Mean occupancy time: linking mechanistic movement models, population dynamics and landscape ecology to population persistence

Abstract

Reaction–diffusion models for the dynamics of a biological population in a fragmented landscape can incorporate detailed descriptions of movement and behavior, but are difficult to analyze and hard to parameterize. Patch models, on the other hand, are fairly easy to analyze and can be parameterized reasonably well, but miss many details of the movement process within and between patches. We develop a framework to scale up from a reaction–diffusion process to a patch model and, in particular, to determine movement rates between patches based on behavioral rules for individuals. Our approach is based on the mean occupancy time, the mean time that an individuals spends in a certain area of the landscape before it exits that area or dies. We illustrate our approach using several different landscape configurations. We demonstrate that the resulting patch model most closely captures persistence conditions and steady state densities as compared with the reaction–diffusion model.

Appendices

Appendix A: Core and buffer habitat

We consider a core habitat, say a reserve, of length $2L_1,$ with a buffer zone of length $L_2$ attached to the core on either side (Cantrell and Cosner 1999). Since the one-dimensional model is symmetric with respect to the middle of the core habitat, we use this symmetry to reduce the model to the core ($\varOmega _1=[0,L_1]$) and the buffer ($\varOmega _2=[-L_2,0]$). The linearized equations for the density of individuals in the core and buffer are

$$\begin{aligned} u_t=D_i u_{xx}-m_i u +b_i u,\quad x\in \varOmega _i \end{aligned}$$

(70)

with $r_i=b_i-m_i$ and $r_1>r_2.$ We have a hostile boundary at $x=-L_2$ and, due to symmetry, no-flux conditions at $x=L_1,$ i.e.

$$\begin{aligned} u_x(L_1,t)=0,\quad u(-L_2,t)=0. \end{aligned}$$

(71)

At the interface between core and buffer, we require the flux to be continuous, but the density can be discontinuous if either there is movement preference by the individuals or if the diffusion rates differ between core and buffer (Ovaskainen and Cornell 2003), i.e.

$$\begin{aligned} D_1 u_x(0^+,t)=D_2 u_x(0^-,t),\quad u(0^+,t)=k u(0^-,t), \end{aligned}$$

(72)

where superscripts $\pm$ stand for right- and left-sided limits. Parameter $k$ summarizes bias and changes in movement. If individuals prefer the core habitat with probability $p,$ then either (I) $k=\frac{p\sqrt{D_2}}{(1-p) \sqrt{D_1}}$ or (II) $k=\frac{p D_2}{(1-p)D_1},$ see (38).

We denote $\mathcal M$ as the operator defined by (70–72) with $b_1=b_2=0.$ To determine the adjoint operator, we use the definition

$$\begin{aligned} \int \limits _{-L_2}^{L_1}v(x)\mathcal M u(\cdot ) dx=\int \limits _{-L_2}^{L_1}u(x)\mathcal M ^* v(\cdot ) dx \end{aligned}$$

(73)

We find that $\mathcal M ^*$ is given by $D_iv_{xx}-m_iv$ for $x\in \varOmega _i$ with interface and boundary conditions,

$$\begin{aligned} kD_1 v_x(0^+,t)&= D_2 v_x(0^-,t),\quad v(0^+,t)=v(0^-,t),\end{aligned}$$

(74)

$$\begin{aligned} v_x(L_1,t)&= 0,\quad v(-L_2,t)=0. \end{aligned}$$

(75)

Occupancy time in the core ($T_{\varOmega _1}$) and buffer ($T_{\varOmega _2}$) satisfy (20), which yields the following ODEs

$$\begin{aligned}&D_1\frac{d^2T_{\varOmega _1}}{dy^2}-m_1T_{\varOmega _1}=-1,\quad y\in \varOmega _1;\quad D_2\frac{d^2T_{\varOmega _1}}{dy^2}-m_2T_{\varOmega _1}=0,\quad y\in \varOmega _2;\qquad \quad \end{aligned}$$

(76)

$$\begin{aligned}&D_1\frac{d^2T_{\varOmega _2}}{dy^2}-m_1T_{\varOmega _2}=0,\quad y\in \varOmega _1;\quad D_2\frac{d^2T_{\varOmega _2}}{dy^2}-m_2T_{\varOmega _2}=-1,\quad y\in \varOmega _2.\qquad \quad \end{aligned}$$

(77)

