tx_{n(k)}, Tu)\leq \psi(M(x_{n(k)},u)) \end{equation} where \begin{equation}\label{10aks} \begin{split} &M(x_{n(k)},u)\\ &=\max\{\sigma(x_{n(k)},u),\sigma(x_{n(k)},Tx_{n(k)}),\sigma(u,Tu), \frac{\sigma(x_{n(k)},Tu)+\sigma(u,Tx_{n(k)})}{4}\}\\ &=\max\{\sigma(x_{n(k)},u),\sigma(x_{n(k)},x_{n(k)+1}),\sigma(u,Tu), \frac{\sigma(x_{n(k)},Tu)+\sigma(u,x_{n(k)+1})}{4}\}. \end{split} \end{equation} By \eqref{e7EKC1} and \eqref{b10}, we have \begin{equation} \lim_{k\to \infty} M (x_{n(k)},u)=\sigma(u,Tu). \label{b11} \end{equation} Letting $k\to \infty$ in \eqref{h1}) \begin{equation}\label{e11} \sigma(u,Tu) \leq \psi(\sigma(u,Tu)) <\sigma(u,Tu), \end{equation} which is a contradiction. Hence, we obtain that $u$ is a fixed point of $T$, that is, $Tu=u$. \end{proof} \begin{definition} \rm Let $(X,\sigma)$ be a metric-like space and $T: X\to X$ be a given mapping. We say that $T$ is a generalized $\alpha-\psi$ contractive mapping of type B if there exist two functions $\alpha: X\times X \to [0, \infty)$ and $\psi\in\Psi$ such that \begin{equation} \label{e1gN} \alpha(x,y)\sigma(Tx,Ty)\leq\psi(M_0(x,y)), \text{ for all } \,x,y\in X, \end{equation} where \begin{equation} \label{Nxy} M_0(x,y)=\max\{\sigma(x,y),\sigma(x,Tx),\sigma(y,Ty)\}. \end{equation} \end{definition} \begin{theorem}\label{thm2c} Let $(X, d)$ be a complete metric-like space and $T: X\to X$ be a generalized $\alpha-\psi$ contractive mapping of type B. Suppose that \begin{itemize} \item[(i)] $T$ is $\alpha$-admissible; \item[(ii)] there exists $x_{0}\in X$ such that $\alpha(x_{0},Tx_{0})\geq 1$; \item[(iii)] $T$ is continuous. \end{itemize} Then there exists a $u\in X$ such that $\sigma(u,u)=0$. If in addition {\rm (H1)} holds, then such $u$ is a fixed point of $T$, that is, $Tu=u$. \end{theorem} \begin{proof} Along the lines of the proof of Theorem \ref{thm1}, we get the desired result. Because of the analogy, we skip the details of the proof. \end{proof} \begin{theorem}\label{thm2d} Let $(X, d)$ be a complete metric-like space and $T: X\to X$ be a generalized $\alpha-\psi$ contractive mapping of type B. Suppose that \begin{itemize} \item[(i)] $T$ is $\alpha$-admissible; \item[(ii)] there exists $x_{0}\in X$ such that $\alpha(x_{0},Tx_{0})\geq 1$; \item[(iii)] if $\{x_n\}$ is a sequence in $X$ such that $\alpha(x_n,x_{n+1})\geq 1$ for all $n$ and $x_n \to x\in X$ as $n\to \infty$, then there exists a subsequence $\{x_{n(k)}\}$ of $\{x_n\}$ such that $\alpha(x_{n(k)},x)\geq 1$ for all $k$. \end{itemize} Then, there exists $u\in X$ such that $Tu=u$. \end{theorem} We omit the proof because of the similarity to Theorem \ref{thm2}. \begin{definition} \rm Let $(X,\sigma)$ be a metric-like space and $T: X\to X$ be a given mapping. We say that $T$ is a $\alpha-\psi$ contractive mapping if there exist two functions $\alpha: X\times X \to [0, \infty)$ and $\psi\in\Psi$ such that \begin{equation} \label{e1gNg} \alpha(x,y)\sigma(Tx,Ty)\leq\psi(\sigma(x,y)), \quad \text{for all } x,y\in X, \end{equation} \end{definition} \begin{theorem}\label{thm2e} Let $(X, d)$ be a complete metric-like space and $T: X\to X$ be a $\alpha-\psi$ contractive mapping. Suppose that \begin{itemize} \item[(i)] $T$ is $\alpha$-admissible; \item[(ii)] there exists $x_{0}\in X$ such that $\alpha(x_{0},Tx_{0})\geq 1$; \item[(iii)] $T$ is continuous. \end{itemize} Then there exists a $u\in X$ such that $\sigma(u,u)=0$. If in addition {\rm (H1)} holds, then such $u$ is a fixed point of $T$, that is, $Tu=u$. \end{theorem} The above theorem is a simple consequence of Theorem \ref{thm1}. \begin{theorem}\label{thm2f} Let $(X, d)$ be a complete metric-like space and $T: X\to X$ be a $\alpha-\psi$ contractive mapping. Suppose that \begin{itemize} \item[(i)] $T$ is $\alpha$-admissible; \item[(ii)] there exists $x_{0}\in X$ such that $\alpha(x_{0},Tx_{0})\geq 1$; \item[(iii)] if $\{x_n\}$ is a sequence in $X$ such that $\alpha(x_n,x_{n+1})\geq 1$ for all $n$ and $x_n \to x\in X$ as $n\to \infty$, then there exists a subsequence $\{x_{n(k)}\}$ of $\{x_n\}$ such that $\alpha(x_{n(k)},x)\geq 1$ for all $k$. \end{itemize} Then, there exists $u\in X$ such that $Tu=u$. \end{theorem} The above theorem follows from Theorem \ref{thm2}. \section{Consequences of the main results} In the following, we present some illustrated consequences of our obtained results given by Theorem \ref{thm1} and Theorem \ref{thm2}. \subsection{Standard fixed point results in metric-like spaces} \label{con1} \begin{corollary} \label{Cor1} Let $(X,\sigma)$ be a complete metric-like space and $T: X\to X$ be such that \[ \sigma(Tx,Ty)\leq \psi(M(x,y))\quad\text{for all }x,y\in X \] where $M(x,y)$ is defined by \eqref{Mxy}. Then, $T$ has a fixed point. \end{corollary} To prove the above corollary it suffices to take $\alpha(x,y)=1$ in Theorem \ref{thm2}. \begin{corollary}\label{Cor2} Let $(X,\sigma)$ be a complete metric-like space and $T: X\to X$ be such that \[ \sigma(Tx,Ty)\leq \lambda\,M(x,y)\quad\text{for all}\,\,x,y\in X \] where $\lambda\in [0,1)$. Then, $T$ has a fixed point. \end{corollary} \begin{proof} To prove the above corollary it suffices to take $\psi(t)=\lambda t$ in Corollary \ref{Cor1}. \end{proof} \begin{corollary} \label{Cor1b} Let $(X,\sigma)$ be a complete metric-like space and $T: X\to X$ be such that \[ \sigma(Tx,Ty)\leq \psi(M_0(x,y))\quad\text{for all}\,\,x,y\in X \] where $M_0(x,y)$ is defined by \eqref{Nxy}. Then, $T$ has a fixed point. \end{corollary} To prove the above corollary it suffices to take $\alpha(x,y)=1$ in Theorem \ref{thm2c}. \begin{corollary}\label{Cor2b} Let $(X,\sigma)$ be a complete metric-like space and $T: X\to X$ be such that \[ \sigma(Tx,Ty)\leq \lambda\,M_0(x,y)\quad\text{for all}\,\,x,y\in X \] where $\lambda\in [0,1)$. Then, $T$ has a fixed point. \end{corollary} To prove the