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\begin{document}

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\title[Nonlinear first order system of differential equation]{Some properties of solutions of the nonlinear first order system of
differential equation}
\author{Veronica Ilea}%$^{*}$ and Diana Otrocol$^{**}$}
\address{$^{*}$``Babe\c{s}-Bolyai'' University, Faculty of Mathematics and Computer
Science, M. Kog\u{a}lniceanu St. 1, RO-400084 Cluj-Napoca, Romania}
\email{vdarzu@math.ubbcluj.ro}
\author{Diana Otrocol}
\address{$^{\ast\ast}$Technical University of Cluj-Napoca, Memorandumului St. 28, 400114, Cluj-Napoca, Romania} 
\email{Diana.Otrocol@math.utcluj.ro}
\address{$^{\ast\ast}$``T. Popoviciu'' Institute of Numerical Analysis, Romanian
Academy, P.O.Box. 68-1, 400110, Cluj-Napoca, Romania}

\subjclass{34A12, 34C10, 34A34.}
\keywords{Nonlinear second order differential system,
extremal principle, zeros of solutions, Sturm-type theorem, Nicolescu-type
theorem, Butlewski-type theorem.}

\begin{abstract}
In this paper we present for the solutions of the nonlinear first order system
of differential equation, extremal principle, Nicolescu-type and
Butlewski-type separation theorems. Some applications and examples are given.
\end{abstract}
\date{}
\maketitle


\section{Introduction}

Let $F,G\in C([a,b]\times\mathbb{R}^{3}).$ We consider the following first
order system of differential equation%
\begin{equation}
\left\{
\begin{array}
[c]{l}%
F(x,y,z,y^{\prime})=0\\
G(x,y,z,z^{\prime})=0.
\end{array}
\right.  \label{ec1.1}%
\end{equation}


In this paper by a solution of the system (\ref{ec1.1}) we understand a
function $(y,z)\in C^{1}([a,b],\mathbb{R}^{2})$ which satisfies (\ref{ec1.1}).

For a function $u:[a,b]\rightarrow\mathbb{R}$ we denote by $Z_{u}$ the zero
set of $u$, $Z_{u}:=\{x\in\lbrack a,b]|\ u(x)=0\}$.

Let us recall now some essential definitions and fundamental results.

\begin{definition}
\label{def-homm} A function $f:D\rightarrow\mathbb{R}\ (D\subset\mathbb{R}%
^{2})$ is called homogeneous of degree $n$ if $f(tu,tv)=t^{n}f(u,v),$ for each
$(u,v)\in D$ and $t>0$.
\end{definition}

The linear case of (\ref{ec1.1}) is the following system%
\begin{equation}
\left\{
\begin{array}
[c]{c}%
y^{\prime}+p_{1}(x)y+q_{1}(x)z=0\\
z^{\prime}+p_{2}(x)y+q_{2}(x)z=0
\end{array}
\right.  \label{ec1.2}%
\end{equation}
with $p_{i},q_{i}\in C[a,b]$, $i=1,2$.

For the system (\ref{ec1.2}) the following properties of the solution are well
known (see \cite{Rus96}, \cite{Sansone1}, \cite{Sansone2}, \cite{Reid},
\cite{Hartman}, \cite{Swanson}).

\begin{theorem}
\label{th1.1rus} If $(y,z)\neq 0$ is a solution of (\ref{ec1.2}) then
we have:

\begin{itemize}
\item[(i)] $Z_{y}\cap Z_{z}=\emptyset;$

\item[(ii)] if $q_{1}(x)\neq0,\ p_{2}(x)\neq0,\ \forall x\in\lbrack a,b],$
then the zeros of $y$ and $z$ are simple and isolated on $[a,b]$.
\end{itemize}
\end{theorem}

\begin{theorem}
\label{th1.3} (Nicolescu's theorem \cite{Nicolescu}) We suppose that
$q_{1}(x)p_{2}(x)<0,$ $\forall x\in\lbrack a,b].$ If $(y,z)\neq 0$ is a solution of
(\ref{ec1.2}), then the zeros of $y$ and $z$ separate each other on $[a,b]$.
\end{theorem}

\begin{theorem}
\label{th1.4} (Butlewski's theorem \cite{Butlewski}) We suppose that
$p_{i},q_{i}\in C[a,b]$, $i=1,2$ and $q_{1}(x)\neq0$ $(p_{2}(x)\neq0),$
$\forall x\in\lbrack a,b].$ If $(y_{1},z_{1})$ and $(y_{2},z_{2})$ are two
linear independent solutions of (\ref{ec1.2}), then the zeros of $y_{1}$ and
$y_{2}$ $(z_{1}\ $and $z_{2})$ separate each other on $[a,b]$.
\end{theorem}

The aim of this paper is to extend the above results to the solutions of
(\ref{ec1.1}). For some results in this directions see \cite{Rus63},
\cite{Rus85}, \cite{Muresan} and \cite{IOR2015}.

