\documentclass[reqno]{amsart}
\usepackage{hyperref}

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2010(2010), No. 83, pp. 1--7.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
\newline ftp ejde.math.txstate.edu}
\thanks{\copyright 2010 Texas State University - San Marcos.}
\vspace{9mm}}

\begin{document}
\title[\hfilneg EJDE-2010/83\hfil Existence of weak solutions]
{Existence of weak solutions for mixed problems
of parabolic systems}

\author[P. T. Duong, D. V. Loi\hfil EJDE-2010/83\hfilneg]
{Pham Trieu Duong, Do Van Loi}  % in alphabetical order

\address{Pham Trieu Duong \newline
Department of Mathematics, Hanoi National University of Education,
Hanoi, Vietnam}
\email{duongptmath@hnue.edu.vn}

\address{Do Van Loi \newline
Department of Mathematics, Hong Duc University, Thanhhoa, Vietnam}
\email{37loilinh@gmail.com}

\thanks{Submitted May 20, 2010. Published June 18, 2010.}
\thanks{Supported by National Foundation for Science and
Technology Development (Nafosted)}
\subjclass[2000]{35K35}
\keywords{Parabolic system; generalized solution; asymptotic
behavior; \hfill\break\indent Garding's inequality}

\begin{abstract}
 The purpose of this paper is to investigate the existence of
 generalized solutions for strongly parabolic systems in a
 cylindrical domain. The decay  of the solution at infinity,
 depending on the right-hand of the equation, is also studied in
 this article.
\end{abstract}

\maketitle
\numberwithin{equation}{section}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{proposition}[theorem]{Proposition}


\section{Introduction}

The study of boundary value problems for non-stationary systems of
PDE's in arbitrary domains differs from the study of stationary
systems. In  previous works of the first author
\cite{HD1,HD2}, the existence of solutions for parabolic
system has been obtained by a Galerkin's approximation scheme.
Following the method in \cite{Lad}, which have been successfully
applied for studying  the Stokes equation, we prove in this
article the  existence and uniqueness of  solutions for general
parabolic systems. This approach allows us to study the decay of
solutions for large time for bounded and for unbounded domains.
However, from these proofs it is not clear how to find the solutions.

This paper is organized as follows. In the second section we
introduce the necessary notation and functional spaces for our
problem. The third section is devoted to formulate and prove the
main theorem. The last section is intended to
conduct the research on the asymptotic behavior of solution at
infinity.

 \section{Notation}
Let $\Omega$ be a domain in ${\mathbb R}^n$ and $T$ be arbitrary,
$0< T\leq \infty$. We denote by $Q_T$  the cylinder $\Omega\times
(0,T)$. We consider the  differential operator
$$
L(x,t,D)=\sum_{|p|,|q|=0}^m D^p(a_{pq}D^q),
$$
where $a_{pq}$ are bounded functions in $C^\infty (Q_T)$ together with $\frac{\partial a_{pq}}{\partial t}$. For $|p|=|q|=m$ we assume
the condition $a_{pq}=a_{qp}^{\star}$, where the asterisk denotes
the transposed complex conjugate of a matrix.

The operator $L$ is strongly elliptic. Then there exists a
constant $C>0$ such that for all real vectors $\xi \in
\mathbb{R}^n$ and all complex vectors $\eta\in \mathbb{C}^s$
\begin{equation}\label{eE}
 \sum_{|p|,|q|=m}a_{pq}\xi^p\xi^q\eta\overline{\eta} \ge
C|\xi|^{2m}|\eta|^2 \quad \forall (x,t)\in Q_T \,.
\end{equation}
The spaces $H^{m}(\Omega), H^{m,k}(Q_T)$ are the usual Sobolev spaces,
where $m, k$ denote the order of derivatives with respect to
 $x$ and $t$ respectively. In  addition $ H^{m,k}_0(Q_T)$
is the completion with respect to $H^{m,k}(Q_T)$ norm of functions
from $C^\infty (Q_T) $ which vanish near the lateral surface
$S_T=\partial \Omega\times (0,T)$.

In this study, we  consider the initial-boundary value
problem (IBVP) in the  cylinder $Q_T$ for  the system of PDE's:
 \begin{gather}
 Pu\equiv u_t +(-1)^m L(t,x, D_x) u = f(x,t)\quad \text{in }Q_T,
\label{e1}\\
 \frac{\partial ^j u}{\partial \nu^j}=0,\quad 0\leq j\leq m-1 \quad
\text{on } S_T, \label{e2}\\
u(x,0)=\varphi (x) \label{e3}
\end{gather}
where $\nu$ is the unit outer normal vector  to the
lateral surface $S_T$. Here $f(x,t)$ and $\varphi (x)$ are given
functions.

