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\markboth{\hfil Uniform controllability \hfil EJDE--1999/03}
{EJDE--1999/03\hfil Miguel Angel Moreles \hfil}
\begin{document}
\title{\vspace{-1in}\parbox{\linewidth}{\footnotesize\noindent
{\sc  Electronic Journal of Differential Equations},
Vol. {\bf 1999}(1999), No.~03, pp. 1--13. \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)}
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Uniform controllability for Kirchhoff and Mindlin-Timoshenko elastic systems 
\thanks{ {\em 1991 Mathematics Subject Classifications:} 93B05, 35L20, 34H05.
\hfil\break\indent
{\em Key words and phrases:} exact controllability, elastic plates,
hyperbolic operators.
\hfil\break\indent
\copyright 1999 Southwest Texas State University  and University of
North Texas. \hfil\break\indent
Submitted November 30, 1998. Published February 10, 1999. \hfil\break\indent
Partially supported by CONACYT Project 28540-A.} }
\date{}
%
\author{Miguel Angel Moreles}
\maketitle

\begin{abstract} 
The Mindlin-Timoshenko operator is a perturbation of the Kirchhoff operator
and it is well known that there exist solutions to the exact controllability 
problem for their associated systems. This article shows that the
solution of the controlled problem of the Mindlin-Timoshenko system
converges to that of Kirchhoff; that is, we show uniform controllability.
\end{abstract}

\newtheorem{theorem}{Theorem}
\newtheorem{corollary}{Corollary}[theorem]


\section{Introduction}

We are concerned with the boundary controllability problem known as the
{\em Exact-Controllability Problem} (ECP). Suppose we have a
well-posed initial-boundary value problem for an  evolution
equation $Lw=0$ in a cylindrical domain $Q=\Omega \times \lbrack 0,+\infty
) $, where $\Omega $ is a bounded domain in ${\mathbb R}^n$. The ECP deals
with the following question: Given Cauchy data in $\Omega $ at $t=0$, can
this data be supplemented with appropriate inhomogeneous time-dependent
boundary data (boundary controls), prescribed in the lateral boundary of 
$Q$, such that the solution of the initial-boundary problem will vanish for 
$t\geq T_{1}$~?

In this work, we are interested in the evolution equations associated with the
Kirchhoff operator 
\[
L=\rho h\partial _t^2-\frac{\rho h^3}{12}\Delta \partial _t^2+\Delta ^2 
\]
and with the Mindlin-Timoshenko operator 
\[
L_K=\rho h\partial _t^2-\frac{\rho h^3}{12}\Delta \partial _t^2+\Delta ^2+%
\frac{\rho h}K\left( \frac{\rho h^3}{12}\partial _t^4-\Delta \partial
_t^2\right)\,,
\]
where  $\Delta $ is the Laplacian Operator acting in the space variables, i.e.,
$$
\Delta =\sum\limits_{i=1}^n\partial _{x_i}^2\,.
$$ 


When $n=2$, the equations $Lw=0$ and $L_{K}w_{K}=0$, in a suitable
domain with appropriate initial and boundary data, model vibrations of
elastic plates. In such a case $w(x,t)$ represents the vertical displacement
of the {\em middle surface}. The physical constants are: $\rho \equiv $
density, $h\equiv $ plate thickness, $K\equiv $ shear modulus. The Mindlin-Timoshenko
equation is obtained by uncoupling the Reissner-Mindlin Plate. See
Lagnese \cite{LJ}, Lagnese \& Lions \cite{LL} and Lagnese, Leugering \&
Schmidt \cite{LLS}.

Komornik \cite{KV} studies the Reissner-Mindlin plate equation and established 
that the control time is independent of $K$. A Similar result
follows for the corresponding Timoshenko beam equation. To our knowledge,
there are no results about convergence of the solution of the control
problem for the Reissner-Mindlin system to that of the Kirchhoff operator.

For the Timoshenko beam equation, a solution to this problem is presented in
Moreles \cite{MM}. In addition to uniform control time, it is shown that
the solution of the control problem for the Timoshenko system, converges in
a {\em strong} sense to the solution of the control problem of the Rayleigh
system. The latter is the one dimensional version of the Kirchhoff system.
In this work, we generalize this result to several space dimensions, in
particular for plates.\smallskip

The Mindlin-Timoshenko operator is hyperbolic and the solution of the ECP is a 
direct application of Theorem 2 in Littman \cite{LW}. The method of solution
therein was successfully generalized to elastic operators (beams and plates)
in Littman \cite{LW1}. It is observed that the Kirchhoff operator does not
correspond to those of the cited works. However, we show that Littman's
method still applies and solve the ECP. Moreover the control time for both
systems is the same and independent of $K$.

Once the ECP is solved for both operators we prove uniformity, that is, we show
that the solution of the controlled problem for the Mindlin-Timoshenko operator
converges to that of Kirchhoff when $K\rightarrow +\infty $.  Our proof rests on
the results in Moreles \cite{MM}.

The outline of this article is as follows:
Section 2 presents the statement of the main result. The proof is carried
out in the remaining sections. To illustrate our result we consider a weak
version of the ECP in Lasiecka \& Triggiani \cite{LT}.
In Section 3 we show the first step of Littman's method. More precisely, we
show that the solutions of the Cauchy problems for the Mindlin-Timoshenko and Kirchhoff operators
are smooth away from the origin if Cauchy data with compact support is
provided. This is accomplished by studying the {\em singular support} 
 of the fundamental solutions. For the Mindlin-Timoshenko operator there is
nothing to do: the information about the singular support contained in the
principal part follows from the theory of hyperbolic operators. 
The Kirchhoff operator, however, does require some analysis. 
We construct the fundamental solution
as a Fourier integral operator in the sense of H\"{o}rmander \cite{HL}
and apply the results therein to determine the singular support.
In Section 4 we deal with two perturbation problems that arise on Littman's 
method, and prove the analogue results to those in Moreles \cite{MM}. 
Consequently we deduce uniform controllability.

