Appendix A: Calculation of non-trivial limits and derivatives

Prey dependent Type II

The prey-dependent Holling Type II model is specified as (Holling 1959; Abrams & Ginzburg 2000):
\begin{equation} f\left(N,\ P\right)=\frac{\text{αN}}{N+b}\nonumber \\ \end{equation}\begin{equation} g\left(N,P\right)=\frac{\text{αPN}}{N+b}\nonumber \\ \end{equation}
Therefore:
\begin{equation} \frac{\text{dg}}{\text{dN}}=\frac{\text{αP}\left(N+b\right)-\alpha PN}{\left(N+b\right)^{2}}=\frac{\text{αPb}}{\left(N+b\right)^{2}}\nonumber \\ \end{equation}
Since all variables are nonnegative numbers,\(\frac{\text{Pb}}{\left(N+b\right)^{2}}\) is also a nonnegative number. Given\(\alpha\in\left.\ \left[0,\ \ \infty\right.\ \right)\),\(\frac{\text{dg}}{\text{dN}}=\alpha\frac{\text{Pb}}{\left(N+b\right)^{2}}\)is therefore in\(\left.\ \left[0,\ \ \infty\right.\ \right)\).

Prey dependent Type III

The prey-dependent Holling Type III model is specified as (Holling 1959; Abrams & Ginzburg 2000):
\begin{equation} f\left(N,\ P\right)=\frac{\alpha N^{2}}{N^{2}+b}\nonumber \\ \end{equation}\begin{equation} g\left(N,P\right)=\frac{\text{αP}N^{2}}{N^{2}+b}\nonumber \\ \end{equation}
Therefore:
\begin{equation} \frac{\text{dg}}{\text{dN}}=\frac{2\alpha PN*\left(N^{2}+b\right)-2N\left(\text{αP}N^{2}\right)}{\left(N^{2}+b\right)^{2}}=\frac{2\alpha PNb}{\left(N^{2}+b\right)^{2}}\nonumber \\ \end{equation}
Since all variables are nonnegative numbers,\(\frac{2PNb}{\left(N^{2}+b\right)^{2}}\) is also a nonnegative number. Given\(\alpha\in\left.\ \left[0,\ \ \infty\right.\ \right)\),\(\frac{\text{dg}}{\text{dN}}=\alpha\frac{2PNb}{\left(N^{2}+b\right)^{2}}\)is therefore in\(\left.\ \left[0,\ \ \infty\right.\ \right)\).

Ratio dependent Type II

The ratio-dependent Holling Type II model is specified as (Arditi & Ginzburg 1989; Abrams & Ginzburg 2000):
\begin{equation} f\left(N,\ P\right)=\frac{\alpha\frac{N}{P}}{\frac{N}{P}+b}\nonumber \\ \end{equation}\begin{equation} g\left(N,P\right)=\frac{\text{αN}}{\frac{N}{P}+b}\nonumber \\ \end{equation}
Therefore:
\begin{equation} \operatorname{}\left(f\right)=\operatorname{}\left[\frac{\text{αN}}{N+bP}\right]=0\nonumber \\ \end{equation}\begin{equation} \operatorname{}\left(f\right)=\operatorname{}\left[\frac{\frac{\alpha}{P}}{\frac{1}{P}+\frac{b}{N}}\right]=\frac{\frac{\alpha}{P}}{\frac{1}{P}}=\alpha\nonumber \\ \end{equation}\begin{equation} \operatorname{}\left(f\right)=\frac{0}{0+b}=0\nonumber \\ \end{equation}\begin{equation} \operatorname{}\left(g\right)=\frac{\text{αN}}{0+b}=\frac{\text{αN}}{b}\nonumber \\ \end{equation}\begin{equation} \operatorname{}\left(g\right)=\frac{\text{αN}}{\infty}=0\nonumber \\ \end{equation}\begin{equation} \operatorname{}\left(g\right)=\operatorname{}\left[\frac{\alpha}{\frac{1}{P}+\frac{b}{N}}\right]=\frac{\alpha}{\frac{1}{P}}=\alpha P\nonumber \\ \end{equation}\begin{equation} \operatorname{}\left(g\right)=\frac{0}{0+b}=0\nonumber \\ \end{equation}\begin{equation} \frac{\text{dg}}{\text{dN}}=\frac{\alpha\left[\frac{N}{P}+b\right]-\alpha N\left[\frac{1}{P}\right]}{\left[\frac{N}{P}+b\right]^{2}}=\frac{\text{αb}}{\left[\frac{N}{P}+b\right]^{2}}\nonumber \\ \end{equation}
Since all variables are nonnegative numbers,\(\frac{b}{\left[\frac{N}{P}+b\right]^{2}}\) is also a nonnegative number. Given\(\alpha\in\left.\ \left[0,\ \ \infty\right.\ \right)\),\(\frac{\text{dg}}{\text{dN}}=\alpha\frac{b}{\left[\frac{N}{P}+b\right]^{2}}\)is therefore in\(\left.\ \left[0,\ \ \infty\right.\ \right)\).

