#  Simultaneous controllability of wave sound propagations

Alexis Rodriguez Carranza<EMAIL_ADDRESS>Luis J. Caucha
<EMAIL_ADDRESS>Obidio Rubio Mercedes<EMAIL_ADDRESS>Universidad
Nacional de Trujillo, Perú  Universidad Nacional de Tumbes, Perú

###### Abstract

In this work we study one problem of mathematical interest for their
applications in several topics in Applied Science. We study simultaneous
controllability of a pair of systems which model the evolution of sound in a
compressible flow considered as a transmission problem. We show the well posed
of the problem. Furthermore provided appropriate conditions in the geometry of
the domain are valid and suitable assumptions on the fluid, is possible to
conduce the pair of systems to the equilibrium in a simultaneous way using
only one control.

###### keywords:

Controllability, wave sound, H.U.M

,

##  1 Introduction

In this work, we considered an equations system to describe an evolution of
the wave sound or compressible fluids. A linear model well know is given by a
system [12]

${}\left\\{\begin{aligned} &\frac{\partial u}{\partial t}+\alpha\nabla
p=0,\quad\mbox{in}\quad\Omega\times(0,T)\\\ &\frac{\partial p}{\partial
t}+\beta\mbox{div}(u)=0,\quad\mbox{in}\quad\Omega\times(0,T)\\\
&u.\eta=Q,\quad\mbox{in}\quad S_{0}\times(0,T)\\\ &p=0,\quad\mbox{in}\quad
S_{1}\times(0,T)\\\ &u(x,0)=u_{0}(x),p(x,0)=p_{0}(x)\end{aligned}\right.$ (1)

where $p=p(x,t)$ is acoustic precision, $u=(u_{1},u_{2},u_{3})$ and
$u_{j}=u_{j}(x,t)$ are fluid velocity field, $\alpha>0$ is the density of
equilibrium and $\beta>0$ is the compressibility factor of fluid. Here
$\Omega$ is an open subset of $I\\!\\!R^{3}$ with regularity boundary
conditions $S_{0}\cup S_{1}=\partial\Omega$ and $S_{0}\cap S_{1}=\emptyset$.
To solve the simultaneous controllability we considered a system given by

${}\left\\{\begin{aligned} &\frac{\partial v}{\partial t}+\gamma\nabla
q=0,\quad\mbox{in}\quad\Omega\times(0,T)\\\ &\frac{\partial q}{\partial
t}+\tau\mbox{div}(v)=0,\quad\mbox{in}\quad\Omega\times(0,T)\\\
&q=P,\quad\mbox{in}\quad S_{0}\times(0,T)\\\ &q=0,\quad\mbox{in}\quad
S_{1}\times(0,T)\\\ &v(x,0)=v_{0}(x),q(x,0)=q_{0}(x)\end{aligned}\right.$ (2)

where $\gamma>0$ and $\tau>0$. $Q$ and $P$ in (1) and (2), respectively; these
are control functions. In $1986$, D.L. Russell[20] and J.L.Lions [13] proposed
to solve a exact controllability problem for an evolution model, using only
one control function. They called that problem as simultaneous
controllability. The absences of dissipative effects as in (1) and (2), the
problem present difficulties for the solution, see the examples [7], [8], [10]
and [13], where they perturbed the multipliers used for the controllability.
The problem of simultaneous controllability for the systems (1) and (2) is to
take a control for both of system using only one control function, i.e., given
$T>0$ any initial condition, $(u_{0},p_{0},v_{0},q_{0})$, and final
$(\tilde{u}_{0},\tilde{p}_{0},\tilde{v}_{0},\tilde{q}_{0})$ in appropriate
functional space, find $P(x,t)$ and $Q(x,t)$ such that

  1. a)

A solution $\\{u,p,v,q\\}$ of (1) and (2) satisfied in $T$

$(u(.,T),p(.,T),v(.,T),q(.,T))=(\tilde{u}_{0},\tilde{p}_{0},\tilde{v}_{0},\tilde{q}_{0})$

  2. b)

The control function, $P(x,t)$, for (2) was given in terms of $Q(x,t)$.

A method to solve the controllability problem is Hilbert Uniqueness Method
(H.U.M) proposed by J.L.Lions, it is a construction of an appropriate
structure for the Hilbert space in the initial conditions space.
These structure are connected by uniqueness properties. An important
contribution to the controllability problems (1) and (2) were made by
Kapitonov et. G. Perla Menzala [8], [10]. In [10] and [8] the author answered
positively for a simultaneous control and They showed that the control
$P=-\frac{\beta}{\gamma}Q$ could be use to solve a problem. In this work we
study a controllability problem of these systems with a perspective for
applications as a problem of transmission; this is described below.

Given $\sigma_{0}$ and $\sigma_{1}$ open limited subset and conexo in
$i\\!\\!r^{3}$, with $\bar{\sigma}_{1}\subseteq\sigma_{0}$. Also
$\omega=\sigma_{0}\setminus\bar{\sigma}_{1}$, we denoted
$\partial\sigma_{0}=s_{0},\quad\partial\sigma_{1}=s_{1}$. And fixed an integer
$m>1$ and $k=1,2,\dots,m$. For each $k$, $b_{k}$ is an open subset and conexo,
with regularity in the boundary such that, $\bar{\sigma}_{1}\subseteq
b_{k}\subseteq\sigma_{0}$, $\bar{b}_{k}\subseteq b_{k+1}$. We put
$\omega_{0}=b_{1}\setminus\bar{\sigma}_{1}$,
$\omega_{k}=b_{k+1}\setminus\bar{b}_{k}$, $k=1,2,\dots,m-1$ and
$\omega_{m}=\sigma_{0}\setminus\bar{b}_{m}$. and,
$\omega=\cup^{m}_{j=0}\omega_{j}$, for $i\neq j$, we take
$\omega_{i}\cap\omega_{j}=\emptyset$ and $\partial\omega=s_{0}\cup s_{1}$.
Examples for this decomposition is showed in

Figure 1: Case $m=0$ and $m=3$

We need a solution defined by part on each sub domain; for that, we considered
the systems (1) and (2) rewrite on sub domains $\Omega_{k}$, and

${}\left\\{\begin{array}[]{ccc}\frac{\partial u^{k}}{\partial
t}+\alpha^{k}\nabla p^{k}&=&0,\quad\mbox{in}\quad\Omega_{k}\times(0,T)\\\
\frac{\partial p^{k}}{\partial
t}+\beta^{k}\mbox{div}(u^{k})&=&0,\quad\mbox{in}\quad\Omega_{k}\times(0,T)\\\
u^{k}(x,0)=&u^{k}_{0}(x),&p^{k}(x,0)=p^{k}_{0}(x)\end{array}\right.$ (3)

${}\left\\{\begin{array}[]{ccc}\frac{\partial v^{k}}{\partial
t}+\gamma^{k}\nabla q^{k}&=&0,\quad\mbox{in}\quad\Omega_{k}\times(0,T)\\\
\frac{\partial q^{k}}{\partial
t}+\tau^{k}\mbox{div}(v^{k})&=&0,\quad\mbox{in}\quad\Omega_{k}\times(0,T)\\\
v^{k}(x,0)=&v^{k}_{0}(x),&q^{k}(x,0)=q^{k}_{0}(x)\end{array}\right.$ (4)

$k=0,1,2,\dots,m$.

with boundary conditions (1) and (2). The interfaces of transmission
conditions $\Gamma_{k}=\partial\Omega_{k}$, given by

${}\left\\{\begin{array}[]{ccc}\alpha^{k-1}p^{k-1}&=&\alpha^{k}p^{k}\\\
\beta^{k-1}(u^{k-1}.\eta)&=&\beta^{k}(u^{k}.\eta)\\\
k=2,\dots,m,&(x,t)\in\Gamma_{k}\times(0,T)\end{array}\right.$ (5)

${}\left\\{\begin{array}[]{ccc}\gamma^{k-1}q^{k-1}&=&\gamma^{k}q^{k}\\\
\tau^{k-1}(v^{k-1}.\eta)&=&\tau^{k}(v^{k}.\eta)\\\
k=2,\dots,m,&(x,t)\in\Gamma_{k}\times(0,T)\end{array}\right.$ (6)

for the systems (3) and (4), respectively.
The functions $\alpha^{k},\beta^{k},\gamma^{k}$ and $\tau^{k}$ are the
restriction for the functions $\alpha,\beta,\gamma,\tau$ on the systems (1)
and (2), we assumed that those functions were constant by parts, strictly
positive and we lost the continuity only in $\Gamma_{k}$, $k=1,2,\dots,m$.
The objective in this section is to get the estimation of

$\displaystyle(T-T_{0})\sum_{k=0}^{m}\int_{\Omega_{k}}\left[\beta^{k}\mid
u^{k}\mid^{2}+\alpha^{k}(p^{k})^{2}+\tau^{k}\mid
v^{k}\mid^{2}+\gamma^{k}(q^{k})^{2}\right]dx$ (7) $\displaystyle\leq
C\int_{0}^{T}\int_{S_{0}}\left[\alpha p-\tau(v.\eta)\right]^{2}\frac{\partial
h}{\partial\eta}dS_{0}dt.$

For some $T_{0}>0$, $C>0$ and $T>T_{0}$. The inequality (7) is named from an
inequality of observation which is in the theorem 3.12 assuming geometrical
properties on domain $\Omega$ and in the interfaces $\Gamma_{k}$. Such that,
to prove (7) we assumed monotonicity conditions in the coefficients of the
systems (3) and (4). The requirement necessary were found by Lions[13] in his
study of transmission problem. Lagnese[7] used the same hypothesis to prove
the result of controllability for a hyperbolic problem.
Then this will be done.

  1. 1.

We showed that (3)-(6) is well posed problem, we used the semigroup theory
[17]

  2. 2.

We obtained an inequality of simultaneous observability, for both systems (3)
and (4), these we solved using a multipliers technique[6]

  3. 3.

We applied the H.U.M (Hilbert Uniqueness Method) to obtain the simultaneous
controllability [13].

##  2 Functional spaces

Given the Hilbert space
$X_{1}=\left[L^{2}(\Omega)\right]^{3}\times\left[L^{2}(\Omega)\right]$,
associate to (3). We define an scalar product in $X_{1}$, given by
$(\tilde{u},\tilde{p}),(u,p)\in X_{1}$, then:

${}\left<(u,p),(\tilde{u},\tilde{p})\right>_{X_{1}}=\sum^{m}_{k=0}\int_{\Omega_{k}}\left\\{\beta^{k}u^{k}\Large\textbf{.}\tilde{u}^{k}+\alpha^{k}p^{k}\tilde{p}^{k}\right\\}dx$
(8)

Consequently, we considered
$X_{2}=\left[L^{2}(\Omega)\right]^{3}\times\left[L^{2}(\Omega)\right]$
associate to (4). We define a scalar product in $X_{2}$, as
$(\tilde{v},\tilde{q}),(v,q)\in X_{2}$, then:

${}\left<(v,q),(\tilde{v},\tilde{q})\right>_{X_{2}}=\sum^{m}_{k=0}\int_{\Omega_{k}}\left\\{\tau^{k}v^{k}\Large\textbf{.}\tilde{v}^{k}+\gamma^{k}q^{k}\tilde{q}^{k}\right\\}dx$
(9)

We have considered a total energy to the problem (3), (4), (5), (6) and the
boundary conditions in (1), (2), as

${}E(t)=\frac{1}{2}\sum_{k=0}^{m}\int_{\Omega_{k}}\left\\{\beta_{k}\mid
u^{k}\mid^{2}+\alpha^{k}(p^{k})^{2}+\tau_{k}\mid
v^{k}\mid^{2}+\gamma^{k}(q^{k})^{2}\right\\}dx$ (10)

Making a rigorously way for the interfaces conditions, we can see a lemma 2.1;
for more details see Perla et al.[10].

######  Lemma 2.1.

Given $\Omega$ bounded region in $I\\!\\!R^{3}$, with regularity in the
boundary $\partial\Omega$. The application

$\begin{array}[]{ccc}\left[C^{1}(\bar{\Omega})\right]^{3}&\rightarrow&C^{1}(\partial\Omega)\\\
u=(u_{1},u_{2},u_{3})&\rightarrow&u\Large\textbf{.}\eta\end{array}$

where $\eta=\eta(x)$ is as exterior unit normal vector in
$x\in\partial\Omega$. We can extend by continuity application

$\tilde{H}\longrightarrow H^{-1/2}(\partial\Omega)$

where $\tilde{H}=\left\\{u\in\left[L^{2}(\Omega)\right]^{3},\quad\mbox{such
that},\quad\mbox{div}(u)\in L^{2}(\Omega)\right\\}$ and
$H^{-1/2}(\partial\Omega)$ is dual space of $H^{1/2}(\partial\Omega)$

To simplify the notation we write $u^{k}$ as $u$, $\beta^{k}$ as $\beta$, the
same way for all symbols in the region $\Omega_{k}$. by the lemma 2.1 is
clearly that the spaces

$\begin{array}[]{ccc}H_{1}=\left\\{(u,p)\in X_{1},\quad\mbox{such that
},\quad\left(-\alpha\nabla p,-\beta\mbox{div}(u)\right)\in
X_{1}\right\\}&\subseteq&X_{1}\\\ H_{2}=\left\\{(v,q)\in X_{2},\quad\mbox{such
that},\quad\left(-\gamma\nabla q,-\tau\mbox{div}(v)\right)\in
X_{2}\right\\}&\subseteq&X_{2}\end{array}$

we can define the sub spaces:

$Z_{1}=\left\\{(u,p)\in H_{1},\quad\mbox{such that
}\begin{array}[]{cc}\alpha^{k-1}p^{k-1}=&\alpha^{k}p^{k}\\\
\beta^{k-1}(u^{k-1}\Large\textbf{.}\eta)=&\beta^{k}(u^{k}\Large\textbf{.}\eta)\\\
u\Large\textbf{.}\eta=0\in S_{0},&\quad p=0\in
S_{1}\end{array},\quad\mbox{in}\quad\Gamma_{k},k=2,\dots,m.\right\\}$

and,

$Z_{2}=\left\\{(v,q)\in H_{2},\mbox{such
that}\begin{array}[]{cc}\gamma^{k-1}q^{k-1}=&\gamma^{k}q^{k}\\\
\tau^{k-1}(v^{k-1}\Large\textbf{.}\eta)=&\tau^{k}(v^{k}\Large\textbf{.}\eta)\\\
q=0,&\mbox{in}\quad\partial\Omega=S_{0}\cup
S_{1}\end{array},\quad\mbox{in}\quad\Gamma_{k},\quad k=2,\dots,m\right\\}$

Observe that $\left[C^{1}(\bar{\Omega})\right]\subset Z_{j},\quad j=1,2$. Also
$Z_{1}$ and $Z_{2}$ are dense in $X_{1}$ and $X_{2}$, respectively.
Considering the bounded operator

$A_{j}:Z_{j}=D(A_{j})\subseteq X_{j}\longrightarrow X_{j}$

defined as

  1. 1.

