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\markboth{\hfil Resolvent estimates for Couette flow \hfil EJDE--2002/92}
{EJDE--2002/92\hfil Pablo Braz e Silva \hfil}
\begin{document}
\title{\vspace{-1in}\parbox{\linewidth}{\footnotesize\noindent
{\sc  Electronic Journal of Differential Equations},
Vol. {\bf 2002}(2002), No. 92, pp. 1--15. \newline
ISSN: 1072-6691. URL: http://ejde.math.swt.edu or http://ejde.math.unt.edu
\newline ftp  ejde.math.swt.edu  (login: ftp)}
 \vspace{\bigskipamount} \\
 %
  Resolvent estimates for 2-dimensional perturbations of plane Couette flow
 %
\thanks{ {\em Mathematics Subject Classifications:} 76E05, 47A10.
\hfil\break\indent
{\em Key words:} Couette flow, resolvent estimates.
\hfil\break\indent
\copyright 2002 Southwest Texas State University. \hfil\break\indent
Submitted September 24, 2002. Published October 27, 2002.
\hfil\break\indent
Partially supported by grant BEX1901/99-0 from CAPES -- Bras\'\i lia - Brasil} }
\date{}
%
\author{Pablo Braz e Silva}
\maketitle

\begin{abstract}
  We present results concerning resolvent estimates for the linear
  operator associated with the system of differential equations
  governing 2 dimensional perturbations of plane Couette flow.
  We prove estimates on the $L_2$ norm of the resolvent of this
  operator showing this norm to be proportional to the Reynolds
  number $R$ for a region of the unstable half plane.
  For the remaining region, we show that the problem can be
  reduced to estimating the solution of a homogeneous ordinary
  differential equation with non-homogeneous boundary conditions.
  Numerical approximations indicate that the norm of the resolvent
  is proportional to $R$ in the whole region of interest.
\end{abstract}

\numberwithin{equation}{section}
\newtheorem{theorem}{Theorem}
\newtheorem{lemma}[theorem]{Lemma}

\section{Introduction}
We consider the initial boundary-value problem
\begin{equation} \label{eq1.1}
    \begin{gathered}
    w_t + (w\cdot \nabla )W + (W\cdot \nabla )w + \nabla p
    = \frac{1}{R}\Delta w + H\\
         \nabla \cdot w = 0 \\
        w(x,0,t) = w(x,1,t) = (0 , 0) \\
        w(x,y,t) = w(x + 1 ,y,t) \\
        w(x,y,0) = (0 , 0)
     \end{gathered}
\end{equation}
where $w : \mathbb{R}\times [0 , 1] \times [0 , \infty) \to
\mathbb{R}^2$ is the unknown function $w(x,y,t) = (u(x,y,t) ,
v(x,y,t))$. $W$ is The vector field $W = (y,0)$ and the Reynolds
number $R$ is a positive parameter. The forcing $H(x,y,t) =
(F(x,y,t),G(x,y,t))$ is a given $C^{\infty}$ function satisfying
satisfying $\int_{0}^{\infty} ||H(\cdot , \cdot , t)||^2 dt <
\infty$ and $\nabla \cdot H = 0 $ for all $(x , y , t) \in
\mathbb{R}\times [0,1] \times [0 , \infty)$. These equations are
the linearization of the equations governing 2 dimensional
perturbations $w(x,y,t)$ of $W = (y,0)$, known as Couette flow,
which is a steady solution of
\begin{equation}\label{eq1}
\begin{gathered}
         U_t + (U\cdot \nabla )U +  \nabla P = \frac{1}{R}\Delta U \\
         \nabla \cdot U = 0 \\
        U(x,0,t) = (0 , 0) \\ U(x,1,t)= (1 , 0) \\
        U(x,y,t) = U(x + 1 ,y,t) . \\
     \end{gathered}
\end{equation}
The pressure term $p(x,y,t)$ in (\ref{eq1.1}) is determined in
terms of $w$ by the linear elliptic equation
\begin{equation} \label{eq1.2}
\begin{gathered}
     \Delta p = -\nabla \cdot ((w\cdot \nabla) W)
     - \nabla \cdot ((W\cdot \nabla) w) \\
     p_y(x,0,t) = \frac{1}{R}v_{yy}(x,0,t) \\
     p_y(x,1,t) = \frac{1}{R}v_{yy} (x,1,t).
\end{gathered}
\end{equation}
We note that $p$ depends linearly on $w$, and is determined up to
a constant. The estimates derived in this paper are independent of
$p$. With $p$ given by the above equation, the solution $w$ of
(\ref{eq1.1}) remains divergence free for all $t\geq 0$. Therefore
we drop the continuity equation and write the problem as
\begin{equation} \label{eq1.5}
 \begin{gathered}
w_t = \frac{1}{R}\Delta w - (w\cdot \nabla )W - (W\cdot \nabla )w
- \nabla p+ H =: \mathcal{L}_R w + H \\
w(x,y,0) = (0,0)
\end{gathered}
\end{equation}
with $w$ satisfying the boundary conditions described in
(\ref{eq1.1}). Note that $ \mathcal{L}_R$ is a linear operator on
$L_2 \big( \Omega \big)$, $\Omega = [0 , 1]\times [0 , 1]$. The
$L_2$ inner product and norm over $\Omega$ are denoted by
$$
 \langle w_1 , w_2 \rangle = \int_{\Omega} \overline{w}_1 \cdot w_2\,dx\,dy\,,
 \quad \|w\|^2 = \langle w , w \rangle .
$$
As motivation, consider a general linear evolution
equation
\begin{equation}\label{eq1.3}
\begin{gathered}
   w_t = \mathcal{L} w + H \\
   w(x,y,0) = 0 .
\end{gathered}
\end{equation}
Formally, after Laplace transformation with respect to
$t$, the transformed equation is $s \widetilde{w} = \mathcal{L}
\widetilde{w} + \widetilde{H}$ for $\mathop{\rm Re}  (s)\geq 0$, where
$\widetilde{w}$ is the Laplace transform of $w$. The solution of
the transformed equation is given by
\begin{equation} \label{eq1.6}
   \widetilde{w} = (s\mathcal{I} - \mathcal{L} )^{-1} \widetilde{H} .
\end{equation}
where $\mathcal{I}$ is the identity operator, and then we
get the solution of (\ref{eq1.3}) by applying the inverse Laplace
transform to $\widetilde{w}$. Moreover, from (\ref{eq1.6}),
 \begin{equation}\label{eq1.7}
    \|\widetilde{w} ( \cdot , s)\|^2 \leq \|(s\mathcal{I} - \mathcal{L} )^{-1}\|^2 \|
   \widetilde{H} (\cdot, s)\|^2 .
\end{equation}
We recall the following definition.

