#  Revisiting the Direct Fourier Filtering Technique for the Maximal Decay
Rate of Boundary-damped Wave Equation by Finite Differences and Finite
Elements

A. Ö. Özer  Department of Mathematics, Western Kentucky University,
Bowling Green, KY 42101, USA.
E-mail<EMAIL_ADDRESS>Rafi Emran  Department of Mathematics, Western
Kentucky University,
Bowling Green, KY 42101, USA.
E-mail<EMAIL_ADDRESS>
(Date: January 1, 2004)

###### Abstract.

The one-dimensional PDE model of the wave equation with a state feedback
controller at its boundary, which describes wave dynamics of a wide-range of
controlled mechanical systems, has exponentially stable solutions. However, it
is known that the reduced models of the wave equation by the standard Finite
Differences and Finite Elements suffer from the lack of exponential stability
(and exact observability without a state feedback controller) uniformly as the
discretization parameter tends to zero. This is due to the loss of uniform gap
among the high-frequency eigenvalues as the discretization parameter tends to
zero. One common remedy to overcome this discrepancy is the direct Fourier
filtering of the reduced models, where the high-frequency spurious eigenvalues
are filtered out. After filtering, besides from the strong convergency, the
exponential decay rate, mimicking the one for the partial differential
equation counterpart, can be retained uniformly. However, the existing results
in the literature are solely based on an observability inequality of the
control-free model, to which the filtering is implemented. Moreover, the decay
rate as a function of the filtering parameter is implicit. In this paper,
exponential stability results for both filtered Finite Difference and Finite
Element reduced models are established directly by a Lyapunov-based approach
and a thorough eigenvalue estimation.The maximal decay rate is explicitly
provided as a function of the feedback gain and filtering parameter. Our
results, expectedly, mimic the ones of the PDE counterpart uniformly as the
discretization parameter tends to zero. Several numerical tests are provided
to support our results.

###### Key words and phrases:

wave equation $\cdot$ boundary feedback stabilization $\cdot$ computational
issues $\cdot$ model reductions $\cdot$ finite differences $\cdot$ finite
elements $\cdot$ numerical Fourier filtering $\cdot$ maximal decay rate

###### 1991 Mathematics Subject Classification:

35Q60, 35Q93; 74F15; 35Q74, 93B52.

##  1\. Introduction

Let dots denote the derivatives with respect to the time variable $t$.
Consider the standard one-dimensional wave equation clamped at the left. The
boundary feedback is injected at the free end:

$\displaystyle\left\\{\begin{array}[]{ll}\ddot{v}-c^{2}v_{xx}=0,\quad\left(x,t\right)\in\left(0,L\right)\times\mathbb{R}^{+}&\\\
v\left(0,t\right)=0,\quad c^{2}v_{x}\left(L,t\right)=-\xi
v_{t}\left(L,t\right),~{}~{}t\in\mathbb{R}^{+}&\\\
\left[v,v_{t}\right]\left(x,0\right)=\left[v_{0},v_{1}\right]\left(x\right),\quad
x\in\left[0,L\right]\end{array}\right.$ (4)

where $c>0$ is the wave propagation speed and $\xi>0$ is the feedback gain.
The closed-loop system (4) is widely used in the literature to demonstrate
transverse vibrations on a string [11], longitudinal vibrations on a
piezoelectric beam [17], or sound vibrations in a duct [26], etc.

The natural energy of the solutions for (4) is defined as

$\displaystyle
E(t)=\frac{1}{2}\int^{L}_{0}\left[\left|\dot{v}\right|^{2}+c^{2}|v_{x}|^{2}\right]dx.$
(5)

Defining the state $\vec{\psi}=(\psi^{1},\psi^{2})^{\rm T}=(v,\dot{v})^{\rm
T},$ the system (4) can be formed into the first-order form

$\left\\{\begin{array}[]{ll}{\dot{\vec{\psi}}}=\mathcal{A}\vec{\psi}=\begin{bmatrix}0&I\\\
c^{2}\frac{\partial^{2}}{\partial x^{2}}&0\\\ \end{bmatrix}\vec{\psi},\quad
t\in\mathbb{R}^{+}&\\\ \psi^{1}(0)=0,\quad\psi^{1}_{x}(L)=-\xi\psi^{2}(L),\\\
\vec{\psi}(x,0)=\vec{\psi}_{0}(x).\end{array}\right.$ (6)

In fact, the eigenvalues $\\{\lambda(\xi)\\}_{k\in\mathbb{Z}}$ of the non-
self-adjoint system operator $\mathcal{A}$ in (6) are calculated explicitly
[2] as the following

$\displaystyle\lambda_{k}(\xi)=\begin{array}[]{ll}-\frac{c}{2L}ln\left|\frac{\xi+c}{\xi-c}\right|+\frac{(2k+1)\pi
c}{2L}i,&~{}~{}\xi<c.\end{array}$ (8)

Therefore, the fed-back sensor measurement,$\dot{v}(L,t)$ is sufficient to
push all the eigenvalues to the left-half plane, uniformly bounded away from
the imaginary axis. This leads to the known exponential stability result [2,
10].

The same exponential stability result can be also proved directly by a
Lyapunov approach. A Lyapunov function $E_{\delta}(t)$ is constructed by a
$\delta-$perturbation of the energy $E(t)$ by a standard energy-like
functional $F(t)$ as the following

$\displaystyle F(t):=\int_{0}^{L}2\dot{u}xu_{x}dx,\quad
E_{\delta}(t):=E(t)+\delta F(t).$ (9)

######  Theorem 1.

[15, Chap. 5] Let $H^{1}_{L}(0,L):=\\{v\in H^{1}(0,L)~{}:~{}v(0)=0\\}$. There
exist constants $M,\delta,\omega>0$ such that for all initial data
$(v^{0},v^{1})\in H^{1}_{L}(0,L)\times L^{2}(0,L)$, the solutions
$(v,\dot{v})$ of the controlled system (4) with $0<\xi<c$ are exponentially
stable, i.e.

$E(v,\dot{v};t)\leq M(\delta)E(v_{0},v_{1};0)e^{-\sigma(\delta(\xi))t}$

where

$\displaystyle\begin{array}[]{ll}&\delta(\xi)=\frac{1}{2}\min\left\\{\frac{c}{2L},\quad\frac{\xi
c^{2}}{L(c^{2}+\xi^{2})}\right\\},\quad\sigma(\delta)=2\delta\left(1-\frac{2L\delta}{c}\right),\quad
M(\delta)=\frac{c+2L\delta}{c-2L\delta}.\end{array}$ (11)

Hence, the maximal decay rate $\sigma_{max}=\frac{c}{4L}$ is attained with the
optimal feedback gain $\xi=c.$ Indeed, at $\xi=c,$ the spectral abscissa
approaches $-\infty.$ Therefore, the solutions disappear in finite time.

It is also known that the system (4) with $\xi=0$ is exactly observable in the
energy space $H^{1}_{L}(0,L)\times L^{2}(0,L)$. Indeed, proving Theorem 1 is
mathematically equivalent to proving the so-called an observability inequality
for the control-free problem, i.e. $\xi=0.$

######  Theorem 2.

[10, Chap. 3] Consider the solutions $(v,\dot{v})\in H^{1}_{L}(0,L)\times
L^{2}(0,L)$ of the control-free system (4), i.e. $\xi=0$. For any
$T>\frac{2L}{c}$, there exists a constant $C(T)>0$ such that

$\displaystyle\int^{T}_{0}\left|\dot{v}(L,t)\right|^{2}~{}dt\geq C(T)E(0)$
(12)

for all initial data $(v_{0},v_{1})\in H^{1}_{L}(0,L)\times L^{2}(0,L).$

It is widely known that Theorem 1 above fails to hold true as the
discretization parameter $h$ in Finite Difference and Finite Element-based
approximations tends to zero. This discrepancy is first observed in the
pioneer work of [2], where the widely used model reductions failed to mimic
the PDE counterpart (4). The lack of exponential stability of Finite
Difference and Finite Element-based model reductions as the discretization
parameter approaches zero is shown rigorously in [21] by thorough spectral
estimates. Indeed, it is observed that the high-frequency of eigenvalues tend
to the imaginary axis as the discretization parameter approaches zero. Later,
it is shown by a multipliers approach that the lack exponential stability is
due to the the lack of exact observability of the control-free model, i.e.
$\xi\equiv 0$, as the discretization parameter in both Finite Differences and
Finite Element approximations tend to zero. In other words, the observability
constant $C(T)$ in (12) of Theorem 2 turns out to blow up to infinity as the
discretization parameter approaches zero. Investigating this discrepancy
further shows that high-frequency eigenvalues tend to a specific value as the
discretization parameter approaches zero. To remedy this issue, a “direct
Fourier filtering technique”, controlling only the low-frequency part of the
solution in order to eliminate the short wave-length (high-frequency)
components of the solutions, is proposed for the first time in [9] for fully-
clamped boundary conditions. With this approach, the observability results is
recovered fully, see also the detailed review paper on this issue [25]. Since
the numerical approximation is proved to converge to the PDE counterpart, the
high-frequency components of the solutions are retained by choosing the
discretization small enough.

The first attempt of a Finite-Difference-based numerical scheme for (4) is
reported in [23]. The proof of the exponential stability result is solely
based on an observability result, and a numerical viscosity term artificially
added to the system (4), which is referred to the “indirect filtering”
technique in the literature. The proof of exponential stability uses the
decomposition of the PDE model (4) into a control-free problem with non-zero
initial conditions and the controlled problem with zero initial conditions.
The techniques in the proof prevents the rigorous investigation of the maximal
decay rate in terms of the numerical filtering parameter. Later on, with the
direct Fourier filtering technique, as in [9], the exponential stability for
the Finite Difference-based model is shown to be retained [5] as the
discretization parameter approaches zero. The proof outlined in [5] is based
off of the similar decomposition technique as in [23] yet the “direct Fourier
filtering” is implemented to the control-free model, $\xi=0$ in (4). The major
drawback of the exponential stability results in both [5] and [23] is that the
proof of the exponential stability result solely relies on an observability
result of the control-free model. This together with the decomposition
argument make the maximal decay rate analysis, and therefore the analysis of
finding the optimal feedback gain to achieve the maximal decay rate, more
complicated.

