\documentclass[reqno]{amsart}
\usepackage{hyperref}

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2014 (2014), No. 47, 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 2014 Texas State University - San Marcos.}
\vspace{8mm}}

\begin{document}
\title[\hfilneg EJDE-2014/47\hfil Fuzzy differential equations]
{Fuzzy differential equations under dissipative and
compactness type conditions}

\author[T. Donchev, A. Nosheen \hfil EJDE-2014/47\hfilneg]
{Tzanko Donchev, Ammara Nosheen}  % in alphabetical order

\address{Tzanko Donchev \newline
Department of Mathematics, "Al. I. Cuza" University,
Ia\c{s}i 700506, Romania}
\email{tzankodd@gmail.com}

\address{Ammara Nosheen \newline
Abdus Salam School of Mathematical Sciences, 68-B, New Muslim Town,
Lahore, Pakistan}
\email{hafiza\_amara@yahoo.com}

\thanks{Submitted July 27, 2012. Published February 19, 2014.}
\subjclass[2000]{34A07, 34G20}
\keywords{Lyapunov-like function; compact perturbations; fuzzy sets}

\begin{abstract}
 Fuzzy differential equation with right-hand side defined as a sum of two almost
 continuous functions is studied. The first function satisfies dissipative-type
 condition with respect to Lyapunov-like function. The second maps
 bounded sets into relatively compact sets. The existence of solution is proved
 with aid of Schauder's fixed point theorem.
\end{abstract}

\maketitle
\numberwithin{equation}{section}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{example}[theorem]{Example}
\newtheorem{remark}[theorem]{Remark}
\allowdisplaybreaks


\section{Introduction}
Starting from \cite{KB}, the theory of fuzzy differential equations
is rapidly developed due to many applications in the real world problems.
Notice only the basic work in this direction \cite{K,LM,PH,PS}.
As it is shown in \cite{K}, the set of fuzzy numbers is not locally compact.
It means that the classical Peano theorem is (probably) no longer valid and
some extra conditions along with continuity of right-hand side are needed.

In \cite{WS} the existence of solutions of fuzzy differential equation with
uniformly continuous right-hand side is proved under compactness-type condition.
The existence and uniqueness of solution under dissipative-type conditions
when the right-hand side is continuous is studied in \cite{GS,Lu,SW}.
In this paper we study fuzzy differential equation whose right-hand side is
a sum of two almost continuous functions, one satisfies dissipative-type
condition, and another maps bounded sets into relatively compact sets. To the
authors knowledge there are not  related results in the literature.

We study the  fuzzy differential equation
\begin{equation}\label{4.1}
\dot x(t) = f(t,x) + g(t,x);\ x(0) = x_0,\  t \in I,
\end{equation}
where $f: I \times \mathbb{E}\to \mathbb{E}$ satisfies dissipative-type
condition and $g: I \times \mathbb{E}\to \mathbb{E}$ satisfies 
compactness-type assumption.
Here and further in the paper $I=[0,1]$.
$\mathbb{E}= \{ x: \mathbb{R}^n \to [0,1]; x\text{ satisfies (1)--(4)}\}$ 
is the space of fuzzy numbers:
\begin{itemize}
\item[(1)] $x$ is normal i.e. there exists $y_0\in \mathbb{R}^n$ such
that $x(y_0)=1$,

\item[(2)] $x$ is fuzzy convex i.e. $x(\lambda y + (1 - \lambda )z)\ge
\min \{ x(y) , x(z)\}$ whenever $y,z \in \mathbb{R}^n$ and
$\lambda \in [0,1]$,

\item[(3)] $x$ is upper semicontinuous i.e. for any $y_0 \in
\mathbb{R}^n$ and $\varepsilon >0$ there exists $\delta
(y_0,\varepsilon)>0$ such that $x(y) <
x(y_0)+\varepsilon$ whenever $|y-y_0|<\delta,\ y\in
\mathbb{R}^n$,

\item[(4)] The closure of the set $\{ y\in \mathbb{R}^n;\ x(y)>0\}$ is
compact.
\end{itemize}
The set $[x]^\alpha= \{ y\in \mathbb{R}^n;\ x(y)\geq \alpha\}$ is
called $\alpha$-level set of $x$.

It follows from (1)--(4) that the $\alpha$-level sets $[x]^\alpha$
are convex compact subsets of $\mathbb{R}^n$ for all $\alpha\in (0,1]$.
The fuzzy zero is defined by
\[
\hat{{0}}(y)=\begin{cases}
    0 &\text{if } y\neq 0, \\
    1 &\text{if }  y=0.
\end{cases}
\]
The metric in $\mathbb{E}$ is defined by 
$ D(x,y)=\sup _{\alpha\in (0,1]} D_H([x]^\alpha, [y]^\alpha)$,
where 
\[
 D_H(A,B)= \max\{ \max_{a\in A}\min_{b\in B}|a-b|,
\max_{b\in B}\min_{a\in A}|a-b|\}
\]
is the Hausdorff distance between the
convex compact subsets of $\mathbb{R}^n$.

