text:
\section{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:
\\
\hspace*{0.2cm} 1) The linear viscous damping or "external damping", it does mostly model an external frictional force, such that the auto-mobile shock absorber.
\\
\hspace*{0.2cm} 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 \cite{liu-rao2,tebou}). Nonetheless, when the damping is  arbitrary localized with singular coefficient, it's not the case any more (see \cite{AHR2,liu-rao1}). Actually, in one-dimensional case we can consider the following problem
\begin{equation} \label{intro}
\left\lbrace\begin{array}{ll}
u_{tt}-[u_{x}(x,t)+b(x)u_{xt}]_x=0 &-1<x<1,\; t\geq0,
\\
u(t,-1)=u(t,1)=0& t\geq 0,
\\
u(0,x)=u_0(x) , u_t(0,x)=u_1(x) &-1\le x \le 1,
\end{array} \right.
 \end{equation}
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 \cite{liu-zhang} 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 \cite{liu-rao1} showed that system \eqref{intro} is polynomially stable with an order equal to $2$ where few years ago Liu and Liu \cite{liu-liu} 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 \cite{F.P} the coupled waves with partial frictional damping
\begin{equation*}
\left\lbrace \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.
\end{equation*}
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 \cite{AHR2,hassine1,hassine2,hassine4,hassine3,ALS} (see also \cite{ammari-niciase}) 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
\medskip
\begin{equation}\label{sc}
\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.
\end{equation}
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 \eqref{sc} at an instant $t$ is given by
$$
E(t)= \frac{1}{2} \int_{-1}^{1} \left(|u_t (x,t)|^2+|v_t(x,t)|^2 +|u_{x}(x,t)|^2 + c \, |v_x(x,t)|^2 \right)\,\ud x, \forall \, t > 0.
$$
Multiplying the first equation of \eqref{sc} by $\bar{u}$, the second by $\bar{v}$, integrating over (-1,1) and then taking the real part leads to
\begin{equation*}
E^\prime (t) = -\int_{-1}^1 a(x) \, |u_{xt}(x,t)|^2 \,\ud x,\;\forall\,t > 0.
\end{equation*}
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 \ref{CWKVWP}, we prove that system \eqref{sc} is well-posed. In section \ref{CWKVSS}, we show that the energy of the system is the strong stability. In section \ref{CWKVLES}, we prove the lack of exponential stability. In section \ref{CWKVPS}, we prove a polynomial stability decay of the energy.
\section{Well-posedness}\label{CWKVWP}
In this section, we discuss the well-posesness of the problem (\ref{sc}) using the semigroup theory.
\medskip
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
$$
\left\langle U_1,U_2\right\rangle _{\mathcal{H}}= \left\langle
u^1_x,u^2_x\right\rangle _{L^{2}(-1,1)} + \left\langle
\sqrt{c} v^1_x,\sqrt{c} v^2_x \right\rangle _{L^{2}(-1,1)}
+ \left\langle w^1,w^2\right\rangle _{L^{2}(-1,1)}
+\left\langle
z^1,z^2\right\rangle_{L^{2}(-1,1)}.
$$

\medskip

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 (\ref{sc}) as a first order differential equation as follow
\begin{equation}\label{damped}
\dot{y}(t) =  {\mathcal A} y(t), \qquad y(0) = y_0,
\end{equation}
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
\begin{align*}
(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,
\\
v^1 \in H^2(-1,1) \cap H^1_0 (-1,1), \left(u^1_x + a u^2_x \right)_x \in L^2(-1,1) \big\}.
\end{align*}
For the well-posedness of system (\ref{damped}) we have the following proposition:
\begin{proposition}\label{exist}
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 (\ref{damped}). Moreover, if $y_0 \in \mathcal{D}(\mathcal{A})$, then
$$
y = (u,v,u_t,v_t) \in C([0,\,+\infty),\, \mathcal{D}(\mathcal{A}))\cap C^{1}([0,\,+\infty),\, \mathcal{H}).
$$
\end{proposition}
\begin{proof}
By Lumer-Phillips' theorem (see \cite{Pazy}), it suffices to show that $\mathcal{A}$ is dissipative and  maximal.
\medskip

(1) We first prove that $\mathcal{A}$ is dissipative. Take $Z = (u,v,w,z) \in \mathcal{D}(\mathcal{A})$.
Then
\begin{align*}
\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)}
\\
+ \left\langle
c v_{xx} + w,z\right\rangle_{L^{2}(-1,1)}.
\end{align*}
By integration by parts and using the boundary conditions, it holds:

\begin{equation}\label{res1}
(\mathcal{A} Z,Z)_{\mathcal{H}} = - \left\langle a w_{x},w_{x}\right\rangle_{L^2(-1,1)}= - \int_{-1}^{1} a|w_x|^2 \, \ud x \leq 0.
\end{equation}
This shows that $\mathcal{A}$ is the dissipative.

\medskip
(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
\begin{equation}\label{WPwave}
\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.
\end{equation}
It is well known that by Lax-Milgram's theorem the system \eqref{WPwave} 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 (\ref{WPwave}) by $\overline{u}, \overline{v}$ respectively and integrating over $(-1,1)$ and using Poincar\'e 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 \cite[Theorem 1.4.6]{Pazy}. Then thanks to Lumer-Phillips Theorem (see \cite[Theorem 1.4.3]{Pazy}), the operator $\mathcal{A}$ generates a $C_{0}$-semigroup of contractions on the Hilbert $\mathcal{H}$ denoted by $(e^{t \mathcal{A}})_{t \geq0}$.

\medskip
\end{proof}
\section{Strong stability}\label{CWKVSS}
\begin{thm} \label{th1}
The semigroup $(e^{t \mathcal{A}})_{t \geq 0}$ is strongly stable in the energy space $\mathcal{H}$ i.e.,
$$
\lim_{t\to+\infty}\|e^{t \mathcal{A}}y_{0}\|=0,\qquad \forall\,y_{0}\in\mathcal{D}(\mathcal{A}).
$$
\end{thm}

\begin{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 \cite{liu-rao2}) 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.
\medskip
\subparagraph*{Step 1} Let $(u,v,w,z)\in\mathcal{D}(\mathcal{A})$ such that
\begin{equation}\label{ICwave}
\mathcal{A}(u,v,w,z) = i\mu(u,v,w,z).
\end{equation}
or equivalently,
\begin{equation}\label{waveIC}
\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.
\end{equation}
Then taking the real part of the scalar product
of \eqref{ICwave} 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
\begin{equation*}
w_{x} =0 \qquad \text{ in }\,(0,1).
\end{equation*}
This implies that from the first equation \eqref{waveIC} that
\begin{equation*}
u_{x} =0 \qquad \text{ in }\,(0,1),
\end{equation*}
which means that $u$ is a constant in $(0,1)$ and since $u(1)=0$ we obtain that
\begin{equation*}
u=w=0 \qquad \text{ in }\,(0,1),
\end{equation*}
Hence, from the third and the second equation of \eqref{waveIC} one gets
\begin{equation}\label{DCwave}
u=w=v=z=0 \qquad \text{ in }\,(0,1),
\end{equation}
Using \eqref{DCwave} then \eqref{waveIC} is reduced to the following problem
\begin{equation}\label{waveIC1}
\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.
\end{equation}
\medskip
Let $y=(u,v,u_x,v_x)$ and $y_x=(u_x,v_x,u_{xx},v_{xx})$ then \eqref{waveIC1} is recast as follow
\begin{equation} \label{Cwave3}
\left\lbrace \begin{array}{cc}
y_x=A_{\mu}y & \text{in} (-1,0)
\\
Y(0)=0
\end{array} \right.
\end{equation}
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}

\end{pmatrix}.
$$
Since $A_{\mu} $ is a bounded operator then the unique solution of \eqref{Cwave3} is $y=0$ therefore $u=v=0$ in $(-1,0)$. Moreover, from the fist and the second equation of \eqref{waveIC1} we have $w=z=0$ in $(-1,1)$. Combining all this with\eqref{DCwave}, we deduce that $u=v=w=z=0$ in $(-1,1)$. This conclude the fist part of this proof.
\medskip
\subparagraph*{Step 2} Now given $(f,g)\in\mathcal{H}$, we solve the equation
$$
(i\mu I-\mathcal{A})(u,v,w,z)=(f,g,f_1,g_1).
$$
Or equivalently,
\begin{equation}\label{Swave}
\left\{\begin{array}{llcc}
w=i\mu u-f \\
z =i \mu v - g \\
\mu^{2}u + u_{xx} + i\mu\,(a u_x)_x - i \mu v = (a f_x)_x -i\mu f - f_1 - g=F
\\
\mu^2 v + c \, v_{xx} + i \mu u = - \mu g + f - g_1=G.
\end{array}\right.
\end{equation}
Let's define the operator
\begin{align*}
\begin{array}{rrrl}
 A :& (H_0^{1}(-1,1))^2 &\longrightarrow &(H^{-1}(-1,1))^2
\\
&(u,v)&\longmapsto&(-u_{xx}-i\mu (au_x)_x+i\mu v , -cv_{xx} -i \mu u).
\end{array}
\end{align*}
First we are going to show that $A$ is an  isomorphism. For this purpose we consider the two operator $\tilde{A}$ and $C$ such that
$$
\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}
$$
and $C$ such that $A = C+ \tilde{A}$.
\medskip
It's easy to show that $\tilde{ A}$ is an isomorphism, then we could rewrite $A = \tilde{A} (Id - \tilde{ A}^{-1} (-C))$.
\medskip
To begin with, thanks to  the compact embedding
$$
 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.
\medskip

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
\begin{equation}\label{SCwave11}
\left\{\begin{array}{ll}
u_{xx} + i\mu\,(a u_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.
\end{equation}
Multiplying the first equation of \eqref{SCwave11} by $\bar{u}$ and the conjugate of the second by $v$, after integration over $(-1,1)$, it follows
$$
- \int _{-1}^{1} \vert u_x \vert ^2 \ud x +c \int _{-1}^{1} \vert v_x \vert ^2 \ud x -i \mu \int _{-1}^{1} a \vert u_x \vert ^2 \ud x =0
$$
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 \eqref{SCwave11} we obtain $v=0$ in $(0,1)$, which implies that \eqref{SCwave11} that
\begin{equation}\label{Swave111}
\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.
\end{equation}
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:
\begin{equation*}
\left\lbrace \begin{array}{ll}
y_x=D_{\mu}y&\text{in }(-1,0)
\\
y(0)=0,&
\end{array} \right.
\end{equation*}
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}

\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).

