# Stability for coupled waves with locally disturbed Kelvin-Voigt damping

Fathi Hassine  UR Analysis and Control of PDEs, UR 13ES64, Department of
Mathematics, Faculty of Sciences of Monastir, University of Monastir, Tunisia
<EMAIL_ADDRESS>and  Nadia Souayeh  UR Analysis and Control of PDEs,
UR 13ES64, Department of Mathematics, Faculty of Sciences of Monastir,
University of Monastir, Tunisia<EMAIL_ADDRESS>
###### Abstract.

We consider a coupled wave system with partial Kelvin-Voigt damping in the
interval $(-1,1)$, where one wave is dissipative and the other does not. When
the damping is effective in the whole domain $(-1,1)$ it was proven in [17]
that the energy is decreasing over the time with a rate equal to
$t^{-\frac{1}{2}}$. In this paper, using the frequency domain method we show
the effect of the coupling and the non smoothness of the damping coefficient
on the energy decay. Actually, as expected we show the lack of exponential
stability, that the semigroup loses speed and it decays polynomially with a
slower rate then given in [17], down to zero at least as $t^{-\frac{1}{12}}$.

###### Key words and phrases:

Coupled system, Kelvin-Voigt damping, frequency domain approach

###### 2010 Mathematics Subject Classification:

35B35, 35B40, 93D20

###### Contents

  1. 1 Introduction
  2. 2 Well-posedness
  3. 3 Strong stability
  4. 4 Lack of exponential stability
  5. 5 Polynomial stabilization

##  1\. Introduction

When a vibrating source disturbs the first particular of a medium, a wave is
created. This phenomena begins to travel from particle to particle along the
medium, which is typically modelled by a wave equation. In order to suppress
those vibrations, the most common approach is adding damping. It’s more likely
to use one of two types:
1) The linear viscous damping or ”external damping”, it does mostly model an
external frictional force, such that the auto-mobile shock absorber.
2) The Kelvin-Voigt damping, it’s also called the ”internal damping” or the
”material damping”, which is originated from the extension or compression of
the vibrating particles.
In the recent years, many researchers showed interest in problems involving
this kind of damping. In control theory for instance it was shown that when
the Kelvin-Voigt damping coefficient is satisfying some geometrical control
conditions the semigroup corresponding to this system is exponential stable
(see [14, 19]). Nonetheless, when the damping is arbitrary localized with
singular coefficient, it’s not the case any more (see [2, 13]). Actually, in
one-dimensional case we can consider the following problem

(1.1)
$\left\\{\begin{array}[]{ll}u_{tt}-[u_{x}(x,t)+b(x)u_{xt}]_{x}=0&-1<x<1,\;t\geq
0,\\\ u(t,-1)=u(t,1)=0&t\geq 0,\\\ u(0,x)=u_{0}(x),u_{t}(0,x)=u_{1}(x)&-1\leq
x\leq 1,\end{array}\right.$

with $b\in L^{\infty}(-1,1)$ And

$b(x)=\left\\{\begin{array}[]{ll}0&\text{for }x\in[0,1)\\\ a(x)&\text{for
}x\in(-1,0).\end{array}\right.$

Under the assumption that the damping coefficient has a singularity at the
interface of the damped and undamped regions, and behaves like $x^{\alpha}$
near the interface, it was proven by Liu abd Zhang [15] that the semigroup
corresponding to the system is polynomially or exponentially stable and the
decay rate depends on the parameter $\alpha\in(0,1]$. When $\alpha=0$, Liu and
Rao [13] showed that system (1.1) is polynomially stable with an order equal
to $2$ where few years ago Liu and Liu [12] proved the lack of the exponential
stability.

When dealing with systems involving quantities described by several
components, pretending to control or observe all the state variables it turns
out that certain systems possess an internal structure that compensates the
lack of control variables. Such a phenomenon is referred to as indirect
stabilization or indirect control. For instance Alabo et al. did study in [1]
the coupled waves with partial frictional damping

$\left\\{\begin{array}[]{ll}u_{tt}-\Delta u+\alpha v=0&x\in\Omega,\;t\geq
0,\\\ v_{tt}-\Delta v+\alpha u+\beta v_{t}=0&x\in\Omega,\;t\geq
0,\end{array}\right.$

subjected to Dirichlet boundary conditions. It was proven then the semigroup
corresponding to this system is not exponentially stable, but it’s
polynomially with the rate $t^{\frac{-1}{2}}$. In 2016, Oquendo and Pacheco
studied the wave equation with internal coupled terms where the Kelvin-Voigt
damping is global in one equation and the second equation is conservative.
Although the damping is stronger than the frictional one, they had shown that
the semigroup loses speed with a slower rate of $t^{-\frac{1}{4}}$. For this
kind of coupled visco-elastic models we distinguish what is called the
transmission problems which have been intensively studied by the first author,
Ammari and their collaborators in [2, 6, 7, 8, 9, 3] (see also [4]) where the
systems studied in these papers are the wave or the plate equation or a
coupled wave-plate equation. Assuming a non smooth and singular damping
coefficient it was shown in these works a uniform and a non-uniform decay
rates of the energy. In this work, we examine the behaviour of a coupled waves
system with a partial Kelvin-Voigt damping, namely we consider the following
system where the first wave is dissipative and the second one is conservative

(1.2)
$\left\\{\begin{array}[]{ll}u_{tt}(x,t)-[u_{x}(x,t)+a(x)u_{xt}(x,t)]_{x}+v_{t}(x,t)=0&(x,t)\in(-1,1)\times(0,+\infty),\\\
v_{tt}(x,t)-c\,v_{xx}(x,t)-u_{t}(x,t)=0&(x,t)\in(-1,1)\times(0,+\infty),\\\
u(0,t)=v(0,t)=0,u(1,t)=v(1,t)=0&\forall\,t>0,\\\
u(x,0)=u_{0}(x),u_{t}(x,0)=u_{1}(x)&\forall\,x\in(-1,1),\\\
v(x,0)=v_{0}(x),v_{t}(x,0)=v_{1}(x)&\forall\,x\in(-1,1),\end{array}\right.$

where $c>0$ and $a\in L^{\infty}(-1,1)$ is non-negative function. In this
paper we assume that the damping coefficient is piecewise function in
particular we suppose that $a$ have the following form
$a=d\,\mathbb{1}_{[0,1]}$, where $d$ is a strictly positive constant. Since
the damping is singular, this system can be seen as a coupling of a
conservative wave equation and a transmission wave equation.

The natural energy of (u,v) solution of (1.2) at an instant $t$ is given by

$E(t)=\frac{1}{2}\int_{-1}^{1}\left(|u_{t}(x,t)|^{2}+|v_{t}(x,t)|^{2}+|u_{x}(x,t)|^{2}+c\,|v_{x}(x,t)|^{2}\right)\,{\rm
d}x,\forall\,t>0.$

Multiplying the first equation of (1.2) by $\bar{u}$, the second by $\bar{v}$,
integrating over (-1,1) and then taking the real part leads to

$E^{\prime}(t)=-\int_{-1}^{1}a(x)\,|u_{xt}(x,t)|^{2}\,{\rm
d}x,\;\forall\,t>0.$

Therefore, the energy is a non-increasing function of the time variable $t$.
We show the lack of the exponential stability and prove that the semigroup
corresponding to this system is polynomially stable for regular initial data
and with a slower rate, down to $t^{-\frac{1}{12}}$.

This paper is organized as follows. In section 2, we prove that system (1.2)
is well-posed. In section 3, we show that the energy of the system is the
strong stability. In section 4, we prove the lack of exponential stability. In
section 5, we prove a polynomial stability decay of the energy.

##  2\. Well-posedness

In this section, we discuss the well-posesness of the problem (1.2) using the
semigroup theory. Let
$\mathcal{H}=(H_{0}^{1}(-1.1))^{2}\times(L^{2}(-1,1))^{2}$ be the Hilbert
space endowed with the inner product define, for
$U_{1}=(u^{1},v^{1},w^{1},z^{1})\in\mathcal{H}$ and
$U_{2}=(u^{2},v^{2},w^{2},z^{2})\in\mathcal{H}$, by

$\left\langle U_{1},U_{2}\right\rangle_{\mathcal{H}}=\left\langle
u^{1}_{x},u^{2}_{x}\right\rangle_{L^{2}(-1,1)}+\left\langle\sqrt{c}v^{1}_{x},\sqrt{c}v^{2}_{x}\right\rangle_{L^{2}(-1,1)}+\left\langle
w^{1},w^{2}\right\rangle_{L^{2}(-1,1)}+\left\langle
z^{1},z^{2}\right\rangle_{L^{2}(-1,1)}.$

By setting $y(t)=(u(t),v(t),u_{t}(t),v_{t}(t))$ and
$y_{0}=(u_{0},v_{0},u_{1},v_{1})$ we can rewrite system (1.2) as a first order
differential equation as follow

(2.1) $\dot{y}(t)={\mathcal{A}}y(t),\qquad y(0)=y_{0},$

where

${\mathcal{A}}(u^{1},v^{1},u^{2},v^{2})=\left(u^{2},v^{2},\left(u_{x}^{1}+au_{x}^{2}\right)_{x}-v^{2},c\,v^{1}_{xx}+u^{2}\right),$

with

$\displaystyle(u^{1},v^{1},u^{2},v^{2})\in{\mathcal{D}}({\mathcal{A}})=\big{\\{}(u^{1},v^{1},u^{2},v^{2})\in\mathcal{H},\,(u^{2},v^{2})\in(H^{1}_{0}(-1,1))^{2},$
$\displaystyle v^{1}\in H^{2}(-1,1)\cap
H^{1}_{0}(-1,1),\left(u^{1}_{x}+au^{2}_{x}\right)_{x}\in
L^{2}(-1,1)\big{\\}}.$

For the well-posedness of system (2.1) we have the following proposition:

######  Proposition 2.1.

For an initial datum $y_{0}=(u_{0},v_{0},u_{1},v_{1})\in\mathcal{H}$, there
exists a unique solution $y=(u,v,u_{t},v_{t})\in
C([0,\,+\infty),\,\mathcal{H})$ to problem (2.1). Moreover, if
$y_{0}\in\mathcal{D}(\mathcal{A})$, then

$y=(u,v,u_{t},v_{t})\in C([0,\,+\infty),\,\mathcal{D}(\mathcal{A}))\cap
C^{1}([0,\,+\infty),\,\mathcal{H}).$

###### Proof.

By Lumer-Phillips’ theorem (see [16]), it suffices to show that $\mathcal{A}$
is dissipative and maximal.

