Patchy Invasion of Stage-Structured Alien Species with Short-Distance and Long-Distance Dispersal

Abstract

Understanding of spatiotemporal patterns arising in invasive species spread is necessary for successful management and control of harmful species, and mathematical modeling is widely recognized as a powerful research tool to achieve this goal. The conventional view of the typical invasion pattern as a continuous population traveling front has been recently challenged by both empirical and theoretical results revealing more complicated, alternative scenarios. In particular, the so-called patchy invasion has been a focus of considerable interest; however, its theoretical study was restricted to the case where the invasive species spreads by predominantly short-distance dispersal. Meanwhile, there is considerable evidence that the long-distance dispersal is not an exotic phenomenon but a strategy that is used by many species. In this paper, we consider how the patchy invasion can be modified by the effect of the long-distance dispersal and the effect of the fat tails of the dispersal kernels.

Appendices

Appendix 1: Details of numerical integration

1.1 Boundary Conditions, Stationary Case

For the purposes of this paper, we need a boundary condition as non-intrusive as possible in order to minimize the boundary effect on the population dynamics in the interior of the domain. Since the kernel-based model is non-local, the relevant boundary condition is expected to be non-local as well.

Consider the normally distributed symmetrical kernel in the 1D domain $\Omega$:

$$\begin{aligned} \displaystyle {k(x,y)=\frac{1}{\sqrt{2\pi \alpha ^2}}\exp \left( -\frac{(x-y)^2}{2\alpha ^2}\right) }. \end{aligned}$$

(33)

where $(x,y) \in \Omega$. In the context of individual organism’s movement, the dispersal kernel k(x, y) gives the probability density of the event that an individual located at the position y before the dispersal will be found at the position x after the dispersal, and parameter $\alpha$ quantifies the spatial scale of the dispersal. We therefore require that the total probability is

$$\begin{aligned} P(x)=\int \limits _{\Omega }k(x,y)\mathrm{d}x = \int \limits _{\Omega }k(x,y)\mathrm{d}y \equiv 1. \end{aligned}$$

(34)

The boundary can be regarded as non-intrusive when the requirement (34) holds at any point in the computational domain $\Omega$. However, it is obviously not so when x is sufficiently close to the domain’s boundary regardless the size of the domain, see Fig. 11.

Fig. 11

Validation of the condition (34) in the domain $\Omega$. The kernel (33) has the variance $\alpha =3$. The domain size is $\Omega =[-L,L]$, where a $L=10$ and b $L=50$

In order to understand how the problem should be modified in order to make sure that condition (34) holds everywhere in the computational domain, we now consider the 1D domain $\Omega =[-L,L]$. From (33) and (34), we obtain:

$$\begin{aligned} P(x) = \int \limits _{\Omega }k(x,y)\mathrm{d}y = \int \limits _{-L}^L k(x,y)\mathrm{d}y = \frac{1}{2}\left[ \text{ erf }\left( \frac{L-x}{\sqrt{2}\alpha }\right) +\text{ erf }\left( \frac{L+x}{\sqrt{2}\alpha }\right) \right] , \end{aligned}$$

where $\text{ erf }(\xi )$ is the error function. Clearly, in order to satisfy (34), we need to ensure that $\text{ erf }\left( (L-x)/(\sqrt{2}\alpha )\right) \approx 1$ and $\text{ erf }\left( (L+x)/(\sqrt{2}\alpha )\right) \approx 1$ with sufficient precision.