The organization of this paper is as follows. In Section 2 we prove some
extremal principles for nonlinear first order system of differential equations
and in Section 3 we discussed some properties of the zeros of the components
of the solutions for such systems and in the end we prove Nicolescu-type and
Butlewski-type separation theorems, by using Tonelli's Lemmas. The results
presented in this paper generalize the main results in \cite{IOR2015}.

\section{Extremal principles}

We consider the system (\ref{ec1.1}) with $F,G\in C([a,b]\times\mathbb{R}%
^{3})$. We have the following extremal principle for the solutions of
(\ref{ec1.1}).

\begin{theorem}
\label{th.extr.princ.} (Extremal principle) Let $(y,z)\in C^{1}%
([a,b],\mathbb{R}^{2})$ be a solution of (\ref{ec1.1}) and we suppose that:

\begin{itemize}
\item[(i)] $F(x,\cdot,\cdot,0)$ and $G(x,\cdot,\cdot,0)$ are homogeneous for
all $x\in\lbrack a,b]$;

\item[(ii)] $F(x,y,\cdot,0)\ $and$\ G(x,\cdot,z,0)$ are increasing$,\ \forall
x\in\lbrack a,b];$

\item[(iii)] $F(x,1,1,0)<0$,$\ G(x,1,1,0)<0,\ \forall x\in\lbrack a,b]$.
\end{itemize}

Then:

\begin{itemize}
\item[(a)] If there exists $x_{0}\in\lbrack a,b]$ such that $\max\left\{
\underset{x\in\lbrack a,b]}{\max}y(x),\underset{x\in\lbrack a,b]}{\max
}z(x)\right\}  =\max\left\{  y(x_{0}),z(x_{0})\right\}  >0,$ then $x_{0}%
\in\{a,b\};$

\item[(b)] If there exists $x_{0}\in\lbrack a,b]$ such that $\min\left\{
\underset{x\in\lbrack a,b]}{\max}y(x),\underset{x\in\lbrack a,b]}{\max
}z(x)\right\}  =\max\left\{  y(x_{0}),z(x_{0})\right\}  <0,$ then $x_{0}%
\in\{a,b\}.$
\end{itemize}
\end{theorem}

\begin{proof}
(a) We suppose that $x_{0}\in]a,b[.\ $Let $\max\left\{  \underset{x\in\lbrack
a,b]}{\max}y(x),\underset{x\in\lbrack a,b]}{\max}z(x)\right\}  =\max\left\{
y(x_{0}),z(x_{0})\right\}  =y(x_{0})>0.\ $We shall show that this leads to a contradiction.

Since $y\in C^{1}[a,b]$ we have that $y^{\prime}(x_{0})=0.$ From (\ref{ec1.1})
we have%
\[
F(x_{0},y(x_{0}),z(x_{0}),0)=0.
\]
Using (i) and (ii) we obtain%
\begin{align*}
0  &  =F(x_{0},y(x_{0}),z(x_{0}),0)\leq F(x_{0},y(x_{0}),y(x_{0}),0)=\\
&  =y(x_{0})F(x_{0},1,1,0)<0.
\end{align*}
So, $\ x_{0}\in\{a,b\}.$

Now let $\max\left\{  \underset{x\in\lbrack a,b]}{\max}y(x),\underset
{x\in\lbrack a,b]}{\max}z(x)\right\}  =\max\left\{  y(x_{0}),z(x_{0})\right\}
=z(x_{0})>0$.