(1) We first prove that $\mathcal{A}$ is dissipative. Take
$Z=(u,v,w,z)\in\mathcal{D}(\mathcal{A})$. Then

$\displaystyle\left\langle\mathcal{A}Z,Z\right\rangle_{\mathcal{H}}=\left\langle
w_{x},u_{x}\right\rangle_{L^{2}(-1,1)}+c\,\left\langle
z_{x},v_{x}\right\rangle_{L^{2}(-1,1)}+\left\langle(u_{x}+aw_{x})_{x},w\right\rangle_{L^{2}(-1,1)}$
$\displaystyle+\left\langle cv_{xx}+w,z\right\rangle_{L^{2}(-1,1)}.$

By integration by parts and using the boundary conditions, it holds:

(2.2) $(\mathcal{A}Z,Z)_{\mathcal{H}}=-\left\langle
aw_{x},w_{x}\right\rangle_{L^{2}(-1,1)}=-\int_{-1}^{1}a|w_{x}|^{2}\,{\rm
d}x\leq 0.$

This shows that $\mathcal{A}$ is the dissipative.

(2) Let us now prove that $\mathcal{A}$ is maximal, i.e., that $\lambda
I-\mathcal{A}$ is surjective for some $\lambda>0$. So, for any given
$(f,g,f_{1},g_{1})\in\mathcal{H}$, we solve the equation
$\mathcal{A}(u,v,w,z)=(f,g,f_{1},g_{1})$, which is recast on the following way

(2.3) $\left\\{\begin{array}[]{l}w=f\\\ z=g\\\ u_{xx}+(af_{x})_{x}=f_{1}+g\\\
c\,v_{xx}=g_{1}-f.\end{array}\right.$

It is well known that by Lax-Milgram’s theorem the system (2.3) admits a
unique solution $(u,v)\in H_{0}^{1}(-1,1)\times H^{1}_{0}(-1,1)$. Moreover by
multiplying the second and the third lines of (2.3) by
$\overline{u},\overline{v}$ respectively and integrating over $(-1,1)$ and
using Poincaré inequality and Cauchy-Schwarz inequality we find that there
exists a constant $C>0$ such that

$\int_{-1}^{1}\left(|u_{x}(x)|^{2}+|v_{x}(x)|^{2}\right)\,dx\leq
C\,\int_{-1}^{1}\left(|f_{x}(x)|^{2}+|g_{x}(x)|^{2}+|f_{1}(x)|^{2}+|g_{1}(x)|^{2}\right)\,dx.$

It follows that $(u,v,w,z)\in\mathcal{D}(\mathcal{A})$ and we have

$\|(u,v,w,z)\|_{\mathcal{H}}\leq C\|(f,g,f_{1},g_{1})\|_{\mathcal{H}}.$

This imply that $0\in\rho(\mathcal{A})$ and by contraction principle, we
easily get $R(\lambda\mathrm{I}-\mathcal{A})=\mathcal{H}$ for sufficient small
$\lambda>0$. The density of the domain of $\mathcal{A}$ follows from [16,
Theorem 1.4.6]. Then thanks to Lumer-Phillips Theorem (see [16, Theorem
1.4.3]), the operator $\mathcal{A}$ generates a $C_{0}$-semigroup of
contractions on the Hilbert $\mathcal{H}$ denoted by
$(e^{t\mathcal{A}})_{t\geq 0}$.

∎

##  3\. Strong stability

######  Theorem 3.1.

The semigroup $(e^{t\mathcal{A}})_{t\geq 0}$ is strongly stable in the energy
space $\mathcal{H}$ i.e.,

$\lim_{t\to+\infty}\|e^{t\mathcal{A}}y_{0}\|=0,\qquad\forall\,y_{0}\in\mathcal{D}(\mathcal{A}).$

###### Proof.

To show that the semigroup $(e^{t\mathcal{A}})_{t\geq 0}$ is strongly stable
we only have to prove that the intersection of $\sigma(\mathcal{A})$ with
$i\mathbb{R}$ is an empty set. Since the resolvent of the operator
$\mathcal{A}$ is not compact (see [14]) but $0\in\rho(\mathcal{A})$ we only
need to prove that $(i\mu I-\mathcal{A})$ is a one-to-one correspondence in
the energy space $\mathcal{H}$ for all $\mu\in\mathbb{R}^{*}$. The proof will
be done in two steps: In the first step we prove the injective property of
$(i\mu I-\mathcal{A})$ and in the second step we prove the surjective property
of the same operator.

### Step 1

Let $(u,v,w,z)\in\mathcal{D}(\mathcal{A})$ such that

(3.1) $\mathcal{A}(u,v,w,z)=i\mu(u,v,w,z).$

or equivalently,

(3.2) $\left\\{\begin{array}[]{ll}w=i\mu u&\text{in }(-1,1),\\\ z=i\mu
v&\text{in }(-1,1),\\\ (u_{x}+aw_{x})_{x}-z=i\mu w&\text{in }(-1,1),\\\
c\,v_{xx}+w=i\mu z&\text{in }(-1,1),\\\
u(-1)=u(1)=0,\,v(-1)=v(1)=0.\end{array}\right.$

Then taking the real part of the scalar product of (3.1) with $(u,v,w,z)$ we
get

$\mathrm{Re}(i\mu\|(u,v,w,z)\|_{\mathcal{H}}^{2})=\mathrm{Re}\left\langle\mathcal{A}(u,v,w,z),(u,v,w,z)\right\rangle_{\mathcal{H}}=-d\int_{0}^{1}|w_{x}|^{2}dx=0.$

Which implies that

$w_{x}=0\qquad\text{ in }\,(0,1).$

This implies that from the first equation (3.2) that

$u_{x}=0\qquad\text{ in }\,(0,1),$

which means that $u$ is a constant in $(0,1)$ and since $u(1)=0$ we obtain
that

$u=w=0\qquad\text{ in }\,(0,1),$

Hence, from the third and the second equation of (3.2) one gets

(3.3) $u=w=v=z=0\qquad\text{ in }\,(0,1),$

Using (3.3) then (3.2) is reduced to the following problem

(3.4) $\left\\{\begin{array}[]{ll}w=i\mu u&\text{in }(-1,0),\\\ z=i\mu
v&\text{in }(-1,0),\\\ \mu^{2}u+u_{xx}-i\mu v=0&\text{in }(-1,0),\\\
\mu^{2}v+c\,v_{xx}+i\mu u=0&\text{in }(-1,0),\\\
u(-1)=u(0)=0,\,v(-1)=v(0)=0.&\end{array}\right.$

Let $y=(u,v,u_{x},v_{x})$ and $y_{x}=(u_{x},v_{x},u_{xx},v_{xx})$ then (3.4)
is recast as follow

(3.5) $\left\\{\begin{array}[]{cc}y_{x}=A_{\mu}y&\text{in}(-1,0)\\\
Y(0)=0\end{array}\right.$

where

$A_{\mu}=\begin{pmatrix}\begin{array}[]{cccc}0&0&1&0\\\ 0&0&0&1\\\
-\mu^{2}&i\mu&0&0\\\
-i\frac{\mu}{c}&-\frac{\mu^{2}}{c}&0&0\end{array}\par\end{pmatrix}.$

Since $A_{\mu}$ is a bounded operator then the unique solution of (3.5) is
$y=0$ therefore $u=v=0$ in $(-1,0)$. Moreover, from the fist and the second
equation of (3.4) we have $w=z=0$ in $(-1,1)$. Combining all this with(3.3),
we deduce that $u=v=w=z=0$ in $(-1,1)$. This conclude the fist part of this
proof.

### Step 2

Now given $(f,g)\in\mathcal{H}$, we solve the equation

$(i\mu I-\mathcal{A})(u,v,w,z)=(f,g,f_{1},g_{1}).$

Or equivalently,

(3.6) $\left\\{\begin{array}[]{llcc}w=i\mu u-f\\\ z=i\mu v-g\\\
\mu^{2}u+u_{xx}+i\mu\,(au_{x})_{x}-i\mu v=(af_{x})_{x}-i\mu f-f_{1}-g=F\\\
\mu^{2}v+c\,v_{xx}+i\mu u=-\mu g+f-g_{1}=G.\end{array}\right.$

Let’s define the operator

$\displaystyle\begin{array}[]{rrrl}A:&(H_{0}^{1}(-1,1))^{2}&\longrightarrow&(H^{-1}(-1,1))^{2}\\\
&(u,v)&\longmapsto&(-u_{xx}-i\mu(au_{x})_{x}+i\mu v,-cv_{xx}-i\mu
u).\end{array}$

First we are going to show that $A$ is an isomorphism. For this purpose we
consider the two operator $\tilde{A}$ and $C$ such that

$\begin{array}[]{rrrl}\tilde{A}:&(H_{0}^{1}(-1,1))^{2}&\longrightarrow&(H^{-1}(-1,1))^{2}\\\
&(u,v)&\longmapsto&(-u_{xx}-i\mu(au_{x})_{x},-cv_{xx}),\end{array}$

and $C$ such that $A=C+\tilde{A}$. It’s easy to show that $\tilde{A}$ is an
isomorphism, then we could rewrite $A=\tilde{A}(Id-\tilde{A}^{-1}(-C))$. To
begin with, thanks to the compact embedding

$H_{0}^{1}(-1,1)^{2}\hookrightarrow
L^{2}(-1,1)^{2}\;\hbox{and}\;L^{2}(-1,1)^{2}\hookrightarrow H^{-1}(-1,1)^{2},$

we see that $\tilde{A}^{-1}$ is a compact operator. Secondly, it’s clear that
$C$ is a bounded operator, therefore, thanks to Fredholm’s alternative, we
only need to prove that $(Id-\tilde{A}^{-1}(-C))$ is injective.