The map $F:I\times \mathbb{E} \to \mathbb{E}$ is said to be
continuous at $(s,y)$ when for every $\varepsilon>0$ there exists
$\delta > 0$ such that $D(F(s,y),F(t,x))< \varepsilon$ for
every $t\in I$ and $x\in \mathbb{E}$ with $|t-s|+ D(x,y)< \delta$.
The map $F:I\times \mathbb{E} \to \mathbb{E}$  is said to be
almost continuous if there
exists a sequence $\{ I_k \}_{k =1}^\infty$ of pairwise
disjoint compact sets with $\operatorname{meas}(I_k)> 0$ and
$\operatorname{meas}\big(\cup_{k = 1}^\infty  {I_k}\big)
= \operatorname{meas}(I)$ such
that $F:I_k\times \mathbb{E} \to \mathbb{E}$ is continuous for every $k$.

Since $I_k$ is compact for every $k$, one has that 
$\cup_{k=1}^nI_k$ is also compact and hence 
$ (0,1)\setminus \cup_{k=1}^n I_k= \cup_{i=1}^\infty (a_i,b_i)$ is open,
because every open set in $\mathbb{R}$ is a union of countable sets of
pairwise disjoint open intervals.

 Throughout this paper both $f:I\times \mathbb{E} \to \mathbb{E}$
and $g:I\times \mathbb{E} \to \mathbb{E}$ are assumed to be almost continuous.

\begin{remark}\label{rem4.1}
Due to Lusin's theorem  (see e.g.~\cite{LT} for short proof) 
$\Lambda:I\to \mathbb{E}$ is strongly measurable if and only if it
satisfies Lusin property, i.e.
for all $\varepsilon > 0$ there exists $I_\varepsilon \subset I$ with
$ \operatorname{meas}(I \backslash I_\varepsilon)\le \varepsilon$ such that $
\Lambda:I_\varepsilon \to \mathbb{E}$ is continuous.
\end{remark}

A mapping $\Upsilon: I\to \mathbb{E}$ is said to be differentiable at
$t\in I$ if for sufficiently small $h>0$ the differences 
$\Upsilon(t + h) - \Upsilon(t)$, $\Upsilon(t) - \Upsilon(t-h)$ 
(in sense of Hukuhara) exist and
there exists $\dot \Upsilon(t) \in \mathbb{E}$ such that the limits 
$ \lim_{h \to 0^ + } \frac{\Upsilon(t + h) - \Upsilon(t)}{h}$ and
$ \lim_{h \to 0^ + } \frac{\Upsilon(t) - \Upsilon(t-h)}{h}$ exist, and are
equal to $\dot \Upsilon(t)$. At the end points of $I$ we consider only the
one sided derivative.

The integral of fuzzy function $\Upsilon: I \to \mathbb{E}$ is defined
levelwise, i.e. there exists $\Lambda: I \to \mathbb{E}$ such that
$ [\Lambda(t)]^\alpha = \int_0^t [\Upsilon(s)]^\alpha ds$, where
the integral is in Auman sense.
Every such function $\Lambda(\cdot)$ is absolutely continuous (AC).

The sequence of strongly measurable functions
$ \{ y_n(\cdot)\}_{n=1}^\infty$ is said to be integrally bounded if 
there exists $\lambda(t) \in L_1(I,\mathbb{R}^+)$ 
(non negative valued integrable function) such that 
$D(y_n(t),\hat 0)\leq \lambda(t)$ for every
$n$ and a.a. $t\in I$.

The Caratheodory function $v: I\times \mathbb{R}^+\to \mathbb{R}^+$ is
said to be Kamke function if it is integrally bounded on the bounded
sets, $v(t,0)=0$ and the unique solution of $\dot r(t)= v(t,r(t))$
with $r(0)=0$ is $r(t)\equiv 0$.

\section{Fuzzy differential equation under dissipative-type condition}

In this section we consider the  fuzzy differential equation
\begin{equation}\label{4e}
\dot x(t) = f(t,x),\quad x(0) = x_0,
\end{equation}
where $f: I\times \mathbb{E}\to \mathbb{E}$ satisfies dissipative-type
condition. We extend the results of \cite{PS} to the case of fuzzy
differential equations with almost continuous right-hand side.
We need the following hypothesis:
\begin{itemize}
\item[(F1)]  $D(f(t,x),\hat 0) \le \lambda(t)(1 + D(x,\hat 0))$ for
some $\lambda(t) \in L_1(I,\mathbb{R}^+)$.

