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\title[Population Dynamics]{Modeling Population Dynamics}
\author{Raul Guajardo}
\institute{University of Texas - Pan-American}
\date{03/25/2014}
\begin{document}
\begin{frame}
\titlepage
\end{frame}
% Uncomment these lines for an automatically generated outline.
%\begin{frame}{Outline}
% \tableofcontents
%\end{frame}
\section{Population Dynamics}
\begin{frame}{Introduction}
\begin{itemize}
\item Population dynamics can be represented as Ordinary Differential Equations, although it is not exactly precise, it can be extremely useful in dealing with problems like population control, medical treatment etc.
\item For the following slides we will let \texttt{p(t)} represent the total number of members of a population at time \texttt{t}.
\item Below we first look at a steady production model.
\end{itemize}
\begin{block}{Steady Production}
If population increase is determined by a constant. Our ODE model is very simple and shown below.
$$\frac{dp}{dt} = c;\space p(0) = p_0\Rightarrow p(t)=ct+p_0.$$
\end{block}
\end{frame}
\subsection{Malthusian model}
\begin{frame}{Malthusian model}
\begin{itemize}
\item The malthusian model assumes that the \texttt{birth rate} and \texttt{death rate} of a population are proportional to the current size of the population.
\item To include it in your document, use the \texttt{includegraphics} command (see the comment below in the source code).
\end{itemize}
\begin{block}{Malthusian Model}
In this case we will let \texttt{b} represent birth rate and \texttt{d} represent death rate.
$$\frac{dp}{dt} = bp - dp = rp; \space p(0) = p_0\Rightarrow p(t)=p_0exp(rt).$$
Where r will be referred to as the \texttt{growth parameter} of the population.
\end{block}
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% Commands to include a figure:
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%\caption{\label{fig:your-figure}Caption goes here.}
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\subsection{Logistic Model}
\begin{frame}{Logistic Model}
\begin{itemize}
\item A problem with the previous model is that it assumes that there is no maximum limit for the size of the population.
\item In reality that is not true, the environment can't sustain a population that is always increasing.
\item Therefore we introduce \texttt{K}, the \texttt
{"carrying capacity"}.
\item In other words \texttt{K} is the greatest population sustainable by the environment.
\end{itemize}
\begin{block}{Logistic Model}
$$\frac{dp}{dt} = rp(1-\frac{p}{K}); \space p(0) = p_0\Rightarrow p(t)=\frac{p_0K}{(K-p_0)exp(-rt)+p_0}.$$
\end{block}
%Let $X_1, X_2, \ldots, X_n$ be a sequence of independent and identically distributed random variables with $\text{E}[X_i] = %\mu$ and $\text{Var}[X_i] = \sigma^2 < \infty$, and let
%$$S_n = \frac{X_1 + X_2 + \cdots + X_n}{n}
% = \frac{1}{n}\sum_{i}^{n} X_i$$
%denote their mean. Then as $n$ approaches infinity, the random variables $\sqrt{n}(S_n - \mu)$ converge in distribution to a %normal $\mathcal{N}(0, \sigma^2)$.
\end{frame}
\begin{frame}{Logistic Model Analysis}
We will analyze the logistic ODE model and determine the functionality of its critical points.
In order to solve for the critical points we must set $$\frac{dp}{dt} = 0$$
In this case we are left with the equation...
$$rp_e(1-\frac{p_e}{K}) = 0 \Rightarrow p_e = 0,K.$$
Where 0 and K are the equilibrium points of the logistic model.
\end{frame}
\begin{frame}{Logistic Model Analysis}
We can see how the critical points behave by checking the sign of $\frac{dp}{dt}$ on either side of the critical points.
\begin{itemize}
\item For $p_e= K$
\item We can see that at $p_e>K, \frac{dp}{dt}<0$. In other words when the population is greater than \texttt{K}, the \texttt{carrying capacity}, we see that the population decreases.
\item As for when $p_e<K, \frac{dp}{dt}>0$. Which means that when the population is less than K , the population increases towards K.
\item Therefore K is a stable critical point, since the population always approaches K from either side.
\end{itemize}
\end{frame}
\begin{frame}{Logistic Model Analysis}
\begin{itemize}
\item For $p_e= 0$
\item We can see that at $p_e>0, \frac{dp}{dt}>0$. In other words when the population is greater than \texttt{0}, we see that the population will increase away from the 0.
\item As for $p_e<0$, we ignore it because we cannot have a negative population.
\item Therefore 0 is an unstable critical point, since the population can only increase from 0.
\end{itemize}
\end{frame}
\begin{frame}
\begin{block}{Resources}
Modeling with ODE\par
P. Howard\par
Fall 2009
\url{<http://www.math.tamu.edu/~phoward/m442/modode.pdf>}
\end{block}
\end{frame}
\begin{frame}
\Huge \center Any Questions?
\end{frame}
\end{document}
```