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\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:
\begin{equation} \label{e1.1} 
x'(t)=f(t, x(t)), \quad   x(t_0)=x_0 
\end{equation}
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
\begin{equation} \label{e1.2}
t'(x)=g(x, t(x)), \quad
         t(x_0)=t_0  
\end{equation}
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 $0<\epsilon<a$;

\item[(iii)]  there is a number $\alpha $  such that for all $(t,x)$  
and $(s,x)$  in $U$, 
\[
|f(t,x)- f(s,x)| \leq\alpha|t-s|.
\]
\end{itemize}
Then there is a unique solution to the initial value problem \eqref{e1.1}  
in some interval  $(t_0-\nu, t_0+\eta)$. 
\end{theorem}

\begin{proof}
   Let $x$  be a solution to \eqref{e1.1} where $f$ satisfies the conditions 
of the theorem.  Define the closed set $D=\{t\in[t_0,t_0+a]:x'(t)=0\}$.  
Suppose, for contradiction, that $x$  is not strictly monotone in any interval 
$[t_0,t_0+\varepsilon]$.   Then $D$  is infinite and $t_0 \in D$.  
Since $D$  is closed, $\sup D\equiv t_1\in D$.  
The set $(t_0,t_1)-D$   is open, and, by (ii), non-empty.
Therefore, there is an interval $(u,v) \subseteq (t_0,t_1)-D$  with $u$  
and $v$ both in $D$.
Thus $x'(u) =x'(v)=0$.  Then by condition (i), $x(u)=x(v)=x_0$  and it follows
by Rolle's Theorem that there is a $\xi \in (u,v)$   such that $x'(\xi)=0 $.
But this leads to the contradiction that $\xi \in D \cap (u,v) = \emptyset $.
Thus every solution of \eqref{e1.1} is strictly monotone 
(and therefore invertible) on some interval $[t_0,t_0+\delta_1)$.   
If $t(x)$  is the inverse of an increasing solution of \eqref{e1.1} 
then $t(x)$ satisfies
\begin{equation} \label{e2.1}
t'(x)= \frac{1}{f(t,x)}, \quad  t(x_0)=t_0  
\end{equation}
for $x>x_0$.

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
\begin{equation} \label{e3.1}
x'(t)=g(t) + h(t)|x(t)|^r, \quad x(0)= 0  
\end{equation}
where $g$ and $h$  are non-negative Lipschitz continuous functions and $0<r<1$.
The theorem given here can be applied to show that \eqref{e3.1} has a 
unique solution in a neighborhood of 0  provided $h(0) \neq 0$  and $0<r<1/2$.
However, any theorem which relies only on the difference $f(t,x)-f(t,y)$ 
--such as those mentioned in the introduction--  would evidently not apply 
to \eqref{e3.1}.  For if such a theorem did apply to \eqref{e3.1}  
it would also have to apply to the example  $x'(t)=h(t)|x(t)|^r$, 
$x(0)=0 $, since $f(t,x)-f(t,y)$ is the same in this example as in \eqref{e3.1}. 
But this example has the two solutions
$x(t) = [((1-r) \int_ 0^t h(s)ds)]^{1/(1-r)}$ and  $x(t)\equiv 0$ .


Example \eqref{e3.1} is a generalization of an example which appears in [2], 
where a theorem is given which also does not rely on the difference 
$f(t,x)-f(t,y)$.  But the theorem in [2] does not apply to \eqref{e3.1} 
unless $g(0)\neq 0$.


\subsection*{Remarks}

Conditions (i) and (ii) in Theorem \ref{mainthm} replace the stronger 
condition that $f(t_0,x_0)\neq 0 $  as discussed in the Introduction.  
Neither of these conditions can be dropped.  Consider these two choices for $f$: 
\begin{itemize}
\item[(a)] $f(t,x) = x^{3/5} + \frac{1}{100} t^{3/2}$  

\item[(b)] $f(t,x) = x^{1/3}$. 
\end{itemize}
With $t_0 = x_0 = 0$, both of these functions satisfy  the conditions 
of  Theorem \ref{mainthm},  except that example (a)  does not satisfy 
condition (i),  and example (b)  does not satisfy condition (ii).  
The non-uniqueness of solutions of the corresponding initial value problem 
\eqref{e1.1} is shown below.
\begin{itemize}
\item[(a)]  $x(t) = kt^{5/2}$  is a solution where $k$ is any of the three 
real numbers which satisfy the equation  
$(\frac{5k}{2} - \frac{1}{100})^5 = k^3$.

\item[(b)]  $x(t)\equiv 0$  and $x(t) = (\frac{2t}{3})^{3/2}$  
   are both solutions.
\end{itemize}


\begin{thebibliography}{0}

\bibitem{AVL1}  R. P. Agarwal, V. Lakshmikantham;
\emph{Uniqueness and nonuniqueness criteria  for ordinary differential equations},
 World Scientific Publishing,  1993.

\bibitem{JMB} J. M. Bownds;
\emph{A uniqueness theorem for $y'(x) = f(x,y)$} using a certain factorization 
of $f$, Journal of Differential Equations, 
Vol 7 (1970), 227-231

\bibitem{MN1} M. Nagumo;
\emph{Eine hinreichende bedingung f\"ur die unit\"at der l\"osung von
     differentialgleichungen erster ordnung}, 
Japan J. Math. 3 (1926), 107-112.

 \bibitem{OG1}  W. Osgood;
\emph{Beweis der existenz einer l\"osung der differentialgle ichung 
$dy/dx = f(x,y)$ ohne hinzunahme der Cauchy-Lipshitzchen bedingung}, 
Monatah. Math.  Phys., Vol 9 (1898), 331-345.

\bibitem{OP1} O. Perron;
\emph{Eine hinreichende bedingung f\"ur die unit\"at der l\"osung von
      Differentialgleichungen erster ordnung}, Math. Z., Vol 28 (1928), 216-219.

\end{thebibliography}

\end{document}