\documentclass[twoside]{article}
\usepackage{amsfonts}  % font used for R in Real numbers
\pagestyle{myheadings} \markboth{\hfil Non-degenerate implicit
evolution inclusions \hfil EJDE--2000/34} {EJDE--2000/34\hfil
Kenneth Kuttler \hfil}
\begin{document}
\title{\vspace{-1in}\parbox{\linewidth}{\footnotesize\noindent
{\sc  Electronic Journal of Differential Equations}, Vol.~{\bf
2000}(2000), No.~34, pp.~1--20. \newline ISSN: 1072-6691. URL:
http://ejde.math.swt.edu or http://ejde.math.unt.edu
\newline ftp  ejde.math.swt.edu \quad ftp ejde.math.unt.edu (login: ftp)}
 \vspace{\bigskipamount} \\
%
  Non-degenerate implicit evolution inclusions
\thanks{ {\em Mathematics Subject Classifications:}
34A60, 35R70, 35A05, 34A09, 35K20, 47H15. \hfil\break\indent {\em
Key words:} Nonlinear evolution inclusion,
pseudo-monotone mapping, implicit inclusions.
\hfil\break\indent \copyright 2000 Southwest Texas State
University  and University of North Texas. \hfil\break\indent
Submitted April 12, 2000. Published May 12, 2000.} }
\date{}
%
\author{ Kenneth Kuttler }
\maketitle

\begin{abstract}
 We prove the existence of solutions for the implicit evolution inclusion
 $$
  ( B(t) u( t) ) ' + A(t,u(t)) \ni f(t)
 $$
 under conditions that are easy to verify on the set valued
 operator $A( t,\cdot)$ and that do not imply the operator is monotone.
 We also present an example where our existence theorem applies to a time
 dependent implicit inclusion.
\end{abstract}


\newtheorem{theorem}{Theorem}[section]
\newtheorem{corollary}[theorem]{Corollary}
\newtheorem{definition}[theorem]{Definition}
\newtheorem{example}[theorem]{Example}
\newtheorem{lemma}[theorem]{Lemma}
\renewcommand{\theequation}{\arabic{section}.\arabic{equation}}


\section{Introduction} \label{introduction}

There are many works which deal in the theory of implicit
evolution equations of the form
\[
( Bu) '+Au=f.
\]
In the case where $A$ is monotone and $B$ is linear, the book by
Carroll and Showalter \cite{car76}, gives many of the best
theorems allowing the equation to be replaced by $\ni $ thus
including evolution inclusions. Equations of this form have also
been discussed by many authors in the case where $A$ is some sort
of single valued operator from a Banach space to its dual which
may fail to be monotone. See for example, Lions \cite{lio69},
Bardos and Brezis \cite{bar69}, \cite{bre68}, or \cite{Ken}. More
recently these theorems have been generalized to include the case
where $A$ may be non monotone and set valued, a recent paper being
\cite{kut99}.

The paper \cite{kut99} includes as a special case the situation
where $V$ and $W$ are separable reflexive Banach spaces satisfying
\begin{equation}
V\hookrightarrow W\hookrightarrow W'\hookrightarrow V'
\label{3.05}
\end{equation}
and $V$ is dense in $W$ along with a family of linear operators,
$B( t) \in \mathcal{L}( W,W') $ satisfying
\begin{eqnarray}
&\langle B( t) u,v\rangle =\langle B( t) v,u\rangle ,& \label{3.1}
\\ &\langle B( t) u,u\rangle \geq 0, & \label{3.2} \\ &B( t) =B(
0) +\int_{0}^{t}B'(s)\,ds, & \label{3.3}
\end{eqnarray}
for all $u,v\in W$, and $B'\in L^{\infty }( 0,T;\mathcal{L}( W,W')
) $.

In the above formulae, $\langle \cdot ,\cdot \rangle $ denotes the
duality pairing of the Banach space, $W$, with its dual space. We
will use this notation in the present paper, the exact
specification of which Banach space being determined by the
context in which this notation occurs. Thus in the above, it is
clear from context, since $B( t) \in \mathcal{L}( W,W') $,
 that the Banach space is $W$.  We use this
notation throughout the present paper to make the presentation
less cluttered with symbols. Occasionally, when it is desired to
emphasize which Banach space is meant, we will write the symbol in
the form $\langle v,u\rangle _{X',X}$, thus indicating the duality
pairing between $X$ and $X'$.  We will also use the notation, $\|
u\|_Y$ to denote the norm of $u$ in the space $Y$ where $Y$ is a
Banach space or in the form $\| u\| $ when it is clear which space
is meant. The symbol, $L^p( 0,T;Y) $ denotes the space of strongly
measurable $Y$ valued functions, $f$, for which
\[
\int_{0}^{T}\| f( t) \| _Y^p\,dt<\infty
\]
in the case where $p\geq 1$, and the essential supremum of the
function, $t\to \| f( t) \| $ in the case where $p=\infty $. For
further discussion of these spaces we refer to \cite{die77}.

In \cite{kut99}, there is also a set valued operator, $A$,
pseudo-monotone in the sense of \cite{nan95}, which maps elements
of a solution space, $X$, defined below, to $\mathcal{P}( X') $
the power set of $X'$. The function, $f\in L^{p'}( 0,T;V') \equiv
\mathcal{V}'$ is also given. The paper, \cite{kut99} includes
existence theorems for the following implicit inclusion,
\begin{equation}
( Bu) '+Au\ni f\mbox{ in }\mathcal{V}',\; Bu( 0) =B( 0)
u_{0},\;u_{0}\in W,  \label{pseudo.1}
\end{equation}
where the prime denotes differentiation in the sense of $V'$
valued distributions. Thus, for $\phi \in C_{c}^{\infty }( 0,T) $,
\[
( Bu) '( \phi ) \equiv -\int_{0}^{T}B( t) u( t) \phi '( t) \,dt
\]

The solution, $u$, is found in the space of solutions, $X$ where
\begin{equation}
X\equiv \left\{ u\in L^p( 0,T;V) \equiv \mathcal{V}:( Bu) '\equiv
Lu\in L^{p'}( 0,T;V^{\prime }) \equiv \mathcal{V}'\right\}
\label{pseudo.2}
\end{equation}
and $p\geq 2$ is always assumed. Thus, $u$ is a solution to
(\ref{pseudo.1})
if $u\in \mathcal{V}$, $( Bu) '\in \mathcal{V}'$%
, and there exists $\xi \in Au\subseteq \mathcal{V}'$ such that
along with the initial condition, whose precise meaning is given
below, we have the following equation.
\[
( Bu) '+\xi =f\,.
\]
The meaning of the initial condition is dependent on the following
theorem about the space of solutions, $X$, a special case of one
proved in \cite {kut99}. It is less general because in
\cite{kut99}, it is not assumed $V$ is dense in $W$ and more
general function spaces are considered. See also \cite{Ken} for a
slightly less general version of the same theorem.

\begin{theorem}
\label{spaceofsolutions} Let $u,v\in X$, then the following hold.

\begin{enumerate}
\item  \label{b1} $t\to \langle B( t) u( t)
,v( t) \rangle _{W',W}$ equals an absolutely continuous function
a.e. $t$, denoted by $\langle Bu,v\rangle ( \cdot ) $.

\item  \label{b2} $\langle Lu( t) ,u( t) \rangle =%
\frac{1}{2}\left[ \langle Bu,u\rangle '( t) +\langle B'( t) u( t)
,u( t) \rangle \right] $ a.e. $t$.

\item  \label{b3} $\left| \langle Bu,v\rangle ( t) \right| \leq
C\| u\| _X\| v\| _X$ for some $C>0$ and for all $t\in [0,T]$.

\item  \label{b4} $t\to B( t) u( t) $ equals a
function in $C( 0,T;W') $, a.e. $t$, denoted by $Bu( \cdot ) $.

\item  \label{b5} $\sup \{\| Bu( t) \|
_{W'} ,t\in \left[ 0,T\right] \}\leq C||u||_X$ for some $C>0$.

\item  \label{b6}  Let $K:X\to X'$ be given by
\[
\langle Ku,v\rangle _{X',X}\equiv \int_0^T\langle Lu( t) ,v ( t)
\rangle dt+\langle Bu,v\rangle ( 0)\,.
\]
Then  $K$ is linear, continuous and weakly continuous.

\item  \label{b7} $\langle Ku,u\rangle =\frac{1}{2}[\langle Bu,u\rangle
( T) +\langle Bu,u\rangle ( 0) ]+\frac{1}{2}%
\int_{0}^{T}\langle B'( t) u( t) ,u( t) \rangle dt$.
\end{enumerate}
\end{theorem}

In proving this theorem it is shown that $C^{\infty }( \left[
0,T\right] ;V) $, the space of infinitely differentiable functions
having values in $V$ is dense in $X$.  Therefore, we also obtained
the following formula which is valid for all $u\in X$.
\begin{equation}
Bu( t) =Bu( 0) +\int_{0}^{t}( Bu) ^{\prime }( s) ds
\label{extraconditiontheorem1}
\end{equation}
Here $Bu( 0) \in W'$.