above corollary it suffices to take $\psi(t)=\lambda t$ in Corollary \ref{Cor1b}. \begin{corollary}\label{Cor1c} Let $(X,\sigma)$ be a complete metric-like space and $T: X\to X$ be such that \[ \sigma(Tx,Ty)\leq \psi(\sigma(x,y))\quad\text{for all}\,\,x,y\in X. \] Then, $T$ has a fixed point. \end{corollary} To prove the above corollary it suffices to take $\alpha(x,y)=1$ in Theorem \ref{thm2f}. \begin{corollary}\label{Cor2c} Let $(X,\sigma)$ be a complete metric-like space and $T: X\to X$ be such that \[ \sigma(Tx,Ty)\leq \lambda\,\sigma(x,y)\quad\text{for all}\,\,x,y\in X \] where $\lambda\in [0,1)$. Then, $T$ has a fixed point. \end{corollary} To prove the above corollary it suffices to take $\psi(t)=\lambda t$ in Corollary \ref{Cor1c}. \subsection{Standard fixed point results in partial metric spaces} \label{con2} The partial metric spaces were introduced by Matthews \cite{Ma} as a part of the study of denotational semantics of data for networks. \begin{definition}[\cite{Ma}] \label{partial}\rm A partial metric on a nonempty set $X$ is a function $p : X\times X\to [0, +\infty)$ such that for all $x, y, z \in X$: \begin{itemize} \item[(P1)] $x =y\Longleftrightarrow p(x,x)=p(x,y)= p(y,y)$, \item[(P2)] $p(x,x) \leq p(x,y)$, \item[(P3)] $p(x,y)= p(y,x)$, \item[(P4)] $p(x,y)\leq p(x,z)+ p(z,y)-p(z,z)$. \end{itemize} A partial metric space is a pair $(X,p)$ such that $X$ is a nonempty set and $p$ is a partial metric on $X$. \end{definition} If $p$ is a partial metric on $X$, then the function $d_p : X\times X \to \mathbb{R}^+_0$ given by \begin{equation} \label{ps} d_p(x,y)=2p(x,y)-p(x,x)-p(y,y), \end{equation} is a metric on $X$. \begin{lemma} \label{le-s} Let $(X,p)$ be a partial metric space. Then, (a) $\{x_{n}\}$ is a Cauchy sequence in $(X,p)$ if and only if it is a Cauchy sequence in the metric space $(X,d_p)$, (b) $X$ is complete if and only if the metric space $(X,d_p)$ is complete. \end{lemma} \begin{corollary} \label{Cor3} Let $(X,\sigma)$ be a complete partial space and $T: (X,p)\to (X,p)$ be such that \[ \alpha(x,y) p(Tx,Ty)\leq \psi(N(x,y))\quad\text{for all}\,\,x,y\in X \] where $N(x,y)$ is defined as \[ N(x,y)=\max\{p(x,y),p(x,Tx),p(y,Ty),\frac{p(x,Ty)+p(y,Tx)}{2}\}. \] Suppose that \begin{itemize} \item[(i)] $T$ is $\alpha$-admissible; \item[(ii)] there exists $x_{0}\in X$ such that $\alpha(x_{0},Tx_{0})\geq 1$; \item[(iii)] $T$ is continuous. \end{itemize} Then there exists a $u\in X$ such that $p(u,u)=0$. If in addition {\rm (H1)} holds, then such $u$ is a fixed point of $T$, that is, $Tu=u$. \end{corollary} \begin{proof} It suffices to replace the metric-like $\sigma$ in Theorem \ref{thm1} by the partial metric $p$ which itself a metric-like. Note that we considered in $N(x,y)$ the fourth term $\frac{p(x,Ty)+p(y,Tx)}{2}$ instead of $\frac{p(x,Ty)+p(y,Tx)}{4}$ due to the inequality $p(x,x)\leq p(x,y)$. Its proof is evident. \end{proof} Similar