The coercivity   of the bilinear form associated with the operator
$\mathcal{L}$ is a direct consequence of  Garding's inequality
which is formulated as follows.

\begin{proposition} \label{prop2.1}
Assume that the coefficients of $L$ satisfy the condition
\eqref{eE} on $Q_T$. Moreover, we suppose that $a_{pq}$ are
continuous functions on the $x$ variable, uniformly with respect to
 $t\in (0,T)$.  Then there exist  constants $\mu>0, \lambda\geq 0$
such that for all $u(x,t)\in  H^{m,k}_0(Q_T)$,
$$
\int_{\Omega}\sum_{|p|,|q|=0}^m (-1)^{|p|+m}a_{pq}D^q u
\overline{D^p u}dx  \geq \mu \Vert u\Vert^2_{H^{m}(\Omega)}
-\lambda \Vert u\Vert^2_{L_2(\Omega)}\,.
$$
\end{proposition}

We note that the constant $\lambda$ can be  equal  $0$, since
by a substitution $v=e^{-\lambda t}u$  the initial problem
can be transformed to a problem with constant $\lambda=0$.

\section{The existence result}

We obtained previously in  \cite{HD1,HD2}  some results
on the solvability of above problem.
Although, in these works we studied this problem only
in non-smooth cylinders, the assertion there
is still valid  for arbitrary domains; since  there we
 considered only  the generalized solutions possessing
weak derivatives of orders less or equal $m$.
In this  section we  present another proof of this
result by introducing an operator associating to
the initial- boundary problem and its properties.

\begin{theorem} \label{thm3.1}
The problem \eqref{e1}--\eqref{e3} is uniquely solvable
for arbitrary functions $f(x,t)\in L_2(Q_T)$ and  $\varphi (x,t)\in V$.
The solution $u(x,t)$ has the derivative $u_t$ and  the value
$\mathcal{L}u$ (in distributional sense) from the space
$L_2(Q_T)$ and it satisfies  system \eqref{e1}--\eqref{e3}
in the classical sense in each subset $\Omega'\times (0,T)$,
where $\Omega'$ is a compact subset of $\Omega$.
\end{theorem}

\begin{proof}
We will use the Lax-Milgram's theorem. First, we will use the
coercivity and  the self-adjoint property of $-\mathcal{L}$  to
show the solvability of  the elliptic problem.

\subsection*{The Lax-Milgram procedure}
Let
\[
a(t,u,v)=\int_{\Omega}\sum_{|p|,|q|=0}^m
(-1)^{|p|+m}a_{pq}D^q u \overline{D^p v}dx
\]
which is a bilinear form
defined on the space $V={\mathaccent"7017 H}^m(\Omega)$.
Since $a$ is coercive, by virtue of Lax-Milgram's theorem, for
each function $f(x)\in L_2(\Omega)$ there exists a solution
$u\in V $ of the variational problem
$$
a(t,u,v)=(f,v), \forall v\in V
$$
It is easy to show that $-\mathcal{L}$ is a closed, symmetric
operator with the range dense in space $H=L_2(\Omega)$; therefore,
$-\mathcal{L}$  is a self-adjoint operator.

\textit{Operator $A$:} Let us introduce the space
\[
H^t=\{u(x,t)|f \in L_2(\Omega)\text{ for almost }t\in (0,T);
        u=0 \text{ near } \partial\Omega\}.
\]
We  define an operator $A$ which  associates the function $u(x,t)$
with the right-hand $f$ and the initial value $\varphi$ by the
formula:
$$
Au=\big(Pu,u(x,0)\big).
$$
This operator is defined in the space $D(A)$ consisting of
functions $u(x,t)$ which are represented by the formula
$$
u = \varphi_0(x) +\int_0^T \varphi_1 (x,t) \,dx\,dt,
$$
where $\varphi_0(x)\in V$ and $\varphi_1(x,t)\in V$
for a.e. $t\in (0,T)$.

We can easily verify  that $D(A)$ is a dense subset in
${\mathaccent"7017 H}^m(Q_T)$. The image of $A$ is considered to be
the elements in a certain Hilbert space $W$, which consists of the
pairs $(f(x,t); \varphi(x)$ with $f(x,t)\in H^t$ and $\varphi
(x)\in V$. In $W$ the scalar product is defined as
$$
\{(f_1;\varphi_1),(f_2;\varphi_2)\}
= \int_0^T (f_1,f_2)+ [\varphi_1,\varphi_2] dt
$$
where $(,)$ and $[,]$ are the inner products in the spaces $H$ and
$V$ respectively.