Given $(u,p)\in D(A_{1})$, then, $A_{1}(u,p)=\left(-\alpha\nabla
p,-\beta\mbox{div}(u)\right)$

  2. 2.

Given $(v,q)\in D(A_{2})$, then , $A_{2}(v,q)=\left(-\gamma\nabla
q,-\tau\mbox{div}(v)\right)$

The adjoint operator of $A_{1}$ is denoted by $A_{1}^{*}$; It is calculated
and given as :

$A^{*}_{1}(\tilde{u},\tilde{p})=\left(\alpha\nabla\tilde{p},\beta\mbox{div}(\tilde{u})\right)$

and

$D(A_{1}^{*})=\left\\{(\tilde{u},\tilde{p})\in H_{1},\quad\mbox{such that
},\begin{array}[]{cc}\alpha^{k-1}\tilde{p}^{k-1}=&\alpha^{k}\tilde{p}^{k}\\\
\beta^{k-1}(\tilde{u}^{k-1}.\eta)=&\beta^{k}(\tilde{u}^{k}.\eta)\\\
p=0\quad\mbox{in}\quad S_{1},&\tilde{u}.\eta=0\quad\mbox{in}\quad
S_{0}\end{array},\mbox{in}\quad\Gamma_{k},\quad k=2,\dots,m\right\\}$

Perla et al. [10] showed that operator $A_{1}$ is skew-adjoint, i.e,
$A_{1}^{*}=-A_{1}$, the same result was proved for $A_{2}$ . Using the Stone’s
theorem, we have proved that $A_{1}$ and $A_{2}$ generate infinitesimally a
group of strongly continuous unit operators $\\{U_{j}(t)\\}_{t\in I\\!\\!R}$,
in $X_{1}$ and $X_{2}$, respectively. Moreover, $U_{j}(t)w_{j}$ is strongly
differentiable in relation to $t$ and for any $w_{j}\in D(A_{j})$,

$\frac{d}{dt}U_{j}(t)w_{j}=A_{j}U(t)w_{j}$

Now, we study some properties of the solutions (3), these are used in

######  Lemma 2.2.

Given $V_{j}=\left[\mbox{Ker}(A^{*}_{j})\right]^{\bot}$, considering the
orthogonality in relation with the scalar product defined in $X_{1}$ and
$X_{2}$, respectively. Then, the following results are valid:

  1. 1.

$U_{j}(t)\left(V_{j}\cap D(A_{j})\right)\subset V_{j}\cap D(A_{j})$

  2. 2.

Fixing $t\in I\\!\\!R$, and $(u,p)\in V_{1}\cap D(A_{1})$, then, in the
distributions manner

    1. (a)

$\mbox{curl}(u^{k})=0$, in $\Omega_{k}$, $k=0,1,\dots,m$

    2. (b)

$u\times\eta=0$ in $S_{1}$

    3. (c)

$u^{k-1}\times\eta=u^{k}\times\eta$ in $\Gamma_{k}$, $k=2,\dots,m$

where $\times$ denoted a vectorial product of $I\\!\\!R^{3}$ and $\eta(x)$ is
an exterior normal vector $\Gamma_{k}$

  3. 3.

Fixing $t\in I\\!\\!R$, $(v,q)\in V_{2}\cap D(A_{2})$, then, in the
distribution manner

    1. (a)

$\mbox{curl}(v^{k})=0$, in $\Omega_{k}$, $k=0,1,\dots,m$

    2. (b)

$v\times\eta=0$ in $\Gamma$

    3. (c)

$v^{k-1}\times\eta=v^{k}\times\eta$ in $\Gamma_{k}$, $k=2,\dots,m$

###### Proof.

The proof of $A_{1}$ , $A_{2}$ are similar. we make the proof for the first
one. Given $(u,p)\in V_{1}\cap D(A_{1})$, then $(u,p)\in V_{1}$ and $(u,p)\in
D(A_{1})$. As consequence of semigroup theory, we know that $U(t)(u,p)\in
D(A_{1})$, $\forall t\in I\\!\\!R$, Now we need to prove that $U(t)(u,p)\in
V_{1}$. See that, $\mbox{Ker}(A^{*}_{1})\neq\emptyset$ has elements of the
form $(\beta^{-1}\mbox{Curl}(v),0)$ where
$v\in\left[H^{2}(\Omega)\right]^{3},\quad v=0$ in $S_{0}$ Given
$w=(w_{1},w_{2})\in\mbox{Ker}(A_{1}^{*})$, then $A_{1}^{*}(w_{1},w_{2})=0$.
And,

$\frac{d}{dt}\left(U_{1}(t)(u,p),(w_{1},w_{2})\right)=\left<A_{1}U_{1}(t)(v,p),(w_{1},w_{2})\right>_{X_{1}}=\left<U_{1}(t)(u,p),A^{*}_{1}(w_{1},w_{2})\right>=0$

we have that,

$U_{1}(t)(u,p),(w_{1},w_{2})=\mbox{C},\quad\mbox{C=constant}\quad\forall t\in
I\\!\\!R.$

In particular, for $t=0$, $\left<(u,p),(w_{1},w_{2})\right>_{X_{1}}=\mbox{C}$,
and , $(u,p)\in\left[\mbox{Ker}(A^{*}_{1})\right]^{\perp}$ and
$(w_{1},w_{2})\in\mbox{Ker}(A^{*}_{1})$, this implied that $C=0$. In the same
form, $\left<U_{1}(t)(u,p),(w_{1},w_{2})\right>=0,\quad\forall t\in I\\!\\!R$.
Given $U_{1}(t)(u,p)\in\left[\mbox{Ker}(A^{*}_{1})\right]^{\perp}=V_{1}$ and
the item $1.)$ was proved. Now, we prove the first item $2)$. Given
$v\in\left[H^{2}(\Omega)\right]^{3}$ with support in $\Omega_{k}$ and
considering the elements
$\left(\beta^{-1}\mbox{Curl}(v),0\right)\in\mbox{Ker}(A^{*}_{1})$. Then, for
all $(u,p)\in V_{1}\cap D(A_{1})$, we have;

$0=\left<(u,p),\left(\beta^{-1}\mbox{Curl}(v),0\right)\right>_{X_{1}}=\sum^{m}_{k=0}\int_{\Omega_{k}}u{\Large\textbf{.}}\mbox{Curl}(v)dx$

because the support of $v$ is in $\Omega_{k}$. The same form,
$\mbox{Curl}(v)=0$ in $\Omega_{k}$, $k=0,1,\dots,m$ in the distributions way.
For proving the item $b)$ of $2)$ We use the identities such as,

${}\int_{\Omega}\mbox{Curl}(u){\Large\textbf{.}}vdx=\int_{\Omega}u{\Large\textbf{.}}\mbox{Curl}(v)dx-\int_{\partial\Omega}v{\Large\textbf{.}}(u\times\eta)d\Gamma$
(11)

Given $v\in\left[H^{2}(\Omega)\right]^{3}$ and
$\left(\beta^{-1}\mbox{Curl}(v),0\right)\in\mbox{Ker}(A^{*}_{1})$, as, $v=0$
in $\bigcup^{j=1}_{m}\bar{\Omega_{j}}$. Using (11) we have

$\displaystyle{}0$ $\displaystyle=$
$\displaystyle\left<(u,p),\left(\beta^{-1}\mbox{Curl}(v),0\right)\right>_{X_{1}}=\sum^{m}_{k=0}\int_{\Omega_{k}}u^{k}{\Large\textbf{.}}\mbox{Curl}(v^{k})dx$
(12) $\displaystyle=$
$\displaystyle\sum^{m}_{k=0}\int_{\Omega_{k}}\mbox{Curl}(u^{k})\Large\textbf{.}v^{k}dx+\sum^{m}_{k=0}\int_{\partial\Omega_{k}}v^{k}\Large\textbf{.}(u^{k}\times\eta)d\Gamma_{k}$
$\displaystyle 0$ $\displaystyle=$
$\displaystyle\sum^{m}_{k=0}\int_{\partial\Omega_{k}}v^{k}\Large\textbf{.}(u^{k}\times\eta)d\Gamma_{k}=\int_{S_{1}}v\Large\textbf{.}(u\times\eta)dS_{1}$

The last prove is for the item $c)$ of $2)$. Given
$v\in\left[H^{2}(\Omega)\right]^{3}$ and
$\left(\beta^{-1}\mbox{Curl}(v),0\right)\in\mbox{Ker}(A^{*}_{1})$, using the
identities (12), we have

$\displaystyle{}0$ $\displaystyle=$
$\displaystyle\sum^{m}_{k=0}\int_{\Omega_{k}}\mbox{Curl}(u^{k}).v^{k}dx+\sum^{m}_{k=0}\int_{\partial\Omega_{k}}v^{k}\Large\textbf{.}(u^{k}\times\eta)d\Gamma_{k}$
$\displaystyle 0$ $\displaystyle=$
$\displaystyle\sum^{m}_{k=0}\int_{\Omega_{k}}v^{k}\Large\textbf{.}(u^{k}\times\eta)d\Gamma_{k}$
(13)

Now, we chose $v$ such that $v=0$ in $S_{0}$ and $v=0$ in
$\begin{array}[]{c}m\\\ \bigcup\\\ j=1\\\ j\neq k\end{array}\Gamma_{j}$, by
(2) we have that

$0=\int_{\Gamma_{k}}v\Large\textbf{.}\left\\{u^{k-1}\times\eta-u^{k}\times\eta\right\\}d\Gamma_{k}$

that is the expected result.
The item $3)$ is proved in the same way. ∎

The theorem 2.3 has a summary of the results .

######  Theorem 2.3.

Given $V_{j}$ an orthogonal complement of the subset $\mbox{Ker}(A^{*}_{j})$,
$j=1,2$, in $X_{j}$. Consider the problems (3), (4), (1), (2) and the initial
conditions $(u_{0},p_{0})\in V_{1}\cap D(A_{1})$ and $(v_{0},q_{0})\in
V_{2}\cap D(A_{2})$. Then $(u,p)=U_{1}(t)(u_{0},p_{0})$ and
$(v,q)=U_{2}(t)(v_{0},q_{0})$ are the uniqueness solutions, respectively. That
is,

$\begin{array}[]{cc}(u,p)\in&C(I\\!\\!R;V_{1}\cap D(A_{1}))\cap
C(I\\!\\!R,X_{1})\\\ (v,q)\in&C(I\\!\\!R;V_{2}\cap D(A_{2}))\cap
C(I\\!\\!R,X_{2})\end{array}$

In addition, these solutions satisfies the properties in the lemma 2.2.

Before to show the inequality of observability, we prove some important
properties.
The energy associate to the systems (3) and (4), with null boundary
conditions, are given by:

$E_{1}(t)=\frac{1}{2}\sum_{k=0}^{m}\int_{\Omega_{k}}\left\\{\beta_{k}\mid
u^{k}\mid^{2}+\alpha^{k}(p^{k})^{2}\right\\}dx$

and

$E_{2}(t)=\frac{1}{2}\sum_{k=0}^{m}\int_{\Omega_{k}}\left\\{\tau_{k}\mid
v^{k}\mid^{2}+\gamma^{k}(q^{k})^{2}\right\\}dx$

respectively. To prove that these are not dependent of the time $t$, In fact,
we multiplied the first equation of (3) by $\beta^{k}u^{k}$ and integrating in
$\Omega_{k}$ and adding in $k=0,1,\dots,m$, we have

${}\frac{1}{2}\frac{d}{dt}\sum_{k=0}^{m}\int_{\Omega_{k}}\beta^{k}\mid
u^{k}\mid^{2}dx-\sum_{k=0}^{m}\int_{\Omega_{k}}\beta^{k}\alpha^{k}p^{k}\mbox{div}(u^{k})dx+\sum_{k=0}^{m}\int_{\partial\Omega_{k}}\beta^{k}\alpha^{k}p^{k}(u^{k}{\Large\textbf{.}}\eta)dx.$
(14)

Multiplying the second equation of (3) by $\alpha^{k}p^{k}$, after integrating
in $\Omega_{k}$ and adding in $k$, we have that

${}\frac{1}{2}\frac{d}{dt}\sum_{k=0}^{m}\int_{\Omega_{k}}\alpha^{k}(p^{k})^{2}dx+\sum_{k=0}^{m}\int_{\Omega_{k}}\alpha^{k}\beta^{k}p^{k}\mbox{div}(u^{k})dx=0$
(15)

adding (14) and (15), we have that

$\frac{1}{2}\frac{d}{dt}\sum_{k=0}^{m}\int_{\Omega_{k}}\left\\{\beta^{k}\mid
u^{k}\mid^{2}+\alpha^{k}(p^{k})^{k}\right\\}dx+\sum_{k=0}^{m}\int_{\partial\Omega_{k}}\beta^{k}\alpha^{k}p^{k}(u^{k}\Large\textbf{.}\eta)dx=0$

moreover ,

$\displaystyle\sum_{k=0}^{m}\int_{\partial\Omega_{k}}\beta^{k}\alpha^{k}p^{k}(u^{k}\Large\textbf{.}\eta)dx$
$\displaystyle=\int_{S_{1}}\alpha\beta
p(u\Large\textbf{.}\eta)dS_{1}+\sum_{k=1}^{m}\int_{\Gamma_{k}}\left\\{\alpha^{k-1}\beta^{k-1}p^{k-1}(u^{k-1}\Large\textbf{.}\eta)-\right.$
$\displaystyle\left.\alpha^{k}\beta^{k}p^{k}(u^{k}\Large\textbf{.}\eta)\right\\}d\Gamma_{k}+\int_{S_{0}}\alpha\beta
p(u.\eta)dS_{0}$

Using the contour and interface conditions (5), we have that
$\sum_{k=0}^{m}\int_{\partial\Omega_{k}}\alpha^{k}\beta^{k}p^{k}(u^{k}\Large\textbf{.}\eta)d\Gamma_{k}=0$.
Follows the affirmation. The case $E_{2}(t)$ is similar .

