# INTERNAL EXACT CONTROLLABILITY FOR NAGDHI SHELL

Alexis Rodriguez Carranza  A. Rodriguez. National University of Trujillo
<EMAIL_ADDRESS>,  Jose Luis Ponte Bejarano  J.Ponte. National
University of Trujillo<EMAIL_ADDRESS>and  Juan Carlos Ponte Bejarano
C. Ponte. National University of Trujillo<EMAIL_ADDRESS>
###### Abstract.

In this work we study the exponential stability of the energy associated with
a Naghdi shell model with localized internal dissipation. Using several tools
from Riemannian Geometry we show the well possedness of the model, via
semigroup theory, and obtain observability inequalities which allows to prove
the exponential decay of the total energy. As a consequence then we use
Russell Principle for obtain exact controllability.

###### Key words and phrases:

Internal controllability, Naghdi shell

###### Contents

  1. 1 Introduction and summary
    1. 1.1 Preliminaries and stationary problem
    2. 1.2 Equation of motion
  2. 2 The stabilization result
  3. 3 Some comments on scape vector fields:
  4. 4 Naghdi shell stabilization with internal dissipation
  5. 5 Controllability via Stability
  6. 6 conclusions

##  1\. Introduction and summary

Let us denote by $S$ a two dimensional smoooth Riemannian Manifold with the
induced metric of $I\\!\\!R^{3}$ and inner product denoted by $g$ or simply
$<.,.>$. Recall that this means that for each $p\in S$ we have an inner
product $<.,.>$ on the tangent space $T_{p}S$ and this relation is
$C^{\infty}$. We will consider $S$ as the middle surface of a thin shell.

Suppose we consider a bounded open region $M$ of $S$ with a smooth boundary
$\Gamma=\partial M$. This paper is devoted to consider $M$ as a Naghdi type
thin shell. In this dynamical model we study exponential stabilization of the
total energy provided we assume an internal localized dissipation acting on
$M$.

In the literature, mos of the authors prefer to use the classical geometrical
approach while working on properties of the solutions of evolution thin shell.
In those situations the middle surface is instead, the image under a smooth
map of a two dimensional connected domain de $I\\!\\!R^{2}$ and therefore
described under just a one coordinate path. Several interesting model for
plates or shells with a variable coefficients and other hyperbolic type
systems obtained by using traditional geomentry became very difficult to treat
mainly due to the explicit presence of Chritoffel symbols
$\Gamma^{k}_{ij}(p)$. Interesting aternative was given in the work due to peng
fei Yao [35] and collaborators about twenty years ago. In [35], used and
intrinsic model of the middle surface of a shallow shell as a two dimensional
Riemannianan manifold. This approach allows to get satisfactory results using
multipliers. The basic idea was iniciated by S. Bochner in [8]. In order to
very an identity or a pointwise estimate, it sufficies to do so at each point
$p$ relative to a coordinate frame field wich give as more simplifications.
The best coordinate system in our case would be the one in wich the symbols
$\Gamma^{k}_{ij}(p)$ vanish at the given point $p$. As in [8] the best frame
will be given by the so called coordinate system normal at $p$.

This paper is devoted to study exponential stabilization of the total energy
associated with a dynamic thin shell equation of Naghdi type in the presence
of localized internal dissipation.Using several Yao ideas we show the well
posedness of the model, via semigroup theory, and obtain observability
inequalities wich allo us to prove the exponential decay of the total energy.
As a consequence then we use Russell Principle for obtain exact
controllability.

###  1.1. Preliminaries and stationary problem

Let $S$ and $M$ be as in the introduction. In addition we will assume $S$
orientable and a normal field at each $x\in M$ will be denote by $N(x)$. The
shell, a body of $I\\!\\!R^{3}$ is definied as

$S=\\{p\in I\\!\\!R^{3},\quad p=x+zN(x),\quad x\in M,\quad|x|<\frac{h}{2}\\}$

where $h>0$ denotes the, small, thickness off the shell and $z\in I\\!\\!R$.
In Naghdi model, the displacement vector $\xi(p)$ at a point $p\in S$ can be
approximated by

(1) $\xi(p)=\xi_{1}(p)+z\Psi(x),\quad x\in M,$

where $p=x+zN(x)\xi_{1}(x)\in I\\!\\!R^{3}$ denotes the displacement vector of
the middle surface and $\Psi(x)$ captures the rotations of the normal $N(x)$
at each $x\in M$. Following Naghdi description [27] we assume that a normal
after a deformation may not be again a normal, but, the distance from a point
on the normal to the surface remains invariant. As a consequence, the
deformations in the direction of the normal can be neglected. This implies in
particular that $\xi_{1}(x)$ and $\Psi(x)$ in (1) can be decompose as

(2) $\xi_{1}(x)=W_{1}(x)+w_{1}(x)N(x)$

and

(3) $\Psi_{1}(x)=V(x)+w_{2}(x)N(x)$

where $W_{1}$ and $V\in\chi(M)$, and $w_{1}$, $w_{2}\in C^{\infty}(M)$. Here
$\chi(M)$ denotes the set of all vector fieds on $M$. We recall that a vector
field is a map wich associates to each point $p\in M$ a vectr $X(p)\in
T_{p}(M)$ wich is the tangential plane of $M$ at $p$.
In order to find a model describing deformations of the middle surface we need
to analyse the tensor field (of variation of the metric) using the second and
thrid-fundamental forms on the surface $M$. In other words, we consider

(4) $\Upsilon=\frac{1}{2}(\tilde{g}-g)$

where $g$ and $\tilde{g}$ denote the metric induced on the middle surface
before and after the deformation, respectively. $\Upsilon$ is called the
strain tensor of the middle surface. Given $x\in M$, we choose a coordinare
system normal at $x$, say $\\{E_{1}(x),E_{2}(x),E_{3}(x)=N(x)\\}$ wich is a
basis of $I\\!\\!R^{3}$. Now, we calculate
$\tilde{g}(E_{i},E_{j})-g(E_{i},E_{j})$ to find, after linearization

(5) ${}\Upsilon(\xi)=\frac{1}{2}(DW_{1}+D^{*}W_{1})+w_{1}\Pi$

where $\xi=(W_{1},V,w_{1},w_{2})$, $DW_{1}$ is the covariant differential of
$W_{1}$ of $W_{1}$ and $D^{*}W_{1}$ is the transpose of $DW_{1}$ and $\Pi$ is
the second fundamental form of $M$ wich is a $2-$ covariant tensor.
$\Upsilon(\xi)$ is called the tensor change of metric(linearized). In a
similar way we can deduce the change of curvature tensor(linearized)

(6)
${}\chi_{0}(\xi)=\frac{1}{2}\\{DV+D^{*}V+\Pi(.,DW_{1})+\Pi(DW_{1},.)\\}+w_{2}\Pi+w_{1}c$

where $c$ is the third fundamental form on the surface $M$. Also, the tensor
wich captures rotations of the normal is given by:

(7) ${}\varphi_{0}(\xi)=\frac{1}{2}[Dw_{1}+V-i(W_{1})\Pi]$

where $i(W_{1})\Pi$ is the interior product of the tensro field $\Pi$ by the
vector field $W_{1}$. It is convenient to write (5), (6), (7) in a more
concise way. For example, consider the change of variable,

(8) ${}W_{2}=V+i(W_{1})\Pi$

for $x\in M$. As above we consider the system normal at $x$,
$\\{E_{1}(x),E_{2}(x),E_{3}(x)=N(x)\\}$. Direct calculations give us,

$DW_{2}(E_{i},E_{j})=DV(E_{i},E_{j})+D\Pi(W_{1},E_{i},E_{j})+\Pi(E_{i},D_{E_{j}}W_{1})$

because $D_{E_{j}}E_{i}(x)=0.$ Thus $DW_{2}=DV+\Pi(.,DW_{1})+i(W_{1})D\Pi$ for
$x\in M.$ Substitution into (6) gives

(9) ${}\chi_{0}(\xi)=\frac{1}{2}(DW_{2}+D^{*}W_{2})+K_{ol}(\xi)$

substitution of (8) into (7) give us

(10) ${}\varphi_{0}(\xi)=\frac{1}{2}Dw_{1}+\varphi_{ol}(\xi)$

where $K_{ol}(\xi)=-i(W_{1})D\Pi+w_{1}c+w_{2}\Pi$ and
$\varphi_{ol}(\xi)=-i(W_{1})\Pi+\frac{W_{2}}{2}$ Assuming the material of the
shell is homogeneous and isotropic we find in the literature (see for instance
[15]) the stress-strain relations of the $3-$ dimensional shell on the middle
surface $M$ and express energy as an integral over
$M\times[\frac{-h}{2},\frac{h}{2}]$. Let $R>0$ be the smallest principal
radius of curvature of the undeformed middle surface (see [15], pg 166). As
usual, for a thin shell it is assumed that $\frac{h}{R}<<1$ (see [15], pg 18).
Using this assumption, the following approximation of the strain energy of the
shell is obtain (see [15], pg 253)

(11) $\displaystyle I(\xi)=$ $\displaystyle\alpha
h\int_{M}\left\\{\left.|\Upsilon(\xi)\right|^{2}+2\left|\varphi_{0}(\xi)\right|^{2}+\omega_{2}^{2}+\beta\left(tr(\Upsilon(\xi))+w_{2}\right)^{2}\right.$
$\displaystyle\left.+\gamma\left[|\chi_{0}(\xi)|^{2}+\frac{\left|Dw_{2}\right|^{2}}{2}+\beta\left(tr(\chi_{0}(\xi))\right)^{2}\right]\right\\}dM$

where $\alpha=\frac{E}{1+\mu}$, $\beta=\frac{\mu}{1-2\mu}$ and
$\gamma=\frac{h^{2}}{12}$. Here, $E$ denotes Young modules and $\mu$ is
Poisson ratio $(0<\mu<\frac{1}{2})$.

The above expression (11) of $I(\xi)$ allows us to consider the following
symetric, bilinear form $\tilde{B_{0}}$ associated to the strain energy,
definied on the space $Z=[H^{1}(M,\wedge)]^{2}\times[H^{1}(M)]^{2}:$

(12) ${}\tilde{B_{0}}(\xi,\theta)=\frac{\alpha
h}{2}\int_{M}B_{0}(\xi,\theta)dM$

where $\xi=(W_{1},W_{2},w_{1},w_{2})\in Z$,
$\theta=(\theta_{1},\theta_{2},u_{1},u_{2})\in Z$ and

(13) $\displaystyle
B_{0}(\xi,\theta)=2\langle\Upsilon(\xi),\Upsilon(\theta)\rangle$
$\displaystyle\left.+4<\varphi_{0}(\xi),\varphi_{0}(\theta)\right\rangle+2\omega_{2}u_{2}+$
$\displaystyle+2\beta\cdot\left(\operatorname{tr}(\Upsilon(\xi))+\omega_{2}\right)\left(t_{2}\Upsilon(\theta)+u_{2}\right)$
$\displaystyle\left.\left.+2\gamma<\chi_{0}(\xi),\chi_{0}(\theta)\right)\right\rangle+\gamma\left\langle
D\omega_{2},Du_{2}\right\rangle$
$\displaystyle+2\gamma\beta\operatorname{tr}(\chi_{0}(\xi)).\operatorname{tr}(\chi_{0}(\theta))$

In order to obtain Green identify we consider the Hodge-Laplace type operator
$\Delta_{\beta}$ given by:

(14) ${}\Delta_{\beta}=-[\delta d+2(1+\beta)d\delta]$

where $\beta=\frac{\mu}{1-2\mu}$, $d$ is the exterior derivative and $\delta$
is its formal adjoint. The operator $\Delta_{\beta}$ acts taking a $p$-form to
another $p$-form. In our case, we will need only $\Delta_{\beta}$ operating on
$1$-forms. In [35](theorem 5.1) the following result was prove: Let us
consider the bilinear form $\tilde{B_{0}}(.,.)$ given by(12). Then, for any
$\xi=(W_{1},W_{2},w_{1},w_{2})\in Z$ and
$\theta=(\theta_{1},\theta_{2},u_{1},u_{2})\in Z$, the identity:

(15) ${}\widetilde{B}_{0}(\xi,\theta)=\frac{\alpha h}{2}\left\langle
A_{0}\xi,\theta\right\rangle L^{2}+\frac{\alpha h}{2}\int_{\Gamma=\partial
M}\partial(A_{0}\xi,\theta)d\Gamma$

holds, where $L^{2}=[L^{2}(M,\wedge)]^{2}\times[L^{2}(M)]^{2}$ and

(16) $\displaystyle A_{0}(\xi)=-$
$\displaystyle\left(\Delta\omega_{1}+F_{1}(\xi),\gamma\Delta_{\beta}W_{2}+F_{2}(\xi),\right.$
$\displaystyle\left.\Delta
w_{1}+f_{1}(\xi),\gamma\Delta\omega_{2}+f_{2}(\xi)\right)$
$\displaystyle\partial\left(A_{0}\xi,\theta\right)$
$\displaystyle=\left\langle
c_{1}(\xi),\theta_{1}\right\rangle+\gamma\left\langle\operatorname{c}_{2}(\xi),\theta_{2}\right\rangle$
$\displaystyle+2\left\langle\varphi_{0}(\xi);\eta\right\rangle
u_{1}+\gamma\frac{\partial\omega_{2}}{\partial\eta}u_{2}$

with

(17) $\displaystyle
c_{1}(\xi)=2i(\eta)\Upsilon(\xi)+2\beta\left(\operatorname{tr}\Upsilon(\xi)+w_{2}\right)\eta$
$\displaystyle c_{2}(\xi)=2\gamma
i(\eta)\chi_{0}(\xi)+2\beta(\operatorname{tr}(\chi_{0}(\xi))\eta$

here $\eta$ denotes the exterior normal vector along the curve
$\Gamma=\partial M,$ $\Delta$ is the usual Laplace-Beltrami operator on the
Riemannian manifold $M$ and $F_{j}(\xi)$ and $f_{j}(\xi)$ are first order
terms, i.e, of order $\leq 1$ for $j=1\,\,\mbox{or}\,\,2$

The above description was shown in [35]. In fact, the variable
$\xi=(W_{1},W_{2},w_{1},w_{2})$ satisfies the following system

$\begin{cases}W_{1}^{\prime\prime}-\Delta W_{\beta}+F_{1}(\xi)=0&\text{ on
}M\times(0,\infty)\\\ W_{2}^{\prime\prime}-\Delta W_{2}+F_{2}(\xi)=0&\text{ on
}M\times(0,+\infty)\\\ w_{1}^{\prime\prime}-\Delta w_{1}+f_{1}(\xi)=0&\text{
on }M\times(0,+\infty)\\\ w_{2}^{\prime\prime}-\Delta
w_{2}+f_{2}(\xi)=0&\text{ on }M\times(0,+\infty)\end{cases}$

with

$\left\\{\begin{array}[]{l}\xi=0\quad\text{ on }\Gamma(M)\times(0,+\infty)\\\
\xi(0)=\xi_{0},\xi_{t}(0)=\xi\quad\text{on}\quad\mbox{M}\end{array}\right.$

Here $\Gamma(M)$ denotes the boundary of $M$. Let us consider the following
spaces

$\displaystyle H_{\Gamma}^{1}(M)=\left\\{u\in H^{1}(M),u\equiv 0\text{ on
}\Gamma=\Gamma(M)\right\\}$ $\displaystyle
H_{\Gamma}^{1}(M,\Lambda)=\left\\{z\in H^{1}(M,\Lambda),z\equiv 0\text{ on
}\Gamma=\Gamma(M)\right\\}$

and

$X_{\Gamma}=\left[H_{\Gamma}^{1}(M,N)\right]^{2}\times\left[H_{\Gamma}^{\prime}(M)\right]^{2}$

Next, we can prove that the symetric bilinear form $\tilde{B}(\xi,\theta)$
defined in (15) for any $\bar{\xi},\theta\in Z$ is coercive, that is, there
exits a positive constant $c_{3}$ such that

(18) ${}\widetilde{B}(\xi,\xi)\geqslant c_{3}\|\xi\|_{X_{\Gamma}}^{2}\text{
for any }\xi\epsilon X_{\Gamma}$

In fact, substitution of (5), (9) and (10) into (13) followed by integration
over $M$ give us

(19) ${}\tilde{B}_{0}(\xi,\xi)+K_{2}\|\xi\|_{L^{2}}^{2}\geqslant
K_{1}\|\xi\|_{H_{\Gamma}^{1}}^{2}(M)$

for some positive constants $K_{1}$ and $K_{2}$.

In order to use (19) to obtain (18) we can use the following uniqueness
results: Let $\xi=\left(W_{1},W_{2},w_{1},w_{2}\right)$ belonging to
$X_{\Gamma}$ such that $\Upsilon(\xi)=0,\chi_{0}(\xi)=0,\varphi_{0}(\xi)=0$
and $w_{2}=0$. Then $\xi=0$ for all $x\in M$. Using this uniqueness result
together with the method called compactness-uniqueness we can “adsorb" the
term $K_{2}\|\xi\|_{L^{2}}^{2}$ into the term of the right hand side of (19)
to conclude the validity of (18). The expression (15) for the bilinear form
$\widetilde{B}_{0}(\xi,\theta)$ is known. Using the above discusiion we deduce
that the variational problem associated to the bilinear form
$\widetilde{B}_{0}$ it is equivalent to the following boundary value problem:
To find
$\xi=\left(\boldsymbol{W}_{1},\mathbf{W}_{2},\mathbf{w}_{1},\mathbf{w}_{2}\right)$
such that

(20) ${}\left\\{\begin{array}[]{l}\frac{\alpha h}{2}A_{0}\xi=\tilde{F}\\\
\left.W_{1}\right|_{\Gamma}=\left.W_{2}\right|_{\Gamma}=0,\left.w_{1}\right|_{\Gamma}=\left.w_{2}\right|_{\Gamma}=0\end{array}\right.$

for a given
$\tilde{F}\in\mathbf{L}^{2}=\left[L^{2}(M,\Lambda)\right]^{2}\times\left[L^{2}(M)\right]^{2}$

###  1.2. Equation of motion

In this Section we will consider the equations of motion of Naghdi model. We
assume that there are no external loads on the shell and the shell is clamped
along $\Gamma=\Gamma(M)$. In this situation $\tilde{\xi}=\tilde{\xi}(x,t),$
$x\in M$ and $t$ is time. In onder to include the kinetic energy to our
problem it is convevient to consider the change of variables $t\rightarrow
tc_{1}^{-1}$ where $c_{1}^{2}=\frac{2}{\alpha}$ with $\alpha$ as in (11).
Also, denoting by

$R=\left[\begin{array}[]{llll}1&0&0&0\\\ 0&\gamma&0&0\\\ 0&0&1&0\\\
0&0&0&\gamma\end{array}\right]$

and $\xi=R^{1/2}\cdot\widetilde{\xi}$ we write the equations of evolution of a
Naghdi shell for
$\xi=\left(\boldsymbol{W}_{1},\mathbf{W}_{2}\mathbf{w}_{1},\mathbf{w}_{2}\right)$
as

(21) ${}\left\\{\begin{array}[]{l}\xi+A\xi=0\text{ on
}M\times\mathbb{R}^{+}\\\ \xi=0\text{ on }\Gamma(M)\times\mathbb{R}^{+}\\\
\xi(x,0)=\xi_{0}(x),\xi_{t}(x,0)=\xi_{1}(x)\text{ on }M\end{array}\right.$

where $\mathbf{A}=R^{-1/2}\mathbf{A}_{0}R^{-1/2}$. The bilinear form $J$
associated to operator $\mathbf{A}$ is given by

(22) $\displaystyle\mathbf{J}(\xi,u)$
$\displaystyle=2\langle\Upsilon(\xi),\Upsilon(u)\rangle+4\left\langle\boldsymbol{\varphi}_{0}(\xi),\boldsymbol{\varphi}_{0}(u)\right\rangle\gamma$
$\displaystyle+2\beta\left[Tr(\Upsilon(\xi))+\frac{w_{2}}{\sqrt{\gamma}}\right]\left[Tr(\Upsilon(u))+\frac{u_{2}}{\sqrt{\gamma}}\right]$
$\displaystyle+2\beta
Tr(\chi_{0}(\xi))Tr(\chi_{0}(u))+2\left\langle\chi_{0}(\xi),\chi_{0}(u)\right\rangle$
$\displaystyle+\left(Dw_{2},Du_{2}\right\rangle+\frac{2}{\gamma}w_{2}u_{2}$

for all $\xi=\left(W_{1},W_{2},w_{1},w_{2}\right)$ and
$u=\left(U_{1},U_{2},u_{1},u_{2}\right)$ with $\xi\text{ and }u\in Z$. Also
$\Upsilon,\chi_{0}$ and $\varphi_{0}$ are as in (5), (6) and (7) respectively.
The bilinear form associated with the operator $A\text{ is }\mathbf{J}(\xi,u)$
and the coresponding Green formula would be

(23)
${}\widetilde{J}(\xi,u)=\langle\mathbf{A}\xi,u\rangle_{Y}+\int_{\Gamma=\partial
M}\partial\left(\mathbf{A}\xi,u\right)dr$

where $\tilde{J}(\xi,u)=\int_{M}\mathbf{J}(\xi,u)dM$. Now, we consider a
localized perturbation model (21): To find
$\xi=\left(W_{1},W_{2},w_{1},w_{2}\right)\in Z$ satisfying

(24) ${}\left\\{\begin{array}[]{l}\xi_{tt}+A\xi+a(x)\,\,\xi_{t}=0\quad\text{
on }M\times(0,+\infty)\\\ \xi=0\text{ on }\Gamma(M)\times(0,+\infty)\\\
\xi(x,0)=\xi_{0}(x),\xi_{t}(x,0)=\xi_{1}(x)\text{ on }M\end{array}\right.$

where $a(x)$ is a real valued function defined for all $x\in M$ such that has
support in a small interior region of $M$. Well posedness of problem (24) it
follows using standar tools, for example, the semigroup theory [29] and omitt
the proof here. Before we present a proof of the uniform stabilization of the
total energy of model (24) we need some preliminaries

Let us denote by $T^{2}(M)$ the set of all tensor fields on $M$ of rank $2$.
We define the bilinear form $b(\cdot,\cdot):\quad T^{2}(M)\times
T^{2}(M)\mapsto\mathbb{R}$ gives by

(25) ${}b\left(T_{1},T_{2}\right)=\left\langle
T_{1},T_{2}\right\rangle+\beta(trace(T_{1}))(trace(T_{2}))$

for any $T_{1},T_{2}\in T^{2}(M)$. Here, for any $T\in T^{2}(M)$, the trace of
$T$ at $x\in M$ is given by $trace(T)=\sum_{i=1}^{2}T(e_{i},e_{i})$ where
$\\{e_{1},e_{2}\\}$ is an ortonormal basis of $T_{x}M$. For any $W\in
H^{1}(M,\Lambda)$ we define

(26) ${}\widetilde{S}(W)=\frac{1}{2}\left(DW+D^{*}W\right)$

It is known that there exist a positive constant $\lambda$ such that

(27)
${}2\|\tilde{S}(W)\|_{L^{2}\left(M,T^{2}\right)}=\left\|DW+D^{*}W\right\|_{{L^{2}\left(M,T^{2}\right)}}{\geq\lambda\|W\|_{H^{1}(M,\Lambda)}}$

for any $w\in H^{1}(M,\Lambda)$. See lemma $(4.5)$ in [35]. We claim that
there exist $\lambda_{0}>0$ such that

(28)
${}\lambda_{0}\int_{M}\left[b(\widetilde{S}(w),\tilde{S}(w))+|W|^{2}\right]dM\geqslant\|Dw\|_{{}_{L^{2}\left(M,\Lambda\right)}}^{2}$

holds for any $W\in H_{\Gamma}^{1}(M,\Lambda)$. In fact,

$b(\tilde{S}(w),\tilde{S}(w))=\frac{1}{4}\left|Dw+D^{*}w\right|^{2}+\beta\left(Trace\tilde{S}(w)\right)^{2}$

consequently,

(29)
${}b(\widetilde{S}(w),\widetilde{S}(w))+|w|^{2}\geq\frac{1}{4}\left|DW+D^{*}W\right|^{2}$

integration of (29) over $M$ and using (27) we obtain,

$\displaystyle\int_{m}\left[b(\tilde{S}(w),\widetilde{S}(W))+\mid
Wi^{2}\right]dM$
$\displaystyle\geqslant\frac{1}{4}\left\|DW+D^{*}w\right\|_{L^{2}(M,T^{2})}^{2}$
$\displaystyle\geqslant\frac{\lambda}{4}\|w\|_{H^{1}(M,\Lambda)}^{2}$
$\displaystyle\geqslant\frac{\lambda}{4}\|DW\|_{L^{2}(M,\Lambda)}^{2}$

wich prove our claim. Next, was will use the tedmiage of multiplies to stain
appropiatte identitis and inequalities. wet us assume that given $V\in\chi(M)$
there exists a function $v(x)\in C^{\infty}(M)$ such that

(30) ${}DV(x)(X,X)=v(x)|X|^{2}$

for all $X\in T_{x}(M),x\in M$. Given
$\xi=\left(W_{1},W_{2},w_{1},w_{2}\right)\in Z$ we consider

(31)
${}m(\xi)=\left(D_{V}W_{1},D_{V}W_{2},V\left(w_{1}\right),V\left(w_{2}\right)\right)$

we recall that
$DV(X,X)=\operatorname{DV}_{X}(X)=\left\langle\operatorname{DV}_{X},X\right\rangle$,
for all $X\in T_{x}(M),x\in M$.

