\documentclass[reqno]{amsart}
\usepackage{hyperref}

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2010(2010), No. 172, pp. 1--9.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
\newline ftp ejde.math.txstate.edu}
\thanks{\copyright 2010 Texas State University - San Marcos.}
\vspace{9mm}}

\begin{document}
\title[\hfilneg EJDE-2010/172\hfil Uniqueness for differential systems]
{Uniqueness for n-th order differential systems with strong
 singularities}

\author[Y. Pan, M. Wang\hfil EJDE-2010/172\hfilneg]
{Yifei Pan, Mei Wang}  % in alphabetical order

\address{Yifei Pan \newline
Department of Mathematical Sciences \\
Indiana University - Purdue University Fort Wayne \\
Fort Wayne, IN 46805-1499, USA.\newline
School of Mathematics and Informatics \\
Jiangxi Normal University, Nanchang, China}
\email{pan@ipfw.edu}

\address{Mei Wang \newline
Department of Statistics\\
University of Chicago\\
Chicago, IL 60637, USA}
\email{meiwang@galton.uchicago.edu}

\thanks{Submitted July 26, 2010. Published December 6, 2010.}
\subjclass[2000]{34A12, 65L05}
\keywords{Unique continuation; uniqueness; Carath\'eodory theorem;
\hfill\break\indent  Gronwall inequality}

\begin{abstract}
 Using a Lipschitz type condition, we obtain the uniqueness
 of solutions for a system of n-th order nonlinear ordinary
 differential  equations where the coefficients are allowed
 to have singularities.
\end{abstract}

\maketitle
\numberwithin{equation}{section}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{corollary}[theorem]{Corollary}

\section{Introduction and main results}

 Lipschitz condition was a key part in proving classical results
on the existence and uniqueness of ordinary differential equations, as
extensively surveyed and summarized in \cite{AL}.  In this paper, we use a
Lipschitz type condition to obtain the uniqueness of solutions of n-th
order nonlinear ordinary differential systems where the coefficients are
allowed to have singularities.

 Our results are in the spirit of the
Carath\'eodory theorem on the existence of ordinary differential
equations \cite[Chapter 2]{CL},  which gives a Lipschitz condition in
first order differential equations.  The conditions in this paper are
in terms of absolute continuity, thus the uniqueness result is on
solutions in weaker or more general sense.  For first order
differential equations, Nagumo's Theorem \cite{Nagumo} and its
generalizations \cite{Athanassov90, Constantin10} give precise
coefficients and sharp order for an isolated singularity in the
Lipschitz condition.  A natural generalization of the classical
Carath\'eodory condition is to higher order linear and nonlinear
differential systems (e.g. \cite{Bartusek97, Nth2004}).  Our
results are on higher order differential equations with coefficients
of singularities under integrability conditions.

  The main results are stated below.  The proofs are
provided in the next section.  In the last section, we provide two
applicable forms of the main theorem as corollaries, and we give an example
to illustrate the sharpness of the singularity order allowed in the main
condition of the Lipschitz type in an $n$ order differential equation.

  Let $L^1(a,b)$ denote the set of real Lebesgue
integrable functions on the interval $(a,b)$, $\|\cdot\|$ the Euclidean
norm in $\mathbb{R}^d$, and $\mathcal{C}^{k}(a,b)$ the set of
$d$-dimensional functions with $k$-th continuously differentiable
components on $(a,b)$.

 The following theorem is on the uniqueness of solutions
of differential systems.