Ratio dependent Type III

The ratio-dependent Holling Type III model is specified as (Arditi & Ginzburg 1989; Abrams & Ginzburg 2000):
\begin{equation} f\left(N,\ P\right)=\frac{\alpha\left(\frac{N}{P}\right)^{2}}{\left(\frac{N}{P}\right)^{2}+b}\nonumber \\ \end{equation}\begin{equation} g\left(N,\ P\right)=\frac{\alpha\frac{N^{2}}{P}}{\left(\frac{N}{P}\right)^{2}+b}\nonumber \\ \end{equation}\begin{equation} \operatorname{}\left(f\right)=\operatorname{}\left[\frac{\alpha N^{2}}{N^{2}+bP^{2}}\right]=\frac{\alpha N^{2}}{N^{2}}=\alpha\nonumber \\ \end{equation}\begin{equation} \operatorname{}\left(f\right)=\operatorname{}\left[\frac{\frac{\alpha}{P^{2}}}{\frac{1}{P^{2}}+\frac{\alpha}{N^{2}}}\right]=\frac{\frac{\alpha}{P^{2}}}{\frac{1}{P^{2}}}=\alpha\nonumber \\ \end{equation}\begin{equation} \operatorname{}\left(f\right)=\frac{0}{0+b}=0\nonumber \\ \end{equation}\begin{equation} \operatorname{}\left(g\right)=\operatorname{}\left[\frac{\alpha N^{2}}{D\left[\left(\frac{N}{P}\right)^{2}+b\right]}\right]=0\nonumber \\ \end{equation}\begin{equation} \operatorname{}\left(g\right)=\operatorname{}\left[\frac{\alpha N^{2}}{\left[\left(\frac{N^{2}}{P}\right)+bP\right]}\right]=0\nonumber \\ \end{equation}\begin{equation} \operatorname{}\left(g\right)=\operatorname{}\left[\frac{\frac{\alpha}{P}}{\left(\frac{1}{P}\right)^{2}+\frac{b}{N^{2}}}\right]=\frac{\frac{\alpha}{P}}{\left(\frac{1}{P}\right)^{2}}=\alpha P\nonumber \\ \end{equation}\begin{equation} \operatorname{}\left(g\right)=\frac{0}{0+b}=0\nonumber \\ \end{equation}\begin{equation} \frac{\text{dg}}{\text{dN}}=\frac{2\alpha\left(\frac{N}{P}\right)\left[\left(\frac{N}{P}\right)^{2}+b\right]-\alpha\left(\frac{N^{2}}{P}\right)\frac{2N}{P^{2}}}{\left[\left(\frac{N}{P}\right)^{2}+b\right]^{2}}=\frac{2\alpha b\left(\frac{N}{P}\right)}{\left[\left(\frac{N}{P}\right)^{2}+b\right]^{2}}\nonumber \\ \end{equation}
Since all variables are nonnegative numbers,\(\frac{b\left(\frac{N}{P}\right)}{\left[\left(\frac{N}{P}\right)^{2}+b\right]^{2}}\)is also a nonnegative number. Given\(\alpha\in\left.\ \left[0,\ \ \infty\right.\ \right)\),\(\frac{\text{dg}}{\text{dN}}=2\alpha\frac{b\left(\frac{N}{P}\right)}{\left[\left(\frac{N}{P}\right)^{2}+b\right]^{2}}\)is therefore in\(\left.\ \left[0,\ \ \infty\right.\ \right)\).