Using our assumption (30), we take the inner product of equation (24) with
$m(\xi)$ and integrate over $M$

(32)
$\displaystyle\left\langle\xi_{tt},m(\xi)\right\rangle_{L^{2}(M,\Lambda)}+\langle
A\xi,m(\xi)\rangle+\langle a(x)\xi_{t},m(\xi)\rangle=0$

we can use identity (23) to deduce from (32)

(33)
$\displaystyle\left\langle\xi_{tt},m(\xi)\right\rangle_{L^{2}(M,\Lambda)}+\tilde{J}(\xi,m(\xi))-\int_{\Gamma=\Gamma(M)}\partial(A\xi,m(\xi))d\Gamma=-\left\langle
a(x)\xi_{t},m(\xi)\right\rangle.$

Using similar calculations as the ones given in Lemma $5.2$ in [35] we deduce
the identity

(34) $\displaystyle 2\widetilde{J}(\xi,m(\xi))=$
$\displaystyle\int_{\Gamma}J(\xi,\xi)\langle V,\eta\rangle
d\Gamma-2\int_{M}vJ(\xi,\xi)dM+2\int_{M}K(\xi,\xi)dM$
$\displaystyle+l_{0}(\xi)$

where $\eta$ is the outside normal vector field along $\Gamma=\Gamma(M)$,

(35) $\displaystyle K(\xi,\xi)$
$\displaystyle=2b(\tilde{S}(W_{1}),G\left(V,DW_{1}\right))+2b(\tilde{S}\left(W_{2}\right),G\left(V,DW_{2}\right))+$
$\displaystyle+4v\left|\Psi_{0}(\xi)\right|^{2}+v\left|DW_{2}\right|^{2}.$

Here $b(\cdot,\cdot)$ is as in (25) and $G$ is a map defined by,

(36) $\displaystyle G:\chi(M)\times T^{2}(M)\longmapsto T^{2}(M)$
$\displaystyle G(W,T)=\frac{1}{2}\left[T(\cdot,\nabla\cdot
W)+T^{*}(\cdot,W\cdot W)\right]$

Finally, in (34), $l_{0}(\xi)$ denotes lower order terms with respect to the
energy in the sense that for any $\epsilon>0$ there exists $C_{\epsilon}>0$
such that

(37) ${}\left|l_{0}(\xi)\right|^{2}\leqslant\varepsilon
d(\xi)+C_{\varepsilon}h(\xi)\text{ for all }x\in M$

where $d(\xi)$ is the density energy $e(t)=\frac{1}{2}\int_{M}d(\xi)dM$ y
$h(\xi)$ has partial derivatives only up to order $1$.

##  2\. The stabilization result

Let $\xi$ be the displacement field of Naghdi shell. The total energy of the
model is

(38)
${}E(t)=\frac{1}{2}\left\|\xi_{t}\right\|_{L^{2}(M)}^{2}+\frac{1}{2}\tilde{J}(\xi,\xi)$

where $\tilde{J}$ is given as in(23). In order to study the stabilization
result for the solutions of model (24) we will consider the following
assumption on $a(x)$

Assumption 1 a:M $\longmapsto\mathbb{R}^{+}$is a real valued function
nonnegative and with support in a small interior region of $M$.

######  Theorem A.

Consider the solution of problem (24), $\text{ (with }\Gamma_{0}=\Gamma\text{
) }$ and initial data $\xi_{0}\in
Z=\left[\mathbf{H}^{1}(M,1)\right]^{2}\times\left[\mathbf{H}^{1}(M)\right]^{2}$,
$\left.\xi_{1}\in
Y=\left[L^{2}(M,\Lambda)\right]^{2}\times\left[L^{2}(M)\right]\right]^{2}$.
Assume that condition (30) holds. Then, the identity:

(39)
$\displaystyle\int_{0}^{T}\int_{\Gamma}\left[2\partial(A\xi,m(\xi))+\left(\left|\xi_{t}\right|^{2}-J(\xi,\xi)\right)\langle
V,\eta\rangle\right]drdt$
$\displaystyle\left.\quad=2\left(\xi_{t},m(\xi)\right)_{L^{2}}\right]_{0}^{T}+2\int_{0}^{T}\int_{M}v\left[|\xi_{t}|^{2}-J(\xi\xi)\right]dMdt$
$\displaystyle+2\int_{0}^{T}\int_{M}^{T}k(\xi,\xi)dMdt-\int_{0}^{T}\left(a(x)\xi_{t},2m(\xi)\right)dt$
$\displaystyle\quad+l_{0}(\xi)$

holds.

holds. Here $m(\xi)$ is a in (31) and $K$ given in (35). Furthermore, if we
consider $P:M\longrightarrow\mathbb{R}$ any smooth function, then, the
identity

(40) $\displaystyle\int_{0}^{T}\int_{\Gamma}\partial(A\xi,p\xi)d\Gamma dt$
$\displaystyle=\int_{0}^{T}\int_{M}p\left[J(\xi,\xi)-|\xi|^{2}\right]dMdt$
$\displaystyle-\int_{0}^{T}\left(a(x)\xi_{t},p\xi\right)_{L^{2}}dt+l_{0}(\xi)$

holds.

Proof: As before we can use (32) and (23) to obtain from equation (24) the
identity

(41)
$\displaystyle\left\langle\xi_{tt},2m(\xi)\right\rangle_{L^{2}(M,\lambda)}+\tilde{J}(\xi,2m(\xi)\rangle$
$\displaystyle-\iint_{\Gamma}\partial(A\xi,2m(\xi))d\Gamma=-\left\langle
a(x)|\xi_{t}|,2m(\xi)\right\rangle.$

Using (30), (31) we obtain after taking inner product, integrate over $M$ and
use Green formula

(42)
${}-2\left\langle\xi_{t},m\left(\xi_{t}\right)\right\rangle_{L^{2}}=2\int_{M}v\left|\xi_{t}\right|^{2}dM-\int_{\Gamma=\partial
M}\left|\xi_{t}\right|^{2}\langle V,\eta\rangle d\Gamma$

Substitution of (42) into (41) gives us

(43) $\displaystyle\left\langle\xi_{tt},2m(\xi)\right\rangle_{L^{2}}$
$\displaystyle=2\frac{\partial}{\partial
t}\left\langle\xi_{t},m(\xi)\right\rangle_{L^{2}}+2\int_{M}v\left|\xi_{t}\right|^{2}dM$
$\displaystyle-\int_{\Gamma=\partial M}\left|\xi_{t}\right|^{2}\langle
V,\eta\rangle d\Gamma$

Using Green formula for operator $A$ and relation (34) we deduce

(44) $\displaystyle\langle A\xi,2m(\xi)\rangle_{L^{2}}$
$\displaystyle=\int_{\Gamma=\partial M}[\boldsymbol{J}(\xi,\xi)\langle
V,\eta\rangle-2\partial(\boldsymbol{A}\xi,m(\xi))]d\Gamma$
$\displaystyle-2\int_{M}v\boldsymbol{J}(\xi,\xi)dM+2\int_{M}K(\xi,\xi)dM+l_{0}(\xi)$

Using identities (43), (44) follow (40). We need some geometric hypothesis in
orde to obtain the desired stabilization result in case of localized internal
dissipation.

######  Definition B.

Let $V$ a vector field on $M$ that is $V\in X(M)$. We say that $V$ is an
escape vector field for the Naghdi shell if the following conditions are
satisfy:

  * a)

There exists a function $v$ on $M$ such that

$DV(X,X)=v(x)|X|^{2}$

for all $X\in T_{x}(M),x\in M$

  * b)

Let $\varepsilon(x)$ denote the “volume element of the middle surface M",
consider

$f(x)=\frac{\langle DV,\varepsilon\rangle}{2}\quad\text{for}\,\,\,x\in M$

The functions $v(x)$ and $l(x)$ are assume to satisfy the inequality

$2\min_{x\in M}r(x)>\lambda_{0}(1+2\beta)\max_{x\in M}|\ell(x)|$

where $\lambda_{0}\geqslant 1$ satisfies (28) with
$\lambda_{0}^{-1}=\frac{c}{4}$ and $\beta=\frac{\mu}{1-2\mu}$

##  3\. Some comments on scape vector fields:

  * 1.

It is well known that on a $2$-dimensional middle surface $M$ there always
exists a vector freed $V$ satisfying assumption a). To obtain assumption b)
may be the difficult part. It is known that a necessary condition to hold
condition b) is that there is no closed geodesics inside the middle surface
$M$.

  * 2.

Condition b) says in a sense that function $l(x)$ is related to the symmetry
of the covariant differential DV. In fact, if $DV$ is symmetric then $l(x)=0$
for all $x\in M$.

  * 3.

In our case $M$ is an oriented Riemsannian manifold of
$\operatorname{din}M=2$. Let $\left\\{e_{1},e_{2}\right\\}\in T_{x}(M)$ be
linearly independent and let $\varepsilon$ be a differential form of degree 2
defined as

$\displaystyle\varepsilon\left(e_{1},e_{2}\right)(x)$
$\displaystyle=\pm\sqrt{\operatorname{det}\left(\left\langle
e_{i}e_{j}\right\rangle\right)}$
$\displaystyle=\text{orient.vol}\\{e_{1},e_{2}\\}$

$x\in M$. The oriented volume is affected by the sign $+$ or $-$ depending on
wheter or not a basis $\\{e_{1},e_{2}\\}$ belongs to the orientation on $M$.
$\varepsilon=\varepsilon(x)$ is called the volume element of $M$. Some texts
define the volume element as a $2-$ form on $M$ if for any frame field
$\\{e_{1},e_{2}\\}$, $|\varepsilon(e_{1},e_{2})|=1$. In chapter $4$ of [35]
several examples are presented where we can assure the construction of escape
vector fields for a shallow shells.

Next, we define the motion of scape region a piece $\overline{M}$ of the
middle surface $M$ wich will be convenient in order to obtain the desired
stabilization result. We are interested in case where $\overline{M}$ is not
the whole $M$ and maybe small as possible (in some sense). The notion was used
by several authors in related work (see [35], [30], and references therein)

######  Definition C.

A region $G\subset\Omega$ is called a scape region for the Naghdi shell, if

  1. $1)$

There is a finite number of sub-regions
$\left\\{\Omega_{i}\right\\}_{i=1}^{J}$, with boundary $\Gamma_{i}$, $J$ a
natural positive, such that

$\Omega_{i}\cap\Omega_{j}=\phi\quad\mbox{para todo}\quad 1\leq i<j\leq J.$

  2. $2)$

For each $\Omega_{i}$ there is a vector field $V_{i}$ and a function $v_{i}$
such that,

$\displaystyle DV_{i}(X,X)$ $\displaystyle=$ $\displaystyle
v_{i}(x)|X|^{2}\quad\mbox{on}\quad\Omega_{i}$ $\displaystyle
2\min_{x\in\Omega_{i}}v_{i}(x)$ $\displaystyle>$
$\displaystyle\lambda_{0}(1+2\beta)\max_{x\in\Omega_{i}}\frac{|l_{i}(x)|}{2},$

where $l_{i}(x)=\frac{\left<DV_{i},E\right>}{2}$ for all $1\leq i\leq J$;

  3. $3)$

$G\supset\bar{\Omega}\cap
N_{\epsilon}\left[\cup_{i=1}^{J}\Gamma_{i0}\cup\left(\Omega\setminus\cup_{i=1}^{J}\Omega_{i}\right)\right]$

where $\epsilon>0$, small, and:

$\displaystyle N_{\epsilon}(S)$ $\displaystyle=$ $\displaystyle\cup_{x\in
S}\left\\{y\in\Omega/d_{g}(y,x)<\epsilon\right\\}\quad{para}\quad S\subset M$
$\displaystyle\Gamma_{i0}$ $\displaystyle=$
$\displaystyle\left\\{x\in\Gamma_{i},\left<V_{i}(x),\nu(x)\right>>0\right\\},$

$\nu_{i}$ is the normal to $\Omega_{i}$ pointing out.

In general, there is no defined scape vector field over the entire median
surface $\Omega$. However, such fields can be defined on small geodesic balls.
Then, considering $\Omega=\cup_{n\in I\\!\\!N}B(x_{n},\delta)$ with
$x_{n}\in\Omega$ and $\delta>0$ small enough, an escape vector field can be
defined in $B(x_{n},\delta)$. Then
$\mu(\Omega)=\lim_{k\rightarrow\infty}\sum_{n=1}^{k}\mu(B(x_{n},\delta))$,
where $\mu$ is the two-dimensional Lebesgue measure on the surface $\Omega$.
So, given $\epsilon>0$, there is $N\in I\\!\\!N$ big enough that

$\sum_{\infty}^{n=N+1}\mu(B(x_{n},\delta))<\epsilon.$

Then considering $\Omega_{i}=B(x_{i},\delta)$ con $1\leq i\leq N$, we have
proved the following

######  Theorem D.

For $\epsilon>0$ given, an escape region can be chosen $G\subset\bar{\Omega}$
such that

$\mu(G)<\epsilon$

where $\mu(G)$ is the two-dimensional Lebesgue measure of $G$

Now we will give some examples.