to Corollary \ref{Cor3}, from Theorem \ref{thm2} we deduce the following result. \begin{corollary} \label{Cor4} Let $(X,\sigma)$ be a complete partial space and $T: (X,p)\to (X,p)$ be such that \[ \alpha(x,y) p(Tx,Ty)\leq \psi(N(x,y))\quad\text{for all}\,\,x,y\in X. \] Suppose that \begin{itemize} \item[(i)] $T$ is $\alpha$-admissible; \item[(ii)] there exists $x_{0}\in X$ such that $\alpha(x_{0},Tx_{0})\geq 1$; \item[(iii)] if $\{x_n\}$ is a sequence in $X$ such that $\alpha(x_n,x_{n+1})\geq 1$ for all $n$ and $x_n \to x\in X$ as $n\to \infty$, then there exists a subsequence $\{x_{n(k)}\}$ of $\{x_n\}$ such that $\alpha(x_{n(k)},x)\geq 1$ for all $k$. \end{itemize} Then, there exists $u\in X$ such that $Tu=u$. \end{corollary} \begin{remark} \label{rmk3.11} \rm It is clear that one can easily state the analog of Theorem \ref{thm2c}, Theorem \ref{thm2d}, Theorem \ref{thm2e} and Theorem \ref{thm2f} in the setting of partial metric spaces. \end{remark} \subsection{Fixed point results with a partial order} \label{con3} The study of the existence of fixed points on metric spaces endowed with a partial order can be considered as one of the very interesting improvements in the field of fixed point theory. This trend was initiated by Turinici \cite{T} in 1986, but it became one of the core research subject after the publications of Ran and Reurings in \cite{RR} and Nieto and Rodr\'iguez-L\'opez \cite{NR}. \begin{definition} \rm Let $(X,\preceq)$ be a partially ordered set and $T: X\to X$ be a given mapping. We say that $T$ is nondecreasing with respect to $\preceq$ if $$ x,y\in X,\,\,x\preceq y \Longrightarrow Tx\preceq Ty. $$ \end{definition} \begin{definition} \rm Let $(X,\preceq)$ be a partially ordered set. A sequence $\{x_n\}\subset X$ is said to be nondecreasing with respect to $\preceq$ if $x_n\preceq x_{n+1}$ for all $n$. \end{definition} \begin{definition} Let $(X,\preceq)$ be a partially ordered set and $\sigma$ be a metric-like on $X$. We say that $(X,\preceq,\sigma)$ is regular if for every nondecreasing sequence $\{x_n\}\subset X$ such that $x_n\to x\in X$ as $n\to \infty$, there exists a subsequence $\{x_{n(k)}\}$ of $\{x_n\}$ such that $x_{n(k)}\preceq x$ for all $k$. \end{definition} \begin{corollary}\label{CT7} Let $(X,\preceq)$ be a partially ordered set and $\sigma$ be a metric-like on $X$ such that $(X,\sigma)$ is complete. Let $T: X\to X$ be a nondecreasing mapping with respect to $\preceq$. Suppose that there exists a function $\psi\in \Psi$ such that $$ \sigma(Tx,Ty)\leq \psi(M(x,y)), $$ for all $x,y\in X$ with $x\succeq y$. Suppose also that the following conditions hold: \begin{itemize} \item[(i)] there exists $x_0\in X$ such that $x_0\preceq Tx_0$; \item[(ii)] ($T$ is continuous and the property $(H)$ holds) or ($(X,\preceq,\sigma)$ is regular). \end{itemize} Then, $T$ has a fixed point. \end{corollary} \begin{proof} Define