The Theory of Distributions of L. Schwartz and the 
the Theory of Hyperbolic Operators will be assumed throughout this article. 
As a classical reference we have H\"{o}rmander's book \cite{HL1}.

\section{Main Result}

Let $\Omega $ be a bounded domain in ${\mathbb R}^n$. Let $W_{2}^{l}(\Omega )$
be the Sobolev space of functions $f\in L_{2}(\Omega)$ for which the 
distributional derivatives $\partial _{x}^{\alpha }f$ are in $L_{2}(\Omega )$ 
for $| \alpha | \leq l$, and let the
norm be denoted by $\left\| f\right\| _{l}$. 
 When $\Omega ={\mathbb R}^n$ we regard $W_{2}^{l}$ as $H_{l}$ and 
\[
\left\| f\right\| _{l}=\left( \int \Lambda _{2l}\left| \widehat{f}\right|
^{2}\right) ^{1/2}\,, 
\]
where $\widehat{f}$ is the Fourier Transform of $f$, and 
\[
\Lambda _{s}(\xi )=\left( 1+\left| \xi \right| ^{2}\right) ^{\frac{s}{2}%
}\,,\ s\in {\mathbb R} \,.
\]
Let
$w^{0}$ and $w^{1}$ be functions in $W_{2}^{m+l}(\Omega )$, and let 
$w^{2}$ and $w^{3}$ be functions in $W_{2}^{m+l-2}(\Omega )$, 
where $m$ and $l$ are nonnegative integers with $l>\frac{n}{2}$.
Let $\varepsilon$ be a positive constant. Extend the Cauchy Data to have 
compact support in an $\varepsilon $-neighborhood of $\Omega $ which we call 
again $\Omega$. For $b\geq 0$ we denote by $Q_b$ the set 
$\overline{\Omega }\times [b,+\infty )$.


\begin{theorem} \begin{description}
 \item{(i)} Let $T_{0}=d\sqrt{\rho h^3/12}$, 
where $d$  is the diameter of $\Omega $. Then for
each $T_{1}>2T_{0}$  there exist solutions $w(x,t)$ and $w_{K}(x,t)$
to the Cauchy problems 
\begin{eqnarray*}
&Lw=0,\quad \mbox{in }\overline{\Omega }\times \{t\geq 0\}\,,& \\ 
&w(x,0)=w^{0}(x)\,,\quad w_{t}(x,0)=w^{1}(x)\quad \mbox{in }\overline{\Omega }&
\end{eqnarray*}
and 
\begin{eqnarray*}
&L_{K}w_{K}=0\,,\quad \mbox{in }\overline{\Omega }\times \{t\geq 0\}\,,& \\ 
&\partial _{t}^{j}w_{K}(x,0)=w^{j}(x)\,, \quad j=0,1,2,3\quad 
\mbox{ in }\overline{\Omega }
\end{eqnarray*}
both vanishing in $\overline{\Omega }\times \{t\geq T_{1}\}$. 
Moreover, If $| \alpha | \leq m$,  then 
$\partial_{x}^{\alpha }w_{K}$ converges to $\partial _{x}^{\alpha }w$
when $K\rightarrow +\infty $ in the $L^{\infty }$-norm in
bounded subsets of $Q_{0}$.

\item{(ii)} If we restrict further $w^{0}\in W_{2}^{m+l+1}(\Omega )$, 
$w^{2}\in W_{2}^{m+l-1}(\Omega )$,  then 
$\partial _{x}^{\alpha }\partial _{t}w_{K}$ converges to 
$\partial _{x}^{\alpha }\partial_{t}w$  when 
$K\rightarrow +\infty ${\it \ in the }$L^{\infty }$-norm in bounded subsets 
of $Q_{0}$ also for $\left| \alpha \right|\leq m$. 
\end{description}
\end{theorem}

The functions $w(x,t)$ and $w_{K}(x,t)$ with the alleged properties are
constructed in the proof. Hence to solve the ECP for both equations we just
need to read off appropriate boundary conditions to have uniqueness of the
resulting initial-boundary-value problem. 
For instance, the following boundary conditions guarantee uniqueness
\begin{eqnarray*}
&f_{K}=w_{K}\,,\quad g_{K}=\Delta w_{K}\quad \mbox{in }\Gamma \times
\lbrack 0,T]\,; & \\
&f=w\,,\quad g=\Delta w\quad \mbox{in } \Gamma \times \lbrack 0,T]\,. 
\end{eqnarray*}
Here $\Gamma =\partial \Omega $ is the (smooth) boundary of $\Omega $.
Observe that these functions are bounded in $Q_{0}$. Moreover 
$f_{K}\rightarrow f$ and $g_{K}\rightarrow g$ in the $L^{\infty}$-norm.


Observe that the control time $T_{1}$ in the theorem is independent of $K$,
that is, the control time is uniform. We shall see that the functions 
$w(x,t)$ and $w_{K}(x,t)$ are smooth in $Q_{T_{0}}$. Hence, in this set the
convergence is uniform.\smallskip

Now the drawbacks. From the point of view of Exact Controllability we have a
weak result. For instance, in Lasiecka \& Triggiani \cite{LT} the Kirchhoff 
system is considered with the following boundary conditions 
\begin{eqnarray*}
&w=0\quad \mbox{in } \Gamma \times \lbrack 0,T]& \\ 
&\Delta w=0\quad \mbox{in } \Gamma _{0}\times \lbrack 0,T]& \\ 
&\Delta w=u\quad \mbox{in } \Gamma _{1}\times \lbrack 0,T]\,,&
\end{eqnarray*}
where $\Gamma $ is the disjoint union of $\Gamma _{0}$ and $\Gamma _{1}$.
Thus, only one control in part of the boundary is required to drive the
system to rest.

