%% ****** Start of file template.aps ****** %
%%
%%
%%   This file is part of the APS files in the REVTeX 4 distribution.
%%   Version 4.0 beta 3 of REVTeX, November 30, 1999.
%%
%%
%%   Copyright (c) 1999 The American Physical Society.
%%
%%   See the REVTeX 4 README file for restrictions and more information.
%%
%%
% This is a template for producing files for use with REVTEX 4.0 beta
% Copy this file to another name and then work on that file.
% That way, you always have this original template file to use.
%
% Group addresses by affiliation. Use superscriptaddress for long
% author lists or if there are many overlapping affiliations
% For Phys. Rev. look and feel change preprint to twocolumn
% showpacs will print out PACS numbers
%\documentclass[aps,prl,reprint,twocolumn,showpacs,epsf]{revtex4}
\documentclass[aps,pre,preprint,superscriptaddress]{revtex4}
%\documentclass[aps,twocolumn,groupedaddress]{revtex4}

\usepackage{graphicx}
\usepackage{psfrag}
\usepackage{color}

\def\dd{\mbox{d}}

\begin{document}
% You should use BibTeX and revtex.bst for references
\bibliographystyle{apsrev}



% Use the \preprint command to place your local institutional report
% number on the title page in preprint mode.
% Multiple \preprint commands are allowed.
%\preprint{}



%Title of paper
\title{Notes on the LSA}
% Optional argument for running titles on pages
%\title[]{}
% repeat the \author .. \affiliation  etc. as needed
% \email, \thanks, \homepage, \altaffiliation all apply to the current
% author. Explanatory text should go in the []'s, actual e-mail
% address or url should go in the {}'s for \email and \homepage.
% Please use the appropriate macro for the type of information



% \affiliation command applies to all authors since the last
% \affiliation command. The \affiliation command MUST follow the
% other information
%\author{Maria}


%astrobiomathematicalphysicochemicalhydrodynamicist
%\email[]{Your e-mail address}
%\homepage[]{Your web page}
%\thanks{}
%\altaffiliation{}
%\affiliation{Department of Mathematics, UCLA, Los Angeles, CA 90095}
%\affiliation{}


%Collaboration name if desired (requires use of superscriptaddress
%option in \documentclass). \noaffiliation is required (may also be
%used with the \author command).
%\collaboration{}
%\noaffiliation



\date{\today}



\begin{abstract}
Some notes on the semi-2D linear stability analysis
\end{abstract}



%insert suggested PACS numbers in braces on next line
\pacs{05.65.+b, 45.50.-j, 45.40.Ln, 87.18.Ed}
%\maketitle must follow title, authors, abstract and PACS

\maketitle
\noindent
From the Linear Stability Analysis related to the $2D$ potentials
we find that the possible unstable wavelengths $k$ are given by
the ones that satisfy the following

\begin{eqnarray}
\label{omega}
\omega(k) = - 2 \pi k^2 \left[\frac{s_0 R_s r_s^2}{(1+k^2 r_s^2)^{3/2}} +
\frac{g_0 R_g r_g^2}{(1+k^2 r_g^2)^{3/2}} -
\frac{g_0 A_g a_g^2}{(1+k^2 a_g^2)^{3/2}}
\right] \geq 0
\end{eqnarray}

\noindent
and where $s_0$ and $g_0$ are the steady state, equilibrium
concentrations of solitary and gregarious locusts, repectively and
given as

\begin{eqnarray}
s_0 &=& \frac{\rho_0 \delta_1 k_1^2 (k_2^2 + \rho_0^2)}{\delta_1 k_1^2
  k_2^2 + (\delta_1 k_1^2 + \delta_2 k_2^2) \rho_0^2 + \delta_2
  \rho_0^4} \\
\nonumber
\\
g_0 &=& \frac{\delta_2 \rho_0^3 (k_1^2 + \rho_0^2)}{\delta_1 k_1^2
  k_2^2 + (\delta_1 k_1^2 + \delta_2 k_2^2) \rho_0^2 + \delta_2
  \rho_0^4} 
\end{eqnarray}

\noindent
In order to make analytical progress we set $k_1 = k_2 = h$ and
$\delta_1 = \delta_2 = \delta$. The use of $h$ is to avoid confusion
with the wavelength $k$. We also use the fact that $r_s=a_g$. Later
it will be useful to recall that within our model $r_g < r_s$.
Inserting these simplifications into Eq.\,\ref{omega}
we find that $\omega(k)$ can be
written as

\begin{eqnarray}
\label{omega2}
\omega(k) = \frac {2 \pi R_g \rho_0^2 r_g^2 k^2}{(1+k^2 r_s^2)^{3/2}} 
\left[M - m(k)
%frac {(1+k^2 r_s^2)^{3/2}}{(1+k^2 r_g^2)^{3/2}}
\right] 
\end{eqnarray}

\noindent
where

\begin{eqnarray}
M = \frac{\rho_0^2 A_g r_s^2 - h^2 R_s r_s^2}{R_g \rho_0^2 r_g^2}
\end{eqnarray}

\noindent
and

\begin{eqnarray}
m(k) = \frac {(1+k^2 r_s^2)^{3/2}}{(1+k^2 r_g^2)^{3/2}}.
\end{eqnarray}