\medskip

Now, rewriting the third and the fourth lines of \eqref{Swave} one gets
\begin{equation*}
(u,v)-\mu ^2 A^{-1} (u,v)=A^{-1}(F,G).
\end{equation*}
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:
\begin{equation}\label{sWAVE10}
\left\{\begin{array}{ll}
\mu^{2}u + u_{xx} + i\mu\,(a u_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.
\end{equation}
Furthermore, multiplying the first equation of \eqref{sWAVE10} by $ \bar{u}$ and the conjugate of the second by $v$, after integration over $(-1,1)$ and taking the imaginary part, we deduce that
\begin{equation*}
\int _{-1}^{1} a \vert u_x \vert ^2 \ud x =d \int _{0}^{1} \vert u_x \vert ^2 \ud x =0.
\end{equation*}
So, we get the same system as in the first step (see \eqref{waveIC}). Thus, $\ker ( I-\mu^2 A^{-1})= \lbrace 0_{(H^{-1}(-1,1))^2} \rbrace$.
\\
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 \eqref{Swave} have a unique solution in $ H_0^{1}(-1,1)^2$.
This completes the proof.
\end{proof}
\section{Lack of exponential stability}\label{CWKVLES}
\medskip
Now, we prove the lack of  exponential stability given by the following theorem
\begin{thm}\label{th3}
The semigroup $(e^{t\mathcal{A}})_{t \geq 0}$, is not exponentially stable in the energy space provided that $c>1$ and that
\begin{equation}\label{thetaAssum}
\sin(2\sqrt{c}n\pi)\neq O(n^{-\frac{1}{2}}),
\end{equation}
\end{thm}
Noting that the assumption $c>1$ is made here just to make the calculation readable. The second assumption \eqref{thetaAssum} can be fulfilled for instance by taking $c$ such that $2\sqrt{c}$ is an integer number. To prove \eqref{th3} we mainly use the following theorem
\begin{thm}(see \cite{huang1, pruss}) \label{3.4}
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
$$
\Vert e^{t\mathcal{B}}\Vert_{\mathcal{L}(\mathcal{H})}  \le M e^{-at} , \forall t \geq 0
$$
if and only if
$$\limsup_{\omega \in \mathbb{R}, \, |\omega| \rightarrow \infty} \Vert (i\omega I-\mathcal{B})^{-1}\Vert_{\mathcal{L}(\mathcal{H})} < \infty.$$
 \end{thm}
Now, based on the Theorem \ref{3.4} we prove the Theorem \ref{th3}.
\begin{proof}
Our main objective is to show that:
\begin{equation} \label{Objective}
\Vert (\lambda I-\mathcal{A})^{-1} \Vert_{\mathcal{L}(\mathcal{H})} \text{ is unbounded on the imaginary axis}.
\end{equation}
For $n \in \mathbb{N}$ large enough let  $\lambda= \lambda_n = i\omega_{n}$, where
\begin{equation}\label{omegan}
\omega_n =  \sqrt{\frac{8c(c+1)n^2\pi^2+2c +\sqrt{\Delta}}{4c}}\qquad\text{with}\qquad\Delta=(8c(c-1)\pi^2n^2)^2+32(c+1)(c\pi n)^2+4c^2.
\end{equation}
It's clear that  $\omega_n \longrightarrow +\infty$ and in particular we have
\begin{equation}\label{omeganBH}
\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)
\end{equation}
and
\begin{equation}\label{BHomega}
\frac{1}{\omega_{n}}=\frac{1}{2n\pi\sqrt{c}}-\frac{1}{16\sqrt{c}(c-1)(\pi n)^{3}}+o(n^{-4}).
\end{equation}

Define $(F_1 , G_1, F_2 ,G_2)$ $\in (H_0^1(0,1))^2 \times (L^2(0,1) )^2$, such that
\begin{align*}
F_1= F_1(x,n)= 0 \qquad \forall\,x \in (-1,1),
\end{align*}
 \begin{align*}
 G_1=G_1(x,n)= \left\lbrace \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.
\end{align*}
\begin{align*}
F_2=F_2(x,n)=0 \qquad \forall\,x \in (-1,1),
\end{align*}
\begin{align*}
 G_2=G_2(x,n)= \left\lbrace \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.
\end{align*}
A straight forward calculation leads to
\begin{equation}\label{fg}
\Vert(F_1,G_1,F_2,G_2)\Vert_{\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.
\end{equation}
Our goal is to prove that $\displaystyle\lim_{|\lambda| \rightarrow \infty} \Vert(\lambda I - \mathcal{A})^{-1} \Vert_{\mathcal{L}(\mathcal{H})} = \infty$. That's why, we  solve the resolvent equation
\begin{equation}\label{resol}
(\lambda I-\mathcal{A})(u^1,v^1,u^2,v^2)=(F_1,G_1,F_2,G_2).
\end{equation}
\textbf{Step 1.} For all  $x \in (0,1)$ , we have
\begin{align} \label{eq01}
\left\lbrace \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.
\end{align}
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
\begin{equation*}
\displaystyle r=\sqrt{a^2+b^2},\quad \cos(\phi)= \frac{a}{r}  \quad\text{ and }\quad   \sin(\phi)= \frac{b}{r},
\end{equation*}
with
\begin{align*}
\displaystyle a&=-(1-c)^2+d^2\omega^{2}-\frac{4c}{\omega_n^{2}}
\\
\displaystyle b&=-2d\left((1-c)\omega_n+\frac{2c}{\omega_n}\right).
\end{align*}
It is important to note that
\begin{equation*}
\sqrt{a}=d\omega-\frac{(c-1)^2}{2d}\omega^{-1}-\frac{(c-1)^{4}+16cd^{2}}{8d^{3}}\omega^{-3}+o(\omega^{-3}),
\end{equation*}
\begin{equation*}
\frac{b}{a}=\frac{2(c-1)}{d}+\frac{2(c-1)-4cd}{d}\omega^{-3}+o(\omega^{-4})
\end{equation*}
and
\begin{align*}
i\frac{\sqrt{r}}{d}\mathrm{e}^{i\frac{\phi}{2}}&=\lambda-\frac{c-1}{d}-\frac{(c-1)^{3}-d^{2}(c-1)+2cd^{2}}{d^{3}}\lambda^{-2}
\\
&+\frac{d^{2}(c-1)^{2}-(c-1)^{4}-2cd^{3}(c-1)-2cd^{2}}{d^{4}}\lambda^{-3}+o(\omega^{-3}).
\end{align*}
Then we obtain
\begin{align}\label{eta+}
\eta_{+}&=-\lambda+\frac{c}{d}-\frac{c}{d^{2}}\lambda^{-1}+\frac{(c-1)^{3}+d^{2}(c+1)+2c}{2d^{3}}\lambda^{-2}
\\
&+\frac{(c-1)^{4}-(c-1)^{3}-d^{2}(c-1)(c-2)-2c}{2d^{4}}+o(\omega^{-3})\nonumber
\end{align}
and
\begin{align}\label{eta-}
\eta_{-}&=-\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})
\end{align}
\medskip
A straightforward calculation leads to
\begin{align} \label{u eta v}
(u^1+\eta_+ v^1)_{xx}=  ( \beta_+)^2(u^1+\eta_+ v^1)
\\
(u^1+\eta_- v^1)_{xx}=  ( \beta_-)^2(u^1+\eta_- v^1),
\end{align}
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
\begin{equation*}
\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}}
\end{equation*}
with
\begin{align*}
\displaystyle a_{\pm}&=-(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 b_{\pm}&=cd\omega_n \pm \sqrt{r}\left(-\cos\left(\frac{\phi}{2}\right)-d\omega_n\sin\left(\frac{\phi}{2}\right)\right).
\end{align*}
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
\begin{equation}\label{beta+}
\beta_{+}=\frac{\lambda}{\sqrt{c}}-\frac{(c-1)(c-2)}{8\sqrt{2}d^{2}}+o(\omega^{-1}),
\end{equation}
and
\begin{equation}\label{beta+2}
\beta_{+}^{2}=\frac{\lambda^{2}}{c}+o(1).
\end{equation}
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
\begin{equation}\label{beta-}
\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}),
\end{equation}
and
\begin{align}\label{beta-2}
\beta_{-}^{2}&=\frac{\lambda}{d}-\frac{1}{d^{2}}+\frac{(c-1)^{3}+d(c-1)+2c}{2cd^{3}}\lambda^{-1}
\\
&-\frac{(c-1)^{3}(2-c)+d(c-1)(c-2-2cd)+2c}{2cd^{4}}\lambda^{-2}+o(\omega^{-2})\nonumber
\end{align}
Next, from  \eqref{u eta v}, we get
\begin{equation*}
(u^1+\eta_+v^1)= c_1 e^{x\beta_+}+c_2 e^{-x \beta_+}
\end{equation*}
and
\begin{equation*}
(u^1+\eta_- v^1)= c_3 e^{x \beta_-}+c_4 e^{-x \beta_-}.
\end{equation*}
Recalling that $u ^1(1)=v^1(1)=0$ we can rewrite the last two equations as follow
\begin{equation} \label{eta-2}
(u^1+\eta_+ v^1)= c_1( e^{x\beta_+}- e^{(2-x)\beta_+}),
\end{equation}
\begin{equation} \label{eta+2}
(u^1+\eta_- v^1)= c_3(e^{x\beta_-}- e^{(2-x)\beta_-}).
\end{equation}
Hence by combining \eqref{eta-2} and \eqref{eta+2} we obtain
\begin{equation}\label{solu1+}
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),
\end{equation}
and
\begin{equation}\label{solv1+}
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).
\end{equation}
\textbf{Step 2.} For all $x \in (-1,0)$ we have
\begin{align}\label{eq(-1.0)}
\left\lbrace \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.
\end{align}
 \medskip

Following to the third and the fourth equation of \eqref{eq01} and of \eqref{eq(-1.0)} we can deduce, thanks to the regularity of the stats, that
\begin{align}
(1+ \lambda d)u_{x}^{1}(0^+)&=u_{x}^{1}(0^-) \label{cond1u0},
\\
v_{x}^{1}(0^+)&=v_{x}^{1}(0^-)\label{cond1v0}.
\end{align}
and
\begin{align}
(1+ \lambda d)u_{xx}^{1}(0^+)&=u_{xx}^{1}(0^-),\label{cond2u0}
\\
v_{xx}^{1}(0^+)&=v_{xx}^{1}(0^-)\label{cond2v0}.
\end{align}
We denote by
\begin{equation}\label{alpha+}
\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}),
\end{equation}
and
\begin{equation}\label{alpha-}
\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})
\end{equation}
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 \eqref{omegan} one get
$$
\displaystyle\mu_{\pm}^2= \frac{\lambda}{\lambda-\frac{\alpha_{\pm}}{c}}.
$$
Besides, we have
\begin{equation}\label{mu+}
\mu_{+}=\sqrt{c}\left(1-\frac{c}{2(c-1)}\lambda^{-2}+o(\omega^{-2})\right),
\end{equation}
\begin{equation}\label{mu-}
\mu_{-}=1+\frac{\lambda^{-2}}{2(c-1)}+o(\omega^{-2}),
\end{equation}
and
\begin{equation}\label{mu+/mu-}
\frac{\mu_{+}}{\mu_{-}}=\sqrt{c}\left(1-\frac{c+1}{2(c-1)}\lambda^{-2}+o(\omega^{-2})\right).
\end{equation}
We set
\begin{align}
\omega^+_1(x)&= (u^2+\alpha_+v^2+  \mu_+(u^1_x + \alpha_+v^1_x),\label{omega1+}
\\
\omega ^-_1(x)&=(u^2+\alpha_+v^2-\mu_+(u^1_x + \alpha_+v^1_x)),\label{omega1-}
\\
\omega ^+_2(x)&= (u^2+\alpha_ -v^2 + \mu_-(u^1_x + \alpha_- v^1_x)),\label{omega2+}
\\
\omega ^-_2(x)&= (u^2+\alpha_- v ^2-\mu_-(u^1_x + \alpha_- v^1_x)).\label{omega2-}
\end{align}