\paragraph{Definition.} % definition 1
Let $\mathcal{L}$ be a linear operator in a Banach space $X$ and
$\cal{I}$ the identity operator. The resolvent set $\rho
(\mathcal{L} ) \subseteq \mathbb{C}$ of $\mathcal{L}$ is the set of all
complex numbers $s$ such that the operator $(s\mathcal{I} -
\mathcal{L})^{-1}$ exists, is bounded and has a dense domain in
$X$. The linear operator $(s\mathcal{I} - \mathcal{L})^{-1}$, for
$s \in \rho (\mathcal{L})$, is said to be the resolvent of
$\mathcal{L}$. The set $\sigma (\mathcal{L}) = \mathbb{C} \setminus \rho
(\mathcal{L})$ is the spectrum of $\mathcal{L}$. \smallskip


Therefore, if the complex number $s$ is such that $s \in \rho
(\mathcal{L})$, the formal argument above to express the
transformed function $\widetilde{w}$ in terms of the transformed
forcing $\widetilde{H}$ by (\ref{eq1.6}) is valid and we have the
estimate (\ref{eq1.7}). According to Romanov \cite{R}, the
unstable half plane $\mathop{\rm Re}  s \geq 0$ is contained in the
resolvent set of the linear operator $ \mathcal{L}_R$ defined in
(\ref{eq1.5}) , for all positive Reynolds numbers $R$, and the
eigenvalue of $\mathcal{L}_R$ with largest real part is at a
distance from the imaginary axis proportional to $1/R$.
Our aim in this paper is to estimate the dependence on $R$ of the
$L_2$ norm of the resolvent $ \sup_{Re\, s\geq0}\|(s\mathcal{I} -
\mathcal{L}_R)^{-1}\|$. For each $R$, we derive estimates for the
region $|s|\geq 2\sqrt{2}(1+\sqrt{R})$, showing the norm of the
resolvent to be proportional to $R$. For the remaining region, we
prove that we can reduce the problem to estimating the norm of the
solution of a homogeneous ordinary differential equation with
non-homogeneous boundary conditions. This is the main contribution
of the paper, since in principle it simplifies the problem either
if one wants to prove the estimates analytically or use numerical
computations. We perform numerical computations indicating the
norm of the resolvent to be proportional to $R$ also for
$|s|<2\sqrt{2}(1+\sqrt{R})$. There are results in Liefvendahl \&
Kreiss\cite{L2}, Reddy \& Henningson\cite{RE}, Trefethen {\it et
al.}\cite{T}, Kreiss {\it et al.}\cite{K}, also partly based in
computations, indicating the $L_2$ norm of the resolvent to be
proportional to $R^2$ in the case of 3 space dimensions. In the
papers above, the computations are performed using different
methods than the one used here.

\section{Resolvent estimates}
First we study the case when $s$ bounded away from 0.

\begin{theorem} \label{teorema1}
If $|s| \geq 2\sqrt{2}( 1+\sqrt{R})$, then $
\|(s\mathcal{I} - \mathcal{L}_R)^{-1}\|^2 \leq
\frac{8}{|s|^2}(1+\sqrt{R})^2 \leq 1$.
\end{theorem}

\paragraph{Proof:}
After applying the Laplace transformation in $t$, the first equation in
(\ref{eq1.1}) is transformed to
\begin{equation} \label{eq2.1}
 s \widetilde{w}
  = \frac{1}{R}\Delta \widetilde{w} - (\widetilde{w}\cdot \nabla )W - (W\cdot \nabla )\widetilde{w} - \nabla \tilde{p}+ \widetilde{H} =
  \mathcal{L}_R \widetilde{w} + \widetilde{H}
\end{equation}
with $\widetilde{w}$ satisfying the boundary conditions in the
space variables described in (\ref{eq1.1}). Taking the inner
product of (\ref{eq2.1}) with $\widetilde{w}$, we have
\begin{equation}  \label{eq2.2}
 \langle\widetilde{w},s\widetilde{w}\rangle =
\langle\widetilde{w}, \frac{1}{R}\Delta\widetilde{w}\rangle -
\langle\widetilde{w},(\widetilde{w} \cdot \nabla )W\rangle -
\langle\widetilde{w},(W\cdot \nabla)\widetilde{w} \rangle -
\langle\widetilde{w} , \nabla \widetilde{p}\rangle +
\langle\widetilde{w} ,\widetilde{H}\rangle .
\end{equation}
Integrating by parts and using the divergence free and
boundary conditions, one can prove that $\langle\widetilde{w} ,
\nabla \widetilde{p}\rangle = 0$ and $\langle\widetilde{w},(W\cdot
\nabla)\widetilde{w} \rangle$ is purely imaginary. Therefore,
\begin{equation}  \label{eq2.3}
s\|\widetilde{w}\|^2 + \frac{1}{R}\|D\widetilde{w} \|^2 =
-\langle\widetilde{w},(\widetilde{w} \cdot \nabla )W\rangle  -
\langle\widetilde{w},(W\cdot \nabla)\widetilde{w} \rangle +
\langle\widetilde{w} , \widetilde{H}\rangle =: P
\end{equation}
where $D\widetilde{w}$ denotes the derivative of $\widetilde{w}$
with respect to the space variables, and $\|D\widetilde{w}\|$ its
Frobenius norm. $P$ satisfies
\begin{equation} \label{eq2.4}
 | P | \leq  \|\widetilde{w}\| \|(\widetilde{w} \cdot
\nabla)W\| + \|\widetilde{w}\|\|(W\cdot \nabla )\widetilde{w}\| +
\|\widetilde{w}\| \|\widetilde{H}\|
 \end{equation}
which implies, since $\|D W\| = 1$,
\begin{equation}\label{eq2.5}
  | P | \leq  \|\widetilde{w}\|^2 +
\|\widetilde{w}\|\|D\widetilde{w}\| + \|\widetilde{w}
\|\|\widetilde{H}\| .
\end{equation}
 Since
$\langle\widetilde{w},(W\cdot\nabla)\widetilde{w} \rangle$ is
purely imaginary, taking the real part of (\ref{eq2.3}) gives
\begin{equation}\label{equacao1}
      \mathop{\rm Re}  s \|\widetilde{w}\|^2 + \frac{1}{R}\|D\widetilde{w} \|^2 \leq
|\langle\widetilde{w},(\widetilde{w} \cdot \nabla )W\rangle| +
|\langle\widetilde{w} , \widetilde{H}\rangle |
              \leq \|\widetilde{w}\|^2 + \|\widetilde{w} \|\|\widetilde{H}\|
\end{equation}
and therefore
\begin{equation} \label{eq2.6}
  \mathop{\rm Re} s \|\widetilde{w}\|^2 \leq \|\widetilde{w}\|^2 + \|\widetilde{w}
\|\|\widetilde{H}\|.
\end{equation}
Note that (\ref{equacao1}) and (\ref{eq2.6}) are valid for all $s$
satisfying $\mathop{\rm Re} s \geq 0$.
To complete the proof, we consider two separate cases:

