\documentclass[twoside]{article}
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\markboth{\hfil High regularity for a nonlinear parabolic problem
 \hfil EJDE--2002/48}
{EJDE--2002/48\hfil L. Barbu, Gh. Moro\c sanu, \& W. L. Wendland \hfil}
\begin{document}
\title{\vspace{-1in}\parbox{\linewidth}{\footnotesize\noindent
{\sc  Electronic Journal of Differential Equations},
Vol. {\bf 2002}(2002), No. 48, pp. 1--12. \newline
ISSN: 1072-6691. URL: http://ejde.math.swt.edu or http://ejde.math.unt.edu
\newline ftp  ejde.math.swt.edu  (login: ftp)}
 \vspace{\bigskipamount} \\
 %
  High regularity of the solution of a nonlinear parabolic
  boundary-value problem
 %
\thanks{ {\em Mathematics Subject Classifications:}
35K60, 35K05, 35K20, 34G20, 47H05, 47H20.
\hfil\break\indent
{\em Key words:} Parabolic equation, nonlinear
 boundary conditions, \hfil\break\indent
 maximal monotone operator, subdifferential,
 compatibility conditions.
\hfil\break\indent
\copyright 2002 Southwest Texas State University. \hfil\break\indent
Submitted April 11, 2002.  Published May 29, 2002.
\hfil\break\indent
Gh. Moro\c sanu was supported
by the German Research Foundation DFG \hfil\break\indent
(Deutsche Forschungsgemeinschaft) under the project  DFG We 659/35-2.
} }
\date{}
%
\author{Lumini\c{t}a Barbu, Gheorghe Moro\c sanu, \& Wolfgang L. Wendland}
\maketitle

\begin{abstract}
  The aim of this paper is to report some results concerning
  high regularity of the solution  of a nonlinear parabolic
  problem with a linear parabolic differential equation in
  one spatial dimension and nonlinear boundary conditions.
  We show that any regularity can be reached provided that
  appropriate smoothness of the data and compatibility
  assumptions are required.
\end{abstract}

\newtheorem{theorem}{Theorem}[section]
\newtheorem{remark}{Remark}[section]
\numberwithin{equation}{section}

\section{Introduction}\label{s:1}

We consider the parabolic boundary value problem (BVP):
\begin{gather}
y_t-y_{xx}+ gy=f(x,t), \quad \mbox{in} \ D_T, \label{E}\\[2pt]
 \begin{gathered}
y_x(0,t)\in {\beta}_1(y(0,t)),\\
-y_x(1,t)\in {\beta}_2(y(1,t)), \quad 0<t<T,
\end{gathered} \label{BC} \\[2pt]
y(x,0)=y_0(x), \quad  0<x<1, \label{IC}
\end{gather}
where $D_T:= \{ (x,t)\in \mathbb{R}^2; 0<x<1, 0<t<T\}$ for a fixed
$T\in (0,\infty )$, where $g \geq 0$ is a given constant,
$f:\overline{D_T}\to \mathbb{R}$ and $y_0:[0,1]\to \mathbb{R}$ are
given functions, and
${\beta}_i: D({\beta_i})\subseteq\mathbb{R}\to \mathbb{R}$, $i=1,2$, are
nonlinear mappings that might possibly be multivalued. So, this
BVP is a nonlinear problem.

 Notice that the BVP is a model
for heat conduction and diffusion processes. Also, it can be
viewed as a reduced model for a singular perturbation problem
associated with an electrical circuit, in which the specific
inductance is negligible and is set equal to zero \cite{BWCM}. In
\cite{BWCM} some nonhomogeneous boundary conditions appear, but
they can easily be homogenized by a simple change of the unknown
function:
\begin{equation}\label{eq:1.1}
\tilde y(x,t) = y(x,t)+x(1-x)[a(t)x+b(t)]\,.
\end{equation}
%
In the case of integrated circuits with negligible specific
inductances, we can set them equal to zero and thus arrive at a
similar BVP, but with $n$ equations instead of (\ref{E})
\cite{GK,MN}.

An existence theory for the BVP in the case in
which $\beta _1, \; \beta _2$ are monotone mappings can be found in
\cite[Chapter 3, \S 3]{M}, even in a more general framework. In
the present paper we are concerned with higher order regularity of
the solution. As we shall see, this requires higher regularity of
$f, \; \beta _1, \; \beta _2, \;  y_0$ and higher order compatibility
conditions. The main difficulty in this problem is due to the
nonlinearity of (\ref{BC}) whereas the linear case is well-known from
\cite[Chapter VII, \S 10]{LSU}. There the compatibility
conditions, corresponding to our particular BVP with linear
conditions, consist in the fact that the time derivatives
$(\partial ^ky/\partial t^k)(x,0)$ ($k\ge 0$), which can be
calculated from (\ref{E}) and (\ref{IC}), for $x=0$ and $x=1$ must satisfy
the boundary conditions and the relations obtained from their
differentiation with respect to $t$. Of course, the case of the
nonlinear condition (\ref{BC}) is more delicate.

