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\chapter{Sensitivity Analysis}
The first step in our method to obtain the sensitivity of each parameter value is to differentiate the right hand side of each model with respect to each model parameter. The partial derivatives for the right hand side of our linear response model (\ref{ivplinear}) are contained in Table 7.1 . Table 7.2 lists the associated partials for our stage structured model (\ref{ivp22}). We note that each resulting partial in Table 7.1 and Table 7.2 are continuous with respect to the independent variable $t$, and all population densities $P, C,$ and $R$ for all positive parameter values. Thus by Theorem 3 and Theorem 4 we can differentiate the solution to each model with respect to each parameter.
%When looking at our models designed for omnivory, one might ask the question, why do we care about sensitivity analysis? Since our models are used to approximate natural systems over time, it can be very useful to know what causes change and how these changes affect the rest of the populations involved. The way our differential equations are set up makes us aware that changes in one trophic level directly affects every other trophic level. Hence, it could be very useful to understand how small changes in parameter values affect the solution to our model's as a whole. Recall that our parameter values represent numerical values collected by field biologist, research scientist, and other mathematical techniques to approximate and monitor wildlife. Thus, an important aspect of our research ... Specifically, when comparing parameters in each model, it could be very useful to know which value has the highest "sensitivity" such that it could drastically change the entire ecosystem. Hence, we are interested in determining which parameter values affect our systems the most. This information can be a valuable resource for biologist in field or even conservationist. Our goal is to mathematically describe how small changes in parameter values affect the solution to each respective model.
%Now that we have some intuition of how our differential equations in each system are setup and know what it means for a parameter to be "sensitive" in our models, we start to analyze our models. Our sensitivities involve the rate of change of each population density with respect to each parameter value. We have 3 population densities and 10 parameters in our linear model, while our stage structured model has 4 population densities and 15 parameters. \section{Deriving Sensitivity Equations}
Recall from equations (\ref{ivplinear}) and (\ref{ivp22}) that both our systems take the form,
\begin{equation} \label{diff} \begin{aligned}
\der{y_{1}}=f_{1}( y_{1},\hdots ,y_{j}) \\
\der{y_{2}}=f_{2} (y_{1},\hdots ,y_{j}) \\
\vdots \\
\der{y_{j}}=f_{j} ( y_{1},\hdots ,y_{j}). \\
\end{aligned}
\end{equation}
where $y_{j}$, $j=1,2,3$, represent the population densities $(P,C,R)$ for our linear response model. For our stage structure model we have $j=1,2,3,4$ for our population densities $(P_{2}, P_{1}, C, R)$. We represent each parameter value by $\alpha_{i}$, where $i=1,2,\hdots ,10$ for our linear response model, and $i=1,2,\hdots , 15$ for our stage structure model. We define the sensitivity of the state variable $y_{j}$ with respect to parameter $\alpha_{i}$ as the partial derivative of $y_{j}$ with respect $\alpha_{i}$. Using Definition 6, we can define the sensitivity of each population with respect to the parameter $\alpha_{i}$ as,
\begin{equation}\label{def1} S_{y_{j},\alpha_i} = \pder{y_{j}}{\alpha_{i}} \end{equation}
for $j=1,2,3$ with $i=1,2, \hdots, 10$ (linear response model) and $j=1,2,3,4$ with $i=1,2,\hdots, 15$ (stage sturcture model).
Since our sensitivities involve the rate of change in population densities with respect to model parameters, we need to differentiate each trophic level with respect to each model parameter. To justify these calculations we look back at Theorems 3 and 4. In chapters 5 and 6 we proved by means of Theorem 1 and Lemma 2, that each respective model satisfied the conditions of existence and uniqueness of solutions and that the solution for each model can be continued for $t\geq t_{0}$ (Theorems 5 and 6). We also showed, by means of Theorem 3 that the solution to each system depends continuously upon the parameter vector
\begin{align}\label{lipp}
\bfa &= \begin{bmatrix}
\alpha_{1}\\
\alpha_{2}\\
\vdots\\
\alpha_{i} \\
\end{bmatrix}^T,
\end{align}
on the premise that the partial derivatives in Tables 5.1 and 6.1 are continuous with respect to $t$ and $\bfy$ for all $t\geq0$. The conclusion of Theorem 3 guarantees us that the solution $\bfy(t, \bfa)$ satisfying,
\begin{equation} \bfy (t, \bfa_{0})=\bfy_{0} \end{equation}
is continuous with respect to $\bfa$ for all $\bfa \in D_{\alpha}$ and $t\geq t_{0}$.
Hence we have we have satisfied one of the two conditions for Theorem 4. The other condition is satisfied by the calculations contained in Tables 7.1 and 7.2. That is, the right hand sides of (\ref{ivplinear}) and (\ref{Stage1}) have continuous partial derivatives in D with respect to the vectors $\bfy$ and $\bfa$. Hence by Theorems 3 and 4, we can differentiate the solution to each model with respect to model parameters. \\
\section{Sensitivity Equations}
Our next step is to obtain the sensitivity equations. We use Theorems 4 and 5 for this task. Our goal is to calculate the rate of change of the population densities with respect to each parameter value. Recall that the hypothesis of Theorem 5 is satisfied by meeting the conditions of Theorem 4. Theorem 5 allows us to differentiate each differential equation (in each model) with respect to each model parameter. It then allows us to interchange the order of differentiation to derive a linear system of equations for the sensitivities that solves,
\begin{equation}\label{derek15} \der{}S_{{y_{j}},\alpha_{i}} = \sum_{k=1}^{n} (\pder{f_{j}}{y_{k}}S_{y_{j},\alpha_{i}}) + \pder{f_{j}}{\alpha_{i}}. \end{equation}
Where $S_{y_{j}, \alpha_{i}}$ represent the sensitivity of the population density $y_j$ with respect to the parameter $\alpha_i$. Thus, we are looking to solve each system for,
\begin{equation}\label{morgan} S_{y_{j},\alpha_{i}}(t)=\pder{y_{j}(t)}{\alpha_{i}}, \end{equation}
where $y_{j}$ is the j-th component of the state, $j=1,2,3$ (linear response model) or $j=1,2,3,4$ (for stage structured model), and $\alpha_{i}$ are the model parameters.