##  3 Inequality of observability

In this section we show the inequality of observability. This inequality
satisfies the systems (3) and (4) simultaneously. Using the multiplier’s
theory (see Komornik[6]), we make the proof. The multiplier was modified to
get a good estimates in the boundary. These multiplier were used in several
works. The invariant of the systems (1) and (2), in relations to dilatations
groups in all variables, see [8] and [10]. Given $h:C(\Omega)\cap
C^{1}(\bar{\Omega})\longrightarrow I\\!\\!R$ an auxiliary function, it will be
chosen in the next steps; and, given $(u,p)\in V_{1}\cap D(A_{1})$ a solution
of the system (1). Considering the multiplier given by:

$\left\\{\begin{aligned} M_{1}&=2\left(\alpha tp-u\Large\textbf{.}\nabla
h+\alpha\int_{0}^{t}p(x,s)ds\right)\\\ M_{2}&=2\left(\beta tu-p\nabla
h\right)\\\ M_{3}&=2\beta u\end{aligned}\right.$

and $(u,p)$ solution of (1), we have the identities

$0=M_{1}\left\\{p_{t}+\beta\mbox{div}u\right\\}+M_{2}\Large\textbf{.}\left\\{u_{t}+\alpha\nabla
p\right\\}+M_{3}\Large\textbf{.}\left\\{\int_{0}^{t}(u_{s}+\alpha\nabla
p)ds\right\\}$

The expression above, we make rewrite as

${}0=\frac{\partial A}{\partial t}-\mbox{div}(\vec{B})-J$ (16)

where

$\displaystyle A$ $\displaystyle=t\left[\beta\mid u\mid^{2}+\alpha
p^{2}\right]-2p(u\Large\textbf{.}\nabla h)+2\alpha
p\int_{0}^{t}p(x,s)ds-2\beta u(x,0)\Large\textbf{.}\int_{0}^{t}u(x,s)ds$
$\displaystyle\vec{B}$ $\displaystyle=-2\alpha\beta tpu+\alpha p^{2}\nabla
h-\beta\mid u\mid^{2}\nabla h+2\beta(u\Large\textbf{.}\nabla
h)u-2\alpha\beta\left(\int_{0}^{t}p(x,s)ds\right)u$ $\displaystyle J$
$\displaystyle=\beta(\Delta-1)\mid
u\mid^{2}-2\beta\sum_{i,j=1}^{3}\frac{\partial^{2}h}{\partial x_{i}\partial
x_{j}}u_{i}u_{j}-\alpha(\Delta h-3)p^{2}$

######  Remark 3.1.

We have considered $h(x)=\frac{1}{2}\mid x-x_{0}\mid^{2}$ for some $x_{0}\in
I\\!\\!R$ fixed, then $J=0$. In this case (16) represents a conservation law.
Integrating (16) and $\Omega_{k}$ , we observe that: in the expression $B$
need to fix the $\frac{\partial h}{\partial\eta}$ to get good estimations;
then, we chose $h(x)$ as a small perturbation of $\frac{1}{2}\mid
x-x_{0}\mid^{2}$ for some $x_{0}\in I\\!\\!R^{3}$.

Integrating the identity (16) in $\Omega_{k}\times(0,s)$ and adding in $k$, we
have

${}0=\sum_{k=0}^{m}\int_{\Omega_{k}}\left[A^{k}(x,s)-A^{k}(x,0)\right]dx-\sum_{k=0}^{m}\int_{0}^{s}\int_{\partial\Omega_{k}}\vec{B}^{k}{\Large\textbf{.}}\eta
d\Gamma-\sum_{k=0}^{m}\int_{\Omega_{k}}\int_{0}^{s}J^{k}dtdx$ (17)

replacing the expression $A$ in (17), we have that

$\displaystyle 0$
$\displaystyle=\sum_{k=0}^{m}\int_{\Omega_{k}}\left\\{s\left[\beta^{k}\mid
u^{k}\mid^{2}+\alpha^{k}(p^{k})^{2}\right]-2p^{k}(x,s)(u^{k}\Large\textbf{.}\nabla
h)+2\alpha^{k}p^{k}\int_{0}^{s}p^{k}(x,\tau)d\tau\right.$
$\displaystyle-\left.2\beta^{k}u_{0}^{k}(x)\Large\textbf{.}\int_{0}^{s}u^{k}(x,r)dr+2p_{0}^{k}(u_{0}^{k}\Large\textbf{.}\nabla
h)\right\\}dx-\sum_{k=0}^{m}\int_{0}^{s}\int_{\partial\Omega_{k}}\vec{B}^{k}\Large\textbf{.}\eta
d\Gamma_{k}-\sum_{k=0}^{m}\int_{\Omega_{k}}\int_{0}^{s}J^{k}dtdx$

thus,

$\displaystyle s\sum_{k=0}^{m}\int_{\Omega_{k}}\left[\beta^{k}\mid
u^{k}\mid^{2}+\alpha^{k}(p^{k})^{2}\right]$
$\displaystyle=2\sum_{k=0}^{m}\int_{\Omega_{k}}2p^{k}(x,s)(u^{k}{\Large\textbf{.}}\nabla
h)dx-2\sum_{k=0}^{m}\int_{\Omega_{k}}\alpha^{k}p^{k}\int_{0}^{s}p^{k}(x,\tau)d\tau$
(18)
$\displaystyle+2\sum_{k=0}^{m}\int_{\Omega_{k}}\beta^{k}u_{0}^{k}(x){\Large\textbf{.}}\int_{0}^{s}u^{k}(x,r)drdx-2\sum_{k=0}^{m}\int_{\Omega_{k}}p_{0}^{k}(u_{0}^{k}{\Large\textbf{.}}\nabla
h)dx$
$\displaystyle+\sum_{k=0}^{m}\int_{0}^{s}\int_{\partial\Omega_{k}}\vec{B}^{k}{\Large\textbf{.}}\eta
d\Gamma_{k}+\sum_{k=0}^{m}\int_{\Omega_{k}}\int_{0}^{s}J^{k}dtdx$

where $p^{k}_{0}=p^{k}(x,0)$, $u^{k}_{0}=u^{k}(x,0)$. The proof of the main
result is to get the inequality of the observability; we obtain this
estimation using the right side of (18). Making good presentation of the
proof, we show many lemmas

######  Lemma 3.2.

Given $\\{u,p\\}$ regular solution for the problem (3)-(5), this was given by
the theorem 2.3. Then

$\sum_{k=0}^{m}\int_{\Omega_{k}}2p^{k}(x,s)(u^{k}.\nabla h)dx\leq
C_{1}\sum_{k=0}^{m}\left\\{\beta^{k}\mid
u^{k}\mid^{2}+\alpha^{k}(p^{k})^{2}\right\\}dx$

where
$C_{1}=3\max_{k=0,1,\dots,m}\left\\{(\alpha^{k})^{-1},(\beta^{k})^{-1}\right\\}\max_{x\in\Omega}\mid\nabla
h\mid$

###### Proof.

Using the Holder inequality in the first term in the right-hand side of (18),
we have that

$\displaystyle 2\sum_{k=0}^{m}\int_{\Omega_{k}}p^{k}(x,s)(u^{k}.\nabla h)dx$
$\displaystyle=2\sum_{k=0}^{m}\sum_{i=1}^{3}\int_{\Omega_{k}}p^{k}u_{i}^{k}\frac{\partial
h}{\partial x_{i}}dx\leq 2\max_{x\in\Omega}\mid\nabla
h\mid\sum_{k=0}^{m}\sum_{i=1}^{3}\int_{\Omega_{k}}p^{k}u_{i}^{k}dx$ (19)
$\displaystyle\leq 2\max_{x\in\Omega}\mid\nabla
h\mid\sum_{k=0}^{m}\sum_{i=1}^{3}\left(\int_{\Omega_{k}}(p^{k})^{2}\right)^{1/2}\left(\int_{\Omega_{k}}(u_{i}^{k})^{2}\right)^{1/2}$
$\displaystyle\leq\max_{x\in\Omega}\mid\nabla
h\mid\sum_{k=0}^{m}\left\\{3\int_{\Omega_{k}}(p^{k})^{2}dx+\int_{\Omega_{k}}\mid
u^{k}\mid^{2}dx\right\\}$ $\displaystyle\leq
3\max_{k=0,1,\dots,m}\left\\{(\alpha^{k})^{-1},(\beta^{k})^{-1}\right\\}\max_{x\in\Omega}\mid\nabla
h\mid\sum_{k=0}^{m}\left\\{\beta^{k}\mid
u^{k}\mid^{2}+\alpha^{k}(p^{k})^{2}\right\\}dx$
$\displaystyle=C_{1}\sum_{k=0}^{m}\left\\{\beta^{k}\mid
u^{k}\mid^{2}+\alpha^{k}(p^{k})^{2}\right\\}dx$

where,

$C_{1}=3\max_{k=0,1,\dots,m}\left\\{(\alpha^{k})^{-1},(\beta^{k})^{-1}\right\\}\max_{x\in\Omega}\mid\nabla
h\mid$

∎

The second term in the right-hand side of (18),
$2\sum_{k=0}^{m}\int_{\Omega_{k}}\alpha^{k}p^{k}\int_{0}^{s}p^{k}(x,\tau)d\tau$,
we may write as

$\displaystyle
2\sum_{k=0}^{m}\int_{\Omega_{k}}\alpha^{k}p^{k}\int_{0}^{s}p^{k}(x,\tau)d\tau=-\frac{\partial}{\partial
s}\sum_{k=0}^{m}\int_{\Omega_{k}}\alpha^{k}\left[\int_{0}^{s}p^{k}(x,r)dr\right]dx.$
(20)

To estimate the fourth term in the right-hand side of (18),
$-2\sum_{k=0}^{m}\int_{\Omega_{k}}p_{0}^{k}(u_{0}^{k}{\Large\textbf{.}}\nabla
h)dx$, we used the assumption the independence of the energy with the time the
estimation is (19), Thus:

$\displaystyle-2\sum_{k=0}^{m}\int_{\Omega_{k}}p_{0}^{k}(u_{0}^{k}{\Large\textbf{.}}\nabla
h)dx\leq C_{1}\sum_{k=0}^{m}\int_{\Omega_{k}}\left[\beta^{k}\mid
u^{k}\mid^{2}+\alpha^{k}(p^{k})^{2}\right]^{2}dx.$ (21)

The fifth term of right-hand side of (18),
$\sum_{k=0}^{m}\int_{0}^{s}\int_{\partial\Omega_{k}}\vec{B}^{k}.\eta
d\Gamma_{k}$, we need to make an analysis more carefully.

######  Lemma 3.3.

Given $\\{u,p\\}$ a regular solution of the problem (1)-(5) by theorem 2.3.
For $k=1,2,\dots,m$, we have the following identity

$\vec{B}^{k-1}{\Large\textbf{.}}\eta-\vec{B}^{k}{\Large\textbf{.}}\eta=-\frac{\partial
h}{\partial\eta}\left\\{\frac{(\alpha^{k-1}-\alpha^{k})}{\alpha^{k-1}}\alpha^{k}(p^{k})^{2}+(\beta^{k-1}-\beta^{k})\frac{\beta^{k}}{\beta^{k-1}}\mid
u^{k}{\Large\textbf{.}}\eta\mid^{2}+(\beta^{k-1}-\beta^{k})\mid
u^{k}\times\eta\mid^{2}\right\\}$

is validated in $\Gamma_{k}$, $k=1,2,\dots,m$.

###### Proof.

Using the boundary conditions (5), we have that $x\in S_{1}$ , then

$\displaystyle\vec{B}{\Large\textbf{.}}\eta$ $\displaystyle=-\beta\mid
u\mid^{2}\frac{\partial h}{\partial\eta}+2\beta(u{\Large\textbf{.}}\nabla
h)(u.\eta)$ (22) $\displaystyle=-\beta\mid u\mid^{2}\frac{\partial
h}{\partial\eta}+2\beta\left\\{\mid u\mid^{2}\frac{\partial
h}{\partial\eta}+(u\times\eta)(\nabla h\times u)\right\\}$
$\displaystyle=-\beta\mid u\mid^{2}\frac{\partial h}{\partial\eta}+2\beta\mid
u\mid^{2}\frac{\partial h}{\partial\eta}$ $\displaystyle=\beta\mid
u\mid^{2}\frac{\partial h}{\partial\eta}.$

and $x\in S_{0}$, then

$\vec{B}{\Large\textbf{.}}\eta=\alpha p^{2}\frac{\partial
h}{\partial\eta}-\beta\mid u\mid^{2}\frac{\partial h}{\partial\eta}.$

Using the interface conditions (5), for $x\in\Gamma_{k}$ we have the following
identity

$\displaystyle\vec{B}^{k-1}{\Large\textbf{.}}\eta-\vec{B}^{k}{\Large\textbf{.}}\eta$
$\displaystyle=-2\alpha^{k-1}\beta^{k-1}+p^{k-1}(u^{k-1}.\eta)+\alpha^{k-1}(p^{k-1})^{2}\frac{\partial
h}{\partial\eta}-\beta^{k-1}\mid u^{k-1}\mid^{2}\frac{\partial
h}{\partial\eta}+$ (23) $\displaystyle+2\beta^{k-1}(u^{k-1}.\nabla
h)(u^{k-1}{\Large\textbf{.}}\eta)-2\alpha^{k-1}\beta^{k-1}\left(\int_{0}^{t}p^{k-1}(x,s)ds\right)(u^{k-1}{\Large\textbf{.}}\eta)+$
$\displaystyle-\alpha^{k}(p^{k})^{2}\frac{\partial
h}{\partial\eta}+\beta^{k}\mid u^{k}\mid^{2}\frac{\partial
h}{\partial\eta}-2\beta^{k}(u^{k}{\Large\textbf{.}}\nabla
h)(u^{k}{\Large\textbf{.}}\eta)+2\alpha^{k}\beta^{k}\left(\int_{0}^{t}p^{k}(x,s)ds\right)(u^{k}{\Large\textbf{.}}\eta)$
$\displaystyle=\alpha^{k-1}(p^{k-1})^{2}\frac{\partial
h}{\partial\eta}-\beta^{k-1}\mid u^{k-1}\mid^{2}\frac{\partial
h}{\partial\eta}+2\beta^{k-1}(u^{k-1}{\Large\textbf{.}}\nabla
h)(u^{k-1}{\Large\textbf{.}}\eta)-$
$\displaystyle\alpha^{k}(p^{k})^{2}\frac{\partial
h}{\partial\eta}+\beta^{k}\mid u^{k}\mid^{2}\frac{\partial
h}{\partial\eta}-2\beta^{k}(u^{k}{\Large\textbf{.}}\nabla
h)(u^{k}{\Large\textbf{.}}\eta)$

else,

$\displaystyle\alpha^{k-1}(p^{k-1})^{2}\frac{\partial
h}{\partial\eta}-\alpha^{k}(p^{k})^{2}\frac{\partial h}{\partial\eta}$
$\displaystyle=\frac{\left(\alpha^{k-1}p^{k-1}\right)^{2}}{\alpha^{k-1}}\frac{\partial
h}{\partial\eta}$ (24)
$\displaystyle=\left(\frac{1}{\alpha^{k-1}}-\frac{1}{\alpha^{k}}\right)(\alpha^{k}p^{k})^{2}\frac{\partial
h}{\partial\eta}$
$\displaystyle=-(\alpha^{k-1}-\alpha^{k})\frac{\alpha^{k}}{\alpha^{k-1}}(p^{k})^{2}\frac{\partial
h}{\partial\eta}$