Now, define $Y=(\omega^+_1 , \omega^-_1,\omega^+_2,\omega_2^-)^t$ and $Z=(g_{1x},g_2)^t$. Then we have
\begin{equation}\label{w}
Y_x = AY + BZ
\end{equation}
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
\begin{equation}\label{LES1}
\omega_1^+(-1)=-\omega_1^-(-1) \; \hbox{and} \; \omega_2^+(-1)=-\omega_2^-(-1),
\end{equation}
Taking into account of \eqref{LES1} then the solution of \eqref{w} is written as follow
\begin{align}
\omega_1^+(x)&=\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],\label{sol1+}
\\
\omega_1^-(x)&=-\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],\label{sol1-}
\\
\omega_2^+(x)&=\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],\label{sol2+}
\\
\omega_2^-(x)&=-\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].\label{sol2-}
\end{align}
Taking the trace of $\omega_1^+$ and $\omega_1^-$ in \eqref{sol1+}-\eqref{sol1-} and  in \eqref{omega1+}-\eqref{omega1-} on the boundary $0$ and using the continuity of the states $u_{2}$ and $v_{2}$ we obtain
 \begin{align*}
(\omega_1^+ + \omega_1^-)(0^-)&=\alpha_+\left(\frac{\mu_+}{\mu_-}-1\right)= 2u^2(0^-)+2\alpha_+v^2(0^{-})\nonumber
\\
&= 2\lambda( u^1(0^-) + \alpha_+ v^1(0^-))=2\lambda( u^1(0^+) + \alpha_+ v^1(0^+))\nonumber
\\
&=\frac{2\lambda}{\eta_+-\eta_-}\left(c_1(1-e^{2\beta_+})(\alpha_+-\eta_-)+ c_3(1-e^{2 \beta_-})(\eta_+-\alpha_+)\right),
\end{align*}
where we have used the the expressions of $u^{1}$ and $v^{1}$ in \eqref{solu1+} and \eqref{solv1+}.
 \medskip
This implies that
\begin{equation}\label{c3=f(c1)}
c_3=\frac{1-e^{2\beta_+}}{1-e^{2\beta_-}}A_nc_1+\frac{B_n}{1-e^{2\beta_{-}}}
\end{equation}
where
\begin{align} \label{An}
A_n= \frac{\eta_{-}-\alpha_{+}}{\eta_{+}-\alpha_{+}}&=\frac{c-1}{c}\left(1+\frac{\lambda^{-1}}{d}-\frac{\lambda^{-2}}{c-1}+o(\omega^{-2})\right)\nonumber
\\
&=\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),
\end{align}
and
\begin{align}\label{Bn}
B_n&=\frac{\alpha_{+}(\eta_{+}-\eta_{-})\left(\frac{\mu_{+}}{\mu_{-}}-1\right)}{2\lambda(\eta_{+}-\alpha_{+})}\nonumber
\\
&=\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)\nonumber
\\
&=\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).
\end{align}
where we used here \eqref{eta+}, \eqref{eta-}, \eqref{alpha+}, \eqref{alpha-}, \eqref{mu+/mu-} and \eqref{omegan}.
\\
Using \eqref{sol1+}-\eqref{sol1-} and \eqref{cond2u0}-\eqref{cond2v0}, one gets
\begin{align*}
(\omega_1^+-\omega_1^-)'(0^-)&=2i n \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^+)
\\
&=\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_{-}}.
\end{align*}
Then we obtain
\begin{equation}\label{c1=f(c3)}
c_{1}=\frac{1-e^{2\beta_-}}{1-e^{2\beta_+}}A_{n}'c_{3}+\frac{B_{n}'}{1-e^{2\beta_{+}}}
\end{equation}
where
\begin{align}\label{An'}
A_{n}'&=\frac{\beta_{-}^{2}(\alpha_{+}-(1+\lambda d)\eta_{+})}{\beta_{+}^{2}(\alpha_{+}-(1+\lambda d)\eta_{-})}\nonumber
\\
&=\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)\nonumber
\\
&=\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),
\end{align}
and
\begin{align}\label{Bn'}
B_{n}'&=\frac{in\pi\alpha_{+}(\eta_{+}-\eta_{-})\left(\frac{\mu_{+}}{\mu_{-}}-1\right)}{\mu_{+}\beta_{+}^{2}(\alpha_{+}-(1+\lambda d)\eta_{-})}\nonumber
\\
&=\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})\nonumber
\\
&=\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).
\end{align}
where we used here \eqref{eta+}, \eqref{eta-}, \eqref{beta+2}, \eqref{beta-2}, \eqref{alpha+}, \eqref{alpha-}, \eqref{mu+}, \eqref{mu+/mu-} and \eqref{omegan}.
\\
Combining \eqref{c3=f(c1)} and \eqref{c1=f(c3)} then we find that
\begin{equation}\label{c1}
c_{1}=\frac{1}{1-e^{2\beta_{+}}}\times\frac{A_{n}'B_{n}+B_{n}'}{1-A_{n}A_{n}'}=\frac{c_{1}'}{1-e^{2\beta_{+}}}
\end{equation}
and
\begin{equation}\label{c3}
c_{3}=\frac{1}{1-e^{2\beta_{-}}}\times\frac{A_{n}B_{n}'+B_{n}}{1-A_{n}A_{n}'}=\frac{c_{3}'}{1-e^{2\beta_{-}}},
\end{equation}
where following to \eqref{An}, \eqref{Bn}, \eqref{An'} and \eqref{Bn'} we have
\begin{equation}\label{c'}
c_{1}'=O(1)\qquad \text{ and }\qquad c_{3}'=O(1).
\end{equation}
\medskip
In another hand, by denoting $\theta=-i\mu_{-}\left(\lambda-\frac{\alpha_{-}}{c}\right)$ and by using the same argument as previously, one gets
\begin{align*}
(\omega_2^++\omega_2^-)(0^-)&= 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^+)
\\
&=\frac{2\lambda}{\eta_+-\eta_-}\left(c_{1}'(\alpha_{-}-\eta_{-})+ c_{3}'(\eta_+-\alpha_-)\right).
\end{align*}
It's clear that $\theta \neq 0[\pi]$ then we can write
\begin{align}\label{omega2}
\omega_2^+(-1)&= \frac{\lambda}{i\sin(\theta)(\eta_{+}-\eta_{-})}\left[c_{1}'(\alpha_--\eta_-)+ c_{3}'(\eta_+-\alpha_-)\right]-\frac{\alpha_{-}}{2in\pi+i\theta}.
\end{align}
Noting that from \eqref{omegan}, \eqref{omeganBH}, \eqref{alpha-} and \eqref{mu-} we have
\begin{equation}\label{theta}
\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)
\end{equation}
Then from \eqref{omeganBH}, \eqref{eta+}, \eqref{eta-}, \eqref{alpha-}, \eqref{mu-} and \eqref{theta} we deduce that
\begin{equation}\label{omega2BH}
\omega_{2}^{+}(-1)\sim \frac{2\pi n\sqrt{c}\,c_{3}'}{\sin(\theta)
\end{equation}
\medskip
Using \eqref{cond1u0}-\eqref{cond1v0}, \eqref{omega1+}-\eqref{omega1-} and \eqref{sol1+}-\eqref{sol1-} we get
\begin{align}\label{omega1}
\omega_{1}^{+}(-1)&=\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^+)\nonumber
\\
&=\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].
\end{align}
Then from  \eqref{omeganBH}, \eqref{eta+}, \eqref{eta-}, \eqref{beta+}, \eqref{beta-}, \eqref{alpha-} and \eqref{mu+} we deduce that
\begin{equation}\label{omega1BH}
\omega_{1}^{+}(-1)\sim c_{3}'\sqrt{\frac{c}{d}}\mathrm{e}^{-i\frac{\pi}{4}}(2\pi\sqrt{c}n)^{\frac{3}{2}}.
\end{equation}
Next, for all $x \in (-1,0)$ we have
\begin{align}\label{v1x}
v_x^{1}(x)&=\frac{1}{2 \mu_- \mu_+ (\alpha_+-\alpha_-)}\left[\alpha_{-}(\omega_1^+(x)-\omega_1^-(x))-\alpha_{+}(\omega_2^+(x)-\omega_2^-(x))\right]\nonumber
\\
&=\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)
\\
&-\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],\nonumber
\end{align}
where we have used \eqref{omega1+}-\eqref{omega2-} and \eqref{sol1+}-\eqref{sol2-}. Thus further leads to
\begin{align}\label{v1xN}
\|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}}
\\
-\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}}.\nonumber
\end{align}
Since,
$$
\sin(\theta)\neq O(n^{-\frac{1}{2}}),
$$
as $n$ goes to the infinity (by \eqref{theta} assumption \eqref{thetaAssum}) then by using \eqref{omegan}, \eqref{mu+}, \eqref{mu-}, \eqref{alpha+}, \eqref{alpha-}, \eqref{omega2BH} and \eqref{omega1BH} we can show that the second and the fourth terms of the right hand side of \eqref{v1xN} are bounded while the sum of the fist and the third terms  tends to the infinity as $n$ goes to $+\infty$, therefore we obtain
\begin{equation}\label{vx1BH}
\|v_{x}^{1}\|_{L^{2}(-1,0)}^{2}\,\,\quad\text{as}\quad n\nearrow+\infty.
\end{equation}
Last but not least, we have
 \begin{equation} \label{fin}
 \Vert (i \omega_n I- \mathcal{A})^{-1}(F_1,G_1,F_2,G_2)\Vert_{\mathcal{H}}=\Vert (u^1,v^1,u^2,v^2)\Vert_{\mathcal{H}}^2\geq
\int_{-1}^0 \vert v_{x}^{1}(x) \vert^2 \ud x \longrightarrow+\infty,\quad\text{as}\quad n\nearrow+\infty.
\end{equation}
Finally we conclude, using \eqref{fin} and \eqref{fg} that
$$
\limsup_{\omega \in \mathbb{R}, |\omega| \rightarrow \infty}\Vert ( i\omega I- \mathcal{A})^{-1}\Vert_{\mathcal{L}(\mathcal{H})} =+\infty.
$$
So, $e^{t\mathcal{A}}$ is not exponentially stable in the energy space. This completes the proof.
\end{proof}
\section{Polynomial stabilization}\label{CWKVPS}
\medskip
This subsection aims to prove the polynomial stability given by the following theorem:
\begin{thm} \label{th2}
The semigroup of contraction $(e^{T\mathcal{A}})_{t\geq 0}$ is polynomially stable of order $\displaystyle\frac{1}{12}$.
\end{thm}
 Our method is based on the Borichev and Tomilov result given by the following:
\begin{thm}\label{SMKV19}\cite[Theorem 2.4]{BT}
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 $i\R\subset\sigma(\mathcal{B})$ then $e^{t\mathcal{B}} $ is polynomially stable with order $\frac{1}{\gamma}, \gamma > 0$ i.e. there exists $C>0$ such that
$$
\|e^{t\mathcal{B}}U_{0}\|_{\mathcal{X}}\leq\frac{C}{(1+t)^{\frac{1}{\gamma}}}\|U_{0}\|_{\mathcal{D}(\mathcal{B})},\qquad\forall\,t\geq0,\;\forall\,U_{0}\in\mathcal{D}(\mathcal{B}),
$$
if and only if
$$
\limsup_{\beta\in\R, \, |\beta| \rightarrow\infty}\|\beta^{-\gamma}(i\beta-\mathcal{B})^{-1}\|_{\mathcal{L}(\mathcal{X})}<+\infty.
$$
\end{thm}
Based on Theorem \ref{SMKV19} we are able now to prove our main result given in Theorem \ref{th2} of this section. For this purpose, let's consider the following:
\begin{prop} The operator $\mathcal{A}$ defined in \eqref{damped} satisfies:
\begin{equation}\label{SMKV16}
\limsup_{\beta\in\R, \, |\beta| \rightarrow \infty}\|\beta^{-12}(i\beta-\mathcal{A})^{-1}\|_{\mathcal{L}(\mathcal{H})}<+\infty.
\end{equation}
\end{prop}
\begin{proof}
To prove \eqref{SMKV16}  we use an argument of contradiction. In fact, if \eqref{SMKV16} is false, then, there exist $\beta_{n}\in \R_{+}$ and $Y_{n}=(u_n^1,v_n ^1,u_n^2,v_n^2)\in\mathcal{D}(\mathcal{A})$ such that
\begin{equation}\label{SMKV4}