In contrast we require controls in the whole boundary. Also, these controls
need not be unique, and we offer no criteria for comparison. Furthermore,
for solving the initial-boundary-value problem we require on the Cauchy 
data more regularity than
usual. Our offering is convergence in a rather strong sense.\smallskip

A natural continuation to this work, is to consider the problem under weaker
assumptions.

\section{The Cauchy Problems}

\begin{theorem}  Let $u_{K}$ and $u$ be the solutions of the Cauchy problems 
\begin{eqnarray} \label{c1}
&Lu=0\quad \mbox{in }{\mathbb R}^n\times \{t\geq 0\}\,,& \\ 
&u(x,0)=w^{0}(x)\,,\quad u_{t}(x,0)=w^{1}(x)\quad \mbox{in }{\mathbb R}^n&
\nonumber\end{eqnarray}
and 
\begin{eqnarray}\label{c2}
&L_{K}u_{K}=0\quad \mbox{in }{\mathbb R}^n\times \{t\geq 0\}\,,&\\ 
&\partial _{t}^{j}u_{K}(x,0)=w^{j}(x)\,,\quad j=0,1,2,3\quad \mbox{in }
{\mathbb R}^n \,.& \nonumber
\end{eqnarray}
Then $u_{K}$ and $u$ are smooth in $Q_{T_{0}}$.
\end{theorem}

As remarked before, this theorem is obtained as a consequence of the study
of the singular support of the fundamental solutions.

\subsection*{The Mindlin-Timoshenko Operator}
Since the Mindlin-Timoshenko Operator is hyperbolic the problem is settled. 
Indeed, the principal part is 
\[
L_{K4}=\frac{\rho ^2h^4}{12K}\left( \partial _t^2-\frac K{\rho h}\Delta
\right) \left( \partial _t^2-\frac{12}{\rho h^3}\Delta \right) \,. 
\]
It has two characteristic cones, namely 
\begin{eqnarray*}
\Gamma & = & \left\{ (x,t)\in {\mathbb R}^{n+1}:t^2=\frac{12}{\rho h^3}
|x|^2\,,\; t\geq 0\right\} \\ 
\Gamma _1 & = & \left\{ (x,t)\in {\mathbb R}^{n+1}:t^2=\frac K{\rho
h}|x|^2\,,\; t\geq 0\right\}
\end{eqnarray*}
with dual cones 
\begin{eqnarray}\label{c5}
\Gamma ^0 & = & \left\{ (x,t)\in {\mathbb R}^{n+1}:t^2=\frac{\rho h^3}{12}
|x|^2\,,\; t\geq 0\right\} \\ 
\Gamma _1^0 & = & \left\{ (x,t)\in {\mathbb R}^{n+1}:t^2=\frac{\rho h}
K|x|^2\,,\; t\geq 0\right\}\,. \nonumber
\end{eqnarray}
Thus if $G_K$ is the fundamental solution of $L$ then 
\[
\mbox{sing\, supp}\;G_K\subset \Gamma ^0\cup \Gamma _1^0\,. 
\]

\subsection*{The Kirchhoff Operator}
The principal part of the Kirchhoff Operator is 
\[
L_4=\frac{\rho h^3}{12}\Delta \left( \partial _t^2-\frac{12}{\rho h^3}\Delta
\right) . 
\]
Notice that the wave operator $\partial _t^2-\frac{12}{\rho h^3}\Delta $ has
characteristic cone $\Gamma $. We shall see that this hyperbolic part of 
$L_4$ determines the singularities of the fundamental solution of $L$, that
is, we show that 
\[
\mbox{sing supp}\;G\subset \Gamma ^0\,. 
\]

For convenience we use the following notation 
\[
D_{x_j}\equiv -i\partial _{x_j}\,,\quad 
D_x=D_{x_1},\ldots ,D_{x_n}\,, \quad 
D_t\equiv -i\partial _t\,, \quad \overline{\Delta }=-\Delta \,. 
\]
and write the Kirchhoff Operator in the form 
\begin{equation}
P(D_x,D_t)=\left( 1+\frac{h^2}{12}\overline{\Delta }\right) D_t^2-\frac
1{\rho h}\overline{\Delta }^2\,.
\label{c10}
\end{equation}

We shall construct a distribution $G(x,t)$ supported in the half space $%
\left\{ (x,t)\in {\mathbb R}^{n+1}:t\geq 0\right\} $ such that 
\[
\begin{array}{l}
\left[ \left( 1+\frac{h^2}{12}\overline{\Delta }\right) D_t^2-\frac 1{\rho h}%
\overline{\Delta }^2\right] G(x,t)=\delta (x,t)\,.
\end{array}
\]
Let 
\[
S(\xi ,t)  = \frac{1}{2\lambda (\xi )}\left( e^{i\lambda (\xi)t}-e^{-i\lambda (\xi )t}\right)  
=  i\frac{\sin \lambda (\xi )t}{\lambda (\xi )}\,,
\]
where 
\begin{equation}
\lambda (\xi )=\frac{\left| \xi \right| ^{2}}{\sqrt{\rho h+\frac{\rho h^{3}}{%
12}\,\left| \xi \right| ^{2}}}\mbox{ .}  \label{c13}
\end{equation}
it satisfies 
\begin{eqnarray}\label{c14}
&\left[ \left( 1+\frac{h^{2}}{12}\left| \xi \right| ^{2}\right) D_{t}^{2}-
\frac{1}{\rho h}\left| \xi \right| ^{4}\right] S(\xi ,t)=0\,, \quad
\mbox{for } t>0\,.& \\ 
&S(\xi ,0)=0\,,\quad D_{t}S(\xi ,0)=1\,.  &  \nonumber
\end{eqnarray}
Observe that 
\[
P(\xi ,D_{t})=\left( 1+\frac{h^{2}}{12}\left| \xi \right| ^{2}\right)
D_{t}^{2}-\frac{1}{\rho h}\left| \xi \right| ^{4} 
\]
is obtained from (\ref{c10}) after applying the Fourier Transform with
respect to $x$.