\noindent
Note that eq.\,\ref{omega2} holds for all paramter
choices, provided that $k_1= k_2$, $r_s = a_g$ and $\delta_1 = \delta_2$.
Also note that since $m(k) > 0$
for instability to occur we must
require $M > 0$, otherwise $\omega(k)$ will always be negative. This
is not a sufficient condition, but it is a necesary one and translates
to

\begin{eqnarray}
\rho_0 > \rho_c = \sqrt{\frac{R_s}{A_g}} h,
\end{eqnarray}

\noindent
which in our case for $R_s = 41.5, A_g= 13.3$ and $h=65$ implies that
if $\rho_0 < \rho_c = 115$ the system will always be stable. Let us now look
at the terms inside the square brackets of Eq.\,\ref{omega2}.  We know
that $r_g < r_s$. This implies that the function $m(k)$ is
monotonically increasing. Hence, for $\omega(k)$ to be positive, it is
also necessary that $M$ be greater that the minimum value of $m(k)$
which is attained at $k=0$. Thus, a more stringent condition is that

\begin{eqnarray}
M > m(k=0) = 1
\end{eqnarray}

\noindent
Using the fact that $m(k \to \infty) = r_s^3/ r_g^3 > 1$
we can now distinguish three cases

\begin{itemize}
\item {if $M < 1$ \, then $\omega(k)$ always stable} 
\item {if $\displaystyle{1 \leq M \leq \frac{r_s^3}{r_g^3}}$ 
\, then
instabilities will arise only at small wavelengths, $k \to 0$}
\item{if $M > \displaystyle{\frac{r_s^3}{r_g^3}}$ \, then
the system is always unstable for all wavelengths.}
\end{itemize}

\noindent
The condition $M > 1$, to guarantee instability can be rewritten as

\begin{eqnarray}
\rho_0 > \rho_{c,2} = \sqrt{\displaystyle{
\frac{R_s}{A_g - R_g \frac{r_g^2}{r_s^2}}}}\, h
\end{eqnarray}

\noindent
so that, we are now guaranteed instability as long as $\rho_0 >
\rho_{c,2} = 116.7$ which is exactly what we see
from numerical plots. This condition is now sufficient for
instability to set in.
Note that for the parameters at hand, the third
case cannot be ever verified, since it can be shown that $M <
r_s^3/r_g^3$ for the numbers we use in the paper.  It may however, be
a viable occurrence for other parameter choices.
So, it seems that for the parameters we have chosen,
we get instabilities for $\rho > 146$.


\section{Maximum Instability}
\noindent
Finally, if we are interested in the most unstable wavelength, in the
case where $M > 1$, we can calculate $\omega'(k)$ to find

\begin{eqnarray}
\omega'(k) = \frac{2 \pi R_g \rho_0^2 r_g^2 k}{(1 + r_s^2 k^2)^{5/2}} 
\left[M \frac{2 - r_s^2 k^2}{(1 + r_s^2 k^2)^{5/2}}- 
\frac{2 - r_g^2 k^2}{(1+r_g^2 k^2)^{5/2}} \right].
\end{eqnarray}

\noindent
note that the two terms in the parenthesis contain the function

\begin{eqnarray}
g(y) = \frac{2 - y^2}{(1 + y^2)^{5/2}}
\end{eqnarray}

\noindent
so that finding the maximum corresponds
to setting the term in parenthesis to zero and
finding values of $k=k^*$ so that

\begin{eqnarray}
\label{tosolve}
M g(r_s k^*) = g(r_g k^*)
\end{eqnarray}

\noindent
The $g(y)$ function is such that $g(0) = 2$, $g(y \to \infty) = 0$, it
intersects the $y$ axis at $y = \sqrt{2}$ initially decreases and
attains a minimum at $y = 2$.  We note that this means, specifically,
that at $k=0$ the function $M g(r_s k=0) > g(r_g k=0)$ since $M > 1$
in this case.  Hence, $omega'(k)$ starts out positive for small $k$.
On the other hand, the function $M g(r_s k)$ will be zero at the
location $k_s =\sqrt{2}/r_s$, whereas the function $g(r_g k)$ will be
zero at the location $k_g = \sqrt{2}/r_g$. We also know that $k_g >
k_s$, due to our choice $r_s< r_g$. In particular, this means that for
$k=k_s$ $M g(r_s k =r_s k_s) = 0$ but $g(r_g k=r_g k_s)$ is still
positive, because it will become zero for larger $k=k_g$.

These facts lead us to conclude that $M g(r_s k = r_sk_s) = 0 
< g(r_g k = k_s)$, and that there must exist a point
$0 < k^*< k_s$  where the two terms are equal. In particular,
for the choice of $r_s = 0.14$ this leads to $k^* < 10$.
The value of $k^*$ is the value of $k$ that leads to the fastest
growing perturbation. There is another solution to
Eq.\,\ref{tosolve},but that will represent a minimum.


\begin{figure}[t]
\label{Fig1}
\begin{center}
\includegraphics[width=5.3in, angle=0]{Chad.pdf}
\caption{The magenta curve is the function $g(r_g k)$
while the blue curve is the $M g(r_s k)$ curve, where $M > 1$.
Here, we set $M=5$, $r_g = 0.04$ and $r_s = 0.14$. Note that
as predicted, the two meet at a value $k^* < \sqrt{2}/r_s$
which is the point that makes the curve $M g(r_s k) = 0$.
In this case, $k^* \simeq 6.02$}
\end{center}
\end{figure}










\end{document}