\begin{theorem} \label{f3-thm-AC}
Consider the
system of differential equations of $y: (a,b)\to \mathbb{R}^d$,
%
\begin{equation} \label{f3-odesys}
\begin{gathered}
y^{(n)}= f(x,y,y',\dots,y^{(n-1)}), \quad x\in(a,b),\; a<0<b, \\
y(0) = a_0\\
y'(0) = a_1\\
\dots \\
y^{(n-1)}(0) = a_{n-1}\\
\end{gathered}
\end{equation}
where $f: (a,b)\times \mathbb{R}^{nd}\to \mathbb{R}^d$ satisfies the
Lipschitz type condition
%
\begin{equation} \label{f3-lip}
\|f(x, s_0, \dots, s_{n-1})
- f(x, r_0,\dots, r_{n-1}) \|
\le \sum_{k=0}^{n-1} \lambda_k(x)
 \frac{\|s_k - r_k\|}{|x|^{n-k-1}},  \quad\text{a. e. }x\in(a,b)
\end{equation}
%
for $\lambda_k(x)\ge 0$, $\lambda_k\in L^1(a,b)$ and $(s_0,\dots,
s_{n-1})$, $(r_0,\dots, r_{n-1})$ in a domain in $\mathbb{R}^{nd}$
containing $(a_0,\dots,a_{n-1})$.  If $u,v\in\mathcal{C}^{n-1}(a,b)$ are
solutions of \eqref{f3-odesys} and $u^{(n-1)}, v^{(n-1)}$ are
absolutely continuous with respect to the Lebesgue measure on $(a,b)$,
then
\[
v(x) \equiv u(x) \quad for \quad x\in(a,b).
\]
\end{theorem}
%

 The following theorem gives the condition for a
function that satisfies a differential inequality to be identically
zero, which can be considered as an $n$-th order Gronwall
\cite{Gronwall} uniqueness theorem.


\begin{theorem} \label{f3-thm-global}
 Let $y:(a,b)\to \mathbb{R}^d$,
  $y\in\mathcal{C}^{n-1}(a,b)$ and $y^{(n-1)}$ absolute continuous with
  respect to the Lebesgue measure on $(a,b), ~a<0<b$.  If
\[
y(0)=y^{(1)}(0)=\dots=y^{(n-1)}(0)=\vec 0
\]
and
\begin{equation}\label{f3-thm-globalinq}
\|y^{(n)}(x)\| \le
\sum_{k=0}^{n-1}\lambda_k(x)\frac{\|y^{(k)}(x)\|}{|x|^{n-k-1}}
\quad\text{a. e. } x\in(a,b)
\end{equation}
%
with $\lambda_k\in L^1(a,b)$, $\lambda_k\ge 0$, for $k=0,1,\dots, n-1$,
then
$y (x)\equiv \vec 0$ for all $x\in(a,b)$.
%
\end{theorem}


\section{Proofs of the main results}


 The absolute continuity assumption in the theorems is needed in
applying the Fundamental Theorem of Calculus to prove the following
lemma.

\begin{lemma} \label{f3-keylem}
 Assume $\phi: (a,b) \to \mathbb{R}^d$,
$\phi\in \mathcal{C}^{n-1}(a,b)$, $\phi^{(n-1)}$ absolutely
continuous on   $(a,b)$ for $a<0<b$.  If
\[
\phi^{(k)}(0) =\vec 0 \in \mathbb{R}^d, \quad k=0,1,\dots, n-1,
\]
then for any $\lambda(x)\in L^1(a,b)$, $\lambda(x)\ge 0$ and
$k=0,1,\dots,n-1$,
\begin{equation} \label{f3-keyleminq}
\int_0^x \lambda(t)\frac{\|\phi^{(k)}(t)\|}{|t|^{n-k-1}} dt
\le \int_0^x \lambda(t)dt \int_0^x \|\phi^{(n)}(t)\| dt,
\quad \forall x\in(a,b).
\end{equation}
%
\end{lemma}