It is worthwhile to mention that an alternate semi-discretized Finite-
Difference based model reduction, based on reducing the order of the system
(4) in time and space variables, is reported [14, 22]. This type of model
reduction, similar to the one by mixed-finite elements [2, 3, 16], does not
need any numerical filtering. There are several other remedies proposed in the
literature for interested audience worth to read for, e.g. Tichonoff
regularization [8], non-uniform meshes [6], etc.

In this paper, following the work in [21], the spectral investigation of the
non-self-adjoint system operator $\mathcal{A}$ in (6) is extended for the
estimation of the maximum modulus of the eigenvalues of the system matrices
for each Finite Difference and Finite Element-based model reductions of (4).
It is proved that both reduced models lack uniform observability with respect
to mesh parameter $h\to 0$. By the implementation of the direct Fourier
filtering technique to the closed-loop model reductions directly, the
exponential stability of the model reductions with respect to mesh parameter
is recovered uniformly. The maximal decay rate for each model reduction is
established as a function of the filtering parameter. The exponential
stability results mimic the ones in Theorem 1, and the proofs are solely based
on a Lyapunov approach, discrete multipliers, and through spectral estimates.
Maximal decay rates for both approximations as a function of the filtering
parameter are also provided for each approximation technique.

The overall methodology presented here not only extends the results in [7, 5,
23] but also provides better insights to understand the overall exponential
stability of the Fourier-filtered solutions of (4) by Finite Differences and
Finite Elements [2, 21]. To the best of our knowledge, the exponential
stability of the Finite Elements with direct Fourier filtering is not reported
at all. More importantly, our analysis is applicable to large collections of
wave and beam models.

##  2\. Semi-discretizations of (4) in the $x-$variable

Let $N\in\mathbb{N}$ be given, and define the mesh size $h:=\frac{1}{N+1}$.
Consider a uniform discretization of the interval $[0,L]$:
$0=x_{0}<x_{1}<...<x_{N-1}<x_{N}<x_{N+1}=L.$

###  2.1. Finite Differences with $\xi\equiv 0$

Let $v_{j}=v_{j}(t)\approx v(x_{j},t)$ the approximation of the solution
$v(x,t)$ of (4) at the point space $x_{j}=j\cdot h$ for any $j=0,1,...,N,N+1$,
and $\vec{v}=[v_{1},v_{2},...,v_{N}]^{T}$. Now consider the central
differences for $z^{\prime\prime}(x_{j})\approx(-A_{h}\vec{v})_{j}$ with the
matrix $A_{h}$ defined by

$A_{h}^{FD}:=\frac{c^{2}}{h^{2}}\begin{bmatrix}2&-1&0&\dots&\dots&\dots&0\\\
-1&2&-1&0&\dots&\dots&0\\\ &\ddots&\ddots&\ddots&\ddots&\ddots&\\\
0&\dots&\dots&0&-1&2&-1\\\ 0&\dots&\dots&\dots&0&-1&1\\\
\end{bmatrix}_{N\times N}$ (13)

whose eigen-pairs $(\mu^{FD}_{k}(h),\vec{\phi}_{k}(h))$ are [5]:

$\left\\{\begin{array}[]{ll}\mu_{k}^{FD}(h)=\frac{4c^{2}}{h^{2}}\sin^{2}\left(\frac{(2k-1)\pi
h}{2(2-h)L}\right),&\\\ \phi_{k,j}=\sin\left(\frac{(2k-1)j\pi
h}{L(2-h)}\right),\quad k,j=1,2,...,N.&\end{array}\right.$ (14)

Considering no feedback $\xi\equiv 0$ in (4), the following discretization is
obtained

$\left\\{\begin{array}[]{ll}\vec{\ddot{v}}+A_{h}^{FD}\vec{v}=0,\quad
t\in\mathbb{R}^{+}&\\\ v_{0}=0,\quad v_{N+1}-v_{N}=0,&\\\
v_{j}(0)=v^{0}_{j},~{}\dot{v}_{j}(0)=v^{1}_{j},\quad
j=0,...,N+1.\end{array}\right.$ (15)

The discretized energy corresponding to (5) is

$\displaystyle\begin{array}[]{ll}&E_{h,0}^{FD}(t):=\frac{h}{2}\sum\limits_{j=0}^{N}\rho\left|\dot{v}_{j}\right|^{2}+c^{2}\left|\frac{v_{j+1}-v_{j}}{h}\right|^{2}.\end{array}$
(17)

Defining $\mathcal{A}_{h}^{FD}$ by
$\mathcal{A}_{h}^{FD}(\vec{u}_{1},\vec{u}_{2}):=\left(\vec{u}_{2},-A_{h}^{FD}\vec{u}_{1}\right),$
and calling $\vec{y}_{h}=(\vec{v},\dot{\vec{v}})$, (15) can be reformulated as

$\frac{d}{dt}\vec{y}_{h}=\mathcal{A}_{h}^{FD}\vec{y}_{h}.$ (18)

######  Lemma 1.

[5] For $K:=\\{-N,...,-1,1,...N\\},$ the eigen-pairs
$\\{(\lambda_{k}^{FD}(h),\vec{\varphi}_{k}(h))\\}_{K}$ of
$\mathcal{A}_{h}^{FD}$ are

$\begin{array}[]{ll}(\lambda_{k}^{FD},\vec{\varphi}_{k})=\left(i\sqrt{\mu^{FD}_{k}(h)},\begin{bmatrix}\frac{1}{i\sqrt{\mu^{FD}_{k}(h)}}\vec{\phi}_{k}(h)\\\
\vec{\phi}_{k}(h)\end{bmatrix}\right),\end{array}$ (19)

and $\sqrt{\lambda_{k}^{FD}(h)}:=-\sqrt{\lambda_{-k}^{FD}(h)}$ and
$\vec{\varphi}_{k}:=\vec{\varphi}_{-k}$ for $k=-1,-2,...,-N.$ Therefore, the
solutions to (15) can be expressed as

$\vec{v}(t)=\sum_{k\in
K}(a_{k}e^{i\sqrt{\lambda_{k}^{FD}(h)}t})\vec{\phi}_{k}(h).$ (20)

###  2.2. Finite Elements with $\xi\equiv 0$

First, multiply both sides of the equation in (4) by a continuously
differentiable test function $\phi(x)\in C_{0}^{\infty}[0,L]$, and integrate
both sides of the equation over $[0,L]$ to get

$\displaystyle\int_{0}^{L}u_{tt}\phi~{}dx+\int_{0}^{L}u_{x}\phi_{x}~{}dx=0.$
(21)

At each node $\left\\{x_{i}\right\\}_{i=1}^{N}$, the following linear splines
are defined

$\displaystyle\phi_{i}(x)=\left\\{\begin{array}[]{ll}\frac{1}{h}(x-x_{i}),&x_{i-1}<x<x_{i}\\\
\frac{1}{h}(x-x_{i+1}),&x_{i}<x<x_{i+1}\\\
0,&{\textrm{o}therwise,}\end{array}\right.$ (25)
$\displaystyle\phi_{N+1}(x)=\left\\{\begin{array}[]{ll}\frac{1}{h}(x-x_{N}),&x_{N}<x<x_{N+1}\\\
0,&{\textrm{o}therwise}.\end{array}\right.$ (28)

Defining the $(N+1)\times(N+1)$ matrices $A_{h}$ and $M$ by

$A_{h}^{FEM}:=\frac{c^{2}}{h^{2}}\begin{bmatrix}2&-1&0&\dots&\dots&\dots&0\\\
-1&2&-1&0&\dots&\dots&0\\\ &\ddots&\ddots&\ddots&\ddots&\ddots&\\\
0&\dots&\dots&0&-1&2&-1\\\ 0&\dots&\dots&\dots&0&-1&1\\\ \end{bmatrix},$ (29)
$\displaystyle M:=\left(\begin{array}[]{cccccc}2/3&1/6&0&0&\ldots&0\\\
1/6&2/3&1/6&0&\ldots&0\\\ 0&1/6&2/3&1/6&\ldots&0\\\
\vdots&\vdots&\vdots&\ddots&\vdots&\vdots\\\ 0&0&0&1/6&2/3&1/6\\\
0&0&\ldots&0&1/6&1/3\\\ \end{array}\right),$ (36)

and seeking solutions to (21) with $\xi\equiv 0$ of the form
$v(x,t)=\sum\limits_{i=0}^{N+1}v_{i}(t)\phi_{i}(x)$ leads to

$\left\\{\begin{array}[]{ll}\vec{\ddot{v}}+M^{-1}A^{FEM}_{h}\vec{v}=0,\quad
t\in\mathbb{R}^{+}&\\\ v_{0}=0,\quad
h\frac{2\ddot{v}_{N+1}+\ddot{v}_{N}}{6}+c^{2}\frac{v_{N+1}-v_{N}}{h}=0,&\\\
v_{j}(0)=v^{0}_{j},~{}\dot{v}_{j}(0)=v^{1}_{j},\quad
j=0,...,N+1.\end{array}\right.$ (37)

The discretized energy corresponding to (37) is defined by

$\displaystyle\begin{array}[]{ll}E_{h}^{FEM}(t):=\frac{h}{12}\left[|\dot{v}_{N+1}|^{2}+\sum_{j=1}^{N}\left(2\left|\dot{v}_{j}\right|^{2}+\left|\dot{v}_{j}+\dot{v}_{j+1}\right|^{2}+6c^{2}\left|\frac{v_{j+1}-v_{j}}{h}\right|^{2}\right)\right].\end{array}$
(39)