We recall that $\text{ erf }(-\xi )=-\text{ erf }(\xi )$, and $\text{ erf }(\xi )$ is a monotone function of its argument $\xi$, $\text{ erf }(\xi ) \rightarrow 1$ as $\xi \rightarrow \infty$. It is well known that $\text{ erf }(\xi )$ is very close to 1 for $\xi \ge 3$, as we have $\text{ erf }(3)=0.99998$. Hence, we require that

$$\begin{aligned} \frac{L-x}{\sqrt{2}\alpha } \ge 3 \quad \text{ and } \quad \frac{L+x}{\sqrt{2}\alpha } \ge 3 \end{aligned}$$

(35)

in order to make $P \approx 1$ with sufficient precision. That can be achieved by performing the integration on a smaller domain, i.e., $x \in [-L+3\sqrt{2}\alpha , L-3\sqrt{2}\alpha ]$. Alternatively, however, if our domain of interest is $[-L,L]$, we can consider an extended domain $\Omega _\mathrm{ext}$ where the integration is performed. From the conditions (35), it is obvious that the extended domain preserving the condition (34) with sufficient accuracy can be defined as follows:

$$\begin{aligned} \Omega _\mathrm{ext}=[-L-3\sqrt{2}\alpha , L+3\sqrt{2}\alpha ]. \end{aligned}$$

(36)

Note that, apart from the size of the extended domain, parameter $\alpha$ also gives us a rough estimate of the grid step size in the problem, as we require that the interval of the length $\alpha$ should contain at least one grid point. For instance, if $L=10$ and $\alpha =0.1$, then the minimum size of the domain $\Omega _\mathrm{ext}$ is $\Omega _\mathrm{ext}=[-10.425, 10.425]$ and the minimum sensible number of grid points should be $n_{\min }=210$. For $n<n_{\min }$, the poor approximation will result in P being considerably less than 1 or may lead to $P>1$ which is senseless.

The analysis similar to that performed above for the normal distribution can be carried out for a different type of the kernel. Consider now the Cauchy-distributed kernel,

$$\begin{aligned} \displaystyle {k(x,y)=\frac{\beta }{\pi ((x-y)^2+\beta ^2)}}, \end{aligned}$$

(37)

where $\beta$ is a parameter. Again, we require that the condition (34) holds. Let us fix the value of x in (37) and consider it as the Cauchy distribution of the variable y. Integration over the domain $\Omega =[-L,L]$ gives

$$\begin{aligned} P(x)=\int \limits _{-L}^Lk(y)\mathrm{d}y=\frac{1}{\pi }\left[ \arctan \left( \frac{L-x}{\beta }\right) +\arctan \left( \frac{L+x}{\beta }\right) \right] . \end{aligned}$$

(38)

In order to meet the requirement $P=1$, we need to ensure that $\arctan \left( (L-x)/\beta \right) =\pi /2$ and $\arctan \left( (L+x)/\beta \right) =\pi /2$.

Let $v^*$ be a parameter that provides the required accuracy of the integration, such that $\arctan (v^*) \approx \pi /2$ with the desired precision. We then require

$$\begin{aligned} \frac{L-x}{\beta } \ge v^* \quad \text{ and }\quad \frac{L+x}{\beta } \ge v^*, \end{aligned}$$

(39)

in order to approximate $P \approx 1$ in the expression (38). That will give us the necessary range of x as

$$\begin{aligned} x \in [-L+\beta v^*, L-\beta v^*]. \end{aligned}$$

Therefore, the extended domain $\Omega _\mathrm{ext}$ can be defined as

$$\begin{aligned} \Omega _\mathrm{ext}=[-L-\beta v^*, L+\beta v^*]. \end{aligned}$$

(40)

Clearly, for any chosen accuracy $v^*$, the size of the domain is fully controlled by the value of the parameter $\beta$.

We notice here that the asymptotic convergence of the function $\arctan (v)$ is much slower than the convergence of the function $\text{ erf }(v)$. Correspondingly, in simulations with the Cauchy kernel, the domain extension has to be considerably larger than in simulations with the normally distributed kernel. By the way of example, several relevant values of $\arctan (v)$ are given in Table 1. Considering, for instance, the minimum accuracy of 0.2 % (i.e., at most 0.002 of the total population is lost because of its dispersal through the domain boundary), we observe that factor $v^*\approx 200$. For a hypothetical value $\beta =0.1$, it leads to the requirement that the margin separating the spreading population from the domain boundary should be about 20 or larger.