Note that $i=1,2,\hdots, 10$ for our linear model and $i=1,2,\hdots 15$ for our Stage Structured Model. We must also differentiate the initial conditions with respect to each parameter, hence we have the initial conditions,
\begin{equation}\label{morgan1} S_{y_{j},\alpha_{i}(0)} = 0, \mbox{ for each } y_{j},\alpha_{i}. \end{equation}
We call the equations in (\ref{morgan}) the sensitivity equations and note that they require that,
\begin{equation} \pder{\bff_{j}}{y_{k}}, \end{equation}
be continuous with respect to each state variable $y_j$, where $j=1,2,3$ (for our linear model) and $j=1,2,3,4$ (for our stage structure model) and the independent variable $t$.
%For our linear response model we define,
%\begin{equation}\begin{split}
%S_{P}= S_{1,i} = \pder{P}{\alpha_{i}}\\
%S_{C}= S_{2,i} = \pder{C}{\alpha_{i}}\\
%S_{R}= S_{3,i} = \pder{R}{\alpha_{i}}
%\end{split}.\end{equation}\vspace{.02in}
%For our stage structure model we define,
%\begin{equation}\begin{split}
%S_{P_{2}}= S_{1,i} = \pder{P_{2}}{\alpha_{i}}\\
%S_{P_{1}}= S_{2,i} = \pder{P_{1}}{\alpha_{i}}\\
%S_{C}= S_{3,i} = \pder{C}{\alpha_{i}}\\
%S_{R}= S_{4,i} = \pder{R}{\alpha_{i}}
%\end{split}.\end{equation}\vspace{.02in}
\section{Calculating Linear Response Model Sensitivity Equations}
\noindent We have already established from the initial value problem in Chapter 5, particularly in (\ref{ivplinear}), that we can represent our linear response model as,
\begin{equation}\label{useforrhs} \begin{aligned}
\der{P}= f_{1}(P,C,R)\\
\der{C}= f_2(P,C,R), \\
\der{R}= f_3(P,C,R),
\end{aligned}\end{equation}
where $f_{1},f_{2},$ and $f_{3}$ are continuous functions of more than one variable. Thus using the notation above and the result of Theorem 4 we can differentiate each differential equation with respect to our ten parameters. We start by taking the partial derivatives of each differential equation with respect to $\alpha _1$ then proceed until we reach $\alpha_ {10}$. We show the process for each row vector denoted $f_1, f_2$, and $f_3$.
\noindent For $f_1$ we have,
\begin{equation}\label{uva}\begin{aligned}
\pder{}{\alpha_{1}} \der{P}&=\pder{f_{1}(P,C,R)}{\alpha_{1}}\\
\pder{}{\alpha_{2}} \der{P}&=\pder{f_{1}(P,C,R)}{\alpha_{2}}\\
\vdots \\
\pder{}{\alpha_{10}} \der{P}&=\pder{f_{1}(P,C,R)}{\alpha_{10}}, \end{aligned}\end{equation} \\
for $f_2$ we have,
\begin{equation}\label{uva1}\begin{aligned}
\pder{}{\alpha_{1}} \der{C}&=\pder{f_{2}(P,C,R)}{\alpha_{1}}\\
\pder{}{\alpha_{2}} \der{C}&=\pder{f_{2}(P,C,R)}{\alpha_{2}}\\
\vdots \\
\pder{}{\alpha_{10}} \der{C}&=\pder{f_{2}(P,C,R)}{\alpha_{10}}, \end{aligned}\end{equation}\\
and, finally, for $f_3$ we have,
\begin{equation}\label{uva2}\begin{aligned}
\pder{}{\alpha_{1}} \der{R}&=\pder{f_{3}(P,C,R)}{\alpha_{1}}\\
\pder{}{\alpha_{2}} \der{R}&=\pder{f_{3}(P,C,R)}{\alpha_{2}}\\
\vdots \\
\pder{}{\alpha_{10}} \der{R}&=\pder{f_{3}(P,C,R)}{\alpha_{10}}. \end{aligned}\end{equation}
\subsection{Chain Rule Differentiation}
Given the notation in equations, (\ref{uva}), (\ref{uva1}), and (\ref{uva2}) we can now formally differentiate the right hand side of each. We use the multi-variable chain rule for differentiation of the right hand sides of (\ref{uva}), (\ref{uva1}), and (\ref{uva2}) with respect to $\alpha_{1}, \hdots, \alpha_{10}$ to obtain our linear response sensitivity equations. The general form of the chain rule given by Stewart \cite{Stewart} states; \\
``If $u$ is a differentiable function of $n$ variables $y_1, y_2, y_3, \hdots, y_n$ and each $y_n$ is a differentiable function of $m$ variables $t_1, t_2, t_3, \hdots, t_m$. Then $u$ is a function of $t_1, t_2, t_3,\hdots,t_m$ and \\
\begin{eqnarray*}
\pder{u}{t_i} = \pder{u}{y_1}\pder{y_1}{t_i}+\pder{u}{y_2}\pder{y_2}{t_i}+\hdots +\pder{u}{y_n}\pder{y_n}{t_i}
\end{eqnarray*}for each $ i= 1, 2,\hdots , m$ ."\cite{Stewart}\\In our differential equations $u=f_{i}, i\in {1,2,3}$. Our variables, $n$ are as follows: $y_1= P, y_2=C, y_3=R$. Our parameters are represented by, $t_{i}=\alpha_{i}, i\in {1,2,3,\hdots,10}$.\\
Applying the chain rule to the right hand side of each equation in (\ref{uva}), (\ref{uva1}), and (\ref{uva2}) gives us, \\
\vspace{.3cm}
\centerline{\emph{Predator}}
\begin{equation}\label{resource1}\begin{aligned}
\der{}\pder{P}{\alpha _1}&= \pder{f_1}{P}\pder{P}{\alpha _1}+\pder{f_1}{C}\pder{C}{\alpha _1}+\pder{f_1}{R}\pder{R}{\alpha _1}+\pder{f_1}{\alpha _1}\\
\der{}\pder{P}{\alpha _2}&= \pder{f_1}{P}\pder{P}{\alpha _2}+\pder{f_1}{C}\pder{C}{\alpha _2}+\pder{f_1}{R}\pder{R}{\alpha _2}+\pder{f_1}{\alpha _2}\\
&=\vdots\\
\der{}\pder{P}{\alpha _{10}}&= \pder{f_1}{P}\pder{P}{\alpha _{10}}+\pder{f_1}{C}\pder{C}{\alpha _{10}}+\pder{f_1}{R}\pder{R}{\alpha _{10}}+\pder{f_1}{\alpha _{10}},
\end{aligned}\end{equation}
\vspace{1.cm}\centerline{\emph{Consumer}}
\begin{equation}\label{consumer1}\begin{aligned}
\der{}\pder{C}{\alpha _1}&= \pder{f_2}{P}\pder{P}{\alpha _1}+\pder{f_2}{C}\pder{C}{\alpha _1}+\pder{f_2}{R}\pder{R}{\alpha _1}+\pder{f_2}{\alpha _1}\\
\der{}\pder{C}{\alpha _2}&= \pder{f_2}{P}\pder{P}{\alpha _2}+\pder{f_2}{C}\pder{C}{\alpha _2}+\pder{f_2}{R}\pder{R}{\alpha _2}+\pder{f_2}{\alpha _2}\\
&=\vdots\\
\der{}\pder{C}{\alpha _{10}}&= \pder{f_2}{P}\pder{P}{\alpha _{10}}+\pder{f_2}{C}\pder{C}{\alpha _{10}}+\pder{f_2}{R}\pder{R}{\alpha _{10}}+\pder{f_2}{\alpha _{10}},
\end{aligned}\end{equation}
\vspace{1.cm}\centerline{\emph{Resource}}
\begin{equation}\label{resource12}\begin{aligned}
\der{}\pder{R}{\alpha _1}&= \pder{f_3}{P}\pder{P}{\alpha _1}+\pder{f_3}{C}\pder{C}{\alpha _1}+\pder{f_3}{R}\pder{R}{\alpha _1}+\pder{f_3}{\alpha _1}\\
\der{}\pder{R}{\alpha _2}&= \pder{f_3}{P}\pder{P}{\alpha _2}+\pder{f_3}{C}\pder{C}{\alpha _2}+\pder{f_3}{R}\pder{R}{\alpha _2}+\pder{f_3}{\alpha _2}\\
&=\vdots\\
\der{}\pder{R}{\alpha _{10}}&= \pder{f_3}{P}\pder{P}{\alpha _{10}}+\pder{f_3}{C}\pder{C}{\alpha _{10}}+\pder{f_3}{R}\pder{R}{\alpha _{10}}+\pder{f_3}{\alpha _{10}}.