Defining
$\mathcal{A}_{h}^{FEM}(\vec{u}_{1},\vec{u}_{2}):=\left(\vec{u}_{2},-M^{-1}A^{FEM}_{h}\vec{u}_{1}\right),$
and calling $\vec{y}_{h}=(\vec{v},\dot{\vec{v}})$, (37) can be reformulated as

$\frac{d}{dt}\vec{y}_{h}=\mathcal{A}_{h}^{FEM}\vec{y}_{h}.$ (40)

Introduce an $(N+1)\times(N+1)$ diagonal matrix $K$ by
$K={\mathrm{d}iag}(2,\ldots,2,1)$ so that

$\displaystyle\begin{array}[]{ll}\lambda_{j}\left(M^{-1}A_{h}^{FEM}\right)=\lambda_{j}\left(M^{-1}KK^{-1}A_{h}^{FEM}\right)=\frac{\lambda_{j}\left(K^{-1}A_{h}^{FEM}\right)}{\lambda_{j}\left(K^{-1}M\right)},~{}j=1,\ldots,N+1,\end{array}$
(42)

where matrices $K^{-1}A_{h}$ and $K^{-1}M$ are band-matrices:

$\displaystyle\begin{array}[]{ll}K^{-1}A_{h}^{FEM}=\frac{c^{2}}{2h^{2}}\left(\begin{array}[]{cccccc}2&-1&0&0&\ldots&0\\\
-1&2&-1&0&\ldots&0\\\ \vdots&\vdots&\vdots&\ddots&\vdots&\vdots\\\
0&0&0&-1&1&-1\\\ 0&0&\ldots&0&-2&2\\\ \end{array}\right),\\\
K^{-1}M=\left(\begin{array}[]{cccccc}\frac{1}{3}&\frac{1}{12}&0&0&\ldots&0\vskip
3.0pt plus 1.0pt minus 1.0pt\\\
\frac{1}{12}&\frac{1}{3}&\frac{1}{12}&0&\ldots&0\vskip 3.0pt plus 1.0pt minus
1.0pt\\\ \vdots&\vdots&\vdots&\ddots&\vdots&\vdots\\\
0&0&0&\frac{1}{12}&\frac{1}{3}&\frac{1}{12}\vskip 3.0pt plus 1.0pt minus
1.0pt\\\ 0&0&\ldots&0&\frac{1}{6}&\frac{1}{3}\\\
\end{array}\right).\end{array}$ (55)

The next two lemmas are necessary for finding the eigenvalues of matrices
(55).

######  Lemma 2.

The eigenvalues and eigenfunctions of the matrix $K^{-1}A_{h}^{FEM}$,
respectively, are

$\left\\{\begin{array}[]{ll}\lambda_{j}(h)=\frac{2c^{2}}{h^{2}}\sin^{2}\left(\frac{(2j-1)\pi}{4N}\right),&\\\
u_{j,k}=\sin\left(\frac{(2j-1)k\pi}{2N}\right),~{}j,k=1,2,...,N+1.\end{array}\right.$
(56)

Proof: Letting $u=[u_{1},u_{2},\ldots,u_{N}]^{T}$, the eigenvalue problem for
$k=1,2,\ldots,N+1$ is

$\displaystyle\left\\{\begin{array}[]{ll}-u_{k-1}+(2-2h^{2}\lambda)u_{k}-u_{k+1}=0,&\\\
u_{0}=0,~{}~{}u_{N+1}=u_{N}.\end{array}\right.$ (59)

With $u_{k}=z^{k}$, $u_{k-1}=z^{k-1}$ and $u_{k+1}=z^{k+1}$, both $z^{k-1}\neq
0$ and $(-1+(2-2h^{2}\lambda)z-z^{2})=0.$ Thus, the general solution to (59)
is

$\displaystyle u_{k}=c_{1}z^{k}(\lambda)+c_{2}z^{-k}(\lambda).$ (60)

By the boundary conditions,

$\displaystyle\begin{array}[]{ll}&z_{j}=e^{\frac{i(2j-1)k\pi}{2N}},~{}j=1,\ldots,N+1.\\\
\end{array}$ (62)

Therefore, substituting $z_{j}$ into (60) leads to

$\displaystyle\begin{array}[]{ll}u_{j,k}=sin\left(\frac{(2j-1)k\pi
h}{2(L-h)}\right),\quad j,k=1,2,\ldots,N+1.\end{array}$ (64)

Lastly, the eigenvalues are solved by using $z_{1}z_{2}=1$ and
$z_{1}+z_{2}=2-2h^{2}\lambda_{j}$:

$\displaystyle
2\left(1-2\sin^{2}\left({\frac{(2j-1)\pi}{4N}}\right)\right)=2-2h^{2}\lambda_{j}.$
(65)

Therefore, noting $N=\frac{L-h}{h}$, (56) is obtained. $\square$

######  Lemma 3.

For $j=1,\ldots,N+1,$ the eigenvalues $M^{-1}A_{h}^{FEM}$ of (40) are given by

$\displaystyle{\lambda}_{j}(M^{-1}A_{h}^{FEM})=\frac{1}{h^{2}}\frac{6-6\cos{\left(\frac{(2j-1)\pi
h}{2(L-h)}\right)}}{2+\cos{\left(\frac{(2j-1)\pi h}{2(L-h)}\right)}}.$ (66)

Proof: Define the $(N+1)\times(N+1)$ matrix $J$ by
$J:={\mathrm{tridiag}(1,0,1)}.$ One can readily verify that
$h^{2}K^{-1}A=I-K^{-1}J$ and $6K^{-1}M=2I+K^{-1}J.$ Since
$h^{2}K^{-1}A_{h}^{FEM}$ and $6K^{-1}M$ are diagonalizable, there exists an
invertible matrix P such that $6P^{-1}K^{-1}MP=2I+P^{-1}K^{-1}JP.$ Lastly,
letting $P^{-1}K^{-1}AP=D$ where D is a diagonal matrix,
$6P^{-1}K^{-1}MP=3I-h^{2}D$,
$P^{-1}K^{-1}MP=\frac{1}{6}\left(3I-h^{2}D\right).$ Therefore, the eigenvalues
of the matrix $K^{-1}M$ lie on the diagonal of matrix $P^{-1}K^{-1}MP$,

$\displaystyle\lambda_{j}\left(K^{-1}M\right)=\frac{1}{6}\left(3-h^{2}\lambda_{j}\left(K^{-1}A_{h}^{FEM}\right)\right).~{}$
(67)

Therefore,

$\displaystyle\begin{array}[]{ll}\lambda_{j}\left(M^{-1}A_{h}^{FEM}\right)&=\frac{\lambda_{j}\left(K^{-1}A_{h}^{FEM}\right)}{\lambda_{j}\left(K^{-1}M\right)}=\frac{\frac{12}{h^{2}}\sin^{2}{\left(\frac{(2j-1)\pi
h}{4(L-h)}\right)}}{2\left(\frac{3}{2}-\sin^{2}{\left(\frac{(2j-1)\pi
h}{4(L-h)}\right)}\right)}.\end{array}$ (69)

Following the sub-eigenvalue problem (42), (66) is obtained. $\square$

######  Theorem 3.

It can be shown that the eigen-pairs
$\\{\lambda_{k}^{FEM}(h),\vec{\varphi}_{k}(h)\\}_{K}$ of $\mathcal{A}_{h}$ are

$\begin{array}[]{ll}\left(i\sqrt{\lambda_{k}(M^{-1}A_{h}^{FEM})},\begin{bmatrix}\frac{1}{i\sqrt{\lambda_{k}(M^{-1}A_{h}^{FEM})}}\vec{\phi}_{k}\\\
\vec{\phi}_{k}\end{bmatrix}\right),\end{array}$ (70)

where $\sqrt{\lambda_{k}^{FEM}(h)}:=-\sqrt{\lambda_{-k}^{FEM}(h)}$ and
$\vec{\varphi}_{k}:=\vec{\varphi}_{-k}$ for $k=-1,-2,...,-N.$ The solutions to
(15) can be expressed as

$\vec{v}(t)=\sum_{k\in
K}(a_{k}e^{i\sqrt{\lambda_{k}^{FEM}(h)}t})\vec{\phi}_{k}(h).$ (71)

##  3\. Lack of Observability and Exponential Stability as $h\to 0$

It can be shown easily the observability result stated in Theorem 2 does not
hold uniformly for the discretized models since

$\displaystyle\left\\{\begin{array}[]{ll}\frac{h^{2}|\lambda_{N+1}^{FD}(h)|^{2}}{c^{2}}\to
4,\qquad\frac{h^{2}|\lambda_{N+1}^{FEM}(h)|^{2}}{c^{2}}\to 12&{\rm
as}~{}N\to\infty.\end{array}\right.$ (73)

######  Lemma 4.

Considering either (15) or (37), for any $T>0,$ respectively,

$\displaystyle\lim\limits_{h\to
0}\sup\frac{E_{h,0}^{FD}(0)}{\int_{0}^{T}\left|\frac{\dot{v}_{N}}{h}\right|^{2}},\qquad\lim\limits_{h\to
0}\sup\frac{E^{FEM}_{h}(0)}{\int_{0}^{T}\left|\frac{\dot{v}_{N}}{h}\right|^{2}}\to\infty.$
(74)

Proof: The first limit is proved in [5]. The second limit follows from the
same argument together with Theorem 3. $\square$

###  3.1. Finite Differences with $\xi>0$

First, observe that the absorbing boundary conditions
$c^{2}v_{x}(L,t)=-\xi\dot{v}(L,t)$ in (4) can be approximated by