\|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.
\end{equation}
Equivalently, we have
\begin{equation} \label{eq1}
\beta_{n}^{\gamma}\left(i\beta_{n}u_{n}^{1}-u_{n}^{2}\right)=f_{n}^{1}\,\longrightarrow 0 \hspace{0.3cm} \text{in} \hspace{0.3cm} H_{0}^{1}(-1,1),
\end{equation}
\begin{equation} \label{eq2}
\beta_{n}^{\gamma}\left(i\beta_{n}v_{n}^{1}-v_{n}^{2}\right)=g_{n}^{1}\,\longrightarrow 0 \hspace{0.3cm} \text{in} \hspace{0.3cm} H_{0}^{1}(-1,1),
\end{equation}
 \begin{equation} \label{eq3}
\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 \hspace{0.3cm} \text{in} \hspace{0.3cm} L^{2}(-1,1),
\end{equation}
\begin{equation}\label{eq4}
\beta_{n}^{\gamma}\left(i\beta_{n}v_{n}^{2}-cv_{nxx}^{1}-u_{n}^{2}\right)=g_{n}^{2}\,\longrightarrow 0 \hspace{0.3cm} \text{in}\hspace{0.3cm}L^{2}(-1,1).
\end{equation}
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
$$
\beta_{n}^{\gamma} \int_{0}^{1}d.\vert u_{nx}^{2}\vert^{2}\,\ud x\,\longrightarrow 0,
$$
which leads to
\begin{equation}\label{ux2}
\beta_{n}^{\frac{\gamma}{2}}\|u_{nx}^{2}\|_{L^2(0,1)}\longrightarrow 0.
\end{equation}
Now thanks to \eqref{eq1} and \eqref{ux2}, we obtain
\begin{equation}\label{unx1}
\beta_{n}^{\frac{\gamma}{2}+1}\|u_{nx}^{1}\|_{L^{2}(0,1)}\longrightarrow 0.
\end{equation}
From \eqref{ux2} and \eqref{unx1}, it follows
\begin{equation}\label{Tn}
\beta_{n}^{\frac{\gamma}{2}}\|T_{n}\|_{L^{2}(0,1)} \longrightarrow 0.
\end{equation}
Taking the inner product of \eqref{eq3} with $u_n^2$ in $L^{2}(0,1)$ we get
\begin{equation}\label{SMKV1}
\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).
\end{equation}
Thanks to \eqref{SMKV4}, \eqref{ux2} and \eqref{Tn}, it's clear that the second and the last terms converge to zero. Furthermore, we have
$$
\beta_{n}^{\frac{3\gamma}{4}}T_{n}(0^{+})\overline{u_n^2}(0^+)\leq C\beta_{n}^{\frac{\gamma}{2}}\left(\|T_{n}\|_{L^2(0,1)}^{\frac{1}{2}}.\|u_{nx}^{2}\|_{L^2(0,1)}^{\frac{1}{2}}.\|T_{n}'\|_{L^{2}(0,1)}^{\frac{1}{2}}.\|u_{n}^{2}\|_{L^2(0,1)}^{\frac{1}{2}}\right).
$$
From \eqref{eq3} we can see that $\|\beta_{n}u_{n}^{2}+v_{n}^{2}\|_{L^2(0,1)}\sim\|T_{n}'\|_{L^2(0,1)}$ which implies that
\begin{align}\label{SMKV2}
\beta_{n}^{\frac{3\gamma}{4}}|T_{n}(0^{+})|.|\overline{u_{n}^{2}}(0^{+})|&\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\nonumber
\\
&\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}}\nonumber
\\
&\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\nonumber
\\
&\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)\nonumber
\\
&\leq\left(1+\beta_{n}^{\frac{1}{2}+\frac{\gamma}{4}}.\|u_{n}^{2}\|_{L^2(0,1)}\right)o(1).
\end{align}
Combining \eqref{SMKV1} and \eqref{SMKV2}, one follows
\begin{equation}\label{un2}
\beta_{n}^{\frac{1}{2}+\frac{3\gamma}{8}}\|u_{n}^{2}\|_{L^{2}(0,1)}\,\longrightarrow 0.
\end{equation}
Moreover, multiplying \eqref{eq3} by $\beta_{n}^{-\frac{\gamma}{2}}(1-x)T_{n}$ and integrating over the interval $(0,1)$ then by taking account of \eqref{Tn}, an integration by parts leads to
\begin{align}\label{SMKV5}
\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)
\\
+\beta_{n}^{\frac{\gamma}{2}}\mathrm{Re}\langle v_{n}^{2},(1-x)T_{n}\rangle_{L^2(0,1)}=o(1).\nonumber
\end{align}
We suppose that $\displaystyle\gamma\geq\frac{4}{3}$. It's clear from \eqref{SMKV4}, \eqref{Tn} and \eqref{un2} that the first, the third and the last terms of \eqref{SMKV5} converge to zero then one gets
\begin{equation}\label{Tn0}
\beta_{n}^{\frac{\gamma}{4}}.|T_{n}(0^{+})|\,\longrightarrow 0.
\end{equation}
Taking into account to \eqref{unx1} then the trace formula gives
\begin{equation}\label{u10}
\beta_{n}^{\frac{\gamma}{2}+1}.|u_{n}^{1}(0^{+})|\,\longrightarrow 0 .
\end{equation}
Substituting \eqref{eq2} into \eqref{eq3} 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
\begin{equation}\label{SMKV3}
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).
\end{equation}
Taking $\displaystyle\gamma\geq12$ and using \eqref{SMKV4}, \eqref{Tn}, \eqref{un2} and \eqref{Tn0} we can see that the first, the second and the fourth terms of \eqref{SMKV3} converge to zero, therefore
\begin{equation}\label{v1}
\beta_{n}^{2}.\|v_{n}^{1}\|_{L^{2}(0,1)}\,\longrightarrow 0.
\end{equation}
From \eqref{eq2} and \eqref{v1} it follows
\begin{equation}\label{v2}
\beta_{n}\|v_{n}^{2}\|_{L^2(0,1)}\,\longrightarrow 0.
\end{equation}
Multiplying \eqref{eq4} with $\beta_{n}^{-\gamma}(1-x)\overline{v_{nx}^{1}}$ and integrating over $(0,1)$ then by taking the real part we find
\begin{align*}
\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)}
\\
-\mathrm{Re}\langle i\beta_{n}v_{n}^{2},(1-x)v_{nx}^{1}\rangle_{L^{2}(0,1)}+o(1).
\end{align*}
Using \eqref{SMKV4}, \eqref{un2} and \eqref{v2} leads to
\begin{equation}\label{SMKV17}
|v_{nx}^{1}(0^{+})|^{2}-\|v_{nx}^{1}\|_{L^{2}(0,1)}^{2}\,\longrightarrow 0.
\end{equation}
We take the inner product of \eqref{eq4} with $\beta_{n}^{-\gamma}xv_{n}^{1}$ in $L^{2}(0,1)$ then we have
\begin{equation*}
c\left(\int_{0}^{1}x|v_{nx}^{1}(x)|^{2}\,\ud 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).
\end{equation*}
Using \eqref{SMKV4}, \eqref{un2} and \eqref{v2} we deduce that
\begin{equation*}
\int_{0}^{1}x|v_{nx}^{1}(x)|^{2}\,\ud x\,\longrightarrow 0.
\end{equation*}
This implies in particular that for every $\varepsilon$ in $(0,1)$ we have
\begin{equation}\label{SMKV6}
\|v_{nx}^{1}\|_{L^{2}(\varepsilon,1)}\,\longrightarrow 0\;\text{ as }n\nearrow+\infty.
\end{equation}
Multiplying \eqref{eq4} with $\beta_{n}^{-\gamma}(1-x)\overline{v_{nx}^{1}}$ and integrating over $(0,\varepsilon)$ then by taking the real part we find
\begin{align*}
\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).
\end{align*}
Besides,  from \eqref{SMKV4}, \eqref{un2}, \eqref{v2} and \eqref{SMKV6} we follow
\begin{equation*}
|v_{nx}^{1}(\varepsilon)|\,\longrightarrow 0\;\text{ as }n\nearrow+\infty.
\end{equation*}
Then we deduce that
\begin{equation}\label{SMKV18}
v_{nx}^{1}(x)\,\longrightarrow 0\;\text{ a.e. in } [0,1]\text{ as }n\nearrow+\infty.
\end{equation}
Now, \eqref{SMKV4} and \eqref{SMKV18} allows  the use of the dominated convergence theorem and lead to
\begin{equation}\label{SMKV9}
\|v_{nx}^{1}\|_{L^{2}(0,1)}\,\longrightarrow 0.
\end{equation}
Therefore, we obtain
\begin{equation}\label{SMKV10}
|v_{n}^{1}(0^{+})|\,\longrightarrow 0.
\end{equation}
By combining \eqref{SMKV17} and \eqref{SMKV9} we find
\begin{equation}\label{SMKV8}
|v_{nx}^{1}(0^{+})|\,\longrightarrow 0.
\end{equation}
Furthermore, taking the inner product of \eqref{eq2} with $\beta_n^{1-\gamma}(1-x)v_{nx}^1$ and then considering the imaginary part one gets
\begin{align*}
\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)
\\
=\frac{1}{2}( \beta_n^2 \vert v_n^1(0^+) \vert ^2 - \beta_n^2 \Vert v_n^1 \Vert^2) - \beta_n Im \langle v_n^2 , (1-x) v_{nx}^1\rangle
\end{align*}
Adding to this \eqref{SMKV10}, \eqref{v1} and \eqref{v2} we can deduce that :
\begin{equation}\label{ vn01}
\beta _n  \vert v_n^1 (0^+) \vert\longrightarrow 0
\end{equation}
Thanks to \eqref{Tn0}, \eqref{u10}, \eqref{SMKV10} and \eqref{SMKV8} one gets
\begin{align}
\beta_{n}^{\frac{\gamma}{2}+1}.u_{n}^{1}(0^{-})\,\longrightarrow 0,\label{0-1}
\\
\beta^{\frac{\gamma}{4}}.u_{nx}^{1}(0^{-})\,\longrightarrow 0,
\\
\beta_n v_{n}^{1}(0^{-})\,\longrightarrow 0,
\\
v_{nx}^{1}(0^{-})\,\longrightarrow 0.\label{0-2}
\end{align}
Next, inserting \eqref{eq1} into \eqref{eq3} and inserting \eqref{eq2} into \eqref{eq4} and consider both equations in the interval $(0,1)$, leads to
\begin{equation}\label{SMKV11}
-\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},
\end{equation}
 and
\begin{equation}\label{SMKV12}
 -\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}.
\end{equation}
A straightforward calculation shows that the real part of the inner product of \eqref{SMKV11} with $(x+1).u_{nx}^{1}$ and that the real part of the inner of \eqref{SMKV12} with $(x+1).v_{nx}^{1}$ leads to
\begin{align}\label{-104}
\frac{1}{2}\int_{-1}^{0}\left(|\beta_{n}u_{n}^{1}|^{2}+|u_{nx}^{1}|^{2}\right)\,\ud x=\frac{1}{2}\left(|u_{nx}^{1}(0^{-})|^{2}+\beta_{n}^{2}|u_{n}^{1}(0^{-})|^{2}\right)
\\
-\mathrm{Re}\langle v_{n}^{2},(x+1)u_{nx}^{1}\rangle_{L^{2}(-1,0)}+o(1),\nonumber
\end{align}
and
\begin{align}\label{-105}
\frac{1}{2}\int_{-1}^{0}\left(|\beta_{n}v_{n}^{1}|^{2}+c|v_{nx}^{1}|^{2}\right)\,\ud x=\frac{1}{2}\left(c|v_{nx}^{1}(0^{-})|^{2}+\beta_{n}^{2}|v_{n}^{1}(0^{-})|^{2}\right)
\\
+\mathrm{Re}\langle u_{n}^{2},(x+1)v_{nx}^{1}\rangle_{L^{2}(-1,0)}+o(1).\nonumber
\end{align}
Where we have used \eqref{SMKV4}-\eqref{eq4}. In another hand, from \eqref{SMKV4}, \eqref{un2}, \eqref{v2} and \eqref{0-1}-\eqref{0-2} we get
\begin{equation}\label{SMKV13}
\int_{-1}^{0}\left(|\beta_{n}u_{n}^{1}|^{2}+|u_{nx}^{1}|^{2}\right)\,\ud x\,\longrightarrow 0,
\end{equation}
and
\begin{equation}\label{SMKV14}
\int_{-1}^{0}\left(|\beta_{n}v_{n}^{1}|^{2}+c|v_{nx}^{1}|^{2}\right)\,\ud x\,\longrightarrow 0.
\end{equation}
Now by summing \eqref{unx1} \eqref{un2}, \eqref{v1}, \eqref{v2}, \eqref{SMKV13} and \eqref{SMKV14} we can see that
\begin{equation}\label{SMKV15}
\|Y_{n}\|_{\mathcal{H}}\,\longrightarrow 0.
\end{equation}
This contradicts \eqref{SMKV4} and so \eqref{SMKV16} holds true with $\gamma\geq12$. This completes the proof.
\end{proof}