 The regularity question for
the BVP is also important as an intermediate step in developing
an asymptotic analysis of the telegraph system with small specific
inductance (see \cite{BWCM,BM1,BM2}).

\section{Preliminaries}\label{s:2}

Let us first recall the following result (see
\cite[Appendix]{Br2}), which will be used later:

\begin{theorem}\label{th: 2.1.}
Let $X$ be a reflexive real Banach space and let
$u\in L^p(a,b;\,X)$ with $-\infty <a<b<\infty $ and $1<p<\infty $.
Then, the following two conditions are equivalent:
\begin{enumerate}
\item[(i)] $u\in W^{1,p}(a,b;\, X)$;
\item[(ii)] There exists a constant $C>0$ such that
$\int_a^{b-\delta}\Vert u(t+\delta)-u(t) \Vert^p_X\le C{\delta}^p $
for every $\delta\in(0, b-a]$.
\end{enumerate}
The implication $(i)\Rightarrow (ii)$ is
still valid for $p=1$. Moreover, $(ii)$ holds if $u$ is of bounded
variation on $[a,b]$ and $X$ is not necessarily reflexive.
\end{theorem}

Here and in what follows, $L^p$ and $W^{k,p}$ denote the usual
function and Sobolev spaces, respectively. Now, in order to
state the next result, let us consider two real Hilbert spaces $V$
and $H$, such that $V$ is densely and continuously embedded into
$H$. If $H$ is identified with its own dual, then we have
$V\subset H\subset V'$, algebraically and topologically. We have
denoted by $V'$ the dual of $V$. Denote also by $\langle \cdot ,
\cdot \rangle$ the pairing between $V$ and $V'$, i.e., $\langle
v,v^*\rangle := v^*(v), \quad \mbox{for} \quad v\in V, \quad v^*\in V'$.
For $v^*\in H' \equiv H , \quad \langle v, v^*\rangle $ reduces to
the scalar product in $H$ of $v$ and $v^*$. Following
\cite[Chapter 1]{LM}, for some $-\infty < a < b < \infty$ we set
$$
W(a,b):=\{u\in L^2(a,b;V);\, u'\in L^2(a,b;V')\}\,,
$$
where $u'$ is the distributional derivative of $u$, as a
$V$-valued distribution.

 \begin{theorem}\label{th:2.2.}
Every $u\in W(a,b)$ has a representative $u_1\in C([a,b];H)$ and
so $u$ can be identified with such a continuous function.
Furthermore, if $u,\, \tilde u\in W(a,b)$, then the function
$t\mapsto \langle u(t),\tilde u(t)\rangle$
 is absolutely continuous on $[a,b]$ and
$$
{\frac{d}{dt}}\langle u(t), \tilde u(t)\rangle = \langle u(t),
\tilde u'(t)\rangle + \langle \tilde u(t), u'(t)\rangle \quad \mbox{for a.a.} \quad t\in (a,b),
$$
and, hence, in particular,
$$
{\frac{d}{dt}}\Vert u(t)\Vert ^2_H = 2\langle u(t), u'(t)\rangle \quad \mbox{for a.a.} \quad t\in (a,b)\,.
$$
\end{theorem}

Finally, we recall the following theorem due to H. Attouch and A.
Damlamian \cite{AD} which will be needed for the derivation of
higher regularity of the solution of the BVP:

\begin{theorem}\label{th:2.3.} Let
 $A(t)=\partial\phi(t,\cdot)$ for $0\let\le T$,
where $\phi(t,\cdot)\,:\,H\to (-\infty,+\infty]$ are proper,
convex, and lower semi-continuous functions, with a domain of
definition $D(\phi(t,\cdot))=D$ which is independent of $t$. Here
$H$ is a real Hilbert space. Assume further that there exists a
nondecreasing function $\gamma$$:$$[0,T]$$\to $$\mathbb{R}$ and
some real constants $C_1$, $C_2$ such that
%
\begin{equation}\label{eq:2.1}
\phi(t,v)\le\phi(s,v)+[\gamma (t) - \gamma (s)]\cdot [\phi (s,v) +
C_1\Vert v\Vert ^2_H +C_2]
\end{equation}
%
for all $v\in D$ and  $0\le s\le t\le T$. Then, for every $u_0\in
D$ and $f\in L^2(0,T; H)$, there exists a unique solution $u\in
W^{1,2}(0,T; H)$ of the equation  $u'(t)+A(t)u(t)=f(t)$ for a.a.
 $t\in (0,T)$ with the initial condition $u(0)=u_0$. Moreover, there
exists a function $h\in L^1(0,T)$ such that
%
\begin{equation}\label{eq:2.2}
 \phi (t,u(t))\le \phi (s,u(s))+\int _s^t h(r)dr \quad \mbox{for all} \
 \ 0\le s\le t\le T\,.
\end{equation}
\end{theorem}

Here, we denote by $\partial \phi (t, \cdot )$ the subdifferential
of the function $\phi (t, \cdot )$. In what follows, we shall use
the theory of evolution equations associated with monotone
operators in Hilbert spaces. For details we refer to
\cite{Ba,Br2,M}.