Hassel-Varley Type I

The Hassel-Varley Holling Type I model is specified as (Hassell & Varley 1969; Abrams & Ginzburg 2000):
\begin{equation} f\left(N,\ P\right)=\alpha\left(\frac{N}{P^{m}}\right)\nonumber \\ \end{equation}\begin{equation} g\left(N,\ P\right)=\alpha NP^{1-m}\nonumber \\ \end{equation}
Therefore, \(\frac{\text{dg}}{\text{dN}}=\alpha P^{1-m}\), which is in the range \(\left.\ \left[0,\ \ \infty\right.\ \right)\)given\(\alpha\in\left.\ \left[0,\ \ \infty\right.\ \right)\).

Hassel-Varley Type II

The Hassel-Varley Holling Type II model is specified as (Hassell & Varley 1969; Abrams & Ginzburg 2000):
\begin{equation} f\left(N,\ P\right)=\frac{\alpha\left(\frac{N}{P^{m}}\right)}{\left(\frac{N}{P^{m}}\right)+b}\nonumber \\ \end{equation}\begin{equation} g\left(N,\ P\right)=\frac{\text{αN}P^{1-m}}{\left(\frac{N}{P^{m}}\right)+b}\nonumber \\ \end{equation}
Therefore,
\begin{equation} \operatorname{}\left(f\right)=\operatorname{}\left[\frac{\alpha\left(N\right)}{N+bP^{m}}\right]=\left\{\begin{matrix}\alpha N,\ \ \&m=0\\ 0,\ \ \&m>0\\ \end{matrix}\right.\ \nonumber \\ \end{equation}\begin{equation} \operatorname{}\left(f\right)=\operatorname{}\left[\frac{\alpha\frac{1}{P^{m}}\ }{\frac{1}{P^{m}}+\frac{b}{N}}\right]=\frac{\alpha\frac{1}{P^{m}}\ }{\frac{1}{P^{m}}}=\alpha\nonumber \\ \end{equation}\begin{equation} \operatorname{}\left(f\right)=\frac{0}{0+b}=0\nonumber \\ \end{equation}\begin{equation} \operatorname{}\left(g\right)=\operatorname{}\left[\frac{\text{αN}P^{1-m}}{\left(\frac{N}{P^{m}}\right)+b}\right]=\left\{\begin{matrix}\text{αN}\frac{\infty}{N+b}=\infty,\ \ \&m=0\\ \text{αN}\frac{\infty}{\frac{N}{\infty}+b}=\infty,\ \ \&0<m<1=0.5\\ \frac{\text{αN}}{\frac{N}{\infty}+b}=\frac{\text{αN}}{b},\ \ \&m=1\\ \end{matrix}\right.\ =\left\{\begin{matrix}\infty,\ \ \&0\leq m<1\\ \frac{\text{αN}}{b},\ \ \&m=1\\ \end{matrix}\right.\ \nonumber \\ \end{equation}

Hassel-Varley Type III

The Hassel-Varley Holling Type III model is specified as (Hassell & Varley 1969; Abrams & Ginzburg 2000):
\begin{equation} f\left(N,\ P\right)=\frac{\alpha\left(\frac{N}{P^{m}}\right)^{2}}{\left(\frac{N}{P^{m}}\right)^{2}+b}\nonumber \\ \end{equation}\begin{equation} g\left(N,\ P\right)=\frac{\alpha\left(\frac{N^{2}}{P^{2m-1}}\right)}{\left(\frac{N}{P^{m}}\right)^{2}+b}\nonumber \\ \end{equation}
Therefore,
\begin{equation} \operatorname{}\left(f\right)=\operatorname{}\left[\frac{\alpha N^{2}}{N^{2}+b{*P}^{2m}}\right]=\left\{\begin{matrix}\frac{\alpha N^{2}}{N^{2}+b},\ \ \&m=0\\ 0,\ \ \&0.5<m\leq 1\\ \end{matrix}\right.\ \nonumber \\ \end{equation}\begin{equation} \operatorname{}\left(g\right)=\operatorname{}\left[\frac{\alpha N^{2}}{\left(\frac{N^{2}}{P}\right)+b{*P}^{2m-1}}\right]=\left\{\begin{matrix}\frac{\alpha N^{2}}{\frac{N^{2}}{\infty}+b*0}=\infty,\ \ \&0\leq m<0.5\\ \frac{\alpha N^{2}}{\frac{N^{2}}{\infty}+b*1}=\frac{\alpha N^{2}}{b},\ \ \&m=0.5\\ \frac{\text{αN}}{\frac{N^{2}}{\infty}+b*\infty}=0,\ \ \&0.5<m\leq 1\\ \end{matrix}\right.\ \nonumber \\ \end{equation}