\paragraph{Case 1: $\mathop{\rm Re}  s \geq |\mathop{\rm Im} s|$.}
In this case, $ \mathop{\rm Re}  s \geq |s|/\sqrt{2}$, and we
choose $s$ such that $ |s|/(2\sqrt{2}) \geq 1$. Using
(\ref{eq2.6}),
$$
     \frac{|s|}{2\sqrt{2}} \|\widetilde{w}\|^2
 = \big( \frac{|s|}{\sqrt{2}} - \frac{|s|}{2\sqrt{2}}\big)
 \| \widetilde{w} \|^2 \leq ( \mathop{\rm Re} s - 1 ) \|\widetilde{w}\|^2
 \leq \|\widetilde{w}\|\|\widetilde{H}\|
$$
which implies
\begin{equation}\label{eq2.7}
 |s|^2 \|\widetilde{w}\|^2 \leq 8\|\widetilde{H}\|^2.
\end{equation}

\paragraph{Case 2: $|\mathop{\rm Im} s| \geq \mathop{\rm Re}  s \geq 0$.}
In this case, $ |\mathop{\rm Im} s| \geq |s|/\sqrt{2}$. Taking
the imaginary part of equation (\ref{eq2.3}) gives
\begin{equation} \label{eq2.9}
     |\mathop{\rm Im} s| \|\widetilde{w}\|^2 \leq  |P|
\leq \|\widetilde{w}\|^2 + \|\widetilde{w}\|\|D\widetilde{w}\| +
\|\widetilde{w} \|\|\widetilde{H}\| .
\end{equation}
By (\ref{equacao1}),
$$
 \frac{1}{R}\|D\widetilde{w}\|^2 \leq
\|\widetilde{w}\|^2 + \|\widetilde{w}\|\|\widetilde{H}\|\leq
\left(\|\widetilde{w}\| + \|\widetilde{H}\|\right)^2
$$
and then
\begin{equation}\label{equacao2}
  \|D\widetilde{w}\| \leq \sqrt{R} \left( \|\widetilde{w}\| +
\|\widetilde{H}\|\right).
\end{equation}
Using this estimate in (\ref{eq2.9}), we get
\begin{equation} \label{equacao3}
 |\mathop{\rm Im} s| \|\widetilde{w}\|^2 \leq
(1+\sqrt{R})\|\widetilde{w}\|^2 + (1+\sqrt{R})\|\widetilde{w}
\|\|\widetilde{H}\|.
\end{equation}
Choosing $s$ such that $ \frac{|s|}{2\sqrt{2}} \geq 1+\sqrt{R}$,
we have
$$
\frac{|s|}{2\sqrt{2}} \|\widetilde{w}\|^2 \leq \left( | \mathop{\rm Im} s |-
(1+\sqrt{R}) \right) \|\widetilde{w}\|^2 \leq (1+\sqrt{R})\|\widetilde{w}\|
\|\widetilde{H}\|
$$
and this implies
\begin{equation}\label{eq2.8}
 |s|^2 \|\widetilde{w}\|^2 \leq
8(1+\sqrt{R})^2\|\widetilde{H}\|^2 .
\end{equation}
From the two cases, we conclude that
\begin{equation} \label{eq2.12}
|s| \geq 2\sqrt{2}(1+\sqrt{R})  \Rightarrow |s|^2 \| \widetilde{w}
\|^2 \leq 8(1+\sqrt{R})^2\|\widetilde{H}\|^2
\end{equation}
which implies the desired estimate
\begin{equation} \label{eq2.13}
   \|(s\mathcal{I} - \mathcal{L}_R)^{-1}\|^2 \leq
\frac{8}{|s|^2}(1+\sqrt{R})^2 \leq 1,
\end{equation}
valid in the region $|s| \geq 2\sqrt{2}(1+\sqrt{R})$ of the
unstable half plane $\mathop{\rm Re} s \geq 0$. \hfil$\square$
\smallskip

Now we study the case when $|s| < 2\sqrt{2}(1+\sqrt{R})$. For this case,
write the problem (\ref{eq1.1}) componentwise:
\begin{equation} \label{eq3.1}
    \begin{gathered}
         u_t + y u_x + v + p_x = \frac{1}{R}\Delta u + F\\
         v_t + y v_x  + p_y = \frac{1}{R}\Delta v + G\\
         u_x + v_y  = 0 \\
        u(x,0,t) = u(x,1,t) = v(x,0,t) = v(x,1,t) = 0  \\
        u(x,y,t) = u(x+1 ,y,t) \\
        v(x,y,t) = v(x+1,y,t) \\
        u(x,y,0) = v(x,y,0) = 0
     \end{gathered}
\end{equation}
Introduce the stream function $\psi$ by
\begin{equation}
\label{eq3.2}
     \psi_x = v, \quad
     \psi_y = -u .
\end{equation}
Apply the Laplace transform to problem (\ref{eq3.1}) with respect to $t$,
expand in a Fourier series in the periodic direction $x$. The
equations for the transformed functions $\widehat{u}(k,y,s)$,
$\widehat{v}(k,y,s)$, $\widehat{p}(k,y,s)$, $\widehat{F} (k,y,s)$,
$\widehat{G} (k,y,s)$ are
\begin{equation}\label{eq3.3}
\begin{gathered}
s \widehat{u} + i k y  \widehat{u} + \widehat{v} + ik\widehat{p}
  = - \frac{k^2}{R}\widehat{u} +
        \frac{1}{R}\widehat{u}_{yy}+ \widehat{F}\\
s \widehat{v} + i k y  \widehat{v} + \widehat{p}_y
  = - \frac{k^2}{R}\widehat{v} +\frac{1}{R}
        \widehat{v}_{yy}+ \widehat{G}\\
         i k u + \widehat{v}_y  = 0 .
\end{gathered}
\end{equation}
Differentiating the first equation in (\ref{eq3.1}) with respect
to $y$, the second with respect to $x$ and subtracting the second
from the first, we eliminate the pressure $p$, and using the
divergence free condition, we see that the transformed stream
function $\widehat{\psi}$ satisfies the following boundary value
problem for an ordinary differential equation with three
parameters $s$, $R$, $k$:
\begin{equation} \label{eq3.4}
\begin{gathered}
  \frac{1}{R} \widehat{\psi} ''''
  - \Big(s + \frac{2k^2}{R} +iky\Big)\widehat{\psi} '' +
 \Big(sk^2 +\frac{k^4}{R} +ik^3 y\Big)\widehat{\psi} = I \\
  \widehat{\psi} (k,0,s) = \widehat{\psi} (k,1,s)
 = \widehat{\psi} ' (k,0,s) = \widehat{\psi} ' (k,1,s) = 0
\end{gathered}
\end{equation}
where $I := \widehat{F}_y - ik\widehat{G}$ and $'$ denotes the
derivative with respect to $y$. We also use the notation $
\mathcal{D} = \frac{\partial}{\partial y}$ , and write the problem
as
\begin{equation} \label{eq3.5}
  \begin{gathered}
   T T_0 \widehat{\psi}
= \Big( \frac{1}{R} \mathcal{D}^2 - (s + \frac{k^2}{R} + i k y) \Big)
(\mathcal{D}^2 - k^2)  \widehat{\psi} = I\\
    \widehat{\psi} (k,0,s) = \widehat{\psi} (k,1,s) = \widehat{\psi}_y (k,0,s)
 = \widehat{\psi}_y (k,1,s) = 0
\end{gathered}
\end{equation}
where $ T = \frac{1}{R} \mathcal{D}^2 - \big( s +
\frac{k^2}{R} +ik y\big)$ and $ T_0 = \mathcal{D}^2 - k^2$ are
differential operators depending on the parameters $R$, $k$ and
$s$. To derive the resolvent estimates, we use the following
Lemma.