From this theorem we see that we can define a continuous function
which equals $B( t) u( t) $ a.e. and therefore, we can give a
meaning to the expression $Bu( 0) =B( 0) u$.  The
operator $A$ is assumed to be a set valued pseudo-monotone map from $X$ to $%
\mathcal{P}( X') $.  Following \cite{nan95} these operators are
given according to the following definition.

\begin{definition}
\label{d2.1} We say $A:X\to \mathcal{P}( X') $ is
\textit{pseudo-monotone} if the following hold.

\begin{enumerate}
\item  \label{a1} The set $Au$ is non-empty, bounded, closed and convex for
all $u\in X$.

\item  \label{a2} If $F$ is a finite dimensional subspace of $X$, $u\in F$,
and if $U$ is a weakly open set in $V'$ such that $Au\subseteq U$,
then there exists a $\delta >0$ such that if $v\in B_{\delta
}(u)\cap F$ then $Av\subseteq U$.

\item  \label{a3} If $u_{i}\to u$ weakly in $X$ and $u_{i}^{\ast
}\in Au_{i}$ is such that
\begin{equation}
\lim \sup_{i\to \infty }\langle u_{i}^{\ast },u_{i}-u\rangle \leq
0, \label{2.1}
\end{equation}
then, for each $v\in X$, there exists $u^{\ast }( v) \in Au$ such
that
\begin{equation}
\lim \inf_{i\to \infty }\langle u_{i}^{\ast },u_{i}-v\rangle \geq
\langle u^{\ast }(v),u-v\rangle .  \label{2.2}
\end{equation}
\end{enumerate}
\end{definition}

The following lemma shows the second condition in the above list
follows as a special case of something which can be proved if it
is assumed that $A$ is bounded in addition to Conditions
(\ref{a3}) and (\ref{a1}). First we give a definition of what we
mean by bounded.

\begin{definition}
Let $A:X\to \mathcal{P}( X') $ be a set valued map. We say $A$ is
bounded if for each bounded set, $G\subseteq X$,
\[
\sup \left\{ \| z\| :z\in Ax,x\in G\right\} <\infty .
\]
Thus $A$ is bounded if the norms of all possible elements of $Ax$
for $x\in G $ are bounded above.
\end{definition}

\begin{lemma}
\label{2isredundant} Let $A:X\to \mathcal{P}( X^{\prime }) $
satisfy conditions (\ref{a1}) and (\ref{a3}) above and suppose $A$
is bounded. Then if $x_{n}\to x$ in $X$, and if $U$ is a weakly
open set containing $Ax$, then $Ax_{n}\subseteq U$ for all $n$
large enough.
\end{lemma}

\paragraph{Proof:} If this is not true, there exists $x_{n}\to x$, a
weakly open set, $U$, containing $Ax$ and $z_{n}\notin Ax_{n}$,
but $\ z_{n}\notin U$. Taking a subsequence if necessary, we
obtain a sequence which satisfies $z_{n}\rightharpoonup z\notin U$
in addition to this. Then
\[
\lim \sup_{n\to \infty }\langle z_{n},x_{n}-x\rangle =0
\]
so if $y\in X$ there exists $z( y) \in Ax$ such that $\langle
z,x-y\rangle =\lim \inf_{n\to \infty }\langle z_{n},x_{n}-y\rangle
\geq \langle z( y) ,x-y\rangle $. Letting $w=x-y$, this shows,
since $y\in X$ is arbitrary, that the following inequality holds
for every $w\in X$.
\[
\langle z,w\rangle \geq \langle z( x-w) ,w\rangle .
\]
In particular, we may replace $w$ with $-w$ and obtain
\[
\langle z,-w\rangle \geq \langle z( x+w) ,-w\rangle ,
\]
which implies
\[
\langle z( x-w) ,w\rangle \leq \langle z,w\rangle \leq \langle z(
x+w) ,w\rangle .
\]
Therefore, there exists
\[
z_{\lambda }( y) \equiv \lambda z( x-w) +( 1-\lambda ) z( x+w) \in
Ax
\]
such that $\langle z,w\rangle =\langle z_{\lambda }( y) ,w\rangle
$. But this is a contradiction to $z\notin Ax$ because if $z\notin
Ax$ there exists $w\in X$ such that  $\langle z,w\rangle >\langle
z_{1},w\rangle$ for all $z_{1}\in Ax$. Therefore, $z\in Ax$ which
contradicts the assumption that $z_{n}$ and consequently $z$ are
not contained in $U$.

This condition which implies Condition \ref{a2} above is known as
Upper semi-continuity with respect to the strong topology on $X$
and the weak topology on $X'$.  If $A$ were single valued, this
would be called demicontinuity because it would imply that the
image of strongly convergent sequences converges weakly. The
importance of this lemma is that it shows Condition \ref{a2} is
redundant if it is assumed that $A$ satisfies the other two
conditions and is bounded.

For $B( t) $ defined above, we may consider the operators, $B'$
and $B$ as maps from $\mathcal{V}$ to $\mathcal{V}'$ according to
the definition,
\[
\langle B'u,v\rangle \equiv \int_{0}^{T}\langle B'( t) u( t) ,v(
t) \rangle dt,\;\langle Bu,v\rangle \equiv \int_{0}^{T}\langle B(
t) u( t) ,v( t) \rangle dt
\]
and we shall follow this convention throughout the paper. Note the
same
notation also applies to these operators considered as maps from $%
L^{2}( 0,T;W) $ to $L^{2}( 0,T;W') $ since $%
p\geq 2$ and $V$ is assumed to be dense in $W$.

In \cite{kut99} an existence theorem is given which contains the
following result as a special case.

\begin{theorem}
\label{mainexistencetheorem} Let $B,X,W$, and $V$ be as defined
above, $f\in \mathcal{V}'$,  $u_{0}\in W$, and suppose
$A:\mathcal{V}\rightarrow \mathcal{P}( \mathcal{V}') $ is such
that $A$ is set
valued, bounded as a map from $\mathcal{V}$  to $\mathcal{P}( \mathcal{%
V}') $, and pseudo-monotone when considered as a map from $X$ to
$\mathcal{P}( X') $. Then if $A+\frac{1}{2}B' $ is coercive,
\[
\lim_{\| u\| _{\mathcal{V}}\to \infty }\frac{%
\langle Au,u\rangle +\frac{1}{2}\langle B'u,u\rangle } {\| u\|
_{\mathcal{V}}}=\infty ,
\]
it follows there exists a solution, $u\in X$ to
\begin{equation}
( Bu) '+Au\ni f,\;Bu( 0) =Bu_{0}. \label{pseudo10}
\end{equation}
Furthermore, $u$ is a solution to (\ref{pseudo10}) if and only if
$u$ is a solution to the following equation which holds for all
$v\in X$.
\begin{equation}
\langle Ku,v\rangle +\langle u^{\ast },v\rangle =\langle
f,v\rangle +\langle Bv( 0) ,u_{0}\rangle   \label{pseudo10.3}
\end{equation}
for some $u^{\ast }\in $ $Au$.
\end{theorem}

While this theorem is very general, even allowing $A$ to depend on
the history of the function, $u$, the hypotheses are sometimes
difficult to verify. Therefore, it is important to consider the
special case in which the operator, $A$, is of the form $Au( t)
=A( t,u( t)
) $ and to determine easy to verify conditions on the operators, $%
A( t,\cdot ) $ which will imply the above conditions on $A$.

A paper by Bian and Webb, \cite{bia99} gives such convenient
conditions in
the special case where $B=I$ and $W=H$, a Hilbert space with $H=H'$%
. Conditions are given on the operators, $A( t,\cdot ) $ which
make it possible to obtain an existence theorem for the evolution
inclusion, (\ref{pseudo.1}) in the case where $u_{0}\in V$. We
will demonstrate that under appropriate conditions on $B$,
including the case when $B=I$, their
conditions actually imply the operator, $A$ is Pseudo-monotone on the space, $%
X$, which makes possible the consideration of more general initial
data. In the context of evolution inclusions they considered, it
will mean we can take the initial data in $H$ rather than only in
$V$.  When this theorem has been proved, we apply it to some
existence theorems which follow from it and Theorem
\ref{mainexistencetheorem}. Next we consider a model problem for
an implicit inclusion in which the nonlinear operators are
obtained as a sum of operators considered by Browder \cite{bro77}
and a set valued operator. We conclude by giving a proof of a
measurability result. Throughout the paper the symbol
$\rightharpoonup $ will mean weak or weak $\ast $ convergence and
the symbol $\to $ will mean strong convergence.

\section{Pointwise pseudo-monotone maps}
\label{pointwisepseudomonotone} \setcounter{equation}{0}

We will need to consider some sort of measurability condition for
set valued operators, $S( t,\cdot ) $ mapping the Banach space,
$V$, described above, to $\mathcal{P}( V') $. There is quite a
well developed theory of set valued maps found in \cite{aub90},
but for our purposes, we will say the operators, $S( t,\cdot ) $
are measurable if the following condition holds.

