Consequences of Dispersal Heterogeneity for Population Spread and Persistence

Abstract

Dispersal heterogeneity is increasingly being observed in ecological populations and has long been suspected as an explanation for observations of non-Gaussian dispersal. Recent empirical and theoretical studies have begun to confirm this. Using an integro-difference model, we allow an individual’s diffusivity to be drawn from a trait distribution and derive a general relationship between the dispersal kernel’s moments and those of the underlying heterogeneous trait distribution. We show that dispersal heterogeneity causes dispersal kernels to appear leptokurtic, increases the population’s spread rate, and lowers the critical reproductive rate required for persistence in the face of advection. Wavespeed has been shown previously to be determined largely by the form of the dispersal kernel tail. We qualify this by showing that when reproduction is low, the precise shape of the tail is less important than the first few dispersal moments such as variance and kurtosis. If the reproductive rate is large, a dispersal kernel’s asymptotic tail has a greater influence over wavespeed, implying that estimating the prevalence of traits which correlate with long-range dispersal is critical. The presence of multiple dispersal behaviors has previously been characterized in terms of long-range versus short-range dispersal, and it has been found that rare long-range dispersal essentially determines wavespeed. We discuss this finding and place it within a general context of dispersal heterogeneity showing that the dispersal behavior with the highest average dispersal distance does not always determine wavespeed.

Appendix: Moments and Wavespeed

In this appendix, we will show that a dispersal kernel with faster tail decay can result in larger wavespeeds than a kernel with a more slowly decaying tail as long as the reproductive rate is sufficiently small.

Let $M_1(s)$ and $M_2(s)$ be two moment-generating functions for mean-zero, exponentially bounded, symmetric dispersal kernels $f_1(x)$ and $f_2(x)$, respectively. We assume that $f_2(x)>f_1(x)$ for $x>y>0$ and that $f_2(x)<f_1(x)$ for some $x<y$. Only a finite number of moments of $f_2$ can be less than those of $f_1$ (see 7.1 below).

The $2n$th moment of a distribution $f$ is defined as $m_{i,2n}=\displaystyle \int _{-\infty }^\infty x^{2n} f_i(x)\mathrm{d}x$. We further assume that the first $k-1$ even moments of the distributions are the same (all odd moments are zero), but that the $2k$th moment of $f_1$ is larger than that for $f_2$. We drop the first subscript on the first $k-1$ moments since they are equivalent among the two kernels to get

$$\begin{aligned} M_1(s)&= 1+m_2\frac{s^2}{2!}+m_4\frac{s^4}{4!}+\cdots +m_{2(k-1)}\frac{s^{2(k-1)}}{({2(k-1)})!}+m_{1,2k}\frac{s^{2k}}{(2k)!}\nonumber \\&+\sum _{j=k+1}^\infty m_{1,2j}\frac{s^{2j}}{(2j)!}\end{aligned}$$

(36)

$$\begin{aligned} M_2(s)&= 1+m_2\frac{s^2}{2!}+m_4\frac{s^4}{4!}+\cdots +m_{2(k-1)}\frac{s^{2(k-1)}}{({2(k-1)})!}+m_{2,2k}\frac{s^{2k}}{(2k)!}\nonumber \\&+\sum _{j=k+1}^\infty m_{2,2j}\frac{s^{2j}}{(2j)!}. \end{aligned}$$

(37)

We know that $m_{1,2k}>m_{2,2k}$, and the rest of the moments can have any arbitrary relationship as long as some $K$ exists such that $m_{1,2j}<m_{2,2j}$ for all $j>K$.

$$\begin{aligned} M_1(s)-M_2(s)=\left( (m_{1,2k}-m_{2,2k})+\sum _{j=k+1}^\infty (m_{1,2j}-m_{2,2j})(2k)!\frac{s^{2j-2k}}{(2j)!}\right) \frac{s^{2k}}{(2k)!} \end{aligned}$$

(38)

For $s$ small enough, $M_1(s)-M_2(s)>0$ because the dominant term becomes that with the $2k$th moment $m_{i,2k}$, which is larger for $f_1$. However, $M_1(s)<M_2(s)$ for $s$ large enough, since the higher-order moments of $f_2$ are larger. Thus, $\hat{s}$ exists such that $M_1(\hat{s}) = M_2(\hat{s})$, $M_1(s) > M_2(s)$ for $0 < s < \hat{s}$ and $M_1(s) < M_2(s)$ for $\hat{s} < s < \tilde{s}$ (for some $\tilde{s}$—depending on the precise relationship between the moments—it may be possible for the moment-generating functions to cross at multiple points).

For any given reproductive rate, $R$:

$$\begin{aligned} c(\hat{s};R)=c_1(\hat{s};R)=\frac{1}{\hat{s}}\ln { \left( RM _1(\hat{s}) \right) }=\frac{1}{\hat{s}}\ln { \left( RM _2(\hat{s}) \right) }=c_2(\hat{s};R), \end{aligned}$$

(39)

and $c_1(s;R)>c_2(s;R)$ for $s<\hat{s}$. The wavespeed is defined as

$$\begin{aligned} c_i^*=c(s_i^*;R)=\min { \left\{ \frac{1}{s}\ln \left( RM _i(s) \right) \right\} }. \end{aligned}$$

(40)

As $R$ goes down to one, both $s_1^*$ and $s_2^*$ go to zero since $R=1$ corresponds to a wavespeed of zero with a shape parameter $s^*=0$. So for $R$ sufficiently close to one, we can get $s_1^*$ and $s_2^*$ as close to zero as we like (see Appendix 7.2) and hence both smaller than $\hat{s}$. Because both critical shape parameters are in the region where $c_1(s;R)>c_2(s;R)$, $c_1^*>c_2^*$. This shows that a dispersal kernel with a “thinner” tail can give a larger wavespeed than a dispersal kernel with a “fatter tail” as long as the former dominates the latter in lower-order moments and that the net reproductive rate is sufficiently close to one.

