\documentclass[reqno]{amsart}
\usepackage{hyperref}
\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2007(2007), No. 160, 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 (login: ftp)}
\thanks{\copyright 2007 Texas State University - San Marcos.}
\vspace{9mm}}
\begin{document}
\title[\hfilneg EJDE-2007/160\hfil Resonant problem]
{Resonant problem for some second-order differential equation
on the half-line}
\author[K. Szyma\'nska\hfil EJDE-2007/160\hfilneg]
{Katarzyna Szyma\'nska}
\address{Katarzyna Szyma\'nska \newline
Institute of Mathematics\\
Technical University of Lodz\\
90-924 Lodz, ul. W\'olcza\~nska 215, Poland}
\email{grampa@zb.net.lodz.pl}
\thanks{Submitted June 20, 2007. Published November 21, 2007.}
\thanks{Supported by grant 1 PO3A 001 28 from MNiI}
\subjclass[2000]{34B15, 34B40}
\keywords{Asymptotic boundary value problem; resonant problem;
\hfill\break\indent nonlinear boundary value problem; set-valued maps}
\begin{abstract}
We prove the existence of at least one solution to a nonlinear
second-order differential equation on the half-line,
with the boundary conditions $x'(0)=0$ and with the first derivative
vanishing at infinity. Our main tool is the multi-valued version
of the Miranda Theorem.
\end{abstract}
\maketitle
\numberwithin{equation}{section}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{proposition}[theorem]{Proposition}
\section{Introduction}
Most nonlinear differential, integral or, more generally, functional equations
have the form $Lx=N(x)$, where $L$ is a linear and $N$ nonlinear operator,
in appropriate Banach spaces.
We have no problem if $L$ is a linear Fredholm operator of index $0$.
Then the kernel of the linear part of the above equation is trivial.
It means that there exists an integral operator and we can apply known
topological methods to prove the existence theorems
\cite{Agarwal,Jackson,Lipovan}.
If kernel $L$ is nontrivial then the equation is called resonant and one can
manage the problem by using the coincidence degree in that case \cite{Mawhin}.
But, if the domain is unbounded (for example the half-line) the operator
is usually non-Fredholm
(the range of $L$ is not a closed subspace in any reasonable Banach space)
Such problems have
been studied by different methods in many papers. We mention only
\cite{Andres,Const, Furi,Przera,Rabier,Regan}.
For instant, the perturbation method
was developed in a series of papers \cite{Karp,Karp,Karpi,Sta,St}.
We consider the following asymptotic boundary-value problem (BVP)
on the half-line
\[
x''=f(t,x,x') , \quad x'(0)=0 , \quad \lim_{t\to \infty }{x'(t)}=0.
\]
The problem is resonant, since the corresponding homogeneous linear problem:
$x''=0,\quad x'(0)=\lim _{t\to \infty }x'(t)=0$ has nontrivial
solutions - constant functions.
Similar problem was considered in \cite{Rabier}. There, the asymptotic boundary condition
$\lim _{t\to \infty }{x'(t)}=0$ is replaced by $x\in H^{2}(\mathbb{R}_{+})$
that is close but not the same as our one and assumptions are of
completely different kind.
Our problem has been already studied in \cite{Szym}.
In this paper, we have obtained the existence result in a completely different way
than by using standard methods for resonant problems. This enables us to get it under
weak assumptions: a linear growth condition and
a sign condition for the nonlinear term $f$. Similar assumptions appear also
for other boundary-value problems.
But there we assumed also that function $f$ is Lipschitz continuous. It
was an artificial condition. We needed it only to show that defined
there mapping $g$ is a function and then to apply the Miranda
Theorem \cite[p. 124]{PSV}. In this paper we
omit the assumption of Lipschitz continuity. In this way the
mapping $g$ becomes a multifunction. Consequently, to get the
Theorem about the existence of at least one solution to the
resonant problem we have to prove the multi-valued version of
Miranda Theorem (see Appendix, Theorem \ref{miranda}).
In this paper we apply Theorem \ref{miranda} only in case $k=1$.
But as we know the Theorems of the type of Miranda's theorem has many
applications for instance in control theory \cite{Marek}. That's why it
seams that Theorem \ref{miranda} can be very useful as well.
\section{The main result}
Let us consider an asymptotic BVP
\begin{equation}
\label{eq:pr}
x''=f(t,x,x'), \quad x'(0)=0 , \quad \lim_{t\to \infty } ,
{x'(t)}=0
\end{equation}
where $f:\mathbb{R}_{+}\times \mathbb{R} \times \mathbb{R}
\to \mathbb{R} $ is continuous.