redpajama_set_name: RedPajamaArXiv


text:
Join the Pegasus Alzheimer's clinical trial
Helpful links about ALS
Helpful links about Alzheimer's
About AMX0035
Amylyx Announces Appointment of Debra Canner as Global Head of Human Resources
molinger2019-12-12T19:04:15+00:00
Passionate HR Leader Brings 20+ Years of Experience to Solidify the Growth of the Company's Team
CAMBRIDGE, Mass. (BUSINESS WIRE)–Amylyx Pharmaceuticals, Inc., today announced the appointment of Debra Canner as the company's first Global Head of Human Resources. Deb will serve on Amylyx' Executive Team, overseeing the company's human resource efforts, including talent management, organizational leadership, culture and corporate responsibility.
Throughout her distinguished career as a human resources leader, including chief human resources officer experience at three preeminent life science and technology organizations, Deb brings a strategic yet practical approach to her work.Deb will play a large role in shaping and fostering a culture focused on attracting, retaining and developing bold, collaborative talent that aims to improve the lives those of suffering from disease.
"We're thrilled Deb has joined us to continue building a strong core team that is eager to dedicate themselves to changing patients' lives," said Joshua Cohen, CEO, Chairman, and Co-Founder of Amylyx. "Deb's tremendous experience in building and scaling companies will add great value to Amylyx and we look forward to her leadership in ensuring our culture and key growth initiatives are aligned."
"Becoming part of the Amylyx team at this central moment in time is tremendously exciting. We have an opportunity to bring together a team dedicated to fearless innovation and that believes we can be better together to achieve our goal of helping patients."
Prior to her CHRO roles at Akamai Technologies, Joslin Diabetes Center and Haemonetics Corporation, Deb was part of Genzyme, now a fully owned subsidiary of Sanofi, and rose to become the number two HR leader there during a period of significant growth. Since 2013, Deb has been a human resources consultant to leaders of public and private small and emerging growth life science companies with a focus on HR leadership across all areas of Human Resources. Past clients include: Gen9 Bio, CeQur, Joslin Diabetes Center, Juniper Pharmaceuticals, Navitor Pharmaceuticals, and CANbridge Pharmaceuticals.
Deb received a B.A. from the University of Massachusetts, Amherst. She holds certification in executive leadership development from the Center for Creative Leadership.
About Amylyx Pharmaceuticals
Amylyx Pharmaceuticals, Inc. is a pharmaceutical company developing a novel therapeutic for Amyotrophic Lateral Sclerosis (ALS), Alzheimer's disease and other neurodegenerative diseases. For more information, visit www.amylyx.com and follow us on LinkedIn and Twitter.
Merissa Muller
Merissa.muller@finnpartners.com
Amylyx Pharmaceuticals was founded in 2013 with the mission of improving the lives of patients with neurodegenerative diseases. The company's lead therapeutic, AMX0035, is designed to simultaneously target pathways originating in the mitochondria and endoplasmic reticulum which lead to neuronal death and degradation. AMX0035 is currently in clinical trials in both Alzheimer's disease and ALS.
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redpajama_set_name: RedPajamaCommonCrawl


Please visit https://jth30th-productshowcase.eventbrite.com to register for our 30th Anniversary Celebration & Product Showcase!
JTH and 12 of our manufacturer partners will be at the 2018 Upper Midwest Electrical Expo at the Minneapolis Convention Center. We will once again be hosting an Expo Celebration following the first day of the expo.
Three members from the JTH sales team hosted a group of customers at the 3rd annual ice fishing event, held at Ballard's Resort.
Enjoy product demonstrations, lunch and fantastic fall scenery with your favorite JTH and Pinnacle crew aboard the Paradise Lady.
Please join us for the 2nd Annual JTH Decorative Lighting Show!
JTH Lighting Alliance will be showing one of our manufacturer's products at the 10th Annual Minnecon at Target Field, 12:00-7:00 p.m..
JTH Lighting Alliance and Bega will be at Brit's Pub to showcase products from the LIMBURG Collection Catalog-5.
JTH Lighting Alliance and ConTech Lighting host 2nd Annual Golf and Fishing Outing.
JTH Lighting Alliance will be hosting the 2nd Annual Contractor Ice Fishing Trip at Ballard's Resort, which will be sponsored by Elite Lighting, H.E. Williams and Leviton.


I think I see where you are going with this. Clever design. Can't wait to see this finished.
Nice job…I do like free plans.
Very nifty engineering goin on there Alex. Comin along nicely. Look forward to pt. 2.


--From a video shot from a streetcar traveling down Market Street in San Francisco in 1905. (super cool) I can see why they had to sit down and say, "Hmmm, thinkin' some rules for the road might be a good idea with all these new street cars we got here..."
--Ah! I discovered this kickin' new site. 12Most. I am in love with lists, and the ones featured here daily are uniquely charming. Like, 12 Most Reliable Ways to Build Trust, 12 Most Honest Ways to Break Off a Friendship, 12 Most Clever Ways to Ask Someone on a Date. You get the idea. Go check them out.
--These tips via BuzzFeed really work. Especially #17. "27 Ways To Make Your Groceries Last As Long As Possible." Do 'em, because how dumb is it to keep buying well intended fruits and vegetables, then throwing them out because they get slimy and brown? Pretty dumb. So here's how to keep them fresher, longer.
Loved the streetcar video! Amazing.


text:
Author Archives: Philip Peters
March 30, 2010Philip Peters
President Obama and Cuba Palabra Nueva Article by Philip Peters in the March 2010 issue of Palabra Nueva, the magazine of the Archdiocese of Havana. Follow these links to the original article in Spanish, the article in English translation, and the Palabra Nueva website [Read More...]
February 23, 2010Philip Peters
Hapless in Havana Published in ForeignPolicy.com, February 23 2010 Alan P. Gross of Maryland recently had the rare experience of being thrown in jail for doing his job. On Dec. 4, Cuban authorities arrested the U.S. Agency for International Development (USAID) contractor at [Read More...]
November 19, 2009Philip Peters
Time to Lift the Ban on Travel to Cuba Statement before the Committee on Foreign Affairs United States House of Representatives Mr. Chairman, Congresswoman Ros-Lehtinen: I applaud you for convening this hearing and I'm pleased to state my strong belief that an end to Cuba travel restrictions is squarely [Read More...]
September 24, 2009Philip Peters
Hugonomics: Venezuela's fight against poverty, and against the private sector Research Study Hugo Chavez was elected President of Venezuela in 1998 and has proceeded to lead and symbolize a new variant of the Latin American political left. His Bolivarian Revolution and his push toward socialism have been characterized by strong [Read More...]
July 8, 2009Philip Peters
Raulonomics: Tough Diagnosis and Partial Prescriptions in Raul Castro's Economic Policies Cuban President Raul Castro has been at the head of Cuba's government for nearly three years, as acting President since July 2006 and as President since February 2008. His time in office has been marked by stark assessments of Cuba's [Read More...]
June 17, 2009Philip Peters
TV Marti: An Idea Whose Time Has Come – and Gone Mr. Chairman, members of the Subcommittee: I appreciate the opportunity to join you today to discuss TV Marti in the context of U.S. policy toward Cuba. I support public diplomacy in the large sense of that term, encompassing information and [Read More...]
April 29, 2009Philip Peters
Diplomacy with Cuba and U.S. National Security Mr. Chairman, members of the Subcommittee: Thank you for this opportunity to discuss the security aspects of American policy toward Cuba. I believe that a shift toward a policy of engagement with Cuba would serve U.S. interests at a time [Read More...]
The Administration's Cuba Family Sanctions: Time for Repeal Mr. Chairman, members of the Subcommittee: Thank you for inviting me to address the issue of U.S. sanctions that limit Cuban American visits and aid to their family members in Cuba. I oppose all restrictions on American travel to Cuba, [Read More...]
August 2, 2007Philip Peters
Congress Moving to Increase UN Support Issue Brief Conflicts over the war in Iraq and turmoil at the Justice Department are obscuring a rare case of cooperation between the Administration and Capitol Hill. The White House has asked for greater support for the United Nations' budget [Read More...]
"True Believer" Leaves Many Questions Unanswered The 16-year career of Ana Belen Montes as an agent of Cuban intelligence came to a prosaic end the morning of September 21, 2001. Her supervisor at the Defense Intelligence Agency (DIA), where she was the top Cuba analyst, directed [Read More...]