When the mean dispersal location is negative, the critical reproductive rate is greater than one and the above argument does not apply since $s^*$ does not approach zero as $R$ goes down to $R^*$. However, as long as the mean dispersal location is close enough to zero, $R^*$ is close to one, and for a fixed $R$ close enough to $R^*$, $f_1$ gives larger wavespeeds than $f_2$.

1.1 Miscellaneous Moment Calculations

High-order moments tend toward infinity Here, we show that the magnitude of a probability distribution’s moments grows unboundedly as we look at higher and higher-order moments.

To see that the $2n$th moment goes to infinity as $n\rightarrow \infty$ for a symmetric kernel $f$ which has support for $x>1$, find $j>1$ such that $f(x)>C_j$ on the interval $(j,j+1)$. The $2n$th moment is $m_{2n}=2\int _0^\infty x^{2n}f(x)\mathrm{d}x$. This integral is then bounded from below by $2 j^{2n} C_j$. As $n\rightarrow \infty$, $m_{2n}>2 j^{2n} C_j \rightarrow \infty$ which proves the result.

High-order moments are larger for slower decaying tail In this section, we show that if one dispersal kernel is eventually above another $f_2(x)>f_1(x)$ for all $x$ greater than some $y$, then the higher-order moments of $f_2$ are all larger than those of $f_1$ (for sufficiently large order). This is a somewhat looser requirement than $f_2$ having a slower tail decay rate, which may involve showing that the ratio $f_1(x)/f_2(x)$ goes to zero as $x$ goes to infinity.

Assume that symmetric, mean-zero, exponentially bounded kernels satisfy $f_2(x)>f_1(x)$ for all $x>y$. Also, assume that the minimum of $f_2(x)-f_1(x)=-C<0$ for $0<x<y$. For the interval $0<x<y$, $x^{2n}(f_2(x)-f_1(x))>-Cy^{2n}$, and for $x>y$, $x^{2n}(f_2(x)-f_1(x))>\tilde{C}z^{2n}$ assuming that $f_2(x)-f_1(x)>\tilde{C}$ for $z<x<z+1$ given some $z>y$.

$$\begin{aligned} \frac{1}{2}(m_{2,2n}-m_{1,2n})&= \int _0^y x^{2n}(f_2(x)-f_1(x))\mathrm{d}x+\int _y^\infty x^{2n}(f_2(x)-f_1(x))\mathrm{d}x \\&>-Cy^{2n{+1}}+\tilde{C}z^{2n} \end{aligned}$$

Because $z$ is greater than $y$, $n$ being large enough ensures that $\tilde{C}z^{2n}-Cy^{2n{+1}}$ is positive, demonstrating that all higher-order moments of $f_2(x)$ are larger than those of $f_1(x)$.

1.2 As $R\searrow 1$, $c^*\rightarrow 0$ and $s^*\rightarrow 0$

As the net reproductive rate $R$ goes down to one, we will show that the wavespeed $c^*$ goes to zero and the corresponding critical shape parameter $s^*$ also goes to zero.

Let $R_n$ be a decreasing sequence of reproductive rates which converges to one. The moment-generating function for some mean-zero, symmetric, and exponentially bounded dispersal kernel is

$$\begin{aligned} M(s)=1+m_2\frac{s^2}{2}+\cdots . \end{aligned}$$

When $s$ is near zero, the moment-generating function is near one and is thus $M(s)=1+O(s^2)$. The power series expansion of the natural logarithm is $\ln (1+x)=x-\frac{1}{2}x^2+\cdots$, thus $\ln (M(s))=O(s^2)$. Now, we look at the sequence of functions

$$\begin{aligned} c_n(s)=\frac{1}{s}\ln \left( R_nM(s)\right) . \end{aligned}$$

It is clear that $c_n(s)>c_{n+1}(s)$ for any $n$ and for any $s$ because $R_n$ is a decreasing sequence and the natural logarithm is a monotone function.

Following a standard analytical technique, for any $\epsilon >0$, choose an $s$ close enough to zero such that $\frac{1}{s}\ln (M(s))=O(s)<\frac{\epsilon }{2}$. Because $R_n\rightarrow 1$ and for the fixed value of $s$ we just chose, a natural number $N$ can be chosen such that $\frac{1}{s}\ln (R_n)<\frac{\epsilon }{2}$ for any $n>N$ since $\ln (R_n)$ can be made as close to zero as we like.

Now, we have established that, for the sequence of functions $c_n(s)$, we can choose an $N$ and a fixed $s$ such that for any $n>N$,

$$\begin{aligned} c_n(s)=\frac{1}{s}\ln (R_n)+\frac{1}{s}\ln (M(s))<\frac{\epsilon }{2}+ \frac{\epsilon }{2}=\epsilon . \end{aligned}$$

Because $c_n(s)<\epsilon$ at some $s$, its minimum is also less than $\epsilon$. Thus, we can choose $R$ close enough to one to make the wavespeed as close to zero as we like.

Using the parametric representation (6) and letting $R$ go down to one

$$\begin{aligned} c^*=\frac{M'(s^*)}{M(s^*)}\rightarrow 0. \end{aligned}$$

Since the moment-generating function $M(s)$ is positive and monotonically increasing in $s$ (hence $M'(s)>0$ for any $s>0$), as $R$ goes down to one, $M'(s^*)$ must go to zero. This happens only if $s^*\rightarrow 0$.

Now, we have established that for any exponentially bounded, symmetric dispersal kernel with mean-zero, we can choose a $R$ close enough to one to give $s^*$ as close to zero as we like.