# Some $L^{q}(\mathbb{R})$-norm decay estimates ($q\in[1,+\infty]$) for two
Cauchy systems of type Rao-Nakra sandwich beam with a frictional damping or an
infinite memory

Aissa Guesmia

###### Abstract.

In this paper, we consider two systems of type Rao-Nakra sandwich beam in the
whole line $\mathbb{R}$ with a frictional damping or an infinite memory acting
on the Euler-Bernoulli equation. When the speeds of propagation of the two
wave equations are equal, we show that the solutions do not converge to zero
when time goes to infinity. In the reverse situation, we prove some
$L^{2}(\mathbb{R})$-norm and $L^{1}(\mathbb{R})$-norm decay estimates of
solutions and theirs higher order derivatives with respect to the space
variable. Thanks to interpolation inequalities and Carlson inequality, these
$L^{2}(\mathbb{R})$-norm and $L^{1}(\mathbb{R})$-norm decay estimates lead to
similar ones in the $L^{q}(\mathbb{R})$-norm, for any $q\in[1,+\infty]$. In
our both $L^{2}(\mathbb{R})$-norm and $L^{1}(\mathbb{R})$-norm decay
estimates, we specify the decay rates in terms of the regularity of the
initial data and the nature of the control.

Institut Elie Cartan de Lorraine, UMR 7502, Université de Lorraine

3 Rue Augustin Fresnel, BP 45112, 57073 Metz Cedex 03, France

Keywords. Rao-Nakra sandwich beam, frictional damping, infinite memory,
unbounded domain,
asymptotic behavior, $L^{2}(\mathbb{R})$-norm and $L^{1}(\mathbb{R})$-norm
decay estimates, energy method, Fourier analysis. AMS classification. 34B05,
34D05, 34H05, 35B40, 35L45, 74H40, 93D20, 93D15.

††footnotetext: E-mail addresses<EMAIL_ADDRESS>ORCID: 0000
0003 4782 5053

##  1\. Introduction

Let $\rho_{1},\,\rho_{2},\,\rho_{3},\,k_{0},\,k_{1},\,k_{2},\,k_{3},\,\gamma$
and $l$ be real positive constants and
$g:\mathbb{R}_{+}:=[0,+\infty)\to\mathbb{R}_{+}$ satisfying $g\in
C^{1}(\mathbb{R}_{+})$,

(1.1) $0<g_{0}:=\displaystyle\int_{0}^{+\infty}g(s)\,ds<k_{3}.$

and, for some real positive constants $\beta_{1}$ and $\beta_{2}$,

(1.2) $-\beta_{2}g\leq g^{\prime}\leq-\beta_{1}g.$

Condition (1.2) implies that $g^{\prime}\leq 0$, $g$ is integrable over
$\mathbb{R}_{+}$ and, by integrating,

(1.3) $g(0)e^{-\beta_{2}s}\leq g(s)\leq g(0)e^{-\beta_{1}s},\quad
s\in\mathbb{R}_{+}.$

So (1.1) is valid if $g(0)>0$ and $\beta_{1}$ is big enough. For example, one
can take $g(s)=d_{1}e^{-d_{2}s}$ with $d_{1},\,d_{2}>0$ satisfying
$\frac{d_{1}}{d_{2}}<k_{3}$, so (1.1) and (1.2) hold, for
$\beta_{1}=\beta_{2}=d_{2}$. This paper deals with the stability of two
systems of type Rao-Nakra sandwich beam in the whole line $\mathbb{R}$ with a
control acting only on the Euler-Bernoulli equation. These systems consist of
two wave equations for the longitudinal displacements of the top and bottom
layers, and one Euler-Bernoulli equation for the transversal displacement. The
considered control is provided through a frictional damping of size $\gamma$:

(1.4)
$\left\\{\begin{array}[]{ll}\rho_{1}\varphi_{tt}-k_{1}\varphi_{xx}+k_{0}(\varphi+\psi+lw_{x})=0,\vspace{0.2cm}\\\
\rho_{2}\psi_{tt}-k_{2}\psi_{xx}+k_{0}(\varphi+\psi+lw_{x})=0,\vspace{0.2cm}\\\
\rho_{3}w_{tt}+k_{3}w_{xxxx}-lk_{0}(\varphi+\psi+lw_{x})_{x}+\gamma
w_{t}=0,\vspace{0.2cm}\\\
\varphi(x,0)=\varphi_{0}(x),\,\psi(x,0)=\psi_{0}(x),\,w(x,0)=w_{0}(x),\vspace{0.2cm}\\\
\varphi_{t}(x,0)=\varphi_{1}(x),\,\psi_{t}(x,0)=\psi_{1}(x),\,w_{t}(x,0)=w_{1}(x)\end{array}\right.$

or an infinite memory of kernel $g$:

(1.5)
$\left\\{\begin{array}[]{ll}\rho_{1}\varphi_{tt}-k_{1}\varphi_{xx}+k_{0}(\varphi+\psi+lw_{x})=0,\vspace{0.2cm}\\\
\rho_{2}\psi_{tt}-k_{2}\psi_{xx}+k_{0}(\varphi+\psi+lw_{x})=0,\vspace{0.2cm}\\\
\rho_{3}w_{tt}+k_{3}w_{xxxx}-lk_{0}(\varphi+\psi+lw_{x})_{x}-\displaystyle\int_{0}^{+\infty}g(s)\,w_{xxxx}(x,t-s)ds=0,\vspace{0.2cm}\\\
\varphi(x,0)=\varphi_{0}(x),\,\psi(x,0)=\psi_{0}(x),\,w(x,-t)=w_{0}(x,t),\vspace{0.2cm}\\\
\varphi_{t}(x,0)=\varphi_{1}(x),\,\psi_{t}(x,0)=\psi_{1}(x),\,w_{t}(x,0)=w_{1}(x),\end{array}\right.$

where $x\in\mathbb{R}$ and $t\in\mathbb{R}_{+}$ are, respectively, the spacial
and time variables, $\varphi_{0},\,\varphi_{1},\,\psi_{0},\,\psi_{1},\,w_{0}$
and $w_{1}$ are fixed initial data, and

$(\varphi,\,\psi,\,w):\,\mathbb{R}\times\mathbb{R}_{+}\to\mathbb{R}^{3}$

is the unknown of (1.4) and (1.5). A subscript $r$ denotes the derivative with
respect to $r$. Also, we use $\partial_{r}^{m}$ or $\frac{d^{m}}{dr^{m}}$ to
denote the differential operator of order $m$ with respect to $r$; i.e.
$\frac{\partial^{m}}{\partial r^{m}}$. When a function has only one variable,
its derivative is noted by $\prime$. Since the works [23, 31, 36], several
layer laminated beam and plate models have been introduced during the last
sixty years. The known generalized Rao-Nakra beam (composed of a top and a
bottom face plate), presented in [21], takes into account the shear effect of
the bottom and top layers, and it is gien by

(1.6)
$\left\\{\begin{array}[]{ll}\rho_{1}h_{1}u_{tt}-E_{1}h_{1}u_{xx}-\tau=0,\vspace{0.2cm}\\\
\rho_{3}h_{3}v_{tt}-E_{3}h_{3}v_{xx}+\tau=0,\vspace{0.2cm}\\\ \rho
hw_{tt}+EIw_{xxxx}-G_{1}h_{1}(w_{x}+\phi_{1})_{x}-G_{3}h_{3}(w_{x}+\phi_{3})_{x}-h_{2}\tau_{x}=0,\vspace{0.2cm}\\\
\rho_{1}I_{1}\phi_{1,tt}-E_{1}I_{1}\phi_{1,xx}-\frac{1}{2}h_{1}\tau+G_{1}h_{1}(w_{x}+\phi_{1})=0,\vspace{0.2cm}\\\
\rho_{3}I_{3}\phi_{3,tt}-E_{3}I_{3}\phi_{3,xx}-\frac{1}{2}h_{3}\tau+G_{3}h_{3}(w_{x}+\phi_{3})=0,\end{array}\right.$

where $\rho_{j},\,h_{j},\,E_{j},\,G_{j},\,I_{j},\,EI$ and $\rho h$ are real
positive constants representing some physical parameters and satisfying some
relationships, $u,\,\phi_{1},\,v$ and $\phi_{3}$ are, respectively, the
longitudinal displacement and shear angle of the top and bottom layers, $w$ is
the transverse displacement of the beam and $\tau$ is the shear stress in the
core layer defined by

$\tau=-u-\frac{1}{2}h_{1}\phi_{1}+h_{2}w_{x}+v-\frac{1}{2}h_{3}\phi_{3}.$

By neglecting some components and/or parameters and/or considering some
connections between them, several models were derived from (1.6) and studied
in the literature like the Rao-Nakra sandwich beam [29]:

(1.7)
$\left\\{\begin{array}[]{ll}\rho_{1}h_{1}u_{tt}-E_{1}h_{1}u_{xx}-k(-u+v+\alpha
w_{x})=0,\vspace{0.2cm}\\\ \rho_{3}h_{3}v_{tt}-E_{3}h_{3}v_{xx}+k(-u+v+\alpha
w_{x})=0,\vspace{0.2cm}\\\ \rho hw_{tt}+EIw_{xxxx}-\alpha k(-u+v+\alpha
w_{x})_{x}=0,\end{array}\right.$

the laminated beam [17]:

(1.8) $\left\\{\begin{array}[]{ll}\rho w_{tt}+G(\psi-
w_{x})_{x}=0,\vspace{0.2cm}\\\ I_{\rho}(3s-\psi)_{tt}-D(3s-\psi)_{xx}-G(\psi-
u_{x})=0,\vspace{0.2cm}\\\ 3I_{\rho}s_{tt}-3Ds_{xx}+3G(\psi-w_{x})+4\mu
s+4\delta s_{t}=0,\end{array}\right.$

the Bresse model [4]:

(1.9)
$\left\\{\begin{array}[]{ll}\rho_{1}u_{tt}-k_{1}(u_{x}+\psi+lw)_{x}-lk_{3}(w_{x}-lu)=0,\vspace{0.2cm}\\\
\rho_{2}\psi_{tt}-k_{2}\psi_{xx}+k_{1}(u_{x}+\psi+lw)=0,\vspace{0.2cm}\\\
\rho_{3}w_{tt}-k_{3}(w_{x}-lu)_{x}+lk_{1}(u_{x}+\psi+lw)=0,\end{array}\right.$

the Mead-Markos model [23]:

(1.10) $\left\\{\begin{array}[]{ll}\rho
hw_{tt}+EIw_{xxxx}-\alpha_{1}(-u+v+\alpha w_{x})=0,\vspace{0.2cm}\\\
(-u+v+\alpha w_{x})_{xx}-\alpha_{2}(-u+v+\alpha w_{x})-\alpha
w_{xxxx}=0\end{array}\right.$

and the Timoshenko beam [35]:

(1.11)
$\left\\{\begin{array}[]{ll}\rho_{1}u_{tt}-k_{1}(u_{x}+\psi)_{x}=0,\vspace{0.2cm}\\\
\rho_{2}\psi_{tt}-k_{2}\psi_{xx}+k_{1}(u_{x}+\psi)=0,\end{array}\right.$

where all the coefficients are real postive constants. The study of long time
behavior of (1.7)-(1.11) has been the subject of an active research and
mathematical endeavor. Thereby, a huge number of research articles have been
appeared. In order to highlight the main contribution and the novelty of the
present paper, we will merely point out the articles whose content concerns
the Rao-Nakra sandwich beam (1.7), and so their content is closely related to
our problems (1.4) and (1.5). For (1.8)-(1.11) (in different contexts), we
refer the reader, for example, to [1, 2, 5, 8, 9, 10, 11, 14, 21, 24, 32, 33]
and the references therein. In [20], the authors considered the Rao-Nakra
sandwich beam with $x\in(0,1)$ and an internal frictional damping acting
either on the beam equation or one of the wave equations, and proved that the
polynomial stability occurs. Similar polynomial stability results were proved
in [19] under internal damping or Kelvin-Voigt damping working on two of the
three equations. The authors of [30] proved the well-posedness and exponential
stability of Rao-Nakra sandwich beam in $(0,L)$, $L>0$, under a heat
conduction of secound sound acting on the second wave equation (see [6, 22,
27] for details on the second sound mechanism). The same well-posedness and
exponential stability results of [30] were proved in [25] using boundary
feedbacks at $x=L$. The (global or local) boundary controllability problems of
the Rao-Nakra beam were also the subject of several studies in the literature;
see, for example, [12, 13, 15, 16, 26, 28] and the references therein. The
subject of this paper is to treat the stability of the Rao-Nakra sandwich beam
(1.7) in the whole line $\mathbb{R}$ under a frictional damping (1.4) or an
infinite memory (1.5) (where we denoted $(u,v)$ by $(-\varphi,\psi)$ and
simplified the notation of the coefficients). The frictinal damping and
infinite memory terms generate the unique dissipation for (1.4) and (1.5) (see
(2.15) and Remark 1 below). This dissipation is acting on the third equation
of (1.4) and (1.5) (transversal displacement). To the best of our knowledge,
these situations have never been considered in the literature. The main
objective of this paper is to get decay estimates of

(1.12)
$t\mapsto\left\|\partial_{x}^{j}U(\cdot,t)\right\|_{2}\quad\hbox{and}\quad
t\mapsto\left\|\partial_{x}^{j}U(\cdot,t)\right\|_{1},$

where $U$ is defined in (2.7) below, $j\in\mathbb{N}$ and $\|\cdot\|_{q}$
denotes the norm of $L^{q}(\mathbb{R})$, for $q\in[1,+\infty]$. We will prove
the instability of (1.4) and (1.5) when

(1.13) $\frac{k_{1}}{\rho_{1}}=\frac{k_{2}}{\rho_{2}}.$

However, when

(1.14) $\frac{k_{1}}{\rho_{1}}\neq\frac{k_{2}}{\rho_{2}},$

we prove that the functions (1.12) satisfy some polynomial stability estimates
with decay rates depending on the regularity of the initial data. Our
$L^{2}(\mathbb{R})$-norm and $L^{1}(\mathbb{R})$-norm decay estimates cover
all the $L^{q}(\mathbb{R})$-norm decay estimates of $\partial_{x}^{j}U$, for
any $q\in[1,+\infty]$. Indeed, using interpolation inequalities, the decay
estimates of the $L^{\infty}$-norm is obtained immediately from the inequality
($\delta_{q}$ denotes a real positive constant, which does not depend neither
on $U$ nor on $j$)

(1.15)
$\|\partial_{x}^{j}U\|_{\infty}\leq\delta_{\infty}\|\partial_{x}^{j}U\|_{2}^{1/2}\|\partial_{x}^{j+1}U\|_{2}^{1/2}.$

This eventually lead to the decay rate of the $L^{q}$-norm, $2<q<+\infty$,
through the interpolation inequality

(1.16)
$\|\partial_{x}^{j}U\|_{q}\leq\delta_{q}\|\partial_{x}^{j}U\|_{\infty}^{1-2/q}\|\partial_{x}^{j}U\|_{2}^{2/q}.$

To fill the gap $1<q<2$, we use the interpolation inequality

(1.17)
$\left\|\partial_{x}^{j}U\right\|_{{q}}\leq\delta_{q}\left\|\partial_{x}^{j}U\right\|_{{2}}^{2(q-1)/q}\left\|\partial_{x}^{j}U\right\|_{1}^{(2-q)/q}.$

To get the $L^{1}(\mathbb{R})$-norm decay estimate, we treat the asymptotic
behavior of $\|x\partial_{x}^{j}U\|_{{2}}$, and then we use the Carlson
inequality (see [3] for instance)

(1.18)
$\|\partial_{x}^{j}U\|_{{1}}\leq\|\partial_{x}^{j}U\|_{{2}}^{1/2}\|x\partial_{x}^{j}U\|_{{2}}^{1/2}.$

The proof is based on the energy method combined with the Fourier analysis (by
using the transformation in the Fourier space). The paper is organized as
follows. In section 2, we formulate (1.4) and (1.5) in a first order Cauchy
system. Section 3 will be devoted to the proof of the asymptotic behavior of
$|\widehat{U}|$. In section 4, we prove our $L^{2}(\mathbb{R})$-norm decay
estimate. The asymptotic behavior of $|\partial_{\xi}\widehat{U}|$ will be
treated in section 5. Finally, in section 6, we prove our
$L^{1}(\mathbb{R})$-norm decay estimate. The last section presents some
concluding remarks.