Let $(u,v)\in(H_{0}^{1}(-1,1))^{2}$ such that
$(Id-\tilde{A}^{-1}(-C))(u,v)=0$, which implies that

$(\tilde{A}-(-C))(u,v)=0.$

Or equivalently

(3.7) $\left\\{\begin{array}[]{ll}u_{xx}+i\mu\,(au_{x})_{x}-i\mu v=0&\text{in
}(-1,1)\\\ c\,v_{xx}+i\mu u=0&\text{in }(-1,1)\\\
u(-1)=u(1)=0,\,v(-1)=v(1)=0.\end{array}\right.$

Multiplying the first equation of (3.7) by $\bar{u}$ and the conjugate of the
second by $v$, after integration over $(-1,1)$, it follows

$-\int_{-1}^{1}|u_{x}|^{2}{\rm d}x+c\int_{-1}^{1}|v_{x}|^{2}{\rm
d}x-i\mu\int_{-1}^{1}a|u_{x}|^{2}{\rm d}x=0$

Next, by taking the imaginary part, we can deduce that $u_{x}=0$ in $(0,1)$
then $u$ is constant in $(0,1)$ where with the boundary condition $u(1)=0$ we
have $u=0$ in (0,1). Moreover, using the second equation of (3.7) we obtain
$v=0$ in $(0,1)$, which implies that (3.7) that

(3.8) $\left\\{\begin{array}[]{ll}u_{xx}=i\mu v&\text{in}(-1,1)\\\
v_{xx}=-i\frac{\mu}{c}u&\text{in }(-1,1)\\\
u(0)=u(-1)=0,\,v(0)=v(-1)=0\end{array}\right.$

Let $y=(u,v,u_{x},v_{x})$ and $y_{x}=(u_{x},v_{x},u_{xx},v_{xx})$, using the
trace theorem we have:

$\left\\{\begin{array}[]{ll}y_{x}=D_{\mu}y&\text{in }(-1,0)\\\
y(0)=0,&\end{array}\right.$

where

$D_{\mu}=\begin{pmatrix}\begin{array}[]{cccc}0&0&1&0\\\ 0&0&0&1\\\
0&i\mu&0&0\\\ -i\frac{\mu^{2}}{c}&0&0&0\end{array}\par\end{pmatrix}.$

With a same approach as in the first step, we can have the result that we are
looking for (i.e. A is an isomorphism).

Now, rewriting the third and the fourth lines of (3.6) one gets

$(u,v)-\mu^{2}A^{-1}(u,v)=A^{-1}(F,G).$

Let $(u,v)\in\ker(Id-\mu^{2}A^{-1})$, i.e. $\mu^{2}(u,v)-A(u,v)=0$, so we can
see that:

(3.9) $\left\\{\begin{array}[]{ll}\mu^{2}u+u_{xx}+i\mu\,(au_{x})_{x}-i\mu
v=0&\text{in }(-1,1)\\\ \mu^{2}v+c\,v_{xx}+i\mu u=0&\text{in
}(-1,1).\end{array}\right.$

Furthermore, multiplying the first equation of (3.9) by $\bar{u}$ and the
conjugate of the second by $v$, after integration over $(-1,1)$ and taking the
imaginary part, we deduce that

$\int_{-1}^{1}a|u_{x}|^{2}{\rm d}x=d\int_{0}^{1}|u_{x}|^{2}{\rm d}x=0.$

So, we get the same system as in the first step (see (3.2)). Thus,
$\ker(I-\mu^{2}A^{-1})=\\{0_{(H^{-1}(-1,1))^{2}}\\}$.
In another hand, thanks to the compact embeddings
$H_{0}^{1}(-1,1)^{2}\hookrightarrow L^{2}((-1,1))^{2}$ and
$L^{2}(-1,1)^{2}\hookrightarrow H^{-1}(-1,1)^{2}$, we see that $A^{-1}$ is a
compact operator. Now, thanks to the Fredholm’s alternative, the operator
($Id-\mu^{2}A^{-1}$) is bijective in $(H_{0}^{1}(-1,1))^{2}$.Finally, the
equation (3.6) have a unique solution in $H_{0}^{1}(-1,1)^{2}$. This completes
the proof. ∎

##  4\. Lack of exponential stability

Now, we prove the lack of exponential stability given by the following theorem

######  Theorem 4.1.

The semigroup $(e^{t\mathcal{A}})_{t\geq 0}$, is not exponentially stable in
the energy space provided that $c>1$ and that

(4.1) $\sin(2\sqrt{c}n\pi)\neq O(n^{-\frac{1}{2}}),$

Noting that the assumption $c>1$ is made here just to make the calculation
readable. The second assumption (4.1) can be fulfilled for instance by taking
$c$ such that $2\sqrt{c}$ is an integer number. To prove (4.1) we mainly use
the following theorem

######  Theorem 4.2.

(see [10, 18]) Let $e^{t\mathcal{B}}$ be a bounded $C_{0}$-semigroup on a
Hilbert space $H$ with generator $\mathcal{B}$ such that
$i\mathbb{R}\subset\rho(\mathcal{B})$. Then $e^{t\mathcal{B}}$ is
exponentially stable if and only if There exist $a>0$ and $M>0$, such that

$\|e^{t\mathcal{B}}\|_{\mathcal{L}(\mathcal{H})}\leq Me^{-at},\forall t\geq 0$

if and only if

$\limsup_{\omega\in\mathbb{R},\,|\omega|\rightarrow\infty}\|(i\omega
I-\mathcal{B})^{-1}\|_{\mathcal{L}(\mathcal{H})}<\infty.$

Now, based on the Theorem 4.2 we prove the Theorem 4.1.

###### Proof.

Our main objective is to show that:

(4.2) $\|(\lambda I-\mathcal{A})^{-1}\|_{\mathcal{L}(\mathcal{H})}\text{ is
unbounded on the imaginary axis}.$

For $n\in\mathbb{N}$ large enough let $\lambda=\lambda_{n}=i\omega_{n}$, where

(4.3)
$\omega_{n}=\sqrt{\frac{8c(c+1)n^{2}\pi^{2}+2c+\sqrt{\Delta}}{4c}}\qquad\text{with}\qquad\Delta=(8c(c-1)\pi^{2}n^{2})^{2}+32(c+1)(c\pi
n)^{2}+4c^{2}.$

It’s clear that $\omega_{n}\longrightarrow+\infty$ and in particular we have

(4.4)
$\omega_{n}=\sqrt{c}\left(2n\pi+\frac{n^{-1}}{4\pi(c-1)}-\frac{cn^{-3}}{32\pi^{3}(c-1)^{3}}+o(n^{-4})\right)$

and

(4.5)
$\frac{1}{\omega_{n}}=\frac{1}{2n\pi\sqrt{c}}-\frac{1}{16\sqrt{c}(c-1)(\pi
n)^{3}}+o(n^{-4}).$

Define $(F_{1},G_{1},F_{2},G_{2})$
$\in(H_{0}^{1}(0,1))^{2}\times(L^{2}(0,1))^{2}$, such that

$\displaystyle F_{1}=F_{1}(x,n)=0\qquad\forall\,x\in(-1,1),$ $\displaystyle
G_{1}=G_{1}(x,n)=\left\\{\begin{array}[]{ll}0&\text{in }(0,1)\\\ \displaystyle
g_{1}=\frac{\sin(2n\pi x)}{2n\pi}&\text{in }(-1,0),\end{array}\right.$
$\displaystyle F_{2}=F_{2}(x,n)=0\qquad\forall\,x\in(-1,1),$ $\displaystyle
G_{2}=G_{2}(x,n)=\left\\{\begin{array}[]{ll}0&\text{in }(0,1)\\\ \displaystyle
g_{2}=\frac{c\sin(2n\pi
x)}{i\sqrt{\frac{2c}{c+1+\sqrt{(c-1)^{2}+\frac{4c}{\omega_{n}^{2}}}}}}&\text{in
}(-1,0).\end{array}\right.$

A straight forward calculation leads to

(4.6)
$\|(F_{1},G_{1},F_{2},G_{2})\|_{\mathcal{H}}^{2}=\frac{1}{2}+\frac{1}{2\mu_{-}}\longrightarrow\frac{1}{2}\left(1+\frac{1}{\sqrt{c}}\right)\quad\text{as}\quad
n\nearrow+\infty.$

Our goal is to prove that
$\displaystyle\lim_{|\lambda|\rightarrow\infty}\|(\lambda
I-\mathcal{A})^{-1}\|_{\mathcal{L}(\mathcal{H})}=\infty$. That’s why, we solve
the resolvent equation

(4.7) $(\lambda
I-\mathcal{A})(u^{1},v^{1},u^{2},v^{2})=(F_{1},G_{1},F_{2},G_{2}).$

Step 1. For all $x\in(0,1)$ , we have

(4.13) $\displaystyle\left\\{\begin{array}[]{llll}\lambda u^{1}-u^{2}=0\\\
\lambda v^{1}-v^{2}=0\\\ \lambda u^{2}-(1+\lambda d)u^{1}_{xx}+v^{2}=0\\\
\lambda v^{2}-cv_{xx}^{1}-u^{2}=0\\\ v^{1}(1)=u^{1}(1)=0\end{array}\right.$