\begin{proof}
By the absolute continuity assumption of $\phi^{(n-1)}$,
the Fundamental Theorem of Calculus grants the form
\[
 \phi^{(k)}(x)=\int_0^x \int_0^{t_{n-k-1}} \dots \int_0^{t_1}
\phi^{(n)}(t) dt ~dt_1 \dots dt_{n-k-1},\quad x\in [0,b)
\]
for $k=0,1,\dots, n-1$.  Define
\begin{equation*}
g(x)=\int_0^x \int_0^{t_{n-1}} \dots \int_0^{t_1}
\|\phi^{(n)}(t)\| dt ~dt_1 \dots dt_{n-1},\quad x\in[0,b).
\end{equation*}
Then $g\in\mathcal{C}^{n-1}[0,b)$,
$g^{(n)}=\|\phi^{(n)}\|$ a. e. in $(a,b)$ and for
$k=0,1,\dots,n-1$,
%
\begin{gather*}
g^{(k)}(x)
=\int_0^x \int_0^{t_{n-k-1}}  \dots \int_0^{t_1}
\|\phi^{(n)}(t)\| dt \,dt_1 \dots dt_{n-k-1}
\quad \nearrow \quad \text{w. r. t. } x\in(0,b) \\
g^{(k)}(0)=0.
\end{gather*}
%
By Taylor's formula, for any $t\in(0,b)$ and $k=0,\dots,n-2$,
%
\begin{align*}
 g^{(k)}(t)
&= g^{(k)}(0)+g^{(k+1)}(c)(t-0)  \quad c\in(0,t) \\
& \le g^{(k+1)}(t) ~t \le g^{(k+2)}(t) ~t^2  \le \dots \\
& \le  g^{(n-1)}(t) ~t^{n-k-1}
\end{align*}
%
By $\|\phi^{(k)}(x)\| \le g^{(k)}(x)$ and the monotonicity
of $g^{(k)}$ on $[0,b)$,
%
\begin{align*}
\int_0^x  \lambda(t)\frac{\|\phi^{(k)}(t)\|}{t^{n-k-1}} dt
& \le \int_0^x  \lambda(t)\frac{g^{(k)}(t)}{t^{n-k-1}} dt
\quad x\in[0,b),\; k=0,1,\dots,n-1\\
&\le \int_0^x \lambda(t) g^{(n-1)}(t) dt
\le  g^{(n-1)}(x) \int_0^x \lambda(t) dt \\
&= \int_0^x  g^{(n)}(t) dt \int_0^x \lambda(t) dt
=\int_0^x \|\phi^{(n)}(t)\| dt \int_0^x \lambda(t) dt.
\end{align*}
%
Thus the result holds for $x\in[0,b)$.  For $x\in(a,0]$,
consider $x^*=-x$ and define
%
\[
\phi^*(x^*)=\phi(-x^*), \quad
g^*(x^*)= \int_0^{x^*} \int_0^{t_{n-1}} \dots \int_0^{t_1}
\|{\phi^*}^{(n)}(t)\| dt \,dt_1 \dots dt_{n-1},
\]
where $x^*\in[0,-a)$.
Repeating the above derivation for $x\in[0,b)$ leads to
\[
\int_0^{x^*}  \lambda(-t)\frac{\|{\phi^*}^{(k)}(t)\|}{t^{n-k-1}} dt
\le  \int_0^{x^*} \|{\phi^*}^{(n)}(t)\| dt
\int_0^{x^*} \lambda(-t) dt, \quad \forall x^*\in[0,-a).
\]
Subsisting $t$ by $-t$ in the integrands and replacing $x^*$
by $x=-x^*\in(a,0]$,
\[
\int_{x}^0  \lambda(t)\frac{\|{\phi}^{(k)}(t)\|}{|t|^{n-k-1}} dt
\le  \int_{x}^0 \|{\phi}^{(n)}(t)\| dt \int_{x}^0 \lambda(t) dt,
\quad \forall x\in(a,0].
\]
Therefore, \eqref{f3-keyleminq} holds for all $x\in(a,b)$.
This completes the proof.
\end{proof}

The above lemma is used in the proof of Theorem
\ref{f3-thm-AC} below.