The following assumption will be needed in this paper:
\begin{itemize}
\item[(i)] $|f(t,x,y)|\leq b(t)|y|+c(t)$,
where $b, c \in L^{\infty}(0, \infty)$;
\item[(ii)] there exists $M> 0$ such that $xf(t,x,y)\geq 0$ for
$t\geq 0$, $y \in \mathbb{R}$ and $|x|\geq M$.
\end{itemize}
First, we consider problem
\begin{equation}
\label{eq:zpy}
y'=f(t,c+\int_{0}^{t}{y},y), \quad y(0)=0 ,
\end{equation}
for fixed $c\in \mathbb{R}$. Observe that \eqref{eq:zpy} is equivalent to an
initial value problem
\begin{equation}
\label{eq:zpx}
x''=f(t,x,x'), \quad x(0)=c, \quad x'(0)=0 .
\end{equation}
Since $f$ is continuous, then by assumption (i) and the Local Existence
Theorem we get that problem
\eqref{eq:zpx} has at least one local solution.
We can write \eqref{eq:zpy} as
\begin{equation}
\label{eq:yc}
y_{c}(t)=\int_{0}^{t}{f\Big(s,c+\int_{0}^{s}{y_{c}(u)}du,y_{c}(s)\Big)}ds
\end{equation}
Set
\begin{equation}
\label{eq:abc}
B:= \int_{0}^{\infty }{b(s)ds}, \quad
C:= \int_{0}^{\infty }{c(s)ds}.
\end{equation}
By (i) and \eqref{eq:abc} we get
\begin{equation}
\label{eq:wspogr a}
|y_{c}(t)|\leq \int_{0}^{t}{(b(s)|y_{c}(s)|+c(s))}ds
\leq C + \int_{0}^{t}{b(s)|y_{c}(s)|}ds .
\end{equation}
Now, due to Gronwall's Lemma \cite[p. 17]{PSV}, we have
\[
|y_{c}(t)|\leq C{\rm \exp}\int_{0}^{t}{b(s)}ds
\]
Hence, by the Theorem on a Priori Bounds \cite[ p. 146]{PSV},
\eqref{eq:zpx} has a global solution for $t\geq 0$.
We obtain that \eqref{eq:zpy} has a global solution for $t\geq 0$.
Moreover, by assumption (i) and \eqref{eq:abc}, we have
\begin{equation}
\label{eq:wspogr}
|y_{c}(t)|\leq C{\rm \exp}\int_{0}^{t}{b(s)}ds
\leq C{\rm \exp}\int_{0}^{\infty}{b(s)}ds = C{\rm \exp} B < \infty
\end{equation}
Hence all global solutions are bounded for $t\geq 0$.
The function $t \mapsto f(t,c+\int_{0}^{t}{y(u)}du,y(t))$
is absolutely integrable; i.e.,
\[
\forall_{\varepsilon > 0} \quad \exists_{M > 0} \quad
\big|\int_{M}^{\infty}{f\Big(t,c+\int_{0}^{t}{y(u)}du,y(t)\Big)}dt\big|
< \varepsilon .
\]
In particular, there exists a limit
$\lim_{t\to\infty}{y_{c}(t)}$, for every $c$.
Thus all solutions of \eqref{eq:zpy} have finite limits at $+\infty$.
Denote by $BC(\mathbb{R}_{+},\mathbb{R})$ the space of continuous and
bounded functions with supremum norm and by $BCL(\mathbb{R}_{+},\mathbb{R})$
its closed subspace of continuous and bounded functions
which have finite limits at $+\infty$.
Let us consider the nonlinear operator
$F:BCL(\mathbb{R}_{+},\mathbb{R})\to BCL(\mathbb{R}_{+},\mathbb{R})$ given by
\begin{equation}\label{eq:fyc}
F(c,x)(t)=\int_{0}^{t}{f(s,c+\int_{0}^{s}{x(u)du},x(s))ds},
\end{equation}
where $c\in \mathbb{R}$ is fixed.
It is easy to see that $F$ is well defined. By using
the Lebesgue Dominated Convergence Theorem one can prove the continuity
of $F$.