For $\mid\eta\mid=1$ is validated

${}\mid u\mid^{2}=\mid(u.\eta)\mid^{2}+\mid u\times\eta\mid^{2}$ (25)

Substituting (25) in (23), we have that

$\displaystyle\beta^{k}\mid u^{k}\mid^{2}\frac{\partial
h}{\partial\eta}-\beta^{k-1}\mid u^{k-1}\mid^{2}\frac{\partial
h}{\partial\eta}$
$\displaystyle=\beta^{k}\left(\mid(u^{k}.\eta)\mid^{2}+\mid(u^{k}\times\eta)\mid^{2}\right)\frac{\partial
h}{\partial\eta}-\beta^{k-1}\left(\mid(u^{k-1}{\Large\textbf{.}}\eta)\mid^{2}+\right.$
(26) $\displaystyle+\left.\mid(u^{k-1}\times\eta)\mid^{2}\right)\frac{\partial
h}{\partial\eta}$ $\displaystyle=\left(\beta^{k}\mid
u^{k}{\Large\textbf{.}}\eta\mid^{2}-\beta^{k-1}\mid
u^{k-1}{\Large\textbf{.}}\eta\mid^{2}\right)\frac{\partial
h}{\partial\eta}+\left(\beta^{k}\mid u^{k}\times\eta\mid^{2}-\right.$
$\displaystyle\left.\beta^{k-1}\mid
u^{k-1}\times\eta\mid^{2}\right)\frac{\partial h}{\partial\eta}$
$\displaystyle=\left[\frac{(\beta^{k}\mid
u^{k}{\Large\textbf{.}}\eta\mid)^{2}}{\beta^{k}}-\frac{(\beta^{k-1}\mid
u^{k-1}{\Large\textbf{.}}\eta\mid)^{2}}{\beta^{k-1}}\right]\frac{\partial
h}{\partial\eta}+$ $\displaystyle(\beta^{k}-\beta^{k-1})\mid
u^{k}\times\eta\mid^{2}\frac{\partial h}{\partial\eta}$
$\displaystyle=\left(\frac{1}{\beta^{k}}-\frac{1}{\beta^{k-1}}\right)(\beta^{k}\mid
u^{k}{\Large\textbf{.}}\eta\mid)^{2}+(\beta^{k}-\beta^{k-1})\mid
u^{k}\times\eta\mid^{2}\frac{\partial h}{\partial\eta}$
$\displaystyle=(\beta^{k-1}-\beta^{k})\frac{\beta^{k}}{\beta^{k-1}}\mid
u^{k}.\eta\mid^{2}\frac{\partial h}{\partial\eta}-(\beta^{k-1}-\beta^{k})\mid
u^{k}\times\eta\mid^{2}\frac{\partial h}{\partial\eta}$

Finally,

$\displaystyle 2\beta^{k-1}(u^{k-1}{\Large\textbf{.}}\nabla
h)(u^{k-1}.\eta)-2\beta^{k}(u^{k}.\nabla h)(u^{k}.\eta)$
$\displaystyle=2\beta^{k-1}\mid
u^{k-1}{\Large\textbf{.}}\eta\mid^{2}-2\beta^{k}\mid
u^{k}{\Large\textbf{.}}\eta\mid^{2}\frac{\partial h}{\partial\eta}$ (27)
$\displaystyle=2\frac{\left(\beta^{k-1}\mid
u^{k-1}{\Large\textbf{.}}\eta\mid\right)^{2}}{\beta^{k-1}}\frac{\partial
h}{\partial\eta}-2\frac{\left(\beta^{k}\mid
u^{k}{\Large\textbf{.}}\eta\mid\right)^{2}}{\beta^{k}}\frac{\partial
h}{\partial\eta}$
$\displaystyle=2\left(\frac{1}{\beta^{k-1}}-\frac{1}{\beta^{k}}\right)(\beta^{k}\mid
u^{k}{\Large\textbf{.}}\eta\mid)^{2}\frac{\partial h}{\partial\eta}$
$\displaystyle=-2(\beta^{k-1}-\beta^{k})\frac{\beta^{k}}{\beta^{k-1}}\mid
u^{k}{\Large\textbf{.}}\eta\mid^{2}\frac{\partial h}{\partial\eta}.$

Substituting (24), (25) and (27) in (23), we have that

$\displaystyle\vec{B}^{k-1}.\eta-\vec{B}^{k}.\eta$
$\displaystyle=-(\alpha^{k-1}-\alpha^{k})\frac{\alpha^{k}}{\alpha^{k-1}}(p^{k})^{2}\frac{\partial
h}{\partial\eta}-(\beta^{k-1}-\beta^{k})\frac{\beta^{k}}{\beta^{k-1}}\mid
u^{k}{\Large\textbf{.}}\eta\mid^{2}\frac{\partial h}{\partial\eta}-$ (28)
$\displaystyle(\beta^{k-1}-\beta^{k})\mid
u^{k}\times\eta\mid^{2}\frac{\partial h}{\partial\eta}$
$\displaystyle=-\frac{\partial
h}{\partial\eta}\left\\{\frac{(\alpha^{k-1}-\alpha^{k})}{\alpha^{k-1}}\alpha^{k}(p^{k})^{2}+(\beta^{k-1}-\beta^{k})\frac{\beta^{k}}{\beta^{k-1}}\mid
u^{k}{\Large\textbf{.}}\eta\mid^{2}+(\beta^{k-1}-\beta^{k})\mid
u^{k}\times\eta\mid^{2}\right\\}.$

∎

Now, we estimate the sixth term,
$\sum_{m}^{k=0}\int_{\Omega_{k}}\int_{0}^{s}J^{k}dtdx$. Remember that,

${}\sum_{m}^{k=0}\int_{\Omega_{k}}\int_{0}^{s}J^{k}dtdx=\sum_{k=0}^{m}\int_{\Omega_{k}}\int_{0}^{s}\left\\{\beta^{k}(\Delta
h-1)\mid u^{k}\mid^{2}-2\beta^{k}\sum_{i,j=1}^{m}\frac{\partial^{2}h}{\partial
x_{i}\partial x_{j}}u_{i}^{k}u_{j}^{k}-\alpha^{k}(\Delta
h-3)(p^{k})^{2}\right\\}$ (29)

To estimate (29), we choose a function $h$ as:

${}h(x)=\frac{1}{2}\mid x-x_{0}\mid^{2}+\delta_{0}\Phi(x)$ (30)

where $x_{0}\in\sigma_{1}$ and $\Phi$ satisfy

${}\left\\{\begin{aligned} \Delta\Phi&=1\quad\mbox{en}\quad\Omega\\\
\frac{\partial\Phi}{\partial\eta}&=2\frac{\mbox{Vol}(\Omega)}{\mbox{area}(S_{0})},\quad\mbox{in},\quad
S_{0}\\\
\frac{\partial\Phi}{\partial\eta}&=-\frac{\mbox{Vol}(\Omega)}{\mbox{area}(S_{1})},\quad\mbox{in},\quad
S_{1}\end{aligned}\right.$ (31)

######  Remark 3.4.

Given $\mu=\mu(\Omega)$, by

${}\mu(\Omega)=\inf_{\begin{aligned} x\in\Omega\\\
\mid\xi\mid=1\end{aligned}}2\sum_{i,j=1}^{3}\frac{\partial^{2}\Phi(x)}{\partial
x_{i}\partial x_{j}}\xi_{i}\xi_{j}$ (32)

we may observe that, considering $\xi=(1,0,0),\xi=(0,1,0),\xi=(0,0,1)$ , have
that

$\mu(\Omega)\leq 2\frac{\partial^{2}\Phi(x)}{\partial
x_{1}^{2}},\quad\mu(\Omega)\leq 2\frac{\partial^{2}\Phi(x)}{\partial
x_{2}^{2}},\quad\mu(\Omega)\leq 2\frac{\partial^{2}\Phi(x)}{\partial
x_{3}^{2}}$

that, adding the last expressions we have that

$3\mu(\Omega)\leq 2\Delta\Phi,\Longrightarrow\mu(\Omega)\leq\frac{2}{3}$

the expression of $h(x)$, then

$\frac{\partial^{2}h(x)}{\partial x_{i}\partial
x_{j}}=\delta_{ij}+\delta_{0}\frac{\partial^{2}\Phi(x)}{\partial x_{i}\partial
x_{j}}$

and $\Delta h=3+\delta_{0}$.

######  Lemma 3.5.

Given $\\{u,p\\}$ a regular solution of the problem (1)-(5) by theorem 2.3.
Choosing $h$ as (30), we have that

$\sum_{k=0}^{m}\int_{\Omega_{k}}\int_{0}^{s}J^{k}dtdx\leq\delta_{0}(1-\mu(\Omega))\sum_{k=0}^{m}\int_{\Omega_{k}}\int_{0}^{s}\left\\{\beta^{k}\mid
u^{k}\mid^{2}+\alpha^{k}(p^{k})^{2}\right\\}dtdx$

for any $\delta_{0}>0$

###### Proof.

Using (30), (31), (32) and the observation (3.4), we have that

$\displaystyle\sum_{k=0}^{m}\int_{\Omega_{k}}\int_{0}^{s}J^{k}dtdx$
$\displaystyle=\sum_{k=0}^{m}\int_{\Omega_{k}}\int_{0}^{s}\left\\{\beta^{k}(2+\delta_{0})\mid
u^{k}\mid^{2}-2\beta^{k}\sum_{i,j=1}^{m}\left(\delta_{ij}+\delta_{0}\frac{\partial^{2}\Phi}{\partial
x_{i}\partial
x_{j}}\right)u_{i}^{k}u_{j}^{k}-\alpha^{k}\delta_{0}(p^{k})^{2}\right\\}$ (33)
$\displaystyle\leq\sum_{k=0}^{m}\int_{\Omega_{k}}\int_{0}^{s}\left\\{\beta^{k}(2+\delta_{0})\mid
u^{k}\mid^{2}-2\beta^{k}\mid u^{k}\mid^{2}-\beta^{k}\mu(\Omega)\delta_{0}\mid
u^{k}\mid^{2}-\alpha^{k}\delta_{0}(p^{k})^{2}\right\\}$
$\displaystyle\leq\sum_{k=0}^{m}\int_{\Omega_{k}}\int_{0}^{s}\left\\{\delta_{0}\beta^{k}\mid
u^{k}\mid^{2}-\mu(\Omega)\delta_{0}\beta^{k}\mid
u^{k}\mid^{2}-\alpha^{k}\delta_{0}(p^{k})^{2}\right\\}dtdx$
$\displaystyle\leq\sum_{k=0}^{m}\int_{\Omega_{k}}\int_{0}^{s}\left\\{\delta_{0}(1-\mu(\Omega))\beta^{k}\mid
u^{k}\mid^{2}\right\\}dtdx$
$\displaystyle\leq\delta_{0}(1-\mu(\Omega))\sum_{k=0}^{m}\int_{\Omega_{k}}\int_{0}^{s}\left\\{\beta^{k}\mid
u^{k}\mid^{2}+\alpha^{k}(p^{k})^{2}\right\\}dtdx$

∎

To estimate the rest of the terms of (18),
$2\sum_{k=0}^{m}\int_{\Omega_{k}}\beta^{k}u_{0}^{k}(x){\Large\textbf{.}}\int_{0}^{s}u^{k}(x,r)drdx$,
considering a hypothesis about initial condition $u_{0}(x)$, we assumed that,
it satisfied the following system:

${}\left\\{\begin{aligned}
u_{0}^{k-1}.\eta&=u_{0}^{k}.\eta\quad\mbox{in}\quad\Gamma_{k}\\\
u_{0}^{k-1}\times\eta&=u_{0}^{k}\times\eta\quad\mbox{in}\quad\Gamma_{k}\\\
u_{0}^{k}&=\nabla l^{k}(x)\quad\mbox{in}\quad\Gamma_{k}\\\ l^{k}&\in
H^{2}(\Omega_{k})\quad\mbox{and}\quad h=0\quad\mbox{in}\quad
S_{1}\end{aligned}\right.$ (34)

######  Remark 3.6.

The hypothesis have been made in (34), inclusive the solution of the problem
satisfy the properties of the lemma (2.2), these are necessaries because the
domain is conexo.