Operator $A$ acts from $V^t$ to $W$. Now we will show that
$A$ can be extended to its closure $\overline{A}$, with
range $R(\overline{A})$ exhausting entire the space $W$. It is
equivalent to conclude  that the problem \eqref{e1}--\eqref{e3}
 has a solution $u$ which belongs to $D(\overline{A})$ for
arbitrary functions $f\in H^t, \varphi \in V$.

\subsection*{The existence of $\overline{A}$}
Let $\{u_n(x,t)\} \subset D(A)$ is such sequence of functions
that $u_n$ tend to $0$ in $H^t$ and $A(u_n)$ tend to $(f,\varphi)$
in $W$ for $n\to\infty$. We have to verify that
$ (f,\varphi)=(0,0)$.

To do that, let us  multiply the equation $P u_n=f_n$ by
an arbitrary smooth function $\Phi (x,t)$ from $D(A)$ which
equal $0$ for $t=T$, and then  integrate the result over $Q_T$.
After that, we will pass all the derivatives from $u$ to
$\Phi$, by integration by parts.
In the end we arrive at
\begin{align*}
\int_{Q_T} f_n \Phi \,dx\,dt
&= \int_{Q_T} (u_{nt}-\mathcal{L}u)\Phi \,dx\,dt\\
&=\int_{Q_T} u_n (-\Phi_t-\mathcal{L}\Phi )\,dx\,dt
-\int_{\Omega}\varphi_n(x) \Phi(x,0) dx
\end{align*}
Passing $n\to\infty$, by our assumption, we obtain
$$
\int_{Q_T} f \Phi \,dx\,dt = -\int_{\Omega}\varphi \Phi(x,0) dx.
$$
However, noting that the smooth functions $\Phi(x,t)$
from $D(A)$ which vanish for $t=T$ and $t=0$
form a dense subset in $H^t$, we can conclude that $f(x,t)\equiv 0$.
Since the values $\Phi(x,0)$ are dense in $V$, we have
$\varphi\equiv 0$. From these facts we can guarantee the
existence of  the closure $\overline{A}$.

\subsection*{The domain of $\overline{A}$}
We consider for $u\in D(A)$ the expression $\int_0^t (P u,P u) dt$.
We can transform this  by the integration by parts
\begin{equation} \label{e4}
\begin{aligned}
&\int_0^t\int_{\Omega} (Pu,Pu) \,dx\,dt\\
&= \int_{Q_t}[u_t.\overline{ u_t}+ \mathcal{L}u.
\overline{\mathcal{L}u}+2 Re u_t \mathcal{L}u)]\,dx\,dt\\
&=\int_{Q_t}[u_t^2+(\mathcal{L}u)^2]\,dx\,dt
- 2\int_{Q_t}\sum_{|p|,|q|=0}^m (-1)^{|p|+m-1}a_{pq}D^q u
\overline{D^p u_t}\,dx\,dt\,.
\end{aligned}
\end{equation}
The  second term in the right-hand side can be rewritten as
\begin{equation} \label{e5}
\begin{aligned}
&-2\mathop{\rm Re}\int_{Q_{t_0}}\sum_{|p|,|q|=0}^m (-1)^{|p|+m-1}
a_{pq}D^p u \overline{D^q u_t}\,dx\,dt\\
& =-\int_{Q_{t_0}}\big[\frac{\partial}{\partial t}
(\sum_{|p|,|q|=0}^m (-1)^{|p|+m-1}a_{pq}D^q u \overline{D^p u})\\
&\quad -\sum_{|p|,|q|=0}^m (-1)^{|p|+m-1}
\frac{\partial a_{pq}}{\partial t} D^q u \overline{D^p u}\big].
\end{aligned}
\end{equation}
      Therefore, from Garding's inequality and  the boundedness
of $\frac{\partial a_{pq}}{\partial t}$ we can show that
     \begin{equation}\label{eq5}c
\int_{0}^{t_0}(|u_t|^2+ |\mathcal{L}u|^2 )dt+\mu \Vert u (.,t_0)\Vert_V^2 \leq \int_{0}^{t_0} |Pu|^2 dt + C \Vert (u.,0) \Vert_V^2+ C\int_0^{t_0} \Vert u\Vert_V^2 dt.
     \end{equation}
with the constant $C$ depends on the bound of
$a_{pq}$ and $\frac{\partial a_{pq}}{\partial t}$.