\end{aligned}\end{equation}
\vspace{1.3cm}
Visually analyzing the differential equations above, we see the term on the end of each differential equation changes for each respective parameter value. The term we are looking at is denoted as $\pder{\bff_{j}}{\alpha_{i}}$ and we call this term the \textbf{particular part of the equations}. This term is located on the right hand side of (\ref{derek15}) after the addition sign. We call this the particular part of the equations because it changes for every parameter $\alpha_{i}$.
We note that the term $\pder{\bff_{j}}{y_{k}}$ with $j$ and $k \in {1,2,3}$ does not depend on the parameter $\alpha_{i}$, hence this term does not change for each parameter. This can be seen in equations (\ref{resource1}) and (\ref{consumer1}). We use this observation to define the \textbf{general sensitivity equations}.
Knowing that $\pder{\bff_j}{y_k}$ remains the same for each parameter we can use the notation used in equation (\ref{morgan}) to form our general sensitivity equations. We call the following the \textbf{general sensitivity equations} for the system of linear differential equations,
\begin{equation}\label{morgan3} \der{}S_{y_{j},\alpha_{i}} = \sum_{k=1}^{n} (\pder{f_{j}}{y_{k}}S_{y_{j},\alpha_{i}}). \end{equation}
The partial derivatives computed from the \textbf{particular part of the sensitivity equations} are given in Tables 7.1 and 7.2 for each respective model parameter. We note that each partial derivative in the tables noted are continuous with respect to $t$ and all respective population densities for all positive values. If we include the particular part of the equations, we have the exact form on the right hand side as provided in equation (\ref{derek15}). \subsection{Changing Order of Differentiation}
Having already noted that Theorem 5 is satisfied by meeting the conditions of Theorem 4, we can use the result of Theorem 5 which allows us to interchange the order of differentiation on the left hand side of each differential equation. We note that we have already calculated the right hand side of each equation in the previous section.
\noindent Starting with the differential equation related to the predator, we have\\
\begin{eqnarray*}
\der{P}=f_1(P, C, R)\end{eqnarray*}
Applying the partial derivative to both sides with respect to the parameter value $\alpha_i$ gives,\\
$$\pder{}{\alpha _1}\der{P}=\pder{}{\alpha _1}f_1(P, C, R).$$
Now concentrating on the left hand side of the equation, we interchange the order of differentiation,
\begin{eqnarray*}
\der{} \pder{P}{\alpha _i}= \pder{f_1(P,C,R)}{\alpha_i} .\\
\end{eqnarray*}
The same follows for the $C$ and $R$:\\
\centerline{Consumer}
\begin{equation*}\begin{aligned}
\der{C} \pder{}{\alpha _i}&= \pder{}{\alpha_i}f_2(P,C,R), \hspace{1.2cm} \mbox{is equivalent to}\\\\
\der{} \pder{C}{\alpha _i} &= \pder{f_2(P, C, R)}{\alpha_i},
\end{aligned} \end{equation*}
\centerline{Resource}
\begin{equation*}\begin{aligned}
\der{R}\pder{}{\alpha_i} &= \pder{}{\alpha_i}f_3(P,C,R), \hspace{1.2cm} \mbox{is equivalent to}\\\\
\der{} \pder{R}{\alpha _i} &= \pder{f_3(P, C, R)}{\alpha_i}.
\end{aligned} \end{equation*}
We continue this process for each population and all ten parameter values. Recalling from equation (\ref{250}) that our sensitivity for parameter $\alpha_{i}$ is defined as,
\begin{equation*}\label{251} S_{j,i} = \pder{y_{j}(t, \bfa)}{\alpha_{i}} . \end{equation*}
where $S_{j,i}$ represents the sensitivity of population $j$ with respect to parameter $i$. Hence we can replace the term $\pder{y_{j}}{\alpha_{i}}$, with the notation of $S_{j,\alpha_i}$, where $j=P,C,R$ and $i=1,2,\hdots, 10$.