\begin{proof}[Proof of Theorem \ref{f3-thm-AC}]
  Let $\phi(x)=u(x)-v(x)$, $x\in(a,b)$.  Then $\phi$ satisfies the
conditions in Lemma \ref{f3-keylem}: $\phi\in\mathcal{C}^{n-1}(a,b)$,  
$\phi^{(n-1)}$ is absolutely
continuous on $(a,b)$, and $\phi^{(k)}(0)=\vec 0$ for
$k=0,1,\dots, n-1$.
  By the Lipschitz property \eqref{f3-lip},
\begin{equation} \label{f3-lipdiff}
\begin{aligned}
\|\phi^{(n)}(t)\|
&= \|f(t,u(t),u'(t),\dots,u^{(n-1)}(t))
- f(t,v(t),v'(t),\dots,v^{(n-1)}(t)) \|   \\
&\le \sum_{k=0}^{n-1}
\lambda_k(t)\frac{\|u^{(k)}(t)-v^{(k)}(t)\|}{t^{n-k-1}}
=  \sum_{k=0}^{n-1}
\lambda_k(t)\frac{\|\phi^{(k)}(t)\|}{t^{n-k-1}}.
\end{aligned}
\end{equation}
We assume $f$ is well defined so that for solutions $u,v$ on
$(a,b)$, $(u,u',\dots,u^{(n-1)})$, $(v,v',\dots, v^{(n-1)})$ are in the
domain for which \eqref{f3-lip} holds.  For any $\varepsilon\in(0,b)$,
define
\[
A(\varepsilon) = \int_0^\varepsilon \Big(\sum_{k=0}^{n-1}
\lambda_k(t)\frac{\|\phi^{(k)}(t)\|}{t^{n-k-1}}\Big) dt
\]
Applying inequalities \eqref{f3-keyleminq} within the sum and then
using \eqref{f3-lip}, we have
%
\begin{align*}
A(\varepsilon)
& = \sum_{k=0}^{n-1}  \int_0^\varepsilon
\lambda_k(t)\frac{\|\phi^{(k)}(t)\|}{t^{n-k-1}} dt\\
&\le \sum_{k=0}^{n-1}  \int_0^\varepsilon \lambda_k(t) dt
\int_0^\varepsilon \|\phi^{(n)}(t)\| dt\\
&\le \sum_{k=0}^{n-1} \int_0^\varepsilon  \lambda_k(t) dt
\int_0^\varepsilon \Big(\sum_{k=0}^{n-1} \lambda_k(t)
 \frac{\|\phi^{(k)}(t)\|}{t^{n-k-1}}\Big) dt\\
&=\Big(\int_0^\varepsilon \sum_{k=0}^{n-1} \lambda_k(t) dt\Big)
A(\varepsilon).
\end{align*}
%
If $A(\varepsilon)\not=0$, then
\[
\int_0^{\varepsilon} \sum_{k=0}^{n-1} \lambda_k(t) dt \ge 1.
\]
However, $\lambda_k\in L^1(a,b)$ means
\[
\int_0^{\varepsilon} \sum_{k=0}^{n-1} \lambda_k(t) dt \to 0
\quad as \quad \varepsilon\to 0.
\]
This contradiction implies that there exists $\varepsilon\in(0,b)$ such
that $A(\tilde\varepsilon)=0, ~\forall\tilde\varepsilon\le\varepsilon$.
Integrating \eqref{f3-lipdiff} on both sides,
\[
\int_0^\varepsilon \|\phi^{(n)}(t)\| dt
 \le \int_0^\varepsilon  \Big(\sum_{k=0}^{n-1}
\lambda_k(t)\frac{\|\phi^{(k)}(t)\|}{t^{n-k-1}}\Big) dt
=A(\varepsilon)  =0
\]
Hence
\[
\|\phi^{(n)}(t)\| =0
\text{ a. e. } t\in(0,\varepsilon)
\quad \Longrightarrow  \quad \phi^{(n)}(t) =\vec 0
\text{ a. e. } t\in(0,\varepsilon).
\]
Recall $\phi^{(n-1)}(0)=u^{(n-1)}(0)-v^{(n-1)}(0)=\vec 0$.  Thus
\[
 \phi^{(n-1)}(\varepsilon) =\int_0^\varepsilon \phi^{(n)}(t)dt=\vec 0
\quad \Longrightarrow \quad
\phi^{(n-1)}(t)= \vec 0 \text{ a. e. }t\in(0,\varepsilon).
 \]
Repeating the argument results in
\[
\phi(t) = u(t)-v(t)= \vec 0 \quad a.~e. ~~  ~t\in(0,\varepsilon).
\]
Since $u,v\in\mathcal{C}^{n-1}(a,b)$ and $u(0)=v(0)$, we obtain
\[
\phi(t)=u(t)-v(t)\equiv \vec 0, \quad \forall t\in[0,\varepsilon).
\]
Let $\varepsilon'=\max\{\varepsilon: \phi(t) = \vec 0,\; \forall
t\in(0,\varepsilon)\}$.  If $\varepsilon'<b$, then
$\phi(t) = \vec 0$ for
$t\in(0,\varepsilon']$ by the continuity of $u$ and $v$.
Then applying the above derivation to functions
$u(x-\varepsilon')$, $v(x-\varepsilon')$ on the
interval $[\varepsilon',b)$ would yield $u(t)-v(t) = \vec 0$ for
$t\in(\varepsilon',\varepsilon'+\varepsilon'')$ for
some $\varepsilon''>0$,
which contradicts the definition of $\varepsilon'$.
Therefore we must have
\[
u(t)-v(t)= \vec 0, \quad \forall t\in[0,b).
\]
To obtain the results on $(a,0]$, replacing $u(x)$, $v(x)$, $x\in(a,b)$ by
$u^*(x)=u(-x)$, $v^*(x)=v(-x)$, $x\in(-b,-a)$ to obtain
\[
u^*(t)-v^*(t)= \vec 0 \quad \forall t\in[0,-a).
 \]
Combining the results we arrived at
\[
u(t)-v(t)= \vec 0 \quad \forall t\in(a,b).
\]
This concludes the proof for Theorem \ref{f3-thm-AC}.
\end{proof}