If
\begin{equation}
\emptyset \neq F( t) \equiv \left\{ w\in S( t,x( t) ) :\langle
w,x( t) -y( t) \rangle \leq \alpha ( t) \right\}
\label{measurabilitycondition}
\end{equation}
for $\alpha $ measurable and $x,y\in \mathcal{V}$, then there
exists $z\in \mathcal{V}'$ such that $z( t) \in F( t) $ a.e.

The next three conditions are modifications of conditions proposed
by Bian and Webb, \cite{bia99}. We let $V$ and $W$ be the
reflexive Banach spaces defined in (\ref{3.05}).

\begin{enumerate}
\item  \label{limitpseudomonotone}$v\to A( t,v) $ is a
set valued pseudo-monotone map from $V$ to $\mathcal{P}( V) $
satisfying conditions 1 and 3 in the definition of pseudo-monotone
given in Section \ref{introduction}.

\item  \label{boundedestimate}There exists $b_{1}\geq 0$ and $b_{2}\in
L^{p'}( 0,T) $ such that
\[
\| z\| _{V'}\leq b_{1}\| u\| _V^{p-1}+b_{2}( t) \mbox{ for all
}z\in A( t,u)
\]

\item  \label{coerciveestimate}There exists $b_{3}>0$, $\alpha \in (
0,p) $, $b_{4}\geq 0$, and $b_{5}\in L^{1}( 0,T) $ such that
\[
\inf_{z\in A( t,u) }\langle z,u\rangle +\frac{1}{2}\langle B'( t)
u,u\rangle \geq b_{3}\| u\|_V^p-b_{4}\| u\| _V^{\alpha}-b_{5}( t)
\]
and
\[
\inf_{z\in A( t,u) }\langle z,u\rangle \geq b_{3}\| u\|
_V^p-b_{4}\| u\| _V^{\alpha }-b_{5}( t) .
\]

To these three conditions, we append the following.

\item  \label{measurableestimate} The operators, $A( t,\cdot ) $
are measurable in the sense of (\ref{measurabilitycondition}).
\end{enumerate}

In Section \ref{measurability} we show how Condition
\ref{measurableestimate} follows from standard definitions of
measurability, in particular, the assumptions in Bian and Webb,
\cite{bia99}. The first inequality of 3. implies we are assuming
that $B'( t) $ cannot be too negative and we require both
inequalities to hold.

In the case that $A( t,\cdot ) $ is single valued, Condition 4
would be satisfied in the context of conditions
(\ref{limitpseudomonotone})
- (\ref{coerciveestimate}) if we assumed, as it is reasonable to do, that $%
t\to A( t,v) $ is measurable. This is because the operators, $A(
t,\cdot ) $, being pseudo-monotone and bounded would be
demicontinuous also. Therefore, if $x\in \mathcal{V}$ we could
obtain $x$ as a pointwise limit of simple functions, $s_{n}$ for which $%
t\to A( t,s_{n}( t) ) $ is measurable and use the demicontinuity
of $A( t,\cdot ) $ to conclude $t\rightarrow A( t,x( t) ) $ is
measurable. Then $F( t)
=A( t,x( t) ) $ and the given estimates would imply $%
t\to A( t,x( t) ) $ is in $\mathcal{V}^{\prime }$.

We will need the following definition.

\begin{definition}
\label{definitionofA}We define an operator, $\widehat{A}:\mathcal{V}%
\to \mathcal{P}( \mathcal{V}') $ by
\[
\widehat{A}( u) \equiv \left\{ z\in \mathcal{V}':z(
t) \in A( t,u( t) ) \mbox{ a.e.}t\in \left[ 0,T%
\right] \right\}
\]
\end{definition}

The following is the main theorem in this section.

\begin{theorem}
\label{Apseudomonotone}Suppose conditions
\ref{limitpseudomonotone} - \ref
{measurableestimate} hold and $B( t) $ is one-to-one for a.e. $t$%
. Then $\widehat{A}$ is pseudo-monotone as a map from $X$ to $\mathcal{P}%
( X') .\;$
\end{theorem}

\paragraph{Proof:} First we need to verify $\widehat{A}u$ is nonempty closed
and convex. It is clear that this is convex and closed. We need to
verify this set is nonempty. Let $u,v\in \mathcal{V}$ and let
\begin{eqnarray*}
\lefteqn{ F( t)}\\
 &\equiv& \Big\{ w\in A( t,u( t) ) :\langle w,u( t) -v( t) \rangle
\leq \left[b_{1}\| u( t) \| _V^{p-1}+b_{2}(t) \right] \| u( t) -v(
t) \|
   \Big\} .
\end{eqnarray*}
Note that by \ref{boundedestimate}, $F( t) $ is nonempty. Letting
\[
\alpha ( t) \equiv \left[ b_{1}\| u( t) \| _V^{p-1}+b_{2}( t)
\right] \| u( t) -v( t) \| ,
\]
it follows from Condition \ref{measurableestimate} that there
exists $z\in \mathcal{V}'$ such that $z( t) \in F( t) $ a.e. Thus
$z\in \widehat{A}u$.

Next we must verify the pseudo-monotone limit condition for
$\widehat{A}$ on $X$.  Let $u_{n}\rightharpoonup u$ in $X$,
$z_{n}\in \widehat{A}u_{n}$, and suppose
\begin{equation}
\lim \sup_{n\to \infty }\langle z_{n},u_{n}-u\rangle _{\mathcal{V}%
',\mathcal{V}}\leq 0.  \label{pseudo3.5}
\end{equation}
We note there is a set of measure zero, $\Sigma _{1}$ such that
for $t\notin \Sigma _{1}$, we have the following holding for all
$n$. $$\displaylines{ z_{n}( t) \in A( t,u_{n}( t) ) ,\;Bu( t) =B(
t) u( t) , \cr Bu_{n}( t) =B( t) u_{n}( t) ,\;\mbox{and } B( t)
\mbox{ is one-to-one.} }$$ First we verify the following claim.

\paragraph{Claim:} Let $u_{n}\rightharpoonup u$ in $X$ and let $t\notin \Sigma
_{1}$.  Then
\[
\lim \inf_{n\to \infty }\langle z_{n}( t) ,u_{n}( t) -u( t)
\rangle \geq 0.
\]

\paragraph{Proof of the claim:} Fix $t\notin \Sigma _{1}$ and suppose to the
contrary that
\begin{equation}
\lim \inf_{n\to \infty }\langle z_{n}( t) ,u_{n}( t) -u( t)
\rangle <0.  \label{pseudo3.55}
\end{equation}
Then there exists a subsequence, $n_k$ such that
\begin{equation}
\lim_{k\to \infty }\langle z_{n_k}( t) ,u_{n_k}( t) -u( t) \rangle
=\lim \inf_{n\to \infty }\langle z_{n}( t) ,u_{n}( t) -u( t)
\rangle <0. \label{pseudo3.6}
\end{equation}
Therefore, for all $k$ large enough, the second formula in \ref
{coerciveestimate} implies
\begin{eqnarray*}
b_{3}\| u_{n_k}( t) \| _V^p-b_{4}\| u_{n_k}( t) \| _V^{\alpha
}-b_{5}( t) &<&\| z_{n_k}( t) \| _{V'}\| u( t) \|_V \\ &\leq& (
b_{1}\| u_{n_k}( t) \| _V^{p-1}+b_{2}( t) ) \| u( t) \| _V
\end{eqnarray*}
which implies $\| u_{n_k}( t) \| _V$ and consequently $\| z_{n_k}(
t) \| _{V'}$ are bounded. ( $\| z_{n_k}( t) \| _{V'}$ is bounded
independent of $n_k$ because of the assumption that $A( t,\cdot )
$ is bounded and we just showed $\| u_{n_k}( t) \| _V$ is
bounded.) Now by (\ref{extraconditiontheorem1}),
\begin{equation}
Bu_{n_k}( t) =Bu_{n_k}( 0) +\int_{0}^{t}( Bu_{n_k}) '( s)
ds,\;\;Bu( t) =Bu( 0) +\int_{0}^{t}( Bu) '( s) ds \label{pseudo2}
\end{equation}
where the initial values $Bu_{n_k}( 0) $ and $Bu( 0)
$ are in $W'$. From Theorem \ref{spaceofsolutions} $%
Bu_{n_k}( 0) $ is bounded in $W'$.  Also from this theorem, the
mapping, $w\to Bw( 0) $ is a continuous and linear map from $X$ to
$W'$ and so, taking a further subsequence if necessary, we may
obtain
\[
Bu_{n_k}( 0) \rightharpoonup Bu( 0) \mbox{ in }W'
\]
and also $u_{n_k}( t) \rightharpoonup \xi$  in $V$.
 From (\ref{pseudo2}), and the assumption that $u_{n}\rightharpoonup u$ in
$X$,
\[
B( t) u_{n_k}( t) =Bu_{n_k}( t) \rightharpoonup Bu( t) =\mbox{ }B(
t) u( t) \mbox{ in }W'.
\]
Since $B( t) $ is continuous, it is also closed and hence weakly
closed. Therefore, the above implies
\[
B( t) u( t) =B( t) \xi \mbox{ in }W^{\prime }
\]
and since $B( t) $ is one-to-one, $\xi =u( t)$. Now from
(\ref{pseudo3.6}), and (\ref{pseudo3.55}),
\[
\lim_{k\to \infty }\langle z_{n_k}( t) ,u_{n_k}( t) -u( t) \rangle
<0,
\]
and so the pseudo-monotone limit condition for $A( t,\cdot ) $
implies that there exists $z_{\infty }$ in $A( t,u( t) ) $ such
that
\begin{eqnarray*}
\lim \inf_{k\to \infty }\langle z_{n_k}( t) ,u_{n_k}( t) -u( t)
\rangle &\geq &\langle z_{\infty },u( t) -u( t) \rangle =0 \\
&>&\lim_{k\to \infty }\langle z_{n_k}( t) ,u_{n_k}( t) -u( t)
\rangle ,
\end{eqnarray*}
a contradiction. This proves the claim.

