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Analysis of spread and persistence for stream insects with winged adult stages

Abstract

Species such as stoneflies have complex life history details, with larval stages in the river flow and adult winged stages on or near the river bank. Winged adults often bias their dispersal in the upstream direction, and this bias provides a possible mechanism for population persistence in the face of unidirectional river flow. We use an impulsive reaction–diffusion equation with non-local impulse to describe the population dynamics of a stream-dwelling organism with a winged adult stage, such as stoneflies. We analyze this model from a variety of perspectives so as to understand the effect of upstream dispersal on population persistence. On the infinite domain we use the perspective of weak versus local persistence, and connect the concept of local persistence to positive up and downstream spreading speeds. These spreading speeds, in turn are connected to minimum travelling wave speeds for the linearized operator in upstream and downstream directions. We show that the conditions for weak and local persistence differ, and describe how weak persistence can give rise to a population whose numbers are growing but is being washed out because it cannot maintain a toe hold at any given location. On finite domains, we employ the concept of a critical domain size and dispersal success approximation to determine the ultimate fate of the populations. A simple, explicit formula for a special case allows us to quantify exactly the difference between weak and local persistence.

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Acknowledgments

Funding for this work came from the Alberta Water Research Institute (MAL, OV), NSERC Discovery (FL, MAL) and Accelerator (MAL) grants, and a Canada Research Chair and Killam Research Fellowship (MAL).

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Correspondence to Frithjof Lutscher.

Appendix

Appendix

Proof of Lemma 4.2

Proof

In Eq. (4.3), we have the iteration

$$\begin{aligned} u_{n+1}(x,0)=\rho e^{-\alpha \tau } (K*\Gamma _{q\tau ,2d\tau })*u_n(x,0). \end{aligned}$$

Let us denote \(\nu =\rho e^{-\alpha \tau }\) and \(L=K*\Gamma _{q\tau ,2d\tau }.\) Then \(u_n=\nu ^nL^{*n}u_0,\) where \(^{*n}\) denotes the n-fold convolution. We assume that original number of individuals released is \(||u_0||_1=n_0\) on a bounded set of measure b. Without loss of generality we may choose the set to be \(-b/2< x< b/2\).

By assumption, K and therefore L have finite mean and variance. We denote the mean and variance of L by \(\mu \) and \(\sigma ^2,\) respectively. To show weak persistence for \(\nu >1\) we demonstrate that there exists an \(x_n\) such that \(u_n(x_n,0)=\nu ^nL^{*n}u_0(x_n,0)\) grows for n sufficiently large. To start, we calculate the distance between \(L^{*n-1}L\) and the related Gaussian distribution \(\Gamma _{n\mu ,n\sigma ^2}\). We expect this to become small because of the Central Limit Theorem.

$$\begin{aligned}&\sup _x |L^{*n-1}L(x)-\Gamma _{n\mu ,n\sigma ^2}(x)|\nonumber \\&\quad =\frac{1}{\sigma \sqrt{n}} \sup _y |\sigma \sqrt{n}L^{*n-1}L(\sigma \sqrt{n}y+\mu )-\Gamma _{0,1}(y)| \le \frac{1}{\sigma \sqrt{n}}\frac{c}{\sigma \sqrt{n}}=\frac{c}{\sigma ^2n}. \end{aligned}$$

The convergence estimate was established by Petrov (1975), Theorem 10, Chapter VII. Next we use Hölder’s Inequality to calculate the distance between \(L^{*n-1}L\) and the corresponding Gaussian distribution \(\Gamma _{n\mu ,n\sigma ^2}\) convolved with the initial condition \(u_0(x,0)\).

$$\begin{aligned}&\sup _x |L^{*n}u_0(x,0)-\Gamma _{n\mu ,n\sigma ^2}*u_0(x,0)|\nonumber \\&\quad =||(L^{*n}-\Gamma _{n\mu ,n\sigma ^2}*)u_0(x,0)||_{\infty }\nonumber \\&\quad \le ||L^{*n-1}L(x)-\Gamma _{n\mu ,n\sigma ^2}(x)||_{\infty } ||u_0(x,0)||_{1} =\frac{cn_0}{\sigma ^2n}. \end{aligned}$$

This arises from Hölder’s inequality (see, for example, Kuptsov 2001) and the translation invariance of the Lebesgue measure.