The following theorem gives a sufficient condition for compactness
in the space $BC(\mathbb{R}_{+},\mathbb{R})$, (\cite{Przerad}).
\begin{theorem}\label{kzbc}
If $A \subset BC(\mathbb{R}_{+},\mathbb{R})$
satisfies following conditions:
\begin{enumerate}
\item There exists $M > 0$, that for every $x\in A$ i
$t\in [0 , \infty )$ we have $|x(t)|\leq L$;
\item for each $t_{0} \geq 0$, the family $A$ is equicontinuous at $t_{0}$;
\item for each $\varepsilon > 0$ there exists $T> 0$ and $\delta >0$
such that if $|x(T)-y(T)|\leq \delta $, then
$|x(t)-y(t)|\leq \varepsilon $ for $t\geq T$ and all
$x,y\in A$.
\end{enumerate}
Then $A$ is relatively compact in $BC(\mathbb{R}_{+},\mathbb{R})$.
\end{theorem}
Now, we can prove that operator $F$ is completely continuous.
\begin{proposition}\label{pelopf}
Under assumption (i) operator $F$ is completely continuous.
\end{proposition}
\begin{proof}
We shall show that the image of $A:= \{x\in
BCL(\mathbb{R}_{+},\mathbb{R}) \mid
\|x\|_{BC(\mathbb{R}_{+},\mathbb{R})}\leq M\}$
under $F$ is relatively compact.
Condition (1) of Theorem \ref{kzbc} is satisfied, since
$|F(c,x)(t)|\leq C+M B$.
Now, we prove condition (2). We show that for any $t_{0}\geq 0 $ and
$\varepsilon > 0$ there exists $\delta > 0$ that for each $x\in B$
if $|t-t_{0}|< \delta$, then
$|F(c,x)(t)-F(c,x)(t_{0})| < \varepsilon $.
Let us choose an arbitrary $\varepsilon > 0$.
By (i) there exist
$\delta_{1}$, $\delta_{2} > 0$ such that
\begin{gather*}
\text{if } |t-t_{0}|< \delta_{1}, \text{ then }
\int_{\min \{t_{0},t\}}^{\max \{t_{0},t\}}
{b(s)ds} < \frac{\varepsilon }{2 M}, \\
\text{if } |t-t_{0}|< \delta_{2}, \text{ then }
\int_{\min \{t_{0},t\}}^{\max \{t_{0},t\}}
{c(s)ds} < \frac{\varepsilon }{2}.
\end{gather*}
Let $\delta =\min \{\delta _{1}, \delta _{2}\} $. Then, for
$|t-t_{0}|< \delta $, we get
\begin{align*}
|F(c,x)(t) -F(c,x)(t_{0})|
&\leq \int_{\min \{t_{0},t\}}^{\max \{t_{0},t\}}
{|f(s,c+\int_{0}^{s}{x(u)du},x(s))|ds}\\
&\leq M \int_{\min \{t_{0},t\}}^{\max \{t_{0},t\}}
{b(s)ds} + \int_{\min \{t_{0},t\}}^{\max
\{t_{0},t\}} {c(s)ds}\\
& < M\frac{\varepsilon }{2 M}+ \frac{\varepsilon }{2} =\varepsilon .
\end{align*}
It remains to prove condition (3). By assumption
(i) for every $\varepsilon > 0$ there exist $t_{1}$,
$t_{2}$ large enough that
\[
\int_{t_{1}}^{\infty }{b(s)ds} < \frac{\varepsilon }{6 M}, \quad
\int_{t_{2}}^{\infty }{c(s)ds} < \frac{\varepsilon }{6}.
\]
Let $T = \max \{t_{1}, t_{2}\}$
and $\delta := \varepsilon/3$.
If $|F(c,x)(T)-F(c,y)(T)|\leq \delta $, then for $t\geq T$ we get
\begin{align*}
&|F(c,x)(t)-F(c,y)(t) | \\
&\leq |F(c,x)(T)-F(c,y)(T)|
+\int_{T}^{\infty} {\big|f(s,c+\int_{0}^{s}{x(u)}du,x(s))\big|}ds\\
&\quad + \int_{T}^{\infty}{\big|f(s,c+\int_{0}^{s}{y(u)}du,y(s))\big|}ds \\
&\leq |F(c,x)(T)-F(c,y)(T)|+
2\int_{T}^{\infty}{M b(s)}ds+
2\int_{T}^{\infty}{c(s)}ds \\
&\leq \frac{\varepsilon }{3 }+2\frac{\varepsilon }{6 M}
+2\frac{\varepsilon }{6} =\varepsilon.