Table 1 Convergence of $\arctan (v)$ to its limiting value $\pi /2$ for large argument v

The above approach readily applies to the 2D problem as well, with the obvious modification that it should be used in both directions x and y.

1.2 Accumulation of Integration Error with Time

We now investigate how fast the numerical error is accumulated with time when the kernel is defined in either the original domain $\Omega$ or the extended domain $\Omega _\mathrm{ext}$. For this purpose, we consider

$$\begin{aligned} N_t(x)=\int \limits _{\Omega }k(x,y) N_{t-1}(y)\mathrm{d}y, \end{aligned}$$

(41)

where, at each generation t, we take into account only dispersal but not reproduction. Consider the case where the dispersal kernel is normally distributed; see (33). Assuming for the sake of simplicity that the initial condition is given by the normal distribution as well, i.e.,

$$\begin{aligned} N_0(y)=\frac{1}{\sqrt{2\pi \alpha _0^2}}\exp \left( -\frac{(y-\mu )^2}{2\alpha _0^2}\right) , \end{aligned}$$

(42)

the population density after t generations is given by the following normal distribution:

$$\begin{aligned} N_t(x)=\frac{1}{\sqrt{2\pi \alpha _t^2}}\exp {-\frac{(x-\mu )^2}{2\alpha _t^2}}, \end{aligned}$$

(43)

where the variance is

$$\begin{aligned} \alpha _t^2=\alpha _0^2+t\alpha ^2, \end{aligned}$$

(44)

$t=1,2,3\ldots$.

Let us now compute the function $\tilde{N}_t(x)$ by numerical integration in the domain $\Omega$ and compare it with the exact solution $N_t(x)$ given by (43). For any fixed t, the error $e_i=|N_t(x_i)-\tilde{N}_t(x_i)|$ is computed at every point $i=1,2,\ldots ,K$ of a uniform computational grid where K is the total number of grid nodes. The error norm is then defined as

$$\begin{aligned} ||e|| = \max \limits _{i=1,\ldots ,K}e_i. \end{aligned}$$

(45)

The graph of the error norm (45) as a function of t is shown in Fig. 12 by the dashed curve. The parameters of this test case are $\alpha =3.0$, $\alpha _0=1.0$, $\mu =0$, $t=10$ and $\Omega =[-20,20]$. The number of grid nodes on a uniform computational grid is $K=2049$. Here and below, the error is shown on the logarithmic scale. It is readily seen that the error increases rapidly as the time progresses. Further refinement of the grid does not result in any significant improvement in accuracy. Hence, we conclude that poor accuracy of numerical integration for $t>10$ is related to inaccurate kernel computation at the domain boundaries as discussed in the previous section.

Fig. 12

Computation of the convolution by numerical integration. The test case parameters are $\alpha =3.0$, $\alpha _0=1.0$, $\mu =0$, $t=10$ and $\Omega =[-20,20]$. The graph of the error norm (45) as a function of t. The computation is made in the domain $\Omega _\mathrm{ext}$ (solid line, open square) and the domain $\Omega$ (dashed line, closed circle)

We now make use of the findings in the previous section and compute the solution of Eq. (41), which we denote as $\tilde{N}_t(x)$, by numerical integration in the extended domain $\Omega _\mathrm{ext}$. The numerical solution $\tilde{N}_t(x)$ is then compared with the exact solution $N_t(x)$ in the domain $\Omega$. We emphasize that the domain $\Omega _\mathrm{ext}$ should be thought of as an auxiliary domain only used for accurate computation of the kernel. The resulting function $\tilde{N}_t(x)$ is still considered in the domain $\Omega$ where we assume the species population exists in the framework of our model.

