Sensitivity Analysis model for omnivory systems (coupled system of ode's)

In summary, sensitivity analysis is a method used to determine the effect of small changes in input parameters on the output of a mathematical model. It can be applied to omnivory systems to understand the influence of key parameters on their dynamics and stability. There are different types of sensitivity analysis that can be applied to coupled systems of ODEs, and it can also be used to predict the behavior of omnivory systems under different scenarios.
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DrWahoo
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\chapter{Stage Structured Omnivory Model}

%\indent For our initial computations we  will be using a standard Lotka-Volterra omnivory model which was developed to analyze the population and community level dynamics. The model depicted below (figure 1.4)  is a standard Lotka-Volterra model of omnivory.Also for mathematical computations, we will say this model of omnivory is a model consisting of only one predator [P] , one consumer [C], and one resource [R]. The model is given below in Figure 1.4.\\

To better model our complex system of wolves, elk, and resources we use a stage structured approach with non-linear functional responses. Vance \cite{Vance} uses a similar model in,``Permanent Coexistence for Omnivory Models." However, in his model he assumes the mortality rate of both predator classes to be the same. In our case, the mortality rate of our juvenile wolf is much greater than that of an adult wolf. Separating the mortality rates into two different parameters will likely give us better results. This model separates the predator into two classes, adults and juveniles.  As Gotelli \cite{Got} notes, this can make a huge difference in the accuracy of a model.  By separating the predator into two classes, we have to include the maturation rate $\mu p$, the time it takes a juvenile wolf to become an adult wolf. We assume the juvenile predator, denoted $P_{1}$, feeds solely on the resource, while the adult predator, $P_{2}$, feeds on both the consumer and resource.   The model is given below with the following set of differential equations:

\centerline{\bf{Predator Stage Structured Omnivory Model}}
\begin{equation}\label{Stage1}
\begin{aligned}
\der{P_{2}}=&\mu_{P}P_{1} - m_{P_{2}}P_{2}.\\
\der{P_{1}}=&\frac{e_{RP}\lambda_{RP}R + e_{CP}\lambda_{CP}C}{1+\lambda_{RP}h_{RP}R+\lambda_{CP}h_{CP}C}P_{2}-(\mu_{P} + m_{P_{1}})P_{1}.\\
\der{C}=&C[\frac{e_{RC}\lambda_{RC}R}{1+\lambda_{RC}h_{RC}R}- \frac{\lambda_{CP}P_{2}}{1+\lambda_{RP}h_{RP}R+\lambda_{CP}h_{CP}C}-m_{C}].\\
\der{R}=&R[r(1-\frac{R}{K})-\frac{\lambda_{RC}C}{1+\lambda_{RC}h_{RC}R}-\frac{\lambda_{RP}P_{1}}{1+\lambda_{RP}h_{RP}R}- \frac{\lambda_{RP}P_{2}}{1+\lambda_{RP}h_{RP}R+\lambda_{CP}h_{CP}C}].\\
\end{aligned}
\end{equation}
We assume that the initial population densities of all species ($P_{2}$, $P_{1}$, $C$, $R$) are such that,
\begin{equation} P_{2}(0)= c_{1}>0, P_{2}(0)=c_{2}>0, C(0)=c_{3}>0, R(0)=c_{4}>0.\end{equation}
Then we have the following initial value problem:
\begin{equation}\label{ivp22}
\begin{aligned}
\der{P_{2}}=&f_{1}(t, P_{2}, P_{1}, C,R)\\
\der{P_{1}}=&f_{2}(t, P_{2}, P_{1}, C,R)\\
\der{C}=&f_{3}(t, P_{2}, P_{1}, C,R)\\
\der{R}=&f_{4}(t, P_{2}, P_{1}, C,R)\\
P_{2}(0)=c_{1}, P_{1}(0)=c_{2},& \hspace{.1cm}C(0)=c_{3}, R(0)=c_{4}.\\
\end{aligned}\end{equation} defined on $D = \Rone \times \Rfour$.  

