\documentclass[10pt]{article}
\usepackage{url}
\usepackage{sober}
\usepackage{color}
\usepackage{times}
\usepackage{multicol}
\usepackage{amsmath}
\usepackage{fullpage}
\newcommand{\qq}[1]{\color{blue} #1 \color{black}}
\title{Linear discrete-time models}
\author{\copyright\ Ben Bolker: \today}
\date{}
\begin{document}
\maketitle

\enlargethispage{20pt}
\thispagestyle{empty}
\begin{multicols}{2}

Univariate (one-dimensional), discrete-time,
deterministic model: $N(t+1) = f(N(t))$.
(Typically, state variable is continuous.)
(Units of time, stock? When does discrete time
make sense?)

\section*{Geometric growth (decay)}
Simplest possible model.
$f(N)=RN$ (sometimes stated as $f(N)=(1+r)N$).
Solve recursion analytically.
Now we know everything about the dynamics.
Suppose $N(0)>0$, $R>0$ (note: the R language indexes
vectors starting from 1).
(What happens if $N(0)<0$ ? model of debt?)

\qq{What happens if $R<0$?}

(Even this ridiculously simple rule --- %
or generalizations of it --- is the basis of serious modeling 
in conservation biology.)

The limiting set of points as $t\to\infty$ is
called an \emph{attractor}. If $R<1$, $N \to 0$
but $N=0$ only in the limit, unless it starts there.
A value of $N$ such that $f(N^*)=N^*$ is called
an \emph{equilibrium} (or a \emph{fixed point}).

\emph{Stability}: what happens for perturbations
in the neighborhood of the fixed point?
Consider displacing the population away from
$N^*$ by $\delta$, where $\delta \ll 1$;
what happens? 
\begin{equation*}
\begin{split}
f(N^*+\delta) & = 
f(N^*)+\delta f'(N)|_{N=N^*} \\
\qquad \mbox{} & +\delta^2/2 f''(N)|_{N=N^*} + \ldots
\end{split}
\end{equation*}
Therefore the deviation $\delta \to \delta f'(N^*)$.
\textbf{If $|g| \equiv |f'(N^*)|<1$ then the deviation from
the equilibrium decreases geometrically with time}:
\emph{stable} fixed point.
Behavior for $g < (-1)$, $-1<g<0$, $0<g<1$, $g>1$ \ldots

$N=0$ is always an equilibrium, stable iff $|R|<1$.

\section*{Affine models}

Now suppose (as in the example in the book) we
are adding or subtracting a fixed amount per time 
step: $N(t+1) = a + b N(t)$.
As before we can work out the recursion.
% $N(t+2) = a + b (a + b N(t)) = a (1+b) + b^2 N(t)$
% $N(t+3) = a + b(a(1+b) + b^2 N(t)) = a (1 + b + b^2) + b^3 N(t)$

Summing the series for $t$ steps gives
$a(1-b^t)/(1-b) + b^t N$; the limit is
$a/(1-b) + \lim_{t\to\infty} b^t N(0)$.
If $|b|>1$ this is a bit boring.
If $|b|<1$ we get a stable equilibrium at $a/(1-b)$.
(For $b<0$ (``bucket model''): 
$a$ is the supply rate, $1/(1-b)$ is
the average \emph{residence time}.)

Useful component for larger models.
(Autoregressive model in time series analysis; sometimes used
as the bottom level in food chain modeling.)

In general, things are not so simple. The general approach is
(1) solve for equilibria (directly); 
(2) evaluate stability of equilibria;
(3) possibly evaluate for small $N$ (near system boundaries);
(4) if possible solve for time-dependent solution.

\section*{Graphical approaches: cobwebbing}

\begin{center}
 \SweaveOpts{height=3,width=3}
 \setkeys{Gin}{width=3in}
<<echo=FALSE,fig=TRUE>>=
source("../R/cobweb.R")
par(mfrow=c(2,2))
par(mar=c(2,4,1,1),las=1,mgp=c(2.5,1,0))
par(cex=0.5)
cobweb(0.75*x,start=20,N=10,xlab="")
cobweb(1.25*x,start=1,N=10,xlab="")
cobweb(10+0.75*x,start=2,N=10,xlab="",xlim=c(0,70),ylim=c(0,70))
cobweb(10+0.75*x,start=70,add=TRUE,scol=2)
abline(a=10,b=0.75) ## cheat
cobweb(10-1.25*x,start=8.1,N=10,xlab="")
cobweb(10-1.25*x,start=7.9,N=10,add=TRUE,scol=2)
@ 
\end{center}

\section*{Multiple lags}
What if $N(t+1)$ depends on previous time steps $N(t-1)$ etc. as
well as $N(t)$?  Homogeneous linear equations: $\sum_{i=0}^m a_i N(t-i)=0$.
Plug in $N(t)=C \lambda^t$.  Solve characteristic equation \ldots
get a linear combination of geometric growth/decay, $\sum C_i \lambda_i^n$:
largest \emph{eigenvalue} dominates long-term behavior.

\end{multicols}
\end{document}