Since $z\in C^{1}[a,b]$ we have that $z^{\prime}(x_{0})=0.$ From (\ref{ec1.1})
we have%
\[
G(x_{0},y(x_{0}),z(x_{0}),0)=0.
\]
Using (i) and (ii) we obtain%
\begin{align*}
0  &  =G(x_{0},y(x_{0}),z(x_{0}),0)\leq G(x_{0},z(x_{0}),z(x_{0}),0)=\\
&  =z(x_{0})G(x_{0},1,1,0)<0.
\end{align*}
So, $\ x_{0}\in\{a,b\}.$

(b) Let $\min\left\{  \underset{x\in\lbrack a,b]}{\max}y(x),\underset
{x\in\lbrack a,b]}{\max}z(x)\right\}  =\max\left\{  y(x_{0}),z(x_{0})\right\}
=y(x_{0})<0.\ $We suppose that $x_{0}\in]a,b[$. Analogous, we prove that this
leads to a contradiction.
\end{proof}

\begin{corollary}
\label{cor1}Let $(y,z)\in C^{1}([a,b],\mathbb{R}^{2})$ be a solution of the
following system%
\[
\left\{
\begin{array}
[c]{c}%
p_{1}y+q_{1}z-y^{\prime}=0\\
p_{2}y+q_{2}z-z^{\prime}=0
\end{array}
\right.
\]
and we suppose that $p_{2}>0,q_{1}>0,\ p_{1}+q_{1}<0$ and $p_{2}+q_{2}<0.$ Then:

\begin{itemize}
\item[(a)] If there exists $x_{0}\in\lbrack a,b]$ such that $\max\left\{
\underset{x\in\lbrack a,b]}{\max}y(x),\underset{x\in\lbrack a,b]}{\max
}z(x)\right\}  =\max\left\{  y(x_{0}),z(x_{0})\right\}  >0,$ then $x_{0}%
\in\{a,b\};$

\item[(b)] If there exists $x_{0}\in\lbrack a,b]$ such that  $\min\left\{
\underset{x\in\lbrack a,b]}{\max}y(x),\underset{x\in\lbrack a,b]}{\max
}z(x)\right\}  =\max\left\{  y(x_{0}),z(x_{0})\right\}  <0,$ then $x_{0}%
\in\{a,b\}.$
\end{itemize}
\end{corollary}

\begin{example}
\label{ex.cor32} We consider on $[0,1]$ the system%
\[
\left\{
\begin{array}
[c]{c}%
-2y+z-y^{\prime}=0\\
y-4z-z^{\prime}=0
\end{array}
\right.
\]
with initial conditions $y(0)=z(0)=1.$ We have $q_{1}=p_{2}=1,\ p_{1}%
+q_{1}<0,\ p_{2}+q_{2}<0.\ $From Figure 1 one can see that the conditions of
Corollary \ref{cor1} hold.
\end{example}

%

%TCIMACRO{\FRAME{ftbpF}{3.7559in}{2.3013in}{0pt}{}{}{fig1.eps}%
%{\special{ language "Scientific Word";  type "GRAPHIC";  display "USEDEF";
%valid_file "F";  width 3.7559in;  height 2.3013in;  depth 0pt;
%original-width 5.8219in;  original-height 4.3708in;  cropleft "0";
%croptop "1";  cropright "1";  cropbottom "0";
%filename 'fig1.eps';file-properties "NPEU";}}}%
%BeginExpansion
\begin{figure}
[ptb]
\begin{center}
\includegraphics[
height=2.3013in,
width=3.7559in
]%
{fig1.eps}%
\end{center}
\caption{Plot of $\max\left\{  y(x),z(x)\right\} $ as function of $x$}
\end{figure}
%EndExpansion


\begin{theorem}
\label{th.1(PE)}Let $(y,z)\in C^{1}([a,b],\mathbb{R}^{2})$ be a solution of
the following system%
\[
\left\{
\begin{array}
[c]{c}%
f(x,y,z)-y^{\prime}=0\\
g(x,y,z)-z^{\prime}=0
\end{array}
\right.
\]
and we suppose that

\begin{itemize}
\item[(i)] $f$ and $g$ are homogeneous with respect to the last two arguments;

\item[(ii)] $f(x,y,\cdot)\ $and$\ g(x,\cdot,z)$ are increasing$,\ \forall
x\in\lbrack a,b];$

\item[(iii)] $f(x,1,1)<0$,$\ g(x,1,1)<0,\ \forall x\in\lbrack a,b]$.
\end{itemize}