From (\ref{Stage1}) and (\ref{ivp22}), we have
\begin{equation}\label{forp}
\begin{split}
f_{1}(t, P_{2}, P_{1}, C,R)&=\mu_{P}P_{1} - m_{P_{2}}P_{2}\\
f_{2}(t, P_{2}, P_{1}, C,R)&=\frac{e_{RP}\lambda_{RP}R + e_{CP}\lambda_{CP}C}{1+\lambda_{RP}h_{RP}R+\lambda_{CP}h_{CP}C}P_{2}-(\mu_{P} + m_{P_{1}})P_{1}\\
f_{3}(t, P_{2}, P_{1}, C,R)&=C[\frac{e_{RC}\lambda_{RC}R}{1+\lambda_{RC}h_{RC}R}- \frac{\lambda_{CP}P_{2}}{1+\lambda_{RP}h_{RP}R+\lambda_{CP}h_{CP}C}-m_{C}]\\
f_{4}(t, P_{2}, P_{1}, C,R)&=R[r(1-\frac{R}{K})-\frac{\lambda_{RC}C}{1+\lambda_{RC}h_{RC}R}-\frac{\lambda_{RP}P_{1}}{1+\lambda_{RP}h_{RP}R}- \frac{\lambda_{RP}P_{2}}{1+\lambda_{RP}h_{RP}R+\lambda_{CP}h_{CP}C}].\\
\end{split}
\end{equation}
Thus we can represent our system by the following initial value problem,
\begin{equation}\label{IVPP} \der{\bff}=\bff( t, \bfy) \hspace{.7cm} \bff(0)=\bfc \end{equation} 
defined on D = $\rm I\!R$ $\times$ $\Rfour$, with the understood definitions for $\bff$, $\bfy$, and $\bfc$ where boldface letters represent our vectors. 

\section{Solution to system}
Investigating long-term survival of species requires that a unique solution exists for all $t\geq 0$. We use a similar approach as in the previous chapter for our linear response model. We first show that the $f$ and its partial derivatives are all continuous with respect to population densities $P_{2},P_{1},C,R$ for all positive t, P(t), C(t), R(t).  We calculate the derivatives below and provide them in Table 6.1We show the components of $\bff$ with respect to population densities below:

\begin{displaymath}
\pder{\mathbf{f}}{P_{2}} =
\left( \begin{array}{c}
\pder{f_1}{P_{2}}\vspace{.2cm} \\
\pder{f_2}{P_{2}}\vspace{.2cm}\\
\pder{f_3}{P_{2}}\vspace{.2cm}\\
\pder{f_4}{P_{2}}\vspace{.2cm}
\end{array} \right)=
\left( \begin{array}{c}
 -m_{P_{2}} \\
 \frac{e_{RP}\lambda_{RP}R+e_{CP}\lambda_{CP}C}{1+\lambda_{RP}h_{RP}R+\lambda_{CP}h_{CP}C}\\
-\frac{\lambda_{CP}C}{1+\lambda_{RP}h_{RP}R+\lambda_{CP}h_{CP}C}\\
-\frac{\lambda_{RP}R}{1+\lambda_{RP}h_{RP}R+\lambda_{CP}h_{CP}C}
\end{array} \right)
\end{displaymath}\begin{displaymath}
\pder{\mathbf{f}}{P_{1}} =
\left( \begin{array}{c}
\pder{f_1}{P_{1}}\vspace{.2cm} \\
\pder{f_2}{P_{1}}\vspace{.2cm}\\
\pder{f_3}{P_{1}}\vspace{.2cm}\\
\pder{f_4}{P_{1}}\vspace{.2cm}
\end{array} \right)=
\left( \begin{array}{c}
\mu_{P} \\
 -(\mu_{P}+m_{P_{1}}) \\
0\\
-\frac{\lambda_{RP}R}{1+\lambda_{RP}h_{RP}R}
\end{array} \right)
\end{displaymath}