\section{High Regularity of Solutions}\label{s:3}
\setcounter{equation}{0} If $\beta _1, \ \beta _2$ are maximal
monotone mappings, then existence and uniqueness of the solution
to the BVP is well known. The most important results, even for
$n$ dimensions, were established by H. Brezis \cite{Br1,Br2}. Our
problem can be expressed as a Cauchy problem in $L^2(0,1)$,
associated with a subdifferential and, hence, existence is
available (see, e.g., Brezis's theorem in \cite[p. 56]{M}, where
the regularizing effect of the subdifferential on the initial data
is pointed out). For a more general problem than ours see
\cite[Chapter 3, \S 3]{M}, where the existence to a higher order,
one-dimensional, parabolic equation is discussed.

 \begin{theorem}\label{th:3.1.}
Assume that
%
\begin{gather}\label{eq:3.1}
f\in W^{1,1}(0,T;L^2(0,1))\,;\\
\label{eq:3.2}
{\beta}_1, \ {\beta}_2  \quad  \mbox{are both maximal monotone}\,;  \\
\label{eq:3.3} y_0\in H^2(0,1)\,;
\end{gather}
and that the following first-order compatibility conditions are
fulfilled:
%
\begin{equation}\label{eq:3.4}
 y'_0(0)\in {\beta}_1(y_0(0)), \quad -y'_0(1)\in
 {\beta}_2(y_0(1))\,.
\end{equation}
%
 Then the BVP (\ref{E}), (\ref{BC}), (\ref{IC}) has a unique strong solution
%
\begin{equation}\label{eq:3.5}
y\in W^{1,\infty}(0,T;L^2(0,1))\cap L^\infty(0,T;H^2(0,1))\cap
W^{1,2}(0,T;H^1(0,1))\,.
\end{equation}
 \end{theorem}

\paragraph{Proof:} Let $H=L^2(0,1)$ with
$$ \langle p,q\rangle = \int_0^1 p(x)q(x)\, dx, \quad \Vert p
{\Vert}_H^2 = \int_0^1 p(x)^2dx\, .
$$
We define the operator $A\,:\,D(A)\subseteq H\to  H$ as
$$
Ap = -p'' + gp
$$
on the domain of definition
%
\begin{eqnarray*}
D(A)&=&
\big\{p\in H^2(0,1) ;  p(0)\in D(\beta _1), \\
&&  p(1)\in D(\beta _2),  p'(0)\in {\beta}_1(p(0)),
-p'(1)\in {\beta}_2(p(1))\big\}\,.
\end{eqnarray*}
%
Then the BVP may be written as the Cauchy problem in $H$,
%
\begin{equation}\label{eq:3.6}
\begin{gathered}
y'(t)+Ay(t)=f(t), \quad \ 0<t<T,\\
y(0)=y_0,
\end{gathered}
\end{equation}
%
where $y(t):=y(\cdot ,t)$ and $f(t):= f(\cdot ,t)$. The operator
$A$ is maximal monotone. Moreover, $A$ is the subdifferential of
an appropriate proper, convex, lower semi-continuous function (see
\cite[Chapter 3, \S 3]{M}). By the existence and uniqueness result
[see, e.g., \cite[Theorem 2.1, p. 48]{M}), it follows that there
exists a unique strong solution of problem $(3.6)$, $ y \in
W^{1,\infty}(0,T;H)\cap L^\infty(0,T;H^2(0,1))\,. $
\bigskip

Now, using $(3.6)_1$, i.e. Eq. (\ref{E}),  we deduce that
\begin{align*}
y_t(x,t+\delta)-y_t(x,t)-[y_{xx}(x,t+\delta)-y_{xx}(x,t)]&\\
+g[y(x,t+\delta )-y(x,t)]&=f(x,t+\delta)-f(x,t),
\end{align*}
where $\delta \in (0,T)$.  If we multiply the last equality by
$y(\cdot ,t+\delta)-y(\cdot ,t)$ scalarly and then integrate on
$[0,T-\delta ]$, we arrive at the inequality
%
\begin{equation}\label{eq:3.7}
\begin{aligned}
  \int_0^{T-\delta}& \Vert y_x(\cdot,t+\delta)-y_x(\cdot,t) \Vert_H^2 dt
   \\
\leq&
 \frac 12 \Vert y(\cdot,\delta)-y_0{\Vert}_H^2
+\int_0^{T-\delta} \Vert f(\cdot,t+\delta)-f(\cdot,t) {\Vert}_H
\Vert y(\cdot,t+\delta) -y(\cdot,t) {\Vert}_H dt\,. \notag
\end{aligned}
\end{equation}
%
Since $y\in W^{1,\infty}(0,T;\, H)$, there exists a positive
constant $C_1$ such that for all $t\in [0,T-\delta ]$:
%
\begin{equation}\label{eq:3.8}
\Vert y(\cdot ,t+\delta ) - y(\cdot ,t){\Vert}_H\le C_1\delta \,.
\end{equation}
%
Now, combining (3.7), (3.8), (3.1), and Theorem 2.1 for $p=1$, we
can easily see that
%
\begin{equation}\label{eq:3.9}
\int_0^{T-\delta} {\Vert y_x(\cdot,t+\delta)-y_x(\cdot,t)
{\Vert}}_H^2 dt\leq C_2\delta^2\,,
\end{equation}
%
with some positive constant $C_2$. Taking again into account
Theorem 2.1, it follows by (3.9) that $y_x\in W^{1,2}(0,T;H)$,
which concludes the proof of the theorem.
%