We now continue with the proof of the theorem. It follows from
this claim that for a.e. $t$, in fact any $t\notin \Sigma _{1}$,
we have
\begin{equation}
\lim \inf_{n\to \infty }\langle z_{n}( t) ,u_{n}( t) -u( t)
\rangle \geq 0.  \label{pseudo3}
\end{equation}
Now also from the coercivity condition, \ref{coerciveestimate}, if
$y\in \mathcal{V}$,
\begin{eqnarray*}
\langle z_{n}( t) ,u_{n}( t) -y( t) \rangle &\geq &b_{3}\| u_{n}(
t) \|_V^p-b_{4}\| u_{n}( t) \|_V^{\alpha }-b_{5}( t) \\ &&-(
b_{1}\| u_{n}( t) \|^{p-1}+b_{2}( t) ) \| y( t) \| _V\,.
\end{eqnarray*}
Using $p-1=\frac{p}{p'}$, where $\frac{1}{p}+\frac{1}{p'}%
=1$, the right side of the above inequality equals
\[
b_{3}\| u_{n}( t) \| _V^p-b_{4}\| u_{n}( t) \| _V^{\alpha }-b_{5}(
t) -b_{1}\| u_{n}( t) \| ^{p/p'}\| y( t) \| _V-b_{2}( t) \| y( t)
\| _V.
\]
Now using Young's inequality, we can obtain a constant, $C(
b_{3},b_{4}) $, depending on $b_{3}$ and $b_{4}$ such that
\[
b_{4}\| u_{n}( t) \| _V^{\alpha }\leq \frac{b_{3}}{2}\| u_{n}( t)
\| _V^p+C( b_{3},b_{4})
\]
and another constant, $C( b_{1},b_{3}) $ depending on $b_{1}$ and
$b_{3}$ such that
\[
b_{1}\| u_{n}( t) \| ^{p/p^{\prime
}}\| y( t) \| _V\leq \frac{b_{3}}{2}%
\| u_{n}( t) \| _V^p+C( b_{1},b_{3}) \| y( t) \| _V^p.
\]
Letting $k( t) =b_{5}( t) +C( b_{3},b_{4}) $ and $C=C(
b_{1},b_{3}) $, it follows $k\in L^{1}( 0,T) $ and
\begin{equation}
\langle z_{n}( t) ,u_{n}( t) -y( t) \rangle \geq -k( t) -C\| y( t)
\| _V^p.  \label{pseudo4}
\end{equation}
Letting $y=u$, we may use Fatou's lemma to write
\begin{eqnarray*}
\lefteqn{ \lim \inf_{n\to \infty }\int_{0}^{T}( \langle z_{n}( t)
,u_{n}( t) -u( t) \rangle +k( t)+C\| y( t) \| _V^p) \,dt }\\
&\geq& \int_{0}^{T}\lim \inf_{n\to \infty }\langle z_{n}( t)
,u_{n}( t) -u( t) \rangle +( k( t)+C\| y( t) \| _V^p) \,dt \\
&\geq & \int_{0}^{T}( k( t) +C\| y( t)\| _V^p) \,dt\,.
\end{eqnarray*}
Consequently,
\begin{eqnarray*}
0 &\geq &\lim \sup_{n\to \infty }\langle z_{n},u_{n}-u\rangle _{%
\mathcal{V}',\mathcal{V}}\geq \lim \inf_{n\to \infty
}\int_{0}^{T}\langle z_{n}( t) ,u_{n}( t) -u( t) \rangle dt \\
&=&\lim \inf_{n\to \infty }\langle z_{n},u_{n}-u\rangle _{\mathcal{V}%
',\mathcal{V}}\geq \int_{0}^{T}\lim \inf_{n\to \infty }\langle
z_{n}( t) ,u_{n}( t) -u( t) \rangle dt\geq 0
\end{eqnarray*}
showing that
\begin{equation}
\lim_{n\to \infty }\langle z_{n},u_{n}-u\rangle _{\mathcal{V}%
',\mathcal{V}}=0\,.  \label{pseudo5}
\end{equation}
We need to show that for all $y\in X$ there exists $z( y) \in
\widehat{A}u$ such that
\[
\lim \inf_{n\to \infty }\langle z_{n},u_{n}-y\rangle _{\mathcal{V}%
',\mathcal{V}}\geq \langle z( y) ,u-y\rangle _{\mathcal{V%
}',\mathcal{V}}.
\]
Suppose to the contrary that for some $y\in X$,
\[
\lim \inf_{n\to \infty }\langle z_{n},u_{n}-y\rangle _{\mathcal{V}%
',\mathcal{V}}<\langle z,u-y\rangle _{\mathcal{V}',\mathcal{V}}
\]
for all $z\in \widehat{A}u$.  Taking a subsequence if necessary,
we assume
\[
\lim \inf_{n\to \infty }\langle z_{n},u_{n}-y\rangle _{\mathcal{V}%
',\mathcal{V}}=\lim_{n\to \infty }\langle z_{n},u_{n}-y\rangle
_{\mathcal{V}',\mathcal{V}}
\]
 From (\ref{pseudo4}),
\[
0\leq \langle z_{n}( t) ,u_{n}( t) -u( t) \rangle ^{-}\leq k( t)
+C\| u( t) \| _V^p\,.
\]
Thanks to (\ref{pseudo3}), we know that for a.e. $t$, $\langle
z_{n}( t) ,u_{n}( t) -u( t) \rangle \geq -\varepsilon $ for all
$n$ large enough. Therefore, for such $n,\langle z_{n}( t) ,u_{n}(
t) -u( t) \rangle ^{-}\leq \varepsilon $ if $\langle z_{n}( t)
,u_{n}( t) -u( t) \rangle <0$ and $\langle z_{n}( t) ,u_{n}( t)
-u( t) \rangle ^{-}=0$ if $\langle z_{n}( t) ,u_{n}( t) -u( t)
\rangle \geq 0$.  Therefore, $\lim_{n\to \infty }\langle z_{n}( t)
,u_{n}( t) -u( t) \rangle ^{-}=0$ and so we may apply the
dominated convergence theorem and conclude
\[
\lim_{n\to \infty }\int_{0}^{T}\langle z_{n}( t) ,u_{n}( t) -u( t)
\rangle ^{-}dt=\int_{0}^{T}\lim_{n\to \infty }\langle z_{n}( t)
,u_{n}( t) -u( t) \rangle ^{-}dt=0
\]
from (\ref{pseudo3}). Now by (\ref{pseudo5}) and the above
equation,
\begin{eqnarray*}
\lefteqn{ \lim_{n\to \infty }\int_{0}^{T}\langle z_{n}( t) ,u_{n}(
t) -u( t) \rangle ^{+}\,dt }\\ &=&\lim_{n\to \infty
}\int_{0}^{T}\langle z_{n}( t) ,u_{n}( t) -u( t) \rangle +\langle
z_{n}( t) ,u_{n}( t) -u( t) \rangle ^{-}\,dt \\
&=&\lim_{n\to \infty }\langle z_{n},u_{n}-u\rangle _{\mathcal{V}%
',\mathcal{V}}=0\,.
\end{eqnarray*}
Therefore,
\[
\lim_{n\to \infty }\int_{0}^{T}\left| \langle z_{n}( t) ,u_{n}( t)
-u( t) \rangle \right| \,dt=0
\]
so there exists a subsequence, $n_k$ such that
\begin{equation}
\langle z_{n_k}( t) ,u_{n_k}( t) -u( t) \rangle \to 0\mbox{ a.e.}
\label{pseudo6}
\end{equation}
Therefore, by the pseudo-monotone limit condition for $A( t,\cdot
) $, there exists $w_{t}$ in $A( t,u( t) ) $ such that for a.e.
$t$,
\[
\alpha ( t) \equiv \lim \inf_{k\to \infty }\langle z_{n_k}( t)
,u_{n_k}( t) -y( t) \rangle \geq \langle w_{t},u( t) -y( t)
\rangle .
\]
Let
\[
F( t) \equiv \left\{ w\in A( t,u( t) ) :\langle w,u( t) -y( t)
\rangle \leq \alpha (t) \right\}
\]
By the condition on measurability, \ref{measurableestimate}, there exists $%
z\in \mathcal{V}'$ such that $z( t) \in F( t) $ a.e. Therefore,
$z\in \widehat{A}u$ and for a.e. $t$,
\[
\lim \inf_{k\to \infty }\langle z_{n_k}( t) ,u_{n_k}( t) -y( t)
\rangle \geq \langle z(t) ,u( t) -y( t) \rangle .
\]
From (\ref{pseudo4}) and Fatou's lemma,
\begin{eqnarray*}
\lefteqn{ \int_{0}^{T}\lim \inf_{k\to \infty }\left\{ \langle
z_{n_k}( t) ,u_{n_k}( t) -y( t) \rangle +k(t) +C\| y( t) \|
_V^p\right\} dt }\\ &\leq& \lim \inf_{k\to \infty
}\int_{0}^{T}\left\{ \langle z_{n_k}( t) ,u_{n_k}( t) -y( t)
\rangle +k( t) +C\| y( t) \| _V^p\right\} dt
\end{eqnarray*}
which implies
\begin{eqnarray*}
\lim \inf_{k\to \infty }\langle z_{n_k},u_{n_k}-y\rangle _{%
\mathcal{V}',\mathcal{V}} &\geq &\int_{0}^{T}\lim \inf_{k\to
\infty }\langle z_{n_k}( t) ,u_{n_k}( t) -y( t) \rangle dt \\
&\geq &\int_{0}^{T}\langle z( t) ,u( t) -y( t) \rangle dt \\
&=&\langle z,u-y\rangle _{\mathcal{V}',\mathcal{V}} \\
&>&\lim_{k\to \infty }\langle z_{n_k},u_{n_k}-y\rangle _{%
\mathcal{V}',\mathcal{V}},
\end{eqnarray*}
a contradiction that completes the present proof.

