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\begin{document}

\newpage


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\title{\ifnum\TeacherEdition=1 TEACHER VERSION
%\footnote{DELETE THIS FOOTNOTE BEFORE SUBMISSION.  In the \TeX \, source code there is a toggle/option to produce a Teacher Version or a Student Version by ``carving'' out with  $\backslash$teacher\{\} that material which should only appear in the Teacher version. See the source code.}
\else STUDENT VERSION \fi\\Climate Change Effects on Insect Outbreaks}





\markboth{Climate Change Effects on Insect Outbreaks}{Climate Change Effects on Insect Outbreaks}

\author{Jacob Duncan\\
Winona State University\\
Winona MN 55987 USA\\
jacob.duncan@winona.edu
}

\makeStitlePDFLaTex





\begin{abstract}

\noindent The mountain pine beetle (MPB, \textit{Dendroctonus ponderosae}), a tree-killing
bark beetle, has historically been part of the normal disturbance regime in lodgepole
pine (\textit{Pinus contorta}) forests. In recent years, warmer weather has allowed
MPB populations to achieve synchronous emergence and successful attacks, resulting
in widespread population outbreaks and resultant tree mortality across western
North America. Here we will use a system of difference equations that incorporates a
temperature-dependent MPB population growth rate to model the outbreak and recovery cycle in MPB-infested forests (see Figure \ref{cycle}).
We will explore analytical methods, applied to the model,
to predict outbreak severity in terms of the climate-sensitive growth rate parameter.
These model-based predictions elucidate the connection between warming climate and increased severity
of recent MPB outbreaks.
\end{abstract}








\section*{SCENARIO DESCRIPTION}

\begin{figure}[h]
    \centering
    \includegraphics[width=150mm]{mpb1.png}\\
    %\caption{} \label{}
\end{figure}


\section*{Motivation}

\begin{enumerate}
    \item Before class, students should
    \begin{enumerate}
        \item Watch the following short videos about mountain pine beetle (MPB) outbreaks:
            \begin{itemize}
                \item \textit{National Geographic}: \href{https://www.youtube.com/watch?v=vR30qlK0-Cw}{https://www.youtube.com/watch?v=vR30qlK0-Cw}  \cite{NatlGeo}
                \item \textit{The YEARS Project}: \href{https://www.youtube.com/watch?v=FzrfzQG6YD0}{https://www.youtube.com/watch?v=FzrfzQG6YD0}  \cite{years}
            \end{itemize}

         \item Read Sections 1 and 2 of this project.

        \item Read Sections 1 and 2 (Subsections 2.1 and 2.2 only) of \cite{Duncan2015}.\\
          * The main purpose of this reading is to acquaint students with an academic paper.  However, this step may be omitted since the readings are summarized in Sections 1 and 2 of this project.

    \end{enumerate}

    \item In class, show students \textit{Matlab} simulations of MPB outbreaks.\\
         \teacher{* \textit{Matlab} code included.  }

\end{enumerate}



\begin{figure}[h]
    \centering
    \includegraphics[width=150mm]{cycle.png}
    \caption{Relationship between MPB life-cycle and lodgepole pine forests.  } \label{cycle}
\end{figure}

\section{Introduction}
The mountain pine beetle (MPB, \textit{Dendroctonus ponderosae}), a tree-killing
bark beetle, has historically been part of the normal disturbance regime in lodgepole
pine (\textit{Pinus contorta}) forests. In recent years, however, warmer weather has allowed
MPB populations to achieve synchronous emergence and successful attacks, resulting
in widespread population outbreaks and resultant tree mortality across western
North America \cite{Bentz2016}.

In late spring and summer, MPB emerge from infested trees and search for new trees to colonize.  MPB attack adult pine trees \emph{en masse} in order to overcome the tree's defensive mechanisms \cite{Berryman1985}.  Once colonized, MPB lay eggs beneath the bark of the tree.  In spring, eggs hatch and larvae feed on the tree's phloem layer, which kills the tree.  Once larvae have molted into adult MPB, they emerge from the tree in summer and the cycle begins again \cite{Amman1988} (see Figure \ref{cycle}).