######  Exemplo 3.1.

In the case of an escape vector field, $V$, defined on all $\Omega$, then in
the definition (C) we have to $J=1$. By condition($3$) a region escape is
supported in the boundary region $\Gamma_{0}$, where

$\Gamma_{0}=\left\\{x\in\partial\Omega/\left<V(x),\nu(x)\right>>0\right\\}.$

That escape region was already used by many authors[11], [6], [25]. The escape
field considered, in the case $I\\!\\!R^{n}$, was $V=x-x_{0}$

######  Exemplo 3.2.

consider now

$\Omega=C=\left\\{x=(x_{1},x_{2},x_{3})\in
I\\!\\!R^{3}/x_{1}^{2}+x_{2}^{2}=1,\quad-1\leq x_{3}\leq 1\right\\},$

the cylinder limited in $I\\!\\!R^{3}$. it is known that is not possible to
define a vanishing vector field over all $\Omega$. To construct an escape
region, let $x_{0}=\left(x_{01},x_{02},x_{03}\right)\in C$, with $x_{03}=0$
and either $L_{0}$ the generating line containing $x_{0}$. Be $x_{1}\in C$ the
antipode of $x_{0}$. Since the interior of the cut-locus of $x_{1}$ is
$C\setminus\setminus L_{0}$, there exists an escape vector field defined over
$C\setminus L_{0}$. Thus, an escape region for $\Omega$ is supported in a
neighborhood of the edge of $\Omega$ and the $L_{0}$.

##  4\. Naghdi shell stabilization with internal dissipation

To continue with the resolution of the problem of the exponential decay of
energy, we need the following lemmas.

######  Lema E.

Let $V\in\chi(\Omega)$ satisfying the first condition of the definition (B).
So the tensor field $DV$ can be decomposed as

$DV=v(x)g+l(x)E\quad\mbox{para}\quad x\in\Omega$

###### Proof.

decomposing $DV$ in its symmetric and antisymmetric part, by

(45)
${}DV=\frac{1}{2}\left(DV+D^{*}V\right)+\frac{1}{2}\left(DV-D^{*}V\right)$

Given that $\frac{1}{2}\left(DV-D^{*}V\right)$ is a $2$-antisymmetric shape
and $\Omega$ is $2$-dimension, there is a function $q$ such that

(46) ${}\frac{1}{2}\left(DV-D^{*}V\right)=q(x)E\quad\mbox{para}\quad x\Omega.$

Because the dimension of the $2$-shapes defined over spaces $2$-dimensionais
is $1$. substituting (46) in the expression of $l(x)$, we have

$\displaystyle l(x)$ $\displaystyle=$
$\displaystyle\frac{1}{2}\left<DV,E\right>=\frac{1}{2}\left<2q(x)E+D^{*}V,E\right>$
$\displaystyle=$
$\displaystyle\left<q(x)E,E\right>+\frac{1}{2}\left<D^{*}V,E\right>=2q(x)+\frac{1}{2}\left<D^{*}V,E\right>$
$\displaystyle=$ $\displaystyle 2q(x)-\frac{1}{2}\left<DV,E\right>$
$\displaystyle=$ $\displaystyle 2q(x)-l(x),$

de dónde,

(47) ${}l(x)=q(x).$

substituting (47) in (46) and using that $DV=vg$ we have the result. ∎

Another notation to fix is the following. Given $W\in\chi(\Omega)$ e $T\in
T^{2}(\Omega)$, be $G(W,T)\in T^{2}(\Omega)$ given by

(48) ${}G(W,T)=\frac{1}{2}\left[T(.,\nabla_{.}W)+T^{*}(.,\nabla_{.}W)\right]$

Now we prove some necessary lemmas

######  Lema F.

There exists a constant $c>0$ such that

(49) ${}||DW+D^{*}W||_{L^{2}(\Omega,T^{2})}\geq
c||W||_{H^{1}(\Omega,\Lambda)}\quad\forall W\in
H_{\Gamma_{0}}^{1}(\Omega,\Lambda)$

We note that, for $W\in H_{\Gamma_{0}}^{1}(\Omega,\Lambda)$

$\displaystyle b(S(W),S(W))$ $\displaystyle=$
$\displaystyle\left<S(W),S(W)\right>+\left(\mbox{Tra\c{c}}(S(W))\right)^{2}$
$\displaystyle=$
$\displaystyle\frac{1}{4}|DW+D^{*}W|^{2}+\left(\mbox{Tra\c{c}}(S(W))\right)^{2}$

Then,

$b(S(W),S(W))+|W|^{2}\geq\frac{1}{4}|DW+D^{*}W|^{2}.$

From where, integrating in $\Omega$ and using the lemma (49)

$\displaystyle{}\int_{\Omega}\left[b(S(W),S(W))+|W|^{2}\right]dx$
$\displaystyle=$
$\displaystyle\frac{1}{4}||DW+D^{*}W||^{2}_{L^{2}(\Omega,T^{2})}\geq\frac{c}{4}||W||^{2}_{H^{1}(\Omega,\Lambda)}$
$\displaystyle\geq$
$\displaystyle\frac{c}{4}||DW||^{2}_{L^{2}(\Omega,\Lambda)}$ (51)
$\displaystyle\lambda_{0}\int_{\Omega}\left[b(S(W),S(W))+|W|^{2}\right]dx$
$\displaystyle\geq$ $\displaystyle||DW||^{2}_{L^{2}(\Omega,\Lambda)}$

with $\lambda_{0}=\frac{4}{c}$.

######  Lema G.

Be $V$ an escape vector field for the Naghdi shell model and let $G(V,DW)\in
T^{2}(\Omega)$ given by (48), for $W\in H^{1}(\Omega,\Lambda)$. Then,

$\sigma_{1}\int_{\Omega}b\left(S(W),S(W)\right)dx\leq\int_{\Omega}b\left(S(W),G(V,DW)dx+Lo(\xi)\right).$

where
$\sigma_{1}=\min_{x\in\Omega}v(x)-(1+2\beta)\max_{x\in\Omega}\frac{|l(x)|}{2}$.

###### Proof.

Remembering that, for $T_{1},T_{2}\in T^{2}(\Omega)$, we have:

$b(T_{1},T_{2})=\left<T_{1},T_{2}\right>+\beta\mbox{Tra\c{c}}T_{1}\mbox{Tra\c{c}}T_{2}.$

Then,

(52)
${}b(S(W),G(V,DW))=\left<S(W),G(V,DW)\right>+\mbox{Tra\c{c}}(S(W))\mbox{Tra\c{c}}(G(V,DW))$

Now let estimate each term of (52). For this, given $x\in\Omega$, be
$\\{e_{1},e_{2}\\}$ an orthonormal basis of $T_{x}\Omega$ such that

(53) ${}DW(e_{1},e_{2})+D^{*}W(e_{1},e_{2})=0\quad\mbox{em}\quad x$

This is possible since the order tensor $2$, $DW+D^{*}W$, it is symmetrical.
It follows that,

(54) ${}S(W)(e_{1},e_{2})=0.$

Considering $W_{ij}=DW(e_{i},e_{j})$, have

(55) $\displaystyle{}\mbox{Tra\c{c}}(S(W))$ $\displaystyle=$ $\displaystyle
S(W)(e_{1},e_{2})+S(W)(e_{1},e_{2})$ $\displaystyle=$ $\displaystyle
DW(e_{1},e_{2})+DW(e_{2},e_{1})$ $\displaystyle=$ $\displaystyle
W_{1}1+W_{2}2$ (56) $\displaystyle{}\mbox{Tra\c{c}}G(V,DW)$ $\displaystyle=$
$\displaystyle G(V,DW)(e_{1},e_{2})+G(V,DW)(e_{2},e_{2})$ $\displaystyle=$
$\displaystyle DW(e_{1},\nabla_{e_{1}}V)+DW(e_{2},\nabla_{e_{2}}V)$

Now, using the lemma (E), have

(57) $\displaystyle{}\nabla_{e_{1}}V$ $\displaystyle=$
$\displaystyle\left<\nabla_{e_{1}}V,e_{1}\right>e_{1}+\left<\nabla_{e_{1}}V,e_{2}\right>e_{2}$
$\displaystyle=$ $\displaystyle DV(e_{1},e_{2})e_{1}+DV(e_{2},e_{1})e_{2}$
$\displaystyle=$ $\displaystyle v(x)e_{1}-l(x)e_{2}$

(58) $\displaystyle{}\nabla_{e_{2}}V$ $\displaystyle=$
$\displaystyle\left<\nabla_{e_{2}}V,e_{1}\right>e_{1}+\left<\nabla_{e_{2}}V,e_{2}\right>e_{2}$
$\displaystyle=$ $\displaystyle DV(e_{1},e_{2})e_{1}+DV(e_{2},e_{2})e_{2}$
$\displaystyle=$ $\displaystyle l(x)e_{1}+v(x)e_{2}$

substituting (58), (57) en (56), have

$\displaystyle\mbox{Tra\c{c}}G(V,DW)$ $\displaystyle=$ $\displaystyle
DW(e_{1},v(x)e_{1}-l(x)e_{2})+DW(e_{2},l(x)e_{1}+v(x)e_{2})$ $\displaystyle=$
$\displaystyle
v(x)DW(e_{1},e_{1})-l(x)DW(e_{1},e_{2})+l(x)DW(e_{2},e_{1})+v(x)DW(e_{2},e_{2})$

que, por (53), tenemos

(59) $\displaystyle{}\mbox{Tra\c{c}}G(V,DW)$ $\displaystyle=$ $\displaystyle
v(x)(W_{11}+W_{22})+2l(x)W_{21}$ $\displaystyle=$ $\displaystyle
v(x)\mbox{Tra\c{c}}(DW)+2l(x)W_{21}$

substituting (54), (55) and (59) in (52), obtain:

$\displaystyle b(S(W),G(V,DW))$ $\displaystyle=$
$\displaystyle\beta\mbox{Tra\c{c}}DW\left(v(x)\mbox{Tra\c{c}}DW+2l(x)W_{21}\right)$
$\displaystyle=$ $\displaystyle\beta
v(x)\left(\mbox{Tra\c{c}}DW\right)^{2}+2\beta
l(x)\left(W_{11}+W_{22}\right)W_{21}$ $\displaystyle=$ $\displaystyle
v(x)b(S(W),S(W))+2\beta l(x)\left(W_{11}+W_{22}\right)W_{21}$
$\displaystyle\geq$
$\displaystyle\min_{x\in\Omega}v(x)b(S(W),S(W))-(1+2\beta)\max_{x\in\Omega}\frac{|l(x)|}{2}|DW|^{2}+Lo(\xi),$

Integrating this last equation into $\Omega$, have:

(60)
${}\int_{\Omega}b(S(W),G(V,DW))dx\geq\min_{x\in\Omega}\int_{\Omega}b(S(W),S(W))dx-(1+2\beta)\max_{x\in\Omega}\frac{|l(x)|}{2}\int_{\Omega}|DW|^{2}dx$

Finally, using (4) in (60), have:

$\displaystyle\int_{\Omega}b(S(W),G(V,DW))dx$ $\displaystyle\geq$
$\displaystyle\min_{x\in\Omega}\int_{\Omega}b(S(W),S(W))dx$ $\displaystyle-$
$\displaystyle\lambda_{0}(1+2\beta)\max_{x\in\Omega}\frac{|l(x)|}{2}\int_{\Omega}b(S(W),S(W))dx+Lo(\xi)$
$\displaystyle=$ $\displaystyle\sigma_{1}\int_{\Omega}b(S(W),S(W))dx+Lo(\xi)$

∎

######  Lema H.

$2\mathsf{B}(\xi,m(\xi))=\int_{\Gamma}\mathbf{B}(\xi,\xi)\left<V,\nu\right>d\Gamma-2\int_{\Omega}v\mathbf{B}(\xi,\xi)+2\int_{\Omega}e(\xi,\xi)dx+Lo(\xi)$

where $\mathsf{B}$ is the bilinear form given in (15) and

$e(\xi,\xi)=2b(S(W_{1}),G(V,DW_{1}))+2b(S(W_{2}),G(V,DW_{2}))+4v|\varphi(\xi)|^{2}+v|Dw_{2}|^{2}$

###### Proof.

by the formula (25) we have to estimate
$\Upsilon(\xi),\chi(m(\xi)),\varphi(m(\xi))$ e
$\left<Dw_{2},D(V(w_{2}))\right>$. We start with the first term,