the mapping $\alpha: X\times X\to [0,\infty)$ by \[ \alpha(x,y)=\begin{cases} 1 &\text{if } x\preceq y \text{ or } x\succeq y,\\ 0 &\text{otherwise}. \end{cases} \] Clearly, $T$ is a generalized $\alpha-\psi$ contractive mapping of type A; that is, $$ \alpha(x,y)\sigma(Tx,Ty)\leq \psi(M(x,y)), $$ for all $x,y\in X$. From condition (i), we have $\alpha(x_0,Tx_0)\geq 1$. Moreover, for all $x,y\in X$, from the monotone property of $T$, we have $$ \alpha(x,y)\geq 1 \Longrightarrow x\succeq y \text{ or } x\preceq y \Longrightarrow Tx\succeq Ty \text{ or } Tx\preceq Ty \Longrightarrow \alpha(Tx,Ty)\geq 1. $$ Thus, $T$ is $\alpha$-admissible. Now, if $T$ is continuous and the hypothesis (H1) holds, the existence of a fixed point follows from Theorem \ref{thm1}. Suppose now that $(X,\preceq,d)$ is regular. Let $\{x_n\}$ be a sequence in $X$ such that $\alpha(x_n,x_{n+1})\geq 1$ for all $n$ and $x_n \to x\in X$ as $n\to \infty$. From the regularity hypothesis, there exists a subsequence $\{x_{n(k)}\}$ of $\{x_n\}$ such that $x_{n(k)}\preceq x$ for all $k$. This implies from the definition of $\alpha$ that $\alpha(x_{n(k)},x)\geq 1$ for all $k$. In this case, the existence of a fixed point follows from Theorem \ref{thm2}. \end{proof} \begin{remark} \label{rmk} \rm Notice that we may obtain the analog of Theorem \ref{thm2c}, Theorem \ref{thm2d}, Theorem \ref{thm2e}, Theorem \ref{thm2f} and the results of Subsection \ref{con1} and Subsection \ref{con2} in the setting of partially ordered metric-like spaces. \end{remark} \subsection{Fixed point results for cyclic contractions} Kirk, Srinivasan and Veeramani \cite{Kirk} proved very interesting generalizations of the Banach Contraction Mapping Principle by introducing a cyclic contraction. This remarkable paper \cite{Kirk} has been appreciated by many several researchers (see, for example, \cite{CC3, CC4, CC3P, Petric, CC2} and the related reference therein). In this subsection, we derive some fixed point theorems for cyclic contractive mappings in the setting of metric-like spaces. \begin{corollary}\label{CT14} Let $ \{A_i\}_{i=1}^2$ be nonempty closed subsets of a complete metric-like space $(X,\sigma)$ and $T:Y \to Y$ be a given mapping, where $Y=A_1\cup A_2$. Suppose that the following conditions hold: \begin{itemize} \item[(I)] $T(A_1)\subseteq A_2$ and $T(A_2)\subseteq A_1$; \item[(II)] there exists a function $\psi\in \Psi$ such that $$ \sigma(Tx,Ty)\leq \psi(M(x,y)), \text{ for all } (x,y)\in A_1\times A_2. $$ \end{itemize} Then $T$ has a fixed point that belongs to $A_1\cap A_2$. \end{corollary} \begin{proof} Since $A_1$ and $A_2$ are closed subsets of the complete metric-like space $(X,\sigma)$, then $(Y,\sigma)$ is complete. Define the mapping $\alpha: Y\times Y\to [0,\infty)$ by \[ \alpha(x,y)=\begin{cases} 1 &\text{if } (x,y)\in (A_1\times A_2)\cup (A_2\times A_1),\\ 0 &\text{otherwise.