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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\\ How many sheep may safely graze?}




\markboth{Modeling a Grazing System}{Modeling a Grazing System}

\author{Mary Vanderschoot\\
Wheaton College\\
Wheaton IL 60187 USA\\
mary.vanderschoot@wheaton.edu
}

\makeStitlePDFLaTex


\begin{abstract}

One of the most well-known mathematical models in ecology is the Lotka-Volterra predator-prey system of differential equations.  Initially, this model was used to analyze interactions between two animal populations. But ecologists discovered that it could also be applied to plant (`prey') and herbivore (`predator') interactions.   A grazing system (such as sheep in a pasture) is a special type of plant-herbivore system in which the herbivore population is controlled by humans.  Because the number of herbivores does not change, the model consists of a single differential equation for the vegetation.   In this activity, students will apply graphical analysis (such as phase lines) to determine the long-term predictions of a differential equation model for pasture grass using two different formulas for the herbivore consumption rate.
\end{abstract}


\section*{SCENARIO DESCRIPTION}

One of the most well-known mathematical models in ecology is the Lotka-Volterra predator-prey system of differential equations.  Though the model was originally intended for two animal populations, ecologists discovered that the model could also be used to describe plant (`prey') and herbivore (`predator') interactions.  Imanuel Noy-Meir was one of the first ecologists to apply the predator-prey model to a type of plant-herbivore system called a grazing system \cite{IMN}.   In a grazing system (sheep in a pasture, for instance) the number of herbivores is controlled by humans.  Of these systems Noy-Meir wrote, ``\ldots grazing systems are one of the type of ecosystems which are of greatest importance to man; if theoretical ecology could contribute to their understanding and to the solution of their practical management problems, this would be a very useful contribution indeed'' \cite{IMN}. In this scenario you'll use graphical analysis to analyze a mathematical model for a grazing system.






\section*{SUBMISSION VERSION}

\section{Grazing Model Design}

In order to apply the predator-prey model to herbivore (`predator') and plant (`prey') interactions, ecologists needed to find a useful way to represent the `plant population' using a single variable.  This is often done by letting $P$ represent the plant biomass (in $kg/m^2$).  The quantity $P$ can be thought of as the amount of vegetation that is available, both for consumption by the herbivore and for producing plant growth.

Suppose that a farmer places $H_0$ sheep to graze in a pasture with biomass $P$.   What is the long-term impact that this level of grazing will have on the vitality of the pasture grass?  The standard grazing model is based on the following assumptions about the rate of growth of the pasture grass and the consumption rate of the sheep:  
     
\begin{itemize} 

\item In the absence of sheep $P$ exhibits logistic growth with intrinsic growth rate $r$ and carrying capacity $K$.  That is, $P$ has growth rate $G  = rP(1-P/K)$.  

\item  The per sheep consumption rate of grass, denoted $c(P)$, depends only on the quantity of available vegetation.  The function $c(P)$ is called the \emph{functional response}. Thus the total consumption rate of vegetation by the sheep population is $C=H_0 c(P)$. 

\end{itemize}

\noindent The net change in plant biomass $P$ is the growth $G$ minus consumption $C$, that is, $$\frac{dP}{dt}=G-C=G(P)-C(P),$$ or 
\begin{eqnarray}
\frac{dP}{dt} & = & r P \left(1 - \frac{P}{K} \right) - H_0 c(P).
\label{Noy}
\end{eqnarray}  

Ecologists have found a number of useful formulas for $c(P)$ to model the consumption rates of different herbivores.  Functional responses that satisfy the following assumptions are called \emph{saturation} functional responses:  The herbivore's consumption rate is limited by foraging availability.  The consumption rate increases as more edible plant biomass becomes available.  But there is an upper limit (based on digestion) on the rate at which an herbivore can feed.  

\begin{enumerate}

\item  The following two saturation functional responses (where $a, b >0$) are often used in plant-herbivore models:  
\begin{eqnarray}
c_{1}(P)&=&\left(\frac{aP}{b+P}\right)
\label{func1}
\end{eqnarray}
\begin{eqnarray}
c_2(P)&=&\left(\frac{aP^2}{b^2+P^2}\right) 
\label{func2}
\end{eqnarray} 
\noindent Sketch the graphs of $c_1$ and $c_2$ for $a=1$ and $b=10$ on one set of axes and include any horizontal asymptotes.  Use your sketch to answer the following:  
\begin{enumerate}
\item Briefly explain why $c_1$ and $c_2$ are saturation functional responses.  
\item In each case, what is the ``saturation constant'', that is, what is the value of the maximum rate at which vegetation can be consumed?    In your own words, explain what $b$ signifies.  
\item For which functional response is the sheep more efficient at finding food at low vegetation densities?  For which is the sheep more efficient at reaching satiation?  
\end{enumerate}
\end{enumerate}

\section{Analyzing a Model}

Suppose that for a pasture with biomass $P$ and $H_0$ grazing sheep, the following parameters were known:   $r=0.1$, $K =100$, $a=1$ and $b=10$.   And suppose that, in the long-run, the per sheep consumption rate must be at least 75\% of the saturation constant.  (Below this level the sheep risk malnourishment.)  In the following exercises, you will  use graphical analysis to analyze the grazing model in   (\ref{Noy}) with the functional responses in  (\ref{func1}) and (\ref{func2}).  Your task is to determine the maximum number of sheep that the pasture can support.  

 
\begin{enumerate}

\item[2.] Suppose the farmer allows one sheep to graze in this pasture, that is, assume $H_0=1$.     Sketch the growth rate $G(P)$ and the consumption rate $C_1(P)=H_0 c_1(P)$ as functions of $P$ on the same set of axes.    Use technology to determine the value(s) of $P$ where the two curves intersect.  Note that $\frac{dP}{dt}$ is zero at these values of $P$, that is, these values of $P$ are the equilibrium solutions to  (\ref{Noy}).   Sketch the corresponding phase line and briefly explain what the model predicts about the vegetation in the long-run and at what percent of the saturation constant the sheep will be grazing.