\end{align*}
The proof is complete.
\end{proof}
Note that the solutions of \eqref{eq:zpy} are fixed points of operator
$F$ defined by \eqref{eq:fyc}. Let $\mathop{\rm fix} F(c,\cdot )$ denote the
set of fixed points of operator $F$, where $c$ is given.
Let us consider the multifunction $g:\mathbb{R}\to 2^{\mathbb{R}}$ given by
\[
g(c):=\{\lim_{t\to\infty}{y_{c}(t)}:y_{c}\in\mathop{\rm fix} F(c,\cdot )\}.
\]
We will show that $g$ is upper semicontinuous map from $\mathbb{R}$
into its compact connected subsets.
To prove this we will define two maps: $\varphi$ and $\Phi $.
Let us consider the function
$\varphi : BCL(\mathbb{R}_{+},\mathbb{R})\to\mathbb{R}$
given by
\[
\varphi (y_{c})=\lim_{t\to\infty}{y_{c}(t)}.
\]
It is easily seen that function $\varphi $ is continuous.
Set
\[
\Phi :\mathbb{R}\ni c\to \mathop{\rm fix} F(c,\cdot )
\subset BCL(\mathbb{R}_{+},\mathbb{R}).
\]
We can now formulate the following result.
\begin{proposition}\label{stw1}
Let assumption {\rm (i)} hold. Then the
multi-valued map $\Phi$ is upper semicontinuous, that is for each
$c_{0}\in \mathbb{R}$ and for any neighborhood $U\subset
{BCL(\mathbb{R}_{+},\mathbb{R})} $ of $\Phi (c_{0})$ there
exists a neighborhood $V$ of $c_{0}$ such that $\Phi (c) \subset
U$, for all $c\in V$.
\end{proposition}
\begin{proof}
Suppose, contrary to our claim, that $\Phi$ is not upper semicontinuous;
i. e., for some $c_{0}\in \mathbb{R}$
there exist neighborhood $U\subset BCL(\mathbb{R}_{+},\mathbb{R})$
of $\mathop{\rm fix}F(c_{0},\cdot )$ and sequence
$c_{n} \to c_{0}$ and
$x_{n}\in\mathop{\rm fix}F(c_{n},\cdot )\setminus U$.
By \eqref{eq:wspogr} we get that the solutions of \eqref{eq:zpy} are
equibounded for any $c$. Hence the sequence $(x_{n})$ is bounded.
Moreover, we have
\begin{equation}
\label{eq:wzf} x_{n}=F(c_{n},x_{n}). \end{equation}
Proposition \ref{pelopf} yields
that operator $F$ is completely continuous.
Then, by \eqref{eq:wzf},
$(x_{n})$ is relatively compact. Hence,
from the sequence $(x_{n})$, we can extract a
subsequence $(x_{n_{l}})$ which is convergent to some $x_{0}$ in
$BCL(\mathbb{R}_{+},\mathbb{R})$. Moreover, $c_{n_{k}}\to c_{0}$.
Letting $k\to\infty$ in the equality
\[
x_{n_{k}}=F(c_{n_{k}},x_{n_{k}}),
\]
we get
\[
x_{0}=F(c_{0},x_{0}).
\]
Hence, $x_{0}\in \mathop{\rm fix}F(c_{0},\cdot )\subset U$.
This contradicts the fact that
$x_{n_{k}}\in\mathop{\rm fix}F(c_{n_{k}},\cdot )\setminus U$, for every
$n_{k}$, and completes the proof.
\end{proof}
\begin{proposition}\label{stw2}
Let assumption (i) hold. Then the set-valued map $\Phi$
has compact and connected values.
\end{proposition}
\begin{proof}
By \eqref{eq:wspogr} we know that the set of solutions of IVP \eqref{eq:zpy}
is equibounded for any $c$. Now, the proof that $\mathop{\rm fix} F(c,\cdot )$
is relatively compact in $BCL(\mathbb{R}_{+},\mathbb{R})$ follows
by the same method as in Proposition \ref{pelopf}.
We next show that $\mathop{\rm fix} F(c,\cdot )$ is closed.
Let $(y_{n})$ be an arbitrary convergent sequence such that
$y_{n}\in \mathop{\rm fix} F(c,\cdot )$,
and let $y_{n}\to y$. We have
\begin{equation}
\label{eq:prawastrona}
F(c,y_{n})(t)=\int_{0}^{t}{f(s,c+\int_{0}^{s}{y_{n}(u)du},y_{n}(s))ds}.