\section{Existence theorems}
\label{existencetheorems} \setcounter{equation}{0}

In this section we generalize the main existence theorem of Bian
and Webb, \cite{bia99} to the case of implicit evolution
inclusions and more general initial data. With the result of
Section \ref{pointwisepseudomonotone}, we may state the following
corollary of Theorem \ref{mainexistencetheorem} which generalizes
the main result of \cite{bia99}.

\begin{theorem}
\label{existenceBonetoone}Let $B,\widehat{A},X,W$, and $V$ be as
defined above, $f\in \mathcal{V}'$, $u_{0}\in W$, and suppose
$\widehat{A}$ is given by Definition \ref{definitionofA} where $A(
t,\cdot ) $ satisfies the conditions \ref{limitpseudomonotone} -
\ref{measurableestimate} of Section \ref{pointwisepseudomonotone}.
Then if $B( t) $ is one to one a.e. t, there exists a solution,
$u\in X$ to
\[
( Bu) '+\widehat{A}u\ni f,\;Bu( 0) =Bu_{0}
\]
\end{theorem}

\paragraph{Proof:} Theorem \ref{Apseudomonotone} implies $\widehat{A}$ is
pseudo-monotone as a map from $X$ to $\mathcal{P}( X') $ and so
the estimates \ref{coerciveestimate} and \ref{boundedestimate} of
Section \ref{pointwisepseudomonotone} apply to give $\widehat{A}+\frac{1}{2}%
B'$ is bounded and coercive. Therefore, Theorem \ref
{mainexistencetheorem} implies the desired conclusion.

There is an easy generalization to Theorem
\ref{existenceBonetoone} which we state next. Suppose $B$
satisfies (\ref{3.1}) - (\ref{3.3}) and we modify the conditions
\ref{limitpseudomonotone} - \ref{measurableestimate} of Section
\ref{pointwisepseudomonotone}. For some $\lambda \geq 0$, we have

\begin{enumerate}
\item  $u\to \lambda B_{\lambda }( t) u+A(
t,u) $ is a set valued pseudo-monotone map satisfying Conditions 1
and 3 in the definition of pseudo-monotone given in Section
\ref{introduction} where
\[
B_{\lambda }( t) u=B( t) e^{-\lambda t}u.
\]

\item  There exists $b_{1}\geq 0$ and $b_{2}\in L^{p'}(
0,T) $ such that
\[
\| z\| _{V'}\leq b_{1}\| u\| _V^{p-1}+b_{2}( t) \mbox{ for all
}z\in A( t,u)
\]

\item  There exists $b_{3}>0$, $\alpha \in ( 0,p) $, $b_{4}\geq
0, $ and $b_{5}\in L^{1}( 0,T) $ such that
\[
\inf_{z\in \lambda B_{\lambda }( t) u+A( t,u) }\langle z,u\rangle
+\frac{e^{-\lambda t}}{2}\langle B'( t) u,u\rangle \geq b_{3}\|
u\| _V^p-b_{4}\| u\| _V^{\alpha }-b_{5}( t) .
\]
and
\[
\inf_{z\in \lambda B_{\lambda }( t) u+A( t,u) }\langle z,u\rangle
\geq b_{3}\| u\| _V^p-b_{4}\| u\| _V^{\alpha }-b_{5}( t)
\]

\item  The operators, $\lambda B+A( t,\cdot ) $ are measurable in
the sense of the condition on measurability,
(\ref{measurabilitycondition}).
\end{enumerate}

\begin{corollary}
\label{easycorollary} Let $B,\widehat{A},X,W$, and $V$ be as defined above, $%
f\in \mathcal{V}'$, $u_{0}\in W$, and suppose $\widehat{A}$ is
given by Definition \ref{definitionofA} where $A( t,\cdot ) $ and
$B$ satisfy the conditions 1 - 4 above. Then if $B$ is one-to-one,
there exists a solution, $u\in X$ to
\begin{equation}
( Bu) '+\widehat{A}u\ni f,\;Bu( 0) =Bu_{0}. \label{pseudo17.5}
\end{equation}
\end{corollary}

\paragraph{Proof:} We define a new dependent variable, $y$, by $u(
t) =e^{\lambda t}y( t) $.  Then the evolution inclusion of this
corollary, written in terms of $y$ becomes
\begin{equation}
( By) '+\lambda By+\widehat{A}_{\lambda }( y) \ni g,By( 0)
=Bu_{0},  \label{pseudo18}
\end{equation}
where $g( t) =e^{-\lambda t}f( t) $ and
\[
A_{\lambda }( t,v) \equiv e^{-\lambda t}A( t,e^{\lambda t}v) .
\]
It is almost immediate that Condition \ref{boundedestimate} of
Section \ref {pointwisepseudomonotone} holds for $\lambda
B+A_{\lambda }$.  Condition 3 above implies Condition
\ref{coerciveestimate} of Section \ref {pointwisepseudomonotone}
for $\lambda B+\widehat{A}_{\lambda }$ with modified $b_{i}$.
\begin{eqnarray*}
\lefteqn{ \left\{ w\in \lambda Bx( t) +A_{\lambda }( t,x( t) )
:\langle w,x( t) -y( t) \rangle \leq \alpha ( t) \right\} }\\ & =&
\left\{ w\in \lambda Bv( t) +A( t,v( t) ) :\langle w,v( t)
-e^{\lambda t}y( t) \rangle \leq e^{\lambda t}\alpha ( t)
\right\}\,,
\end{eqnarray*}
where $v( t) =\exp(\lambda t)x( t)$.  So the measurability
condition (\ref{measurabilitycondition}) holds for $\lambda
B+A_{\lambda }$.  It is also clear that $\lambda B+A_{\lambda }$
is set valued pseudo-monotone if Condition 1 above holds.
Therefore, we apply Theorem \ref{existenceBonetoone} to the
inclusion (\ref{pseudo18}).

