SUPPLEMENTAL INFORMATION
A sufficient condition for the critical road density
Here we derive Eq. 2 using the following theorem by Senn (1983) and used by Cantrell and Cosner (2001):
Theorem: Suppose that r(x,y) is continuous in a finite two-dimensional region, , and is positive at some point (x0,y0). If
A1.1
then the population described by Eq. 1 (without logistic growth) grows exponentially.
If
then the integral in A1.1 is simply A0r0+ A1r1, where A0 and A1 are the road area and the non-road area, respectively, and condition A1.1 becomes A0r0+A1r10. Defining h=A0/(A0+A1), we then obtain Eq. 2.
The important point is that A1.1 is only a sufficient condition, because it is possible that and for the population to still persist in the region (Cantrell and Cosner 2001). Notice that the growth rate in this version does not follow into the premises of the above theorem since it only takes two values, r0 and r1, hence it is necessarily discontinuous. We consider, however, the theorem to be a good first approximation to our problem.
Solution of Eq. 1 (formulas for minimum patch size and relaxation time)
Here, for completeness, we solve Eq. 1 under the assumptions of case 2, that is, exponential growth and lethal roads, hence r0=-. The model is then
,
subject to the initial condition N(x,y,0)=N0(x,y) and, because roads are a lethal habitat, the following boundary conditions: N(0,y,t)= N(Lx,y,t)= N(x,0,t)= N(x,Ly,t)=0, where Lx and Lyare the distance between roads along the x and y directions (see Fig. 1), and to simplify the notation we use D=2/2 .
The methods used to solve Eq. 1 are well known, see for example, Pipes and Harvill (1971) or Cosner (2008). To solve the top branch of this equation we use the method of separation of variables twice, first to separate the temporal from the spatial components, and then to separate the x and y spatial components. The first separation of variables assumes N(x,y,t)=V(x,y)U(t), thus we have
.
If V(x,y)0 and U(t)0, diving both terms of the previous equation by V(x,y)U(t) leads to
.
Because the right hand side contains only terms with x and y and the left hand side only terms with t, the equality is only possible if both sides are equal to a constant, that we call -k2. Thus, we have
(A2.1)
and
. (A2.2)
The solution of Eq. A2.1 with the initial condition U(0)=N0(x,y) is
. (A2.3)
To solve Eq. A2.2 we repeat the method of separation of variables, using now V(x,y)=W(x)Z(y), which leads to
.
Again, this equality is only valid if both terms are equal to a constant, that we now call , and we obtain the following equations
and
,
with , whose solutions are of the form (e.g., Pipes and Harvil1, 1971)
and
with A1, A2, B1 and B2 being constants.
The roads being lethal, the density must be equal to 0 when x=0 and y=0 which implies that the terms with cosines should be null, hence, B1=B2=0. Applying the same reasoning, the density also has to be equal to 0 for x=Lx and y=Ly, hence with n=1,2,…, with m=1,2,…, and
(n=m=0, are trivial solutions that we ignore). Thus
(A2.4)
and
. (A2.5)
Combining Eq. A2.3, A2.4 and A2.5 we finally obtain
, A2.6
where Cn,m are constants that depend on the initial density N0(x,y).
As in case 1, we are interested in finding under which conditions a population persists. Now, the exponential term in Eq. A2.6 shows that a population will increase or remain constant over time when -Dk2+r1 ≥ 0. On the basis of this inequality we can determine the minimum distance between roads such that the population does not become extinct. Substituting k2 by , we obtain
,
or, using L=Lx and Ly=Lx, with ≥ 1,
.
The minimum value of the smallest side of the rectangle, Lm, is then obtained when n=m=1,
, A2.7
and in the case of a square, because =1,
. A2.8
We can also use Eq. A2.6 to determine the time to extinction of a population in a nonviable patch, that is, one with area, A, smaller than that of the minimum patch size, Am. Here we use the relaxation time, trel, defined as the time a population takes to reach 1/e of its original size. If roads were built at time t=0, and the population starts to decline, then, according to exponential term in Eq. A2.6 the time it will take N(x,y,0) to reach 1/e of its size original abundance at t=0 is (-Dk2+r1)trel = -1, or
.
Numerical approximation of Eq. 1 using Euler explicit method
In order to approximate the partial differential Eq. 1 by a difference equation we used Euler’s explicit method that consists of transforming the partial derivatives for time and space according to
,
,
.
Making these substitutions into Eq. 1 we obtain Eq. 6.
It is important to assess the domain of stability of a difference equation. A heuristic derivation of the stability condition for Eq. 6 consists of observing that the density, ni,j,t, cannot be negative. If we consider the extreme case when adjacent cells do not contribute to the central cell density, that is, , then we have to guarantee in the top branch of Eq. 6 that , or . We obtain the worst case when ni,j,t=K and then <0.25. This condition is not surprising because the fraction of the density lost to the adjacent cells (the term -4) cannot be larger than 1. A similar analysis applied to the lower branch of Eq. 6 leads to <0.25-|r0|/4 which is more demanding then the previous one.
References
Cantrell RS, Cosner C (2001) Spatial heterogeneity and critical patch size: area effects via diffusion in closed environments. Journal of Theoretical Biology 209:161-171.
Pipes LA, Harvill LR (1971) Applied mathematics of engineers and physicists, 3rd edn. Mc Graw Hill, Tokyo.
Senn S (1983) On a linear eigenvalue problem with Neumann boundary condition, with an application to population genetics. Commun. Partial Differential Equations 8:1199-1228.
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