You can get in touch with the Odeon Lincoln by calling 0871 224 4007. The cinema is located in the city centre right next to the Brayford Pool and River Witham.
Odeon cinemas are a chain of cinemas that are located throughout the UK. The Odeon Lincoln is located on the Brayford Pool and offers great views from its large windows not that you would need to the views and they show the latest and best new movies available. The cinema is also conveniently located next to a variety of restaurants giving you a lot of options to eat out before or after your movie showing. There is also a bus stop on Newland (A57) where a few different buses stop making it convenient to access via public transport. There are also two car parks on Lucy Tower Street very close to the cinema, a convenient place to park if you do not want to walk far after your movie as well as a cash point on the same road if you do not want to pay with your card.
There are a total of 9 screens in the Odeon Lincoln all of which can show movies in 'RealD 3D', there is also premier seats for you to watch your film comfortably. They have a Costa Coffee and a Ben & Jerry's stand on site as well as refreshments and snacks available at the counter. If you want to skip the lines they have two fast ticket machines where you can purchase tickets (card only) or collect tickets you bought online. Odeon wants everyone to be able to enjoy movies, the Odeon Lincoln location is fully accessible including all tills having hearing induction hoops, accessible toilets as well as fully accessible screens with wheelchair spaces and infrared enabling audio description or amplification headsets. If you want to know more about the accessible facilities in the cinema call the helpline on 0800 138 3315.
The Odeon Lincoln is open 24 hours a day but they don't usually show movies in the early hours of the morning, they mostly stop at around 12 am and the films usually all finish at around 3 am. The Costa Coffee closes Monday to Sunday at 21:30, it usually opens at 8.30 but on Saturday it opens at 9.00 and on Sunday it opens at 10.00. The Ben and Jerry's stand opens and closes at the same times the Costa Coffee does.
You can get in touch with the Odeon Lincoln via phone call or email. To get in touch with the cinema via phone call all you need to do is dial 0871 224 4007 and make your enquiry directly to a member of their staff. To email the cinema you need to go to to the website and click either live chat or guest feedback, with live chat you can talk directly to a member of their staff live or with the guest feedback you will receive a reply within a few days.
This site is a directory for Lincoln businesses, attractions, schools and organisations. You can find contact information and company information for each business on this site.


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HOME | Monitor | USDA Invests Nearly $19 Million in Rural Broadband for Alaskan Families
USDA Invests Nearly $19 Million in Rural Broadband for Alaskan Families
Dec 4, 2019 | Monitor
CORDOVA—US Department of Agriculture (USDA) Alaska State Director for Rural Development Jerry Ward announced USDA has invested $18.88 million in high-speed broadband infrastructure that will create or improve rural e-Connectivity for nearly 270 rural households in Yakutat. This is one of many funding announcements in the first round of USDA's ReConnect Pilot Program investments.
"Geographic isolation should no longer be a barrier to economic prosperity, especially here in Alaska," Ward said. "With ReConnect, we are bridging the so-called 'digital divide' and bringing modern communications to rural Alaskan communities like Yakutat. Under the leadership of President Trump and Agriculture Secretary Sonny Perdue, USDA is committed to being a strong partner to rural communities in deploying this critical infrastructure, because we know when rural America thrives, all of America thrives."
Nathan Moulton, Executive Director of Yakutat Tlingit Tribe, added "The Yakutat Tlingit Tribe and our members are excited to work with Cordova Telecom to bring high-speed broadband to Yakutat. The USDA grant investment in Yakutat will positively impact lives and opportunities for everyone in our community."
The project, dubbed NICEY (New Internet Communications for Everyone in Yakutat), will deliver broadband internet service to all 662 year-round residents and businesses in Yakutat via fiber-to-the-home (FTTH) connections. Yakutat will be connected to Cordova's submarine fiber optics via a new microwave middle-mile network spanning 230 miles between the two communities.
Like seafood processing, hospitality, and the oil industry, trucking has traditionally been an Alaska industry that depends heavily on out-of-state workers. According to the Alaska Department of Labor and Workforce Development's most recent report on nonresident workers, out-of-state drivers represented 18.5 percent of the workforce of 3,602 heavy and tractor-trailer drivers in 2016. That category of driver made an average quarterly wage of almost $14,000 that year. The nonresident count includes both seasonal workers and those who live elsewhere and regularly fly in for rotations.
Crum says that since 2016, oil and gas companies have flown in fewer commercial truck drivers from out of state. But he says the practice remains common in the seafood processing industry where companies frequently struggle to find qualified drivers at their remote plants.


