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\usepackage[english]{babel}
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\title{The Logistic Flow}
\subtitle{(continuous)}

\author{Tom Carter}

\institute{Complex Systems Summer School}

\date{June, 2009}

\subject{Logistic}

\def\defn#1{{\color{red} #1}}

\begin{document}

\begin{frame}
  \titlepage
\end{frame}

%\begin{frame}
%  \frametitle{Outline}
%  \tableofcontents
%\end{frame}

%\section{Section 1}

\begin{frame}
\frametitle{Discrete logistic map}

    We all know that the discrete logistic map 
          \begin{eqnarray*}
            P_{n+1} & = & r  P_n  (1 - P_n) 
         \end{eqnarray*}
       exhibits interesting behavior of various sorts for various values of the parameter $r$, including chaos, etc.
\end{frame}
\begin{frame}
\frametitle{Discrete logistic map -- bifurcation diagram}
     \centerline{\includegraphics[width = 4.5in]{logistic-pic}}
\end{frame}
\begin{frame}
\frametitle{Continuous logistic flow}
    What kind of behavior can we expect from a continuous version of a logistic flow:
     \begin{eqnarray*}
          \frac{dP}{dt} & = & r   P   (1 - P)  \ \ \   ?
     \end{eqnarray*}

\end{frame}
\begin{frame}
\frametitle{Continuous logistic flow - solving}
  Note that this is a non-linear ODE, but fortunately we can actually integrate \ldots

            \begin{eqnarray*}
               \frac{dP}{dt} & = & r   P   (1 - P) \\
               \\
              \frac{dP}{P   (1 - P)} & = & r   dt \\
            \end{eqnarray*}

%\end{frame}
%\begin{frame}
%\frametitle{Continuous logistic flow - solving}
            
            Thus: 
            \begin{eqnarray*}
               \int \frac{dP}{P   (1 - P)} & = & \int r   dt \\
               \\
               \int \frac{dP}{P   (1 - P)} & = & r   t + c_1 \\
            \end{eqnarray*}


\end{frame}
\begin{frame}
\frametitle{Continuous logistic flow - solving}
            
 By partial fractions, we have:
        
             \begin{eqnarray*}
               \int \frac{dP}{P} + \int \frac{dP}{(1 - P)} & = & r   t + c_1 \\
               \\
               \log (P) - \log (1 - P) & = & r   t + c_1 \\
               \\
               \log ( \frac{P}{1 - P}) & = & r   t + c_1 \\
               \\
               \frac{P}{1 - P} & = & e^{r   t + c_1} \\
               \\
               \frac{P}{1 - P} & = & c_2   e^{r   t} \\
            \end{eqnarray*}

\end{frame}
\begin{frame}
\frametitle{Continuous logistic flow - solving}
            
This gives us:

           $$    P  =  ( 1 - P )   c_2   e^{r   t} $$
            
 And thus: 
             \begin{eqnarray*}
               P & = & c_2   e^{r   t} - P   c_2   e^{r   t} \\
               \\
               P + P   c_2   e^{r   t} & = & c_2   e^{r   t} \\
               \\
               P   ( 1 +  c_2   e^{r   t}) & = & c_2   e^{r   t} \\
            \end{eqnarray*}

\end{frame}
\begin{frame}
\frametitle{Continuous logistic flow - solving}
            
    From this we get:
       \begin{eqnarray*}
               P & = & \frac { c_2   e^{r   t}} { 1 +  c_2   e^{r   t} } \\
            \end{eqnarray*}


    Finally, dividing top and bottom by $c_2   e^{r   t}$ and simplifying, we have:
         \begin{eqnarray*}
               P & = & \frac {1} { 1 +  c   e^{-r   t} } \\
            \end{eqnarray*}
            

\end{frame}

\begin{frame}
\frametitle{The classic logistic/sigmoid curve}

      \centerline{\includegraphics[width = 3in]{Logistic-curve}}
      
      and changes in $c$ and $r$ make minor changes in the behavior near $0$ \ldots
            

\end{frame}
\begin{frame}
\frametitle{Discrete vs. Continuous}

           The difference between the behavior of the discrete and continuous logistic functions
      can give us some idea of the significance of working in the discrete regime . . .


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\frametitle{Fin}
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\begin{center}
{\bf  \ldots}
\end{center}

\vskip10pt

\begin{center}
Slides for this talk will be available at:\\
\url{http://csustan.csustan.edu/~tom/SFI-CSSS/2009}
\vskip12pt
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\titlepage
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\end{document}