##  2\. Abstract formulation of (1.4) and (1.5)

To formulate (1.4) and (1.5) in an abstract first order system, we consider
the following variables:

(2.1) $u=\varphi_{t},\quad y=\psi_{t},\quad\theta=w_{t},\quad
v=\varphi_{x},\quad z=\psi_{x},\quad\phi=w_{xx}\quad\hbox{and}\quad
p=\varphi+\psi+lw_{x}.$

We observe that (1.4)1-(1.4)3 can be written in the form

(2.2) $\displaystyle\begin{cases}v_{t}-u_{x}=0,\\\
\rho_{1}u_{t}-k_{1}v_{x}+k_{0}p=0,\\\ z_{t}-y_{x}=0,\\\
\rho_{2}y_{t}-k_{2}z_{x}+k_{0}p=0,\\\ \phi_{t}-\theta_{xx}=0,\\\
\rho_{3}\theta_{t}+k_{3}\phi_{xx}-lk_{0}p_{x}+\gamma\theta=0,\\\
p_{t}-u-y-l\theta_{x}=0.\end{cases}$

In case (1.5), we consider the additional variable introduced in [7]

(2.3) $\displaystyle\eta(x,t,s)=w(x,t)-w(x,t-s)$

with its initial data $\eta_{0}(x,s)=\eta(x,0,s)$. The variable $\eta$
satisfies

(2.4) $\displaystyle\eta_{t}(x,t,s)+\eta_{s}(x,t,s)=w_{t}(x,t)$

and

(2.5)
$\displaystyle\displaystyle\int_{0}^{+\infty}\,g(s)\eta_{xxxx}(x,t,s)\,ds=g_{0}w_{xxxx}(x,t)-\displaystyle\int_{0}^{+\infty}\,g(s)w_{xxxx}(x,t-s)\,ds.$

Then (1.5)1-(1.5)3 can be formulated in the form

(2.6) $\displaystyle\begin{cases}v_{t}-u_{x}=0,\\\
\rho_{1}u_{t}-k_{1}v_{x}+k_{0}p=0,\\\ z_{t}-y_{x}=0,\\\
\rho_{2}y_{t}-k_{2}z_{x}+k_{0}p=0,\\\ \phi_{t}-\theta_{xx}=0,\\\
\rho_{3}\theta_{t}+(k_{3}-\tau_{2}g_{0})\phi_{xx}-lk_{0}p_{x}+\displaystyle\int_{0}^{+\infty}\,g(s)\eta_{xxxx}\,ds=0,\\\
p_{t}-u-y-l\theta_{x}=0,\\\ \eta_{t}+\eta_{s}-\theta=0.\end{cases}$

We define the variable $U$ and its initial data $U_{0}$ by

(2.7) $\displaystyle
U(x,t)=\begin{cases}(v,\,u,\,z,\,y,\,\phi,\,\theta,\,p)^{T}(x,t)\quad&\hbox{in
case \eqref{S1}},\\\
(v,\,u,\,z,\,y,\,\phi,\,\theta,\,p,\eta)^{T}(x,t)\quad&\hbox{in case
\eqref{S2}}\end{cases}\quad\hbox{and}\quad U_{0}(x)=U(x,0).$

Therefore, the systems (2.2) and (2.6) lead to

(2.8) $\displaystyle\begin{cases}U_{t}(x,t)+AU(x,t)=0,\\\
U(x,0)=U_{0}(x),\end{cases}$

where

$AU=A_{4}U_{xxxx}+A_{2}U_{xx}+A_{1}U_{x}+A_{0}U$

and the operators $A_{j}$ are defined in case (2.2) by

$\displaystyle\begin{cases}A_{4}U_{xxxx}=(0,0,0,0,0,0,0)^{T},\\\
A_{2}U_{xx}=\left(0,0,0,0,-\theta_{xx},\frac{k_{3}}{\rho_{3}}\phi_{xx},0\right)^{T},\\\
A_{1}U_{x}=\left(-u_{x},-\frac{k_{1}}{\rho_{1}}v_{x},-y_{x},-\frac{k_{2}}{\rho_{2}}z_{x},0,-\frac{lk_{0}}{\rho_{3}}p_{x},-l\theta_{x}\right)^{T},\\\
A_{0}U=\left(0,\frac{k_{0}}{\rho_{1}}p,0,\frac{k_{0}}{\rho_{2}}p,0,\frac{\gamma}{\rho_{3}}\theta,-u-y\right)^{T},\end{cases}$

and $A_{j}$ in case (2.6) are defined by

$\displaystyle\begin{cases}A_{4}U_{xxxx}=\left(0,0,0,0,0,\frac{1}{\rho_{3}}\displaystyle\int_{0}^{+\infty}\,g(s)\eta_{xxxx}\,ds,0,0\right)^{T},\\\
A_{2}U_{xx}=\left(0,0,0,0,-\theta_{xx},\frac{k_{3}-g_{0}}{\rho_{3}}\phi_{xx},0,0\right)^{T},\\\
A_{1}U_{x}=\left(-u_{x},-\frac{k_{1}}{\rho_{1}}v_{x},-y_{x},-\frac{k_{2}}{\rho_{2}}z_{x},0,-\frac{lk_{0}}{\rho_{3}}p_{x},-l\theta_{x},0\right)^{T},\\\
A_{0}U=\left(0,\frac{k_{0}}{\rho_{1}}p,0,\frac{k_{0}}{\rho_{2}}p,0,0,-u-y,\eta_{s}-\theta\right)^{T}.\end{cases}$

For a given function $h:\mathbb{R}\to\mathbb{C}$, we use the notations
$Re\,h$, $Im\,h$, $\bar{h}$ and $\widehat{h}$ to denote, respectively, the
real part of $h$, the imaginary part of $h$, the conjugate of $h$ and the
Fourier transformation of $h$ given by

(2.9) $\widehat{h}(\xi)=\displaystyle\int_{-\infty}^{+\infty}\,e^{-i\xi
x}h(x)\,dx,\quad\xi\in\mathbb{R}.$

Applying the Fourier transformation (with respect to the space variable $x$)
to (2.8), we obtain the following first-order Cauchy system in the Fourier
space:

(2.10)
$\displaystyle\begin{cases}\widehat{U}_{t}(\xi,\,t)+{\tilde{A}}(\xi)\widehat{U}(\xi,\,t)=0,\\\
\widehat{U}(\xi,\,0)=\widehat{U}_{0}(\xi),\end{cases}$

where
${\tilde{A}}(\xi)=\xi^{4}\,{A}_{4}-\xi^{2}\,{A}_{2}+i\,\xi\,{A}_{1}+A_{0}$.
The solution of (2.10) is given by

(2.11) $\widehat{U}(\xi,\,t)=e^{-{\tilde{A}}(\xi)\,t}\ \widehat{U}_{0}(\xi).$

Computing the term $e^{-{\tilde{A}}(\xi)t}$ is a challenging problem, and in
many situations, this cannot be done. Consequently, in order to show the
asymptotic behavior of $\widehat{U}$, it suffices to find a non negative
function $f(\xi)$ and two positive constants ${\tilde{c}}$ and $c$ such that,
for each $(\xi,t)\in\mathbb{R}\times\mathbb{R}_{+}$,

(2.12) $\left|e^{-{\tilde{A}}(\xi)t}\right|\leq{\tilde{c}}e^{-cf(\xi)t}.$

Let $\widehat{E}$ be the energy associated with (2.10) given by

(2.13) $\displaystyle\widehat{E}(\xi,t)$ $\displaystyle=$
$\displaystyle\frac{1}{2}\left[k_{1}\,|\widehat{v}|^{2}+\rho_{1}|\widehat{u}|^{2}+k_{2}\,|\widehat{z}|^{2}+\rho_{2}|\widehat{y}|^{2}+(k_{3}-{\tau}_{0}g_{0})|\widehat{\phi}|^{2}+\rho_{3}|\widehat{\theta}|^{2}+k_{0}|\widehat{p}|^{2}\right](\xi,\,t)$
$\displaystyle+\frac{{\tau}_{0}}{2}\xi^{4}\displaystyle\int_{0}^{+\infty}\,g(s)|\widehat{\eta}(\xi,\,t)|^{2}\,ds,$

where

(2.14) $\displaystyle{\tau}_{0}=\begin{cases}0\quad&\hbox{in case
\eqref{e10}},\\\ 1\quad&\hbox{in case \eqref{e20}}.\end{cases}$

######  Lemma 2.1.

The energy functional $\widehat{E}$ satisfies, for each
$(\xi,t)\in\mathbb{R}\times\mathbb{R}_{+}$,

(2.15)
$\frac{d}{dt}\widehat{E}(\xi,\,t)=-(1-{\tau}_{0})\gamma\,|\widehat{\theta}(\xi,\,t)|^{2}+\frac{{\tau}_{0}}{2}\xi^{4}\displaystyle\int_{0}^{+\infty}\,g^{\prime}(s)|\widehat{\eta}(\xi,\,t)|^{2}\,ds.$

###### Proof.

The equation (2.10)1 in case (2.2) is equivalent to

(2.16) $\displaystyle\begin{cases}\widehat{v}_{t}-i\xi\widehat{u}=0,\\\
\rho_{1}\widehat{u}_{t}-ik_{1}\xi\,\widehat{v}+k_{0}\widehat{p}=0,\\\
\widehat{z}_{t}-i\xi\widehat{y}=0,\\\
\rho_{2}\widehat{y}_{t}-ik_{2}\xi\,\widehat{z}+k_{0}\widehat{p}=0,\\\
\widehat{\phi}_{t}+\xi^{2}\widehat{\theta}=0,\\\
\rho_{3}\widehat{\theta}_{t}-k_{3}\xi^{2}\widehat{\phi}-ilk_{0}\,\xi\widehat{p}+\gamma\widehat{\theta}=0,\\\
\widehat{p}_{t}-\widehat{u}-\widehat{y}-il\xi\widehat{\theta}=0.\end{cases}$

Multiplying the equations in (2.16) by $k_{1}\bar{\widehat{v}}$,
$\bar{\widehat{u}}$, $k_{2}\bar{\widehat{z}}$, $\bar{\widehat{y}}$,
$k_{3}\bar{\widehat{\phi}}$, $\bar{\widehat{\theta}}$ and
$k_{0}\bar{\widehat{p}}$, respectively, adding the obtained equations, taking
the real part of the resulting expression and using the identity, for
$h,\,d:\mathbb{R}\to\mathbb{C}$ two differentiable functions,

(2.17)
$\frac{d}{dt}Re\,\left(h{\bar{d}}\right)=Re\,\left(h_{t}{\bar{d}}+d_{t}{\bar{h}}\right),$

we obtain (2.15) with ${\tau}_{0}=0$. Similarily, (2.10)1 in case (2.6) is
reduced to

(2.18) $\displaystyle\begin{cases}\widehat{v}_{t}-i\xi\widehat{u}=0,\\\
\rho_{1}\widehat{u}_{t}-ik_{1}\xi\,\widehat{v}+k_{0}\widehat{p}=0,\\\
\widehat{z}_{t}-i\xi\widehat{y}=0,\\\
\rho_{2}\widehat{y}_{t}-ik_{2}\xi\,\widehat{z}+k_{0}\widehat{p}=0,\\\
\widehat{\phi}_{t}+\xi^{2}\widehat{\theta}=0,\\\
\rho_{3}\widehat{\theta}_{t}-(k_{3}-g_{0})\xi^{2}\widehat{\phi}-ilk_{0}\,\xi\widehat{p}+\xi^{4}\displaystyle\int_{0}^{+\infty}\,g(s)\widehat{\eta}\,ds=0,\\\
\widehat{p}_{t}-\widehat{u}-\widehat{y}-il\xi\widehat{\theta}=0,\\\
\widehat{\eta}_{t}+\widehat{\eta}_{s}-\widehat{\theta}=0.\end{cases}$

Multiplying (2.18)1-(2.18)7 by $k_{1}\bar{\widehat{v}}$, $\bar{\widehat{u}}$,
$k_{2}\bar{\widehat{z}}$, $\bar{\widehat{y}}$,
$(k_{3}-g_{0})\bar{\widehat{\phi}}$, $\bar{\widehat{\theta}}$ and
$k_{0}\bar{\widehat{p}}$, respectively, multiplying (2.18)8 by
$\xi^{4}g(s)\bar{\widehat{\eta}}$ and integrating on $\mathbb{R}_{+}$ with
respect to $s$, adding all the obtained equations, taking the real part of the
resulting expression and using (2.17), we get

$\frac{d}{dt}\widehat{E}(\xi,\,t)=-\frac{1}{2}\xi^{4}\displaystyle\int_{0}^{+\infty}\,g(s)\frac{d}{ds}|\widehat{\eta}|^{2}\,ds,$

therefore, by integrating with respect to $s$, we obtain

$\frac{d}{dt}\widehat{E}(\xi,\,t)=-\frac{1}{2}\left[\xi^{4}\displaystyle\int_{0}^{+\infty}\,g(s)|\widehat{\eta}|^{2}\,ds\right]_{s=0}^{s=+\infty}+\frac{1}{2}\xi^{4}\displaystyle\int_{0}^{+\infty}\,g^{\prime}(s)|\widehat{\eta}|^{2}\,ds.$

Because (1.3) and (2.3) imply that

(2.19)
$\displaystyle\lim_{s\to+\infty}g(s)=0\quad\hbox{and}\quad\widehat{\eta}(\xi,t,0)=0,$

then we find (2.15) with ${\tau}_{0}=1$. ∎

######  Remark 1.

Notice that (2.15) implies that

$\frac{d}{dt}\widehat{E}(\xi,\,t)\leq 0,$

since $\gamma>0$ and $g^{\prime}\leq 0$, so (2.10) is dissipative. If the
frictional damping and infinite memory are not considered (i.e. $\gamma=g=0$),
then (2.10) is conservative; that is

$\widehat{E}(\xi,\,t)=\widehat{E}(\xi,\,0).$

On the other hand, we put

$\displaystyle|\widehat{U}(\xi,\,t)|^{2}=\left[|\widehat{v}|^{2}+|\widehat{u}|^{2}+|\widehat{z}|^{2}+|\widehat{y}|^{2}+|\widehat{\phi}|^{2}+|\widehat{\theta}|^{2}+|\widehat{p}|^{2}\right](\xi,\,t)+\tau_{0}\xi^{4}\displaystyle\int_{0}^{+\infty}\,g(s)|\widehat{\eta}(\xi,\,t)|^{2}\,ds.$

So we deduce that, thanks to the right inequality in (1.1),

(2.20) $|\widehat{U}|^{2}\sim\widehat{E}.$

##  3\. Estimation of $|\widehat{U}|$

This section is dedicated to the study of the asymptotic behavior of
$\widehat{U}(\xi,\,t)$, when time $t$ goes to infinity. We prove the next
theorem.

######  Theorem 3.1.

Let $\widehat{U}$ be the solution of (2.10). 1\. In case (1.13) and for every
$\xi\in\mathbb{R}$, $|\widehat{U}(\xi,\,t)|$ doesn’t converge to zero when $t$
goes to infinity. 2\. In case (1.14), there exist $c,\,\widetilde{c}>0$
(independent on $\xi$ and $t$) such that

(3.1)
$|\widehat{U}(\xi,\,t)|\leq\widetilde{c}\,e^{-\,c\,f(\xi)\,t}\,|\widehat{U}_{0}(\xi)|,\quad\forall(\xi,t)\in\mathbb{R}\times\mathbb{R}_{+},$

where

(3.2) $f(\xi)=\frac{\xi^{4+2\tau_{0}}}{\xi^{10}+1}.$

###  3.1. Case (1.13)

We prove here, under (1.13) and for every $\xi\in\mathbb{R}$, that
$|\widehat{U}(\xi,\,t)|$ doesn’t converge to zero when time $t$ goes to
infinity. It suffices to prove that, for any $\xi\in\mathbb{R}$,
$-{\tilde{A}}(\xi)$ has at least a pure imaginary eigenvalue (see [34]); that
is

(3.3)
$\forall\xi\in\mathbb{R},\,\,\exists\lambda\in\mathbb{R}^{*},\,\,\exists\widehat{U}\neq
0:\quad i\lambda\widehat{U}+{\tilde{A}}(\xi)\widehat{U}=0.$

System (2.16). To get (3.3), it is enough to prove that

(3.4) $\forall\xi\in\mathbb{R},\,\,\exists\lambda\in\mathbb{R}^{*}:\quad
det\left(i\lambda I+{\tilde{A}}(\xi)\right)=0,$

where $I$ denotes the identity matrix. We see that (1.13) implies, in case
(2.16), that

$i\lambda I+{\tilde{A}}(\xi)=\begin{pmatrix}i\lambda&-i\xi&0&0&0&0&0\\\
-i\frac{k_{1}}{\rho_{1}}\xi&i\lambda&0&0&0&0&\frac{k_{0}}{\rho_{1}}\\\
0&0&i\lambda&-i\xi&0&0&0\\\
0&0&-i\frac{k_{1}}{\rho_{1}}\xi&i\lambda&0&0&\frac{k_{0}}{\rho_{2}}\\\
0&0&0&0&i\lambda&\xi^{2}&0\\\
0&0&0&0&-\frac{k_{3}}{\rho_{3}}\xi^{2}&i\lambda+\frac{\gamma}{\rho_{3}}&-i\frac{lk_{0}}{\rho_{3}}\xi\\\
0&-1&0&-1&0&-il\xi&i\lambda\end{pmatrix}.$

A direct computation shows that

(3.5) $\displaystyle det\left(i\lambda I+{\tilde{A}}(\xi)\right)$
$\displaystyle=$ $\displaystyle
i\lambda\left(\lambda^{2}-\frac{k_{1}}{\rho_{1}}\xi^{2}\right)\left[\lambda^{2}-k_{0}\left(\frac{1}{\rho_{1}}+\frac{1}{\rho_{2}}\right)-\frac{k_{1}}{\rho_{1}}\xi^{2}\right]\left[i\lambda\left(i\lambda+\frac{\gamma}{\rho_{3}}\right)+\frac{k_{3}}{\rho_{3}}\xi^{4}\right]$
$\displaystyle+i\frac{l^{2}k_{0}}{\rho_{3}}\lambda\xi^{2}\left(\lambda^{2}-\frac{k_{1}}{\rho_{1}}\xi^{2}\right)^{2}.$

It is clear that $det\left(i\lambda I+{\tilde{A}}(\xi)\right)=0$ for

(3.6)
$\displaystyle\lambda=\begin{cases}{\sqrt{\frac{k_{1}}{\rho_{1}}}}\xi\quad\quad&\hbox{if}\,\,\xi\neq
0,\\\
{\sqrt{k_{0}\left(\frac{1}{\rho_{1}}+\frac{1}{\rho_{2}}\right)}}\quad\quad&\hbox{if}\,\,\xi=0.\end{cases}$

Consequently, (3.4) holds. System (2.18). For $\lambda\in\mathbb{R}^{*}$, we
put

(3.7) $\widehat{\eta}(s)=\frac{1}{i\lambda}\left(1-e^{-i\lambda
s}\right)\widehat{\theta}\quad\hbox{and}\quad{\tilde{g}}(\lambda)=\int_{0}^{+\infty}\left(1-e^{-i\lambda
s}\right)g(s)ds.$