\begin{displaymath}
\pder{\mathbf{f}}{C} =
\left( \begin{array}{c}
\pder{f_1}{C}\vspace{.2cm} \\
\pder{f_2}{C}\vspace{.2cm}\\
\pder{f_3}{C}\vspace{.2cm}\\
\pder{f_4}{C}\vspace{.2cm}
\end{array} \right)=
\left( \begin{array}{c}
0\\
\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} \\
\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}   \\
-\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}
\end{array} \right)
\end{displaymath}

\begin{displaymath}
\pder{\mathbf{f}}{R} =
\left( \begin{array}{c}
\pder{f_1}{R}\vspace{.2cm} \\
\pder{f_2}{R}\vspace{.2cm}\\
\pder{f_3}{R}\vspace{.2cm}\\
\pder{f_4}{R}\vspace{.2cm}
\end{array} \right)=
\left( \begin{array}{c}
 0 \\
 \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} \\
 \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}          \\
 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}- \frac{\lambda_{RP}P_{2}(1+\lambda_{CP}h_{CP}C)}{(1+\lambda_{RP}h_{RP}R+\lambda_{CP}h_{CP}C)^2}
\end{array} \right)
\end{displaymath}\begin{table}[!h]

\centering % used for centering table
\begin{tabular}{|l|l|l|} % centered columns (2 columns)
\hline %inserts horizontal lines
Pop. & Partial of $f_1$  \\
%heading
\hline % inserts single horizontal line \\
$P_{2}$ & $ -m_{P_{2}}$  \\
$P_{1}$ &  $\mu_{p}$ \\
$C$ & $0$   \\ 
$R$ & $0$  \\
 \hline
Pop. & Partial of $f_2$ \\
\hline

$P_{2}$ &  $\frac{e_{RP}\lambda_{RP}R+e_{CP}\lambda_{CP}C}{1+\lambda_{RP}h_{RP}R+\lambda_{CP}h_{CP}C}$ \\
$P_{1}$ &   $-(\mu_{p}+m_{p_{1}})$\\
$C$  & $\frac{e_{CP}\lambda_{CP}P_{2}(1+\lambda_{RP}h_{RP}R)-e_{RP}\lambda_{RP}\lambda_{CP}h_{CP}RP_{2}}{(1+\lambda_{RP}h_{RP}R+\lambda_{CP}h_{CP}C)^2}$ \\ 
$R$ & $\frac{e_{RP}\lambda_{RP}P_{2}(1+\lambda_{CP}h_{CP}C)-e_{CP}\lambda_{RP}\lambda_{CP}h_{RP}CP_{2}}{(1+\lambda_{RP}h_{RP}R+\lambda_{CP}h_{CP}C)^2}$ \\

 \hline
Pop. & Partial of $f_3$ \\
\hline
$P_{2}$ & $-\frac{\lambda_{CP}C}{1+\lambda_{RP}h_{RP}R+\lambda_{CP}h_{CP}C}$\\
$P_{1}$ & $0$ \\
$C$ & $\frac{e_{RC}\lambda_{RC}R}{1+\lambda_{RC}h_{RC}R}-\frac{\lambda_{CP}P_{2}(1+\lambda_{RP}h_{RP}R)}{(1+\lambda_{RP}h_{RP}R+\lambda_{CP}h_{CP}C)^2}-m_{C}$  \\
$R$ &  $\frac{e_{RC}\lambda_{RC}C}{(1+\lambda_{RP}h_{RP}R)^2} + \frac{\lambda{CP}\lambda_{RP}h_{RP}CP_{2}}{(1+\lambda_{RP}h_{RP}R+\lambda_{CP}h_{CP}C)^2}$ \\
\hline
Pop. & Partial of $f_4$\\
\hline
$P_{2}$ &  $-\frac{\lambda_{RP}R}{1+\lambda_{RP}h_{RP}R+h_{CP}\lambda_{CP}C}$\\
$P_{1}$ & $-\frac{\lambda_{RP}R}{1+\lambda_{RP}h_{RP}R}$\\
$C$ & $ -\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}$  \\
$R$  & $r(1-\frac{2R}{K})-\frac{\lambda_{RC}C}{(1+\lambda_{RC}h_{RC}R)^2}-\frac{\lambda_{RP}P_{1}}{(1+\lambda_{RP}h_{RP}R)^2} -  \frac{\lambda_{RP}P_{2}(1+\lambda_{CP}h_{CP}C)}{(1+\lambda_{RP}h_{RP}R+\lambda_{CP}h_{CP}C)^2}$ \\
\hline
\end{tabular}
\caption{Partials with respect to Population Densitites for Predator Stage Structure Model} % title of Table
\end{table}