 The proof of Theorem \ref{f3-thm-global} is
similar to the proof of Theorem \ref{f3-thm-AC}.

\begin{proof}[Proof of Theorem \ref{f3-thm-global}]
Notice that $y$ satisfies the assumptions in Lemma
\ref{f3-keylem}.   Define
\[
B(\varepsilon) = \int_0^\varepsilon \Big(\sum_{k=0}^{n-1}
\lambda_k(t)\frac{\|y^{(k)}(t)\|}{t^{n-k-1}}\Big) dt
\]
for any $\varepsilon\in(0,b)$.  Applying \eqref{f3-keyleminq} in Lemma
\ref{f3-keylem} and then \eqref{f3-thm-globalinq},
\begin{align*}
B(\varepsilon) & = \sum_{k=0}^{n-1}  \int_0^\varepsilon
\lambda_k(t)\frac{\|y^{(k)}(t)\|}{t^{n-k-1}} dt
\le \sum_{k=0}^{n-1}  \int_0^\varepsilon \lambda_k(t) dt
\int_0^\varepsilon \|y^{(n)}(t)\| dt\\
&\le
\sum_{k=0}^{n-1} \int_0^\varepsilon  \lambda_k(t) dt
\int_0^\varepsilon \Big(\sum_{k=0}^{n-1} \lambda_k(t)
 \frac{\|y^{(k)}(t)\|}{t^{n-k-1}}\Big) dt\\
&=\Big(\int_0^\varepsilon \sum_{k=0}^{n-1} \lambda_k(t) dt\Big)
B(\varepsilon).
\end{align*}
%
$B(\varepsilon)\not=0$ would imply
\[
\int_0^{\varepsilon} \sum_{k=0}^{n-1} \lambda_k(t) dt \ge 1.
\]
On the other hand, $\lambda_k\in L^1(a,b)$ implies
\[
\int_0^{\varepsilon} \sum_{k=0}^{n-1} \lambda_k(t) dt \to 0
\quad \text{as } \varepsilon\to 0.
\]
Thus there must exist $\varepsilon\in(0,b)$ such that
$B(\tilde\varepsilon)=0, \forall \tilde\varepsilon \le \varepsilon$.
Integrating
\eqref{f3-thm-globalinq} on both sides,
\[
\int_0^\varepsilon \|y^{(n)}(t)\| dt
 \le \int_0^\varepsilon  \Big(\sum_{k=0}^{n-1}
\lambda_k(t)\frac{\|y^{(k)}(t)\|}{t^{n-k-1}}\Big) dt
=B(\varepsilon)  =0
\]
which leads to
\[
\|y^{(n)}(t)\| =0
\text{ a. e. }  t\in(0,\varepsilon)
\quad \Longrightarrow  \quad y^{(n)}(t) =\vec 0\quad 
\text{a. e. } t\in(0,\varepsilon).
\]
Consequently,
\[
y^{(n-1)}(\varepsilon) =\int_0^\varepsilon y^{(n)}(t)dt=\vec 0
\quad \Longrightarrow \quad y^{(n-1)}(t)= \vec 0
\quad\text{a. e. }t\in(0,\varepsilon).
\]
This argument leads to
\[
y(t)= \vec 0 \quad \text{a. e. }t\in(0,\varepsilon).
\]
where ``a. e.'' can be removed by the continuity of $y$.  The
interval $(0,\varepsilon)$ on which $y\equiv \vec 0$ can be extended to
$(0,b)$ and $(a,0)$ using arguments analogous to the ones used in the
proof of Theorem \ref{f3-thm-AC}.
This concludes the proof of Theorem
\ref{f3-thm-global}.
\end{proof}