We observe that ${\tilde{g}}(\lambda)$ is well defined (according to (1.1))
and $\widehat{\eta}$ is the unique function satisfying (3.3)8 and
$\widehat{\eta}(0)=0$. On the other hand, (1.13) implies that the first seven
equations of (3.3) are equivalent to
${\tilde{B}}(\xi)\left(\widehat{v},\widehat{u},\widehat{z},\widehat{y},\widehat{\phi},\widehat{p}\right)^{T}=0$,
where

${\tilde{B}}(\xi)=\begin{pmatrix}i\lambda&-i\xi&0&0&0&0&0\\\
-i\frac{k_{1}}{\rho_{1}}\xi&i\lambda&0&0&0&0&\frac{k_{0}}{\rho_{1}}\\\
0&0&i\lambda&-i\xi&0&0&0\\\
0&0&-i\frac{k_{1}}{\rho_{1}}\xi&i\lambda&0&0&\frac{k_{0}}{\rho_{2}}\\\
0&0&0&0&i\lambda&\xi^{2}&0\\\
0&0&0&0&-\frac{k_{3}-g_{0}}{\rho_{3}}\xi^{2}&i\lambda+\frac{{\tilde{g}}(\lambda)}{\rho_{3}}\xi^{4}&-i\frac{lk_{0}}{\rho_{3}}\xi\\\
0&-1&0&-1&0&-il\xi&i\lambda\end{pmatrix}.$

Then the problem (3.3) is reduced to prove that

(3.8) $\forall\xi\in\mathbb{R},\,\,\exists\lambda\in\mathbb{R}^{*}:\quad
det\,{\tilde{B}}(\xi)=0.$

A direct computation shows that (as for (3.5) with $k_{3}-g_{0}$ and
$\xi^{4}{\tilde{g}}(\lambda)$ instead of $k_{3}$ and $\gamma$, respectively)

$\displaystyle det{\tilde{B}}(\xi)$ $\displaystyle=$ $\displaystyle
i\lambda\left(\lambda^{2}-\frac{k_{1}}{\rho_{1}}\xi^{2}\right)\left[\lambda^{2}-k_{0}\left(\frac{1}{\rho_{1}}+\frac{1}{\rho_{2}}\right)-\frac{k_{1}}{\rho_{1}}\xi^{2}\right]\left[i\lambda\left(i\lambda+\frac{{\tilde{g}}(\lambda)}{\rho_{3}}\xi^{4}\right)+\frac{k_{3}-g_{0}}{\rho_{3}}\xi^{4}\right]$
$\displaystyle+i\frac{l^{2}k_{0}}{\rho_{3}}\lambda\xi^{2}\left(\lambda^{2}-\frac{k_{1}}{\rho_{1}}\xi^{2}\right)^{2}.$

We remark that $det\,{\tilde{B}}(\xi)=0$ for $\lambda$ given by (3.6). Thus
(3.8) holds. Consequently, the proof of the first result in Theorem 3.1 is
ended.

###  3.2. Some differential equations

We start the proof of (3.1) by proving some useful differential equations. To
simplify the computations, we put

(3.9)
$G=(1-\tau_{0})\gamma\widehat{\theta}+\tau_{0}\xi^{4}\int_{0}^{+\infty}g(s)\widehat{\eta}ds,$

where $\tau_{0}$ is defined in (2.14). Let consider the system

(3.10) $\displaystyle\begin{cases}\widehat{v}_{t}-i\xi\widehat{u}=0,\\\
\rho_{1}\widehat{u}_{t}-ik_{1}\xi\,\widehat{v}+k_{0}\widehat{p}=0,\\\
\widehat{z}_{t}-i\xi\widehat{y}=0,\\\
\rho_{2}\widehat{y}_{t}-ik_{2}\xi\,\widehat{z}+k_{0}\widehat{p}=0,\\\
\widehat{\phi}_{t}+\xi^{2}\widehat{\theta}=0,\\\
\rho_{3}\widehat{\theta}_{t}-(k_{3}-\tau_{0}g_{0})\xi^{2}\widehat{\phi}-ilk_{0}\,\xi\widehat{p}+G=0,\\\
\widehat{p}_{t}-\widehat{u}-\widehat{y}-il\xi\widehat{\theta}=0,\\\
\widehat{\eta}_{t}+\widehat{\eta}_{s}-\widehat{\theta}=0.\end{cases}$

It is evident that (2.16) is identical to (3.10)1-(3.10)7 if $\tau_{0}=0$, and
(2.18) coincides with (3.10) if $\tau_{0}=1$. Multiplying (3.10)4 and (3.10)3
by $i\,\xi\,\overline{\widehat{z}}$ and
$-i\rho_{2}\,\xi\,\overline{\widehat{y}}$, respectively, adding the resulting
equations, taking the real part and using (2.17), we obtain

(3.11)
$\frac{d}{dt}Re\left(i\rho_{2}\,\xi\,\widehat{y}\,\overline{\widehat{z}}\right)=\xi^{2}\left(\rho_{2}|\widehat{y}|^{2}-k_{2}|\widehat{z}|^{2}\right)+k_{0}\,Re\left(i\,\xi\,\widehat{z}\,\overline{\widehat{p}}\right).$

Similarily, multiplying (3.10)2 and (3.10)1 by
$i\,\xi\,\overline{\widehat{v}}$ and
$-i\rho_{1}\,\xi\,\overline{\widehat{u}}$, respectively, adding the resulting
equations, taking the real part and using (2.17), we find

(3.12)
$\frac{d}{dt}Re\left(i\rho_{1}\,\xi\,\widehat{u}\,\overline{\widehat{v}}\right)=\xi^{2}\left(\rho_{1}|\widehat{u}|^{2}-k_{1}|\widehat{v}|^{2}\right)+k_{0}Re\left(i\,\xi\,\widehat{v}\,\overline{\widehat{p}}\right).$

Also, multiplying (3.10)6 and (3.10)5 by $-\overline{\widehat{\phi}}$ and
$-\rho_{3}\,\overline{\widehat{\theta}}$, respectively, adding the resulting
equations, taking the real part and using (2.17), we get

(3.13)
$\frac{d}{dt}Re\left(-\rho_{3}\,\widehat{\theta}\,\overline{\widehat{\phi}}\right)=\xi^{2}\left(\rho_{3}|\widehat{\theta}|^{2}-(k_{3}-\tau_{0}g_{0})|\widehat{\phi}|^{2}\right)+lk_{0}\,Re\left(i\,\xi\,\widehat{\phi}\,\overline{\widehat{p}}\right)+\,Re\left(\overline{\widehat{\phi}}\,G\right).$

Multiplying (3.10)4 and (3.10)7 by $\xi^{2}\,\overline{\widehat{p}}$ and
$\rho_{2}\xi^{2}\,\overline{\widehat{y}}$, respectively, adding the resulting
equations, taking the real part and using (2.17), we obtain

(3.14)
$\frac{d}{dt}Re\left(\rho_{2}\xi^{2}\,\widehat{p}\,\overline{\widehat{y}}\right)=\xi^{2}\left(\rho_{2}\,|\widehat{y}|^{2}-k_{0}|\widehat{p}|^{2}\right)+\rho_{2}\xi^{2}\,Re\left(\widehat{y}\,\overline{\widehat{u}}\right)+l\rho_{2}\xi^{2}\,Re\left(i\xi\,\widehat{\theta}\,\overline{\widehat{y}}\right)+k_{2}\xi^{2}\,Re\left(i\,\xi\,\widehat{z}\,\overline{\widehat{p}}\right).$

After, multiplying (3.10)3 and (3.10)6 by
$\rho_{3}\,\overline{\widehat{\theta}}$ and $\overline{\widehat{z}}$,
respectively, adding the resulting equations, taking the real part and using
(2.17), we entail

(3.15)
$\frac{d}{dt}Re\left(\rho_{3}\widehat{z}\,\overline{\widehat{\theta}}\right)=-\rho_{3}\,Re\left(i\xi\widehat{\theta}\,\overline{\widehat{y}}\right)+(k_{3}-\tau_{0}g_{0})\,\xi^{2}\,Re\left(\widehat{\phi}\,\overline{\widehat{z}}\right)-lk_{0}\,Re\left(i\,\xi\,\widehat{z}\,\overline{\widehat{p}}\right)-\,Re\left(\overline{\widehat{z}}\,G\right).$

Multiplying (3.10)5 and (3.10)4 by $i\rho_{2}\xi\,\overline{\widehat{y}}$ and
$-i\xi\overline{\widehat{\phi}}$, respectively, adding the resulting
equations, taking the real part and using (2.17), it follows that

(3.16)
$\frac{d}{dt}Re\left(i\rho_{2}\xi\,\widehat{\phi}\,\overline{\widehat{y}}\right)=-\rho_{2}\xi^{2}\,Re\left(i\xi\widehat{\theta}\,\overline{\widehat{y}}\right)+k_{2}\xi^{2}\,Re\left(\widehat{\phi}\,\overline{\widehat{z}}\right)-k_{0}\,Re\left(i\xi\widehat{\phi}\,\overline{\widehat{p}}\right).$

Next, multiplying (3.10)2 and (3.10)3 by $i\xi\overline{\widehat{z}}$ and
$-i\rho_{1}\xi\,\overline{\widehat{u}}$, respectively, adding the resulting
equations, taking the real part and using (2.17), it appears that

(3.17)
$\frac{d}{dt}Re\left(i\rho_{1}\xi\,\widehat{u}\,\overline{\widehat{z}}\right)=-k_{1}\xi^{2}\,Re\left(\widehat{v}\,\overline{\widehat{z}}\right)+\rho_{1}\xi^{2}\,Re\left(\widehat{y}\,\overline{\widehat{u}}\right)+k_{0}\,Re\left(i\xi\widehat{z}\,\overline{\widehat{p}}\right).$

Multiplying (3.10)2 and (3.10)5 by $i\xi\overline{\widehat{\phi}}$ and
$-i\rho_{1}\xi\,\overline{\widehat{u}}$, respectively, adding the resulting
equations, taking the real part and using (2.17), we see that

(3.18)
$\frac{d}{dt}Re\left(i\rho_{1}\xi\,\widehat{u}\,\overline{\widehat{\phi}}\right)=-k_{1}\xi^{2}\,Re\left(\widehat{\phi}\,\overline{\widehat{v}}\right)+\rho_{1}\,\xi^{2}\,Re\left(i\xi\widehat{\theta}\,\overline{\widehat{u}}\right)+k_{0}\,Re\left(i\xi\widehat{\phi}\,\overline{\widehat{p}}\right).$

Also, multiplying (3.10)1 and (3.10)4 by
$i\rho_{2}\xi\,\overline{\widehat{y}}$ and $-i\xi\overline{\widehat{v}}$,
respectively, adding the resulting equations, taking the real part and using
(2.17), we find

(3.19)
$\frac{d}{dt}Re\left(i\rho_{2}\xi\widehat{v}\,\overline{\widehat{y}}\right)=-\rho_{2}\xi^{2}\,Re\left(\widehat{y}\,\overline{\widehat{u}}\right)+k_{2}\xi^{2}\,Re\left(\widehat{v}\,\overline{\widehat{z}}\right)-k_{0}\,Re\left(i\xi\widehat{v}\,\overline{\widehat{p}}\right).$

Multiplying (3.10)1 and (3.10)6 by $\rho_{3}\,\overline{\widehat{\theta}}$ and
$\overline{\widehat{v}}$, respectively, adding the resulting equations, taking
the real part and using (2.17), we get

(3.20)
$\frac{d}{dt}Re\left(\rho_{3}\,\widehat{v}\,\overline{\widehat{\theta}}\right)=-\rho_{3}\,Re\left(i\xi\widehat{\theta}\,\overline{\widehat{u}}\right)+(k_{3}-\tau_{0}g_{0})\,\xi^{2}\,Re\left(\widehat{\phi}\,\overline{\widehat{v}}\right)-lk_{0}\,Re\left(i\xi\widehat{v}\,\overline{\widehat{p}}\right)-\,Re\left(\overline{\widehat{v}}\,G\right).$

Now, multiplying (3.10)6 by
$-\xi^{2}\displaystyle\int_{0}^{+\infty}\,g(s)\overline{\widehat{\eta}}\,ds$,
multiplying (3.10)8 by $-\rho_{3}\xi^{2}g(s)\overline{\widehat{\theta}}$ and
integrating over $\mathbb{R}_{+}$ with respect to $s$, adding the resulting
equations, taking the real part and using (2.17), we infer that

$\displaystyle\frac{d}{dt}Re\left(-\rho_{3}\xi^{2}{\widehat{\theta}}\,\displaystyle\int_{0}^{+\infty}\,g(s)\overline{\widehat{\eta}}\,ds\right)$
$\displaystyle=$
$\displaystyle-\rho_{3}g_{0}\xi^{2}|\widehat{\theta}|^{2}-(k_{3}-\tau_{0}g_{0})\xi^{4}Re\left(\widehat{\phi}\displaystyle\int_{0}^{+\infty}\,g(s)\overline{\widehat{\eta}}\,ds\right)$
$\displaystyle-
lk_{0}\xi^{2}Re\left(i\xi\widehat{p}\displaystyle\int_{0}^{+\infty}\,g(s)\overline{\widehat{\eta}}\,ds\right)-\rho_{3}\xi^{2}Re\left(\overline{\widehat{\theta}}\displaystyle\int_{0}^{+\infty}\,g^{\prime}(s){\widehat{\eta}}\,ds\right)$
$\displaystyle+\xi^{2}Re\left(G\displaystyle\int_{0}^{+\infty}\,g(s)\overline{{\widehat{\eta}}}\,ds\right).$

Similarily, multiplying (3.10)4 by
$i\xi\displaystyle\int_{0}^{+\infty}\,g(s)\overline{\widehat{\eta}}\,ds$,
multiplying (3.10)8 by $-i\rho_{2}\xi g(s)\,\overline{\widehat{y}}$ and
integrating over $\mathbb{R}_{+}$ with respect to $s$, adding the resulting
equations, taking the real part and using (2.17), we get

(3.22)
$\displaystyle\frac{d}{dt}Re\left(i\rho_{2}\xi{\widehat{y}}\,\displaystyle\int_{0}^{+\infty}\,g(s)\overline{\widehat{\eta}}\,ds\right)$
$\displaystyle=$
$\displaystyle-\rho_{2}g_{0}Re\left(i\xi{\widehat{\theta}}\overline{\widehat{y}}\right)-k_{2}\xi^{2}Re\left(\widehat{z}\displaystyle\int_{0}^{+\infty}\,g(s)\overline{\widehat{\eta}}\,ds\right)$
$\displaystyle-
k_{0}Re\left(i\xi\widehat{p}\displaystyle\int_{0}^{+\infty}\,g(s)\overline{\widehat{\eta}}\,ds\right)-\rho_{2}Re\left(i\xi\overline{\widehat{y}}\displaystyle\int_{0}^{+\infty}\,g^{\prime}(s){\widehat{\eta}}\,ds\right).$

Finally, multiplying (3.10)2 by
$i\xi\displaystyle\int_{0}^{+\infty}\,g(s)\overline{\widehat{\eta}}\,ds$,
multiplying (3.10)8 by $-\rho_{1}\xi g(s)\overline{\widehat{u}}$ and
integrating over $\mathbb{R}_{+}$ with respect to $s$, adding the resulting
equations, taking the real part and using (2.17), we see that

(3.23)
$\displaystyle\frac{d}{dt}Re\left(i\rho_{1}\xi{\widehat{u}}\,\displaystyle\int_{0}^{+\infty}\,g(s)\overline{{\widehat{\eta}}}\,ds\right)$
$\displaystyle=$ $\displaystyle-
k_{1}\xi^{2}Re\left(\widehat{v}\displaystyle\int_{0}^{+\infty}\,g(s)\overline{\widehat{\eta}}\,ds\right)-k_{0}Re\left(i\xi\widehat{p}\displaystyle\int_{0}^{+\infty}\,g(s)\overline{\widehat{\eta}}\,ds\right)$
$\displaystyle-\rho_{1}Re\left(i\xi\overline{\widehat{u}}\displaystyle\int_{0}^{+\infty}\,g^{\prime}(s){\widehat{\eta}}\,ds\right)-\rho_{1}g_{0}Re\left(i\xi{\widehat{\theta}}\overline{\widehat{u}}\right).$

Let $\lambda_{1},\cdots,\lambda_{13}$ be real numbers to be chosen later
(which do not depend on time $t$ but may depend on $\xi$ and the parameters of
(1.4) and (1.5)). We define the functionals ${F}_{0},\,F_{1},\,F_{2}$ and
${F}_{3}$ as follows:

$\displaystyle{F}_{0}(\xi,\,t)$ $\displaystyle=$ $\displaystyle
Re\left(i\rho_{2}\lambda_{1}\xi\widehat{y}\overline{\widehat{z}}+i\rho_{1}\lambda_{2}\xi\widehat{u}\overline{\widehat{v}}-\rho_{3}\lambda_{3}\widehat{\theta}\overline{\widehat{\phi}}+\rho_{2}\lambda_{4}\xi^{2}\widehat{p}\overline{\widehat{y}}+\rho_{3}\lambda_{5}\widehat{z}\overline{\widehat{\theta}}\right)$
$\displaystyle+Re\left(i\rho_{2}\lambda_{6}\xi\widehat{\phi}\overline{\widehat{y}}+i\rho_{1}\lambda_{7}\xi\widehat{u}\overline{\widehat{z}}+i\rho_{1}\lambda_{8}\xi\widehat{u}\overline{\widehat{\phi}}+i\rho_{2}\lambda_{9}\xi\widehat{v}\overline{\widehat{y}}+\rho_{3}\lambda_{10}\widehat{v}\overline{\widehat{\theta}}\right)$
$\displaystyle+\tau_{0}Re\left[\left(-\rho_{3}\lambda_{11}\xi^{2}\widehat{\theta}+i\rho_{2}\lambda_{12}\xi\widehat{y}+i\rho_{1}\lambda_{13}\xi\widehat{u}\right)\displaystyle\int_{0}^{+\infty}\,g(s)\overline{\widehat{\eta}}ds\right],$
(3.25)
${F}_{1}(\xi,\,t)=-\xi^{2}\left(B_{1}|\widehat{z}|^{2}+B_{2}|\widehat{v}|^{2}+B_{3}|\widehat{\phi}|^{2}+B_{4}|\widehat{u}|^{2}+B_{5}|\widehat{p}|^{2}+B_{6}|\widehat{y}|^{2}+B_{7}|\widehat{\theta}|^{2}\right),$
(3.26)
${F}_{2}(\xi,\,t)=Re\left[i\xi\left(A_{1}\widehat{z}+A_{2}\widehat{v}+A_{3}\widehat{\phi}\right)\overline{\widehat{p}}+A_{4}\widehat{\phi}\overline{\widehat{z}}+A_{5}\widehat{v}\overline{\widehat{z}}+A_{6}\widehat{\phi}\overline{\widehat{v}}+A_{7}\xi^{2}\widehat{y}\overline{\widehat{u}}+iA_{8}\xi\widehat{\theta}\overline{\widehat{u}}+iA_{9}\xi\widehat{\theta}\overline{\widehat{y}}\right]$

and

(3.27) $\displaystyle{F}_{3}(\xi,\,t)$ $\displaystyle=$ $\displaystyle
Re\left[\left(\lambda_{3}\overline{\widehat{\phi}}-\lambda_{5}\overline{\widehat{z}}-\lambda_{10}\overline{\widehat{v}}\right)G\right]-\tau_{0}Re\left[\left(\rho_{3}\lambda_{11}\xi^{2}\overline{\widehat{\theta}}+i\rho_{2}\lambda_{12}\xi\overline{\widehat{y}}+i\rho_{1}\lambda_{13}\xi\overline{\widehat{u}}\right)\displaystyle\int_{0}^{+\infty}\,g^{\prime}(s){\widehat{\eta}}ds\right]$
$\displaystyle+\tau_{0}Re\left[\lambda_{11}\left[-(k_{3}-\tau_{0}g_{0})\xi^{4}\widehat{\phi}-ilk_{0}\xi^{3}\widehat{p}+\xi^{2}G\right]\displaystyle\int_{0}^{+\infty}\,g(s)\overline{\widehat{\eta}}ds\right]$
$\displaystyle-\tau_{0}Re\left[\left[\lambda_{12}\left(k_{2}\xi^{2}\widehat{z}+ik_{0}\xi\widehat{p}\right)+\lambda_{13}\left(k_{1}\xi^{2}\widehat{v}+ik_{0}\xi\widehat{p}\right)\right]\displaystyle\int_{0}^{+\infty}\,g(s)\overline{\widehat{\eta}}ds\right],$

where

$B_{1}=k_{2}\lambda_{1},\quad B_{2}=k_{1}\lambda_{2},\quad
B_{3}=(k_{3}-\tau_{0}g_{0})\lambda_{3},\quad B_{4}=-\rho_{1}\lambda_{2},$
$B_{5}=k_{0}\lambda_{4},\quad B_{6}=-\rho_{2}(\lambda_{1}+\lambda_{4}),\quad
B_{7}=\rho_{3}(\tau_{0}g_{0}\lambda_{11}-\lambda_{3}),$
$A_{1}=k_{0}\lambda_{1}+k_{2}\lambda_{4}\xi^{2}-lk_{0}\lambda_{5}+k_{0}\lambda_{7},\quad
A_{2}=k_{0}(\lambda_{2}-\lambda_{9}-l\lambda_{10}),\quad
A_{3}=k_{0}(l\lambda_{3}-\lambda_{6}+\lambda_{8})$
$A_{4}=(k_{3}-\tau_{0}g_{0})\lambda_{5}\xi^{2}+k_{2}\lambda_{6}\xi^{2},\quad
A_{5}=-k_{1}\lambda_{7}\xi^{2}+k_{2}\lambda_{9}\xi^{2},\quad
A_{6}=-k_{1}\lambda_{8}\xi^{2}+(k_{3}-\tau_{0}g_{0})\lambda_{10}\xi^{2},$
$A_{7}=\rho_{1}\lambda_{7}+\rho_{2}(\lambda_{4}-\lambda_{9}),\quad
A_{8}=\rho_{1}(\lambda_{8}\xi^{2}-\tau_{0}g_{0}\lambda_{13})-\rho_{3}\lambda_{10}\quad\hbox{and}\quad
A_{9}=\rho_{2}(l\lambda_{4}-\lambda_{6})\xi^{2}-\rho_{3}\lambda_{5}-\tau_{0}\rho_{2}g_{0}\lambda_{12}.$

Multiplying (3.11)-(3.23) by
$\lambda_{1},\cdots,\lambda_{10},\,\tau_{0}\lambda_{11},\,\tau_{0}\lambda_{12}$
and $\tau_{0}\lambda_{13}$, respectively, and adding the obtained equations,
we deduce that

(3.28)
$\frac{d}{dt}{F}_{0}(\xi,\,t)={F}_{1}(\xi,\,t)+{F}_{2}(\xi,\,t)+{F}_{3}(\xi,\,t).$

According to the notations (2.1), we have $p_{x}=v+z+l\phi$. Because (2.9)
implies that ${\widehat{p_{x}}}=i\xi{\widehat{p}}$, then

$i\xi\overline{\widehat{p}}=-\overline{\widehat{v}}-\overline{\widehat{z}}-l\overline{\widehat{\phi}},$

this identity allows to formulate the first term in $F_{2}$ as

$\displaystyle
Re\left[i\xi\left(A_{1}\widehat{z}+A_{2}\widehat{v}+A_{3}\widehat{\phi}\right)\overline{\widehat{p}}\right]$
$\displaystyle=$
$\displaystyle-\left(A_{1}|\widehat{z}|^{2}+A_{2}|\widehat{v}|^{2}+lA_{3}|\widehat{\phi}|^{2}\right)-\left(A_{1}+A_{2}\right)Re\left({\widehat{v}}\overline{\widehat{z}}\right)$
$\displaystyle-\left(A_{3}+lA_{1}\right)Re\left({\widehat{\phi}}\overline{\widehat{z}}\right)-\left(lA_{2}+A_{3}\right)Re\left({\widehat{\phi}}\overline{\widehat{v}}\right).$

By combining (3.28) and (3.2), we get

(3.30)
$\frac{d}{dt}{F}_{0}(\xi,\,t)={F}_{5}(\xi,\,t)+{F}_{4}(\xi,\,t)+{F}_{3}(\xi,\,t),$

where

$\displaystyle{F}_{5}(\xi,\,t)$ $\displaystyle=$
$\displaystyle-\left[\left(B_{1}\xi^{2}+A_{1}\right)|\widehat{z}|^{2}+\left(B_{2}\xi^{2}+A_{2}\right)|\widehat{v}|^{2}+\left(B_{3}\xi^{2}+lA_{3}\right)|\widehat{\phi}|^{2}\right]$
$\displaystyle-\xi^{2}\left[B_{4}|\widehat{u}|^{2}+B_{5}|\widehat{p}|^{2}+B_{6}|\widehat{y}|^{2}+B_{7}|\widehat{\theta}|^{2}\right]$

and

$\displaystyle{F}_{4}(\xi,\,t)$ $\displaystyle=$ $\displaystyle
Re\left[\left(A_{4}-A_{3}-lA_{1}\right)\widehat{\phi}\overline{\widehat{z}}+\left(A_{5}-A_{1}-A_{2}\right)\widehat{v}\overline{\widehat{z}}+\left(A_{6}-lA_{2}-A_{3}\right)\widehat{\phi}\overline{\widehat{v}}\right]$
$\displaystyle+Re\left[A_{7}\xi^{2}\widehat{y}\overline{\widehat{u}}+iA_{8}\xi\widehat{\theta}\overline{\widehat{u}}+iA_{9}\xi\widehat{\theta}\overline{\widehat{y}}\right].$

Now, we choose the different real numbers $\lambda_{1},\cdots,\lambda_{15}$ in
order to have

(3.31) $A_{4}-A_{3}-lA_{1}=A_{5}-A_{1}-A_{2}=A_{6}-lA_{2}-A_{3}=A_{7}=0,$
(3.32) $B_{1}\xi^{2}+A_{1}={\tilde{B}}_{1}\xi^{2},\quad
B_{2}\xi^{2}+A_{2}={\tilde{B}}_{2}\xi^{2},\quad
B_{3}\xi^{2}+lA_{3}={\tilde{B}}_{3}\xi^{2},$ (3.33)
${\tilde{B}}_{1}>0,\quad{\tilde{B}}_{2}>0,\quad{\tilde{B}}_{3}>0,\quad
B_{4}>0,\quad B_{5}>0,\quad B_{6}>0$

and

(3.34) $\displaystyle
A_{8}=A_{9}=0\,\,\,\hbox{and}\,\,\,B_{7}>0\quad\hbox{if}\,\,\tau_{0}=1.$

We start by choosing $\lambda_{7},\,\lambda_{8}$ and $\lambda_{9}$ as follows:

(3.35)
$\lambda_{7}=k_{2}\xi^{2}+l\lambda_{5}-\lambda_{1},\quad\lambda_{8}=-\frac{2(k_{3}-\tau_{0}g_{0})}{l}\xi^{2}+\lambda_{6}-l\lambda_{3}\quad\hbox{and}\quad\lambda_{9}=-k_{1}\xi^{2}+\lambda_{2}-l\lambda_{10}.$

The choices (3.35) guarantee (3.32) with

(3.36)
${\tilde{B}}_{1}=B_{1}+k_{2}\lambda_{4}+k_{0}k_{2},\quad{\tilde{B}}_{2}=B_{2}+k_{0}k_{1}\quad\hbox{and}\quad{\tilde{B}}_{3}=B_{3}-2k_{0}(k_{3}-\tau_{0}g_{0}).$

Also, we select $\lambda_{1},\,\lambda_{5},\,\lambda_{6}$ and $\lambda_{10}$
by

(3.37)
$\lambda_{6}=\frac{1}{k_{2}}\left[lk_{2}\lambda_{4}-(k_{3}-\tau_{0}g_{0})\lambda_{5}+k_{0}\left(lk_{2}-\frac{2(k_{3}-\tau_{0}g_{0})}{l}\right)\right],$
(3.38)
$\lambda_{10}=\frac{1}{lk_{2}}\left[-2k_{1}k_{2}\xi^{2}+k_{1}\lambda_{1}+k_{2}\lambda_{2}-k_{2}\lambda_{4}-lk_{1}\lambda_{5}-k_{0}(k_{1}+k_{2})\right],$
(3.39)
$\lambda_{1}=-\frac{k_{2}}{k_{1}}\lambda_{2}-\frac{l^{2}k_{2}}{k_{3}-\tau_{0}g_{0}}\lambda_{3}+\left(\frac{l^{2}k_{2}}{k_{3}-\tau_{0}g_{0}}+\frac{k_{2}}{k_{1}}\right)\lambda_{4}+k_{0}\left(\frac{2l^{2}k_{2}}{k_{3}-\tau_{0}g_{0}}-1-\frac{k_{2}}{k_{1}}\right)$

and

(3.40)
$\lambda_{5}=-\frac{k_{2}}{l}\xi^{2}+\frac{1}{l}\lambda_{1}-\frac{k_{0}\rho_{2}(k_{1}+k_{2})}{l(k_{1}\rho_{2}-k_{2}\rho_{1})}$

($\lambda_{5}$ and $\lambda_{1}$ are well defined thanks to (1.14) and the
right inequality in (1.1), respectively). According to (3.35), the selections
(3.37) and (3.38) imply $A_{4}-A_{3}-lA_{1}=0$ and $A_{5}-A_{1}-A_{2}=0$,
respectively, (3.37)-(3.39) lead to $A_{6}-lA_{2}-A_{3}=0$, and (3.38) and
(3.40) guarantee $A_{7}=0$, so (3.31) is satisfied. We put

(3.41)
$B_{0}=\min\left\\{{\tilde{B}}_{1},{\tilde{B}}_{2},{\tilde{B}}_{3},B_{4},B_{5},B_{6}\right\\}.$

Then, multiplying (3.30) by $\xi^{2}$ and exploiting (3.31) and (3.32), we
obtain

$\displaystyle\frac{d}{dt}\left[\xi^{2}{F}_{0}(\xi,\,t)\right]$
$\displaystyle\leq$ $\displaystyle-
B_{0}\xi^{4}\left[|\widehat{z}|^{2}+|\widehat{v}|^{2}+|\widehat{\phi}|^{2}+|\widehat{u}|^{2}+|\widehat{p}|^{2}+|\widehat{y}|^{2}\right]-B_{7}\xi^{4}|\widehat{\theta}|^{2}$
$\displaystyle+\xi^{2}Re\left[iA_{8}\xi\widehat{\theta}\overline{\widehat{u}}+iA_{9}\xi\widehat{\theta}\overline{\widehat{y}}\right]+\xi^{2}F_{3}(\xi,t).$

In order to estimate $\xi^{2}F_{3}(\xi,t)$, first, we notice the following
evident inequality:

(3.43)
$|\xi|^{m_{2}}\leq|\xi|^{m_{1}}+|\xi|^{m_{3}},\quad\forall\xi\in\mathbb{R},\,\,\forall
0\leq m_{1}\leq m_{2}\leq m_{3}.$

Second, using Hölder’s inequality and the right inequality in (1.2), we see
that, for
${\tilde{\eta}}\in\left\\{{\widehat{\eta}},\overline{\widehat{\eta}}\right\\}$,

$\displaystyle\left|\displaystyle\int_{0}^{+\infty}\,g(s){\tilde{\eta}}(\xi,s)\,ds\right|^{2}$
$\displaystyle=$
$\displaystyle\left|\displaystyle\int_{0}^{+\infty}\,{\sqrt{g(s)}}{\sqrt{g(s)}}{\tilde{\eta}}(\xi,s)\,ds\right|^{2}$
$\displaystyle\leq$
$\displaystyle\left(\displaystyle\int_{0}^{+\infty}\,g(s)\,ds\right)\displaystyle\int_{0}^{+\infty}\,g(s)|{\tilde{\eta}}(\xi,s)|^{2}\,ds$
$\displaystyle\leq$
$\displaystyle-\frac{g_{0}}{\beta_{1}}\displaystyle\int_{0}^{+\infty}\,g^{\prime}(s)|{\tilde{\eta}}(\xi,s)|^{2}\,ds.$

Similarly, using the limit in (2.19), we have

$\displaystyle\left|\displaystyle\int_{0}^{+\infty}\,g^{\prime}(s){\tilde{\eta}}(\xi,s)\,ds\right|^{2}$
$\displaystyle=$
$\displaystyle\left|\displaystyle\int_{0}^{+\infty}\,{\sqrt{-g^{\prime}(s)}}{\sqrt{-g^{\prime}(s)}}{\tilde{\eta}}(\xi,s)\,ds\right|^{2}$
$\displaystyle\leq$
$\displaystyle\left(-\displaystyle\int_{0}^{+\infty}\,g^{\prime}(s)\,ds\right)\displaystyle\int_{0}^{+\infty}\,(-g^{\prime}(s))|{\tilde{\eta}}(\xi,s)|^{2}\,ds$
$\displaystyle\leq$
$\displaystyle-g(0)\displaystyle\int_{0}^{+\infty}\,g^{\prime}(s)|{\tilde{\eta}}(\xi,s)|^{2}\,ds.$

Then, using these two inequalities and Young’s inequality, we get, for any
$h:\mathbb{R}\to\mathbb{C}$ and $\varepsilon>0$,

(3.44)
$Re\,\left(h(\xi)\displaystyle\int_{0}^{+\infty}\,g(s){\tilde{\eta}}(\xi,s)\,ds\right)\leq\varepsilon|h(\xi)|^{2}-\frac{g_{0}}{4\varepsilon\beta_{1}}\displaystyle\int_{0}^{+\infty}\,g^{\prime}(s)|{\tilde{\eta}}(\xi,s)|^{2}\,ds$

and

(3.45)
$Re\,\left(h(\xi)\displaystyle\int_{0}^{+\infty}\,g^{\prime}(s){\tilde{\eta}}(\xi,s)\,ds\right)\leq\varepsilon|h(\xi)|^{2}-\frac{g(0)}{4\varepsilon}\displaystyle\int_{0}^{+\infty}\,g^{\prime}(s)|{\tilde{\eta}}(\xi,s)|^{2}\,ds.$

Third, in the sequel, we use $C$ (sometimes $C_{1},\,C_{2},\,\cdots$) to
denote a generic real positive constant, and $C_{\varepsilon}$ to denote a
generic real positive constant depending on some real positive constant
$\varepsilon$, where $C$ and $C_{\varepsilon}$ may be different from step to
step. The constants $C$ ($C_{1},\,C_{2},\,\cdots$), $\varepsilon$ and
$C_{\varepsilon}$ are independent on $x$, $\xi$ and $t$. Applying Young’s
inequality for all the terms in $\xi^{2}{F}_{3}(\xi,\,t)$ and using (3.43),
(3.44), (3.45) and the fact that $\tau_{0}^{2}=\tau_{0}$ and
$\tau_{0}(1-\tau_{0})=0$, we get, for any $\varepsilon>0$,

$\displaystyle\xi^{2}{F}_{3}(\xi,\,t)$ $\displaystyle\leq$
$\displaystyle\varepsilon\xi^{4}\left[|\widehat{z}|^{2}+|\widehat{v}|^{2}+|\widehat{\phi}|^{2}+\tau_{0}\left(|\widehat{u}|^{2}+|\widehat{y}|^{2}+|\widehat{\theta}|^{2}+|\widehat{p}|^{2}\right)\right]$
$\displaystyle+C_{\varepsilon}\left(\lambda_{3}^{2}+\lambda_{5}^{2}+\lambda_{10}^{2}\right)|{G}|^{2}+\tau_{0}|\lambda_{11}|\xi^{4}|G|\left|\displaystyle\int_{0}^{+\infty}\,g(s)\overline{\widehat{\eta}}(\xi,s)\,ds\right|$
$\displaystyle-\tau_{0}C_{\varepsilon}\left[\lambda_{11}^{2}\xi^{4}\left(\xi^{4}+1\right)+\left(\lambda_{12}^{2}+\lambda_{13}^{2}\right)\xi^{2}\left(\xi^{2}+1\right)\right]\displaystyle\int_{0}^{+\infty}\,g^{\prime}(s)|{\widehat{\eta}}(\xi,s)|^{2}\,ds$

Now, we select
$\lambda_{2},\,\lambda_{3},\,\lambda_{4},\,\lambda_{11},\,\lambda_{12}$ and
$\lambda_{13}$ in order to get (3.33) and (3.34) (and then, in particular,
$B_{0}>0$ since (3.41)). To do so, we distinguish the two cases related to
(2.14).