\end{equation}
The sequence $(y_{n})$ is bounded, since it is convergent.
Moreover $y_{n}$ is uniformly convergent to $y$ on $[0,\infty)$.
By
(i) and the Lebesgue Dominated Convergence Theorem,
letting $n\to\infty$, we get
\[
\lim_{n\to\infty}F(c,y_{n})(t)=\int_{0}^{t}{f(s,c+\int_{0}^{s}{y(u)du},y(s))ds}.
\]
On the other hand, $y_{n}= F(c,y_{n} )$ and consequently
\[
y(t)=\lim_{n\to\infty}y_{n}(t)=\lim_{n\to\infty}F(c,y_{n})(t)=F(c,y)(t)
\]
Hence $y\in \mathop{\rm fix} F(c,\cdot )$. From the above it follows
that $\mathop{\rm fix} F(c,\cdot )$ is compact.
It is left is to show that $\mathop{\rm fix} F(c,\cdot )$ is connected in
$BCL(\mathbb{R}_{+},\mathbb{R})$.
On the contrary, suppose that the set is not connected.
Since $\mathop{\rm fix} F(c,\cdot )$ is compact, there exist compact sets
$A$ and $B$ such that $A,B\neq \emptyset $, $A\cap B=\emptyset $ and
$A\cup B=\mathop{\rm fix} F(c,\cdot )$.
Let $\varepsilon :=\mathop{\rm dist} ( A,B)$, $\varepsilon > 0$. Then
\begin{equation}
\label{eq:sp1}
\forall _{y\in A, z\in B} \quad \|y-z\|\geq \varepsilon .
\end{equation}
By \eqref{eq:wspogr}
there exists $T> 0$ such that for any $y\in \mathop{\rm fix} F(c,\cdot )$ we get
\begin{equation}
\label{eq:sp3}
\int_{T}^{\infty}{\big|f\big(t,c+\int_{0}^{t}{y(u)}du,y(t)\big)\big|}dt<
\frac{1}{3} \varepsilon .
\end{equation}
Let $y\in A$, i $z\in B$. Now, consider the functions $y$ and $z$
cut to the compact set $[0,T]$ and set $y|_{[0,T]}$ and
$z|_{[0,T]}$. By Kneser's Theorem \cite[p. 413]{krasnosielski},
the set $\mathop{\rm fix} F(c,\cdot )|_{[0,T]}$ is connected in
$C([0,T],\mathbb{R})$. From this, there exist $x_{1}, \dots
,x_{k}$ in $\mathop{\rm fix} F(c,\cdot )|_{[0,T]}$ such that
$x_{1}=y\mid _{[0,T]}$, $x_{k}=z\mid _{[0,T]}$ and
\begin{equation}
\label{eq:sp2}
\|x_{i}-x_{i+1}\|_{C([0,T],\mathbb{R})} <\frac{1}{3}\varepsilon .
\end{equation}
Hence, at least two sequel $x_{i}$, $x_{i+1}$ in
$\mathop{\rm fix} F(c,\cdot )|_{[0,T]}$ are such that
$x_{i} \in A$ i $x_{i+1} \in B$.
Moreover, $x_{i}, x_{i+1}\in \mathop{\rm fix}F(c,\cdot )$.
By \eqref{eq:sp1}, \eqref{eq:sp3} and \eqref{eq:sp2} we get a contradiction.
Indeed
\begin{align*}
\varepsilon &\leq \|x_{i}-x_{i+1}\|_{BCL(\mathbb{R}_{+},\mathbb{R})}\\
&\leq \max \{\|x_{i}-x_{i+1}\|_{C([0,T],\mathbb{R})},
\sup _{t\geq T}|x_{i}(t)-x_{i+1}(t)| \} \\
&\leq \|x_{i}-x_{i+1}\|_{C([0,T],\mathbb{R})} + \sum_{j=i}^{i+1}{
\int_{T}^{\infty}{|f(t,c+\int_{0}^{t}{x_{j}(u)}du,x_{j}(t))|}dt} \\
&< \frac{1}{3}\varepsilon +\frac{1}{3}\varepsilon +\frac{1}{3}\varepsilon
=\varepsilon .
\end{align*}
Hence, $\mathop{\rm fix} F(c,\cdot )$ is connected in
$BCL(\mathbb{R}_{+},\mathbb{R})$, which proves the Proposition.