######  Lemma 3.7.

Given $\\{u,p\\}$ regular solution of the problem (3)-(5) by theorem 2.3, and
the initial condition $u_{0}$ that satisfied (34). Then

$2\sum_{k=0}^{m}\int_{\Omega_{k}}\beta^{k}u_{0}^{k}(x)\int_{0}^{s}u^{k}(x,r)drdx=2\sum_{k=1}^{m}\int_{\Gamma_{k}}l^{k}(p^{k}(x,s)-p_{0}^{k}(x))dx$

###### Proof.

Using the hypothesis $l$ and the boundary condition in $S_{0}$, we have that

$\displaystyle
2\sum_{k=0}^{m}\int_{\Omega_{k}}\beta^{k}u_{0}^{k}(x)\int_{0}^{s}u^{k}(x,r)drdx$
$\displaystyle=2\sum_{k=0}^{m}\int_{\Omega_{k}}\int_{0}^{s}\beta^{k}\nabla
l^{k}{\Large\textbf{.}}u^{k}$ (35)
$\displaystyle=2\sum_{k=0}^{m}\int_{0}^{s}\left\\{-\int_{\Omega_{k}}\beta^{k}l^{k}\mbox{div}(u^{k})+\int_{\partial\Omega_{k}}\beta^{k}l^{k}(u^{k}{\Large\textbf{.}}\eta)\right\\}$
$\displaystyle=2\sum_{k=0}^{m}\int_{0}^{s}\int_{\Omega_{k}}l^{k}\frac{\partial
p^{k}}{\partial
t}+2\sum_{k=0}^{m}\int_{0}^{s}\int_{\partial\Omega_{k}}\beta^{k}l^{k}(u^{k}{\Large\textbf{.}}\eta)$
$\displaystyle=2\sum_{k=0}^{m}\int_{0}^{s}\int_{\Omega_{k}}l^{k}\frac{\partial
p^{k}}{\partial t}+2\int_{0}^{s}\int_{S_{1}}\beta l(u{\Large\textbf{.}}\eta)+$
$\displaystyle
2\sum_{k=1}^{m}\int_{0}^{s}\int_{\Gamma_{k}}\left[\beta^{k-1}l^{k-1}(u^{k-1}{\Large\textbf{.}}\eta)-\beta^{k}l^{k}(u^{k}.\eta)\right]+2\int_{0}^{s}\int_{S_{0}}\beta
l(u{\Large\textbf{.}}\eta)$
$\displaystyle=2\sum_{k=1}^{m}\int_{\Gamma_{k}}l^{k}(p^{k}(x,s)-p_{0}^{k}(x))+$
$\displaystyle
2\sum_{k=1}^{m}\int_{0}^{s}\int_{\Gamma_{k}}\left[\beta^{k-1}l^{k-1}(u^{k-1}{\Large\textbf{.}}\eta)-\beta^{k}l^{k}(u^{k}{\Large\textbf{.}}\eta)\right]$

by the hypothesis made in the initial condition (34), we have that

$\displaystyle\mbox{De},\quad
u^{k-1}_{0}{\Large\textbf{.}}\eta=u^{k}_{0}.\eta\quad\mbox{in}\quad\Gamma_{k}$
$\displaystyle\Longrightarrow\nabla l^{k-1}{\Large\textbf{.}}\eta=\nabla
l^{k}.\eta\quad\mbox{in}\quad\Gamma_{k}$
$\displaystyle\Longrightarrow\nabla(l^{k-1}-l^{k}){\Large\textbf{.}}\eta=0\quad\mbox{in}\quad\Gamma_{k}$
$\displaystyle\Longrightarrow\nabla(l^{k-1}-l^{k}){\Large\textbf{.}}\eta\perp\eta\quad\mbox{in}\quad\Gamma_{k}.$
$\displaystyle\mbox{De},\quad
u^{k-1}_{0}\times\eta=u^{k}_{0}\times\eta\quad\mbox{in}\quad\Gamma_{k}$
$\displaystyle\Longrightarrow\nabla l^{k-1}\times\eta=\nabla
l^{k}\times\eta\quad\mbox{in}\quad\Gamma_{k}$ (36)
$\displaystyle\Longrightarrow\nabla(l^{k-1}-l^{k})\times\eta=0\quad\mbox{in}\quad\Gamma_{k}$
$\displaystyle\Longrightarrow\nabla(l^{k-1}-l^{k})\times\eta\mbox{//}\eta\quad\mbox{in}\quad\Gamma_{k}.$

thereby, $\nabla(l^{k-1}-l^{k})=0$ in $\Gamma_{k}$, it implies that
$l^{k-1}-l^{k}=\mbox{C}$ in $\Gamma_{k}$, $\mbox{C}=\mbox{constant}$.
Substituting (36) in the second tern on the right-hand side of (35), then

$\displaystyle
2\sum_{k=1}^{m}\int_{0}^{s}\int_{\Gamma_{k}}\left[\beta^{k-1}l^{k-1}(u^{k-1}.\eta)-\beta^{k}l^{k}(u^{k}.\eta)\right]$
$\displaystyle=2\sum_{k=1}^{m}\int_{0}^{s}\int_{\Gamma_{k}}(l^{k-1}-l^{k})\beta^{k}(u^{k}.\eta)$
(37)
$\displaystyle=2\mbox{C}\sum_{k=1}^{m}\int_{0}^{s}\int_{\Gamma_{k}}\beta^{k}(u^{k}.\eta)$
$\displaystyle=2\int_{0}^{s}\int_{S_{0}}\beta(u.\eta)=0$

and following substituting (37) in (35). ∎

Substituting the obtained estimations, (19), (20), (21), (22),(29), (37) in
(18), then ,

$\displaystyle s\sum_{k=0}^{m}\int_{\Omega_{k}}\left[\beta^{k}\mid
u^{k}\mid^{2}+\alpha^{k}(p^{k})^{2}\right]dx$
$\displaystyle=2C_{1}s\sum_{k=0}^{m}\int_{\Omega_{k}}\left[\beta^{k}\mid
u^{k}\mid^{2}+\alpha^{k}(p^{k})^{2}\right]dx-$ (38)
$\displaystyle\frac{\partial}{\partial
s}\sum_{k=0}^{m}\alpha^{k}\left[p^{k}(x,r)dr\right]^{2}dx+\int_{0}^{s}\beta\mid
u\mid^{2}\frac{\partial h}{\partial\eta}+$
$\displaystyle\sum_{k=0}^{m}\int_{0}^{s}\int_{\Gamma_{k}}\left[\vec{B}^{k-1}.\eta-\vec{B}^{k}.\eta\right]+\int_{0}^{s}\int_{S_{0}}\left(\alpha
p^{2}\frac{\partial h}{\partial\eta}-\beta\mid u\mid^{2}\frac{\partial
h}{\partial\eta}\right)+$
$\displaystyle\delta_{0}(1-\mu(\Omega))\sum_{k=0}^{m}\int_{0}^{s}\int_{\Gamma_{k}}\left[\beta^{k}\mid
u^{k}\mid^{2}+\alpha^{k}(p^{k})^{2}\right]+$ $\displaystyle
2\sum_{k=0}^{m}\int_{\Omega_{k}}l^{k}(p^{k}(x,s)-p^{k}_{0}(x))dx$

Integrating (38) in $(0,T)$ and using the independence of energy of the model
with the time, we have that

$\displaystyle\frac{T}{2}\left[1-\delta_{0}(1-\mu(\Omega))\right]\sum_{k=0}^{m}\int_{\Omega_{k}}\left[\beta^{k}\mid
u^{k}\mid^{2}+\alpha^{k}(p^{k})^{2}\right]dx\leq
2C_{1}T\sum_{k=0}^{m}\int_{\Omega_{k}}\left[\beta^{k}\mid
u^{k}\mid^{2}+\alpha^{k}(p^{k})^{2}\right]dx-$ (39)
$\displaystyle\sum_{k=0}^{m}\int_{\Omega_{k}}\alpha^{k}\left[\int_{0}^{T}p^{k}(x,r)\right]^{2}dx+\int_{0}^{T}\int_{0}^{s}\int_{S_{1}}\beta\mid
u\mid^{2}\frac{\partial
h}{\partial\eta}+\sum_{k=0}^{m}\int_{0}^{T}\int_{0}^{s}\int_{\Gamma_{k}}\left[\vec{B}^{k-1}.\eta-\vec{B}^{k}.\eta\right]+$
$\displaystyle\int_{0}^{T}\int_{0}^{s}\int_{S_{0}}\left(\alpha
p^{2}\frac{\partial h}{\partial\eta}-\beta\mid u\mid^{2}\frac{\partial
h}{\partial\eta}\right)+2\sum_{k=0}^{m}\int_{0}^{T}\int_{\Omega_{k}}l^{k}(p^{k}(x,s)-p^{k}_{0}(x))$

Now, we need the hypothesis in the domain $\Omega$. Given $\delta_{0}>0$ such
that, some $x_{0}\in\sigma_{1}$, we have

${}\left\\{\begin{aligned} &\delta_{0}(1-\mu(\Omega))<1,\\\
&(x-x_{0}).\eta\geq-2\delta_{0}\frac{\mbox{Vol}(\Omega)}{\mbox{area}(S_{0})},\quad\mbox{for}\quad
x\in S_{0}\\\
&(x-x_{0}).\eta\leq\delta_{0}\frac{\mbox{Vol}(\Omega)}{\mbox{area}(S_{1})},\quad\mbox{for}\quad
x\in S_{1}\\\ &(x-x_{0}).\eta+\delta_{0}\frac{\partial\Phi}{\partial\eta}\geq
0,\quad\forall x\in\Gamma_{k},\quad k=1,2,\dots,m\end{aligned}\right.$ (40)

######  Remark 3.8.

The hypothesis made in (40), These are true when $\delta_{0}=0$ satisfied the
surf of kind ”star-shaped”.

Using (40) in $S_{1}$, we have

${}\frac{\partial h}{\partial\eta}=\nabla
h.\eta=(x-x_{0}).\eta+\delta_{0}\frac{\partial\Phi}{\partial\eta}=(x-x_{0}).\eta-\delta_{0}\frac{\mbox{Vol}(\Omega)}{\mbox{area}S_{1}}\leq
0$ (41)

and, for $x\in S_{0}$,

${}\frac{\partial h}{\partial\eta}=\nabla
h.\eta=(x-x_{0}).\eta+\delta_{0}\frac{\partial\Phi}{\partial\eta}=(x-x_{0}).\eta+2\delta_{0}\frac{\mbox{Vol}(\Omega)}{\mbox{area}S_{0}}\geq
0$ (42)

Substituting (42) and (41) in (39), then

$\int_{0}^{T}\int_{0}^{s}\int_{S_{1}}\beta\mid u\mid^{2}\frac{\partial
h}{\partial\eta}\leq 0$ (43)

and

$-\int_{0}^{T}\int_{0}^{s}\int_{S_{0}}\beta\mid u\mid^{2}\frac{\partial
h}{\partial\eta}\leq 0$ (44)

moreover, if the coefficients $\alpha^{k},\beta^{k}$ satisfied

$\displaystyle{}\alpha^{k-1}$ $\displaystyle\leq$ $\displaystyle\alpha^{k}$
(45) $\displaystyle\beta^{k-1}$ $\displaystyle\leq$ $\displaystyle\beta^{k}$

then, the fourth condition in (40), (45) and the lemma (3.3) we have that

${}\sum_{k=1}^{m}\int_{0}^{T}\int_{0}^{s}\int_{\Gamma_{k}}\left[\vec{B}^{k-1}.\eta-\vec{B}^{k}.\eta\right]\leq
0$ (46)

Using that the associate energy to the system (1)-(3); with not dependency of
the time, we prove the next lemma

######  Lemma 3.9.

Given a regular solution $\\{u,p\\}$ of the problem (1) by theorem 2.3, and
the initial condition $u_{0}$ that satisfied (34). Then

$\displaystyle
2\sum_{k=0}^{m}\int_{0}^{T}\int_{\Omega_{k}}l^{k}(p^{k}(x,s)-p^{k}_{0}(x))dxds$
$\displaystyle\leq\sum_{k=0}^{m}\int_{\Omega_{k}}\left[\int_{0}^{T}p^{k}(x,s)\right]^{2}dx+$
$\displaystyle(C_{3}+C_{4}T)\sum_{k=0}^{m}\int_{\Omega_{k}}\left[\beta^{k}\mid
u^{k}\mid^{2}+\alpha^{k}(p^{k})^{2}\right]dx$

where $C_{3}=C_{2}\max_{k}\left\\{(\alpha^{k}\beta^{k})^{-1}\right\\}$ and
$C_{4}=\max_{k}\left\\{C_{2}(\beta^{k})^{-1},(\alpha^{k})^{-1}\right\\}$

###### Proof.

$\displaystyle
2\sum_{k=0}^{m}\int_{0}^{T}\int_{\Omega_{k}}l^{k}(p^{k}(x,s)-p^{k}_{0}(x))=2\sum_{k=0}^{m}\int_{0}^{T}\int_{\Omega_{k}}\mid
l^{k}p^{k}(x,s)\mid+2\sum_{k=0}^{m}\int_{0}^{T}\int_{\Omega_{k}}\mid
l^{k}p^{k}_{0}(x)\mid$ (47) $\displaystyle
2\leq\sum_{k=0}^{m}\int_{\Omega_{k}}\mid l^{k}(x)\mid\int_{0}^{T}\mid
l^{k}p^{k}(x,t)\mid+2\sum_{k=0}^{m}\int_{0}^{T}\left(\int_{\Omega_{k}}\mid
l^{k}(x)\mid^{2}\right)^{1/2}\left(\int_{\Omega_{k}}\mid
p^{k}(x,s)\mid^{2}\right)^{1/2}$ $\displaystyle\leq
2\sum_{k=0}^{m}\left[\int_{\Omega_{k}}(l^{k}(x))^{2}\right]^{1/2}\left[\int_{\Omega_{k}}\left[p^{k}(x,s)\right]^{2}\
dx\right]^{1/2}+\sum_{k=0}^{m}\int_{0}^{T}\int_{\Omega_{k}}\left\\{(l^{k})^{2}+(p^{k})^{2}\right\\}$
$\displaystyle\leq\sum_{k=0}^{m}\left\\{(\alpha^{k-1})^{-1}\int_{\Omega_{k}}(l^{k}(x))^{2}+\alpha^{k}\int_{\Omega_{k}}\left[\int_{0}^{T}p^{k}(x,s)\right]^{2}dx\right\\}+\sum_{k=0}^{m}\int_{0}^{T}\int_{\Omega_{k}}\left\\{(l^{k})^{2}+(p^{k})^{2}\right\\}$
$\displaystyle\leq
C_{2}\sum_{k=0}^{m}\int_{\Omega_{k}}(\alpha^{k})^{-1}\mid\nabla
l^{k}\mid^{2}+\sum_{k=0}^{m}\alpha^{k}\left[\int_{0}^{T}p^{k}(x,s)\right]^{2}dx+C_{2}\sum_{k=0}^{m}\int_{0}^{T}\int_{\Omega_{k}}\mid\nabla
l^{k}\mid^{2}+\sum_{k=0}^{m}\int_{0}^{T}\int_{\Omega_{k}}(p^{k})^{2}$
$\displaystyle\leq C_{2}\sum_{k=0}^{m}\int_{\Omega_{k}}(\alpha^{k})^{-1}\mid
u_{0}^{k}(x)\mid^{2}+\sum_{k=0}^{m}\alpha^{k}\left[\int_{0}^{T}p^{k}(x,s)\right]^{2}dx+C_{2}\sum_{k=0}^{m}\int_{0}^{T}\int_{\Omega_{k}}\mid\nabla
l^{k}\mid^{2}+\sum_{k=0}^{m}\int_{0}^{T}\int_{\Omega_{k}}(p^{k})^{2}$
$\displaystyle\leq
C_{2}\max_{k}\\{(\alpha^{k}\beta^{k})^{-1}\\}\sum_{k=0}^{m}\int_{\Omega_{k}}\left[\beta^{k}\mid
u^{k}\mid^{2}+\alpha^{k}(p^{k})^{2}\right]+\sum_{k=0}^{m}\alpha^{k}\left[\int_{0}^{T}p^{k}(x,s)\right]^{2}dx+$
$\displaystyle
C_{2}\sum_{k=0}^{m}\int_{\Omega_{k}}(\beta^{k})^{-1}\int_{0}^{T}\int_{\Omega_{k}}\left[\beta^{k}\mid
u^{k}\mid^{2}+\alpha^{k}(p^{k})^{2}\right]+\sum_{k=0}^{m}(\alpha^{k})^{-1}\int_{0}^{T}\int_{\Omega_{k}}\left[\beta^{k}\mid
u^{k}\mid^{2}+\alpha^{k}(p^{k})^{2}\right]$ $\displaystyle\leq
C_{2}\max_{k}\\{(\alpha^{k}\beta^{k})^{-1}\\}\sum_{k=0}^{m}\int_{\Omega_{k}}\left[\beta^{k}\mid
u^{k}\mid^{2}+\alpha^{k}(p^{k})^{2}\right]+\sum_{k=0}^{m}\alpha^{k}\left[\int_{0}^{T}p^{k}(x,s)\right]^{2}dx+$
$\displaystyle\max_{k}\\{C_{2}(\beta^{k}))^{-1},(\alpha^{k})^{-1}\\}T\sum_{k=0}^{m}\int_{\Omega_{k}}\left[\beta^{k}\mid
u^{k}\mid^{2}+\alpha^{k}(p^{k})^{2}\right]dx$
$\displaystyle\leq\sum_{k=0}^{m}\int_{\Omega_{k}}\left[\int_{0}^{T}p^{k}(x,s)\right]^{2}dx+(C_{3}+C_{4}T)\sum_{k=0}^{m}\int_{\Omega_{k}}\left[\beta^{k}\mid
u^{k}\mid^{2}+\alpha^{k}(p^{k})^{2}\right]dx$