Then:

\begin{itemize}
\item[(a)] If there exists $x_{0}\in\lbrack a,b]$ such that $\max\left\{
\underset{x\in\lbrack a,b]}{\max}y(x),\underset{x\in\lbrack a,b]}{\max
}z(x)\right\}  =\max\left\{  y(x_{0}),z(x_{0})\right\}  >0,$ then $x_{0}%
\in\{a,b\};$

\item[(b)] If there exists $x_{0}\in\lbrack a,b]$ such that  $\min\left\{
\underset{x\in\lbrack a,b]}{\max}y(x),\underset{x\in\lbrack a,b]}{\max
}z(x)\right\}  =\max\left\{  y(x_{0}),z(x_{0})\right\}  <0,$ then $x_{0}%
\in\{a,b\}.$
\end{itemize}
\end{theorem}

\begin{proof}
The system satisfies the condition from Theorem \ref{th.extr.princ.}.
\end{proof}

\begin{corollary}
\label{cor2} Let $(y,z)\in C^{1}([a,b],\mathbb{R}^{2})$ be a solution of the
following system%
\[
\left\{
\begin{array}
[c]{c}%
p_{1}y^{3}+q_{1}z^{3}-y^{\prime}=0\\
p_{2}y^{3}+q_{2}z^{3}-z^{\prime}=0
\end{array}
\right.
\]
and we suppose that $p_{2}>0,q_{1}>0,\ p_{1}+q_{1}<0$ and $p_{2}+q_{2}<0.$

Then:

\begin{itemize}
\item[(a)] If there exists $x_{0}\in\lbrack a,b]$ such that $\max\left\{
\underset{x\in\lbrack a,b]}{\max}y(x),\underset{x\in\lbrack a,b]}{\max
}z(x)\right\}  =\max\left\{  y(x_{0}),z(x_{0})\right\}  >0,$ then $x_{0}%
\in\{a,b\};$

\item[(b)] If there exists $x_{0}\in\lbrack a,b]$ such that $\min\left\{
\underset{x\in\lbrack a,b]}{\max}y(x),\underset{x\in\lbrack a,b]}{\max
}z(x)\right\}  =\max\left\{  y(x_{0}),z(x_{0})\right\}  <0,$ then $x_{0}%
\in\{a,b\}.$
\end{itemize}
\end{corollary}

\begin{example}
\label{ex2} We consider on $[0,1]$ the system%
\[
\left\{
\begin{array}
[c]{c}%
-5y^{3}+2z^{3}-y^{\prime}=0\\
2y^{3}-6z^{3}-z^{\prime}=0
\end{array}
\right.
\]
with initial conditions $y(0)=z(0)=1.$ We have $q_{1}=2,p_{2}=2,\ p_{1}%
+q_{1}<0,\ p_{2}+q_{2}<0.\ $From Figure 2 one can see that the conditions of
Corollary \ref{cor2} hold.
\end{example}

%

%TCIMACRO{\FRAME{ftbpF}{3.5475in}{2.4431in}{0pt}{}{}{fig2.eps}%
%{\special{ language "Scientific Word";  type "GRAPHIC";  display "USEDEF";
%valid_file "F";  width 3.5475in;  height 2.4431in;  depth 0pt;
%original-width 5.8219in;  original-height 4.3708in;  cropleft "0";
%croptop "1";  cropright "1";  cropbottom "0";
%filename 'fig2.eps';file-properties "NPEU";}}}%
%BeginExpansion
\begin{figure}
[ptb]
\begin{center}
\includegraphics[
height=2.4431in,
width=3.5475in
]%
{fig2.eps}%
\end{center}
\caption{Plot of $\max\left\{  y(x),z(x)\right\} $ as function of $x$}
\end{figure}
%EndExpansion


In the end of this section, we consider the following functional-differential
system%
\begin{equation}
\left\{
\begin{array}
[c]{l}%
F(x,y,y(g),z,y^{\prime})=0\\
G(x,y,z,z(h),z^{\prime})=0,
\end{array}
\right.  \label{ex.EPD}%
\end{equation}


\begin{theorem}
\label{th.2(PE)}Let $(y,z)\in C^{1}([a,b],\mathbb{R}^{2})$ be a solution of
the system (\ref{ex.EPD}), where $g,h\in C[a,b]),g(x)\leq x,h(x)\leq x,\ a\leq
g(x)\leq b,\ a\leq h(x)\leq b,\ \forall x\in\lbrack a,b]$ and we suppose that:

\begin{itemize}
\item[(i)] $F(x,\cdot,\cdot,\cdot,0)$ and $G(x,\cdot,\cdot,\cdot,0)$ are
homogeneous, for all $x\in\lbrack a,b]$;