We see that $\bff$ and all our derivatives contained in Table 6.1 are continuous with respect to our four population densities.  Applying Lemma 1 to our system, we see that our vector $\bff$ satisfies, in $D = \rm I\!R$ $\times$ $\Rfour$, a local Lipschitz condition with respect to $\bfy$ where:\\
\begin{align}\label{lip23}
\bfy &= \begin{bmatrix}
P_{2}\\
P_{1}\\
C\\
R\\
\end{bmatrix}.
\end{align} 
We now use Theorem 1 to show existence and uniqueness.  Recall the hypothesis of Theorem 1 is satisfied if $\pder{\bff}{\bfy}$ is continuous in D. We already have this hypothesis satisfied by the calculations (found in Table 6.1) hence, we conclude the initial value problem
\begin{equation} \bfy'=\bff(t,\bfy), \hspace{.7cm} \bfy(t_{0})=\bfc \end{equation} 
has exactly one solution in D.  This solution can be extended to the left and right up to the boundary of D.  \\

%\der{P_{2}}=&\mu_{P}P_{1} - m_{P_{2}}P_{2}.\\
%\der{p_{1}}=&\frac{e_{RP}\lambda_{RP}R + e_{CP}\lambda_{CP}C}{1+\lambda_{RP}h_{RP}R+\lambda_{CP}h_{CP}C}P_{2}-(\mu_{P} + m_{P_{1}})P_{1}.\\
%\\Since we are dealing with time and populations over time, we also need to show that a unique solution is available for all $t \geq 0$.  We provide the proof in the form of a theorem below. We credit Vance for this approach and proof. 
\begin{lemma} \cite{Vance} The initial value problem (\ref{ivp22}) has a unique solution in $\Rfour$ for all $t\geq 0$. \end{lemma}
\textit{Proof.} Assume the population densities are such that, 
\begin{equation}\label{64} P_{2}(t), P_{1}(t), C(t), R(t) \geq 0  \end{equation}
for all  $t\geq 0.$\\
In model (\ref{Stage1}), we have that, 
\begin{equation}\der{R}=R[r(1-\frac{R}{K})-\frac{\lambda_{RC}C}{1+\lambda_{RC}h_{RC}R}-\frac{\lambda_{RP}P_{1}}{1+\lambda_{RP}h_{RP}R}- \frac{\lambda_{RP}P_{2}}{1+\lambda_{RP}h_{RP}R+\lambda_{CP}h_{CP}C}].\end{equation}