$\displaystyle\begin{array}[]{ll}c^{2}v_{xx}(x_{N+1})&\approx
c^{2}\frac{v_{x}(x_{N+1})-v_{x}(x_{N})}{h}=\frac{-\xi\dot{v}_{N+1}-c^{2}\left(\frac{v_{N+1}-v_{N}}{h}\right)}{h}.\end{array}$

Then, the following Finite-Difference approximation for (4) can be considered

$\left\\{\begin{array}[]{ll}\left\\{\ddot{v}_{j}-c^{2}\left(\frac{v_{j+1}+v_{j-1}-2v_{j}}{h^{2}}\right)=0\right\\}_{j=1}^{N},\\\
v_{0}=0,\ddot{v}_{N+1}+c^{2}\frac{v_{N+1}-v_{N}}{h^{2}}+\xi\frac{\dot{v}_{N+1}}{h}=0,\\\
v_{j}(0)=v^{0}_{j},~{}\dot{v}_{j}(0)=v^{1}_{j},\quad
j=0,...,N+1.\end{array}\right.$ (76)

The energy $E_{h,0}^{FD}(t)$, defined in (17) for the case $\xi\neq 0$, is now
redefined as the following

$\displaystyle\begin{array}[]{ll}&E_{h}^{FD}(t):=\frac{h}{2}\sum\limits_{j=0}^{N+1}\rho\left|\dot{v}_{j}\right|^{2}+c^{2}\frac{h}{2}\sum\limits_{j=0}^{N}\left|\frac{v_{j+1}-v_{j}}{h}\right|^{2}.\end{array}$
(78)

Letting $\vec{u}_{1,h}=(v_{1},v_{2},...,v_{N+1})^{\rm T}$,
$\vec{u}_{2,h}=(\dot{v}_{1},...,\dot{v}_{N+1})^{\rm T},$
$\vec{y}_{h}=(\vec{u}_{1,h},\vec{u}_{2,h})^{\mathrm{T}},$ the system (76) is
written in the first-order form

${\dot{\vec{y}}_{h}}=\mathcal{A}_{h}^{FD}(\xi)\vec{y}_{h}=(\vec{y}_{2,h},-A_{h}\vec{y}_{1,h}+B_{h}\vec{y}_{2,h})^{\mathrm{T}}$
(79)

where $A_{h}$ is the matrix defined by (29),
and$B_{N+1,N+1}=\frac{-\xi}{h}\neq 0$ and otherwise $B_{i,j}\equiv 0$ for any
other $i,j.$

Consider the eigenvalue problem for (79):

$\displaystyle\mathcal{A}_{h}^{FD}(\xi)\vec{y}=\lambda^{FD}(\xi,h)\vec{y}.$
(80)

For $0<\xi<c,$ it can be shown that the real part of the eigenvalues are
negative, e.g. ${\rm
Re}\lambda^{FD}(\xi,h)=\frac{-L\xi|\lambda^{FD}(\xi,h)y_{1,N+1}|^{2}}{hc\left(|\lambda^{FD}(\xi,h)\vec{y}_{1,h}|^{2}-\vec{y}_{1,h}^{\rm
T}A\vec{y}_{1,h})\right)}<0.$

Seeking the solution of (80) of the form $y_{1,k}=z^{2k}-z^{-2k}$ with
$k=1,2,\ldots,N+1$ and $z\in\mathbb{C}$ leads to
$\lambda^{FD}(\xi,h)=c(N+1)(z-z^{-1})$ where $z$ satisfies $z^{2}\neq\mp 1$
and

$p(z)=z^{4N+6}+\frac{\xi}{c}z^{4N+5}-\frac{\xi}{c}z+1=0.$ (81)

Note that the eigenvalues $\lambda^{FD}(\xi,h)$ of $\mathcal{A}_{h}^{FD}(\xi)$
solve (81), but some of the solutions of (81) are not related to the
eigenvalues $\lambda^{FD}(\xi,h)$. In fact, since $\mathcal{A}_{h}^{FD}(\xi)$
is a real matrix, it is sufficient to identify all eigenvalues of
$\mathcal{A}_{h}^{FD}(\xi)$ with positive imaginary part.

Moreover, observe that $z=\mp i$ are the roots of $p.$ On the other hand, if
$z$ is a root of $p(z),$ then $-z$ and $z^{-1}$ are both the roots of $p(z).$
As $z\neq-\frac{\xi}{c},$ (81) is equivalent to

$z^{4N+5}=\frac{\xi z-c}{\xi+cz}.$ (82)

The following list of results are proved in [21].

######  Lemma 5.

[21, Prop. 3.3] The polynomial $p(z)$ satisfies the following:

  * 1.

The only roots of $p(z)$ on the unit circle are $z=\mp i.$ Moreover, $z=\mp i$
are the only root of $p(z)$ with purely imaginary parts.

  * 2.

All roots of $p(z)$ with positive real parts must be within the unit circle.
All roots of $p(z)$ with negative real parts must be outside the unit circle.

It is inferred from these results that each root $z$ of $p(z)$ in the first
quadrant of the complex plane leads to three other roots of $p(z):$
$\bar{z},z^{-1},{\bar{z}}^{-1}.$ Therefore, the eigenvalue of
$\mathcal{A}_{h}^{FD}(\xi)$ corresponding to the roots of $p(z)$ has positive
imaginary part if and only if $z$ has positive imaginary part. As a result, we
only estimate the roots of $p(z)$ in the first quadrant.

Following the clever discussion in [21], the analysis of the distribution of
the roots of (81) is solely based on the analysis of the mapping defined by
$\mathcal{T}(z):={\mathcal{G}}^{\frac{1}{4N+5}}(z)$ where
${\mathcal{G}}(z)=\frac{\xi z-c}{zc+\xi}.$ Indeed, a fixed point $z$ of
$\mathcal{T}$ is a root of the polynomial (81). Therefore, the root of $p$
satisfies

$z=[\mathcal{G}(z)]^{\frac{1}{4N+5}}=\left(\frac{\xi
z-c}{zc+\xi}\right)^{\frac{1}{4N+5}}.$

Define the following sector in the first quadrant of $\mathbb{C}:$

$S=\\{z\in\mathbb{C}~{}|~{}Re(z)\geq 0,Im(z)\geq
0,|z|\leq\frac{\xi+c}{2\xi}\\}.$

Since $T$ is multi-valued, considering the branches ${\mathcal{T}}_{j}$ of
$\mathcal{T}$ as the following

$\displaystyle\begin{array}[]{ll}&{\mathcal{T}}_{j}(z)=\mathcal{G}(z)]^{\frac{1}{4N+5}}e^{\frac{i(\theta(z)+2j\pi)}{4N+5}},~{}~{}1<j\leq
4N+4,\\\ &\theta(z)={\textrm{A}rg}({\mathcal{G}}(z)).\end{array}$ (85)

######  Lemma 6.

[21, Prop. 3.4 & 3.5, Cor. 3.1] The following results hold:

  * 1.

The complex functions ${\mathcal{T}}_{j},$ $j=1,\ldots 4N+4,$ are analytic
over $S,$ and ${\mathcal{T}}_{j}$ is a contraction mapping over $S$ for $N$
large enough.

  * 2.

Since $G$ is analytic over $S,$ there exists a constant $M_{G}>0$ such that
$|G(z)|<M_{G}$ for all $z\in S.$

  * 3.

As well, for every $j\in[0,N],$ the subsections $S_{j}$ of $S$ defined by

$S_{j}:=\left\\{z\in
S,{\textrm{A}rg}(z)\in\left[\frac{2j\pi}{4N+5},\frac{(2j+1)\pi}{4N+5}\right]\right\\}$

is invariant under ${\mathcal{T}}_{j}$ for large enough $N$.

  * 4.

The fixed point $z$ of $\mathcal{T}_{j}$ in $S_{j}$ satisfies

$\displaystyle|z|\geq(\frac{\xi c-\xi^{2}}{2\xi^{2}+\xi
c+c^{2}})^{\frac{1}{4N+5}}-\sqrt{2}(1+\frac{c}{\xi})\frac{{M_{G}}^{\frac{1}{4N+5}}}{4N+5},$
(86)
$\displaystyle|z|\leq\left(\frac{c}{\xi}\right)^{\frac{1}{4N+5}}+\left(1+\frac{c}{\xi}\right)\frac{{M_{G}}^{\frac{1}{8N+10}}}{4N+5}.$
(87)

Now, notice that

$\displaystyle\begin{array}[]{ll}{\mathrm{R}e}\lambda_{j}^{FD}(\xi,h)=\frac{c(N+1)}{L}(z_{j}-\frac{1}{z_{j}})\cos{({\textrm{A}rg}~{}z_{j})},\\\
{\mathrm{I}m}\lambda_{j}^{FD}(\xi,h)=\frac{c(N+1)}{L}(z_{j}+\frac{1}{z_{j}})\sin{({\textrm{A}rg}~{}z_{j})},\end{array}$
(90)

where
$Arg(z_{j})=Arg(T_{j}(z_{k}))=\frac{\theta(z_{j})+2j\pi}{4N+5},j=0,1,\ldots
4N+4,$ and

$\begin{array}[]{ll}&\theta(z_{j})=\pi-\arctan\left(\frac{Im~{}\mathcal{G}(z_{j})}{Re~{}\mathcal{G}(z_{j})}\right),\qquad\frac{\pi}{2}\leq\theta(z_{j})\leq\frac{3\pi}{2}.\end{array}$
(91)

Finally, the following results describe the eigenvalues of
$\mathcal{A}_{h}^{FD}(\xi)$ in terms of the roots of (81).