\item[(ii)] $F(x,y,\cdot,\cdot,0)\ $and$\ G(x,\cdot,z,\cdot,0)$ are
increasing$,\ \forall x\in\lbrack a,b];$

\item[(iii)] $F(x,1,1,1,0)<0$,$\ G(x,1,1,1,0)<0,\ \forall x\in\lbrack a,b]$.
\end{itemize}

Then:

\begin{itemize}
\item[(a)] If there exists $x_{0}\in\lbrack a,b]$ such that $\max\left\{
\underset{x\in\lbrack a,b]}{\max}y(x),\underset{x\in\lbrack a,b]}{\max
}z(x)\right\}  =\max\left\{  y(x_{0}),z(x_{0})\right\}  >0,$ then $x_{0}%
\in\{a,b\};$

\item[(b)] If there exists $x_{0}\in\lbrack a,b]$ such that $\min\left\{
\underset{x\in\lbrack a,b]}{\max}y(x),\underset{x\in\lbrack a,b]}{\max
}z(x)\right\}  =\max\left\{  y(x_{0}),z(x_{0})\right\}  <0,$ then $x_{0}%
\in\{a,b\}.$
\end{itemize}
\end{theorem}

\begin{proof}
(a) We suppose that $x_{0}\in]a,b[.\ $Let $\max\left\{  \underset{x\in\lbrack
a,b]}{\max}y(x),\underset{x\in\lbrack a,b]}{\max}z(x)\right\}  =\max\left\{
y(x_{0}),z(x_{0})\right\}  =y(x_{0})>0.\ $We shall show that this leads to a contradiction.

Since $y\in C^{1}[a,b]$ we have that $y^{\prime}(x_{0})=0.$ From (\ref{ec1.1})
we have%
\[
F(x_{0},y(x_{0}),y(g(x_{0})),z(x_{0}),0)=0.
\]
Using (i) and (ii) we obtain%
\begin{align*}
0  &  =F(x_{0},y(x_{0}),y(g(x_{0})),z(x_{0}),0)\leq F(x_{0},y(x_{0}%
),y(x_{0}),y(x_{0}),0)=\\
&  =y(x_{0})F(x_{0},1,1,1,0)<0.
\end{align*}
So, $\ x_{0}\in\{a,b\}.$

Now let $\max\left\{  \underset{x\in\lbrack a,b]}{\max}y(x),\underset
{x\in\lbrack a,b]}{\max}z(x)\right\}  =\max\left\{  y(x_{0}),z(x_{0})\right\}
=z(x_{0})>0$.

Since $z\in C^{1}[a,b]$ we have that $z^{\prime}(x_{0})=0.$ From (\ref{ec1.1})
we have%
\[
G(x_{0},y(x_{0}),z(x_{0}),z(h(x_{0})),0)=0.
\]
Using (i) and (ii) we obtain%
\begin{align*}
0  &  =G(x_{0},y(x_{0}),z(x_{0}),z(h(x_{0})),0)\leq G(x_{0},z(x_{0}%
),z(x_{0}),z(x_{0}),0)=\\
&  =z(x_{0})G(x_{0},1,1,1,0)<0.
\end{align*}
So, $\ x_{0}\in\{a,b\}.$

(b) Let $\min\left\{  \underset{x\in\lbrack a,b]}{\max}y(x),\underset
{x\in\lbrack a,b]}{\max}z(x)\right\}  =\max\left\{  y(x_{0}),z(x_{0})\right\}
=y(x_{0})<0.\ $We suppose that $x_{0}\in]a,b[$. Analogous, we prove that this
leads to a contradiction.
\end{proof}

\begin{corollary}
\label{cor3} Let $(y,z)\in C^{1}([a,b],\mathbb{R}^{2})$ be a solution of the
following system%
\[
\left\{
\begin{array}
[c]{l}%
p_{1}y+q_{1}z-y^{\prime}=0\\
p_{2}y+q_{2}z+q_{3}z(h)-z^{\prime}=0
\end{array}
\right.
\]
and we suppose that $q_{1}>0,\ p_{2}>0,\ q_{3}>0,\ p_{1}+q_{1}<0$ and
$p_{2}+q_{2}+q_{3}<0,h(x)\leq x,a\leq h(x)\leq b,\ \forall x\in\lbrack a,b].$