\item[(F2)] There exists a Lyapunov-like function $ W: \mathbb{E}\times
\mathbb{E}\to \mathbb{R}^+$ for \eqref{4e}.
\end{itemize}
A continuous map $ W: \mathbb{E}\times \mathbb{E}\to \mathbb{R}^+$
is said to be Lyapunov-like function for \eqref{4e} if the following conditions
hold (cf.~\cite{LL}):
\begin{itemize}
\item[(1)] $W(x,x)= 0$, $W(x,y)>0$ for $x\neq y$ and $ \lim_{m\to
\infty} W(x_m,y_m)=0$ implies $ \lim_{m\to \infty}D(x_m,y_m)=0$,

\item[(2)] There exists a constant $L>0$ such that
$$
|W(x_1,y_1)-W(x_2,y_2)|\leq L\left(D(x_1,x_2)+ D(y_1,y_2)\right),
$$

\item[(3)] There exists a Kamke function $v:I\times \mathbb{R}^+ \to \mathbb{R}^+$ such that
$$
\lim_{h\to 0^+} h^{-1} \left[ W \left( x + h f(t,x),y
+ h f(t,y) \right) - W(x,y)\right] \leq v(t,W(x,y))
$$
for any $x,y \in \mathbb{E}$.
\end{itemize}

\begin{lemma}\label{l4.3}
Let {\rm (F1)} holds, then for $\varepsilon > 0$ and $\delta > 0$ there
exists an AC function $x_\varepsilon(t)$ such that 
$D(\dot x_\varepsilon(t),f(t,x_\varepsilon(t))) \leq \varepsilon$
for all $t \in I_\varepsilon \subset I$, where $I_\varepsilon$
is a compact set with measure greater than $1- \delta$.
\end{lemma}

\begin{proof}
Since $f:I \times \mathbb{E} \to \mathbb{E}$ is almost continuous
there exists a sequence $\{ I_k \}_{k = 1}^\infty$ of
pairwise disjoint compact sets such that 
$\operatorname{meas}\big(\cup_{k=1}^\infty I_k\big) = \operatorname{meas}(I)$ 
and $f:I_k \times \mathbb{E} \to \mathbb{E}$
is continuous for every $k$. For large $n$ we have
$ \operatorname{meas}(I_\delta)> 1 - \delta$, where 
$ I_\delta= \big( \cup_{k = 1}^n I_k \big)$.
Let the needed solution $x_\varepsilon(\cdot)$ be defined on
$[0,\tau]$ where $\tau \leq 1$ ($\tau = 0$ is possible).
If $\tau = 1$ then we have done, otherwise two cases would be possible:

(i) $\tau \in (a_l,b_l) \text{ where } (0,1)\setminus{I_\delta}=
\cup _{l=1}^\infty (a_l,b_l)$. In this case we extend
$x_\varepsilon(\cdot)$ on $[\tau,b_l)$ by $x_\varepsilon(t) = x_\varepsilon(\tau)$
and denote $\tau_1 = b_l > \tau$,

(ii) $\tau \notin \cup_{i=1}^\infty [a_i,b_i)$ then we define
$$
x_\varepsilon(t) = x_\varepsilon(\tau)+(t-\tau)f(\tau,x_\varepsilon(\tau)),\quad 
t \in [\tau,\tau_1]\cap I_\delta.
$$
Since $f(\cdot, x_\varepsilon(\cdot))$ is continuous on $I_\delta$,
then $ D(\dot x_\varepsilon(t) = f(t,x_\varepsilon(t)),f(\tau,x_\varepsilon(\tau)))
\leq \varepsilon,\ \forall t \in [\tau, \tau_1] \cap I_\delta$.

One can continue by induction. Suppose the largest interval on which
$x_\varepsilon(\cdot)$ satisfies lemma conditions is $[0,\bar\tau)$.
Since $ D(f(t,x_\varepsilon), \hat 0) \leq \lambda(t)(1+D(x_\varepsilon(t),\hat 0))$,
one has that
$$
D(\dot x_\varepsilon(t),\hat 0) \leq \lambda(t)(1+D(x_\varepsilon(t),\hat0)) +
\varepsilon \quad \text{for } t \in [0,\bar\tau).
$$
Consequently,
\begin{gather*}
D(x_\varepsilon(t),\hat0) \leq e^{\int _0^{\bar\tau} \lambda(s)ds}
D(x_0,\hat0)+\varepsilon, \\
D(\dot x_\varepsilon(t),\hat0)
\leq \lambda(t)(1+N_\varepsilon)+\varepsilon,
\end{gather*}
where
$$
N_\varepsilon =  e^{\int _0^{\bar\tau} \lambda(s)ds}\big(D(x_0,\hat0) + 2\big).
$$
Therefore, $D(\dot x_\varepsilon(t),\hat0) \in L_1(I,\mathbb{R}^+)$.
Furthermore, since $x_\varepsilon(\cdot)$ is AC, then one can conclude that
$x_\varepsilon(\cdot)$ is uniformly continuous on $[0,\bar\tau)$. Thus
$\lim_{t\uparrow \bar\tau}x_\varepsilon(t)=x(\bar\tau)$ exists, which is a
contradiction to the fact that $[0,\tau]$ is maximum interval of existence.
If $\bar\tau=1$ then the proof is complete.