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Cinema TDOG
By Bill and TDOG. Discovered by Player FM and our community — copyright is owned by the publisher, not Player FM, and audio is streamed directly from their servers. Hit the Subscribe button to track updates in Player FM, or paste the feed URL into other podcast apps.
2Films 2Curious « »
The Gamers: Dorkness Rising and Dungeons and Dragons
What is a Dungeons and Dragons movie? Is it a movie in the Dungeons and Dragons world, like Solo A Star Wars Story is a Star Wars movie; or is it a movie about Dungeons and Dragons, like Fanboys is a Star Wars movie?
The Gamers: Dorkness Rising, sequel to the oft Youtube linked The Gamers, is a movie about a bunch of RPG nerds playing Dungeons and Dragons. Dungeons and Dragons, starring Riff Raff from The Rocky Horror Picture Show, is a movie with a couple of mysterious dungeons and a ton of huge Dragons.
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Outdoor Idaho "Salmon River Lodges & Legacies" This special travels the Salmon River, stopping each night at a historic lodge to meet the people, who continue in the footsteps of hearty pioneers in the wild heart of Idaho. Each stop reveals a rich history. Rafters, guides, boat operators and residents all reveal how they are drawn to the river repeatedly - or permanently.G
Oregon Field Guide "Sea Otter Specials" Sea otters lived on Oregon's coast for thousands of years until fur traders wiped them out. But they may be making a comeback.G
Roadtrip Nation "In-Studio Interview: Veronica Belmont" Past Roadtrippers Mariana and Hannah join RTN staffer Jason on the Roadtrip Nation Winnebago stage to interview Veronica Belmont, Gamer and Host of Tekzilla, a technology news podcast. The audience hears about Veronica's love for technology and the moment of serendipity that led her from humble beginnings to internet fame. Veronica urges the audience to "Think big," and fields questions from her fans about gaming, technology, and the Road she's paved to arrive at a meaningful job. D
Live From Lincoln Center "One Singular Sensation: Celebrating Marvin Hamlisch" The New York Philharmonic rings in 2013 with this celebration of the life and work of Marvin Hamlisch. The late composer-conductor was scheduled to conduct this concert before his death in August. Guest star appearances include violinist Joshua Bell, Raul Esparza (Company), Michael Feinstein, Maria Friedman, Josh Groban, Megan Hilty, Audra McDonald, Kelli O'Hara and Frederica Von Stade.G
Music Voyager "From D.C. to Cajun Country" Music Voyager host Jacob Edgar continues his Road to a GRAMMY journey by meeting up with GRAMMY winner India. Arie in Washington D.C. Dobet Gnahore, a stunning singer from west Africa, who shares a GRAMMY nomination with India. D
Wild! "Elephants: The Last Mahout" A unique look into the plight of the Asian Elephants, exploring the tragic collisions of man and beast and speaking plainly about the steps necessary to save this remarkable animal. D
Globe Trekker "Caribbean Island: St. Lucia, Martinique & Montserrat" Trekker Zoe Palmer visits the islands of St. Lucia, Martinique and Montserrat. With rainforests, mountains and volcanoes and a hybrid of English, African and French heritage, these islands are among the top vacation spots in the world.G
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Biz Kid$ "What Can You Do with Money?" Decisions, decisions! Join the Biz Kids and you'll ponder the merits of spending, saving, donating, and investing money. Visit the New York Stock Exchange, get tips from the author of Not Buying It, and learn to avoid compulsive shopping. You can also learn how to create a financial diary and track spending. D
Oregon Field Guide "Cavity Nesting Bird Habitat, Zumwalt Prairie, Sensor Fish/Dam Research" NULL
Cook's Country from America's Test Kitchen "Dixie Chicken" Test cook Julia Collin Davison shows host Christopher Kimball how to make the best Moravian Chicken Pie. And finally, test cook Bridget Lancaster revives a classic recipe for Chicken and Slicks. D
Rick Steves' Europe "Istanbul" Rick loses his way in the Grand Bazaar, munches through the famously fragrant Spice Market and follows the fall of the ancient capital of Byzantium and the rise of Islam at the city's wall.G
NOVA "Doomsday Volcanoes" The eruption of Iceland's Eyjafjallajkull volcano in 2010 turned much of the northern hemisphere into an ash-strewn no-fly zone. But Eyjafjallajkull may be just the start. Katla, an Icelandic volcano 10 times bigger, has begun to swell and grumble.G
Independent Lens "Men Who Swim" In 2009, Sweden's all-male synchronized swimming team, inspired by classical techniques used by 1950s-era swimmer Esther Williams, entered the first international championship for the sport. The move would require increased training for what started as a weekly escape from the daily grind - and in a year several of them turned 40.G
Chatsworth House The Duke and Duchess of Devonshire are at home in the 300-room Chatsworth House on the estate which has been in the Duke's family for 450 years, and are preparing the house to be open to the public. Everyone has a role to play, and the duke and duchess join the annual litter clean up around the estate. Part 1 of 3G
Reagan Presidency This detailed analysis features interviews with notable Reagan staff members, congressional leaders, scholars, journalists and presidential biographers. Part 1 of 3G
Not In Our Town: Light in the Darkness In 2008 in Patchogue, New York, a series of attacks against Latino residents ended with the killing of 37-year-old Marcelo Lucero, an Ecuadorian immigrant who had lived there for 13 years. Seven local high school students arrested for the crime admitted they were "looking for a Mexican" to beat up. Over a two-year period, the film followed Patchogue Mayor Paul Pontieri as he led a diverse group of residents to confront the anti-immigrant bias in their town and repair the fabric of their community life.G
Rough Cut - Woodworking with Tommy Mac "Laminated Coat Rack" NULL
P. Allen Smith's Garden Home "Keeping It Local" In this episode, Allen discusses the importance of buying locally in your community or growing your own. He shares tips on planting gourds and when to harvest garlic. Plus, instead of using an artificial tree, we talk to an expert about buying from a local Christmas Tree Farm. D
Katie Brown Workshop "Who Needs A Wedding Planner?" Katie's ideas for invitations and one-flower arrangements and recipes for quiche cups with shrimp, chicken pinwheels and polenta sandwiches will help keep a wedding day simple. D
Christina "Eating Well for Life" If you think it's a diet, you'll fail. But if you think of eating well for life, it becomes a part of how you live and who you are. Eating well is required for your body to function at its best. D
The Jazzy Vegetarian "Pasta with Pizazz!" Host Laura Theodore prepares pasta favorites starting with easy "Peanut Noodles." Next, "Penne with Garbanzo Beans and Broccoli Rabe" prepares quickly, but is packed with zingy flavor. A surprise choice for noodles is showcased in "Pea and White Bean Puttanesca," followed by "Arugula, Artichoke, and Tomato Pasta" which features a no-cook sauce, making it a breeze to prepare for a summertime meal. D
Christina "Italian Light" Think you can't lose weight and enjoy an Italian feast? Sophia Loren famously said she owed her great looks to pasta and olive oil. The natural deliciousness of Italian cuisine can help you achieve the body of your dreams with its great balance of carbs, protein and fat. And with delicious dishes like Penne fra Diavolo, dieting has never been so yummy. D
The Jazzy Vegetarian "Lunch Bunch" Host Laura Theodore prepares lunch, starting with simple "Four-Ingredient Leek Soup." A trio of main dish salads include: tantalizing "Avocado and Black Bean Confetti Salad," tasty "Fancy Chickpea Salad" and satisfying "Quinoa Tabbouleh," served with crusty bread on the side. We'll finish our lunch with a refreshing and elegant "Banana Raspberry Sorbet. D
In Search of the Novel "Who Owns The Novel?" This workshop probes the living nature of the novel by illustrating how each reader makes a novel his or her own. It shows how the interpretation of a novel can change, depending on the reader's culture, class, generation, gender, and personality. D
Primary Sources: Workshops In American History "Disease and History: Typhoid Mary and the Search For Perfect Control" This workshop looks at the history of infectious disease in America, particularly typhoid, diphtheria, and polio, and their "conquest" by medical research and public health regulation. With the aid of contemporary medical journal articles and New York City health records, the onscreen participants investigate the medical and civil liberties issues exemplified by the case of "Typhoid Mary" Mallon. Facing off as either Board of Health officials or friends of Mary Mallon, workshop participants debate the typhoid carrier's fate.G
Rediscovering Biology: Molecular to Global Perspectives "Session 13. Genetically Modified Organisms" While genetic modification of organisms has occurred for millennia, we now have the tools to insert specific genes from one organism into cells of unrelated species. This session illustrates the processes used and how such genetically transformed organisms are increasingly common in agriculture, industry, and medicine, and introduces the ethical considerations of GMO research. D
Rediscovering Biology: Molecular to Global Perspectives "Session 1. Genomics" Having determined the complete DNA nucleotide sequence of humans and several other organisms, today. D
Critical Issues In School Reform "Reflecting On Teaching Practice: Student Work, Teacher Work, and Standards, Part 1 Math" A tenth-grade math teacher from San Bruno, California, presents a sample of student work from her classroom to a group of teachers, administrators, and a facilitator. The group uses a "tuning protocol" to examine this work, give the teacher feedback, and discuss its implications for her teaching practice. D
Critical Issues In School Reform "Reflecting On Teaching Practice: Student Work and Teacher Work, Part II Science" A fourth-grade teacher from Worcester, Massachusetts, shares a sample of work from her science classroom with a group of teachers and administrators in a facilitated "consultancy" that focuses on a question posed by the teacher about the student work and her teaching practice.G
Critical Issues In School Reform "Stories of Public Engagement: Patrick O'hearn School" A Boston elementary school has enhanced student success through close cooperation with families. D
Hands On Crafts for Kids "My Workspace" This episode features crafts to fix up your workspace -whether it's the kitchen table or a desk in your room - make the supplies you use every day. We begin with a recycled portfolio with monster clips. Next, there's no excuse for having a mess on your desk with a great organizer made from old CD's. D
Hey Kids, Let's Cook "Breakfast for Mom" Whether it's Mother's Day, your mom's birthday, or just simply just because she's special, you should set your alarm, ask your dad to meet you in the kitchen at 6:00 A.m., and go for it! Make your mom that special breakfast and take it to her before she gets up! And today we're going to learn how to make scones, a delicious bread brought to us from the English and Scottish. And we'll serve it up with an omelet, and fresh fruit. D
Jacques Pepin: More Fast Food My Way "Smooth Food" Daughter Claudine joins Jacques in the kitchen. He uses local blue cheese for mini savory cheesecakes on arugula, and makes tapioca banana coconut pudding.G
America's Test Kitchen from Cook's Illustrated "Cool and Creamy Desserts" Host Christopher Kimball visits Toscanini's to learn the secrets behind professional ice-cream making. Next, test cook Bridget Lancaster shows Chris how to make the Best Homemade Ice Cream. Then, equipment expert Adam Ried reveals his pick for the top ice-cream makers, and tasting expert Jack Bishop challenges Chris to a tasting of vanilla beans. D
Cuisine Culture "Arnaud Lallement, Reims, France" Reims is the capital of Champagne, and it's also where the culinary adventure begins in chef Arnaud Lallement's Two-Star Michelin restaurant, L'Assiette Champenoise. Lallement demonstrates how to grill langoustines, braise lamb and make an explosive dessert, a Raspberry Vacheron. Sweets await in one of the world's oldest cookie-making factories. D
Pati's Mexican Table "Lip-Smacking Mexican Meal" By adding a few key Mexican ingredients to what you'd normally find in an all-American pantry and fridge, you get these to-die-for, lip-smacking dishes. Chicken a la Trash; Everyday Green Salad; Impossible Chocoflan. D
Ask This Old House "Restoring Terrazzo Floors/Repairing a Cracked Floor Joist" Host Kevin O'Connor heads to Del Ray Beach, Florida, to restore terrazzo floors. Then, Kevin, along with landscape contractor Roger Cook, plumbing and heating expert Richard Trethewey and general contractor Tom Silva, asks, "What Is It?" Afterward, Tom helps a homeowner repair a cracked floor joist. D
Rick Steves' Europe "Vienna and the Danube" Rick takes in the city's wealth of elegant gardens, great art and fine music. He travels along the Danube River and hikes through the Baroque Melk Abbey.G
Equitrekking "Spain" Equitrekking host Darley Newman rides at Cortijo El Esparragal, an estate and working farm outside of Seville that dates back to the Bronze Age. On prize-winning Andalusians, Darley learns the purpose and history of the Spanish Vaquera style of riding. Next, she enjoys a sunset ride outside of Segovia and visits the castles and Roman aqueducts in this central region. D
Mexico -- One Plate at a Time with Rick Bayless "From Lobster to Chocolate Clams: A Delicious Feast in Magdalena Bay" Rick and local guide Luis Garduno of REDTours venture out on a skiff to learn about sustainable fishing on Magdalena Bay. They catch up with local lobster fishermen and learn about sizing lobsters. Then they watch as clam divers haul up the local chocolate clams from the bottom of the bay. D
Caprial and John's Kitchen: Cooking for Family and Friends "Decadent Organic Dairy" On a visit to an Oregon organic dairy, Caprial and John come across products that lend amazing flavor and aroma to their creations: mushroom dip with truffle oil, basil sour cream dressing and grilled lamb chops with cumin yogurt sauce.G
Theater Talk "The Mystery of Edwin Drood" The Mystery of Edwin Drood's lyricist/librettist Rupert Holmes and star player Will Chase discuss the current and highly successful revival of Holmes' 1985 musical mystery on Broadway. In this show, based on Charles Dickens' last, unfinished novel, the audience is assigned the task of picking the murderer, because although Dickens laid out the case in the novel, he died before revealing the killer's identity. D
Closer to Truth "Asking Ultimate Questions" I like pushing boundaries, trying to discern existence, searching the foundations of reality, knowing all that can be known. Overly ambitious? Sophomoric? I don't care. I do it anyway. D
Dialogue "Doris Kearns Goodwin: Conversations from the Sun Valley Writers' Conference" The Pulitzer Prize-winning author and historian Doris Kearns Goodwin talks with host Marcia Franklin about the lessons we can learn from Abraham Lincoln.G
Intelligence Squared U.S. "Ration End-Of-Life Care" Just because we can extend life, should we? The U.S. is expected to spend $2.8 trillion on health care in 2012. Medicare alone will cost taxpayers $590 billion, with over 25% going toward patients in their last year of life. D
Intelligence Squared U.S. "The Rich Are Taxed Enough" How do we fix the economy? The U.S. government's budget deficit is nearing a trillion dollars for the fourth straight year and unemployment remains high. With the Bush-era tax cuts that are set to expire at the end of 2012, what is the best move for continued economic recovery? President Obama says we should raise taxes on those making more than $250,000 to reduce the deficit. D
America ReFramed "Beyond Belief" Susan Retik and Patti Quigley are two ordinary soccer moms living in the affluent suburbs of Boston until tragedy strikes. Rather than turning inwards, grief compels these women to focus on the country where the terrorists who took their husbands' lives were trained: Afghanistan. Over the course of two years, as they cope with loss and struggle to raise their families as single mothers, these extraordinary women dedicate themselves to empowering Afghan widows whose lives have been ravaged by decades of war, poverty and oppression - factors they consider to be the root causes of terrorism. D
Global Voices "Arusi Persian Wedding	" Set against the turbulent relationship between the United States and Iran, Iranian-American filmmaker Marjan Tehrani captures the struggle and excitement of her brother Alex and his bride Heather as they plan a Persian Islamic wedding in Iran. But when Alex's Iranian-born parents and Heather's conservative American father meet for the first time, cultures clash and test the couple to their limits. D
Global Voices "Carmen Meets Borat" In "Carmen Meets Borat," director Mercedes Stalenhoef follows the life of 17-year-old Carmen. She lives in a gypsy village in Romania, where the men spend their days tying one on and exchanging coarse words in her father's bar. In the evenings, she watches a Spanish soap opera and dreams of a better life in Spain, where the men are romantic and decent. D
Idaho Reports 2013 Host Greg Hahn with news reporters and political analysts looks at the week's activities in the Idaho Legislature and examines what to expect in the weeks ahead. Hahn also talks with and profiles legislative personalities and reports on events at the Statehouse.G
Scientific American Frontiers "Dead Men's Tales" Alda meets scientists reconstructing mysterious past events from the evidence of excavated remains. Stories include the mysterious destruction of the Jamestown colony and the unconfirmed hanging of outlaw Wild Bill Longley. D
Arthur "Get Smart/Baby Steps" Get Smart - Everyone loves HUGO-the new interactive whiteboard that claims to be 100% accurate. If only Mr. Ratburn knew how to use it. After a crash course from Brain, Mr. Ratburn embraces the high tech teaching tool; that is, until HUGO corrects him in front of the class. D
Wild Kratts "Termites Vs. Tongues" When Aviva and Koki accidentally miniaturize themselves and get carried away by termites, Martin and Chris must split up to search for them. D
WordGirl "Hello New Year, Goodbye Moon/Art in the Park" HELLO NEW YEAR, GOODBYE MOON - It's New Years Eve! While the Botsfords celebrate with the greatest party ever, Dr. Two Brains is planning to turn the moon into cheese. Will the new year begin with Two Brains acting on his resolution to feast on a cheesy moon, or will WordGirl be able to stop him in time? Vocabulary Words: Massive, Resolution. ART IN THE PARK - Becky agrees to help out Violet with her poetry/painting/music performance. D
Cyberchase "Fortress of Attitude" The kids use their skills at measuring length to climb a massive, booby-trapped statue of Hacker.G
Sesame Street "Telly The Tiebreaker" Elmo, Zoe, Rosita and some friends are making a beautiful block house together but can't agree whether to put a square or circle on top. Alan helps them solve the problem by having them raise their hands for the one they like best. After counting up the votes, the square wins by one. D
Clifford The Big Red Dog "Clifford On Parade/Follow The Leader" Clifford on Parade - Emily and Charley are having a hard time deciding what kind of float to make for the annual Birdwell Parade. With a little help from Clifford, they learn the importance of compromise. Follow the Leader - Cleo has a hard time giving up her role as " leader" in a game of Follow The Leader, but soon learns that playing with friends is more fun when everyone cooperates and takes turns with one another. D
Arthur "Francine's Big Top Trouble/George Blows His Top" Francine's Big Top Trouble - Francine is sure she will be a star at circus camp. Since Catherine did so well there - how hard could it be? Well, very, it turns out. Will Francine ever get the hang of tumbling? And, more importantly, will she ever step out of her sister's shadow? George Blows His Top - George is always more than happy to lend his pals whatever they need, especially Buster. D
Martha Speaks "Eyes on the Skies/Camp Truman" Eyes on the Skies - Astrophysicist Neil DeGrasse Tyson (Truman's #1 hero), is holding a seminar at the library. Truman becomes more and more certain that his activity partner T.D. is going to embarrass him in front of his idol. D
Curious George "Auctioneer George/Sock Monkey Opera" Auctioneer George - George has one dollar to buy anything he wants at the Annual Corn Roast and Attic Auction, and he sets his sights on a pair of red elephant mittens. When the Man with the Yellow Hat leaves George alone, he bids.and bids. D
Cat in the Hat Knows A Lot About That! "Bounce/Timmy Tippy Toe" Bounce - Nick and Sally are wondering why their ball is so bouncy. Fish tells them it's because it's made from rubber, and rubber comes from trees. Cat doesn't believe him so it's time to ask the expert, Dr. Twiggles! It turns out Fish is right! And Dr. Twiggles shows them all just how rubber is made, and all the wonderful things rubber can make! Timmy Tippy Toe - Nick and Sally are playing on a climbing frame but just can't manage to get to the top. D
Super Why! "The Three Billy Goats Gruff" Gramma will not let Red into the kitchen to get some extra special Red Velvet Cake - and it's making Red extremely mad! The Super Readers head into the story of the Three Billy Goats Gruff, who really want to cross the Trolls bridge#only this time the Troll gets to tell his side of the story. The Super Readers wonder if anyone can make those rude Billy Goats Gruff behave, and Red figures out what Gramma needs to hear. Educational Objectives: To learn the importance of good behavior, discover the alphabet and identify the letters P-O-L-E. D
Dinosaur Train "The Old Bird/Diamond Don" The Old Bird - Buddy, Tiny and Mom make a play date with Petey Peteinosaurus to ride the Dinosaur Train to meet Arlene Archeopteryx, a bird from the family of the oldest birds ever! The kids are excited to learn that Arlene is a bird and a dinosaur. Home With Shiny & Don - Shiny and Don stay at home and we get to see that just because they don't always ride with Buddy and Tiny on the Dinosaur Train doesn't mean that they are not having fun. An adventure into a local caves leads to a marvelous and brilliant discovery. D
Sid The Science Kid "Sid's Super Kick!" In his backyard, Sid is pretending to be "Super Sid!" To show his super strength, he kicks a soccer ball really hard and it sails across the yard. But when he tries to kick it to a nearby tree with the same might, the ball once again sails really far. On a special trip to the Science Center, Sid and his friends investigate how using lots of force makes something move really far and fast-just like kicking a soccer ball! (Educational objective - Force). D
WordWorld "Shark's First Day of School/Shark's First Loose Tooth" Shark's First Day of School - It's the first day of school for the youngest WordFriends, but Shark is afraid to go. With the help of his good buddy Duck, and some gentle encouragement from their teacher Cat, Shark's fear turns to confidence, and by day's end he's at the head of the class. Shark's First Loose Tooth - When Shark loses his first baby tooth at a sleepover party with Duck, he's got to get to sleep fast so the tooth fairy can come and leave him a present. D
Caillou "Grown-Ups and Me" Caillou meets a lot of new people today. He's invited to spend the day with Clementine's family and is nervous until he discovers her family is a lot like his. Caillou's first day at day care is scary until he meets his teacher Miss Martin, and makes two new friends, Leo and Clementine. D
Barney & Friends "The Misbegotten Moon, A Space Adventure" To her dismay, Baby Bop sees a half-moon and thinks the moon is broken in half! Through comic events, she goes on a search to find the missing pieces. Barney takes her and the kids on a magical adventure to the moon where she meets the Man in the Moon who eases her concerns. D
Thomas & Friends "If A Job's Worth Doing" Thomas and the Garbage Train - Thomas volunteers to take the Duchess to tea. Diesel tells Thomas that his work with Whiff will make him too stinky to take the Duchess. As Whiff and Thomas puff along, Thomas uncouples from the Garbage train and Thomas takes 'sweeter smelling' tracks. D
The Electric Company "The Wordball Is My Oyster" Lisa gets a case of stage fright when she's cast as the lead in Hector's play. Can Lisa overcome her fears in time to save the play, or will she have to give her part to Annie?. D
WordGirl "The Masked Meat Marauder/Sandwich World" The Masked Meat Marauder - While the Botsfords and the Mings go head to head in a charity car wash, the Butcher battles his new rival--the Masked Meat Marauder. Can this town handle two meat-flinging villains or will WordGirl have to send one of them packing? Sandwich World - Chuck is fed up with living at home in his mom's basement. So he builds an evil lair, Sandwich World! Will WordGirl and Captain Huggy Face escape Chuck's crazy sandwich theme park? And will Chuck be strong enough to avoid getting homesick?. D
Fetch! With Ruff Ruffman "Ruff Fetches The Fetchers" Season Five is about to start but Ruff has been so busy trying to find his long-lost parents, he's forgotten to choose his Season 5 FETCHers! In this casting special, viewers get a behind-the-scenes look at how Ruff goes about picking his contestants. Unfortunately, shortly after selecting his intrepid kids, Ruff loses them on the way to Studio G!. D