\item[3.] Repeat Exercise 2 for the values of $H_0=2,3$ and $4$.   

\item[4.] Sketch a bifurcation diagram in the $H_0-P$ plane that includes the phase lines for $H_0=0, 1,2,3$ and $4$.  

\item[5.] What does the model predict is the maximum number of sheep that can be placed in the pasture without the biomass of the pasture grass going to zero?  What initial amount of plant biomass is necessary to maintain this level of grazing?  

\item[6.] What number of sheep results in overexploitation of the vegetation, regardless of the initial plant biomass?  
 
\item[7.] Repeat Exercises 2--6 with the function $c_2(P)$ as the functional response.  What are the differences in the long-term predictions between the two models?   

\end{enumerate}




\begin{thebibliography}{98}

\bibitem{Feng} Feng Z., DeAngelis. D. 2018. {\it Mathematical Models of Plant-Herbivore Interactions}. Boca Raton FL USA: CRC Press.

\bibitem{IMN} Noy-Meir, I. 1975. Stability of Grazing Systems: An Application of Predator-Prey Graphs.  {\it Journal of Ecology\/}. 63: 459-579. 



\end{thebibliography}



\teacher{
\section*{NOTES FOR TEACHERS}
This scenario can be assigned as a group project early in the course.  Students should have some experience using phase lines and graphical analysis to analyze autonomous differential equations.   Following are some comments on the exercises:
\begin{enumerate}
\item The two functional responses, $c_1(P)$ and $c_2(P)$, are called Type-I and Type-2 Holling responses.   The questions in the first exercise are intended to help students develop their understanding of a functional response and should be discussed in small groups.  (The constant $b$ is called the half-saturation constant.)

\item[2-6.] Figure~\ref{sheep} shows the graphs of $C_1(P)$ and $G(P)$ for $H_0 = 1, 2, 3$ and $4$.  

\begin{figure}[!h]
\centering
\includegraphics[width=16cm]{sheep.png}
\vspace*{-1. true cm}
\caption{Growth and Consumption Rates for $c_1(P)$.}
\label{sheep}
\end{figure}

\begin{enumerate}
\item  If $H_0=1$, there are two equilibria, at $P=0$ and $P=90$.  For any $P(0)>0$ the model predicts that the pasture grass biomass will approach 90.  The sheep will be consuming grass at 90\% of the saturation constant.
\item If $H_0 =2$, there are three equilibria, at $P=0$, $P \approx 13$ and $P \approx 77$.  The equilibrium at $P=0$ is now a sink,  $P \approx 13$ is a source, and $P \approx 77$ is a sink.  If $P(0)<13$, the biomass will approach zero.  If $P(0)>13$, then $P(t)$ will approach 77, and each sheep will be consuming grass at about 88.5\% of the saturation constant.
\item The case of $H_0=3$ is similar to $H_0=2$, but the nonzero equilibrium points are $P=40$ and $P=50$.  If $P(0)<40$, then the biomass will approach zero.  If $P(0)>40$, then biomass will approach 50 in the long-run and each sheep will be consuming grass at 83\% of the saturation constant.
\item When $H_0=4$ we see a sudden overexploitation of the vegetation.  There are no nonzero equilibria and the model predicts that the biomass will approach zero regardless of $P(0)$.
\end{enumerate} 
\end{enumerate}
\noindent The corresponding bifurcation diagram is shown in Figure~\ref{bifurcation}.  

\begin{figure}[h]
\centering
\includegraphics[width=8cm]{bifurcation.png}
\caption{Bifurcation Diagram for $c_1(P)$.}
\label{bifurcation}
\end{figure}

\begin{enumerate}
\item[7.] Figure~\ref{herbivores} shows the graphs of $C_2(P)$ and $G(P)$ for $H_0 = 1, 2, 3$ and $4$.  
\begin{figure}[h]
\centering
   \includegraphics[width=16cm]{herbivores.png}
   \vspace*{-1. true cm}
   \caption{Growth and Consumption Rates for $c_2(P)$.}
      \label{herbivores}
   \end{figure}
\begin{enumerate}
\item  If $H_0=1$, there are two equilibria, at $P=0$ and $P\approx 88.89$.  For any $P(0)>0$ the model predicts that the pasture grass biomass will approach 88.89.  (This is very similar to what the model predicted using $c_1(P)$.)  The sheep will be consuming grass at about 98.75\% of the saturation constant.
\item If $H_0 =2$, there are four equilibria, at $P=0$, $P\approx6.83$, $P=20$, and $P\approx73.17$.  The equilibria $P\approx6.83$ and $P\approx73.17$ are sinks, and the equilibria $P=0$ and $P=20$ are sources. If $P(0)>20$, then $P(t)$ will approach 73.17, and the sheep will be consuming grass at 98\% of the saturation constant.   If $0<P(0)<20$, the biomass will approach 6.83 but the the sheep will only be consuming grass at about 32\% of the saturation constant.
\item If $H_0=3$, there are two equilibria, at $P=0$ and $P\approx 3.64$.  For any $P(0)>0$, the model predicts that the biomass will approach 3.64  but the the sheep will only be consuming grass at about 12\% of the saturation constant.
\item The case of $H_0=4$ is similar to $H_0=3$, with only one nonzero equilibrium point at $P \approx 2.6$.  For any $P(0)>0$, the biomass will approach 2.6 but the sheep consumption rate will be only 6\% of the saturation constant.
\end{enumerate} 
\end{enumerate}
}



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