Then:

\begin{itemize}
\item[(a)] If there exists $x_{0}\in\lbrack a,b]$ such that $\max\left\{
\underset{x\in\lbrack a,b]}{\max}y(x),\underset{x\in\lbrack a,b]}{\max
}z(x)\right\}  =\max\left\{  y(x_{0}),z(x_{0})\right\}  >0,$ then $x_{0}%
\in\{a,b\};$

\item[(b)] If there exists $x_{0}\in\lbrack a,b]$ such that $\min\left\{
\underset{x\in\lbrack a,b]}{\max}y(x),\underset{x\in\lbrack a,b]}{\max
}z(x)\right\}  =\max\left\{  y(x_{0}),z(x_{0})\right\}  <0,$ then $x_{0}%
\in\{a,b\}.$
\end{itemize}
\end{corollary}

\begin{example}
\label{exTh34} We consider on $[0,1]$ the system%
\[
\left\{
\begin{array}
[c]{l}%
-4y^{6}(x)+z^{6}(x)-y^{\prime}(x)=0\\
3y(x)-5z(x)+z(x^{2})-z^{\prime}(x)=0
\end{array}
\right.
\]
with initial conditions $y(0)=z(0)=1.$ We have $q_{1}=1,\ p_{2}=3,\ q_{3}%
=1,\ p_{1}+q_{1}<0,\ p_{2}+q_{2}+q_{3}<0,h(x)=x^{2}.\ $From Figure 3 one can
see that the conditions of Theorem \ref{th.2(PE)} hold.
\end{example}

%

%TCIMACRO{\FRAME{ftbpF}{3.4437in}{2.5201in}{0pt}{}{}{fig3.eps}%
%{\special{ language "Scientific Word";  type "GRAPHIC";  display "USEDEF";
%valid_file "F";  width 3.4437in;  height 2.5201in;  depth 0pt;
%original-width 5.8219in;  original-height 4.3708in;  cropleft "0";
%croptop "1";  cropright "1";  cropbottom "0";
%filename 'fig3.eps';file-properties "NPEU";}}}%
%BeginExpansion
\begin{figure}
[ptb]
\begin{center}
\includegraphics[
height=2.5201in,
width=3.4437in
]%
{fig3.eps}%
\end{center}
\caption{Plot of $\max\left\{  y(x),z(x)\right\} $ as function of $x$}
\end{figure}
%EndExpansion


\section{Zeros of the components of the solutions of the system (\ref{ec1.1})}

Let us consider the following conditions on the system (\ref{ec1.1}):

\begin{itemize}
\item[(C$_{1}$)] $F(x,0,0,0)=0,G(x,0,0,0)=0,\forall x\in\lbrack a,b].$

\item[(C$_{2}$)] If $(y,z)$ is a solution of (\ref{ec1.1}) and $y(x_{0}%
)=z(x_{0})=0$ for some $x_{0}\in\lbrack a,b],$ then $y=z=0$.

\item[(C$_{3}$)] The Cauchy problem for (\ref{ec1.1}) has at most a solution.

\item[(C$_{4}$)] Let $(y,z)$ a solution of (\ref{ec1.1}), then the problem
\[
\left\{
\begin{array}
[c]{l}%
F(x,y,z,y^{\prime})=0\\
G(x,y,z,z^{\prime})=0\\
y(x_{0})=y_{0},y^{\prime}(x_{0})=y_{0}^{\prime}%
\end{array}
\right.
\]
where $x_{0}\in\lbrack a,b],$ $y_{0},y_{0}^{\prime}\in\mathbb{R}$ has at most
a solution.