######  Theorem 4.

[21, Thm 3.1] For sufficiently large $N$, all roots of $p(z)$ are simple.
There are exactly $N+1$ roots in the first quadrant of the complex plane. Each
subset $S_{k}$ contains exactly one root $z_{j}$ of $p(z)$ for $j=1,2,\ldots
N.$ As a result, the matrix $\mathcal{A}_{h}^{FD}(\xi)$ has $2N+2$ eigenvalues
$\lambda_{j}^{FD}(\xi,h).$ Moreover, this implies that
$Re~{}\lambda_{j}^{FD}(\xi,h)\rightarrow 0$ as $N\to\infty$. Hence, the system
(76) lacks of exponential stability uniformly as $N\to\infty$.

Now we are ready to state one of our main results, which helps estimate the
eigenvalues further:

######  Theorem 5.

For all $j=1,2,\ldots,2N+2$ and sufficiently large $N$ (or small enough $h$),
the eigenvalues $\lambda_{j}^{FD}(\xi,h)$ of $\mathcal{A}_{h}^{FD}(\xi)$
satisfy

$\displaystyle h~{}|Re\lambda_{j}^{FD}(\xi,h)|=O(h),$ (92) $\displaystyle
h~{}|Im\lambda_{j}^{FD}(\xi,h)|=2c\left|\sin\left(\frac{2j\pi}{4N+5}\right)\right|+O(h).$
(93)

Hence,

$h^{2}|\lambda_{j}^{FD}(\xi,h)|^{2}\leq
4c^{2}\sin^{2}\left(\frac{2j\pi}{4N+5}\right)+O(h).$ (94)

Proof: First observe that (92) holds by (91) since

$\displaystyle\begin{array}[]{ll}|h~{}Re\lambda_{j}^{FD}(\xi,h)|&=ch\frac{N+1}{L}|(|z_{j}|-|z_{j}|^{-1})\cos(Arg(z_{j}))|\\\
&=ch\frac{1}{h}\left|\left(\frac{|z_{j}|^{2}-1}{|z_{j}|}\right)\cos\left(\frac{2j\pi}{4N+5}+\frac{\theta(z_{j})}{4N+5}\right)\right|\\\
&=c\left|\frac{-1-|z_{j}|}{|z_{j}|}\right||1-|z_{j}||\left|\cos\left(\frac{2j\pi}{4N+5}\right)\cos\left(\frac{\theta(z_{j})}{4N+5}\right)-\sin\left(\frac{2j\pi}{4N+5}\right)\sin\left(\frac{\theta(z_{j})}{4N+5}\right)\right|,\end{array}$

and by utilizing (86) and (87) leading to (92)

$\displaystyle\begin{array}[]{ll}|h~{}Re\lambda_{j}^{FD}(\xi,h)|&=ch\frac{N+1}{L}|(|z_{j}|-|z_{j}|^{-1})\cos(Arg(z_{j}))|\\\
&=ch\frac{1}{h}\left|\left(\frac{|z_{j}|^{2}-1}{|z_{j}|}\right)\cos\left(\frac{2j\pi}{4N+5}+\frac{\theta(z_{j})}{4N+5}\right)\right|\\\
&=c\left|\frac{-1-|z_{j}|}{|z_{j}|}\right||1-|z_{j}||\left|\cos\left(\frac{2j\pi}{4N+5}\right)\cos\left(\frac{\theta(z_{j})}{4N+5}\right)-\sin\left(\frac{2j\pi}{4N+5}\right)\sin\left(\frac{\theta(z_{j})}{4N+5}\right)\right|\\\
&\leq c\left|-1-\left(\frac{\xi c-\xi^{2}}{2\xi^{2}+\xi
c+c^{2}}\right)^{\frac{1}{4N+5}}+\sqrt{2}(1+\frac{c}{\xi})\frac{M^{\frac{1}{4N+5}}}{4N+5}\right|\\\
&\quad\times\left|1-\left(\frac{\xi c-\xi^{2}}{2\xi^{2}+\xi
c+c^{2}}\right)^{\frac{1}{4N+5}}+\sqrt{2}(1+\frac{c}{\xi})\frac{M^{\frac{1}{4N+5}}}{4N+5}\right|\times|1+O(h)|\\\
&=c\left|-1-1+O(h)\right|O(h)\\\ &=O(h).\end{array}$

On the other hand, analogously, by (86)-(91)

$\displaystyle\begin{array}[]{ll}\left|h~{}Im\lambda_{j}^{FD}(\xi,h)\right|&=h\frac{c(N+1)}{L}|(|z_{j}|+|z_{j}|^{-1})\sin(Arg(z_{j}))|\\\
&=c||z_{j}|+|z_{j}|^{-1}|\left|\sin\left(\frac{2j\pi}{4N+5}+\frac{\theta(z_{j})}{4N+5}\right)\right|\\\
&=c||z_{j}|+|z_{j}|^{-1}|\left|\sin\left(\frac{2j\pi}{4N+5}\right)\cos\left(\frac{\theta(z_{j})}{4N+5}\right)+\cos\left(\frac{2j\pi}{4N+5}\right)\sin\left(\frac{\theta(z_{j})}{4N+5}\right)\right|\\\
&=c||z_{j}|+|z_{j}|^{-1}|\left|\sin\left(\frac{2j\pi}{4N+5}\right)+O(h)\right|\\\
&\leq
c\left|\left(\frac{c}{\xi}\right)^{\frac{1}{4N+5}}+\left(1+\frac{c}{\xi}\right)\frac{M^{\frac{1}{8N+10}}}{4N+5}+\frac{1}{\left(\frac{\xi
c-\xi^{2}}{2\xi^{2}+\xi
c+c^{2}}\right)^{\frac{1}{4N+5}}-\sqrt{2}(1+\frac{c}{\xi})\frac{M^{\frac{1}{4N+5}}}{4N+5}}\right|\\\
&\qquad\qquad\times\left|\sin\left(\frac{2j\pi}{4N+5}\right)+O(h)\right|\\\
&\leq
c\left|\sin\left(\frac{2j\pi}{4N+5}\right)+O(h)\right|\left|\left(\frac{c}{\xi}\right)^{\frac{1}{4N+5}}+O(h)+\frac{1}{\left(\frac{\xi
c-\xi^{2}}{2\xi^{2}+\xi c+c^{2}}\right)^{\frac{1}{4N+5}}+O(h)}\right|\\\ &\leq
c\left|\sin\left(\frac{2j\pi}{4N+5}\right)+O(h)\right|\left|1+O(h)+\frac{1}{1+O(h)}\right|\\\
&\leq 2c\left|\sin\left(\frac{2j\pi}{4N+5}\right)\right|+O(h).\end{array}$

Finally, (94) follows from (92) and (93). $\square$

###  3.2. Finite Elements with $\xi>0$

Considering the discretized model by Finite Elements

$\left\\{\begin{array}[]{ll}\left\\{\frac{\ddot{v}_{j+1}+4\ddot{v}_{j}+\ddot{v}_{j-1}}{6}+c^{2}\frac{v_{j+1}-2v_{j}+v_{j-1}}{h^{2}}=0\right\\}_{j=1}^{N},&\\\
v_{0}=0,~{}\frac{2\ddot{v}_{N+1}+\ddot{v}_{N}}{6}+c^{2}\frac{v_{N+1}-v_{N}}{h^{2}}=-\frac{\xi}{h}\dot{v}_{N+1},&\\\
v_{j}(0)=v^{0}_{j},~{}\dot{v}_{j}(0)=v^{1}_{j},\quad
j=0,...,N+1,\end{array}\right.$ (98)

the analysis and arguments above can be replicated analogously, see [21] for
more in-depth discussion, to get the following result.

######  Theorem 6.

For all $j=1,2,\ldots,2N+2$ and sufficiently large enough $N$ (or small enough
$h$ ), the eigenvalues $\lambda_{j}^{FEM}(\xi,h)$ of
$\mathcal{A}_{h}^{FEM}(\xi)$ satisfy

$h^{2}|\lambda_{j}^{FEM}(\xi,h)|^{2}\leq
12c^{2}\sin^{2}\left(\frac{2j\pi}{4N+5}\right)+O(h).$ (99)

##  4\. Exponential Stability as $h\to 0$

###  4.1. Finite Differences with $\xi\neq 0$

######  Lemma 7.

The system (76) is dissipative, i.e.

$\displaystyle\frac{dE_{h}^{FD}}{dt}+\xi|\dot{v}_{N+1}|^{2}=0.$ (100)

Proof: Multiply both sides of (76) by $h\dot{v}_{j}$ and take sum from $j=1$
to $N:$

$\displaystyle\begin{array}[]{ll}\sum\limits_{j=1}^{N}h\dot{v}_{j}\ddot{v}_{j}-c^{2}\sum\limits_{j=1}^{N}\frac{v_{j+1}-2v_{j}+v_{j-1}}{h}\dot{v}_{j}=0.\end{array}$
(102)

Since

$\displaystyle\begin{array}[]{ll}&-hc^{2}\sum\limits_{j=1}^{N}\frac{v_{j+1}-2v_{j}+v_{j-1}}{h^{2}}\dot{v}_{j}=\frac{c^{2}}{h}(v_{N}-v_{N+1})\dot{v}_{N+1}+\frac{c^{2}}{h}\sum\limits_{j=0}^{N}(v_{j+1}-v_{j})(\dot{v}_{j+1}-\dot{v}_{j}),\end{array}$
(104)

substituting (104) into (102) yields

$\displaystyle\begin{array}[]{ll}\frac{dE_{h}^{FD}}{dt}-\frac{c^{2}}{h}(v_{N+1}-v_{N})\dot{v}_{N+1}-h\dot{v}_{N+1}\ddot{v}_{N+1}=0,\end{array}$
(106)

and this together with the boundary conditions (76), (100) follows. $\square$

Following (9), define the following Lyapunov functional

$\displaystyle L_{h}^{FD}(t):=E_{h}^{FD}+\delta F_{h}^{FD}(t)$ (107)

where the auxiliary function $F_{h}^{FD}$ is defined by

$\displaystyle\begin{array}[]{ll}F_{h}^{FD}(t)=\sum\limits_{j=1}^{N}jh\dot{v}_{j}\left(\frac{v_{j+1}-v_{j-1}}{2}\right)+\frac{L}{2}(v_{N+1}-v_{N})\dot{v}_{N+1}-\frac{L\xi
h}{4c^{2}}|\dot{v}_{N+1}|^{2}.\end{array}$ (109)