$C_{3}=C_{2}\max_{k}\left\\{(\alpha^{k}\beta^{k})^{-1}\right\\}$ and
$C_{4}=\max_{k}\left\\{C_{2}(\beta^{k})^{-1},(\alpha^{k})^{-1}\right\\}$ where
$C_{3}=C_{2}\max_{k}\left\\{(\alpha^{k}\beta^{k})^{-1}\right\\}$ and
$C_{4}=\max_{k}\left\\{C_{2}(\beta^{k})^{-1},(\alpha^{k})^{-1}\right\\}$ ∎

Substituting in (39), and using the $\frac{\partial h}{\partial\eta}\geq 0$ in
$S_{0}$, we have that

$\displaystyle\frac{T}{2}\left[1-\delta_{0}(1-\mu(\Omega))\right]\sum_{k=0}^{m}\int_{\Omega_{k}}\left[\beta^{k}\mid
u^{k}\mid^{2}+\alpha^{k}(p^{k})^{2}\right]dx$ (48)
$\displaystyle\leq\left\\{(2C_{1}+C_{4})T+C_{3}\right\\}\sum_{k=0}^{m}\int_{\Omega_{k}}\left[\beta^{k}\mid
u^{k}\mid^{2}+\alpha^{k}(p^{k})^{2}\right]dx+T\int_{0}^{T}\int_{S_{0}}\alpha
p^{2}\frac{\partial h}{\partial\eta}$

thus,

$\displaystyle\frac{T}{2}\left[1-\delta_{0}(1-\mu(\Omega))\right]\sum_{k=0}^{m}\int_{\Omega_{k}}\left[\beta^{k}\mid
u^{k}\mid^{2}+\alpha^{k}(p^{k})^{2}\right]dx-2C_{5}\sum_{k=0}^{m}\int_{\Omega_{k}}\left[\beta^{k}\mid
u^{k}\mid^{2}+\alpha^{k}(p^{k})^{2}\right]dx\leq$ (49) $\displaystyle
2\int_{0}^{T}\int_{S_{0}}\alpha p^{2}\frac{\partial h}{\partial\eta}dS_{0}dt$

where $C_{5}=C_{3}+(2C_{1}+C_{4})T$ and considered $T>\max{1,(2C_{1}+C_{4})}$.
We have proved the

######  Theorem 3.10.

Taking $\Phi$ as in (31), the geometry properties (40), and the hypothesis of
monotony of coefficients (45) and the hypothesis (34); these were made for the
initial condition. Then, $\exists C_{5}>0$, with independence of
$t,u,u_{0},p_{0}$, such that

$\frac{T}{2}\left[1-\delta_{0}(1-\mu(\Omega))-2C_{5}\right]\sum_{k=0}^{m}\int_{\Omega_{k}}\left[\beta^{k}\mid
u^{k}\mid^{2}+\alpha^{k}(p^{k})^{2}\right]dx\leq
2\int_{0}^{T}\int_{S_{0}}\alpha p^{2}\frac{\partial h}{\partial\eta}dS_{0}dt.$

The same manner, we obtain the inequality of observability for the system
(2)-(4) with their interface conditions and the monotonicity of the
coefficients, given by:

$\displaystyle{}\gamma^{k-1}$ $\displaystyle\leq\gamma^{k}$ (50)
$\displaystyle\tau^{k-1}$ $\displaystyle\leq\tau^{k}$

######  Theorem 3.11.

Assuming $\Phi$ as in (31), the monotonicity for the coefficients (50) and the
hypothesis of the theorem 3.10 with $h(x)=\frac{1}{2}\mid
x-x_{0}\mid^{2}+\delta_{0}\Phi(x)$ and $(v_{0},q_{0})\in V_{2}\cap D(A_{2})$,
$v^{k}_{0}=\nabla m^{k}$, with $m^{k}\in H^{2}(\Omega_{k})$, $m=0$ in $S_{1}$.
Then, there is a constant $C_{6}>0$, with independence of $t,v_{0},q_{0}$ such
that

$\displaystyle
T\left[1-\delta_{0}(1-\mu(\Omega))\right]\sum_{k=0}^{m}\int_{\Omega_{k}}\left[\tau^{k}\mid
v^{k}\mid^{2}+\gamma^{k}(q^{k})^{2}\right]dx-2C_{6}\sum_{k=0}^{m}\int_{\Omega_{k}}\left[\tau^{k}\mid
v^{k}\mid^{2}+\gamma^{k}(q^{k})^{2}\right]dx$ $\displaystyle\leq
2\int_{0}^{T}\int_{S_{0}}\tau\mid v.\eta\mid^{2}\frac{\partial
h}{\partial\eta}dS_{0}dt.$

###### Proof.

The proof was obtained using the results in [10] and the estimations that were
made in the proof of the theorem 3.10 ∎

Assuming the hypothesis of the theorems 3.10 and 3.11, we obtain the
inequalities of the observability:

$\displaystyle(T-T_{0})\sum_{k=0}^{m}\int_{\Omega_{k}}\left[\beta^{k}\mid
u^{k}\mid^{2}+\alpha^{k}(p^{k})^{2}+\tau^{k}\mid
v^{k}\mid^{2}+\gamma^{k}(q^{k})^{2}\right]dx$ (51) $\displaystyle\leq
C_{7}\int_{0}^{T}\int_{S_{0}}\left[\alpha p^{2}+\tau\mid
v.\eta\mid^{2}\right]\frac{\partial h}{\partial\eta}dS_{0}dt.$

For any $T\geq
T_{0}=\max\left\\{1,\frac{C_{5}+C_{6}}{1-\delta_{0}(1-\mu(\Omega))}\right\\}$.
That was made in [10] and [8], (51). This is an inequality of observability,
moreover, it is not convenient to use the H.U.M technique. Starting of (51),
we obtain an appropriate inequality.

######  Theorem 3.12.

Assuming the hypothesis of the theorem 3.10, 3.11. Moreover, we suppose that :

$\displaystyle\alpha^{k}\beta^{k}$ $\displaystyle=\gamma^{k}\tau^{k}$ (52)
$\displaystyle\beta^{k-1}\tau^{k}$ $\displaystyle=\beta^{k}\tau^{k-1}.$

Then, There is a positive constant $C>0$ such that

$\displaystyle(T-T_{0})\sum_{k=0}^{m}\int_{\Omega_{k}}\left[\beta^{k}\mid
u^{k}\mid^{2}+\alpha^{k}(p^{k})^{2}+\tau^{k}\mid
v^{k}\mid^{2}+\gamma^{k}(q^{k})^{2}\right]dx$ $\displaystyle\leq
C\int_{0}^{T}\int_{S_{0}}\left[\alpha p-\tau(v.\eta)\right]^{2}\frac{\partial
h}{\partial\eta}dS_{0}dt.$

$\forall
T>\max\left\\{1,\frac{C_{5}+C_{6}}{1-\delta_{0}(1-\mu(\Omega))}\right\\}$.

###### Proof.

Using the theorems (4) and (52), we have that

$\displaystyle\frac{d}{dt}\int_{\Omega}(\tau u.v+\alpha pq)dx$
$\displaystyle=\frac{d}{dt}\sum_{k=0}^{m}\int_{\Omega_{k}}(\tau^{k}u^{k}.v^{k}+\alpha^{k}p^{k}q^{k})dx=\sum_{k=0}^{m}\int_{\Omega_{k}}\left\\{\tau^{k}\frac{\partial
u^{k}}{\partial t}.v^{k}+\tau^{k}u^{k}.\frac{\partial v^{k}}{\partial
t}+\right.$ (53) $\displaystyle\left.\alpha^{k}\frac{\partial p^{k}}{\partial
t}q^{k}+\alpha^{k}p^{k}\frac{\partial q^{k}}{\partial t}\right\\}$
$\displaystyle=\sum_{k=0}^{m}\int_{\Omega_{k}}\left\\{-\alpha^{k}\tau^{k}\nabla
p^{k}.v^{k}-\tau^{k}\gamma^{k}u^{k}\nabla
q^{k}-\alpha^{k}\beta^{k}\mbox{div}(u^{k})q^{k}-\alpha^{k}\tau^{k}p^{k}\mbox{div}(v^{k})\right\\}$
$\displaystyle=-\sum_{k=0}^{m}\int_{\partial\Omega_{k}}\alpha^{k}\tau^{k}p^{k}(v^{k}.\eta)+\sum_{k=0}^{m}\int_{\Omega_{k}}\left\\{\alpha^{k}\tau^{k}p^{k}\mbox{div}(v^{k})-\tau^{k}\gamma^{k}u^{k}\nabla
q^{k}-\right.$
$\displaystyle\left.\alpha^{k}\beta^{k}\mbox{div}(u^{k})q^{k}-\alpha^{k}\tau^{k}p^{k}\mbox{div}(v^{k})\right\\}dx$
$\displaystyle=-\sum_{k=0}^{m}\int_{\partial\Omega_{k}}\alpha^{k}\tau^{k}p^{k}(v^{k}.\eta)-\sum_{k=0}^{m}\int_{\partial\Omega_{k}}\tau^{k}\gamma^{k}q^{k}(u^{k}.\eta)+$
$\displaystyle\sum_{k=0}^{m}\int_{\partial\Omega_{k}}\left\\{\gamma^{k}\tau^{k}\mbox{div}(u^{k})q^{k}-\alpha^{k}\beta^{k}\mbox{div}(u^{k})q^{k}\right\\}$
$\displaystyle=-\sum_{k=0}^{m}\int_{\partial\Omega_{k}}\alpha^{k}\tau^{k}p^{k}(v^{k}.\eta)-\sum_{k=0}^{m}\int_{\partial\Omega_{k}}\tau^{k}\gamma^{k}q^{k}(u^{k}.\eta)$
$\displaystyle=-\int_{S_{1}}\alpha\tau
p(v.\eta)-\sum_{k=0}^{m}\int_{\Gamma_{k}}\left[\alpha^{k-1}\tau^{k-1}p^{k-1}(v^{k-1}.\eta)-\alpha^{k}\tau^{k}p^{k}(v^{k}.\eta)\right]-$
$\displaystyle\int_{S_{0}}\alpha\tau p(v.\eta)-\int_{S_{1}}\tau\gamma
q(u.\eta)-\sum_{k=0}^{m}\int_{\Gamma_{k}}\left[\tau^{k-1}\gamma^{k-1}q^{k-1}(u^{k-1}.\eta)-\tau^{k}\gamma^{k}q^{k}(u^{k}.\eta)\right]-$
$\displaystyle\int_{S_{0}}\tau\gamma q(u.\eta)$

and, using the boundary condition and (52) in the interfaces, we have that

${}\frac{d}{dt}\sum_{k=0}^{m}\int_{\Omega_{k}}(\tau^{k}u^{k}.v^{k}+\alpha^{k}p^{k}q^{k})dx=-\int_{S_{0}}\alpha\tau
p(v.\eta)dS_{0}$ (54)

This is clearly that the first, second, fourth and sixth term of (53) are null
only substituting directly the interface conditions. The fifth term is made
using (52)