(61)
${}\Upsilon(m(\xi))=\frac{1}{2}\left[D(\nabla_{V}W)+D^{*}(\nabla_{V}W)\right]+V(w_{1})\Pi$

We will use the Bochner technique. Be $x\in\Omega$ e
$\left\\{E_{i}\right\\}_{i=1}^{2}$ a normal referential in $x$, we have:

$\displaystyle D(\nabla_{V}W_{1})(E_{i},E_{j})$ $\displaystyle=$
$\displaystyle E_{j}\left(\left<\nabla_{V}W_{1},E_{i}\right>\right)$
$\displaystyle=$ $\displaystyle
E_{j}\left(DW_{1}(E_{i},V)\right)=D^{2}W_{1}\left(E_{i},V,E_{j}\right)+DW_{1}(E_{i},\nabla_{E_{j}}V)$
$\displaystyle=$ $\displaystyle
D^{2}W_{1}\left(E_{i},E_{j},V\right)+\left<R_{VE_{j}}W_{1},E_{i}\right>+DW_{1}\left(E_{i},\nabla_{E_{j}}V\right)$
$\displaystyle=$
$\displaystyle\nabla_{V}DW_{1}(E_{i},E_{j})+R(V,E_{j},W_{1},E_{i})+DW(E_{i},\nabla_{E_{j}}V)$

Where from,

(62)
${}D(\nabla_{V}W_{1})=\nabla_{V}DW_{1}+R(V,.,W_{1},.)+DW_{1}(.,\nabla_{.}V)$

Similarly,

(63)
${}D^{*}(\nabla_{V}W_{1})=\nabla_{V}D^{*}W_{1}+R(V,.,W_{1},.)+DW_{1}(.,\nabla_{.}V)$

and,

$\displaystyle V(w_{1}\Pi)$ $\displaystyle=$ $\displaystyle
V(w_{1})\Pi+w_{1}\nabla_{V}\Pi$

So

(64) ${}V(w_{1})\Pi=V(w_{1}\Pi)-w_{1}\nabla_{V}\Pi$

substituting (64), (63), (62) in (61), have

(65) $\displaystyle{}\Upsilon(m(\xi))$ $\displaystyle=$
$\displaystyle\frac{1}{2}\left\\{\nabla_{V}\left(DW_{1}+D^{*}W_{1}\right)+DW_{1}(.,\nabla_{.}V)+DW_{1}(\nabla_{.}V,.)\right\\}$
$\displaystyle+$ $\displaystyle R(V,.,W_{1},.)+V(w_{1}\Pi)-w_{1}\nabla_{V}\Pi$
$\displaystyle=$ $\displaystyle\nabla_{V}\Upsilon(\xi)+G(V,DW_{1})+Lo(\xi)$

where, $Lo(\xi)=R(V,.,W_{1},.)-w_{1}\nabla_{V}\Pi$. Continuing with the
following terms,

(66)
${}\chi(m(\xi))=\frac{1}{2}\left(D(\nabla_{V}W_{2})+D^{*}(\nabla_{V}W_{2})\right)+V(w_{2})\Pi-\sqrt{\gamma}\left(i(\nabla_{V}W_{1})D\Pi-V(w_{1})c\right)$

now estimating the terms of (66),

$\displaystyle\nabla_{V}\left(i(W_{1})D\Pi\right)(E_{i},E_{j})$
$\displaystyle=$ $\displaystyle
D\left(i(W_{1})D\Pi\right)(E_{i},E_{j},V)=V\left(i(W_{1})D\Pi(E_{i},E_{j})\right)$
$\displaystyle=$ $\displaystyle V\left(D\Pi(W_{1},E_{i},E_{j})\right)$
$\displaystyle=$ $\displaystyle
D(D\Pi)(W_{1},E_{i},E_{j},V)+D\Pi(\nabla_{V}W_{1},E_{i},E_{j})$
$\displaystyle=$ $\displaystyle
i(W_{1})\nabla_{V}D\Pi(E_{i},E_{j})+i(\nabla_{V}W_{1})D\Pi(E_{i},E_{j})$

Therefore,

(67)
${}i(\nabla_{V}W_{1})D\Pi=\nabla_{V}\left(i(W_{1})D\Pi\right)-i(W_{1})\nabla_{V}D\Pi$

substituting (62), (63), (64) y (67) in (66), have

(68) $\displaystyle{}\chi(m(\xi))$ $\displaystyle=$
$\displaystyle\frac{1}{2}\left\\{\nabla_{V}\left(DW_{2}+D^{*}W_{2}\right)+DW_{2}(.,\nabla_{.}V)+DW_{2}(\nabla_{.}V,.)\right\\}$
$\displaystyle+$ $\displaystyle
R(V,.,W_{2},.)+V(w_{2}\Pi)-w_{2}\nabla_{V}\Pi-\sqrt{\gamma}\nabla_{V}(i(W_{1})D\Pi)-\sqrt{\gamma}i(W_{1})\nabla_{V}D\Pi$
$\displaystyle-$
$\displaystyle\sqrt{\gamma}V(w_{1}c)-\sqrt{\gamma}w_{1}\nabla_{V}c$
$\displaystyle=$ $\displaystyle\nabla_{V}\chi(\xi)+G(V,DW_{2})+Lo(\xi)$

continuing with $\varphi(\xi)$,

(69)
${}\varphi(m(\xi))=\frac{1}{2}D(V(w_{1}))-i(\nabla_{V}W_{1})\Pi+\frac{1}{\sqrt{\gamma}}\nabla_{V}W_{2}$

Estimating the terms of (69),

$\displaystyle\left<D(V(w_{1})),E_{i}\right>$ $\displaystyle=$ $\displaystyle
E_{i}(V(w_{1}))=E_{i}(\left<Dw_{1},V\right>)$ $\displaystyle=$
$\displaystyle\left<\nabla_{E_{i}}Dw_{1},V\right>+\left<Dw_{1},\nabla_{E_{i}}V\right>$
$\displaystyle=$
$\displaystyle\left<\nabla_{V}Dw_{1},E_{i}\right>+Dw_{1}\left(\nabla_{E_{i}}V\right)$

Therefore,

(70) ${}D(V(w_{1}))=\nabla_{V}Dw_{1}+Dw_{1}\left(\nabla_{.}V\right)$

continuing,

$\displaystyle\left<\nabla_{V}\left(i(W_{1})\Pi,E_{i}\right)\right>$
$\displaystyle=$ $\displaystyle
D(i(W_{1})\Pi)(E_{i},V)=V(i(W_{1})\Pi(E_{i}))=V(\Pi(W_{1},E_{i}))$
$\displaystyle=$ $\displaystyle
D\Pi(W_{1},E_{i},V)+\Pi(\nabla_{V}W_{1},E_{i})$ $\displaystyle=$
$\displaystyle\nabla_{V}\Pi(W_{1},E_{i})+\left<i(\nabla_{V}W_{1})\Pi,E_{i}\right>$
$\displaystyle=$
$\displaystyle\left<i(W_{1})\nabla_{V}\Pi,E_{i}\right>+\left<i(\nabla_{V}W_{1})\Pi,E_{i}\right>$

Therefore,

(71)
${}i(\nabla_{V}W_{1})\Pi=\nabla_{V}\left(i(W_{1})\Pi\right)-i(W_{1})\nabla_{V}\Pi$

substituting (70), (71) in (69), have,

(72) $\displaystyle{}\varphi(m(\xi))$ $\displaystyle=$
$\displaystyle\frac{1}{2}\left(\nabla_{V}Dw_{1}+Dw_{1}\left(\nabla_{.}V\right)\right)-\nabla_{V}\left(i(W_{1})\Pi\right)+i(W_{1})\nabla_{V}\Pi+\frac{1}{\sqrt{\gamma}}\nabla_{V}W_{2}$
$\displaystyle=$
$\displaystyle\nabla_{V}\varphi(\xi)+\varphi(\xi)\left(\nabla_{.}V\right)+Lo(\xi)$

Now writing the equation (16), with $\eta=m(\xi)$, have

$\displaystyle{}\mathbf{B}(\xi,m(\xi))$ $\displaystyle=$ $\displaystyle
2\left<\Upsilon(\xi),\Upsilon(m(\xi))\right>+4\left<\varphi(\xi),\varphi(m(\xi))\right>+2\beta\left(\mbox{Tra\c{c}}(\Upsilon(\xi)+\frac{1}{\sqrt{\gamma}})w_{2}\right)$
$\displaystyle\left(\mbox{Tra\c{c}}(\Upsilon(m(\xi))+\frac{1}{\sqrt{\gamma}})V(w_{2})\right)+2\left<\chi(\xi),\chi(m(\xi))\right>+2\beta\mbox{Tra\c{c}}(\chi(\xi))\mbox{Tra\c{c}}(\chi(m(\xi)))$
$\displaystyle+$
$\displaystyle\left<Dw_{2},D(V(w_{2}))\right>+\frac{1}{\gamma}w_{2}V(w_{2})$

using (65), (68) and (72) Let estimate each term of (4).

$\displaystyle 2\left<\varphi(\xi),\varphi(m(\xi))\right>$ $\displaystyle=$
$\displaystyle
2\left<\varphi(\xi),\nabla_{V}\varphi(\xi)+\varphi(\xi)\left(\nabla_{.}V\right)+Lo(\xi)\right>$
$\displaystyle=$ $\displaystyle
2\left<\varphi(\xi),\nabla_{V}\varphi(\xi)\right>+2\left<\varphi(\xi),\varphi(\xi)\left(\nabla_{.}V\right)\right>+Lo(\xi)$
$\displaystyle=$ $\displaystyle
V(|\varphi(\xi)|^{2})+2\left<\varphi(\xi),\nabla_{\varphi(\xi)}V\right>+Lo(\xi)$
$\displaystyle=$ $\displaystyle
V(|\varphi(\xi)|^{2})+2DV(\varphi(\xi),\varphi(\xi))+Lo(\xi).$

From where, using (B), we have

(74)
${}2\left<\varphi(\xi),\varphi(m(\xi))\right>=V(|\varphi(\xi)|^{2})+2v|\varphi(\xi)|^{2}+Lo(\xi)$

Continuing with the terms of (4),

(75) $\displaystyle{}2\left<Dw_{2},D(V(w_{2}))\right>$ $\displaystyle=$
$\displaystyle
2\left<Dw_{2},\nabla_{V}Dw_{2}+Dw_{2}\left(\nabla_{.}V\right)\right>$
$\displaystyle=$ $\displaystyle
2\left<Dw_{2},\nabla_{V}Dw_{2}\right>+2\left<Dw_{2},\nabla_{Dw_{2}}V\right>$
$\displaystyle=$ $\displaystyle V(|Dw_{2}|^{2})+2v|Dw_{2}|^{2}$

and,

(76) $\displaystyle{}2\left<\Upsilon(\xi),\Upsilon(m(\xi))\right>$
$\displaystyle=$
$\displaystyle\left<\Upsilon(\xi),\nabla_{V}\Upsilon(\xi)+G(V,DW_{1})+Lo(\xi)\right>$
$\displaystyle=$ $\displaystyle
V(|\Upsilon(\xi)|^{2})+2\left<\Upsilon(\xi),G(V,DW_{1})\right>+Lo(\xi)$