\subsection{Sensitivity Differential Equations}
Since we know our sensitivities are $\pder{y_j}{\alpha _i}$ where $y_{j}=P, C, R$, $j=1,2,3$ and $ i=1, \hdots,10$ we can replace $\pder{y_{j}}{\alpha _i}$ with the notation $S_{y_{j}, \alpha _i}$. This gives us the following sets of sensitivity differential equations.
\begin{eqnarray*}
\der{}S_{y_{1},\alpha _{1}} = \pder{f_1}{P}S_{y_{1},\alpha _{1}}+\pder{f_1}{C}S_{y_{2},\alpha _{1}}+\pder{f_1}{R}S_{y_{3},\alpha _{1}}+\pder{f_1}{\alpha _1}\\
\der{}S_{y_{1},\alpha _{2}} = \pder{f_1}{P}S_{y_{1},\alpha _{2}}+\pder{f_1}{C}S_{y_{2},\alpha _{2}}+\pder{f_1}{R}S_{y_{3},\alpha _{2}}+\pder{f_1}{\alpha _2}\\
\der{}S_{y_{1},\alpha _{3}} = \pder{f_1}{P}S_{y_{1},\alpha _{3}}+\pder{f_1}{C}S_{y_{2},\alpha _{3}}+\pder{f_1}{R}S_{y_{3},\alpha _{3}}+\pder{f_1}{\alpha _3}\\
\vdots\\
\der{}S_{y_{1},\alpha _{10}} = \pder{f_1}{P}S_{y_{1},\alpha _{10}}+\pder{f_1}{C}S_{y_{2},\alpha _{10}}+\pder{f_1}{R}S_{y_{3},\alpha _{10}}+\pder{f_1}{\alpha _{10}}\\
\end{eqnarray*}
\begin{eqnarray*}
\der{}S_{y_{2},\alpha _{1}} = \pder{f_2}{P}S_{y_{1},\alpha _{1}}+\pder{f_2}{C}S_{y_{2},\alpha _{1}}+\pder{f_2}{R}S_{y_{3},\alpha _{1}}+\pder{f_2}{\alpha _1}\\
\der{}S_{y_{2},\alpha _{2}} = \pder{f_2}{P}S_{y_{1},\alpha _{2}}+\pder{f_2}{C}S_{y_{2},\alpha _{2}}+\pder{f_2}{R}S_{y_{3},\alpha _{2}}+\pder{f_2}{\alpha _2}\\
\der{}S_{y_{2},\alpha _{3}} = \pder{f_2}{P}S_{y_{1},\alpha _{3}}+\pder{f_2}{C}S_{y_{2},\alpha _{3}}+\pder{f_2}{R}S_{y_{3},\alpha _{3}}+\pder{f_2}{\alpha _3}\\
\vdots\\
\der{}S_{y_{2},\alpha _{10}} = \pder{f_2}{P}S_{y_{1},\alpha _{10}}+\pder{f_2}{C}S_{y_{2},\alpha _{10}}+\pder{f_2}{R}S_{y_{3},\alpha _{10}}+\pder{f_2}{\alpha _{10}}\\
\end{eqnarray*}
\begin{eqnarray*}
\der{}S_{y_{3},\alpha _{1}} = \pder{f_3}{P}S_{y_{1},\alpha _{1}}+\pder{f_3}{C}S_{y_{2},\alpha _{1}}+\pder{f_3}{R}S_{y_{3},\alpha _{1}}+\pder{f_3}{\alpha _1}\\
\der{}S_{y_{3},\alpha _{2}} = \pder{f_3}{P}S_{y_{1},\alpha _{2}}+\pder{f_3}{C}S_{y_{2},\alpha _{2}}+\pder{f_3}{R}S_{y_{3},\alpha _{2}}+\pder{f_3}{\alpha _2}\\
\der{}S_{y_{3},\alpha _{3}} = \pder{f_3}{P}S_{y_{1},\alpha _{3}}+\pder{f_3}{C}S_{y_{2},\alpha _{3}}+\pder{f_3}{R}S_{y_{3},\alpha _{3}}+\pder{f_3}{\alpha _3}\\
\vdots\\
\der{}S_{y_{3},\alpha _{10}} = \pder{f_3}{P}S_{y_{1},\alpha _{10}}+\pder{f_3}{C}S_{y_{2},\alpha _{10}}+\pder{f_3}{R}S_{y_{3},\alpha _{10}}+\pder{f_3}{\alpha _{10}}\\
\end{eqnarray*}
\section{Chain rule differentiation for stage structure model}
Recall that the differential equations in our stage structured model contains $f_{j}, j\in$ {$1,2,3,4$}, $y_1= P_{2}, y_2=P_{1}, y_3=C, y_4=R$, and $t=$ $\alpha_{i}, i\in ${$1,2,\hdots, 15$}. Hence, using the chain rule we have the following differential equations for our stage structure model;
\centerline{\emph{Adult Predator}}
\begin{eqnarray*}
\der{}\pder{P_{2}}{\alpha _1}= \pder{f_1}{P_{2}}\pder{P_{2}}{\alpha _1}+\pder{f_1}{P_{1}}\pder{P_{1}}{\alpha _1}+\pder{f_1}{C}\pder{C}{\alpha _1}+\pder{f_1}{R}\pder{R}{\alpha _1}+\pder{f_1}{\alpha _1}\\
\der{}\pder{P_{2}}{\alpha _2}= \pder{f_1}{P_{2}}\pder{P_{2}}{\alpha _2}+\pder{f_1}{P_{1}}\pder{P_{1}}{\alpha _2}+\pder{f_1}{C}\pder{C}{\alpha _2}+\pder{f_1}{R}\pder{R}{\alpha _2}+\pder{f_1}{\alpha _2}\\
\vdots\\
\der{}\pder{P_{2}}{\alpha _{15}}= \pder{f_1}{P_{2}}\pder{P_{2}}{\alpha _{15}}+\pder{f_1}{P_{1}}\pder{P_{1}}{\alpha _{15}}+\pder{f_1}{C}\pder{C}{\alpha _{15}}+\pder{f_1}{R}\pder{R}{\alpha _{15}}+\pder{f_1}{\alpha _{15}}\\
\end{eqnarray*}
\centerline{\emph{Juvenile Predator}}
\begin{eqnarray*}
\der{}\pder{P_{1}}{\alpha _1}= \pder{f_2}{P_{2}}\pder{P_{2}}{\alpha _1}+\pder{f_2}{P_{1}}\pder{P_{1}}{\alpha _1}+\pder{f_2}{C}\pder{C}{\alpha _1}+\pder{f_2}{R}\pder{R}{\alpha _1}+\pder{f_2}{\alpha _1}\\
\der{}\pder{P_{1}}{\alpha _2}= \pder{f_2}{P_{2}}\pder{P_{2}}{\alpha _2}+\pder{f_2}{P_{1}}\pder{P_{1}}{\alpha _2}+\pder{f_2}{C}\pder{C}{\alpha _2}+\pder{f_2}{R}\pder{R}{\alpha _2}+\pder{f_2}{\alpha _2}\\
\vdots\\
\der{}\pder{P_{1}}{\alpha _{15}}= \pder{f_2}{P_{2}}\pder{P_{2}}{\alpha _{15}}+\pder{f_2}{P_{1}}\pder{P_{1}}{\alpha _{15}}+\pder{f_2}{C}\pder{C}{\alpha _{15}}+\pder{f_2}{R}\pder{R}{\alpha _{15}}+\pder{f_2}{\alpha _{15}}\\