One year following an infestation, a tree's needles turn red (the tree is then called a red snag).  Two years following an infestation, the needles turn gray (gray snag).  Finally, three years after infestation, the needles fall off of the tree completely.  This opens up the forest floor (which will typically be littered with heat-dependent serotinous pine cones) to the sun's thermal radiation, generating new juvenile tree growth.  Juvenile trees are not susceptible to MPB attacks until they have a big enough phloem layer to support a colonization (around 50 to 80 years of age) \cite{Amman1988}.

In this project, we will use a system of difference equations that incorporates
temperature-dependent MPB population growth rate to model the outbreak and recovery cycle in MPB-infested forests. We then explore analytical methods, applied to the model, to predict outbreak severity in terms of the climate-sensitive growth rate parameter. These model-based predictions elucidate the connection between warming climate and increased severity of recent MPB outbreaks.






\section{The SIJ Model}

\begin{figure}[h]
    \centering
    \includegraphics[width=150mm]{model.png}
    \caption{The SIJ model.  %Natural juvenile mortality, $d$, opens up forest floor space to new seedling growth.
%Likewise, infestation mortality translates to seedling growth (after a once infested tree spends $2$ years as a snag). Juvenile age class survivorship $s = 1 - d$ is constant.   A tree spends $NJ$ years as a juvenile before graduating (maturing) to the susceptible class.  Susceptible trees in the forest become infested at a rate proportional to the MPB population growth rate, $R$, as well as the rate at which MPB are able to find susceptible trees, given by $e^{-\beta(T-S_{n+1})}$.
} \label{model}
\end{figure}


\begin{figure}[h]
    \centering
    \includegraphics[width=150mm]{table1.png}
    \caption{Model variables, parameters, and estimated parameter values from \cite{Powell2009}.}
    \label{table}
\end{figure}

The SIJ model of \cite{Duncan2015} is a system of difference equations that tracks yearly populations of susceptible trees $S_n$, infected trees $I_n$, and juvenile trees $J_n$ (see Figures. \ref{model} and \ref{table}).
Natural juvenile mortality, $d$, opens up forest floor space to new seedling growth.
Likewise, infestation mortality translates to seedling growth (after a once infested tree spends $2$ years as
a snag). Juvenile age class survivorship $s = 1 - d$ is constant.   A tree spends $NJ$ years as a juvenile before graduating (maturing) to the susceptible class.  Susceptible trees in the forest become infested at a rate proportional to last year's population of infested trees.  The constant of proportionality is the MPB population growth rate.  The rate of infestation also depends on how easily MPB are able to find susceptible trees and can be characterized by $e^{-\beta(T-S_{n+1})}$, where $\beta$ is a parameter measuring the efficiency with which MPB search out and find new host trees to colonize.  The total population of trees in the forest is denoted by $T$.
The equation modeling infested trees is
\begin{equation}\label{infEqn}
I_{n+1}=RI_{n}e^{-\beta(T-S_{n+1})},
\end{equation}
where $R$ represents the MPB population growth rate parameter.  Note that $R$ implicitly depends on temperature since MPB developmental rates are driven by ambient thermal energy as they progress through their life stages \cite{Bentz2016}.

The current number of \emph{nonsusceptible} trees in the forest is given by $(T-S_{n+1})$.  Hence, as MPB infest more and more trees each year, the number of nonsusceptible trees increases, causing the exponential term in (1) to approach $0$.  This significantly reduces the spread of infestation as MPB no longer have a sufficient number of susceptibles trees left to colonize.  Eventually, this leads to a decrease in infested trees and an increase in susceptible trees.  When the forest has recovered from an outbreak, it contains mostly susceptible trees.  In this case, the exponential term approaches $1$, reducing the infestation equation to $I_{n+1} = RI_n$.  This yields exponential growth of infestation, triggering another outbreak.  Thus, we see that MPB outbreaks are periodic in nature.  Figure \ref{sim} illustrates this through simulation of the model.

\begin{figure}[h]
    \centering
    \includegraphics[width=150mm]{sim2.png}
    \vspace*{-.1 true cm}\caption{The outbreak-recovery cycle: Simulation output showing tree populations over two full periods. } \label{sim}
\end{figure}