} \end{cases} \] From (II) and the definition of $\alpha$, we can write $$ \alpha(x,y)d(Tx,Ty)\leq \psi(M(x,y)), $$ for all $x,y\in Y$. Thus $T$ is a generalized $\alpha-\psi$ contractive mapping of type A. Let $(x,y) \in Y\times Y$ such that $\alpha(x,y)\geq 1$. If $(x,y)\in A_1\times A_2$, from (I), $(Tx,Ty)\in A_2\times A_1$, which implies that $\alpha(Tx,Ty)\geq 1$. If $(x,y)\in A_2\times A_1$, from (I), $(Tx,Ty)\in A_1\times A_2$, which implies that $\alpha(Tx,Ty)\geq 1$. Hence, in all cases, we conclude that $\alpha(Tx,Ty)\geq 1$ which yields that that $T$ is $\alpha$-admissible. Notice also that, from (I), for any $a\in A_1$, we have $(a,Ta)\in A_1\times A_2$, which implies that $\alpha(a,Ta)\geq 1$. Now, let $\{x_n\}$ be a sequence in $X$ such that $\alpha(x_n,x_{n+1})\geq 1$ for all $n$ and $x_n \to x\in X$ as $n\to \infty$. This implies from the definition of $\alpha$ that $$ (x_n,x_{n+1})\in (A_1\times A_2)\cup (A_2\times A_1), \quad\text{for all } n. $$ Since $(A_1\times A_2)\cup (A_2\times A_1)$ is a closed set with respect to the metric-like $\sigma$, we get that $$ (x,x)\in (A_1\times A_2)\cup (A_2\times A_1), $$ which implies that $x\in A_1\cap A_2$. Thus we get immediately from the definition of $\alpha$ that $\alpha(x_n,x)\geq 1$ for all $n$. Now, all the hypotheses of Theorem \ref{thm2} are satisfied and $T$ has a fixed point in $Y$.\bigskip \end{proof} Note that Corollary \ref{CT14} is a generalization of \cite[Corollary 1.10]{Erdal1}. \section{Examples} We present the following two concrete examples to support our results. \begin{example} \label{examp4.1} \rm Consider $X=\{0,1,2\}$. Take the metric-like $\sigma:X\times X\to \mathbb{R}^+_0$ defined by \begin{gather*} \sigma(0,0)=\sigma(1,1)=0,\quad \sigma(2,2)=\frac{9}{20}, \\ \sigma(0,2)=\sigma(2,0)=\frac{2}{5},\quad \sigma(1,2)=\sigma(2,1)=\frac{3}{5},\\ \sigma(0,1)=\sigma(1,0)=\frac{1}{2}. \end{gather*} Note that $\sigma(2,2)\neq 0$, so $\sigma$ is not a metric and $\sigma(2,2)>\sigma(0,2)$, so $\sigma$ is not a partial metric. Clearly, $(X,\sigma)$ is a complete metric-like space. Given $T:X\to X$ as $T0=T1=0$ and $T2=1$. Take $\psi(t)=5t/6$ for each $t\geq 0$. Define the mapping $\alpha: X\times X\to [0,\infty)$ by \[ \alpha(x,y)=\begin{cases} 1 & \text{if } x=0,\\ 0 & \text{otherwise}. \end{cases} \] First, let $x,y\in X$ such that $\alpha(x,y)\geq 1$. By the definition of $\alpha$, this implies that $x=0$ and since $T0=0$, so $\alpha(Tx,Ty)=1$ for each $y\in X$, that is, $T$ is $\alpha$-admissible. We distinguish two cases: \noindent Case 1: If ($x=0$ and $y=0$) or ($x=0$ and $y=1$), we have \[ \alpha(Tx,Ty)\sigma(Tx,Ty)=\sigma(Tx,Ty)=0. \] Case 2: If $x=0$ and $y=2$, we have \begin{align*} \alpha(Tx,Ty)\sigma(Tx,Ty)&=\sigma(Tx,Ty)=\sigma(0,1)=\frac{1}{2}=\frac{5}{6}\sigma(2,1)\\ &=\psi(\sigma(y,Ty))\\ &\leq \psi(M(x,y)) \end{align*} where $M(x,y)$ is defined by \eqref{Mxy}. It is also