The error norm for the function $\tilde{N}(t,x)$ when the computation is performed in the domain $\Omega _\mathrm{ext}$ is shown in Fig. 12 by the solid curve. The problem parameters remain the same as in the previous test case. According to the analysis done in the previous section, see (36), the size of the extended domain is $\Omega _\mathrm{ext}=[-32.7279, 32.7279]$. It is therefore clear that computation in the extended domain $\Omega _\mathrm{ext}$ provides very good accuracy for the solution evaluation in the original domain $\Omega$.

Note that the error only becomes large when the integrand function $\tilde{N}_{t-1}(x)$ has relatively large values close to the endpoints of the domain. Let us fix the time t and vary the parameter $\alpha$ in the formula (33). The corresponding graphs of the function $N_t(x)$ are shown in Fig. 13a. For each value of $\alpha$, we compute the error norm shown in Fig. 13b, where the results of computation are presented in the domain $\Omega$ and the domain $\Omega _\mathrm{ext}$.

Fig. 13

Computation of the convolution by numerical integration when the kernel parameter $\alpha$ is varied. The other parameters are $\alpha _0=1.0$, $\mu =0$, $\Omega =[-20,20]$ and $t=25$. a The graph of the function N(t, x) given by (43) for $\alpha =0.05$, $\alpha =0.5$ and $\alpha =2.0$. b The graph of the error norm (45) as a function of $\alpha$. The computation is made in the domain $\Omega$ (dashed line, closed circle) and the domain $\Omega _\mathrm{ext}$ (solid line, open square)

We therefore conclude that the accuracy of computation will deteriorate with time, provided that the support of the integrand $N_{t-1}(x)$ gets bigger as the time progresses. The ‘critical’ time $t_c$ when the error becomes unacceptably large can be roughly estimated from the condition $3\alpha _t=L$, where $\alpha _t$ is given by the Eq. (44). For $L=20$, $\alpha _0=1.0$ and $\alpha =3.0$, we have $t_c \approx 4$ and this estimate appears to be in a good agreement with the results of Fig. 13a.

1.3 Grid Convergence Test

Now we consider a nonlinear integro-difference equation that takes into account both dispersal and reproduction:

$$\begin{aligned} N_t(x)=\int \limits _{\Omega }k(x,y)f(N_{t-1}(y))\mathrm{d}y. \end{aligned}$$

(46)

In order to test the quality of our numerical approach, we need a function f that could provide a non-trivial spatiotemporal dynamics such as pattern formation. Correspondingly, we consider

$$\begin{aligned} \displaystyle {f(N)=rN\exp {(-N)}}. \end{aligned}$$

(47)

As for the dispersal, we consider the normally distributed kernel given by (33).

Fig. 14

Comparison of the solution on a fine grid of 8193 nodes (a bold line) with the solution on a coarse grid of 129 nodes at the fixed time $t=20$. The domain size is $[-L,L]$, where $L=15.0$. The other parameters are $\alpha =0.1$ and $r=7.0$

Equation (46) is solved numerically in the domain $[-L,L]$ to obtain the solution $N_t(x)$ at generation t from the solution $N_{t-1}(x)$ at the previous generation $t-1$. We use a regular grid, so that the location of each grid node in the domain is given by $x_{i+1}=x_i+\Delta$, where the grid step size $\Delta =2L/K$ and K is the number of grid nodes. For any fixed time t, the accuracy of the solution depends on the total number of nodes in the spatial grid used for numerical integration. The example of numerical solution on a coarse grid of $K=129$ nodes and a fine grid of $K=8193$ nodes at the fixed time $t=20$ is shown in Fig. 14. It is readily seen that the solution accuracy is lost on the coarse grid where the grid step size is not sufficiently small to resolve the solution oscillations.