\vspace*{-.5 true cm}

%\subsection*{Move from Discrete-time to Continuous-time Model}
\section{Convert the Difference Equation to a Differential Equation Model}

Here we move from a discrete-time to a continuous-time model by transforming the difference equation governing infested trees into an approximating differential equation.


\begin{enumerate}
  \item Take the natural logarithm of both sides of (\ref{infEqn}) and solve for $\beta S_{n+1}$.  Label the result as (2).
    \begin{mdframed}[backgroundcolor=gray!20]
          \noindent\verb"Solution":
              \begin{equation}
                \beta S_{n+1} = \beta T - \ln R - \ln I_n + \ln I_{n+1}
              \end{equation}
    \end{mdframed}

    \mbox{}

\vspace*{-.3 true cm}
  \item Re-index the equation by subtracting $1$ from all variable indices.  Label the result as (3).
    \begin{mdframed}[backgroundcolor=gray!20]
          \noindent\verb"Solution":
              \begin{equation}
                \beta S_{n} = \beta T - \ln R - \ln I_{n-1} + \ln I_{n}
              \end{equation}
    \end{mdframed}

    \mbox{}


  \item Subtract (3) from (2).  Label the result as (4).
    \begin{mdframed}[backgroundcolor=gray!20]
          \noindent\verb"Solution":
              \begin{equation}
                \beta(S_{n+1} - S_{n}) = \ln I_{n-1} - 2\ln I_{n} + \ln I_{n+1}
              \end{equation}
    \end{mdframed}

    \mbox{}
\vspace*{-.3 true cm}

  \item Since the number of juvenile trees is small during an outbreak, we may neglect the term containing $J_{NJ,n}$ in the equation governing susceptible trees, $S_{n+1} = S_n - I_n + sJ_{NJ,n}$.  This results in $S_{n+1} = S_n - I_n$.
  Substitute the right side of this equation in for $S_{n+1}$ in (4).  Label the result as (5).
    \begin{mdframed}[backgroundcolor=gray!20]
          \noindent\verb"Solution":
              \begin{equation}
                -\beta I_{n} = \ln I_{n-1} - 2\ln I_{n} + \ln I_{n+1}
              \end{equation}
    \end{mdframed}

    \mbox{}
\vspace*{-.3 true cm}
  \item We are  ready to make the transition from a discrete-time to a continuous-time model.  Let $y(t)=\ln I_n$ be some continuous function of time $t$ years where $t=n$ for whole number values of $t$.  Note that $t$ is a real variable, as opposed to $n$ which is an integer variable.  Write (5) in terms of $y$.  Label  result as (6).
    \begin{mdframed}[backgroundcolor=gray!20]
          \noindent\verb"Solution":
              \begin{equation}
                -\beta e^{y(t)} = y(t-1) - 2y(t) + y(t+1)
              \end{equation}
    \end{mdframed}

    \mbox{}
\vspace*{-.3 true cm}

  \item Use  second finite difference approximation for  second derivative, $y'' \approx y(t+1) - 2y(t) + y(t-1)$,
  %\[
  %  y'' \approx y(t+1) - 2y(t) + y(t-1),
  %\]
  to convert  (6) to a second order differential equation (solved for $y''$).  Label  result as (7).
    \begin{mdframed}[backgroundcolor=gray!20]
          \noindent\verb"Solution":
              \begin{equation}
                y''(t) = -\beta e^{y(t)}
              \end{equation}
    \end{mdframed}

\end{enumerate}


\section{Solve the Differential Equation}

\begin{enumerate}
  \item Multiply both sides of the second order differential equation (7) by $y'(t)$.
        %\begin{mdframed}[backgroundcolor=gray!20]
        %      \noindent\verb"Solution":
        %          \[
        %            y'(t)y''(t) = -\beta e^{y(t)}y'(t)
        %          \]
        %\end{mdframed}
\vspace*{-.3 true cm}
        \begin{enumerate}
          \item Integrate the left side of the equation with respect to time (omit the constant of integration).   Hint: Integration by parts!
            \begin{mdframed}[backgroundcolor=gray!20]
              \noindent\verb"Solution":
                  \[
                    \int y'(t)y''(t) \mbox{ } dt  = \frac{1}{2}\left(y'(t)\right)^2
                  \]
            \end{mdframed}