Let

$\eta_{+}=\frac{-\lambda(1+\lambda
d-c)+\omega_{n}\sqrt{r}e^{i\frac{\phi}{2}}}{2(1+\lambda d)}\qquad\text{ and
}\qquad\eta_{-}=\frac{-\lambda(1+\lambda
d-c)-\omega_{n}\sqrt{r}e^{i\frac{\phi}{2}}}{2(1+\lambda d)}$

where

$\displaystyle r=\sqrt{a^{2}+b^{2}},\quad\cos(\phi)=\frac{a}{r}\quad\text{ and
}\quad\sin(\phi)=\frac{b}{r},$

with

$\displaystyle\displaystyle a$
$\displaystyle=-(1-c)^{2}+d^{2}\omega^{2}-\frac{4c}{\omega_{n}^{2}}$
$\displaystyle\displaystyle b$
$\displaystyle=-2d\left((1-c)\omega_{n}+\frac{2c}{\omega_{n}}\right).$

It is important to note that

$\sqrt{a}=d\omega-\frac{(c-1)^{2}}{2d}\omega^{-1}-\frac{(c-1)^{4}+16cd^{2}}{8d^{3}}\omega^{-3}+o(\omega^{-3}),$
$\frac{b}{a}=\frac{2(c-1)}{d}+\frac{2(c-1)-4cd}{d}\omega^{-3}+o(\omega^{-4})$

and

$\displaystyle i\frac{\sqrt{r}}{d}\mathrm{e}^{i\frac{\phi}{2}}$
$\displaystyle=\lambda-\frac{c-1}{d}-\frac{(c-1)^{3}-d^{2}(c-1)+2cd^{2}}{d^{3}}\lambda^{-2}$
$\displaystyle+\frac{d^{2}(c-1)^{2}-(c-1)^{4}-2cd^{3}(c-1)-2cd^{2}}{d^{4}}\lambda^{-3}+o(\omega^{-3}).$

Then we obtain

(4.14) $\displaystyle\eta_{+}$
$\displaystyle=-\lambda+\frac{c}{d}-\frac{c}{d^{2}}\lambda^{-1}+\frac{(c-1)^{3}+d^{2}(c+1)+2c}{2d^{3}}\lambda^{-2}$
$\displaystyle+\frac{(c-1)^{4}-(c-1)^{3}-d^{2}(c-1)(c-2)-2c}{2d^{4}}+o(\omega^{-3})$

and

(4.15) $\displaystyle\eta_{-}$
$\displaystyle=-\frac{(c-1)^{3}+d^{2}(c+1)}{2d^{3}}\lambda^{-2}+\frac{(c-1)^{3}(2-c)+d^{2}(c-1)(c-2-2cd)}{2d^{4}}\lambda^{-3}+o(\omega^{-3})$

A straightforward calculation leads to

(4.16)
$\displaystyle(u^{1}+\eta_{+}v^{1})_{xx}=(\beta_{+})^{2}(u^{1}+\eta_{+}v^{1})$
(4.17)
$\displaystyle(u^{1}+\eta_{-}v^{1})_{xx}=(\beta_{-})^{2}(u^{1}+\eta_{-}v^{1}),$

where

$(\beta_{\pm})^{2}=\frac{c\lambda^{2}-\lambda\eta_{\pm}(1+\lambda
d)}{c(1+\lambda d)}.$

So, for $n$ large enough we get

$\beta_{\pm}=\frac{\omega_{n}}{\sqrt{2c(1+(d\omega_{n})^{2})}}\sqrt{r_{\pm}}e^{i\frac{\phi_{\pm}}{2}},$

where

$\displaystyle
r_{\pm}=\sqrt{a_{\pm}^{2}+b_{\pm}^{2}},\quad\cos(\phi_{\pm})=\frac{a_{\pm}}{r_{\pm}}\quad\text{
and }\quad\sin(\phi_{\pm})=\frac{b_{\pm}}{r_{\pm}}$

with

$\displaystyle\displaystyle a_{\pm}$
$\displaystyle=-(1+c)-(d\omega_{n})^{2}\pm\sqrt{r}\left(-d\omega_{n}\cos\left(\frac{\phi}{2}\right)+\sin\left(\frac{\phi}{2}\right)\right)$
$\displaystyle\displaystyle b_{\pm}$
$\displaystyle=cd\omega_{n}\pm\sqrt{r}\left(-\cos\left(\frac{\phi}{2}\right)-d\omega_{n}\sin\left(\frac{\phi}{2}\right)\right).$

Noting that

$|a_{+}|=2(d\omega)^{2}+\frac{c^{2}-3c+6}{2}+o(\omega^{-1}),$
$\sqrt{|a_{+}|}=\sqrt{2}d\omega+\frac{c^{2}-3c+6}{4\sqrt{2}d}\omega^{-1}+o(\omega^{-1}),$
$b_{+}=\left(d(2cd+1-c)+\frac{(c-1)^{3}}{d}\right)\omega^{-1}+o(\omega^{-1}),$

and

$\frac{b_{+}}{a_{+}}=o(\omega^{-2}).$

Then we obtain

(4.18)
$\beta_{+}=\frac{\lambda}{\sqrt{c}}-\frac{(c-1)(c-2)}{8\sqrt{2}d^{2}}+o(\omega^{-1}),$

and

(4.19) $\beta_{+}^{2}=\frac{\lambda^{2}}{c}+o(1).$

Similarly we have

$b_{-}=2cd\omega+\left(d(c-1-2cd)-\frac{(c-1)^{3}}{d}\right)\omega^{-1}+o(\omega^{-1}),$
$\sqrt{b_{-}}=\sqrt{2cd\omega}\left(1+\left(\frac{c-1-2cd}{4c}-\frac{(c-1)^{3}}{4cd^{2}}\right)\omega^{-2}\right)+o(\omega^{-2})$
$a_{-}=-2c+o(\omega^{-1}),$

and

$\frac{a_{-}}{b_{-}}=-\frac{\omega^{-1}}{d}+o(\omega^{-2}),$

then consequently we obtain

(4.20)
$\beta_{-}=\sqrt{\frac{\omega}{d}}\mathrm{e}^{\frac{i\pi}{4}}-\frac{\omega^{-\frac{1}{2}}}{2d^{\frac{3}{2}}}\mathrm{e}^{-i\frac{\pi}{4}}+o(\omega^{-1}),$

and

(4.21) $\displaystyle\beta_{-}^{2}$
$\displaystyle=\frac{\lambda}{d}-\frac{1}{d^{2}}+\frac{(c-1)^{3}+d(c-1)+2c}{2cd^{3}}\lambda^{-1}$
$\displaystyle-\frac{(c-1)^{3}(2-c)+d(c-1)(c-2-2cd)+2c}{2cd^{4}}\lambda^{-2}+o(\omega^{-2})$

Next, from (4.16), we get

$(u^{1}+\eta_{+}v^{1})=c_{1}e^{x\beta_{+}}+c_{2}e^{-x\beta_{+}}$

and

$(u^{1}+\eta_{-}v^{1})=c_{3}e^{x\beta_{-}}+c_{4}e^{-x\beta_{-}}.$

Recalling that $u^{1}(1)=v^{1}(1)=0$ we can rewrite the last two equations as
follow

(4.22) $(u^{1}+\eta_{+}v^{1})=c_{1}(e^{x\beta_{+}}-e^{(2-x)\beta_{+}}),$
(4.23) $(u^{1}+\eta_{-}v^{1})=c_{3}(e^{x\beta_{-}}-e^{(2-x)\beta_{-}}).$

Hence by combining (4.22) and (4.23) we obtain

(4.24)
$u^{1}(x)=-\frac{c_{1}\eta_{-}}{\eta_{+}-\eta_{-}}\left(\mathrm{e}^{\beta_{+}x}-\mathrm{e}^{\beta_{+}(2-x)}\right)+\frac{c_{3}\eta_{+}}{\eta_{+}-\eta_{-}}\left(\mathrm{e}^{\beta_{-}x}-\mathrm{e}^{\beta_{-}(2-x)}\right),$

and

(4.25)
$v^{1}(x)=\frac{c_{1}}{\eta_{+}-\eta_{-}}\left(\mathrm{e}^{\beta_{+}x}-\mathrm{e}^{\beta_{+}(2-x)}\right)-\frac{c_{3}}{\eta_{+}-\eta_{-}}\left(\mathrm{e}^{\beta_{-}x}-\mathrm{e}^{\beta_{-}(2-x)}\right).$

Step 2. For all $x\in(-1,0)$ we have

(4.31) $\displaystyle\left\\{\begin{array}[]{llll}\lambda u^{1}-u^{2}=0\\\
\lambda v^{1}-v^{2}=g_{1}\\\ \lambda u^{2}-u^{1}_{xx}+v^{2}=0\\\ \lambda
v^{2}-cv_{xx}^{1}-u^{2}=g_{2}\\\ v^{1}(-1)=u^{1}(-1)=0.\end{array}\right.$

Following to the third and the fourth equation of (4.13) and of (4.31) we can
deduce, thanks to the regularity of the stats, that

(4.32) $\displaystyle(1+\lambda d)u_{x}^{1}(0^{+})$
$\displaystyle=u_{x}^{1}(0^{-}),$ (4.33) $\displaystyle v_{x}^{1}(0^{+})$
$\displaystyle=v_{x}^{1}(0^{-}).$

and

(4.34) $\displaystyle(1+\lambda d)u_{xx}^{1}(0^{+})$
$\displaystyle=u_{xx}^{1}(0^{-}),$ (4.35) $\displaystyle v_{xx}^{1}(0^{+})$
$\displaystyle=v_{xx}^{1}(0^{-}).$