 \begin{remark}\label{r:3.1} \rm
In fact, the results similar to those above are known \cite{Br1}.
However, it seems that, for multivalued $\beta _1$ and $\beta _2$,
no improvement of the above regularity is possible, although we
could not find an appropriate counter-example yet. But an
improvement of regularity is possible for single-valued and smooth
$\beta _1$ and $\beta _2$.
 \end{remark}

\begin{theorem}\label{th: 3.2.}
Assume that
%
\begin{equation}\label{eq:3.10}
f\in W^{2,1}(0,T;L^2(0,1)), \quad f(\cdot,0)\in H^2(0,1)\,;
\end{equation}
%
$\beta _1$, $\beta _2$ are both defined on $\mathbb{R}$,
single-valued, and satisfy
%
\begin{equation}\label{eq:3.11}
\beta _1, \ \beta _2\in W_{loc}^{2,\infty}\mathbb{R}\,,  \quad {\beta}'_1 \ge 0, \ {\beta}'_2 \ge 0\,.
\end{equation}
%
Moreover,
%
\begin{equation}\label{eq:3.12}
y_0\in H^4(0,1)\,.
\end{equation}
%
In addition,  we require (\ref{eq:3.4}) (where the inclusions must be
replaced by equalities) as well as the following second order
compatibility conditions:
%
\begin{equation}\label{eq:3.13}
z'_0(0)=\beta'_1(y_0(0))z_0(0), \quad -z'_0(1)=\beta'_2(y_0(1))z_0(1)\,,
 \end{equation}
%
where $z_0$ is defined as
%
\begin{equation}\label{eq:3.14}
z_0(x) = f(x,0) + y''_0(x) - gy_0(x), \quad 0\le x \le
1\,.
\end{equation}
%
Then the solution $y$ of the BVP belongs to
$$W^{2,2}(0,T;H^1(0,1))\cap W^{2,\infty}(0,T;L^2(0,1))$$
\end{theorem}
%

