Homogenization analysis of invasion dynamics in heterogeneous landscapes with differential bias and motility

Abstract

Animal movement behaviors vary spatially in response to environmental heterogeneity. An important problem in spatial ecology is to determine how large-scale population growth and dispersal patterns emerge within highly variable landscapes. We apply the method of homogenization to study the large-scale behavior of a reaction–diffusion–advection model of population growth and dispersal. Our model includes small-scale variation in the directed and random components of movement and growth rates, as well as large-scale drift. Using the homogenized model we derive simple approximate formulas for persistence conditions and asymptotic invasion speeds, which are interpreted in terms of residence index. The homogenization results show good agreement with numerical solutions for environments with a high degree of fragmentation, both with and without periodicity at the fast scale. The simplicity of the formulas, and their connection to residence index make them appealing for studying the large-scale effects of a variety of small-scale movement behaviors.

Appendix A: Homogenization

We apply the method of homogenization, substituting the expansion (8) into the two-scale model (9), solving at each order of $\varepsilon$. We seek the leading order solution $\rho _0$.

1.1 Order $\varepsilon ^{-2}$

The governing equation at order $\varepsilon ^{-2}$ is

$$\begin{aligned} 0=\partial _y\left[ I^{-1}\partial _y(I\eta \rho _0)\right] . \end{aligned}$$

(51)

Note that $I^{-1}$ denotes the reciprocal of I, not its inverse. Integrating twice results in

$$\begin{aligned} I\eta \rho _0=p_1(x,t)\gamma (y)+p(x,t), \end{aligned}$$

(52)

where $p_1$ and p are unknown functions of x and t, and

$$\begin{aligned} \gamma (y)=\int _{y_0}^yI(s)\,ds. \end{aligned}$$

(53)

The integral, $\gamma (y)$, is unbounded as $y\rightarrow \infty$. Thus, we require $p_1(x,t)=0$, so that $\rho _0$ remains bounded. Consequently,

$$\begin{aligned} I(y)\rho _0=\frac{p(x,t)}{\eta (y)}, \end{aligned}$$

(54)

which results in Eq. (12).

1.2 Order $\varepsilon ^{-1}$

The governing equation at order $\varepsilon ^{-1}$ is

$$\begin{aligned} \partial _y(u\rho _0)=\partial _y\left[ I^{-1}\partial _y(I\eta \rho _1) \right] +\partial _y\left[ I^{-1}\partial _x(I\eta \rho _0)\right] +\partial _x\left[ I^{-1}\partial _y(I\eta \rho _0)\right] . \end{aligned}$$

(55)

Recall that $u=u(x)$ does not depend on y. Substituting $p(x,t)=I\eta \rho _0$, from Eq. (54), and integrating twice yields

$$\begin{aligned} I\eta \rho _1=u\nu (y)p-y\partial _x p+q_1(x,t)\gamma (y)+q(x,t). \end{aligned}$$

(56)

Here, $q_1$ and q are unknown functions of x and t, and

$$\begin{aligned} \nu (y)=\int _{y_0}^y\eta (s)^{-1}\,ds. \end{aligned}$$

(57)

Note that the first three terms on the right side of Eq. (56) are O(y). So that $\rho _1$ remains bounded as $y\rightarrow \infty$, we require

$$\begin{aligned} q_1(x,t)=\frac{1}{\langle I \rangle }\partial _xp-\frac{\langle \eta ^{-1}\rangle }{\langle I\rangle }up, \end{aligned}$$

(58)

where,

$$\begin{aligned} \langle \eta ^{-1}\rangle =\lim _{y\rightarrow \infty }\frac{\nu (y)}{y}, \quad \text {and} \quad \langle I\rangle =\lim _{y\rightarrow \infty }\frac{\gamma (y)}{y}, \end{aligned}$$