Continuing with the rest of the terms,

(77) $\displaystyle
2\beta\left(\mbox{Tra\c{c}}\Upsilon(\xi)+\frac{1}{\sqrt{\gamma}}w_{2}\right)\left(\mbox{Tra\c{c}}\Upsilon(m(\xi))+\frac{1}{\sqrt{\gamma}}V(w_{2})\right)=2\beta\mbox{Tra\c{c}}\Upsilon(\xi)\mbox{Tra\c{c}}\Upsilon(m(\xi))$
$\displaystyle+$
$\displaystyle\frac{2\beta}{\sqrt{\gamma}}\mbox{Tra\c{c}}\Upsilon(\xi)V(w_{2})+\frac{2\beta}{\sqrt{\gamma}}w_{2}\mbox{Tra\c{c}}\Upsilon(m(\xi))+\frac{2\beta}{\sqrt{\gamma}}w_{2}V(w_{2})$
$\displaystyle=$ $\displaystyle
2\beta\mbox{Tra\c{c}}\Upsilon(\xi)\left(\mbox{Tra\c{c}}\nabla_{V}\Upsilon(\xi)+\mbox{Tra\c{c}}G(V,DW_{1})+Lo(\xi)\right)+\frac{2\beta}{\sqrt{\gamma}}\mbox{Tra\c{c}}\Upsilon(\xi)V(w_{2})$
$\displaystyle+$
$\displaystyle\frac{2\beta}{\sqrt{\gamma}}w_{2}\mbox{Tra\c{c}}\Upsilon(m(\xi))+\frac{\beta}{\gamma}V(w_{2}^{2})$
$\displaystyle=$ $\displaystyle
2\beta\mbox{Tra\c{c}}\Upsilon(\xi)\mbox{Tra\c{c}}\nabla_{V}\Upsilon(\xi)+2\beta\mbox{Tra\c{c}}\Upsilon(\xi)\mbox{Tra\c{c}}G(V,DW_{1})+\frac{2\beta}{\sqrt{\gamma}}\mbox{Tra\c{c}}\Upsilon(\xi)V(w_{2})$
$\displaystyle+$
$\displaystyle\frac{2\beta}{\sqrt{\gamma}}w_{2}\mbox{Tra\c{c}}\Upsilon(m(\xi))+\frac{\beta}{\gamma}V(w_{2}^{2})$
$\displaystyle=$ $\displaystyle\beta
V\left(\left(\mbox{Tra\c{c}}\Upsilon(\xi)+\frac{1}{\gamma}w_{2}\right)^{2}\right)+2\beta\mbox{Tra\c{c}}\Upsilon(\xi)\mbox{Tra\c{c}}G(V,DW_{1})+Lo(\xi)$

and,

(78) $\displaystyle{}2\left<\chi(\xi),\chi(m(\xi))\right>$ $\displaystyle=$
$\displaystyle
2\left<\chi(\xi),\nabla_{V}\chi(\xi)+G(V,DW_{2})+Lo(\xi)\right>$
$\displaystyle=$ $\displaystyle
2\left<\chi(\xi),\nabla_{V}\chi(\xi)\right>+2\left<\chi(\xi),G(V,DW_{2})\right>+Lo(\xi)$
$\displaystyle=$ $\displaystyle
V\left(|\chi(\xi)|^{2}\right)+2\left<\chi(\xi),G(V,DW_{2})\right>+Lo(\xi)$

and,

$\displaystyle{}2\beta\mbox{Tra\c{c}}(\chi(\xi))\beta\mbox{Tra\c{c}}(\chi(m(\xi)))$
$\displaystyle=$ $\displaystyle
2\beta\mbox{Tra\c{c}}(\chi(\xi))\left(\mbox{Tra\c{c}}\nabla_{V}\chi(\xi)+\mbox{Tra\c{c}}G(V,DW_{2})+Lo(\xi)\right)$
$\displaystyle=$ $\displaystyle\beta
V\left(\left(\mbox{Tra\c{c}}\chi(\xi)\right)^{2}\right)+2\beta\mbox{Tra\c{c}}\chi(\xi)\mbox{Tra\c{c}}G(V,DW_{2})+Lo(\xi)$

substituting (4), (78), (LABEL:trauptramup), (76), (75) and (74) in (4), we
have

$\displaystyle\mathbf{B}(\xi,m(\xi))$ $\displaystyle=$ $\displaystyle
V\left(|\Upsilon(\xi)|^{2}\right)+2\left<\Upsilon(\xi),G(V,DW_{1})\right>+2V\left(|\varphi(\xi)|^{2}\right)+4v|\varphi(\xi)|^{2}$
$\displaystyle+$ $\displaystyle\beta
V\left(\left(\mbox{Tra\c{c}}\Upsilon(\xi)+\frac{1}{\sqrt{\gamma}}w_{2}\right)^{2}\right)+2\beta\mbox{Tra\c{c}}\Upsilon(\xi)\mbox{Tra\c{c}}G(V,DW_{1})$
$\displaystyle+$ $\displaystyle
2\beta\mbox{Tra\c{c}}\Upsilon(\xi)\mbox{Tra\c{c}}G(V,DW_{1})+V\left(|\chi(\xi)|^{2}\right)+2\left<\chi(\xi),G(V,DW_{2})\right>+\beta
V\left(\left(\mbox{Tra\c{c}}\chi(\xi)\right)^{2}\right)$ $\displaystyle+$
$\displaystyle
2\beta\mbox{Tra\c{c}}\chi(\xi)\mbox{Tra\c{c}}G(V,DW_{2})+\frac{1}{2}V\left(|Dw_{2}|^{2}\right)+v|Dw_{2}|^{2}$
$\displaystyle+$ $\displaystyle\frac{1}{\gamma}V(w_{2}^{2})+Lo(\xi)$
$\displaystyle=$
$\displaystyle\frac{1}{2}V\left(2|\Upsilon(\xi)|^{2}+4|\varphi(\xi)|^{2}+2\beta\left(\mbox{Tra\c{c}}\Upsilon(\xi)+\frac{1}{\sqrt{\gamma}}w_{2}\right)^{2}+2|\chi(\xi)|^{2}+2\beta\left(\mbox{Tra\c{c}}\chi(\xi)\right)^{2}\right)$
$\displaystyle+$
$\displaystyle\frac{1}{2}\left(|Dw_{2}|^{2}+\frac{1}{\gamma}w_{2}^{2}\right)+2\left<\Upsilon(\xi),G(V,DW_{1})\right>+4v|\varphi(\xi)|^{2}+2\beta\mbox{Tra\c{c}}\Upsilon(\xi)\mbox{Tra\c{c}}G(V,DW_{1})$
$\displaystyle+$ $\displaystyle
2\left<\chi(\xi),G(V,DW_{2})\right>+2\beta\mbox{Tra\c{c}}\chi(\xi)\mbox{Tra\c{c}}G(V,DW_{2})+v|Dw_{2}|^{2}+Lo(\xi)$
$\displaystyle=$
$\displaystyle\frac{1}{2}V\left(B(\xi,\xi)\right)+2b\left(\Upsilon(\xi),G(V,DW_{1})\right)+2b\left(\chi(\xi),G(V,DW_{2})\right)$
$\displaystyle+$ $\displaystyle 4v|\varphi(\xi)|^{2}+v|Dw_{2}|^{2}+Lo(\xi)$

Then,

(80) $\displaystyle{}2\mathsf{B}(\xi,m(\xi))$ $\displaystyle=$
$\displaystyle\int_{\Omega}\left[V\left(B(\xi,xi)\right)+4b\left(\Upsilon(\xi),G(V,DW_{1})\right)+b\left(\chi(\xi),G(V,DW_{2})\right)\right]dx$
$\displaystyle+$
$\displaystyle\int_{\Omega}\left[8v|\varphi(\xi)|^{2}+2v|Dw_{2}|^{2}+Lo(\xi)\right]$

Now, being $V=\sum_{i=1}^{2}V_{i}E_{i}$, where $\\{E_{1},E_{2}\\}$ is a normal
referential that varies with $x$. have

(81) $\displaystyle{}\int_{\Omega}V\left(B(\xi,\xi)\right)dx$ $\displaystyle=$
$\displaystyle\sum_{i=1}^{2}\int_{\Omega}V_{i}E_{i}(B(\xi,\xi))dx$
$\displaystyle=$
$\displaystyle\sum_{i=1}^{2}\int_{\Omega}E_{i}\left(V_{i}B(\xi,\xi)\right)dx-\sum_{i=1}^{2}\int_{\Omega}E_{i}(V_{i})B(\xi,\xi)dx$
$\displaystyle=$
$\displaystyle\int_{\Omega}\mbox{div}\left(B(\xi,\xi)V\right)dx-\int_{\Omega}\mbox{div}(V)B(\xi,\xi)dx$
$\displaystyle=$
$\displaystyle\int_{\Omega}B(\xi,\xi)\left<V,\nu\right>dx-2\int_{\Omega}vB(\xi,\xi)dx$

In the last line of (81) we will use (B). In fact,

$\mbox{div}(V)=\mbox{Tra\c{c}}DV=\sum_{i=1}^{2}DV(E_{i},E_{i})=\sum_{i=1}^{2}v|E_{i}|^{2}=2v$

Substituting (81) into (80), we have

$2\mathsf{B}(\xi,m(\xi))=\int_{\Gamma}B(\xi,\xi)dx-2\int_{\Omega}vB(\xi,\xi)dx+2\int_{\Omega}e(\xi,\xi)dx+Lo(\xi)$

with
$e(\xi,\xi)=2b\left(\Upsilon(\xi),G(V,DW_{1})\right)+b\left(\chi(\xi),G(V,DW_{2})\right)+4v|\varphi(\xi)|^{2}+v|Dw_{2}|^{2}$
and the lemma is . ∎

Other result need is the following

######  Theorem I.

Let $\xi=\left(W_{1},W_{2},w_{1},w_{2}\right)\in\mathsf{H}^{1}(\Omega)$
problem solution

(82) ${}\xi_{tt}+A\xi+a(x)\xi_{t}=0,\quad\mbox{em}\quad(0,T)\times\Omega$

Then

(83)
$\displaystyle\int_{\Sigma}\left[2\partial(A\xi,m(\xi))+\left(|\xi_{t}|^{2}-B(\xi,\xi)\right)\left<V,\nu\right>\right]d\Sigma$
$\displaystyle=2\left(\xi_{t},m(\xi)\right)|_{0}^{T}+2\int_{Q}v\left[|\xi_{t}|^{2}-B(\xi,\xi)\right]dQ$
$\displaystyle+2\int_{Q}e(\xi,\xi)dQ-\int_{0}^{T}\left(a\xi_{t},2m(\xi)\right)+L(\xi)$

where $\Sigma=(0,T)\times\Gamma$ and
$e(\xi,\xi)=e(\xi,\xi)=2b(S(W_{1}),G(V,DW_{1}))+2b(S(W_{2}),G(V,DW_{2}))+4v|\varphi(\xi)|^{2}+v|Dw_{2}|^{2}$
. $L(\xi)$denotes the terms of lower order with respect to energy.
Additionally, if $p$ is a function in $\Omega$, then

(84)
${}\int_{\Sigma}\partial\left(A\xi,p\xi\right)dQ=\int_{Q}p\left[B(\xi,\xi)-|\xi_{t}|^{2}\right]dQ-\int_{0}^{T}\left(a\xi_{t},p\xi\right)+L(\xi)$

###### Proof.

Multiplying the equation (82) by $2m(\xi)$ and integrating in $\Omega$ we
have:

$\displaystyle{}\left(\xi_{tt},2m(\xi)\right)_{\mathsf{L}^{2}(\Omega)}+\left(A\xi,2m(\xi)\right)$
$\displaystyle=$ $\displaystyle-\left(a\xi_{t},2m(\xi)\right)$ (85)
$\displaystyle\left(\xi_{tt},2m(\xi)\right)_{\mathsf{L}^{2}(\Omega)}+B(\xi,2m(\xi))-\int_{\Gamma}\partial\left(\xi,2m(\xi)\right)$
$\displaystyle=$ $\displaystyle-\left(a\xi_{t},2m(\xi)\right)$

Estimating the first term on the left hand side of (4), since the second term
was estimated in the lemma (H).

(86)
${}\left(\xi_{tt},2m(\xi)\right)_{\mathsf{L}^{2}(\Omega)}=2\left[\left(\xi_{t},m(\xi)\right)_{\mathsf{L}^{2}(\Omega)}\right]_{t}-2\left(\xi_{t},m(\xi_{t})\right)_{\mathsf{L}^{2}(\Omega)}$

and,

(87) $\displaystyle{}2\left(\xi_{tt},2m(\xi)\right)_{\mathsf{L}^{2}(\Omega)}$
$\displaystyle=$ $\displaystyle
2\left(\left(W_{1t},W_{2t},w_{1t},w_{2t}\right),\left(\nabla_{V}W_{1t},\nabla_{V}W_{2t},V(w_{1t}),V(w_{2t})\right)\right)_{\mathsf{L}^{2}(\Omega)}$
$\displaystyle=$ $\displaystyle
V\left(||W_{1t}||_{L^{2}(\Omega,\Lambda)}^{2}+||W_{2t}||_{L^{2}(\Omega,\Lambda)}^{2}+||w_{1t}||_{L^{2}(\Omega)}^{2}+||w_{2t}||_{L^{2}(\Omega)}^{2}\right)$
$\displaystyle=$ $\displaystyle
V(||\xi_{t}||_{L^{2}(\Omega)}^{2})=\int_{\Omega}V(|\xi_{t}|^{2})dx=\sum_{i=1}^{2}\int_{\Omega}V_{i}E_{i}(|\xi_{t}|^{2})dx$
$\displaystyle=$
$\displaystyle\sum_{i=1}^{2}\int_{\Omega}E_{i}\left(V_{i}|\xi_{t}|^{2}\right)dx-\sum_{i=1}^{2}\int_{\Omega}E_{i}(V_{i})|\xi_{t}|^{2}dx$
$\displaystyle=$
$\displaystyle\int_{\Gamma}|\xi_{t}|^{2}\left<V,\nu\right>d\Gamma-2\int_{\Omega}v|\xi_{t}|^{2}dx$

substituting (87) en (86), have

(88)
${}\left(\xi_{tt},2m(\xi)\right)_{\mathsf{L}^{2}(\Omega)}=2\left[\left(\xi_{t},m(\xi)\right)_{\mathsf{L}^{2}(\Omega)}\right]_{t}+2\int_{\Omega}v|\xi_{t}|^{2}dx-\int{\Gamma}|\xi_{t}|^{2}\left<V,\nu\right>d\Gamma$

substituting (88) and using the lema (H) in (4), have:

(89)
$\displaystyle{}2\left[\left(\xi_{tt},m(\xi)\right)_{\mathsf{L}^{2}(\Omega)}\right]_{t}$
$\displaystyle+$ $\displaystyle
2\int_{\Omega}v|\xi_{t}|^{2}dx-\int_{\Gamma}|\xi_{t}|^{2}\left<V,\nu\right>d\Gamma+\int_{\Gamma}B(\xi,\xi)\left<V,\nu\right>d\Gamma$
$\displaystyle-$ $\displaystyle
2\int_{\Omega}vB(\xi,\xi)dx+2\int_{\Omega}e(\xi,\xi)dx-\int_{\Gamma}\partial\left(\xi,2m(\xi)\right)d\Gamma+Lo(\xi)=$
$\displaystyle-$ $\displaystyle\left(a\xi_{t},2m(\xi)\right)$

integrating (89) of $0$ to $T$, have

$\displaystyle
2\left[\left(\xi_{tt},m(\xi)\right)_{\mathsf{L}^{2}(\Omega)}\right]|^{T}_{0}$
$\displaystyle+$ $\displaystyle
2\int_{Q}v|\xi_{t}|^{2}dx-\int_{\Sigma}|\xi_{t}|^{2}\left<V,\nu\right>d\Sigma+\int_{\Sigma}B(\xi,\xi)\left<V,\nu\right>d\Sigma$
$\displaystyle-$ $\displaystyle
2\int_{Q}vB(\xi,\xi)dx+2\int_{Q}e(\xi,\xi)dx-\int_{\Sigma}\partial\left(\xi,2m(\xi)\right)d\Sigma+Lo(\xi)=$
$\displaystyle\int_{0}^{T}\left(a\xi_{t},2m(\xi)\right)$

where,

$\displaystyle\int_{\Sigma}\left[2\partial(\xi,m(\xi))+\left(|\xi_{t}|^{2}-B(\xi,\xi)\right)\left<V,\nu\right>\right]d\Sigma$
$\displaystyle=$ $\displaystyle
2\left[\left(\xi_{t},m(\xi_{t})\right)_{\mathsf{L}^{2}(\Omega)}\right]|_{0}^{T}+2\int_{Q}v\left[|\xi_{t}|^{2}-B(\xi,\xi)\right]$
$\displaystyle+$ $\displaystyle
2\int_{Q}e(\xi,\xi)dQ-\int_{0}^{T}\left(a\xi_{t},2m(\xi)\right)+Lo(\xi)$

And the theorem is shown. ∎

In this section we state and prove the main result, the stabilization of the
evolution equation of the Naghdi model. We will need the following result.

######  Theorem J.

Let $V$ an escape vector field for the Naghdi shell. Let $\xi$ problem
solution.

(90) ${}\xi_{tt}+A\xi+a(x)\xi_{t}=0.$

Then for $T>0$,

(91)
${}\int_{\Sigma}SBd\Sigma+\lambda_{0}\sigma_{0}\left[E(0)+e(T)\right]-\int_{0}^{T}\int_{\Omega}a\left<\xi,\xi_{t}\right>\geq
2\sigma_{1}\int_{0}^{T}E(t)dt+L(\xi)$

where

(92) $\displaystyle SB$
$\displaystyle=\partial\left(A\xi,2m(\xi)+\rho\xi\right)+\left[\mid\xi_{t}\mid^{2}-B(\xi,\xi)\right]\left<V,\nu\right>$
$\displaystyle m(\xi)$
$\displaystyle=\left(\nabla_{V}W_{1},\nabla_{V}W_{2},V(w_{1}),V(w_{2})\right),\quad\rho=2v-\sigma_{1}$

###### Proof.

Let $p=\rho$ in the identity (84) and adding with the identity (84), obtain

(93) $\displaystyle\int_{\Sigma}SBd\Sigma$
$\displaystyle=2\left(\xi_{t},m(\xi)\right)_{(L^{2}(\Omega))}\mid_{0}^{T}+\sigma_{1}\int_{Q}\left[\mid\xi_{t}\mid^{2}+B(\xi,\xi)\right]dQ+$
$\displaystyle 2\int_{Q}e(\xi,\xi)dQ+L(\xi)$

and, by the expression of $B$, we have to

(94) $\displaystyle B(\xi,\xi)$ $\displaystyle=2b(S(W_{1}),S(W_{1}))+2\beta
b(S(W_{2}),S(W_{2}))+$ $\displaystyle 4\mid\varphi(\xi)\mid^{2}+\mid
Dw_{2}\mid+L(\xi)$

Using the lemma (G), have

(95) ${}\int_{Q}e(\xi,\xi)dQ\geq\sigma_{1}\int_{Q}B(\xi,\xi)dQ+L(\xi)$

using the identity (94), the coercivity of $b$, have

(96) $\displaystyle 2\left(\xi_{t},m(\xi)\right)_{L^{2}(\Omega)}$
$\displaystyle\leq\sigma_{0}\left[\parallel\xi_{t}\parallel^{2}_{L^{2}(\Omega)}+\sum_{i=1}^{2}\left(\parallel
DW_{i}\parallel^{2}_{L^{2}(\Omega,T^{2})}+\parallel
Dw_{i}\parallel^{2}_{L^{2}(\Omega,\Lambda)}\right)\right]$ $\displaystyle\leq
2\lambda_{0}\sigma_{0}E(t)+L(\xi)$

finally, substituting (96) and (95), in (93), we have inequality (91). ∎

We now state and prove our main result, the stabilization of the Naghdi
evolution model.

######  Theorem K.

Let be the evolution equation for the Naghdi shell model, with internal
dissipation

(97) $\displaystyle\xi_{tt}+A\xi+a(x)\xi_{t}$
$\displaystyle=0\quad\mbox{em}\quad\Omega\times(0,T)$ $\displaystyle\xi$
$\displaystyle=0\quad\mbox{em}\quad\partial\Omega\times(0,T)$

where the function $a=a(x)$ is supported in an escape region
$w\subset\bar{\Omega}$. Let $a_{0}>0$ such that

(98) ${}a(x)\geq a_{0}>0\quad\mbox{para todo}\quad x\in w$

So, there are constants $c_{1},c_{2}$ such that

(99) ${}E(t)\leq c_{1}E(0)\exp^{(-c_{2}t)}$

where $E(t)$ is the total energy of the system (97)

###### Proof.

multiplying the equation (97) for $\xi_{t}$, integrating into $\Omega$ and
considering the boundary conditions, we have that

(100) ${}\frac{d}{dt}E(t)=-\int_{\Omega}a(x)\mid\xi_{t}\mid^{2}dx$

from where, integrating into $(0,T)$, we have

(101) ${}E(T)=E(0)-\int_{\Omega}a(x)\mid\xi_{t}\mid^{2}dx.$

By (101) it is enough to prove that there is a $T>0$ e $C>0$, independent of
the solutions of the problem (97), such that

$E(T)\leq C\int_{Q}a\mid\xi_{t}\mid^{2}dQ,$

for in this case, substituting in (101), we have to

(102) ${}E(T)\leq\frac{C}{C+1}E(0)$

We affirm that (99) if follow from (102). In fact, notice that (102) is
equivalent

(103) ${}E(T)\leq\gamma E(0)$

where $\gamma=\frac{C}{C+1}<1$. Since the system is invariant by translations
in time, we have that (103)is valid in $[(m-1)T,mT]$, so

(104) $\displaystyle E(mT)$ $\displaystyle\leq\gamma E((m-1)T)$
$\displaystyle\leq\gamma^{2}E((m-2)T)$ $\displaystyle\vdots$
$\displaystyle\leq\gamma^{m}E(0)$ $\displaystyle\leq e^{-\omega mT}E(0)$

where $\omega=\frac{1}{T}\ln(\frac{1}{\gamma})>0$. For $t>0$ arbitrary, exists
$m=1,2,\dots$, such that $(m-1)T<t\leq mT$. Finally, using that the energy of
the system is decreasing, we have:

$\displaystyle E(t)$ $\displaystyle\leq$ $\displaystyle E((m-1)T)$
$\displaystyle\leq$ $\displaystyle e^{-\omega(m-1)T}E(0)$ $\displaystyle\leq$
$\displaystyle\frac{1}{\gamma}e^{-\omega mT}E(0)$ $\displaystyle\leq$
$\displaystyle\frac{1}{\gamma}e^{-\omega t}E(0)$

Which proves our statement. So, let try (102)

By the definition of escape fields and regions discussed, we can assume that
there are subsets of $\Omega$, $\\{\Omega_{i}\\}_{i=1}^{N}$, satisfying (B).
Then the identities (92) and (84) can be used on each $\Omega_{i}$, since
vectorial escape fields are defined on each of them.

The idea is to estimate the total energy of the system, first inside $\Omega$
using that on the $\Omega_{i}$ we have vanishing vector fields defined on them
and in the plugin use the property of the dissipation function, $a$.

Now, since (91) is valid only in $\Omega_{i}$, We are going to restrict
ourselves, first, to making estimates on them.

be then $0<\epsilon_{2}<\epsilon_{1}<\epsilon_{0}<\epsilon$ and be

$V_{j}=N_{\epsilon_{j}}\left\\{\cup_{i=1}^{N}\Gamma_{0}^{i}\cup\left(\Omega\setminus\cup_{i=1}^{N}\Omega_{i}\right)\right\\},\quad
j=0,1,2.$

notice that $V_{2}\subset V_{1}\subset V_{0}\subset\bar{V_{0}}$. be

$\phi^{i}=\left\\{\begin{array}[]{ccc}1,&\Omega_{i}\setminus
V_{i},&i=1,2,\dots,N\\\ 0,&V_{2}\end{array}\right.$

consider now $V^{i}=\phi^{i}H^{i}$, $p^{i}=\phi^{i}q$ and
$\xi^{i}=\phi^{i}\xi$ where $H^{i}$ is an escape vector field in $\Omega_{i}$
and $q$ a function defined in $\Omega$, respectively. So, in each $\Omega_{i}$
it is valid (91) and we have

(105) $\displaystyle
2\sigma_{1}\int_{0}^{T}E^{i}(t)\leq\int_{\Sigma_{i}}SB_{i}d\Sigma_{i}+\lambda_{0}\sigma_{0}\left[E^{i}(0)+E^{i}(T)\right]-\int_{0}^{T}\int_{\Omega_{i}}a\left<\xi^{i},\xi^{i}_{t}\right>dxdt$

where,

(106)
${}SB_{i}=\partial\left(A\xi^{i},2m(\xi^{i})+\rho\xi^{i}\right)+\left[\mid\xi^{i}_{t}\mid^{2}-B(\xi^{i},\xi^{i})\right]\left<V^{i},\nu\right>$

By the definition of $V_{2}$, we have to $\Gamma^{i}_{0}\subset V_{2}$ and
what $\phi^{i}=0$ in $V_{2}$, we have to $x\in\Gamma^{i}_{0}$, the terms at
the border are annulled. So

(107) ${}SB_{i}=0\quad\quad\mbox{para}\quad x\in\Gamma^{i}_{0}.$

If $x\in\Omega_{i}\cap\Gamma$, using boundary conditions, $\xi=0$, replacing
in (106) have

(108) $\displaystyle SB_{i}$
$\displaystyle=2\partial\left(A\xi_{i},m(\xi^{i})\right)-B(\xi^{i},\xi^{i})\left<V^{i},\nu\right>$
$\displaystyle=B(\xi^{i},\xi^{i})\left<V^{i},\nu\right>$ $\displaystyle\leq 0$

where we use the coercivity of $B$. It is concluded that

(109) ${}\int_{\Sigma_{i}}SB_{i}\leq 0\quad\mbox{para todo}\quad i$

Using the estimate (107) and (109) in (105) have

$\displaystyle
2\sigma_{1}\int_{0}^{T}\int_{\Omega_{i}}\mid\xi_{t}\mid^{2}+2\sigma_{1}\int_{0}^{T}\int_{\Omega_{i}}B(\xi,\xi)\leq-\int_{0}^{T}\left(a\xi_{t},2m(\xi)\right)+\lambda_{0}\sigma_{0}\left[E(T)+E(0)\right]+L(\xi)$

where,

(110) $\displaystyle\int_{0}^{T}\int_{\Omega_{i}\setminus V_{1}}B(\xi,\xi)$
$\displaystyle\leq C_{1}\int_{T}^{0}\int_{\Omega_{i}\cap
V_{1}}B(\xi,\xi)+C_{\beta}\int_{0}^{T}\int_{\Omega_{i}}\mid\xi\mid^{2}+\beta\int_{0}^{T}\int_{\Omega_{i}}B(\xi,\xi)dQ+$
$\displaystyle\lambda_{0}\sigma_{0}\left[E(T)+E(0)\right]+L(\xi)$

where $\beta>0$ is small enough that
$\beta\int_{0}^{T}\int_{\Omega_{i}}B(\xi,\xi)dxdt\leq\int_{0}^{T}\int_{\Omega_{i}\cap
V_{1}}B(\xi,\xi)dxdt$.