Setting $m_1=0,$ and using the boundary condition at $x=L_1,$ we obtain

$$\begin{aligned} T_{\varOmega _1}(y)=\frac{1}{D_1}\left(-\frac{y^2}{2}+L_1 y+C\right),\quad y\in \varOmega _1. \end{aligned}$$

(78)

A convenient representation of the solution on $\varOmega _2$ that takes the boundary at $x=-L_2$ into account, is

$$\begin{aligned} T_{\varOmega _1}(y)=A\sinh \left(\sqrt{\frac{m_2}{D_2}}(y+L_2)\right),\quad y\in \varOmega _2. \end{aligned}$$

(79)

The interface condition for $T_{\varOmega _1}$ gives

$$\begin{aligned} C=D_1 A \sinh \left(\sqrt{\frac{m_2}{D_2}}L_2\right), \end{aligned}$$

(80)

whereas the condition on $T_{\varOmega _1}^{\prime }$ gives

$$\begin{aligned} A=\frac{kL_1}{\sqrt{D_2 m_2}\cosh \left(\sqrt{\frac{m_2}{D_2}}L_2\right)}. \end{aligned}$$

(81)

The calculations for $T_{\varOmega _2}$ are similar.

The definitions in (23) now give the elements of the matrix $\mathcal T .$ Finally, the average population growth rate is given by the eigenvalues of the matrix

$$\begin{aligned} \begin{bmatrix} b_1&\quad 0 \\ 0&\quad b_{2} \end{bmatrix}-\mathcal T ^{-1}. \end{aligned}$$

(82)

Appendix B: MFPT of a rectangle

We consider the rectangular domain $\varOmega =[0,a]\times [0,b]$ and the diffusion equation

$$\begin{aligned} \frac{\partial u}{\partial t}=D\left(\frac{\partial ^2 u}{\partial x^2}+\frac{\partial ^2 u}{\partial y^2}\right), \quad (x,y)\in \varOmega \end{aligned}$$

(83)

with hostile boundary conditions $u=0$ for $x\in \{0,a\}$ and $y\in \{0,b\}.$

The dominant eigenvalue can be calculated by standard separation of variables as

$$\begin{aligned} \lambda =-\pi ^2 D\frac{a^2+b^2}{a^2 b^2}. \end{aligned}$$

(84)

When we keep the area of the rectangle constant, we set $b=1/a.$ In that case, the dominant eigenvalue is maximal when $a=1.$

The MFPT is given by the equation

$$\begin{aligned} D\left(\frac{\partial ^2 T}{\partial x^2}+\frac{\partial ^2 T}{\partial y^2}\right)=-1\quad \mathrm{on}\quad \varOmega . \end{aligned}$$

(85)

The ansatz

$$\begin{aligned} T(x,y)=\sum _{m,n} B_{m,n}\sin \left(\frac{m\pi x}{a}\right)\sin \left(\frac{n\pi x}{b}\right) \end{aligned}$$

(86)

leads to the condition

$$\begin{aligned} \sum _{m,n} \hat{B}_{m,n}\sin \left(\frac{m\pi x}{a}\right)\sin \left(\frac{n\pi x}{b}\right)=1,\quad \hat{B}_{m,n}= D\pi ^2 B_{m,n}\left(\frac{m^2}{a^2}+\frac{n^2}{b^2}\right). \end{aligned}$$

(87)

The coefficients must satisfy $\hat{B}_{m,n}=16/(nm\pi ^2)$ when $n$ and $m$ are odd and $\hat{B}_{m,n}=0$ otherwise. Accordingly, we obtain

$$\begin{aligned} T(x,y)=\sum _{m,n \;\mathrm{odd}} \frac{16}{nmD\pi ^2}\left(\frac{m^2}{a^2}+\frac{n^2}{b^2}\right)^{-1}\sin \left(\frac{m\pi x}{a}\right)\sin \left(\frac{n\pi x}{b}\right). \end{aligned}$$

(88)

After integrating, we find

$$\begin{aligned} \overline{T}=\sum _{m,n \;\mathrm{odd}} \frac{16}{nmD\pi ^2}\frac{a^2 b^2}{a^2 n^2+ b^2 m^2}. \end{aligned}$$

(89)

We evaluate this expression numerically with $b=1/a$ and plot the result as a function of $a\in [0,1]$ in Fig. 8.

Fig. 8

Plot of MOT equation (89) for a rectangular domain with hostile boundary conditions. The area is fixed to 1, while the dimensions of the domain are varied, so illustrate how shape can effect MOT. The parameter $D=1$