\section{Corollaries and an example}

 In Corollary \ref{f3-cor-jacobian} below, we give an
explicit form of the $L^1$ functions in the Lipschitz condition
\eqref{f3-lip} in Theorem \ref{f3-thm-AC} in terms of the Jacobians,
under stronger differentiability conditions on the function $f$ in the
differential system \eqref{f3-odesys} in Theorem \ref{f3-thm-AC}.
Recall that
\[
f(x,s_0,s_1,\dots,s_{n-1}):
(a,b)\times \mathbb{R}^d \times \dots\times \mathbb{R}^d
\quad \to \quad  \mathbb{R}^d
\]
where
$ f=(f_1,\dots,f_d)\in\mathbb{R}^d$,
$s_k = \left(s_{k1},\dots,s_{kd}\right)\in \mathbb{R}^d$,
$ k=0,\dots,n-1$.
For each $x\in(a,b)$, denote the Jacobian
\[
J_k = J_k(x) = \det\Big(
\frac{\partial f(x,s_0,\dots,s_{n-1})}{\partial s_k}\Big)
= \left| \begin{matrix}
\frac{\partial f_1}{\partial s_{k1}} & \dots &
\frac{\partial f_1}{\partial s_{kd}} \\
\vdots && \vdots \\
\frac{\partial f_d}{\partial s_{k1}} & \dots &
\frac{\partial f_d}{\partial s_{kd}}  \\
\end{matrix}
\right|,
\]
for $k=0,1,\dots,n-1$.

\begin{corollary}\label{f3-cor-jacobian}
  If $f(x, s_0,\dots,s_{n-1})$ is differentiable on
  $(s_0,\dots,s_{n-1})\in\mathbb{R}^{nd}$, a. e. $x\in(a,b)$, and
\[
 J_k(x) ~x^{n-k-1}\in L^1(a,b), \quad k=0,\dots,n-1,
\]
then there are $\lambda_k(x) \in L^1(a,b)$ such that the Lipschitz
condition \eqref{f3-lip} holds in Theorem \ref{f3-thm-AC}.
\end{corollary}