obvious that hypothesis (iii) of Theorem \ref{thm2} is satisfied. Thus, we map apply Theorem \ref{thm2} and so $T$ has a fixed point, which is $u=0$. \end{example} \begin{example} \label{examp4.2} \rm Let $X=[0,\infty)$ be endowed with the metric-like $\sigma$ given as $\sigma(x,y)=\max\{x,y\}$. Define the mapping $T:X\to X$ by \begin{equation*} Tx=\begin{cases} \frac{1}{2} x^2 & \text{if }x\in [0,1], \\ 3x-1 & \text{otherwise}. \end{cases} \end{equation*} Consider $\psi:[0,\infty)\to [0,\infty)$ defined by \begin{equation*} \psi(t)=\begin{cases} \frac{1}{2} t^2 & \text{if } 0\leq t<1, \\ \frac{1}{2} & \text{otherwise}. \end{cases} \end{equation*} Obviously, $\psi\in \Psi$. Consider $\alpha: X\times X\to [0,\infty)$ as \[ \alpha(x,y)=\begin{cases} 1 &\text{if } x,y\in [0,1],\\ 0 &\text{otherwise}. \end{cases} \] First, let $x,y\in X$ such that $\alpha(x,y)\geq 1$, so $x,y\in [0,1]$. In this case, \[ \alpha(Tx,Ty)=\alpha(\frac{1}{2} x^2,\frac{1}{2} y^2)=1; \] that is, $T$ is $\alpha$-admissible. Here we also have \begin{align*} \alpha(Tx,Ty)\sigma(Tx,Ty) &=\sigma(Tx,Ty)=\sigma(\frac{1}{2} x^2,\frac{1}{2} y^2)\\ &=\sigma(\psi(x),\psi(y))=\max(\psi(x),\psi(y))\\ &=\psi(\max(x,y))=\psi(\sigma(x,y))\\ &\leq \psi(M(x,y)). \end{align*} Note that hypothesis (iii) of Theorem \ref{thm2} is also satisfied. Applying Theorem \ref{thm2}, $T$ has a fixed point in $X$, which is $u=0$. \end{example} \section{Applications} Here, we consider the following two-point boundary-value problem for the second-order differential equation \begin{equation} \label{ap1} \begin{gathered} -\frac{d^2x}{dt^2}=f(t,x(t)),\quad t\in [0,1]\\ x(0)=x(1)=0, \end{gathered} \end{equation} where $f:[0,1]\times \mathbb{R}\to \mathbb{R}$ is a continuous function. Recall that the Green's function associated to \eqref{ap1} is \begin{equation}\label{ap2} G(t,s)= \begin{cases} t(1-s) &\text{if } 0\leq t\leq s\leq1\\ s(1-t) &\text{if } 0\leq s\leq t\leq1. \end{cases} \end{equation} Let $X=\mathcal{C}(I) (I = [0, 1])$ be the space of all continuous functions defined on $I$. We consider on $X$, the metric-like $\sigma$ given by \[ \sigma(x,y)=\|x-y\|_{\infty}+\|x\|_{\infty}+\|y\|_{\infty}\quad\text{for all } x,y\in X, \] where $\|u\|_{\infty}=\max_{t\in[0,1]} |u(t)|$ for each $u\in X$. Note that $\sigma$ is also a partial metric on $X$ and since \[ d_{\sigma} (x,y):=2\sigma(x,y)-\sigma(x,x)-\sigma(y,y)=2\|x-y\|_{\infty}, \] so by Lemma \ref{le-s}, $(X,\sigma)$ is complete since the metric space $(X,\|\cdot\|_\infty)$ is complete. It is well known that $x\in C^2(I)$ is a solution of \eqref{ap1} is equivalent to that $x\in X=C(I)$ is a solution of the integral equation \begin{equation}\label{ap4} x(t) =\int_0^ 1 G(t, s) f (s, x(s)) ds, \quad\text{for all } t \in I. \end{equation} \begin{theorem}\label{t-ap} Suppose the following conditions hold:\\ \begin{itemize} \item there exists a continuous function $p: I\to \mathbb{R}^+_0$ such that \[ |f (s, a)-f (s, b)|\leq 8\, p(s)\,|a-b|, \] for each $s\in I$ and $a,b\in \mathbb{R}$; \item there exists a continuous function $q: I\to \mathbb{R}^+_0$ such that \[ |f (s, a)|\leq 8\, q(s)\,|a|, \] for each $s\in I$ and $a\in \mathbb{R}$; \item $\sup_{s\in I} p(s)=\lambda_1<\frac{1}{3}$; \item $\sup_{s\in I} q(s)=\lambda_2<\frac{1}{3}$. \end{itemize} Then problem \ref{ap1} has a solution $u\in X=\mathcal{C}(I,\mathbb{R})$. \end{theorem} \begin{proof} Consider the mapping $T:X\to X$ defined by \[ Tx(t)=\int_0^ 1 G(t, s) f (s, x(s)) ds. \] for all $x\in X$ and $t\in I$. Then, problem \eqref{ap1} is equivalent to finding $u\in X$ that is a fixed point of $T$. Now, let $x,y\in X$. We have \begin{align*} |Tx(t)-Ty(t)| &=|\int_0^ 1 G(t, s) f (s, x(s)) ds-\int_0^ 1 G(t, s) f (s, y(s)) ds|\\ &\leq \int_0^ 1 G(t, s) |f (s, x(s))-f (s, y(s))|\, ds\\ &\leq 8\int_0^ 1 G(t, s) p(s)\,|x(s)-y(s)|\, ds\\ &\leq 8\lambda_1 \|x-y\|_{\infty} \sup_{t\in I}\int_0^ 1 G(t, s)\,ds\\ &= \lambda_1 \|x-y\|_{\infty}. \end{align*} In the above equality, we used that for each $t\in I$, we have $\int_0^1 G(t,s)\,ds=-\frac{t^2}{2}+\frac{t}{2}$, and so $\sup_{t\in I}\int_0^ 1 G(t, s)\,ds=\frac{1}{8}$. Therefore, \begin{equation} \label{ap5} \|Tx-Ty\|_{\infty}\leq \lambda_1 \|x-y\|_{\infty}. \end{equation} Again, we have \begin{align*} |Tx(t)|&=|\int_0^ 1 G(t, s) f (s, x(s)) ds|\\ &\leq \int_0^ 1 G(t, s)\,|f (s, x(s))|\, ds\\ &\leq 8\int_0^ 1 G(t, s)\, q(s)\,|x(s)|\, ds\\ &\leq 8\lambda_2 \|x\|_{\infty} \sup_{t\in I}\int_0^ 1 G(t, s)\,ds\\ &\leq \lambda_2 \|x\|_{\infty}. \end{align*} Thus \begin{equation} \label{ap6} \|Tx\|_{\infty}\leq \lambda_2 \|x\|_{\infty}. \end{equation} Proceeding similarly, \begin{equation}\label{ap7} \|Ty\|_{\infty}\leq \lambda_2 \|y\|_{\infty}. \end{equation} Take $\lambda=\lambda_1+2\lambda_2$. Under assumptions in Theorem \ref{t-ap}, we have $\lambda<1$. Summing \eqref{ap5} to \eqref{ap7}, we find \begin{align*} \sigma(Tx,Ty) &=\|Tx-Ty\|_{\infty}+\|Tx\|_{\infty}+\|Ty\|_{\infty}\\ &\leq \lambda_1 \|x-y\|_{\infty}+\lambda_2 \|x\|_{\infty} +\lambda_2 \|y\|_{\infty}\\ &\leq (\lambda_1+2\lambda_2)(\|x-y\|_{\infty}+\|x\|_{\infty}+\|y\|_{\infty})\\ &= \lambda \sigma(x,y)\leq \lambda M(x,y). \end{align*} So all hypotheses of Corollary \ref{Cor2} are satisfied, and so $T$ has a fixed point $u\in X$, that is, the problem \eqref{ap1} has a solution $u\in C^2(I)$. \end{proof} \subsection*{Conclusion} All fixed point results presented in this article are also valid for metric spaces. Consequently, our results extend and unify several results from the literature. \begin{thebibliography}{99} \bibitem{Aage} C. T. Aage, J. N. Salunke; \emph{Some results of fixed point theorem in dislocated quasi metric space}. Bull. Marathadawa Math. Soc. 9 (2008),1--5 . \bibitem{Aage2} C. T. Aage, J. N. 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