If $\bar\tau< 1$ then we can continue this process by defining
$$
x_\varepsilon(t)=x_\varepsilon(\bar \tau)+(t-\bar \tau)f(\bar \tau,
x_\varepsilon(\bar \tau)),\ t \in [\bar{\tau},\tilde \tau]
$$
for $\bar \tau \notin \cup _{l=1}^\infty [a_l,b_l)$
or $\tilde \tau = b_l$ if $\bar{\tau} \in [a_l,b_l)$ for some
$l$, therefore there exists a $\tilde \tau_1 >\tilde \tau$ such
that $x_\varepsilon(\cdot)$ satisfies the conclusion of the lemma on
$[0, \tilde \tau_1]$. Continuing in the same way the so defined
$x_\varepsilon(\cdot)$ will satisfy the conclusion of the lemma on $[0,1]$
\end{proof}

\begin{theorem} \label{th1}
Let {\rm (F1)} and {\rm (F2)} hold, then  \eqref{4e} admits
unique solution.
\end{theorem}

\begin{proof}
Denote $ \chi_n(t) = \lambda(t)(1 + N_\varepsilon) + \frac{\varepsilon}{2^n}$,
where $N_\varepsilon$ is from Lemma \ref{l4.3}. 
Let $ I_{\delta_n} = \cup_{n=1}^{k_{\delta_n}} I_n$ be such that
$ \operatorname{meas}(I_{\delta_n}) >1-\frac{\delta}{2^n}$ and 
$f: I_n \times \mathbb{E} \to \mathbb{E}$ is
continuous. Consider the sequence of  approximate solutions 
$\{ x_n(\cdot)\}_{n = 0}^\infty$ where $x_n(\cdot)$ is the AC 
function defined in Lemma \ref{l4.3} when $\varepsilon$ is replaced by 
$\frac{\varepsilon}{2^n}$.
Therefore $ D(\dot x_n(t),f(t,x_n(t)))\leq \eta_n(t)$, where
$$
 \eta_n(t) = \begin{cases}
\varepsilon/2^n &  \text{if }t \in I_{\delta_n},\\
\chi_n(t) & \text{if }t \notin I_{\delta_n}.
\end{cases}
$$
We have to prove that $ \{x_n(\cdot)\}_{n=0}^\infty$ is a Cauchy sequence.
To this end we take $\{x_n(\cdot)\}$, $\{x_m(\cdot)\}$, where $n<m$.
Without loss of generality we can assume that $\dot x_n(\cdot)$,
 $\dot x_m(\cdot)$ and $f(\cdot,x(\cdot))$ are continuous on $J_n$, 
where $J_n \subset I_{\delta_n}$ with 
$ \operatorname{meas}(J_n) > 1- \frac{\delta}{2^n}$. If $t\in J_n$, then
\begin{align*}
&D^+ W(x_n(t),x_m(t)) \\
&= \lim_{h \to 0^ + } \frac{W(x_n(t+h),x_m(t+h))-
W(x_n(t),x_m(t))}{h}\\
&\leq\lim_{h \to 0^ + } \frac{W(x_n(t)+h\dot x_n(t),
x_m(t)+h\dot x_m(t))- W(x_n(t),x_m(t))+o(h)}{h} \\
&\leq \lim_{h \to 0^ + } \frac{W(x_n(t)+h\dot x_n(t),x_m(t)+
h\dot x_m(t))- W(x_n(t),x_m(t))}{h} \\
&\leq \lim_{h \to 0^ + } \frac{W(x_n(t)+ h f(t,x_n(t)),
x_m(t)+ h f(t,x_m(t)))-  W(x_n(t),x_m(t))}{h} \\
&\quad + \lim_{h \to 0^ + } \frac{Lh\left[ D(\dot x_n(t),
f(t,x_n(t)))+ D(\dot x_m(t),f(t,x_m(t)))\right]}{h}\\
&\leq v(t,D(x_n(t),x_m(t)))+  \frac{2L\varepsilon}{2^n}.
\end{align*}
For almost all $t \notin J_n$, we have
\begin{align*}
&D^+ W(x_n(t),x_m(t))\\
&=\lim_{h \to 0^ + } \frac{W(x_n(t+h),x_m(t+h))-
W(x_n(t),x_m(t))}{h} \\
&\leq \lim_{h \to 0^ + } \frac{W(x_n(t)+h\dot x_n(t),
x_m(t)+h\dot x_m(t))- W(x_n(t),x_m(t))+o(h)}{h} \\
&\leq \lim_{h \to 0^ + } \frac{W(x_n(t)+h\dot x_n(t),x_m(t)+h
\dot x_m(t))- W(x_n(t),x_m(t))}{h} \\
&\leq \lim_{h \to 0^ + } \frac{W(x_n(t)+ h f(t,x_n(t)),
x_m(t)+ h f(t,x_m(t)))-  W(x_n(t),x_m(t))}{h} \\
&\quad + \lim_{h \to 0^ + } \frac{Lh\left[ D(\dot x_n(t),
f(t,x_n(t)))+ D(\dot x_m(t),f(t,x_m(t)))\right]}{h}\\
&\leq v(t,D(x_n(t),x_m(t)))+  2L \chi_n(t).
\end{align*}
Consequently, $ D^+ W(x_n(t),x_m(t)) \leq v(t,D(x_n(t),x_m(t)))+
2L\eta_n(t)$, because $n< m$.