\end{eqnarray*}\centerline{\emph{Consumer}}
\begin{eqnarray*}
\der{}\pder{C}{\alpha _1}= \pder{f_3}{P_{2}}\pder{P_{2}}{\alpha _1}+\pder{f_3}{P_{1}}\pder{P_{1}}{\alpha _1}+\pder{f_3}{C}\pder{C}{\alpha _1}+\pder{f_3}{R}\pder{R}{\alpha _1}+\pder{f_3}{\alpha _1}\\
\der{}\pder{C}{\alpha _2}= \pder{f_3}{P_{2}}\pder{P_{2}}{\alpha _2}+\pder{f_3}{P_{1}}\pder{P_{1}}{\alpha _2}+\pder{f_3}{C}\pder{C}{\alpha _2}+\pder{f_3}{R}\pder{R}{\alpha _2}+\pder{f_3}{\alpha _2}\\
\vdots\\
\der{}\pder{C}{\alpha _{15}}= \pder{f_3}{P_{2}}\pder{P_{2}}{\alpha _{15}}+\pder{f_3}{P_{1}}\pder{P_{1}}{\alpha _{15}}+\pder{f_3}{C}\pder{C}{\alpha _{15}}+\pder{f_3}{R}\pder{R}{\alpha _{15}}+\pder{f_3}{\alpha _{15}}\\
\end{eqnarray*}
\newpage
\centerline{\emph{Resource}}
\begin{eqnarray*}
\der{}\pder{R}{\alpha _1}= \pder{f_4}{P_{2}}\pder{P_{2}}{\alpha _1}+\pder{f_4}{P_{1}}\pder{P_{1}}{\alpha _1}+\pder{f_4}{C}\pder{C}{\alpha _1}+\pder{f_4}{R}\pder{R}{\alpha _1}+\pder{f_4}{\alpha _1}\\
\der{}\pder{R}{\alpha _2}= \pder{f_4}{P_{2}}\pder{P_{2}}{\alpha _2}+\pder{f_4}{P_{1}}\pder{P_{1}}{\alpha _2}+\pder{f_4}{C}\pder{C}{\alpha _2}+\pder{f_4}{R}\pder{R}{\alpha _2}+\pder{f_4}{\alpha _2}\\
\vdots\\
\der{}\pder{R}{\alpha _{15}}= \pder{f_4}{P_{2}}\pder{P_{2}}{\alpha _{15}}+\pder{f_4}{P_{1}}\pder{P_{1}}{\alpha _{15}}+\pder{f_4}{C}\pder{C}{\alpha _{15}}+\pder{f_4}{R}\pder{R}{\alpha _{15}}+\pder{f_4}{\alpha _{15}}\\
\end{eqnarray*}
\section{Sensitivity Differential Equations for Stage Structured Model}
Since we know our sensitivities are $\pder{y_j}{\alpha _i}$ where $y_{j}= j=P_{2}, P_{1}, C, R$, $j=1,2,3,4$ and $ i=1, \hdots,15$, we can replace $\pder{y_{j}}{\alpha _i}$ with the notation $S_{y}, \alpha _i$. Thus we have:\\
\centerline{\emph{Adult Predator}}
\begin{eqnarray*}
\der{}S_{y_{1},\alpha _{1}} = \pder{f_1}{P_{2}}S_{y_{1},\alpha _{1}}+\pder{f_1}{P_{1}}S_{y_{2},\alpha_{1}}+\pder{f_1}{C}S_{y_{3},\alpha _{1}}+\pder{f_1}{R}S_{y_{4},\alpha _{1}}+\pder{f_1}{\alpha _1}\\
\der{}S_{y_{1},\alpha _{2}} = \pder{f_1}{P_{2}}S_{y_{1},\alpha _{2}}+\pder{f_1}{P_{1}}S_{y_{2},\alpha_{2}}+\pder{f_1}{C}S_{y_{3},\alpha _{2}}+\pder{f_1}{R}S_{y_{4},\alpha _{2}}+\pder{f_1}{\alpha _2}\\
\vdots\\
\der{}S_{y_{1},\alpha _{15}} = \pder{f_1}{P_{2}}S_{y_{1},\alpha _{15}}+\pder{f_1}{P_{1}}S_{y_{2},\alpha_{15}}+\pder{f_1}{C}S_{y_{3},\alpha _{15}}+\pder{f_1}{R}S_{y_{4},\alpha _{1}}+\pder{f_1}{\alpha _{15}}\\
\end{eqnarray*}
\centerline{\emph{Juvenile Predator}}
\begin{eqnarray*}
\der{}S_{y_{2},\alpha _{1}} = \pder{f_2}{P_{2}}S_{y_{1},\alpha _{1}}+\pder{f_2}{P_{1}}S_{y_{2},\alpha_{1}}+\pder{f_2}{C}S_{y_{3},\alpha _{1}}+\pder{f_2}{R}S_{y_{4},\alpha _{1}}+\pder{f_2}{\alpha _1}\\
\der{}S_{y_{2},\alpha _{2}} = \pder{f_2}{P_{2}}S_{y_{1},\alpha _{2}}+\pder{f_2}{P_{1}}S_{y_{2},\alpha_{2}}+\pder{f_2}{C}S_{y_{3},\alpha _{2}}+\pder{f_2}{R}S_{y_{4},\alpha _{2}}+\pder{f_2}{\alpha _2}\\
\vdots\\
\der{}S_{y_{2},\alpha _{15}} = \pder{f_2}{P_{2}}S_{y_{1},\alpha _{15}}+\pder{f_2}{P_{1}}S_{y_{2},\alpha_{15}}+\pder{f_2}{C}S_{y_{3},\alpha _{15}}+\pder{f_2}{R}S_{y_{4},\alpha _{1}}+\pder{f_2}{\alpha _{15}}\\
\end{eqnarray*}
\centerline{\emph{Consumer}}
\begin{eqnarray*}
\der{}S_{y_{3},\alpha _{1}} = \pder{f_3}{P_{2}}S_{y_{1},\alpha _{1}}+\pder{f_3}{P_{1}}S_{y_{2},\alpha_{1}}+\pder{f_3}{C}S_{y_{3},\alpha _{1}}+\pder{f_3}{R}S_{y_{4},\alpha _{1}}+\pder{f_3}{\alpha _1}\\
\der{}S_{y_{3},\alpha _{2}} = \pder{f_3}{P_{2}}S_{y_{1},\alpha _{2}}+\pder{f_3}{P_{1}}S_{y_{2},\alpha_{2}}+\pder{f_3}{C}S_{y_{3},\alpha _{2}}+\pder{f_3}{R}S_{y_{4},\alpha _{2}}+\pder{f_3}{\alpha _2}\\
\vdots\\
\der{}S_{y_{3},\alpha _{15}} = \pder{f_3}{P_{2}}S_{y_{1},\alpha _{15}}+\pder{f_3}{P_{1}}S_{y_{2},\alpha_{15}}+\pder{f_3}{C}S_{y_{3},\alpha _{15}}+\pder{f_3}{R}S_{y_{4},\alpha _{1}}+\pder{f_3}{\alpha _{15}}\\
\end{eqnarray*}
\centerline{\emph{Resource}}
\begin{eqnarray*}
\der{}S_{y_{4},\alpha _{1}} = \pder{f_4}{P_{2}}S_{y_{1},\alpha _{1}}+\pder{f_4}{P_{1}}S_{y_{2},\alpha_{1}}+\pder{f_4}{C}S_{y_{3},\alpha _{1}}+\pder{f_4}{R}S_{y_{4},\alpha _{1}}+\pder{f_4}{\alpha _1}\\
\der{}S_{y_{4},\alpha _{2}} = \pder{f_4}{P_{2}}S_{y_{1},\alpha _{2}}+\pder{f_4}{P_{1}}S_{y_{2},\alpha_{2}}+\pder{f_4}{C}S_{y_{3},\alpha _{2}}+\pder{f_4}{R}S_{y_{4},\alpha _{2}}+\pder{f_4}{\alpha _2}\\