Therefore, we can bound the true solution above and below by expressions involving the Gaussian distribution:

$$\begin{aligned} \Gamma _{n\mu ,n\sigma ^2}*u_0(x,0) -\frac{cn_0}{\sigma ^2n} \le L^{*n}u_0(x,0) \le \Gamma _{n\mu ,n\sigma ^2}*u_0(x,0) +\frac{cn_0}{\sigma ^2n} \end{aligned}$$

for all x. Multiplying by \(\nu ^n\) allows us to rewrite the left hand inequality as

$$\begin{aligned} u_n(x) \ge \nu ^n\left( \Gamma _{n\mu ,n\sigma ^2}*u_0(x,0) -\frac{cn_0}{\sigma ^2n}\right) \end{aligned}$$

for all x. To evaluate weak persistence, we choose \(x=x_n=n\mu \) so it tracks the mean displacement of L. We observe that over the interval (\(n\mu -b/2\le x \le n\mu +b/2\)) the quantity \(\Gamma _{n\mu ,n\sigma ^2}(x)\ge \Gamma _{n\mu ,n\sigma ^2}(n\mu +b/2)\). Hence

$$\begin{aligned} \Gamma _{n\mu ,n\sigma ^2}*u_0(n\mu )\ge \Gamma _{n\mu ,n\sigma ^2}(b/2) n_0 = \frac{n_0e^{-\frac{b^2}{8\sigma ^2 n}}}{\sqrt{2\pi \sigma ^2 n}} \end{aligned}$$

and so

$$\begin{aligned} u_n(n\mu ) \ge n_0\nu ^n \left( \frac{e^{-\frac{b^2}{8\sigma ^2 n}}}{\sqrt{2\pi \sigma ^2 n}} -\frac{c}{\sigma ^2n}\right) \ge n_0\nu ^n \left( \frac{e^{-\frac{b^2}{8\sigma ^2}}}{\sqrt{2\pi \sigma ^2 n}} -\frac{c}{\sigma ^2n}\right) \end{aligned}$$

The right hand quantity is positive and bounded below for all \(n>n^*\) where \(n^*>2c\pi \frac{e^{b^2/(8\sigma ^2)}}{\sigma ^2}\). Consequently, there exists some \(\varepsilon >0\) such that for all \(n>n^*\) there exists \(x_n=\mu n\in \mathbb {R}\) such that \(u_n(x_n,0)>\varepsilon \). This makes the population weakly persistent by definition (3.2). \(\square \)

Proof of Lemma 5.1

Proof

Consider a sequence \(v_n\rightarrow v\) in B. We show that \(K*v_n\) converges to \(K*v\) uniformly on compact subsets. By linearity, we may assume \(v=0.\)

Consider \(M>0\) and \(\varepsilon >0\). We need to find \(N>0\) such that for any \(x\in [-M,M]\) and \(n>N\) we have \(0\le (K*v_n)(x)<\varepsilon \). Since K is integrable, we can find some \(L>0\) such that

$$\begin{aligned} \int _{L}^{\infty }K(z)dz<\frac{\varepsilon }{3U^*}\quad \mathrm{and}\quad \int _{-\infty }^{-L} K(z)dz<\frac{\varepsilon }{3U^*}. \end{aligned}$$

By convergence, we can choose \(N>0\) be such that \(0\le v_n(x)<\frac{\varepsilon }{3}\) for \(n>N\) and any \(x\in [-M-L,M+L]\). Now, let \(x\in [-M,M]\) and \(n>N\). Then

$$\begin{aligned} \int _{-\infty }^\infty K(x-y)v_n(y)dy= & {} \int _{-\infty }^{-M-L}K(x-y)v_n(y)dy+\int _{-M-L}^{M+L}K(x-y)v_n(y)dy\nonumber \\&+\int _{M+L}^\infty K(x-y)v_n(y)dy, \end{aligned}$$

where

$$\begin{aligned} \int _{-\infty }^{-M-L}K(x-y)v_n(y)dy\le & {} U^*\int _{x+M+L}^\infty K(z)dz\le U^*\int _L^\infty K(z)dz<\frac{\varepsilon }{3},\\ \int _{M+L}^\infty K(x-y)v_n(y)dy\le & {} U^*\int _{-\infty }^{x-M-L} K(z)dz\le U^*\int _{-\infty }^{-L} K(z)dz<\frac{\varepsilon }{3}, \end{aligned}$$

and

$$\begin{aligned} \int _{-M-L}^{M+L}K(x-y)v_n(y)dy\le \frac{\varepsilon }{3}\int _{-M-L}^{M+L}K(x-y) dy\le \frac{\varepsilon }{3}\int _{-\infty }^\infty K(x-y)dy=\frac{\varepsilon }{3}. \end{aligned}$$

Thus,

$$\begin{aligned} (K*v_n)(x)=\int _{-\infty }^\infty K(x-y)v_n(y)dy<\varepsilon , \end{aligned}$$

as needed. \(\square \)

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Vasilyeva, O., Lutscher, F. & Lewis, M. Analysis of spread and persistence for stream insects with winged adult stages. J. Math. Biol. 72, 851–875 (2016). https://doi.org/10.1007/s00285-015-0932-x

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  • DOI: https://doi.org/10.1007/s00285-015-0932-x

Keywords

  • Drift paradox
  • Non-local impulsive reaction–diffusion equation
  • Spreading speed
  • Persistence condition

Mathematics Subject Classification

  • 92B05
  • 35K57
  • 34A38
  • 37L15