\end{proof}
\begin{proposition} \label{fung}
Let assumption (i) hold. Then the multifunction $g$ is upper semicontinuous
map from $\mathbb{R}$ into its compact intervals.
\end{proposition}
\begin{proof}
By Proposition \ref{stw2} we know that $\Phi$ has compact and connected
values. Now, by continuity of $\varphi$, the set
\[
\{\lim_{t\to\infty}y_{c}(t)\mid y_{c}\in \mathop{\rm fix}F(c,\cdot)\}
\]
is compact and connected subset of $\mathbb{R}$ for
every $c$. From this, we conclude that multifunction $g(c)=(\varphi
\circ \Phi )(c)$ maps $\mathbb{R}$ into its compact intervals.
Moreover, $g$ is upper semicontinuous as a superposition of
set-valued map with compact values and continuous function
\cite[p. 47]{AF}.
\end{proof}
We can now formulate our main result.
\begin{theorem}
Under assumption (i)-(ii) problem \eqref{eq:pr} has at least one solution.
\end{theorem}
\begin{proof}
Let $y_{c}\in\mathop{\rm fix} F(c,\cdot )$ be the bounded global
solution of \eqref{eq:zpy} and
\[
g(c):=\{\lim_{t\to\infty}{y_{c}(t)}\mid y_{c}\in\mathop{\rm fix}
F(c,\cdot )\}.
\]
Observe that $x(t)=c+\int_{0}^{t}{y_{c}(s)}ds$ is a solution of
\eqref{eq:pr} if there
exists an $c\in \mathbb{R}^{k}$ such that $0\in g(c)$.
We shall show that $g$ satisfies assumptions of
the multi-valued version of Miranda Theorem for $k=1$ (see Appendix).
By Proposition \ref{fung} we get that
multifunction $g$ is upper semicontinuous map from $\mathbb{R}$
into its compact (so convex) intervals.
Let $c=M+1$, where $M> 0$ is the constant from assumption (ii).
Set $\widehat{M}:=M+1$.
We will show that $y_{\widehat{M}}(t)\geq 0$ for $t\geq 0$ and all
$y_{\widehat{M}}\in \mathop{\rm fix}F(\widehat{M},\cdot )$.
By \eqref{eq:zpy} we have $y_{\widehat{M}}(0)=0$. Assume that for some t
and $y_{\widehat{M}}\in \mathop{\rm fix}F(\widehat{M},\cdot )$ we have $y_{\widehat{M}}(t)< 0$.
Then there exist $t_{\ast} :=\inf \{{t\mid y_{\widehat{M}}(t)<
0}\}$ such that $y_{\widehat{M}}(t_{\ast})=0$ and $y_{\widehat{M}}(t)\geq 0$
for $t< t_{\ast}$ (if $t_{\ast}\neq 0$).
By continuity of $y_{\widehat{M}}(t)$ there exists
$t_{1} > t_{\ast}$ such that
$ \int_{t_{\ast}}^{t_{1}}{|y_{\widehat{M}}(t)|}dt\leq 1$.
Hence, we get
\begin{equation}
x(t)=c+\int_{t_{\ast}}^{t}{y_{\widehat{M}}(s)}ds =
\widehat{M}+\int_{t_{\ast}}^{t}{y_{\widehat{M}}(s)}ds\geq M \quad \text{for}
\quad t\in \left[t_{\ast}, t_{1}\right]. \nonumber
\end{equation}
Now, by
(ii) we have
\[
x(t)f(t,x(t),y(t))=x(t)y'_{M+1}(t)\geq 0 .
\]
Hence $y'_{\widehat{M}}(t)\geq 0$ for $t\in [t_{\ast},t_{1}]$.
It means that $y_{\widehat{M}}(t)$ is nondecreasing on
$[t_{\ast}, t_{1}]$.
Since $y_{\widehat{M}}(t_{\ast})=0$ we get
a contradiction. Hence $y_{\widehat{M}}(t)\geq 0$ for $t\geq 0$. In
consequence, if $d\in g(\widehat{M})=\lim_{t\to \infty}{y_{M+1}(t)}$, then
$d\geq 0$.
To prove that if $d\in g(-\widehat{M})$, then $d\leq 0$ we proceed
analogously.
Hence, by multi-valued version of Miranda Theorem, there exists an
$c\in [-\widehat{M},\widehat{M}]$ such that, $0 \in g(c)$. This completes the
proof.