$\displaystyle\sum_{k=0}^{m}\int_{\Gamma_{k}}\left[\tau^{k-1}\gamma^{k-1}q^{k-1}(u^{k-1}.\eta)-\tau^{k}\gamma^{k}q^{k}(u^{k}.\eta)\right]$
$\displaystyle=\sum_{k=0}^{m}\int_{\Gamma_{k}}\left[\frac{\tau^{k-1}}{\beta^{k-1}}\beta^{k-1}\gamma^{k-1}q^{k-1}(u^{k-1}.\eta)-\right.$
(55)
$\displaystyle\left.\frac{\tau^{k}}{\beta^{k}}\beta^{k}\gamma^{k}q^{k}(u^{k}.\eta)\right]$
$\displaystyle=\sum_{k=0}^{m}\int_{\Gamma_{k}}\left(\frac{\tau^{k-1}}{\beta^{k-1}}-\frac{\tau^{k}}{\beta^{k}}\right)\gamma^{k}q^{k}\beta^{k}(u^{k}.\eta)$
$\displaystyle=\sum_{k=0}^{m}\int_{\Gamma_{k}}\frac{\gamma^{k}}{\beta^{k-1}}(\beta^{k-1}\tau^{k-1}-\beta^{k-1}\tau^{k})q^{k}(u^{k}.\eta)$
$\displaystyle=0$

and ,

${}\int_{0}^{T}\int_{S_{0}}\left[\alpha
p-\tau(v.\eta)\right]^{2}=\int_{0}^{T}\int_{S_{0}}\alpha^{2}p^{2}+\int_{0}^{T}\int_{S_{0}}\tau^{2}(v.\eta)^{2}-2\tau\alpha\int_{0}^{T}\int_{S_{0}}p(v.\eta)$
(56)

Integrating (54) in $(0,T)$ and substituting in (56), we have that

$\int_{0}^{T}\int_{S_{0}}\left[\alpha
p+\tau(v.\eta)\right]^{2}=\int_{0}^{T}\int_{S_{0}}\alpha^{2}p^{2}+\int_{0}^{T}\int_{S_{0}}\tau^{2}(v.\eta)^{2}+\sum_{k=0}^{m}\int_{\Omega_{k}}(\tau^{k}u^{k}.v^{k}+\alpha^{k}p^{k}q^{k})\mid_{0}^{T}$

Now, we have that

${}\int_{0}^{T}\int_{S_{0}}\left(\alpha^{2}p^{2}+\tau^{2}(v.\eta)^{2}\right)=\int_{0}^{T}\int_{S_{0}}\left[\alpha
p-\tau(v.\eta)\right]^{2}-\sum_{k=0}^{m}\int_{\Omega_{k}}(\tau^{k}u^{k}.v^{k}+\alpha^{k}p^{k}q^{k})\mid_{0}^{T}$
(57)

using (57) to estimate the term of right-hand side of the inequality (51)

$\displaystyle\int_{0}^{T}\int_{S_{0}}\left[\alpha
p^{2}+\tau(v.\eta)^{2}\right]$
$\displaystyle\leq\max\\{(\alpha^{-1}),(\tau^{-1})\\}\int_{0}^{T}\int_{S_{0}}\left[\alpha^{2}p^{2}+\tau^{2}(v.\eta)^{2}\right]$
(58) $\displaystyle\leq C_{8}\int_{0}^{T}\int_{S_{0}}\left[\alpha
p-\tau(v.\eta)\right]^{2}-C_{8}\sum_{k=0}^{m}\int_{\Omega_{k}}(\tau^{k}u^{k}.v^{k}+\alpha^{k}p^{k}q^{k})\mid_{0}^{T}$

Substituting in (51), we have that

$\displaystyle(T-T_{0})\sum_{k=0}^{m}\int_{\Omega_{k}}\left[\beta^{k}\mid
u^{k}\mid^{2}+\alpha^{k}(p^{k})^{2}+\tau^{k}\mid
v^{k}\mid^{2}+\gamma^{k}(q^{k})^{2}\right]dx+C_{8}\sum_{k=0}^{m}\int_{\Omega_{k}}(\tau^{k}u^{k}.v^{k}+\alpha^{k}p^{k}q^{k})\mid_{0}^{T}$
(59) $\displaystyle\leq C_{7}C_{8}\int_{0}^{T}\int_{S_{0}}\left[\alpha
p-\tau(v.\eta)\right]^{2}$

and the proof of the theorem follows the inequalities:

$\displaystyle\sum_{k=0}^{m}\int_{\Omega_{k}}(\tau^{k}u^{k}.v^{k}+\alpha^{k}p^{k}q^{k})\mid_{0}^{T}\leq
C_{9}\left[\beta^{k}\mid u^{k}\mid^{2}+\alpha^{k}(p^{k})^{2}+\tau^{k}\mid
v^{k}\mid^{2}+\gamma^{k}(q^{k})^{2}\right]dx$ (60)

where
$C_{9}=\max\left\\{\tau^{k}(\beta^{k})^{-1},\alpha^{k}(\tau^{k})^{-1}\right\\}$
∎

As a corollary , of uniqueness of the theorem 3.12, we have that:

######  Corollary 3.13.

with the hypothesis of the theorem 3.12, given $\\{u,p\\}$ and $\\{v,q\\}$
solutions of the problem (3) and (2), respectively. Making that

$\alpha p(x,t)=\tau v(x,t).\eta\quad\quad\forall(x,t)\in\Gamma_{0}\times(0,T)$

then, in $T>T_{0}$, $u=v=0$ and $p=q=0$ for all $(x,t)\in\Omega\times(0,T)$

##  4 Exact controllability

As a consequence of the corollary 3.13 , we have that: for $T>T_{0}$, the
expression

${}\left[\int_{0}^{T}\int_{\Gamma_{0}}\left[\alpha
p-\tau(v.\eta)\right]\right]^{1/2}$ (61)

define a norm in a space of initial data $(u_{0},p_{0})$ and $(v_{0},q_{0})$
the problems (3) and (2). We denote by $Y$ the Hilbert space defined as
closure of $V_{1}\cap D(A_{1})\times V_{1}\cap D(A_{1})$ in $X=X_{1}\times
X_{2}$ with the norm (61). The obtained number in (61) is denoted by
$\parallel(u_{0},p_{0},v_{0},v_{0},q_{0})\parallel_{Y}$. Now, $Y\subset X$ and

$\displaystyle\parallel(u_{0},p_{0},v_{0},q_{0})\parallel^{2}_{X}$
$\displaystyle=\parallel(u_{0},p_{0})\parallel^{2}_{X_{1}}+\parallel(v_{0},q_{0})\parallel^{2}_{X_{2}}$
(62) $\displaystyle\leq C\parallel(u_{0},p_{0},v_{0},q_{0})\parallel^{2}_{Y}$

for some positive constant $C$. The dual space of $Y$ respect to $X$ is
denoted by $Y^{{}^{\prime}}$. In $\Omega\times(0,T)$ consider the systems (3)
and (4) with initial condition $(u_{0},p_{0},v_{0},q_{0})\in Y^{{}^{\prime}}$.
Using the transportation method, the solution to the problems (3) and (4) with
no homogeneous contour conditions.

######  Definition 4.1.

Given $(u(x,t),p(x,t),v(x,t),q(x,t))\in C(O,T;Y^{{}^{\prime}})$, is a solution
of (3) and (4), if

${}\left<(u,p,v,q),(\tilde{u},\tilde{p},\tilde{v},\tilde{q})\right>_{X}=\left<(u_{0},p_{0},v_{0},q_{0}),(\tilde{u}_{0},\tilde{p}_{0},\tilde{v}_{0},\tilde{q}_{0})\right>_{X}-\int_{0}^{T}\int_{\Gamma_{0}}\left[\alpha\beta
Q\tilde{p}+\gamma\tau P(\tilde{n}.\eta)\right]d\Gamma_{0}ds$ (63)

for all $(\tilde{u}_{0},\tilde{p}_{0},\tilde{v}_{0},\tilde{q}_{0})\in Y$ and
$0<t<T$. Here $(\tilde{u}_{0},\tilde{p}_{0})$ and
$(\tilde{v}_{0},\tilde{q}_{0})$ are the solutions of (3) and (4),
respectively, for the functions $P,Q\in C(0,T;L^{2}(\Gamma_{0}))$. In (63) is
given by

$\left<(u,p,v,q),(\tilde{u},\tilde{p},\tilde{v},\tilde{q})\right>_{X}=\left<(u,p),(\tilde{u},\tilde{p}\right>_{X_{1}}+\left<(v,q),(\tilde{v},\tilde{q})\right>_{X_{2}}$

######  Definition 4.2.

A solution of (3) and (4); This is null in the time $t=T$. The function
$(u(x,t),p(x,t),v(x,t),q(x,t))(\ref{Par_Ecu_Ondas_Multilayer_1})$ and
$(\ref{Par_Ecu_Ondas_Multilayer_2})inC(0,T;Y^{{}^{\prime}})$ such that

${}\left<(u,p,v,q),(\tilde{u},\tilde{p},\tilde{v},\tilde{q})\right>_{X}=\int_{0}^{T}\int_{\Gamma_{0}}\left[\alpha\beta
Q\tilde{p}+\gamma\tau P(\tilde{n}.\eta)\right]d\Gamma_{0}ds$ (64)

for all $(\tilde{u},\tilde{p},\tilde{v},\tilde{q})\in Y$ and $0<t<T$

Given the lineal and reversible systems (3) and (4) in the time; it is clearly
to solve the problem of exact controllability, it is sufficient to prove that,
for all initial condition in $Y^{{}^{\prime}}$, and their solutions, it can be
take in the equilibrium of time $T$.
Given $G_{1}=(w_{0},k_{0})$ and $G_{2}=(m_{0},l_{0})$ arbitrary elements of
$Y$. We denote by

$\displaystyle(w(x,t),k(x,t))=U_{1}(t)(w_{0},k_{0})$
$\displaystyle(m(x,t),l(x,t))=U_{2}(t)(m_{0},l_{0})$

consider the following functions

$\displaystyle Q=\beta\left(\alpha k(x,t)-\tau m(x,t).\eta\right)$ (65)
$\displaystyle P=-\frac{\beta}{\gamma}Q$

and given $(u,p)$ and $(v,q)$ the solution of (3) and (4), these are null in
the instant $T$, $(T>T_{0})$ and the contour conditions (65).
Considering the map

$\displaystyle\wedge:$ $\displaystyle Y\longrightarrow Y^{{}^{\prime}}$
$\displaystyle(G_{1},G_{2})\longrightarrow\wedge(G_{1},G_{2})=(u,p,v,q)\mid_{t=0}$

Using (64) in $t=0$ and substituting $P$ and $Q$ given by (65), we have

$\displaystyle\left<\wedge(G_{1},G_{2}),(\tilde{u}_{0},\tilde{p}_{0},\tilde{v}_{0},\tilde{q}_{0})\right>_{X}$
$\displaystyle=\int_{0}^{T}\int_{\Gamma_{0}}\left[\alpha\beta
Q\tilde{p}+\gamma\tau P\tilde{v}.\eta\right]d\Gamma_{0}ds$ (66)
$\displaystyle=\int_{0}^{T}\int_{\Gamma_{0}}(\gamma k-\tau
m.\eta)(\alpha\tilde{p}-\tau\tilde{v}.\eta)d\Gamma_{0}ds$
$\displaystyle=\left<(G_{1},G_{2}),(\tilde{u}_{0},\tilde{p}_{0},\tilde{v}_{0},\tilde{q}_{0})\right>_{Y}$

of (66), we can conclude that $\wedge$ is an isomorphism of $Y$ in
$Y^{{}^{\prime}}$. Putting

$\displaystyle(G_{1},G_{2})$
$\displaystyle=\wedge^{-1}((u_{0},p_{0}),(v_{0},q_{0}))$ (67) $\displaystyle
Q$ $\displaystyle=\beta^{-1}\left(\alpha k(x,t)-\tau m(x,t).\eta\right)$
$\displaystyle P$ $\displaystyle=-\frac{\beta}{\gamma}Q$

Using (63) with $t=T>T_{0}$, we have that

$\displaystyle\left<(u(x,T),p(x,T),v(x,T),q(x,T)),(U_{1}(t)(\tilde{u}_{0},\tilde{p}_{0}),U_{2}(t)(\tilde{v},\tilde{q}))\right>_{X}=$
$\displaystyle=\left<\wedge(G_{1},G_{2}),(\tilde{u}_{0},\tilde{p}_{0},\tilde{v}_{0},\tilde{q}_{0})\right>_{X}-\left<(G_{1},G_{2}),(\tilde{u}_{0},\tilde{p}_{0},\tilde{v}_{0},\tilde{q}_{0})\right>_{Y}$

for all $(\tilde{u}_{0},\tilde{p}_{0},\tilde{v}_{0},\tilde{q}_{0})\in Y$.
Using (66), we have that $(u(x,T),p(x,T),v(x,T),q(x,T))$ is a functional null
on $Y$. In conclusion we prove the following theorem

######  Theorem 4.3.

Assuming the hypothesis of theorem 3.12. If given $T>T_{0}$ and initial
condition $(u_{0},p_{0},v_{0},q_{0})\in Y^{{}^{\prime}}$ the problem (1), (2),
(5), (5). Then, there is exist an control $Q(x,t)\in C(0,T;L^{2}(\Gamma_{0}))$
such that the corresponding solution $(u,p,v,q)$ with the boundary condition

$\left\\{\begin{aligned} &u.\eta=Q,\quad\mbox{in}\quad S_{0}\times(0,T)\\\
&p=0,\quad\mbox{in}\quad S_{1}\times(0,T)\\\
&u(x,0)=u_{0}(x),p(x,0)=p_{0}(x)\end{aligned}\right.$ (68)

and

$\left\\{\begin{aligned} &q=P,\quad\mbox{in}\quad S_{0}\times(0,T)\\\
&q=0,\quad\mbox{in}\quad S_{1}\times(0,T)\\\
&v(x,0)=v_{0}(x),q(x,0)=q_{0}(x)\end{aligned}\right.$ (69)

with $P=-\beta\gamma^{-1}Q$, satisfy , for $x\in\Omega$

$\left\\{\begin{aligned} &u(x,T)=0\\\ &p(x,T)=0\\\ &v(x,T)=0\\\
&q(x,T)=0\end{aligned}\right.$ (70)

##  5 Acknowledgment

We thank to professors G. Perla Mezala and Boris Kapitonov for the suggestions
in this article.