Fig. 15

The error norm as a function of the number K of grid nodes

The above observations can be summarized by computing the solution error on a sequence of spatial grids when the time t is fixed. Namely, we first compute a numerical solution on a very fine grid of $K_f=8193$ nodes. We consider this numerical solution as this ‘exact’ solution and denote it $N^\mathrm{exact}(x)$. We then generate a sequence of uniformly refined grids where the ‘exact’ solution obtained on the fine grid should be available on each grid in the sequence. Hence, we consider a projection of fine grid onto a uniform coarse grid of K nodes. The number K is defined as $K=sK_0+1$, where $K_0=32$ is the number of grid subintervals on the initial coarse grid and the scaling coefficient is $s=2^{p}, p=0,1,2,\ldots ,7$. The nodal coordinates $x_i^{c}$, $i=1,\ldots ,K$ and $x_i^{f}$, $k=1,\ldots ,K_f$, considered on the coarse and fine grid, respectively, are related as $x_i^{c}=x_k^{f}$, where $k=si$. Once the grid projection has been made, the ‘exact’ solution is readily available at nodes $x_i$ of a coarse grid and the solution error

$$\begin{aligned} e_i=|N(t,x_i)-N^\mathrm{exact}(t,x_i)| \end{aligned}$$

(48)

is computed at each node. The error norm is defined accordingly as $||e||=\max _i e_i$.

The graph of the error norm as a function of the number K of grid nodes is shown in Fig. 15. It is seen from the figure that very good accuracy of computation is approached when the number of grid nodes is $K \ge 513$, i.e., the grid step size is $\Delta <0.0586$. Meanwhile, the solution is very poorly resolved on coarse grids with $K \le 65$ where the maximum error is $||e|| \sim 1$.

Appendix 2: Details of the FFT Numerical Technique

Let the function f(x) be defined in the domain $x \in (-\infty , +\infty )$. The Fourier transform $\widehat{f}(s)$ of the function f(x) is given by

$$\begin{aligned} \widehat{f}(s)=\int \limits _{-\infty }^{+\infty }f(x)e^{-2\pi isx}\mathrm{d}x, \end{aligned}$$

(49)

and the inverse Fourier transform is

$$\begin{aligned} f(x)=\int \limits _{-\infty }^{+\infty }\widehat{f}(s)e^{2\pi isx}\mathrm{d}s. \end{aligned}$$

Consider now two functions f(x) and g(x) defined for $x \in (-\infty , +\infty )$. Their convolution denoted ${f *g}$ is defined as

$$\begin{aligned} {f *g}=\int \limits _{-\infty }^{+\infty }f(y)g(x-y)\mathrm{d}y. \end{aligned}$$

(50)

Obviously, the convolution ${f *g}$ is a function of x, ${f *g}\equiv {f *g}(x)$ and we can apply (49) to ${f *g}(x)$ to obtain the Fourier transform $\widehat{f *g}(s)$ of the convolution.

Let $\widehat{f}(s)$ be the Fourier transform of a function f(x) and $\widehat{g}(s)$ be the Fourier transform of a function g(x). The convolution theorem states that

$$\begin{aligned} \widehat{f *g}(s)=\frac{1}{2\pi }\widehat{f}(s) \widehat{g}(s), \end{aligned}$$

(51)

i.e., the Fourier transform of the convolution of two functions is equal to the product of their Fourier transforms (e.g., see Champeney 1973). Thus, the convolution ${f *g}(x)$ of two functions can be found by calculating and inverting the Fourier transform $\widehat{f *g}(s)$ rather than by performing straightforward integration in (50).

It is important to note that the convolution theorem (51) can be applied in the multi-dimensional case where $f(\mathbf {x})$ and $g(\mathbf {x})$ are functions of the vector argument $\mathbf {x}$. The following discussion refers to the one-dimensional case as all basic results can be readily extended to a two-dimensional problem.