(59)

assuming that these limits exist. Here and throughout the analysis, we assume that these limits also have the same values as $y\rightarrow -\infty$. Then, Eq. (56) becomes

$$\begin{aligned} I\eta \rho _1=-\left( \frac{\langle \eta ^{-1}\rangle \gamma (y)}{\langle I\rangle }-\nu (y)\right) u p+\left( \frac{\gamma (y)}{\langle I\rangle }-y\right) \partial _xp+q(x,t). \end{aligned}$$

(60)

1.3 Order $\varepsilon ^{0}$

The governing equation at order $\varepsilon ^{0}$ is

$$\begin{aligned} \partial _t\rho _0 + \partial _y(u\rho _1)+\partial _x(u\rho _0)= & {} \partial _y\left[ I^{-1}\partial _y(I\eta \rho _2)\right] +\partial _y \left[ I^{-1}\partial _x(I\eta \rho _1)\right] \nonumber \\&+\,\partial _x\left[ I^{-1}\partial _y(I\eta \rho _1)\right] +\partial _x\left[ I^{-1}\partial _x(I\eta \rho _0)\right] \nonumber \\&+\,f(\rho _0,y), \end{aligned}$$

(61)

where we f is given by Eq. (4), so that $f(\rho _0,y)=(\lambda (y)-\mu (y)\rho _0)\rho _0$. Substituting Eqs. (60) and (54) into (61), integrating twice, and simplifying yields

$$\begin{aligned} I\eta \rho _2= & {} \left[ \int _{y_0}^yI(s)\beta (s)\,ds\right] \partial _tp\nonumber \\&+\,\left[ \int _{y_0}^y\left( \frac{\langle \eta ^{-1}\rangle }{\langle I\rangle }\gamma (s)+\frac{\langle \eta ^{-1}\rangle }{\langle I\rangle }sI(s)-\nu (s)\right) \,ds\right] \partial _x(up)\nonumber \\&-\,\left[ \int _{y_0}^y\left( \frac{\gamma (s)}{\langle I\rangle }+\frac{sI(s)}{\langle I\rangle }-s\right) \,ds\right] \partial _x^2p-\left[ \int _{y_0}^yI(s) \omega (s)\,ds\right] p\nonumber \\&+\,\left[ \int _{y_0}^yI(s)\kappa (s)\,ds\right] p^2 +\left[ \int _{y_0}^y\eta ^{-1}(s)\left( \nu (s)-\frac{\langle \eta ^{-1}\rangle }{\langle I\rangle }\gamma (s)\right) \,ds\right] u^2p\nonumber \\&+\,\left[ \int _{y_0}^y\eta ^{-1}(s)\left( \frac{\gamma (s)}{\langle I\rangle }-s\right) \,ds\right] u\partial _x p +\left\{ \nu (y)\right\} uq-\left\{ y\right\} \partial _xq\nonumber \\&-\,\left\{ \gamma (y)\right\} w_1(x,t) -w(x,t), \end{aligned}$$

(62)

where $w_1$ and w are unknown functions of x and t, and $\beta (y)$, $\omega (y)$, and $\kappa (y)$ are defined by

$$\begin{aligned} \beta (y)=\int _{y_0}^yI(s)^{-1}\eta (s)^{-1}\,ds, \quad \omega (y)=\int _{y_0}^y\lambda (s)I(s)^{-1}\eta (s)^{-1}\,ds, \end{aligned}$$

(63)

and

$$\begin{aligned} \kappa (y)=\int _{y_0}^y\mu (s)I(s)^{-2}\eta (s)^{-2}\,ds. \end{aligned}$$

(64)

In Eq. (62), the terms with coefficients contained in square brackets, $[\cdot ]$, are $O(y^2)$, the terms with coefficients contained in curly braces, $\{\cdot \}$, are $O(y^1)$, and the remaining terms are $O(y^0)$. For $\rho _2$ to remain bounded terms of the same order are required to balance each other as $y\rightarrow \infty$. In particular at $O(y^2)$

$$\begin{aligned} \partial _tp+U\partial _x(up)=D\partial _x^2p+(\varLambda -Mp)p+Ru^2p+Vu\partial _xp, \end{aligned}$$