Choose $\chi (\xi )\in {\cal D}$ such that $\chi (\xi )=1$ in a neighborhood
of $0$. Let 
\begin{eqnarray}
P_{0}u(x,t) & = & \frac{1}{\left( 2\pi \right) ^{n}}\int e^{i\left(
x-y\right) \cdot \xi }\frac{1}{1+\frac{h^{2}}{12}\left| \xi \right| ^{2}}%
S(\xi ,t)\chi (\xi )u(y)\,d\xi\,dy   \nonumber \\ 
Au(x,t) & = & \frac{1}{\left( 2\pi \right) ^{n}}\int e^{i\varphi (x,y,\xi
)}a(\xi )u(y)\,d\xi\,dy      \label{c16} \\ 
Bu(x,t) & = & \frac{1}{\left( 2\pi \right) ^{n}}\int e^{i\psi (x,y,\xi
)}a(\xi )u(y)\,d\xi\,dy\,, \nonumber
\end{eqnarray}
where 
\begin{eqnarray}
a(\xi )=\frac{1}{2\lambda (\xi )}\cdot \frac{1}{1+\frac{h^{2}}{12}\left| \xi
\right| ^{2}}\cdot \left( 1-\chi (\xi )\right)  \label{c17} \\
\varphi (x,y,\xi ) & = & \left( x-y\right) \cdot \xi +\lambda (\xi )t \nonumber \\ 
\psi (x,y,\xi ) & = & \left( x-y\right) \cdot \xi -\lambda (\xi )t\,. \nonumber
\end{eqnarray}
Since $\frac{\partial \varphi }{\partial x_i}=\xi _i$ it follows that 
\begin{equation}
\left| \nabla _x\varphi \right| ^2+\left| \xi \right| ^2\left| \nabla _\xi
\varphi \right| ^2\geq \left| \xi \right| ^2\mbox{ .}  \label{c19}
\end{equation}
Moreover, $\lambda (\xi )\in S^1$ and so is $\varphi (x,y,\xi )$. Recall
that for any real $m$ $S^m$ is the set of symbols of order $m$.

Therefore, according to Definition 2.3 in H\"{o}rmander \cite{HL}, $\varphi $
is a {\em phase function}. Similar argument applies to $\psi $. 
Also, it is readily seen that $a(\xi )$ in (\ref{c17}) is in $S^{-3}$. 
Then from Theorem 2.4 in H\"{o}rmander \cite{HL}, $A$ and $B$ are Fourier integral
operators with symbol 
$a(\xi )$ and phase functions $\varphi (x,y,\xi )$, $\psi (x,y,\xi )$
respectively.

To study singularities of the fundamental solution, we need the following
definitions.
Let $\Omega _\varphi $ be the set of all $((x,t),y)\in {\mathbb R}^{n+1}\times 
{\mathbb R}^n\,$ such that for some constant $C$, depending on $((x,t),y),$%
\[
1\leq C\left| \nabla \varphi \right| ^2\,,\quad \left| \xi \right| >C\,. 
\]
Then the set $\Omega _\varphi $ is open, hence its complement, 
denoted by  $F_\varphi$, is closed. Define similarly $\Omega_\psi $ and 
$F_\psi$. 
For subsets $X$ of ${\mathbb R}^n$ define 
\[
F_\varphi X=\left\{ (x,t)\in {\mathbb R}^{n+1}:((x,t),y)\in F_\varphi 
\mbox{
for some }y\in {\mathbb R}^n\right\} \,. 
\]

Our claim about the fundamental solution will be a consequence of the
following theorem.

\begin{theorem} Let $\Gamma ^{0}$ be the cone in (\ref{c5}), and let 
$u=\delta $ in ( \ref{c16}). Define 
\[
E\delta (x,t)=P_{0}\delta (x,t)+A\delta (x,t)-B\delta (x,t)
\]
Then $E\delta $  is a distribution such that 
the singular support of $E\delta$ is a subset of $\Gamma ^{0}$
\end{theorem}

\paragraph{Proof.}
We shall prove that in the complement of $\Gamma ^{0}$ the distribution $E\delta
(x,t)$ is smooth. Observe that $P_{0}\delta (x,t)$ is $C^{\infty }$ in 
${\mathbb R}^{n+1}$; hence we need to consider only $A\delta (x,t)$ and 
$B\delta (x,t)$.

In light of Corollary 2.7 in H\"{o}rmander \cite{HL}, it suffices to show
that if $\left| x\right| ^2\neq \frac{12}{\rho h^3}t^2$ for $t\geq 0$ then 
$(x,t)\notin F_\varphi \,\mbox{sing supp}\;\delta $ and 
$(x,t)\notin F_\psi\,\mbox{ sing supp}\;\delta $. 
We prove the assertion for $\varphi $. For $\psi $, the proof is similar.