          \mbox{}


          \item Integrate the right side of the resulting equation with respect to time (label the constant of integration $C$).   Hint: $u$-substitution!
          \begin{mdframed}[backgroundcolor=gray!20]
              \noindent\verb"Solution":
                  \[
                    \int -\beta e^{y(t)}y'(t) \mbox{ } dt  = -\beta e^{y(t)} + C
                  \]
            \end{mdframed}

          \mbox{}

        \end{enumerate}

  \item Solve for $y'$ (take only the positive square-root).  Label the result as (8).  Note that the result is a \textit{first order} differential equation.
        \begin{mdframed}[backgroundcolor=gray!20]
          \noindent\verb"Solution":
              \begin{equation}
                y' =  \sqrt{-2\beta e^y + C}
              \end{equation}
        \end{mdframed}

          \mbox{}


  \item We can solve for the constant of integration, $C$, in (8) using the fact that $t=0$ corresponds to the peak of the outbreak.  That is, $y'(0)=0$ and $y(0)=\ln I_{max}$ where $I_{max}$ represents the maximum infestation (the value of which will be determined later).  Solve for $C$ and substitute it back into equation (8).  Factor out any common factors.  Label the result as Equation (9).
        \begin{mdframed}[backgroundcolor=gray!20]
          \noindent\verb"Solution":
              \begin{equation}
                y' =  \sqrt{2\beta}\sqrt{I_{max} - e^y}
              \end{equation}
        \end{mdframed}

          \mbox{}


  \item Using the change of variables, $\hat{I}(t)=e^{y(t)}$, write (9) in terms of $\hat{I}(t)$ (solved for $\hat{I}^{\prime}(t)$).  Note that $\hat{I}(t)$ is our continuous function approximation for the discrete function $I_n$.  Label the result as (10).
        \begin{mdframed}[backgroundcolor=gray!20]
          \noindent\verb"Solution":\\
              Note that $y'(t) = \frac{\hat{I}'(t)}{\hat{I}(t)}$.
              \begin{equation}
                \hat{I}' =  \sqrt{2\beta}\text{ } \hat{I} \sqrt{I_{max} - \hat{I}}
              \end{equation}
        \end{mdframed}

          \mbox{}


  %\item Use a graphical ODE solver, such as the Java applet \verb"DFIELD", to numerically solve the first-order equation (10).  \verb" DFIELD" can be obtained at \url{https://www.cs.unm.edu/~joel/dfield/} (you may have to retype the tilde in a web browser).

  \item Solve the first order (10) using the separation of variables method.  First, separate the variables.  Then integrate both sides.  You will need the integral formula
  \[
    \int \frac{1}{x\sqrt{k-x}}dx = \frac{-2}{\sqrt{k}}\tanh^{-1}\left( \frac{\sqrt{k-x}}{\sqrt{k}} \right)
  \]
  where $k$ is a positive constant.  Do not solve for $\hat{I}(t)$ yet.

        \begin{mdframed}[backgroundcolor=gray!20]
          \noindent\verb"Solution":
              \[
                 \frac{2}{\sqrt{I_{max}}} \tanh^{-1} \left( \frac{\sqrt{I_{max} - \hat{I}}}{\sqrt{I_{max}}} \right) =  \sqrt{2\beta}t + C
              \]
        \end{mdframed}

          \mbox{}



  \item Find the value of the constant of integration using the fact that $\hat{I}(0)=I_{max}$.
        \begin{mdframed}[backgroundcolor=gray!20]
          \noindent\verb"Solution":
              \[
                C = 0
              \]
        \end{mdframed}

          \mbox{}



  \item Now solve for $\hat{I}(t)$.  You will need the identities $\tanh^2(x) + \textrm{sech}^2(x) = 1$ and $\textrm{sech}(-x)=\textrm{sech}(x)$.
  %\[
  %  \tanh^2(x) + \textrm{sech}^2(x) = 1
  %\]
  %and
  %\[
  %  \textrm{sech}(-x)=\textrm{sech}(x).
  %\]
  Label the result as (11).    (This is the \textit{main result} of the project.)