We denote by

(4.36)
$\alpha_{+}=\frac{\lambda}{2}\left(c-1+\sqrt{(1-c)^{2}+\frac{4c}{\omega_{n}^{2}}}\right)=(c-1)\lambda-\frac{c}{c-1}\lambda^{-1}-\frac{c^{2}}{(c-1)^{3}}+o(\omega^{-3}),$

and

(4.37)
$\alpha_{-}=\frac{\lambda}{2}\left(c-1-\sqrt{(1-c)^{2}+\frac{4c}{\omega_{n}^{2}}}\right)=\frac{c}{c-1}\lambda^{-1}+o(\omega^{-1})$

and we define for $n$ large enough $\mu_{\pm}$ as follow

$\mu_{\pm}=\frac{\sqrt{2c}}{\sqrt{c+1-\left(\pm\sqrt{(c-1)^{2}+\frac{4c}{\omega_{n}^{2}}}\right)}},$

in particular with the chose of $\omega_{n}$ in (4.3) one get

$\displaystyle\mu_{\pm}^{2}=\frac{\lambda}{\lambda-\frac{\alpha_{\pm}}{c}}.$

Besides, we have

(4.38)
$\mu_{+}=\sqrt{c}\left(1-\frac{c}{2(c-1)}\lambda^{-2}+o(\omega^{-2})\right),$
(4.39) $\mu_{-}=1+\frac{\lambda^{-2}}{2(c-1)}+o(\omega^{-2}),$

and

(4.40)
$\frac{\mu_{+}}{\mu_{-}}=\sqrt{c}\left(1-\frac{c+1}{2(c-1)}\lambda^{-2}+o(\omega^{-2})\right).$

We set

(4.41) $\displaystyle\omega^{+}_{1}(x)$
$\displaystyle=(u^{2}+\alpha_{+}v^{2}+\mu_{+}(u^{1}_{x}+\alpha_{+}v^{1}_{x}),$
(4.42) $\displaystyle\omega^{-}_{1}(x)$
$\displaystyle=(u^{2}+\alpha_{+}v^{2}-\mu_{+}(u^{1}_{x}+\alpha_{+}v^{1}_{x})),$
(4.43) $\displaystyle\omega^{+}_{2}(x)$
$\displaystyle=(u^{2}+\alpha_{-}v^{2}+\mu_{-}(u^{1}_{x}+\alpha_{-}v^{1}_{x})),$
(4.44) $\displaystyle\omega^{-}_{2}(x)$
$\displaystyle=(u^{2}+\alpha_{-}v^{2}-\mu_{-}(u^{1}_{x}+\alpha_{-}v^{1}_{x})).$

Now, define
$Y=(\omega^{+}_{1},\omega^{-}_{1},\omega^{+}_{2},\omega_{2}^{-})^{t}$ and
$Z=(g_{1x},g_{2})^{t}$. Then we have

(4.45) $Y_{x}=AY+BZ$

where

$A=\left(\begin{array}[]{cccc}\mu_{+}(\lambda-\frac{\alpha_{+}}{c})&0&0&0\\\
0&\mu_{+}(-\lambda+\frac{\alpha_{+}}{c})&0&0\\\
0&0&\mu_{-}(\lambda-\frac{\alpha_{-}}{c})&0\\\
0&0&0&\mu_{-}(-\lambda+\frac{\alpha_{-}}{c})\end{array}\right)$

and

$B=\left(\begin{array}[]{cc}-\alpha_{+}&-\mu_{+}\frac{\alpha_{+}}{c}\\\
-\alpha_{+}&\mu_{+}\frac{\alpha_{+}}{c}\\\
-\alpha_{-}&-\mu_{-}\frac{\alpha_{-}}{c}\\\
-\alpha_{-}&\mu_{-}\frac{\alpha_{-}}{c}\end{array}\right).$

Then, a straightforward calculation leads to:

$\mu_{+}(\lambda-\frac{\alpha_{+}}{c})=2in\pi.$

Using the boundary condition at $-1$ we get

(4.46)
$\omega_{1}^{+}(-1)=-\omega_{1}^{-}(-1)\;\hbox{and}\;\omega_{2}^{+}(-1)=-\omega_{2}^{-}(-1),$

Taking into account of (4.46) then the solution of (4.45) is written as follow

(4.47) $\displaystyle\omega_{1}^{+}(x)$
$\displaystyle=\omega_{1}^{+}(-1)\mathrm{e}^{2in\pi
x}-\frac{\alpha_{+}}{2}\left[\left(1-\frac{\mu^{+}}{\mu^{-}}\right)(x+1)\mathrm{e}^{2in\pi
x}+\frac{1}{2n\pi}\left(1+\frac{\mu^{+}}{\mu^{-}}\right)\sin(2n\pi x)\right],$
(4.48) $\displaystyle\omega_{1}^{-}(x)$
$\displaystyle=-\omega_{1}^{+}(-1)\mathrm{e}^{-2in\pi
x}-\frac{\alpha_{+}}{2}\left[\left(1-\frac{\mu^{+}}{\mu^{-}}\right)(x+1)\mathrm{e}^{-2in\pi
x}+\frac{1}{2n\pi}\left(1+\frac{\mu^{+}}{\mu^{-}}\right)\sin(2n\pi x)\right],$
(4.49) $\displaystyle\omega_{2}^{+}(x)$
$\displaystyle=\omega_{2}^{+}(-1)\mathrm{e}^{\mu_{-}\left(\lambda-\frac{\alpha_{-}}{c}\right)(x+1)}+\frac{\alpha_{-}}{2in\pi+\mu_{-}\left(\lambda-\frac{\alpha_{-}}{c}\right)}\left[\mathrm{e}^{-2in\pi
x}+\mathrm{e}^{\mu_{-}\left(\lambda-\frac{\alpha_{-}}{c}\right)(x+1)}\right],$
(4.50) $\displaystyle\omega_{2}^{-}(x)$
$\displaystyle=-\omega_{2}^{+}(-1)\mathrm{e}^{-\mu_{-}\left(\lambda-\frac{\alpha_{-}}{c}\right)(x+1)}-\frac{\alpha_{-}}{2in\pi+\mu_{-}\left(\lambda-\frac{\alpha_{-}}{c}\right)}\left[\mathrm{e}^{2in\pi
x}+\mathrm{e}^{-\mu_{-}\left(\lambda-\frac{\alpha_{-}}{c}\right)(x+1)}\right].$

Taking the trace of $\omega_{1}^{+}$ and $\omega_{1}^{-}$ in (4.47)-(4.48) and
in (4.41)-(4.42) on the boundary $0$ and using the continuity of the states
$u_{2}$ and $v_{2}$ we obtain

$\displaystyle(\omega_{1}^{+}+\omega_{1}^{-})(0^{-})$
$\displaystyle=\alpha_{+}\left(\frac{\mu_{+}}{\mu_{-}}-1\right)=2u^{2}(0^{-})+2\alpha_{+}v^{2}(0^{-})$
$\displaystyle=2\lambda(u^{1}(0^{-})+\alpha_{+}v^{1}(0^{-}))=2\lambda(u^{1}(0^{+})+\alpha_{+}v^{1}(0^{+}))$
$\displaystyle=\frac{2\lambda}{\eta_{+}-\eta_{-}}\left(c_{1}(1-e^{2\beta_{+}})(\alpha_{+}-\eta_{-})+c_{3}(1-e^{2\beta_{-}})(\eta_{+}-\alpha_{+})\right),$

where we have used the the expressions of $u^{1}$ and $v^{1}$ in (4.24) and
(4.25). This implies that

(4.51)
$c_{3}=\frac{1-e^{2\beta_{+}}}{1-e^{2\beta_{-}}}A_{n}c_{1}+\frac{B_{n}}{1-e^{2\beta_{-}}}$

where

$\displaystyle A_{n}=\frac{\eta_{-}-\alpha_{+}}{\eta_{+}-\alpha_{+}}$
$\displaystyle=\frac{c-1}{c}\left(1+\frac{\lambda^{-1}}{d}-\frac{\lambda^{-2}}{c-1}+o(\omega^{-2})\right)$
(4.52) $\displaystyle=\frac{c-1}{c}\left(1+\frac{n^{-1}}{2i\pi
d\sqrt{c}}-\frac{n^{-2}}{4\pi^{2}c(c-1)}+o(\omega^{-2})\right),$

and

$\displaystyle B_{n}$
$\displaystyle=\frac{\alpha_{+}(\eta_{+}-\eta_{-})\left(\frac{\mu_{+}}{\mu_{-}}-1\right)}{2\lambda(\eta_{+}-\alpha_{+})}$
$\displaystyle=\frac{(c-1)(\sqrt{c}-1)}{2c}\left(1-\frac{c-1}{d}\lambda^{-1}-\left(\frac{1}{(c-1)^{2}}+\frac{\sqrt{c}(c+1)}{2(\sqrt{c}-1)(c-1)}\right)\lambda^{-2}+o(\omega^{-2})\right)$
(4.53) $\displaystyle=\frac{(c-1)(\sqrt{c}-1)}{2c}\left(1-\frac{c-1}{2i\pi
d\sqrt{c}}n^{-1}-\left(\frac{1}{(c-1)^{2}}+\frac{\sqrt{c}(c+1)}{2(\sqrt{c}-1)(c-1)}\right)\times\frac{n^{-2}}{4\pi^{2}c}+o(n^{-2})\right).$

where we used here (4.14), (4.15), (4.36), (4.37), (4.40) and (4.3).
Using (4.47)-(4.48) and (4.34)-(4.35), one gets

$\displaystyle(\omega_{1}^{+}-\omega_{1}^{-})^{\prime}(0^{-})$
$\displaystyle=2in\pi\alpha_{+}\left(\frac{\mu_{+}}{\mu_{-}}-1\right)=2\mu_{+}(u^{1}+\alpha_{+}v^{1})_{xx}(0^{-})=2\mu_{+}((1+\lambda
d)u^{1}+\alpha_{+}v^{1})_{xx}(0^{+})$
$\displaystyle=\frac{2\mu_{+}\left[c_{1}\beta_{+}^{2}(1-\mathrm{e}^{2\beta_{+}})(\alpha_{+}-(1+\lambda
d)\eta_{-})+c_{3}\beta_{-}^{2}(1-\mathrm{e}^{2\beta_{-}})((1+\lambda
d)\eta_{+}-\alpha_{+})\right]}{\eta_{+}-\eta_{-}}.$