\paragraph{Proof:}  Obviously, all the
conditions of Theorem 3.1 are fulfilled and so the BVP has a
unique solution $y$ satisfying $(3.5)$. Let us denote $V=H^1(0,1)$
and $V'= (H^1(0,1))'$ (the dual space). We will first show that
$y_t\in W^{1,2}(0,T; V')$. To this end, it suffices to prove (cf.
Theorem 2.1) that there exists a positive constant $C$ such that
for every $\delta \in (0, T]$:
%
\begin{equation}\label{eq:3.15}
\int _0^{T-\delta}\big\Vert y_t(\cdot ,t+\delta )- y_t(\cdot
,t)\big\Vert ^2_{V'}dt\le C\delta ^2\,.
\end{equation}
%
Indeed, we have for a.a. $t\in (0,T-\delta )$ and all $\phi \in V$ the equation
%
\begin{eqnarray}\label{eq:3.16}
\lefteqn{\hspace*{-.5cm}
  \big\langle y_t(\cdot , t+\delta )-y_t(\cdot ,t), \phi\big\rangle} \notag\\
 & =&
\big\langle y_{xx}(\cdot , t+\delta ) - y_{xx}(\cdot ,t),\phi
\big\rangle\notag\\
&&\mbox{}
-g\langle y(\cdot ,t+\delta )-y(\cdot ,t), \phi \rangle + \langle f(\cdot ,t+\delta ) -f(\cdot ,t), \phi \rangle \notag\\
& =&
 -\big\langle y_x(\cdot , t+\delta ) - y_x(\cdot ,t), \phi' \big\rangle -
[\beta _2 (y(1,t+\delta )) -\beta _2(y(1,t))]\phi (1)\notag\\
&&\mbox{}
- [\beta _1 (y(0,t+\delta )) -\beta _1(y(0,t))]\phi (0)\notag\\
&&\mbox{}
 - g\langle y(\cdot ,t+\delta )-y(\cdot ,t), \phi\rangle +
\langle f(\cdot ,t+\delta ) - f(\cdot , t), \phi \rangle \,,
\end{eqnarray}
%
where $\langle \cdot , \cdot \rangle $ denotes the $L^2$-duality
pairing between $V$ and $V'$. Taking into account $(3.5)$ and
$(3.11)$, which in particular implies that $\beta _1$, $\beta _2$
are Lipschitzian on bounded sets, one obtains from $(3.16)$ the
estimate
%
\begin{equation}\label{eq:3.17}
\Vert y_t(\cdot , t+\delta )-y_t(\cdot ,t)\Vert ^2_{V'}\le C_3\{
\Vert y(\cdot ,t+\delta ) - y(\cdot ,t)\Vert ^2_V + \Vert f(\cdot
, t+\delta )-f(\cdot ,t)\Vert ^2_H\}\,,
\end{equation}
%
where $C_3$ is some positive constant. Now, $(3.5)$, $(3.10)$,
$(3.17)$ and Theorem 2.1 lead us to the desired estimate $(3.15)$.
Therefore $z:=y_t\in W^{1,2}(0,T;V')$ and, thus, we can
differentiate with respect to $t$ the equation
%
\begin{align*}
\langle& y_t(\cdot ,t), \phi \rangle + \langle y_x(\cdot ,t),
{\phi}' \rangle + \beta _1(y(0,t))\phi (0)+ \beta
_2(y(1,t))\phi (1)+
g\langle y(\cdot ,t), \phi \rangle \\
& =
 \langle f(\cdot ,t), \phi \rangle
\end{align*}
%
to obtain
 %
\begin{gather}\label{eq:3.18}
  \langle z_t(\cdot ,t), \phi \rangle + \langle z_x(\cdot ,t), \phi'\rangle +
g_1(t)z(0,t)\phi (0) + g_2(t)z(1,t)\phi (1)
+g\langle z(\cdot ,t), \phi \rangle\notag\\
 = \langle f_t(\cdot ,t),\phi )
 \mbox{for all} \quad \phi \in V, \quad \mbox{and a.a.} \quad t\in (0,T)\,,
\end{gather}
%
where
$$
g_1(t):= \beta _1' (y(0,t)), \quad \ g_2(t):= \beta
_2' (y(1,t)).
$$
In addition,
%
\begin{equation}\label{eq:3.19}
z(\cdot ,0) = z_0\,,
\end{equation}
%
with $z_0$ defined by $(3.14)$. Consequently, $z$ is the unique
solution of the problem $(3.18)$, $(3.19)$ in the class of $y_t$.
Indeed, if we take in $(3.18)$, $(3.19)$  $f_t \equiv 0$, $z_0
\equiv 0$, and $\phi :=z(\cdot ,t)$, then, according to Theorem
2.2, we have
$$
{\frac{d}{dt}}\Vert z(\cdot ,t)\Vert _H^2 \le 0 \quad \mbox{for
a.a.} \quad t\in (0,T),
$$
which clearly implies that $z\equiv 0$. Note that $z=y_t$
satisfies the linear problem
%
\begin{equation}\label{eq:3.20}
\begin{gathered}
z_t-z_{xx}+gz=f_t, \quad \mbox{in} \ D_T,\\[2mm]
z(x,0)=z_0(x), \quad 0<x<1,\\
z_x(0,t)=g_1(t)z(0,t), \quad -z_x(1,t)=g_2(t)z(1,t), \quad 0<t<T
\end{gathered}
\end{equation}
%
formally. By Theorem 3.1 and the assumption $(3.11)$ it follows
that $g_1, \ g_2 \in  H^1(0,T)$, and that $g_1\ge 0, \quad g_2\ge 0$
in $[0,1]$. Actually, problem $(3.20)$ has a unique strong
solution. To show this, let us use the fact that this problem can
be expressed as the following time-dependent Cauchy problem in
$H=L^2(0,1)$:
%
\begin{equation}\label{eq:3.21}
\begin{gathered}
z'(t)+B(t)z(t)=f_t(\cdot, t) \quad \mbox{for} \quad 0<t<T, \\
z(0)=z_0,
\end{gathered}
\end{equation}
%
 where $z(t):=z(\cdot,t)$ and  $B(t)\,:\,D(B(t))\subset H\to  H$ is defined by
$$
B(t)p = -p^{\prime\prime} + gp
$$
on the domain of definition
$$
D(B(t))=\left\{p\in H^2(0,1);~
p'(0)=g_1(t)p(0),~-p'(1)=g_2(t)p(1)\right\}.
$$
$B(t)$ is maximal monotone for every $t$$\in$$[0,T]$ and, even
more, $B(t)$ is the sub-\linebreak differential of the function
$\varphi(t,\cdot):H\to (-\infty,+\infty]$, given by
$$
\varphi(t,p)= \begin{cases}
(1\slash{2})\big\{\int_0^1{p'(x)}^2dx+g\int_0^1p(x)^2dx\\[1mm]
 +g_1(t)p(0)^2 + g_2(t)p(1)^2\big\} \quad & \mbox{for }
 p\in H^1(0,1)\,,\\[2mm]
+\infty & \mbox{for }p\in H\backslash H^1(0,1)
\end{cases}
$$
(see, e. g., \cite[Chapter 3, \S 3]{M}). For every $t \in
[0,T],$ the effective domain is
$D(\varphi(t,\cdot))$$=$$H^1(0,1),$ i.e. it is independent on $t.$
Now, we are going to show that the condition $(2.1)$ of Theorem
2.3 is satisfied. To this end, let $p$$\in$$H^1(0,1)$ and
$0$$\leq$$s$$\leq$$t$$\leq$$T.$ We have
%
\begin{eqnarray*}
\varphi(t,p)-\varphi(s,p) &=& \frac{g_1(t)-g_1(s)}{2}p(0)^2+
\frac{g_2(t)-g_2(s)}{2}p(1)^2\\
&=& \tfrac{1}{2}p(0)^2\int_s^tg_1'(\tau)d\tau+
\tfrac{1}{2}p(1)^2\int_s^tg_2'(\tau)d\tau\\
&\leq& \tfrac{K}{2} {(\|p\|}_H^2 +\|p'\|_H^2) \int_s^t\big(\mid
g_1'(\tau)\mid+\mid g_2'(\tau)\mid\big)d\tau\,,
\end{eqnarray*}
%
where $K$ is a positive constant due to the continuous embedding
of $H^1(0,1)$ into $C[0,1]$. Therefore,
$$
\varphi(t,p)\leq\varphi(s,p)+[\gamma(t)-\gamma(s)]\cdot
[\varphi(s,p)+{\frac{1}{2}}{\|p\|}_H^2],
$$
where
$$\gamma(t)=K\int_0^t\big(\mid g_1'(\tau)\mid+\mid g_2'(\tau)\mid\big)d\tau .
$$
By Theorem 2.3 one obtains that problem $(3.21)$ has a unique solution
$$
z\in W^{1,2}(0,T;H)\cap L^\infty(0,T;H^1(0,1)).
$$
This solution $z$ satisfies problem $(3.18)$, $(3.19)$ and, hence,
it coincides with $y_t$. So, we have already proved that
$$
y\in W^{2,2}(0,T;H)\cap W^{1,\infty}(0,T;H^1(0,1)).
$$
We are now going to show that
$$
z_t = y_{tt}\in L^{\infty}(0,T;H)\cap L^2(0,T;H^1(0,1)).
$$
To this end, we proceed in a standard manner by starting from the
estimates
 %
\begin{eqnarray*}
\lefteqn{
 {\frac{1}{2}}{\frac{d}{dt}}\Vert z(\cdot ,t+h) - z(\cdot ,t)\Vert _H^2
 + \Vert z_x(\cdot , t+h)-
z_x(\cdot ,t)\Vert _H^2 }\\
\lefteqn{+[g_1(t+h)z(0,t+h)-g_1(t)z(0,t)]\cdot [z(0,t+h)-z(0,t)]  }\\
\lefteqn{+[g_2(t+h)z(1,t+h)-g_2(t)z(1,t)]\cdot [z(1,t+h)-z(1,t)] }\\
&\le&
 \Vert f_t(\cdot ,t+h) -f_t(\cdot ,t)\Vert _H\cdot \Vert z(\cdot ,t+h)
 - z(\cdot ,t)\Vert _H \hspace{10mm}
\end{eqnarray*}
for a.a. $0\le t<t+h\le T$, and
\begin{eqnarray*}
 \lefteqn{ {\frac{1}{2}}{\frac{d}{dh}}\Vert z(\cdot ,h) -z_0\Vert _H^2
 + \Vert z_x(\cdot ,h) - z_0' \Vert _H^2 +[g_1(h)z(0,h)} \\
\lefteqn{-g_1(0)z_0(0)]\cdot [z(0,h)-z_0(0)]
 +[g_2(h)z(1,h -g_2(0)z_0(1)]\cdot [z(1,h)-z_0(1)] }\\
&\le&
 \Vert f_h(\cdot ,h)-B(0)z_0\Vert _H\cdot \Vert z(\cdot ,h)
 - z_0\Vert _H \quad \mbox{for a.a.} \quad h\in (0,T)\,.\hspace{17mm}
\end{eqnarray*}
%