###  3.3. Case 1: $\tau_{0}=0$

Notice that, because $\tau_{0}=0$, the numbers $\lambda_{11},\,\lambda_{12}$
and $\lambda_{13}$ are not used. On the other hand, we select
$\lambda_{2},\,\lambda_{3}$ and $\lambda_{4}$ as follows:

(3.47) $-k_{0}<\lambda_{2}<0,$ (3.48)
$0<\lambda_{4}<\frac{lk_{2}}{k_{1}k_{3}+k_{2}\left(l^{2}k_{1}+k_{3}\right)}\left[\frac{k_{3}}{l}\lambda_{2}+lk_{1}\lambda_{3}+k_{0}\left(\frac{k_{1}k_{3}}{lk_{2}}+\frac{k_{3}}{l}-2lk_{1}\right)\right],$
(3.49)
$\lambda_{4}>\frac{k_{1}k_{3}}{k_{1}k_{3}+k_{2}\left(l^{2}k_{1}+k_{3}\right)}\left[\frac{k_{2}}{k_{1}}\lambda_{2}+\frac{l^{2}k_{2}}{k_{3}}\lambda_{3}+k_{0}\left(\frac{k_{2}}{k_{1}}-\frac{2l^{2}k_{2}}{k_{3}}\right)\right]$

and

(3.50)
$\lambda_{3}>\max\left\\{2k_{0},\frac{1}{l^{2}k_{1}k_{2}}\left[-k_{2}k_{3}\lambda_{2}+k_{0}\left(2l^{2}k_{1}k_{2}-k_{1}k_{3}-k_{2}k_{3}\right)\right]\right\\}.$

We go back in the reverse order to select the numbers in the following manner:
1\. Choose $\lambda_{2}$ by (3.47). 2\. After, select $\lambda_{3}$ large
enough in such a way that (3.50) holds. 3\. Take $\lambda_{4}$ so that (3.48)
and (3.49) are true. According to (3.50), $\lambda_{4}$ exists. 4\. Now, it is
possible to find $\lambda_{1}$ through (3.39). 5\. After, take $\lambda_{5}$
as in (3.40). 6\. Next, it is time to pick $\lambda_{6}$ and $\lambda_{10}$
verifying (3.37) and (3.38), respectively. 7\. Finally, we can select
$\lambda_{7}$, $\lambda_{8}$ and $\lambda_{9}$ by (3.35).

We observe that the left inequality in (3.47), (3.50), the right inequality in
(3.47) and the left one in (3.48) imply that
${\tilde{B}}_{2}>0,\,{\tilde{B}}_{3}>0,\,B_{4}>0$ and $B_{5}>0$, respectively.
Moreover, (3.49) and the right inequality in (3.48) combined with (3.39) imply
${\tilde{B}}_{1}>0$ and $B_{6}>0$, respectively. Therefore (3.33) holds. Thus
(3.31)-(3.33) are satisfied. On the other hand, because
$\lambda_{1},\,\lambda_{2},\,\lambda_{3}$ and $\lambda_{4}$ do not depend on
$\xi$,

(3.51) $|\lambda_{j}|\leq
C\left(\xi^{2}+1\right),\,\,j=5,\cdots,10,\quad\hbox{and}\quad|A_{j}|\leq
C\left(\xi^{4}+1\right),\,\,j=8,9$

(since (3.43)), then, applying Young’s inequality and using (3.43), we get,
for any $\varepsilon>0$,

(3.52)
$\xi^{2}Re\left[iA_{8}\xi\widehat{\theta}\overline{\widehat{u}}+iA_{9}\xi\widehat{\theta}\overline{\widehat{y}}\right]\leq\varepsilon\xi^{4}\left(|\widehat{u}|^{2}+|\widehat{y}|^{2}\right)+C_{\varepsilon}\xi^{2}\left(\xi^{8}+1\right)|\widehat{\theta}|^{2}.$

Moreover, we conclude from (3.9) and (3.2) (with $\tau_{0}=0$) that

(3.53)
$\xi^{2}{F}_{3}(\xi,\,t)\leq\varepsilon\xi^{4}\left[|\widehat{z}|^{2}+|\widehat{v}|^{2}+|\widehat{\phi}|^{2}\right]+C_{\varepsilon}\left(\xi^{4}+1\right)|\widehat{\theta}|^{2}.$

By combining (3.2), (3.52) and (3.53), choosing $0<\varepsilon<B_{0}$
($\varepsilon$ exists thanks to (3.33) and (3.41)), and using (2.13) (with
$\tau_{0}=0$) and (3.43), we deduce that there exist real positive constants
$c_{1}$ and $c_{2}$ such that (notice that $B_{7}$ does not depend on $\xi$)

(3.54) $\frac{d}{dt}\left[\xi^{2}{F}_{0}(\xi,\,t)\right]\leq-
c_{1}\xi^{4}\widehat{E}(\xi,\,t)+c_{2}\left(\xi^{10}+1\right)|\widehat{\theta}|^{2}.$

Now, let $\lambda$ be a real positive constant and

(3.55)
$F(\xi,\,t)=\lambda\,\widehat{E}(\xi,\,t)+\frac{\xi^{2}}{\xi^{10}+1}\,{F}_{0}(\xi,\,t).$

From (2.15) (with $\tau_{0}=0$), (3.54) and (3.55), we find

(3.56) $\frac{d}{dt}{F}(\xi,\,t)\leq-
c_{1}f(\xi)\widehat{E}(\xi,\,t)-\left(\gamma\,\lambda-
c_{2}\right)|\widehat{\theta}|^{2},$

where $f$ is defined by (3.2) with $\tau_{0}=0$. Moreover, using the
definitions of $\widehat{E},\,{F}_{0}$ and ${F}$, and exploiting (3.43) and
(3.51), we observe that there exists $c_{3}>0$ (not depend on $\lambda$) such
that

(3.57)
$|{F}(\xi,\,t)-\lambda\,\widehat{E}(\xi,\,t)|\leq\frac{c_{3}\,\xi^{2}(|\xi|^{3}+1)}{\xi^{10}+1}\,\widehat{E}(\xi,\,t)\leq
2c_{3}\,\widehat{E}(\xi,\,t).$

Therefore, for $\lambda$ satisfying
$\lambda>\max\left\\{\frac{c_{2}}{\gamma},\,2c_{3}\right\\}$, we deduce from
(3.56) and (3.57) that

(3.58) $\frac{d}{dt}{F}(\xi,\,t)\leq-c_{1}\,f(\xi)\,\widehat{E}(\xi,\,t)$

and $F\sim\widehat{E}$, since

(3.59)
$(\lambda-2c_{3})\,\widehat{E}(\xi,\,t)\leq{F}(\xi,\,t)\leq(\lambda+2c_{3})\,\widehat{E}(\xi,\,t).$

Consequently, a combination of (3.58) and the second inequality in (3.59)
leads to, for $c=\frac{c_{1}}{2(\lambda+2c_{3})}$,

(3.60) $\frac{d}{dt}{F}(\xi,\,t)\leq-2c\,f(\xi)\,F(\xi,\,t).$

Multiplying (3.60) by $e^{2cf(\xi)t}$, we find

(3.61) $\frac{d}{dt}\left(e^{2cf(\xi)t}{F}(\xi,\,t)\right)\leq 0,$

then, by integration (3.61) with respect to $t$,

(3.62) ${F}(\xi,\,t)\leq e^{-2cf(\xi)t}{F}(\xi,\,0).$

Finally, using (2.20) and (3.59), (3.1) in case $\tau_{0}=0$ follows from
(3.62).

###  3.4. Case 2: $\tau_{0}=1$

We choose $\lambda_{1},\cdots,\,\lambda_{10}$ as in case $1$ but with
$k_{3}-g_{0}$ instead of $k_{3}$ ($k_{3}-g_{0}>0$ thanks to the right
inequality in (1.1)), and we get (3.31)-(3.33). Next, we select

(3.63)
$\lambda_{13}=\frac{1}{g_{0}}\lambda_{8}\xi^{2}-\frac{\rho_{3}}{g_{0}\rho_{1}}\lambda_{10},$
(3.64)
$\lambda_{12}=\frac{1}{g_{0}}(l\lambda_{4}-\lambda_{6})\xi^{2}-\frac{\rho_{3}}{g_{0}\rho_{2}}\lambda_{5}$

and

(3.65) $\lambda_{11}>\frac{1}{g_{0}}\lambda_{3}$

($g_{0}>0$ according to the left inequality in (1.1)). The choices
(3.63)-(3.65) imply tha $A_{8}=A_{9}=0$ and $B_{7}>0$, respectively, thus
(3.34) is satisfied. Therefore, (3.2) implies

(3.66)
$\frac{d}{dt}\left[\xi^{2}{F}_{0}(\xi,\,t)\right]\leq-\min\\{B_{0},B_{7}\\}\xi^{4}\left[|\widehat{z}|^{2}+|\widehat{v}|^{2}+|\widehat{\phi}|^{2}+|\widehat{u}|^{2}+|\widehat{p}|^{2}+|\widehat{y}|^{2}+|\widehat{\theta}|^{2}\right]+\xi^{2}F_{3}(\xi,t).$

On the other hand, $\lambda_{1},\,\lambda_{2},\,\lambda_{3},\lambda_{4}$ and
$\lambda_{11}$ do not depend on $\xi$, then

(3.67) $|\lambda_{j}|\leq
C\left(\xi^{2}+1\right),\,\,j=5,\cdots,10,\quad\hbox{and}\quad|\lambda_{j}|\leq
C\left(\xi^{4}+1\right),\,\,j=12,13,$

therefore, applying Young’s inequality and using (3.9) (with $\tau_{0}=1$) and
(3.43), we conclude from (3.2), for any $\varepsilon>0$, that

(3.68) $\displaystyle\xi^{2}{F}_{3}(\xi,\,t)$ $\displaystyle\leq$
$\displaystyle\varepsilon\xi^{4}\left[|\widehat{z}|^{2}+|\widehat{v}|^{2}+|\widehat{\phi}|^{2}+|\widehat{p}|^{2}+|\widehat{u}|^{2}+|\widehat{y}|^{2}+|\widehat{\theta}|^{2}\right]$
$\displaystyle-
C_{\varepsilon}\xi^{2}\left(\xi^{10}+1\right)\displaystyle\int_{0}^{+\infty}\,g^{\prime}(s)|{\widehat{\eta}}(\xi,s)|^{2}\,ds.$

By combining (3.66) and (3.68), choosing
$0<\varepsilon<\min\\{B_{0},B_{7}\\}$, and using (2.13) and (3.43), we deduce
that there exist real positive constants $c_{1}$ and $c_{2}$ such that

(3.69) $\frac{d}{dt}\left[\xi^{2}{F}_{0}(\xi,\,t)\right]\leq-
c_{1}\xi^{4}\widehat{E}(\xi,\,t)-c_{2}\xi^{2}\left(\xi^{10}+1\right)\displaystyle\int_{0}^{+\infty}\,g^{\prime}(s)|{\widehat{\eta}}(\xi,s)|^{2}\,ds.$

Now, let $\lambda$ be a real positive constant and

(3.70)
$F(\xi,\,t)=\lambda\,\widehat{E}(\xi,\,t)+\frac{\xi^{4}}{\xi^{10}+1}\,{F}_{0}(\xi,\,t).$

From (2.15) (with $\tau_{0}=1$), (3.69) and (3.70), we find

(3.71) $\frac{d}{dt}{F}(\xi,\,t)\leq-
c_{1}f(\xi)\widehat{E}(\xi,\,t)+\left(\frac{\lambda}{2}-c_{2}\right)\xi^{4}\displaystyle\int_{0}^{+\infty}\,g^{\prime}(s)|{\widehat{\eta}}(\xi,s)|^{2}\,ds,$

where $f$ is defined in (3.2) with $\tau_{0}=1$. Also, using the definitions
of $\widehat{E},\,{F}_{0}$ and ${F}$, and exploiting (3.43) and (3.67), we see
that there exists $c_{3}>0$ (not depend on $\lambda$) such that

(3.72)
$|{F}(\xi,\,t)-\lambda\,\widehat{E}(\xi,\,t)|\leq\frac{c_{3}\,\xi^{4}\left(|\xi|^{5}+1\right)}{\xi^{10}+1}\,\widehat{E}(\xi,\,t)\leq
2c_{3}\,\widehat{E}(\xi,\,t).$

Therefore, choosing $\lambda>\max\left\\{2c_{2},\,2c_{3}\right\\}$ and using
(3.71) and (3.72) and the fact that $g^{\prime}\leq 0$, we find (3.58) and
(3.59). Consequently, (3.58) and the second inequality in (3.59) lead to
(3.60). Finally, by integration (3.60) with respect to $t$ and using (2.20)
and (3.59), we obtain (3.1) in case $\tau_{0}=1$. The proof of the second
result in Theorem 3.1 is finished.

##  4\. Estimation of $\left\|\partial_{x}^{j}U\right\|_{2}$

In this section, we use the second result in Theorem 3.1 to get some decay
estimates on $\|\partial_{x}^{k}U\|_{2}$ when (1.14) is valid.

######  Theorem 4.1.

Assume that (1.14) is satisfied. Let $N\,\in\mathbb{N}^{*}$,

(4.1) $U_{0}\in H^{N}(\mathbb{R})\cap L^{1}(\mathbb{R})$

and $U$ be the solution of (2.8). Then, for any
$\ell\in\\{1,\,2,\,\ldots,\,N\\}$ and $j\in\\{0,\,1,\,\ldots,\,N-\ell\\}$,
there exists $c_{0}>0$ such that

(4.2) $\|\partial_{x}^{j}U\|_{2}\leq
c_{0}\,(1+t)^{-1/8-j/4}\,\|U_{0}\|_{1}+c_{0}\,(1+t)^{-\ell/6}\,\|\partial_{x}^{j+\ell}U_{0}\|_{2},\quad
t\in\mathbb{R}_{+}$

if $\tau_{0}=0$, and

(4.3) $\|\partial_{x}^{j}U\|_{2}\leq
c_{0}\,(1+t)^{-1/12-j/6}\,\|U_{0}\|_{1}+c_{0}\,(1+t)^{-\ell/4}\,\|\partial_{x}^{j+\ell}U_{0}\|_{2},\quad
t\in\mathbb{R}_{+}$

if $\tau_{0}=1$.