\end{proof}
\section{Appendix}
The theorem below is a generalization of the well known Miranda Theorem
\cite[p. 214]{PSV} which gives zeros of point-valued continuous maps.
\begin{theorem} \label{miranda}
Let $g$ be an upper semicontinuous map from the hypercube
$[-\widehat{M},\widehat{M}]^{k}$
into convex and compact subsets of $\mathbb{R}^{k}$ and satisfying for
$d=(d_{1},\dots ,d_{k})\in \mathbb{R}^{k}$ the conditions
\begin{equation}
\label{eq:wieksze}
\text{if } d\in g(x_{1},\dots ,
x_{i-1},\widehat{M},x_{i+1},\dots ,x_{k}), \text{ then }
d_{i}\geq 0 \end{equation}
and
\begin{equation}
\label{eq:mniejsze}
\text{if }d\in g(x_{1},\dots , x_{i-1},-\widehat{M},x_{i+1},\dots ,x_{k}),
\text{ then } d_{i}\leq 0 ,
\end{equation}
for every $i=1,\dots,k$. Then there exists
$\widetilde{x}\in [-\widehat{M},\widehat{M}]^{k}$ such that
$0\in g(\widetilde{x})$.
\end{theorem}
\begin{proof}
To prove this theorem, first of all suppose that the inequalities
in \eqref{eq:wieksze} and \eqref{eq:mniejsze} hold in the strict
sense. Let $g_{i}=\text{P}_{i}g$ for $i=1,\dots ,k$, where
$\text{P}_{i}$ is the projection of multifunction $g$ on i-th
axis. By \eqref{eq:wieksze} and \eqref{eq:mniejsze} for
$i=1,\dots ,k$ we have
\begin{gather*}
g_{i}(x_{1},\dots ,
x_{i-1},\widehat{M},x_{i+1},\dots ,x_{k}) \subset (0,\infty),\\
g_{i}(x_{1},\dots , x_{i-1},-\widehat{M},x_{i+1},\dots ,x_{k}) \subset
(-\infty ,0).
\end{gather*}
It is easy to see that $g_{i}$ is upper semicontinuous map from
$[-\widehat{M},\widehat{M}]^{k}$
into compact convex intervals for every $i$.
By \eqref{eq:wieksze} and the fact that $g_{i}$ is upper
semicontinuous there exists $\gamma _{i}> 0$ such that for any
$x\in [-\widehat{M},\widehat{M}]^{k}$, where
$x_{i}\in (\widehat{M}-\gamma _{i},\widehat{M}]$, we get
$g_{i}(x)\subset (0 , \infty)$, for every $i=1,\dots ,k$.
Similarly, by \eqref{eq:mniejsze} and the fact that $g_{i}$ is
upper semicontinuous there exists $\beta _{i}> 0$ such that for
any $x\in [-\widehat{M},\widehat{M}]^{k}$, where
$x_{i}\in [-\widehat{M}, -\widehat{M}+\beta _{i})$, we
have $g_{i}(x)\subset (-\infty , 0)$, for every $i=1,\dots ,k$.
An upper semicontinuous map with compact values from a compact space
has compact graph (\cite{Aubin}). Hence
\[
\widehat{g}:=\sup \{|d|\mid d\in g_{i}(x),
x\in [-\widehat{M},\widehat{M}]^{k}, i=1, \dots, k\}<\infty .
\]
Let $\delta := \min \{\beta _{1}, \dots ,\beta _{k},
\gamma _{1}, \dots ,\gamma _{k}, \widehat{M}\}$.
Set $\varepsilon := \frac{\delta }{\widehat{g}}$ and consider the
set-valued mapping given by
\[
F_{i}(x)=x_{i}-\varepsilon g_{i}(x).
\]
Then, for any $x_{i}\in[-\widehat{M}+\delta ,\widehat{M}-\delta ]$ and
$y\in g_{i}(x)$ we have
\[-\widehat{M}=-\widehat{M}+\delta - \delta =-\widehat{M} +\delta
-\varepsilon \widehat{g} \leq x_{i}-\varepsilon y\leq \widehat{M}-\delta +
\varepsilon \widehat{g} =\widehat{M}-\delta +\delta =\widehat{M}.
\]
For $x_{i}\in [-\widehat{M},-\widehat{M} +\delta)$,
if $y \in g_{i}(x)$, then $y<0$ and from this
$-\varepsilon y> 0$. We get
\[-\widehat{M}\leq x_{i} \leq
x_{i}-\varepsilon y\leq -\widehat{M} +\delta +\varepsilon \widehat{g}\leq
-\widehat{M}+2\delta \leq \widehat{M}.