\begin{proof}
By the differentiability of $f$ and the mean value theorem,
\[
 f(x,s_0,\dots,s_{n-1}) - f(x,r_0,\dots,r_{n-1})
= Df(x,s') (s_0-r_0,\dots,s_{n-1}-r_{n-1}),
\]
a. e. $x\in(a,b)$, where
\[
 Df(x,s') = \Big(\frac{\partial f(x, s_0,\dots,s_{n-1})}{\partial s_0}, \dots,
\frac{\partial f(x, s_0,\dots,s_{n-1})}{\partial s_{n-1}} \Big)
\Big|_{(x,s')}
\]
and $(x,s')=(x,s'_0,\dots,s'_{n-1})\in(a,b)\times \mathbb{R}^{nd}$ is
on the line connecting the two points $(x,s_0,\dots,s_{n-1})$ and
$(x,r_0,\dots,r_{n-1})$.  By matrix multiplication,
\[
 f(x,s_0,\dots,s_{n-1}) - f(x,r_0,\dots,r_{n-1})
= \sum_{k=0}^{n-1} \frac{\partial f(x,s_0,\dots,s_{n-1})}
{\partial s_k}(x,s') ~(s_k - r_k) ,
\]
 a. e. $x\in(a,b)$.
Taking the norm, we have
\[
 \|f(x,s_0,\dots,s_{n-1}) - f(x,r_0,\dots,r_{n-1})\|
\le \sum_{k=0}^{n-1}
| J_k(x,s'_k)| \,\|s_k - r_k\|,
\]
a. e. $x\in(a,b)$.
Therefore, the functions
\[
 \lambda_k(x) = J_k(x,s') ~x^{n-k-1} \in L^1(a,b),\quad k=0,1,\dots,n-1
\]
satisfy the Lipschitz condition \eqref{f3-lip}.
\end{proof}

 When $f$ in Theorem \ref{f3-thm-AC} is linear, the
results can be stated as the corollary below.

\begin{corollary}
Let $a<0<b$ and $y:(a,b)\to\mathbb{R}^d$ be a solution of
\[
y^{(n)}+a_{n-1}(x,y) y^{(n-1)} + \dots + a_o(x,y) y = 0, \quad
  x\in(a,b),
\]
  where $y^{(n-1)}$ is absolutely continuous on $(a,b)$, and the
  coefficient functions
\[
 |a_k(x,y)| \le \frac{|\lambda_k(x)|}{|x|^{n-k-1}},
\quad \lambda_k\in L^1(a,b), \; k=0,1,\dots, n-1.
\]
If
\[
y(0)=y^{(1)}(0)=\dots=y^{(n-1)}(0)=\vec 0,
\]
then $y\equiv \vec 0$ on $(a,b)$.
\end{corollary}

 Theorem \ref{f3-thm-AC} is on uniqueness of $n$-th
order differential systems where the $n$th derivative of the solution
only needs to exist almost everywhere, which is in the spirit of the
classical Carath\'eodory theorem on the existence of ordinary
differential equations.  Theorem \ref{f3-thm-AC} for $n=1$ can be
stated as  follows.

\medskip
{\it
 Consider the differential equation
\[
 y' = f(x,y(x)),\quad  x\in(a,b),  \quad y(x_o)=y_o.
\]
If $f: (a,b)\times\mathbb{R}^d \to\mathbb{R}^d$ satisfies
\[
 \|f(x,y_1)-f(x,y_2)\|\le\lambda(x) \|y_1-y_2\|,
\quad\text{a. e. } x\in(a,b)
\]
where $\lambda\in L^1(a,b)$, $\lambda\ge 0$, then the solution of the
differential equation is unique.
%
}

\bigskip
 We conclude by an example to show the
sharpness of the orders in \eqref{f3-lip} and \eqref{f3-thm-globalinq}.