Thus $W(x_n(t),x_m(t)) \leq r_n(t)$, where $r_n(t)$ is the maximal
solution of $ \dot r(t)= v(t,r(t))+ 2L\eta_n(t)$.

Clearly $\eta_n(\cdot)$ is integrally bounded (as a sequence of
real valued functions), and
$\lim_{n \to \infty}\eta_n(t)=0$ for almost all $t \in I$.
Since $v(\cdot,\cdot)$ is Kamke function, then
$\lim_{n \to \infty}r_n(t)=0$ uniformly on $I$. Therefore
there exists a sequence of continuous real valued functions $S_n(t)$
with $\lim_{n \to \infty}D(x_n(t),x_m(t)) \leq S_n(t)$
for all $m \geq n$ and $\lim_{n \to \infty}S_n(t)=0$ uniformly
on $I$. Thus the sequence $\{x_n(\cdot)\}_{n=1}^\infty$ is a Cauchy sequence
and hence $\lim_{n \to \infty}x_n(t)=x(t)$ uniformly on $I$. Consequently
$ f(t,x_n(t))\to f(t,x(t))$ for a.a. $t\in I$. Furthermore, $
D(f(t,x_n(t)),\hat 0)\leq \chi_n(t)\leq \chi_1(t)$. Due to dominated
convergence theorem we get
\begin{equation} \label{4e1}
x(t)=x_0+\int_0^t f(s,x(s))ds.
\end{equation}
The proof is complete thanks to Lemma \ref{l4.4} given below.
\end{proof}

\begin{lemma}\label{l4.4}
If $f: I \times \mathbb{E} \to \mathbb{E}$ is almost continuous and
integrally bounded
then every solution of \eqref{4e} is a solution of \eqref{4e1} and vice versa.
\end{lemma}

\begin{proof}
The space $\mathbb{E}$ can be embedded as a closed convex cone in a Banach
space $\mathbb{X}$. The embedding map $j: \mathbb{E} \to \mathbb{X}$ is an
isometry and isomorphism. From (cf\cite{DK}) we know that $ j(\dot x(t))=
\frac{d}{dt}j(x(t))$. The fact that every solution of \eqref{4e1} is at the
same time a solution of \eqref{4e} is tautology because $ \int
_0^t \dot x(s)ds =\int_0^t f(s, x(s))ds$. Let $x:I \to \mathbb{E}$
be a solution of \eqref{4e1}. Since $x:I \to \mathbb{E}$ is
continuous, therefore $f: I \times \mathbb{E} \to \mathbb{E}$ satisfies
Lusin property and hence
$$
g(t) = \frac{d}{dt}\Big(\int_0^t g(s)ds \Big)
$$
for a.a. $t \in I$. i.e. $\dot x(t) = g(t) = f(t, x(t))$.

Evidently, $x:I \to \mathbb{E}$ is AC, i.e.
$x(\cdot)$ is a solution of \eqref{4e}.
\end{proof}

\begin{remark} \label{r4.1} \rm
Let us consider the equation
\begin{equation}\label{4.2}
\dot x_n = f(t,x_n(t))+\varphi_n(t),\ x_n(0)=x_0.
\end{equation}
If $\{\varphi_n(\cdot)\}_{n=1}^\infty$ is integrally bounded and
$\lim_{n \to \infty} \varphi_n(t)=0$,
then $\lim_{n \to \infty} x_n(t)=x(t)$,
where $\dot x(t) = f(t,x(t)),\ x(0)=x_0.$ Therefore the solution of
\eqref{4.2} depends continuously on the right-hand side.
\end{remark}

\section{Compact perturbations of dissipative fuzzy system}

In this section we prove the existence of solution of the differential
equation \eqref{4.1}. We will use the additional hypotheses:
\begin{itemize}
\item[(F3)] $W(x+z,y+z) = W(x,y)$ for any fuzzy number $z$.