\section{An example}
\label{example} \setcounter{equation}{0}

We give a simple example in this section, a modification of that
in \cite {bia99}. Let $\Omega $ be a bounded open set in
$\mathbb{R}^{3}$ having Lipschitz boundary and let $V$ be a closed
subspace of $H^{1}( \Omega ) ,W\equiv L^{6}( \Omega ) $ and
$b:\left[ 0,T\right] \times \Omega \to \mathbb{R}$ is Borel
measurable and for a.e. $t$, $$\displaylines{ b( t,x) >0\mbox{ a.e
}x, \cr b( t,x) =b( 0,x) +\int_{0}^{t}b_{t}( s,x)\, ds }$$ where
$b( 0,x) \in L^{3/2}( \Omega ) $ and
\[
\sup_{s\in \left[ 0,T\right] }\int_{0}^{T}\left| b_{t}( s,x)
\right| ^{3/2}dx<\infty
\]
Then we define an operator, $B( t) :W\to W'$ by $B( t) u( x)
\equiv b( t,x) u( x)$. Thus $B( t) $ is one-to-one for a.e. $t$.
We define a time dependent operator, $A( t,\cdot ) $ mapping $V$
to $V'$ as follows.
\[
\langle A( t,u) ,v\rangle \equiv \int_{\Omega }(
\sum_{i=1}^{3}a_{i}( t,x,u,\nabla u) \partial _{i}v) +a_{0}(
t,x,u,\nabla u) v\,dx\,.
\]
We make the following assumptions on the functions $a_{i}$.

\begin{enumerate}
\item  $( t,x) \to a_{i}( t,x,\mathbf{z}) $ is
measurable while $\mathbf{z}\to a_{i}( t,x,\mathbf{z}) $ is
continuous.

\item  There exist constants, $C_{1},C_{2},C_{3}>0$ and functions $k_{1}\in
L^{2}( \left[ 0,T\right] \times \Omega ) $ and $k_{2}\in L^{1}(
\left[ 0,T\right] \times \Omega ) ,k_{3}\in L^{1}( \Omega ) $,
such that
\[
\left| a_{i}( t,x,u,\mathbf{p}) \right| \leq C( \left|
\mathbf{p}\right| +\left| u\right| ) +k_{1}( t,x) ,
\]
and for some $\lambda $,
\begin{eqnarray*}
\lefteqn{ \lambda e^{-\lambda t}b( t,x) u^{2}+\sum_{i=1}^{3}a_{i}(
t,x,u,\mathbf{p}) p_{i}+a_{0}( t,x,u,\mathbf{p}) u+
\frac{e^{-\lambda t}}{2}b_{t}( t,x) u^{2}  }\\ &\geq&  C_{2}(
\left|\mathbf{p}\right| ^{2}+\left| u\right| ^{2}) -k_{2}( t,x)
\hspace{5cm}
\end{eqnarray*}
and
\[
\lambda e^{-\lambda t}b( t,x) u^{2}+\sum_{i=1}^{3}a_{i}(
t,x,u,\mathbf{p}) p_{i}+a_{0}( t,x,u,\mathbf{p}) u\geq C_{3}\left|
\mathbf{p}\right| ^{2}-k_{3}( x)
\]

\item  For all $( u,\mathbf{p}) $ and $( u,\widehat{\mathbf{p%
}}) $, if $\mathbf{p\neq }\widehat{\mathbf{p}}$
\[
\sum_{i=1}^{3}( a_{i}( t,x,u,\mathbf{p}) -a_{i}( t,x,u,%
\widehat{\mathbf{p}}) ) ( p_{i}-\widehat{p}_{i}) >0\,.
\]
\end{enumerate}

Now we define a set valued operator, $G:\left[ 0,T\right] \times
\Omega \times \mathbb{R}\to \mathcal{P}( \mathbb{R}) \setminus
\left\{ \emptyset \right\} $ as follows.

\begin{enumerate}
\item  $G( t,x,u) =\left[ g_{1}( t,x,u) ,g_{2}(t,x,u) \right] $
where $( t,u) \to G(t,x,u) $ is upper semi-continuous, meaning
that for each $x$, $$G(s,x,v) \subseteq \left[ g_{1}( t,x,u)
-\varepsilon,g_{2}( t,x,u) +\varepsilon \right] $$ whenever
$\left| v-u\right|+\left| s-t\right|$ is small enough. Also assume
$g_{i}:\left[ 0,T\right] \times \Omega \times \mathbb{R}\to
\mathbb{R}$ is Borel measurable.

\item  There exist $C_{3}>0$ and $C_{4}\in \lbrack 0,C_{2}),k_{3}\in
L^{2}( \Omega ) $, and $k_{4}\in L^{1}( \Omega ) $ such that $G$
satisfies the  estimates $$\displaylines{ \left| G( t,x,u) \right|
=\max \left\{ \left| g_{1}( t,x,u) \right| ,\left| g_{2}( t,x,u)
\right| \right\} \leq C_{3}\left| u\right| +k_{3}( x) , \cr \inf
\left\{ ug_{1}( t,x,u) ,ug_{2}( t,x,u) \right\} \geq -C_{4}\left|
u\right| ^{2}-k_{4}( x) . }$$ We define $w\in G( t,u) $ if and
only if $w( x) \in
G( t,x,u( x) ) $ a.e. so $G:V\to \mathcal{P}%
( H) \subseteq \mathcal{P}( V') $ where $%
H\equiv L^{2}( \Omega ) $.
\end{enumerate}

\noindent We will consider the abstract evolution inclusion
\begin{equation}
( Bu) '+\widehat{A+G}\ni 0,\;Bu( 0) =Bu_{0},\;u_{0}\in L^{6}(
\Omega )\,.  \label{pseudo19}
\end{equation}
Solutions to (\ref{pseudo19}) yield weak solutions to the
nonlinear differential inclusion $$\displaylines{ ( b( t,x) u)
_{t}-\sum_{i=1}^{3}\partial _{i}( a_{i}( t,x,u,\nabla u) ) +a_{0}(
t,x,u,\nabla u)\in -G( t,x,u) \cr u( 0,x) = u_{0}( x) , }$$ along
with appropriate boundary conditions depending on the choice of
$V$. We need to verify the hypotheses of the main existence
theorem. First we deal with the measurability issue for the
function, $G$.

\begin{lemma} \label{closed}
If conditions 1 and 2 hold for $G$, then for all closed convex
subset $P$ of $H$,
\[
G^{-}( P) \equiv \left\{ ( t,u) \in \left[ 0,T\right] \times V:G(
t,u) \cap P\neq \emptyset \right\}
\]
is a closed set.
\end{lemma}

\paragraph{Proof:} Let $( t_{n},u_{n}) \in G^{-}( P) $
and suppose $( t_{n},u_{n}) \to ( t,u) $ in $\left[ 0,T\right]
\times H$.  Taking a subsequence, still denoted by $(t_{n},u_{n})
$, we may also assume $u_{n}( x) \rightarrow u( x) $ for a.e. $x$.
By the assumption that $( t_{n},u_{n}) \in G^{-}( P) $, there
exists $w_{n}\in H\cap P $ such that for a.e. $x$,
\[
g_{1}( t_{n},x,u_{n}( x) ) \leq w_{n}( x) \leq g_{2}(
t_{n},x,u_{n}( x) ) .
\]
By the given estimates, we see $w_{n}$ is bounded in $H$ and so we
may take
a further subsequence, still denoted by $w_{n}$ such that $%
w_{n}\rightharpoonup w\in H$.  Now let
\[
\mathfrak{B}_{\varepsilon }\equiv \left\{ x\in \Omega :w( x) \leq
g_{1}( t,x,u( x) ) -\varepsilon \right\} .
\]
We know from Estimate 2 given above for $G$ that
\begin{eqnarray*}
\lefteqn{ \int_{\mathfrak{B}_{\varepsilon }}-C_{3}\left| u_{n}( x)
\right| -k_{3}( x) dx }\\ &\geq& \int_{\mathfrak{B}_{\varepsilon
}}\left[ ( g_{1}( t_{n},x,u_{n}( x) ) -( C_{3}\left| u_{n}( x)
\right| +k_{3}( x) ) ) -w_{n}( x) \right] dx\,.
\end{eqnarray*}
Adding $3\int_{\mathfrak{B}_{\varepsilon }}C_{3}\left| u_{n}( x)
\right| +k_{3}( x) dx$ to both sides in order to get both
integrands positive, we obtain
\begin{eqnarray*}
\lefteqn{ 2\int_{\mathfrak{B}_{\varepsilon }}C_{3}\left| u_{n}( x)
\right| +k_{3}( x) dx }\\ &\geq& \int_{\mathfrak{B}_{\varepsilon
}}\left[ ( g_{1}( t_{n},x,u_{n}( x) ) -w_{n}( x) ) +2( C_{3}\left|
u_{n}( x) \right| +k_{3}( x) ) \right] dx\,.
\end{eqnarray*}
Taking the $\lim \inf $ of both sides and using Fatou's lemma,
\begin{eqnarray*}
\lefteqn{2\int_{\mathfrak{B}_{\varepsilon }}C_{3}\left| u( x)
\right| +k_{3}( x) dx }\\ &\geq& \int_{\mathfrak{B}_{\varepsilon
}}\lim \inf_{n\to \infty }( g_{1}( t_{n},x,u_{n}( x)) -w_{n}( x) )
dx+ 2\int_{\mathfrak{B}_{\varepsilon }}( C_{3}\left| u( x) \right|
+k_{3}( x) )
\end{eqnarray*}
and so from the upper semi-continuity of $G$,
\begin{eqnarray*}
0&\geq& \int_{\mathfrak{B}_{\varepsilon }}\lim \inf_{n\to \infty
}( g_{1}( t_{n},x,u_{n}( x) ) -w_{n}( x) ) dx \\
&=&\int_{\mathfrak{B}_{\varepsilon }}( g_{1}( t,x,u( x) ) -w( x) )
dx\geq \int_{\mathfrak{B}_{\varepsilon }}\varepsilon dx
\end{eqnarray*}
showing that $m( \mathfrak{B}_{\varepsilon }) =0$.  Thus, since $%
\varepsilon >0$ is arbitrary, it follows $w( x) \geq g_{1}( t,x,u(
x) ) $ a.e. Similar reasoning shows $w( x) \leq g_{2}( t,x,u( x) )
$ a.e. Since $P$ is\ a closed and convex subset of $H$, $w\in P$
as well. Thus $( t,u) \in G^{-}( P) $, and this shows this set is
closed as claimed.