###### Proof.

For the proof of (4.2), we see that (3.2) with $\tau_{0}=0$ implies that (low
and high frequences)

(4.4)
$f(\xi)\geq\left\\{\begin{array}[]{cc}\frac{1}{2}\,\xi^{4}&\mbox{if}\quad|\xi|\leq
1,\\\ \\\ \frac{1}{2}\,\xi^{-6}&\mbox{if}\quad|\xi|>1.\end{array}\right.$

Applying Plancherel’s theorem and (3.1), we entail

(4.5)
$\|\partial_{x}^{j}U\|_{2}^{2}=\left\|\widehat{\partial_{x}^{j}U}(x,\,t)\right\|_{2}^{2}=\int_{\mathbb{R}}\xi^{2\,j}\,|\widehat{U}(\xi,\,t)|^{2}d\xi$
$\displaystyle\leq$
$\displaystyle\widetilde{c}\int_{\mathbb{R}}\xi^{2\,j}\,e^{-\,c\,f(\xi)\,t}\,|\widehat{U}_{0}(\xi)|^{2}d\xi$
$\displaystyle\leq$ $\displaystyle\widetilde{c}\int_{|\xi|\leq
1}\xi^{2\,j}\,e^{-\,c\,f(\xi)\,t}\,|\widehat{U}_{0}(\xi)|^{2}d\xi+\widetilde{c}\int_{|\xi|>1}\xi^{2\,j}\,e^{-\,c\,f(\xi)\,t}\,|\widehat{U}_{0}(\xi)|^{2}\,d\xi:=J_{1}+J_{2}.$

Applying Lemma 2.3 of [8] and (4.4)1, it follows, for the low frequency
region,

(4.6) $J_{1}\leq C\,\|\widehat{U}_{0}\|_{\infty}^{2}\int_{|\xi|\leq
1}\xi^{2\,j}\,e^{-\,\frac{c}{2}\,t\,\xi^{4}}\ d\xi\leq
C\left(1+t\right)^{-\,\frac{1}{4}(1+2\,j)}\|{U}_{0}\|_{1}^{2}.$

For the high frequency region, using (4.4)2, we observe that

$\displaystyle J_{2}$ $\displaystyle\leq$ $\displaystyle
C\int_{|\xi|>1}|\xi|^{2\,j}\,e^{-\frac{c}{2}\,t\,\xi^{-6}}\,|\widehat{U}(\xi,\,0)|^{2}\,d\xi$
$\displaystyle\leq$ $\displaystyle C\
\sup_{|\xi|>1}\left\\{|\xi|^{-2\,\ell}\,e^{-\frac{c}{2}\,t\,|\xi|^{-6}}\right\\}\int_{\mathbb{R}}|\xi|^{2\,(j+\ell)}\,|\widehat{U}(\xi,\,0)|^{2}\
d\xi,$

then, using Lemma 2.4 of [8],

(4.7) $J_{2}\leq
C\left(1+t\right)^{-\frac{1}{3}\,\ell}\,\|\,\partial_{x}^{j+\ell}U_{0}\,\|_{2}^{2},$

and so, by combining (4.5)-(4.7), we get

(4.8) $\|\partial_{x}^{j}U\|_{2}^{2}\leq
C\left[\left(1+t\right)^{-\,\frac{1}{4}(1+2\,j)}\|{U}_{0}\|_{1}^{2}+\left(1+t\right)^{-\frac{1}{3}\,\ell}\,\|\,\partial_{x}^{j+\ell}U_{0}\,\|_{2}^{2}\right].$

Finally, by combining (4.8) and the inequality

(4.9) ${\sqrt{a_{1}+a_{2}}}\leq{\sqrt{a_{1}}}+{\sqrt{a_{2}}},\quad\forall
a_{1},a_{2}\in\mathbb{R}_{+},$

we find (4.2). The proof of (4.3) is identical to the one of (4.2) where,
instead of (4.4), we have (according to (3.2) with $\tau_{0}=1$)

(4.10)
$f(\xi)\geq\left\\{\begin{array}[]{cc}\frac{1}{2}\,\xi^{6}&\mbox{if}\quad|\xi|\leq
1,\\\ \\\ \frac{1}{2}\,\xi^{-4}&\mbox{if}\quad|\xi|>1.\end{array}\right.$

∎

##  5\. Estimation of $|\partial_{\xi}\widehat{U}|$

In this section, we study the asymptotic behavior (with respect to $t$) of
$\partial_{\xi}\widehat{U}$. In order to simplify the computations, let us
denoting $\partial_{\xi}\widehat{U}=\widehat{\mathbf{U}}$,
$\partial_{\xi}\widehat{U}_{0}=\widehat{\mathbf{U}}_{0}$ and

$\left(\partial_{\xi}\widehat{v},\partial_{\xi}\widehat{u},\partial_{\xi}\widehat{z},\partial_{\xi}\widehat{y},\partial_{\xi}\widehat{\phi},\partial_{\xi}\widehat{\theta},\partial_{\xi}\widehat{p},\partial_{\xi}\widehat{\eta}\right)=\left(\widehat{\mathbf{v}},\widehat{\mathbf{u}},\widehat{\mathbf{z}},\widehat{\mathbf{y}},\widehat{\Phi},\widehat{\Theta},\widehat{\mathbf{p}},\widehat{\Lambda}\right).$

As (2.13), the energy associated to $\widehat{\mathbf{U}}$ is define by

(5.1) $\displaystyle\widehat{\mathbf{E}}(\xi,t)$ $\displaystyle=$
$\displaystyle\frac{1}{2}\left[k_{1}\,|\widehat{\mathbf{v}}|^{2}+\rho_{1}|\widehat{\mathbf{u}}|^{2}+k_{2}\,|\widehat{\mathbf{z}}|^{2}+\rho_{2}|\widehat{\mathbf{y}}|^{2}+(k_{3}-{\tau}_{0}g_{0})|\widehat{\Phi}|^{2}+\rho_{3}|\widehat{\Theta}|^{2}+k_{0}|\widehat{\mathbf{p}}|^{2}\right](\xi,\,t)$
$\displaystyle+\frac{{\tau}_{0}}{2}\xi^{4}\displaystyle\int_{0}^{+\infty}\,g(s)|\widehat{\Lambda}(\xi,\,t)|^{2}\,ds,$

where $\tau_{0}$ is defined by (2.14). Applying the operator $\partial_{\xi}$
to (3.10), we obtain the system

(5.2)
$\displaystyle\begin{cases}\widehat{\mathbf{v}}_{t}-i\xi\widehat{\mathbf{u}}=i\widehat{u},\\\
\rho_{1}\widehat{\mathbf{u}}_{t}-ik_{1}\xi\,\widehat{\mathbf{v}}+k_{0}\widehat{\mathbf{p}}=ik_{1}\widehat{v},\\\
\widehat{\mathbf{z}}_{t}-i\xi\widehat{\mathbf{y}}=i\widehat{y},\\\
\rho_{2}\widehat{\mathbf{y}}_{t}-ik_{2}\xi\,\widehat{\mathbf{z}}+k_{0}\widehat{\mathbf{p}}=ik_{2}\widehat{z},\\\
\widehat{\Phi}_{t}+\xi^{2}\widehat{\Theta}=-2\xi\widehat{\theta},\\\
\rho_{3}\widehat{\Theta}_{t}-(k_{3}-\tau_{0}g_{0})\xi^{2}\widehat{\Phi}-ilk_{0}\,\xi\widehat{\mathbf{p}}+{\mathbf{G}}=G_{0},\\\
\widehat{\mathbf{p}}_{t}-\widehat{\mathbf{u}}-\widehat{\mathbf{y}}-il\xi\widehat{\Theta}=il\widehat{\theta},\\\
\widehat{\Lambda}_{t}+\widehat{\Lambda}_{s}-\widehat{\Theta}=0,\end{cases}$

where

(5.3)
${\mathbf{G}}=(1-\tau_{0})\gamma\widehat{\Theta}+\tau_{0}\xi^{4}\int_{0}^{+\infty}g(s)\widehat{\Lambda}ds\,\,\hbox{and}\,\,G_{0}=2(k_{3}-\tau_{0}g_{0})\xi\widehat{\phi}+ilk_{0}\widehat{p}-4\tau_{0}\xi^{3}\displaystyle\int_{0}^{+\infty}\,g(s){\widehat{\eta}}(\xi,s)\,ds.$

As for the proof of (2.15), multiplying (5.2)1-(5.2)7 by
$k_{1}\bar{\widehat{\mathbf{v}}}$, $\bar{\widehat{\mathbf{u}}}$,
$k_{2}\bar{\widehat{\mathbf{z}}}$, $\bar{\widehat{\mathbf{y}}}$,
$(k_{3}-\tau_{0}g_{0})\bar{\widehat{\Phi}}$, $\bar{\widehat{\Theta}}$ and
$k_{0}\bar{\widehat{\mathbf{p}}}$, respectively, multiplying (5.2)8 by
$\tau_{0}\xi^{4}g(s)\bar{\widehat{\Lambda}}$ and integrating on
$\mathbb{R}_{+}$ with respect to $s$, adding all the obtained equations,
taking the real part of the resulting expression and using (2.17), we have

$\displaystyle\frac{d}{dt}\widehat{\mathbf{E}}(\xi,\,t)$ $\displaystyle=$
$\displaystyle-(1-\tau_{0})\gamma\,|\widehat{\Theta}|^{2}+\frac{\tau_{0}}{2}\xi^{4}\displaystyle\int_{0}^{+\infty}\,g^{\prime}(s)|\widehat{\Lambda}|^{2}\,ds$
$\displaystyle+Re\left[ik_{1}\widehat{u}\bar{\widehat{\mathbf{v}}}+ik_{1}\widehat{v}\bar{\widehat{\mathbf{u}}}+ik_{2}\widehat{y}\bar{\widehat{\mathbf{z}}}+ik_{2}\widehat{z}\bar{\widehat{\mathbf{y}}}-2(k_{3}-\tau_{0}g_{0})\xi\widehat{\theta}\bar{\widehat{\Phi}}+G_{0}\bar{\widehat{\Theta}}+ilk_{0}\widehat{\theta}\bar{\widehat{\mathbf{p}}}\right].$

This identity shows that $\widehat{\mathbf{E}}$ is not necessarily
nonincreasing with respect to $t$. Because

$|\widehat{\mathbf{U}}(\xi,\,t)|^{2}=\left[|\widehat{\mathbf{v}}|^{2}+|\widehat{\mathbf{u}}|^{2}+|\widehat{\mathbf{z}}|^{2}+|\widehat{\mathbf{y}}|^{2}+|\widehat{\Phi}|^{2}+|\widehat{\Theta}|^{2}+|\widehat{\mathbf{p}}|^{2}\right](\xi,\,t)+\tau_{0}\xi^{4}\displaystyle\int_{0}^{+\infty}\,g(s)|\widehat{\Lambda}(\xi,\,t)|^{2}\,ds,$

then we see that, according to the right inequality in (1.1),

(5.5) $|\widehat{\mathbf{U}}|^{2}\sim\widehat{\mathbf{E}}.$

######  Theorem 5.1.

Assume that (1.14) is satisfied. Let $\widehat{U}$ be the solution of (2.10).
Then there exist ${\mathbf{c}},\,\widetilde{{\mathbf{c}}}>0$ such that, for
any $t\in\mathbb{R}_{+}$ and for any $\xi\in\mathbb{R}^{*}$,

(5.6)
$|\widehat{\mathbf{U}}(\xi,\,t)|\leq\widetilde{{\mathbf{c}}}e^{-\,{\mathbf{c}}\,{f}(\xi)\,t}\left[|\widehat{\mathbf{U}}_{0}(\xi)|+\left(\xi^{7-2\tau_{0}}+\xi^{-(4+2\tau_{0})}\right)\,|\widehat{U}_{0}(\xi)|\right],$

where $f$ is defined in (3.2).

###### Proof.

We observe that the left hand sides of (5.2) are identical to the ones of
(3.10) if we replace $\widehat{v}$, $\widehat{u},\,\widehat{z},\,\widehat{y}$,
$\widehat{\phi}$, $\widehat{\theta}$, $\widehat{p}$, $\widehat{\eta}$ and $G$
by $\widehat{\mathbf{v}}$,
$\widehat{\mathbf{u}},\,\widehat{\mathbf{z}},\,\widehat{\mathbf{y}}$,
$\widehat{\Phi}$, $\widehat{\Theta}$, $\widehat{\mathbf{p}}$,
$\widehat{\Lambda}$ and ${\mathbf{G}}$, respectively. So, first, we use the
same arguments to get similar differential identities to (3.11)-(3.23), and
second, we treat the additional terms generated by the ones in the right hand
sindes of (5.2). Multiplying (5.2)4 and (5.2)3 by
$i\,\xi\,\overline{\widehat{\mathbf{z}}}$ and
$-i\rho_{2}\,\xi\,\overline{\widehat{\mathbf{y}}}$, respectively, adding the
resulting equations, taking the real part and using (2.17), we obtain

(5.7)
$\frac{d}{dt}Re\left(i\rho_{2}\,\xi\,\widehat{\mathbf{y}}\,\overline{\widehat{\mathbf{z}}}\right)=\xi^{2}\left(\rho_{2}|\widehat{\mathbf{y}}|^{2}-k_{2}|\widehat{\mathbf{z}}|^{2}\right)+k_{0}\,Re\left(i\,\xi\,\widehat{\mathbf{z}}\,\overline{\widehat{\mathbf{p}}}\right)+Re\left(\rho_{2}\,\xi\widehat{y}\,\overline{\widehat{\mathbf{y}}}-k_{2}\,\xi\widehat{z}\,\overline{\widehat{\mathbf{z}}}\right).$

Similarily, multiplying (5.2)2 and (5.2)1 by
$i\,\xi\,\overline{\widehat{\mathbf{v}}}$ and
$-i\rho_{1}\,\xi\,\overline{\widehat{\mathbf{u}}}$, respectively, adding the
resulting equations, taking the real part and using (2.17), we find

(5.8)
$\frac{d}{dt}Re\left(i\rho_{1}\,\xi\,\widehat{\mathbf{u}}\,\overline{\widehat{\mathbf{v}}}\right)=\xi^{2}\left(\rho_{1}|\widehat{\mathbf{u}}|^{2}-k_{1}|\widehat{\mathbf{v}}|^{2}\right)+k_{0}Re\left(i\,\xi\,\widehat{\mathbf{v}}\,\overline{\widehat{\mathbf{p}}}\right)+Re\left(\rho_{1}\,\xi\widehat{u}\,\overline{\widehat{\mathbf{u}}}-k_{1}\,\xi\widehat{v}\,\overline{\widehat{\mathbf{v}}}\right).$

Also, multiplying (5.2)6 and (5.2)5 by $-\overline{\widehat{\Phi}}$ and
$-\rho_{3}\,\overline{\widehat{\Theta}}$, respectively, adding the resulting
equations, taking the real part and using (2.17), we get

$\displaystyle\frac{d}{dt}Re\left(-\rho_{3}\,\widehat{\Theta}\,\overline{\widehat{\Phi}}\right)$
$\displaystyle=$
$\displaystyle\xi^{2}\left(\rho_{3}|\widehat{\Theta}|^{2}-(k_{3}-\tau_{0}g_{0})|\widehat{\Phi}|^{2}\right)+lk_{0}\,Re\left(i\,\xi\,\widehat{\Phi}\,\overline{\widehat{\mathbf{p}}}\right)$
$\displaystyle+\,Re\left(\overline{\widehat{\Phi}}\,{\mathbf{G}}\right)+Re\left(2\rho_{3}\xi\widehat{\theta}\,\overline{\widehat{\Theta}}-G_{0}\,\overline{\widehat{\Phi}}\right).$

Multiplying (5.2)4 and (5.2)7 by $\xi^{2}\,\overline{\widehat{\mathbf{p}}}$
and $\rho_{2}\xi^{2}\,\overline{\widehat{\mathbf{y}}}$, respectively, adding
the resulting equations, taking the real part and using (2.17), we obtain

$\displaystyle\frac{d}{dt}Re\left(\rho_{2}\xi^{2}\,\widehat{\mathbf{p}}\,\overline{\widehat{\mathbf{y}}}\right)$
$\displaystyle=$
$\displaystyle\xi^{2}\left(\rho_{2}\,|\widehat{\mathbf{y}}|^{2}-k_{0}|\widehat{\mathbf{p}}|^{2}\right)+\rho_{2}\xi^{2}\,Re\left(\widehat{\mathbf{y}}\,\overline{\widehat{\mathbf{u}}}\right)+l\rho_{2}\xi^{2}\,Re\left(i\xi\,\widehat{\Theta}\,\overline{\widehat{\mathbf{y}}}\right)$
$\displaystyle+k_{2}\xi^{2}\,Re\left(i\,\xi\,\widehat{\mathbf{z}}\,\overline{\widehat{\mathbf{p}}}\right)+Re\left(ik_{2}\xi^{2}\widehat{z}\,\overline{\widehat{\mathbf{p}}}+il\rho_{2}\,\xi^{2}\widehat{\theta}\,\overline{\widehat{\mathbf{y}}}\right).$

After, multiplying (5.2)3 and (5.2)6 by
$\rho_{3}\,\overline{\widehat{\Theta}}$ and $\overline{\widehat{\mathbf{z}}}$,
respectively, adding the resulting equations, taking the real part and using
(2.17), we entail

$\displaystyle\frac{d}{dt}Re\left(\rho_{3}\widehat{\mathbf{z}}\,\overline{\widehat{\Theta}}\right)$
$\displaystyle=$
$\displaystyle-\rho_{3}\,Re\left(i\xi\widehat{\Theta}\,\overline{\widehat{\mathbf{y}}}\right)+(k_{3}-\tau_{0}g_{0})\,\xi^{2}\,Re\left(\widehat{\Phi}\,\overline{\widehat{\mathbf{z}}}\right)-lk_{0}\,Re\left(i\,\xi\,\widehat{\mathbf{z}}\,\overline{\widehat{\mathbf{p}}}\right)$
$\displaystyle-\,Re\left(\overline{\widehat{\mathbf{z}}}\,{\mathbf{G}}\right)+Re\left(i\rho_{3}\widehat{y}\,\overline{\widehat{\Theta}}+G_{0}\,\overline{\widehat{\mathbf{z}}}\right).$

Multiplying (5.2)5 and (5.2)4 by
$i\rho_{2}\xi\,\overline{\widehat{\mathbf{y}}}$ and
$-i\xi\overline{\widehat{\Phi}}$, respectively, adding the resulting
equations, taking the real part and using (2.17), it follows that

$\displaystyle\frac{d}{dt}Re\left(i\rho_{2}\xi\,\widehat{\Phi}\,\overline{\widehat{\mathbf{y}}}\right)$
$\displaystyle=$
$\displaystyle-\rho_{2}\xi^{2}\,Re\left(i\xi\widehat{\Theta}\,\overline{\widehat{\mathbf{y}}}\right)+k_{2}\xi^{2}\,Re\left(\widehat{\Phi}\,\overline{\widehat{\mathbf{z}}}\right)-k_{0}\,Re\left(i\xi\widehat{\Phi}\,\overline{\widehat{\mathbf{p}}}\right)$
$\displaystyle+Re\left(k_{2}\,\xi\widehat{z}\,\overline{\widehat{\Phi}}-2i\rho_{2}\,\xi^{2}\widehat{\theta}\,\overline{\widehat{\mathbf{y}}}\right).$

Next, multiplying (5.2)2 and (5.2)3 by $i\xi\overline{\widehat{\mathbf{z}}}$
and $-i\rho_{1}\xi\,\overline{\widehat{\mathbf{u}}}$, respectively, adding the
resulting equations, taking the real part and using (2.17), it appears that