Our interest is $u=\delta $,  the Dirac's Delta distribution. In
this case 
\[
\mbox{sing\,supp}\;\delta =\{0\}\,. 
\]

Thus $(x,t)\notin F_\varphi \,\mbox{sing supp}\;\delta $ is equivalent to 
$((x,t),0)\notin F_\varphi $ that is $((x,t),0)\in \Omega _\varphi $.

It can be easily shown that
\[
\frac{\partial \varphi }{\partial \xi _i}=x_i+\frac{\frac 1{\sqrt{\rho h}}t}{%
\left( 1+\frac{h^2}{12}\left| \xi \right| ^2\right) ^{3/2}}\left( 2+\frac{h^2%
}{12}\left| \xi \right| ^2\right) \xi _i\,; 
\]
thus 
\[
\nabla _\xi \varphi ((x,t),0,\xi )=x+\frac{\frac 1{\sqrt{\rho h}}t}{\left( 1+%
\frac{h^2}{12}\left| \xi \right| ^2\right) ^{3/2}}\left( 2+\frac{h^2}{12}%
\left| \xi \right| ^2\right) \xi 
\]
and 
\[
\big| \nabla _\xi \varphi ((x,t),0,\xi )\big| \geq \bigg| \left| x\right|
-\frac{\frac 1{\sqrt{\rho h}}t}{\left( 1+\frac{h^2}{12}\left| \xi \right|
^2\right) ^{3/2}}\left( 2+\frac{h^2}{12}\left| \xi \right| ^2\right) \left|
\xi \right| \bigg| \,. 
\]
Consider the function 
\[
f(r)=\frac{2+\frac{h^2}{12}r^2}{\left( 1+\frac{h^2}{12}r^2\right) ^{3/2}}%
\,r\,,\quad r\geq 0\,,\quad r=\left| \xi \right| \,. 
\]

Then $f(r)\searrow \frac{\sqrt{12}}{h}$ when $r\rightarrow \infty $ and $f$
attains its maximum value at $r=\frac{\sqrt{24}}{h}$ with $f\left( \frac{
\sqrt{24}}{h}\right) =\frac{8}{3\sqrt{6}}\frac{\sqrt{12}}{h}$.

If $\left| x\right| <\sqrt{\frac{12}{\rho h^3}}\,t$ then $\left| x\right|
=\left( 1-\varepsilon \right) \sqrt{\frac{12}{\rho h^3}}\,t$ for some $
\varepsilon $, $0<\varepsilon <1$, thus 
\[
\left| \nabla _\xi \varphi ((x,t),0,\xi )\right| \geq \frac t{\sqrt{\rho h}
}f(r)-\left| x\right| 
\]
but $f(r)\geq \frac{\sqrt{12}}h$ for $r\geq $ $\frac{\sqrt{24}}h$, so 
\[
\left| \nabla _\xi \varphi ((x,t),0,\xi )\right| \geq \varepsilon \sqrt{%
\frac{12}{\rho h^3}}\,t 
\]
or 
\[
\left| \nabla _\xi \varphi ((x,t),0,\xi )\right| C\geq 1\,,\quad %
\mbox{if\quad }\left| \xi \right| >C 
\]
with $C=\max \big\{ \frac 1{\varepsilon \sqrt{\frac{12}{\rho h^3}}\,t},%
\frac{\sqrt{24}}h\big\} $ , i.e. $(x,t)\in \Omega _\varphi $.

For the case $\left| x\right| >\sqrt{\frac{12}{\rho h^3}}\,t$ we have $%
\left| x\right| =\left( 1+\varepsilon \right) \sqrt{\frac{12}{\rho h^3}}\,t$
for some $\varepsilon >0$ , thus 
\[
\left| \nabla _\xi \varphi ((x,t),0,\xi )\right| \geq \left| x\right| -\frac
t{\sqrt{\rho h}}f(r) 
\]

Choose $\widetilde{\varepsilon }$ small enough so that 
\[
\varepsilon -\widetilde{\varepsilon }\left( \frac 8{3\sqrt{6}}-1\right) >0 
\]
there is $r_{\varepsilon ,\widetilde{\varepsilon }}\geq $ $\frac{\sqrt{24}}h$
so that, if $r>r_{\varepsilon ,\widetilde{\varepsilon }}$ then $f(r)\leq %
\left[ 1+\widetilde{\varepsilon }\left( \frac 8{3\sqrt{6}}-1\right) \right] 
\frac{\sqrt{12}}h$ . We have 
\[
\left| \nabla _\xi \varphi ((x,t),0,\xi )\right| \geq \sqrt{\frac{12}{\rho
h^3}}\,t\left( \varepsilon -\widetilde{\varepsilon }\left( \frac 8{3\sqrt{6}%
}-1\right) \right) 
\]
Again 
\[
\left| \nabla _\xi \varphi ((x,t),0,\xi )\right| C\geq 1\,,\quad %
\mbox{if\quad }\left| \xi \right| >C 
\]
with $C=\max \big\{ \left( \sqrt{\frac{12}{\rho h^3}}\,t\left( \varepsilon -%
\widetilde{\varepsilon }\left( \frac 8{3\sqrt{6}}-1\right) \right) \right)
^{-1},r_{\varepsilon ,\widetilde{\varepsilon }}\big\} $ . Therefore $%
(x,t)\in \Omega _\varphi $.

