\documentclass[reqno]{amsart} \usepackage{hyperref} \usepackage{amssymb} \AtBeginDocument{{\noindent\small \emph{Electronic Journal of Differential Equations}, Vol. 2013 (2013), No. 124, pp. 1--3.\newline ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu \newline ftp ejde.math.txstate.edu} \thanks{\copyright 2013 Texas State University - San Marcos.} \vspace{9mm}} \begin{document} \title[\hfilneg EJDE-2013/124\hfil Existence and uniqueness of a local solution] {Existence and uniqueness of a local solution for $x' = f(t,x)$ using inverse functions} \author[J. T. Hoag \hfil EJDE-2013/124\hfilneg] {Jeffrey T. Hoag} % in alphabetical order \address{Jeffrey T. Hoag \newline Mathematics Department, Providence College, Providence, RI 02918, USA} \email{jhoag@providence.edu} \thanks{Submitted January 13, 2013. Published May 20, 2013.} \subjclass[2000]{34A12} \keywords{Existence; uniqueness; ordinary differential equation} \begin{abstract} A condition on the function $f$ is given such that the scalar ordinary differential equation $x' = f(t,x)$ with initial condition $x(t_0) = x_0$ has a unique solution in a neighborhood of $t_0$. An example illustrates that this result can be used when other theorems which put conditions on the difference $f(t,x)-f(t,y)$ do not apply. \end{abstract} \maketitle \numberwithin{equation}{section} \newtheorem{theorem}{Theorem}[section] \allowdisplaybreaks \section{Introduction} Consider the differential equation with initial condition: $$\label{e1.1} x'(t)=f(t, x(t)), \quad x(t_0)=x_0$$ where $f$ is a scalar-valued function which is continuous in a neighborhood $N$ of $(t_0,x_0)$. The continuity of $f$ guarantees that there is at least one solution to this initial value problem. There are various other conditions that can be imposed on $f$ which will ensure that \eqref{e1.1} has a unique solution. Over twenty such uniqueness conditions are collected in [1]. Most of these, including results by Nagumo [3], Osgood [4] and Perron [5], rely on restrictions on $f(t,x)-f(t,y)$ and can be considered generalizations of the Lipschitz condition in the second argument. In this article, a uniqueness theorem for \eqref{e1.1} is given which instead puts the Lipschitz condition on the first argument of $f$. That is, the condition is on the difference $f(t,x)-f(s,x)$ for $(t,x)$ and $(s,x)$ in $N$. It is easy to see that this is possible when $f(t_0,x_0)\neq 0$ because in this case a solution of \eqref{e1.1} is invertible in a neighborhood of $(t_0,x_0)$ and so if $t(x)$ is the inverse of a solution to \eqref{e1.1}, it satisfies $$\label{e1.2} t'(x)=g(x, t(x)), \quad t(x_0)=t_0$$ where we define $g(x,t) = 1/f(t,x)$. If $f$ is Lipschitz in its first argument in a neighborhood $N$ of $(t_0,x_0)$ then there is a neighborhood $M$ of $(x_0,t_0)$ where $g$ is Lipschitz in its second argument. From this it follows that \eqref{e1.2} has a unique solution in a neighborhood of $(x_0,t_0)$ and therefore \eqref{e1.1} has a unique solution in a neighborhood of $(t_0,x_0)$. The theorem that follows extends this approach to include cases when $f(t_0,x_0)=0$. It will be followed by an example for which this theorem applies but other uniqueness theorems do not. \section{Main result} \begin{theorem} \label{mainthm} For $(t_0,x_0)\in\mathbb{R}^2$ and positive numbers $a$ and $b$, define $U = [t_0 - a,t_0+a]\times[x_0-b,x_0+b].$ Let $f:U\to\mathbb{R}$ be a continuous function satisfying the following three conditions: \begin{itemize} \item[(i)] there are constants $c>0$ and $r\in(0,1/2)$ such that $|f(t,x)|\geq c|x-x_0|^r\quad\text{for all }(t,x)\in U;$ \item[(ii)] $f(t,x_0)$ is not identically zero on any interval $(t_0-\varepsilon, t_0+\varepsilon)$ for $0<\epsilonx_0$. Now let $x$ and $\tilde x$ be any two increasing solutions of \eqref{e1.1} with inverses $t$ and $\tilde t$. Since $t$ and $\tilde t$ are both solutions to \eqref{e2.1}, $$|t(x) -\tilde t(x)| \le |t(y) -\tilde t(y)| + \int_y^x \frac{|f(t(s),s) -f(\tilde t(s), s)|} {|f(t(s),s)|\,|f(\tilde t(s), s)|} ds$$ for $x \geqslant y > x_0$. Then, using conditions (i) and (iii), $$|t(x) -\tilde t(x)| \le |t(y) -\tilde t(y)| + \frac{\alpha}{c^2} \int_y^x\frac{|t(s) -\tilde t(s)| }{ |s-x_0|^{2r}} ds\,.$$ Applying the Gronwall-Reid Lemma to this inequality yields $$|t(x) -\tilde t(x)| \le |t(y) -\tilde t(y)| \exp\Big\{ \frac{\alpha}{c^2} \int_y^x\frac{1 }{ |s-x_0 |^{2r}} ds\Big\}.$$ Now take the limit as $y\to x_0 +$. Since $2r < 1$, the improper integral converges. Also $|t(y) -\tilde t(y)|\to |t(x_0) -\tilde t(x_0)|=0$. Therefore, $t(x) =\tilde t(x)$ in some interval $[x_0,x(t_0+\delta_1)]$ and so $x(t) = \tilde x(t)$ for $t\in [t_0,t_0 + \delta_1)$. Thus there is at most one increasing solution to \eqref{e1.1} on an interval $[t_0,t_0 + \delta_1)$. A similar arguments shows that there is at most one decreasing solution to \eqref{e1.1} on an interval $[t_0,t_0 + \delta_2)$. Since it is well-known that \eqref{e1.1} has either one solution or infinitely many solutions, and since every solution of \eqref{e1.1} is monotone, it follows that \eqref{e1.1} has a unique solution on some interval $[t_0,t_0 + \eta)$. A similar argument shows that there is also a unique solution on some interval $(t_0-\nu,t_0]$ . \end{proof} \subsection*{Examples} Consider the initial-value problem $$\label{e3.1} x'(t)=g(t) + h(t)|x(t)|^r, \quad x(0)= 0$$ where $g$ and $h$ are non-negative Lipschitz continuous functions and \$0