Given that $\Omega\subset\left(\cup\Omega_{i}\right)\cup V_{1}$, then

(111) $\displaystyle\int_{0}^{T}\int_{\Omega\setminus V_{1}}B(\xi,\xi)$
$\displaystyle\leq\sum_{i=1}^{N}\int_{0}^{T}\int_{\Omega_{i}\setminus
V_{1}}B(\xi,\xi)$ $\displaystyle\leq C_{2}\int_{0}^{T}\int_{\Omega\cap
V_{1}}B(\xi,\xi)dQ$
$\displaystyle+C_{3}\int_{0}^{T}a\parallel\xi_{t}\parallel^{2}dt+\lambda_{0}\sigma_{0}\left[E(T)+E(0)\right]+L(\xi)$

Now let us estimate in the complement of the union of the $\Omega_{i}$. For
this, be $\psi\in C^{\infty}(I\\!\\!R^{3})$ given by

(112) ${}\psi(x)=\left\\{\begin{aligned} 0&,\quad x\in I\\!\\!R^{3}\setminus
V_{0}\\\ 1&,\quad x\in V_{1}\end{aligned}\right.$

Considering $p=\psi$ en (84) have

$\int_{Q_{i}}B(\xi,\xi)dQ_{i}=\int_{Q_{i}}\mid\xi_{t}\mid^{2}dQ_{i}+\int_{0}^{T}\left(a\xi_{t},\psi\xi\right)dt+L(\xi)$

where from,

$\displaystyle\int_{0}^{T}\int_{\Omega\cap
V_{1}}B(\xi,\xi)dQ+\int_{0}^{T}\int_{\Omega\cap V_{0}}B(\xi,\xi)dQ$
$\displaystyle\leq\int_{0}^{T}\int_{\Omega\cap
V_{0}}B\parallel\xi_{t}\parallel^{2}dQ+\int_{0}^{T}\int_{\Omega\cap
V_{0}}\left<a\xi_{t},\xi\right>+L(\xi)$

And what $\epsilon_{0}<\epsilon$, then $w\supset\bar{\Omega}\cap V_{0}$. Using
the function hypothesis $a$ in $w$, have

$\displaystyle\int_{0}^{T}\int_{\Omega\cap V_{1}}B(\xi,\xi)dQ$
$\displaystyle\leq\int_{0}^{T}\int_{\Omega\cap
V_{0}}\parallel\xi_{t}\parallel^{2}dQ+\int_{0}^{T}\int_{\Omega\cap
V_{0}}\left<a\xi_{t},\xi\right>+L(\xi)$
$\displaystyle\leq\frac{1}{a_{0}}\int_{0}^{T}\int_{\Omega\cap
V_{0}}a\mid\xi\mid^{2}dt+L(\xi)+\beta\int_{0}^{T}\parallel\xi_{t}\parallel^{2}dt+L(\xi)$

where $\beta$ will be chosen later. so, of (111), (4), have

(113) $\displaystyle\int_{Q}B(\xi,\xi)$
$\displaystyle=\int_{0}^{T}\int_{\Omega\setminus
V_{1}}B(\xi,\xi)dxdt+\int_{0}^{T}\int_{\Omega\cap V_{1}}B(\xi,\xi)dxdt$
$\displaystyle\leq C_{4}\int_{0}^{T}\int_{\Omega\cap
V_{1}}B(\xi,\xi)dxdt+C_{5}\int_{0}^{T}a\parallel\xi_{t}\parallel^{2}dt+L(\xi)$
$\displaystyle\leq
C_{6}\int_{Q}a\mid\xi_{t}\mid^{2}dt+\beta\int_{0}^{T}\parallel\xi_{t}\parallel^{2}dt+\lambda_{0}\sigma_{0}\left[E(T)+E(0)\right]+L(\xi)$

now considering $p=\frac{1}{2}$ in (84), have

(114)
$\displaystyle\frac{1}{2}\int_{Q}\left[\mid\xi_{t}\mid^{2}-B(\xi,\xi)\right]dQ$
$\displaystyle=\frac{1}{2}\int_{0}^{T}\left(a\xi_{t},\xi\right)dt+L(\xi)$
$\displaystyle\leq C_{8}\int_{0}^{T}a\parallel\xi_{t}\parallel^{2}dt+L(\xi)$

grouping (113) and (114), have

(115) $\displaystyle\int_{0}^{T}E(t)dt$
$\displaystyle=\int_{Q}B(\xi,\xi)dQ+\frac{1}{2}\int_{Q}\left[\mid\xi_{t}\mid^{2}-B(\xi,\xi)\right]dQ$
$\displaystyle\leq
C\int_{Q}a\mid\xi_{t}\mid^{2}dQ+\beta\int_{0}^{T}\parallel\xi_{t}\parallel^{2}dt+\lambda_{0}\sigma_{0}\left[E(T)+E(0)\right]+C_{8}\int_{Q}a\mid\xi_{t}\mid
dQ+L(\xi)$ $\displaystyle\leq C_{9}\int_{Q}a\mid\xi_{t}\mid
dQ+\beta\int_{0}^{T}E(t)dt+C_{9}\left[E(T)+E(0)\right]+L(\xi)$

Taking $\beta=\frac{1}{2}$, considering that the energy is decreasing, this
is, $E(t)\geq E(T)$ for $t\in[0,T]$ and
$E(T)=E(0)-\int_{Q}a\mid\xi\mid^{2}dQ$, have

(116) $\displaystyle\frac{1}{2}\int_{0}^{T}E(t)dt$ $\displaystyle\leq
C_{9}\int_{Q}a\mid\xi_{t}\mid dQ+C_{9}\left[E(T)+E(0)\right]+L(\xi)$
$\displaystyle\frac{T}{2}E(T)$ $\displaystyle\leq
C_{9}\int_{Q}a\mid\xi_{t}\mid
dQ+\lambda_{0}\sigma_{0}\left[E(T)+E(T)+\int_{Q}a\mid\xi_{t}\mid
dQ\right]+L(\xi)$
$\displaystyle\left(\frac{T}{2}-2\lambda_{0}\sigma_{0}\right)$
$\displaystyle\leq C_{10}\int_{Q}a\mid\xi_{t}\mid dQ+L(\xi)$

For $T>4\lambda_{0}\sigma_{0}$ of (116) and the uniqueness-compactness
argument [35], have

$E(t)\leq C\int_{Q}a\mid\xi_{t}\mid dQ$

This proves the theorem. ∎

##  5\. Controllability via Stability

Russell Principle[32] provides a method for testing exact controllability
using the uniform stabilization result.

In this section we are going to study the exact controllability for the Naghdi
evolution system. For this, we apply Russell Principle using the result of
energy decay proved in the theorem K

The exact controllability problem, with controls inside, consists of finding a
vector function $F=F(x,t)$, control call, such that for some $T>0$ the
following problem has a solution:

(117) ${}\left\\{\begin{aligned}
&\eta_{tt}+A\eta=F(x,t)\quad\mbox{em}\quad\Omega\times(0,T)\\\
&\eta=0\quad\mbox{em}\quad\partial\Omega\times(0,T)\\\
&\eta(0)=\eta_{0},\quad\eta_{t}(0)=\eta_{1}\\\
&\eta(T)=\tilde{\eta}_{0},\quad\eta_{t}(T)=\tilde{\eta}_{1}\end{aligned}\right.$

for any initial data $(\eta_{0},\eta_{1})$ and endings
$(\tilde{\eta}_{0},\tilde{\eta}_{1})$ in the appropriate functional spaces and
$F$ acting in a subregion of $\Omega$.

We will prove that the function $F$ needs to act only in a sub-region of
$\Omega$ arbitrarily small, precisely in the escape region for the Naghdi
shell.

We consider $T>0$ such that

(118) ${}c_{1}e^{-c_{2}T}<1$

where $c_{1}>0$ and $c_{2}>0$ are the constants that appear in the theorem K.

In this section we will prove the following result:

######  Theorem L.

Let $\Omega$ the median surface of the Naghdi shell and $w\subset\Omega$ the
vanishing region given in the proof of the theorem K. Then, if $T>0$ satisfies
(118) the system (117) is exactly controllable at time $T$ with controls
located at $w$

###### Proof.

Since the system is linear and reversible in time, it is enough to consider
the controllability at zero, that is,
$(\tilde{\eta}_{0},\tilde{\eta}_{1})=(0,0)$

We have to, for $(\eta_{0},\eta_{1})\in V\times H$, there is only one solution
$\eta\in C\left([0,\infty);V\right)\times C^{1}\left([0,\infty);H\right)$ of
the problem.

(119) ${}\left\\{\begin{aligned}
&\eta_{tt}+A\eta+a(x)\eta_{t}=0\quad\mbox{em}\quad\Omega\times(0,T)\\\
&\eta=0\quad\mbox{em}\quad\partial\Omega\times(0,T)\\\
&\eta(0)=\eta_{0},\quad\eta_{t}(0)=\eta_{1}\\\ \end{aligned}\right.$

Additional, for the initial data $(-\eta(T),\eta_{t}(T))$ there is only one
solution to the problem.

(120) ${}\left\\{\begin{aligned}
&\theta_{tt}+A\theta+a(x)\theta_{t}=0\quad\mbox{em}\quad\Omega\times(0,T)\\\
&\theta=0\quad\mbox{em}\quad\partial\Omega\times(0,T)\\\
&\theta(0)=-\eta(T),\quad\theta_{t}(T)=\eta_{t}(T)\\\ \end{aligned}\right.$

where $a$ acts in the escape region given in the theorem K.

let define $\xi(x,t)=\eta(x,t)+\theta(x,T-t)$. The field $\xi$ satisfies

(121) ${}\left\\{\begin{aligned}
&\xi_{tt}+A\xi=a(x)(\eta_{t}+\theta_{t})\quad\mbox{em}\quad\Omega\times(0,T)\\\
&\xi=0\quad\mbox{em}\quad\partial\Omega\times(0,T)\\\
&\xi(0)=\eta_{0}+\theta(T),\quad\xi_{t}(0)=\eta_{1}-\theta_{t}(T)\\\
&\xi(T)=0,\quad\xi_{t}(T)=0\end{aligned}\right.$

Of (121) we have that the initial data that are brought to equilibrium have
the form

(122)
${}\left(\xi_{0},\xi_{1}\right)=\left(\eta_{0}+\theta(T),\eta_{1}-\theta_{t}(T)\right)$

Thus it is enough to prove that for each initial data
$\left(\xi_{0},\xi_{1}\right)\in V\times H$, exists
$\left(\eta_{0},\eta_{1}\right)$ satisfying (122). Equivalently, show that the
application:

(123) $\displaystyle L:$ $\displaystyle V\times H\longrightarrow V\times H$
$\displaystyle\left(\eta_{0},\eta_{1}\right)\longrightarrow\left(\eta_{0}+\theta(T),\eta_{1}-\theta_{t}(T)\right)$

is surjective. As $L=I-K$ where $K$ is the application given by

$K(\eta_{0},\eta_{1})=\left(-\theta(T),\theta_{t}(T)\right)$

it is enough to show that $\parallel K\parallel_{V\times H}<1$. Applying the
theorem twice K have

(124) $\displaystyle\parallel K(\eta_{0},\eta_{1})\parallel_{V\times H}$
$\displaystyle=\parallel\left(-\theta(T),\theta_{t}(T)\right)\parallel_{V\times
H}$ $\displaystyle\leq
c_{1}e^{-c_{2}T}\parallel\left(-\eta(T),\eta_{t}(T)\right)\parallel_{V\times
H}$ $\displaystyle\leq
c_{3}e^{-c_{4}T}\parallel\left(\eta(0),\eta_{1}\right)\parallel_{V\times H}.$

choosing $T>0$ such that $c_{3}e^{-c_{4}T}<1$ we have to $\parallel
K\parallel_{V\times H}<1$. Thus $L=I-K$ is a surjective application and
therefore the theorem L it is demonstrated. ∎

##  6\. conclusions

  * (a)

The dissipation can be considered in an arbitrarily small region of the shell
and obtain stabilization and controllability.

  * (b)

The existence of flight regions are verifiable for Naghdi shells, this implies
that localized dissipation in the complement of the union of said regions
generates stabilization and controllability.

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