        \begin{mdframed}[backgroundcolor=gray!45]
          \noindent\verb"Solution  (Main result!)":
              \begin{equation}
                \hat{I}(t) = I_{max} \text{ } \textrm{sech}^2\left( \sqrt{ \frac{\beta I_{max}}{2} } t \right)
              \end{equation}
              \mbox{}
        \end{mdframed}




  \noindent Note that this model (11) has $t$ as the \textit{independent variable} and $\hat{I}(t)$ as the \textit{dependent variable}.  \\
  The \textit{parameters} of the model are $\beta$ and $I_{max}$.

\end{enumerate}


\section{Model Parameterization}

We now fit (11) to data taken from a recent outbreak in central Idaho given in Table 1 (data from \cite{Duncan2017}).  This will entail adjusting the parameter values of the model ($\beta$ and $I_{max}$) until the graph of the solution, (11), best fits a scatter plot of the data.
  %\begin{center}
    \begin{table}[h!]%{ |c|c| }
    \centering
        \begin{tabular}{ |c|c| }
         \hline
         Year & Infested tree density (trees/hectare) \\
         \hline
         1995 & 3.1 \\
         \hline
         1996 & 3.2 \\
         \hline
         1997 & 2.9 \\
         \hline
         1998 & 3.0 \\
         \hline
         1999 & 4.5 \\
         \hline
         2000 & 5.5 \\
         \hline
         2001 & 7.4 \\
         \hline
         2002 & 39.1 \\
         \hline
         2003 & 27.0 \\
         \hline
         2004 & 11.8 \\
         \hline
         2005 & 6.5 \\
         \hline
         2006 & 4.6 \\
         \hline
         2007 & 5.7 \\
         \hline
         2008 & 5.2 \\
         \hline
    \end{tabular}
    \caption{Infestation data taken from a MPB outbreak in the Sawtooth National Recreation Area, Idaho. }
    \end{table}
%\end{center}

\noindent Note: A \textit{hectare} (ha) is equal to 10,000 square meters (approximately 2.47 acres).

\begin{enumerate}
  \item Plot the data in \textit{Desmos} \cite{Desmos} (or similar graphing software). \\ \textit{Desmos} can be found at \href{https://www.desmos.com/calculator}{https://www.desmos.com/calculator}.
       \begin{mdframed}[backgroundcolor=gray!20]
       \newpage
          \noindent\verb"Solution":\\ \\
                \includegraphics[width=140mm]{desmos1.png}\\ \\
                \includegraphics[width=140mm]{desmos2.png}
            \begin{center}
                Figure 5. \textit{Desmos} plot of infestation data from Table 1.
            \end{center}
        \end{mdframed}

\newpage
  \item Graph (11) in \textit{Desmos} along with the data.  Create sliders for the parameters $\beta$ and $I_{max}$.  Since the peak of the graph of (11) is centered around zero, and the peak in the data is centered somewhere in the early 2000s, we must horizontally shift the graph of (11) to align the peaks.  Replace the time variable with $(t-t_{peak})$ and create a slider for $t_{peak}$ (the time at which the outbreak peaks).
      \begin{mdframed}[backgroundcolor=gray!20]
        \noindent\verb"Solution":\\ \\
                \includegraphics[width=140mm]{desmos3.png}
            \begin{center}
                Figure 6. \textit{Desmos} graph of (11) along with infestation data from Table 1.
            \end{center}
      \end{mdframed}

    \mbox{}


  \item Adjust the sliders until the graph of (11) fits the data best by visual inspection of the graph.
  \item Now calculate (in \textit{Desmos} or similar) the sum of the squared errors ($SSE$) between the model and the data:
  \[
  SSE = \sum(\textrm{observed infestation} - \textrm{predicted infestation})^2
  \]
  \[
  = \sum\limits_{n=1995}^{2008}\left(I(n) - \hat{I}(n)\right)^2.
  \]

  \newpage
    \begin{mdframed}[backgroundcolor=gray!20]
        \noindent\verb"Solution":\\ \\
                \includegraphics[width=140mm]{desmos4.png}
            \begin{center}
                Figure 7. \textit{Desmos} graph of (11) along with infestation data from Table 1.  Parameter sliders have been added for the parameterization process.  Parameters are varied until $S$ (SSE) is minimized.
            \end{center}
      \end{mdframed}