Finally we show that the singular support is contained in $\Gamma ^0$. Now we
use the phase function $\psi .$ We have 
\[
\nabla _\xi \psi ((x,t),0,\xi )=x-\frac{\frac t{\rho h}}{\left( 1+\frac{h^2}{%
12}\left| \xi \right| ^2\right) ^{3/2}}\left( 2+\frac{h^2}{12}\left| \xi
\right| ^2\right) \xi 
\]

In the cone $\Gamma ^0$, $
\left| x\right| =\sqrt{\frac{12}{\rho h^3}}\,t$
so 
\[
\nabla _\xi \psi ((x,t),0,\xi )=x-\frac{h\left| x\right| }{\sqrt{12}}\frac{%
f(r)}r\xi\,. 
\]
Let $\xi =\mu x$, with $\mu >0$ to be chosen later. Then 
\[
\left| \nabla _\xi \psi ((x,t),0,\xi )\right| =\left| 1-\frac h{\sqrt{12}%
}f\left( \mu \left| x\right| \right) \right| \left| x\right| 
\]

Since $f(r)\searrow \frac{\sqrt{12}}h$ when $r\nearrow \infty $ it follows
that for any $C>0$ we may choose $\mu >>0$ so that 
\[
\left| 1-\frac h{\sqrt{12}}f\left( \mu \left| x\right| \right) \right|
\left| x\right| <\frac 1C 
\]
that is if $\left| x\right| =\sqrt{\frac{12}{\rho h^3}}\,t\left| x\right| $
then $(x,t)\notin \Omega _\psi $ , i.e., $(x,t)\in F_\psi $ and 
the singular support of $E\delta$ is a subset of $\Gamma ^0$ as asserted.


Let $H(t)$\ be the Heaviside function. From $\left( \ref{c14}\right) $ we
see that 
\[
\left[ \left( 1+\frac{h^2}{12}\left| \xi \right| ^2\right) D_t^2-\frac
1{\rho h}\left| \xi \right| ^4\right] \left( \frac i{1+\frac{h^2}{12}\left|
\xi \right| ^2}S(\xi ,t)H(t)\right) =\delta (t) 
\]

This fact together with the previous theorem give us

\begin{corollary} If $G(x,t)=iE\delta (x,t)H(t)$, 
then $G(x,t)$ is a fundamental solution of the Kirchhoff Operator supported
in the half space $t\geq 0$.  Moreover the singular support of 
$G$ is a subset of $\Gamma^{0}$. 
\end{corollary}

\paragraph{Remark.} Existence of fundamental solutions for differential
operators with constant coefficients is a classical result. 
In applications some times
a more explicit expression is necessary as illustrated in the present work.
The Kirchhoff operator is quasi-hyperbolic in the sense of Ortner \& Wagner \cite
{OW, OW1}. They are able to find explicit expressions for some of such
operators. However, their approach does not seem to apply here.

\section{Perturbation Problems}

>From the Cauchy problems (\ref{c1}) and (\ref{c2}), we obtain the first 
perturbation problem.

After applying the Fourier Transform on the space variables, we obtain 
\begin{eqnarray*}
&L(i\xi ,t)U=0 &\\ 
&U(\xi ,0)=W^{0}(\xi )\,, \quad U_{t}(\xi ,0)=W^{1}(\xi )&
\end{eqnarray*}
and
\begin{eqnarray*}
&L_{K}(i\xi ,t)U_{K}=0 &\\ 
&\partial _{t}^{j}U_{K}(\xi ,0)=W^{j}(\xi )\,, j=0,1,2,3&
\end{eqnarray*}

Let $R=U_{k}-U$. We will establish the following inequalities
\begin{eqnarray} 
\left| R(\xi ,t)\right|  & \leq  & \frac{c(t)}{K}\left[ \left| W^{0}\right|
+\left| W^{1}\right| +\Lambda _{-2}\left( \left| W^{2}\right| +\left|
W^{3}\right| \right) \right]  
\nonumber \\ 
\left| \partial _{t}R(\xi ,t)\right|  & \leq  & \frac{c(t)}{\sqrt{K}}\left(
\Lambda _{1}\left| W^{0}\right| +\left| W^{1}\right| +\Lambda _{-1}\left|
W^{2}\right| +\Lambda _{-2}\left| W^{3}\right| \right)   
\label{d4} \\ 
\left| \partial _{t}^{2}R(\xi ,t)\right|  & \leq  & c(t)\left( \Lambda
_{2}\left| W^{0}\right| +\Lambda _{1}\left| W^{1}\right| +\left|
W^{2}\right| +\Lambda _{-1}\left| W^{3}\right| \right) \,, \nonumber
\end{eqnarray}
where $c(t)$ is a polynomial on $t$ independent of $K$. From these estimates, 
using the fact that $\int \Lambda _{-s}$ converges for 
$s>n$, and H\"{o}lder's inequality,  for any multi-index $\alpha $ we have that 
 \begin{eqnarray*}
\left| \partial _{x}^{\alpha }\left( u_{K}(x,t)-u_{0}(x,t)\right) \right|  & 
\leq  & \frac{c(t)}{K}\bigg( \left\| w^{0}\right\| _{\left| \alpha \right| +%
\frac{s}{2}}+\left\| w^{1}\right\| _{\left| \alpha \right| +\frac{s}{2}%
}  \\ 
&  & + \left\| w^{2}\right\| _{\left| \alpha \right| +\frac{s}{2}%
-2}+\left\| w^{3}\right\| _{\left| \alpha \right| +\frac{s}{2}-2}\bigg) 
\end{eqnarray*}
and 
\begin{eqnarray*}
\left| \partial _{x}^{\alpha }\partial _{t}\left(
u_{K}(x,t)-u_{0}(x,t)\right) \right|  & \leq  & \frac{c(t)}{\sqrt{K}}%
\,\bigg( \left\| w^{0}\right\| _{\left| \alpha \right| +\frac{s}{2}%
+1}+\left\| w^{1}\right\| _{\left| \alpha \right| +\frac{s}{2}}\\
&& +\left\| w^{2}\right\| _{\left| \alpha \right| +\frac{s}{2}%
-1}+\left\| w^{3}\right\| _{\left| \alpha \right| +\frac{s}{2}-2}\bigg)\,.
\end{eqnarray*}
Consequently, we have the following result.

