Neutral Genetic Patterns for Expanding Populations with Nonoverlapping Generations

Abstract

We investigate the inside dynamics of solutions to integrodifference equations to understand the genetic consequences of a population with nonoverlapping generations undergoing range expansion. To obtain the inside dynamics, we decompose the solution into neutral genetic components. The inside dynamics are given by the spatiotemporal evolution of the neutral genetic components. We consider thin-tailed dispersal kernels and a variety of per capita growth rate functions to classify the traveling wave solutions as either pushed or pulled fronts. We find that pulled fronts are synonymous with the founder effect in population genetics. Adding overcompensation to the dynamics of these fronts has no impact on genetic diversity in the expanding population. However, growth functions with a strong Allee effect cause the traveling wave solution to be a pushed front preserving the genetic variation in the population. In this case, the contribution of each neutral fraction can be computed by a simple formula dependent on the initial distribution of the neutral fractions, the traveling wave solution, and the asymptotic spreading speed.

Appendix

1.1 Proof of Lemma 1

Proof

For simplicity in notation we focus on a single neutral fraction and drop the superscript i notation. By assumption, $x^2v_0(x)\mathrm{e}^{sx} \in L^{1}({\mathbb {R}})\cap L^{\infty }({\mathbb {R}})$. Thus, we have

$$\begin{aligned} x^2v_0(x)\mathrm{e}^{sx} \le (1+x^2)v_0(x)\mathrm{e}^{sx} \le C \end{aligned}$$

(93)

for all $x \in {\mathbb {R}}$ where C is a positive constant. Rearranging the previous inequality,

$$\begin{aligned} v_0(x) \le \frac{C\mathrm{e}^{-sx}}{1+x^2} \end{aligned}$$

(94)

for all $x \in {\mathbb {R}}$. Thus, there exists a positive constant C such that the function $w_0(x)$ defined by

$$\begin{aligned} w_0(x) := \frac{C\mathrm{e}^{-sx}}{1+x^2} \end{aligned}$$

(95)

satisfies $v_0(x) \le w_0(x)$ for all $x \in {\mathbb {R}}$. It is easy to see that $w_0(x)\mathrm{e}^{sx}\in L^{1}({\mathbb {R}})\cap L^{\infty }({\mathbb {R}})$. Hence, the Fourier transform of $w_0(x)\mathrm{e}^{sx} \in L^1({\mathbb {R}})$. To calculate the Fourier Transform of $w_0(x)\mathrm{e}^{sx}$, note that

$$\begin{aligned} {\mathcal {F}}\left[ \mathrm{e}^{-|x|}\right] (\omega )&=\int _{-\infty }^{\infty }\mathrm{e}^{-|x|}\mathrm{e}^{-i\omega x}\,\mathrm{d}x \end{aligned}$$

(96)

$$\begin{aligned}&= \int _{-\infty }^0 \mathrm{e}^{(1-i\omega )x}\,\mathrm{d}x+ \int _{0}^\infty \mathrm{e}^{-(1+i\omega )x}\,\mathrm{d}x \end{aligned}$$

(97)

$$\begin{aligned}&= \lim _{b\rightarrow \infty } \left[ \frac{\mathrm{e}^{(1-i\omega )x}}{(1-i\omega )}\bigg |_{-b}^0 - \frac{ \mathrm{e}^{-(1+i\omega )x}}{(1+i\omega )}\bigg |_0^b\right] \end{aligned}$$

(98)

$$\begin{aligned}&= \lim _{b\rightarrow \infty } \left[ \frac{1}{(1-i\omega )}-\frac{\mathrm{e}^{-(1-i\omega )b}}{(1-i\omega )} - \frac{ \mathrm{e}^{-(1+i\omega )b}}{(1+i\omega )}+\frac{1}{(1+i\omega )}\right] \end{aligned}$$

(99)

$$\begin{aligned}&=\left[ \frac{1}{(1-i\omega )} +\frac{1}{(1+i\omega )}\right] \end{aligned}$$

(100)

$$\begin{aligned}&= \frac{2}{1+\omega ^2}. \end{aligned}$$

(101)

From the inverse Fourier transform,

$$\begin{aligned} \pi \mathrm{e}^{-|x|}&= \frac{\pi }{2\pi } \int _{-\infty }^\infty \frac{2}{1+\omega ^2}\mathrm{e}^{i\omega x}\,\mathrm{d}\omega \end{aligned}$$

(102)

$$\begin{aligned}&= \int _{-\infty }^\infty \frac{1}{1+\omega ^2}\mathrm{e}^{i\omega x}\,\mathrm{d}\omega . \end{aligned}$$

(103)

Using the above result,

$$\begin{aligned} {\mathcal {F}}\left[ \frac{C}{1+x^2}\right] (\omega )&={\mathcal {F}}\left[ \frac{C}{1+(-x)^2}\right] (\omega ) \end{aligned}$$

(104)

$$\begin{aligned}&=C \int _{-\infty }^{\infty }\frac{1}{1+(-x)^2}\mathrm{e}^{-i\omega (-x)}\,\mathrm{d}x \end{aligned}$$

(105)

$$\begin{aligned}&=C \int _{-\infty }^{\infty }\frac{1}{1+x^2}\mathrm{e}^{i\omega x}\,\mathrm{d}x \end{aligned}$$

(106)

$$\begin{aligned}&= C\pi \mathrm{e}^{-|\omega |}. \end{aligned}$$

(107)

The proof of the lemma is complete. $\square$