\begin{lemma} \label{lema1}
If for all $k \in \mathbb{Z}$ and for all $s \in \mathbb{C}$ such that $|s| <
2\sqrt{2}(1+\sqrt{R})$ the solution $\widehat{\psi} (k,y,s)$ of
(\ref{eq3.4}) satisfies
\begin{equation}\label{equacao}
\begin{gathered}
    k^2 \|\widehat{\psi} (k, \cdot , s)\|^2 \leq
C R^2 (\|\widehat{F}(k, \cdot , s)\|^2 + \|\widehat{G}(k, \cdot , s)\|^2) \\
    \|\widehat{\psi} ' (k , \cdot , s)\|^2 \leq
C R^2 (\|\widehat{F}(k, \cdot , s)\|^2 + \|\widehat{G}(k, \cdot ,
s)\|^2)
\end{gathered}
\end{equation}
where $C$ is a constant independent of $s$, $R$, $k$,
$\widehat{F}$, $\widehat{G}$, then
$$
\|(s\mathcal{I} - \mathcal{L}_R)^{-1}\|^2 \leq 2C R^2
$$
for $|s| < 2\sqrt{2}(1+\sqrt{R})$.
\end{lemma}

\paragraph{Proof:}
If for all $k \in \mathbb{Z}$ and for all $s \in \mathbb{C}$ such that $|s| <
2\sqrt{2}(1+\sqrt{R})$
\begin{equation} \label{eq3.6}
 \begin{gathered}
    k^2 \|\widehat{\psi} (k, \cdot , s)\|^2 \leq
C R^2 (\|\widehat{F}(k, \cdot , s)\|^2 + \|\widehat{G}(k, \cdot , s)\|^2) \\
    \|\widehat{\psi} ' (k , \cdot , s)\|^2 \leq
C R^2 (\|\widehat{F}(k, \cdot , s)\|^2 + \|\widehat{G}(k, \cdot ,
s)\|^2)
\end{gathered}
\end{equation}
then by the definition (\ref{eq3.2}) of the stream function,
\begin{equation} \label{eq3.7}
\begin{gathered}
     \|\widehat{v} (k , \cdot , s) \|^2  = k^2 \|\widehat{\psi}
(k , \cdot , s)\|^2 \leq
C R^2(\|\widehat{F}(k, \cdot , s)\|^2 + \|\widehat{G}(k, \cdot , s)\|^2) \\
     \|\widehat{u} (k , \cdot , s) \|^2 = \|\widehat{\psi} ' (k , \cdot , s)\|^2 \leq
C R^2 (\|\widehat{F}(k, \cdot , s)\|^2 + \|\widehat{G}(k, \cdot ,
s)\|^2)\,.
\end{gathered}
\end{equation}
Since
\begin{gather*}
 \| \widetilde{u} (\cdot , \cdot , s) \|^2
 = \sum_{k=-\infty}^{\infty}\| \widehat{u} (k , \cdot,s)\|^2\\
 \| \widetilde{v} (\cdot , \cdot , s) \|^2
 = \sum_{k=-\infty}^{\infty} \| \widehat{v} (k ,\cdot,s) \|^2 \\
 \|\widetilde{F} ( \cdot , \cdot , s) \|^2 +
\|\widetilde{G} ( \cdot , \cdot , s) \|^2 =\sum_{k =
-\infty}^{\infty}(\|\widehat{F} ( k , \cdot , s) \|^2 +
\|\widehat{G} ( k ,\cdot, s) \|^2) ,
\end{gather*}
(\ref{eq3.7}) implies
 \begin{equation} \label{eq3.8}
\begin{gathered}
     \| \widetilde{u} (\cdot , \cdot , s) \|^2\leq
C R^2 ( \|\widetilde{F} ( \cdot , \cdot , s) \|^2 + \|\widetilde{G}
( \cdot , \cdot , s) \|^2)\\
     \| \widetilde{v} (\cdot , \cdot , s) \|^2
\leq  C R^2 ( \|\widetilde{F} ( \cdot , \cdot , s) \|^2 +
\|\widetilde{G} ( \cdot , \cdot , s) \|^2)
\end{gathered}
\end{equation}
and this implies $ \|(s\mathcal{I} -\mathcal{L}_R)^{-1}\|^2 \leq 2 C R^2$
for $|s| < 2\sqrt{2}(1+\sqrt{R}).$ \hfil $\square$ \smallskip