\item[(G1)] $g(t,\cdot)$ maps the bounded subsets of $\mathbb{E}$
into relatively compact subsets of $\mathbb{E}$ for a.a. $t\in I$.

\item[(G2)] $D(g(t,x),\hat 0) \le \nu(t)(1 + D(x,\hat 0))$,
where  $\nu(\cdot) \in L_1(I,\mathbb{R}^+)$.

\end{itemize}
Condition (F3) is essential here.
Notice that it holds automatically if $W(x,y) = \zeta (D(x,y))$, where $\zeta$
is some continuous function such that $W(x,y)$ is Lyapunov-like function for
\eqref{4e}.

If $x(\cdot)$ is a solution of \eqref{4.1} then 
$D(\dot x(t), \hat 0) \leq \big(\lambda(t) + \nu(t)\big) ( 1+D(x(t) ,\hat 0))$. 
Therefore,
$$
D(x(t), \hat 0) \leq D(x_0, \hat 0) + e^{\int _0^t(\lambda(s) + \nu(s))ds}
\Big(D(x_0, \hat 0)+\int_0^t [\lambda(s) + \nu(s)]ds\Big).
$$
We can assume without loss of generality that $D(x(t), \hat 0) \le N$
and $D(\dot x(t), \hat 0) \leq \gamma(t)$, where 
$\gamma(t) =  (\lambda(t) + \nu(t))(1+N)$ is Lebesgue integrable. 
Let $A = \{y \in \mathbb{E}:D(y,x_0) \leq N\}$. 
It follows from (G1) that $g(t,A)\subset K(t)$, where 
$K(t)\subset \mathbb{E}$ is a convex compact set for a.a. $t\in I$.

\begin{theorem}\label{th4.3}
Let {\rm (F1), (F2), (F3), (G1), (G2)} hold, then the
differential equation \eqref{4.1} admits a solution.
\end{theorem}

We need the following lemma for proving Theorem \ref{th4.3}.

\begin{lemma} \label{l4.5}
Let $\{\varphi_n(\cdot)\}_{n=1}^\infty$ be an integrally bounded 
(by an integrable function $c(\cdot)$) sequence of
strongly measurable functions from $I$ to $\mathbb{E}$ such that 
\[
\overline{\rm co} \big\{ \cup_{i=1}^\infty \{ \varphi_i(t)\}\big\}= K(t)
\]
is compact for a.a. $t\in I$ and
\begin{equation}\label{4*}
 \dot x_n(t) = f(t,x_n(t))+\varphi_n(t),\ x_n(0)=x_0.
\end{equation}
Passing to subsequence, if necessarily, $x_n(\cdot)$
converges uniformly to $x(\cdot)$, such that
$$
\dot x(t) \in f(t,x(t))+ K(t).
$$
\end{lemma}

\begin{proof}
Clearly $D(\varphi_n(t), \hat 0)\leq c(t)$ implies that 
$ z_n(t) = \int_0^t \varphi_n(s)ds$ is equicontinuous sequence. 
Furthermore,
\[
 \int_0^t \big[ \cup_{n=1}^\infty \varphi_n(s)\big]ds
\subset \int_0^t K(s)ds= R(t),
\]
where  $ \overline{\cup_{t\in [0,1]}\{R(t)\}}$ is a compact subset of $\mathbb{E}$. 
Then the sequence $ z_n(t) = \int_0^t \varphi_n(s)ds$
is $C(I,\mathbb{E})$ precompact. By Arzela Ascoli theorem, passing to
subsequence we have $z_n(t) \to z(t)$ uniformly on $I$.

As we pointed out, $\mathbb{E}$ can be embedded as a closed convex cone
in a Banach space $\mathbb{X}$ with a continuous embedding map $j:
\mathbb{E} \to \mathbb{X}$. Thus $j(K)\subset \mathbb{X}$ is compact.
Then due to Diestel criterion (see proposition 9.4 of \cite{D}) the set
$\{j(\varphi_n(\cdot))\}_{n=1}^\infty$ is weakly precompact in
$L_1(I, \mathbb{X})$. Thus passing to subsequence in $L_1(I, \mathbb{X})$
we have $j(\varphi_n(t)) \rightharpoonup s(t)$.
Since $s(t) \in j(K)$, then there exists
$\varphi(t)$ such that $j(\varphi(t))=s(t)$ and $ z(t)= \int_0^t
\varphi(s)ds$.