(65)

where,

$$\begin{aligned} U= & {} \lim _{y\rightarrow \infty }\frac{\int _{y_0}^y\left( \frac{\langle \eta ^{-1}\rangle }{\langle I\rangle }\gamma (s)+\frac{\langle \eta ^{-1}\rangle }{\langle I\rangle }sI(s) -\nu (s)\right) \,ds}{\int _{y_0}^yI(s)\beta (s)\,ds}, \end{aligned}$$

(66)

$$\begin{aligned} D= & {} \lim _{y\rightarrow \infty }\frac{\int _{y_0}^y\left( \frac{\gamma (s)}{\langle I\rangle }+\frac{sI(s)}{\langle I\rangle }-s\right) \,ds}{\int _{y_0}^yI(s)\beta (s)\,ds}, \end{aligned}$$

(67)

$$\begin{aligned} \varLambda= & {} \lim _{y\rightarrow \infty }\frac{\int _{y_0}^yI(s)\omega (s)\,ds}{\int _{y_0}^yI(s)\beta (s)\,ds}, \end{aligned}$$

(68)

$$\begin{aligned} M= & {} \lim _{y\rightarrow \infty }\frac{\int _{y_0}^yI(s)\kappa (s)\,ds}{\int _{y_0}^yI(s) \beta (s)\,ds}, \end{aligned}$$

(69)

$$\begin{aligned} R= & {} \lim _{y\rightarrow \infty }\frac{\int _{y_0}^y\eta ^{-1}(s)\left( \frac{\langle \eta ^{-1}\rangle }{\langle I\rangle }\gamma (s)-\nu (s)\right) \,ds}{\int _{y_0}^yI(s)\beta (s)\,ds}, \end{aligned}$$

(70)

and

$$\begin{aligned} V=\lim _{y\rightarrow \infty }\frac{\int _{y_0}^y\eta ^{-1}(s) \left( s-\frac{\gamma (s)}{\langle I\rangle }\right) \,ds}{\int _{y_0}^yI(s)\beta (s)\,ds}, \end{aligned}$$

(71)

if these limits exist.

Assuming that the limit

$$\begin{aligned} \lim _{y\rightarrow \infty }\frac{1}{y^2}\int _{y_0}^yI(s)\beta (s)\,ds \end{aligned}$$

(72)

exists, the coefficients (66–67) and (70–71) may be simplified to obtain

$$\begin{aligned} U= & {} \left( \lim _{y\rightarrow \infty }\int _{y_0}^y\eta (s)^{-1}\,ds\right) \left( 2\lim _{y\rightarrow \infty }\frac{1}{y^2}\int _{y_0}^yI(s)\beta (s)\,ds\right) ^{-1}, \end{aligned}$$

(73)

$$\begin{aligned} D= & {} \left( 2\lim _{y\rightarrow \infty }\frac{1}{y^2}\int _{y_0}^yI(s)\beta (s)\,ds\right) ^{-1}, \end{aligned}$$

(74)

$$\begin{aligned} R= & {} 0, \end{aligned}$$

(75)

$$\begin{aligned} V= & {} 0. \end{aligned}$$

(76)

For any integrable function, h(s), provided that h(s) / s is bounded for large s, it can be shown that,

$$\begin{aligned} \lim _{y\rightarrow \infty }\frac{1}{y^2}\int _{y_0}^yh(s)I(s)\,ds=\frac{1}{2}\left( \lim _{y\rightarrow \infty }\frac{h(y)}{y}\right) \left( \lim _{y\rightarrow \infty }\frac{1}{y}\int _{y_0}^yI(s)\,ds\right) , \end{aligned}$$

(77)

provided that the limits exist. Consequently, the coefficients (68–69) and (73–74) may be simplified to obtain the homogenized coefficients (13–16), and, thus, we arrive at the homogenized model (11).