  This Lemma shows
that to estimate the norm of the resolvent, it is enough to prove
estimates of the form (\ref{eq3.6}) for the solution
$\widehat{\psi}$ of (\ref{eq3.4}). To prove those estimates for
$\widehat{\psi}$, take the inner product of the differential
equation in (\ref{eq3.4}) with $\widehat{\psi}$ and obtain
\begin{equation} \label{eq3.9}
   \frac{1}{R} \langle \widehat{\psi} , \widehat{\psi} '''' \rangle -
  \langle \widehat{\psi} , (s + \frac{2k^2}{R} +iky)\widehat{\psi} '' \rangle
  + \langle \widehat{\psi} ,  (sk^2 +\frac{k^4}{R} +ik^3 y)\widehat{\psi}\rangle  = \langle \widehat{\psi} , I \rangle .
\end{equation}
Through integration by parts,
\begin{equation} \label{eq3.10}
 \langle \widehat{\psi} , \widehat{\psi} '''' \rangle
 = \| \widehat{\psi} '' \|^2 \quad\mbox{and}\quad
 - \langle \widehat{\psi} , \widehat{\psi} '' \rangle
 = \| \widehat{\psi} '\|^2 .
\end{equation}
Therefore, equation (\ref{eq3.9}) becomes
$$
   \frac{1}{R}  \| \widehat{\psi} '' \|^2 + \big( \frac{2k^2}{R}+s\big) \|\widehat{\psi} '\|^2 +
  \big( s k^2+\frac{k^4}{R}\big) \| \widehat{\psi} \|^2
  -ik\langle \widehat{\psi} , y\widehat{\psi} '' \rangle
  + ik^3\langle \widehat{\psi} ,   y\widehat{\psi}\rangle
  = \langle \widehat{\psi} , I \rangle .
$$
Again through integration by parts, we have $\langle
\widehat{\psi} , y\widehat{\psi} '' \rangle = - \langle
\widehat{\psi} , \widehat{\psi} ' \rangle - \langle
y\widehat{\psi} ' , \widehat{\psi} ' \rangle  $, and since $\psi$
is a real function, $\langle \widehat{\psi} , \widehat{\psi} '
\rangle$ is purely imaginary, $\langle y\widehat{\psi} ' ,
\widehat{\psi} ' \rangle$ and $\langle \widehat{\psi} ,
y\widehat{\psi}\rangle$ are real numbers. Therefore, taking the
real part of the previous equation we get
$$
   \frac{1}{R}  \| \widehat{\psi} '' \|^2
+ \big( \frac{2k^2}{R}+\mathop{\rm Re} s\big) \|\widehat{\psi} '\|^2 +
  \big( (\mathop{\rm Re} s) k^2+\frac{k^4}{R}\big) \| \widehat{\psi} \|^2
   = \mathop{\rm Re} \langle \widehat{\psi} , I \rangle - \mathop{\rm Re}
   (ik \langle \widehat{\psi} , \widehat{\psi} ' \rangle )
$$
and since $\mathop{\rm Re}  s \geq 0$ , this equation implies
\begin{equation} \label{eq3.13}
   \frac{1}{R}  \| \widehat{\psi} '' \|^2 +  \frac{2k^2}{R} \|\widehat{\psi} '\|^2 +  \frac{k^4}{R} \| \widehat{\psi} \|^2
    - |k| |\langle \widehat{\psi} , \widehat{\psi} ' \rangle |
    \leq | \langle \widehat{\psi} , I \rangle | .
\end{equation}
We separate the analysis into three cases for $k \in
\mathbb{Z}$: $ |k|> \sqrt{R/\sqrt{2}}$ , $ 0<|k|\leq\sqrt{R/\sqrt{2}}$,
and the special case $k = 0$.

\paragraph{Case 1: $ |k| > \sqrt{R/\sqrt 2}$}
For this case, we prove the following theorem.

\begin{theorem}
If $ |k| >\sqrt{R/\sqrt{2}}$, then
$$
 \begin{gathered}
    k^2 \|\widehat{\psi} (k, \cdot , s)\|^2 \leq 16 R^2 (\|\widehat{F}(k, \cdot , s)\|^2
   + \|\widehat{G}(k, \cdot , s)\|^2) \\
    \|\widehat{\psi} ' (k , \cdot , s)\|^2 \leq 16 R^2 (\|\widehat{F}(k, \cdot , s)\|^2 +
\|\widehat{G}(k, \cdot , s)\|^2).
\end{gathered}
$$
\end{theorem}

\paragraph{Proof:}
Since $ |\langle \widehat{\psi} , \widehat{\psi} ' \rangle | \leq
\|\widehat{\psi}\| \|\widehat{\psi} '\| \leq \frac{R}{4 |k|}
\|\widehat{\psi}\|^2 + \frac{|k|}{ R} \|\widehat{\psi} '\|^2$,
inequality (\ref{eq3.13}) implies
\begin{equation} \label{eq3.14}
    \frac{1}{R}  \| \widehat{\psi} '' \|^2 +
 \big( \frac{2k^2}{R}  - \frac{k^2}{R}\big) \|\widehat{\psi} '\|^2 +
 \big( \frac{k^4}{R} - \frac{R}{4}\big) \| \widehat{\psi} \|^2
 \leq | \langle \widehat{\psi} , I \rangle |
\end{equation}
and since  $ |k| > \sqrt{R/\sqrt{2}}$ , we have
$ \frac{k^4}{R} - \frac{R}{4} > \frac{k^4}{2 R}$ and
(\ref{eq3.14}) implies \begin{equation} \label{eq3.15}
    \frac{1}{R}  \| \widehat{\psi} '' \|^2 + \frac{k^2}{R} \|\widehat{\psi} '\|^2 +
  \frac{k^4}{2R} \| \widehat{\psi} \|^2 \leq | \langle \widehat{\psi} , I \rangle | .
\end{equation}
Since equation (\ref{eq3.4}) is linear and the forcing is $I =
\widehat{F}_y - ik\widehat{G}$, it suffices, to get the desired
estimates for the solution $\widehat{\psi}$, to prove estimates
for $\widehat{\psi}_1$ and $\widehat{\psi}_2$, solutions of the
boundary-value problems
\begin{equation}
\begin{gathered}
  \frac{1}{R} \widehat{\psi}_1 '''' - \big(s + \frac{2k^2}{R} +iky\big)
  \widehat{\psi}_1 '' +
 \big(sk^2 +\frac{k^4}{R} +ik^3 y\big)\widehat{\psi}_1 = \widehat{F}_y \\
  \widehat{\psi}_1 (k,0,s) = \widehat{\psi}_1 (k,1,s)
  = \widehat{\psi}_1 ' (k,0,s) = \widehat{\psi}_1 ' (k,1,s) = 0
\end{gathered}
\end{equation}
and
\begin{equation}
\begin{gathered}
  \frac{1}{R} \widehat{\psi}_2 ''''
  - \big(s + \frac{2k^2}{R} +iky\big)\widehat{\psi}_2 '' +
 \big(sk^2 +\frac{k^4}{R} +ik^3 y\big)\widehat{\psi}_2 =  ik\widehat{G}\\
  \widehat{\psi}_2 (k,0,s) = \widehat{\psi}_2 (k,1,s)
  = \widehat{\psi}_2 ' (k,0,s) = \widehat{\psi}_2 ' (k,1,s)
  = 0 ,
\end{gathered}
\end{equation}
since $\widehat{\psi}$ is given by $\widehat{\psi} =
\widehat{\psi}_1 - \widehat{\psi}_2$.

