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Figure 8.5. (a) The Alabama Beach Mouse, and projections (in 2002) of population size based on Eq. (8.29) at two different sites: (b) the site BPSU, and (c) the site GINS. Photo courtesy of US Fish and Wildlife Service. I show the mean and the upper and lower 95% confidence intervals.

Suppose that r = 0. Show that the solution of Eq. (8.30) is

so that if the population starts at the ceiling (n = K) the mean persistence time is T(K) = K2/v. Interpret its shape and compare it with the MacArthur-Wilson result (Figure 8.3).

When r > 0, we rewrite Eq. (8.30) using subscripts to denote derivatives as

and we now recognize that the left hand side is the same as

Tn exp( — n so that Eq. (8.32) can be rewritten as d dn

Tiexp( 2rn

2 2r

which we integrate once to obtain

1 f2r

where c1 is a constant of integration. We now apply the boundary condition that the first derivative of T(n) is 0 when N = K to conclude that c1 = (1/r)exp((2r/v)n) and we can thus write that

Tn —---Y -exp — K exp--n r r v v and we now integrate this equation once again to obtain n v 2r 2r

0 0

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