######  Lemma 8.

For $0<\delta<\frac{c}{L},$ $L_{h}^{FD}(t)$ is equivalent to $E_{h}^{FD}(t),$

$\left(1-\frac{L\delta}{c}\right)E_{h}^{FD}\leq
L_{h}^{FD}\leq\left(1+\frac{L\delta}{c}\right)E_{h}^{FD}.$ (110)

Proof: First of all, by Hölders, and Cauchy-Schwartz inequalities for sums,
the first and second terms in (109) are estimated as the following

$\displaystyle\begin{array}[]{ll}&\left|\sum\limits_{j=1}^{N}jh\dot{v}_{j}\left(\frac{v_{j+1}-v_{j-1}}{2}\right)\right|\leq\frac{Lh}{2c}\left[\sum\limits_{j=0}^{N+1}|\dot{v}_{j+1}|^{2}+\sum\limits_{j=0}^{N}c^{2}\left|\frac{v_{j+1}-v_{j}}{h}\right|^{2}-\frac{c^{2}}{2}\left|\frac{v_{N+1}-v_{N}}{h}\right|^{2}-|\dot{v}_{N+1}|^{2}\right],\end{array}$
(112)
$\displaystyle\begin{array}[]{ll}&\frac{L}{2}\left|(v_{N+1}-v_{N})\dot{v}_{N+1}\right|\leq\frac{Lh}{4c}\left[c^{2}\left|\frac{v_{N+1}-v_{N}}{h}\right|^{2}+\dot{v}_{N+1}^{2}\right].\end{array}$
(114)

Considering (112) and (114), the following is immediate

$\displaystyle\begin{array}[]{ll}|F_{h}^{FD}|&\leq\frac{Lh}{c}E_{h}(t)-\frac{Lh}{4c}|\dot{v}_{N+1}|^{2}-\frac{L\xi
h}{4c^{2}}|\dot{v}_{N+1}|^{2}\leq\frac{Lh}{c}E_{h}(t).\end{array}$ (116)

Therefore, (110) follows from (116). $\square$

######  Lemma 9.

The function $F_{h}$ satisfies

$\displaystyle\begin{array}[]{ll}&\frac{dF_{h}^{FD}(t)}{dt}\leq-\left(1-\frac{\kappa^{FD}}{4c^{2}}\right)E_{h}^{FD}(t)+\frac{L}{2}\left(1+\frac{1}{2N+2}+\frac{\xi^{2}}{c^{2}}\right)|\dot{v}_{N+1}|^{2},\end{array}$
(118)

where
$\kappa^{FD}:=\max\left\\{h^{2}\left|\lambda_{j}^{FD}(\xi,h)\right|^{2}\right\\}_{j=1}^{N+1},$
and the upper bound for the term $h^{2}\left|\lambda_{j}^{FD}(\xi,h)\right|$
is given by (94).

Proof: Finding the derivative of $F_{h}(t)$ along the solutions of (76) leads
to

$\displaystyle\begin{array}[]{ll}\frac{dF_{h}^{FD}}{dt}&=h\left[\sum\limits_{j=1}^{N}j\ddot{v}_{j}(\frac{v_{j+1}-v_{j-1}}{2})+\dot{v}_{j}\frac{{\dot{v}}_{j+1}-{\dot{v}}_{j-1}}{2})\right]+\frac{L}{2}(\dot{v}_{N+1}-\dot{v}_{N})(\dot{v}_{N+1})\\\
&\quad+\frac{L}{2}(v_{N+1}-v_{N})(\ddot{v}_{N+1})-\frac{L\xi
h}{2c^{2}}\dot{v}_{N+1}\ddot{v}_{N+1}\\\
&=-\frac{hc^{2}}{2}\sum\limits_{j=0}^{N}|\frac{v_{j+1}-v_{j}}{h}|^{2}-\frac{h}{2}\sum\limits_{j=0}^{N}|{\dot{v}}_{j}|^{2}+\frac{h^{3}}{4}\sum\limits_{j=0}^{N}|\frac{\dot{v}_{j+1}-\dot{v}_{j}}{h}|^{2}+\frac{Lc^{2}}{2}\left|\frac{v_{N+1}-v_{N}}{h}\right|^{2}-\frac{h}{4}|{\dot{v}}_{N+1}|^{2}\\\
&\quad-\frac{Lh^{2}}{4}\frac{|{\dot{v}}_{N+1}-{\dot{v}}_{N}|^{2}}{h^{2}}+\frac{L(|{\dot{v}}_{N+1}|^{2}+|{\dot{v}}_{N}|^{2})}{4}+\frac{L}{2}(\dot{v}_{N+1}-\dot{v}_{N})(\dot{v}_{N+1})\\\
&\quad+\frac{L}{2}(v_{N+1}-v_{N})\ddot{v}_{N+1}-\frac{L\xi
h}{2c^{2}}\dot{v}_{N+1}\ddot{v}_{N+1}\\\
&=-E_{h}^{FD}(t)+\frac{h}{2}|\dot{v}_{N+1}|^{2}+\frac{h^{3}}{4}\sum\limits_{j=0}^{N}|\frac{\dot{v}_{j+1}-\dot{v}_{j}}{h}|^{2}-\frac{h}{4}|{\dot{v}}_{N+1}|^{2}\\\
&\quad+\frac{L}{2}|\dot{v}_{N+1}|^{2}-\frac{L\xi}{2h}(v_{N+1}-v_{N})\dot{v}_{N+1}-\frac{L\xi
h}{2c^{2}}\dot{v}_{N+1}\ddot{v}_{N+1}\\\
&=-E_{h}^{FD}(t)+\frac{h^{3}}{4}\sum\limits_{j=0}^{N}|\frac{\dot{v}_{j+1}-\dot{v}_{j}}{h}|^{2}+\frac{h}{4}|{\dot{v}}_{N+1}|^{2}-\frac{L\xi
h}{2c^{2}}\dot{v}_{N+1}\ddot{v}_{N+1}\\\
&\quad+\frac{L}{2}\dot{v}_{N+1}\left[\dot{v}_{N+1}+\frac{\xi
h}{c^{2}}\ddot{v}_{N+1}+\frac{\xi^{2}}{c^{2}}\dot{v}_{N+1}\right]\\\
&=-E_{h}^{FD}(t)+\frac{h^{3}}{4}\sum\limits_{j=0}^{N}|\frac{\dot{v}_{j+1}-\dot{v}_{j}}{h}|^{2}+\frac{L}{2}\left(1+\frac{1}{2N+2}+\frac{\xi^{2}}{c^{2}}\right)|\dot{v}_{N+1}|^{2}.\end{array}$
(129)

Defining the higher-order energy
$\tilde{E}^{FD}_{h}(t):=\frac{h}{2}\sum\limits_{j=0}^{N}\rho\left|\ddot{v}_{j}\right|^{2}+c^{2}\left|\frac{\dot{v}_{j+1}-\dot{v}_{j}}{h}\right|^{2},$
the inequality above is reduced to

$\displaystyle\begin{array}[]{ll}\frac{dF_{h}^{FD}}{dt}&\leq-
E_{h}^{FD}(t)+\frac{h^{2}}{4c^{2}}{\tilde{E}}_{h}^{FD}(t)+\frac{L}{2}\left(1+\frac{1}{2N+2}+\frac{\xi^{2}}{c^{2}}\right)|\dot{v}_{N+1}|^{2}\end{array}$
(131)

It follows from the definition of $\kappa_{FD}$ that
$h^{2}\tilde{E}^{FD}_{h}(t)\leq\kappa^{FD}E^{FD}_{h}(t)$. Hence, (118)
follows. $\square$

######  Theorem 7.

Suppose that there exists a constant $0<\delta<\frac{2\xi
c^{2}}{L\left[c^{2}\left(1+\frac{1}{2N+2}\right)+\xi^{2}\right]}<\frac{2\xi
c^{2}}{L\left[c^{2}+\xi^{2}\right]}$ such that for all initial conditions
$\vec{v}^{0},\vec{v}^{1}\in\mathbb{R}^{N+2},$ the energy $E_{h}^{FD}(t)$
corresponding to (76) satisfies

$\displaystyle E_{h}^{FD}(t)\leq\frac{c+\delta L}{c-\delta
L}e^{-\delta(1-\frac{L\delta}{c})(1-\frac{\kappa^{FD}}{4c^{2}})t}E_{h}^{FD}(0),\quad\forall
t>0.$ (132)

Proof: Since
$\frac{L_{h}^{FD}(t)}{dt}=\frac{dE^{FD}(t)}{dt}+\delta\frac{dF_{h}^{FD}(t)}{dt},$
by Lemma 8

$\displaystyle\begin{array}[]{ll}\frac{dL_{h}^{FD}(t)}{dt}&=-\delta\left(1-\frac{\kappa^{FD}}{4c^{2}}\right)E_{h}^{FD}(t)-\left(\xi-\frac{\delta
L}{2}\left[1+\frac{1}{2N+2}+\frac{\xi^{2}}{c^{2}}\right]\right)|\dot{v}_{N+1}|^{2}\\\
&\leq-\delta\left(1-\frac{L\delta}{c}\right)\left(1-\frac{\kappa^{FD}}{4c^{2}}\right)L_{h}^{FD}(t).\end{array}$
(135)