\paragraph{Estimates for $\widehat{\psi}_1$:} Through
integration by parts,
$$
 | \langle \widehat{\psi}_1 , \widehat{F}_y \rangle |
 = | \langle \widehat{\psi}_1 ' , \widehat{F} \rangle | .
$$
Using the Cauchy-Schwarz inequality, (\ref{eq3.15}) with forcing
$\widehat{F}_y$ yields
\begin{equation} \label{eq3.16}
 \frac{1}{R}  \| \widehat{\psi}_1 '' \|^2 +
\frac{k^2}{R} \|\widehat{\psi}_1 '\|^2 +
  \frac{k^4}{2R} \| \widehat{\psi}_1 \|^2
  \leq \|\widehat{\psi}_1 '\|\|\widehat{F}\|\,.
\end{equation}
Since all the terms on the left hand side of this equation are
positive, we have
\begin{gather} \label{eq3.17}
    \frac{k^2}{R}\|\widehat{\psi}_1 '\|^2 \leq \|\widehat{\psi}_1 '\|\|\widehat{F}\|
   \Rightarrow k^4 \|\widehat{\psi}_1 '\|^2 \leq R^2
   \|\widehat{F}\|^2\,,\\
\label{eq3.18}
     \frac{k^4}{2R} \| \widehat{\psi}_1 \|^2 \leq \|\widehat{\psi}_1 '\|\|\widehat{F}\|
     \Rightarrow
    k^6 \|\widehat{\psi}_1 \|^2 \leq 2 R^2 \|\widehat{F}\|^2\,,\\
\label{3.19}
   \frac{1}{R}\|\widehat{\psi}_1 '' \|^2 \leq \|\widehat{\psi}_1 '\|
\|\widehat{F}\|
  \Rightarrow
  k^2\|\widehat{\psi}_1 '' \|^2 \leq R^2 \|\widehat{F}\|^2 .
\end{gather}

\paragraph{Estimates for $\widehat{\psi}_2$:}
By (\ref{eq3.15}) with forcing $i k \widehat{G}$ and the Cauchy-Schwarz
inequality, we have
\begin{equation} \label{eq3.20}
 \frac{1}{R}  \| \widehat{\psi}_2 '' \|^2 +
\frac{k^2}{R} \|\widehat{\psi}_2 '\|^2 +
  \frac{k^4}{2R} \| \widehat{\psi}_2 \|^2 \leq |k| \|\widehat{\psi}_2 \|
\|\widehat{G}\|\,.
\end{equation}
Then, as in the estimates for $\widehat{\psi}_1$ above, we obtain
\begin{equation} \label{eq3.21}
\begin{gathered}
   k^6 \|\widehat{\psi}_2 \|^2 \leq 4 R^2 \|\widehat{G}\|^2 \\
      k^4 \|\widehat{\psi}_2 '\|^2 \leq 2 R^2 \|\widehat{G}\|^2 \\
      k^2 \|\widehat{\psi}_2 ''\|^2 \leq 2 R^2 \|\widehat{G}\|^2 .
\end{gathered}
\end{equation}
Therefore, $\widehat{\psi}$, the solution of problem (\ref{eq3.4}),
satisfies
$$
k^2 \|\widehat{\psi} ''\|^2 + k^4 \|\widehat{\psi} '\|^2 + k^6
\|\widehat{\psi}\|^2
    \leq 16 R^2 (\|\widehat{F}\|^2 + \|\widehat{G}\|^2) = 16 R^2
\|\widehat{H}\|^2 .$$

\paragraph{Case 2: $ 0 < |k| \leq \sqrt{R/\sqrt 2}$}
For this case, we prove that we can reduce the problem to the
study of the solutions of a homogenous 4th order ordinary
differential equation with non-homogenous boundary conditions. We
compute the norms of those functions numerically. The computations
indicate that the norm of the solutions of these simplified
problems do not grow as $R$ grows. This implies the norm of the
resolvent of the operator $\mathcal{L}_R$ to be proportional to
$R$. We begin by noting that we can restrict ourselves to the case
when $s = i \xi$ is purely imaginary. This is a consequence of the
following theorem for holomorphic mappings in Banach spaces (Chae
\cite{C}).

\begin{theorem}[Maximum Modulus Theorem] \label{the1}
 Let $U$ be a connected, open subset of $\mathbb{C}$ and $f : U \to E$
 a holomorphic
 mapping, where $E$ is a Banach space. If $\|f(z)\|$ has a maximum
 at a point in $U$, then $\|f(z)\|$ is constant on $U$.
\end{theorem}