Using the fact that (\ref{64}) and all parameter values are positive we can say, 
\begin{equation}\label{spl}\begin{split}
 -\frac{\lambda_{RC}C}{1+\lambda_{RC}h_{RC}R} & \leq 0,\\
-\frac{\lambda_{RP}P_{1}}{1+\lambda_{RP}h_{RP}R} & \leq 0, \mbox{and}\\
-\frac{\lambda_{RP}P_{2}}{1+\lambda_{RP}h_{RP}R+\lambda_{CP}h_{CP}C} & \leq 0 .\end{split}\end{equation}  
Likewise, from  (\ref{spl}), we know the quantity,
\begin{equation} [-\frac{\lambda_{RC}C}{1+\lambda_{RC}h_{RC}R}-\frac{\lambda_{RP}P_{1}}{1+\lambda_{RP}h_{RP}R}- \frac{\lambda_{RP}P_{2}}{1+\lambda_{RP}h_{RP}R+\lambda_{CP}h_{CP}C}] \leq 0. \end{equation} 
Hence we can form the differential inequality, 
\begin{equation}\der{R(t)} \leq R(t)r(1-R(t)/K).\end{equation}
Equivalently,
\begin{equation}\label{ident}\der{R(t)}\leq R(t)[r(1-R(t)/K]. \end{equation} 
Note this differential inequality (\ref{ident}) is identical to (\ref{tg}) in the proof of Lemma 5. Thus from (\ref{49}), we conclude that,
\begin{equation}\label{rrr} R(t)\leq K_{max} \hspace{.5cm} for \hspace{.5cm} 0\leq t\leq \infty .\end{equation}
We use the same approach as above for the following differential equation,
\begin{equation} \der{C}=C[\frac{e_{RC}\lambda_{RC}R}{1+\lambda_{RC}h_{RC}R}- \frac{\lambda_{CP}P_{2}}{1+\lambda_{RP}h_{RP}R+\lambda_{CP}h_{CP}C}-m_{C}].\end{equation} 
Once again using (\ref{64}) and knowing all parameter values are positive we can say, 
\begin{equation}\label{067}\begin{split}
- \frac{\lambda_{CP}P_{2}}{1+\lambda_{RP}h_{RP}R+\lambda_{CP}h_{CP}C} &\leq 0, \mbox{and}\\
-m_{C}&\leq 0. \end{split}\end{equation} 
Thus we know the quantity,
\begin{equation} 
(- \frac{\lambda_{CP}P_{2}}{1+\lambda_{RP}h_{RP}R+\lambda_{CP}h_{CP}C}-m_{C}) \leq 0 .\end{equation} 
Hence we can form the differential inequality,
\begin{equation}\label{alll} \der{C} \leq C \frac{e_{RC}\lambda_{RC}R}{1+\lambda_{RC}h_{RC}R}. \end{equation} 
Noticing that denominator on the right hand side of (\ref{alll}) is always greater than 1 and that the numerator is always positive, we have
\begin{equation} C \frac{e_{RC}\lambda_{RC}R}{1+\lambda_{RC}h_{RC}R} < Ce_{RC}\lambda_{RC}R.\end{equation}
This gives us the following differential inequality, 
\begin{equation}\label{09} \der{C} \leq C[e_{RC}\lambda_{RC}R]. \end{equation}
Then from the proof of Lemma 5, particularly differential inequality (5.37), we replace $R$ with $K_{max}$ giving us,
\begin{equation}\label{rer} \der{C} \leq C[e_{RC}\lambda_{RC}K_{max}] .\end{equation}
Now we let the positive constant $\Gamma$  represent the quantity of $[e_{RC}\lambda_{RC}K_{max}]$. 
That is, 
\begin{equation} 
\Gamma = e_{RC}\lambda_{RC}K_{max} \end{equation} 
 Substituting $\Gamma$ into equation (\ref{rer}) gives us the exact form we need to satisfy the hypothesis of Lemma 4 Part A on the interval \hspace{.5cm}$0\leq t< \infty$, 
\begin{equation}\label{sats} \der{C(t)} \leq  C(t)\Gamma  \hspace{.5cm} \mbox{with} \hspace{.5cm}  0\leq t <\infty .\end{equation} 
Note that our independent variable t is bounded below. That is, $0\leq t< \infty$. Hence Lemma 4 Part A with our interval, $0\leq t< \infty$  (In our case $0=a$ is our lower bound) yields the following conclusion from (\ref{220}), 
\begin{equation}\label{365}  C(t)\leq C(0)e^{\Gamma (t-0)}  \mbox{   for   } 0\leq t < \infty .\end{equation} 
Recall that our initial value problem (\ref{ivp22}) gives us that $C(0)=c_{3}$.  Plugging this value into (\ref{365}) and simplifying yields,
\begin{equation}\label{3643}\begin{split}
  C(t)&\leq C(0)e^{\Gamma (t)}\\  
  &=c_{3}e^{\Gamma (t)}. 
\end{split}\end{equation}
This exponential function does not approach infinity in finite time.  
Now we focus on our predator population. Taking the negative terms away from the adult population $P_{2}$, and using the numerator of the positive term, juvenile predator population, $P_{1}$, we can form the following differential inequalities,
\begin{align} \der{P_{1}(t)}&\leq (e_{RP}\lambda_{RP}R(t)+e_{CP}\lambda)_{CP}C(t))P_{2}(t)\\
\der{P_{2}}(t)&\leq \mu P_{1}(t). \end{align}
Recall that $R(t)\leq K_{max}$ from equation (\ref{rrr}), and $C(t)\leq c_{3}e^{\Gamma t}$ from equation (\ref{3643}). 
\\Using that information we can say,
\begin{equation}\der{P_{1}}(t)\leq (\Gamma +e_{CP}\lambda_{CP}c_{3}e^{\Gamma t})P_{2}(t), \end{equation} 
and 
\begin{equation} \der{P_{2}}(t)\leq \mu P_{1}(t),\end{equation} 
for $0\leq t\leq \infty$.\\
We now denote the product $e_{CP}\lambda_{CP}c_{3}$ with $\theta$ to simplify our inequalities. \\
That is,
\begin{equation} e_{CP}\lambda_{CP}c_{3}=\theta .\end{equation}
Rewriting our inequalities for both predator populations in matrix form using $\theta$ gives us, 