We just showed that
\begin{equation}
\left\{ ( t,v) :G( t,v) \cap P\neq \emptyset \right\}
\label{pseudomonotone19}
\end{equation}
is a Borel set whenever $P$ is convex and closed.

\begin{lemma}\label{measrabilitycondition}
For $G$ defined above, $G$ is measurable in the sense of
(\ref{measurabilitycondition}).
\end{lemma}

\paragraph{Proof:} This follows from Lemma \ref{sufficientforcondition3} which
is proved in the next section because we have just shown the
hypothesis of Lemma \ref{sufficientforcondition3} are satisfied in
\ref{pseudomonotone19}.
Since $A( t,\cdot ) $ is single valued, it follows the sum $%
A( t,\cdot ) +G( t,\cdot ) $ satisfies the measurability
condition, (\ref{measurabilitycondition}).

Next we must consider the question of whether the operators are
pseudo-monotone. That $\lambda B_{\lambda }( t) +A( t,\cdot ) $ is
pseudo-monotone follows from results of Browder \cite{bro77}.
Therefore, we must verify $G( t,\cdot ) $ is set valued
pseudo-monotone. It remains to verify Conditions 1 and 3 in the
list of conditions for Pseudo-monotone because the assumed
estimates imply $G( t,\cdot ) $ is bounded. Since weak convergence
in $V$ implies strong convergence in $L^{2}( \Omega ) \equiv H$,
it is easy to verify that $G( t,u) $ is a closed and convex subset
of $H\subseteq V'$. From estimates on $g_{i}$ it follows $g_{i}(
t,\cdot
,u( \cdot ) ) \in L^{2}( \Omega ) $ and so $%
G( t,u) \neq \emptyset $.  It remains to verify condition 3, the
limit condition for pseudo-monotone. Suppose $u_{n}\rightharpoonup
u$ in $V$. We show for each $v\in V$, there exists $u_{\infty
}^{\ast }\in G( t,u) $ such that
\[
\lim \inf_{n\to \infty }\langle u_{n}^{\ast },u_{n}-v\rangle \geq
\langle u_{\infty }^{\ast },u-v\rangle .
\]
The weak convergence of $u_{n}$ in $V$ implies $u_{n}\to u$ in $H$
and taking a subsequence, we may assume that $u_{n}( x) \to u( x)
$ a.e. Let $u_{n}^{\ast }\in G( t,u_{n}) \subseteq H$.  By the
estimates, $u_{n}^{\ast }$ is bounded in $H\equiv L^{2}( \Omega )
$ and so, taking a subsequence, we may assume that $u_{n}^{\ast }$
converges weakly in $H$ to $u_{\infty }^{\ast }$. By the
compactness of the embedding of $V$ into $H$, it follows that the
embedding of $H$ into $V'$ is also compact and so $u_{n}^{\ast }$
converges strongly to $u_{\infty }^{\ast }$ in $V'$. We need to
verify $u_{\infty }^{\ast }\in G( t,u) $.  Define
\[
\mathfrak{B}_{\varepsilon }\equiv \left\{ x:u_{\infty }^{\ast }(
x) \geq g_{2}( t,x,u( x) ) +\varepsilon \right\} .
\]
We know
\begin{eqnarray*}
\lefteqn{ -\int_{\mathfrak{B}_{\varepsilon }}C_{3}\left| u_{n}( x)
\right| +k_{3}( x) dx }\\ &\leq& \int_{\mathfrak{B}_{\varepsilon
}}-u_{n}^{\ast }( x) dx-\int_{\mathfrak{B} _{\varepsilon }}\left[
C_{3}\left| u_{n}( x) \right| +k_{3}( x) -g_{2}( t,x,u_{n}( x) )
\right] dx
\end{eqnarray*}
so taking the lim sup of both sides using the assumed weak convergence of $%
u_{n}^{\ast }$ to $u_{\infty }^{\ast }$,
\begin{eqnarray*}
\lefteqn{ -\int_{\mathfrak{B}_{\varepsilon }}C_{3}\left| u( x)
\right| +k_{3}( x) dx }\\ &\leq& \int_{\mathfrak{B}_{\varepsilon
}}-u_{\infty }^{\ast }( x) dx-\lim \inf
\int_{\mathfrak{B}_{\varepsilon }}\left[ C_{3}\left| u_{n}( x)
\right| +k_{3}( x) -g_{2}( t,x,u_{n}( x) ) \right] dx\,.
\end{eqnarray*}
Using Fatou's lemma as in the proof of Lemma \ref{closed} and the
upper semi-continuity of $( t,u) \to G( t,x,u) $, we obtain
\begin{eqnarray*}
\lefteqn{ -\int_{\mathfrak{B}_{\varepsilon }}C_{3}\left| u( x)
\right| +k_{3}( x) dx }\\ &\leq& \int_{\mathfrak{B}_{\varepsilon
}}-u_{\infty }^{\ast }( x) dx -\int_{\mathfrak{B}_{\varepsilon
}}\left[ C_{3}\left| u( x) \right| +k_{3}( x) -\lim \sup g_{2}(
t,x,u_{n}( x)) \right] dx \\ &\leq
&\int_{\mathfrak{B}_{\varepsilon }}-u_{\infty }^{\ast }( x)
dx-\int_{\mathfrak{B}_{\varepsilon }}\left[ C_{3}\left| u( x)
\right| +k_{3}( x) -g_{2}( t,x,u( x) ) \right] dx,
\end{eqnarray*}
which implies
\[
0\leq \int_{\mathfrak{B}_{\varepsilon }}g_{2}( t,x,u( x) )
-u_{\infty }^{\ast }( x) dx\leq -\varepsilon m( \mathfrak{B}%
_{\varepsilon })
\]
so $m( \mathfrak{B}_{\varepsilon }) =0$.  Therefore, letting $%
\varepsilon $ be replaced by $\frac{1}{n}$, taking the union of $\mathfrak{B}%
_{1/n}$, we see that $u_{\infty }^{\ast }( x) \leq g_{2}( x,u( x)
) $ a.e. and a similar argument shows $u_{\infty }^{\ast }( x)
\geq g_{1}( x,u( x) ) $ a.e. We have shown that if
$u_{n}\rightharpoonup u$ in $V$, there exists a subsequence,
$u_{n_k}$ such that $u_{n_k}^{\ast }\rightharpoonup u_{\infty
}^{\ast }\in G( t,u) $ in $H$. Therefore, by the strong
convergence of $u_{n}^{\ast }$ to $u_{\infty }^{\ast }$ in $V'$
and
the weak convergence of $u_{n_k}$ to $u$ in $V$, it follows that for any $%
v\in V$,
\[
\lim_{k\to \infty }\langle u_{n_k}^{\ast },u_{n_k}-v\rangle
_{V',V}=\langle u_{\infty }^{\ast },u-v\rangle _{V',V}
\]
If for some $v\in V$, there exists a sequence,
$u_{n}\rightharpoonup u$ in $V $ such that
\[
\lim \inf_{n\to \infty }\langle u_{n}^{\ast },u_{n}-v\rangle
_{V',V}<\langle u^{\ast },u-v\rangle _{V',V},
\]
for all $u^{\ast }\in G( u) $, this is a contradiction to what was
just shown. Therefore, we have verified the pseudo-monotone limit
condition.

It follows $\lambda B_{\lambda }( t) +A( t,\cdot ) +G( t,\cdot )
$, being a sum of pseudo-monotone operators is pseudo-monotone
\cite{nan95}. The assumed estimates give the rest of the
hypotheses of Corollary \ref{easycorollary}. Therefore, there
exists a solution to (\ref{pseudo19}).