  From Gronwall-Bellman's inequality, as  was carried out
in \cite{HD1}, from \eqref{eq5}, it is obvious that by  assuming
 the convergence of $u_n$ to $u$ in $H^t$ and $Au_n$ to $w$ in $W$,
we attain immediately:
\begin{gather*}
\frac{\partial u_n}{\partial t}\to \frac{\partial u}{\partial t}
 \quad \text{in } H \\
\mathcal{L}u_n \to  \mathcal{L}u \quad \text{in }H\\
D^p u_n \to D^p u  \quad \text{uniformly w.r.t to $t$ in }
  L_2(\Omega), \forall p: 0\leq |p|\leq m
\end{gather*}
Therefore, the function $u$ in domain $D(\overline{A})$
possesses  derivatives
$\frac{\partial u}{\partial t}, D^p u, 0\leq |p|\leq 2m$
in $L_2(Q_T)$. Meanwhile, all $D^p u, 0<\leq p\leq m$ depend on
$t$ continuously which are functions  in the space $H$
for all $t\in (0,T)$. According to this fact, the image of $u$
under $\overline{A}$  is computed as usually
      \begin{equation}\label{eq6}
\overline{A}u= (u_t-\mathcal{L}u; u(x,0))
      \end{equation}
  From \eqref{eq5} it is also valid that if $Au_n$ converges in
$W$ then the sequence $u_n$  also converges  in $H^t$
(even in some stronger sense). It says about that the closure
of the operator $A$ implies the closure of the range $R(A)$; i.e.,
 $R(\overline{A})=\overline{R(A)}$.

To complete the proof of the existence result,
it is sufficient to show that  the orthogonal complement
to $\overline{R(A)}$ in space $W$ consists of an element $0$ solely.
Let us assume the reverse; i.e., there exists an element
$(f;\varphi)\ne (0;0)$ which is orthogonal to all elements
from $\overline{R(A)}$ , or equivalently, from $R(A)$:
\begin{equation} \label{eq7}
     0\equiv\{(f;\varphi),(Au;u(x,0))\}
=\int_{Q_T}(f(u_t-\mathcal{L}u))\,dx\,dt
+ \int_{\Omega}\sum_{|p|=0}^m D^p \varphi\overline{D^p u(x,0)}dx\,.
\end{equation}
  From the solvability of the variational problem, we can find
 basing on $f$ the function  $(\mathcal{L})^{-1}f$.
 Put
$$
u(x,t)=-\int_{0}^{t}(\mathcal{L})^{-1}f(x,\tau)d\tau
$$
in the identity $(\ref{eq7})$ (this function obviously belongs
to $D(A)$). Therefore
$$
0=\int_{Q_T}(f\big((\mathcal{L})^{-1}f \,dx\,dt
+\int_{0}^{t}f d\tau\big).
$$
 Since $\mathcal{L}^{-1}$ is  a negatively defined and self-adjoint
operator, we can rewrite the last equality in the form:
$$
0= \int_{Q_T}|-\mathcal{L}^{\frac{1}{2}}f| \,dx\,dt
 +\frac{1}{2}\int_{\Omega}\big(\int_0^t f d\tau\big)^2,
$$
from which it is clear that ${-\mathcal{L}}^{\frac{1}{2}}f\equiv 0$
and thus $f\equiv 0$.

Consider again the equality \eqref{eq7} which now takes a simple form
$$
0\equiv [u(x,0),\varphi]_V, \forall u\in D(A).
$$
 From this formula we must have $\varphi\equiv 0$, since for
above selected  functions $u$ the values $u(x,0)$ form a dense
subset in $V$.

\subsection{Unbounded domains}
If the domain $\Omega$ is unbounded then the operator
$\mathcal{L}$ may not have the bounded inverse.
In this case the spectrum of $\mathcal{L}^{-1}$ may contain
the element $0$, but in the whole it is non-positive.
By the substitution $v=u e^{-\gamma_0 t}, \gamma_0>0$ we
will arrive to the system:
\begin{equation} \label{eq8}
     v_t+(-1)^m\mathcal{L}_1v=fe^{-\gamma_0 t}
\end{equation}
 with the operator $\mathcal{L}_1=\mathcal{L}-\gamma_0 I$
having the negative spectrum, that means the operator
$\mathcal{L}^{-1}_1$ exists. The solvability for \eqref{eq8}
 follows in the same manner as it was carried out for the bounded domain.
     Note that by this argument we also have proved the uniqueness
of the weak solution. The proof is complete.
 \end{proof}