Since $y\in W^{1,\infty}(0,T;H^1(0,1))$ and $\beta _1, \ \beta _2
\in W_{loc}^{2,\infty}(\mathbb{R})$, both the functions $g_1, \
g_2$ are Lipschitz continuous. So, taking also into account that
$g_1\ge 0$ and $g_2\ge 0$ in $[0,1]$, we get from the above
inequalities
%
\begin{eqnarray}\label{eq:3.22}
\lefteqn{
  {\frac{1}{2}}{\frac{d}{dt}}\Vert z(\cdot ,t+h) - z(\cdot ,t)\Vert _H^2
  + \Vert z_x(\cdot ,t+h) - z_x(\cdot ,t)\Vert _H^2  }\notag\\
& \le&
 \Vert f_t(\cdot ,t+h) -f_t(\cdot ,t)\Vert _H\cdot \Vert z(\cdot ,t+h)
 - z(\cdot ,t)\Vert _H \\
&&+L_1h\vert z(0,t+h)-z(0,t)\vert +
L_2h\vert z(1,t+h)-z(1,t)\vert \,,\notag
\end{eqnarray}
and
\begin{eqnarray} \label{eq:3.23}
 \lefteqn{ \frac{1}{2} \frac{d}{dh}
 \Vert z(\cdot ,h) - z_0\Vert _H^2 + \Vert z_x(\cdot ,h)-z_0'\Vert _H^2 }
 \notag \\
 &\le& \Vert f_h(\cdot ,h)-B(0)z_0\Vert _H\cdot \Vert z(\cdot ,h) -
 z_0\Vert _H  \\
&&+L_1h\vert z(0,h)-z_0(0)\vert + L_2h \vert z(1,h)-z_0(1)\vert \,.\notag
\end{eqnarray}
%
Now, taking into account the continuous embedding of $H^1(0,1)$
into $C[0,1]$ and the inequality $ab\le \varepsilon a^2 +
b^2/(4\varepsilon )$ for all $a, \, b\ge 0, \ \varepsilon >0$, we
obtain from $(3.22)$ and $(3.23)$ the two estimates
%
\begin{eqnarray}\label{eq:3.24}
\lefteqn{
\frac{1}{2}{\frac{d}{dt}}\Vert z(\cdot , t+h)-z(\cdot ,t)\Vert _H^2 + \Vert z_x(\cdot ,t+h) - z_x(\cdot ,t)\Vert _H^2}\notag\\
&\le&
 \Vert f_t(\cdot ,t+h) -f_t(\cdot ,t)\Vert _H\cdot \Vert z(\cdot ,t+h)-z(\cdot ,t)\Vert _H  \\
 &&+
{\frac{1}{2}}(\Vert z_x(\cdot , t+h)-z_x(\cdot ,t)\Vert _H^2 +
\Vert
z(\cdot ,t+h) - z(\cdot ,t)\Vert _H^2) + {\frac{1}{2}}C_4h^2\,,\notag
\end{eqnarray}
and
\begin{eqnarray} \label{eq:3.25}
  \lefteqn{
  {\frac{1}{2}}{\frac{d}{dh}}\Vert z(\cdot ,h) - z_0\Vert _H^2 +
  \Vert z_x(\cdot ,h)-z_0'\Vert _H^2}\notag\\
& \le&  \Vert f_h(\cdot ,h)
 -B(0)z_0\Vert _H\cdot \Vert z(\cdot ,h)-z_0\Vert _H\\
&&+{\frac{1}{2}}(\Vert z_x(\cdot ,h) - z_0'\Vert _H^2 + \Vert z(\cdot ,h) - z_0\Vert _H^2) +
{\tfrac{1}{2}}C_5h^2\,, \notag
\end{eqnarray}
%
where $C_4$ and $C_5$ are some positive constants. Therefore,
%
\begin{eqnarray}\label{eq:3.26}
  \lefteqn{ {\tfrac{d}{dt}}(e^{-t}\Vert z(\cdot ,t+h)-z(\cdot ,t)\Vert _H^2) +e^{-t}\Vert z_x(\cdot ,t+h)-z_x(\cdot ,t)\Vert _H^2 }\notag\\
&\le&
 2e^{-t}\Vert f_t(\cdot ,t+h)-f_t(\cdot ,t)\Vert _H\cdot \Vert z(\cdot
 ,t+h)-z(\cdot ,t)\Vert _H + C_4h^2\,,
 \end{eqnarray}
%
 and
%
\begin{equation}\label{3.27}
 {\tfrac{d}{dh}}(e^{-h}\Vert z(\cdot ,h)-z_0\Vert _H^2)\le
 C_5h^2+2e^{-h}\Vert f_h(\cdot ,h)-B(0)z_0\Vert _H \cdot \Vert z(\cdot
 ,h)-z_0\Vert _H\,.
\end{equation}
%