We denote for convenience $y(t) = j(x(t))$, $y_n(t) = j(x_n(t))$,
$p(t) = j(z(t))$, $\psi(t) = j(\varphi(t))$,
$y(t)-p(t)= u(t)$, $y_n(t)- p_n(t)=u_n(t)$ and $q(t,y) = j(f(t,x))$.

Consider the functions $y_n(t) - p_n(t)=u_n(t)$. We have
\begin{align*}
&W(u(t)+h\dot u(t),u_n(t)+h\dot u_n(t)) \\
&=  W(u(t)+ h q(t,y(t)), u_n(t)+ h q(t,y_n(t)))+o(h)\\
&= W(u(t)+ p_n(t)+h q(t,y(t)),y_n(t)+h q(t,y_n(t)))+o(h) \\
&= W(u(t)+p_n(t)+h q(t,y(t)-p(t)+p_n(t)),y_n(t)+h q(t,y_n(t)))\\
&\quad + h |q(t,y(t))-q(t,y(t)-p(t)+p_n(t))| + o(h).
\end{align*}
Consequently,
\begin{align*}
&\lim_{h \to 0^ + } \frac {W(u(t)+h(\dot u(t), u_n(t)+ h\dot u_n(t))
- W(u(t),u_n(t))}{h}\\
&= \lim_{h \to 0^+ } \frac{W(u(t+h),
u_n(t+h))- W(u(t), u_n(t))}{h}\\
&= \lim_{h \to 0^ + } \frac{W(u(t)+h\dot u(t),
u_n(t)+ h\dot u_n(t))-W(u(t),u_n(t))}{h}\\
&\leq v(t,|u(t)- u_n(t)|)+ |q(t,y(t)-p(t)+p_n(t))-q(t,y(t))|.
\end{align*}
Thus
\begin{align*}
D^+ W(y(t)-p(t),y_n(t)-p_n(t))\leq v(t,|y(t)-p(t)-(y_n(t)-p_n(t))|)\\
+  |q(t,y(t))-q(t,y(t)-p(t)+p_n(t))|.
\end{align*}
The latter implies that
$$
W(y(t)-p(t),y_n(t)-p_n(t)) \leq r_n(t),
$$
where
$$
\dot r_n(t) = v(t,r_n(t))+|q(t,y(t)-p(t)+p_n(t))-q(t,y(t))|,\ r_n(0)=0.
$$
Since $v(\cdot,\cdot)$ is Kamke function and since
$$
\lim _{n \to \infty} |q(t,y(t)-p(t)+p_n(t))-q(t,y(t))|=0
\text{ for a.a } t \in I,
$$
one has that $\lim _{n \to \infty} r_n(t)=0$, which implies that
$\lim _{n \to \infty} W(y(t)-p(t),y_n(t)-p_n(t))=0$.
Thus $y_n(t) \to y(t)$ uniformly on $I$, where
$\dot y(t)= q(t,y(t))+\psi(t)$, i.e $\dot x(t)= f(t,x(t))+\varphi(t)$.
\end{proof}

\begin{proof}[Proof of Theorem \ref{th4.3}]
Consider the set
$$
Q=\{ z(\cdot) \in C(I,K):\ D(\dot z(t), \hat 0) \leq \gamma(t),\ z(0)=x_0\}.
$$
It is easy to see that $Q \subset C(I, \mathbb{E})$ is closed, bounded and
convex. Consider the map $\xi: z(\cdot) \to x_z(\cdot)$, where
$x_z(\cdot)$ is the unique solution of
$$
\dot x_z(t)= f(t,x_z(t))+g(t,z(t));\ x_z(0)=x_0,\, t \in I.
$$
Due to Remark \ref{r4.1} the map $\xi: Q\to Q$ is continuous. Furthermore,
$ \overline{\xi(Q)} \subset Q$ is compact by Lemma \ref{l4.5}.
It follows from Schauder's theorem that there exist a fixed point $z(\cdot)
\in Q$ such that $\xi(z)=z$. This function $z(\cdot)$ is a solution of \eqref{4.1}.
\end{proof}

Notice that the linear growth conditions \textbf{(F1), (G2)} can be relaxed
in order to prove only local existence, i.e. we can assume that 
$f:I \times \mathbb{E} \to \mathbb{E}$ is integrally bounded on the bounded sets.
In that case, Theorem \ref{th1} is formulated as follows.