\medskip
\noindent\textbf{Example.}
For $p\in(0,1/2)$, let
\[
 y= e^{-1/{|x|^p}}, \quad x\in(-1,1),
\quad y^{(k)}(0)=0, \quad k=0,\dots,n-1.
\]
Then $y$ and its derivatives are even functions.
For $x\in(0,1)$, we have
\begin{align*}
y'&=p\frac{y}{x^{p+1}} \\
y''&=p\frac{y'}{x^{p+1}} - p(p+1)\frac{y}{x^{p+2}}\\
y'''& =p\frac{y''}{x^{p+1}} - 2p(p+1)\frac{y'}{x^{p+2}}
+ p(p+1)(p+2)\frac{y'}{x^{p+3}}
\end{align*}
%
The general form can be written as
%
\begin{equation} \label{f3-ex}
y^{(m)}(x) = \sum_{k=0}^{m-1} C^n_k \frac{y^{(k)}}{|x|^{m-k+p}},
\quad x\in(-1,1)\setminus\{0\}, \; m=1,\dots, n.
\end{equation}
%
where $C^m_k$ are constants depending on $n,k$ and $p$.
Formula \eqref{f3-ex} can be verified by induction.
Taking derivative of \eqref{f3-ex} for $x\in(0,1)$,
%
\begin{align*}
y^{(m+1)}(x)
&= \sum_{k=0}^{m-1} C^m_k \frac{y^{(k+1)}}{x^{m-k+p}}
+ \sum_{k=0}^{m-1} C^m_k (-m+k-p)\frac{y^{(k)}}{x^{m-k+p+1}} \quad
(k'=k+1) \\
&= \sum_{k'=1}^{m}
C^m_{k'-1} \frac{y^{(k')}}{x^{m+1-k'+p}}
+ \sum_{k=0}^{m-1} C^m_k (-m+k-p)\frac{y^{(k)}}{x^{m+1-k+p}}  \\
& = \sum_{k=0}^{m} C^{m+1}_k \frac{y^{(k)}}{x^{m+1-k+p}}
\end{align*}
%
where $C^{m+1}_k$ are constants depending on $n,k$ and $p$.  Therefore
\eqref{f3-ex} holds for all $m=1,\dots, n$.  We may write the case of
$m=n$ as
\[
|y^{(n)}(x)| \le  \sum_{k=0}^{n-1}
\Big(\frac{C^{n}_k}{|x|^{1-p}}\Big) \frac{y^{(k)}(x)}{|x|^{n-k-1+2p}}
=\sum_{k=0}^{n-1}
\lambda_k(x) \frac{y^{(k)}(x)}{|x|^{n-k-1+2p}},
\quad \forall x\in(-1,1)\setminus \{0\}
\]
where $\lambda_k\in L^1(-1,1)$ for $p\in(0,1/2)$.
In \eqref{f3-lip} and
\eqref{f3-thm-globalinq}, the order of singularity corresponding to
$y^{(k)}$ is $n-k-1$. In this example, the corresponding order is
$n-k-1+2p$, where $p\in(0,1/2)$ can be arbitrarily small.
However it is enough to make $y\not\equiv 0$ on $(-1,1)$.
Alternatively, we may write
\[
|y^{(n)}(x)| \le  \sum_{k=0}^{n-1}
\Big(\frac{C^{n}_k}{|x|^{1+p}}\Big) \frac{y^{(k)}(x)}{|x|^{n-k-1}}
=\sum_{k=0}^{n-1}
\lambda^*_k(x) \frac{y^{(k)}(x)}{|x|^{n-k-1}},
\quad \forall x\in(-1,1)\setminus \{0\}
\]
Now the order of singularity corresponding to $y^{(k)}$ is $n-k-1$
as in \eqref{f3-lip} and \eqref{f3-thm-globalinq}, however
\[
 \lambda^*_k(x)= \frac{C^{n}_k}{|x|^{1+p}} \quad \in L^q(-1,1),
\quad \forall q< \frac{1}{1+p},\quad p\in(0,1/2).
\]
Thus we do not have $y\equiv 0$ as in the conclusions of the theorems.

  We are interested in the uniqueness of solutions
ordinary differential systems when the coefficients are allowed to
have singularities \cite{PW1,PW2}.  In this paper, We give a Lipschitz
type condition for the uniqueness of weak solutions in the style of
the Carath\'eodory theorem for nonlinear $n$th order nonlinear
ordinary differential systems.  We also give a unique continuation
condition for functions satisfying an $n$th order Gronwall
differential inequality.  We use an example to show the sharpness of
the order of singularity required in the conditions.



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\end{document}