\section{Behavior of solutions at infinity}

We begin with the inequality:
     \begin{align}\label{eq9}
\frac{1}{2}\frac{\partial}{\partial t} \Vert u(x,t)\Vert ^2
+ \mu \Vert u(x,t)\Vert_V ^2 \leq (f(x,t),u(x,t))
     \end{align}
     which follows from \eqref{e1} by multiplying both sides
of the system by $u$ and using the Garding's inequality.
     From the last relation it follows  that
$$
u\frac{\partial}{\partial t} \Vert u\Vert
\leq \Vert f \Vert \Vert u\Vert,
$$
     where  $\Vert . \Vert$ denotes the norm in $H$.
 In other words, the above statement confirms that
$\Vert u\Vert =0$ or $\frac{\partial}{\partial t} \Vert u\Vert
\leq \Vert f\Vert$.

However, we know beforehand that $u$ is a continuous function
with respect to $t$ (see the proof of Theorem \ref{thm3.1}),
thus for all $t$ and $\tau$ we  have
\begin{equation} \label{eq10}
     \Vert u(x,t)\Vert \leq \Vert u(x,\tau)\Vert
+ \int_{\tau}^t \Vert f(x,\xi)\Vert d\xi\,.
\end{equation}
Differentiating both sides of inequality \eqref{eq9} with respect to
$t$ from $\tau$ to $t$ and using \eqref{eq10} we can conclude that
\begin{equation} \label{eq11}
\begin{aligned}
      \Vert u(x,t)\Vert^2 + 2\mu \int_{\tau}^t dt
&\leq \Vert u(x,t)\Vert^2+ 2\int_{\tau}^t\Vert f\Vert dt
 \big(\Vert u(x,\tau)\Vert+\int_{\tau}^t\Vert f\Vert dt\big) \\
&\leq 2\Vert u(x,\tau)\Vert^2+3\big(\int\limits_{\tau}^t (f(.,\xi) d\xi\big)^2\,.
\end{aligned}
\end{equation}
We consider two cases:

\subsection*{Bounded $\Omega$}
If the integral $\int_0^\infty \Vert f\Vert dt$ converges,
then from \eqref{eq9} it follows that the integral
$\int_0^\infty \Vert u\Vert_V dt$ also converges according
to the Gronwall-Bellman's inequality. The direct consequence
from it is the existence of a sequence $\{t_k\}$ tending
to $\infty$, for which  the norm $\Vert u(x,t_k)\Vert_V$ tends to zero.

However, for the bounded domains:
$\Vert u(x,t)\Vert \leq C\Vert u(x,t)\Vert_V$,
hence the norm $\Vert u(x,t_k)\Vert$ also tends to $0$
when $\{t_k\} \to \infty$. Using $(\ref{eq10})$ with
$t\geq \tau= t_k$ we get that
$\Vert u(x,t)\Vert\to 0$ as $t\to\infty$.

\subsection*{Unbounded $\Omega$}
 We cannot apply the Friedrichs's inequality now,
therefore we have to use both inequalities \eqref{eq5}, \ref{eq11}
 simultaneously.
 If we assume the convergence of
 $$
\int_{0}^\infty ( \Vert f\Vert +\Vert f\Vert^2) dt,
$$
then from \eqref{eq11}   there exists a sequence
$\{t_k\}\to\infty$, for which $\Vert u(x,t_k)\Vert_V\to 0$.
By  \eqref{eq5} and  the Gronwall-Bellman's inequality,
\[
\Vert u(x,t)\Vert_V^2 \leq \Vert u(x,t_k)\Vert^2_V
 + \frac{1}{\mu}\int_{\tau}^t \Vert f(x,\xi)\Vert^2 d\xi, \quad
\text{for all} t\geq t_k\,.
\]
Consequently,   the norm $\Vert u(x,t)\Vert_V$ can be made
 arbitrarily small  for sufficiently large $t$.
 By this argument we   have proved the following result.

\begin{theorem} \label{thm4.1}
If $\Omega$ is a bounded domain and the integral
$\int_0^\infty \Vert f\Vert dt$ converges then
$\Vert u(x,t)\Vert\to 0$ when $t\to\infty$.
If $\Omega$ is an arbitrary domain then $\Vert u(x,t)\Vert_V\to 0$
when $t\to\infty$ if  the integral
$\int_{0}^\infty ( \Vert f\Vert +\Vert f\Vert^2) dt$  converges.
\end{theorem}


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 \end{document}