  \item Adjust the sliders so that the $SSE$ is minimized (as small as possible).\\


   \item Write down the parameter values that minimize the $SSE$ as well as the value of the $SSE$.\\
  \begin{mdframed}[backgroundcolor=gray!20]
        \noindent\verb"Solution (answers may vary)":
        \[
            \beta \approx 0.033
        \]
        \vspace*{-.5 true cm}
        \[
            I_{max} \approx 38.6
        \]

        \[
            SSE =  236.9
        \]

        \mbox{}

      \end{mdframed}


\end{enumerate}

\noindent * Here we offer a link to a live \textit{Desmos} plot in which the reader can adjust parameter values in order to minimize the SSE: \href{https://www.desmos.com/calculator/l4avq5n5f7}{https://www.desmos.com/calculator/l4avq5n5f7}.









\section{Effect of Global Warming on Outbreak Severity}

One measure of the severity of an outbreak is the \textit{outbreak footprint} -- the total number of trees (per hectare of forest) killed over the course of an outbreak.
A formula for approximating the outbreak footprint derived in \cite{Duncan2015} is given by
\begin{equation}
    F = \frac{2\ln R}{\beta}.
\end{equation}
This approximation is in terms of the MPB search efficiency $\beta$, and population growth rate $R$.   Since $R$ depends heavily on temperature, $F$ is used to assess the effect of climate change on the severity of outbreaks. \\

\vspace*{-.3 true cm}
\begin{enumerate}
    \item At the onset of an ongoing outbreak, it was determined that the MPB population growth rate was approximately 4.6.  Predict the total number of trees killed in this outbreak (total infestation).  (Use the fitted value of $\beta$ from the previous section.)  
        \begin{mdframed}[backgroundcolor=gray!20]
            \noindent\verb"Solution (answers may vary)":
            \[
                F = \frac{2\ln(4.6)}{0.033} \approx 92.5 \text{ trees/ha }
            \]

       \end{mdframed}

        \mbox{ }

    \item Graph the footprint formula (12) in \textit{Desmos} using  fitted value of $\beta$, with $R$ as the independent variable. \\
        \begin{mdframed}[backgroundcolor=gray!20]
             \includegraphics[width=140mm]{desmos5.png}
             \begin{center}
                Figure 8. Desmos graph of the outbreak footprint formula (12) along with the predicted total infestation for a growth rate of $R=4.6$.
            \end{center}
        \end{mdframed}


        \mbox{ }

    \item Given that increasing temperatures lead to higher MPB population growth rates, assess the impact of climate change on the severity of MPB outbreaks using the graph of $F$. \\
        \begin{mdframed}[backgroundcolor=gray!20]
            \noindent\verb"Solution (answers may vary)":\\
                 As global temperature rises, MPB population growth rate rises; as the growth rate rises, so does the outbreak footprint (according to (12)).  Hence, we see more clearly the nature by which warming climate leads to increased severity of outbreaks.
        \end{mdframed}


\end{enumerate}



\teacher{
\section*{NOTES FOR TEACHERS}
\begin{itemize}
  \item The overall aims of this project are to
    \begin{itemize}
      \item show how the discrete mathematics world and continuous mathematics world can intersect in an applied context;
      \item acquaint students with real mathematical biology/ecology research and journal articles;
      \item illustrate how mathematics can be a bridge in understanding the connections between climate change and local ecosystems.
    \end{itemize}

  \item The project is best run in the second half of a first semester differential equations or mathematical modeling course and is designed to be completed in approximately 3 (50 min) class periods as follows:
   \begin{center}
        \begin{tabular}{ |c|l| }
         \hline
         Day & Activities \\
         \hline
         1 & Watch National Geographic video: \href{https://www.youtube.com/watch?v=vR30qlK0-Cw}{https://www.youtube.com/watch?v=vR30qlK0-Cw} \cite{NatlGeo} \\
           & Discuss the SIJ model (Sections 1 and 2)\\
           & Show model simulations of MPB outbreaks \\
           & Students work through section: \\
           & \mbox{      } \emph{3. Convert the Difference Equation to a Differential Equation Model} \\
         \hline
         2 & Students work through section: \\
           & \mbox{      } \emph{4. Solve the Differential Equation} \\
         \hline
         3 & Students work through sections: \\
           & \mbox{      } \emph{5. Model Parameterization} and \\
           & \mbox{      } \emph{6. Effect of Global Warming on Outbreak Severity} \\
         \hline
        \end{tabular}
   \end{center}