Applying the theorem to the holomorphic function $f(s) =
(s\mathcal{I} - \mathcal{L}_R)^{-1}$ defined on $\mathop{\rm Re} s >
0$, taking values in the Banach space of bounded linear operators
on $L_2(\Omega)$ and noting that (\ref{eq2.13}) implies
$\lim_{|s|\to \infty}\|f(s)\| = 0$, we conclude
\begin{equation}
    \sup_{Re s \geq 0} \|(s\mathcal{I} - \mathcal{L}_R)^{-1}\|
    = \sup_{\xi\in\mathbb{R}} \| (i\xi\mathcal{I} -
   \mathcal{L}_R)^{-1}\| .
\end{equation}
We note that since we just need to consider $s = i\xi$, $\xi \in
\mathbb{R}$, and in this case $|s| = |\xi | = |\mathop{\rm Im}
s|$, the proof of Theorem \ref{teorema1} is valid for $|s| =
|\mathop{\rm Im} s|> 2(1+\sqrt{R})$. This restricts the parameter
range for the numerical calculations. We now reduce our problem to
the study of solutions of a homogenous ordinary differential
equation. To this end, first consider the second order system
$$
\begin{gathered}
   T h  = \big( \frac{1}{R} \mathcal{D}^2 - (s + \frac{k^2}{R} + i k y) \big) h
   = I = \widehat{F}_y - ik\widehat{G}, \quad  h(0) = h(1) = 0 \\
    T_0 g = (\mathcal{D}^2 - k^2) g = h , \quad  g(0) = g(1) = 0
\end{gathered}
$$
for $s = i \xi$, $\xi \in \mathbb{R}$. Taking the inner product
with $h$ in the first equation and with $g$ in the second, and
since the boundary conditions for both equations imply that the
boundary terms after integration by parts vanish, we get
\begin{equation}\label{eq3.24}
    \begin{gathered}
         \|g ''\|^2 + k^2 \| g '\|^2 + k^4 \|g\|^2 \leq C_1\|h\|^2 \\
         \|h'\|^2 + k^2\|h\|^2 \leq C_2 R^2 (\|\widehat{F}\|^2 + \|\widehat{G}\|^2)
    \end{gathered}
\end{equation}
where $C_1$, $C_2$ are constants independent of $R$, $s$, $k$,
$\widehat{F}$, $\widehat{G}$. Combining those two inequalities, it
follows that
\begin{equation}\label{eq3.25}
   k^2\|g ''\|^2 + k^4 \| g '\|^2 + k^6 \|g\|^2 \leq C R^2 (\|\widehat{F}\|^2
   + \|\widehat{G}\|^2)
\end{equation}
where $C$ is a constant independent of $R$, $s$, $k$,
$\widehat{F}$, $\widehat{G}$. Note that $g$ satisfies
\begin{equation}\label{eq3.26}
\begin{gathered}
      T T_0 g = \Big( \frac{1}{R} \mathcal{D}^2
      - \big( s + \frac{k^2}{R} + i k y\big) \Big) (\mathcal{D}^2 - k^2) g =
            \widehat{F}_y - ik\widehat{G} \\
      g(0) = g(1) = 0 .\\
    \end{gathered}
\end{equation}
Let $g'(0) = \alpha$ and $g'(1) = \beta$. The numbers
$\alpha$ and $\beta$ can be estimated using a 1-dimensional
Sobolev type inequality. Indeed, $ |g'|^2_{\infty} \leq \|g'\|^2 +
\|g'\|\|g''\|$. Using the estimates (\ref{eq3.25}) and that $k \in
\mathbb{Z}$ satisfies $1 \leq |k| $, we conclude that
\begin{equation} \label{eq 3.29}
      |g'|^2_{\infty} \leq C R^2 (\|\widehat{F}\|^2 + \|\widehat{G}\|^2 )\,.
\end{equation}
Therefore, \begin{equation} \label{eq3.30}
\begin{gathered}
      |\alpha |^2 = |g'(0)|^2 \leq C R^2 (\|\widehat{F}\|^2
      + \|\widehat{G}\|^2 ) \\
      |\beta |^2  = |g'(1)|^2 \leq C R^2 (\|\widehat{F}\|^2
      + \|\widehat{G}\|^2 ),
\end{gathered}
\end{equation}
where $C$ represents a constant independent of $s$, $R$, $k$,
$\widehat{F}$, $\widehat{G}$. Now, let $ \delta (k , y , s)$ be
the solution of the problem
\begin{equation}\label{eq3.27}
\begin{gathered}
T T_0 \delta = \Big( \frac{1}{R} \mathcal{D}^2
- \big( s + \frac{k^2}{R} + i k y\big) \Big) (\mathcal{D}^2 - k^2)
\delta = 0 \\
      \delta (0) = \delta (1) = 0 \\
      \delta ' (0) = \alpha \\
      \delta ' (1) = \beta .
\end{gathered}
\end{equation}
Then, $ \widehat{\psi} (k , y , s)  = g (k , y , s)-
\delta (k , y , s) $ is the solution of (\ref{eq3.5}). Indeed,
$$
  T T_0 \widehat{\psi} = T T_0 (g - \delta) = T T_0 g - T T_0 \delta
  = \widehat{F}_y - ik\widehat{G}
$$
and
\begin{gather*}
 \widehat{\psi} (k , 0 , s)  = g (k , 0 , s)- \delta (k , 0 , s) = 0 \\
 \widehat{\psi} (k , 1 , s)  = g (k ,1,s)- \delta (k,1,s) =0 \\
 \widehat{\psi} ' (k , 0 , s)  = g' (k,0,s)- \delta' (k,0,s) = \alpha - \alpha = 0 \\
 \widehat{\psi} ' (k , 1 , s)  = g' (k,1,s)- \delta'
(k,1,s) = \beta - \beta = 0 .
\end{gather*}
Since we already have suitable estimates for $g$, our problem is
reduced to deriving estimates for the solution of
\begin{equation}\label{eq3.28}
\begin{gathered}
T T_0 \delta = \Big( \frac{1}{R} \mathcal{D}^2 -
\big( s + \frac{k^2}{R} + i k y\big) \Big) (\mathcal{D}^2 - k^2) \delta= 0 \\
\delta (0) = \delta (1) = 0 \\
      \delta ' (0) = \alpha \\
      \delta ' (1) = \beta
\end{gathered}
\end{equation}
where $|\alpha |^2 \leq C R^2 (\|\widehat{F}\|^2 +
\|\widehat{G}\|^2 ) $ and $|\beta |^2 \leq C R^2
(\|\widehat{F}\|^2 + \|\widehat{G}\|^2 ) $. Therefore, it remains
to estimate $k^2 \|\delta(k , \cdot , s)\|^2$ and $
\|\delta^\prime (k , \cdot , s)\|^2$ for the parameter range $(k ,
\xi) \in \mathbb{Z} \times \mathbb{R}$,
$1\leq|k|\leq\sqrt{R/\sqrt{2}}$ , $ |\xi | <2(1+\sqrt{R})$, where
$s = i\xi$. To study this problem numerically, we simplify it as
follows: let $\delta_1$ be the solution of
\begin{equation}\label{eq3.31}
\begin{gathered}
     T T_0 \delta_1 = 0 \\
      \delta_1 (0) = \delta_1 (1) = 0 \\
      \delta_1 ' (0) = 1 \\
      \delta_1 ' (1) = 0
\end{gathered}
\end{equation}
and $\delta_2$ be the solution of
\begin{equation}\label{eq3.32}
\begin{gathered}
     T T_0 \delta_2 =  0 \\
      \delta_2 (0) = \delta_2 (1) = 0 \\
      \delta_2 ' (0) = 0 \\
      \delta_2 ' (1) = 1 .
\end{gathered}
\end{equation}
Then $\delta = \alpha \delta_1 + \beta \delta_2$ is the solution
of (\ref{eq3.28}). Therefore,
\begin{gather*}
k^2 \|\delta(k , \cdot , s)\|^2 \leq 2 |\alpha |^2 k^2\|\delta_1
(k , \cdot , s)\|^2 +2 |\beta |^2 k^2 \|\delta_2 (k , \cdot ,
s)\|^2 \\
\|\delta ' (k , \cdot , s)\|^2 \leq 2 |\alpha |^2 \|\delta_1 ' (k
, \cdot , s)\|^2 +2 |\beta |^2  \|\delta_2 ' (k , \cdot , s)\|^2.
\end{gather*}
Thus we can restrict ourselves to the dependence on $R$ of $k^2 \|
\delta_j (k , \cdot , s)\|^2$ and $\|\delta_j ' (k , \cdot ,
s)\|^2$, for $j =1 , 2$. Moreover, since $\|\psi (k , \cdot ,
s)\|^2 = \| \psi (-k , \cdot , -s)\|^2$, where $\psi$ is the
solution of (\ref{eq3.4}), we can restrict ourselves to $0 \leq
\xi < 2(1+\sqrt{R})$.