By the Grönwall’s inequality,

$\displaystyle\begin{array}[]{ll}L_{h}^{FD}(t)\leq
e^{-\delta\left(1-\frac{L\delta}{c}\right)\left(1-\frac{\kappa^{FD}}{4c^{2}}\right)t}L_{h}^{FD}(0),\end{array}$
(137)

and Lemma 8, (132) is obtained. $\square$

###  4.2. Finite Elements with $\xi\neq 0$

Letting $\vec{u}_{1,h}=(v_{1},v_{2},...,v_{N+1})^{\rm T}$,
$\vec{u}_{2,h}=(\dot{v}_{1},...,\dot{v}_{N+1})^{\rm T},$
$\vec{y}_{h}=(\vec{u}_{1,h},\vec{u}_{2,h})^{\mathrm{T}},$ the system (98) is
written in the first-order form

${\dot{\vec{y}}_{h}}=\mathcal{A}_{h}^{FEM}(\xi)\vec{y}_{h}=(\vec{y}_{2,h},-M^{-1}A_{h}^{FEM}\vec{y}_{1,h}+B_{h}\vec{y}_{2,h})^{\mathrm{T}}$
(138)

where $A_{h}$ is defined by (13), and $B_{N+1,N+1}=\frac{-\xi}{h}\neq 0$ and
otherwise $B_{i,j}\equiv 0$ for any other $i,j.$

Consider the eigenvalue problem for (79):

$\displaystyle\mathcal{A}_{h}^{FEM}(\xi)\vec{y}=\lambda^{FEM}(\xi,h)\vec{y}.$
(139)

For $0<\xi<c,$ it can be shown that the real part of the eigenvalues are
negative, e.g. ${\rm
Re}\lambda^{FEM}(\xi,h)=\frac{-L\xi|\lambda^{FEM}(\xi,h)y_{1,N+1}|^{2}}{hc\left(|\lambda^{FEM}(\xi,h)\vec{y}_{1,h}|^{2}-\vec{y}_{1,h}^{\rm
T}A\vec{y}_{1,h})\right)}<0.$

######  Lemma 10.

The system (37), or (79), is dissipative, i.e.

$\displaystyle\frac{dE_{h}^{FEM}}{dt}+\xi|\dot{v}_{N+1}|^{2}=0.$ (140)

Proof: Multiply both sides of (98) by $h\dot{v}_{j}$ and take sum from $j=1$
to $N:$

$\displaystyle\begin{array}[]{ll}\sum\limits_{j=1}^{N}\frac{\ddot{v}_{j+1}+4\ddot{v}_{j}+\ddot{v}_{j-1}}{6}h\dot{v}_{j}+c^{2}\sum\limits_{j=1}^{N}\frac{v_{j+1}-2v_{j}+v_{j-1}}{h}\dot{v}_{j}=0,\end{array}$
(142)

where

$\displaystyle\begin{array}[]{ll}&h\sum\limits_{j=1}^{N}\frac{\ddot{v}_{j+1}+4\ddot{v}_{j}+\ddot{v}_{j-1}}{6}\dot{v}_{j}=-\frac{h}{6}(\ddot{v}_{N+1}+\ddot{v}_{N})\dot{v}_{N+1}+\frac{h}{3}\sum\limits_{j=0}^{N}\ddot{v}_{j}\dot{v}_{j}+\frac{h}{6}\sum\limits_{j=0}^{N}(\dot{v}_{j+1}+\dot{v}_{j})(\ddot{v}_{j+1}+\ddot{v}_{j}),\end{array}$
(144)
$\displaystyle\begin{array}[]{ll}&-hc^{2}\sum\limits_{j=1}^{N}\frac{v_{j+1}-2v_{j}+v_{j-1}}{h^{2}}\dot{v}_{j}=\frac{c^{2}}{h}(v_{N}-v_{N+1})\dot{v}_{N+1}+\frac{c^{2}}{h}\sum\limits_{j=0}^{N}(v_{j+1}-v_{j})(\dot{v}_{v+1}-\dot{v}_{j}).\end{array}$
(146)

Substituting (144) and (146) into (142) yields (140). $\square$

Following (9), define the following Lyapunov functional

$\displaystyle L_{h}^{FEM}(t):=E_{h}^{FEM}+\delta F_{h}^{FEM}(t)$ (147)

where the auxiliary function $F_{h}(t)$ is defined by

$\displaystyle\begin{array}[]{ll}F_{h}^{FEM}(t)=h\sum\limits_{j=1}^{N}\frac{\dot{v}_{j+1}+4\dot{v}_{j}+\dot{v}_{j-1}}{6}j\frac{v_{j+1}-v_{j-1}}{2}+\frac{L}{6}\left(2\dot{v}_{N+1}+\dot{v}_{N}\right)(v_{N+1}-v_{N}).\end{array}$
(149)

######  Lemma 11.

For $0<\delta<\frac{c}{L},$ $L_{h}$ is equivalent to $E_{h}^{FEM},$ i.e.

$\left(1-\frac{L\delta}{c}\right)E_{h}^{FEM}\leq
L_{h}^{FEM}\leq\left(1+\frac{L\delta}{c}\right)E_{h}^{FEM}.$ (150)

Proof: Applying the Cauchy-Schwartz inequality leads to

$\displaystyle\begin{array}[]{ll}|F_{h}^{FEM}(t)|&\leq\frac{Lh}{6}\left|\frac{2\dot{v}_{N+1}+\dot{v}_{N}}{2}\right|^{2}+\frac{Lh}{6}\left|\frac{v_{N+1}-v_{N}}{h}\right|^{2}\\\
&~{}~{}+\frac{Lh}{2}\sum\limits_{j=1}^{N}\left|\frac{v_{j+1}-v_{j}}{2h}\right|^{2}+\frac{Lh}{2}\sum\limits_{j=1}^{N}\left|\frac{\dot{v}_{j+1}+4\dot{v}_{j}+\dot{v}_{j-1}}{6}\right|^{2}\\\
&\leq\frac{Lh}{9}\left|\frac{\dot{v}_{N+1}+\dot{v}_{N}}{2}\right|^{2}+\frac{Lh}{4}\left|\frac{v_{N+1}-v_{N}}{h}\right|^{2}+\frac{Lh}{18}|\dot{v}_{N+1}|^{2}+\frac{Lh}{4}\sum\limits_{j=1}^{N}\left|\frac{v_{j+1}-v_{j}}{h}\right|^{2}\\\
&~{}~{}+\frac{Lh}{4}\sum\limits_{j=1}^{N}\left|\frac{v_{j}-v_{j-1}}{h}\right|^{2}+\frac{Lh}{9}\sum\limits_{j=1}^{N}\left|\frac{\dot{v}_{j+1}+\dot{v}_{j}}{2}\right|^{2}+\left|\frac{\dot{v}_{j}+\dot{v}_{j-1}}{2}\right|^{2}+\left|\dot{v}_{j}\right|^{2}\\\
&\leq\frac{L}{c}E_{h}^{FEM}(t).\end{array}$ (156)

Therefore, (147) follows. $\square$

######  Lemma 12.

The function $F_{h}^{FEM}$ satisfies

$\begin{array}[]{ll}\frac{dF_{h}^{FEM}}{dt}\leq&-\left(1-\frac{\kappa^{FEM}}{12}\right)E_{h}^{FEM}(t)+\frac{L}{2}\left(1+\frac{\xi^{2}}{c^{2}}\right)|\dot{v}_{N+1}|^{2}\end{array}$
(157)

where
$\kappa_{FEM}=\max\left\\{h^{2}\left|\lambda_{j}^{FEM}(\xi,h)\right|^{2}\right\\}_{j=1}^{N+1}.$

Proof: Finding the derivative of $F^{FEM}_{h}(t)$ along the solutions of (98)
leads to

$\displaystyle\begin{array}[]{ll}F_{h}^{FEM}(t)=h\sum\limits_{j=1}^{N}\frac{\ddot{v}_{j+1}+4\ddot{v}_{j}+\ddot{v}_{j-1}}{6}j\frac{v_{j+1}-v_{j-1}}{2}+h\sum\limits_{j=1}^{N}\frac{\dot{v}_{j+1}+4\dot{v}_{j}+\dot{v}_{j-1}}{6}j\frac{\dot{v}_{j+1}-\dot{v}_{j-1}}{2}&\\\
\quad+\frac{L}{6}\left(2\ddot{v}_{N+1}+\ddot{v}_{N}\right)(v_{N+1}-v_{N})+\frac{L}{6}\left(2\dot{v}_{N+1}+\dot{v}_{N}\right)(\dot{v}_{N+1}-\dot{v}_{N})\\\
=-\frac{c^{2}h}{2}\sum\limits_{j=0}^{N}\left|\frac{v_{j+1}-v_{j}}{h}\right|^{2}-\frac{h}{12}\sum\limits_{j=0}^{N}\left|\dot{v}_{j+1}+\dot{v}_{j}\right|^{2}-\frac{h}{6}\sum\limits_{j=1}^{N}\dot{v}_{j+1}\dot{v}_{j}+\frac{c^{2}L}{2}\left|\frac{v_{N+1}-v_{N}}{h}\right|^{2}&\\\
\quad-c^{2}L\left|\frac{v_{N+1}-v_{N}}{h}\right|^{2}-\frac{L\xi}{h}\dot{v}_{N+1}(v_{N+1}-v_{N})+L\frac{(\dot{v}_{N+1}+\dot{v}_{N})^{2}}{12}+\frac{L}{6}\dot{v}_{N+1}\dot{v}_{N}&\\\
\quad+L\frac{2\dot{v}_{N+1}^{2}-\dot{v}_{N+1}\dot{v}_{N}-\dot{v}_{N}^{2}}{6}&\\\
=-E_{h}^{FEM}(t)+\frac{h}{6}\sum\limits_{j=0}^{N}\left[\left|\dot{v}_{j}\right|^{2}-\dot{v}_{j+1}\dot{v}_{j}\right]-\frac{c^{2}L}{2}\left|\frac{v_{N+1}-v_{N}}{h}\right|^{2}&\\\
\quad-\frac{L\xi}{h}\dot{v}_{N+1}(v_{N+1}-v_{N})+L\frac{5\dot{v}_{N+1}^{2}+2\dot{v}_{N+1}\dot{v}_{N}-\dot{v}_{N}^{2}}{12}&\\\
\leq-
E_{h}^{FEM}(t)+\frac{h}{6}\sum\limits_{j=0}^{N}\left[\left|\dot{v}_{j}\right|^{2}-\dot{v}_{j+1}\dot{v}_{j}\right]+\frac{L}{2}(1+\frac{\xi^{2}}{c^{2}})|\dot{v}_{N+1}|^{2}.\end{array}$
(166)