(5.13)
$\frac{d}{dt}Re\left(i\rho_{1}\xi\,\widehat{\mathbf{u}}\,\overline{\widehat{\mathbf{z}}}\right)=-k_{1}\xi^{2}\,Re\left(\widehat{\mathbf{v}}\,\overline{\widehat{\mathbf{z}}}\right)+\rho_{1}\xi^{2}\,Re\left(\widehat{\mathbf{y}}\,\overline{\widehat{\mathbf{u}}}\right)+k_{0}\,Re\left(i\xi\widehat{\mathbf{z}}\,\overline{\widehat{\mathbf{p}}}\right)+Re\left(\rho_{1}\,\xi\widehat{y}\,\overline{\widehat{\mathbf{u}}}-k_{1}\,\xi\widehat{v}\,\overline{\widehat{\mathbf{z}}}\right).$

Multiplying (5.2)2 and (5.2)5 by $i\xi\overline{\widehat{\Phi}}$ and
$-i\rho_{1}\xi\,\overline{\widehat{\mathbf{u}}}$, respectively, adding the
resulting equations, taking the real part and using (2.17), we see that

$\displaystyle\frac{d}{dt}Re\left(i\rho_{1}\xi\,\widehat{\mathbf{u}}\,\overline{\widehat{\Phi}}\right)$
$\displaystyle=$ $\displaystyle-
k_{1}\xi^{2}\,Re\left(\widehat{\Phi}\,\overline{\widehat{\mathbf{v}}}\right)+\rho_{1}\,\xi^{2}\,Re\left(i\xi\widehat{\Theta}\,\overline{\widehat{\mathbf{u}}}\right)+k_{0}\,Re\left(i\xi\widehat{\Phi}\,\overline{\widehat{\mathbf{p}}}\right)$
$\displaystyle+Re\left(2i\rho_{1}\,\xi^{2}\widehat{\theta}\,\overline{\widehat{\mathbf{u}}}-k_{1}\,\xi\widehat{v}\,\overline{\widehat{\Phi}}\right).$

Also, multiplying (5.2)1 and (5.2)4 by
$i\rho_{2}\xi\,\overline{\widehat{\mathbf{y}}}$ and
$-i\xi\overline{\widehat{\mathbf{v}}}$, respectively, adding the resulting
equations, taking the real part and using (2.17), we find

(5.15)
$\frac{d}{dt}Re\left(i\rho_{2}\xi\widehat{\mathbf{v}}\,\overline{\widehat{\mathbf{y}}}\right)=-\rho_{2}\xi^{2}\,Re\left(\widehat{\mathbf{y}}\,\overline{\widehat{\mathbf{u}}}\right)+k_{2}\xi^{2}\,Re\left(\widehat{\mathbf{v}}\,\overline{\widehat{\mathbf{z}}}\right)-k_{0}\,Re\left(i\xi\widehat{\mathbf{v}}\,\overline{\widehat{\mathbf{p}}}\right)+Re\left(k_{2}\xi\widehat{z}\,\overline{\widehat{\mathbf{v}}}-\rho_{2}\,\xi\widehat{u}\,\overline{\widehat{\mathbf{y}}}\right).$

Multiplying (5.2)1 and (5.2)6 by $\rho_{3}\,\overline{\widehat{\Theta}}$ and
$\overline{\widehat{\mathbf{v}}}$, respectively, adding the resulting
equations, taking the real part and using (2.17), we get

$\displaystyle\frac{d}{dt}Re\left(\rho_{3}\,\widehat{\mathbf{v}}\,\overline{\widehat{\Theta}}\right)$
$\displaystyle=$
$\displaystyle-\rho_{3}\,Re\left(i\xi\widehat{\Theta}\,\overline{\widehat{\mathbf{u}}}\right)+(k_{3}-\tau_{0}g_{0})\,\xi^{2}\,Re\left(\widehat{\Phi}\,\overline{\widehat{\mathbf{v}}}\right)-lk_{0}\,Re\left(i\xi\widehat{\mathbf{v}}\,\overline{\widehat{\mathbf{p}}}\right)$
$\displaystyle-\,Re\left(\overline{\widehat{\mathbf{v}}}\,{\mathbf{G}}\right)+Re\left(i\rho_{3}\widehat{u}\,\overline{\widehat{\Theta}}+G_{0}\,\overline{\widehat{\mathbf{v}}}\right).$

Now, multiplying (5.2)6 by
$-\xi^{2}\displaystyle\int_{0}^{+\infty}\,g(s)\overline{\widehat{\Lambda}}\,ds$,
multiplying (5.2)8 by $-\rho_{3}\xi^{2}g(s)\overline{\widehat{\Theta}}$ and
integrating over $\mathbb{R}_{+}$ with respect to $s$, adding the resulting
equations, taking the real part and using (2.17), we infer that

$\displaystyle\frac{d}{dt}Re\left(-\rho_{3}\xi^{2}{\widehat{\Theta}}\,\displaystyle\int_{0}^{+\infty}\,g(s)\overline{\widehat{\Lambda}}\,ds\right)$
$\displaystyle=$
$\displaystyle-\rho_{3}g_{0}\xi^{2}|\widehat{\Theta}|^{2}-(k_{3}-\tau_{0}g_{0})\xi^{4}Re\left(\widehat{\Phi}\displaystyle\int_{0}^{+\infty}\,g(s)\overline{\widehat{\Lambda}}\,ds\right)$
$\displaystyle-
lk_{0}\xi^{2}Re\left(i\xi\widehat{\mathbf{p}}\displaystyle\int_{0}^{+\infty}\,g(s)\overline{\widehat{\Lambda}}\,ds\right)-\rho_{3}\xi^{2}Re\left(\overline{\widehat{\Theta}}\displaystyle\int_{0}^{+\infty}\,g^{\prime}(s){\widehat{\Lambda}}\,ds\right)$
$\displaystyle+\xi^{2}Re\left({\mathbf{G}}\displaystyle\int_{0}^{+\infty}\,g(s)\overline{{\widehat{\Lambda}}}\,ds\right)-\xi^{2}Re\left(G_{0}\displaystyle\int_{0}^{+\infty}\,g(s)\overline{{\widehat{\Lambda}}}\,ds\right).$

Similarily, multiplying (5.2)4 by
$i\xi\displaystyle\int_{0}^{+\infty}\,g(s)\overline{\widehat{\Lambda}}\,ds$,
multiplying (5.2)8 by $-i\rho_{2}\xi g(s)\,\overline{\widehat{\mathbf{y}}}$
and integrating over $\mathbb{R}_{+}$ with respect to $s$, adding the
resulting equations, taking the real part and using (2.17), we arrive at

$\displaystyle\frac{d}{dt}Re\left(i\rho_{2}\xi{\widehat{\mathbf{y}}}\,\displaystyle\int_{0}^{+\infty}\,g(s)\overline{\widehat{\Lambda}}\,ds\right)$
$\displaystyle=$
$\displaystyle-\rho_{2}g_{0}Re\left(i\xi{\widehat{\Theta}}\overline{\widehat{\mathbf{y}}}\right)-k_{2}\xi^{2}Re\left(\widehat{\mathbf{z}}\displaystyle\int_{0}^{+\infty}\,g(s)\overline{\widehat{\Lambda}}\,ds\right)$
$\displaystyle-
k_{0}Re\left(i\xi\widehat{\mathbf{p}}\displaystyle\int_{0}^{+\infty}\,g(s)\overline{\widehat{\Lambda}}\,ds\right)-\rho_{2}Re\left(i\xi\overline{\widehat{\mathbf{y}}}\displaystyle\int_{0}^{+\infty}\,g^{\prime}(s){\widehat{\Lambda}}\,ds\right)$
$\displaystyle-
k_{2}\,Re\left(\xi\,\widehat{z}\displaystyle\int_{0}^{+\infty}\,g(s)\overline{{\widehat{\Lambda}}}\,ds\right).$

Finally, multiplying (5.2)2 by
$i\xi\displaystyle\int_{0}^{+\infty}\,g(s)\overline{\widehat{\Lambda}}\,ds$,
multiplying (5.2)8 by $-\rho_{1}\xi g(s)\overline{\widehat{\mathbf{u}}}$ and
integrating over $\mathbb{R}_{+}$ with respect to $s$, adding the resulting
equations, taking the real part and using (2.17), we see that

$\displaystyle\frac{d}{dt}Re\left(i\rho_{1}\xi{\widehat{\mathbf{u}}}\,\displaystyle\int_{0}^{+\infty}\,g(s)\overline{{\widehat{\Lambda}}}\,ds\right)$
$\displaystyle=$ $\displaystyle-
k_{1}\xi^{2}Re\left(\widehat{\mathbf{v}}\displaystyle\int_{0}^{+\infty}\,g(s)\overline{\widehat{\Lambda}}\,ds\right)-k_{0}Re\left(i\xi\widehat{\mathbf{p}}\displaystyle\int_{0}^{+\infty}\,g(s)\overline{\widehat{\Lambda}}\,ds\right)$
$\displaystyle-\rho_{1}Re\left(i\xi\overline{\widehat{\mathbf{u}}}\displaystyle\int_{0}^{+\infty}\,g^{\prime}(s){\widehat{\Lambda}}\,ds\right)-\rho_{1}g_{0}Re\left(i\xi{\widehat{\Theta}}\overline{\widehat{\mathbf{u}}}\right)$
$\displaystyle-
k_{1}\,Re\left(\xi\,\widehat{v}\displaystyle\int_{0}^{+\infty}\,g(s)\overline{{\widehat{\Lambda}}}\,ds\right).$

Let ${\mathbf{F}}_{0},\cdots,{\mathbf{F}}_{5}$ as ${F}_{0},\cdots,{F}_{5}$,
respectively, with $\widehat{\mathbf{v}}$,
$\widehat{\mathbf{u}},\,\widehat{\mathbf{z}},\,\widehat{\mathbf{y}}$,
$\widehat{\Phi}$, $\widehat{\Theta}$, $\widehat{\mathbf{p}}$,
$\widehat{\Lambda}$ and ${\mathbf{G}}$ instead of $\widehat{v}$,
$\widehat{u},\,\widehat{z},\,\widehat{y}$, $\widehat{\phi}$,
$\widehat{\theta}$, $\widehat{p}$, $\widehat{\eta}$ and $G$, respectively.
Exploiting (5.7)-(5), we find (instead of (3.30))

(5.20)
$\frac{d}{dt}{\mathbf{F}}_{0}(\xi,\,t)={\mathbf{F}}_{5}(\xi,\,t)+{\mathbf{F}}_{4}(\xi,\,t)+{\mathbf{F}}_{3}(\xi,\,t)+R(\xi,t),$

where $R$ gathers the mixed products of the components of $U$ and
${\mathbf{U}}$, and it is given by

(5.22) $\displaystyle R(\xi,\,t)$ $\displaystyle=$ $\displaystyle
Re\left[-k_{2}\lambda_{1}\xi\widehat{z}\bar{\widehat{\mathbf{z}}}+\rho_{2}\lambda_{1}\xi\widehat{y}\bar{\widehat{\mathbf{y}}}-k_{1}\lambda_{2}\xi\widehat{v}\bar{\widehat{\mathbf{v}}}+\rho_{1}\lambda_{2}\xi\widehat{u}\bar{\widehat{\mathbf{u}}}-\lambda_{3}G_{0}\bar{\widehat{\Phi}}+2\rho_{3}\lambda_{3}\xi\widehat{\theta}\bar{\widehat{\Theta}}\right]$
$\displaystyle+Re\left[ik_{2}\lambda_{4}\xi^{2}\widehat{z}\bar{\widehat{\mathbf{p}}}+i\rho_{2}l\lambda_{4}\xi^{2}\widehat{\theta}\bar{\widehat{\mathbf{y}}}+i\rho_{3}\lambda_{5}\widehat{y}\bar{\widehat{\Theta}}+\lambda_{5}{G_{0}}\bar{\widehat{\mathbf{z}}}-2i\rho_{2}\lambda_{6}\xi^{2}\widehat{\theta}\bar{\widehat{\mathbf{y}}}+k_{2}\lambda_{6}\xi\widehat{z}\bar{\widehat{\Phi}}\right]$
$\displaystyle+Re\left[-k_{1}\lambda_{7}\xi\widehat{v}\bar{\widehat{\mathbf{z}}}+\rho_{1}\lambda_{7}\xi\widehat{y}\bar{\widehat{\mathbf{u}}}-k_{1}\lambda_{8}\xi\widehat{v}\bar{\widehat{\Phi}}+2i\rho_{1}\lambda_{8}\xi^{2}\widehat{\theta}\bar{\widehat{\mathbf{u}}}-\rho_{2}\lambda_{9}\xi\widehat{u}\bar{\widehat{\mathbf{y}}}+k_{2}\lambda_{9}\xi\widehat{z}\bar{\widehat{\mathbf{v}}}\right]$
$\displaystyle+Re\left[i\rho_{3}\lambda_{10}\widehat{u}\bar{\widehat{\Theta}}+\lambda_{10}G_{0}\bar{\widehat{\mathbf{v}}}+\tau_{0}\left(-\lambda_{11}\xi^{2}G_{0}-k_{2}\lambda_{12}\xi\widehat{z}-k_{1}\lambda_{13}\xi\widehat{v}\right)\int_{0}^{+\infty}g(s)\bar{\widehat{\Lambda}}ds\right].$

Considering the same choices of $\lambda_{1}\,\cdots,\,\lambda_{13}$ and using
(3.44) and (3.45) for
${\tilde{\eta}}\in\left\\{{\widehat{\Lambda}},\overline{{\widehat{\Lambda}}}\right\\}$,
we get (instead of (3.54) and (3.69))

$\displaystyle\frac{d}{dt}\left[\xi^{2+2\tau_{0}}{\mathbf{F}}_{0}(\xi,\,t)\right]$
$\displaystyle\leq$ $\displaystyle-
c_{1}\xi^{4+2\tau_{0}}\widehat{\mathbf{E}}(\xi,\,t)+\xi^{2+2\tau_{0}}\,R(\xi,t)$
$\displaystyle+c_{2}\xi^{2\tau_{0}}\left(\xi^{10}+1\right)\left[(1-\tau_{0})|\widehat{\Theta}|^{2}-\tau_{0}\xi^{2}\displaystyle\int_{0}^{+\infty}\,g^{\prime}(s)|{\widehat{\Lambda}}(\xi,s)|^{2}\,ds\right].$

Using Young’s inequality and exploiting (3.43), (3.51) and (3.67), we obtain,
for any $\epsilon>0$,

$\displaystyle\xi^{2+2\tau_{0}}|R(\xi,\,t)|$ $\displaystyle\leq$
$\displaystyle\epsilon\xi^{4+2\tau_{0}}\left(|{\widehat{\mathbf{z}}}|^{2}+|{\widehat{\mathbf{y}}}|^{2}+|{\widehat{\mathbf{v}}}|^{2}+|{\widehat{\mathbf{u}}}|^{2}+|{\widehat{\Phi}}|^{2}+|{\widehat{\Theta}}|^{2}+|{\widehat{\mathbf{p}}}|^{2}\right)$
$\displaystyle+\tau_{0}\epsilon\xi^{6+4\tau_{0}}\displaystyle\int_{0}^{+\infty}\,g(s)|{\widehat{\Lambda}}(\xi,s)|^{2}\,ds+C_{\epsilon}\xi^{2\tau_{0}}\left(\xi^{8}+1\right)|{\widehat{U}}|^{2},$

thus, using (5.5), we obtain in both cases $\tau_{0}=0$ and $\tau_{0}=1$

(5.24) $\xi^{2+2\tau_{0}}|R(\xi,\,t)|\leq\epsilon
C_{1}\xi^{4+2\tau_{0}}{\widehat{\mathbf{E}}}(\xi,\,t)+C_{\epsilon}\xi^{2\tau_{0}}\left(\xi^{8}+1\right)|\widehat{U}(\xi,\,t)|^{2},$

therefore, chosing $\epsilon=\frac{c_{1}}{2C_{1}}$, we deduce from (5) and
(5.24) that

$\displaystyle\frac{d}{dt}\left[\xi^{2+2\tau_{0}}{\mathbf{F}}_{0}(\xi,\,t)\right]$
$\displaystyle\leq$
$\displaystyle-\frac{c_{1}}{2}\,\xi^{4+2\tau_{0}}\,{\widehat{\mathbf{E}}}(\xi,\,t)+C\xi^{2\tau_{0}}\left(\xi^{8}+1\right)|\widehat{U}(\xi,\,t)|^{2}$
$\displaystyle+c_{2}\xi^{2\tau_{0}}\left(\xi^{10}+1\right)\left[(1-\tau_{0})|\widehat{\Theta}|^{2}-\tau_{0}\xi^{2}\displaystyle\int_{0}^{+\infty}\,g^{\prime}(s)|{\widehat{\Lambda}}(\xi,s)|^{2}\,ds\right].$

Now, let $\lambda>0$ and ${\mathbf{F}}$ defined as $F$ in (3.55) and (3.70);
that is

${\mathbf{F}}(\xi,\,t)=\lambda\,{\widehat{\mathbf{E}}}(\xi,\,t)+\frac{\xi^{2+2\tau_{0}}}{\xi^{10}+1}{\mathbf{F}}_{0}(\xi,\,t).$