The functions f and g in the theorem (51) are generally supposed to be complex functions. Clearly, real functions in the generic population dynamics model (2) present a particular case of complex functions and the theorem (51) can therefore be employed in our problem to compute

$$\begin{aligned} {f *k}=\int _{-L}^{L}f(y)k(x-y)\mathrm{d}y, \end{aligned}$$

(52)

where ${f *k}$ is required to obtain population distributions and the definition of the kernel k(x) is given in the text. The interval L has to be chosen large enough so that in the time considered in the simulation, there is no boundary effects on the solution and we can assume that $y \in [-L,L]$ is a good approximation of the infinite interval (see also the discussion in Appendix 1).

The convolution theorem gives us the theoretical background for finding the values of a continuous function ${f *k}$ in the formula (52). However, when the problem (2) is solved numerically, see Sect. 3.3, both the population density and the kernel are only defined at nodes of a computational grid. Thus, the continuous Fourier transform has to be replaced with the discrete Fourier transform (DFT).

Let a continuous function f(x) be discretized over the interval $x \in [0,1]$ so that only the values $f_k \equiv f(x_k)$ are considered, where $x_k=k\Delta x$, $k=0,\ldots K-1$, $\Delta x=1/(K-1)$ is the grid step size and K is the number of grid nodes chosen in the problem. We denote $[f_k]$ the discrete function given by the set of numbers $f_0,f_1,\ldots ,f_{K-1}$. The DFT of the function $[f_k]$ denoted $F_s$ is defined as

$$\begin{aligned} F_s=\displaystyle {\frac{1}{K}\sum _{k=0}^{K-1} f_{k}e^{2\pi iks/K}}. \end{aligned}$$

(53)

The corresponding inverse transform is

$$\begin{aligned} f_k=\displaystyle {\sum _{s=0}^{K-1} F_{s}e^{-2\pi iks/K}}. \end{aligned}$$

(54)

The discrete Fourier transform can be loosely thought of as approximation of the integral (49) by the finite sum (53).

One important consequence of the definition (53) is that the convolution theorem is still valid in the discrete case stating that the product of the two individual DFTs will give the DFT of the discrete convolution (e.g., see Nussbaumer 1982). Thus, the task of computing the convolution (52) can be decomposed as computing $\widehat{k}$ and $\widehat{f}$ to produce $\widehat{k *f}=\widehat{k} \widehat{f}$ and then computing the inverse DFT of the product.

Computing and inverting the DFT can be done efficiently with help of the fast Fourier transform (FFT) numerical algorithms. The key idea behind any FFT computational routine is to reduce the number of operations required to compute the DFT and its inverse transform.

While the number of operations in a straightforward DFT computation using the formula (53) is $O(K^2)$, an FFT algorithm reduces that number to $O(K\log _2(K))$. It is worth noting here that the FFT is also superior to methods of numerical integration. For instance, numerical integration of (52) by a composite trapezoidal rule can be done in $O(K^2)$ operations.

The significant reduction in the number of operations requires a sophisticated algorithm incorporating a number of computational tricks, e.g., choosing the number K to be $K=2^m$ for some integer m and interchanging the first and second parts of the output vector in a computer program. The detailed explanation of the FFT algorithm can be found elsewhere (e.g., Press et al. 2007). In our problem, there also are some small modifications to the standard FFT routine dictated by the problem statement: We have to multiply the result by 2L (as the standard FFT algorithm assumes that the functions are defined on the interval [0, 1]) and remove the imaginary vector components created during the calculations as we deal with real functions only.

In the two-dimensional case, the FFT can be split into a series of one-dimensional FFTs resulting in the total number of operations $O(K^2\log _2(K))$. We use internal Mathematica routines to calculate the two-dimensional Fourier transform by FFT and, hence, obtain the population distributions over the lattice.

The results of the FFT computation performed for several test cases have been verified by direct numerical integration of (52), and very good agreement between the results of the two methods (i.e., the FFT and the trapezoidal rule of integration) was demonstrated in all test cases.