\begin{equation}\label{matrix1}
   \frac{d}{dt}\begin{bmatrix}
         P_{1}(t) \\
    P_{2}(t) 
     \end{bmatrix}\leq 
    \begin{bmatrix}
        0       & \Gamma +\theta e^{\Gamma t}\\
    \mu p       & 0
    \end{bmatrix}
   \begin{bmatrix}
         P_{1}(t) \\
    P_{2}(t) 
     \end{bmatrix}    
.\end{equation}
Note that 

\begin{equation}\label{matrix2}
   2(P_{1} \hspace{.2cm} P_{2})\hspace{.17cm} \frac{d}{dt} \begin{bmatrix}
         P_{1} \\
    P_{2}
     \end{bmatrix}=
    \frac{d}{dt} 
\left| \left| \begin{array}{cc}
P_{1}\\
P_{2} \end{array} \right| \right|^2    
.\end{equation}
Thus we can multiply both sides of (\ref{matrix1}) by $2$ and the row vector $(P_{1} \hspace{.17cm} P_{2})$  to get

\begin{equation}\label{dad} \frac{d}{dt}
\left| \left| \begin{array}{cc}
P_{1}\\
P_{2} \end{array} \right| \right|^2 
\leq 2(\Gamma +\theta e^{\Gamma t})(P_{1}\hspace{.13cm} P_{2}). \end{equation} 

Since $2(P_{1} \hspace{.13cm} P_{2}) \leq ((P_{1})^2 + (P_{2})^2)$ we can rewrite (\ref{dad}) as
\begin{equation}\label{daddy} \frac{d}{dt}
\left| \left| \begin{array}{cc}
P_{1}\\
P_{2} \end{array} \right| \right|^2 
\leq (\Gamma +\theta e^{\Gamma t})((P_{1})^2 + (P_{2})^2). \end{equation} 

Note, 
\begin{equation}\label{because}
\left| \left| \begin{array}{cc}
P_{1}\\
P_{2} \end{array} \right| \right|^2 = (P_{1})^2 + (P_{2})^2 \end{equation}. 