It was a beautiful day in late October when Prue and Paul exchanged vows at the Caversham House in Swan Valley. Prue and her bridemaids arrived in style in a strikingly red Chevrolet stretched limousine. The ceremony was held at the magnificent grounds of the Caversham house.
The portrait session was also performed at the Caversham House. Caversham house is well known for its perfectly manicured English and European gardens thus it was a great place for portrait shoots.
The bridal party proceed to the Stirling Square Gardens at Guildford for some fun photo shoots before finally arriving at the Rosehill Estate for the reception.
We managed to sneak some more photo opportunities while the guests were enjoying some pre-dinner drinks at the reception. Situated at South Guildford, the Rosehill Estate consists of an impressive federation style manor and an old style stables building in an enchanting country setting.
Prue and Paul, thank you for sharing your special day with us.
Mandi and Pierre were married in June at the Caversham House. The ceremony was held on the ground by the river. It was a beautiful ceremony witnessed by their dearest and closest family and friends. The Caversham House also has some of the most beautiful settings and thus we had the bridal portrait photo session there after the ceremony.
The reception was held in the Marquee. It was filled with lots of fun, emotions and laughters. Traditional Croatian food, songs and dances were some the best parts of reception as well as the traditional toasts done in full Croatian fashion.


Winton is a town and locality in the Shire of Winton in Central West Queensland, Australia. Australian Age of Dinosaurs and Qantilda Museum are good opportunities to explore the local culture. Spend the 21st (Sun) exploring nature at Bladensburg National Park. Next up on the itinerary: see the interesting displays at Waltzing Matilda Centre, get your game on at Ben's Chicken Stampede, learn more about the world around you at Corfield & Fitzmaurice General Merchants, and make a trip to Arno's Wall.
To find traveler tips and more tourist information, refer to the Winton trip app .
You'll set off for home on the 3rd (Fri).


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1939 Sylvia 2021
Sylvia Campbell
It is with our deepest sorrow, that we announce the passing of our beloved Mother, Ms. Sylvia Joyce Sutherland Campbell, of Lovelock, Nevada. Sylvia passed away peacefully on Monday, March 22, 2021. Sylvia was a former resident of Eureka, Utah where she was raised. She was Valedictorian of her high school graduating class. After graduating from Tintic High School, Eureka, Utah she attended the University of Utah until her marriage to John B. Campbell in 1959. After a brief time living in Salt Lake City, Sylvia moved back to Eureka to raise her family. Sylvia was one of the first of many who helped organize the committee that created the Tintic Historical Society of Eureka. Donating many old antiques from her grandmother's house which are now displayed in the Depot Museum. Sylvia was a member of the Tintic Ladies of Elks, and a member of the Eureka United Methodist Church. Sylvia moved to Reno, Nevada where she returned to college at the age of 50, attending University Nevada Reno, and received her Bachelor's Degree in Elementary Education. She taught Special Education, English as a second language, and tutored adults studying for their GEDs in Lovelock, NV, for the Pershing County School District for 15 years. Sylvia retired in 2007. Sylvia was an avid reader, and sat on the library board in Lovelock, Nevada. She loved gardening in her yard, and doing crossword puzzles. She enjoyed attending luncheons at the local Senior Citizen Center after she retired, where she got her daily dose of socializing.
Sylvia is survived by her son, Shawn Campbell, (Michelle Joiner Olivas) of Lovelock, Nevada, and her two daughters, Nicole (Stanley) Green of Winnemucca, NV, and Traci (Kevin) Schow of Layton, Utah. Her sister, Ann (James) Anderson of Scottsdale Arizona, 3 brothers, John (Brenda) Sutherland Eureka, Utah, Craig (Betty) Sutherland of Peach Spring, Arizona, and Kerry Sutherland (Cathy Gardelli) of Denver, Colorado. She has 8 grandchildren, and several great-grandchildren. She was proceeded in death, by her son, Anthony (Tony) Campbell, her husband, John Campbell, and her parents, Walt and Ila Sutherland.
Viewing will be on Saturday, March 27, 2021 at Brown Family Mortuary, 66 S. 300 E. Santaquin, Utah from 10 am– 12pm. Burial will take place at the Eureka City Cemetery at 1:00 pm following viewing.
The Family would like to thank the staff at Apple Village Assisted Living, Layton, Utah and Bristol Hospice Ogden, Utah for their compassion and care.
To order memorial trees or send flowers to the family in memory of Sylvia Campbell, please visit our flower store.
24 Hour Viewing
Brown Family Mortuary
66 S 300 E, Santaquin, UT 84655
Eureka City Cemetery
, Eureka, UT 84628