\begin{theorem} Let $s>n$ and $\alpha $ be a
multi-index such that $|\alpha| \leq m$.  Assume that
$w^{0}$ and are in $w^{1}\in H^{m+\frac{s}{2}}$, and that
$w^{2}$ and $w^{3}$ in $H^{m+\frac{s}{2}-2}$.
Then \begin{description}
\item{(i)} $\partial _{x}^{\alpha }u_{K}$ converges to $\partial
_{x}^{\alpha }u$  when $K\rightarrow +\infty $  in the norm 
$L^{\infty }\left( {\mathbb R}^n\times \lbrack 0,\tau ]\right) $  with 
$\tau <+\infty $. 
\item{(ii)} If we restrict further $w^{0}\in H^{m+\frac{s}{2}+1}$  and 
$w^{2}\in H^{m+\frac{s}{2}-1}$, then 
$\partial _{x}^{\alpha }\partial_{t}u_{K}$  converges to 
$\partial _{x}^{\alpha }\partial _{t}u$ when $K\rightarrow +\infty $ 
in the norm $L^{\infty }\left( {\cal R}^{n}\times \lbrack 0,\tau ]\right)$
with $\tau <+\infty $. 
\end{description} \end{theorem} 

\paragraph{Remarks.} \begin{description}
\item{(i)} Notice that in Theorem 1 $l\geq \lceil s/2\rceil$. 
\item{(ii)} Thanks to Estimate (\ref{d4}) we require less
regularity of the Cauchy Data than in Theorem 3.2 in Moreles \cite{MM}. Thus,
Theorem 4 lead us to a slight improvement of the main result therein.
\item{(iii)} Observe that the last estimate in (\ref{d4}) implies
weaker convergence when taking second derivatives with respect to $t$.
\end{description}

Let us discuss the proof of the estimates in (\ref{d4}). 
It is not difficult to see that 
\[
U_{0}(\xi ,t)=W^{0}\cos \lambda t+\frac{1}{\lambda }W^{1}\sin \lambda t
\]
with 
\[
\lambda \equiv \lambda (\xi )=\frac{\left| \xi \right| ^{2}}{\sqrt{\rho h+%
\frac{\rho h^{3}}{12}\left| \xi \right| ^{2}}}\,,
\]
as in (\ref{c13}). The remainder $R$ satisfies 
\begin{eqnarray*}
&L_{K}(i\xi ,t)R=\frac{\rho ^{2}h^{4}}{12K}F \,&\\ 
&\partial _{t}^{j}R(\xi ,0)=R^{j}(\xi )\,,&
\end{eqnarray*}
where 
$F(\xi ,t)=-\partial _{t}^{4}U_{0}-\frac{12}{\rho h^{3}}\left| \xi \right|
^{2}\partial _{t}^{2}U_{0}$
and 
\begin{eqnarray*}
&R^{0}(\xi )=R^{1}(\xi )=0\, &\\ 
&R^{2}(\xi )=W^{2}(\xi )-\partial _{t}^{2}U_{0}\left( \xi ,0\right) \,,& \\ 
&R^{3}(\xi )=W^{3}(\xi )-\partial _{t}^{3}U_{0}\left( \xi ,0\right) \,.&
\end{eqnarray*}

Then we obtain 
\[
F(\xi ,t)=-\left( \lambda ^{2}-\frac{12}{\rho h^{3}}\left| \xi \right|
^{2}\right) \left( \lambda ^{2}W^{0}\cos \lambda t+\lambda W^{1}\sin \lambda
t\right)\,. 
\]
Notice that 
\begin{equation}
\big| \lambda ^{2}-\frac{12}{\rho h^{3}}\left| \xi \right| ^{2}\big| \leq
c\,;  \label{d14}
\end{equation}
hence, in terms of $\Lambda _{s}$ we obtain 
\begin{equation}
\left| F(\xi ,t)\right| \leq c\left( \Lambda _{2}\left| W^{0}\right|
+\Lambda _{1}\left| W^{1}\right| \right)\,.   \label{d15}
\end{equation}

Now we mimic the proof of Theorem 3.2 in Moreles \cite{MM} to obtain the
estimates (\ref{d4}).

\paragraph{Remark.} Estimate (\ref{d15}) follows from
the explicit form of $F$ and ( \ref{d14}). Compare (\ref{d15}) with the
estimate after (3.10) in Moreles \cite{MM}.\medskip 

The second perturbation problem follows from the proof of Theorem 2 in
Littman \cite{LW}. Let us rework such a proof to derive this perturbation
problem as well as the uniform time of controllability.