Using (\ref{because}) we can rewrite (\ref{daddy}) as

\begin{equation}\label{dad1} \frac{d}{dt}
\left| \left| \begin{array}{cc}
P_{1}(t)\\
P_{2}(t) \end{array} \right| \right| ^2 
\leq  (\Gamma +\theta e^{\Gamma t}) \left| \left| \begin{array}{cc}
 P_{1}(t)\\
P_{2}(t) \end{array} \right| \right| ^2 
.\end{equation} 

If we define
\begin{equation*}  \Lambda =\frac{\theta }{\Gamma } \end{equation*}
then we can apply Lemma 4 part C to the scalar equation (\ref{dad1}). We note that the left hand side of (\ref{dad1}) matches the left hand side of the hypothesis of Lemma 4 part C because the norm of $P_1$ and $P_2$ are functions of $t$. The constant terms $\Gamma$ and $\theta$ represent  $M_{1}$ and $M_{2}$ from the right hand side of Lemma 4 Part C.  Finally, because the norm on the right hand side of (\ref{dad1}) is a function of $t$. Thus apply Lemma 4 part C to the scaler equation (\ref{dad1}) gives us the desired result of   

\begin{equation}\label{dad3} \frac{d}{dt}
\left| \left| \begin{array}{cc}
P_{1}(t)\\
P_{2}(t) \end{array} \right| \right| ^2 
\leq  ((c_{2})^2 + (c_{1})^2)e^{\Gamma t+\Lambda (e^{\Gamma t-1)}}.\end{equation} 

This exponential function does not approach infinity in finite time. The inequality above also ensures that $P_{1}(t)$ and $P_{2}(t)$ do not reach infinity in finite time \cite{Vance}.   Hence, by Lemma 2 and inequalities (\ref{rrr}), (\ref{3643}), and (\ref{dad3}) a unique solution for the initial value problem (\ref{ivp22}) exists for all $t\geq 0$.  This completes the proof and allows us to say that a unique solution is available to (\ref{ivp22}) for all $t\geq 0$.
 

1. What is sensitivity analysis?

Sensitivity analysis is a method used to determine the effect of small changes in the input parameters on the output of a mathematical model. It helps to identify which parameters have the greatest impact on the model's behavior and can be used to evaluate the robustness and reliability of the model.

2. How does sensitivity analysis apply to omnivory systems?

In the context of omnivory systems, sensitivity analysis is used to determine the influence of key parameters on the dynamics of the system. This can include parameters such as the rate of consumption, growth rates of species, and carrying capacities. By varying these parameters, we can understand how changes in the system's environment can affect the population dynamics of omnivorous species.

3. What types of sensitivity analysis can be applied to coupled systems of ODEs?

There are several types of sensitivity analysis that can be applied to coupled systems of ODEs, including local sensitivity analysis, global sensitivity analysis, and rank-based sensitivity analysis. Each type has its own advantages and limitations, and the choice of method depends on the specific research questions and goals of the study.

4. How can sensitivity analysis help in understanding the stability of omnivory systems?

Sensitivity analysis can provide insights into the stability of omnivory systems by identifying the key parameters that contribute to the system's stability or instability. By varying these parameters, we can understand how changes in the environment or management practices can affect the system's stability and make informed decisions to maintain or improve the system's stability.

5. Can sensitivity analysis be used to predict the behavior of omnivory systems under different scenarios?

Yes, sensitivity analysis can be used to predict the behavior of omnivory systems under different scenarios by varying the input parameters and observing the changes in the model's output. This can help in making predictions about the system's response to environmental changes, management strategies, and other external factors.

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