Beddington-DeAngelis

The Beddington-DeAngelis model is specified as (Beddington 1975; DeAngelis et al. 1975):
\begin{equation} f\left(N,P\right)=\frac{\text{αN}}{N+cP+b}\nonumber \\ \end{equation}\begin{equation} g\left(N,\ P\right)=\frac{\text{αNP}}{N+cP+b}\nonumber \\ \end{equation}
Therefore,
\begin{equation} \operatorname{}\left(f\right)=\frac{0}{cP+b}=0\nonumber \\ \end{equation}\begin{equation} \operatorname{}\left(f\right)=\operatorname{}\left[\frac{\alpha}{1+\frac{\left(cP+b\right)}{N}}\right]=\frac{\alpha}{1}=\alpha\nonumber \\ \end{equation}\begin{equation} \operatorname{}\left(g\right)=\frac{0}{N+0+b}=0\nonumber \\ \end{equation}\begin{equation} \operatorname{}\left(g\right)=\operatorname{}\left[\frac{\text{αN}}{c+\frac{\left(N+b\right)}{P}}\right]=\frac{\text{αN}}{c}\nonumber \\ \end{equation}\begin{equation} \operatorname{}\left(g\right)=\frac{0}{cP+b}=0\nonumber \\ \end{equation}\begin{equation} \operatorname{}\left(g\right)=\operatorname{}\left[\frac{\text{αP}}{1+\frac{\left(cP+b\right)}{N}}\right]=\frac{\text{αP}}{1}=\alpha P\nonumber \\ \end{equation}\begin{equation} \frac{\text{dg}}{\text{dN}}=\frac{\text{αP}\left[N+cP+b\right]-\alpha NP}{\left[N+cP+b\right]^{2}}=\frac{\alpha\left[cP^{2}+bP\right]}{\left[N+cP+b\right]^{2}}\nonumber \\ \end{equation}
Since all variables are nonnegative numbers,\(\frac{cP^{2}+bP}{\left[N+cP+b\right]^{2}}\) is also a nonnegative number. Given\(\alpha\in\left.\ \left[0,\ \ \infty\right.\ \right)\),\(\frac{\text{dg}}{\text{dN}}=\alpha\frac{cP^{2}+bP}{\left[N+cP+b\right]^{2}}\)is therefore in\(\left.\ \left[0,\ \ \infty\right.\ \right)\).