If we drop the second term in the left--hand side of $(3.26)$ and
integrate the resulting inequality over $[0,t]$, then, by using a
Gronwall type lemma (see, e.g., \cite[p. 47]{M}) we arrive at the
estimate
%
\begin{eqnarray}\label{eq:3.28}
  \Vert z(\cdot ,t+h) - z(\cdot ,t)\Vert _H
&\le&
 C_6\{ \Vert z(\cdot ,h)-z_0\Vert _H + h \notag\\
&&\mbox{}
   +\int_0^t\Vert f_s(\cdot ,s+h)-f_s(\cdot ,s)\Vert _Hds\}\,.
\end{eqnarray}
%

Now, from $(3.25)$ we obtain in a similar way
%
\begin{equation}\label{eq:3.29}
\Vert z(\cdot ,h) -z_0\Vert _H \le C_7(h^{3/2} + \int_0^h\Vert
f_s(\cdot ,s) -B(0)z_0\Vert _Hds)\,.
\end{equation}
%

Finally, $(3.28)$ and $(3.29)$ imply that $z_t\in
L^{\infty}(0,T;H)$. Using this conclusion together with $(3.24)$,
we
 get, by using of Theorem 2.1, that $z_x\in W^{1,2}(0,T; H)$.
This concludes the proof.