\vdots\\
\der{}S_{y_{4},\alpha _{15}} = \pder{f_4}{P_{2}}S_{y_{1},\alpha _{15}}+\pder{f_4}{P_{1}}S_{y_{2},\alpha_{15}}+\pder{f_4}{C}S_{y_{3},\alpha _{15}}+\pder{f_4}{R}S_{y_{4},\alpha _{1}}+\pder{f_4}{\alpha _{15}}\\
\end{eqnarray*}
Tables 7.1 and 7.2 contain all the \textbf{particular parts} for the sensitivity equations; the partial derivatives of each population with respect to parameter values. We can now write our initial value problem of sensitivity equations for each model. We suppress the dependence on time for these calculations. The general sensitivity equations (\ref{morgan3}) for our linear response model are given by,
\begin{equation}\label{71}
\begin{split} \der{}(S_{P})&= (e_{RP}\alpha_{RP}R+e_{CP}\alpha_{CP}C-m_{P})S_{P}+ (e_{CP}\alpha_{CP}P)S_{C}+(e_{RP}\alpha_{RP}P)S_{R} \\
\der{}(S_{C})&= (-\alpha_{CP}C)S_{P} + (e_{RC}\alpha_{RC}R-\alpha_{CP}P-m_{C})S_{C}+ (e_{RC}\alpha_{RC}C)S_{R} \\
\der{}(S_{R})&= (-\alpha_{CP}C)S_{P} + (e_{RC}\alpha_{RC}R-\alpha_{CP}P-m_{C})S_{C}+ (e_{RC}\alpha_{RC}C)S_{R} \\
\end{split}\end{equation}
with initial conditions
\begin{equation} S_{P}(0)=0, \hspace{2.cm} S_{C}(0)=0, \hspace{1.6cm} and \hspace{.2cm} S_{R}(0)=0 \end{equation}
where we define
\begin{equation}S_{P}(t) = \pder{}{\alpha_{i}}P(t),\hspace{.3cm} S_{C}(t) = \pder{}{\alpha_{i}}C(t),\hspace{.3cm} S_{R}(t) = \pder{}{\alpha_{i}}R(t).
\end{equation}Similarly, by \ref{morgan3} the general sensitivity equations for our stage structured model are:
\begin{equation}\label{72}
\begin{split} \der{}(S_{P_{2}})&= (-m_{P_{2}})S_{P_{2}}+(\mu_{P})S_{P_{1}} \\
\der{}(S_{P_{1}})&= \frac{e_{RP}\lambda_{RP}R+e_{CP}\lambda_{CP}C}{1+\lambda_{RP}h_{RP}R+\lambda_{CP}h_{CP}C}S_{P_{2}}-(\mu_{P}+m_{P_{1}})S_{P_{1}} \\
&+ \bigg( \frac{e_{CP}\lambda_{CP}P_{2}(1+\lambda_{RP}h_{RP}R)-e_{RP}\lambda_{RP}\lambda_{CP}h_{CP}RP_{2}}{(1+\lambda_{RP}h_{RP}R+\lambda_{CP}h_{CP}C)^2}\bigg) S_{C}\\
&+ \bigg( \frac{e_{RP}\lambda_{RP}P_{2}(1+\lambda_{CP}h_{CP}C)-e_{CP}\lambda_{RP}\lambda_{CP}h_{RP}CP_{2}}{(1+\lambda_{RP}h_{RP}R+\lambda_{CP}h_{CP}C)^2}\bigg) S_{R}\\
\der{}(S_{C})&= \bigg( -\frac{\lambda_{CP}C}{1+\lambda_{RP}h_{RP}R+\lambda_{CP}h_{CP}C}\bigg) S_{P_{2}} \\
& + \bigg( \frac{e_{RC}\lambda_{RC}R}{1+\lambda_{RC}h_{RC}R}- \frac{\lambda_{CP}P_{2}(1+\lambda_{RP}h_{RP}R)}{(1+\lambda_{RP}h_{RP}R+\lambda_{CP}h_{CP}C)^2}-m_{C} \bigg) S_{C} \\
&+ \bigg( \frac{e_{RC}\lambda_{RC}C}{(1+\lambda_{RP}h_{RP}R)^2}+\frac{\lambda_{CP}\lambda_{RP}h_{RP}CP_{2}}{(1+\lambda_{RP}h_{RP}R+\lambda_{CP}h_{CP}C)^2}\bigg) S_{R} \\
\der{}(S_{R})&= \bigg( -\frac{\lambda_{RP}R}{1+\lambda_{RP}h_{RP}R+\lambda_{CP}h_{CP}C} \bigg) S_{P_{2}}\\
&+\bigg( -\frac{\lambda_{RP}R}{1+\lambda_{RP}h_{RP}R}\bigg) S_{P_{1}}\\
&+\bigg( -\frac{\lambda_{RC}R}{1+\lambda_{RC}h_{RC}R}+\frac{\lambda_{RP}\lambda_{CP}h_{CP}RP_{2}}{(1+\lambda_{RP}h_{RP}R+\lambda_{CP}h_{CP}C)^2}\bigg) S_{C}\\
&+ \bigg( r(1-\frac{2R}{K})-\frac{\lambda_{RC}C}{(1+\lambda_{RC}h_{RC}R)^2}- \frac{\lambda_{RP}P_{1}}{(1+\lambda_{RP}h_{RP}R)^2} \bigg) S_{R}\\
&+ \bigg( - \frac{\lambda_{RP}P_{2}(1+\lambda_{CP}h_{CP}C)}{(1+\lambda_{RP}h_{RP}R+\lambda_{CP}h_{CP}C)^2} \bigg) S_{R}\\
\end{split}\end{equation}
with initial conditions
\begin{equation} S_{P_{2}}(0)=0, S_{P_{1}}(0)=0, \hspace{1.1cm} S_{C}(0)=0, \hspace{1.1cm} and \hspace{.2cm} S_{R}(0)=0 \end{equation}
where we define
\begin{equation}
S_{P_{2}}(t) = \pder{}{\alpha_{i}}P_{2}(t),\hspace{.2cm} S_{P_{1}}(t) = \pder{}{\alpha_{i}}P_{1}(t),\hspace{.2cm} S_{C}(t) = \pder{}{\alpha_{i}}C(t),\hspace{.2cm} S_{R}(t) = \pder{}{\alpha_{i}}R(t).\end{equation}\vspace{.02in}
\section{Solving the Sensitivity Equations}
For each parameter that the original system has, we must solve a system of linear sensitivity equations. We note that the number of differential equations in the state system dictates how many differential equations we will have in the linear sensitivity system. For our linear response model, we have ten parameters and three variables. For our stage structured model we have fifteen parameters and four variables. Vance and Eads \cite{Chris} note that, ``Although the sensitivity equations are linear, they are forced by the solution to the state equations." Hence, we will need to solve one hundred and twenty equations in groups of eight (four model and four sensitivity) for our stage structure model.
