Ecological Archives E081-027-A1

Frederick R. Adler, and Julio Mosquera. 2000. Is space necessary? Interference competition and limits to biodiversity. Ecology 81:3226-3232.



Appendix A. Coalition with discontinuous competitiveness function

Suppose the coalition includes all species between m=0 and m=zmax. For stability, we require that

\begin{displaymath}f(m) = 1 - T - m + s \int_{z=0}^{z=m} p(z)dz
- s \int_{z=m}^{z=z_{max}} p(z)dz = 0
\end{displaymath}


for $0 \leq m \leq z_{max}$. Differentiating with respect to mgives that 2s p(m)=1 or $p(m)=\mbox{$\frac{{\textstyle 1}}{{\textstyle 2s}}$ }$ for $0 \leq m \leq z_{max}$. Therefore, $T=\mbox{$\frac{{\textstyle z_{max}}}{{\textstyle 2s}}$ }$.

Next, $\int f(m) p(m) dm = 0$ where the integral is taken over all members of the coalition, because f(m)=0 on this set. However,

\begin{displaymath}\int_0^{z_{max}} f(m) p(m) dm = T(1 - T - \bar{m})
\end{displaymath}


where $\bar{m}$ is the mean of m. The interaction terms cancel because losses by the species with strategy z are gains by the species with strategy m (as can be shown by changing the order of integration in the second term).

We can then solve for $\bar{m}=1-T$. With the additional relation $\bar{m}=z_{max}/2$ for a uniform distribution, we find $z_{max}=\mbox{$\frac{{\textstyle 2s}}{{\textstyle 1+s}}$ }$. This coalition will be stable to invasion because any species with m > zmax has no competitive advantage over zmax itself, but at the cost of higher mortality. In terms of the invasion function f, f(m) < f(zmax) = 0.


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