\begin{remark}\label{r:3.2} \rm
If $f$ in  Theorem 3.2 is assumed to be more regular with respect
to $x$, then $y$ is also more regular with respect to $x$, because
$$
y_{tt}=y_{xxt}-gy_t+f_t = y_{xxxx}-gy_{xx}+f_{xx}-gy_t+f_t.
$$
On the other hand, by reasoning similarly as in the proof of
Theorem 3.2, one can show that $y\in W^{3,2}(0,T;H^1(0,1))\cap
W^{3,\infty}(0,T;L^2(0,1))$ under appropriate assumptions on the
smoothness of $\beta _1, \ \beta _2, \ y_0, \ f$ and
compatibility. The proof needs a slight modification since the
boundary conditions corresponding to $(3.20)$ are now
inhomogeneous. Fortunately, the inhomogeneous terms there are
$H^1$-functions and so Theorem 2.3 is again applicable, with a
slight change of $\phi $. The corresponding $t$-dependent operator
is nonlinear because its domain is an affine subset of $L^2(0,1)$.
Here we will not further present these details. Of course, higher
regularity with respect to $x$ can also be obtained at the expense
of additional regularity and compatibility assumptions. Actually,
our procedure can be repeated as many times as we want to and so
any regularity of the solution with respect to $t$ and $x$ can be
reached under sufficient smoothness of the data and compatibility
conditions. More precisely, the $H^k(D_T)$-regularity can be shown
for every $k$ and, thus, the $C^k(\overline{D_T})$-regularity can
be obtained as well for every $k$, due to Sobolev's embedding
theorem.
\end{remark}
%

\begin{remark}\label{3.3} \rm
Theorem 3.2 is formulated in such a way that the next level of
regularity can be obtained. Note, however, that this is a
parabolic problem so that, in order to get, for instance,
$$
y\in W^{2,2}(0,T;L^2(0,1))\cap W^{1,\infty}(0,T; H^1(0,1)),
$$
we can relax our assumptions. More precisely, it suffices to
assume that
$$
f\in W^{1,2}(0,T;L^2(0,1)), \quad f(\cdot ,0) \in H^1(0,1);
$$
$\beta _1$ and $\beta _2$ satisfy $(3.11)$; $y_0\in H^3(0,1)$, and
$y_0$ satisfies $(3.4)$ with equalities instead of inclusions.
\end{remark}

\begin{remark}\label{r:3.4} \rm
Here,  we do not discuss nonlinear cases of the equation (\ref{E}),
since this would require a special treatment. However, let us
point out some immediate extensions of the above results.
\end{remark}
%
Assume that the linear term $gy$ of Equation (\ref{E}) is replaced by
a nonlinear one, say $g(y)$, where $g:\mathbb{R}\longrightarrow \mathbb{R}$
is a single-valued, continuous and nondecreasing function. Then
Theorem 3.1 is still valid. If, in addition, $g$ is a
$W_{loc}^{2,\infty }(\mathbb{R})$-function, then Theorem 3.2 is
also valid for this nonlinear case with
$$
z_0(x)=f(x,0)+y_0''(x)-g(y_0(x)).
$$
Indeed, $(3.17)$ remains true, because $g$ is Lipschitzian on
bounded sets and $y$ takes values in a bounded set. So, again,
$z=y_t\in W^{1,2}(0,T;V')$. The remainder of the proof of Theorem
3.2 continues with slight modifications. In particular, in Eq.
$(3.20)_1$ the term $gz$ must be replaced by $g'(y)z$ and $B(t)$
becomes
$$
B(t)p=-p'' + g'(y(\cdot ,t))p,
$$
with the same domain of definition as before. This means that in
the definition of the energy function $\phi (t,p)$, the term
$(1/2)g\int _0^1p(x)^2dx$ should be replaced by
$(1/2)\int _0^1g'(y(x,t))p(x)^2dx$. Finally, here $\gamma $ can be
chosen in the specific form
$$
\gamma (t) = K\int _0^t(\vert g_1'(\tau )\vert + \vert
g_2'(\tau )\vert + \Vert g'' (y(\cdot
,\tau))y_{\tau}(\cdot ,\tau)\Vert _{L^{\infty}(0,1)})d\tau .
$$
Then all the conclusions can be derived by similar arguments as
above.

 On the other hand, if the (\ref{BC}) are inhomogeneous and
(\ref{E}) is nonlinear, then a transformation like (\ref{eq:1.1}) would lead
us to an evolution equation where the spatial operator becomes
time-dependent. Instead, we still can keep the inhomogeneous form
of (\ref{BC}) and associate with our problem an energy functional
$\phi (t, \cdot )$ which also includes the inhomogeneous
terms.

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\bigskip

\noindent\textsc{Lumini\c{t}a Barbu}\\
Department of Mathematics and Informatics\\
Ovidius University\\
Blvd. Mamaia 124\\
8700 Constan\c ta, Romania\\
e-mail: lbarbu@univ-ovidius.ro
\medskip

\noindent\textsc{Gheorghe Moro\c sanu}\\ 
Department of Mathematics\\
``Al.~I.~Cuza'' University\\
Blvd. Carol I,  11\\
6600 Ia\c{s}i,
Romania\\
e-mail: gmoro@uaic.ro
\medskip

\noindent\textsc{ Wolfgang L. Wendland} \\
Mathematisches Institut A\\
University of Stuttgart\\
Pfaffenwaldring 57\\
70569  Stuttgart, Germany\\
e-mail: wendland@mathematik.uni-stuttgart.de

 \end{document}