Let $\varphi (t)$ be a cutoff function such that 
\[
\varphi (t)=\left\{ 
\begin{array}{ll}
1 & \mbox{for }t\leq T_0 \\[3pt] 
0 & \mbox{for }t\geq T_0+\varepsilon\,.
\end{array}
\right. 
\]

Let $f=L[u\varphi ]$ and $ f_K=L_K[u\varphi ]$. Then 
$f$ and $f_K$ are smooth and have support in the strip 
$T_0\leq t\leq T_0+\varepsilon$. 

\begin{theorem} There are smooth solutions $V$ and $V_{K}$ for
\begin{equation}
L[V]=f\quad \mbox{and}\quad  L_{K}[V_{K}]=f_{K}\label{d18}
\end{equation}
vanishing in a neighborhood of 
\begin{equation}
\overline{\Omega }\times \{t=0\} \quad \mbox{and}\quad \overline{\Omega }\times
\{t=T_{1}\}\,.
\label{d19}
\end{equation}
Moreover, if $\left| \alpha \right| +l\leq 3$,  then 
$\partial_{x}^{\alpha }\partial _{t}^{l}V_{K}$ converges to 
$\partial_{x}^{\alpha }\partial _{t}^{l}V$ uniformly in compacta.
\end{theorem}

\paragraph{Proof.}
For each unit vector $\omega $ let  
$I_\omega =\left\{ s:s=x\cdot \omega \,,\,x\in \overline{\Omega }\right\}$.
Then we have 
\begin{eqnarray*}
&L(\partial _x,\partial _t)v(x\cdot \omega ,t) =  \left[ \rho h\partial
_t^2-\frac{\rho h^3}{12}\partial _s^2\partial _t^2+\partial _s^4\right]
v(s,t)\,,& \\ 
&L_K(\partial _x,\partial _t)v_K(x\cdot \omega ,t)  =  \left[ \rho
h\partial _t^2-\frac{\rho h^3}{12}\partial _s^2\partial _t^2+\partial _s^4+
\frac{\rho h}K\left( \frac{\rho h^3}{12}\partial _t^4-\partial _s^2\partial
_t^2\right) \right] v_K(s,t)\,. &
\end{eqnarray*}

We obtain the beam operators of Rayleigh and Timoshenko 
\begin{eqnarray*}
&L(\partial _s,\partial _t)  = \rho h\partial _t^2-\frac{\rho h^3}{12}
\partial _s^2\partial _t^2+\partial _s^4 &\\ 
&L_K(\partial _s,\partial _t)  =  \rho h\partial _t^2-\frac{\rho h^3}{12}
\partial _s^2\partial _t^2+\partial _s^4+\frac{\rho h}K\left( \frac{\rho h^3
}{12}\partial _t^4-\partial _s^2\partial _t^2\right)&
\end{eqnarray*}
which are hyperbolic in the $s$-direction. For both operators there are two
fundamental solutions supported respectively in the cones 
\begin{eqnarray*}
&\Gamma _{+}^0=\left\{ (s,t)\in {\mathbb R}^2:t^2=\frac{\rho h^3}{12}%
|s|^2\;,\ s\geq 0\right\}& \\ 
&\Gamma _{-}^0=\left\{ (s,t)\in {\mathbb R}^2:t^2=\frac{\rho h^3}{12}%
|s|^2\;,\ s\leq 0\right\}\,.&
\end{eqnarray*}
Let 
\[
f(x,t)=\int\limits_{\left| \omega \right| =1}f_\omega (s,t)\,d\omega 
\quad\mbox{and}\quad  
f_K(x,t)=\int\limits_{\left| \omega \right| =1}f_{K\omega }(s,t)\,d\omega\,.
\]
be the plane wave decompositions of $f$ and $f_K$.

Let $V_{\omega R}$ , $V_{\omega _{L}}$ , $V_{K\omega R}$, and $V_{K\omega R}$
be the solutions to 
\begin{eqnarray}
&L(\partial _{s},\partial _{t})[V_{\omega R}]=f_{\omega }\,,\ s\geq 0\,;&
\nonumber \\
& L(\partial _{s},\partial _{t})[V_{\omega L}]=f_{\omega }\,,\ s\leq 0\,; 
\label{d25}  \\ 
&L_{K}(\partial _{s},\partial _{t})[V_{K\omega R}]=f_{K\omega }\,, 
\ s\geq 0\,; & \nonumber \\
& L_{K}(\partial _{s},\partial _{t})[V_{K\omega L}]=f_{K\omega}\,, 
\ s\leq 0\,. & \nonumber
\end{eqnarray}
Here we assume that $0\in \Omega $.

We see that the supports of $V_{\omega R}$ , $V_{\omega _L}$ , $V_{K\omega
R} $, and $V_{K\omega R}$ are bounded away from the sets 
\[
\overline{\Omega }\times \{t=0\} \quad \mbox{ and } \quad 
\overline{\Omega }\times \{t=T_1\}
\]
Let 
\begin{equation}
V_{\omega }=V_{\omega R}+V_{\omega _{L}}\,,\quad 
V_{K\omega }=V_{K\omega R}+V_{K\omega R}\,
\label{d27}
\end{equation}
and 
\[
V(x,t)  =  \int\limits_{\left| \omega \right| =1}V_{\omega }\,d\omega \,, \quad 
V_{K}(x,t)  =  \int\limits_{\left| \omega \right| =1}V_{K\omega }\,d\omega \,. 
\]

These functions satisfy (\ref{d18})  and (\ref{d19}).
>From (\ref{d25}), (\ref{d27}), and
Theorem 4.2 in Moreles \cite{MM}, it follow that 
$\partial _{s}^{m}\partial_{t}^{l}V_{K\omega }$ converges to 
$\partial _{s}^{m}\partial_{t}^{l}V_{\omega }$ uniformly in compacta 
independently of $\omega$. This proves the theorem.

Finally let 
\[
w=u\varphi -V \quad\mbox{and}\quad w_K=u_K\varphi -V_K\,.
\]
Then $w$ and $w_K$ satisfy all the requirements of Theorem 1.

\paragraph{Acknowledgment.} We wish to thank the anonymous referees for their
remarks which were of great help.

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\end{thebibliography} 


{\sc Miguel Angel Moreles }\\
Centro de Investigaci\'{o}n en Matem\'{a}ticas (CIMAT)\\
Apdo. Postal 402; 
Guanajuato, GTO 36000, MEXICO\\
Email address: moreles@fractal.cimat.mx


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