\item[(C$_{5}$)] Let $(y,z)$ a solution of (\ref{ec1.1}) then the problem
\[
\left\{
\begin{array}
[c]{l}%
F(x,y,z,y^{\prime})=0\\
G(x,y,z,z^{\prime})=0\\
z(x_{0})=z_{0},z^{\prime}(x_{0})=z_{0}^{\prime}%
\end{array}
\right.
\]
where $x_{0}\in\lbrack a,b],$ $z_{0},z_{0}^{\prime}\in\mathbb{R}$ has at most
a solution.
\end{itemize}

\begin{rem}
\label{remarks}

\begin{itemize}
\item[(1)] If $F(x,u,\cdot,w)=0$ has a solution in $v$, $\forall x_{0}%
\in\lbrack a,b],$ $u,w\in\mathbb{R},$ then (C$_{3}$) implies (C$_{4}$).

\item[(2)] If $F(x,\cdot,v,w)=0$ has a solution in $u$, $\forall x_{0}%
\in\lbrack a,b],$ $v,w\in\mathbb{R},$ then (C$_{3}$) implies (C$_{5}$).

\item[(3)] (C$_{1}$) and (C$_{3}$) imply (C$_{2}$).

\item[(4)] Let us consider the system%
\[
\left\{
\begin{array}
[c]{c}%
p_{1}y+q_{1}z-y^{\prime}=0\\
p_{2}y+q_{2}z-z^{\prime}=0
\end{array}
\right.  ,\ p_{i},q_{i}\in C[a,b],i=1,2.
\]
In this case the conditions (C$_{1}$),(C$_{2}$) and (C$_{3}$) are satisfy.

If $q_{1}(x)\neq0,\forall x_{0}\in\lbrack a,b],$ then the condition (C$_{4}$)
is satisfied.

If $p_{2}(x)\neq0,\forall x_{0}\in\lbrack a,b],$ then the condition (C$_{5}$)
is satisfied.

\item[(5)] (C$_{4}$) and (C$_{5}$) imply (C$_{3}$).
\end{itemize}
\end{rem}

In what follow we also need the following result (see \cite{Tonelli},
\cite{Rus96}, \cite{Muresan}).

\begin{lemma}
\label{lem4.3} (Tonelli's Lemma; see \cite{Tonelli} and \cite{Rus63}) Let
$y_{1},y_{2}\in C^{1}[a,b]$ be two functions which satisfy the following conditions:

\begin{itemize}
\item[(i)] $y_{1}(a)=0,y_{1}(b)=0$ and $y_{1}(x)>0,\forall x\in]a,b[;$

\item[(ii)] $y_{2}(x)>0$ $,$ $\forall$ $x\in\lbrack a,b].$
\end{itemize}

Then there exists $\lambda>0$ and $x_{0}\in]a,b[$ such that:%
\[
y_{2}(x_{0})=\lambda y_{1}(x_{0})\text{ and }y_{2}^{\prime}(x_{0})=\lambda
y_{1}^{\prime}(x_{0}).
\]

\end{lemma}

Next, we use another version of Tonelli's lemma.

\begin{lemma}
\label{lemT2} (Tonelli's Lemma) Let $y_{1},y_{2}\in C[a,b]$ be such that:

\begin{itemize}
\item[(i)] $y_{1}(a)=0,y_{1}(b)=0$ and $y_{1}(x)\neq0,\forall x\in]a,b[;$

\item[(ii)] $y_{2}(x)\neq0$ $,$ $\forall$ $x\in\lbrack a,b].$
\end{itemize}

Then there exists $\lambda\in\mathbb{R}^{\ast}$ and $x_{0}\in\lbrack a,b]$
such that:%
\[
y_{2}(x_{0})=\lambda y_{1}(x_{0})\text{ and }y_{2}^{\prime}(x_{0})=\lambda
y_{1}^{\prime}(x_{0}).
\]

\end{lemma}

\begin{remark}
For Lemma \ref{lem4.3} see also: \cite{IOR2015}, \cite{Muresan} and
\cite{Rus96}.
\end{remark}

Our results are the following.

\begin{theorem}
\label{th4.2} (Nicolescu-type separation theorem) For the system
(\ref{ec1.1}), we suppose that:

\begin{itemize}
\item[(i)] $F$ and $G$ are homogeneous with respect to the last three arguments;

\item[(ii)] $F(x,y,z,\cdot)\ $and$\ G(x,y,z,\cdot)$ are increasing$,\ \forall
x\in\lbrack a,b];$

\item[(iii)] $F(x,1,\lambda,1)\neq0,\ G(x,\lambda,1,1)\neq0,$ for all
$\lambda\in\mathbb{R}^{\ast}.$
\end{itemize}

Then, if $(y,z)$ is a solution of (\ref{ec1.1}), the zeros of $y$ and $z$
separate each other.
\end{theorem}

\begin{proof}
We consider $x_{1}$ and $x_{2}$ two consecutive zeros of $y(x)$. We have to
prove that $z(x)$ has at least one zero in the interval $(x_{1},x_{2})$.