Then we obtain

(4.54)
$c_{1}=\frac{1-e^{2\beta_{-}}}{1-e^{2\beta_{+}}}A_{n}^{\prime}c_{3}+\frac{B_{n}^{\prime}}{1-e^{2\beta_{+}}}$

where

$\displaystyle A_{n}^{\prime}$
$\displaystyle=\frac{\beta_{-}^{2}(\alpha_{+}-(1+\lambda
d)\eta_{+})}{\beta_{+}^{2}(\alpha_{+}-(1+\lambda d)\eta_{-})}$
$\displaystyle=\frac{c}{c-1}\left(1-\frac{\lambda^{-1}}{d}+\left(\frac{(c-1)^{3}}{2cd^{2}}+\frac{c-1}{2cd}+\frac{3-c}{2d^{2}}+\frac{1}{2}\right)\lambda^{-2}+o(\omega^{-2})\right)$
(4.55) $\displaystyle=\frac{c}{c-1}\left(1-\frac{n^{-1}}{2i\pi
d\sqrt{c}}+\left(\frac{(c-1)^{3}}{2cd^{2}}+\frac{c-1}{2cd}+\frac{3-c}{2d^{2}}+\frac{1}{2}\right)\frac{n^{-2}}{4\pi^{2}c}+o(n^{-2})\right),$

and

$\displaystyle B_{n}^{\prime}$
$\displaystyle=\frac{in\pi\alpha_{+}(\eta_{+}-\eta_{-})\left(\frac{\mu_{+}}{\mu_{-}}-1\right)}{\mu_{+}\beta_{+}^{2}(\alpha_{+}-(1+\lambda
d)\eta_{-})}$
$\displaystyle=\frac{n\pi(c-\sqrt{c})}{2\omega}\left(-1+\frac{c}{d}\lambda^{-1}+\left(\frac{c+\sqrt{c}+3}{2(c-1)^{2}}-\frac{c+1+d^{2}}{2d^{2}}\right)\lambda^{-2}\right)+o(\omega^{-2})$
(4.56) $\displaystyle=\frac{\sqrt{c}-1}{2}\left(-1+\frac{\sqrt{c}}{2i\pi
d}n^{-1}+\left(\frac{\sqrt{c}+4}{c-1}-\frac{c+1+d}{d^{2}}\right)\frac{n^{-2}}{8c\pi^{2}}+o(n^{-2})\right).$

where we used here (4.14), (4.15), (4.19), (4.21), (4.36), (4.37), (4.38),
(4.40) and (4.3).
Combining (4.51) and (4.54) then we find that

(4.57)
$c_{1}=\frac{1}{1-e^{2\beta_{+}}}\times\frac{A_{n}^{\prime}B_{n}+B_{n}^{\prime}}{1-A_{n}A_{n}^{\prime}}=\frac{c_{1}^{\prime}}{1-e^{2\beta_{+}}}$

and

(4.58)
$c_{3}=\frac{1}{1-e^{2\beta_{-}}}\times\frac{A_{n}B_{n}^{\prime}+B_{n}}{1-A_{n}A_{n}^{\prime}}=\frac{c_{3}^{\prime}}{1-e^{2\beta_{-}}},$

where following to (4), (4), (4) and (4) we have

(4.59) $c_{1}^{\prime}=O(1)\qquad\text{ and }\qquad c_{3}^{\prime}=O(1).$

In another hand, by denoting
$\theta=-i\mu_{-}\left(\lambda-\frac{\alpha_{-}}{c}\right)$ and by using the
same argument as previously, one gets

$\displaystyle(\omega_{2}^{+}+\omega_{2}^{-})(0^{-})$
$\displaystyle=2i\sin(\theta)\omega_{2}^{+}(-1)+\frac{2\alpha_{-}}{2n\pi-\theta}\sin(\theta)=2\lambda(u^{1}+\alpha_{-}v^{1})(0^{-})=2\lambda(u^{1}+\alpha_{-}v^{1})(0^{+})$
$\displaystyle=\frac{2\lambda}{\eta_{+}-\eta_{-}}\left(c_{1}^{\prime}(\alpha_{-}-\eta_{-})+c_{3}^{\prime}(\eta_{+}-\alpha_{-})\right).$

It’s clear that $\theta\neq 0[\pi]$ then we can write

(4.60) $\displaystyle\omega_{2}^{+}(-1)$
$\displaystyle=\frac{\lambda}{i\sin(\theta)(\eta_{+}-\eta_{-})}\left[c_{1}^{\prime}(\alpha_{-}-\eta_{-})+c_{3}^{\prime}(\eta_{+}-\alpha_{-})\right]-\frac{\alpha_{-}}{2in\pi+i\theta}.$

Noting that from (4.3), (4.4), (4.37) and (4.39) we have

(4.61)
$\theta=\omega\left(1-\frac{3}{2(c-1)}\omega^{-2}+o(\omega^{-2})\right)=\sqrt{c}\left(2n\pi+\frac{c\pi-12}{4c\pi^{2}(c-1)}n^{-1}+o(n^{-1})\right)$

Then from (4.4), (4.14), (4.15), (4.37), (4.39) and (4.61) we deduce that

(4.62) $\omega_{2}^{+}(-1)\sim\frac{2\pi
n\sqrt{c}\,c_{3}^{\prime}}{\sin(\theta)}$

Using (4.32)-(4.33), (4.41)-(4.42) and (4.47)-(4.48) we get

$\displaystyle\omega_{1}^{+}(-1)$
$\displaystyle=\frac{(\omega_{1}^{+}-\omega_{1}^{-})(0^{-})}{2}=\mu_{+}(u^{1}+\alpha_{+}v^{1})_{x}(0^{-})=\mu_{+}((1+\lambda
d)u^{1}+\alpha_{+}v^{1})_{x}(0^{+})$ (4.63)
$\displaystyle=\frac{\mu_{+}}{\eta_{+}-\eta_{-}}\left[c_{1}\beta_{+}(1+e^{2\beta_{+}})(\alpha_{+}-(1+\lambda
d)\eta_{-})+c_{3}\beta_{-}(1+e^{2\beta_{-}})((1+\lambda
d)\eta_{+}-\alpha_{+})\right].$

Then from (4.4), (4.14), (4.15), (4.18), (4.20), (4.37) and (4.38) we deduce
that

(4.64) $\omega_{1}^{+}(-1)\sim
c_{3}^{\prime}\sqrt{\frac{c}{d}}\mathrm{e}^{-i\frac{\pi}{4}}(2\pi\sqrt{c}n)^{\frac{3}{2}}.$

Next, for all $x\in(-1,0)$ we have

$\displaystyle v_{x}^{1}(x)$
$\displaystyle=\frac{1}{2\mu_{-}\mu_{+}(\alpha_{+}-\alpha_{-})}\left[\alpha_{-}(\omega_{1}^{+}(x)-\omega_{1}^{-}(x))-\alpha_{+}(\omega_{2}^{+}(x)-\omega_{2}^{-}(x))\right]$
(4.65)
$\displaystyle=\frac{1}{2\mu_{-}\mu_{+}(\alpha_{+}-\alpha_{-})}\Bigg{[}\mu_{-}\left(2\omega_{1}^{+}(-1)\cos(2n\pi
x)-i\alpha_{+}\left(1-\frac{\mu_{+}}{\mu_{-}}\right)(x+1)\sin(2n\pi x)\right)$
$\displaystyle-\mu_{+}\bigg{(}2\omega_{2}^{+}(-1)\cos\left(\theta(x+1)\right)+\frac{2\alpha_{-}}{2in\pi+i\theta}\left(\cos(2n\pi
x)+\cos\left(\theta(x+1)\right)\right)\bigg{)}\Bigg{]},$

where we have used (4.41)-(4.44) and (4.47)-(4.50). Thus further leads to

(4.66)
$\displaystyle\|v_{x}^{1}\|_{L^{2}(-1,0)}^{2}\geq\max\left\\{\frac{|\omega_{1}^{+}(-1)|^{2}}{2\mu_{+}^{2}|\alpha_{+}-\alpha_{-}|^{2}},\frac{|\omega_{2}^{+}(-1)|^{2}}{\mu_{-}^{2}|\alpha_{+}-\alpha_{-}|^{2}}\right\\}-\frac{|\alpha_{+}|^{2}(\mu_{+}-\mu_{-})^{2}}{4\mu_{-}^{2}\mu_{+}^{2}|\alpha_{+}-\alpha_{-}|^{2}}$
$\displaystyle-\min\left\\{\frac{|\omega_{1}^{+}(-1)|^{2}}{2\mu_{+}^{2}|\alpha_{+}-\alpha_{-}|^{2}},\frac{|\omega_{2}^{+}(-1)|^{2}}{\mu_{-}^{2}|\alpha_{+}-\alpha_{-}|^{2}}\right\\}-\frac{2|\alpha_{-}|^{2}}{\mu_{-}^{2}\mu_{+}^{2}(2n\pi+\theta)^{2}|\alpha_{+}-\alpha_{-}|^{2}}.$

Since,

$\sin(\theta)\neq O(n^{-\frac{1}{2}}),$

as $n$ goes to the infinity (by (4.61) assumption (4.1)) then by using (4.3),
(4.38), (4.39), (4.36), (4.37), (4.62) and (4.64) we can show that the second
and the fourth terms of the right hand side of (4.66) are bounded while the
sum of the fist and the third terms tends to the infinity as $n$ goes to
$+\infty$, therefore we obtain

(4.67) $\|v_{x}^{1}\|_{L^{2}(-1,0)}^{2}\,\,\quad\text{as}\quad
n\nearrow+\infty.$

Last but not least, we have

(4.68)
$\|(i\omega_{n}I-\mathcal{A})^{-1}(F_{1},G_{1},F_{2},G_{2})\|_{\mathcal{H}}=\|(u^{1},v^{1},u^{2},v^{2})\|_{\mathcal{H}}^{2}\geq\int_{-1}^{0}|v_{x}^{1}(x)|^{2}{\rm
d}x\longrightarrow+\infty,\quad\text{as}\quad n\nearrow+\infty.$

Finally we conclude, using (4.68) and (4.6) that

$\limsup_{\omega\in\mathbb{R},|\omega|\rightarrow\infty}\|(i\omega
I-\mathcal{A})^{-1}\|_{\mathcal{L}(\mathcal{H})}=+\infty.$

So, $e^{t\mathcal{A}}$ is not exponentially stable in the energy space. This
completes the proof. ∎

##  5\. Polynomial stabilization

This subsection aims to prove the polynomial stability given by the following
theorem:

######  Theorem 5.1.

The semigroup of contraction $(e^{T\mathcal{A}})_{t\geq 0}$ is polynomially
stable of order $\displaystyle\frac{1}{12}$.