By combining (5) and (5), we arrive at

(5.26) $\displaystyle\frac{d}{dt}{\mathbf{F}}(\xi,\,t)$ $\displaystyle\leq$
$\displaystyle-\frac{c_{1}\xi^{4+2\tau_{0}}}{2\left(\xi^{10}+1\right)}\,{\widehat{\mathbf{E}}}(\xi,\,t)+C\frac{\xi^{2\tau_{0}}\left(\xi^{8}+1\right)}{\xi^{10}+1}|\widehat{U}(\xi,\,t)|^{2}+R_{1}(\xi,\,t)$
$\displaystyle+(1-\tau_{0})\left(c_{2}\xi^{2\tau_{0}}-\gamma\lambda\right)|\widehat{\Theta}|^{2}+\tau_{0}\left(\frac{1}{2}\lambda\xi^{4}-c_{2}\xi^{2+2\tau_{0}}\right)\displaystyle\int_{0}^{+\infty}\,g^{\prime}(s)|{\widehat{\Lambda}}(\xi,s)|^{2}\,ds,$

where

$R_{1}(\xi,\,t)=\lambda\,Re\left[ik_{1}\widehat{u}\bar{\widehat{\mathbf{v}}}+ik_{1}\widehat{v}\bar{\widehat{\mathbf{u}}}+ik_{2}\widehat{y}\bar{\widehat{\mathbf{z}}}+ik_{2}\widehat{z}\bar{\widehat{\mathbf{y}}}-2(k_{3}-\tau_{0}g_{0})\xi\widehat{\theta}\bar{\widehat{\Phi}}+G_{0}\bar{\widehat{\Theta}}+ilk_{0}\widehat{\theta}\bar{\widehat{\mathbf{p}}}\right].$

On the other hand, as (3.57) and (3.72), we have

(5.27) $|{\mathbf{F}}(\xi,\,t)-\lambda\,\widehat{\mathbf{E}}(\xi,\,t)|\leq
2c_{3}\,\widehat{\mathbf{E}}(\xi,\,t).$

Hence, for $\lambda>\left\\{\frac{c_{2}}{\gamma},2c_{3}\right\\}$ in case
$\tau_{0}=0$, and $\lambda>\left\\{2c_{2},2c_{3}\right\\}$ in case
$\tau_{0}=1$, we deduce from (5.26) and (5.27) that

(5.28)
$\frac{d}{dt}{\mathbf{F}}(\xi,\,t)\leq-\frac{c_{1}\xi^{4+2\tau_{0}}}{2\left(\xi^{10}+1\right)}\,{\widehat{\mathbf{E}}}(\xi,\,t)+C\frac{\xi^{2\tau_{0}}\left(\xi^{8}+1\right)}{\xi^{10}+1}|\widehat{U}(\xi,\,t)|^{2}+R_{1}(\xi,\,t)$

and

(5.29) ${\mathbf{F}}\sim{\widehat{\mathbf{E}}}.$

Applying Young’s inequality and using the definitin of $\widehat{U}$,
${\widehat{\mathbf{E}}}$ and $G_{0}$, we see that, for any $\varepsilon>0$,

(5.30)
$\xi^{4+2\tau_{0}}R_{1}(\xi,\,t)\leq\frac{\varepsilon\,\xi^{2(4+2\tau_{0})}}{\xi^{10}+1}{\widehat{\mathbf{E}}}(\xi,\,t)+C_{\varepsilon}\left(\xi^{2}+1\right)\left(\xi^{10}+1\right)|\widehat{U}(\xi,\,t)|^{2},$

then, by multiplying (5.28) by $\xi^{4+2\tau_{0}}$, combining with (5.30),
choosing $\varepsilon=\frac{c_{1}}{4}$ and using (3.43) and (5.29), we obtain,
for some $c_{0}=\frac{c_{1}}{8}$,

(5.31)
$\frac{d}{dt}\left[\xi^{4+2\tau_{0}}{\mathbf{F}}(\xi,\,t)\right]\leq-\frac{2c_{0}\xi^{2(4+2\tau_{0})}}{\xi^{10}+1}\,{\mathbf{F}}(\xi,\,t)+C\frac{\xi^{22}+1}{\xi^{10}+1}|\widehat{U}(\xi,\,t)|^{2},$

therefore, by multiplying (5.31) by $e^{2c_{0}f(\xi)t}$ ($f$ is defined in
(3.2)) and using (3.1), we find

$\frac{d}{dt}\left[\xi^{4+2\tau_{0}}e^{2c_{0}f(\xi)t}{\mathbf{F}}(\xi,\,t)\right]\leq
C\frac{\xi^{22}+1}{\xi^{10}+1}e^{2(c_{0}-c)f(\xi)t}|\widehat{U}_{0}(\xi)|^{2},$

then, by integrating the above inequality with respect to $t$, we find

(5.32) ${\mathbf{F}}(\xi,\,t)\leq
e^{-2c_{0}f(\xi)t}{\mathbf{F}}(\xi,\,0)+C\frac{\xi^{22}+1}{\xi^{2(4+2\tau_{0})}}e^{-2cf(\xi)t}|\widehat{U}_{0}(\xi)|^{2},$

thus, according to (5.5) and (5.29) and using the inequality (4.9), the above
inequality (5.32) implies (5.6). The proof of Theorem 5.1 is now achieved. ∎

##  6\. Estimation of $\left\|\partial_{x}^{j}U\right\|_{1}$

In this section, we show the decay estimate of $\|\partial_{x}^{j}U\|_{1}$ by
exploiting (4.2), (4.3) and (5.6).

######  Theorem 6.1.

Assume that (1.14) is satisfied. Let
$N\,\in\\{5+2\tau_{0},6+2\tau_{0},\cdots\\}$ and $U$ be a solution of (2.8)
corresponding to an initial data $U_{0}$ satisfying

(6.1) $U_{0}\in H^{N+7-2\tau_{0}}(\mathbb{R})\cap
L^{1}(\mathbb{R})\quad\hbox{and}\quad{\tilde{U}}_{0}\in H^{N}(\mathbb{R})\cap
L^{1}(\mathbb{R}),$

where ${\tilde{U}}_{0}(x)=x{U}_{0}(x)$. Then, for any
$\ell\in\\{1,\,2,\,\cdots,\,N-4-2\tau_{0}\\}$ and
$j\in\\{4+2\tau_{0},\,5+2\tau_{0},\,\cdots,\,N-\ell\\}$, there exist
${\mathbf{c}}_{0},\,{\tilde{\mathbf{c}}}_{0}>0$ such that, for any
$t\in\mathbb{R}_{+}$,

$\displaystyle\|\partial_{x}^{j}U\|_{{1}}$ $\displaystyle\leq$
$\displaystyle{\tilde{\mathbf{c}}}_{0}\,\left[(1+t)^{-1/8-j/4}\,\|{\tilde{U}}_{0}\|_{{1}}+(1+t)^{7/8-j/4}\,\|{U}_{0}\|_{{1}}\right]$
$\displaystyle+{\tilde{\mathbf{c}}}_{0}\,(1+t)^{-\ell/6}\,\left[\|\partial_{x}^{j+\ell}{\tilde{U}}_{0}\|_{{2}}+\|\partial_{x}^{j+\ell+7}U_{0}\|_{{2}}+\|\partial_{x}^{j+\ell}U_{0}\|_{{2}}+\|\partial_{x}^{j+\ell-1}U_{0}\|_{{2}}\right].$

if $\tau_{0}=0$, and

$\displaystyle\|\partial_{x}^{j}U\|_{{1}}$ $\displaystyle\leq$
$\displaystyle{\tilde{\mathbf{c}}}_{0}\,\left[(1+t)^{-1/12-j/6}\,\|{\tilde{U}}_{0}\|_{{1}}+(1+t)^{11/12-j/6}\,\|{U}_{0}\|_{{1}}\right]$
$\displaystyle+{\tilde{\mathbf{c}}}_{0}(1+t)^{-\ell/4}\,\left[\|\partial_{x}^{j+\ell}{\tilde{U}}_{0}\|_{{2}}+\|\partial_{x}^{j+\ell+5}U_{0}\|_{{2}}+\|\partial_{x}^{j+\ell}U_{0}\|_{{2}}+\|\partial_{x}^{j+\ell-1}U_{0}\|_{{2}}\right]$

if $\tau_{0}=1$.

###### Proof.

First, using (1.18) and applying Young’s inequality, it follows that

(6.4)
$\|\partial_{x}^{j}U\|_{1}\leq{\frac{1}{2}}\|\partial_{x}^{j}U\|_{2}+{\frac{1}{2}}\|x\partial_{x}^{j}U\|_{2}.$

Because the term $\|\partial_{x}^{j}U\|_{2}$ has yet been estimated in (4.2)
and (4.3), we have only to estimate the term $\|x\partial_{x}^{j}U\|_{2}$.
Using the method in [18] and Plancherel’s theorem, we may write

$\displaystyle\int_{\mathbb{R}}x^{2}|\partial_{x}^{j}U(x,t)|^{2}dx$
$\displaystyle\leq$ $\displaystyle
C\int_{\mathbb{R}}\left|\partial_{\xi}\left(|\xi|^{j}|\widehat{U}(\xi,t)|\right)\right|^{2}d\xi$
$\displaystyle\leq$ $\displaystyle
C\int_{\mathbb{R}}\left(|\xi|^{j-1}|\widehat{U}(\xi,t)|+|\xi|^{j}|\partial_{\xi}\widehat{U}(\xi,t)|\right)^{2}d\xi$
$\displaystyle\leq$ $\displaystyle
C\|\partial_{x}^{j-1}U\|_{2}^{2}+C\int_{\mathbb{R}}\xi^{2j}|\widehat{\mathbf{U}}(\xi,t)|^{2}d\xi.$

It is clear that $\|\partial_{x}^{j-1}U\|_{2}$ can be easily estimated by
using (4.2) and (4.3) (with $j-1$ instead of $j$). To estimate the last
integral in (6), we use (5.6) and apply Plancherel’s theorem, it appears that

$\displaystyle\int_{\mathbb{R}}\xi^{2\,j}\,|\widehat{U}(\xi,\,t)|^{2}d\xi$
$\displaystyle\leq$ $\displaystyle
C\int_{\mathbb{R}^{*}}\xi^{2\,j}\,e^{-2\,{\mathbf{c}}\,f(\xi)\,t}\,\left[|\widehat{\mathbf{U}}_{0}(\xi)|^{2}+\left(\xi^{14-4\tau_{0}}+\xi^{-(8+4\tau_{0})}\right)|\widehat{U}_{0}(\xi)|^{2}\right]\,d\xi$
$\displaystyle\leq$ $\displaystyle C\int_{0<|\xi|\leq
1}\,e^{-2\,{\mathbf{c}}\,f(\xi)\,t}\,\left[\xi^{2\,j}|\widehat{\mathbf{U}}_{0}(\xi)|^{2}+\xi^{2(j-4-2\tau_{0})}|\widehat{U}_{0}(\xi)|^{2}\right]d\xi$
$\displaystyle+C\int_{|\xi|>1}\,e^{-2\,{\mathbf{c}}\,f(\xi)\,t}\,\left[\xi^{2j}|\widehat{\mathbf{U}}_{0}(\xi)|^{2}+\xi^{2\,(j+7-2\tau_{0})}|\widehat{U}_{0}(\xi)|^{2}\right]\,d\xi$
$\displaystyle:=$ $\displaystyle J_{1}+J_{2}.$

Case $\tau_{0}=0$: using (4.4)1 and Lemma 2.3 of [8], we observe that

$\displaystyle J_{1}$ $\displaystyle\leq$ $\displaystyle
C\,\|\widehat{\mathbf{U}}_{0}\|_{{\infty}}^{2}\int_{|\xi|\leq
1}\xi^{2\,j}\,e^{-\,{\mathbf{c}}\,t\,\xi^{4}}\,d\xi+C\,\|\widehat{U}_{0}\|_{{\infty}}^{2}\int_{|\xi|\leq
1}\xi^{2\,(j-4)}\,e^{-\,{\mathbf{c}}\,t\,\xi^{4}}\,d\xi$ $\displaystyle\leq$
$\displaystyle
C\left(1+t\right)^{-\,\frac{1}{4}(1+2\,j)}\|{\tilde{U}}_{0}\|_{{1}}^{2}+C\left(1+t\right)^{-\,\frac{1}{4}[1+2\,(j-4)]}\|{U}_{0}\|_{{1}}^{2}.$

In the high frequency region, using (4.4)2, we entail

$\displaystyle J_{2}$ $\displaystyle\leq$ $\displaystyle
C\int_{|\xi|>1}\xi^{2\,j}\,e^{-{\mathbf{c}}\,t\,\xi^{-6}}\,|\widehat{\mathbf{U}}_{0}(\xi)|^{2}\,d\xi+C\int_{|\xi|>1}\xi^{2\,(j+7)}\,e^{-{\mathbf{c}}\,t\,\xi^{-6}}\,|\widehat{U}_{0}(\xi)|^{2}\,d\xi$
$\displaystyle\leq$ $\displaystyle C\
\sup_{|\xi|>1}\left\\{\xi^{-2\,\ell}\,e^{-{\mathbf{c}}\,t\,\xi^{-6}}\right\\}\left[\int_{\mathbb{R}}\xi^{2\,(j+\ell)}\,|\widehat{\mathbf{U}}_{0}(\xi)|^{2}\
d\xi+\int_{\mathbb{R}}\xi^{2\,(j+\ell+7)}\,|\widehat{U}_{0}(\xi)|^{2}\
d\xi\right]$ $\displaystyle\leq$ $\displaystyle C\
\sup_{|\xi|>1}\left\\{\xi^{-2\,\ell}\,e^{-{\mathbf{c}}\,t\,\xi^{-6}}\right\\}\,\left(\|\,\partial_{x}^{j+\ell}{\tilde{U}}_{0}\,\|_{{2}}^{2}+\|\,\partial_{x}^{j+\ell+7}U_{0}\,\|_{{2}}^{2}\right),$

thus, simple computations (see, for example, Lemma 2.4 of [8]) imply that

(6.8) $J_{2}\leq
C\left(1+t\right)^{-\frac{\ell}{3}}\,\,\left(\|\,\partial_{x}^{j+\ell}{\tilde{U}}_{0}\,\|_{{2}}^{2}+\|\,\partial_{x}^{j+\ell+7}U_{0}\,\|_{{2}}^{2}\right),$

and so, by combining (6)-(6.8), we get

$\displaystyle\int_{\mathbb{R}}\xi^{2\,j}\,|\widehat{U}(\xi,\,t)|^{2}d\xi$
$\displaystyle\leq$ $\displaystyle
C\left(1+t\right)^{-\,\frac{1}{4}(1+2\,j)}\|{\tilde{U}}_{0}\|_{{1}}^{2}+\left(1+t\right)^{-\,\frac{1}{4}[1+2\,(j-4)]}\|{U}_{0}\|_{{1}}^{2}$
$\displaystyle+C\left(1+t\right)^{-\frac{\ell}{3}}\,\,\left(\|\,\partial_{x}^{j+\ell}{\tilde{U}}_{0}\,\|_{{2}}^{2}+\|\,\partial_{x}^{j+\ell+7}U_{0}\,\|_{{2}}^{2}\right)$

thus, by combining (4.2), (6.4) and (6), and using the inequality (4.9), we
get (6.1). Case $\tau_{0}=1$: using (4.3), (4.10) and (5.6), and following the
same arguments as for the proof of (6.1), we obtain (6.1). ∎

##  7\. Concluding discussion

1\. This article is concerned with the stability of two systems of type Rao-
Nakra sandwich beam in the whole line $\mathbb{R}$ under the presence of a
frictional damping or an infinite memory acting on the Euler-Bernoulli
equation. We prove the instability of both systems if the speeds of
propagation of the two wave equations are equal. In the reverse situation, we
are able, despite the presence of only one controle, to obtain the desired
$L^{2}(\mathbb{R})$-norm and $L^{1}(\mathbb{R})$-norm decay estimates (4.2),
(4.3), (6.1) and (6.1). The main ingredient of the proof is the energy method
and the Fourier analysis. 2\. The decay estimates (4.2), (4.3), (6.1) and
(6.1) are still satisfied when (1.13) holds if we add
${\tilde{\gamma}}\varphi_{t}$ to (1.4)1 and (1.5)1, or
${\tilde{\gamma}}\psi_{t}$ to (1.4)2 and (1.5)2, where ${\tilde{\gamma}}>0$.
Indeed, in these cases, we do not need $A_{7}=0$ because either
$|\widehat{u}|^{2}$ or $|\widehat{y}|^{2}$ will be directly controlled via the
derivative of the energy functional $\widehat{E}$. Similarily, (4.2), (4.3),
(6.1) and (6.1) hold even in case (1.13) if the added control is of memory
type:

$\displaystyle\int_{0}^{+\infty}{\tilde{g}}(s)\,\varphi_{xx}(x,t-s)ds\quad\hbox{or}\quad\displaystyle\int_{0}^{+\infty}{\tilde{g}}(s)\,\psi_{xx}(x,t-s)ds$

instead of ${\tilde{\gamma}}\varphi_{t}$ and ${\tilde{\gamma}}\psi_{t}$,
respectively, where ${\tilde{g}}$ is as $g$. 3\. Using the interpolation
inequalities (1.15), (1.16) and (1.17), we see that our
$L^{2}(\mathbb{R})$-norm and $L^{1}(\mathbb{R})$-norm decay estimates lead to
similar $L^{q}(\mathbb{R})$-norm ones, for any $q\in[1,+\infty]$. The
$L^{q}$-norm decay estimates, for $1\leq q<2$, are based on the $L^{1}$-norm
decay estimate, and so they require initial data $U_{0}$ having the regularity
(6.1) with $N\in\\{5,6,\cdots\\}$ in case (1.4), and $N\in\\{7,8,\cdots\\}$ in
case (1.5). However, the $L^{q}$-norm decay estimates, for $q\in[2,+\infty]$
require the weaker regularity (4.1), where $N\in\mathbb{N}^{*}$. 4\. In a
future work, we aspire to treat the case where the control occurs on a one
wave equation of systems. On the other hand, we think that similar results can
be obtained with a control subject to a thermal effect like Fourier law,
Cattaneo law and Gurtin-Pipkin law. These kinds of controls deserve to be
treated and we aspire to do it in a future work. Acknowledgment. The author
thanks Belkacem Said-Houari for useful and fruitful discussions and exchanges
on $L^{q}(\mathbb{R})$-norm decay estimates for Cauchy PDEs.

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