(157) follows from this together with the following inequality

$\displaystyle\begin{array}[]{ll}\frac{h}{6}\sum\limits_{j=0}^{N}|\dot{v}_{j}|^{2}-\dot{v}_{j}\dot{v}_{j+1}&=\frac{h}{12}\sum\limits_{j=0}^{N}\frac{|\dot{v}_{j}-\dot{v}_{j+1}|^{2}}{2}-\frac{h}{12}|\dot{v}_{N+1}|^{2}\\\
&\leq\frac{\kappa^{FEM}h}{12c^{2}}\sum\limits_{j=0}^{N}|\dot{v}_{j}|^{2}\\\
&\leq\frac{\kappa^{FEM}}{12c^{2}}E_{h}^{FEM}(t).~{}~{}\square\end{array}$
(170)

######  Theorem 8.

Suppose that there exists a constant $0<\delta<\frac{2\xi
c^{2}}{L(c^{2}+\xi^{2})}$ such that for all initial conditions
$\vec{v}^{0},\vec{v}^{1}\in\mathbb{R}^{N+2},$ the energy $E_{h}^{FEM}(t)$
corresponding to (98) satisfies $\forall t>0$

$\displaystyle E_{h}^{FEM}(t)\leq\frac{c+\delta L}{c-\delta
L}e^{-\delta\left(1-\frac{L\delta}{c}\right)\left(1-\frac{\kappa^{FEM}}{12c^{2}}\right)t}E_{h}^{FEM}(0).$
(171)

Proof: Since
$\frac{L_{h}^{FEM}(t)}{dt}=\frac{dE^{FEM}(t)}{dt}+\delta\frac{dF_{h}^{FEM}(t)}{dt},$
by Lemma 11

$\displaystyle\begin{array}[]{ll}\frac{dL_{h}^{FEM}(t)}{dt}&=-\left(\xi-\frac{L\delta}{2}\left(1+\frac{\xi^{2}}{c^{2}}\right)\right)|\dot{v}_{N+1}|^{2}-\delta\left(1-\frac{\kappa^{FEM}h^{2}}{12c^{2}}\right)E_{h}^{FEM}(t)\\\
&\leq-\delta\left(1-\frac{L\delta}{c}\right)\left(1-\frac{\kappa^{FEM}}{12c^{2}}\right)L^{FEM}_{h}(t).\end{array}$
(174)

By the Grönwall’s inequality,

$\displaystyle\begin{array}[]{ll}L_{h}^{FEM}(t)\leq
e^{-\delta\left(1-\frac{L\delta}{c}\right)\left(1-\frac{\kappa^{FEM}}{12c^{2}}\right)t}L_{h}^{FEM}(0),\end{array}$
(176)

and Lemma 11, (171) is obtained. $\square$

##  5\. Maximal Decay Rate and Implementation of Direct Fourier Filtering

For $N$ large enough (or small $h$) and $1\leq k\leq 2N+2$, Theorems 5 and 6
leads to

$\displaystyle\begin{array}[]{ll}\frac{h^{2}}{4c^{2}}\max\left\\{\left|\lambda_{k}^{FD}(\xi,h)\right|^{2}\right\\}&=\frac{\kappa^{FD}}{4c^{2}}=:\Gamma^{FD}+O(h),\\\
\frac{h^{2}}{12c^{2}}\max\left\\{\left|\lambda_{k}^{FEM}(\xi,h)\right|^{2}\right\\}&=\frac{\kappa^{FEM}}{12c^{2}}=:\Gamma^{FEM}+O(h).\end{array}$
(178)

where $\Gamma$ is the Fourier filtering parameter. Therefore, for both model
reductions $0<\Gamma^{FD}<1.$ Now, consider the space of filtered solutions
for (79) and (138)

$\mathcal{C}_{h}(\Gamma):=\left\\{\vec{z}_{h}=\sum\limits_{0<\Gamma<1}a_{k}e^{i\sqrt{\lambda_{k}(\xi,h)}t}\vec{\phi}_{k}(h)\right\\}.$

By Theorems 7 and 8, the exponential stability as $h\to 0$ is immediate as
some filtering $0<\Gamma<1$ is considered. If here is no filtering, i.e.
$\Gamma\approx 1$, the exponential stability is at steak since $1-\Gamma\to 0$
as $h\to 0$ e.g. see (137) and (176).

Notice that for each filtered solution in Theorems 7 and 8, the decay rate
$\sigma$ and $\delta$ are functions of $\Gamma$ and $\xi:$

$\displaystyle\begin{array}[]{ll}\sigma(\Gamma,\xi)=\left\\{\begin{array}[]{ll}\delta(1-\frac{L\delta}{c})(1-\Gamma^{FD}),&FD\\\
\delta\left(1-\frac{L\delta}{c}\right)\left(1-\Gamma^{FEM}\right),&FEM\end{array}\right.\\\
\delta(\xi)=\frac{c}{2L}\begin{array}[]{ll}\min\left(1,\frac{2\xi
c}{c^{2}+\xi^{2}}\right),&FD~{}{\text{and}}~{}FEM.\end{array}\end{array}$
(184)

Note that $\delta$ reaches its maximum at $\xi=c:$

$\delta_{max}(\xi=c)=\frac{c}{2L},\qquad\text{FD ~{} and ~{} FEM}$

at which $\sigma$ reaches its maximum

$\sigma_{max}\left(\delta_{max}\right)=\begin{cases}\frac{c}{4L}(1-\Gamma),\quad\text{FD
~{} and ~{} FEM}\end{cases}$ (185)

Our results perfectly mimic the maximal decay rate result (11) in Theorem 1.

##  6\. Numerical Experiments

To show the strength of the Finite Difference (76) and Finite Element-based
(98) model reductions of (4) with and without filtering, we consider high-
enough number of nodes, e.g. $N=30$ with 60 complex eigenvalues of the system
matrix in total. Therefore, $h=1/31\approx 0.0322.$ For simplicity, consider
$c=1$ and $L=1.$ For simulations, the following set of high-frequency initial
conditions is considered

$\displaystyle\left\\{\begin{array}[]{ll}v(x_{j},0)=v_{j}^{0}=10^{-3}\sum_{i=20}^{30}\sin(i\pi
x_{j})&\\\ \dot{v}(x_{j},0)=v_{j}^{1}=10^{-3}\sum_{i=20}^{30}\sin(i\pi
x_{j}).\end{array}\right.$ (188)

The eigenvalues $\lambda_{k}(\xi)$ of the PDE in (8) and the approximated
eigenvalues $\lambda_{k}^{FD}(\xi,h)$ and $\lambda_{k}^{FEM}(\xi,h)$ of the
FEM and FD models, respectively, are simulated in Fig. 1. For $\xi=0.9<c=1,$
the spectral plot shows $40$ high-frequency eigenvalues out of $60$ complex
conjugate eigenvalues in total filtered out. The corresponding filtering
parameters are $\Gamma^{FEM}=1.4133$ and $\Gamma^{FD}=1.017.$ Therefore, the
maximal decay rates in (184) are calculated as $\sigma_{max}^{FEM}=0.2205$ and
$\sigma_{max}^{FD}=0.1864.$ This case can be compared to the maximal decay
rate analysis in Theorem 1 where the optimal decay rate is found to be
$\sigma_{max}=0.25.$

Figure 1.  For $c=1$ and $L=1$, 40 filtered out eigenvalues with $\xi=0.9<c=1$
for $2N=60$ eigenvalues. The filtered eigenvalues converge to the ones of the
PDE.

Next, the simulations of $v(x,t),\dot{v}(x,t),E(t),$ and the tip velocity
feedback $\dot{v}(L,t)$ with FD and FEM algorithms, $\xi=0.9<c=1,$ and for the
unfiltered and filtered algorithms are shown in Figures 2-3 for the set of
initial conditions (188), respectively.

Figure 2.  If neither scheme is filtered, the FEM model can be seen to
stabilize the spurious high-frequency solutions much better than the FD model.
Yet, the solutions are still not stabilized to the rest position.

Figure 3. After filtering $40$ out of $60$ high-frequency solutions, the FEM
model stabilizes the high-frequency solutions much better than the FD model.

Note that the simulations of various initial data (box, sawtooth, sinusoidal,
pinch, square, triangle types) with and without filtering for both FD and FEM
algorithms are real-time demonstrated in the recently-published Wolfram
Demonstrations Project [24].

##  7\. Conclusions

In conclusion, the Lyapunov approach laid out in this paper can be used to
define an explicit decay rate in terms of the filtering parameter $\Gamma$ and
the feedback gain $\xi.$ Our findings are in line with the conclusions in [9]
that FEM provides a more accurate approximation and a better decay rate than
FD to the (4). For most PDE models based off of the wave [17, 18] and beam
equations [1, 13, 12], our approach can be easily adapted. Indeed, an
immediate application under consideration is the obtension of the spectral
estimates for coupled systems where there several branches of eigenvalues.

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