Our method is based on the Borichev and Tomilov result given by the following:

######  Theorem 5.2.

[5, Theorem 2.4] Let $\mathcal{B}$ be a generator of a $C_{0}$-semigroup of
contraction in a Hilbert space $\mathcal{X}$ with domain
$\mathcal{D}(\mathcal{B})$ such that $i\mathbb{R}\subset\sigma(\mathcal{B})$
then $e^{t\mathcal{B}}$ is polynomially stable with order
$\frac{1}{\gamma},\gamma>0$ i.e. there exists $C>0$ such that

$\|e^{t\mathcal{B}}U_{0}\|_{\mathcal{X}}\leq\frac{C}{(1+t)^{\frac{1}{\gamma}}}\|U_{0}\|_{\mathcal{D}(\mathcal{B})},\qquad\forall\,t\geq
0,\;\forall\,U_{0}\in\mathcal{D}(\mathcal{B}),$

if and only if

$\limsup_{\beta\in\mathbb{R},\,|\beta|\rightarrow\infty}\|\beta^{-\gamma}(i\beta-\mathcal{B})^{-1}\|_{\mathcal{L}(\mathcal{X})}<+\infty.$

Based on Theorem 5.2 we are able now to prove our main result given in Theorem
5.1 of this section. For this purpose, let’s consider the following:

######  Proposition 5.1.

The operator $\mathcal{A}$ defined in (2.1) satisfies:

(5.1)
$\limsup_{\beta\in\mathbb{R},\,|\beta|\rightarrow\infty}\|\beta^{-12}(i\beta-\mathcal{A})^{-1}\|_{\mathcal{L}(\mathcal{H})}<+\infty.$

###### Proof.

To prove (5.1) we use an argument of contradiction. In fact, if (5.1) is
false, then, there exist $\beta_{n}\in\mathbb{R}_{+}$ and
$Y_{n}=(u_{n}^{1},v_{n}^{1},u_{n}^{2},v_{n}^{2})\in\mathcal{D}(\mathcal{A})$
such that

(5.2) $\|Y_{n}\|_{\mathbf{H}}=1,\;\beta_{n}\nearrow+\infty\text{ and
}\beta^{\gamma}(i\beta_{n}\mathrm{I}-\mathcal{A})Y_{n}:=(f_{n}^{1},g_{n}^{1},f_{n}^{2},g_{n}^{2})\longrightarrow\,0\text{
in }\mathcal{H}\text{ as }n\nearrow+\infty.$

Equivalently, we have

(5.3)
$\beta_{n}^{\gamma}\left(i\beta_{n}u_{n}^{1}-u_{n}^{2}\right)=f_{n}^{1}\,\longrightarrow
0\hskip 8.5359pt\text{in}\hskip 8.5359ptH_{0}^{1}(-1,1),$ (5.4)
$\beta_{n}^{\gamma}\left(i\beta_{n}v_{n}^{1}-v_{n}^{2}\right)=g_{n}^{1}\,\longrightarrow
0\hskip 8.5359pt\text{in}\hskip 8.5359ptH_{0}^{1}(-1,1),$ (5.5)
$\beta_{n}^{\gamma}\left(i\beta_{n}u_{n}^{2}-\left(u_{nx}^{1}+au_{nx}^{2}\right)_{x}+v_{n}^{2}\right)=f_{n}^{2}\,\longrightarrow
0\hskip 8.5359pt\text{in}\hskip 8.5359ptL^{2}(-1,1),$ (5.6)
$\beta_{n}^{\gamma}\left(i\beta_{n}v_{n}^{2}-cv_{nxx}^{1}-u_{n}^{2}\right)=g_{n}^{2}\,\longrightarrow
0\hskip 8.5359pt\text{in}\hskip 8.5359ptL^{2}(-1,1).$

We denote by

$T_{n}=u_{nx}^{1}+au_{nx}^{2}.$

Taking the real part of
$\displaystyle\left\langle\beta^{\gamma}(i\beta_{n}\mathrm{I}-\mathcal{A})Y_{n},Y_{n}\right\rangle_{\mathcal{H}}$
then by the dissipation property of the semigroup of the operator
$\mathcal{A}$ we get

$\beta_{n}^{\gamma}\int_{0}^{1}d.|u_{nx}^{2}|^{2}\,{\rm d}x\,\longrightarrow
0,$

which leads to

(5.7) $\beta_{n}^{\frac{\gamma}{2}}\|u_{nx}^{2}\|_{L^{2}(0,1)}\longrightarrow
0.$

Now thanks to (5.3) and (5.7), we obtain

(5.8)
$\beta_{n}^{\frac{\gamma}{2}+1}\|u_{nx}^{1}\|_{L^{2}(0,1)}\longrightarrow 0.$

From (5.7) and (5.8), it follows

(5.9) $\beta_{n}^{\frac{\gamma}{2}}\|T_{n}\|_{L^{2}(0,1)}\longrightarrow 0.$

Taking the inner product of (5.5) with $u_{n}^{2}$ in $L^{2}(0,1)$ we get

(5.10)
$\beta_{n}^{\frac{3\gamma}{4}}\left(i\beta_{n}\|u_{n}^{2}\|^{2}_{L^{2}(0,1)}+\langle
T_{n},u_{nx}^{2}\rangle_{L^{2}(0,1)}+T_{n}(0^{+})\overline{u_{n}^{2}}(0^{+})+\langle
v_{n}^{2},u_{n}^{2}\rangle_{L^{2}(0,1)}\right)=o(1).$

Thanks to (5.2), (5.7) and (5.9), it’s clear that the second and the last
terms converge to zero. Furthermore, we have

$\beta_{n}^{\frac{3\gamma}{4}}T_{n}(0^{+})\overline{u_{n}^{2}}(0^{+})\leq
C\beta_{n}^{\frac{\gamma}{2}}\left(\|T_{n}\|_{L^{2}(0,1)}^{\frac{1}{2}}.\|u_{nx}^{2}\|_{L^{2}(0,1)}^{\frac{1}{2}}.\|T_{n}^{\prime}\|_{L^{2}(0,1)}^{\frac{1}{2}}.\|u_{n}^{2}\|_{L^{2}(0,1)}^{\frac{1}{2}}\right).$

From (5.5) we can see that
$\|\beta_{n}u_{n}^{2}+v_{n}^{2}\|_{L^{2}(0,1)}\sim\|T_{n}^{\prime}\|_{L^{2}(0,1)}$
which implies that

$\displaystyle\beta_{n}^{\frac{3\gamma}{4}}|T_{n}(0^{+})|.|\overline{u_{n}^{2}}(0^{+})|$
$\displaystyle\leq
C\beta_{n}^{\frac{3\gamma}{4}}\|T_{n}\|_{L^{2}(0,1)}^{\frac{1}{2}}.\|u_{nx}^{2}\|_{L^{2}(0,1)}^{\frac{1}{2}}\times$
$\displaystyle\left(\|\beta_{n}u_{n}^{2}\|_{L^{2}(0,1)}^{\frac{1}{2}}+\|v_{n}^{2}\|_{L^{2}(0,1)}^{\frac{1}{2}}+o(1)\right).\|u_{n}^{2}\|_{L^{2}(0,1)}^{\frac{1}{2}}$
$\displaystyle\leq
C\|\beta_{n}^{\frac{\gamma}{2}}T_{n}\|_{L^{2}(0,1)}^{\frac{1}{2}}.\|\beta_{n}^{\frac{\gamma}{2}}u_{nx}^{2}\|_{L^{2}(0,1)}^{\frac{1}{2}}\times$
$\displaystyle\left(\|\beta_{n}u_{n}^{2}\|_{L^{2}(0,1)}^{\frac{1}{2}}+\|v_{n}^{2}\|^{\frac{1}{2}}_{L^{2}(0,1)}\right)\|\beta_{n}^{\frac{\gamma}{2}}u_{n}^{2}\|_{L^{2}(0,1)}^{\frac{1}{2}}+o(1)$
(5.11)
$\displaystyle\leq\left(1+\beta_{n}^{\frac{1}{2}+\frac{\gamma}{4}}.\|u_{n}^{2}\|_{L^{2}(0,1)}\right)o(1).$

Combining (5.10) and (5), one follows

(5.12)
$\beta_{n}^{\frac{1}{2}+\frac{3\gamma}{8}}\|u_{n}^{2}\|_{L^{2}(0,1)}\,\longrightarrow
0.$

Moreover, multiplying (5.5) by $\beta_{n}^{-\frac{\gamma}{2}}(1-x)T_{n}$ and
integrating over the interval $(0,1)$ then by taking account of (5.9), an
integration by parts leads to

(5.13) $\displaystyle\mathrm{Re}\langle
i\beta_{n}^{\frac{1}{2}+\frac{3\gamma}{8}}u_{n}^{2},(1-x)\beta_{n}^{\frac{1}{2}-\frac{3\gamma}{8}+\frac{\gamma}{2}}T_{n}\rangle_{L^{2}(0,1)}+\frac{\beta_{n}^{\frac{\gamma}{2}}}{2}\left(|T_{n}(0^{+})|^{2}-\|T_{n}\|^{2}_{L^{2}(0,1)}\right)$
$\displaystyle+\beta_{n}^{\frac{\gamma}{2}}\mathrm{Re}\langle
v_{n}^{2},(1-x)T_{n}\rangle_{L^{2}(0,1)}=o(1).$