Kovai

The Kovai equation for predation is given as follows:
\begin{equation} f\left(N,\ P\right)=\alpha\left(1-e^{\frac{-Nu}{\left(\text{αP}\right)}}\right);\ u=\left(1-\frac{b}{N}e^{\frac{-N}{b}}\right)\nonumber \\ \end{equation}\begin{equation} g\left(N,P\right)=\alpha P\left(1-e^{\frac{-Nu}{\left(\text{αP}\right)}}\right)\nonumber \\ \end{equation}
Limits of the per predator predation rate f for extreme values of N and P can be calculated as follows:
\begin{equation} \operatorname{}\left(f\right)=\alpha\left[1-\operatorname{}\left(e^{\frac{-Nu}{\left(\text{αP}\right)}}\right)\right]=\alpha\left[1-1\right]=0\nonumber \\ \end{equation}\begin{equation} \operatorname{}\left(f\right)=\alpha\left[1-\operatorname{}\left(e^{\frac{\left[-N+b\left(1-e^{-\frac{N}{b}}\right)\right]}{\left(\text{αP}\right)}}\right)\right]=\alpha\left[1-e^{0}\right]=0\nonumber \\ \end{equation}\begin{equation} \operatorname{}\left(f\right)=\alpha\left[1-\operatorname{}\left(e^{\frac{-Nu}{\left(\text{αP}\right)}}\right)\right]=\alpha\left[1-e^{-\infty}\right]=\alpha\nonumber \\ \end{equation}
Limits of the total predation rates for extreme values of P and D are calculated as follows:
\begin{equation} \operatorname{}\left(g\right)=\alpha*0*\left[1-e^{-\infty}\right]=0\nonumber \\ \end{equation}\begin{equation} \operatorname{}\left(g\right)=\operatorname{\alpha P}\left(1-e^{\frac{-Nu}{\left(\text{αP}\right)}}\right)=\alpha P\left(1-0\right)=\alpha P\nonumber \\ \end{equation}
The limit of g as the number of predators become infinitely large is calculated as follows:
\begin{equation} \operatorname{}\left(g\right)=\operatorname{}\left[\text{αP}\left(1-e^{\frac{-Nu}{\left(\text{αP}\right)}}\right)\right]\nonumber \\ \end{equation}
This being in the indeterminate form \(\infty*0\), we reformulate and apply l’Hôpital’s rule:
\begin{equation} \operatorname{}\left(g\right)=\operatorname{}\frac{\alpha\left(1-e^{\frac{-Nu}{\left(\text{αP}\right)}}\right)}{\frac{1}{P}}=\frac{\lim_{P\rightarrow\infty}\left[\frac{\text{Nu}}{P^{2}}e^{-\frac{\text{Nu}}{\left(\text{αP}\right)}}\right]}{\lim_{P\rightarrow\infty}\left[\frac{1}{P^{2}}\right]}=\lim_{P\rightarrow\infty}\left[\text{Nu}{*e}^{-\frac{\text{Nu}}{\left(\text{αP}\right)}}\right]=Nu*1=Nu\nonumber \\ \end{equation}
The derivative of g with regards to N is calculated as follows:
\begin{equation} \frac{\text{dg}}{\text{dN}}=\alpha P\frac{d}{\text{dN}}\left[1-e^{\frac{\left[-N+b\left(1-e^{-\frac{N}{b}}\right)\right]}{\left(\text{αP}\right)}}\right]\nonumber \\ \end{equation}
With \(x=N-b\left(1-e^{-\frac{N}{b}}\right)\), we have
\begin{equation} \frac{\text{dg}}{\text{dN}}=\alpha P\frac{d}{\text{dN}}\left[1-e^{\frac{-x}{\left(\text{αP}\right)}}\right]=\alpha P\frac{d}{\text{dx}}\left[1-e^{\frac{-x}{\left(\text{αP}\right)}}\right]\frac{\text{dx}}{\text{dN}}\nonumber \\ \end{equation}\begin{equation} \frac{\text{dx}}{\text{dN}}=1-b\left(\frac{1}{b}e^{-\frac{N}{b}}\right)=1-e^{-\frac{N}{b}}\nonumber \\ \end{equation}
Therefore
\begin{equation} \frac{\text{dg}}{\text{dN}}=\alpha P\left[\frac{e^{\frac{-x}{\left(\text{αP}\right)}}}{\text{αP}}\right]\left[1-e^{-\frac{N}{b}}\right]=\left[1-e^{-\frac{N}{b}}\right]*e^{\frac{\left[-N+b\left(1-e^{-\frac{N}{b}}\right)\right]}{\left(\text{αP}\right)}}\nonumber \\ \end{equation}
For N, b in \(\left[\left.\ 0,\infty\right)\right.\ \),\(1-e^{-\frac{N}{b}}\) is in the range\(\left[\left.\ 0,1\right)\right.\ \).
The second portion of the derivative may be shown to be similarly bounded as follows:
\begin{equation} e^{\frac{\left[-N+b\left(1-e^{-\frac{N}{b}}\right)\right]}{\left(\text{αP}\right)}}=e^{\frac{-x}{\left(\text{αP}\right)}}\nonumber \\ \end{equation}
We first demonstrate that x is in the range\(\left[\left.\ 0,\infty\right)\right.\ \) for N, b in\(\left[\left.\ 0,\infty\right)\right.\ :\)
\begin{equation} x\left(N=0\right)=0-b\left(1-e^{0}\right)=0\nonumber \\ \end{equation}
As \(\frac{\text{dx}}{\text{dD}}=1-e^{-\frac{D}{b}}\) is in the range \(\left[\left.\ 0,1\right)\right.\ \), x must therefore be a positive number for N, b in\(\left[\left.\ 0,\infty\right)\right.\ \). Therefore,\(e^{\frac{-x}{\left(\text{αP}\right)}}\) is in the range\(\left(\left.\ 0,\ 1\right]\right.\ \), and\(\frac{\text{dg}}{\text{dN}}\) also remains in the range\(\left[\left.\ 0,1\right)\right.\ \).