    \item A basic understanding of the following concepts/skills is strongly recommended:
        \begin{itemize}
          \item Difference equations (recursive functions)
          \item Difference between model variables and parameters
          \item Separation of variables method
          \item Hyperbolic functions
        \end{itemize}

  \item A basic understanding of the following concepts/skills may be beneficial but is not required:
        \begin{itemize}
          \item Second order and nonlinear first order differential equations
          \item Second finite difference approximation
          \item Model parameterization using sum of squared errors
          \item \textit{Desmos}: Free online graphing/computation program: \href{https://www.desmos.com/calculator}{https://www.desmos.com/calculator} \cite{Desmos} \\ Getting started with \emph{Desmos}: \href{https://learn.desmos.com/graphing}{https://learn.desmos.com/graphing}
        \end{itemize}

  \item Students may work in groups (of 3 or 4, for example) or individually (either in class or out of class).

  %\item It is recommended that the instructor briefly discuss the SIJ model from \cite{Duncan2015} before students begin the project.

  \item Solutions have been included to facilitate the use of checkpoints at which students should have their work verified by the instructor before continuing.

\end{itemize}
}


%\section*{Submission Version}



\begin{thebibliography}{98}

\bibitem{Amman1988} Amman, Gene D., Schmitz, Richard F. 1988.  Mountain pine beetle: lodgepole pine interactions and strategies for reducing tree losses. \textit{Ambio} 17:62-68.

\bibitem{Bentz2016} Bentz, Barbara J., Duncan, Jacob P., Powell, James A. 2016. Elevational shifts in thermal suitability for mountain pine beetle
    population growth in a changing climate. \textit{Forestry}. 1–13.

\bibitem{Berryman1985} Berryman, A. A., Dennis, B., Raffa, K. F., Stenseth, N. C. 1985.  Evolution of optimal group attack, with particular reference to bark beetles (Coleoptera: Scolytidae). \textit{Ecology}, 66(3), 898-903.
\bibitem{Desmos} Desmos. 2021. Desmos. \href{https://www.desmos.com/}{https://www.desmos.com/}.

\bibitem{Duncan2015} Duncan, Jacob P., Powell, James A., Gordillo, Luis F., Eason, Joseph. 2015. A model for mountain pine beetle outbreaks in an age-structured forest: Predicting severity and outbreak-recovery cycle period. \textit{Bulletin of Mathematical Biology}. Jul;77(7):1256-84.

\bibitem{Duncan2017} Duncan, Jacob P., Powell, James A. 2017. Analytic Approximation of Invasion Wave Amplitude Predicts Severity of Insect Outbreaks. \textit{SIAM Journal on Applied Mathematics}. 77(1):294-314.

\bibitem{NatlGeo} National Geographic. 2018 March 26.  This Tiny Beetle Is Devastating Forests in the Worst Outbreak Ever | Short Film Showcase. [Video file]. \textit{YouTube}. \href{https://www.youtube.com/watch?v=vR30qlK0-Cw}{https://www.youtube.com/watch?v=vR30qlK0-Cw}.

\bibitem{Powell2009} Powell, James A. and Bentz, Barbara J. 2009. Connecting phenological predictions with population growth rates for mountain pine beetle, an outbreak insect. \textit{Landscape Ecology}. 24(5), pp.657-672.

\bibitem{years} The YEARS Project. 2018 November 27. Attack of the Bark Beetle. [Video file]. \textit{YouTube}. \href{https://www.youtube.com/watch?v=FzrfzQG6YD0}{https://www.youtube.com/watch?v=FzrfzQG6YD0}.

\end{thebibliography}








\end{document}