\section{Measurability}
\label{measurability} \setcounter{equation}{0}

In this section we demonstrate that the conditions on
measurability given by Bian and Webb imply Condition
\ref{measurableestimate} of Section \ref
{pointwisepseudomonotone}. The following definition of
measurability, used in \cite{bia99}, is the one referred to as
strong measurability in \cite {aub90}.

\begin{definition}
\label{measurabledefinition} \rm We say $S:\left[ 0,T\right]
\times V\rightarrow \mathcal{P}( V') $ is measurable if whenever,
$\mathfrak{H} \subseteq V'$ is a closed set, the set $\left\{ (
t,v) :S( t,v) \cap \mathfrak{H}\neq \emptyset\right\}$ is a Borel
set in $\left[ 0,T\right] \times V$.
\end{definition}

The following lemma is an immediate consequence of the above
definition.

\begin{lemma}
If $S$ is measurable as just described and if $x:\left[ 0,T\right]
\to V$ is measurable, then for $\mathfrak{H}$ a closed set in
$V^{\prime }$, the set $\left\{ t\in \left[ 0,T\right] :S( t,x( t)
) \cap\mathfrak{H}\neq\emptyset\right\}$ is measurable.
\end{lemma}

Now we verify the following lemma which is the main result of this
section.

\begin{lemma}
\label{sufficientforcondition3}Suppose $S$ satisfies the condition
of Definition \ref{measurabledefinition} for all $\mathfrak{H}$ a
closed convex set
where $S( t,v) $ equals a closed convex nonempty subset of $%
V'$. Then $S( t,\cdot ) $ is measurable in the sense of
(\ref{measurabilitycondition}).
\end{lemma}

\paragraph{Proof:} We show that if $\alpha $ is real valued and measurable,
and $x,y$ are strongly measurable $V$ valued functions, then $F:\left[ 0,T%
\right] \to \mathcal{P}( V') $ given by
\[
F( t) \equiv \left\{ w\in S( t,x( t) ) :\langle w,x( t) -y( t)
\rangle \leq \alpha ( t) \right\}
\]
has the property that $\left\{ t:F( t) \cap U\neq \emptyset
\right\} $ is measurable whenever $U$ is an open set. We define
\[
F_{n}^{m}( t) \equiv \left\{ w\in S( t,x( t) ) :\langle w,x_{n}(
t) -y_{n}( t) \rangle \leq \alpha _{n}( t) +\frac{1}{m}\right\}
\]
where here $x_{n},y_{n},\alpha _{n}$ are simple functions
converging pointwise to $x,y$, and $\alpha $ respectively. Thus
there exist disjoint measurable subsets of $\left[ 0,T\right] $,
$\left\{ E_{i}^{n}\right\} _{i=1}^{m_{n}}$ such that each of
$x_{n},y_{n}$, and $\alpha _{n}$ are constant on $E_{i}^{n}$. If
$\mathfrak{H}$ is a closed convex set in $V'$
\begin{equation}
\left\{ t\in \left[ 0,T\right] :F_{n}^{m}( t) \cap
\mathfrak{H}\neq \emptyset \right\} =\cup _{i=1}^{m_{n}}\left\{
t\in E_{i}^{n}:F_{n}^{m}( t) \cap \mathfrak{H}\neq \emptyset
\right\} . \label{pseudo7}
\end{equation}
On the set, $E_{i}^{n}$ denote the values of $x_{n},y_{n}$ and
$\alpha _{n}$ as $x_{n}^{i},y_{n}^{i}$, and $\alpha _{n}^{i}$
respectively. Then
\[
F_{n}^{m}( t) \cap \mathfrak{H}=S( t,x( t) ) \cap \mathfrak{H}\cap
C_{n}^{i}
\]
where
\[
C_{n}^{i}\equiv \left\{ w\in V':\langle
w,x_{n}^{i}-y_{n}^{i}\rangle \leq \alpha
_{n}^{i}+\frac{1}{m}\right\}
\]
and is a closed set. Therefore,
\[
\left\{ t\in E_{i}^{n}:F_{n}^{m}( t) \cap \mathfrak{H}\neq
\emptyset \right\} =\left\{ t\in E_{i}^{n}:S( t,x( t) ) \cap
\widetilde{\mathfrak{H}}\neq \emptyset \right\}
\]
where $\widetilde{\mathfrak{H}}\equiv \mathfrak{H}\cap C_{n}^{i}$,
a closed convex set. Therefore, by our hypotheses,  the set
$\left\{ t\in E_{i}^{n}:F_{n}^{m}(t) \cap \mathfrak{H}\neq
\emptyset \right\} $ is measurable and so it follows from
(\ref{pseudo7}) that the set $\left\{ t\in \left[ 0,T\right]
:F_{n}^{m}( t) \cap \mathfrak{H}\neq \emptyset \right\} $ is also
measurable.

\paragraph{Claim:} Let $\mathfrak{H}$ be a closed ball. Then
\begin{equation}
\left\{ t:F( t) \cap \mathfrak{H}\neq \emptyset \right\} =\cap
_{m=1}^{\infty }\cup _{k=1}^{\infty }\cap _{n\geq k}\left\{
t:F_{n}^{m}( t) \cap \mathfrak{H}\neq \emptyset \right\} .
\label{lotsofintersections}
\end{equation}

\paragraph{Proof of the claim:} If $t\in \left\{ t:F( t) \cap
\mathfrak{H}\neq \emptyset \right\} $, then there exists $w\in S(
t,x( t) ) \cap \mathfrak{H}$ such that $\langle w,x( t) -y( t)
\rangle \leq \alpha (t)$. Therefore, for that $w$ it follows that
for each $m$,
\[
\langle w,x_{n}( t) -y_{n}( t) \rangle \leq \alpha _{n}( t)
+\frac{1}{m}
\]
for all $n$ large enough. Therefore, $t$ is an element of the right side of (%
\ref{lotsofintersections}).

Now let $t$ be an element of the right side. Then for all $m$,
there exists $w_{n}^{m}$ in $S( t,x( t) ) \cap \mathfrak{H}$ such
that
\[
\langle w_{n}^{m},x_{n}( t) -y_{n}( t) \rangle \leq \alpha _{n}(
t) +\frac{1}{m}
\]
for all $n$ large enough. Since $S( t,x( t) ) \cap
\mathfrak{H}$ is closed and bounded and convex, we can take a subsequence, $%
w_{n_k}^{m}$ which converges weakly to $w^{m}\in S( t,x( t) ) \cap
\mathfrak{H}$.  Therefore, taking a limit as $k\rightarrow
\infty $, using the strong convergence of $x_{n}( t) $ and $%
y_{n}( t) $ to $x( t) $ and $y( t) $ respectively, we obtain
\[
\langle w^{m},x( t) -y( t) \rangle \leq \alpha ( t) +\frac{1}{m}.
\]
Now take another subsequence, $w^{m_k}$ converging weakly to $w\in
S( t,x( t) ) \cap \mathfrak{H}$ and take a limit as $k\rightarrow
\infty $ to obtain
\[
\langle w,x( t) -y( t) \rangle \leq \alpha ( t) .
\]
It follows $t$ is an element of the left side of (\ref{lotsofintersections}%
), proving the claim.

Now if $U$ is an arbitrary open set, we know that since $V$ and
consequently, $V'$ are separable, $U$ is the union of countably
may closed balls, $U=\cup _{k=1}^{\infty }\mathfrak{H}_k$ and
\[
\left\{ t:F( t) \cap U\neq \emptyset \right\} =\cup _{k=1}^{\infty
}\left\{ t:F( t) \cap \mathfrak{H}_k\neq \emptyset \right\} ,
\]
a measurable set.

We have just verified that for all $U$ open, $\left\{ t:F( t) \cap
U\neq \emptyset \right\} $ is a measurable set. This is a
sufficient condition for the existence of a measurable, selector,
$z( t) \in F( t) $.  \cite{cas77}, \cite{pap97}. From the assumed
estimates, it follows $z\in \mathcal{V}'$ whenever $x\in
\mathcal{V}$.

We did not gain any generality by only requiring the closed set,
$\mathfrak{H}$ to be convex. To see this, note that our argument
is concluded by verifying that the set $\left\{ t:F( t) \cap U\neq
\emptyset \right\} $ is measurable for all $U$ open. It is known
\cite{cas77}, \cite{pap97} this is equivalent under certain
conditions, including the case where Lebesgue measure is used on
the Lebesgue measurable sets of $\left[ 0,T\right] $ to the set
$\left\{ t:F( t) \cap \mathfrak{H}\neq \emptyset \right\} $ being
measurable for all $\mathfrak{H}$ closed or even Borel.
Nevertheless, it is easier to verify the measurability condition
for closed convex sets than for arbitrary closed sets.

\paragraph{Summary.}
The paper has given an existence theorem for implicit evolution
inclusions of the form $( Bu) '+Au\ni f$ under assumptions that
$B( t) $ is one-to-one
 a.e. It would be very interesting to obtain
similar theorems involving reasonable pointwise conditions on the
operators $A( t,\cdot ) $ using different methods, and also to
include the case where $B( t) $ could be a degenerate operator as
in \cite {kut99}.

\paragraph{Acknowledgments.} I am grateful to the anonymous referee for
pointing out several errors in the original manuscript.

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\end{thebibliography}
 \smallskip

\noindent{\sc Kenneth Kuttler }\\ 
Department of Mathematics\\
Brigham Young University\\
Provo, Utah, 84602 USA\\
email: klkuttle@mail.math.byu.edu


\end{document}