We will numerically integrate the linear sensitivity equations and the nonlinear state equations using Matlab's fourth and fifth order adaptive step-size algorithm known as ode45. This is a Runge-Kutta-Fehlberg method that will simultaneously obtain two solutions per step. This helps to monitor the accuracy of the solution and adjusts the step size.
After differentiating our linear response model with respect to our ten parameter values we obtain ten sensitivity equations for each population density. Given our three population densities, we have a total of thirty sensitivity equations to solve. That is, for each parameter in our original system we must solve a system of linear sensitivity equations. We note that the number of differential equations in the state system dictates how many differential equations there will be in the linear sensitivity system. The equations in (\ref{71}) remain the same with only the addition of the particular part to each line. In our linear model, we have ten parameters and three variables. Vance and Eads \cite{Chris} state, "although the sensitivity equations are linear, they are forced by the solution to the state equations". Hence we must solve the initial value problem for the model and the three corresponding sensitivity equations. Hence, to compute all our sensitivities (for each parameter in our linear response model) we solve ten systems of equations with six coupled equations in each system.
Our stage structure model contains four variables and fifteen parameters. Thus, to compute the sensitivities of this model, we are required to solve fifteen systems with eight coupled equations in each. We have eight equations in each group because we must solve the sensitivity equations for each of our four populations and solve the original system simultaneously.
To solve each coupled set of equations in each model we will numerically integrate the linear sensitivity equations and the state equations using Matlab's built-in algorithm known as ode45. Ode45 is a Runge-Kutta-Fehlberg method algorithm which obtains two solutions per step to monitor accuracy of the solution. We note that the Runge-Kutta method we are using allows for adjusting the step size to desired tolerances. We set the the relative error for 1 x $10^{-3}$ and the absolute error at 1 x $10^{-3}$. We set the initial conditions for the state equations as $(1,1,1,1)^T$ and set our initial conditions for the sensitivity equations as $(0,0,0,0)^T$ \cite{Vance} \cite{Chris}.
In order to quantify or give our sensitivities a performance measure we use a weighted Euclidean norm. Our weighted norm for the linear response model will be in three dimensions while our norm for the stage structure model will be in four dimensions. We define the weighted norms for each model below and note that taking the weighted norm will be a function of the parameter value and time only. This will allow us to graph parameter sensitivities against one another to see which of these sensitivity measures rank higher and ones which rank lower. Because we are unsure if any state variable in terms of of performance measure is more important, we simply use a weight of 1 for all our calculations. We use the following weighted Euclidean norm for our linear response model,
\begin{equation} \parallel S_{\alpha_{i}}\parallel =\parallel (S_{{P}, \alpha_{i}}, S_{{C}, \alpha_{i}}, S_{{R}, \alpha_{i}})^T\parallel \sqrt{w_{1}(S_{{P}, \alpha_{i}})^2 + w_{2}(S_{{C}, \alpha_{i}})^2 + w_{3}(S_{{R},\alpha_{i}})^2} \end{equation}
and likewise,
\begin{equation} \parallel S_{\alpha_{i}}\parallel =\parallel (S_{{P_{2}}, \alpha_{i}},S_{{P_{1}}, \alpha_{i}} S_{{C}, \alpha_{i}}, S_{{R}, \alpha_{i}})^T\parallel \sqrt{w_{1}(S_{{P_{2}}, \alpha_{i}})^2 + w_{2}S_{{P_{1}}, \alpha_{i}})^2 w_{3}(S_{{C}, \alpha_{i}})^2 + w_{4}(S_{{R},\alpha_{i}})^2} \end{equation}
as the norm for our stage structure model.