\begin{theorem} \label{th4}
Let $f:I \times \mathbb{E} \to \mathbb{E}$ be integrally bounded
on the bounded sets. Then under
(F2) there exists $a>0$ such that the system \eqref{4e} admits
unique solution on $[0,a]$.
\end{theorem}

\begin{proof} 
Let $M>0$. There exists an integrable function $\zeta: I \to \mathbb{R}^+$ with
$$
\sup_{|x-x_0|\leq M}|f(t,x)|\leq \zeta(t).
$$
Let $a>0$ be such that $ \int_0^a (\zeta(t)+ \varepsilon)dt\leq M$. 
On the interval $[0,a]$ every $\delta$ solution $x_\delta(t)$ satisfies
$ |x_\delta(t)|\leq M$ and $|\dot x_\delta(t)|\leq \zeta(t)+ \varepsilon$.
Therefore, one can continue as in the proof of Theorem \ref{th1}.
\end{proof}

Theorem \ref{th4.3} can be obviously formulated as:

\begin{theorem}\label{th5}
Let $f:I \times \mathbb{E} \to \mathbb{E}$ and 
$g:I \times \mathbb{E} \to \mathbb{E}$ be integrally bounded on the bounded set. 
Then under {\rm (F2),  (F3), (G1)} there exists $a>0$ such that the system
\eqref{4.1} admits a solution on $[0,a]$.
\end{theorem}

\begin{proof}
As in the proof of Theorem \ref{th4} we can see that there exists $a>0$ and
$\varepsilon>0$ such that every $\varepsilon$-solution of \eqref{4.1} is
extendable on $[0,a]$ and $|x_\varepsilon(t)-x_0|\leq M$. 
Let $g(t,x_0+ M \mathbb{B})\subset A(t)$, where 
$A(t)\subset \mathbb{E}$ is a convex compact set. 
It follows from Theorem \ref{th4} that for every strongly
measurable $\varphi(t)\in A(t)$, the fuzzy differential equation
$$
  \dot x(t) = f(t,x(t))+\varphi(t),\quad  x(0)=x_0
$$
admits unique solution on $[0,a]$. One can then continue as in the
proof of Theorem \ref{th4.3}, proving of course the corresponding variant
of Lemma \ref{l4.3}.
\end{proof}

\section{Conclusion}

As it is pointed out in the introduction the space $\mathbb{E}$ is not locally
compact. This implies that it would be very difficult (if it is possible at all)
to prove analogue of the classical Peano theorem, when the right-hand side of
\eqref{4e} is only jointly continuous. On the other hand up to author's knowledge
there is no example of such a system without solutions.

In authors opinion it is very interesting open question to give an example of
fuzzy differential equation without local solution, when the right-hand side
is jointly continuous.

In optimal control problems the controls are measurable functions and it
is one of the main motivation to study differential equations with almost
continuous right-hand sides.

In this paper we proved existence (and uniqueness) of the solution of \eqref{4e}
under as weak as it is possible dissipative-type condition w.r.t.
Lyapunov-like function. We also show the existence of solution when the
right-hand side is the sum of a function satisfying such condition along
with almost continuous function mapping bounded sets into relatively
compact ones. For example such function is $g(t,\cdot)$ which takes
values in a locally compact set $\mathbb{E}_K\subset\mathbb{E}$. It
seems that it is impossible to relax compactness-type assumptions on
$g$ without using stronger dissipative-type conditions on $f$.
We refer the reader to the paper \cite{A}, where it is shown by example that
if $v(\cdot,\cdot)$ is a Kamke function, then it is possible that the function
$w(t,r)= v(t,r)+L(t)r$ is not a Kamke function.

Of course in our proof we essentially used (F3), which is in
general not valid for arbitrary Lyapunov-like function. It is an open
question does the solution exists, when the last condition is dispensed
with?

Now we give a simple example of fuzzy system which satisfies our
conditions.

\begin{example}\label{ex1} \rm
 Consider the  system of crisp first equation and fuzzy second:
\begin{align*}
 \dot x= - \sqrt[3]{x} + f(t,x,y),\quad x(0)=0 \\
\dot y(t)= g(t,x,y),\quad y(0)= y_0.
\end{align*}
Here $x$ is crisp variable, 
$f:I\times \mathbb{R}\times \mathbb{E}\to \mathbb{R}$ 
is continuous and Lipschitzian on $x$ and on $y$. Furthermore
$g:I\times \mathbb{R}\times \mathbb{E}\to \mathbb{E}$   is continuous,
Lipschitzian on $x$ and takes values in a locally compact subset of
$\mathbb{E}$. If some growth condition holds, than the system satisfies
all the conditions of Theorem \ref{th4.3}.
\end{example}

\subsection*{Acknowledgments}
The first author is supported by a grant from  the Romanian National Authority
for Scientific Research, CNCS--UEFISCDI, project number PN-II-ID-PCE-2011-3-0154.
The second author is partially supported by Higher Education
Commission, Pakistan.

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\end{document}