## References

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:
LEAVE UNSET 2021 LEAVE UNSET LEAVE UNSET Machine Learning for Health (ML4H)
2021

# Monte Carlo dropout increases model repeatability

Andreanne Lemay1,2,3<EMAIL_ADDRESS>
1 Martinos Center for Biomedical Imaging  Department of Radiology
Massachusetts General Hospital  Boston  USA
2 NeuroPoly Lab  Institute of Biomedical Engineering  Polytechnique Montreal
Canada
3 Mila  Quebec AI Institute  Canada
Katharina Hoebel1,4<EMAIL_ADDRESS>
4 Massachusetts Institute of Technology  Cambridge  USA
Christopher P. Bridge1,5<EMAIL_ADDRESS>
5 MGH and BWH Center for Clinical Data Science  Boston  USA
Didem Egemen6<EMAIL_ADDRESS>
6 National Cancer Institute  Clinical Genetics Branch  Division of Cancer
Epidemiology & Genetics  Rockville  USA
Ana Cecilia Rodriguez6<EMAIL_ADDRESS>
Mark Schiffman6<EMAIL_ADDRESS>
John Peter Campbell7<EMAIL_ADDRESS>
7 Oregon Health and Science University  Portland  USA
Jayashree Kalpathy-Cramer1<EMAIL_ADDRESS>

###### Abstract

The integration of artificial intelligence into clinical workflows requires
reliable and robust models. Among the main features of robustness is
repeatability. Much attention is given to classification performance without
assessing the model repeatability, leading to the development of models that
turn out to be unusable in practice. In this work, we evaluate the
repeatability of four model types on images from the same patient that were
acquired during the same visit. We study the performance of binary, multi-
class, ordinal, and regression models on three medical image analysis tasks:
cervical cancer screening, breast density estimation, and retinopathy of
prematurity classification. Moreover, we assess the impact of sampling Monte
Carlo dropout predictions at test time on classification performance and
repeatability. Leveraging Monte Carlo predictions significantly increased
repeatability for all tasks on the binary, multi-class, and ordinal models
leading to an average reduction of the 95% limits of agreement by 17% points.

###### keywords:

repeatability, Monte Carlo dropout, cervical screening, breast density,
retinopathy of prematurity, medical classification, computer vision

##  1 Introduction

While the majority of the deep learning (DL) work for medical image
classification focuses on accuracy and classification performance, little is
known about the repeatability of these models. It is of utmost importance and
key to model reliability, that models provide consistent predictions in
clinical settings. Empirically, minor changes in an image can lead to vastly
different predictions by DL models. In clinical practice, this repeatability
issue could lead to dangerous medical errors. Figure 1 illustrates this issue.
Automatic visual assessment of cervix images by artificial intelligence models
has the potential to support cervical cancer screening in low-resource
settings. Two cervical cancer screening images from the same precancerous
cervix acquired during the same visit lead to completely different
predictions. A binary DL model (without dropout layers) trained to distinguish
between a normal cervix and one with precancerous lesions (0: Normal, 1: Pre-
cancer) predicted a normal cervix on one image and classified the second one
as precancerous. This difference is represented by prediction results at each
extreme of the spectrum, i.e., 0.01 and 0.98, suggesting high model certainty
for both outputs. Repeatability is important as it reflects how reliably the
model can achieve a certain classification performance.

[Model prediction: 0.01 (Normal)] [Model prediction: 0.98 (Pre-cancer)]

Figure 1: Illustration of repeatability issues from DL models on different
images of a cervix with precancerous lesions from the same patient taken the
same day. (a) the binary model (non MC) predicts a normal cervix (severity
score: 0.01). (b) the binary model predicts pre-cancer (severity score: 0.98).

tab:dataset

Table 1: Datasets description. Mean image per patient is the mean number of
images available for the same anatomical structure at the same timepoint. img:
image; No.: Normal; Scat.: scattered; Hetero: Heterogeneous.

| # images | Label distribution |  Mean img/patient | Train/Val/Test split
---|---|---|---|---
Cervical | 3509 | No./Gray/Precancer (33%/33%/34%) | 2 | 65%/10%/25%
DMIST | 108,230 | Fatty/Scat./Hetero./Dense (12%/44%/38%/6%) | 5 | 65%/10%/25%
ROP | 5511 | No./Pre-Plus/Plus (82%/15%/3%) | 3 | 78%/13%/8%

###  1.1 Study outline

In the competition to achieve higher and higher classification performance
important aspects like test-retest variability remain overlooked and not all
DL models are equal with respect to their repeatability. To the best of our
knowledge, this is the first study to propose Monte Carlo (MC) dropout (Hinton
et al., 2012; Srivastava et al., 2014) as a method to improve repeatability
and systematically assess it on different tasks, model types, and network
architectures. In this work, we evaluate the repeatability of four types of DL
classification and regression models, binary, multi-class, ordinal, and
regression, each with and without MC dropout. We test the repeatability of
these models’ predictions on three different medical image analysis tasks:
cervical cancer screening, breast density estimation, and retinopathy of
prematurity (ROP) disease severity classification. Based on our results, we
present recommendations for model selection to improve repeatability.

##  2 Methods

fig:bland_altman

Figure 2: Bland-Altman plots on images from the same patient and visit for
cervical classification models. The y-axis of each graph represents the
difference in model prediction for images of the same patient, while the
x-axis refers to the mean of the predicted scores. The 95% limits of agreement
are presented with dashed lines.

tab:metric-table

Table 2: Model performance overview (MEAN $\pm$ 95% CI). Values in bold
indicate statistical difference between MC and non-MC models. LoA: Limits of
agreement; Acc.: Accuracy; CI: Confidence interval.

| Cervical | Breast density | ROP
---|---|---|---
Model |  95% LoA $\downarrow$ |  Acc. $\uparrow$ |  95% LoA $\downarrow$ |  Acc. $\uparrow$ |  95% LoA $\downarrow$ |  Acc. $\uparrow$
Binary | $0.68\pm 0.07$ | $0.73\pm 0.03$ | $0.58\pm 0.01$ | $0.84\pm 0.00$ | $0.88\pm 0.04$ | $0.81\pm 0.02$
MC Binary | $0.33\pm 0.04$ | $0.75\pm 0.03$ | $0.48\pm 0.01$ | $0.85\pm 0.00$ | $0.55\pm 0.05$ | $0.85\pm 0.02$
Multi class | $0.50\pm 0.06$ | $0.47\pm 0.03$ | $0.33\pm 0.00$ | $0.69\pm 0.01$ | $0.48\pm 0.03$ | $0.85\pm 0.02$
MC multi-class | $0.22\pm 0.03$ | $0.52\pm 0.03$ | $0.30\pm 0.00$ | $0.71\pm 0.01$ | $0.39\pm 0.03$ | $0.85\pm 0.02$
Ordinal | $0.51\pm 0.07$ | $0.47\pm 0.03$ | $0.33\pm 0.00$ | $0.68\pm 0.01$ | $0.40\pm 0.04$ | $0.82\pm 0.02$
MC ordinal | $0.29\pm 0.03$ | $0.49\pm 0.03$ | $0.29\pm 0.01$ | $0.69\pm 0.01$ | $0.34\pm 0.03$ | $0.83\pm 0.02$
Regression | $0.29\pm 0.03$ | $0.44\pm 0.03$ | $0.21\pm 0.01$ | $0.70\pm 0.01$ | $0.33\pm 0.03$ | $0.86\pm 0.02$
MC Regression | $0.16\pm 0.02$ | $0.43\pm 0.03$ | $0.21\pm 0.01$ | $0.67\pm 0.01$ | $0.33\pm 0.01$ | $0.79\pm 0.03$

###  2.1 Datasets

Three datasets were used to assess the repeatability of the different models:
cervical photographs for certical cancer screening, mammographies for breast
density prediction, and retinal photographs for ROP severity diagnosis
(summarized in Table LABEL:tab:dataset). All datasets were split on a patient
level to prevent leakage of information. Repeatability was assessed as the
maximum difference between model outputs for all available images for each
patient in the test dataset.

#### Cervical

Cervical cancer is the fourth most common cancer in women worldwide and the
leading cause of cancer-related deaths in western, eastern, middle, and
southern Africa (Arbyn et al., 2020). In addition to HPV testing, the visual
assessment of the cervix using photographs can help to detect precancerous
lesions in low-resource settings (Catarino et al., 2015; Xue et al., 2020; Hu
et al., 2019). Each image, originating from one of two studies (Schiffman and
Solomon, 2003; Bratti et al., 2004), was classified using cytological and
histological data from the patient as one of the following three categories:
Normal, Gray zone, i.e., the presence of precancerous lesions was equivocal,
Pre-cancer/cancer. For the binary models, we utilized only images that were
classified as either normal or pre-cancer/cancer.

#### Breast density

The risk to develop breast cancer rises with increasing breast density (Boyd
et al., 1995). We used images from the Digital Mammographic Imaging Sceening
Trial (DMIST) acquired at 33 institutions (Pisano et al., 2005). Breast
density labels were generated according to the BI-RADS criteria (Liberman and
Menell, 2002): fatty, scattered fibroglandular density, heterogeneously dense,
and extremely dense. The breast density models were trained using all
available views for each patient in the training dataset using either four
labels or a simplified binary labelling system of fatty and scattered as one
class, and dense and heterogeneous as the other class.

#### Retinopathy of Prematurity

ROP is the leading cause of preventable childhood blindness worldwide (IAPB,
). Images from the dataset described in Brown et al. (2018) were classified as
normal, pre-plus, or plus disease following previously published methods (Ryan
et al., 2014). The final label was based on the independent image-based
diagnosis by 3 expert graders. The binary models were trained using normal as
one class and pre-plus and plus as the other class. Following Brown et al.
(2018)’s work, we trained ROP classification models using pre-segmented vessel
maps as input. ROP classification models were trained using only the posterior
field of view and the repeatability was assessed on the posterior, temporal,
and nasal views which represents a slight domain shift (i.e., testing on
unseen views).

#### Classification model training

For each dataset, we trained binary, multi-class and ordinal (Cao et al.,
2019) classification models, as well as regression models each with and
without dropout, resulting in a total of 8 models per dataset. Models with
dropout were trained using spatial dropout with a dropout rate of 0.1 for
cervical images and DMIST, and 0.2 for ROP. For the DenseNet121 architecture,
the dropout was applied after every dense layer while for the ResNets the
dropout layer was applied after each residual block. At test time, the dropout
was enabled to generate $N=50$ slightly different predictions and the final
prediction was obtained by averaging over all the MC samples (Gal and
Ghahramani, 2016). We used the following ImageNet pretrained models for each
dataset based on which performed the best for the conventional multi-class
classification model: DenseNet121 (cervix), ResNet50 (DMIST), and ResNet18
(ROP). Models were trained using binary cross-entropy, cross-entropy, CORAL
(Cao et al., 2019), and mean squared error (MSE) losses for binary, multi-
class, ordinal, and regression models respectively. The code was implemented
using the MONAI framework (MONAI Consortium, 2020) based on the PyTorch
library (Paszke et al., 2017).

###  2.2 Evaluation

#### Severity scores

For direct comparison of a model’s predictions, we summarized each model’s
outputs as a continuous severity score. The methodology and analysis were
chosen based on the consideration that the underlying variable of interest,
i.e. disease severity, of these medical tasks is better represented by a
spectrum rather than clear distinct categories. For the binary and regression
models, the output of the models was directly used without further
modifications (labels from 0 to max class - 1). For the multi-class model, we
utilized the ordinality of all three classification problems and defined the
continuous severity score as a weighted average using softmax probability of
each class as described in Equation 1. For cervical and ROP classification (3
classes), the values lie in the range 0 to 2 while the values extend from 0 to
3 for breast density (4 classes).

$score=\sum_{i=1}^{k}p_{i}\times i-1$ (1)

with $k$ being the number of classes and $p_{i}$ the softmax probability of
class $i$. For the ordinal model, the classification problem of $k$ ranks
(i.e., classes) is modified into $k-1$ binary classification problems (Li and
Lin, 2007) leading to one output unit less than for the classification form.
For instance, for a 3-class problem, the ground truth would be encoded as
follows: class 1 $\rightarrow$ [0, 0]; class 2 $\rightarrow$ [1, 0]; class 3
$\rightarrow$ [1, 1]. The continuous prediction score for ordinal models is
obtained by summing the output neurons leading to values ranging from 0 to 2
and 0 to 3 for 3-class and 4-class problems, respectively.

#### Metrics

Repeatability was evaluated using the 95% limits of agreement (LoA) from the
Bland-Altman plots. Since normality was not reached for the differences for
the LoA, non-parametric LoA were calculated using empirical percentiles (Bland
and Altman, 1999). The LoA was presented as a fraction of the possible value
range. The classification accuracy was also reported. For the regression
models, thresholds to binarize predictions for accuracy calculation were
computed by splitting the range of predictions equally (e.g., 3-class problem
with predictions ranging from 0 to 2: $s\leq 0.67\rightarrow$ class 1;
$0.67<s\leq 1.33\rightarrow$ class 2; $s\geq 1.33\rightarrow$ class 3).

#### Statistics

Statistical difference between models was determined using a two-sided
$t$-test and metric bootstrapping (500 iterations). Models with a p-value
smaller than 0.05 were considered significantly different. The normality of
the distribution was verified using the Shapiro-Wilk test ($\alpha=0.05$).

##  3 Results

Bland-Altman plots of all model types for cervical classification are
summarized in Figure LABEL:fig:bland_altman. Ideally, all cases would lie near
a horizontal line crossing the y-axis at 0 which means the difference between
test-retest score is low. For every task, the MC models showed better test-
retest reliability than their conventional counterparts with the exception of
the regression models. This is illustrated by the narrower 95% LoA and the
highest concentration of differences near 0 on the y-axis. The range of
predicted values remained similar for MC models, indicating that the effect of
the MC model is not simply regressing scores towards the mean.

95% LoA (i.e., repeatability metric) and accuracy for each model can be found
in Table LABEL:tab:metric-table. As observed in Figure LABEL:fig:bland_altman,
repeatability was statistically improved when using MC models for binary,
multi-class and ordinal models for all tasks. On average, across all tasks and
classification models (i.e., excluding regression) the 95% limit of agreement
improved by 17% points. Increased repeatability translates to smaller
disagreement rates during test-retest which is desired for classification
tasks. Accuracy increased for all classification MC models. No consistent
improvement in classification or repeatability performances was shown when
adding MC iterations to regression models over the different medical tasks.
Regression models generally showed better repeatability compared with the
classification models (i.e., 3-class and ordinal), which reduces the potential
to improve repeatability, as regression without MC reaches good repeatability
performances compared to the classification models.

##  4 Conclusion

We evaluated the repeatability of four model types on three medical tasks
using distinct model architectures (ResNet18, ResNet50, DenseNet121). We
demonstrated that MC sampling during test time leads to more reliable
classification models providing more stable and repeatable predictions on
different images from the same patient. Only regression models did not show a
constant improvement when leveraging MC sampling. MC sampling is flexible as
it is applicable to any model type and architecture while being easily
implementable. The code used to train and generate results can be found at
https://github.com/andreanne-lemay/gray_zone_assessment.

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