\]
Next, for $x_{i}\in (\widehat{M}-\delta ,\widehat{M}]$, if
$y\in g_{i}(x)$, then $y> 0$ and in this way $-\varepsilon y< 0$.
We get
\[
-\widehat{M}\leq \widehat{M}-2\delta \leq \widehat{M}-\delta
-\varepsilon \widehat{g}
\leq x_{i}-\varepsilon y\leq x_{i} \leq \widehat{M}.
\]
Now, let us
consider the multi-valued mapping \[F(x)=x-\varepsilon g(x) ,
\]where $\varepsilon := \frac{\delta }{\widehat{g}}$ and $x\in
[-\widehat{M},\widehat{M}]^{k}$. By the assumptions of multifunction $g$, $F$ is upper
semicontinuous with convex and compact values in $\mathbb{R}^{k}$.
However, we know more. We get that $F$ maps from the hypercube
$[-\widehat{M},\widehat{M}]^{k}$ into its compact and convex sets. Indeed, the
projection $\text{P}_{i}F=F_{i}$ of $F$ on i-th axis is compact
interval contained in $[-\widehat{M},\widehat{M}]$ for every $i=1,\dots ,k$.
Hence, by Kakutani's Theorem (\cite{Aubin}), there exists
$\overline{x}\in [-\widehat{M},\widehat{M}]^{k}$ such that
$\overline{x}\in F(\overline{x})$.
On the other hand
\[
F(\overline{x})=\overline{x}-\varepsilon g(\overline{x}).
\]
Thus $0\in F(\overline{x})-\overline{x}=-\varepsilon g(\overline{x})$,
and from this $0\in g(\overline{x})$.
Now, suppose that the inequalities in \eqref{eq:wieksze} and
\eqref{eq:mniejsze} hold in a weak sense. Then for the following
set-valued mapping
\[H_{n}(x)=g(x)+\frac{1}{n}x, \quad
n\in\mathbb{N}
\]
we have sharp inequalities. By the first part of the proof it
follows that there exists $\widetilde{x}_{n}\in[-\widehat{M},\widehat{M}]^{k}$
such that $0\in H_{n}(\widetilde{x}_{n})$, for every $n$. Hence $0\in
g(\widetilde{x}_{n})+\frac{1}{n}\widetilde{x}_{n}$. Since sequence
$(\widetilde{x}_{n})$ is bounded, we can extract a convergent subsequence
of $(x_{n})$. Let $(\widetilde{x}_{n})$ denote
the subsequence and let $\widetilde{x}_{n}\to \widetilde{x}$.
It suffices to show that $0\in g(\widetilde{x})$.
Suppose, contrary to our claim, that $0\notin g(\widetilde{x})$.
Let $\eta :=\mathop{\rm dist} (0,g(\widetilde{x}))$.
Since $g(\widetilde{x})$ is compact, we have that $\eta >0$.
Choose neighborhood $U:=\{y \mid
|y-z|<\frac{1}{3}\eta , z\in g(\widetilde{x}) \}$
of $g(\widetilde{x})$.
For every $n$ we have $0\in
g(\widetilde{x}_{n})+\frac{1}{n}\widetilde{x}_{n}$. From this
there exists $y_{n}\in g(\widetilde{x}_{n})$ such that
$y_{n}+\frac{1}{n}\widetilde{x}_{n}=0$, $n\in\mathbb{N}$.
In particular, $g$ is upper semicontinuous at $\widetilde{x}$.
Hence, there exists $N_{1} $ such that for every $n\geq N_{1}$
we have $g(\widetilde{x}_{n})\subset U$.
Moreover, $\frac{1}{n}\widetilde{x}_{n}\to 0$, as $n\to \infty$.
Hence, there exists $N_{2}$ such that for any $n\geq N_{2}$ we have
$|\frac{1}{n}\widetilde{x}_{n}|<\frac{1}{3}\eta $.
Set $N=\max \{N_{1},N_{2}\}$. Then for $n\geq N$ we get
$|y_{n}|=|-\frac{1}{n}\widetilde{x}_{n}|<\frac{1}{3}\eta
$. On the other hand, $y_{n}\in g(\widetilde{x}_{n})\subset U$,
which is impossible. Hence $0\in g(\widetilde{x})$ and the proof
is complete.
\end{proof}
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