We suppose that $\displaystyle\gamma\geq\frac{4}{3}$. It’s clear from (5.2),
(5.9) and (5.12) that the first, the third and the last terms of (5.13)
converge to zero then one gets

(5.14) $\beta_{n}^{\frac{\gamma}{4}}.|T_{n}(0^{+})|\,\longrightarrow 0.$

Taking into account to (5.8) then the trace formula gives

(5.15) $\beta_{n}^{\frac{\gamma}{2}+1}.|u_{n}^{1}(0^{+})|\,\longrightarrow 0.$

Substituting (5.4) into (5.5) and taking the inner product with
$\beta_{n}^{3-\gamma}v_{n}^{1}$ in $L^{2}(0,1)$ then by integrating by parts
we have

(5.16) $i\beta_{n}^{4}\left\langle
u_{n}^{2},v_{n}^{1}\right\rangle_{L^{2}(0,1)}+\beta_{n}^{3}\left\langle
T_{n},v_{n}^{1}\right\rangle_{L^{2}(0,1)}+i\beta_{n}^{4}\|v_{n}^{1}\|_{L^{2}(0,1)}^{2}+\beta_{n}^{3}T_{n}(0^{+})\overline{v_{n}^{1}(0^{+})}=o(1).$

Taking $\displaystyle\gamma\geq 12$ and using (5.2), (5.9), (5.12) and (5.14)
we can see that the first, the second and the fourth terms of (5.16) converge
to zero, therefore

(5.17) $\beta_{n}^{2}.\|v_{n}^{1}\|_{L^{2}(0,1)}\,\longrightarrow 0.$

From (5.4) and (5.17) it follows

(5.18) $\beta_{n}\|v_{n}^{2}\|_{L^{2}(0,1)}\,\longrightarrow 0.$

Multiplying (5.6) with $\beta_{n}^{-\gamma}(1-x)\overline{v_{nx}^{1}}$ and
integrating over $(0,1)$ then by taking the real part we find

$\displaystyle\frac{c}{2}\left(|v_{nx}^{1}(0^{+})|^{2}-\|v_{nx}^{1}\|_{L^{2}(0,1)}^{2}\right)=\mathrm{Re}\left\langle
u_{n}^{2},(1-x)v_{nx}^{1}\right\rangle_{L^{2}(0,1)}$
$\displaystyle-\mathrm{Re}\langle
i\beta_{n}v_{n}^{2},(1-x)v_{nx}^{1}\rangle_{L^{2}(0,1)}+o(1).$

Using (5.2), (5.12) and (5.18) leads to

(5.19)
$|v_{nx}^{1}(0^{+})|^{2}-\|v_{nx}^{1}\|_{L^{2}(0,1)}^{2}\,\longrightarrow 0.$

We take the inner product of (5.6) with $\beta_{n}^{-\gamma}xv_{n}^{1}$ in
$L^{2}(0,1)$ then we have

$c\left(\int_{0}^{1}x|v_{nx}^{1}(x)|^{2}\,{\rm d}x+\left\langle
v_{nx}^{1},v_{n}^{1}\right\rangle_{L^{2}(0,1)}\right)=\left\langle
u_{n}^{2},xv_{n}^{1}\right\rangle_{L^{2}(0,1)}-i\beta_{n}\langle
v_{n}^{2},xv_{n}^{1}\rangle_{L^{2}(0,1)}+o(1).$

Using (5.2), (5.12) and (5.18) we deduce that

$\int_{0}^{1}x|v_{nx}^{1}(x)|^{2}\,{\rm d}x\,\longrightarrow 0.$

This implies in particular that for every $\varepsilon$ in $(0,1)$ we have

(5.20) $\|v_{nx}^{1}\|_{L^{2}(\varepsilon,1)}\,\longrightarrow 0\;\text{ as
}n\nearrow+\infty.$

Multiplying (5.6) with $\beta_{n}^{-\gamma}(1-x)\overline{v_{nx}^{1}}$ and
integrating over $(0,\varepsilon)$ then by taking the real part we find

$\displaystyle\frac{c}{2}\left(|v_{nx}^{1}(\varepsilon)|^{2}-\|v_{nx}^{1}\|_{L^{2}(\varepsilon,1)}^{2}\right)=\mathrm{Re}\left\langle
u_{n}^{2},(1-x)v_{nx}^{1}\right\rangle_{L^{2}(\varepsilon,1)}-\mathrm{Re}\langle
i\beta_{n}v_{n}^{2},(1-x)v_{nx}^{1}\rangle_{L^{2}(\varepsilon,1)}+o(1).$

Besides, from (5.2), (5.12), (5.18) and (5.20) we follow

$|v_{nx}^{1}(\varepsilon)|\,\longrightarrow 0\;\text{ as }n\nearrow+\infty.$

Then we deduce that

(5.21) $v_{nx}^{1}(x)\,\longrightarrow 0\;\text{ a.e. in }[0,1]\text{ as
}n\nearrow+\infty.$

Now, (5.2) and (5.21) allows the use of the dominated convergence theorem and
lead to

(5.22) $\|v_{nx}^{1}\|_{L^{2}(0,1)}\,\longrightarrow 0.$

Therefore, we obtain

(5.23) $|v_{n}^{1}(0^{+})|\,\longrightarrow 0.$

By combining (5.19) and (5.22) we find

(5.24) $|v_{nx}^{1}(0^{+})|\,\longrightarrow 0.$

Furthermore, taking the inner product of (5.4) with
$\beta_{n}^{1-\gamma}(1-x)v_{nx}^{1}$ and then considering the imaginary part
one gets

$\displaystyle\beta_{n}^{2}Re(v_{nx}^{1},(1-x)v_{n}^{1})-Im\beta_{n}(v_{n}^{2},(1-x)v_{nx}^{1})=o(1)$
$\displaystyle=\frac{1}{2}(\beta_{n}^{2}|v_{n}^{1}(0^{+})|^{2}-\beta_{n}^{2}\|v_{n}^{1}\|^{2})-\beta_{n}Im\langle
v_{n}^{2},(1-x)v_{nx}^{1}\rangle$

Adding to this (5.23), (5.17) and (5.18) we can deduce that :

(5.25) $\beta_{n}|v_{n}^{1}(0^{+})|\longrightarrow 0$

Thanks to (5.14), (5.15), (5.23) and (5.24) one gets

(5.26)
$\displaystyle\beta_{n}^{\frac{\gamma}{2}+1}.u_{n}^{1}(0^{-})\,\longrightarrow
0,$ (5.27)
$\displaystyle\beta^{\frac{\gamma}{4}}.u_{nx}^{1}(0^{-})\,\longrightarrow 0,$
(5.28) $\displaystyle\beta_{n}v_{n}^{1}(0^{-})\,\longrightarrow 0,$ (5.29)
$\displaystyle v_{nx}^{1}(0^{-})\,\longrightarrow 0.$

Next, inserting (5.3) into (5.5) and inserting (5.4) into (5.6) and consider
both equations in the interval $(0,1)$, leads to

(5.30)
$-\beta_{n}^{2}u_{n}^{1}-u_{nxx}^{1}+v_{n}^{2}=\beta_{n}^{-\gamma}f_{n}^{2}+i\beta_{n}^{1-\gamma}f_{n}^{1},$

and

(5.31)
$-\beta_{n}^{2}v_{n}^{1}-cv_{nxx}^{1}-u_{n}^{2}=\beta_{n}^{-\gamma}g_{n}^{2}+i\beta_{n}^{1-\gamma}g_{n}^{1}.$

A straightforward calculation shows that the real part of the inner product of
(5.30) with $(x+1).u_{nx}^{1}$ and that the real part of the inner of (5.31)
with $(x+1).v_{nx}^{1}$ leads to

(5.32)
$\displaystyle\frac{1}{2}\int_{-1}^{0}\left(|\beta_{n}u_{n}^{1}|^{2}+|u_{nx}^{1}|^{2}\right)\,{\rm
d}x=\frac{1}{2}\left(|u_{nx}^{1}(0^{-})|^{2}+\beta_{n}^{2}|u_{n}^{1}(0^{-})|^{2}\right)$
$\displaystyle-\mathrm{Re}\langle
v_{n}^{2},(x+1)u_{nx}^{1}\rangle_{L^{2}(-1,0)}+o(1),$

and

(5.33)
$\displaystyle\frac{1}{2}\int_{-1}^{0}\left(|\beta_{n}v_{n}^{1}|^{2}+c|v_{nx}^{1}|^{2}\right)\,{\rm
d}x=\frac{1}{2}\left(c|v_{nx}^{1}(0^{-})|^{2}+\beta_{n}^{2}|v_{n}^{1}(0^{-})|^{2}\right)$
$\displaystyle+\mathrm{Re}\langle
u_{n}^{2},(x+1)v_{nx}^{1}\rangle_{L^{2}(-1,0)}+o(1).$

Where we have used (5.2)-(5.6). In another hand, from (5.2), (5.12), (5.18)
and (5.26)-(5.29) we get

(5.34)
$\int_{-1}^{0}\left(|\beta_{n}u_{n}^{1}|^{2}+|u_{nx}^{1}|^{2}\right)\,{\rm
d}x\,\longrightarrow 0,$

and

(5.35)
$\int_{-1}^{0}\left(|\beta_{n}v_{n}^{1}|^{2}+c|v_{nx}^{1}|^{2}\right)\,{\rm
d}x\,\longrightarrow 0.$

Now by summing (5.8) (5.12), (5.17), (5.18), (5.34) and (5.35) we can see that

(5.36) $\|Y_{n}\|_{\mathcal{H}}\,\longrightarrow 0.$

This contradicts (5.2) and so (5.1) holds true with $\gamma\geq 12$. This
completes the proof. ∎

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