We suppose that $z(x)\neq0,\ x\in\lbrack x_{1},x_{2}]$. Applying Tonelli's
Lemma \ref{lemT2} there exists $x_{0}\in(x_{1},x_{2})$ and $\lambda
\in\mathbb{R}^{\ast}$ such that%
\[
z(x_{0})=\lambda y(x_{0}),\ z^{\prime}(x_{0})=\lambda y^{\prime}(x_{0}).
\]
From (\ref{ec1.1}) we have%
\[
F(x_{0},y(x_{0}),\lambda y(x_{0}),y^{\prime}(x_{0}))=0.
\]
We suppose that $y(x_{0})\geq y^{\prime}(x_{0}).$ Then%
\begin{align*}
0  &  =F(x_{0},y(x_{0}),\lambda y(x_{0}),y^{\prime}(x_{0}))\leq F(x_{0}%
,y(x_{0}),\lambda y(x_{0}),y(x_{0}))=\\
&  =y(x_{0})F(x_{0},1,\lambda,1)<0,\\
0  &  =G(x_{0},y(x_{0}),\lambda y(x_{0}),\lambda y^{\prime}(x_{0}))\leq
G(x_{0},y(x_{0}),\lambda y(x_{0}),\lambda y(x_{0}))=\\
&  =\lambda y(x_{0})G(x_{0},\tfrac{1}{\lambda},1,1)<0,
\end{align*}
so we have reached a contradiction.

If $y(x_{0})\leq y^{\prime}(x_{0}).$ Then%
\begin{align*}
0  &  =F(x_{0},y(x_{0}),\lambda y(x_{0}),y^{\prime}(x_{0}))\geq F(x_{0}%
,y(x_{0}),\lambda y(x_{0}),y(x_{0}))=\\
&  =y(x_{0})F(x_{0},1,\lambda,1)>0,\\
0  &  =G(x_{0},y(x_{0}),\lambda y(x_{0}),\lambda y^{\prime}(x_{0}))\geq
G(x_{0},y(x_{0}),\lambda y(x_{0}),\lambda y(x_{0}))=\\
&  =\lambda y(x_{0})G(x_{0},\tfrac{1}{\lambda},1,1)>0,
\end{align*}
so we have reached a contradiction.
\end{proof}

\begin{theorem}
\label{th4.3} (Butlewski-type separation theorem) For the homogeneous system
(\ref{ec1.1}), we suppose that it satisfies condition (C$_{2}$). Then, if
$(y_{1},z_{1})$ and $(y_{2},z_{2})$ are two linear independent solutions of
(\ref{ec1.1}), then the zeros of $y_{1}$ and $y_{2}$ ($z_{1}$ and $z_{2}$)
separate each other on $[a,b]$.
\end{theorem}

\begin{proof}
We consider $x_{1}$ and $x_{2}$ two consecutive zeros of $y_{1}(x)$. We have
to prove that $y_{2}(x)$ has at least one zero in $[x_{1},x_{2}]$.

We suppose that $y_{2}(x)\neq0,\ x\in\lbrack x_{1},x_{2}]$. Applying Tonelli's
Lemma \ref{lemT2} there exists $x_{0}\in(x_{1},x_{2})$ and $\lambda
\in\mathbb{R}^{\ast}$ such that%
\[
y_{2}(x_{0})=\lambda y_{1}(x_{0}),\ y_{2}^{\prime}(x_{0})=\lambda
y_{1}^{\prime}(x_{0}).
\]
Taking into account (C$_{2}$) we have that $y_{2}(x)=\lambda y_{1}(x)$ and so
we have reached a contradiction.
\end{proof}

\textbf{Acknowledgement} The work of the first author was partially supported
by a grant of the Romanian National Authority for Scientific Research, CNCS --
UEFISCDI, project number PN-II-ID-PCE-2011-3-0094.
%\end{acknowledgement}


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\end{document}