\begin{table}\label{155}
\newpage
\centering % used for centering table
\begin{tabular}{|l|l|l|l|l|} % centered columns (2 columns)
\hline %inserts horizontal lines
$\alpha _i$ & Parameter & Partial of $f_1$ & Partial of $f_2$ & Partial of $f_3$ \\ [0.1ex] % inserts table
%heading
\hline % inserts single horizontal line
$\alpha _1$ & $e_{RP}$ & $\alpha_{RP}RP$ & $0$ & $0$\\
$\alpha _2$ & $e_{CP}$ & $\alpha _{CP}CP$ & $0$ & $0$ \\
$\alpha _3$ & $e_{RC}$ & $0$ & $\alpha _{RC}RC$ & $0$\\
$\alpha_4$ & $\alpha _{RP}$ & $e_{RP}RP$ & $0$ & $-RP$\\
$\alpha_ 5$ & $\alpha _{CP}$ & $e_{CP}CP$ & $-CP$ & $0$\\
$\alpha _6$ & $\alpha _{RC}$ & $0$ & $e_{RC}RC$ & $-RC$\\
$\alpha _7$ & $m_P$ & $-P$ & $0$ & $0$\\
$\alpha _8$ & $m_C$ & $0$ & $-C$ & $0$\\
$\alpha _9$ & $r$ &$0$ & $0$ & $R(1-R/K)$\\
$\alpha _{10}$ & $K$ & $0$ & $0$ & $rR^2/K^2$\\ \hline
\end{tabular}
\caption{Linear Response Model: Partials with respect to Parameter Values} % title of Table
\end{table}\begin{table}\label{ssp}
\centering
\begin{tabular}{|c|c|c|} % centered columns (2 columns)
\hline %inserts horizontal lines
Parameter & Partial of $f_1$ & Partial of $f_2$ \\ [0.1ex] % inserts table
%heading
\hline % inserts single horizontal line
$r$ & $0$ & $0$\\
$K$ & $0$ & $0$\\
$\mu_{P}$ & $P_{1}$ & $-P_{1}$\\
$m_{P_{2}}$ & $-P_{2}$ & $-P_{1}$\\
$m_{P_{1}}$ & $0$ & $0$\\
$m_{C}$ & $0$ & $0$\\
$e_{RP}$ & $0$ & $\frac{\lambda_{RP}RP_{2}}{1+\lambda_{RP}h_{RP}R+\lambda_{CP}h_{CP}C}$\\
$e_{CP}$ & $0$ & $\frac{\lambda_{CP}CP_{2}}{1+\lambda_{RP}h_{RP}R+\lambda_{CP}h_{CP}C}$\\
$e_{RC}$ & $0$ & $0$\\
$h_{RP}$ & $0$ & $-\frac{\lambda_{RP}RP_{2}(e_{RP}\lambda_{RP}R+e_{CP}\lambda_{CP}C)}{(1+\lambda_{RP}h_{RP}R+\lambda_{CP}h_{CP}C)^2}$\\
$h_{CP}$ & $0$ & $-\frac{\lambda_{CP}CP_{2}(e_{RP}\lambda_{RP}R+e_{CP}\lambda_{CP}C)}{(1+\lambda_{RP}h_{RP}R+\lambda_{CP}h_{CP}C)^2}$\\
$h_{RC}$ & $0$ & $0$\\
$\lambda_{RP}$ & $0$ & $\frac{e_{RP}RP_{2}(1+\lambda_{CP}h_{CP}C)-e_{CP}\lambda_{CP}h_{RP}RCP_{2}}{(1+\lambda_{RP}h_{RP}R+\lambda_{CP}h_{CP}C)^2}$\\
$\lambda_{CP}$ & $0$ & $\frac{e_{CP}CP_{2}(1+\lambda_{RP}h_{RP}R)-e_{RP}\lambda_{RP}h_{CP}RCP_{2}}{(1+\lambda_{RP}h_{RP}R+\lambda_{CP}h_{CP}C)^2}$\\
$\lambda_{RC}$ & $0$ & $0$\\
\hline
Parameter & Partial of $f_3$ & Partial of $f_4$\\ % inserts table
\hline
$r$ & $0$ & $R(1-\frac{R}{K})$ \\
$K$ & $0$ & $\frac{rR^2}{K^2}$\\
$\mu_{P}$ & $0$ & $0$\\
$m_{P_{2}}$ & $0$ & $0$\\
$m_{P_{1}}$ & $0$ & $0$\\
$m_{C}$ & $-C$ & $0$\\
$e_{RP}$ & $0$ & $0$\\
$e_{CP}$ & $0$ & $0$\\
$e_{RC}$ & $\frac{\lambda_{RC}RC}{1+\lambda_{RP}h_{RP}R}$ & $0$\\
$h_{RP}$ & $\frac{\lambda_{RP}\lambda_{CP}RCP_{2}}{(1+\lambda_{RP}h_{RP}R+\lambda_{CP}h_{CP}C)^2}$&$\frac{(\lambda_{RP})^2R^2P_{1}}{(1+\lambda_{RP}h_{RP}R)^2}+\frac{(\lambda_{RP})^2R^2P_{2}}{(1+\lambda_{RP}h_{RP}R+\lambda_{CP}h_{CP}C)^2}$\\%done to here
$h_{CP}$ & $\frac{(\lambda_{CP})^2C^2P_{2}}{(1+\lambda_{RP}h_{RP}R+\lambda_{CP}h_{CP}C)^2}$ & $\frac{\lambda_{RP}\lambda_{CP}RCP_{2}}{(1+\lambda_{RP}h_{RP}R+\lambda_{CP}h_{CP}C)^2}$\\
$h_{RC}$ & $-\frac{e_{RC}(\lambda_{RC})^2R^2C}{(1+\lambda_{RP}h_{RP}R)^2}$ & $\frac{(\lambda_{RC})^2R^2C}{(1+\lambda_{RP}h_{RP}R)^2}$\\
$\lambda_{RP}$ & $\frac{\lambda_{CP}h_{RP}RCP_{2}}{(1+\lambda_{RP}h_{RP}R+\lambda_{CP}h_{CP}C)^2}$ & $-\frac{RP_{1}}{1+\lambda_{RP}h_{RP}R}-\frac{RP_{2}(1+\lambda_{CP}h_{CP}C)}{(1+\lambda_{RP}h_{RP}R+\lambda_{CP}h_{CP}C)^2}$ \\
$\lambda_{CP}$ & $-\frac{CP_{2}(1+\lambda_{RP}h_{RP}R)}{(1+\lambda_{RP}h_{RP}R+\lambda_{CP}h_{CP}C)^2}$ & $\frac{\lambda_{RP}h_{CP}RCP_{2}}{(1+\lambda_{RP}h_{RP}R+\lambda_{CP}h_{CP}C)^2}$\\
$\lambda_{RC}$ & $\frac{e_{RC}RC}{(1+\lambda_{RP}h_{RP}R)^2}$ & $-\frac{RC}{(1+\lambda_{RP}h_{RP}R)^2}$\\
\hline
\end{tabular}
\caption{Stage Structured Model: Partials with respect to Parameter Values} % title of Table
\end{table}