We did numerical computations using
the MATLAB 6.0 built-in boundary value problem solver BVP4C, for
the parameter range $1\leq |k|\leq \sqrt{R/\sqrt{2}}$,
$0\leq \xi \leq 2(1+\sqrt{R})$ and values of $R$ from $1$ up to
$10000$. For $\xi$, we used a mesh with variable number of points
for each $R$. BVP4C makes use of a collocation method, and we
performed computations with different absolute and relative
tolerances, using continuation in the Reynolds number for the
initial guess of the solution. The results were similar for all
cases. Therefore, even though the problem is stiff for some of the
parameter values, the results shown in figures (\ref{fig1}),
(\ref{fig2}), (\ref{fig3}), (\ref{fig4}) in the following pages
should be reliable. For different values of the Reynolds number
$R$, we plot the maximum of $k^2 \| \delta_j (k , \cdot , s)\|^2$
and $\|\delta_j ' (k , \cdot , s)\|^2$ for $1\leq k \leq
 \sqrt{R/\sqrt{2}}$, $0\leq \xi\leq 2(1+\sqrt{R})$.
 The results indicate that
\begin{equation} \label{eq3.33}
  k^2 \| \delta_j (k , \cdot , s)\|^2 \leq 1 \quad\mbox{and}\quad
  \|\delta_j ' (k , \cdot , s)\|^2 \leq 1 \quad j=1,2 .
\end{equation}
Therefore,
\begin{align*}
   k^2 \|\delta(k , \cdot , s)\|^2
\leq & 2 |\alpha |^2 k^2\|\delta_1 (k , \cdot , s)\|^2 +2 |\beta |^2 k^2
\|\delta_2 (k , \cdot , s)\|^2 \\
\leq & 4 C R^2 (\|\widehat{F}\|^2 +\|\widehat{G}\|^2 ),
\end{align*}
that is,
\begin{equation}
k^2 \|\delta(k , \cdot , s)\|^2 \leq \widetilde{C} R^2
(\|\widehat{F}(k , \cdot , s)\|^2 + \|\widehat{G}(k , \cdot ,
s)\|^2 )
\end{equation}
and similarly
\begin{equation}
\|\delta ' (k , \cdot , s)\|^2 \leq \widetilde{C} R^2
(\|\widehat{F}(k , \cdot , s)\|^2 + \|\widehat{G}(k , \cdot ,
s)\|^2 )
\end{equation}
 for $
1\leq|k|\leq\sqrt{R/\sqrt 2}$ , $ |\xi | \leq
2(1+\sqrt{R}) $, where $\widetilde{C}$ is a constant independent
of $R$, $s$, $k$, $\widehat{F}$, $\widehat{G}$. Those inequalities
imply
\begin{gather*}
    k^2 \|\widehat{\psi} (k, \cdot , s)\|^2 \leq C R^2 (\|\widehat{F}(k , \cdot , s)\|^2 +
 \|\widehat{G}(k , \cdot , s)\|^2) \\
    \|\widehat{\psi} ' (k , \cdot , s)\|^2 \leq C R^2 (\|\widehat{F}(k , \cdot , s)\|^2 +
 \|\widehat{G}(k , \cdot , s)\|^2) ,
\end{gather*}
where $C$ is a constant independent of $R$, $s$, $k$,
$\widehat{F}$, $\widehat{G}$. Those are the desired estimates for
$\widehat{\psi}$.

\begin{figure}[th]
\begin{center}
\includegraphics[width=0.5\textwidth]{d1_func.eps}
\end{center}
\caption{ \label{fig1}
$\max_{k,\xi}k^2\|\delta_1(k,\cdot ,i\xi)\|^2$ for
$1\leq k \leq\sqrt{R/\sqrt 2}$,
$-2(1+\sqrt{R})\leq\xi\leq 2(1+\sqrt{R})$.}
\end{figure}

\begin{figure}[th]
\begin{center}
\includegraphics[width=0.5\textwidth]{d1_deriv.eps}
\end{center}
\caption{ \label{fig2}
 $ \max_{k,\xi}\|\delta_1 '(k,\cdot ,i\xi)\|^2$ for $
1\leq k \leq\sqrt{R/\sqrt 2}$,
 $-2(1+\sqrt{R})\leq \xi\leq 2(1+\sqrt{R})$.}
\end{figure}

\begin{figure}[ht]
\begin{center}
\includegraphics[width=0.5\textwidth]{d2_func.eps}
\end{center}
\caption{ \label{fig3}
$\max_{k,\xi}k^2\|\delta_2(k,\cdot ,i\xi)\|^2$ for $
 1\leq k \leq\sqrt{R/\sqrt 2}$,
 $-2(1+\sqrt{R})\leq \xi\leq 2(1+\sqrt{R})$.}
\end{figure}

\begin{figure}[ht]
\begin{center}
\includegraphics[width=0.5\textwidth]{d2_deriv.eps}
\end{center}
\caption{\label{fig4}
$ \max_{k,\xi}\|\delta_2 '(k,\cdot ,i\xi)\|^2$ for $1\leq
k \leq\sqrt{R/\sqrt 2}$, $-2(1+\sqrt{R})\leq \xi\leq
2(1+\sqrt{R})$.}
\end{figure}

\paragraph{Case 3: $ k = 0 $}  For this case,
the differential equation for $\widehat{\psi}$ is reduced to
\begin{equation} \label{eq3.35}
\begin{gathered}
  \frac{1}{R} \widehat{\psi} '''' - i\xi  \widehat{\psi} ''
  = \widehat{F}_y \\
  \widehat{\psi} (0,0,s) = \widehat{\psi} (0,1,s) = \widehat{\psi} ' (0,0,s)
  = \widehat{\psi} ' (0,1,s) = 0 .
\end{gathered}
\end{equation}
By an energy technique, after integration by parts, we get
\begin{equation} \label{eq3.36}
    \frac{1}{R}\|\widehat{\psi} '' \|^2 + i \xi \|\widehat{\psi} '\|^2 = \langle \widehat{\psi} ' , \widehat{F}
\rangle .
\end{equation}
Taking the real part of this equation and using Poincar\'e 's
inequality, we conclude that $ \|\widehat{\psi} ' \|^2 \leq
\frac{R^2}{\pi^4}  \|\widehat{F} \|^2$. Applying the Poincar\'e's
inequality again, we get
\begin{equation} \label{eq3.37}
    \|\widehat{\psi}(0, \cdot , s)\|^2 \leq  \frac{R^2}{\pi^6} \|\widehat{F}(0,\cdot ,s)\|^2 .
\end{equation}
which is the desired estimate. \hfil $\square$ \smallskip

Using Lemma \ref{lema1}, we conclude from the three cases above
that
\begin{equation}
\|(s\mathcal{I} - \mathcal{L}_R)^{-1}\|^2 \leq C R^2
\end{equation}
for $|s| < 2(1+\sqrt{R})$, where $C$ is a constant independent of $s$, $R$,
$\widehat{F}$, $\widehat{G}$.

\subsection*{Conclusions}
The estimates derived for $s$ bounded away from $0$ and computed
numerically  for the remaining region of the unstable half plane
indicate the $L_2$ norm of the resolvent of the operator
$\mathcal{L}_R$ to be proportional to $R$. Deriving the estimates
analytically for the whole unstable half plane is still an open
problem as far as we know. The method proposed here can be helpful
to solve it. It may also be possible to adapt this method for the
3 dimensional problem, and with this approach it may be possible
to give a satisfactory solution to the problem, or at least to
perform more convincing computations. We hope to address these
questions in the future.

\paragraph{Acknowledgement:} The author wish to thank
Professor Jens Lorenz, from the Department of Mathematics and
Statistics of The University of New Mexico, for suggesting the
problem and for the fruitful discussions about it. He also would
like to thank the anonymous referee, for pointing out a mistake in
a previous proof of Theorem \ref{teorema1}, and for his/her suggestions
that improved the presentation of this paper.


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\end{thebibliography}


\noindent \textsc{Pablo Braz e Silva } \\
 The University of New Mexico \\
 Department of Mathematics and Statistics \\
 Albuquerque, NM 87131, USA \\
 e-mail: pablo@math.unm.edu

\end{document}